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EBD_7207, •, , Corporate Office : 45, 2nd Floor, Maharishi Dayanand Marg, Corner Market,, Malviya Nagar, New Delhi-110017, Tel. : 011-26692293 / 26692294, , by, Preetima Bajpai, , Typeset by Disha DTP Team, , DISHA PUBLICATION, All Rights Reserved, © Copyright Publisher, No part of this publication may be reproduced in any form without prior permission of the publisher. The publisher do, not take any legal responsibility for any errors or misrepresentations that might have crept in. We have tried and made, our best efforts to provide accurate up-to-date information in this book., , For further information about the books from DISHA,, Log on to www.dishapublication.com or email to info@dishapublication.com, , (ii)
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www.neetexambooster.in, Contents, 1., , Some Basic Concepts of Chemistry, , 2., , Structure of Atom, , 19-38, , 3., , Classification of Elements and Periodicity in Properties, , 39-52, , 4., , Chemical Bonding and Molecular Structure, , 53-76, , 5., , States of Matter, , 77-94, , 6., , Thermodynamics, , 95-110, , 7., , Equilibrium, , 111-134, , 8., , Redox Reactions, , 135-146, , 9., , Hydrogen, , 147-156, , 10., , The s-Block Elements, , 157-168, , 11., , The p-Block Elements (Group 13 and 14), , 169-180, , 12., , Organic Chemistry – Some Basic Principles and Techniques, , 181-202, , 13., , Hydrocarbons, , 203-226, , 14., , Environmental Chemistry, , 227-238, , 15., , The Solid State, , 239-254, , 16., , Solutions, , 255-274, , 17., , Electrochemistry, , 1-18, , 275-294, (iii)
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18., , Chemical Kinetics, , 295-316, , 19., , Surface Chemistry, , 317-332, , 20., , General Principles and Processes of Isolation of Elements, , 333-344, , 21., , The p-Block Elements (Group 15, 16, 17 and 18), , 345-366, , 22., , The d-and f-Block Elements, , 367-382, , 23., , Coordination Compounds, , 383-404, , 24., , Haloalkanes and Haloarenes, , 405-422, , 25., , Alcohols, Phenols and Ethers, , 423-440, , 26., , Aldehydes, Ketones & Carboxylic acids, , 441-462, , 27., , Amines, , 463-478, , 28., , Biomolecules, , 479-498, , 29., , Polymers, , 499-510, , 30., , Chemistry In Everyday Life, , 511-520, , Mock Test - 1 , , MT-1-6, , Mock Test - 2 , , MT-7-12, , Mock Test - 3 , , MT-13-20, , Mock Test - 4 , , MT-21-26, , Mock Test - 5 , , MT-27-32, , (iv), , EBD_7207, , www.neetexambooster.in
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www.neetexambooster.in, , www.neetexambooster.in, , 1, , SOME BASIC CONCEPTS OF, CHEMISTRY, FACT / DEFINITION TYPE QUESTIONS, 1., , 2., , 3., , 4., , 5., , 6., , 7., , A mixture of sand and iodine can be separated by, (a) crystallisation, (b) distillation, (c) sublimation, (d) fractionation, Difference in density is the basis of, (a) ultrafiltration, (b) molecular sieving, (c) molecular attraction (d) gravity separation, Which of the following is an example of a heterogeneous, substance?, (a) Bottled water, (b) Table salt, (c) Pieces of copper, (d) Candle, Which of the following substances cannot be separated in, to its constituents by physical methods?, (a) Sugar and water solution, (b) Salt and sugar, (c) Solid glucose, (d) Both (a) and (b), Which of the following pair of substances contain, element and compound within a pair ?, (A) O2, CH4, (B) H2, O2, (C) N2, CO2, (D) Na, CO, (a) A, C, D, (b) B only, (c) C and D, (d) All of these, Which of the following statements about a compound is, incorrect?, (a) A molecule of a compound has atoms of different, elements., (b) A compound cannot be separated into its constituent, elements by physical methods of separation., (c) A compound retains the physical properties of its, constituent elements., (d) The ratio of atoms of different elements in a compound, is fixed., Choose the correct combination, Element, Compound Mixture, (a) Ammonia Sodium, Air, (b) Water, Sugar, Aqueous sugar solution, (c) Hydrogen Oxygen, Water, (d) Silver, Water, Air, , 8., , 9., , 10., , 11., , 12., , 13., , 14., , 15., , 16., , Choose the correct statement., (a) The particle s in liquids are more closely held than, gases but less free to move than solids., (b) The particles of solids are arranged in orderly fashion, but they can move as freely as liquids., (c) The particles of gases are far apart as compared to, solids and liquids and their movement is easy and, fast., (d) The particles of gases moves faster than liquids only, when the gases are heated., A mixture contains two or more substances in ________, which are called its ________., (a) fixed ratio, compounds, (b) fixed ratio, elements, (c) any ratio, components, (d) any ratio, elements, Which one of these is not a pure compound?, (a) O3, (b) H2O2, (c) H2O, (d) Sucrose solution, One fermi is, (a) 10–15 cm, (b) 10–13 cm, –10, (c) 10 cm, (d) 10–12 cm, The prefix 1018 is, (a) giga, (b) kilo, (c) exa, (d) nano, The prefix zepto stands for (in m), (a) 109, (b) 10–12, –15, (c) 10, (d) 10–21, The unit J Pa–1 is equivalent to, (a) m3, (b) cm3, 3, (c) dm, (d) None of these, Which has highest weight ?, (a) 1 m3 of water, (b) A normal adult man, (c) 10 litre of Hg, (d) All have same weight, Which one of the following set of units represents the, smallest and largest amount of energy respectively?, (a) J and erg, (b) erg and cal, (c) cal and eV, (d) eV and L-atm
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EBD_7207, SOME BASIC CONCEPTS OF CHEMISTRY, , 2, , 17., , 18., , 19., , 20., , 21., , 22., , 23., , 24., , A measured temperature on Fahrenheit scale, is 200 °F. What will this reading be on Celsius scale ?, (a) 40° C, (b) 94° C, (c) 93.3 °C, (d) 30° C, Which of the following is not a SI unit?, (a) metre, (b) candela, (c) mole, (d) litre, The prefix 10–24 is, (a) yotta, (b) zeta, (c) yocto, (d) zepto, Many countries use Fahrenheit scale for expressing, temperature of atmosphere. If temperature in any such, country is measured 41°F then what is its value in celcius, scale and would you expect hot or cold atmosphere in that, country?, (a) 15°C, cold, (b) 25°C, normal, (c) 5°C, cold, (d) 41°C, hot, A sample was weighted using two different balances. The, results were, (i) 3.929 g, (ii) 4.0 g, How would the weight of the sample be reported?, (a) 3.93 g, (b) 3g, (c) 3.9 g, (d) 3.929 g, Two students performed the same experiment separately, and each one of them recorded two readings of mass which, are given below. Correct reading of mass is 3.0 g. On the, basis of given data, mark the correct option out of the, following statements., Students, Readings, (i) (ii), A, 3.01 2.99, B, 3.05 2.95, (a) Results of both the students are neither accurate nor, precise., (b) Results of student A are both precise and accurate., (c) Results of student B are neither precise nor accurate., (d) Results of student B are both precise and accurate., 0.00016 can be written as ...A... in scientific notaiton. Here,, A refers to, (a) 1.6 × 10–4, (b) 24.50 × 10–9, –8, (c) 2.450 × 10, (d) 24.50 × 10–7, If the true value for an experimental result is 6.23 and the, results reported by three students X, Y and Z are :, X : 6.18 and 6.28, Y : 6.20 and 6.023, Z : 6.22 and 6.24, Which of the following option is correct :, (a) X precise, Y accurate, Z precise and accurate., (b) X precise and accurate, Y not precise, Z precise, (c) Both X & Z precise & accurate, Y not precise., (d) Both X & Y neither precise nor accurate, Z both precise, and accurate., , 25., , In the final answer of the expression, (29.2 20.2) (1.79 105 ), 1.37, , 26., , 27., , 28., , 29., , 30., , 31., , 32., , the number of significant figures is :, (a) 1, (b) 2, (c) 3, (d) 4, The number of significant figures for the three numbers, 161 cm, 0.161 cm, 0.0161 cm are, (a) 3,4 and 5 respectively (b) 3,4 and 4 respectively, (c) 3,3 and 4 respectively (d) 3,3 and 3 respectively, Given P = 0.0030m, Q = 2.40m, R = 3000m, Significant figures, in P, Q and R are respectively, (a) 2, 2, 1, (b) 2, 3, 4, (c) 4, 2, 1, (d) 4, 2, 3, If the density of a solution is 3.12 g mL–1, the mass of 1.5 mL, solution in significant figures is______., (a) 4.7 g, (b) 4680 × 10–3 g, (c) 4.680 g, (d) 46.80 g, In which of the following number all zeros are significant?, (a) 0.0005, (b) 0.0500, (c) 50.000, (d) 0.0050, The correctly reported answer of addition of 29.4406, 3.2, and 2.25 will have significant figures, (a) 3, (b) 4, (c) 2, (d) 5, The number of significant figures in 10.3106 g is, (a) 2, (b) 3, (c) 1, (d) 6, Choose the correct option that represents the result of the, given calculation to the appropriate number of significant, figures:, 43.0 0.0243, 0.340 4, , 33., , 34., , 35., , (a) 0.768, (b) 0.77, (c) 0.76, (d) 0.7683, Arrange the numbers in increasing no. of significant figures., 0.002600, 2.6000, 2.6, 0.260, (a) 2.6 < 0.260 < 0.002600 < 2.6000, (b) 2.6000 < 2.6 < 0.002600 < 0.260, (c) 0.260 < 2.6 < 0.002600 < 2.6000, (d) 0.002600 < 0.260 < 2.6 < 2.6000, Dimension of pressure are same as that of, (a) Energy, (b) Force, (c) Force per unit volume (d) Energy per unit volume, n g of substance X reacts with m g of substance Y to form, p g of substance R and q g of substance S. This reaction, can be represented as, X + Y = R + S. The relation which, can be established in the amounts of the reactants and the, products will be, (a) n – m = p – q, (b) n + m = p + q, (c) n = m, (d) p = q
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SOME BASIC CONCEPTS OF CHEMISTRY, , 36., , 37., , 38., , 20 g of CaCO3 on heating gave 8.8 g of CO2 and 11.2 g of, CaO. This is in accordance with, (a) The law of conservation of mass., (b) The law of constant composition., (c) The law of reciprocal proportion., (d) None of these, Which of the following is the best example of law of, conservation of mass?, (a) 12 g of carbon combines with 32 g of oxygen to form, 44 g of CO2, (b) When 12 g of carbon is heated in a vacuum there is no, change in mass, (c) A sample of air increases in volume when heated at, constant pressure but its mass remains unaltered, (d) The weight of a piece of platinum is the same before, and after heating in air, Which of the following statements is correct about the, reaction given below ?, 4Fe(s), , 39., , 40., , 41., , 3O 2 (g), , 2Fe 2 O 3 (g), , (a) Total mass of iron and oxygen in reactants = total mass, of iron and oxygen in product therefore, it follows law, of conservation of mass., (b) Total mass of reactants = total mass of product;, therefore, law of multiple proportions is followed., (c) Amount of Fe2O3 can be increased by reducing the, amount of any one of the reactants (iron or oxygen)., (d) Amount of Fe2O3 produced will decrease if the amount, of any one of the reactants (iron or oxygen) is taken in, excess., In an experiment 4.2 g of NaHCO3 is added to a solution of, acetic acid weighing 10.0 g, it is observed that 2.2 g of CO2, is released into the atmosphere. The residue left behind is, found to weigh 12.0 g, The above observations illustrate, (a) law of definite proportions., (b) law of conservation of mass, (c) law of multiple proportions, (d) None of these, In one experiment, 4g of H2 combine with 32g of O2 to form, 36g of H2O. In another experiment, when 50g of H2 combine, with 400g of O2 then 450g of H2O is formed. Above two, experiments follow, (a) The law of conservation of mass, (b) The law of constant composition, (c) Both (a) and (b), (d) Neither (a) nor (b), Irrespective of the source, pure sample, of water always yields, 88.89% mass of oxygen and 11.11% mass of hydrogen. This is, explained by the law of, (a) conservation of mass (b) multiple proportions, (c) constant composition (d) constant volume, , 3, , 42. The percentage of copper and oxygen in samples of CuO, obtained by different methods were found to be the same., The illustrate the law of, (a) constant proportions (b) conservation of mass, (c) multiple proportions (d) reciprocal proportions, 43. The law of definite proportions was given by –, (a) John Dalton, (b) Humphry Davy, (c) Proust, (d) Michael Faraday, 44. Which one of the following pairs of compounds illustrate, the law of multiple proportions ?, (a) H2O and Na2O, (b) MgO and Na2O, (c) Na2O and BaO, (d) SnCl2 and SnCl4, 45. Among the following pairs of compounds, the one that, illustrates the law of multiple proportions is, (a) NH3 and NCl3, (b) H2S and SO2, (c) CS2 and FeSO4, (d) CuO and Cu2O, 46. Two samples of lead oxide were separately reduced to, metallic lead by heating in a current of hydrogen. The weight, of lead from one oxide was half the weight of lead obtained, from the other oxide. The data illustrates, (a) law of reciprocal proportions, (b) law of constant proportions, (c) law of multiple proportions, (d) law of equivalent proportions, 47. In compound A, 1.00g of nitrogen unites with 0.57g of, oxygen. In compound B, 2.00g of nitrogen combines with, 2.24g of oxygen. In compound C, 3.00g of nitrogen combines, with 5.11g of oxygen. These results obey the following law, (a) law of constant proportion, (b) law of multiple proportion, (c) law of reciprocal proportion, (d) Dalton’s law of partial pressure, 48. Which of the following statements indicates that law of, multiple proportion is being followed., (a) Sample of carbon dioxide taken from any source will, always have carbon and oxygen in the ratio 1 : 2., (b) Carbon forms two oxides namely CO2 and CO, where, masses fo oxygen which combine with fixed mass of, carbon are in the simple ration 2 : 1., (c) When magnesium burns in oxygen, the amount of, magnesium taken for the reaction is equal to the amount, of magnesium in magnesium oxide formed., (d) At constant temperature and pressure 200 mL of, hydrogen will combine with 100 mL oxygen to produce, 200 mL of water vapour., 49. The molecular weight of O2 and SO2 are 32 and 64, respectively. At 15°C and 150 mm Hg pressure, one litre of, O2 contains ‘N’ molecules. The number of molecules in two, litres of SO2 under the same conditions of temperature and, pressure will be :, (a) N/2, (b) 1N, (c) 2N, (d) 4N
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EBD_7207, SOME BASIC CONCEPTS OF CHEMISTRY, , 4, , 50., , 51., , 52., , 53., , 54., , 55., , dm 3, , 10, of N2 gas and 10, of gas X at the same, temperature contain the same number of molecules, the gas, X is, (a) CO2, (b) CO, (c) H2, (d) NO, One mole of a gas occupies a volume of 22.4 L. This is, derived from, (a) Berzelius’ hypothesis (b) Gay-Lussac’s law, (c) Avogadro’s law, (d) Dalton’s law, One of the following combination which illustrates the law, of reciprocal proportions ?, (a) N2O3, N2O4, N2O5, (b) NaCl, NaBr, NaI, (c) CS2, CO2, SO2, (d) PH3, P2O3, P2O5, Equal volumes of two gases A and B are kept in a container, at the same temperature and pressure. Avogadro’s law is, invalid if, (a) the gases are reactive, (b) the gases are non-reactive, (c) gas A has more number of molecules than gas B., (d) None of these, Molecular mass is defined as the, (a) mass of one atom compared with the mass of one, molecule, (b) mass of one atom compared with the mass of one atom, of hydrogen, (c) mass of one molecule of any substance compared with, the mass of one atom of C-12, (d) None of the above, 1 amu is equal to, (a), , 56., , 57., , dm3, , 1, of O-16, 14, , (b), , 60., , 61., , 62., , 63., , 64., , 65., , 66., , 1, of C-12, 12, , (c) 1 g of H2, (d) 1.66 × 10–23 kg, The modern atomic weight scale is based on, (a) O16, (b) C12, 1, (c) H, (d) C13, The percentage weight of Zn in white vitriol [ZnSO4.7H2O], is approximately equal to ( Zn, , 58., , 59., , 67., , 68., , 65, S 32, O 16 and, , H = 1), (a) 33.65 %, (b) 32.56 %, (c) 23.65 %, (d) 22.65 %, The average atomic mass of neon based on following data, is :, Isotope, Relative abundance, 20 Ne, 0.9051, 21 Ne, 0.0027, 22 Ne, 0.0922, (a) 0.33 u, (b) 20.187 u, (c) 6.729 u, (d) 18.058 u, , 69., , 70., , What is the average atomic mass of bromine from the, following data : (abundance is in %), Isotope, Mass, Abundance, 79Br, 78.9183361, 50.69, 81Br, 80.916289, 49.31, (a) 79.9, (b) 76.6, (c) 75.9, (d) 69.9, What is the mass of an atom of oxygen (in gm)?, (a) 2.656 × 10–23, (b) 1.567 × 10–22, –22, (c) 2.0 × 10, (d) 3.5 × 10–23, If the mass of the one atom is found to be 2.324784×10–23g,, then this atom can be ?, (a) Oxygen, (b) Carbon, (c) Fluorine, (d) Nitrogen, What is the mass of 1 molecule of CO., (a) 2.325 × 10–23, (b) 4.65 × 10–23, –23, (c) 3.732 × 10, (d) 2.895 × 10–23, Calculate the volume at STP occupied by 240 gm of SO2., (a) 64, (b) 84, (c) 59, (d) 73, At S.T.P. the density of CCl4 vapours in g/L will be nearest, to:, (a) 6.87, (b) 3.42, (c) 10.26, (d) 4.57, The number of gram molecules of oxygen in 6.02 × 1024, CO molecules is, (a) 10 gm molecules, (b) 5 gm molecules, (c) 1 gm molecules, (d) 0.5 gm molelcules, The number of oxygen atoms in 4.4 g of CO2 is, (a) 1.2 × 1023, (b) 6 × 1022, 23, (c) 6 × 10, (d) 12 × 1023, Which has maximum number of molecules?, (a) 7 gm N2, (b) 2 gm H2, (c) 16 gm NO2, (d) 16 gm O2, Number of atoms in 558.5 gram Fe (at. wt. of Fe = 55.85, g mol–1) is, (a) twice that in 60 g carbon, (b) 6.023 1022, (c) half that in 8 g He, (d) 558.5 6.023 1023, The number of molecules in 16 g of methane is, (a) 3.0 × 1023, , (b), , 16, 10 23, 6.02, , (c) 6.023 × 1023, , (d), , 16, 10 23, 3.0, , Number of g of oxygen in 32.2 g Na2SO4.10 H2O is, (a) 20.8, (b) 2.24, (c) 22.4, (d) 2.08
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SOME BASIC CONCEPTS OF CHEMISTRY, , 71., , 72., , 73., , 74., , 75., , 76., , 77., , 78., , 79., , 80., , 81., , 82., , The number of moles of oxygen in one litre of air containing, 21% oxygen by volume, under standard conditions are, (a) 0.0093 mole, (b) 0.21 mole, (c) 2.10 mole, (d) 0.186 mole, The number of molecules in 8.96 litre of a gas at 0ºC and 1, atm. pressure is approximately, (a) 6.023 × 1023, (b) 12.04 × 1023, 23, (c) 18.06 × 10, (d) 24.08 × 1022, The mass of a molecule of water is, (a) 3 × 10–25 kg, (b) 3 × 10–26 kg, –26, (c) 1.5 × 10 kg, (d) 2.5 × 10–26 kg, One mole of CO2 contains :, (a) 3 g atoms of CO2, (b) 18.1 × 1023 molecules of CO2, (c) 6.02 × 1023 atoms of O, (d) 6.02 × 1023 atoms of C, Volume of a gas at NTP is 1.12 × 10–7cm3. The number of, molecules in it is :, (a) 3.01 × 1012, (b) 3.01 × 1024, 23, (c) 3.01 × 10, (d) 3.01 × 1020, How many atoms are contained in one mole of sucrose, (C12 H22O11)?, (a) 20 × 6.02 × 1023 atoms/mol, (b) 45 × 6.02 × 1023 atoms/mol, (c) 5 × 6.02 × 1023 atoms/mol, (d) None of these, One litre oxygen gas at S.T.P will weigh :, (a) 1.43 g, (b) 2.24 g, (c) 11.2 g, (d) 22.4 g, Number of moles of NaOH present in 2 litre of 0.5 M NaOH, is :, (a) 1.5, (b) 2.0, (c) 1.0, (d) 2.5, O2, N2 are present in the ratio of 1 : 4 by weight. The ratio of, number of molecules is :, (a) 7 : 32, (b) 1 : 4, (c) 2 : 1, (d) 4 : 1, How many moles of Al2(SO4)3 would be in 50 g of the, substance ?, (a) 0.083 mole, (b) 0.952 mole, (c) 0.481 mole, (d) 0.140 mole, The mass of 1 mole of electrons is, (a) 9.1 × 10–28g, (b) 1.008 mg, (c) 0.55 mg, (d) 9.1 × 10–27 g, 10 g of hydrogen and 64 g of oxygen were filled in a steel, vessel and exploded. Amount of water produced in this, reaction will be:, (a) 3 mol, (b) 4 mol, (c) 1 mol, (d) 2 mol, , 5, , 83. Which has the maximum number of molecules among the, following ?, (a) 44 g CO2, (b) 48 g O3, (c) 8 g H2, (d) 64 g SO2, 84. The weight of one molecule of a compound C 60 H122 is, , 85, , 20, , (a), , 1.2 10, , (c), , 5.025 10 23 gram, , gram, , (b) 1.4 10, (d), , 21, , gram, , 6.023 10 23 gram, , The simplest formula of a compound containing 50% of, element X (atomic mass 10) and 50% of element Y (atomic, mass 20) is, (a) XY, , (b) XY3, , (c) X2Y, , (d), , X2Y3, , 86. Empirical formula of hydrocarbon containing 80% carbon, and 20% hydrogen is :, (a) CH3, , (b) CH4, , (c) CH, , (d) CH2, , 87. The empirical formula of a compound is CH2. One mole of, this compound has a mass of 42 grams. Its molecular formula, is :, (a) C3H6, , (b) C3H8, , (c) CH2, , (d) C2H2, , 88. A compound contains 54.55 % carbon, 9.09% hydrogen ,, 36.36% oxygen. The empirical formula of this compound is, (a), , C3H 5O, , (b), , C 4 H 8O 2, , (c), , C 2 H 4O 2, , (d), , C 2 H 4O, , 89. In a hydrocarbon, mass ratio of hydrogen and carbon is, 1:3, the empirical formula of hydrocarbon is, (a) CH4, , (b) CH2, , (c) C2H, (d) CH3, 90. An organic compound contains carbon, hydrogen and, oxygen. Its elemental analysis gave C, 38.71% and H, 9.67%., The empirical formula of the compound would be :, (a) CH3O, (b) CH2O, (c) CHO, (d) CH4O, 91. A hydrocarbon is composed of 75% carbon. The empirical, formula of the compound is, (a) CH2, (b) CH3, (c) C2H5, (d) CH4, 92. 12 gm of Mg (atomic mass 24) will react completely with, hydrochloric acid to give, (a) One mol of H2, (b) 1/2 mol of H2, (c) 2/3 mol of O2, (d) both 1/2 mol of H2 and 1/2 mol of O2
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EBD_7207, SOME BASIC CONCEPTS OF CHEMISTRY, , 6, , 93., , 94., , 95., , 96., , 20.0 kg of N2(g) and 3.0 kg of H2(g) are mixed to produce, NH3(g). The amount of NH3(g) formed is, (a) 17 kg, (b) 34 kg, (c) 20 kg, (d) 3 kg, 20.0 kg of H2(g) and 32 kg of O2(g) are reacted to produce, H2O(l). The amount of H2O (l) formed after completion of, reaction is, (a) 62 kg, (b) 38 kg, (c) 42 kg, (d) 72 kg, What is the weight of oxygen required for the complete, combustion of 2.8 kg of ethylene ?, (a) 2.8 kg, (b) 6.4 kg, (c) 9.6 kg, (d) 96 kg, In the reaction, 4 NH3 ( g ) 5O2 ( g ) 4 NO( g ) 6H2 O(l ), When 1 mole of ammonia and 1 mole of O2 are made to react, to completion,, (a) 1.0 mole of H2O is produced, (b) 1.0 mole of NO will be produced, (c) all the oxygen will be consumed, (d) all the ammonia will be consumed, , 97., , What is the molarity of 0.2N Na 2 CO 3 solution?, (a) 0.1 M, (c) 0.4 M, , 98., , (b) 0 M, (d) 0.2 M, , The molar solution of H 2SO 4 is equal to :, , (a) N/2 solution, (b) N solution, (c) 2N solution, (d) 3N solution, 99. Volume of water needed to mix with 10 mL 10N HNO3 to get, 0.1 N HNO3 is :, (a) 1000 mL, (b) 990 mL, (c) 1010 mL, (d) 10 mL, 100. One kilogram of a sea water sample contains 6 mg of, dissolved O2. The concentration of O2 in the sample in, ppm is, (a) 0.6, (b) 6.0, (c) 60.0, (d) 16.0, 101. A 5 molar solution of H2SO4 is diluted from 1 litre to a, volume of 10 litres, the normality of the solution will be :, (a) 1N, (b) 0.1N, (c) 5N, (d) 0.5N, 102. With increase of temperature, which of these changes?, (a) Molality, (b) Weight fraction of solute, (c) Molarity, (d) Mole fraction, 20, 103. 6.02 × 10 molecules of urea are present in 100 ml of its, solution. The concentration of urea solution is, (a) 0.02 M, (b) 0.01 M, (c) 0.001 M, (d) 0.1 M, (Avogadro constant, NA = 6.02 × 1023 mol–1), , 104. Two solutions of a substance (non electrolyte) are mixed in, the following manner. 480 ml of 1.5 M first solution + 520 ml, of 1.2 M second solution. What is the molarity of the final, mixture?, (a) 2.70 M, (b) 1.344 M, (c) 1.50 M, (d) 1.20 M, , STATEMENT TYPE QUESTIONS, 105. Which of the following statements are correct ?, (i) Both solids and liquids have definite volume., (ii) Both liquids and gases do not have definite shape., (iii) Both solids and gases take the shape of the container., (a) (i) and (iii), (b) (ii) and (iii), (c) (i) and (ii), (d) (i), (ii) and (iii), 106. Choose correct option based on following statements. Here, T stands for true statement and F for false statement., (i) Homogeneous mixture has uniform composition, throughout., (ii) All components of a heterogeneous mixture are, observable to naked eyes., (iii) All solutions are homogeneous in nature., (iv) Air is an example of heterogeneous mixture., (a) TTFF, (b) TFTF, (c) FFTT, (d) TFFF, 107. Read the following and choose the incorrect statements., (i) Both weight and mass are same quantities used for, measurement of amount of matter present in a, substance, (ii) Mass and weight of a substance vary from one place, to another due to change in gravity., (iii) SI unit of mass is kilogram and while SI unit of weight, is gram., (a) (i) and (iii), (b) (ii) and (iii), (c) (i) and (ii), (d) All of these, 108. Moon takes 27.3 days to complete one orbit around the, Earth. Now read the following statements and choose the, correct code. Here T is for true statement and F is for ‘False, statement’., (i) Moon takes 655.2 hours to complete one orbit around, the Earth., (ii) Moon takes 39312 seconds to complete one orbit, around the earth., (iii) Moon takes 1638 minutes to complete one orbit around, the Earth., (a) F T F, (b) T T T, (c) T F F, (d) T F T
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SOME BASIC CONCEPTS OF CHEMISTRY, , 109. Give the correct order of initials T or F for following, statements. Use T if statement is true and F if it is false., (i) Gay-Lussac’s law of gaseous volumes is actually the, law of definite proportion by volume., (ii) Law of conservation of mass is true for physical, change, but not for chemical change., (iii) The percentage of oxygen in H2O2 is different from, that in H2 O. Hence, it violates law of definite, proportions., (iv) Fixed mass of A reacts with two different masses of B, (say x and y), then the ratio of x/y can be any positive, integer., (v) At STP, 5 mL of N2 and H2 have different no. of, molecules., (a) TTFTF, (b) FTTFT, (c) TFFTF, (d) TFTTF, 110. Consider the following statements., (i) Atoms of H, O, N and C have identical properties but, different mass., (ii) Matter is divisible into atoms which are further, indivisible., (iii) The ratio of N: H in NH3 is 1 : 3 and N : O in nitric oxide, is 2 : 1., (iv) Dalton’s atomic theory support law of conservation, of mass., Which of the following pairs of statements is true according, to Dalton’s atomic theory ?, (a) (i) and (ii), (b) (ii) and (iii), (c) (ii) and (iv), (d) (i) and (iv), 111. Choose the correct option based on following statements., Here ‘T’ stands for true and ‘F’ stands for false statement., (i) Molecular mass of cane sugar (C12H22O11) is 182 amu., (ii) 1 mole of cane sugar contains 6.022 × 1023 molecules, of cane sugar., (iii) 34.20 g of cane sugar contains 6.022 × 1021 molecules, of cane sugar., (a) TTF, (b) TFT, (c) FTF, (d) FTT, , MATCHING TYPE QUESTIONS, 112. Match the items of Column I, II and III appropriately and, choose the correct option from the codes given below., Column I, Column II, Column III, (Multiple), (Prefix), (Symbol), (A) 10–15, (p) Kilo, (i) m, –3, (B) 10, (q) yotta, (ii) f, (C) 103, (r) milli, (iii) k, (D) 1024, (s) femto, (iv) Y, (a) A – (s), (ii); B – (r), (i); C – (p), (iii); D – (q), (iv), (b) A – (p), (ii); B – (q), (iii); C – (r), (i); D – (s), (iv), (c) A – (q), (iv); B – (p), (ii); C – (p), (i); D – (r), (iii), (d) A – (r), (iii); B – (p), (ii); C – (s), (i); D – (q), (iv), , 7, , 113. Match the columns, Column-I, Column-II, (Number), (Significant figures), (A) 29900., (p) 2, (B) 290, (q) 1, (C) 1.23 × 1.331, (r) 4, (D) 20.00, (s) 3, (E) 2.783 – 1, (t) 5, (a) A – (r), B – (q), C – (t), D – (p), E – (s), (b) A – (t), B – (p), C – (s), D – (r), E – (q), (c) A – (p), B – (t), C – (s), D – (r), E – (q), (d) A – (t), B – (s), C – (r), D – (q), E – (p), 114. Match the columns, Column-I, Column-II, (Laws of chemical, (Scientist), combinations), (A) Law of definite, (p) Antoine Lavoisier, proportions, (B) Law of multiple, (q) Gay Lussac, proportions, (C) Law of conservation, (r) Dalton, of mass, (D) Law of gaseous, (s) Joseph Proust, volumes, (a) A – (s), B – (r), C – (p), D – (q), (b) A – (p), B – (r), C – (s), D – (q), (c) A – (r), B – (p), C – (s), D – (q), (d) A – (q), B – (s), C – (r), D – (p), 115. Match the columns, Column-I, Column-II, (A) C6H5NH2, (p) 84, (B) C6H6, (q) 100, (C) C6H12, (r) 93, (D) CaCO3, (s) 78, (a) A – (p), B – (r), C – (q), D – (s), (b) A – (r), B – (s), C – (p), D – (q), (c) A – (r), B – (p), C – (q), D – (s), (d) A – (r), B – (q), C – (s), D – (p), 116. Match the columns., Column-I, Column-II, (A) 88 g of CO2, (p) 0.25 mol, (B) 6.022 × 1023 molecules (q) 2 mol, of H2O, (C) 5.6 litres of O2 at STP (r) 1 mol, (D) 96 g of O2, (s) 6.022 × 1023 molecules, (E) 1 mol of any gas, (t) 3 mol, (a) A – (q), B – (r), C – (p), D – (t), E – (s), (b) A – (r), B – (q), C – (p), D – (t), E – (s), (c) A – (q), B – (p), C – (r), D – (t), E – (s), (d) A – (q), B – (r), C – (p), D – (s), E – (t)
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EBD_7207, 8, , 117. Match the mass of elements given in Column I with the, number of moles given in Column II and mark the appropriate, choice. Choose the correct codes formt he options given, below., Column-I, Column-II, (A) 28 g of He, (p) 2 moles, (B) 46 g of Na, (q) 7 moles, (C) 60 g of Ca, (r) 1 mole, (D) 27 g of Al, (s) 1.5 mole, (a) A – (s), B – (r), C – (q), D – (p), (b) A – (p), B – (r), C – (q), D – (s), (c) A – (r), B – (q), C – (p), D – (s), (d) A – (q), B – (p), C – (s), D – (r), 118. Match the columns., Column-I, Column-II, (Physical quantity), (Unit), (A) Molarity, (p) mol, (B) Mole fraction, (q) Unitless, (C) Mole, (r) mol L–1, (D) Molality, (s) mol kg–1, (a) A – (r), B – (q), C – (s), D – (p), (b) A – (r), B – (p), C – (q), D – (s), (c) A – (r), B – (q), C – (p), D – (s), (d) A – (q), B – (r), C – (p), D – (s), , ASSERTION-REASON TYPE QUESTIONS, Directions : Each of these questions contain two statements,, Assertion and Reason. Each of these questions also has four, alternative choices, only one of which is the correct answer. You, have to select one of the codes (a), (b), (c) and (d) given below., (a) Assertion is correct, reason is correct; reason is a correct, explanation for assertion., (b) Assertion is correct, reason is correct; reason is not a, correct explanation for assertion, (c) Assertion is correct, reason is incorrect, (d) Assertion is incorrect, reason is correct., 119. Assertion : Significant figures for 0.200 is 3 whereas for, 200 it is 1., Reason : Zero at the end or right of a number are significant, provided they are not on the right side of the decimal point., 120. Assertion : 1.231 has three significant figures., Reason : All numbers right to the decimal point are, significant., 121. Assertion : One atomic mass unit is defined as one twelfth, of the mass of one carbon - 12 atom., Reason : Carbon-12 isotope is the most abundunt isotope, of carbon and has been chosen as standard., , SOME BASIC CONCEPTS OF CHEMISTRY, , 122. Assertion : Volume of a gas is inversely proportional to the, number of moles of gas., Reason : The ratio by volume of gaseous reactants and, products is in agreement with their mole ratio., 123. Assertion : Equal moles of different substances contain, same number of constituent particles., Reason : Equal weights of different substances contain, the same number of constituent particles., 124. Assertion : The empirical mass of ethene is half of its, molecular mass., Reason : The empirical formula represents the simplest, whole number ratio of various atoms present in a, compound., , CRITICAL THINKING TYPE QUESTIONS, 125. What are the significant figure(s) in a broken “ruler”, show below?, 0.0, 1.0, 2.0, (B) 2, (A) 1, (C) 3, (D) 0, (a) A, B and C, (b) A, B, D, (c) A only, (d) A and B, 126. Which one of the following sets of compounds correctly, illustrate the law of reciprocal proportions?, (a) P2O3, PH3, H2O, (b) P2O5, PH3, H2O, (c) N2O5, NH3, H2O, (d) N2O, NH3, H2O, 127. If we consider that 1/6, in place of 1/12, mass of carbon, atom is taken to be the relative atomic mass unit, the mass, of one mole of a substance will, (a) decrease twice, (b) increase two fold, (c) remain unchanged, , (d) be a function of the molecular mass of the substance, 128. The maximum number of molecules are present in, (a) 15 L of H2 gas at STP (b) 5 L of N2 gas at STP, (c) 0.5 g of H2 gas, (d) 10 g of O2 gas, 129. How many moles of magnesium phosphate, Mg3(PO4)2 will, contain 0.25 mole of oxygen atoms?, (a) 1.25 × 10–2, (b) 2.5 × 10–2, (c) 0.02, (d) 3.125 × 10–2, 130. Volume occupied by one molecule of water, (density = 1 g cm–3) is : ], (a) 9.0 × 10–23 cm3, (b) 6.023 × 10– 23 cm3, –23, 3, (c) 3.0 × 10 cm, (d) 5.5 × 10– 23 cm3, 131. The number of atoms in 0.1 mol of a triatomic gas is :, (NA = 6.02 ×1023 mol–1), (a) 6.026 × 1022, (b) 1.806 × 1023, (c) 3.600 × 1023, (d) 1.800 × 1022
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SOME BASIC CONCEPTS OF CHEMISTRY, , 132. 1 c.c. N2O at NTP contains :, (a), , 1.8, 10 22 atoms, 224, , (b), , 6.02, 10 23 molecules, 22400, , (c), , 1.32, 10 23 electrons, 224, , (d) All of the above, 133. How much time (in hours) would it take to distribute one, Avogadro number of wheat grains if 10 20 grains are, distributed each second ?, (a) 0.1673, (b) 1.673, (c) 16.73, (d) 167.3, 134. Arrange the following in the order of increasing mass (atomic, mass: O = 16, Cu = 63, N = 14), I. one atom of oxygen, II. one atom of nitrogen, III. 1 × 10–10 mole of oxygen, IV. 1 × 10–10 mole of copper, (a) II < I < III < IV, (b) I < II < III < IV, (c) III < II < IV < I, (d) IV < II < III < I, 135. If 1.5 moles of oxygen combines with Al to form Al 2O3,, the mass of Al in g [Atomic mass of Al = 27] used in the, reaction is, (a) 2.7, (b) 54, (c) 40.5, (d) 81, 136. Which one of the following is the lightest?, (a) 0.2 mole of hydrogen gas, (b) 6.023 × 1022 molecules of nitrogen, (c) 0.1 g of silver, (d) 0.1 mole of oxygen gas, 137. In a compound C, H and N atoms are present in 9 : 1 : 3.5 by, weight. Molecular weight of compound is 108. Molecular, formula of compound is, (a) C2H6N2, (b) C3H4N, (c) C6H8N2, (d) C9H12N3., 138. The empirical formula of an acid is CH2O2, the probable, molecular formula of acid may be :, (a) C3H6O4, (b) CH2O, (c) CH2O2, (d) C2H4O2, 139. A gaseous hydrocarbon gives upon combustion 0.72 g of, water and 3.08 g. of CO2. The empirical formula of the, hydrocarbon is :, (a) C2H4, (b) C3H4, (c) C6H5, (d) C7H8, , 9, , 140. Which of the following is the correct empirical and, molecular formulae of a compound, if the molecular mass, of a compound is 80 and compound contains 60% of C,, 5% of H and 35% of N ?, (a) C2H2N ; C4H4N2, (b) C3H4N2 ; C6H8N4, (c) C2H4N2 ; C4H8N4, (d) C2H2N ; C2H2N, 141. Which of the following is the correct empirical and, molecular formulae of a compound, if the molecular mass, of a compound is 93 and compound containing 77.43% of, C, 7.53% of H and 15.05% of N ?, (a) C3H3.5N1.5 and C6H7N, (b) C6H7N and C6H7N, (c) C3H3N and C6H7N, (d) C3H3N and C6H6N2, 142. Liquid benzene (C6H6) burns in oxygen according to the, equation 2C 6 H 6 (l ) 15O 2 ( g ), 12 CO 2 ( g ) 6 H 2 O ( g ), How many litres of O2 at STP are needed to complete the, combustion of 39 g of liquid benzene?(Mol. wt. of O2 = 32,, C6H6 = 78), (a) 74 L, (b) 11.2 L, (c) 22.4 L, (d) 84 L, 143. Assuming fully decomposed, the volume of CO2 released at, STP on heating 9.85 g of BaCO3 (Atomic mass, Ba = 137), will be, (a) 2.24 L, (b) 4.96 L, (c) 1.12 L, (d) 0.84 L, 144. The mass of BaCO3 produced when excess CO2 is bubbled, through a solution of 0.205 mol Ba(OH)2 is :, (a) 81 g, (b) 40.5 g, (c) 20.25 g, (d) 162 g, 145. For the reaction Fe2O3 + 3CO2 2Fe + 3CO2, the volume, of carbon monoxide required to reduce one mole of ferric, oxide is, (a) 67.2 dm3, (b) 11.2 dm3, (c) 22.4 dm3, (d) 44.8 dm3, 146. How many moles of lead (II) chloride will be formed from a, reaction between 6.5 g of PbO and 3.2 g of HCl ?, (a) 0.044, (b) 0.333, (c) 0.011, (d) 0.029, 147. Fat is an important source of energy and water, this is, important for the desert animals like camel which store fat, in its hump and provide water and energy. How many, grams and moles of H2O are produced from the, combustion of fat C57H110O6 from 450 gram of fat stored, in hump of camel ?, 163, O2, 2, (a) 500.56 , 27.80, (c) 580, 25.0, C57 H110O6, , 57CO 2 55H 2O, , (b) 450, 26.80, (d) 400, 26.6
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SOME BASIC CONCEPTS OF CHEMISTRY, , 11, , FACT / DEFINITION TYPE QUESTIONS, 1., 2., 3., , 4., 5., 6., 7., 8., , 9., , 10., 11., 12., 13., 14., , (c) By sublimation since I2 sublimes., (d) It forms the basis of gravity separation., (d) Candle is a heterogeneous mixture of wax and threads., Copper is an element while bottled water and table salt, are compounds., (c) Glucose is a pure substance hence its constituents, cannot be separated by simple physical method., (a) In case of B, none of O2 and H2 is a compound since, compound consist of two or more different atoms., (c), (d) Silver is an element, water is a compound and air is a, mixture., (c) Attraction between particles in solid is maximum and, hence their movement is minimum amongst the, phases., Attraction between particles in gases is minimum and, hence their movements is maximum amongst the three, phases., Attraction between the particles and their movements, in liquids is intermediate i.e., between solids and gases., (c) A mixture may contain any number of components in, any ratio., e.g. air is a mixture of various gases., (d) It is a solution and is a mixture of sucrose and water., (b) One fermi is 10–13 cm., (c) Exa = 1018, (d) 1 zepto = 10–21, (a) Joule is the unit of work and Pascal is unit of pressure., JPa 1, , 15., , 17., , Work, Pressure, , Nm, Nm, , 2, , 10 6, , kg = 103 kg = 1000 kg, 103, (b) Weight of normal adult man = 65 kg, Weight of 1 m3 of water is highest., (c) Density of Hg = 13.6 g cm–3, Volume of Hg = 10 L = 10 × 1000 = 104 cm3, Weight of Hg = 13.6 × 104 = 136000 g = 136 kg, (d) Smallest and largest amount of energy respectively, are eV and L-atm., 1 eV = 1.6 × 10–19 J, 1L -atm = 101.325 J, (c), , 5, 5, F 32, 41 32 = 5°C, 9, 9, It will be cold., (a) Out of two 3.929 g is more accurate and will be reported, as 3.93 after rounding off., (b), (a) 0.00016 can be written as 1.6 × 10–4 in scientic notation., (d) Both Y and X are neither precise nor accurate as the, two values in each of them are not close. With respect, to X & Y, the values of Z are close & agree with the, true value. Hence, both precise & accurate., (c) On calculation we find, , 20. (c), 21., 22., 23., 24., , 25., , 26., , 27., , m3, , (a) 1 m3 of water 106 cm3 of water, Mass of 106 cm3 water, = 106 cm3 × 1 g cm3 ( density of H2O = 1 g cm3), = 106 g =, , 16., , J, Pa, , 18. (d) Litre (L) is not an SI unit. It is used for measurement of, volume of liquids., 19. (c) yocto = 10–24, , 28., 29., , 30., , 31., , C, , (29.2 – 20.2)(1.79 ´105 ), = 1.17×106, 1.37, As the least precise number contains 3 significant, figures therefore, answers should also contains 3, significant figures., (d) We know that all non-zero digits are significant and, the zeros at the beginning of a number are not, significant. Therefore number 161 cm, 0.161 cm and, 0.0161cm have 3, 3 and 3 significant figures, respectively., (b) Given P = 0.0030m, Q = 2.40m & R = 3000m. In P(0.0030), initial zeros after the decimal point are not significant., Therefore, significant figures in P(0.0030) are 2., Similarly in Q (2.40) significant figures are 3 as in this, case final zero is significant. In R = (3000) all the zeros, are significant hence, in R significant figures are 4, because they come from a measurement., (a), (c) If zero is used to locate the decimal point it is, considered as a significant figure. In 50.000 all zero are, significant., (a) Sum of the figures 29.4406, 3.2 and 2.25 is 34.8906. The, sum should be reported to the first place of decimal as, 3.2 has only one decimal place. After rounding off the, sum is 34.9. Hence number of significant figures is, three., (d) 10.3106 g has 6 significant figures. Since all, non-zero digits are significant and a zero becomes, significant if it appears between two non-zero digits., , 32. (b), , 43.0 0.0243, = 0.7683088, 0.340 4, The least precise term has two significant figures, (leaving the exact number). Hence after rounding off, correct answer is 0.77.
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EBD_7207, SOME BASIC CONCEPTS OF CHEMISTRY, , 12, , 54., , 37., , n + m = p + q by law of conservation of mass., (a) CaCO3 CaO + CO2, 20 g, 8.8 g 11.2 g, mass of reactant = mass of products = 20g., Hence the law of conservation of mass is obeyed., (a), 38. (a), , (b) The number of molecules of N2 and X are same. Hence, they must have the same molecular weights., X is CO., (c), (c) In law of reciprocal proportions, the two elements, combining with the third element, must combine with, each other in the same ratio or multiple of that Ratio of, S and O when combine with C is 2 : 1. Ratio of S and O is, SO2 is 1 : 1, (d) Avogadro’s law is independent of the reactive or, unreactive nature of the gases., According to Avogadro’s law equal volumes of gases, at the same temperature and pressure should contain, equal number of molecules., (c), , 55., , (b) 1 amu =, , 39., , (b), , 56., , 33., , 34., , 35., 36., , 50., , (a) 2.6 has two significant figures., 0.260 has three significant figures., 0.002600 has four significant figures., 2.6000 has five significant figures., Energy, (d), which can be shown, volume, Pressure, = Force, area, X, Y, (b), ng mg, , Work (energy/di stance), Area, R S, , Energy, Volume, , 10.0g, , Residue CO 2, 12.0g, , 2.2g, , Mass of reactants = 4.2 + 10.0 = 14.2 g, Mass of products = 12.0 + 2.2 = 14.2 g, Hence, given reaction illustrate law of conservation, of mass., 40., , 41., 42., , 43., 44., , 45., 46., 47., , 48., 49., , (c) I experiment :, , 53., , pg qg, , NaHCO3 CH3COOH, 4.2g, , 51., 52., , mass of H 2 combined, mass of O 2 combined, , 4, 32, , 1, 8, , mass of H 2 combined 50 1, II experiment :, mass of O 2 combined 400 8, Hence both law of conservation of mass and constant, composition is obeyed., (c) The H : O ratio in water is fixed, irrespective of its, source. Hence it is law of constant composition., (a) Constant proportions according to which a pure, chemical compound always contains same elements, combined together in the same definite proportion of, weight., (c), (d), SnCl2, SnCl4, 119 : 2 × 35.5, 119 : 4 × 35.5, Chlorine ratio in both compounds is, = 2 × 35.5 : 4 × 35.5 = 1 : 2, (d) In CuO and Cu2O the O : Cu is 1 : 1 and 1 : 2 respectively., This is law of multiple proportion., (c), (b) Law of multiple proportion. As the ratio of oxygen, which combine with fix weights of 1 g of nitrogen bears, a simple whole number ratio, 0.57 : 1 : 12 : 1.7031 : 2 : 3, (b), (c) According to Avogadro's law "equal volumes of all, gases contain equal number of molecules under similar, conditions of temperature and pressure". Thus if 1 L, of one gas contains N molecules, 2 L of any other gas, under the same conditions of temperature and pressure, will contain 2N molecules., , 57., , 1, of the mass of C-12., 12, (b) The modern atomic weight scale is based on C12., , (d) Molecular weight of ZnSO 4 .7H 2O, = 65 + 32 + (4 × 16) + 7(2 × 1 + 16) = 287., percentage, , 58., , 59., 60., , 61., , 62., , mass, , zinc, , (d), , 2.824784 10, , 23, , = 14 amu, 1.66056 10–24, Where 1.66056 × 10–24 is equal to one atomic mass, (amu), (b) Gram molecular weight of CO = 12 + 16 = 28 g, 6.023 × 1023 molecules of CO weight 28 g, 28, 6.02 1023, , 4.65 10 23 g, , (b) Molecular weight of SO2 = 32 + 2 × 16 = 64, 64 g of SO2 occupies 22.4 litre at STP, 22.4, 240 84 litre at STP, 64, (a) 1 mol CCl4 vapour = 12 + 4 × 35.5, = 154 g 22.4 L at STP, , 240 g of SO2 occupies =, , 64., , 154, gL 1 6.875 gL 1, 22.4, (b) 6.02 × 1023 molecules of CO =1mole of CO, 6.02 × 1024 CO molecules = 10 moles CO, = 10 g atoms of O = 5 g molecules of O2, , Density =, , 65., , (Zn), , 65, 100 22.65%, 287, (b) Average atomic mass of neon, = 20 × 0.9051 + 21 × 0.0027 + 22 × 0.0922, = 20.187 u, (a) (78.9183361) × (0.5069) + (80.916289) × (0.4931), (a) Mass of oxygen atom is 15.995 amu, becasue 1 amu, = 1.66056 × 10–24 g, hence 15.995 × value of 1 amu, give the value equal to option (a)., , 1 molecule of CO weighs =, 63., , of
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SOME BASIC CONCEPTS OF CHEMISTRY, , 4.4, = 0.1 mol CO2 (mol. wt. of CO2 = 44), 44, 22, = 6 × 10 molecules = 2 × 6 × 1022 atoms of O., 2g of H2 means one mole of H2, hence contains, 6.023 × 1023 molecules. Others have less than one mole,, so have less no. of molecules., 558.5, Fe (no. of moles) =, = 10 moles = 10NA atoms., 55.85, No. of moles in 60 g of C = 60/12 = 5 moles = 5NA, atoms., 16 g CH4 is 1 mol. Hence number of molecules, = Avogadro number = 6.023 × 1023., M. Wt of Na2SO4.10H2O is 322 g which contains 224, g oxygen., 32.2 g will contain 22.4 g oxygen., 21% of 1 litre is 0.21 litre., 22.4 litres = 1 mole at STP, , 13, , 66., , (a) 4.4 g CO2 =, , No. of molecules of O2 =, , 67., , (b), , No. of molecules of N2 =, , 68., , (a), , 69., , (c), , 70., , (c), , 71., , (a), , 0.21, 0.0093 mol, 22.4, 72. (d) At S.T.P. 22.4 litre of gas contains 6.023 × 1023, molecules, molecules in 8.96 litre of gas, , 4, 28, 1 4, :, Ratio of no. of molecules =, 32 28, , 6.023 10 23 8.96, 24.08 10 22, 22.4, (b) Mass of one molecule of Water, , 73., , 74., , 75., , 18, , 3 10 23 g 3 10 26 Kg, 6.023 10 23, (d) 1 molecule of CO2 has one atom of C and two atoms of, oxygen., 1 mole of CO2 has = 6.02 × 1023 atoms of C, = 2 × 6.02 × 1023 atoms of O, (a) Given, V = 1.12 × 10–7 cm3, 22400 cm3 at NTP = 6.02 × 1023 molecules, =, , 6.02 1023, 1.12 10 7, 22400, = 3.01 × 1012 molecules., (b) Total atoms in 1 molecule of C12H22O11, = 12 + 22 + 11 = 45, Total atoms in 1 mole of C12H22O11, = 45 × 6.02 × 1023 atoms/mol., , 81. (c) Mass of 1 electron = 9.11 × 10–28 g, Mass of 1 mole (6.02 × 1023) electrons, = 9.11 × 10–28 × 6.02 × 1023g, = 55 × 10–5 g = 55 × 10–5 × 103 mg = 0.55 mg., 1, O, H2 O, 82. (b), H2, +, 2 2, 10g, , 77., , (a) 22.4 L of O 2 at STP = 32 g, , 78., , 32, 1 = 1.428 g = 1.43 g, 22.4, (c) Given V = 2 L, Molarity = 0.5M, Moles = ?, , 1 L of O2 at STP =, , Molarity, , 79., , Moles, No. of moles of solute or, 0.5, 2, V of solution in L, , Moles = 2 × 0.5 = 1.0, (a) Let mass of O2 = 1 g, Mass of N2 = 4g, , 64g, , 64, 2 mol, 32, , 10, 5 mol, 2, , In this reaction oxygen is the limiting agent. Hence, amount of H2O produced depends on the amount of, O2 taken, 0.5 mole of O2 gives H2O = 1 mol, 2 mole of O2 gives H2O = 4 mol, 83. (c), No. of molecules, 44, 1, NA, Moles of CO2 =, 44, 48, 1, 48, 8, 4, Moles of H2 =, 2, , Moles of O3 =, , NA, 4NA, , 64, 1, NA, 64, (b) Molecular weight of C60H122 = (12 × 60) + 122 = 842., Therefore weight of one molecule, , Moles of SO2 =, , 84., , 1.12 × 10–7 cm3 at NTP =, , 76., , 1, 1, : = 7 : 32, 32 7, , weight, 50, =, = 0.14 mole, 342, mol. wt ., , 80. (d) No. of moles =, , 0.21 litre =, , =, , 1, 32, , =, , Molecular weight of C 60 H 122, Avagadro' s number, 842, , 85., , 1.4 10 21 g, 6.023 10 23, (c) 50% of X (Atomic mass 10), 50% of Y (Atomic mass 20)., , Relative number of atoms of X = 50 5 and than, 10, 50, 2.5, Y=, 20, Simple Ratio 2 : 1. Formula X2Y, 86. (a) Element % Atomic, Relative, Simple ratio, mass no.of atoms, of atoms, C, , 80, , 12, , H, , 20, , 1, , Empirical formula is CH3, , 80, 12, 20, 1, , 6.66, 20.0, , 6.66, 6.66, 20.0, 6.66, , 1, 3
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EBD_7207, SOME BASIC CONCEPTS OF CHEMISTRY, , 14, , 87., , 88., , 89., , 90., , (a) Empirical formula of compound = CH2, Molecular mass of the compound = 42, n = 42/14 = 3, Hence molecular formula = C3H6, (d) C 54.55, 54.55/12 = 4.5 4.5/2.27 = 2, H 9.099.09/1= 9.09 9.09/2.27 = 4, O 36.36, 36.36/16 = 2.27 2.27/2.27= 1, Hence empirical formula of the compound = C2H4O, (a) Mass ratio of H : C = 1 : 12, However, given mass ratio of H : C = 1 : 3, Therefore, for every C atom, there are 4 H atoms, hence, empirical formula = CH4, (a), , Element, , Percentage, , Atomic, weight, , C, , 38.71, , 12, , H, , 9.67, , 1, , O, , 100, (38.71 9.67), 51.62, , 16, , Atomic, ratio, , Simple, ratio, , 38.71, 3.23, 1, 3.23, 3.23, 12, 9.67, 9.67, 9.67, 3, 3.23, 1, 51.62, 3.23, 3.23, 1, 16, 3.23, , Thus empirical formula is CH3O., 91., , (d), , Element %, , At., , Rel.No. of, , (b), , C, , 25, , Mg, 1 mole, 1, mole, 2, , 93., , 94., , 95., , 96., , 2 HCl, , MgCl 2, , = 0.2, , 4, , H2, 1 mole, 1, mole (12g of Mg, 2, , 1, mol ), 2, , (a) We know that, N2 + 3H2, 2NH3, 28 g 6 g, 34 g, 14 g 3 g, 17 g, Here given H2 is 3 kg and N2 is 20 kg but 3 kg of H2 can, only react with 14 g of N2 and thus the obtained NH3, will be of 17 kg., (d) 2H2 + O2, 2H2O, 36g, 4g, 32g, 4kg, 32 kg, 36kg, (c) C2H4 + 3 O2, 2CO2 + 2H2O, 28 g 96 g, 28 g of C2H4 undergo complete combustion by, = 96 g of O2, 2.8 kg of C2H4 undergo complete combustion by, = 9.6 kg of O2., (c) According to stoichiometry they should react as follow, 4NH 3 5O 2, 4NO 6H 2 O, 4 mole of NH3 requires 5 mole of O2., , Equivalent mass, Molecular mass, , M, = 0.1 M, 2 M, , Normality, Replaceable hydrogen atom, , 98., , (a), , 99., , H2SO4 is dibasic acid., Molar solution of H2SO4 = N/2 H2SO4, (b) Given N1 = 10N, V1 = 10 ml, N2 = 0.1N, V2 = ?, , Molarity, , N1V1, , N 2V2, , or 10 × 10 = 0.1 × V2, 10 10, , V2 = 1000 ml, or V2, 0.1, Volume of water to be added, = V2 – V1 = 1000 – 10 = 990 ml., Mass of solute, 106, Mass of solution, , ppm, , Empirical formula is CH4., 92., , (a) Molarity = Normality ×, , Simple, , H, , 25 /1 25, , 97., , 100. (b) ppm, , Ratio, Mass Atoms, 75 12, 75 /12 6.25, 1, , 1, , 5, = 1.25 mole of O2., 4, Hence O2 is consumed completely., , 1 mole of NH3 requires =, , 6 10 –3, 10, 1000, , 6., , 101. (a) 5 M H2SO4 = 10 N H2SO4,, ( Basicity of H2SO4 = 2), N1V1 = N2V2,, 10 × 1 = N2 × 10 or N2 = 1 N, 102. (c) Among all the given options molarity is correct, because the term molarity involve volume which, increases on increasing temperature., 103. (b) Moles of urea present in 100 ml of sol.=, M, , 6.02 1020, 6.02 10 23, , 6.02 10 20 1000, , 0.01M, 6.02 10 23 100, [ M = Moles of solute present in 1L of solution], 104. (b) From the molarity equation., M1V1 + M2V2 = MV, Let M be the molarity of final mixture,, , M=, M, , M1V1 +M V2, , 2, where V = V1 + V2, V, 480 1.5 520 1.2, 1.344 M, 480 520, , STATEMENT TYPE QUESTIONS, 105. (c) Both solids and liquids have definite volume, but, gases do not., Solids have their own shape, but liquids and gases, takes the shape of the container in which they are put, in.
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SOME BASIC CONCEPTS OF CHEMISTRY, , 15, , 106. (b) For statement (ii), it is not necessary that all, components of a h eterogeneous mixture are, observable to naked eyes for example blood is a, heterogeneous mixture whose components are not, visible to naked eyes. For statement (iv) air is a, homogeneous mixture of various gases., 107. (d) Mass of a substance is the amount of matter present, in it while weight is the force exerted by gravity on an, object., Mass is constant while weight may vary from one, place to another due to gravity., SI unit of both mass and weight is kilogram., 108. (c) 27.3 days = 27.3 × 24 hours, = 655.2 hours, 27.3 days = 27.3 × 24 × 60 minutes, = 39312 minutes, 27.3 days = 27.3 × 24 × 60 × 60 seconds, = 2358720 seconds, 109. (c) For statement (i), T = The other name of Gay-Lussac’s, law is law of definite proportions by volume., For statement (ii), F = Law of conservation of mass is, valid for both physical and chemical change., For statement (iii), F = Law of definite proportion is, valid for each compound individually and not for, comparing two different compounds., For statement (iv), T = x/y must be a simple whole, number ratio and must be a positive integer., For statement (v), F = Equal volumes of all gases, under similar conditions of temperature and pressure, contain equal number of molecules., 110. (c) For statement (i) : H, O, C, N = All have different, chemical properties., For statement (ii) : It is true as per Dalton’s postulate., For statement (iii) : N : O = 1 : 1 (NO), For statement (iv) : Dalton’s postulates says, atoms, can neither be created nor destroyed., 111. (c) Molecular mass of cane sugar (C12H22O11), = 12 × 12 + 22 × 1 + 11 × 16, = 342 amu, 1 mole of cane sugar (C12H22O11) = 342 g, (Molecular mass of cane sugar = 342 g), 342 g of cane sugar contain = 6.022 × 1023 molecules, 34.20 g of cane sugar contain =, , 6.022 10 23, 34.20, 342, , = 6.022 × 1022 molecules., , MATCHING TYPE QUESTIONS, 112. (a), 113. (b) Terminal zeros are not significant if there is no, decimal i.e., 290 contains two significant figures, whereas in 29900. there are 5 significant figures;, 1.23 × 1.331 = 1.63713 but keeping the mind the 1.23, has only few significant figures i.e., only three, significant figures, so result should also be reported, in three significant figures only. Thus 1.6373 should, be rounded off to 1.64. Value 1.783 is rounded off to, 2, so has only one significant figure., , 114. (a), , 115. (b), , 116. (a), , 117. (d) A : 28 g of He =, , 28, = 7 mol, 4, , B : 46 g of Na =, , 46, = 2 mol, 23, , C : 60 g of Ca =, , 60, = 1.5 mol, 40, , D : 27 g of Al =, , 27, = 1 mol, 27, , 118. (c), , ASSERTION- REASON TYPE QUESTIONS, 119. (c), 120. (d) 1.231 has four significant figures all no. from left to, right are counted, starting with the first digit that is, not zero for calculating the no. of significant figure., 121. (b), 122. (d) We know that from the reaction H2 + Cl2, 2HCl that, the ratio of the volume of gaseous reactants and, products is in agreement with their molar ratio.The ratio, of H2 : Cl2 : HCl volume is 1: 1: 2 which is the same as, their molar ratio. Thus volume of gas is directly related, to the number of moles. Therefore, the assertion is, false but reason is true., 123. (c) Equal moles of different substances contain same, number of constituent particles but equal weights of, different substances do not contain the same number, of consituent particles., 124. (a), , CRITICAL THINKING TYPE QUESTIONS, 125. (b) For, 0.0 significant figure is zero. For 0.1 to 0.9, significant figure will be 1 whereas from 1.0 to 2.0, significant figures will be 2., 126. (a) In law of reciprocal proportions, the two elements, combining with the third element, must combine with, each other in the same ratio or multiple of that ratio., P2O3, PH3 and H2O correctly illustrate the law of, reciprocal proportions. Ratio in the number of atoms, of hydrogen and oxygen combining with one P is 3 :, 1.5 i.e., 2 : 1., 127. (a) Relative atomic mass, =, or, , Mass of one atom of the element, th, , 1/12 part of the mass of one atom of Carbon 12, , Mass of one atom of the element, , 12, mass of one atom of the C 12, Now if we use 1/6 in place of 1/12 the formula becomes, Relative atomic mass =, , Mass of one atom of element, Mass of one atom of carbon, , Relative atomic mass decrease twice, , 6
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EBD_7207, SOME BASIC CONCEPTS OF CHEMISTRY, , 16, , 128. (a) No. of molecules in different cases, (a), 22.4 litre at STP contains, = 6.023 × 1023 molecules of H2, 15 litre at STP contains =, , 15, 6.023 10 23, 22.4, , 131. (b) The number of atoms in 0.1 mole of a triatomic gas, = 0.1 × 3 × 6.023 × 1023., = 1.806 × 1023, 132. (d) At NTP 22400 cc of N2O = 6.02 × 1023 molecules, 1 cc N2O =, , = 4.03 × 1023 molecules of H2, (b), , 22.4 litre at STP contains, = 6.023×1023 molecules of N2, 5 litre at STP contains =, , (c), , 5, 6.023 10 23, 22.4, , = 1.344 × 1023 molecules of N2, 2 gm of H2= 6.023×1023 molecules of H2, 0.5, 6.023 10 23, 2, , 0.5 gm of H2=, , = 1.505 × 1023 molecules of H2, (d) Similarly 10 g of O2 gas, =, , 10, 6.023 10 23 molecules of O2, 32, , = 1.88 × 1023 molecules of O2, Thus (a) will have maximum number of molecules, 129. (d) 1 Mole of Mg3(PO4)2 contains 8 mole of oxygen atoms, 8 mole of oxygen atoms 1 mole of Mg3(PO4)2, 1, 0.25 mole of, 8, , 0.25 mole of oxygen atom, Mg3(PO4)2, 3.125 10, , 130. (c), , Density, , 2, , Mass, Volume, , 1 gram cm–3 =, Volume =, , mole of Mg3(PO4)2, , 1gram, cm 3, , Mass, Density, , 1gram, , 1gram cm, , 3, , 1cm 3, , Volume occupied by 1 gram water = 1 cm3, or Volume occupied by, , 6.023 1023, molecules of water = 1 cm3, 18, [, , 1g water =, , 1, moles of water], 18, , Thus volume occupied by 1 molecule of water, =, , 1 18, 6.023 10, , 23, , cm 3 = 3.0×10–23 cm3., , 6.02 10 23, molecules, 22400, , 3 6.02 10 23, 1. 8, atoms =, 1022 atoms, 22400, 224, , No. of electrons in a molecule of N 2O = 7 + 7 + 8 = 22, Hence no. of electrons, 23, 6.02 10 23, 22 electrons = 1.32 ´10, 22400, 224, 20, 133. (b) If 10 grains are distributed in one sec, 6.023 × 1023, grains will be distributed in, , 6.023 1023 1, 1020 60 60, , 1.673 hrs, , 134. (a) Mass of 6.023 × 1023 atoms of oxygen = 16 g, Mass of one atom of oxygen, =, , 16, 6.023 10 23, , 2.66 10, , 23, , g, , Mass of 6.023 × 1023 atoms of nitrogen = 14 g, Mass of one atom of nitrogen, 14, , 6.023 1023, , 2.32 10, , 23, , g, , Mass of 1 × 10–10 mole of oxygen = 16 × 10–10, Mass of 1 mole of copper = 63 g, Mass of 1 mole of oxygen = 16 g, Mass of 1 × 10–10 mole of copper = 63 × 1 × 10–10, = 63 × 10–10, So, the order of increasing mass is II < I < III < IV., 135. (b) The equation for the formation of Al 2 O3 can be, represented as, 2Al, , 2 moles, , 3 / 2O 2, 1.5 moles, , Al2 O3, 1 mole, , Thus, 1 mole of alumina is obtained by the reaction of, 1.5 moles of oxygen and 2 moles of aluminium. Thus,, the amount of aluminium, = 2 × 27 g = 54 g. [mol. mass of Al = 27], 136. (c) (a) Weight of H2 = mole × molecular wt., = 0.2 × 2 = 0.4 g, 23, (b) 6.023 × 10 = 1 mole, Thus 6.023 × 1022 = 0.1 mole, Weight of N2 = 0.1 × 28 = 2.8 g, (c) Weight of silver = 0.1 g, (d) Weight of oxygen = 32 × 0.1 = 3.2 g
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SOME BASIC CONCEPTS OF CHEMISTRY, , 137. (c), , Percentage, , 17, , R.N.A, , Simplest ratio, , C, , 9, , 9, 3, =, 12 4, , 3, , H, , 1, , 1, =1, 1, , 4, , N, , 3.5, , 3.5, 1, =, 14, 4, , 1, , Empirical formula = C3H4N, (C3H4N)n = 108, (12 × 3 + 4 × 1 + 14)n = 108, (54)n = 108, 108, n=, =2, 54, molecular formula = C6 H8N2, 138. (c) The acid with empirical formula CH2O2 is formic acid,, H—COOH., 139. (d), 18 gm, H2O contains = 2 gm H, 0.72 gm H2O contains =, , 2, 0.72 gm 0.08 gm H, 18, , 15.05, = 1.075, 14.00g / mol, 1.074 is the smallest value, division by it gives a ratio, of C : H : N, = 5.9 : 6.9 : 1, =6:7:1, Empirical formula = C6H7N, Empirical formula weight = 6 × 12 + 7 + 14 = 93, , Number of moles of N =, , Molecular mass, 1, Empirial formula weight, Molecular formula = 1 × C6H7N = C6H7N, n, , 142. (d), , 2 (78), , C:H=, , 143. (c), , 197 gm, , 197 gm of BaCO3 released carbon dioxide, = 22.4 litre at STP, 22.4, litre, 197, 9.85 gm of BaCO3 released carbon dioxide, , 1 gm of BaCO3 released carbon dioxide =, , Empirical formula = C7H8, 140. (a) Let 100 g of compound be there., , Number of moles of Hydrogen =, , =, , 60, = 4.9, 12.01, Since 2.5 is the smallest value division by it give, ratio, N:H:C, 1 : 1.96 : 1.96, =1:2:2, Empirical formula = C2H2N, Empirical formula weight = 2 × 12 + 2 + 14 = 40, Molecular mass = 80, Molecular formulae = n (C2H2N), , 80, = C4H4N2, 40, 141. (b) Let 100 g of compound be there., = 2 (C2H2N) n, , 77.43g, = 6.44, 12.01g / mol, 7.53g, Number of moles of H =, = 7.47, 1.008g / mol, , Number of moles of C =, , 144. (b), , 22.4, 9.85 = 1.12 litre, 197, , Ba(OH)2 CO2, n mol, , 5, = 4.9, 1.008, , Number of moles of Carbon =, , 15 22.4, litre, 156, , 39 gm of Benzene required oxygen, 15 22.4 39, 84.0 litre, =, 156, BaCO3, BaO CO 2, , 0.84 0.08, :, = 0.07 : 0.08 = 7 : 8, 12, 1, , 35, = 2.5, 14, , 12 CO 2 (g ) 6 H 2 O(g ), , 1 gm of benzene required oxygen =, , 12, 3.08 0.84 gm C, 44, , Number of moles of Nitrogen =, , 15(32 ), , 156 gm of benzene required oxygen = 15 × 22.4 litre, , 44 gm CO2 contains = 12 gm C, 3.08 gm CO2 contains =, , 2C 6 H 6 15O 2 (g ), , 145. (a), , BaCO3 H2 O, n mol, , n mol Ba(OH)2 = n mol BaCO3, 0.205 mol Ba (OH) 2 0.205 mol BaCO3, Wt. of substance = No. of moles × Molecular mass, = 0.205 × 197.3 = 40.5 g, Fe2O3 + 3CO, 2Fe + 3CO2, , 1 vol., 1 mol., , 3 vol., 3 mol., , 2 vol., 2 mol., , 3 vol., 3 mol., , ( vol% = mol%), One gram mol of any gas occupies 22.4 litre at NTP.1, mol of Fe2O3 requires 3 mol of CO for its reduction, i.e., 1 mol of Fe2O3 requires 3 × 22.4 litre or 67.2 dm3, CO to get itself reduced., 146. (d) Writing the equation for the reaction, we get, PbO + 2HCl, PbCl2 + H2O, 207 + 16, = 223 g, , 2 × 36.5, = 73g, , 207 + 71, = 278g, , 6.5, 0.029, 223, 3.2, 0.0877, No. of moles of HCl, 36.5, Thus PbO is the limiting reactant 1 mole of PbO produce, 1 mole PbCl2., 0.029 mole PbO produces 0.029 mole PbCl2., , No. of moles of PbO
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2, STRUCTURE OF ATOM, FACT / DEFINITION TYPE QUESTIONS, 1., , 2., , 3., , 4., , 5., , 6., , Which of the scientist were able to prove that atom is no, longer non-divisible?, (a) Dalton, (b) Michael Faraday, (c) Thomson, (d) Chadwick, Which of the following is never true for cathode rays ?, (a) They possess kinetic energy., (b) They are electromagnetic waves., (c) They produce heat., (d) They produce mechanical pressure., Cathode rays are deflected by, (a) an electric field only (b) magnetic field only, (c) by both, (d) by none, Which of the following statement is not correct about the, characteristics of cathode rays?, (a) They start from the cathode and move towards the, anode., (b) They travel in straight line in the absence of an external, electrical or magnetic field., (c) Characteristics of cathode rays do not depend upon, the material of electrodes in cathode ray tube., (d) Characteristics of cathode rays depend upon the, nature of gas present in the cathode ray tube., Which of the following statements about the electron is, incorrect?, (a) It is negatively charged particle, (b) The mass of electron is equal to the mass of neutron., (c) It is a basic constituent of all atoms., (d) It is a constituent of cathode rays., While performing cathode ray experiments, it was observed, that there was no passage of electric current under normal, conditions. Which of the following can account for this, observation ?, (a) Dust particles are present in air, (b) Carbon dioxide is present in air, (c) Air is a poor conductor of electricity under normal, conditions, (d) None of the above, , 7., , 8., , 9., , 10., , 11., , 12., , 13., , 14., , Which is not true with respect to cathode rays?, (a) A stream of electrons, (b) Charged particles, (c) Move with speed same as that of light, (d) Can be deflected by magnetic fields, What is the optimum conditions required to study the, conduction of electricity through gases., (a) High pressure and low voltage, (b) High pressure and high voltage, (c) Low pressure and high voltage, (d) Low pressure and low voltage, In discharge tube experiment stream of negatively charged, particles travel from, (a) anode to cathode, (b) cathode to anode, (c) Both (a) and (b), (d) Electrons does not travel, Millikan performed an experiment method to determine, which of the following ?, (a) Mass of the electron (b) Charge of the electron, (c) e/m ratio of electron (d) Both (a) and (b), The discovery of neutron became very late because :, (a) neutrons are present in nucleus, (b) neutrons are chargeless, (c) neutrons are fundamental particles, (d) all of the above, Which is correct statement about proton ?, (a) Proton is nucleus of deuterium, (b) Proton is -particle, (c) Proton is ionized hydrogen molecule, (d) Proton is ionized hydrogen atom, The lightest particle is :, (a), -particle, (b) positron, (c) proton, (d) neutron, When beryllium is bombarded with alpha particles, (Chadwick’s experiment) extremely penetrating radiations,, which cannot be deflected by electrical or magnetic field are, given out. These are :, (a) A beam of protons, (b) Alpha rays, (c) A beam of neutrons, (d) A beam of neutrons and protons
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EBD_7207, STRUCTURE OF ATOM, , 20, , 15., , 16., , Neutron is discovered by, (a) Chadwick, (b) Rutherford, (c) Yukawa, (d) Dalton, Suppose beam containing all three fundamental subatomic, particles are allowed to pass through an electric field as, shown in figure. The subatomic particles detected at three, points A, B and C on the screen respectively are ?, +, Beam of, particles, , 17., , 18., , 19., , 20., , 21., , 22., , 23., , 24., , A, B, , –, , (b), , 25., , C, , (a) Protons, neutrons, electrons, (b) Electrons, neutrons, protons, (c) Electrons, protons, neutrons, (d) Neutrons, protons, electrons, Which of the following properties of atom could be explained, correctly by Thomson Model of atom?, (a) Overall neutrality of atom., (b) Spectra of hydrogen atom., (c) Position of electrons, protons and neutrons in atom., (d) Stability of atom., Arrange the following in terms of penetrating power., -rays, -rays, -rays, (a), > >, (b), < <, (c), > <, (b), > >, Which of the rays are not deflected by the electric and, magnetic field ?, (a) -rays, (b) X-rays, (c) -rays, (d) Both (a) and (b), Rutherford’s experiment on the scattering of -particles, showed for the first time that the atom has :, (a) electrons, (b) protons, (c) nucleus, (d) neutrons', When atoms are bombarded with alpha particles, only, a, few in million suffer deflection, others pass out undeflected., This is because, (a) the force of repulsion on the moving alpha particle is, small, (b) the force of attraction between alpha particle and, oppositely charged electrons is very small, (c) there is only one nucleus and large number of electrons, (d) the nucleus occupies much smaller volume compared, to the volume of the atom, Rutherford’s -particle dispersion experiment concludes, (a) all positive ions are deposited at small part, (b) all negative ions are deposited at small part, (c) proton moves around the electron, (d) neutrons are charged particles., Rutherford’s experiment which established the nuclear, model of the atom used a beam of, (a) -particles which impinged on a metal foil and got, absorbed, , -rays which impinged on a metal foil and ejected, electrons, (c) helium atoms, which impinged on a metal foil and got, scattered, (d) helium nuclei, which impinged on a metal foil and got, scattered, Which of the following scientists explained his model on, the basis of centrifugal force ?, (a) Thomson, (b) Dalton, (c) Millikan, (d) Rutherford, The number of neutrons in dipositive zinc ion with mass, number 70 is, (a) 34, (b) 36, (c) 38, (d) 40, 1, 40, 19 K, , 26., , The number of electrons in, , 27., , (b) 40, (a) 20, (c) 18, (d) 19, Which of the following does not contain number of neutrons, equal to that of, , 28., , 30., , 31., , 32., , 33., , 40, 18 Ar ?, , (a), , 41, 19 K, , (b), , 43, 21 Sc, , (c), , 40, 21 Sc, , (d), , 42, 20 Ca, , Number of protons, neutrons and electrons in the element, 89 X, , 29., , is, , 231, , is, (a) 89, 89, 242, (b) 89, 142, 89, (c) 89, 71, 89, (d) 89, 231, 89, An element has atomic number 11 and mass number 24., What does the nucleus contain?, (a) 11 protons, 13 neutrons, (b) 11 protons, 13 neutrons, 13 electrons, (c) 13 protons, 11 neutrons, (d) 13 protons, 11 electrons, The number of electrons and neutrons of an element is 18, and 20 respectively. Its mass number is, (a) 2, (b) 17, (c) 37, (d) 38, ‘A’ represents mass no. and Z represents atomic no. then, - decay is characterized by, (a) Z increases by 2, A decreases by 4, (b) Z decreases by 2, A increases by 4, (c) Z decreases by 2, A decreases by 4, (d) Z increases by 2, A increases by 4., Nucleons are, (a) only neutrons, (b) neutrons + protons, (c) neutrons + protons + electrons, (d) neutrons + electrons, Atoms with same mass number but different atomic numbers, are called, (a) isotopes, (b) isobars, (c) isochores, (d) None of these
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STRUCTURE OF ATOM, , 34., , 35., , 36., , 37., , 38., , 21, , Which of the following pairs will have same chemical, properties ?, (a), , 14, 6 C, , (c), , 40, 18 Ar, , and, , 15, 7 N, , and, , 40, 19 K, , (b), , O2–, , (d), , 35, 17 Cl, , and, , and, , 37, 17 Cl, , What is the difference between two species if one has, atomic mass = 14 and atomic number = 7 whereas the, other has atomic mass = 14 and atomic number = 6 ?, (a) Neutrons, (b) Protons, (c) Electrons, (d) All of these, From the data given below A, B, C and D respectively are,, (A) 10 e–, atomic no. 11 (B) 10 e–, atomic no. 6, (C) 10 e–, atomic no. 10 (D) 10 e–, atomic no. 9, (a) Na+, C4–, Ne, F–, (b) C4–, Ne, Na–, F–, (c) F–, Na+, Ne, C4–, (d) F–, Na+, C4–, Ne, If the wavelength of the electromagnetic radiation is, increased to thrice the digital value, then what will be the, percent ch ange in the value of frequen cy of the, electromagnetic radiation., (a) Increases by 33%, (b) Decreases by 33%, (c) Increases by 66%, (d) Decreases by 66%, Which is the correct schematic representation of the graph, of black body radiation., (a), , T1 > T2, , Intensity, , T2, , T1, , Wavelength, , T1 > T2, (b), , Intensity, , T2, , T1, , Wavelength, , T1 > T2, (c), , T2, Wavelength, , T1, , Intensity, , T2 > T1, , F–, (d), , Wavelength, , T2, T1, , Intensity, , 39. The ideal body, which emits and absorbs radiations of all, frequencies, is called a black body and the radiation emitted, by such a body is called, (a) white body radiation (b) black body radiation, (c) black body emission (d) None of these, 40. Which one of the following is not the characteristic of, Planck’s quantum theory of radiation ?, (a) The energy is not absorbed or emitted in whole number, or multiple of quantum, (b) Radiation is associated with energy, (c) Radiation energy is not emitted or absorbed, continuously but in the form of small packets called, quanta, (d) This magnitude of energy associated with a quantum, is proportional to the frequency., 41. Which of the following is related with both wave nature, and particle nature ?, (a) Interference, (b) E = mc2, (c) Diffraction, (d) E = h, 42. The value of Planck's constant is 6.63 × 10–34 Js. The, velocity of light is 3.0 × 108 m s–1. Which value is closest to, the wavelength in nanometers of a quantum of light with, frequency of 8 × 1015 s–1 ?, (a) 3 × 107, (b) 2 × 10–25, –18, (c) 5 × 10, (d) 4 × 101, 43. In the photo-electron emission, the energy of the emitted, electron is, (a) greater than the incident photon, (b) same as than of the incident photon, (c) smaller than the incident photon, (d) proportional to the intensity of incident photon, 44. When a metal surface is exposed to solar radiations, (a) The emitted electrons have energy less than a maximum, value of energy depending upon frequency of incident, radiations, (b) The emitted electrons have energy less than maximum, value of energy depending upon intensity of incident, radiation, (c) The emitted electrons have zero energy, (d) The emitted electrons have energy equal to energy of, photons of incident light, 45. In photoelectric effect, at which frequency electron will be, ejected with certain kinetic energy ( 0 = threshold, frequency)., (a), > 0, (b) 0 >, (c) 0, (d), 0
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EBD_7207, STRUCTURE OF ATOM, , 22, , 46., , 47., , 48., , 49., , In continous spectrum light of _(i), ___ wavelength is deviated, the ___, ii, (a) (i) = longest, least, (b) (ii) = shortest, least, (c) (i) = shortest, most, (d) (i) = longest, most, Which of the following statements do not form a part of, Bohr’s model of hydrogen atom ?, (a) Energy of the electrons in the orbits are quantized, (b) The electron(s) in the orbit nearest to the nucleus has, the lowest energy, (c) Electrons revolve in different orbits around the nucleus, (d) The position and velocity of the electrons in the orbit, cannot be determined simultaneously, An electron from one Bohr stationary orbit can go to next, higher orbit, (a) by emission of electromagnetic radiation, (b) by absorption of any electromagnetic radiation, (c) by absorption of electromagnetic radiation of particular, frequency, (d) without emission or absorption of electromagnetic, radiation, For a Bohr atom angular momentum M of the electron is, (n = 0, 1, 2, .....) :, (a), , nh 2, 4, , (b), , n2 h2, 4, , nh, nh2, (d), 2, 4, According to Bohr's theory, the angular momentum of an, electron in 5th orbit is, , (c), 50., , (a) 10 h /, 51., , 52., , 53., , 54., , 55., , (b), , 56., , 57., , (a) –, (c) –, , 13.6, n, , 4, , 13.6, 2, , eV, , (b) –, , eV, , (d) –, , 13.6, n3, , eV, , 13.6, eV, n, , n, 58. The energy of second Bohr orbit of the hydrogen atom, is 328 kJ mol 1 ; hence the energy of fourth Bohr orbit, would be:, (a) 41 kJ mol 1, (b) 82 kJ mol 1, 1, (c) 164 kJ mol, (d) 1312 kJ mol 1, 59. In a hydrogen atom, if energy of an electron in ground state, is 13.6. eV, then that in the 2nd excited state is, (a) 1.51 eV, (b) 3.4 eV, (c) 6.04 eV, (d) 13.6 eV., 60. The energy of an electron in second Bohr orbit of hydrogen, atom is :, (a) –5.44 × 10–19eV, (b) –5.44 × 10–19cal, –19, (c) –5.44 × 10 kJ, (d) –5.44 × 10–19J, , 61., , 2.5 h /, , (c) 25 h /, (d) 1.0 h /, In Bohr’s model, atomic radius of the first orbit is y, the, radius of the 3rd orbit, is, (a) y / 3, (b) y, (c) 3y, (d) 9y, The radius of 1st Bohr’s orbit for hydrogen atom is ‘r’. The, radius of second Bohr’s orbit is, (a) 4r, (b) r3, 2, (c) 4r, (d) r1/3, The third line of the Balmer series, in the emission spectrum, of the hydrogen atom, is due to the transition from the, (a) fourth Bohr orbit to the first Bohr orbit, (b) fifth Bohr orbit to the second Bohr orbit, (c) sixth Bohr orbit to the third Bohr orbit, (d) seventh Bohr orbit to the third Bohr orbit, Which one of the following pairs is not correctly matched ?, (a) Rutherford-Proton, (b) J.J. Thomson-Electron, (c) J.H. Chadwick-Neutron, (d) Bohr-Isotopes, If r is the radius of the first orbit, the radius of nth orbit of, H-atom is given by, (a) rn2, (b) rn, (c) r/n, (d) r2 n2, , The radius of hydrogen atom in the ground state is 0.53 Å., The radius of Li2+ ion (atomic number = 3) in a similar state is, (a) 0.17 Å, (b) 0.265 Å, (c) 0.53 Å, (d) 1.06 Å, The energy of an electron in the nth Bohr orbit of hydrogen, atom is, , 62., , 63., , 64., , 65., , The energy of electron in first energy level is 21 .79 10 12, erg per atom. The energy of electron in second energy level, is :, (a), , 54.47 10 12 erg atom 1, , (b), , 5.447 10, , (c), , 0.5447 10 12 erg atom 1, , 12, , erg atom, , 1, , 0.05447 10 12 erg atom 1, (d), The ionisation potential of a hydrogen atom is –13.6 eV., What will be the energy of the atom corresponding to n = 2., (a) – 3.4 eV, (b) – 6.8 eV, (c) – 1.7 eV, (d) –2.7 eV, +, The line spectrum of He ion will resemble that of, (a) hydrogen atom, (b) Li+ ion, (c) helium atom, (d) lithium atom, What does negative sign in the electronic energy for, hydrogen atom convey., (a) Energy of electron when n =, (b) The energy of electron in the atom is lower than the, energy of a free electron in motion, (c) The energy of electron in the atom is lower than the, energy of a free electron of rest, (d) The energy of electron decreases as it moves away, from nucleus, In which of the following Bohr’s stationary state, the, electron will be at maximum distance from the nucleus ?, (a) IInd, (b) Ist, (c) Vth, (d) IIIrd
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STRUCTURE OF ATOM, , 66., , 67., , The wavelength of the radiation emitted, when in a hydrogen, atom electron falls from infinity to stationary state 1, would, be (Rydberg constant = 1.097×107 m–1), (a) 406 nm, (b) 192 nm, (c) 91 nm, (d) 9.1×10–8 nm, The frequency of radiation emitted when the electron falls, from n = 4 to n = 1 in a hydrogen atom will be (Given :, ionization energy of H=2.18 ×10–18J atom–1and h = 6.625 ×, 10–34 J s ), (a), , 68., , 69., , 23, , 1.54 1015 s, , (b) 1.03 1015 s 1, , 1, , (c) 3.08 1015 s 1, (d) 2.00 1015 s 1, Which of the following transitions of electrons in the, hydrogen atom will emit maximum energy ?, (a) n5, n4, (b) n4, n3, (c) n3, n2, (d) all will emit same energy, The first emission line of hydrogen atomic spectrum in the, Balmer series appears is (R = Rydberg constant), 3, 5, R cm 1, R cm 1, (a), (b), 36, 4, 7, 9, R cm 1, R cm 1, (d), 144, 400, According to the Bohr theory, which of the following, transitions in the hydrogen atom will give rise to the least, energetic photon ?, (a) n = 6 to n = 1, (b) n = 5 to n = 4, (c) n = 6 to n = 5, (d) n = 5 to n = 3, The wavelength (in cm) of second line in the Lyman series, of hydrogen atomic spectrum is (Rydberg constant, = R cm–1), , (c), , 70., , 71., , 72., , 73., , (a), , 8R, 9, , (b), , 9, 8R, , (c), , 4, 3R, , (d), , 3R, 4, , The shortest wavelength in hydrogen spectrum of Lyman, series when RH = 109678 cm–1 is, (a) 1002.7Å, (b) 1215.67Å, (c) 1127.30Å, (d) 911.7Å, What is the expression of frequency ( ) associated with, absorption spectra of the photon., (a), , RH 1, h n2, , 1, ni, n 2, , nf, , RH 1, h n2, , 1, nf, n 2, , ni, , i, , (b), , f, , i, , (c), , f, , RH 1, h n2, i, , 1, nf, n 2, f, , (d) All the above are correct, , ni, , 74. Bohr model can explain :, (a) the solar spectrum, (b) the spectrum of hydrogen molecule, (c) spectrum of any atom or ion containing one electron, only, (d) the spectrum of hydrogen atom only, 75. Which of the following statements do not form a part of, Bohr’s model of hydrogen atom ?, (a) Energy of the electrons in the orbits are quantized, (b) The electron in the orbit nearest the nucleus has the, lowest energy, (c) Electrons revolve in different orbits around the nucleus, (d) The position and velocity of the electrons in the orbit, cannot be determined simultaneously., 76. Bohr’s theory can be applied to which of the following, ions., (a) Na+, (b) Be2+, (c) Li+, (d) Li2+, 77. Bohr’s model is not able to account for which of the, following., (a) Stability of atom., (b) Spectrum of neutral helium atom., (c) Energy of free electron at rest., (d) Calculation of radii of the stationary states., 78. If electron, hydrogen, helium and neon nuclei are all moving, with the velocity of light, then the wavelength associated, with these particles are in the order, (a) Electron > hydrogen > helium > neon, (b) Electron > helium > hydrogen > neon, (c) Electron < hydrogen < helium < neon, (d) Neon < hydrogen < helium < electron, 79. The de Broglie wavelength of a tennis ball of mass 60 g, moving with a velocity of 10 metres per second is, approximately, (a) 10–31 metres, (b) 10–16 metres, –25, (c) 10 metres, (d) 10–33 metres, Planck’s constant, h = 6.63 × 10–34 Js, 80. If the energy difference between the ground state of an, atom and its excited state is 4.4 × 10–4 J, the wavelength of, photon required to produce the transition :, (a) 2.26 × 10–12 m, (b) 1.13 × 10–12 m, –16, (c) 4.52 × 10 m, (d) 4.52 × 10–12 m, 81. The mass of a photon with a wavelength equal to, 1.54 × 10–8 cm is, (a) 0.8268 × 10–34 kg, (b) 1.2876 × 10–33 kg, (c) 1.4285 × 10–32 kg, (d) 1.8884 × 10–32 kg, 82. If the Planck’s constant h = 6.6×10–34 Js, the de Broglie, wavelength of a particle having momentum of 3.3 × 10–24, kg ms –1 will be, (a) 0.002 Å, (b) 0.5Å, (c) 2Å, (d) 500Å
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EBD_7207, STRUCTURE OF ATOM, , 24, , 83., , 84., , 85., , 10–34, , The values of Planck's constant is 6.63 ×, Js. The, velocity of light is 3.0 × 108 m s–1. Which value is closest to, the wavelength in nanometres of a quantum of light with, frequency of 8 × 1015 s–1?, (a) 5 × 10–18, (b) 4 × 101, (c) 3 × 107, (d) 2 × 10–25, The de Broglie wavelength of a car of mass 1000 kg and, velocity 36 km/hr is :, (a) 6.626 × l0–34 m, (b) 6.626 × 10–38 m, –31, (c) 6.626 × 10 m, (d) 6.626 × 10– 30 m, Heisenberg uncertainty principle can be explained as, (a), , 87., , 88., , 89., , 90., , 91., , 92., , P h, 4, , x, , P, , h, , (b), , x, , P, , (d), , P, , 94., , 95., , h, 4, , h, x, Heisenberg's uncertainity principle is applicable to, (a) atoms only, (b) electron only, (c) nucleus only, (d) any moving object, The position of both, an electron and a helium atom is, known within 1.0 nm. Further the momentum of the, electron is known within 5.0 × 10–26 kg ms–1 . The, minimum uncertainty in the measurement of the, momentum of the helium atom is, (a) 50 kg ms–1, (b) 80 kg ms–1, –26, –1, (c) 8.0 × 10 kg ms, (d) 5.0 × 10–26 kg ms–1, Uncertainty in the position of an electron, (mass = 9.1 × 10–31 kg) moving with a velocity 300 ms–1,, accurate upto 0.001% will be (h = 6.63 × 10–34 Js), (a) 1.92 × 10–2 m, (b) 3.84 × 10–2 m, –2, (c) 19.2 × 10 m, (d) 5.76 × 10–2 m, The uncertainty in the position of an electron (mass =, 9.1 × 10–28 g) moving with a velocity of 3.0 × 104 cm s–1, accurate upto 0.011% will be, (a) 1.92 cm, (b) 7.68 cm, (c) 0.175 cm, (d) 3.84 cm., The Heisenberg uncertainity principle will be most, significant for which of the following object ?, (a) Object A of mass 9.11 × 10–30 kg, (b) Object B of mass 9.11 × 10–28 g, (c) Object C of mass 9.11 × 10–24 mg, (d) Object D of mass 9.11 × 10–28 kg, The orientation of an atomic orbital is governed by, (a) Spin quantum number, (b) Magnetic quantum number, (c) Principal quantum number, (d) Azimuthal quantum number, For which one of the following sets of four quantum, numbers, an electron will have the heighest energy?, n, l, m, s, (a) 3, 2, 1, 1/2, (b) 4, 2, –1, 1/2, (c) 4, 1, 0, –1/2, (d) 5, 0, 0, –1/2, , (c), , 86., , x, , 93., , 96., , 97., , 98., , 99., , 100., , 101., , 102., , 103., , Which of the following sets of quantum numbers is correct, for an electron in 4f orbital ?, (a) n = 4, = 3, m = + 1, s = + ½, (b) n = 4, = 4, m = – 4, s = – ½, (c) n = 4, = 3, m = + 4, s = + ½, (d) n = 3, = 2, m = – 2, s = + ½, What is the correct orbital designation of an electron with, the quantum number, n = 4, = 3, m = – 2, s = 1/2 ?, (a) 3s, (b) 4 f, (c) 5p, (d) 6s, Which of the following represents correct set of the four, quantum numbers for an electron in a 4d subshell ?, (a) 4, 2, 1, 0, (b) 4, 2, 1, – 1/2, (c) 4, 3, 2, + 1/2, (d) 4, 3, – 2, – 1/2, The total number of electrons that can be accommodated in, all orbitals having principal quantum number 2 and azimuthal, quantum number 1 is, (a) 2, (b) 4, (c) 6, (d) 8, For azimuthal quantum number = 3, the maximum number, of electrons will be, (a) 2, (b) 6, (c) 0, (d) 14, Which of the following is not permissible arrangement of, electrons in an atom?, (a) n = 5, l = 3, m = 0, s = + 1/2, (b) n = 3, l = 2, m = – 3, s = – 1/2, (c) n = 3, l = 2, m = – 2, s = – 1/2, (d) n = 4, l = 0, m = 0, s = – 1/2, Which of the following sets of quantum numbers represents, the highest energy of an atom?, (a) n = 3, l = 0, m = 0, s = +1/2, (b) n = 3, l = 1, m = 1, s = +1/2, (c) n = 3, l = 2, m = 1, s = +1/2, (d) n = 4, l = 0, m = 0, s = +1/2, Which set of quantum numbers are not possible?, n, l, m, s, (a) 3, 2, 0, +1/2, (b) 2, 2, 1, +1/2, (c) 1, 0, 0, –1/2, (d) 3, 2, –2, +1/2, What will be the sum of all possible values of l and m for, n=5?, (a) 12, (b) 13, (c) 4, (d) 9, The following quantum numbers are possible for how many, orbital(s) n = 3, l = 2, m = +2 ?, (a) 1, (b) 3, (c) 2, (d) 4, The orbitals are called degenerate when, (a) they have the same wave functions, (b) they have the same wave functions but different, energies, (c) they have different wave functions but same energy, (d) they have the same energy
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STRUCTURE OF ATOM, , 25, , 104. The number of spherical nodes in 3p orbitals are, (a) one, (b) three, (c) two, (d) None of these, 105. Which of the following graph correspond to one node, , (a), , (b), , a0, , a0, , (c), , (d), , a0, , a0, , 106. If there are five radial nodes, then what can be the correct, representation of the orbital for n = 8., (a) 8d, (b) 8s, (c) 8p, (d) 8f, 107. What can be the representation of the orbital having 3, angular nodes and n = 5., (a) 5d, (b) 5f, (c) 5p, (d) 5s, 108. The number of orbitals present in the fifth shell will be, (a) 25, (b) 10, (c) 50, (d) 20, 109. Arrange the orbital of same shell in the increasing order of, shielding strength of the outer shell of electrons., s, f, d, p, (a) s < p < d < f, (b) s > p < d < f, (c) s > p > d < f, (d) s > p > d > f, 110. Which of the following is not correct for electronic, distribution in the ground state ?, (a) Co, , [Ar], , (b) Ni, , [Ar], , (c) Cu, , [Ar], , (d) All of the above, 111. The electronic configuration of gadolinium (Atomic number, 64) is, (a) [Xe] 4f 8 5d 0 6s2, (b) [Xe] 4f 3 5d 5 6s2, 6, 2, 2, (c) [Xe] 4f 5d 6s, (d) [Xe] 4f 7 5d 1 6s2, , 112. The order of filling of electrons in the orbitals of an atom will, be, (a) 3d, 4s, 4p, 4d, 5s, (b) 4s, 3d, 4p, 5s, 4d, (c) 5s, 4p, 3d, 4d, 5s, (d) 3d, 4p, 4s, 4d, 5s, 113. The number of d-electrons retained in Fe2+, (At. no. of Fe = 26) ion is, (a) 4, (b) 5, (c) 6, (d) 3, 114. The electronic configuration of an element is 1s2, 2s2 2p6,, 3s2 3p6 3d 5, 4s1. This represents its, (a) excited state, (b) ground state, (c) cationic form, (d) anionic form, 115. Number of unpaired electrons in N2+ is, (a) 2, (b) 0, (c) 1, (d) 3, 116. An ion has 18 electrons in the outermost shell, it is, (a) Cu+, (b) Th4+, (c) Cs+, (d) K+, 117. In a given atom no two electrons can have the same values, for all the four quantum numbers. This is called, (a) Hund’s Rule, (b) Aufbau principle, (c) Uncertainty principle (d) Pauli’s exclusion principle, 118. The electronic configuration of Cu (atomic number 29) is, (a), , 1s 2 , 2 s 2 2 p 6 ,3s 2 3 p 6 , 4 s 2 3d 9, , (b) 1s 2 ,2 s 2 2 p 6 ,3s 2 3 p 6 3d 10 , 4 s1, (c), , 1s2 ,2s2 2 p6 ,3 p 2 3 p 6 ,4s2 4 p6 ,5s2 5 p1, , (d) 1s 2 , 2 s 2 2 p 6 ,3 p 2 3 p 6 , 4 s 2 4 p 6 3d 3, 119. The orbital diagram in which the Aufbau principle is violated, is :, 2s, 2p, (a), (b), (c), (d), 120. If n = 6, the correct sequence for filling of electrons will be :, (a) ns (n – 2) f (n – 1) d np, (b) ns (n – 1) d (n – 2) f np, (c) ns (n – 2) f np (n – 1) d, (d) ns np (n – 1) d (n – 2) f, 121. Maximum number of electrons in a subshell of an atom is, determined by the following:, (a) 2 l + 1, (b) 4 l – 2, 2, (c) 2 n, (d) 4 l + 2, 122. The correct order of increasing energy of atomic orbitals is, (a) 5 p < 4 f < 6 s < 5 d, (b) 5 p < 6 s < 4 f < 5 d, (c) 5 p < 5 d < 4 f < 6 s, (d) None of these
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EBD_7207, STRUCTURE OF ATOM, , 26, , 123. For which element, the valence electron will be present in, the highest energy orbital., (a) 3Li, (b) 16 S, (c) 20Ca, (d) 21 Sc, 124. Which of the following electronic configuration of d-orbital, will have highest affinity for gaining an electron?, (a), (b), (c), (d), , wave number of each line is given by v, , STATEMENT TYPE QUESTIONS, 125. On the basis of figure given below which of the following, statement(s) is/are correct ?, , Cathode, , +, , Anode, , –, , 127. Which of the following statement(s) is/are incorrect, regarding photoelectric effect?, (i) The number of electrons ejected is proportional to the, intensity of light., (ii) There is some time lag between the striking of light, beam on the metal surface and the ejection of electrons., (iii) The kinetic energy of ejected electrons depends upon, the brightness of light., (iv) The kinetic energy of the ejected electron is, proportional to the frequency of the incident radiation., (a) (i) and (ii), (b) (ii) and (iii), (c) (ii) only, (d) (ii) and (iv), 128. For Balmer series in the spectrum of atomic hydrogen, the, , Fluorescent screen, Magnet, , (i) At point B, when only electric field is applied., (ii) At point C, when both electric and magnetic field is, applied., (iii) At point B, when both electric and magnetic fields are, balanced., (iv) At point C, when only magnetic field is applied., Which of the following is/are correct?, (a) (i) and (ii), (b) only (iii), (c) (iii) and (iv), (d) (i) and (iii), 126. Which of the following statements are not correct about, electromagnetic radiation ?, (i) Electromagnetic waves require medium to travel., (ii) Different electromagnetic radiations travel at same, speed in vaccum., (iii) The oscillating electric and magnetic fields produced, by oscillating charged particles are perpendicular to, each other, but not to the direction of propagation., (iv) The oscillating electric field and magnetic field are, perpendicular to each other, and also to the direction, of propagation., (a) (i), (ii) and (iii), (b) (ii) and (iii), (c) (i) and (iii), (d) (i) and (iv), , RH, , 1, , n12, , –, , 1, , n 22, , where RH is a constant and n 1 and n 2 are integers. Which, of the following statement(s) is (are) correct?, (i) As wavelength decreases, the lines in the series, converge., (ii) The integer n 1 is equal to 2., (iii) The ionization energy of hydrogen can be calculated, from the wave number of these lines., (iv) The line of longest wavelength corresponds to n 2 = 3., (a) (i), (ii) and (iii), (b) (ii), (iii) and (iv), (c) (i), (i) and (iv), (d) (ii) and (iv), 129. Which of the following statements of quantum mechanics, was in agreement with Bohr’s model?, (i) The path of an electron in an atom can never be, determined accurately., (ii) The energy of electrons in atom is quantized i.e., can, only have specific values., (iii) An orbital cannot contain more than two electrons., (a) Only (i), (b) (i) and (ii), (c) Only (ii), (d) (ii) and (iii), 130. Which of the following statements concerning the quantum, numbers are correct ?, (i) Angular quantum number determines the threedimensional shape of the orbital., (ii) The principal quantum number determines the, orientation and energy of the orbital., (iii) Magnetic quantum number determines the size of the, orbital., (iv) Spin quantum number of an electron determines the, orientation of the spin of electron relative to the chosen, axis., The correct set of option is, (a) (i) and (ii), (b) (i) and (iv), (c) (iii) and (iv), (d) (ii), (iii) and (iv)
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STRUCTURE OF ATOM, , 27, , MATCHING TYPE QUESTIONS, 131. Match the columns., Column-I, (A) 1 H, 2 H and 3 H, 1 1, 1, (B), , 14, 6C, Na+, , and, , 14, 7N, Mg2+, , Column-II, (p) Isobars, (q) Isotopes, , (C), and, (r) Isoelectronic species, (a) A – (p), B – (q), C – (r), (b) A – (q), B – (p), C – (r), (c) A – (r), B – (q), C – (p), (d) A – (p), B – (r), C – (q), 132. Match the columns., Column-I, Column-II, (A) X-rays, (p) v = 100 – 104 Hz, (B) UV, (q) v = 1010 Hz, (C) Long radio waves, (r) v = 1016 Hz, (D) Microwave, (s) v = 1018 Hz, (a) A – (s), B – (r), C – (p), D – (q), (b) A – (r), B – (s), C – (p), D – (q), (c) A – (s), B – (p), C – (r), D – (q), (d) A – (s), B – (r), C – (q), D – (p), 133. Match the columns., Column-I, Column-II, (A), , 2, , z, , (A) d x 2 y2, , (p), x, , y, , z, , (p) Energy can be emitted, , or absorbed in packets, (q) Significant only for, motion of microscopic, objects., (C) Heisenberg, (r) The probability of, finding an electron at a, point within an atom, (D) Planck’s, (s) Every object in motion, has a wave character., (a) A – (q), B – (s), C – (r), D – (p), (b) A – (r), B – (p), C – (q), D – (s), (c) A – (r), B – (s), C – (q), D – (p), (d) A – (s), B – (p), C – (r), D – (q), 134. Match the columns., Column-I, Column-II, (Quantum number), (Information provided), (A) Principal, (p) orientation of the orbital, quantum number, (B) Azimuthal, (q) energy and size of orbital, quantum number, (C) Magnetic, (r) spin of electron, quantum number, (D) Spin quantum, (s) shape of the orbital, number, (a) A – (q), B – (s), C – (p), D – (r), (b) A – (s), B – (q), C – (p), D – (r), (c) A – (q), B – (p), C – (s), D – (r), (d) A – (q), B – (s), C – (r), D – (p), (B) de Brolie, , 135. Match the columns., Column-I, Column-II, Column-III, (Sub shell), (Number of, (Angular/Azimuthal, orbitals), Quantum Number), (A) d, (p) 1, (i) 1, (B) f, (q) 3, (ii) 2, (C) s, (r) 5, (iii) 0, (D) p, (s) 7, (iv) 3, (a) A – (r) – (ii), B – (s) – (iv), C – (p) – (iii), D – (q) – (i), (b) A – (q) – (i), B – (s) – (iv), C – (p) – (iii), D – (r) –(ii), (c) A – (p) – (iii), B – (s) – (iv), C – (r) – (ii), D – (q) – (i), (d) A – (r) – (ii), B – (p) – (iii), C – (s) – (iv), D – (q) – (i), 136. Match the columns., Column-I, Column-II, , (B) d xy, , (q), x, , y, , z, , (C) d yz, , (r), x, , y, z, , (D), , d, , z2, , (s), , y, , (a) A – (p), B – (s), C – (r), D – (q), (b) A – (s), B – (p), C – (r), D – (q), (c) A – (s), B – (p), C – (q), D – (r), (d) A – (s), B – (r), C – (p), D – (q), , x
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EBD_7207, 28, , 137. Match the columns, Column-I, (Rules), (A) Hund’s Rule, , Column-II, (Statements), (p) No two electrons in an atom, can have the same set of four, quantum numbers., (B) Aufbau Principle (q) Half-filled and completely, filled orbitals have extra, stablity., (C) Pault Exclusion, (r) Pairing of electrons in the, Principle, orbitals belonging to the same, subshell does not take place, until each orbital is singly, occupied., (D) Heisenberg’s, (s) It is impossible to determine, Uncertainty, the exact position and exact, Principle, momentum of a subatomic, particle simultaneously., (t) In the ground state of atoms,, orbitals are filled in the order, of their increasing energies., (a) A – (r), B – (p), C – (t), D – (s), (b) A – (r), B – (t), C – (s), D – (p), (c) A – (r), B – (t), C – (p), D – (s), (d) A – (t), B – (r), C – (p), D – (s), 138. Match the columns., Column-I, Column-II, (Atom / Ion), (Electronic configuration), (A) Cu, (p) 1s2 2s2 2p6 3s2 3p6 3d10, 2+, (B) Cu, (q) 1s2 2s2 2p6 3s2 3p6 3d10 4s2, (C) Zn2+, (r) 1s2 2s2 2p6 3s2 3p6 3d10 4s1, 3+, (D) Cr, (s) 1s2 2s2 2p6 3s2 3p6 3d9, (t) 1s2 2s2 2p6 3s2 3p6 3d3, (a) A – (s), B – (r), C – (p), D – (t), (b) A – (r), B – (s), C – (p), D – (t), (c) A – (r), B – (s), C – (t), D – (p), (d) A – (r), B – (s), C – (p), D – (s), , ASSERTION-REASON TYPE QUESTIONS, Directions : Each of these questions contain two statements,, Assertion and Reason. Each of these questions also has four, alternative choices, only one of which is the correct answer. You, have to select one of the codes (a), (b), (c) and (d) given below., (a) Assertion is correct, reason is correct; reason is a correct, explanation for assertion., (b) Assertion is correct, reason is correct; reason is not a, correct explanation for assertion, (c) Assertion is correct, reason is incorrect, (d) Assertion is incorrect, reason is correct., 139. Assertion : The position of an electron can be determined, exactly with the help of an electron microscope., Reason : The product of uncertainty in the measurement of, its momentum and the uncertainty in the measurement of, the position cannot be less than a finite limit., , STRUCTURE OF ATOM, , 140. Assertion : The radius of the first orbit of hydrogen atom is, 0.529Å., Reason : Radius of each circular orbit (rn) - 0.529Å (n2/Z),, where n = 1, 2, 3 and Z = atomic number., 141. Assertion : All isotopes of a given element show the same, type of chemical behaviour., Reason : The chemical properties of an atom are controlled, by the number of electrons in the atom., 142. Assertion : Black body is an ideal body that emits and, absorbs radiations of all frequencies., Reason : The frequency of radiation emitted by a body, goes from a lower frequency to higher frequency with an, increase in temperature., 143. Assertion : It is impossible to determine the exact position, and exact momentum of an electron simultaneously., Reason : The path of an electron in an atom is clearly defined., , CRITICAL THINKING TYPE QUESTIONS, 144. What is the ratio of mass of an electron to the mass of a, proton?, (a) 1 : 2, (b) 1 : 1, (c) 1 : 1837, (d) 1 : 3, 145. The increasing order for the values of e/m (charge/mass) is, (a) e, p, n,, (b) n, p, e,, (c) n, p, , e, (d) n, , p, e, 146. In which of the following the amount of deviation from, their path in the presence of electric and magnetic field will, be maximum?, (a) N2–, (b) N3–, –, (c) N, (d) N, 147. The deflection of the particles from their path in presence, of electric and magnetic field will be maximum in which of, the following., (a) O, (b) N, (c) U, (d) He, 148. Which of the following pairs have identical values of e/m?, (a) A proton and a neutron, (b) A proton and deuterium, (c) Deuterium and an -particle, (d) An electron and -rays, 149. If the alpha-particles are projected against the following, atoms Fe, Be, Mg, Al then increasing order in which the, alpha-particle feel repulsion will be, (a) Be, Mg, Al, Fe, (b) Be, Al, Mg, Fe, (c) Mg, Al, Mg, Fe, (d) Al, Mg, Fe, Be, 150. Chlorine exists in two isotopic forms, C1-37 and C1-35 but, its atomic mass is 35.5. This indicates the ratio of C1-37 and, C1-35 is approximately, (a) 1 : 2, (b) 1 : 1, (c) 1 : 3, (d) 3 : 1, 151. The number of electrons, neutrons and protons in a species, are equal to 10, 8 and 8 respectively. The proper symbol of, the species is, (a) 16 O 8, (b) 18 O 8, (c) 18 Ne10, (d) 16 O82–
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STRUCTURE OF ATOM, , 29, , 152. What will be the difference between electromagnetic, radiation shown in A and B respectively ?, , (A), , He+, , 159. Which of the following levels of H and, have same, energy respectively ?, (A) 1, 2, (B) 3, 4, (C) 2, 4, (D) 3, 6, (a) A and D, (b) A and B, (c) C and D, (d) A, C and D, 160. Bohr radius of nth orbit of an atom is given by the expression:, r, , (c), , r, , (a), , 3e 2, 4 0r, , (B), , (i) Velocity, (ii) Wavelength, (iii) Frequency, (iv) Energy, (a) (ii) only, (b) (ii) and (iv), (c) (ii), (iii) and (iv), (d) (iv) only, 153. Arrange the electromagnetic radiations a, b, c, d and e in, increasing order of energy. Frequencies of a, b and c are, 1015, 1014 and 1017 respectively whereas wavelength of, (d) and (e) are 350 nm and 100 nm respectively ?, (a) a, b, c, d, e, (b) a, b, d, e, c, (c) a, d, b, e, c, (d) b, d, a, e, c, 154. An electron, e1 is moving in the fifth stationary state, and, another electron e2 is moving in the fourth stationary state., The radius of orbit of electron, e1 is five times the radius of, orbit of electron, e2 calculate the ratio of velocity of electron, e1 (v1) to the velocity of electron e2 (v2)., (a) 5 : 1, (b) 4 : 1, (c) 1 : 5, (d) 1 : 4, 2+, 155. The Li ion is moving in the third stationary state, and its, linear momentum is 7.3 × 10–34 kg ms–1. Calculate its angular, momentum., (a) 1.158 × 10–45 kg m2s–1, (b) 11.58 × 10–48 kg m2s–1, (c) 11.58 × 10–47 kg m2s–1, (d) 12 × 10–45 kg m2s–1, 156. The Bohr orbit radius for the hydrogen atom (n = 1) is, approximately 0.530 Å. The radius for the first excited state, (n = 2) orbit is (in Å), (a) 0.13, (b) 1.06, (c) 4.77, (d) 2.12, 157. According to Bohr’s theory the energy required for an electron, in the Li2+ ion to be emitted from n = 2 state is (given that, the ground state ionization energy of hydrogen atom is, 13.6 eV), (a) 61.2 eV, (b) 13.6 eV, (c) 30.6 eV, (d) 10.2 eV, 158. Among species H, Li2+, He+, Be3+ and Al3+ Bohr’s model, was able to explain the spectra of, (a) All of these, (b) None of these, (c) all other species except Be3+, (d) all other species except Al3+, , n2 h2, , (a), , 2, , 4, , me, , n2 h2, , 2, , (b), , r, , (d), , r, , nh, 4, , 2, , mZe2, , n 2 h2, , 4 2 mZ, 4 2 me 2 Z, 161. The ratio between kinetic energy and the total energy of the, electrons of hydrogen atom according to Bohr’s model is, (a) 2 : 1, (b) 1 : 1, (c) 1 : – 1, (d) 1 : 2, 162. The potential energy of electron present in ground state of, Li2+ ion is represented by :, , (c), , (b), , 3e 2, 4, , 0r, , 2, , (d), , 3e, 4, , 0r, , 3e 2, 4 0r, , 163. In hydrogen atomic spectrum, a series limit is found at, 12186.3 cm–1. Then it belong to, (a) Lyman series, (b) Balmer series, (c) Paschen series, (d) Brackett series, 164. Which transition in the hydrogen atomic spectrum will have, the same wavelength as the transition, n = 4 to n = 2 of He+, spectrum?, (a) n = 4 to n = 3, (b) n = 3 to n = 2, (c) n = 4 to n = 2, (d) n = 2 to n = 1, 165. Arrange the following elements in the order of ease of, detection of wave properties, in the de Broglie experiment., H, Li, Be, B, K, (a) H < Be, B < Li < K., (b) H > Li > K > Be > B, (c) H > Li > Be > B > K, (d) H < Li < Be < B < K, 166. A 600 W mercury lamp emits monochromatic rediation of, wavelength 331.3 nm. How many photons are emitted from, the lamp per second ? (h = 6.626 × 10–34 Js; velocity of light, = 3 × 108 ms–1), (a) 1 × 1019, (b) 1 × 1020, 21, (c) 1 × 10, (d) 1 × 1023, 167. Calculate the velocity of ejected electron from the metal, surface when light of frequency 2 × 1015 Hz fall on the, metal surface and the threshold frequency is 7 × 1014 Hz, for metal ?, (a) 1.37 × 106, (b) 1.26 × 106, 7, (c) 1.45 × 10, (d) 1.37 × 107, 168. What is the wavelength (in m) of the electron emitted in, the above question (Q. no. 167) ?, (a) 5.308 × 10–10, (b) 5.89 × 10–11, –13, (c) 4.37 × 10, (d) 3.98 × 10–10
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EBD_7207, STRUCTURE OF ATOM, , 30, , 169., , The velocity of particle A is 0.1 ms–1 and that of particle B is, , 0.05 ms–1. If the mass of particle B is five times that of, particle A, then the ratio of de-Broglie wavelengths, associated with the particles A and B is, (a) 2 : 5, (b) 3 : 4, (c) 6 : 4, (d) 5 : 2, 170. Two fast moving particles X and Y are associated with, de Broglie wavelengths 1 nm and 4 nm respectively. If mass, of X in nine times the mass of Y, the ratio of kinetic energies, of X and Y would be, (a) 3 : 1, (b) 9 : 1, (c) 5 : 12, (d) 16 : 9, 171. Uncertainty in position of a n electron (mass = 9.1 × 10–28 g), moving with a velocity of 3 × 104 cm/s accurate upto 0.001%, will be (use h/4 ) in uncertainty expression where, h = 6.626 ×10–27 erg-second)., (a) 1.93 cm, (b) 3.84 cm, (c) 5.76 cm, (d) 7.68 cm, 172. The measurement of the electron position is associated with, an uncertainty in momentum, which is equal to, 1×10–18 g cm s–1. The uncertainty in electron velocity is,, (mass of an electron is 9 × 10– 28 g), (a) 1 × 109 cm s–1, (b) 1 × 106 cm s–1, 5, –1, (c) 1 × 10 cm s, (d) 1 × 1011 cm s–1, 173. In an atom, an electron is moving with a speed of 600 m/s, with an accuracy of 0.005%. Certainity with which the, position of the electron can be located is ( h = 6.6 × 10 –34 kg, m2s–1, mass of electron, em = 9.1 × 10–31 kg), (a) 5.10 × 10 –3 m, (b) 1.92 × 10 –3 m, –3, (c) 3.84 × 10 m, (d) 1.52 × 10 –4 m, 174. If uncertainty in position and momentum are equal, then, uncertainty in velocity is :, (a), , 1 h, 2m, , (b), , h, 2, , 1 h, h, (d), m, 175. Which of the following sets of quantum numbers is correct?, (a) n = 5, l = 4, m = 0, s = + ½, , (c), , (b) n = 3, l = 3, m = +3, s = + ½, (c) n = 6, l = 0, m = +1, s = – ½, (d) n = 4, l = 2, m = +2, s = 0, 176. Which combinations of quantum numbers, n, , m and s for, the electron in an atom does not provide a permissible, solution of the wave equation ?, (a), , 3, 2,1,, , 1, 2, , (b) 3,1,1,, , 1, 2, , 1, 1, (d) 3, 2, 2,, 2, 2, 177. An electron has principal quantum number 3. The number of, its (i) subshells and (ii) orbitals would be respectively, (a) 3 and 5, (b) 3 and 7, (c) 3 and 9, (d) 2 and 5, 178. The electrons identified by quantum numbers n and :, (A) n = 4, = 1, (B) n = 4, = 0, (C) n = 3, = 2, (D) n = 3, = 1, can be placed in order of increasing energy as :, (a) (C) < (D) < (B) < (A) (b) (D) < (B) < (C) < (A), (c) (B) < (D) < (A) < (C) (d) (A) < (C) < (B) < (D), , (c), , 3, 3,1,, , 179. The five d-orbitals are designated as d xy , d yz , d xz , d, , x2 y2, , and d 2. Choose the correct statement., z, , (a) The shapes of the first three orbitals are similar but, that of the fourth and fifth orbitals are different, (b) The shapes of all five d-orbitals are similar, (c) The shapes of the first four orbitals are similar but that, of the fifth orbital is different, (d) Ths shapes of all five d-orbitals are different, 180. If the nitrogen atom has electronic configuration 1s7, it would, have energy lower than that of the normal ground state, configuration 1s22s22p3, because the electrons would be, closer to the nucleus. Yet 1s7 is not observed because it, violates., (a) Heisenberg uncertainty principle, (b) Hund's rule, (c) Pauli exclusion principle, (d) Bohr postulate of stationary orbits
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STRUCTURE OF ATOM, , FACT / DEFINITION TYPE QUESTIONS, 1., 2., 3., , 4., 8., 9., , 10., 11., 12., 13., , 14., , 15., 16., , 17., 18., , 19., 20., , 21., 22., 23., 24., 25., , (b), (b) Cathode rays are never electromagnetic waves., (c) Cathode rays are made up of negatively charged, particles (electrons) which are deflected by both the, electric and magnetic fields., (d), 5. (b), 6. (c), 7. (c), (c) The electrical discharge through the gases could be, observed only at low pressure and high voltage., (b) The cathode rays (negatively charged particles, stream) originates from cathode and move towards, anode., (b) Millikan determined the value of charge on the, electron by using oil drop experiment., (b), (d) Proton is the nucleus of H-atom (H-atom devoid of its, electron), (b) Positron (positive electron, +1e0) is positively charged, electron without any mass, so it is the lightest particle, among given choices., (c) As the neutron is a chargeless particle, hence, the beam, of neutrons is not deflected by electrical or magnetic, field., (a) James Chadwick in 1932 discovered the neutrons., (b) Since electrons are negatively charged particles they, got deflected toward positively charged electrode, whereas proton being positively charged will get, deflected toward negative electrode. Since neutrons, are neutral, so they went straight., (a), (b), -rays have the least penetrating power, followed by, -rays (100 times that of -rays) and -rays (1000 times, that of -rays)., (d) X- and -rays are not deflected by the electric and, magnetic field., (c) Rutherford’s -ray scattering experiment first showed, the existence of a small positivily charged entity in the, centre of atom, called nucleus., (d) The nucleus occupies much smaller volume compared, to the volume of the atom., (a) All positive ions are deposited at small part. (nucleus, of atom)., (d) Rutherford used doubly charged helium particle., ( - particle), (d) Electrons are revolving around the nucleus, and, centrifugal force is balancing the force of attraction., (d) Number of neutrons = Mass number – Atomic number, = 70 – 30 = 40., , 31, , 26. (a) 19 + 1e– = 20 electrons., 27. (c) 18Ar40 contains 22 neutrons and 21Sc40 contains 19, neutrons. The number of neutrons = (A – Z), 28. (b) Number of p = number of e– = 89 and neutrons, 231 – 89 = 142., 29. (a) Z = 11, A = 24. Hence protons = 11 the neutrons, (24 – 11) = 13., 30. (d) For neutral atom . No. of p = No. of e– = 18 and, A = Z + No. of neutrons = 18 + 20 = 38., 31. (c) When an alpha particle is emitted from a nucleus its, atomic number decreases by two and its atomic mass, decreases by four e.g., Z, , –, , XA, , Z 2X, , A 4, , 32. (b) Nucleons are total number of protons and neutrons., Both of these are collectively known as nucleons., 33. (b) Atoms with mass number but different atomic numbers, are called isobars. Examples; 146C, 147N are isobars., 34. (d), , 17 Cl, , 35, , and, , 17 Cl, , 37, , are isotopes, so they will have, , same chemical properties., 35. (d) Atomic number is equal to number of protons or, number of electrons. Thus if two species have, different atomic number they must contain different, number of protons and electrons. Number of neutrons, = Atomic mass – Atomic number. Therefore due to, difference of atomic numbers two species also have, different number of neutrons., 36. (a), c, 37. (d), 1, l, , 2, , c, , c, 3 1, , 2, , 2, , % change in frequency =, c, 3 1, =, , c, 1, , c, 1, , 100, , 1, , 100, , 1, , 2c, 100, 3 1, c, 1, , = – 66%, 38. (b), 39. (b) The ideal body, which emits and absorbs radiations of, all frequencies, is called a black-body and the radiation, emitted by such a body is called black-body radiation., 40. (a) Energy is always absorbed or emitted in whole number, or multiples of quantum.
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EBD_7207, STRUCTURE OF ATOM, , 32, , 41., , 42., , (d) (i) Interference and diffraction support the wave, nature of electron., (ii) E = mc2 supports the particle nature of electron., hc, (iii) E h, is de-Broglie equation and it, , (d), , supports both wave nature and particle nature of, electron., c, hc, or, E h, , Atomic number of Li (Z) = 3., , 57., , 43., 44., 45., 46., 47., 48., , 49., , nh, ., 2, (b) Angular momentum of an electron in nth orbit is given, by, , =, , 50., , nh, 2, For n = 5, we have, , 58., , 59., , 60., , 51., , (d), , y, , 52., , 53., 55., 56., , (a), , (b), (a), (a), , y, , 9, , So it will 9y., rn = r1 × n2, (for hydrogen atom), rn = r × n2, as r1 = r (given), r2 = r × 22 (n = 2, for second Bohr’s orbit), = 4r, 54. (d), Radius of nth orbit = r1 n2 . (for H-atom), For hydrogen atom (n) = 1 (due to ground state), Radius of hydrogen atom (r) = 0.53 Å., , eV., , 1312, , 13.6, eV 1.51 eV., 9, , eV or E, , 13.6 Z 2, , eV, n2, For second orbit, n = 2, Z = At. no. = 1 (for hydrogen), 13.6 (1) 2, (2), , 13.6, eV, 4, , 2, , 13.6 1.6 10 19, 19, J = 5.44 10 J, 4, 61. (b) If we assume the atom to be hydrogen like, energy of, nth energy level, E, En = 21 where E1 is energy of first energy level, n, =, , E2, , 21.79 10 12, 2, 4, –12, = – 5.447 × 10 erg per atom., , E1, , E1, =, 4, , 2, , E0, , (a) Energy of an electron E, , n2, , For energy level (n = 2), , 64., 65., , 66., , 13.6, 4, , 13.6, , 3.4 eV., 2, (a) H,He+ and Li2+ are single electron species thus show, similar line spectra., (c), (c) Vth stationary state, as radii of stationary state is given, as rn = n2 × a0, n=5, , E, , 63., 32, , n2, , E2, , n2, , For IIIst orbit, , 13.6, , (d) For H atom, En, , 2.5h, , For 1st orbit y = 1, , n2, , kJ mol 1, n2, n = 4 (Fourth Bohr orbit), 1312, = –82 kJ mol–1, Given E4, 42, (a) 2nd excited state will be the 3rd energy level., , En, , 62., , Angular momentum of electron, , 13.6, , (b) We know that En =, , mvr, , 5h, 2, , r1, , relationship. En, , 3 108, , 3.75 10 8 m, 8 1015, In nanometer = 3.75× 10, which is closest to 4 × 101, (c) K.E. of emitted electron, = hv hv0 (i.e. smaller than hv )., (a), (a) At a frequency > 0, the ejected electrons come out, with certain kinetic energy., (a), (d) This statement is known as uncertainty principle which, was given by Heisenberg it is not a Bohr’s postulate., (c) Since the energy difference between two consecutive, Bohr orbits is quantized and the energy of higher orbit, is more than that of lower orbit, so an electron from, one Bohr stationary orbit can go to next higher orbit, by absorption of electromagnetic radiation of particular, wavelength or frequency., (d) For a Bohr atom, angular momentum M of the electron, , (1) 2, n2, 0.53, 0.17, Z, 3, (c) Energy of an electron in Bohr's orbit is given by the, , Radius of Li2+ ion, , (c), , 1, 1, , 2, , R, , 1, , 1, , n12, , n22, , 1.097 107, 91.15 10, , 9, , 1, 1, , 1, , m, , 91nm, , 1.097 107 m 1
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STRUCTURE OF ATOM, , 67., , 1, IE, h, , (c), , 33, , 1, , 1, , n12, , n22, , 18, , 2.18 10, , 68., , 69., , 70., , (c), , (a), , R, , 1, , 1, , n12, , n22, , (a), , 1, , 1, , n12, , n22, , n12, , –, , 21.79, , 1, , 1, , n12, , n22, , J/atom, , 72., , R, , 1, , 1, , 1, , 2, , 2, , 1, , 3, , R, , RH, , RH, , n, , 1, , 2, , 2, , 76., 77., 78., , 1 1, 8R, –, =, 1 9, 9, , 109678, 1, , 6.6 10 34, , 79., , (d), , 80., , (d) Given E = 4.4 × 10–4 j ,, , 60 10 3 10, , hc, =, E, , 81., , 34, , 3.3 10, , 24, , ch, , h, , 8 1015, , Js, , 3 108 m sec, , 1, , kg, , ;and, , 10, , 2 10, , m, , 2Å, , c, , 3.0 108, , 3.0 108, 8 1015, , 0.37 10 7, , 37.5 10 9 m, , 4 101 nm, , h, mv, , 10 33 m, , =?, , 6.6 10 34 3 108, 4.4 10, , 4, , 4.5 10 22 m, , h, h, ; m, mv, v, The velocity of photon (v) = 3 × 108 m sec–1, , (c) We know that, , 1.54 10 8 cm 1.54 10, , 10, , =, , meter, , 36 103, m/sec = 10 m/sec, 60 60, , 6.6 10 34, 10, , 3, , 6.6 10 38 m, , 10, , 85. (b) Heisenberg uncertainity principle can be explained by, the relation, x. P, , (b), (c) Bohr model can explain spectrum of any atom or ion, containing one electron only (that is H-like species), (d) Uncertainty principle which was given by Hiesenberg, and not Bohr’s postulate., (d) Bohr’s model can be applied to one electron system, only., (b) Bohr model can only explain one electron system, (a), = h/ mv ; for the same velocity, varies inversely, with the mass of the particle., h, mv, , 6.6 10, , v = 36 km/hr =, , = 9.117 × 10–6 cm, = 911.7 × 10–10 m = 911.7 Å., , 75., , E, , 32, , 34, , h = 6.6 × 10–34 J/s, m = 1000 kg, , n 22, , (d) The shortest wavelength in hydrogen spectrum of, Lyman series is given by formula :, 1, , 73., 74., , 83. (b), , 84. (b), , For second line in lyman series, n2 = 3, 1, , h, p, , 82. (c), , ;, , 1, 1, , R, , 1.54 10, , 1, , First line in Balmer series results in the transition :, n2 = 3 to n1 = 2., (c) Energy of photon obtained from the transition n = 6 to, n = 5 will have least energy., E 13.6Z 2, , 71., , 3.08 1015 s, , E for two energy levels, , E, , 10 m, , 1.4285 10, , 1 1, 1 16, , 34, , 6.625 10, , 6.626 10, , m, , h, 4, , where x = uncertainity in position, P = uncertainity in momentum, 86. (d) Heisenberg's uncertainty Principle is applicable to any, moving object., 87. (d) By Heisenberg uncertainty Principle, , x, , p, , h, 4, , (which is constant), As x for electron and helium atom is same thus, momentum of electron and helium will also be same, therefore the momentum of helium atom is equal to, 5 × 10–26 kg. m.s–1., 88. (a) Given m = 9.1 × 10–31kg, h = 6.6 × 10–34Js, v=, , 300 .001, 100, , = 0.003ms–1, , From Heisenberg's uncertainity principle, x, , 6.62 10, , 34, , 4 3.14 0.003 9.1 10, , 31, , 1.92 10 2 m
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EBD_7207, 34, , 89., , STRUCTURE OF ATOM, , (c), , x. p, v, , h, 4, , or, , 0.011, 3 10 4, 100, , x.m v, , h, ;, 4, , 3.3 cms 1, , 6.6 10, , 27, , 0.175 cm, 4 3.14 9.1 10 28 3.3, x. v value will be large for object of smallest mass, and is therefore the most significant for calculating, uncertainity., Magnetic quantum no. represents the orientation of, atomic orbitals in an atom. For example px, py & pz, have orientation along X-axis, Y-axis & Z-axis, The sub-shell are 3d, 4d, 4p and 4s, 4d has highest, energy as n + value is maximum for this., The possible quantum numbers for 4f electron are, x, , 90., , (b), , 91., , (b), , 92., , (b), , 93., , (a), , 94., , 1, 2, Of various possiblities only option (a) is possible., (b) n = 4 represents 4th orbit, , 95., , = 3 represents f subshell, m = – 2 represents orientation of f-orbital, s = 1/2 represents direction of spin of electron., The orbital is 4f., (b) For 4d orbitals, n = 4, l = 2, , n = 4, = 3, m = – 3, –2 –1, 0, 1 , 2 , 3 and s, , For s orbital l 0, For p orbital l 1, For d orbital l = 2, , m = –2, –1, 0, +1 or + 2, 1, 1, s = + and, 2, 2, Thus choice b having n = 4, l = 2, d = 1 and s =, 96., , 97., 98., , 99., , 1, is, 2, , correct., (c) n = 2, l = 1 means 2p–orbital. Electrons that can be, accommodated = 6 as p sub-shell has 3 orbital and, each orbital contains 2 electrons., (d), = 3 means f -subshell. Maximum no. of, electrons = 4 + 2 = 4 × 3 + 2 = 14, (b) m = – l to +l, through zero thus for l = 2, values of m, will be – 2, –1, 0, + 1, + 2., Therefore for l = 2, m cannot have the value –3., (c) (a) n = 3,, 0 means 3s-orbital and n + = 3, (b) n = 3,, 1 means 3p-orbital n + = 4, (c) n = 3,, 2 means 3d-orbital n + = 5, (d) n = 4,, 0 means 4s-orbital n + = 4, Increasing order of energy among these orbitals is, 3s < 3p < 4s < 3d, 3d has highest energy.., , 100. (b) Value of l = 0 ..........(n – 1), l cannot be equal to n., 101. (b) For n = 5, l = n – 1 = 5 – 1 = 4, m = 2l + 1 = 2(4) + 1 = 9, Sum of values of l and m = 9 + 4 = 13, 102. (a) Quantum number n = 3, l = 2, m = +2 represent an, orbital with, 1, s, 3d xy or 3d 2 2, x y, 2, which is possible only for one electron., 103. (d) The orbitals which have same energy are called, degenerate orbitals eg. px , p y and pz ., 104. (a) No. of radial nodes in 3p-orbital = (n – – 1), [for p ortbital = 1], =3–1–1=1, 105. (b), 106. (a) As n – l – 1 = 5 or 8 – l – 1 = 5 l = 2., 107. (b) According to given information n = 5 and l = 3., 108. (a) The number of allowed orbitals are given by n 2., Thus when n = 5, (5)2 = 25, 109. (d) Spherical shaped s-orbital shields the outer shell, electrons move effectively than p-orbital, which in turn, shields more effectively than d-orbital and so on., 110. (d) According to Hund’s rule electron pairing in p, d and f, orbitals cannot occur until each orbital of a given, subshell contains one electron each or is singly, occupied., 111. (d) We know that atomic number of gadolinium is 64., Therefore the electronic configuration of gadolinium, is [Xe] 4f 7 5d1 6s2. Because the half filled and fully, filled orbitals are more stable., 112. (b) The sub-shell with lowest value of (n + ) is filled up, first. When two or more sub-shells have same (n + ), value the subshell with lowest value of 'n' is filled up, first therefore the correct order is, orbital 4s, 3d, 4p, 5s, 4d, n+, 4+ 0 3+ 2, 4 +1 5 + 0 4 + 2, value, =4, =5, =5, =5, =6, 113. (c) Fe++ (26 – 2 = 24) = 1s2 2s2 2p6 3s2 3p6 4s0 3d 6 hence, no. of d electrons retained is 6. [Two 4s electron are, removed], 114. (b) This configuration represents ground state electronic, configuration of Cr., 1s2 2s2 2p6 3s23p63d 5 4s1, 115. (c) N(7) = 1s2 2s2 2p3, N2, , 1s 2 , 2 s 2 2 p1x, , Unpaired electrons = 1., 116. (a) Cu+ = 29 – 1 = 28 e–, thus the electronic confingration of Cu+ is, 2, 2, 6, 2, 6, 10, Cu+ (28) = 1s 2s 2 p 3s 3 p 3d, 18e, , 117. (d) This is as per the definition of Pauli’s exclusion, principle.
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STRUCTURE OF ATOM, , 35, , of Cu (29) is 1s2 2s 2 2p6 3s2, , 118. (b) Electronic configuration, 3p6 3d 10 4s1 and not 1s2, 2s2 2p6 3s2 3p6 3d 94s2 due, to extra stability of fully filled orbitals., 119. (b) According to Aufbau principle, the orbital of lower, energy (2s) should be fully filled before the filling of, orbital of higher energy starts., 120. (a) ns (n 2)f, (n 1)d np [n = 6], 121. (d) The number of sub shell is (2 l + 1). The maximum, number of electrons in the sub shell is 2 (2 l + 1), = (4 l + 2)., 122. (b), (n + l), , 5p, , 4f, , 6s, , 5+ 1, , 4+ 3, , 6+ 0, , 6, , 7, , 6, , 5d, 5+ 2, 7, , Hence the order is 5p < 6s < 4f < 5d, 123. (d), 124. (d) The d-orbital represented by option (d) will become, completely filled after gaining an electron. Therefore, option (d) is correct., , 135. (a) For d-subshell Number of orbitals = 5, l = 2, f-subshell Number of orbitals = 7, l = 3, s-subshell Number of orbitals = 1, l = 0, p-subshell Number of orbitals = 3, l = 1, 136. (b), 137. (c), 138. (b), , ASSERTION-REASON TYPE QUESTIONS, 139. (d) The statement-1 is false but the statement-2 is true, exact position and exact momentum of an electron can, never be determined according to Heisenberg’s, uncertainty principle. Even not with the help of electron, microscope because when electron beam of electron, microscope strikes the target electron of atom, the, impact causes the change in velocity and position of, electron ., 140. (a) Both assertion and reason are true and reason is the, correct explanation of assertion., Radius, rn, , MATCHING TYPE QUESTIONS, 131. (b) Isotopes have same atomic number. Isobars have, same mass number, whereas isoelectronic species, have same number of electrons although the (A) has, same number of electrons but the protons they carry, are same while in case of isolelectronic species, number of protons they carry are different., 132. (a), 133. (c), 134. (a), , n2, Z, , 2, , 4 e mZ, , 0.529Å.rn, , For first orbit of H-atom, n=1, , STATEMENT TYPE QUESTIONS, 125. (c) When both electric and magnetic field is applied,, electron strikes at point B, and at point C when only, magnetic field is applied., 126. (c), 127. (b) For statement (ii) there is no time lag between striking, of light beam and the ejection of electrons. For, statement (iii) refer statement (iv)., 128. (c) (i) Beyond a certain wavelength the line spectrum, becomes band spectrum., (ii) For Balmer series n1 = 2, (iii) For calculation of longest wavelength use nearest, value of n 2. Hence for longest wavelength in Balmer, series of hydrogen spectrum,, n1 = 2 & n2 = 3., 129. (c) Statement (i) is related to Heisenberg’s uncertainity, principle. Statement (iii) belongs to Pauli’s exclusion, principle., 130. (b) Angular quantum number determines the 3d shape of, the orbital., Spin quantum number of an electron determines the, orientation of the spin of electron relative to the chosen, axis., , n2 h2, , (1)2, 0.529Å = 0.529Å, 1, 142. (b), 143. (c), r1, , 141. (a), , CRITICAL THINKING TYPE QUESTIONS, 144. (c), 145. (d), , e, for (i) neutron, m, , (ii), , -particle, , 147. (d), 148. (c), 149. (a), , 150. (c), , 0, 2, 4, , 0.5, , 1, 1, 1, 1, (iv) electron, 1837, 1 1837, N3– The amount of deviation depends upon the, magnitude of negative charge on the particle., The lesser is the mass of particle, greater is the, deflection., Deuterium and an -particle have identical values of, e/m., Considering the core of an atom, higher the positive, charge concentrated in the nucleus, greater the, repulsion for an alpha-particle., kq1 ( ze ), Coulombic force of repulsion =, r2, q1 = charge on -particle, (ze) = charge on nucleus of atom, (iii) proton, , 146. (b), , 0, 1
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EBD_7207, STRUCTURE OF ATOM, , 36, , 151. (d) Atomic number = No. of protons = 8, Mass number = No. of protons + No. of neutrons, = 8 + 8 = 16, Since the no. of electrons are two more than the no. of, protons, hence, it is a binegative species. Thus, the, species is 16 O82 ., 152. (c) e/m waves shown in figure A has higher wavelength, in comparison to e/m waves shown in figure B., Thus these waves also differ in frequency and, c, energy., , Also, linear momentum (mv) = 7.3 × 10 –34 kg ms–1, Then angular momentum will be, = (mv) × r, = (7.3 × 10–34 kg ms–1) (158.7 pm), = 7.3 × 10–34 kg ms–1 × (158.7 × 10–12 m), = 11.58 × 10–48 kg m2 s–1, = 11.58 × 10–45 g m2s–1, 156. (d) Given : Radius of hydrogen atom = 0.530 Å, Number of, excited state (n) = 2 and atomic number of hydrogen, atom (Z) = 1. We know that the Bohr radius., n2, Radius of atom, Z, , (r ), 1, , (A), , E1, , E2, , 2, , c, , = 1015, b = 1014,, 17, 15, c = 10 , d = 0.85 × 10, 15, and, e = 10 × 10 ,, 154. (d) From the expression of Bohr’s theory, we know that, , 13.6 (3) 2, , =, , hc, , 153. (d) E = h, and, , (2) 2, , n1, , 159. (d), , m e v1r1, m e v2 r2, , n2, , EnH, , h, 2, , v1, v2, , 5, 4, , 5, 4 5, , 1, 4, , 155. (b) Z = 3 for Li2+ ions, , 52.9 n 2, So rn, Z, n = 3, Z = 3, rn, , 52.9, , 3, , 2, , 3, = 158.7 pm, , 2.18 10 18, , pm, , EnHe, , +, , If nH = 1, If nH = 2, If nH = 3, , 2, h, , Given, r1 = 5 r2, n1 = 5, n2 = 4, me v1 5r2, me v2 r2, , +, , 2.18 10, , n2, , n2, , n2, , 4, , 2, nH, , n2 +, He, , He, , Then nHe+ = 2, Then nHe+ = 4, Then nHe+ = 6, , n2 h 2, , 161. (c), 162. (d) In S.I. units the P.E., For Li2+, Z = 3., , P.E., , 3e 2, ., 4 0r, , Ze 2, ., 4 0r, , +, , nHe+ = 2 × nH, , 4 2 me2 Z, where n = no. of orbit, h = Plank’s constant, e = charge on one electron, m = mass of one electron, Z = atomic number, , 1: 4, , J, , 2.18 10 18 4, , J, , +, , 1, , 160. (d) Radius of nth orbit, , 18, , 2, nH, , Z2, He, , n1 h, n2 2, , z2, , = –30.6 eV, , Z2, 2.18 10 18 2 J, nH, , EnH, , EnHe, , h, 2, , & me v 2 r2, , 13.6, , Energy required = 0 – (–30.6) = 30.6 eV, 158. (d) Except Al3+ all contain one electron and Bohr’s, model could explain the spectra for one electron, system, Bohr’s model was not able to explain the, spectra of multielectron system., , a, , me v1r1, , 2.12 Å, , 157. (c) Energy of electron in 2nd orbit of Li+2, , 1, , 2, , (B), , 4 0.530, , hc, , (2)2, 0.530, 1, , J
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STRUCTURE OF ATOM, , 37, , 163. (c) Series limit is the last line of the series, i.e. n2 = ., 1, , R, , 1, , 1, , n12, , n 22, , 1, , n12, , 2, , R, , B, , n12, , 5mA 0.05, mA 0.1, , 09677.76, , 12186.3, , n12, , 109677.76, 12186.3, , n12, , 1, , R, , A, , 9, , n1, , 1, , 2, , 2, , R, , 1, , 1, , 22, , 42, , Z2 R, , 3R, 4, , R, , 1, , 2, , n12, , n 22, , v2, v1, , 9, 4, , v1, v2, , 4, 9, , KE, , 1, mv 2, 2, , 1, , 1, , n12, , n 22, , or, , R, , 1, , 1, , n12, , n 22, , 1, , 1, , n12, , n 22, , 3, 4, , n1 = 1 and n2 = 2., 165. (c) The wavelengths of elements decreases with increase, , 166. (c) Energy of a photon, E, 6.626 10, , KE1, KE2, , 171. (a), , h, mv, , in their mass., , 34, , 167. (a), , 1, m, 2, , 2, , (Js) 3 108 (ms 1 ), , h, , 1 2, mv, 2, , h(, , v12, , m1, m2, , v 22, , v2, v1, , 9, 1, , 4, 9, , 2, , 16, 9, , Given mass of an electron(v), , 9.1 10, , 28, , g;, , 0.001%, , 0.001, ;, 100, , Actual velocity of the electron, = 6× 10–19 J, , 0.001, 0.3 cm/s ., 100, Planck’s constant (h) = 6.626×10–27erg-sec., ( v) 3 104, , =1021, , h, , 1, 9, , Accuracy in velocity, , hc, , No. of photons emitted per second, 600 (J), , h, mv, , Velocity of electron (v) 3 10 4 cm/s;, , 331.3 10 9 (m), , 6 10 9 (J), , 2.5 5 / 2, , 5:2, , B, , 1, , 1, , 5 0.5, , m 2 v2 1, ;, m1v1 4, , 1, , 3R, 4, , For hydrogen atom,, , :, , m Bv B, m A vA, , 170. (d) de Broglie wavelength, , 3, , The line belongs to Paschen series., 164. (d) For He+ ion,, , h / mA v A, h / mB v B, , A, , Uncertainty in the position of the electron, , ( x), , 0, 0), , 172. (a), , 2h, (, 0), m, 168. (a) According to de-Broglie,, v, , de-Broglie wavelength,, , =1.93 cm, p=m v, Substituting the given values of x and m, we get, 1×10–18 g cm s–1 = 9×10–28 g × v, or, , h, mv, where m = mass of electron, v = velocity, 169. (d) Given, vA = 0.1 ms–1 and vB = 0.05 ms–1 also,, mB = 5mA, h, mv, , h, 6.626´10-27 ´ 7, =, 4 m v 4 ´ 22 ´ (9.1´10-28 )´ 0.3, , v, , 1 10, , 18, , 9 10 28, = 1.1 × 109 cm s–1 1×109 cm s–1, i.e. option (a) is correct., 173. (b) According to Heisenberg uncertainty principle., x.m v, , Here v, , h, 4, , 600 0.005, 100, , x, , 0.03, , h, 4 m v
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EBD_7207, STRUCTURE OF ATOM, , 38, , So, x, , 6.6 10, , 34, , 4 3.14 9.1 10 31 0.03, = 1.92 × 10–3 meter, , h, 4, , h, [, 4, , or m v m v., or, , (b), (c), (d), , p= m v], , h, , 2, , 4 m2, , 1 h, 2m, 4 m, Thus option (a) is the correct option., For n = 5, l may be 0, 1, 2, 3 or 4, For l = 4, m = 2l + 1 = 2 × 4 + 1 = 9, = –4, – 3, – 2, –1, 0, +1, +2, +3, +4, 1, For m = 0, s, 2, Hence, (a) is correct option., For any value of n, the value of l cannot be equal or, greater than value of n, hence it is incorrect., For l = 0, m = 0 hence it is incorrect., The value of s can never be zero. Thus this option is, also incorrect., , or, , 175. (a), , v, , v, , h, , 2, , 1, 1, and, 2, 2, = 3,, Thus for n, may be 0, 1 or 2; but not 3, m may be –2, –1, 0, +1 or +2, 1, 1, s may be, or, 2, 2, , s=, , h, 174. (a) We know p. x, 4, since p = x (given), p. p, , 176. (c) Possible values of and m depend upon the value of n, = 0 to (n – 1), m = – to + through zero, , 177. (c), 178. (b) (A)4 p (B) 4 s, (C) 3 d (D) 3 p, According to Bohr Bury's (n + ) rule, increasing order, of energy will be (D) < (B) < (C) < (A)., Note : If the two orbitals have same value of (n + ), then the orbital with lower value of n will be filled first., 179. (c) First four orbitals contain four lobes, while fifth orbital, consists of only two lobes. The lobes of dxy orbital lie, between x and y axis. Similarly in the case of dyz and, dzx. their lobes lie between yz and zx axis respectively., Four lobes of d 2 2 orbital are lying along x and y, x, , y, , axis while two lobes of d, , z2, , orbital are lying along z-, , axis., 180. (c) As per Pauli exclusion principle "no two electrons in, the same atom can have all the four quantum numbers, equal or an orbital cannot contain more than two, electrons and it can accommodate two electrons only, when their directions of spin are opposite".
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3, CLASSIFICATION OF ELEMENTS AND, PERIODICITY IN PROPERTIES, FACT / DEFINITION TYPE QUESTIONS, 1., , 2., , 3., , 4., , 5., , 6., , 7., , Periodic classification of elements is used to examine the, (a) periodic trends in physical properties of elements, (b) periodic trends in chemical properties of elements, (c) Both (a) and (b), (d) None of the above, Cl, Br, I, if this is Dobereiner’s triad and the atomic masses, of Cl and I are 35.5 and 127 respectively the atomic mass of, Br is, (a) 162.5, (b) 91.5, (c) 81.25, (d) 45.625, If the two members of a Dobereiner triad are phosphorus, and antimony, the third member of this triad is, (a) arsenic, (b) sulphur, (c) iodine, (d) calcium, The law of triads is applicable to a group of, (a) Cl, Br, I, (b) C, N, O, (c) Na, K, Rb, (d) H, O, N, In 1800, only ....X.... elements were known. Here, X refers to, (a) 63, (b) 31, (c) 114, (d) 92, Johann Doberiner gave the idea of trends among physical, and ...X... of several groups of three elements. Here, X refers to, (a) atomic number, (b) atomic mass, (c) chemical properties, (d) None of these, Which of the following is the correct set of elements to, Dobereiner’s triads ?, (a), (c), , (b), , Li Na K, 7, , Fe, , 23, , 39, , Ni, , Co, , Br, 80, , Cl, , I, , 35.5 127, , (d) Data insufficient, , 55.85 58.71 58.93, , 8., , 9., , On which of the following Dobereiner’s Triad law is not, applicable?, (a) Cl, Br, I, (b) Ca, Sr, B, (c) F, Cl, Br, (d) Li, Na, K, Newlands could classify elements only upto –, (a) copper, (b) chlorine, (c) calcium, (d) chromium, , 10. According to Newlands theory, when elements are, arranged in the order of increasing atomic weight which, number element had similar properties to the first element., (a) third, (b) seventh, (c) eighth, (d) sixth, 11. Lothar Meyer plotted the physical properties such as atomic, volume, melting point and ...X... against atomic weight. Here,, X refers to, (a) mass, (b) boiling point, (c) surface tension, (d) None of these, 12. The most significant contribution towards the development, of periodic table was made by, (a) Mendeleev, (b) Avogadro, (c) Dalton, (d) Cavendish, 13. Noble gases were included in Mendeleev’s periodic table in, the, (a) 1st group, (b) 7th group, (c) 8th group, (d) None of these, 14. Mendeleev classified elements in, (a) increasing order of atomic groups, (b) eight periods and eight groups, (c) seven periods and nine groups, (d) eight periods and seven groups, 15. Select the correct chronological order for the discoveries, of the following scientists., Dobereiner, Newlands, Chancourtois, Mendeleev, (a) Chancourtois, Dobereiner, Newlands, Mendeleev, (b) Dobereiner, Chancourtois, Newlands, Mendeleev, (c) Dobereiner, Newlands, Chancourtois, Mendeleev, (d) Chancourtois, Newlands, Dobereiner, Mendeleev, 16. The molecular formula of chloride of Eka-Aluminium and, Eka-Silicon respectively are, (a) GaCl3 and SiO4, (b) GaCl3 and AlCl3, (c) AlCl3 and SiCl4, (d) GaCl3 and GeCl4, 17. Who developed long form of the periodic table?, (a) Lothar Meyer, (b) Neils Bohr, (c) Mendeleev, (d) Moseley, 18. At present, how many elements are known, (a) 110, (b) 112, (c) 113, (d) 118
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EBD_7207, 40, , 19., , 20., , 21., , 22., , 23., , 24., , 25., , 26., , 27., , 28., , 29., , 30., , CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES, , Which of the scientists given below discovered that, periodic table should be based on the atomic number ?, (a) Mendeleev, (b) Newlands, (c) Moseley, (d) Lothar Meyer, How many elements are there in 6th period of periodic table?, (a) 18, (b) 8, (c) 30, (d) 32, Modern periodic table is based on the atomic number of the, elements. The experiment which proved the significance of, the atomic number was, (a) Mulliken’s oil drop experiment, (b) Mosley’s work on X-ray spectra, (c) Bragg’s work on X-ray diffraction, (d) Discovery of X-rays by Rontgen, The period number in the periodic table corresponds to the, ...A... principal quantum number (n) of the elements. Here, A, refers to, (a) lowest, (b) highest, (c) middle, (d) None of these, The symbol and IUPAC name for the element with atomic, number 120, respectively are, (a) Ubn and unbinilium (b) Ubn and unbiunium, (c) Ubn and unnibium, (d) Ubn and unnilium, Element with which of the following atomic number was, named by American Society as Rutherfordium, while by, Soviet Society it was named as Kurchatovium?, (a) 108, (b) 104, (c) 114, (d) 110, What is the IUPAC name of the element with atomic number, 114 ?, (a) Unununnium, (b) Unnilquadium, (c) Ununquadium, (d) Unnilennium., Element with electronic configuration 1s2 2s2 2p6 3s2 3p6, 3d10 4s2 4p6 4d10 5s2 5p3 belongs to the following group of, the periodic table, (a) 2nd, (b) 5th, (c) 3rd, (d) 7th, The long form of periodic table consists of, (a) seven periods and eight groups, (b) seven periods and eighteen groups, (c) eight periods and eighteen groups, (d) eighteen periods and eight groups, All the members in a group in long form of periodic table, have the same, (a) valence, (b) number of valence electrons, (c) chemical properties, (d) All of the above, Elements of which group form anions most readily?, (a) Oxygen family, (b) Nitrogen family, (c) Halogens, (d) Alkali metals, Element having atomic no. of 56 belongs to which of the, following block of periodic table?, (a) p- block, (b) d-block, (c) f- block, (d) s-block, , 31., , 32., , 33., , 34., , 35., , 36., , In the modern periodic table one of the following does not, have appropriate position –, (a) transition elements, (b) inert gases, (c) inner transition elements, (d) halogens, If the atomic number of an element is 33, it will be placed in, the periodic table in the, (a) First group, (b) Third group, (c) Fifth group, (d) Seventh group., An atom has electronic configuration 1s2 2s2 2p6 3s2 3p6, 3d3 4s2, you will place it in which group?, (a) Fifth, (b) Fifteenth, (c) Second, (d) Third, Which of the following is not an actinoid ?, (a) Curium (Z = 96), (b) Californium (Z = 98), (c) Uranium (Z = 92), (d) Terbium (Z = 65), The period number in the long form of the periodic table is, equal to, (a) magnetic quantum number of any element of the period., (b) atomic number of any element of the period., (c) maximum Principal quantum number of any element of, the period., (d) maximum Azimuthal quantum number of any element, of the period., The electronic configuration of four elements are given, below. Which elements does not belong to the same family, as others?, (a), , 37., , 38., , 39., , 40., , [Xe]4f 14 5d10 ls 2, , (b) [Kr]4d10 5s 2, , (c) [Ne]3s23p5, (d) [Ar] 3d10 4s2, The elements with atomic numbers 9, 17, 35, 53 and 85 belong to, (a) alkali metals, (b) alkaline earth metals, (c) halogens, (d) noble gases, Which of the following pairs has both members from the, same period of the periodic table., (a) Na – Ca, (b) Na – Cl, (c) Ca – Cl, (d) Cl – Br, The elements which are characterized by the outer electronic, configuration ns1 to ns2 np6 are collectively called, (a) Transition elements, (b) Representative elements, (c) Lanthanides, (d) Inner transition elements, f-block elements are called inner transition elements because, (a) they have properties similar to those of transition, elements, (b) they exist in between transition elements, (c) the last electron enters into the f-orbital of the, penultimate shell, (d) the last electron enters into any orbital of penultimate, shell
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CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES, , 41., , An element, which belongs to third period and group 16 in, the periodic table has electronic configuration., (a) 1s2, 2s2 2p6, 3s2 3p3 (b) 1s2, 2s2 2p6, 3s2 3p4, (c) 1s2, 2s2 2p6, 3s2 3p5 (d) 1s2, 2s2 2p4, 42. Which of the following is non-metallic ?, (a) B, (b) Be, (c) Mg, (d) Al, 43. Which group of the periodic table contains coinage metal ?, (a) IIA, (b) IB, (c) IA, (d) None of these, 44. The only non-metal which is liquid at ordinary temperature is, (a) Hg, (b) Br2, (c) NH3, (d) None of these, 45. Which is a metalloid?, (a) Pb, (b) Sb, (c) Bi, (d) Zn, 46. In the long form of the periodic table all the non-metals are, placed in, (a) s-block, (b) p-block, (c) f-block, (d) d-block, 47. Arrange the following elements in the order of their, increasing non-metallic character, Li, O, C, Be, F, (a) F < O < C < Be < Li, (b) Li < Be < C < O< F, (c) F < O < C < Be < Li, (d) F < O < Be < C < Li, 48. Which is the correct order of ionic sizes (At. No. : Ce = 58,, Sn = 50, Yb = 70 and Lu = 71) ?, (a) Ce > Sn > Yb > Lu, (b) Sn > Ce > Yb > Lu, (c) Lu > Yb > Sn > Ce, (d) Sn > Yb > Ce > Lu, 49. The order of increasing sizes of atomic radii among the, elements O, S, Se and As is :, (a) As < S < O < Se, (b) Se < S < As < O, (c) O < S < As < Se, (d) O < S < Se < As, 50. In the ions P3–, S2– and Cl–, the increasing order of size is, (a) Cl–, S2–, P3–, (b) P3–, S2–, Cl–, 2–, –, 3–, (c) S , Cl , P, (d) S2–, P3–, Cl–, 51. Which of the following is correct ?, (a) Isoelectronic ions have same nuclear charge, (b) Isoelectronic ions have same neutrons, (c) Isoelectronic ions have same number of electrons, (d) All are correct, 52. On going down a main sub-group in the periodic table, (example Li to Cs in IA or Be to Ra in IIA), the expected, trend of changes in atomic radius is a, (a) continuous increase, (b) continuous decrease, (c) periodic one, an increase followed by a decrease, (d) decrease followed by increase, 53. Why the size of an anion is larger than the parent atom?, (a) Due to increased repulsion among the electrons., (b) Due to decrease in effective nuclear charge., (c) Due to increased in effective nuclear charge., (d) Both (a) and (b), , 41, , 54. Which ionisation potential (IP) in the following equations, involves the greatest amount of energy ?, (b) K, (a) Na, K2, e, Na, e, (c) C 2, (d) Ca, Ca 2, e, C3, e, 55. Arrange S, P, As in order of increasing ionisation energy, (a) S < P < As, (b) P < S < As, (c) As < S < P, (d) As < P < S, 56. The statement that is not correct for periodic classification, of elements is :, (a) The properties of elements are periodic function of, their atomic numbers., (b) Non-metallic elements are less in number than metallic, elements., (c) For transition elements, the 3d-orbitals are filled with, electrons after 3p-orbitals and before 4s-orbitals., (d) The first ionisation enthalpies of elements generally, increase with increase in atomic number as we go along, a period., 57. Consider the following changes, , 58., , 59., , 60., , 61., , 62., , 63., 64., , A A, e : E1 and A, A 2 e : E2, The energy required to pull out the two electrons are E1 and, E2 respectively. The correct relationship between two, energies would be, (a) E1 < E2, (b) E1 = E2, (c) E1 > E2, (d) E1 E2, Of the given electronic configurations for the elements,, which electronic configuration indicates that there will be, abnormally high difference in the second and third ionization, energy for the element?, (a) 1s2 2s2 2p6 3s2, (b) 1s2 2s2 2p6 3s1, (c) 1s2 2s2 2p6 3s2 3p1, (d) 1s2 2s2 2p6 3s2 3p2, Alkali metals are powerful reducing agents because, (a) these are metals, (b) their ionic radii are large, (c) these are monovalent, (d) their ionisation potential is low, Which of the following metals requires the radiation of, highest frequency to cause the emission of electrons ?, (a) Na, (b) Mg, (c) K, (d) Ca, Halogens and chalcogens family have highly ...P.. electron, gain enthalpy. Here, P refers to, (a) negative, (b) positive, (c) zero, (d) infinity, Which of the following represents the correct order of, increasing electron gain enthalpy with negative sign for the, elements O, S, F and Cl ?, (a) Cl < F < O < S, (b) O < S < F < Cl, (c) F < S < O < Cl, (d) S < O < Cl < F, The electron affinity for the inert gases is –, (a) zero, (b) high, (c) negative, (d) positive, The element with positive electron gain enthalpy is, (a) hydrogen, (b) sodium, (c) oxygen, (d) neon
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EBD_7207, CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES, , 42, , 65., , 66., , 67., , 68., , Which of the following will have the least negative electron, gain enthalpy?, (a) P, (b) S, (c) Cl, (d) F, Which is the correct order of electronegativity ?, (a) F > N < O > C, (b) F > N > O > C, (c) F > N > O < C, (d) F < N < O =C, The correct order of decreasing electronegativity values, among the elements I-beryllium, II-oxygen, III-nitrogen and, IV-magnesium is, (a) II > III > I > IV, (b) III > IV > II > I, (c) I > II > III > IV, (d) I > II > IV > III, An element having electronic configuration, , 74., , 75., , 1s 2 2s 2 2p 6 3s 2 3p 6 4s1 forms, , 69., , 70., , 71., , 72., , (a) Acidic oxide, (b) Basic oxide, (c) Amphoteric oxide, (d) Neutral oxide, Diagonal relationship is shown by, (a) All elements with their diagonally downward elements, towards right, (b) Most of the elements of second period, (c) All 3d series elements, (d) None of the above, In any period the valency of an element with respect to, oxygen, (a) Increases one by one from IA to VIIA, (b) Decreases one by one form IA to VIIA, (c) Increases one by one from IA to IVA and then, decreases from VA to VIIA one by one, (d) Decreases one by one from IA to IVA and then, increases from VA to VIIA one by one, What will be the formula of the compound formed by the, silicon and bromine ?, (a) SiBr2, (b) SiBr4, (c) SiBr3, (d) SiBr, Which of the following sequence correctly represents the, decreasing acidic nature of oxides ?, (a) Li2O > BeO > B2O3 > CO2 > N2O3, (b) N2O3 > CO2 > B2O3 > BeO > Li2O, (c) CO2 > N2O3 > B2O3 > BeO > Li2O, (d) B2O3 > CO2 > N2O3 > Li2O > BeO, , STATEMENT TYPE QUESTIONS, 73., , Choose the correct coding for following statements. Here, T stands for True and F stands for False statement., (i) Mendeleev left several gaps in his periodic table for, the undiscovered elements., (ii) The gap under aluminium and a gap under silicon was, left and these elements were called Eka aluminium and, Eka silicon., (iii) Germanium was placed in place of Eka-aluminium and, gallium was placed in place of Eka silicon., (a) TTT, (b) TFF, (c) TTF, (d) TFT, , 76., , Which of the following statement(s) about the modern, periodic table is/are incorrect ?, (i) The elements in the modern periodic table are arranged, on the basis of their decreasing atomic number, (ii) The elements in the modern periodic table are arranged, on the basis of their increasing atomic masses, (iii) Isotopes are placed in adjoining group(s) in the periodic, table, (iv) The elements in the modern periodic table are arranged, on the basis of their increasing atomic number, (a) (i) only, (b) (i), (ii) and (iii), (c) (i), (ii) and (iv), (d) (iv) only, Consider the following statements:, (i) The discovery of inert gases later on did not disturb, Mendeleev's arrangement., (ii) In the present periodic table, periodicity in the, properties of elements is related to the periodicity in, their electronic configurations., Which of these statement(s) is/are correct ?, (a) (i) only, (b) (ii) only, (c) Both (i) and (ii), (d) Neither (i) nor (ii), , Which of the following statements are correct?, (i) The second period (n = 2) starts with lithium and third, electron enters the 2s orbital. The next element,, beryllium has four electrons and has the electronic, configuration 1s22s2. From the next element boron,, the 2p orbitals are filled with electrons when the L, shell is completed at neon (2s22p6). Thus there are 8, elements in the second period., (ii) Successive filling of 3s and 3p orbitals gives rise to, the third period of 8 elements from sodium to argon., (iii) The fourth period (n = 4) starts at potassium and the, added electron fill up the first 4s and 4p orbitals than, 3d orbital is filled., (iv) Fifth period begins with rubidium with the filling of 5s, orbital and ends at xenon with the filling up of the 5p, orbital., (a) (i) and (ii), (b) (i), (ii) and (iii), (c) (iii) and (iv), (d) (i), (ii) and (iv), 77. With reference to the chemical element with atomic number, 17, consider the following statements:, (i) It belongs to second period in the periodic table of, chemical elements., (ii) It forms anion with unit negative charge., Which of the statement(s) given above is/are correct ?, (a) (i) only, (b) (ii) only, (c) Both (i) and (ii), (d) Neither (i) nor (ii), 78. Choose the correct codes for the following statements, related to s-block elements. Here ‘T’ stands for true and F, stands for false statement., (i) They are all reactive metals with low ionization, enthalpies., (ii) Their metallic character and reactivity increase as we, go down the group., (iii) They are found in pure form in nature., (iv) All the compounds of s-block elements are ionic in, nature., (a) TTFF, (b) TTFT, (c) TTTF, (d) TFFF
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CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES, , 79., , 80., , 81., , 82., , 83., , 84., , Consider the following statements:, (i) The elements silicon, germanium and arsenic are called, metalloids., (ii) Metalloids have properties quite different from those, of metals and non-metals., Which of these statement(s) is/are correct ?, (a) (i) only, (b) (ii) only, (c) Both (i) and (ii), (d) Neither (i) nor (ii), Consider the following statements:, (i) Metals will be found on the right side of the periodic, table., (ii) The element P, S and O belong to the same period., Which of these statement(s) is/are correct ?, (a) (i) only, (b) (ii) only, (c) Both (i) and (ii), (d) Neither (i) nor (ii), Consider the following statements:, (i) Atomic radii decreases across a row of the periodic, table when we move from left to right., (ii) Atomic radii increases down the column as we move, from top to bottom., (iii) Although the order of elements is based on atomic, numbers, vertical families share similar chemical, properties., Which of the statement(s) given above is/are correct?, (a) (i) and (ii), (b) (i) and (iii), (c) (ii) and (iii), (d) (i), (ii) and (iii), Consider the following statements:, (i) Fluorine has the highest electron affinity in the periodic, table., (ii) Noble gases are placed extremely left in periodic table., (iii) Magnesium is more metallic in nature than sodium., Which of these statement(s) is/are correct ?, (a) (i) and (ii), (b) (i) and (iii), (c) Only (i), (d) Only (ii), , Which of the following statement(s) is/are incorrect?, (i) Ionization enthalpy is expressed in units of kJmol–1., (ii) Ionization enthalpy is always positive., (iii) Second ionization enthalpy will be higher than the, third ionization enthalpy., (a) Only (ii), (b) Only (iii), (c) (ii) and (iii), (d) None of these, Consider the following statements:, (i) There are 16 groups and 7 periods in the modern, periodic table., (ii) Electro-positive character decreases on moving down, a group., (iii) Electro-negativity in a period increases right from the, alkali metal to the inert gas element., Which of these statement(s) is/are correct ?, (a) (i) and (ii), (b) (i) and (iii), (c) All are correct, (d) All are incorrect, , 43, , 85. Read the following three statements and choose the correct, option. Here T stands for true and F stands for false, statement., (i) Boron has a smaller first ionization enthalpy than, beryllium., (ii) Nitrogen has smaller first ionization enthalpy than, oxygen., (iii) The first ionization enthalpy increases across a, period., (a) FTT, (b) TFT, (c) TFF, (d) FFT, 86. Consider the following statements, (i) The radius of an anion is larger than that of the parent, atom., (ii) The ionization energy generally increases with, increasing atomic number in a period., (iii) The electronegativity of elements increases on moving, down across a group., Which of the above statements is/are correct?, (a) (i) alone, (b) (ii) alone, (c) (i) and (ii), (d) (ii) and (iii), , MATCHING TYPE QUESTIONS, 87. Match the Column-I and Column-II and select correct, answer by given codes., Column-I, Column-II, (Year), (The number of elements, discovered), (A) 1800, (p) 118, (B) 1865, (q) 63, (C) At present, (r) 31, (a) A – (q), B – (p), C – (r), (b) A – (r), B – (p), C – (q), (c) A – (q), B – (r), C – (p), (d) A – (r), B – (q), C – (p), 88. Match the columns., Column-I, Column-II, (A) Newland law, (p) Atomic mass vs, of octaves, Atomic volume, (B) Mendeleev, (q) Li, Na, K, (C) Electronic, (r) One to seven groups subconfiguration, divided into group A and B, (D) Lother Meyer, (s) Periodic repetition of, properties of elements, (E) Dobereiner's triad (t) Only 56 elements known, (a) A–(t); B–(s); C–(r); D–(p); E–(q), (b) A–(t); B–(r); C–(s); D–(p); E – (q), (c) A–(t); B–(r); C–(s); D–(q); E – (p), (d) A–(r); B–(t); C–(s); D–(p); E – (q)
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EBD_7207, 44, , CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES, , 89. Match the columns :, Column-I, Column-II, (A) On arraging in order of (p) Mendeleev, atomic weights, physical, and chemical properties, are repeated at regular, intervals., (B) Elements are arranged in (q) Lothar Meyer, the order of increasing, atomic weights., (C) Elements were arranged (r) Moseley, on the basis of similar, properties ignoring order, of atomic weights, (D) Atomic number is a, (s) Chancourtois, more fundamental, property of an element, than its atomic mass, (a) A – (p), B – (q), C – (s), D – (r), (b) A – (r), B – (s), C – (r), D – (p), (c) A – (q), B – (p), C – (s), D – (r), (d) A – (q), B – (s), C – (p), D – (r), 90. Match Column-I (IUPAC nomenclature of element) with, Column-II (IUPAC official name)., Column-I, Column-II, (A) Unnilhexium, (p) Lawrencium, (B) Unniltrium, (q) Dubnium, (C) Unnilunium, (r) Seaborgium, (D) Unnilpentium, (s) Mendelevium, (a) A – (s), B – (p), C – (r), D – (q), (b) A – (r), B – (p), C – (s), D – (q), (c) A – (r), B – (s), C – (p), D – (q), (d) A – (q), B – (r), C – (p), D – (s), 91. Match the columns., Column-I, Column-II, (Name of element), (Group of element), (A) Nitrogen, (p) 15, (B) Aluminium, (q) 16, (C) Chlorine, (r) 17, (D) Oxygen, (s) 13, (E) Copper, (t) 11, (a) A – (p), B – (s), C – (r), D – (q), E – (t), (b) A – (s), B – (p). C – (r), D – (q), E – (t), (c) A – (p), B–(s), C – (q), D – (r), E – (t), (d) A – (p), B – (s), C – (r), D – (t), E – (q), 92. Match the columns., Column-I, Column-II, (Name of element), (Period of element), (A) Hydrogen, (p) 3, (B) Sodium, (q) 4, (C) Calcium, (r) 6, (D) Barium, (s) 1, (E) Iodine, (t) 5, (a) A – (p), B – (s), C – (q), D – (r), E – (t), (b) A – (s), B – (p), C – (q), D – (r), E – (t), (c) A – (s), B – (q), C – (p), D – (r), E – (t), (d) A – (s), B – (p), C – (q), D – (t), E – (r), , Match the columns., Column-I, Column-II, (A) ‘s’ block elements, (p) Cr, (B) ‘p’ block elements, (q) Na, (C) ‘d’ block elements, (r) Ce, (D) ‘f’ block elements, (s) Si, (a) A – (s), B – (q), C – (p), D – (r), (b) A – (q), B – (s), C – (r), D – (p), (c) A – (q), B – (p), C – (s), D – (r), (d) A – (q), B – (s), C – (p), D – (r), 94. Match the columns., Column-I, Column-II, (A) Element with largest size, (p) Boron, in second period, (B) Element with smallest size, (q) Fluorine, in group 13, (C) Element with maximum, (r) Bromine, non-metallic character, (D) Element with smallest size, (s) Lithium, in fourth period, (E) Element with most metallic, (t) Lead, character in group 14, (a) A – (s), B – (p), C – (q), D – (t), E – (r), (b) A – (p), B – (s), C – (q), D – (r), E – (t), (c) A – (s), B – (q), C – (p), D – (r), E – (t), (d) A – (s), B – (p), C – (q), D – (r), E – (t), 95. Match the columns., Column-I, Column-II, (A) Electronegativity, (p) Isotopes, (B) Lanthanides, (q) increases along a period, (C) Transition elements, (r) f-group of elements, (D) Ionisation energy, (s) d-group of elements, (E) Elements of same, (t) decreases along a group, atomic number, but different, mass number, (a) A – (q), B – (r), C – (s), D – (p), E – (t), (b) A – (r), B – (q), C – (s), D – (t), E – (p), (c) A – (q), B – (r), C – (s), D – (t), E – (p), (d) A – (q), B – (s), C – (r), D – (t), E – (q), 96. Match Column-I with Column-II and select the correct, answer by the given codes., Columnn-I, Column-II, (Atoms), (Properties), (A) He, (p) High electronegative, (B) F, (q) Most electropositive, (C) Rb, (r) Strongest reducing agent, (D) Li, (s) Highest ionisation energy, (a) A – (s), B – (q), C – (r), D – (p), (b) A – (p), B – (s), C – (q), D – (r), (c) A – (s), B – (p), C – (r), D – (q), (d) A – (s), B – (p), C – (q), D – (r), 93.
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CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES, , 97., , Match the Column-I and Column-II and select the correct, answer by given codes., Column-I, Column-II, (Elements), (Properties), (A) Li+ < Al3+ < Mg2+ < K+ (p) DEA (Electron affinity), (B) Li+ > Al3+ > Mg2+ > K+ (q) Ionic radii, (C) Cl > F > Br > I, (r) EN (Electronegativity), (D) F > Cl > Br > I, (s) Effective nuclear charge, (a) A – (q), B – (s), C – (r), D – (p), (b) A – (q), B – (s), C – (p), D – (r), (c) A – (s), B – (q), C – (r), D – (p), (d) A – (s), B – (q), C – (p), D – (r), 98. Match the columns on the basis of diagonal relationship, between elements., Column-I, Column II, (A) Li, (p) Na, (B) Be, (q) Al, (C) B, (r) Si, (s) Mg, (a) (A) – (s), B – (r), C – (p), (b) (A) – (s), B – (q), C – (r), (c) (A) – (s), B – (q), C – (p), (d) (A) – (q), B – (s), C – (p), 99. Match the columns, Column-I, Column-II, (A) [BF4]–, (p) 7, +7, (B) [A1F6]3–, (q) 4, +4, (C) OF2, (r) 6, +3, (D) SiF4, (s) 2, +2, (E) IF7, (t) 4, +3, (a) A – (s), B – (q), C – (t), D – (r), E – (p), (b) A – (t), B – (r), C – (s), D – (q), E – (p), (c) A – (q), B – (r), C– (t), D – (s), E – (p), (d) A – (r), B – (q), C – (s), D – (p), E – (t), , ASSERTION-REASON TYPE QUESTIONS, Directions : Each of these questions contain two statements,, Assertion and Reason. Each of these questions also has four, alternative choices, only one of which is the correct answer. You, have to select one of the codes (a), (b), (c) and (d) given below., (a) Assertion is correct, reason is correct; reason is a correct, explanation for assertion., (b) Assertion is correct, reason is correct; reason is not a, correct explanation for assertion, (c) Assertion is correct, reason is incorrect, (d) Assertion is incorrect, reason is correct., 100. Assertion : In a triad, the three elements present have same, gaps of atomic masses., Reason : Elements in a triad have similar properties., 101. Assertion : According to Mendeleev, periodic properties, of elements is a function of their atomic number., Reason : Atomic number is equal to the number of protons., 102. Assertion : Atomic number of the element ununbium is 112., Reason : Name for digits 1 and 2 is un- and bi-respectively, in latin words., , 45, , 103. Assertion : Second period consists of 8 elements., Reason : Number of elements in each period is four times, the number of atomic orbitals available in the energy level, that is being filled., 104. Assertion : Helium is placed in group 18 along with p-block, elements., Reason : It shows properties similar to p-block elements., 105. Assertion : Hydrogen can be placed in group 1., Reason : Hydrogen can gain an electron to achieve a noble, gas arrangement., 106. Assertion : Atomic size increases along a period., Reason : Effective nuclear charge increases as the atomic, number increases resulting in the increased attraction of, electrons to the nucleus., 107. Assertion : Second ionization enthalpy will be higher the, first ionization enthalpy., Reason : Ionization enthalpy is a quantitative measure of, the tendency of an element to lose electron., 108. Assertion : Alkali metals have least value of ionization energy, within a period., Reason : They precede alkaline earth metals in periodic, table., 109. Assertion : Electron gain enthalpy can be exothermic or, endothermic., Reason : Electron gain enthalpy provides a measure of the, ease with which an atom adds an electron to form anion., 110. Assertion : Smaller the size of an atom greater is the, electronegativity., Reason : Electronegativity refers to the tendency of atom, so share electrons with other atom., , CRITICAL THINKING TYPE QUESTIONS, 111. Which fact is not valid for Dobereiner's triads?, (a) The atomic weight of middle element is roughly average, of the other two elements, (b) The properties of middle element is roughly average, of the other two elements, (c) The elements of triads belong to the same group of, modern periodic table, (d) The elements of triads have same valency electrons., 112. In the Mendeleev periodic table, which of the following, element instead of having lower atomic weight was placed, after the element of higher atomic weight thereby ignoring, the order of increasing atomic weights., (a) Iodine, (b) Antimony, (c) Bromine, (d) Molybdenum, 113. Which of the following is correct about Eka-Aluminium and, Eka-Silicon ?, (a) Oxides of Eka-Aluminium is Al2O3 and Eka-Silicon is, Si2O3, (b) Oxides of Eka-Aluminium is Ga2O3 and Eka-Silicon is, GeO2, (c) Melting point of Eka-Aluminium is lower than the, melting point of Eka-Silicon, (d) Both (a) and (c)
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EBD_7207, 46, , CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES, , 114. Which of the following elements are found in pitch blende?, (a) Actinium and protoactinium, (b) Neptunium and plutonium, (c) Actinium only, (d) Both (a) and (b), 115. Which of the following period contain most of the manmade radioactive elements?, (a) Seventh, (b) Fifth, (c) Sixth, (d) Both (a) and (c), 116. The electronic configuration of an element is, , 117., , 118., , 119., , 120., , 121., , 1s 2 2s 2 2 p 6 3s 2 3 p3 . What is the atomic number of the, element, which is just below the above element in the, periodic table?, (a) 33, (b) 34, (c) 36, (d) 49, Which of the following elements show the given, properties?, (i) All elements are metals., (ii) Most of the elements form coloured ions, exhibit, variable valence and paramagnetism., (iii) Oftently used as catalysts., (a) Chalcogens, (b) Transition elements, (c) Inner transition elements, (d) Representative elements, Which of the given elements A, B, C, D and E with atomic, number 2, 3, 7, 10 and 30 respectively belong to the same, period?, (a) A, B, C, (b) B, C, D, (c) A, D, E, (d) B, D, E, According to Mendeleev's periodic classification, the, electronic configuration of hydrogen atom resembles that, of alkali metals, which are given below as :, H = 1s1, Li = 2s1, Na = 3s1, K = 4s1, On the other hand like halogens, hydrogen also exist as, diatomic molecules, such as : H2, Cl2, Br2, I2, etc., On the basis of above information hydrogen can be placed, with :, (a) Alkali metals, (b) Halogens, (c) Both (a) and (b), (d) None of these, Which of the following statements is incorrect from the, point of view of modern periodic table ?, (a) Elements are arranged in the order of increasing atomic, number, (b) There are eighteen vertical columns called groups, (c) Transition elements fit in the middle of long periods, (d) Noble gases are arbitrarily placed in eighteenth group, Element X forms a chloride with the formula XCl2, which is, a solid with a high melting point. X would most likely be in, the same group of the periodic table as –, (a) Na, (b) Mg, (c) Al, (d) Si, , 122. An element X belongs to fourth period and fifteenth group, of the periodic table. Which one of the following is true, regarding the outer electronic configuration of X ? It has, (a) Partially filled d-orbitals and completely filled s-orbitals, (b) Completely filled s-orbital and completely filled, p-orbitals, (c) Completely filled s-orbital and half-filled p-orbitals, (d) Half-filled d-robitals and completely filled s-orbitals, 123. An element has electronic configuration 1s2 2s2 2p6 3s2 3p4., (a) Period = 3rd, block = p, group = 16, (b) Period = 5th, block = s, group = 1, (c) Period = 3rd, block = p, group = 10, (d) Period = 4th, block = d, group = 12, 124. The periodic table of elements does not, (a) include the inert gases, (b) tell us about the arrangement of atoms in a molecule, (c) allow us to make accurate guess of the properties of, undiscovered elements, (d) reveal regularities in the occurance of elements with, similar properties, 125. The lightest liquid metal is, (a) Hg, (b) Ga, (c) Cs, (d) Fr, 126. The correct sequence which shows decreasing order of the, ionic radii of the elements is, (a), , Al3, , Mg 2, , Na, , F, , (b), , Na, , Mg 2, , Al3, , O2, , (c), , Na, , F, , Mg 2, , O2, , O2, , F, Al3, , O2, F, Na, Mg 2, Al3, 127. The ionic radii (Å) of C 4– and O2– respectively are 2.60 and, 1.40. The ionic radius of the isoelectronic ion N3– would be, (a) 2.6, (b) 1.71, (c) 1.4, (d) 0.95, 128. Which of the following species will have the smallest and, the largest size?, Cl, Na, Cl–, Al3+, Mg2+, Na+, (a) Smallest = Na+, Largest = Cl–, (b) Smallest = Al3+, Largest = Cl–, (c) Smallest = Al3+, Largest = Cl, (d) Smallest = Na, Largest = Cl, 129. Covalent radii of atoms varies in range of 72 pm to 133 pm, from F to I while that of noble gases He to Xe varies from, 120pm to 220pm. This is because in case of noble gases, (a) covalent radius is very large, (b) van der Waal radius is considered, (c) metallic radii is considered, (d) None of these, (d)
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CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES, , 47, , 130. The van der Waal and covalent radii of fluorine atom 135. If ionisation enthalpy of oxygen is lesser than nitrogen, because of two of the four 2p– electrons occupy same 2prespectively from the following figure are., orbital than why such case is not possible with fluorine, which contain greater no of paired electrons because., 294 pm, (a) greater size of atomic orbitals, (b) smaller size of orbitals, (c) nuclear charge overpower electronic repulsions., F, F, F, F, (d) None of these, 136. Which of the following statements is wrong ?, (a) van der Waal’s radius of iodine is more than its covalent, 144 pm, 144 pm, radius, (b) All isoelectronic ions belong to same period of the, (a) 219pm, 72pm, (b) 75pm, 72pm, periodic table, (c) 147pm, 72pm, (d) 147pm, 144pm, (c) I.E.1 of N is higher than that of O while I.E.2 of O is, 131. Arrange the following in increasing order of ionic radii?, higher than that of N, C4–,N3–,F–,O2–, (d) The electron gain enthalpy of N is almost zero while, that of P is 74.3 kJ mol–1, (a) C4– < N3– < O2– < F–, 137. Which one of the following statements is incorrect ?, (b) N3– < C4– < O2– < F–, (a) Greater the nuclear charge, greater is the electron, –, 2–, 3–, 4–, (c) F < O < N < C, affinity, (d) O2– < F– < N3– < C4–, (b) Nitrogen has zero electron affinity, 132. The first ( iH1) and second ( iH2) ionization enthalpies, (c) Electron affinity decreases from fluorine to iodine in, (in kJ mol –1 ) and the electron gain enthalpy ( egH), 17th group, (in kJ mol–1) of the elements I, II, III, IV and V are given, (d) Chlorine has highest electron affinity, below, 138. The elements with zero electron affinity are, (a) Boron and Carbon, Element, iH 1, iH 2, eg H, (b) Beryllium and Helium, I, 520, 7300, – 60, (c) Lithium and Sodium, II, 419, 3051, – 48, (d) Fluorine and Chlorine, III, 1681, 3374, – 328, 139. Which of the following property of element is directly, IV, 1008, 1846, – 295, related to electronegativity?, V, 2372, 5251, + 48, (a) Atomic radius, (b) Ionization enthalpy, (c) Non-metallic character (d) None of these, The most reactive metal and the least reactive non-metal of, 140. Which is not the correct order for the stated property., these are respectively, (a) Ba > Sr > Mg ; atomic radius, (a) I and V, (b) V and II, (b) F > O > N ; first ionization enthalpy, (c) II and V, (d) IV and V, (c) Cl > F > I ; electron affinity, 133. Among the following transition elements, pick out the, (d) O > Se > Te ; electronegativity, element/elements with highest second ionization energy., 141. In which of the following arrangements, the order is NOT, (A) V (At. no = 23), (B) Cr (At. no = 24), according to the property indicated against it?, (C) Mn (At. no = 25), (D) Cu (At. no = 29), (a) Li < Na < K < Rb :, Increasing metallic radius, (E) Zn (At. no = 30), (b), I, < Br < F < Cl :, (a) (A) and (C), (b) (B) and (D), Increasing, electron gain enthalpy, (c) (B) and (E), (d) Only (D), (with negative sign), 134. As we move across the second period from C to F ionisation, (c) B < C < N < O, enthalpy increases but the trend from C to F for ionisation, Increasing first ionization enthalpy, enthalpy is C < O < N < F why it is not C < N < O < F. This is, (d) Al3, Mg 2, Na, F, because, (a) atomic radii of O > atomic radii of N, Increasing ionic size, (b) electronic configuration of N is more stable than 142. The compounds of the s-block elements, with the exception, of lithium and ...X... are predominantly ionic. Here, X refers, electronic configuration of O, to, (c) atomic radii of N > atomic radii of O, (a) hydrogen, (b) helium, (d) None of these, (c) magnesium, (d) beryllium
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EBD_7207, CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES, , 48, , 143.Among Al2O3, SiO2, P2O3 and SO2 the correct order of acid, strength is, (a) Al2O3 < SiO2< SO2 < P2O3, (b) SiO2< SO2 < Al2O3 < P2O3, (c) SO2< P2O3 < SiO2 < Al2O3, (d) Al2O3 < SiO2< P2O3 < SO2, 144. Observe the following periodic table :, He, 2, , H, 1, Li, 2, 1, Na, 2, 8, 1, , Be, 2, 2, Ag, 2, 8, 2, , B, 2, 3, Al, 2, 8, 3, , C, 2, 4, Z, 2, 8, 4, , Y, 2, 5, P, 2, 8, 5, , O, 2, 6, S, 2, 8, 6, , F, 2, 7, , Ne, 2, 8, , Cl, 2, 8, 7, , Ar, 2, 8, 8, , K, X, 2, 8, 8, 1 2, 8, 8, 2, , Arrange the following elements X, Y, Z in increasing order, of their valencies :, (a) X > Z > Y, (b) Y > Z > X, (c) Z > Y > X, (d) X > Y > Z, , 145. Which of the following is the reason for the different, chemical behaviour of the first member of a group of, elements in the s- and p-blocks compared to that of the, subsequent members in the same group?, (i) Small size, (ii) Large charge / radius ratio, (iii) Low electronegativity of the element, (a) (i) and (iii), (b) (i), (ii) and (iii), (c) (i) and (ii), (d) (ii) and (iii), 146. Which of the following statement(s) is/are correct ?, (i) Aluminium react with HCl to form Al3+ and H2 is, liberated, (ii) Aluminium dissolve in NaOH to form NaAl(OH)4 and, H2, (a) (i) and (ii), (b) Only (ii), (c) Only (i), (d) Neither (i) nor (ii)
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CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES, , FACT / DEFINITION TYPE QUESTIONS, 1., , 2., , (c) Periodic classification of elements follow a logical, consequence of electronic configuration of atoms, which is used to examine the physical and chemical, properties of the elements., (c) According to Dobereneir’s triad the atomic mass of Br, will be average of the atomic masses of Cl & I, 35.5 127, 2, , 3., 4., , 5., , 6., , 7., , 81.25, , (a), (a) According to the law of triads the atomic wt of the, middle element is arithmatic mean of I and III., At.wt of Cl At wt of I, At wt of Br, 2, (b) In 1800, only 31 elements were known by 1865 the, number of identified elements had more than doubled, to 63. At present 116 elements are known. Of them the, recently discovered elements are man-made., (c) Johann Dobereiner in early 1800’s was the first to, consider the idea of trends among properties of, elements. By 1829 he noted a similarity among the, physical and chemical properties of several groups of, three elements (triads)., (a) According to law of triad,, Li Na, K, ¯ ¯, ¯, 39 + 7 46, =, = 23, 2, 2, , 8., , (c) Arithmetic mean of atomic mass of F and Br, 19 80, = 49.5., 2, Atomic mass of Cl = 35.5, Arithmetic mean of atomic masses of F and Br, Atomic mass of Cl., , =, , 9., 10., 11., , 12., 15., , (c), (c) Every eighth element had the similar properties to the, first element., (b) Lothar Meyer plotted the physical properties such as, atomic volume, melting point and boiling point against, atomic weight and obtained a periodically repeated, pattern., (a), 13. (d), 14. (c), (b) Correct order is Dobereiner, Chancourtois, Newlands,, Mendeleev., , 49, , 16. (d) Chloride formulas, (i) Eka-Aluminium = GaCl3(ECl3), (ii) Eka-Silicon = GeCl4(ECl4), Mendeleef arranged elements in horizontal rows and, vertical columns of a table in order to their increasing, atomic weights., 17. (b), 18. (d) 118 elements are known at present. The recently, discovered elements are man-made., 19. (c) Moseley discovered that atomic number is more, fundamental property than atomic mass., 20. (d) 6th period consists of 32 elements., 21. (b), 22. (b) The period number corresponds to the highest, principal quantum number (n) of the element., 23. (a) Atomic number (Z) = 120, IUPAC name = Unbinilium, Symbol = Ubn, 24. (b) Element with atomic number 104 was named by, American society as Rutherfordium and Kurchatovium, by soviet society., 25. (c) Digit, Name, 1, un, 4, quad, Using above notation IUPAC name of element 114 is, Ununquadium., 26. (b) Its valence shell has 5 electrons (ns2, np3). It belongs, to 5th group of the periodic table., 27. (b), 28. (d) Because of the presence of same number of valence, electrons the elements of same group have similar, chemical properties., 29. (c) Halogens are most electronegative elements i.e., they, are likely to form anions most readily., 30. (d) Barium has atomic number 56. It is an alkaline earth, metal i.e., found in s-block., 31. (c), 32. (c) Element with Z = 33, , (1s2 2s p6 3s2 p6d10 4s 2 p3 ) lies in fifth (or 15th) group., 33. (a) The electronic configuration clearly suggest that, it is a d-block element (having configuration, (n – 1) d 1– 10 ns 0 – 2 ) which starts from III B and goes, till II B. Hence with d3 configuration it would be, classified in the group., 34. (d), 35. (c)
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EBD_7207, CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES, , 50, , 36., , 37., 38., 39., 41., , 42., 43., 44., 46., 47., 48., 49., , 50., 51., 52., 53., , 54., 55., 57., 58., , 59., 60., 61., , (c) Elements (a), (b) and (d) belong to the same group, since each one of them has two electrons in the s sub, shell. In contrast, element (c) has seven electrons in, the valence shell and hence does not lie in the same, group in which elements (a), (b) and (d) lie., (c), (b) Na and Cl both belongs to III period., (b), 40. (b), 2, , 2, , 6, , 2, , 62., , 1, , 63., 64., , 65., , 4, , (b) 1s , 2s 2p , 3s 3p, ¯, Number of shell = 3, (Principal quantum number), Number of period = 3, Valence electrons = 6 i.e., 2 + 4, Number of group = 16, (a) Metallic ch aracter decreases down group and, increases along a period., (b) Cu, Ag and Au are coinage metals. They belong to group, IB (d-block) of periodic table., (b), 45. (b), (b) Non-metals are mainly placed in p-block elements., (b) Non-metallic character increases on moving from left, to right in a period., (b) Correct order of ionic size is Sn > Ce > Yb > Lu., (c) On moving down in a group atomic radii increases due, to successive addition of extra shell hence, O < S < Se, Further As is in group 15 having one less electron in, its p orbital hence have higher atomic radii than group, 16 elements., i.e., O < S < Se < As, (a), (c) Isoelectronic ions have same number of electrons., (a) Continuous increase as no. of shells increases down, the group., (d) The size of an anion will be larger than that of the, parent atom because the addition of one or more, electron(s) would result in increased repulsion among, the electrons and a decrease in effective nuclear, charge., K, K 2 e . Since e– is to be removed from stable, configuration., (c), 56. (c), (a) IE1 is always less than IE2., , 66., 67., , 68., 69., 71., 72., , (a), (d) Noble gases have positive values of electron gain, enthalpy because the anion is higher in energy than, the isolated atom and electron., (a) Within a group, electron gain enthalpy becomes less, negative down a group. However, adding an electron, to the 2p-orbital leads to greater repulsion than adding, an electron to the larger 3p-orbital. Hence,, phosphorus has the least negative electron gain, enthalpy., (a), (a) Electronegativity values of given elements are as, follows:, Be – 1.5 (I), Mg – 1.2 (IV), O – 3.5 (II), N – 3.0 (III), i.e. II > III > I > IV, (b) It is electronic configuration of alkali metal. Hence it, will form basic oxide., (d), 70. (c), (b) Silicon has valence of 4 and bromine has valence of 1., Hence formula of compound will be SiBr4., (b) On passing from left to right in a period acidic character, of the normal oxides of the elements increases with, increase in electronegativity., , STATEMENT TYPE QUESTIONS, 73., 74., 76., 77., , 78., , (b), , Mg 1s 2 2s 2 2p 6 3s 2, After removing of 2 electron, the magnesium acquired, noble gas configuration hence removing of 3rd, electron will require large amount of energy., (d) ns1 configuration and lesser IE., (b) As I.E. of Mg is more, (a) The halogen (group-17) and the chalcogens (group16) are two groups of elements having highly negative, electron gain enthalpies., , (b) O < S < F < Cl, Electron gain enthalpy –141, –200, –333, – 349 kJ mol–, , 79., 82., , (a), , 83., , 84., 85., 86., , (c) Gallium was placed in place of Eka aluminium and, germanium was placed in place of Eka silicon., (b), 75. (c), (d) In fourth period filling up of 3d orbital becomes, energetically favourable before the 4p orbital is filled., (b) The chemical element with atomic number 17 is, chlorine. It belongs to third period in the periodic table, and forms anion with unit negative charge (Cl–)., (a) For statement (iii) the s-block elements because of, their high reactivity are never found pure in nature., For statement (iv) the compounds of the s-block, elements with the exception of lithium and beryllium, are predominantly ionic., (c), 80. (d), 81. (d), (c) Noble gases are placed extremely right in periodic table., Sodium is more metallic than magnesium as it is more, electropositive and has low ionisation energy., (b) Second ionization enthalpy will be higher than the, first ionization enthalpy but lower than the third, ionization enthalpy., (d), (b) Oxygen has smaller first ionization enthalpy than, nitrogen., (c)
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CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES, , MATCHING TYPE QUESTIONS, 87., 88., 93., 96., , 97., , (d) A. 1800, B. 1865, C. At present, (b), 89. (d), 90., (d), 94. (d), 95., (d) Helium (He) 1s2, , 31 elements were known, 63 elements, 118, (b), 91. (a), 92. (b), (c), Highest ionisation, energy due to noble gas, in nature., Fluorine (F) 1s2, 2s22p3, High electronegativity in, nature due to small size, and –1 oxidation state., Rubidium (Rb), Most electronegative, element due to large, atomic size., Lithium (Li), Strongest reducing, agent due to small size, and positive oxidation, state (+1), (b) A. Li+ < Al2+ < Mg2+ < K+, The cation with the greater positive charge will, have a smaller radius because of the greater, attraction of the electrons to the nucleus. Anion, with the greater negative charge will have the larger, radius., 1, ionic radius, Negative charge ionic radius, B. Greater positive charge, increases effective nuclear, charge in case of isoelectronic species. While for, same group elements effective nuclear charge, decreases down the groups., C. Cl > F > Br > I, electron affinity of Cl is highest in halogen family., D. F > Cl > Br > I, electronegativity of fluorine (F) is higher than Cl,, Br and I., (b), 99. (b), , Positive charge, , 98., , ASSERTION-REASON TYPE QUESTIONS, 100. (d) In a triad, the atomic mass of the middle element is the, mean of the atomic masses of the first and third, elements., 101. (d) According to Mendeleev, periodic properties of, elements is a function of their atomic masses., 102. (a) Both assertion and reason are true and reason is the, correct explanation of assertion., 103. (c) Number of elements in each period is twice the number, of atomic orbitals available in the energy level that is, being filled., 104. (c) He (1s2) should be placed along with s-block elements, because of its electronic configuration but it has a, completely filled valence shell and as a result it exhibits, properties of noble gases, thus it is placed along with, noble gases (ns2, np6)., , 51, , 105. (b) Both the statements are correct but assertion is not, correct explanation for reason., 106. (c) Atomic size generally decreases along a period., 107. (b) Both assertion and reason are true but reason is not, the correct explanation of assertion., It is difficult to remove an electron from a positively, charged ion than a neutral atom., 108. (b) Both assertion and reason are true but reason is not, the correct explanation of assertion., Alkali metals belong to first group and have largest, size in a period and hence low I.E., 109. (b) Both assertion and reason are true but reason is not, the correct explanation of assertion., Depending on the element, the process of adding an, electron to the atom can be either endothermic or, exothermic., 110. (c) Assertion is true but reason is false., Electronegativity refers to the tendency of atom to, attract bonding electrons., , CRITICAL THINKING TYPE QUESTIONS, 111. (b), 112. (a) Iodine with lower atomic weight than that of tellurium, (Group VI) was placed in Group VII along with fluorine,, chlorine, bromine because of similarities in properties., 113. (d) Oxides of Eka-Aluminium = Ga2O3, Oxides of Eka-Silicon = SiO2, Melting point of Eka-Aluminium = Low (302 K), Melting point of Eka-Silicon = High (1231 K), 114. (d) Neptunium and plutonium like actinium and, protoactinium are also found in pitch., 115. (a) Seventh period includes most of the man-made radioactive elements., 116. (a) Atomic number of the given element is 15 and it, belongs to 5th group. Therefore atomic number of the, element below the above element, = 15 + 18 = 33., 117. (b) These are characteristic properties of d-block, elements., 118. (b), 119. (c), 120. (d), 121. (b), 122. (c), 123. (a) By observing principal quantum number (n). Orbital, (s, p, d, f ) and equating no. of e–’s we are able to find, the period, block and group of element in periodic table., 124. (b) Periodic table deals with elements and not molecules., 125. (c) Cs is a metal. It is liquid at room temperature. It is, lighter than Hg (also a liquid metal)., 126. (d) All the given species contains 10 e – each i.e., isoelectronic., For isoelectronic species anion having high negative, charge is largest in size and the cation having high, positive charge is smallest., 127. (b) The ionic radii of isoelectronic ions decrease with the, increase in the magnitude of the nuclear charge., So, decreasing order of ionic radii is C4– > N3– > O2–.
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EBD_7207, 52, , CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES, , 128. (b) Anions will be larger and cations will be smaller than, the parent atoms. Among isoelectronic species (Na+,, Mg2+ and Al3+), the one with the larger positive, nuclear charge will have a smaller radius., Largest = Cl– and smallest = Al3+, 129. (b) In case of halogens covalent radius is considered this, bond is formed by overlapping of electron clouds;, while noble gases remain monoatomic, in this case, only way to obtain radius is through van der Waal, radii., 130. (c) Covalent radius is radius of an atom in its bound state, i.e., in fluorine it is half of distance between two, covalently bonded fluorine atoms; van der Waal radii, is one-half of the distance between the nuclei of two, identical non-bonded isolated atoms. These atoms are, attracted toward each other through weak van der, Waal’s force hence van der Waal radii are very large., 131. (c) All the given species are isoelectronic. In case of, isoelectronic species ionic radii increases with, increase in negative charge on anions., 132. (c) I represents Li, II represents K, III represents Br, IV represents I, V represents He, So, amongst these, II represents most reactive metal, and V represents least reactive non-metal., 133. (b), 134. (b), 135. (c), 136. (b) In the isoelectronic species, all isoelectronic anions, belong to the same period and cations to the next, period., 137. (c) Electron affinity of 9F is less than that of 17Cl, 138. (b) Fully filled electronic configuration., 139. (c) The increase in the electronegativities across a period, is accompanied by an increase in non-metallic, properties (or decrease in metallic properties) of, elements., , 140. (b) On moving along the period, ionization enthalpy, increases., In second period, the order of ionization enthalpy, should be as follows :, F > O > N., But N has half-filled structure, therefore, it is more, stable than O. That is why its ionization enthalpy is, higher than O. Thus, the correct order of IE is, F > N > O., 141. (c) In a period the value of ionisation potential increases, from left to right with breaks where the atoms have, some what stable configuration. In this case N has, half filled stable orbitals. Hence has highest ionisation, energy. Thus the correct order is, B< C< O < N, and not as given in option (c), 142. (d) With the exception of lithium and beryllium compounds, of s-block elements are predominantly ionic., 143. (d) As the size increases the basic nature of oxides, changes to acidic nature i.e., acidic nature increases., SO 2, , P 2O 3, Acidic, , SiO 2, Weak, acidic, , Al 2 O 3, Amphoteric, , SO2 and P2O3 are acidic as their corresponding acids, H2SO3 and H3PO3 are strong acids., , 144. (c), 145. (c) The anomalous behaviour of first member of a group, of element in the s- and p-block element is due to their, small size, large charge/radius ratio and high, electronegativity., 146. (a) Because Al is amphotoric in nature so it dissolve in, both acid and base.
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4, , CHEMICAL BONDING AND, MOLECULAR STRUCTURE, FACT / DEFINITION TYPE QUESTIONS, 1., , 2., , 3., , 4., , 5., , The attractive force which holds various constituents, (atoms, ions etc.) together in different chemical species is, called a, (a) chemical bond, (b) chemical compound, (c) ionic bond, (d) covalent bond, The evolution of various theories of valence and the, interpretation of the nature of chemical bonds have closely, been related to the developments in the understanding of, (a) structure of atom, (b) electronic configuration of elements, (c) periodic table, (d) All of the above, Who provide explanation of valence based on intertness, of noble gases ?, (a) Lewis, (b) Ko ssel-Lewis, (c) Langmuir, (d) Sidgwick & Powell, In the formation of a molecule which of the following take, part in chemical combination?, (a) cation, (b) anion, (c) valence electron, (d) inner shell electron, Which of the following do(es) not represent correct Lewis, symbols?, , 8., , Electrovalence of calcium and chlorine respectively is, (a) + 2, – 1, (b) + 1, – 1, (c) + 1, – 2, (d) + 2, – 2, 9., When a metal atom combines with non-metal atom, the, non-metal atom will, (a) lose electrons and decrease in size, (b) lose electrons and increase in size, (c) gain electrons and decrease in size, (d) gain electrons and increase in size, 10. Who introduced the term covalent bond ?, (a) Lewis, (b) Langmuir, (c) Nyholm and Gillespie (d) Heitler and London, 11. Which of the following is/are not the condition(s) for, Lewis dot structure?, (i) Each bond is formed as a result of sharing of an, electron pair between the atoms., (ii) From the two combining atoms only one atom, contribute electron(s) to the shared pair., (iii) The combining atoms attain the outer shell noble, gas configurations as a result of the sharing of, electrons., (a) (i) and (iii), (b) (ii) and (iii), (c) (ii) only, (d) (iii) only, 12. Which of the following does not represent the correct, Lewis dot structure?, , :C :, , 6., , 7., , Be, O, Ne, B, I, II, III, IV, V, (a) I, IV & V, (b) II, III & IV, (c) II only, (d) II & III, The bond formed as a result of the electrostatic attraction, between the positive and negative ions is termed as ..., (a) Chemical bond, (b) Electrovalent bond, (c) Co-ordinate bond, (d) Covalent bond, Cation and anion combines in a crystal to form following, type of compound, (a) ionic, (b) metallic, (c) covalent, (d) dipole-dipole, , H, , Cl, , Cl, , –, , 8e, , –, , 8e, , –, , 8e, , (B), , O, , C, , –, , 8e, (C), , 8e, , –, , O, 8e, , H, –, , 2e, , (A), , (a) A, (c) C, , O, , –, , (b) B, (d) A and C, , –, , 2e
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EBD_7207, CHEMICAL BONDING AND MOLECULAR STRUCTURE, , 54, , Which of the following statements are correct based on, given Lewis dot structure ?, N, , (i), , H, , H, (ii), , H, , (iii), , (iv), , C, , C, , 8e–, , 8e–, , H, , O, , C, , 8e–, , 8e, , 8e, , H, , C, , C, , O, –, , –, , –, , H, –, , (a), , N, , (c), , N, , 22., , Only II, I, II and III, , N, , (b), , N, , N, , (d), , N, , N, , 23., , 24., , 25., , : : : :, :, : : :, , Which of the following shows the Lewis dot formula for, CO2 ?, (b) :O:C::O:, , : :, , :O ::C:: O:, , :, , (a), 18., , N, , :, , 17., , :O:, III, , In N 2 molecule, the number of electrons shared by each, nitrogen atom is, (a) 1, (b) 2, (c) 3, (d) 5, Which of the following represents the Lewis structure of, N2 molecule?, , :, , 16., , 21., , O=N–O–H, , O, O:, :F:, I, II, Choose the correct option(s)., (a) Only I, (b), (c) Only III, (d), 15., , +, , 20., , : :, , O, , : — :, , :, :, : —, , :F – N – F:, , —, : :, , : :, , : — :, , 8e, 8e, (a) (i) and (iv) represents formation of triple bond, (b) Only (iii) represents formation of double bond, (c) Only (ii) represents formation of single bond, (d) (ii) and (iii) both represents formation of single bond., Which of the following Lewis representation of the, molecules NF3, O3 and HNO3 is correct ?, , : :, , 14., , 19., , N, , : :, , 13., , (a), , :N, , N, , (c), , N, .., , (d) :O:C :O:, (c) :O:: C:O:, Which of the following is the correct electron dot structure, of N2O molecule?, .., , .., , N, , .., , O:, .., , O:, , .., , (b) : N, , N, , (d) : N, , N O, :, .., , O, .. :, .., , 26., , What is X, Y and Z in the following expression of formal, charge., Formal charge (F.C) on an atom in a Lewis structure, 1, = X –Y, (Z ), 2, (a) X = Total number of non bonding electrons, Y = Total number of bonding electrons, Z = Total number of valence electrons in the free, atom, (b) X = Total number of valence electrons in the free, atom, Y = Total number of bonding electrons, Z = Total number of non bonding electrons, (c) X = Total number of valence electrons in the free, atom, Y = Total number of non bonding electrons, Z = Total number of bonding electrons, (d) X = Total number of electrons in the free atom, Y = Total number of non bonding electrons, Z = Total number of valence electrons, The lowest energy structure is the one with the .........., formal charges on the atoms., (a) smallest, (b) highest, (c) zero, (d) negative, 3–, In PO4 ion, the formal charge on each oxygen atom and, P—O bond order respectively are, (a) –0.75, 0.6, (b) – 0.75, 1.0, (c) – 0.75, 1.25, (d) –3, 1.25, In the cyanide ion, the formal negative charge is on, (a) C, (b) N, (c) Both C and N, (d) Resonate between C and N, What are the exceptions of the octet rule ?, (a) The incomplete octet of central atom, (b) An odd number of electrons on central atom., (c) Expanded octet of the central atom, (d) All of these, In which of the following molecules octet rule is not, followed?, (a) NH3, (b) CH4, (c) CO2, (d) NO, In which of the following compounds octet is complete and, incomplete for all atoms :, Al2Cl6, Al2(CH3)6 AlF3 Dimer of Dimer of, BeCl2, BeH2, (a) IC, IC, IC, C, C, (b) C, IC, IC, C, IC, (c) C, IC, C, IC, IC, (d) IC, C, IC, IC, IC, (Note : C for complete octet and IC for incomplete octet.), Which of the following molecule(s) obey the octet rule?, (i) [BF4]–, (ii) [AlCl4]–, (iii) SO2, (iv) CCl4, (a) (i), (ii), (iii), (iv), (b) (ii), (iii), (iv), (c) (i), (iii), (iv), (d) (i), (ii), (iii)
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CHEMICAL BONDING AND MOLECULAR STRUCTURE, , 36., , 37., , 38., , 39., , 40., , 41., , 42., , (a) increases, (b) decreases, (c) remains the same, (d) decreases slowly, Which of the following combination will form an, electrovalent bond ?, (a) P and Cl, (b) NH3 and BF3, (c) H and Ca, (d) H and S, Among the following which compound will show the highest, lattice energy ?, (a) KF, (b) NaF, (c) CsF, (d) RbF, Which of the following bond will have highest ionic, character?, (a) H–I, (b) H–F, (c) H–Cl, (d) H–Br, Which of the following pairs will form the most stable ionic, bond ?, (a) Na and Cl, (b) Mg and F, (c) Li and F, (d) Na and F, Which of the following methods is used for measuring, bond length ?, (a) X-ray diffraction, (b) Electron-diffraction, (c) Spectroscopic techniques, (d) All of these, .......... is measured as the radius of an atom’s core which, is in contact with the core of an adjacent atom in a, bonded situation., (a) van der Waal’s radius, (b) Bond length, (c) Covalent radius, (d) Ionic radius, Following figure represent a chlorine molecule. Identify A, B and C in the given figure., A, , C, , B, pm, , 27., , 360, , Among the following the electron deficient compound is, (a) BCl3, (b) CCl4, (c) PCl5, (d) BeCl2, 28. Which of the following is the electron deficient, molecule?, (a) C2 H6, (b) B2H6, (c) SiH4, (d) PH3, 29. Which of the following compounds does not follow the, octet rule for electron distribution?, (a) PCl5, (b) PCl3, (c) H2O, (d) PH3, 30. A pair of compound which have odd electrons in the group, NO, CO, ClO2, N2O5, SO2 and O3 are, (a) NO and ClO2, (b) CO and SO2, (c) ClO2 and CO, (d) SO2 and O3, 31. Which of the following statements is incorrect ?, (a) The formation of ionic compounds depend upon the, ease of formation of the positive and negative ions, from the respective neutral atoms., (b) Formation of ionic compounds depend upon, arrangement of the positive and negative ions in the, solid., (c) Formation of positive ion involves addition of, electron(s) while that of negative ion involves, removal of electron(s)., (d) None of these, 32. Complete the following statement by choosing the, appropriate option., Ionic bonds will be formed more easily between elements, with comparatively, A, and elements with, comparatively high negative value of, B, ., (a) A = low electronegativity, B = ionization enthalpy, (b) A = low ionization enthalpy, B = electron gain enthalpy, (c) A = high ionization enthalpy, B = electron gain enthalpy, (d) A = high electronegativity, B = ionization enthalpy, 33. In ionic solids how crystal structure get stabilized, (a) By the energy released in the formation of crystal, lattice., (b) By achieving octet of electrons around the ionic, species in gaseous state., (c) By electron gain enthalpy and the ionization enthalpy., (d) None of these, 34. Energy required to completely separate one mole of a, solid ionic compound into gaseous constituent ions is, called .......... ., (a) Ionisation enthalpy, (b) Electron gain enthalpy, (c) Bond dissociation enthalpy, (d) Lattice enthalpy, 35. The effect of more electronegative atom on the strength of, ionic bond, , 55, , (a) A = Bond length,, C = Covalent radius, (b) A = Covalent radius,, C = Ionic radius, (c) A = Ionic radius,, C = Covalent radius, (d) A = Covalent radius,, C = Bond length, , B = van der Waal’s radius, B = Bond length, B = van der Waal’s radius, B = van der Waal’s radius
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EBD_7207, CHEMICAL BONDING AND MOLECULAR STRUCTURE, , 56, , 43., , 44., , 45., , Which of the following statement is correct?, (a) Amount of energy required to break one mole of, bonds of a particular type between two atoms in a, gaseous state is called bond enthalpy., (b) The unit of bond enthalpy is kJ mol–1, (c) Larger the bond dissociation enthalpy, stronger will, be the bond in the molecule, (d) All of these, Complete the following statements., in bond order,, increases, With, A, B, decreases., and, C, (a) A = increase, B = bond length, C = bond enthalpy, (b) A = decrease, B = bond enthalpy, C = bond length, (c) A = increase, B = bond enthalpy, C = bond length, (d) A = increase, B = bond angle, C = bond enthalpy, Which of the following molecules have same bond order ?, , 48., , Which of the following is/are misconception(s) associated, with resonance ?, (i) The molecule exist for a certain fraction of time in, one cannonical form and for other fractions of time, in other cannonical forms., (ii) The cannonical forms have no real existence., (iii) There is no such equilibrium between the cannonical, forms., (a) (i) only, (b) (ii) and (iii), (c) (i) and (iii), (d) (iii) only., , 49., , The number of possible resonance structures for CO32 is, , 50., , H 2 , Cl2 , CO, Br2 , N 2, V, , :O:, , 53., , C, :O:–, , :, , II, , –, , : :, , –, , :O:, , :O, , III, , :, O, :, , pm, , 12, 1, , pm, , 14, 8, , O, , pm, , O, , O, , O, , O, , I, , 54., , 55., , 56., , 8, 14, , 1, 12, , pm, , Choose the correct option., (a) Only A, (b) Only B, (c) Both A and B, (d) None of the above, Which of the following structure represents structure of, O3 more accurately?, O, , 52., , :, , C, , :O:–, , 51., , : :, , :, , –, , :, , C, I, , :, , : :, , :O:, , –, , :O:, , –, , :O:, , :, , B., , 57., , II, 8, 12, , pm, , 58., O, , pm, , 12, 8, , 47., , IV, , :, , :, , 46., , III, , II, , I, , Choose the correct option., (a) I, II and IV have same bond order, (b) III and V have same bond order, (c) Both (a) and (b) are correct, (d) None of the above, Which one of the following is not correct representation of, resonance ?, + .., – .. +, –, A. :O::C::O:, :O::C::O:, :O::C:O:, I, II, III, , O, , O, , 59., , III, , (a) I, (c) III, , (b), (d), , II, I and II, , (b) 3, (a) 2, (c) 6, (d) 9, Which one of the following is not the resonance structure, of CO2?, (a) O = C = O, (b) – O – C O+, +, –, (c), O C–O, (d) O C = O, All the bond lengths of sulphur – oxygen in sulphate ion,, are equal because of:, (a) symmetry, (b) resonance, (c) high electronegativity of oxygen, (d) None of these, Resonance is due to, (a) delocalization of sigma electrons, (b) delocalization of pi electrons, (c) migration of protons, (d) Both (a) and (b), Which one of the following pairs of molecules will have, permanent dipole moments for both members ?, (a) NO2 and CO2, (b) NO2 and O3, (c) SiF4 and CO2, (d) SiF4 and NO2, The molecule which has zero dipole moment is, (a) CH3Cl, (b) NF3, (c) BF3, (d) ClO2, Which of the following has dipole moment?, (a) CO2, (b) p-dichlorobenzene, (c) NH3, (d) CH4, Identify the non polar molecule in the following compounds, (a) H2, (b) HCl, (c) HF and HBr, (d) HBr, A neutral molecule XF3 has a zero dipole moment. The, element X is most likely, (a) chlorine, (b) boron, (c) nitrogen, (d) carbon, Among the following, the molecule of high dipole moment, is, (a) CCl4, (b) NH3, (c) H2O, (d) CHCl3, Which one of the following molecules is expected to have, zero dipole moment?, (a) H2O, (b) CO2, (c) SO2, (d) CaF2
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CHEMICAL BONDING AND MOLECULAR STRUCTURE, , 62., , (c) HF H 2S H 2O, (d) HF H 2 O H 2 S, The most polar bond is, (a) C – F, (b) C – O, (c) C – Br, (d) C – S, Which of the following possess dipole moment SF6(a),, SO2(b), H2S(c) , SF4(d) ?, (a) b and c, (b) a and c, (c) b, c and d, (d) a and b, F, F, , F, , F, , F, , O, , .., S, , :, , O, , S, , F, F, , F, , F, , S, H, , H, , 63., , According to Fajan’s rule, covalent bond is favoured by, (a) Large cation and small anion, (b) Large cation and large anion, (c) Small cation and large anion, (d) Small cation and small anion, 64. Arrange the following in increasing order of covalent, character (i) NaCl, (ii) RbCl, (iii) MgCl2, (iv) AlCl3 ?, (a) (i), (ii), (iii), (iv), (b) (iv), (ii), (i), (iii), (c) (ii), (i), (iii), (iv), (d) (iii), (i), (ii), (iv), 65. The correct sequence of increasing covalent character, is represented by, (a) LiCl < NaCl < BeCl2 (b) BeCl2 < LiCl < NaCl, (c) NaCl < LiCl < BeCl2 (d) BeCl2 < NaCl < LiCl, 66. Which of the following salt shows maximum covalent, character?, (a), , AlCl3, , (c) CsCl, 67., , 68., , 69., , (b), , MgCl 2, , (d), , LaCl3, , Polarisibility of halide ions increases in the order, (a) F –, I – , Br–, Cl–, (b) Cl –, Br – , I–, F–, –, –, –, –, (c) I , Br , Cl , F, (d) F –, Cl – , Br–, l–, The covalent bond length is the shortest in which one of, the following bonds?, (a) C — O, (b) C — C, (c) C N, (d) O — H, Hydrogen chloride molecule contains, (a) polar covalent bond (b) double bond, (c) co-ordinate bond, (d) electrovalent bond, Sodium chloride is an ionic compound whereas hydrogen, chloride is mainly covalent because, (a) sodium is less reactive, (b) hydrogen is non-metal, (c) hydrogen chloride is a gas, (d) electronegativity difference in the case of hydrogen, and chlorine is less than 2.1., , 74., , 75., , 76., , 77., , 78., , 79., , Pair Repulsion (VSEPR) theory will be, (a) planar triangular, (b) pyramidal, (c) tetrahedral, (d) square planar, In BrF3 molecule, the lone pairs occupy equatorial positions, to minimize, (a) lone pair - bond pair repulsion only, (b) bond pair - bond pair repulsion only, (c) lone pair - lone pair repulsion and lone pair - bond pair, repulsion, (d) lone pair - lone pair repulsion only, Which of the correct increasing order of lone pair of electrons, on the central atom?, (a) IF7 < IF5 < CIF3 < XeF2, (b) IF7 < XeF2 < CIF2 < IF5, (c) IF7 < CIF3 < XeF2 < IF5, (d) IF7 < XeF2 < IF5 < CIF3, The number of lone pair and bond pair of electrons on the, sulphur atom in sulphur dioxide molecule are respectively, (a) 1 and 3, (b) 4 and 1, (c) 3 and 1, (d) 1 and 4, A molecule has two lone pairs and two bond pairs around, the central atom. The molecule shape is expected to be, (a) V-shaped, (b) triangular, (c) linear, (d) tetrahedral, Using VSEPR theory, predict the species which has square, pyramidal shape, (a) SnCl2, (b) CCl4, (c) SO3, (d) BrF5, Among the following molecules : SO2, SF4, CIF3, BrF5 and, XeF4, which of the following shapes does not describe any, of the molecules mentioned?, (a) Bent, (b) Trigonal bipyramidal, (c) See-saw, (d) T-shape, Which of the following structure is most stable ?, F, Cl, .., , 70., , 72. The geometry of ClO– ion according to Valence Shell Electron, , 73., , F, , S, , 71. According to VSEPR theory the geometry of a covalent, molecules depends upon, (a) the number of bond pairs of electrons, (b) the number of lone pairs of electrons, (c) the number of electron pairs present in the outer shell, of the central atom, (d) All the above, , .., F, , F, , F, , F, .., , 61., , The correct order of dipole moments of HF, H 2S and H 2 O is, (a) HF H 2 S H 2 O, (b) HF H 2 S H 2 O, , .., , 60., , 57, , Cl, , .., F, I, II, Choose the correct option., (a) Only I, (b) Only II, (c) Only III, (d) All three have same stability, , Cl, F, , F, , .., III, , F
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EBD_7207, CHEMICAL BONDING AND MOLECULAR STRUCTURE, , 58, , 80., , 81., , 82., , 83., , A -bonded molecule MX3 is T-shaped. The number of, non-bonding pairs of electron is, (a) 0, (b) 2, (c) 1, (d) can be predicted only if atomic number of M is known., Shape of methane molecule is, (a) tetrahedral, (b) pyramidal, (c) octahedral, (d) square planar, The shape of stannous chloride molecule is, (a) see-saw, (b) square planar, (c) trigonal pyramidal, (d) bent, Look at the following potential energy curve which of the, following correctly represents the most stable state of, hydrogen molecule., , 89., , 90., , 91., , 92., , D, , A, C, , 93., , Energy, , O, , (a), , (b), , B, , 84., , 85., , 86., , 87., , 88., , The angle between the overlapping of one s-orbital and, one p-orbital is, (a) 180°, (b) 120°, (c) 109 28', (d) 120° 60', The enolic form of a acetone contains, (a) 9 sigma bonds, 1 pi bond and 2 lone pairs, (b) 8 sigma bonds, 2 pi bonds and 2 lone pairs, (c) 10 sigma bonds, 1 pi bond and 1 lone pair, (d) 9 sigma bonds, 2 pi bonds and 1 lone pair, Linear combination of two hybridized orbitals belonging to, two atoms and each having one electron leads to a, (a) sigma bond, (b) double bond, (c) co-ordinate covalent bond, (d) pi bond., Which of the following statements is not correct ?, (a) Double bond is shorter than a single bond, (b) Sigma bond is weaker than a (pi) bond, (c) Double bond is stronger than a single bond, (d) Covalent bond is stronger than hydrogen bond, Which of the following represents zero overlap of atomic, orbitals., , Internuclear distance, (a) A, (b) B, (c) C, (d) D, Which of the following statements is false ?, (a) H2 molecule has one sigma bond, (b) HCl molecule has one sigma bond, (c) Water molecule has two sigma bonds and two lone pairs, (d) Acetylene molecule has three pi bonds and three sigma, bonds, The number of sigma ( ) and pi ( ) bonds present in 1,3,5,7, octatetraene respectively are, (a) 14 and 3, (b) 17 and 4, (c) 16 and 5, (d) 15 and 4, Allyl cyanide molecule contains, (a) 9 sigma bonds, 4 pi bonds and no lone pair, (b) 9 sigma bonds, 3 pi bonds and one lone pair, (c) 8 sigma bonds, 5 pi bonds and one lone pair, (d) 8 sigma bonds, 3 pi bonds and two lone pairs, The molecule not having -bond is, (a) Cl 2, (b) O2, (c) N2, (d) CO2, In hexa-1, 3-diene-5-yne the number of C — C , C — C, and C — H bonds, respectively are, (a) 5, 4 and 6, (b) 6, 3 and 5, (c) 5, 3 and 6, (d) 6, 4 and 5, , z, , pz, , s, , pz, , s, , z, , z, , (c), , s, px, 94., , 95., , (d) All of these, As the s-character of hybridised orbital increases, the bond, angle, (a) increase, (b) decrease, (c) becomes zero, (d) does not change, Which of the following is/are not essential condition(s), for hybridisation?, (i) The orbitals present in the valence shell of the atom, are hybridised., (ii) The orbitals undergoing hybridisation should have, almost equal energy., (iii) Promotion of electron is essential prior to, hybridisation, (iv) Only half filled orbitals participate in hybridisation., (a) (i) only, (b) (iii) only, (c) (iv) only, (d) (iii) and (iv)
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CHEMICAL BONDING AND MOLECULAR STRUCTURE, , 96., , 97., , 98., , 99., , The nature of hybridisation in the ammonia molecule is, (a) sp 2, (b) dp 2, (c) sp, (d) sp 3, The shape of sulphate ion is, (a) square planar, (b) triagonal, (c) trigonal planar, (d) tetrahedral, The strength of bonds formed by s–s and p–p, s–p overlap, in the order of, (a) s–p > s–s > p–p, (b) p–p > s–s > s–p, (c) s–s > p–p > s–p, (d) s–s > s–p > p–p, Which of the following will have sp3 d3 hybridisation?, (a) BrF5, (b) PCl5, (c) XeF6, (d) SF6, , 100. The shape of CO 2 molecule is, (a) linear, (b) tetrahedral, (b) planar, (d) pyramidal, 101. The hybridisation state of carbon in fullerene is, (a) sp, (b) sp2, (c) sp3, (d) sp3 d, 102. Which of the following statements is true for an ion having, sp3 hybridisation?, (a) all bonds are ionic, (b) H-bonds are situated at the corners of a square, (c) all bonds are co-ordinate covalent, (d) H-atoms are situated at the corners of tetrahedron, 103. Which of the following molecule does not have a linear, arrangement of atoms ?, (a) H2S, (b) C2H2, (c) BeH2, (d) CO2, 104. In which one of the following molecules the central atom, said to adopt sp2 hybridization?, (a) BeF2, (b) BF3, (c) C2H2, (d) NH3, 105. Considering the state of hybridization of carbon atoms, find, out the molecule among the following which is linear ?, (a) CH3– CH = CH–CH3, (b) CH3 – C C – CH3, (c) CH2 = CH – CH2 – C CH, (d) CH3 – CH2 – CH2 – CH3, 106. Equilateral shape has, (a) sp hybridisation, (b) sp2 hybridisation, 3, (c) sp hybridisation, (d) None of these, 107. In an octahedral structure, the pair of d orbitals involved in, , 59, , 109. In which of the following species is the underlined carbon, having sp3 - hybridisation ?, (a), 110., , 111., , 112., , 113., , 114., , 115., , 116., , d 2 sp3 hybridization is, , (a), , d, , (c), , d, , d, , x2 y 2 , z 2, , (b), , d xz, d, , d, z 2 , xz, , (d), , d xy, d yz, , x2 y 2, , 108. The trigonal bipyramidal geometry is obtained from the, hybridisation, (a) dsp 3 or sp 3d, (b) dsp 2 or sp 2d, 2, 3, 3, 2, (c) d sp or sp d, (d) None of these, , 117., , CH – COOH, , (b), , CH CH OH, , (c) CH COCH, (d) CH, CH CH, 3, A sp -hybrid orbital contains, (a) 25% s-character, (b) 75% s-character, (c) 50% s-character, (d) 25% p-character, The types of hybridisation of the five carbon atoms from, left to right in the molecule, CH3 — CH == C == CH — CH3 are, (a) sp3, sp2, sp2, sp2, sp3 (b) sp3, sp, sp2, sp2, sp3, (c) sp3, sp2, sp, sp2, sp3 (d) sp3, sp2, sp2, sp, sp3, Pick out the incorrect statement from the following, (a) sp hybrid orbitals are equivalent and are at an angle, of 180° with each other, (b) sp2 hybrid orbitals are equivalent and bond angle, between any two of them is 120°, (c) sp3d2 hybrid orbitals are equivalent and are oriented, towards corners of a regular octahedron, (d) sp3d3 hybrid orbitals are not equivalent, All carbon atoms are sp2 hybridised in, (a) 1, 3-butadiene, (b) CH2 = C = CH2, (c) cyclohexane, (d) 2-butene, Which one of the following is not correct in respect of, hybridization of orbitals?, (a) The orbitals present in the valence shell only are, hybridized, (b) The orbitals undergoing hybridization have almost, equal energy, (c) Promotion of electron is not essential condition for, hybridization, (d) Pure atomic orbitals are more effective in forming stable, bonds than hybrid orbitals, Molecular orbital theory was given by, (a) Kossel, (b) Mosley, (c) Mulliken, (d) Werner, Atomic orbital is monocentric while a molecular orbital is, polycentric. What is the meaning of above statements?, (a) Electron density in atomic orbital is given by the, electron distribution around a nucleus in an atom., While in molecular orbital it is given by the electron, distribution around group of nuclei in a molecule., (b) While an electron in an atomic orbital is influenced, by one nucleus, in a molecular orbital it is influenced, by two or more nuclei depending upon the number, of atoms in the molecule., (c) The electron in an atomic orbital is present in one, nucleus while in molecular orbital electrons are, present on more than one nuclei depending upon, the number of atoms in the molecule., (d) All of these, With increasing bond order, stability of bond, (a) Remain unaltered, (b) Decreases, (c) Increases, (d) None of these
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EBD_7207, CHEMICAL BONDING AND MOLECULAR STRUCTURE, , 60, , 118. The given increasing order of energies of various, molecular orbitals is not true for which of the following, molecule?, 1s <, s < 2s < 2s < ( 2px = 2py) < 2pz < ( 2px, = 2py) < 2pz, (a) B2, (b) C2, (c) N2, (d) O2, 119. Which of the following corresponds unstable molecule?, Here Nb is number of bonding electrons and Na is number, of antibonding electrons., (a) Nb > Na, (b) Nb < Na, (c) Na = Nb, (d) Both (b) and (c), 120. If Nx is the number of bonding orbitals of an atom and Ny is, the number of antibonding orbitals, then the molecule/atom, will be stable if, (a) Nx > Ny, (b) Nx = Ny, (c) Nx < Ny, (d) Nx Ny, 121. In the molecular orbital diagram for O2+ ion, the highest, occupied orbital is, (a), MO orbital, (b), MO orbital, (c) * MO orbital, (d) * MO orbital, 122. The theory capable of explaining paramagnetic behaviour, of oxygen is, (a) resonance theory, (b) V.S.E.P.R. theory, (c) molecular orbital theory, (d) valence bond energy, 123. In an anti-bonding molecular orbital, electron density is, minimum, (a) around one atom of the molecule, (b) between the two nuclei of the molecule, (c) at the region away from the nuclei of the molecule, (d) at no place, 124. When two atomic orbitals combine, they form, (a) one molecular orbital (b) two molecular orbital, (c) three molecular orbital (d) four molecular orbital, 125. Paramagnetism is exhibited by molecules, (a) not attracted into a magnetic field, (b) containing only paired electrons, (c) carrying a positive charge, (d) containing unpaired electrons, 126. The difference in energy between the molecular orbital, formed and the combining atomic orbitals is called, (a) bond energy, (b) activation energy, (c) stabilization energy (d) destabilization energy, 127. The bond order in N2+ is, (a) 1.5, (b) 3.0, (c) 2.5, (d) 2.0, 128. Which molecule has the highest bond order?, (a) N2, (b) Li2, (c) He2, (d) O2, 129. Which one of the following molecules is expected to exhibit, diamagnetic behaviour ?, (a) C2, (b) N2, (c) O2, (d) S2, , 130. The correct statement with regard to H2+ and H2– is, (a) both H2+ and H–2 are equally stable, (b) both H2+ and H2– do not exist, (c) H2– is more stable than H2, (d) H2+ is more stable than H2, 131. Mark the incorrect statement in the following, (a) the bond order in the species O2 , O2+ and O2 –, decreases as O 2, , 132., , 133., , 134., , 135., , O2, , O2, , (b) the bond energy in a diatomic molecule always, increases when an electron is lost, (c) electrons in antibonding M.O. contribute to repulsion, between two atoms., (d) with increase in bond order, bond length decreases, and bond strength increases., According to molecular orbital theory which of the following, statement about the magnetic character and bond order is, correct regarding O 2, (a) Paramagnetic and Bond order < O2, (b) Paramagnetic and Bond order > O2, (c) Diamagnetic and Bond order < O2, (d) Diamagnetic and Bond order > O2, Bond order is a concept in the molecular orbital theory. It, depends on the number of electrons in the bonding and, antibonding orbitals. Which of the following statements is, true about it ? The bond order, (a) can have a negative quantity, (b) has always an integral value, (c) can assume any positive or integral or fractional value, including zero, (d) is a non-zero quantity, Which of the following does not exist on the basis of, molecular orbital theory ?, (b) He2+, (a) H2+, (c) He2, (d) Li2, The paramagnetic property of the oxygen molecule is due, to the presence of unpaired electrons present in, (a), , ( 2 px )1 and ( * 2p x )1, , (b), , ( 2 px )1 and ( 2p y )1, , (c), , ( * 2p y )1 and ( * 2p z )1, , (d), , ( * 2p x )1 and ( * 2p z )1, , 136. In which of the following state of compound the, magnitude of H-bonding will be maximum and in which, case it will be minimum ?, (a) Maximum = Solid, Minimum = Liquid, (b) Maximum = Liquid, Minimum = Gas, (c) Maximum = Solid, Minimum = Gas, (d) Maximum = Gas, Minimum = Solid
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CHEMICAL BONDING AND MOLECULAR STRUCTURE, , 137. Which of the following are correctly classified ?, Intermolecular, Intramolecular, H-bonding, H-bonding, (a) HF, H2O, (b) CH3OH, HF, (c) H2O, o-nitrophenol, (d) HF, p-nitrophenol, 138. Intramolecular hydrogen bond exists in, (a) ortho nitrophenol, (b) ethyl alcohol, (c) water, (d) diethyl ether, 139. The boiling point of p-nitrophenol is higher than that of, o-nitrophenol because, (a) NO2 group at p-position behave in a different way, from that at o-position., (b) intramolecular hydrogen bonding exists in, p-nitrophenol, (c) there is intermolecular hydrogen bonding in, p-nitrophenol, (d) p-nitrophenol has a higher molecular weight than, o-nitrophenol., 140. Which one of the following is the correct order of interactions ?, (a) Covalent < hydrogen bonding < vander Waals < dipoledipole, (b) vander Waals < hydrogen bonding < dipole < covalent, (c) vander Waals < dipole-dipole < hydrogen bonding <, covalent, (d) Dipole-dipole < vander Waals < hydrogen bonding <, covalent., 141. Strongest hydrogen bond is shown by, (a) water, (b) ammonia, (c) hydrogen fluoride, (d) hydrogen sulphide, 142. The low density of ice compared to water is due to, (a) induced dipole-induced dipole interactions, (b) dipole-induced dipole interactions, (c) hydrogen bonding interactions, (d) dipole-dipole interactions, 143. Methanol and ethanol are miscible in water due to, (a) covalent character, (b) hydrogen bonding character, (c) oxygen bonding character, (d) None of these, 144. The hydrogen bond is shortest in, (a) S — H --- S, (b) N — H --- O, (c) S — H --- O, (d) F — H --- F, 145. Hydrogen bonding is maximum in, (a) C2H5OH, (b) CH3OCH3, (c) (CH3)2 C = O, (d) CH3CHO, , 61, , OH, 146. The vapour pressure of, , is higher than, NO 2, , OH, , due to, O 2N, , (a) dipole moment, (b) dipole-dipole interaction, (c) H-bonding, (d) lattice structure, 147. The reason for exceptionally high boiling point of water is, (a) its high specific heat, (b) its high dielectric constant, (c) low ionization of water molecule, (d) hydrogen bonding in the molecules of water, 148. Acetic acid exists as dimer in benzene due to, (a) condensation reaction, (b) hydrogen bonding, (c) presence of carboxyl group, (d) presence of hydrogen atom at -carbon, 149. Hydrogen bonding is formed in compounds containing, hydrogen and, (a) highly electronegative atoms, (b) highly electropositive atoms, (c) metal atoms with d-orbitals occupied, (d) metalloids, , STATEMENT TYPE QUESTIONS, 150. Read the following statements and choose the correct, sequence of T and F from the given codes. Here T, represents true and F represents false statement., (i) The number of dots in Lewis symbol represents the, number of valence electrons., (ii) Number of valence electrons helps to calculate, group valence of element., (iii) Group valence is given as 8 minus the number of, inner shell electrons., (a) T T T, (b) T F F, (c) T T F, (d) F F F, 151. Based on the following Lewis dot structure which of the, given statement(s) is/are correct?, , O, O C, (i), (ii), (iii), (a), (c), , 2–, , O, , There is formation of a double bond and two single, bonds., There are two additional electrons than those, provided by the neutral atoms., The least electropositive atom occupies the central, position in the molecule/ion., (i) and (iii), (b) (i), (ii) and (iii), (iii) only, (d) (i) and (ii)
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EBD_7207, 62, , 152. Choose the correct sequence of T and F for following, statements. Here T stands for true statement and F stands, for false statement., (i) Formal charge in the Lewis structure helps in, keeping track of the valence electrons in the, molecule., (ii) Formal charge indicates the actual charge separation, within the molecule., (iii) Formal charges help in the selection of the lowest, energy structure from a number of possible Lewis, structures., (a) T T F, (b) T F T, (c) T T T, (d) F T T, 153. Read the following statements and choose the correct, option. Here T stands for True and F stands for False, statement., (i) The smaller the size of the cation and the larger the, size of the anion, the greater the covalent character, of an ionic bond., (ii) The smaller the charge on the cation, the greater the, covalent character of the ionic bond., (iii) For cations of the same size and charge, the one,, with electronic configuration (n – 1)d nns0, typical of, transition metals, is more polarising than the one, with a noble gas configuration, ns2 np6, typical of, alkali and alkaline earth metal cations., (a) T T T, (b) T T F, (c) T F T, (d) F T T, 154. Choose the correct sequence of T and F for following, statements. Here T stands for True and F for False, statement., (i) Sigma bond is formed by head on overlap of, bonding orbitals along the internuclear axis., (ii) Pi bond is formed when atomic orbitals overlap in, such a way that their axes remain parallel to each, other and perpendicular to the internuclear axis., (iii) Half-filled s-orbital of one atom and half filled, p-orbitals of another atom forms., bond on, overlapping., (iv) Overlapping in case of pi-bond takes place to a, larger extent as compared to sigma bond., (a) T T T T, (b) T F T F, (c) T T F F, (d) T T F T, 155. Give the correct order of initials T or F for following, statements. Use T if statement is true and F if it is false :, (i) The order of repulsion between different pair of, electrons is lp – lp > lp – bp > bp – bp, (ii) In general, as the number of lone pair of electrons on, central atom increases, value of bond angle from, normal bond angle also increases, (iii) The number of lone pair on O in H2O is 2 while on N in, NH3 is 1, (iv) The structures of xenon fluorides and xenon, oxyfluorides could not be explained on the basis of, VSEPR theory, (a) TTTF, (b) TFTF, (c) TFTT, (d) TFFF, , CHEMICAL BONDING AND MOLECULAR STRUCTURE, , 156. Which of the following statements is/are not correct for, combination of atomic orbitals?, (i) The combining atomic orbitals must have the same, or nearly the same energy., (ii) Greater the extent of overlap, the greater will be the, electron density between the nuclei of a moleculer, orbital., (iii) 2pz orbital of one atom can combine with either of, 2px, 2py or 2pz orbital of other atom as these orbitals, have same energy., (a) (i) and (ii), (b) (iii) only, (c) (i) only, (d) (ii) and (iii), , MATCHING TYPE QUESTIONS, 157. Match the columns, Column-I, (A) BeH2, (B) SF6, (C) NO2, , Column-II, (p) Odd electron molecules, (q) Expanded octet, (r) Incomplete octet of, central atom, (a) A – (p), B – (q), C – (r), (b) A – (q), B – (r), C – (p), (c) A – (r), B – (q), C – (p), (d) A – (r), B – (p), C – (q), 158. Match the columns, Column-I, , Column-II, , (A) HCl, , (p) Covalent compound with, directional bond, , (B) CO2, , (q) Ionic compound with, non-directional bonds, , (C) NaCl, , (r), , Polar molecule, , (D) CCl4, (s) Non-polar molecule, (a) A – (p, q, r), B – (q, r), C – (p, q), D – (r), (b) A – (q), B – (r), C – (p), D – (s), (c) A – (p, r), B – (p, s), C – (q), D – (p, s), (d) A – (q), B – (r), C – (p, q), D – (s), 159. Match Column-I with Column-II and Column-III and choose, the correct option from the given codes., Column-I, Column-II, Column-III, Molecule, (No. of lone, (Shape of molecule), pairs and, bond pairs), (A) NH3, (i) 1, 2, (p) Bent, (B) SO2, (ii) 1, 4, (q) Trigonal pyramidal, (C) SF4, (iii) 2, 3, (r) T-shape, (D) ClF3, (iv) 1, 3, (s) See-Saw, (a) A – (iv, q); B – (ii, p); C – (i, r); D – (iii, s), (b) A – (iv, q); B – (i, p); C – (ii, s); D – (iii, r), (c) A – (i, p); B – (iii, s); C – (iv, r); D – (ii, q), (d) A – (iv, p); B – (i, r); C – (iii, q); D – (ii, s)
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CHEMICAL BONDING AND MOLECULAR STRUCTURE, , 160. Match the columns, Column-I, (A) Trigonal planar, :, , Column-II, (p) PCl5, , 120°, A, :, , :, (B) Tetrahedral, , :, , (q) NH4+, 109.5°, , A, , :, , :, , :, , (r) SF6, , :, , (C) Trigonal bipyramidal, , 90°, , :, , A, , :, , :, , 120°, , :, , (s) BF3, , :, , (D) Octahedral, , 90°, :, , :, , 90°, , :, , :, , :, , A, , (D), , +, , (C), , –, , +, –, , –, , +, , –, , +, , (B), , +, –, , +, , (A), , +, –, , +, , (a) A – (p), B – (q), C – (r), D – (s), (b) A – (s), B – (r), C – (q), D – (p), (c) A – (s), B – (q), C – (p), D – (r), (d) A – (r), B – (p), C – (q), D – (s), 161. Match the columns, Column-I, Column-II, , 63, , (a) A – (p), B – (q), C – (p), D – (r), (b) A – (p), B – (r), C – (q), D – (p), (c) A – (q), B – (q), C – (p), D – (r), (d) A – (r), B – (p), C – (q), D – (p), 162. Match Column-I (molecule) with Column-II (type of, hybridisation) and choose the correct option from the, codes given below., Column-I, Column-II, (Molecule), (Type of hybridisation), (A) SF6, (p) sp 3 d, (B) PF5, (q) sp 3, (C) BCl3, (r) sp 3 d 2, (D) C2H6, (s) sp 2, (a) A – (r), B – (p), C – (s), D – (q), (b) A – (r), B – (p), C – (q), D – (s), (c) A – (p), B – (r), C – (q), D – (s), (d) A – (p), B – (r), C – (s), D – (q), 163. Match the columns, Column-I, Column-II, (A) Valence bond theory (p) Nyholm and Gillespie, (B) Octet rule, (q) F. Hund & R. S Mulliken, (C) Molecular orbital, (r) Heitler and London, theory, (D) The valence shell, (s) Ko ssel and Lewis, electron pair, repulsion theory, (a) A – (p), B – (q),C – (r), D – (s), (b) A – (q), B – (r),C – (s), D – (p), (c) A – (p), B – (s),C – (q), D – (r), (d) A – (s), B – (r), C – (q), D – (p), 164. Match the columns, Column-I, Column-II, Column-III, (A) 1s, , (p), , (q), , (p) negative overlap, , (r), , positive overlap, , +, –, , (i), , *2p, (B) 2pz, , (q) zero overlap, , – + – +, , (ii), , + –, , –, , +, , *1s, (C) 2px, (a), (b), (c), (d), , A–, A–, A–, A–, , (r), , +, –, , –, +, , (q, iii), B – (r, i), C – (p, ii), (q, iii), B – (p, ii), C – (r, i), (p, iii), B – (q, ii), C – (r, i), (p, ii), B – (q, iii), C – (r, i), , (iii), , +, , –
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EBD_7207, CHEMICAL BONDING AND MOLECULAR STRUCTURE, , 64, , ASSERTION-REASON TYPE QUESTIONS, Directions : Each of these questions contain two statements,, Assertion and Reason. Each of these questions also has four, alternative choices, only one of which is the correct answer. You, have to select one of the codes (a), (b), (c) and (d) given below., (a) Assertion is correct, reason is correct; reason is a correct, explanation for assertion., (b) Assertion is correct, reason is correct; reason is not a, correct explanation for assertion, (c) Assertion is correct, reason is incorrect, (d) Assertion is incorrect, reason is correct., 165. Assertion : The correct Lewis structure of O3 may be drawn, as, , :, , +, 1, , O, , 166., , 167., , 168., , 169., , 170., , 171., , 172., , 173., 174., , O, , –, , : :, , : :, , 2, , O: 3, , Reason : The formal charges on atom 1, 2 and 3 are +1, 0, and –1 respectively., Assertion : Atoms can combine either by transfer of, valence of electrons from one atom to another or by, sharing of valence electrons., Reason : Sharing and transfer of valence electrons is, done by atoms to have an octet in their valence shell., Assertion : The lesser the lattice enthalpy more stable is, the ionic compound., Reason : The lattice enthalpy is greater, for ions of highest, charge and smaller radii., Assertion : Sulphur compounds like SF6 and H2SO4 have, 12 valence electrons around S atom., Reason : All sulphur compounds do not follow octet rule., Assertion : BF3 molecule has zero dipole moment., Reason : F is electronegative and B–F bonds are polar in, nature., Assertion : CH2Cl2 is non-polar and CCl4 is polar molecule., Reason : Molecule with zero dipole moment is non-polar in, nature., Assertion : Lone pair-lone pair repulsive interactions are, greater than lone pair-bond pair and bond pair-bond pair, interactions., Reason : The space occupied by lone pair electrons is, more as compared to bond pair electrons., Assertion : In NH3, N is sp3 hybridised, but angle is found, to be 107°., Reason : The decrease in bond angle is due to repulsion, between the lone pair., Assertion : Shape of NH3 molecule is tetrahedral., Reason : In NH3 nitrogen is sp3 hybridized., Assertion : pi bonds are weaker than bonds., Reason : pi bonds are formed by the overlapping of p-p, orbitals along their axes., , 175. Assertion : The bond order of helium is always zero., Reason : The number of electrons in bonding molecular, orbital and antibonding molecular orbital is equal., 176. Assertion : Bonding molecular orbital has greater stability, than corresponding antibonding molecular orbital., Reason : The electron density in a bonding molecular, orbital is located away from the space between the nuclei, while in antibonding molecular orbital it is located, between the nuclei of the bonded atoms., 177. Assertion : Water is one of the best solvent., Reason : H-bonding is present in water molecules., , CRITICAL THINKING TYPE QUESTIONS, 178. What is the correct mode of hybridisation of the central, atom in the following compounds?, NO–2, SF4, PF6 –, 2, (a) sp, sp, sp 3, 2, 3, (b) sp, sp d, sp 3 d 2, 2, 3, (c) sp, sp, d 2 sp 3, 3, 3, (d) sp, sp, sp 3 d 2, 179. Which of the following molecules has trigonal planar, geometry?, (a) BF3, (b) NH3, (c) PCl3, (d) IF3, 180. Which of the following molecules is planar?, (a) SF4, (b) XeF4, (c) NF3, (d) SiF4, 181. Hybridization present in ClF3 is, (a) sp 2, (b) sp 3, (c) dsp 2, (d) sp 3 d, 182. Which of the following represents the given mode of, hybridisation sp2 –sp2 – sp - sp from left to right ?, (a), , H C CH C, , (c), , H C C C, , N, , (b), , HC, , C–C, , CH, , CH2, , 183., , 184., , 185., , 186., , CH, , (d) H2C, Hybridisation states of C in CH3+ and CH4 are, (a) sp2 & sp3, (b) sp3 & sp2, (c) sp2 & sp2, (d) sp3 & sp3, The type of hybridization in xenon atom and the number of, lone pairs present on xenon atom in xenon hexafluoride, molecule are respectively, (a) sp3d3 , one, (b) sp3d3 , two, 3, 3, (c) sp d , two, (d) sp3d2 , zero, In which of the following species, all the three types of, hybrid carbons are present?, (a) CH2 = C = CH2, (b) CH3 – CH= CH – CH+2, +, (d) CH3 – CH= CH – CH–2, (c) CH3 – C C – CH2, If an organic compound contain 92.3% C and 7.7% H, than, number of sp3,sp2 and sp hybridized carbon atoms in all, possible str uctures of compound respectively are, (molecular mass = 52g/mol), (a) 1, 2, 5, (b) 0, 4, 4, (c) 0, 8, 4, (d) None of these
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CHEMICAL BONDING AND MOLECULAR STRUCTURE, , 187. Arrange the following in increasing order of bond length, (i), , N2, , (ii), , N2, , 189., , 190., , 191., , (a) (ii), (i) and (iii), (b) (ii), (iii) and (i), (c) (iii), (ii) and (i), (d) (i), (ii) and (iii), Which of the following molecule exist ?, (a) He2, (b) Be2, (c) Li2, (d) Both (a) and (b), Hybridization and structure of I3– are, (a) sp2 and trigonal planar, (b) sp3d2 and linear, (c) sp3d and linear, (d) sp3 and T-shape, What is the change in hybridization when AlCl3 changes, to [Al2Cl6]–3 ?, (a) sp3 d to sp3d2, (b) sp3 to sp3 d, (c) sp2 to sp3d2, (d) None of these, From the given figure the van der Waal radius and covalent, radius of the hydrogen atom respectively are, 62 pm, , 240 pm, , (a) 151, 31, (b) 120, 31, (c) 31, 100, (d) 30, 120, 192. Which of the following substances has the greatest ionic, character ?, (a) Cl2O, (b) NCl3, (c) PbCl2, (d) BaCl2, 193. Among the following species, identify the pair having same, bond order CN–, O2– , NO+, CN+, (b) O2– and NO+, (a) CN– and O2–, –, +, (c) CN and NO, (d) CN– and CN+, 194. Which of the following is not correct with respect to bond, length of the species ?, (a), (c), , C2, Li 2, , C22, Li 2, , (b), (d), , B2, , B2, , O2, , O2, , 195. The molecule which has the highest bond order is, (a) C2, (b) N2, (c) B2, (d) O2, 196. The compound which cannot be formed is, (a) He, , (b), , He, , (c), , (d), , He, , He 2, , 197. The ground state electronic configuration of valence shell, electrons in nitrogen molecule (N 2 ) is written as, 2s 2 , * 2s 2 , 2p 2x ,, , KK, , (iii) N 22, , 188., , 65, , 2, , molecule is, (a) 0, (c) 2, 198. Bond order in benzene is, (a) 1, (c) 1.5, , 2p 2y 2p 2z Bond order in nitrogen, , (b) 1, (d) 3, (b) 2, (d) None of these, , 199. In O2 , O2 and O 2 2 molecular species, the total number of, antibonding electrons respectively are, (a) 7, 6, 8, (b) 1, 0, 2, (c) 6, 6, 6, (d) 8, 6, 8, 200. N2 and O2 are converted to monopositive cations N2+ and, O2+ respectively. Which is incorrect ?, (a) In N2+ the N–N bond is weakened, (b) In O2+ the bond order increases, (c) In O2+ the paramagnetism decreases, (d) N2+ becomes diamagnetic, 201. Bond order normally gives idea of stability of a molecular, species. All the molecules viz. H2, Li2 and B2 have the same, bond order yet they are not equally stable. Their stability, order is, (a) H2 > B2 > Li2, (b) Li2 > H2 > B2, (c) Li2 > B2 > H2, (d) B2 > H2 > Li2, 202. According to MO theory which of the following lists ranks, the nitrogen species in terms of increasing bond order?, (a), , N 2–, 2, , N 2–, , N2, , (b), , N2, , N 22–, , N 2–, , (c), , N 2–, , N 2–, 2, , N2, , (d), , N 2–, , N2, , N 2–, 2, , 203. Hydrogen bonding would not affect the boiling point of, (a) HI, (b) H2O, (c) NH3, (d) CH3OH, 204. Which one of the following molecules will form a linear, polymeric structure due to hydrogen bonding?, (a) NH3, (b) H2O, (c) HCl, (d) HF, 205. Which among the following can form intermolecular, H – bonding ?, O, H C, , CHO, , CHO, , NO 2, , OH, OH, , (A), , (a) A, (c) B, C and D, , OH, (B), , (C), , (b) B and D, (d) A and C, , OH, (D)
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EBD_7207, CHEMICAL BONDING AND MOLECULAR STRUCTURE, , 66, , FACT / DEFINITION TYPE QUESTIONS, , 4., , 5., , 14., 15., 16., 17., , 6., 7., , 8., 9., , Correct Lewis symbol for Ne = :Ne:, (b) Electrovalent bond is formed as a result of electrostatic, attraction between the positive and negative ions., (a) The electrostatic force that binds the oppositely, charged ions which are formed by transfer of, electron(s) from one atom to another is called ionic, bond. Cation and anion are oppositely charged, particles therefore they form ionic bond in crystal., (a) Calcium is assigned a positive electrovalence of two,, while chlorine has a negative electrovalence of one., (d) When a metal for example Na combines with a non, metal e.g., Cl2. Following reaction occurs, 2Na Cl 2, 2NaCl, In this process Na loses one electron to form Na + and, Cl accepts one electron to form Cl–, , Na, Cl e, , 10., , 11., 12., , Na, , e, Cl, , Therefore, in this process Cl gain electrons and hence, its size increases., (b) Langmuir (1919) refined the Lewis postulations by, abandoning the idea of the stationary cubical, arrangement of the octet, and by introducing the, term covalent bond., (c) Each combining atom contributes at least one, electron to the shared pair., (a) In formation of Cl2 molecule a pair of electrons is, shared between the two chlorine atoms. Each, chlorine atom contribute one electron to the shared, pair., , :, , : :, , Correct Lewis symbol = O, Similarly, Ne is a noble gas having valence electrons, =8, , (a) Lewis structure of N2 is N N, (a) Step I : Skeleton OCO, Step II : A = 1 × 4 for C + 2 × 6 for O = 4 + 12, = 16 electrons, Step III : Total no. of electrons needed to achieve noble, gas configuration (N), N = 1 × 8 + 2 × 8 = 24, Step IV : Shared electrons, S = N – A = 24 – 16, = 8 electrons, Step V : O::C::O, Step VI : :O::C ::O:, , 18., , (b), , :, , 3., , 13., , –, , 8e, 8e, (a) (i) represents : N N :, (iv) represents H – C C – H, (ii) and (iii) represents respectively :, H, H, and, C=C, O=C=O, H, H, (a) All Lewis representation of the molecules NH3, O3, and HNO3 given in question are correct., (c) N2 ; N N, 3 electrons are shared by each nitrogen atom, , :, , 2., , (a) Chemical bond is an attractive force, which holds, various constituents (atoms, ions etc.) together in, different chemical species., (d) The evolution of various theories of valence and the, interpretation of the nature of chemical bonds have, closely been related to the developments in the, understanding of structure of atom, electronic, configuration of elements and Periodic Table., (b) Kossel and Lewis provide some logical explanation, of valence which was based on the intertness of, noble gases., (c) In the formation of a molecule, only the outer shell, electrons take part in chemical combination and they, are known as valence electrons., (d) Valence electrons in O = 6, , Cl, , –, , :, , 1., , Cl, , : O = C = O:, , N, , .., N, , F F F, , H H H, , N –– N– O octet of each atom is complete., 19., , 20., 21., , (c) Formal charge (F.C.) on an atom in a Lewis structure, = [total number of valence electrons in the free atom], – [total number of non bonding (lone pair) electrons], – (1/2) [total number of bonding (shared) electrons], (a) The lowest energy structure is the one with the, smallest formal charges on the atoms., (c) Bond order between P – O, no. of bonds in all possible direction, total no. of resonating structures, –, , O, , O, –, , –, , O P O, , –, , O, –, , –, , O, , –, , O P, O, , –, , O, , –, , O P O, O, , –, , –, , O, , O P O, –, , O, , Formal charge on oxygen =, , 3, 4, , 0.75, , 5, 4, , 1.25
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CHEMICAL BONDING AND MOLECULAR STRUCTURE, , 22., 23., , (b), , In CN– ion formal negative charge is on nitrogen atom, , due to lone pair of electrons., (d) According to octet role, the central atom must have 8, electrons but in some compounds the number of, electrons is more than 8, or less than 8 or an odd number, of electrons is left on the central atom e.g., PCl5, BF5,, NO., , F, F, , F, F, , 10 electrons, around P, , 24., , (d), , 25., , (c) (i), , Cl, Cl, , N=O, , 6 electrons, around B, , Cl, , Al, , : :, , :, , Cl, Cl:B:Cl, , Al, , Cl, , : :, , P, , F, , Odd electrons, around N, , Cl, Cl, , (Complete octet), , (ii), , CH3, , CH3, , Al, , CH3, , CH3, , Al, , CH3, CH3, , (3c – 2e) bond, (incomplete octet), , (iii), , AlF3, , (ionic, compound), , (iv) Cl – Be, , ¾¾, ® Al3+ + 3F, 2s 2 2p6 2s2 2p6, , Cl, Cl, , (octet, complete), , (octet, complete), , Be – C, , (Incomplete octet), , (inomplete octet), , H, (v) H – Be, , Be – H (Incomplete octet), H, , (3c – 2e– ) bond, , 26., , 27., 28., , 29., , (d), , .. .. .. .. .., .. S .. O, O, .., .., , Total no. of valence electron around sulphur in SO 2 is, 10 while in case of other molecules total no. of 8, electrons are present in each., (a) Boron in BCl3 has 6 electrons in outermost shell., Hence BCl3 is a electron deficient compound., (b) The compounds in which octet of central atom is, incomplete are known as electron deficient, compounds. Hence B2H6 is a electron deficient, compound., (a) PCl5 does not follow octet rule, it has 10 electrons in, its valence shell., , 67, , 30. (a), 31. (c) Formation of positive ion involves removal of, electron(s) from neutral atom and that of the, negative ion involves addition of electron(s) to the, neutral atom., 32. (b) Ionic bonds will be formed more easily between, elements with comparatively low ionization enthalpies, and elements with comparatively high negative, value of electron gain enthalpy., 33. (a) In ionic solids, the sum of the electron gain enthalpy, and the ionization enthalpy may be positive but still, the crystal structure gets stabilized due to the, energy released in the formation of the crystal, lattice., 34. (d) Lattice enthalpy is required to completely separate, one mole of a solid ionic compound into gaseous, constituent ions., 35. (a) An ionic bond tightly held the two ions of opposite, charges together, so it is a dipole. More is the, electronegativity of anion higher will be the electron, density and higher will be its charge which, consequently increses the strength of ionic bond., 36. (c) Higher the difference in electronegativity between the, two atoms, more will be electrovalent character of the, bond. Among given choices, calcium and hydrogen, have maximum difference in their electronegativities., 37. (b) For compounds containing cations of same charge,, lattice energy increases as the size of the cation, decrease. Thus, NaF has highest lattice energy. The, size of cations is in the order Na+ < K+ < Rb+ < Cs+, 38. (b) Ionic character of a bond is directly proportional to, the difference of electro negativities of bonded atoms., So, H – F in which electronegativity difference is, highest, will have highest ionic character., 39. (b) The stability of the ionic bond depends upon the lattice, energy which is expected to be more between Mg and, F due to +2 charge on Mg atom., 40. (d) Bond lengths are measured by spectroscopic, X-ray, diffraction and electron diffraction techniques., 41. (c) The covalent radius is measured approximately as, the radius of an atom’s core which is in contact with, the core of an adjacent atom in a bonded situation., 42. (d) A = Covalent radius, B = van der Waal’s radius, C = Bond length, 43. (d) All of the given statements are correct., 44. (c) Bond order, , bond enthalpy, , 1, bond length, , 45. (c) In CO (three shared election pairs between C and O), the bond order is 3. For N2 bond order is 3 and its, Dg H - is 946 kJ mol–1, being one of the highest for a, diatomic molecule, isoelectronic moleculaes and ions, have identical bond order for example F2 and O2–, 2 have, bond ordeer 1, N2, CO and NO+ have bond order 3.
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EBD_7207, CHEMICAL BONDING AND MOLECULAR STRUCTURE, , 68, , 46., 47., , 48., , 49., , (c) Both representation of resonating structures in, molecules of CO2 and CO32– are correct., (c) I and II structure shown above constitute the, cannonical structure. III structure represents the, structure of O3 more accurately. This is also called, resonance hybrid., (a) The molecule does not exist for a certain fraction of, time in one cannonical form and for other fractions, of time in other cannonical forms., (b) There are three resonance structures of CO32 ion., O, , C, –, , (d), , 51., 53., , (b), (b), , 54., , (c), , O, , –, , O–, , C, , C, , O, , O, , O, , F, , F, |, | B, , (b) O = C = O, (a) The correct order of dipole moments of HF, H2S and, H2O us, HF < H2S < H2O, , 61., , (a), , 62., 63., 64., , O–, , (I), (II), (III), Choices (a), (b) and (c) are the resonance structures of, CO2., 52. (b), Both NO2 and O3 have angular shape and hence will, have net dipole moment., The dipole moment of symmetrical molecules is zero., , |, , 50., , O, , –, , O–, , 59., 60., , 65., , 66., , F, , Triangular planar (symmetrical molecule), , 55., , 56., 57., , (c) Dipole moment is a vector quantity, hence the dipole, moment of symmetrical molecules is zero. As CO2,, p-dichlorobenzene and CH4 have regular symmetrical, shape. Hence = 0, Cl, H, .., N H, C H, O=C=O, H H, H H, Cl, However, NH3 has distorted structure due to presence, of lone pair of electrons on N atom and thus has definite, dipole moment., (a) In H2, both atoms are identical, so the molecule is non, polar., (b) BF3 has planar and symmetrical structure thus as a, result the resultant of two bond moments, being equal, and opposite to the third, cancels out and hence, molecule possess zero dipole moment., F, , F, , 68., , 69., , 70., 71., 72., , H, Cl, (b) Hydrogen is non metal and non metal atoms form, covalent bond., (d), (b) Hybridisation is sp3 and shape pyramidal., , ; =0, , B, , F, , 58., , 67., , C F, Because difference between electronegativity of, carbon and flourine is highest., (c) In case of SF6 resultant dipole moment is zero while, all other possess dipole moment., (c), (c) According to Fajan's rule, smaller size and greater, charge on cation favour the formation of partial, covalent character in ionic bonds., (c) As difference of electronegativity increases % ionic, character increases and covalent character decreases, i.e., electronegativity difference decreases covalent, character increases., Further greater the charge on the cation and smaller, the size more will be its polarising power. Hence, covalent character increases., (a) According to Fajan’s rule, as the charge on the cation, increases, and size decreases, its tendency to polarise, the anion increases. This brings more and more, covalent nature to electrovalent compounds. Hence, AlCl3 shows maximum covalent character., (d) In case of anions having same charge as the size of, anion increases, polarisibility of anion also increases., (d) The electronegativity difference is maximum in O—H, bond hence O—H bond length is the smallest among, the given set., (a) A gaseous HCl molecule has hydrogen and chlorine, linked by a covalent bond. Here electronegativity of, chlorine is greater than that of hydrogen. Due to this, the shared pair of electron is more attracted towards, chlorine. Thus, chlorine end of molecule has higher, electron density and becomes slightly negative and, the hydrogen and slightly positive. Hence the covalent, bond in HCl has a polar character as shown below, , (c) CCl4 and BF3 being symmetrical have zero dipole, moment. H2O, CHCl3 and NH3 have dipole moments, of 1.84 D, 1.01 D and 1.46 D respectively. Thus among, the given molecules H2O has highest dipole moment., , :, 73., , (c), , F, , Br, :, , F, , F
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CHEMICAL BONDING AND MOLECULAR STRUCTURE, , 75., , 76., 77., 78., , 79., , 80., , (d), , (a), (d), (b), , (a), , (b), , O, , lp = 1, bp = 4, V-shaped H2O like structure., BrF5 has square pyramidal geometry., SO2 – bent, SF4, – see-saw, ClF3 – T-shape, BrF5 – square pyramidal, XeF4 – square planar., I is the most stable geometry because both the lone, pairs are present at equitorial position. Due to which, repulsion is minimum in molecule as compared to the, repulsion in other molecules where lone pair is in axial, position., X, X, , .., , X, Number of lone pair = 2, (a) In methane molecule C is sp3 hybridised so its shape, will be tetrahedral., .., Sn, , 82., , (d), , 83., , (b) The minimum in the energy curve corresponds to the, most stable state of H2., (d) Structure of acetylene molecule, , 84., , H — C — C— C — C— C — C— C — C — H, H, , H, , Cl, , Cl, , 1, H—, C, , 1, 2, , Bent, , 1, C—, H, , Thus acetylene molecule has 3 bonds and 2 bonds., , H, , H, , H, , H, , H, , H, , 1, 3,5, 7 - octatetraene, 17 and 4, , 86. (b) Allyl cyanide is :, H, H, , C, , H, , H, , C, , C, , C, , ,, , N, , H, , It contains 9 sigma bonds, 3 pi bonds and 1 lone, pair of electrons., 87. (a) Cl2 : Cl – Cl (1 , No ), O 2 : O = O (1 , 1 ), N 2 : N N (1 , 2 ), CO2 : O = C = O (2 , 2 ), 88. (a) The given molecule is, H2, , C, , C, , C, , C, , H, , H, , H, , C, , 2, , C, , H, , The number of C—C ‘ ’ bonds = 5, The number of C—C ‘ ’ bonds = 4, The number of C—C ‘ ’ bonds = 6, , O, , M, , 81., , 85. (b), , .., S, , .., , 74., , In BrF3, both bond pairs as well as lone pairs of, electrons are present. Due to the presence of lone, pairs of electrons (lp) in the valence shell, the bond, angle is contracted and the molecule takes the Tshape. This is due to greater repulsion between two, lone pairs or between a lone pair and a bond pair than, between the two bond pairs., (a) The number of lone pairs of electrons on central atom, in various given species are, Species, Number of lone pairs on, central atom, IF7, nil, IF5, 1, ClF3, 2, XeF2, 3, Thus the correct increasing order is, IF7 < IF5 < ClF3 < XeF2, 0 1, 2, 3, , 69, , 89. (a), , p-orbital, s-orbital, The overlap between s- and p-orbitals occurs along, internuclear axis and hence the angle is 180°., O–H, , 90. (a), , CH3 C, , CH 2 has 9 ,1 and 2lone pairs., , 91. (a) Linear combination of two hybridized orbitals leads to, the formation of sigma bond., 92. (b) Sigma bond is stronger than -bond. The electrons in, the bond are loosely held. The bond is easily broken, and is more reactive than -bond. Energy released, during sigma bond formation is always more than, bond because of greater extent of overlapping, 93. (c) Option (c) represents zero overlapping., 94. (a) Bond angle increases with increase in s-character of, hybridised orbital. The table given below shows the, hybridised orbitals, their % s-chatracter and bond, angles., Hybridised, % s-character, Bond, orbitals, angle, sp 3, 25, 109.5°, sp 2, 33, 120°, sp, 50, 180°, 95. (d) Promotion of electron is not an essential condition, prior to hybridisation. It is not necessary that only, half filled orbitals participate in hybridisation. In, some cases, even filled orbitals of valence shell take, part in hybridisation.
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EBD_7207, CHEMICAL BONDING AND MOLECULAR STRUCTURE, , 70, , 96., , (d) The hybridisation in a molecule is given by, 1, [V M C A], .......(1), 2, where V = no. of valency e– in central atom, M = no. of monovalent atoms around central atom, C = charge on cation, A = charge on anion, For NH3; V = 5, M = 3, C = 0, A = 0, Putting these values in (1), we get, H, , 97., , 1, [5 3 0 0] = 4, H, 2, For H = 4, the hybridisation in molecule is sp3., (d) Hybridisaiton of SO42– ion is given by, 1, [V + M + A– C], 2, Where V = valency of central metal atom, M = no. of monovalent atoms surrounding central metal, atom., A = charge on anion, C = charge on cation, , 100. (a) O C O, C(6) = 1s2 2s2 2p2, ground state, 2, , 2s 2px 2py 2pz, , Excited state, , sp hybridized, , 101. (b), , H=, , 102. (d), 103. (a), , For SO24 ; V = 6, M = 0, A = 2, C = 0, 1, [6 + 0 + 2 – 0] = 4, 2, i.e., sp3 hybridisation and tetrahedral shape., (d) The strength of a bond depends upon the extent of, overlapping. s-s and s-p overlapping results in the, formation of bond but extent of overlapping along, internuclear axis is more in case of s-s overlapping, than in s-p. p-p overlapping may result in bond if, overlapping takes place along internuclear axis or may, result in –bond if sideways overlapping takes place., In any case the extent of overlapping is lesser in p - p, than that of the other two, s-s and s-p. Hence the correct, order is, s - s > s - p > p - p., (c) In case of Xenon compounds, , H=, , 98., , 99., , 1, [No. of valence electrons of, 2, Xe + number of monovalent atoms surrounding Xe –, charge on cation + charge or an ion], , Hybridization =, , 1, [8 + 6 – 0 + 0] = 7 i.e. sp3d3 hybridizaiton., 2, In case of SF6, Sulphur is sp3d2 hybridized., In case of BrF5 .Bromine atom has seven valence, electrons and 5 are contributed by the fluorine atoms., Total number of electron pairs in valence shell of central, , In XeF6 =, , 7 5, 6, 2, Number of shared pairs = 5, Number of one pairs = 1, The molecule is square pyramidal., In PCl5 P is sp3d hybridised. i.e. the correct answer is, XeF6 or option (c), , 104. (b), , two electrons excluded, from hybridization and, partcipate in bonding, with oxygens., In fullerene, carbons are sp2-hybridised like graphite., Fullerenes are also the allotropes of carbon other than, diamond and graphite but have been produced, synthetically, e.g.; C60., Ions having sp3 hybridisation contain hydrogen atoms, at the corners of tetrahedron., For linear arrangement of atoms the hybridisation, should be sp(linear shape, 180° angle). Only H2S has, sp3-hybridization and hence has angular shape while, C2H2, BeH2 and CO2 all involve sp - hybridization and, hence have linear arrangement of atoms., BF3 involves sp2-hybridization., , sp 3 sp sp, sp 3, 105. (b) H3C — C C — CH3, linear, 106. (b) Equilateral or triangular planar shape involves sp2, hybridization., 107. (a) Only those d orbitals whose lobes are directed along, X, Y and Z directions hybridise with s and p orbitals., In other three d orbitals namely dxy, dyz and dxz, the, lobes are at an angle of 45° from both axis, hence the, extent of their overlap with s and p orbitals is much, lesser than d, , x 2 y2, , and d, , z2, , orbitals., , 108. (a) According to VSEPR theory, trigonal bipyramidal, geometry results from sp3d or dsp3 hybridisation., dsp2 hybridisation results in square planar geometry,, while d2sp3 leads to octahedral shape., 109. (b) In CH3 CH2OH underlined C is forming 4 bonds,, hence sp3 hybridisation. In others it is sp2 hybridised, (due to 3 bonds)., 110. (a) Each sp3-hybrid orbital has 25% s-character and 75%, p-character., 111. (c), , H3C CH, sp3, , sp 2, , sp 2, , sp, , C, , sp, , 112. (d), , atom =, , 113. (a), , C, , CH 2, sp3, sp, , sp 2, , CH2, sp3, , 3, , sp3, , sp3, sp3, , CH CH 3, 2, sp, , sp3
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CHEMICAL BONDING AND MOLECULAR STRUCTURE, sp3, , sp 2, , sp, , sp sp, , CH3 CH, HC, sp 2, , CH 2, , C C, sp 2, , sp 2, , 131. (b) The removal of an electron from a diatomic molecule, may increase the bond order as in the conversion, O 2 ( 2 .5) or decrease the bond order as, O2 (2), N 2 ( 2.5), As a, in the conversion, N 2 (3.0), result, the bond energy may increase or decrease., Thus, statement (b) is incorrect., , sp 3, , CH CH 3, sp, , CH, sp 2, , CH CH, 1, 3-butadiene, , sp2, , CH2, , 114. (d) Statement (d) is incorrect., 115. (c) Molecular orbital theory was given by Mulliken., 116. (b) Atomic orbital is monocentric because an electron in, it is influenced by one nucleus. While molecular, orbital is polycentric as it is influenced by two or, more nuclei depending upon the number of atoms in, the molecule., 117. (c), 118. (d) For oxygen correct increasing order is, 1s <, s < 2s < 2s < 2pz < ( 2px = 2py) <, (, px= *2py) < 2pz, 119. (d) Nb < Na or Na = Nb, i.e., a negative or zero bond order corresponds to an, unstable molecule., 120. (a), 121. (c), 122. (c) Paramagnetism of O2 is best explained by molecular, orbital theory., 123. (b), 124. (b) One bonding M.O. and one anti-bonding M.O., 125. (d) Molecules having unpaired electrons show, paramagnetism., 126. (c), 127. (c) N 2 = 7 + 7 – 1 = 13 electrons, Configuration is, 1s 2 , *1s 2 , 2s 2 , * 2s 2 , 2 px 2, = 2 p y 2 , 2 p1z, Bond order =, 1 No. of e s in bonding No. of e s in antibonding, 2 molecular orbital, molecular orbital, 1, 1, 5 2.5, = (9 4), 2, 2, 128. (a), 129. (a, b) The molecular orbital structures of C2 and N2 are, N2, , 1s 2 *1s2 2s 2 *2s 2 2 px2 2 py2 2 pz2, , C2, , 1s 2 *1s 2 2 s 2 * 2 s 2 2 py 2 2 Pz2, , Both N2 and C2 have paired electrons, hence they are, diamagnetic., 130. (d) H 2 ( 1s1), Bond order =, , 1, (1 0), 2, , 1, (2 1), 2, , 132. (b), , O2 : 1s 2 ,, , *, , 1s 2 , 2s 2 ,, , *, , 2 s 2 , 2 p z2 ,, 2 p 2x, 2 p 2y, , ,, , *, , 2 p1x, , *, , 2 p1y, , 10 6, 2, 2, (two unpaired electrons in antibonding molecular orbital), , Bond order, , O 2 : 1s 2 ,, , *, , 1s 2 , 2s 2 ,, , *, , 2s 2 , 2 p 2z ,, , 2 p x2 ,, , *, , 2 p1x, , 2 p 2y ,, , *, , 2 p 0y, , 10 5, 2. 5, 2, (One unpaired electron in antibonding molecular, orbital), , Bond order, , Hence O2 as well as O 2 both are paramagnetic, and, bond order of O 2 is greater than that of O2., 133. (c), 134. (c) Helium molecule does not exist as bond order of, He2 = 0., 135. (c) The paramagnetic property in oxygen came through, unpaired electron which can be explained by molecular, orbital theory., , 2p*x, , 2pz* 2py*, px py pz, , px py pz, , 1, 2, , 2px bonding, , H 2 : ( 1s 2 ) ( *1s1 ), , Bond order =, , 71, , 1, 2, , The bond order of H 2 and H 2 are same but H 2 is more, stable than H 2 . In H 2 the antibonding orbital is filled, with 1 electron so this causes instability., , 2p*y and 2p*z, 136. (c) H-bonding is maximum in the solid state and, minimum in gaseous state., So 2 unpaired of electron present in
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EBD_7207, CHEMICAL BONDING AND MOLECULAR STRUCTURE, , 72, , 137. (c) H2O shows intermolecular hydrogen bonding while, o-nitrophenol shows intramolecular H-bonding., O, , 138. (a), , N, , (Intermolecular, hydrogen bonding), , O, OH, , o-nitrophenol, , 139. (c) The b.p. of p-nitrophenol is higher than that of, o-nitrophenol because in p-nitrophenol there is, intermolecular H-bonding but in o-nitrophenol it is, intramolecular H-bonding., 140. (b) The strength of the interactions follows the order, vander Waal’s < hydrogen – bonding < dipole-dipole, < covalent., 141. (c) H–F shows str ongest H-bonds due to high, electronegativity of F atom., 142. (c) Ice has many hydrogen bonds which give rise to cage, like structure of water molecules. This, structure, possess larger volume and thus makes the density of, ice low., 143. (b) Methanol and ethanol are soluble because of the, hydrogen bonding., 144. (d) F—H----F bond is shortest, because with the increase, of electronegativity and decrease in size of the atom to, which hydrogen is linked, the strength of the hydrogen, bond increases., 145. (a) Hydrogen bonding is possible only in compounds, having hydrogen attached with F, O or N., C2 H 5 OH, (H-bonding possible), , CH 3 O CH 3, (H-bonding not possible), , O, ||, CH3 C CH 3, , O, ||, C H, , CH 3, (H-bonding not possible) (H-bonding not possible), , 146. (c) ortho-Nitrophenol has intramolecular H-bonding, ., , ., , ., ., , N, , O, , and, , para-nitrophenol, , has, , ||, , ., , OH, , O, intermolecular H-bonding., 147. (d) Hydrogen bonding increases the boiling point of, compound., 148. (b), 149. (a) Hydrogen bond is formed when hydrogen is attached, with the atom which is highly electronegative and, having small radius., , 151. (d) The least electronegative atom occupies the central, position in the molecule/ion., 152. (b) Formal charges do not indicate real charge separation, within the molecule. Indicating the charges on the, atoms in the Lewis structure only helps in keeping, track of the valence electrons in the molecule., 153. (c) The greater the charge on the cation, the greater the, covalent character of the ionic bond., 154. (d) Statement (c) and (d) are incorrect., 155. (b) (ii) (F) In general as the number of lone pair of electrons, on central atom increases, value of bond angle from, normal bond angle decreases due to lp – lp > lp – bp., (iv)(F) Structures of xenon fluorides and xenon, oxyfluoride are explained on the basis of VSEPR theory., In SOBr2, S – O bond has minimum bond length in, comparison to S – O bond lengths in SOF2 and SOCl2,, because in SOBr 2, S – O bond has been formed by, hybrid orbital containing less s-character., 156. (b) Atomic orbitals having same or nearly same energy, will not combine if they do not have the same, symmetry. 2pz orbital of one atom cannot combine, with 2px or 2py orbital of other atom because of their, different symmetries., , MATCHING TYPE QUESTIONS, 157. (c) BeH2 : Incomplete octet of central atom., Be has 2 valence electrons, SF6 : Expanded octet, There are 12 electrons around the S atom in SF6, NO2 : Odd electron molecules., In molecules with an odd number of electrons like, NO2, the octet rule is not satisfied., 158. (c), 159. (b) NH3, 1lp, 3bp, Trigonal pyramidal, SO2, 1lp, 2bp, Bent, SF4, 1lp, 4bp, See-saw, ClF3, 2lp, 3bp, T-shape, 160. (c) Trigonal planar = BF3, Tetrahedral = NH +, 4, Trigonal bipyramidal = PCl5, Octahedral = SF6, 161. (c), 162. (a) SF6, sp3d2, PF5, sp3d, BCl3, sp2, C2H6, sp3, 163. (c) Valence bond theory = Heitler and London, , STATEMENT TYPE QUESTIONS, 150. (c) The group valence of the elements is generally either, equal to the number of dots in Lewis symbols or 8, minus the number of dots or valence electrons., , Octet rule = Ko ssel and Lewis, Molecular orbital theory = F. Hund and R.S. Mulliken, VSEPR theory = Nyholm and Gillespie, 164. (b)
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CHEMICAL BONDING AND MOLECULAR STRUCTURE, , 73, , ASSERTION-REASON TYPE QUESTIONS, :, , 1, , : :, , 2, , : :, , 165. (a), , O, , 3, , O:, , O, , Formal charge on O1 = 6 – 2 –, , 1, (6) = +1, 2, , Formal charge on O2 = 6 – 4 –, , 1, (4) = 0, 2, , 1, × 3 = –1, 2, Hence, correct representation of O3 is, Formal charge on O3 = 6 – 6 –, , :, , (+1), , : :, , (0), , : :, , O, , (–1), , O:, , O, , 166. (a) Atoms combine either by transfer of valence, electrons from one atom to another or by sharing of, valence electrons to have an octet in their valence, shell., 167. (d) Assertion is false but reason is true., The greater the lattice enthalpy, more stable is the ionic, compound., 168. (c) Sulphur forms many compounds in which the octet, rule is obeyed. For example SCl2 has an octet of, electrons around it., 169. (b) Both assertion and reason are true but reason is not, the correct explanation of assertion., BF3 is sp2 hybridized. Dipole moment is a vector, quantity. The three bond moments give a net sum of, zero, as the resultant of any two is equal and opposite, to the third., , F, F, , B, (a), , =0, F, , (, , )= 0, , +, (b), , 170. (d) Assertion is false but reason is true., CH2Cl2 is polar while CCl4 is non-polar because in, CCl4 net dipole moment cancels., 171. (a) While the lone pairs are localised on the central, atom, each bonded pair is shared between two, atoms. As a result, the lone pair electrons in a, molecule occupy more space as compared to the, bonding pairs of electrons. This results in greater, repulsion between lone pairs of electrons as compared, to the lone pair -bond pair and bond pair - bond pair, repulsions., , 172. (a) Both assertion and reason are true and reason is the, correct explanation of assertion., lone pair – lone pair repulsion > lone pair – bond pair, repulsion > bond pair – bond pair repulsion. In the, ammonia molecule, NH3 there are three bond pairs and, one lone pair. The three N – H bond pairs are pushed, closer because of the lone pair – bond pair repulsion,, and HNH bond angle gets reduced from 109°23’ (the, tetrahedral angle) to 107°., 173. (d) Assertion is false but reason is true., NH3 molecule is pyramidal is shape, because out of, four electron pairs, three are bonding pairs and one is, lone pair., 174. (a) Both assertion and reason are true and reason is the, correct explanation of assertion., pi bonds are formed by the overlapping of p-p orbitals, perpendicular to their axis i.e., sidewise overlap., 175. (a) Both assertion and reason are true and reason is the, correct explanation of assertion., Helium molecule is formed by linking two helium atoms., Both have 1s orbitals. These will combine to form two, molecular orbitals (1s) and * (1s). Four available, electrons are accommodated as (1s)2 and * (1s)2., 176. (c) The electron density in a bonding molecular orbital, is located between the nuclei of the bonded atoms, because of which the repulsion between the nuclei, is very less while in case of an antibonding, molecular orbital, most of the electron density is, located away from the space between the nuclei., Electrons placed in a bonding molecular orbital tend, to hold the nuclei together and stabilise a molecule., 177. (b) Both assertion and reason are true but reason is not, the correct explanation of assertion., Water is excellent solvent because it has high value of, dielectric constant. Due to high value of dielectric, constant, the electrostatic force of attraction between, the ions decrease and these ions get separated and, ultimately get solvated by the solvent molecules., , CRITICAL THINKING TYPE QUESTIONS, 178. (b) Hybridisation of the central atom in compound is given, by, 1, [V M C A], 2, where V = No. of valency electrons in central metal, atom,, M = No. of monovalent atoms surrounding the central, atom,, C = charge on cation and A = charge on anion, , H=, , 1, [5 0 0 1], 2, sp2 hybridisation, , For NO 2 , H =, , 3
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EBD_7207, CHEMICAL BONDING AND MOLECULAR STRUCTURE, , 76, , 198. (c) Benzene has the following resonance structures–, , Hence, its bond order is, no of possible resonating structures, 1.5., 2, 199. (a) Molecular orbital electronic configuration of these, species are :, , 2 px2, , 2 p2y , *2 px2, , 2 px2, , 2 p 2y , *2 p1x, , 2 px2, , * 2 p1y, , 2 p 2y ,, , *2 px2, , 2, b, , *2 2, a ( b, , 2, 1, b ) b (N 2, , 13electrons), , it contains one unpaired electron hence paramagnetic., 201. (N) None of the given option is correct., The molecular orbital configuration of the given, molecules is, H2 = 1s2 (no electron anti-bonding), Li2 = 1s2 1s2 2s2(two anti-bonding electrons), B2 = 1s2, , 1s2 2s2, , 2s2, , (4 anti-bonding electrons), , 2p1y, , 2p1z, , Bond order =, N 2–, , * 2 p1x, , 2 p2z, , 10 – 6, 2, , * 2 p1y, , 2, , 1s 2 *1s 2 2s 2 * 2s 2, , Bond order =, , *2 p 2y, , Hence number of antibonding electrons are 7, 6 and 8, respectively., *a2, , 2 px2, , 2* p1y, , 1s 2 , *1s 2 , 2 s 2 , * 2 s 2 , 2 p z2 ,, , O22 (18e ), , 1s 2 *1s 2 2 s 2 * 2 s 2 -, , 2 p 2y, , 1s 2 , *1s 2 , 2 s 2 , * 2 s 2 , 2 p z2 ,, , O2 (16e ), , 2, b, , N 2–, 2, , 1s 2 , *1s 2 , 2s 2 , * 2 s 2 , 2 pz2 ,, , O2 (17e ), , 200. (d), , Though the bond order of all the species are same, (B.O = 1) but stability is different. This is due to, difference in the presence of no. of anti-bonding, electron., Higher the no. of anti-bonding electron lower is the, stability hence the correct order is H2 > Li2 > B2, 202. (a) Molecular orbital configuration of, , N2, , 10 5, 2, , 2 px2, 2 p 2y, , * 2 p1x, , 2 pz2, , *2 p 0y, , 2.5, , 1s 2 *1s 2 2 s 2 * 2 s 2, , 2 px2, 2 p 2y, , ,, , 2 pz2, , 10 – 4, 3, 2, The correct order is = N 22– N 2– N 2, 203. (a) Hydrogen bonding is not possible in HI due to low, electronegativity of iodine. So, hydrogen bonding, would not affect boiling point of HI., 204. (d) HF form linear polymeric structure due to hydrogen, bonding., 205. (c) B, C and D form intermolecular hydrogen bonding, while A form intramolecular hydrogen bonding due to, proximity of oxygen and hydrogen., , Bond order =
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5, STATES OF MATTER, FACT / DEFINITION TYPE QUESTIONS, 1., , 2., , 3., , Which of the following is not a type of van der Waal's, forces?, (a) Dipole - dipole forces, (b) Dipole - induced dipole forces, (c) Ion - dipole forces, (d) London forces, Who proposed the concept of dispersion force ?, (a) Heitler and London, (b) van der Waal, (c) Gay Lussac, (d) Fritz London, Which of the following option correctly represents the, relation between interaction energy and distance between, two interacting particles (r) for London forces ?, (a), , (c), 4., , 5., , 6., , 1, r, 1, 12, , r, , (b), , (d), , 7., , 8., , 1, r6, 1, r3, , The interaction energy of London force is inversely, proportional to sixth power of the distance between two, interacting particles but their magnitude depends upon, (a) charge of interacting particles, (b) mass of interacting particles, (c) polarisability of interacting particles, (d) strength of permanent dipoles in the particles., London forces are always ...I... and interaction energy is, inversely proportional to the ...II... power of the distance, between two interacting particles., Here, I and II refer to, (a) I repulsive, II sixth, (b) I attractive, II fourth, (c) I attractive, II sixth, (d) I repulsive, II fourth, Dipole-dipole forces act between the molecules possessing, permanent dipole. Ends of dipoles possess ‘partial charges’., The partial charge is, , 9., , 10., , 11., , (a) more than unit electronic charge, (b) equal to unit electronic charge, (c) less than unit electronic charge, (d) double the unit electronic charge, Dipole-dipole interaction is stronger than the London forces, but is weaker than ion-ion interaction because, (a) only partial charges are involved, (b) only total charges are involved, (c) both (a) and (b), (d) sometimes (a) and sometimes (b), Induced dipole moment depend upon the, I. dipole moment present in the permanent dipole., II polarisability of the electrically neutral molecules., Identify the correct option., (a) I is correct but II is wrong, (b) I is wrong and II is correct, (c) Both I and II are wrong, (d) Both I and II are correct, Dipole-induced dipole interactions are present in which of, the following pairs :, (a) Cl2 and CCl4, (b) HCl and He atoms, (c) SiF4 and He atoms, (d) H2O and alcohol, Which of the following exhibits the weakest intermolecular, forces ?, (a) NH3, (b) HCl, (c) He, (d) H2O, Strength of the hydrogen bond is determined by interaction, between the, I. lone pair of the electronegative atom and the hydrogen, atom of other atom., II. bond pair of the electronegative atom and the hydrogen, atom of other atom., Identify the correct option., (a) Only I is correct, (b) Only II is correct, (c) Both I and II are correct, (d) Neither I nor II are correct
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EBD_7207, STATES OF MATTER, , 78, , 12., , 13., , 14., , 15., , 16., , 17., , 18., , 19., , Which of the following statements regarding thermal energy, is correct ?, (a) Thermal energy is the measure of average kinetic, energy of the particles of the matter and is thus, responsible for movement of particles., (b) Intermolecular forces tend to keep the molecules, together but thermal energy of the molecules tends to, keep them apart., (c) Three states of matter are the result of balance between, intermolecular forces and the thermal energy of the, molecules., (d) All of the above, Which of the following is the correct order of thermal energy, in three states of matter ?, (a) Solid < Liquid < Gas (b) Liquid < Gas < Solid, (c) Liquid < Solid < Gas (d) Gas < Solid < Liquid, Which of the following are arrangement in the correct order?, I. Gas > Liquid > Solid (Thermal energy), II. Solid > Liquid > Gas (Intermolecular force), Select the correct option., (a) I only, (b) II only, (c) Both I and II, (d) None of these, Which one of the following statements is not correct about, the three states of matter i.e., solid, liquid and gaseous ?, (a) Molecules of a solid possess least energy whereas, those of a gas possess highest energy., (b) The density of solid is highest whereas that of gases, is lowest, (c) Gases like liquids possess definite volumes, (d) Molecules of a solid possess vibratory motion, Which of the following is true about gaseous state ?, (a) Thermal energy = Molecular attraction, (b) Thermal energy >> Molecular attraction, (c) Thermal energy << Molecular attraction, (d) Molecular force >> Those in liquids, The first reliable measurement on properties of gases was, made by _____________, (a) Gay Lussac, (b) Jacques charles, (c) Robert Boyle, (d) Avogadro, At constant temperature, for a given mass of an ideal gas, (a) the ratio of pressure and volume always remains, constant., (b) volume always remains constant., (c) pressure always remains constant., (d) the product of pressure and volume always remains, constant., Which of following graph(s) represents Boyle's law, , I., , II., , P, V, , P, V, , III., , IV., , P, , P, , 1/V, , 20., , 1/V, , (a) Only I, (b) II and IV, (c) I and III, (d) Only III, Which of the following represents Boyle's law in terms of, density ?, (a) d . p = k', , (b), , d, p, , k', , dp, k', (d) d k ' p, 2, (Here d = density, p = pressure, k' = constant), Boyle’s law states that the, (a) pressure of a gas is directly proportional to the, temperature at constant volume, (b) pressure of a gas is inversely proportional the volume, at constant temperature, (c) volume is directly proportional to the temperature at, constant pressure, (d) None of the above, 600 c.c. of a gas at a pressure of 750 mm of Hg is compressed, to 500 c.c. Taking the temperature to remain constant, the, increase in pressure, is, (a) 150 mm of Hg, (b) 250 mm of Hg, (c) 350 mm of Hg, (d) 450 mm of Hg, The lowest hypothetical or imaginary temperature at which, gases are supposed to occupy zero volume is called, _________, (a) Kelvin temperature, (b) absolute zero, (c) Charle's temperature (d) constant temperature, 500 ml of nitrogen at 27°C is cooled to –5°C at the same, pressure. The new volume becomes, (a) 326.32 ml, (b) 446.66 ml, (c) 546.32 ml, (d) 771.56 ml, What is the value of X in °C for given volume vs temperature, curve ?, p1 p < p < p < p, p2 1 2 3 4, p3, Volume, p4, , (c), 21., , 22., , 23., , 24., , 25., , 26., , X Temperature (°C), (a) 0° C, (b) 273.15° C, (c) – 273.15° C, (d) 300° C, Which of the following expression at constant pressure, represents Charle’s law?, (a), , V, , 1, T, , (b) V, , (c), , V, , T, , (d) V, , 1, T2, d
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STATES OF MATTER, , 34., , The following graph illustrates, (a) Dalton’s law, , V1 < V2 < V3 < V4, , V, , (b) Charle’s law, , Pressure (bar), , 27., , 79, , (c) Boyle’s law, (d) Gay-Lussac’s law, Temp. (ºC), 28., , 29., , 30., , Air at sea level is dense. This is a practical application of, (a) Boyle’s law, (b) Charle’s law, (c) Kelvin’s law, (d) Brown’s law, Use of hot air balloons in sports and meteorological, observations is an application of, (a) Boyle’s law, (b) Charle’s law, (c) Kelvin’s law, (d) Gay-Lussac’s law, An ideal gas will have maximum density when, (a) P = 0.5 atm, T = 600 K (b) P = 2 atm, T = 150 K, (c) P = 1 atm, T = 300 K (d) P = 1 atm, T = 500 K, , 31., Volume (mL), , P1 < P2 < P 3 < P 4, , 35., , 37., , –273.5, , 32., , 33., , 38., , Arrange the pressures P1, P2, P3 and P4 in the increasing, order which are shown in the graphs., (a) P1 < P2 < P3 < P4, (b) P4 < P3 < P2 < P1, (c) P1 = P2 = P3 = P4, (d) P4 = P3 < P2 = P1, On a ship sailing in pacific ocean where temperature is 23.4°C,, a balloon is filled with 2 L air. What will be the volume of the, balloon when the ship reaches Indian ocean, where, temperature is 26.1°C ?, (a) 2.018 L, (b) 2.8 L, (c) 3.5 L, (d) 1.5 L, Which of the following represents Gay Lussac's law ?, I., , P, T, , III., , P1V1, , constant, , II., , P1T2 = P2T1, , P2V2, , Choose the correct option., (a) I, II and III, (b) II and III, (c) I and III, (d) I and II, , V4, , 100 200, , 300, , 400, , The relationship which is shown in the figure is derived, from, I. Boyle's law., II. Avogadro law., III. Charles' law., Which of the following is the correct option ?, (a) I and II, (b) II and III, (c) I and III, (d) Only I, Which of the following represents Avogadro law ?, m, M, (c) M = kd, (d) All of these, At STP molar volume of an ideal gas or a combination of, ideal gases is ___________, (a) 22.71098 L mol–1, (b) 20.71098 L mol–1, –1, (c) 22.4139 L mol, (d) 24.78 L mol–1, 4.4 g of a gas at STP occupies a volume of 2.24 L, the gas, can be, (a) O2, (b) CO, (c) NO2, (d) CO2, An ideal gas is one which obeys the gas laws under, (a) a few selected experimental conditions, (b) all experimental conditions, (c) low pressure alone, (d) high temperature alone, For an ideal gas, number of moles per litre in terms of its, pressure P, gas constant R and temperature T is, (a) PT/R, (b) PRT, (c) P/RT, (d) RT/P, Select one correct statement. In the gas equation, PV = nRT, (a) n is the number of molecules of a gas, (b) V denotes volume of one mole of the gas, (c) n moles of the gas have a volume V, (d) P is the pressure of the gas when only one mole of gas, is present., Correct gas equation is, , (a) V = kn, , P3, , –300 –200 –100 0, 100, Temperature (°C), , V3, , Temperature (K), , P2, P4, , V2, , 0, , 36., , P1, , V1, , 39., , 40., , 41., , (b) V = k, , (a), , V1T2, P1, , V2T1, P2, , (b), , PV, 1 1, P2V2, , T1, T2, , (c), , PT, 1 2, V1, , P2V2, T2, , (d), , V1V2, T1T2, , P1P2
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EBD_7207, STATES OF MATTER, , 80, , 42., , 43., , The correct value of the gas constant ‘R’ is close to :, (a) 0.082 litre-atmosphere K, (b) 0.082 litre-atmosphere K–1 mol–1, (c) 0.082 litre – atmosphere–1 K mol–1, (d) 0.082 litre –1 atmosphere – 1 K mol, If P, V, M, T and R are pressure, volume, molar mass,, temperature and gas constant respectively, then for an ideal, gas, the density is given by, (a), , RT, PM, , (b), , P, RT, , PM, M, (d), RT, V, Pure hydrogen sulphide is stored in a tank of 100 litre, capacity at 20° C and 2 atm pressure. The mass of the gas, will be, (a) 34 g, (b) 340 g, (c) 282.68 g, (d) 28.24 g, At N.T.P the volume of a gas is found to be 273 ml. What, will be the volume of this gas at 600 mm of Hg and 273°C?, (a) 391.8 ml, (b) 380 ml, (c) 691.6 ml, (d) 750 ml, Pressure of a mixture of 4 g of O2 and 2 g of H2 confined, in a bulb of 1 litre at 0°C is, (a) 25.215 atm, (b) 31.205 atm, (c) 45.215 atm, (d) 15.210 atm, Gas equation PV = nRT is obeyed by, (a) Only isothermal process, (b) Only adiabatic process, (c) Both (a) and (b), (d) None of these, The total pressure of a mixture of two gases is :, (a) the sum of the partial pressures, (b) the difference between the partial pressures, (c) the product of the partial pressures, (d) the ratio of the partial pressures, If three unreactive gases having partial pressures PA, PB, and PC and their moles are 1, 2 and 3 respectively then their, total pressure will be, (a) P = PA + PB + PC, (b) P PA PB PC, 6, , 52., , 53., , 54., , (c), , 44., , 45., , 46., , 47., , 48., , 49., , (c), , 50., , 51., , PA, , PB, , 56., , 57., , 58., , 1, 1, (d), 2, 3, 56 g of nitrogen and 96 g of oxygen are mixed isothermally, and at a total pressure of 10 atm. The partial pressures of, oxygen and nitrogen (in atm) are respectively, (a) 4, 6, (b) 5, 5, (c) 2, 8, (d) 6, 4, If 10–4 dm3 of water is introduced into a 1.0 dm3 flask at 300, K, how many moles of water are in the vapour phase when, equilibrium is established ?, (Given : Vapour pressure of H2O at 300 K is 3170 Pa;, R = 8.314 J K–1 mol–1), (a) 5.56× 10–3 mol, (b) 1.53 × 10–2 mol, (c) 4.46 × 10–2 mol, (d) 1.27 × 10–3 mol, , (c), 59., , PC, , (d) None of these, 3, Dalton’s law of partial pressure will not apply to which of, the following mixture of gases, (a) H2 and SO2, (b) H2 and Cl2, (c) H2 and CO2, (d) CO2 and Cl2, Pressure exerted by saturated water vapour is called, _____________, (a) Aqueous tension, (b) Partial pressure, (c) Total pressure, (d) Both (a) and (b), P, , 55., , The pressure exerted by 6.0g of methane gas in a 0.03 m3, vessel at 129°C is (Atomic masses : C = 12.01, H = 1.01 and, R = 8.314 JK–1 mol –1), (a) 31684 Pa, (b) 215216 Pa, (c) 13409 Pa, (d) 41648 Pa, A gaseous mixture was prepared by taking equal mole of, CO and N2. If the total pressure of the mixture was found 1, atmosphere, the partial pressure of the nitrogen (N2) in the, mixture is, (a) 0.5 atm, (b) 0.8 atm, (c) 0.9 atm, (d) 1 atm, A bubble of air is underwater at temperature 15°C and the, pressure 1.5 bar. If the bubble rises to the surface where the, temperature is 25°C and the pressure is 1.0 bar, what will, happen to the volume of the bubble ?, (a) Volume will become greater by a factor of 1.6., (b) Volume will become greater by a factor of 1.1., (c ) Volume will become smaller by a factor of 0.70., (d) Volume will become greater by a factor of 2.5., A mixture contains 64 g of dioxygen and 60 g of neon at a, total pressure of 10 bar. The partial pressures in bar of, dioxygen and neon are respectively (atomic masses O = 16,, Ne = 20), (a) 4 and 6, (b) 6 and 4, (c) 5 and 5, (d) 8 and 2, 500 mL of air at 760 mm pressure were compressed to 200, mL. If the temperature remains constant, what will be the, pressure after compression?, (a) 1800 mm, (b) 1900 mm, (c) 2000 mm, (d) 1500 mm, Value of universal gas constant (R) depends upon, (a) Number of moles of gas, (b) Volume of gas, (c) Temperature of gas, (d) None of these, Two vessels of volumes 16.4 L and 5 L contain two ideal, gases of molecular existence at the respective temperature, of 27°C and 227°C and exert 1.5 and 4.1 atmospheres, respectively. The ratio of the number of molecules of the, former to that of the later is, (a) 2, (b) 1, , 60.
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STATES OF MATTER, , 61., , 62., , 63., , 64., , 65., , 66., , 67., , The total pressure of a mixture of two gases is, (a) the sum of the partial pressures, (b) the difference between the partial pressures, (c) the product of the partial pressures, (d) the ratio of the partial pressures, Ideal gas equation is the combination of, I. Boyle's law, II. Charles's law, III. Avogadro law, IV. Dalton's law of partial pressure, Choose the correct option., (a) Only I and II, (b) I, II and III, (c) II and III, (d) I, III aor IV, According to the kinetic theory of gases, in an ideal gas,, between two successive collisions a gas molecule travels, (a) in a wavy path, (b) in a straight line path, (c) with an accelerated velocity, (d) in a circular path, Gases consist of large number of identical particles (atoms, or molecules) that are so small and so far apart on the average, that the actual volume of the molecules is negligible in, comparison to the empty space between them., Above given statement explain which property of gases ?, (a) Gases occupy all the space available to them., (b) Gases has fixed shape., (c) Compressibility of gases., (d) None of these., If there were loss of kinetic energy, the motion of gas, molecules will _____A, _______ and gas will ____B, _______., (a) A = increase, B = collide, (b) A = stop, B = settle down, (c) A = increase,, B = exert more pressure on walls of container, (d) A = decrease, B = get liquified, Which of the following assumption of kinetic molecular, theory states that gases do not have fixed shape ?, (a) Particles of a gas move in all possible directions in, straight line., (b) Particles of a gas are always in constant and random, motion., (c) Total energy of molecules before and after the collision, remains same., (d) None of these, Which of the following assumption of kinetic theory if hold, good than the pressure vs volume graph of experimental, data (real gas) and that of theoretically calculated from Boyle's, law (ideal gas) should coincide ?, (i) There is no force of attraction between the molecules, of a gas., (ii) Volume of the molecules of a gas is negligibly small in, comparison to the space occupied by the gas., (a) (i) only, (b) (ii) only, (c) Both (i) and (ii), (d) None of these, , 81, , 68., , 69., , 70., , 71., , 72., , Kinetic theory of gases proves, (a) only Boyle’s law, (b) only Charles’ law, (c) only Avogadro’s law (d) All of these, Which one of the following is the wrong assumption of, kinetic theory of gases ?, (a) Momentum and energy always remain conserved., (b) Pressure is the result of elastic collision of molecules, with the container’s wall., (c) Molecules are separated by great distances compared, to their sizes., (d) All the molecules move in straight line between collision, and with same velocity., When is deviation more in the behaviour of a gas from the, ideal gas equation PV = nRT ?, (a) At high temperature and low pressure, (b) At low temperature and high pressure, (c) At high temperature and high pressure, (d) At low temperature and low pressure, In van der Waal’s equation of state of the gas law,, theconstant ‘b’ is a measure of, (a) volume occupied by the molecules, (b) intermolecular attraction, (c) intermolecular repulsions, (d) intermolecular collisions per unit volume, In van der Waal’s equation of state for a non-ideal gas, the, term that accounts for intermolecular forces is, (a) (V – b), (b) RT, , 73., , 74., , 75., , 76., , a, , (d) (RT)–1, V2, The values of van der Waal’s constant ‘a’ for the gases O2,, N2, NH3 and CH4 are 1.360, 1.390, 4.170 and 2.253 L2 atm, mol–2 respectively. The gas which can most easily be liquified is, (a) O2, (b) N2, (c) NH3, (d) CH4, A gas described by van der Waal’s equation, (i) behaves similar to an ideal gas in the limit of large, molar volume, (ii) behaves similar to an ideal gas in the limit of large, pressure, (iii) is characterised by van der Waal's coefficients that are, dependent on the identity of the gas but are, independent of the temperature, (iv) has the pressure that is lower than the pressure exerted, by the same gas behaving ideally, (a) (i) and (ii), (b) (i) and (iii), (c) (i), (ii) and (iii), (d) (ii) and (iv), The units of constant a in van der Waal’s equation is, (a) dm6 atm mol–2, (b) dm3 atm mol–1, (c) dm atm mol–1, (d) atm mol–1, The van der Waal’s constant ‘a’ for four gases P, Q, R and S, are 4.17, 3.59, 6.71 and 3.8 atm L2 mol–2 respectively., Therefore, the ascending order of their liquefaction is, (a) R < P < S < Q, (b) Q < S < R < P, (c) Q < S < P < R, (d) R < P < Q < S, (c), , P
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EBD_7207, STATES OF MATTER, , 82, , 77., , At low pressure the van der Waal’s equation is reduced to, (a), , 1, , ap, RT, , (b), , Z, , pVm, RT, , 1, , b, p, RT, , pVm, , (a) 1 +, , RT, pb, , RT, , (d), , Gas Tc / K, , Z, , (b) 1, , pb, pb, (d) 1 –, RT, RT, The gas with the highest critical temperature is, (a) H2, (b) He, (c) N2, (d) CO2, A gas is said to behave like an ideal gas when the relation, PV/T = constant. When do you expect a real gas to behave, like an ideal gas ?, (a) When the temperature is low, (b) When both the temperature and pressure are low, (c) When both the temperature and pressure are high, (d) When the temperature is high and pressure is low, An ideal gas can’t be liquefied because, (a) its critical temperature is always above 0°C, (b) Its molecules are relatively smaller in size, (c) it solidifies before becoming a liquid, (d) forces between its molecules are negligible, Gases possess characteristic critical temperature which, depends upon the magnitude of intermolecular forces, between the particles. Following are the critical temperature, of some gases., Gases, H2, He, O2, N2, Critical temperature 33.2 5.3, 154.3 126, in Kelvin, From the above data what would be the order of liquefaction, of these gases ?, Start writing the order from the gas liquefying first, (a) H2, He, O2, N2, (b) He, O2, H2, N2, (c) N2, O2, He, H2, (d) O2, N2, H2, He, Above Boyle point, real gases show ______X, _______ from, ideality and Z values are _______Y, ________ than one., (a) X = Negative deviation, Y = Less, (b) X = Negative deviation, Y = Greater, (c) X = Positive deviation, Y = Less, (d) X = Positive deviation, Y = Greater, Select the one that when used would be considered as best, condition for liquification of a gas., (a) Increasing the temperature., (b) Decreasing the pressure., (c) Increasing the pressure and decreasing the, temperature., (d) Decreasing the pressure and increasing the temperature., , (c) 1 +, , 80., , 81., , 82., , 83., , 84., , 86., , H2, , 33.2, , He, , 5.3, , N2, O2, , 126, 154.3, , (b) H2 < He < N2 < O2, (a) He < N2 < H2 < O2, (c) He < H2 < N2 < O2, (d) O2 < N2 < H2 < He, Choose the correct statement based on the following, isotherms of carbon dioxide at various temperature., G, H, P3, , Pressure, , 79., , Following table represents critical temperature of some, gases. Arrange these gases in their increasing order of, liquifaction., , pVm, a, 1, RT, RT, The compressibility factor for a real gas at high pressure is :, , (c), 78., , pVm, RT, , Z, , 85., , 50°C, D, , 31.1°C, , 73, (Pc ), P2, P1, , E, C, , 21.5°C, , B, , 33.98°C, (Tc ), F, , 13.1°C, , A, , V3 V2 Vc, Volume, , We can move from point A to F vertically by increasing, the temperature., (ii) We can reach the point G by compressing the gas at, constant temperature., (iii) We can move down from G towards D by increasing, the temperature., (iv) As soon as we cross the point D on the critical isotherm, we get liquid., (a) (i) and (ii), (b) (i), (ii) and (iii), (c) (i), (ii) and (iv), (d) (i), (ii), (iii) and (iv), At 1 atm pressure boiling temperature is called _____X, ____., If pressure is 1 bar then the boiling point is called, _____Y, _____ of the liquid., (a) X = Standard boiling point, Y = Normal boiling point, (b) X = Normal boiling point, Y = Standard boiling point, (c) X = Critical boiling point, Y = Normal boiling point, (d) X = Critical boiling point, Y = Standard boiling point, Which of the following statement is incorrect ?, (a) Standard boiling point of liquid is slightly lower than, the normal boiling point., (b) 1 atm pressure is slightly less than 1 bar pressure, (c) The normal boiling point of water is 100°C and its, standard boiling point is 99.6°C, (d) None of the above, , (i), , 87., , 88., , V1
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STATES OF MATTER, , 83, , 89., , A liquid can exist only, (a) between triple point and critical temperature, (b) at any temperature above the melting point, (c) between melting point and critical temperature, (d) between boiling and melting temperature, 90. The kinetic energy of molecules in gaseous state is, (a) more than those in the liquid state, (b) less than those in the liquid state, (c) equal to those in the liquid state, (d) less than those in solid state, 91. A pin or a needle floats on the surface of water, the reason, for this is, (a) surface tension, (b) less weight, (c) upthrust of liquid, (d) None of the above, 92. The spherical shape of rain-drop is due to, (a) density of the liquid (b) surface tension, (c) atmospheric pressure (d) gravity, 93. Which of the following phenomena is caused by surface, tension ?, (a) Particles at the bottom of river remain separated but, they stick together when taken out., (b) A liquid rise in a thin capillary., (c) Small drops of mercury from spherical bead instead of, spreading on the surface., (d) All of the above, 94. A drop of oil is placed on the surface of water. Which of the, following statement is correct ?, (a) It will remain on it as a sphere, (b) It will spread as a thin layer, (c) It will be partly as spherical droplets and partly as thin, film, (d) It will float as a distorted drop on the water surface, 95. When the temperature increases, the viscosity of, (a) gases decreases and viscosity of liquids increases, (b) gases increases and viscosity of liquids decreases, (c) gases and liquids increases, (d) gases and liquids decreases, 96. The surface tension of which of the following liquid is, maximum?, (a) C2H5OH, (b) CH3OH, (c) H2O, (d) C6 H6, 97. In which phenomena water changes into water vapour below, its B.P. ?, , 98., , (a) Evaporation, , (b) Condensation, , (c) Boiling, , (d) No such phenomena exist, , The liquid which has the highest rate of evaporation is, (a) petrol, , (b) nail-polish remover, , (c) water, , (d) alcohol, , 99., , The correct order of viscosity of the following liquids will, be, (a) Water < methyl alcohol < dimethyl ether < glycerol, (b) methyl alcohol < glycerol < water < dimethyl ether, (c) dimethyl ether < methyl alcohol < water < glycerol, (d) glycerol < dimethyl ether < water < methyl alcohol, , STATEMENT TYPE QUESTIONS, 100. Which of the following statement(s) is/are true for London, force ?, (i) These forces are always attractive., (ii) These forces are important for long distance too., (iii) Their magnitude depends on the polarisability of the, particle., (a) (i) and (ii), (b) (i) only, (c) (iii) only, (d) (i) and (iii), 101. Choose the correct sequence of T and F for following, statements. Here T stands for true and F stands for false, statement., (i) Dipole - dipole forces act between the molecules, possessing permanent dipole., (ii) Partial charge possessed by these dipoles is always, equal to unit electronic charge., (iii) Dipole - dipole interaction is weaker than London forces, and ion - ion interaction., (a) TTF, (b) TFF, (c) TTT, (d) TFT, 102. Which of the following statements are correct ?, (i) Hydrogen bonding is a special case of dipole - dipole, interaction., (ii) Energy of hydrogen bond varies between 10 to, 100 kJ mol–1., (iii) Hydrogen bonds are powerful force in determining the, structure and properties of compounds like proteins,, nucleic acids etc., (iv) Strength of the hydrogen bond is determined by the, coulombic interaction between the lone-pair electrons, of the electronegative atom of one molecule and the, hydrogen atom of other molecule., (a) (i) and (ii), (b) (i), (ii) and (iii), (c) (ii), (iii) and (iv), (d) All of these, 103. Which of the following statements are correct regarding, the characteristic of gases ?, (i) Gases are highly compressible., (ii) Gases exert pressure equally in all directions., (iii) Gases have much higher density than the solids and, liquids., (iv) Gases mix evenly and completely in all proportion, without any mechanical aid., Choose the correct option., (a) (i), (ii) and (iii), (b) (ii), (iii) and (iv), (c) (i), (ii) and (iv), (d) (i), (ii), (iii) and (iv)
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EBD_7207, STATES OF MATTER, , 84, , 104. Read the following statements and choose the correct, option. Here T stands for true and F stands for false, statement., (i), , Equation :, , V, , nRT, P, , will be applicable to any gas, under, , those conditions when behaviour of the gas, approaches ideal behaviour., (ii) Value of universal gas constant at 0° C and 1 atm, pressure is 8.20578 × 10–2 L atm K–1 mol–1, (iii) Ideal gas equation describes the state of any gas,, therefore it is also called equation of state., (iv) Value of R in units of Pa m3 K–1 mol–1 is 8.314 × 10–2, (a) T T T T, (b) T T T F, (c) T F T F, (d) T F F T, 105. Choose the correct sequence of T and F for following, statements. Here T stands for true and F stands for false., (i) There may be exchange of energy between colliding, molecules, their individual energies may change, but, the sum of their energies remains constant., (ii) At any particular time, different particles in the gas, have different speeds and hence different kinetic, energies., (iii) Particles of a gas move in all possible directions in, straight lines. During their random motion, they collide, with each other and with the walls of the container., (iv) In kinetic theory it is assumed that average kinetic, energy of the gas molecules is directly proportional to, the absolute temperature., (a) TTTT, (b) TTTF, (c) TFTT, (d) TFFT, 106. Which of the following statements are correct ?, (i) Real gases show deviations from ideal gas law because, molecules interact with each other., (ii) Due to interaction of molecules the pressure exerted, by the gas is given as :, p real = p ideal, , an 2, , V2, (iii) Value of 'a' is measure of magnitude of intermolecular, attractive forces within the gas and depends on, temperature and pressure of gas., (iv) At high pressure volume occupied by the molecules, also becomes significant because instead of moving, in volume V, these are now restricted to volume (V-nb), (a) (i) and (iv), (b) (i), (ii) and (iii), (c) (i), (iii) and (iv), (d) (i) and (iii), 107. Choose the correct sequence of true and false for following, statements. Here T represents true and F represents false, statement., (i) Greater the viscosity, the more slowly the liquid flows., (ii) Glass is an extremely viscous liquid., (iii) Viscosity of liquid increases as the temperature rise., (a) TFF, (b) FFT, (c) TFT, (d) TTF, , MATCHING TYPE QUESTIONS, 108. Match the columns, Column-I, (A) Attractive force that, operates between the, polar molecules having, permanent dipole and, the molecule lacking, permanent dipole, (B) Interaction in which, interaction energy, between stationary, polar molecules is, proportional to, , Column-II, (p) Dipole-dipole, force, , (q) London force, , 1, r3, , (C) Force that are important, (r) Dipole-induced, only at short distances, dipole force, (~ 500 pm), (a) A – (r), B – (p), C – (q), (b) A – (p), B – (r), C – (q), (c) A – (r), B – (q), C – (p), (d) A – (q), B – (r), C – (p), 109. Match the columns, Column-I, Column-II, (A) Volume of a fixed mass, (p) Boyle's Law, of a gas at constant, pressure is directly, proportional to its, absolute temperature, (B) At constant volume,, (q) Avogadro's Law, pressure of a fixed amount, of a gas varies directly with, the temperature., (C) Equal volumes of all, (r) Charle's Law, gases under the same, conditions of temperature, and pressure contain equal, number of molecules., (D) At constant temperature,, (s) Gay Lussac's Law, the pressure of a fixed amount, (i. e., number of moles n) of, gas varies inversely with its, volume., (a) A – (s), B – (r), C – (q), D – (p), (b) A – (r), B – (s), C – (q.), D – (p), (c) A – (r), B – (q), C – (p), D – (s), (d) A – (q), B – (p), C – (s), D – (r)
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STATES OF MATTER, , 110. Match the columns, Column-I, (A) Boyle's law, (B) Charle's law, (C) Dalton's law, (D) Avogadro law, , 85, , Column-II, (p) V n at constant T, and P, (q) ptotal = p1+ p2+ p3+ ...., at constant T, V, pV, = Constant, (r), T, (s) V T at constant n, and p, , 1, at constant n, V, and T, (a) A – (t), B – (s), C – (q), D – (p), (b) A – (s), B – (q), C – (p), D – (t), (c) A – (r), B – (t), C – (q), D – (p), (d) A – (t), B – (q), C – (s), D – (r), 111. Match the graphs between the following variables (Column-I), with their names (Column-II) :, Column-I, Column-II, (Graphs), (Names), (A) Pressure vs temperature (p) Isotherms, graph at constant, molar volume., (B) Pressure vs volume, (q) Constant temperature, graph at constant, curve, temperature, (C) Volume vs temperature (r) Isochores, graph at constant, pressure, (s) Isobars, (a) A – (p), B – (r), C – (s), (b) A – (r), B – (p), C – (s), (c) A – (r), B – (q), C – (p), (d) A – (s), B – (q), C – (r), 112. Match the following graphs of ideal gas (Column-I) with, their co-ordinates (Column-II) :, Column-I, Column-II, (Graphical, (x and y, representation), co-ordinates), , (t), , pµ, , (A), , (p) pV vs. V, , (B), , (q) p vs. V, , (C), , (r) p vs., , 1, V, , (a), (b), (c), (d), , A, A, A, A, , –, –, –, –, , (r), B – (p),, (r), B – (q),, (q), B – (r),, (p), B – (r),, , C, C, C, C, , –, –, –, –, , (q), (p), (p), (q), , ASSERTION-REASON TYPE QUESTIONS, Directions : Each of these questions contain two statements,, Assertion and Reason. Each of these questions also has four, alternative choices, only one of which is the correct answer. You, have to select one of the codes (a), (b), (c) and (d) given below., (a) Assertion is correct, reason is correct; reason is a correct, explanation for assertion., (b) Assertion is correct, reason is correct; reason is not a, correct explanation for assertion, (c) Assertion is correct, reason is incorrect, (d) Assertion is incorrect, reason is correct., 113. Assertion : Three states of matter are the result of balance, between intermolecular forces and thermal energy of the, molecules., Reason : Intermolecular forces tend to keep the molecules, together but thermal energy of molecules tends to keep, them apart., 114. Assertion : Gases expand and occupy all the space available, to them, Reason : There is no force of attraction between the particles, of a gas at ordinary temperature and pressure., 115. Assertion : Gases do not liquefy above their critical, temperature, even on applying high pressure., Reason : Above critical temperature, the molecular speed is, high and intermolecular attractions cannot hold the, molecules together because they escape because of high, speed., 116. Assertion : At critical temperature liquid passes into, gaseous state imperceptibly and continuously., Reason : The density of liquid and gaseous phase is equal, to critical temperature., 117. Assertion : The temperature at which vapour pressure of a, liquid is equal to the external pressure is called boiling, temperature., Reason : At high altitude atmospheric pressure is high., 118. Assertion : Liquids tend to have maximum number of, molecules at their surface., Reason : Small liquid drops have spherical shape., , CRITICAL THINKING TYPE QUESTIONS, 119. Arrange the following in increasing order their, intermolecular interaction, (A) HCl, (B) SF6 and (C) NaCl, (a) A, B, C, (b) A, C, B, (c) B, A, C, (d) B, C, A
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EBD_7207, STATES OF MATTER, , 86, , 120. When a sample of gas is compressed at constant temperature, from 15 atm to 60 atm, its volume changes from 76 cm3 to, 20.5 cm3. Which of the following statements are possible, explanations of this behaviour?, , 124. From the given figure what can be said about the gases, does not deviate much from ideal gases at, Real, , (1) The gas behaves non-ideally, , Ideal, Pressure, , (2) The gas dimerises, (3) The gas is adsorbed into the vessel walls, (a) 1, 2 and 3, , (b) 1 and 2 only, , (c) 2 and 3 only, , (d) 1 only, , 121. Three different gases X, Y and Z of molecular masses 2, 16, and 64 were enclosed in a vessel at constant temperature till, equilibrium is reached. Which of the following statement is, correct?, (a) Gas Z will be at the top of the vessel, (b) Gas Y will be at the top of the vessel, (c) Gas Z will be at the bottom and X will be at the top, (d) Gases will form homogenous mixture, 122. Which of the following volume (V) - temperature (T) plots, represents the behaviour of one mole of an ideal gas at one, atmospheric pressure ?, V(L), , (a), , (22.4 L, 273K), , V(L), , (36.8 L, 373 K), , (b), , (22.4 L, 273K), , T(K), , (26.8 L, 373 K), , Volume, , (a) Higher pressure and low volume., (b) Low pressure and low volume., (c) High pressure and high volume., (d) Low pressure and high volume., 125. 16 g of oxygen and 3 g of hydrogen are mixed and kept at, 760 mm of Hg pressure and 0° C. The total volume occupied, by the mixture will be nearly, (a) 22.4 litres, (b) 33.6 litres, (c) 448 litres, (d) 44800 ml, 126. The density of neon will be highest at, (a) S.T.P., (b) 0ºC, 2 atm, (c) 273ºC, 1 atm., (d) 273ºC, 2 atm., 127. A plot of volume (V) versus temperature (T) for a gas at, constant pressure is a straight line passing through the, origin. The plots at different values of pressure are shown, in figure. Which of the following order pressure is correct, for this gas ?, , T(K), , p1, , (c), , V(L), , (30.6 L, 373 K), , (22.4 L, 273K), , (d), , (22.4 L, 273K), , (14.2 L, 373 K), , p2, , Volume (mL), , V(L), , p3, p4, , T(K), , T(K), , 123. Consider the case of hot air balloon, density of air at 20°, C is 1.2Kg/m3, if the air was heated to 99°C, density of, air becomes 0.94kg/m3. What would be the volume (in m3), at 20°C if the volume at 99°C is 2800 m3 and how much, air (in kg) has been escaped at 99°C , if the air in inflated, balloon was heated to 99°C (if the inflated volume of, balloon was found to be 2800m3) respectively are, (a) 2243, 728, , (b) 3495.3, 596, , (c) 2687, 593, , (d) 2956, 771, , Temperature (K), , (a) p1 > p2 > p3 > p4 (b) p1 = p2 = p3 = p4, (c) p1 < p2 < p3 < p4 (d) p1 < p2 = p3 < p4, 128. At constant temperature, for a given mass of an ideal gas, (a) The ratio of pressure and volume always remains, constant., (b) Volume always remains constant., (c) Pressure always remains constant., (d) The product of pressure and volume always remains, constant.
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STATES OF MATTER, , 87, , 129. If 500 ml of gas A at 400 torr and 666.6 ml of B at 600 torr are, placed in a 3 litre flask, the pressure of the system will be, (a) 200 torr, (b) 100 torr, (c) 550 torr, (d) 366 torr, 130. What is the partial pressure (mmHg) of nitrogen if total, atmospheric pressure is 760mmHg ?, (a) 159, (b) 300, (c) 592.8, (d) 230, 131. Cyclopropane and oxygen at partial pressures 170 torr and, 570 torr respectively are mixed in a gas cylinder. What is the, ratio of the number of moles of cyclopropane to the number, of moles of oxygen (nC3H6/nO2)?, (a), , 170 42, 570 32, , (c), , 170, 740, , 0.39, , 0.23, , (b), , 170, 42, , 170, 42, , (d), , 170, 570, , 0.30, , 570, 32, , 0.19, , 132. An evacuated glass vessel weights 50 g when empty,, 144.0 g when filled with a liquid of density 0.47 g ml–1 and, 50.5 g when filled with an ideal gas at 760 mm Hg at 300, K. The molar mass of the ideal gas is, (Given R = 0.0821 L atm K–1 mol–1), (a) 61.575, (b) 130.98, (c) 123.75, (d) 47.87, 133. The pressure of a 1:4 mixture of dihydrogen and dioxygen, enclosed in a vessel is one atmosphere. What would be the, partial pressure of dioxygen ?, (a) 0.8 × 105 atm, (b) 0.008 Nm–2, (c) 8 × 104 Nm–1, (d) 0.25 atm, 134. Two vessels containing gases A and B are interconected as, shown in the figure. The stopper is opened, the gases are, allowed to mix homogeneously. The partial pressures of A, and B in the mixture will be, respectively, , Gas A, , Gas B, , 12 L, 8 atm, , 8L, 5 atm, , (a) 8 and 5 atom, (b) 9.6 and 4 atm, (c) 4.8 and 2 atm, (d) 6.4 and 4 atm, 135. A neon-dioxygen mixture contains 70.6 g O2 and 167.5 g, neon. If pressure of the mixture of gases in the cylinder is 25, bar. What is the partial pressure of O2 and Ne in the mixture, respectively ?, (a) 5.25 bar, 10 bar, (b) 19.75 bar, 5.25 bar, (c) 19.75 bar, 10 bar, (d) 5.75 bar, 19.75 bar, , 136. 0.5 mole of each H2, SO2 and CH\4 are kept in a container. A, hole was made in the container. After 3 h, the order of partial, pressures in the container will be, (a), , pSO 2, , (c), , pH 2, , pCH 4, pCH 4, , pH 2, , (b), , pH 2, , pSO 2, , pCH 4, , pSO 2, , (d), , pSO 2, , pH 2, , pCH 4, , 137. For a person travelling to the peak of the mountain which, of the following statement(s) is/are correct ?, (i) Oxygen level goes on decreasing., (ii) Gas law can be applied to this situation., (a) Both (i) and (ii), (b) Only (i), (c) Only (ii), (d) Neither (i) nor (ii), 138. Pressure in well inflated tyres of automobiles is almost, constant, but on a hot summer day this increases, considerably and tyre may burst if pressure is not adjusted, properly. During winters, on a cold morning one may find, the pressure in the tyres of a vehicle decreased considerably., Which of the following law explain the above observations?, (a) Charle's Law, (b) Avogadro Law, (c) Boyle's Law, (d) Gay Lussac's Law, 139. What is the ratio of pressure of the 2gm of hydrogen to, that of 4 gm of helium at temperature of 298K, 20ml, volume? (consider the ideal behaviour), (a) 1 : 2, (b) 2 : 1, (c) 1 : 1, (d) 2 : 2, 140. In case of CO and CH4 curve goes to minima then, increases with increase in pressure but in case of H2, and He the curve is linear because:, CO CH4, H2, He, PV, , Ideal gas, , P, (a) Intermolecular interactions for H2 and He are very, low., (b) Molecular size or atomic size for H2 and He is small., (c) Both (a) and (b), (d) Neither (a) nor (b)
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EBD_7207, 88, , 141. Which among the following has lowest surface tension ?, (a) Hexane, (b) Water, (c) CH3OH, (d) CH3CH2OH, 142. Water droplets was not able to maintain its spherical, shape in the presence of gravity but mercury does, why ?, (a) Force of attraction between atoms of mercury is very, high than that of molecules in case of water., (b) Surface tension of mercury is very high., (c) Both (a) and (b), (d) Neither (a) nor (b), , STATES OF MATTER, , 143. Consider the case of honey flowing over a slope for this, situation which of the following statement(s) is/are, correct ?, , (A), , (i), , (B), , Velocity with which honey is flowing is slower in A, than B ( being same in both cases), (ii) Velocity increases with increase in temperature., (a) (i) and (ii), (b) Only (i), (c) Only (ii), (d) Neither (i) nor (ii)
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STATES OF MATTER, , 89, , FACT / DEFINITION TYPE QUESTIONS, 1., , 2., , (c) Attractive forces between an ion and a dipole are, known as ion - dipole forces and these are not van der, Waals forces., (d) Fritz London explained the concept of dispersion force., , 3., , (b) Interaction energy, , 4., 5., , (c), (c) London forces are always attractive and interaction, energy is inversely proportional to the sixth power of, the distance between two interacting particles., (c), (a) Dipole-dipole interaction is stronger than the London, forces but is weaker than ion-ion interaction because, only partial charges are involved e.g., HCl molecules., The attractive force decreases with the increase of, distance between the dipoles., (d) Induced dipole moment depends upon the dipole, moment present in the permanent dipole and the, polarisability of the electrically neutral molecule., Molecules of large size can be easily polarized. High, polarisability increases the strength of attractive, interactions., (b) This type of attractive force operates between the polar, molecules having permanent dipole and the molecules, lacking permanent dipole., HCl is polar (, 0) and He is non polar ( = 0), thus, gives dipole-induced dipole interaction., (c) Nobel gases has no intermolecular forces due to, inertness., (a) Strength of the hydrogen bond is determined by, coulombic interaction between lone pair electrons of, the electronegative atom of one molecule and the, hydrogen atom of the other molecule., (d), 13. (a), (c) Gaseous state of substance has the maximum thermal, energy., (c) Gases do not have definite shape and volume. Their, volume is equal to the volume of the container., (b) It is characteristic of gases i.e., Thermal energy >>, molecular attraction., (c) Robert Boyle made first reliable measurement on, properties of gases., (d) According to Boyle’s law at constant temperature,, , 6., 7., , 8., , 9., , 10., 11., , 12., 14., 15., 16., 17., 18., , V, , 1, r6, , 1, or PV = constant, P, , 19., , (c), , P, , P, , V, Curve I, , 20., , 1/V, Curve III, , Both these graphs represents Boyle's law., (b) According to Boyle's Law, pv = k, p, , m, k, d, , p k, k', d m, , 21., , (b) Boyle’s law, , 1, V, K, P, V, PV K, 22. (a) Given initial volume (V1) = 600 c.c.; Initial pressure, (P1) = 750 mm of Hg and final volume (V2) = 500 c.c., according to Boyle’s law,, P1V1 = P2V2, P, , or 750 × 600 = P2 × 500, 750 600, 900 mm of Hg, 500, Therefore increase of pressure = (900 – 750) = 150 mm, of Hg, (b) The lowest hypothetical or imaginary temperature at, which gases are supposed to occupy zero volume is, called absolute zero., (b) Given initial volume (V1) = 500 ml ; Initial temperature, (T1) = 27ºC = 300 K and final temperature (T2) = –5ºC, = 268 K., From Charle’s law :, , or P2, , 23., , 24., , V1, T1, , V2, 500, or, T2, 300, , V2, 268, , Where V2 = New volume of gas, , 500, 268 446.66ml., 300, (c) At any given pressure, graph of volume vs temperature, (in °C) is a straight line and on extending to zero volume, each line intercepts the temperature axis at – 273.15° C., V2, , 25.
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EBD_7207, STATES OF MATTER, , 90, , k, , (c) According to Charle’s law V, , 27., , (b) Charle’s law V, , 28., , (a), , 29., , more, hence density of air is more., (b) Hot air is lighter due to less density (Charle’s law), , 44., , T at constant P.., , d p, Boyle’s law, d, , d, , T or, , V, T, , 26., , MP, . At sea level pressure is, RT, , MP, RT, MP, RT, , (b) Higher P, lower T, greater the density. d, , 31., 32., , (a) Order of pressure, p1 < p2 < p3 < p4., (a) V1 = 2 L, T2 = (26.1 + 273) K = 299.1 K, V2 = ?, T1 = (23.4 + 273) K = 296.4 K, , V2, , 33., , 34., , (d), , (c), , 35., 36., , (d), (a), , 37., , (d), , P, T, , V1, T1, , 2L 299.1K, 296.4K, = 2.018 L, , V2, T2, , V2, , V1T2, T1, , 2L 1.009, , constant (Gay Lussac's law), , P1 P2, P1T2 P2T1, T1 T2, PV = constant, P1V1 = P2V2 [Boyle's law], This relationship is derived from Boyle's and Charles', law., All of the given relations are true for Avogadro's law., At STP molar volume of an ideal gas or a combination, of ideal gases is 22.71098 L mol–1., 44g at STP occupies volume 22.4 litre which is molecular, mass of CO2. Molecular mass occupies 22.4 litre at, STP., An ideal gas obeys the gas laws under all experimental, conditions., PV = nRT, n/V = P/RT, T., In the equation PV = nRT, n moles of the gas have, volume V., , 38., , (b), , 39., , (c), , 40., , (c), , 41., , (b), , 42., , PV, T1, 1 1, P2V2 T2, (b) R = 0.0082 litre atm K–1 mole–1 ., , 43., , (d) PV = nRT = m RT, , PV, T, , PV, constant or 1 1, T1, , or PM =, , M, m, RT = dRT, V, , P2V2, T2, , d=, , PM, RT, , PV, RT, , n, , m, M, , MPV 34 2, RT, 0.082, PV, T, 760, 1 1 1, (c) V2, T1 P2 600, (a) Total moles, , 100, 282.68gm, 293, 546, 273 691.6 ml, 273, , m, , 45., 46., , =, , 30., , From Charle's law,, , (c), , 4 2, 1.125; PV, 32 2, , nRT, , P 1.125 .0821 273, P = 25.215 atm, 47. (c) PV = nRT is for an ideal gas which follows both, isothermal and adiabatic processes., 48. (a) By Dalton’s law of partial pressures, the total pressure, of a mixture of two gases is the sum of the partial, pressures of components of the mixture., 49. (a), 50. (b) Because H2 & Cl2 gases may react with each other to, produce HCl gas hence Dalton’s law is not applicable., 51. (a) Pressure exerted by saturated water vapour is called, aqueous tension., nRT, V, , 6, 8.314 402, 16.02, 0.03, , 52., , (d) P =, , 53., , (a) Given nCO = n N2, , 41648 Pa, , PCO + PN2 = 1 atm, Partial pressure of a gas = mole fraction of gas × total, pressure, , PN2 =, 54., , n CO, , n N2, , n N2, 1 = 2n, N2, , 1= 1, 2, , 0.5 atm., , (a) Given, P1 = 1.5 bar T1 = 273 + 15 = 288 K V1 = V, P2 = 1.0 bar T1 = 273 + 25 = 298K V2 = ?, P1V1, T1, , P2 V2, T2, , 1.5 V, 288, , 55., , n N2, , 1 V2, 298, , V2 = 1.55 V i.e., volume of bubble will be almost 1.6 time, to initial volume of bubble., (a) Partial pressure = total pressure × mole fraction, Moles of oxygen =, Moles of neon =, , 64, 32, , 60, 20, , 2, 3, , Mole fraction of oxygen =, PO 2, , 2, 10 4 bar, 5, , 2, 2 3, , 2, 5
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STATES OF MATTER, , 91, , Mole fraction of neon =, PNe, , 56., , 58., , 2 3, , 3, 10 6 bar, 5, , 760 500, 200, , 1900mm Hg, , P1V1, , n, , N2, , x, , N2, , 61., , 62., 63., , 56, 28, , 2, n O2, n, , O2, , n, , n, , N2, , N2, , n, , n, , O2, , N2, , n, , 96, 32, 2, , O2, , 2 3, 3, , O2, , 2 3, , or n, , PV, RT, , 74., , a, a, (V – b) = RT, T; Here P, represents, V2, V2, the intermolecular forces., (c) ‘a’ is directly related to forces of attraction. Hence, greater the value of ‘a’, more easily the gas gets, liquified., (b), (i) At very large molar volume, P, , P, , a, , P and Vm b Vm, , Vm2, , (iii) According to van der Waals equation 'a' and 'b' are, independent of temperature., 0.4,, , 75., 76., , 0.6, , (d) From the ideal gas equation :, nRT, , (c), , 68., 69., , 3, , PN = 0.4 × 10 = 4 atm, PO = 0.6 × 10 = 6 atm, 2, 2, , PV, , 72., , 65., , 73., , 2, 1, , (d) On applying Dalton's law,, Partial pressure of a component, = Mole fraction × Total pressure, Given, mass of N2 = 56 g, mass of O2 = 96 g, Total pressure = 10 atm, , x, , 60., , n1T1, n 2 T2, , P1V1T1, P2 V2 T2, 1.5 16.4 500, 4.1 5 300, , 70., 71., , (c) Given statement explain the great compressibility of, gases., (b) If there were loss of kinetic energy, the motion of, molecule will stop and gases will settle down., (b) Particles of a gas are always in constant and random, motion. If the particles were at rest and occupy fixed, positions, then a gas would have a fixed shape which, is not observed., (b) If assumption (ii) is correct, the pressure vs volume, graph of experimental data (real gas) and that of, theoretically calculated for Boyle's law (ideal gas), should coincide., (d) Kinetic theory of gases proves all the given gas laws., (d) Molecules move very fast in all directions in a straight, line by colliding with each other but with different, velocity., (b) At low temperature and high pressure., (a) In van der waals equation ‘b’ is for volume correction, , 67., , (d) Value of gas constant depends only upon units of, measurement., (a) Given conditions, V1 = 16.4 L, V2 = 5 L, P1 = 1.5 atm, P2 = 4.1 atm, T1 = 273 + 27 = 300 K,, T2 = 273 + 227 = 500 K, , n1, n2, , 64., , 66., , Applying gas equation, P V, 2 2, , 59., , 3, 5, , (b) P1V1 = P2V2, 760 × 500 = P2 × 200., P2 =, , 57., , 3, , 3170 10 3, = 1.27 × 10–3, 8.314 300, , (a) By Dalton’s law of partial pressures, the total pressure, of a mixture of two gases is the sum of the partial, pressures., (b) An ideal gas equation is the combination of Boyle's, law, Charles' law and Avogadro law., (b) According to kinetic theory the gas molecules are in a, state of constant rapid motion in all possible directions, colloiding in a random manner with one another and, with the walls of the container and between two, successive collisions molecules travel in a straight line, path but show haphazard motion due to collisions., , 77., , n2a, , PV 2, , atm dm 6 mol-2, V2, n2, (c) Easily liquefiable gases have greater intermolecular, forces which is represented by high value of 'a'. The, greater the value of 'a' more will be liquefiability., So, the order is Q < S < P < R., (a) When pressure is low ‘b’ can be neglected, thus, , (a), , P, , P, , ; a, , a, V2, , PV, , a, V, , PV, , RT, , PV, RT, , RT, RT, , Z, , PV, RT, , V, , RT, , RT, a, V, a, VRT, 1, , a, VRT
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EBD_7207, STATES OF MATTER, , 92, , 78., , (c), , P, , a, , (V, V2, , b), , RT at high pressure, , a, V2, , can be, , neglected, PV – Pb = RT and PV = RT + Pb, PV, RT, , Z, , 79., 80., , 1, 1, , Pb, RT, Pb, ; Z > 1 at high pressure, RT, , (d) CO2 has highest critical temperature of 304.2 K, (d) At low pressure and high temperature: At low pressure, volume correction for 1 mole of a gas in negligible, i.e b, =0, thus the gas equation becomes, æ, ö, çç P + a ÷÷ V = RT, 2 ø÷÷, çè, V, , or Z =, , PVm, a, = 1RT, Vm RT, , of gas in negligible i.e, , a, V2, , =0, , or (P + 0) (V – b) = RT, or P (Vm – b) = RT, or PVm = RT + Pb, , 81., 82., 83., 84., 85., , 86., , 87., , 88., 89., 90., , 97., 98., 99., , (a), 92. (b), (d) All these phenomena are caused by surface tension., (b), 95. (b), (c) Due to intermolecular H-bonding the surface tension, of H2O is more than other liquid. One H2O molecule is, joined with 4 another H2O molecule through H–bond., Hydrogen bonding is in order H2O > C2H5OH >, CH3OH., (a) Boiling point of water is 100°C whereas evaporation of, water into water vapours occurs at room temperature., (a) As intermolecular forces are least in case of petrol., Thus, it has highest rate of evaporation., (c) The correct order of viscosity of the given liquids is, dimethyl ether < methyl alcohol < water < glycerol., , STATEMENT TYPE QUESTIONS, , At higher pressure, the pressure correction for 1 mole, , PVm, Pb, = 1+, RT, RT, (d) In the ideal gas, the intermolecular forces of attraction, are negligible and hence it cannot be liquefied., (d), (d) Above Boyle point, real gases show positive deviation, from ideality and Z values are greater than one., (c), (c) More will be critical temperature easier is the, liquifaction of the gas. Hence correct order will be, He < H2 < N2 < O2, (a) For statement (iii), we can move down from G towards, D by lowering the temperature., For statement (iv), we get liquid as soon as we cross, point H., (b) At 1 atm pressure boiling temperature is called normal, boiling point. If pressure is 1 bar than the boiling point, is called standard boiling point of the liquid., (b) 1 bar pressure is slightly less than 1atm pressure., (d) A substance exists as a liquid above its m. pt. and, below its b. pt., (a) The kinetic energy of molecules in gaseous state is, more than those in the liquid state as the molecules in, gaseous state can move freely (with higher speed) as, compared in liquid state., , or Z =, , 91., 93., 94., 96., , 100. (d) These forces are important only at short distances, (~ 500 pm), 101. (b) For statement (ii), partial charge possessed by these, dipoles is always less than the unit electronic charge., For statement (iii), dipole - dipole interaction is stronger, than London forces but weaker than ion - ion, interaction., 102. (d) All of the given statements are correct for hydrogen, bond., 103. (c) Gases have much lower density than the solids and, liquids., 104. (b) Value of R = 8.314 Pa m3 K–1 mol–1, 105. (a) All the given statements are true., 106. (a) For statement (ii), preal = p ideal, , an 2, V2, , For statement (iii), value of 'a' is independent of, temperature and pressure., 107. (d) Viscosity of liquid decreases as temperature rise., , MATCHING TYPE QUESTIONS, 108. (a), , 109. (b), , 110. (a), , 111. (b), , 112. (c), , ASSERTION-REASON TYPE QUESTIONS, 113. (a), 114. (a) Gases expand and occupy all the space available to, them because there is no force of attraction between, the particles of a gas at ordinary temperature and, pressure., 115. (a), 116. (a), 117. (c) At high altitude atmospheric pressure is low., 118. (d)
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STATES OF MATTER, , 93, , CRITICAL THINKING TYPE QUESTIONS, 119. (c) In case of HCl molecules their is dipole-dipole, interaction which is stronger than London forces as, in case of SF6. Now between HCl and NaCl the ionion interaction present in NaCl is far more stronger, than dipole-dipole interaction of HCl., 120. (d) Given, P1 = 15 atm, P2 = 60 atm, V1 = 76 cm3, V2 = 20.5 cm3., If the gas is an ideal gas, then according to Boyle's law,, it must follow the equation,, P1V1 = P2V2, P1 × V1 = 15 × 76 = 1140, P2 × V2 = 60 × 20.5 = 1230, , It means density of gas is directly proportional to, pressure and inversely proportional to temperature., Density of neon will be maximum at highest pressure, and lowest temperature., 127. (c), 128. (d) According to Boyle’s law at constant temperature,, 1, or PV = constant, P, 129. (a) Applying Boyle's law P1V1 = P2V2 for both gases, V, , 500, 400, 1000, 666.6, 1000, , 600, , P1V1 P2 V2, The gas behaves non-ideally., The given information is not sufficient to comment on, other statements., 121. (d) All the gases occupy the available volume and will, form homogeneous mixture., 122. (c), , V1, T1, , V2, T2, 22.4, 273, , 30.6 litre, , V2, T2, , V1, 298K, , 2800m 3, 372K, , V1 = 2243 m3, 2800 m3 volume of inflated balloon., Mass of air in inflated ballon = 2800 m3 × 0.94 kg m–3, = 2632 kg, Keeping the volume same = 2800 m3, The mass of air, which occupies it with density, (1.2 kg/m3) is 2800 × 1.2 = 3360 kg, Amount of air which had been escaped = 3360 – 2632, = 728 kg, , 125. (d), , n of O 2, , n of H2, , 3, 2, , Total no. of moles, , P1 and n 2, , n1, n2, , V, , 126. (b) d =, , nRT, P, PM, RT, , 1, 2, , 2 .082 273, 1, , 400, 3, , 600, 3, , 200 tor, , P1, P2, , P2, , n1, n2, , 170, 0.30, 570, , PV = nRT =, , w, RT, M, 0.5, 0.0821 300, M, , 0.5 0.0821 300, 200 10–3, , 61.575, , 133. (d) p1 = 1.5 atm, T1 = 15°C = (15 + 273)K = 288 K, p2 = 1 atm, T2 = 25°C = (25 + 273)K = 298 K, , 2, , 44.8lit, , 400, 3, , 132. (a) Given weight of empty glass vessel = 50 g, Weight of vessel filled with liquid = 144 g, Weight of liquid = 144 – 50 = 94 g., Volume of liquid = Mass/density = 94/0.47, = 200 ml = 200 × 10–3 L., Given, pressure of ideal gas = 760 mm Hg = 1 atm, Temperature = 300 K, R = 0.0821 L atm K–1 mol–1, Mass of ideal gas = 50.5 – 50 = 0.5 g, According to ideal gas equation,, , M, , 3, 2, , P, , n1RT, , 1 × 200 × 10–3, , 1, 2, , 3, 200, 3, , P P, , 124. (d), 16, 32, , P, , 200, 3, , P, , 130. (c) Percentage of nitrogen in atmosphere is 78% .Partial, pressure of N2 = 0. 78 × 760, 131. (d) By Ideal gas equation, , n1, V2, , V2, 373, , 123. (a) Since atmospheric pressure remain constant, , V1, T1, , PT, , P1V, , at const. pressure, , P 3, , p1V1, T1, , 44800 ml, p1T2, T1p2, , V2, V1, , V2, V1, , p 2 V2, T2, 1.5 298, 1.55, 288 1
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EBD_7207, STATES OF MATTER, , 94, , 134. (c) Moles of A, (n A), , p A vA, RT, , 8 12, RT, , 96, RT, , pB v B 8 5 40, RT RT, RT, Total pressure × total volume = (nA + nB) × RT, , Moles of B, (n B), , 1, (96 40)RT, RT, p = 6.8, Partial pressure of A = p × mole fraction of A, p (12 8), , 6.8, , 96 96 40, RT, RT, , = 4.8 atm, Partial pressure of B = 6.8 – 4.8 = 2 atm., 135. (d) Number of moles of O2, Number of moles of Ne, , 70.6g, 32g mol, , 1, , 167.5g, 20g mol, , 1, , 2.21 mol, , 8.375 mol, , 2.21, 0.21, 2.21 8.375, Mole fraction of Ne = 1 – 0.21 = 0.79, Partial pressure of a gas = Mole fraction × total pressure, Partial pressure of O2 = 0.21 × 25 = 5.25 bar, Partial pressure of Ne = 0.79 × 25 = 19.75 bar, , Mole fraction of O2, , 136. (a) Extent of diffusion H2 > CH4 > SO2 because rate of, diffusion, , 1, molar mass, , Order of partial pressure after diffusion is, , pSO 2, , pCH 4, , pH 2, , 137. (a) As the height increases, atmospheric pressure, decreases, so now the volume of the gas increases, and gas tends to become less denser, hence the, concentration of oxygen decreases., 138. (d) The mathematical relationship between pressure and, temperature was given by Gay Lussac's law., 139. (c) Number of moles, temperature and volume are same., 140. (c) Due to small size of these species (H2 and He), intermolecular interactions (van der Waal forces) are, very low, therefore it is difficult to compress these ., 141. (a) Since surface tension depends on the attractive, forces between the molecules, and hydrogen bonding, a special type of dipole-dipole interactions in (b), (c), and (d) which is stronger than London forces of, attraction in hexane., 142. (c), 143. (a) Force is required to maintain the flow of layer which, is inversely proportional to the area of contact of, layer therefore flow in B is greater than that in A as, the area of contact is greater in A. Also viscosity of, the fluid decreases with increase in temperature, therefore liquid flow increases.
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6, THERMODYNAMICS, FACT / DEFINITION TYPE QUESTIONS, 1., , 2., , 3., , 4., , 5., , 6., , Thermodynamics is not concerned about____., (a) energy changes involved in a chemical reaction., (b) the extent to which a chemical reaction proceeds., (c) the rate at which a reaction proceeds, (d) the feasibility of a chemical reaction., Which of the following statements is not true regarding the, laws of thermodynamics ?, (a) It deal with energy changes of macroscopic systems., (b) It deal with energy changes of microscopic systems., (c) It does not depends on the rate at which these energy, transformations are carried out., (d) It depends on initial and final states of a system, undergoing the change., A……………... in thermodynamics refers to that part of, universe in which observations are made and remaining, universe constitutes the……………..., (a) surroundings, system (b) system, surroundings, (c) system, surroundings (d) system, boundary, The universe refers to, (a) only system, (b) only surroundings, (c) both system and surroundings, (d) None of these, Which of the following statements is correct?, (a) The presence of reacting species in a covered beaker, is an example of open system., (b) There is an exchange of energy as well as matter, between the system and the surroundings in a closed, system., (c) The presence of reactants in a closed vessel made up, of copper is an example of a closed system., (d) The presence of reactants in a thermos flask or any, other closed insulated vessel is an example of a closed, system., Which of the following is closed system ?, (a) Jet engine, (b) Tea placed in a steel kettle, (c) Pressure cooker, (d) Rocket engine during propulsion, , 7., , An isolated system is that system in which, (a) There is no exchange of energy with the surroundings, (b) There is exchange of mass and energy with the, surroundings, (c) There is no exchange of mass or energy with the, surroundings, (d) There is exchange of mass with the surroundings, 8., The state of a thermodynamic system is described by its, measurable or macroscopic (bulk) properties. These are, (a) Pressure and volume, (b) Pressure, volume, temperature and amount, (c) Volume, temperature and amount, (d) Pressure and temperature, 9., Which of the following are not state functions ?, (I) q + w, (II) q, (III) w, (IV) H - TS, (a) (I) and (IV), (b) (II), (III) and (IV), (c) (I), (II) and (III), (d) (II) and (III), 10. Among the following the state function(s) is (are), (i) Internal energy, (ii) Irreversible expansion work, (iii) Reversible expansion work, (iv) Molar enthalpy, (a) (ii) and (iii), (b) (i), (ii) and (iii), (c) (i) and (iv), (d) (i) only, 11. Enthalpy change ( H) of a system depends upon its, (a) Initial state, (b) Final state, (c) Both on initial and final state, (d) None of these, 12. ………………. is a quantity which represents the total energy, of the system, (a) Internal energy, (b) Chemical energy, (c) Electrical energy, (d) Mechanical energy, 13. Which of the following factors affect the internal energy of, the system ?, (a) Heat passes into or out of the system., (b) Work is done on or by the system., (c) Matter enters or leaves the system., (d) All of the above
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EBD_7207, 96, , 15., , 16., , 17., , 18., , 19., , The system that would not allow exchange of heat between, the system and surroundings through its boundary is, considered as, (a) isothermal, (b) adiabatic, (c) isobaric, (d) isochoric, The enthalpy change of a reaction does not depend on, (a) The state of reactants and products, (b) Nature of reactants and products, (c) Different intermediate reactions, (d) Initial and final enthalpy change of a reaction., The q is ……………. when heat is transferred from the, surroundings to the system and q is……………………….., When heat is transferred from system to the surroundings., (a) positive , negative, (b) negative , positive, (c) high, low, (d) low, high, Adiabatic expansions of an ideal gas is accompanied by, (a) decrease in E, (b) increase in temperature, (c) decrease in S, (d) no change in any one of the above properties, Which of the following statements is incorrect?, (a) q is a path dependent function., (b) H is a state function., (c) Both H and q are state functions., (d) Both (a) and (b), Figure below is showing that one mole of an ideal gas is, fitted with a frictionless piston. Total volume of the gas is, Vi and pressure of the gas inside is p. If external pressure is, pex which is greater than p is applied, piston is moved inward, till the pressure inside becomes equal to pex., , 20., , 21., , 22., , (c) w = nRT ln, , 23., , 24., , 26., Area = pex V, , Vi, , Volume, V, , 27., , pex, , pex, , 28., l, , What does the shaded area represents in the figure ?, (a) Work done, (b) Pressure change, (c) Volume change, (d) Temperature change, , V2, (isothermal reversible expansion of, V1, , an ideal gas), , pex, , Vf, , When 1 mol of a gas is heated at constant volume,, temperature is raised from 298 to 308 K. If heat supplied to, the gas is 500 J, then which statement is correct ?, (a) q = w = 500 J, U = 0 (b) q = U = 500 J, w = 0, (c) q = –w = 500 J, U = 0 (d) U = 0, q = w = –500 J, The work done during the expansion of a gas from a volume, of 4 dm3 to 6 dm3 against a constant external pressure of 3, atm is (1 L atm = 101.32 J), (a) – 6 J, (b) – 608 J, (c) + 304 J, (d) – 304 J, Which of the following statements/relationships is not, correct in thermodynamic changes ?, (a), U = (isothermal reversible expansion of a gas), (b) w = – nRT ln, , 25., , Pressure, p, , 14., , THERMODYNAMICS, , V2, (isothermal reversible expansion of an, V1, , ideal gas), (d) For a system of constant volume heat involved directly, changes to internal energy., An ideal gas expands in volume from 1×10–3 to 1 × 10–2 m3, at 300 K against a constant pressure of 1×105 Nm–2. The, work done is, (a) 270 kJ, (b) – 900 kJ, (c) – 900 J, (d) 900 kJ, The difference between H and U is usually significant, for systems consisting of, (a) only solids, (b) only liquids, (c) both solids and liquids (d) only gases, If a reaction involves only solids and liquids which of the, following is true ?, (a), H< E, (b) H = E, (c), H> E, (d) H = E + RT n, During isothermal expansion of an ideal gas, its, (a) internal energy increases, (b) enthalpy decreases, (c) enthalpy remains unaffected, (d) enthalpy reduces to zero., Assume each reaction is carried out in an open container., For which reaction will H = E ?, (a) C(s) + 2H2O (g) 2H2 (g) + CO2 (g), (b) PCl5 (g) PCl3 (g) + Cl2 (g), (c) 2CO (g) + O2 (g) 2CO2 (g), (d) H2 (g) + Br 2 (g) 2 HBr (g), 1, For the reaction CO (g ), O 2 (g ) CO 2 (g ), 2, Which one of the statement is correct at constant T and P ?, H E, (a), (b), , H, , E, , (c), , H, , E, , (d), , H is independent of physical state of the reactants
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THERMODYNAMICS, , 29., , 30., , 31., , 97, , For a reaction in which all reactants and products are liquids,, which one of the following equations is most applicable ?, (a), H< E, (b) H = S, (c), H= E, (d) H = G, The relationship between enthalpy change and internal, energy change is, (a), H= E+P V, (c), H= E–P V, For the reaction, , C3H8 (g) 5O 2 (g), , (b), (d), , H ( E V P), H= P V– E, , 3CO2 (g) 4H 2O(l), , at constant temperature, H – E is, (a) – RT, (b) + RT, (c) – 3 RT, (d) + 3 RT, 32., , Consider the reaction : N 2, , 3H 2, , 2 NH 3 carried out at, , constant temperature and pressure. If H and U are the, enthalpy and internal energy changes for the reaction, which, of the following expressions is true ?, (a), (b), H, U, H, U, 33., , 34., , 35., , 36., , 37., , 38., , 39., , 40., , H 0, (c), H, U, (d), Among the following, the intensive properties are, (i) molar conductivity, (ii) electromotive force, (iii) resistance, (iv) heat capacity, (a) (ii) and (iii), (b) (i), (ii) and (iii), (c) (i) and (iv), (d) (i) only, Which is an extensive property of the system ?, (a) Volume, (b) Viscosity, (c) Temperature, (d) Refractive index, Which of the following is an example of extensive property?, (a) Temperature, (b) Density, (c) Mass, (d) Pressure, Which of the following factors do not affect heat capacity?, (a) Size of system, (b) Composition of system, (c) Nature of system, (d) Temperature of the system, The heat required to raise the temperature of body by 1 C° is, called, (a) specific heat, (b) thermal capacity, (c) water equivalent, (d) None of these., Equal volumes of two monoatomic gases, A and B, at same, temperature and pressure are mixed. The ratio of specific, heats (Cp/Cv) of the mixture will be :, (a) 0.83, (b) 1.50, (c) 3.3, (d) 1.67, The molar heat capacity of water at constant pressure is 75, JK–1 mol–1. When 1kJ of heat is supplied to 100 g of water,, which is free to expand, the increase in temperature of water, is, (a) 6.6 K, (b) 1.2 K, (c) 2.4 K, (d) 4.8 K, Calorie is equivalent to :, (a) 0.4184 Joule, (b) 4.184 Joule, (c) 41.84 Joule, (d) 418.4 Joule, , 41. Which of the following is not true regarding thermo-chemical, equations?, (a) The coefficients in a balanced thermo-chemical, equation refer to the number of moles of reactants and, products involved in the reaction, (b) The coefficients in a balanced thermo-chemical, equation refer to the number of molecules of reactants, and products involved in the reaction, (c) The numerical value of r H refers to the number of, moles of substances specified by an equation., (d) Standard enthalpy change r H will have units as, kJ mol-1., 42. The enthalpies of elements in their standard states are taken, as zero. The enthalpy of formation of a compound, (a) is always negative, (b) is always positive, (c) may be positive or negative, (d) is never negative, 43. If enthalpies of formation of C 2 H 4 (g) , CO 2 (g) and, H 2 O ( l) at 25°C and 1atm pressure are 52, – 394 and, , – 286 kJ/mol respectively, the change in ethalpy is equal to, (a) – 141.2 kJ/mol, (b) – 1412 kJ/mol, (c) + 14.2 kJ/mol, (d) + 1412 kJ/mol, 44. The enthalpy change for a reaction does not depend upon, (a) use of different reactants for the same product, (b) the nature of intermediate reaction steps, (c) the differences in initial or final temperatures of, involved substances, (d) the physical states of reactants and products, 45. On the basis of thermochemical equations (i), (ii) and (iii),, find out which of the algebric relationships given in options, (a) to (d) is correct., (i) C (graphite) + O2(g) CO2(g); rH = x kJ mol–1, 1, (ii) C(graphite) + O2(g), 2, , (iii) CO (g) +, , 1, O (g), 2 2, , CO (g);, , CO2 (g);, , rH = y kJ mol, , rH = z kJ mol, , –1, , –1, , (a) z = x + y, (b) x = y – z, (c) x = y + z, (d) y = 2z – x, 46. Given that bond energies of H – H and Cl – Cl are 430 kJ mol–, 1 and 240 kJ mol–1 respectively and H for HCl is – 90 kJ, f, mol– 1, bond enthalpy of HCl is, (a) 380 kJ mol–1, (b) 425 kJ mol–1, –1, (c) 245 kJ mol, (d) 290 kJ mol–1, 47. Bond dissociation enthalpy of H2, Cl2 and HCl are 434 , 242, and 431 kJ mol–1 respectively. Enthalpy of formation of HCl, is:, (a) 93 kJ mol–1, (b) – 245 kJmol–1, –1, (c) – 93 kJmol, (d) 245 kJmol–1
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THERMODYNAMICS, , 61., , 62., , 63., , 64., , 65., , 66., , 67., , Pick out the wrong statement, (a) The standard free energy of formation of all elements, is zero, (b) A process accompanied by decrease in entropy is, spontaneous under certain conditions, (c) The entropy of a perfectly crystalline substance at, absolute zero is zero, (d) A process that leads to increase in free energy will be, spontaneous, Identify the correct statement for change of Gibbs energy, for a system ( G system) at constant temperature and, pressure, (a) If Gsystem = 0, the system has attained equilibrium, (b) If Gsystem = 0, the system is still moving in a particular, direction, (c) If Gsystem < 0, the process is not spontaneous, (d) If Gsystem > 0, the process is not spontaneous, Identify the correct statement regarding a spontaneous, process:, (a) Lowering of energy in the process is the only criterion, for spontaneity., (b) For a spontaneous process in an isolated system, the, change in entropy is positive., (c) Endothermic processes are never spontaneous., (d) Exothermic processes are always spontaneous., A chemical reaction will be spontaneous if it is accompanied, by a decrease of, (a) entropy of the system., (b) enthalpy of the system., (c) internal energy of the system., (d) free energy of the system., In which of the following entropy decreases?, (a) Crystallization of sucrose solution, (b) Rusting of iron, (c) Melting of ice, (d) Vaporization of camphor, A spontaneous reaction is impossible if, (a) both, , H and, , S are negative, , (b) both, , H and, , S are positive, , (c), H is negative and S is positive, (d), H is positive and S is negative, For the gas phase reaction,, PCl5(g), , 68., , 99, , Then G for the reaction :, 2ZnS(s) 3O 2 (g), , S2 (s) 2O2 (g), , 2SO2 (g) ; G = – 544 kJ, , 2Zn(s) S2 (s), , 2ZnS(s) ; G = – 293 kJ, , 2ZnO(s) 2SO 2 (g), , will be :, (a) –357 kJ, (b) –731 kJ, (c) –773 kJ, (d) –229 kJ, 69. Identify the correct statement regarding entropy., (a) At absolute zero temperature, entropy of a perfectly, crystalline substance is taken to be zero., (b) At absolute zero temperature, the entropy of a perfectly, crystalline substance is positive., (c) Absolute entropy of a substance cannot be, determined., (d) At 0°C, the entropy of a perfectly crystalline substance, is taken to be zero, 70. Unit of entropy is, (a) JK–1 mol–1, (b) J mol–1, –1, –1, –1, (c) J K mol, (d) JK mol–1, 71. Considering entropy (S) as a thermodynamic parameter,, the criterion for the spontaneity of any process is, Ssystem Ssurroundings 0, (a), , 72., , 73., , 74., , 75., , PCl3(g) + Cl2(g), , which of the following conditions are correct ?, (a), H = 0 and S < 0, (b) H > 0 and S > 0, (c), H < 0 and S < 0, (d) H > 0 and S < 0, The factor of G values is important in metallurgy. The G, values for the following reactions at 800ºC are given as :, , 2ZnO(s) ; G = – 480 kJ, , 2Zn(s) O2 (g), , 76., , (b), , Ssystem, , Ssurroundings, , (c), , Ssystem 0 only, , 0, , Ssurroundings 0 only, (d), In an exothermic reaction (reversible) which of the following, has positive value?, (a) Enthalpy, (b) Entropy, (c) Gibb’s free energy, (d) None of these, A reaction cannot take place spontaneously at any, temperature when, (a) both H and S are positive, (b) both H and S are negative, (c), H is negative and S is positive, (d) H is positive and S is negative, A reaction is spontaneous at low temperature but nonspontaneous at high temperature. Which of the following, is true for the reaction?, (a), H > 0, S > 0, (b) H < 0, S > 0, (c), H > 0, S = 0, (d) H < 0, S < 0, At the sublimation temperature, for the process, CO2(s), CO2(g), (a), H, S and G are all positive, (b) H > 0, S > 0 and G < 0, (c), H < 0, S > 0 and G < 0, (d) H > 0, S > 0 and G = 0, Choose the reaction with negative S value., (a) 2NaHCO3(s) Na2CO3(s) + CO2(g) + H2O(g), (b) Cl2(g) 2Cl(g), (c) 2SO2(g) + O2(g) 2SO3(g), (d) 2KClO3(s) 2KCl(s) + 3O2(g)
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EBD_7207, THERMODYNAMICS, , 100, , 77., , 78., , 79., , 80., , A chemical reaction is spontaneous at 298 K but nonspontaneous at 350 K. Which one of the following is true, for the reaction?, G, H, S, (a) –, –, +, (b) +, +, +, (c) –, +, –, (d) –, –, –, For a particular reversible reaction at temperature T, H and, S were found to be both +ve. If Te is the temperature at, equilibrium, the reaction would be spontaneous when, (a) Te > T, (b) T > Te, (c) Te is 5 times T, (d) T = Te, Which of the following pairs of a chemical reaction is certain, to result in a spontaneous reaction?, (a) Exothermic and increasing disorder, (b) Exothermic and decreasing disorder, (c) Endothermic and increasing disorder, (d) Endothermic and decreasing disorder, In which of the following reactions, standard entropy, change ( S°) is positive and standard Gibb’s energy change, ( G°) decreases sharply with increasing temperature ?, (a) C graphite +, , 1, O (g), 2 2, , (b) CO(g) +, , 1, O (g), 2 2, , CO2(g), , (c) Mg(s) +, , 1, O (g), 2 2, , MgO(s), , 1, 1, 1, C graphite + O2(g), CO2(g), 2, 2, 2, Consider the following reaction occurring in an automobile, 2C 8 H18 (g ) 25 O 2 (g ), , 82., , 83., , 87., , 16CO 2(g) 18H 2O(g ), , the sign of H, S and G would be, (a) +, –, +, (b) –, +, –, (c) –, +, +, (d) +, +, –, A reaction occurs spontaneously if, (a) T S < H and both H and S are + ve, (b) T S > H and H is + ve and S is ve, (c) T S > H and both H and S are + ve, (d) T S = H and both H and S are + ve, The enthalpy of fusion of water is 1.435 kCal/mol., The molar entropy change for the melting of ice at 0°C is :, (a) 10.52 cal / (mol K), (b) 21.04 cal / (mol K), (c) 5.260 cal / (mol K), (d) 0.526 cal / (mol K), , STATEMENT TYPE QUESTIONS, 84., , 86., , CO(g), , (d), , 81., , 85., , Read the following statements carefully and choose the, correct option, (i) The state of the system is specified by state functions, or state variables., , 88., , (ii) Variables like P, V and T are called state variables or, state functions, (iii) Their values depend only on the state of the system, and not on how it is reached., (a) (i) and (ii) are correct, (b) (ii) and (iii) are correct, (c) (i) , (ii) and (iii) are correct, (d) Only (iii) is correct, Read the following statements carefully and choose the, correct option, (i) Internal energy, U, of the system is a state function., (ii) –w shows, that work is done on the system., (iii) +w shows, that work is done by the system, (a) (i) and (ii) are correct (b) (ii) and (iii) are correct, (c) (i) and (iii) are correct (d) Only (i) is correct, Read the following statements carefully and choose the, correct answer, (i) Expansion of a gas in vacuum (pex = 0) is called free, expansion., (ii) Work is done during free expansion of an ideal gas, whether the process is reversible or irreversible, (iii) No work is done during free expansion of an ideal gas, whether the process is reversible or irreversible, (iv) No work is done during free expansion of an ideal gas, when the process is reversible, (a) Only statement (iii) is correct, (b) Statements (i) and (iii) are correct, (c) Statements (ii) and (iv) are correct, (d) Statements (i) and (iv) are correct, Which of the following statement(s) is/are correct ?, (i) In case of expansion maximum amount of work can be, obtained under isothermal conditions by reversibly, carrying out the process rather than through, irreversible route., (ii) In case of compression, minimum amount of work can, be done on system by carrying out the process, irreversibly than reversibly., (a) (i) and (ii), (b) Only (i), (c) Only (ii), (d) Neither (i) nor (ii), Read the following statements carefully and choose the, correct option, (i) In case of diatomic molecules the enthalpy of, atomization is also the bond dissociation enthalpy., (ii) In case polyatomic molecules, bond dissociation, enthalpy is different for different bonds within the same, molecule., (a) Both (i) and (ii) are correct, (b) (i) is correct but (ii) is incorrect, (c) (ii) is correct but (i) is incorrect, (d) Both (i) and (ii) are incorrect
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THERMODYNAMICS, , 89., , 90., , 101, , Read the following statements regarding spontaneity of a, process and mark the appropriate choice., (i) When enthalpy factor is absent than randomness factor, decides spontaneity of a process., (ii) When randomness factor is absent then enthalpy factor, decides spontaneity of a process., (iii) When both the factors take place simultaneously, the, magnitude of both the factors decide spontaneity of a, process., (a) Statements (i) and (ii) are correct and (iii) is incorrect., (b) Statement (iii) is correct, (i) and (ii) are incorrect., (c) Statements (i), (ii) and (iii) are correct., (d) Statements (i), (ii) and (iii) are incorrect., Which of the following statement is incorrect ?, (a) The standard enthalpy of reaction is the enthalpy, change for a reaction when all the participating, substances are in their standard states., (b) The standard state of a substance at a specified, temperature is its pure form at 1 bar., (c) The standard state of solid iron at 298 K is pure iron at, 1 bar, (d) Standard conditions are denoted by adding the, superscript, , to the symbol H e.g.,, , H, , MATCHING TYPE QUESTIONS, 91., , 92., , Match the columns, Column-I, Column-II, (A) Cm, (p) Cv T, (B) q, (q) C/n, (C), (r) Cp T, (D) H, (s) C T, (a) A – (q), B – (s), C – (r), D – (p), (b) A – (q), B – (s), C – (p), D – (r), (c) A – (s), B – (q), C – (p), D – (r), (d) A – (q), B – (p), C – (r), D – (s), Match the columns, Column-I, Column-II, (A) Free expansion at, , Vf, (p) q = – w = nRT n V, i, , V= 0, (B) Isothermal irreversible (q) U = wad, change, (C) Isothermal reversible (r), U = qv, change, (D) For adiabatic change (s) q = – w = Pex (Vf – Vi), (a) A – (s), B – (p), C – (r), D – (q), (b) A – (r), B – (q), C – (p), D – (s), (c) A – (r), B – (s), C – (p), D – (q), (d) A – (q), B – (r), C – (s), D – (p), , 93. Match the columns, Column-I, (A) pext = 0, , Column-II, (p) Free expansion of an, ideal gas, (B) q = pext (Vf – Vi), (q) Adiabatic change, (C) q = 2.303 nRT log (Vf / Vi) (r) Isothermal reversible, change, (D) U = Wad, (s) Isothermal irreversible, change, (a) A – (p), B – (s), C – (r), D – (q), (b) A – (p), B – (q), C – (r), D – (s), (c) A – (p), B – (r), C – (s), D – (q), (d) A – (p), B – (r), C – (q), D – (s), 94. Match the columns, Column-I, Column-II, (A) H 2 (g) Br2 (g), (B) PCl5 (g), (C), , 2HBr (g), , PCl3 (g) Cl 2 (g), , N 2 (g) 3H 2 (g), , (D) 2N 2O5 (g), , 2NH3 (g), , (p), , H = U – 2RT, , (q), , H = U + 3RT, , (r), , H= U, , 4NO 2 (g) O2 (g) (s), , H = U + RT, , (a) A – (r), B – (p), C – (q), D – (s), (b) A – (r), B – (s), C – (p), D – (q), (c) A – (q), B – (p), C – (s), D – (r), (d) A – (s), B – (q), C – (p), D – (r), 95. Match the columns, Column-I, Column-II, (A) C4 H10, , 13, O, 2 2, , (p) Enthalpy of atomisation, , 4CO2 5H 2O; H = – w, (B) CH4 C + 4H ; H = x (q) Enthalpy of formation, (C) H2+ Br2 2HBr ; H = y (r) Enthalpy of combustion, (D) Na–(s) Na(g) ; H = z (s) Enthalpy of sublimation, (a) A – (s), B – (p), C – (q), D – (r), (b) A – (q), B – (r), C – (p), D – (s), (c) A – (r), B – (p), C – (q), D – (s), (d) A – (p), B – (q), C – (s), D – (r), 96. Match the columns, Column-I, Column-II, (A) Exothermic, (p) H = 0, E = 0, (B) Spontaneous, (q) G = 0, (C) Cyclic process, (r) H is negative, (D) Equilibrium, (s) G is negative, (a) A – (q), B – (r), C – (p), D – (s), (b) A – (s), B – (p), C – (r), D – (q), (c) A – (p), B – (q), C – (s), D – (r), (d) A – (r), B – (s), C – (p), D – (q)
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EBD_7207, THERMODYNAMICS, , 102, , 97., , Match the columns, Column-I, (A) H = –ve ; S = –ve, G = –ve, (B), (C), (D), (a), (b), (c), (d), , Column-II, (p) Reaction will be, non-spontaneous at, high temperature, H = –ve ; S = –ve (q) Reaction will be, G = +ve, non-spontaneous at, low temperature, H = +ve ; S = +ve (r) Reaction will be, G = +ve, spontaneous at low, temperature, H = +ve ; S = +ve (s) Reaction will be, G = –ve, spontaneous at high, temperature, A – (q), B – (r), C – (p), D – (s), A – (r), B – (p), C – (q), D – (s), A – (r), B – (q), C – (s), D – (p), A – (q), B – (s), C – (p), D – (r), , ASSERTION-REASON TYPE QUESTIONS, Directions : Each of these questions contain two statements,, Assertion and Reason. Each of these questions also has four, alternative choices, only one of which is the correct answer. You, have to select one of the codes (a), (b), (c) and (d) given below., (a) Assertion is correct, reason is correct; reason is a correct, explanation for assertion., (b) Assertion is correct, reason is correct; reason is not a, correct explanation for assertion, (c) Assertion is correct, reason is incorrect, (d) Assertion is incorrect, reason is correct., 98. Assertion : T, P and V are state variables or state functions., Reason : Their values depend on the state of the system, and how it is reached., 99. Assertion : At constant temperature and pressure whatever, heat absorbed by the system is used in doing work., Reason : Internal energy change is zero., 100. Assertion : For an isothermal reversible process Q = –W, i.e. work done by the system equals the heat absorbed by, the system., Reason : Enthalpy change ( H) is zero for isothermal, process., 101. Assertion : Absolute value of internal energy of a substance, cannot be determined., Reason : It is impossible to determine exact values of, constitutent energies of the substances., 102. Assertion : A process is called adiabatic if the system does, not exchange heat with the surroundings., Reason : It does not involve increase or decrease in, temperature of the system., 103. Assertion : There is exchange in internal energy in a cyclic, process., Reason : Cyclic proces is the one in which the sytem returns, to its initial state after a number of reactions., 104. Assertion : Internal energy is an extensive property., Reason : Internal energy depends upon the amount of the, system., , 105. Assertion : The mass and volume of a substance are the, extensive properties and are proportional to each other., Reason : The ratio of mass of a sample to its volume is an, intensive property., 106. Assertion : First law of thermodynamics is applicable to an, electric fan or a heater., Reason : In an electric fan, the electrical energy is converted, into mechanical work that moves the blades. In a heater,, electrical energy is converted into heat energy., 107. Assertion : The value of enthalpy of neutralization of weak, acid and strong base is numerically less than 57.1 kJ., Reason : All the OH– ions furnished by 1 g equivalent of, strong base are not completely neutralized., 108. Assertion : When a solid melts, decrease in enthalpy is, observed., Reason : Melting of a solid is endothermic., 109. Assertion : Many endothermic reactions that are not, spontaneous at room temperature become spontaneous at, high temperature., Reason : Entropy of the system increases with increase in, temperature., 110. Assertion : An exothermic process which is nonspontaneous at high temperature may become spontaneous, at a low temperature., Reason : There occurs a decrease in entropy factor as the, temperature is decreased., , CRITICAL THINKING TYPE QUESTIONS, 111. In an adiabatic process, no transfer of heat takes place, between system and surroundings. Choose the correct, option for free expansion of an ideal gas under adiabatic, condition from the following., (a) q = 0, T 0, w = 0, (b) q 0, T = 0, w = 0, (c) q = 0, T = 0, w = 0, (d) q = 0, T < 0, w 0, 112. According to the first law of thermodynamics which of the, following quantities represents change in a state function ?, (a) q rev, (b) qrev – Wrev, (c) q rev /W rev, (d) qrev + Wrev, 113. If H is the change in enthalpy and E, the change in, internal energy accompanying a gaseous reaction, then, (a), (b), , H is always greater than E, H < E only if the number of moles of the products, is greater than the number of moles of the reactants, , (c), (d), , H is always less than E, H < E only if the number of moles of products is, less than the number of moles of the reactants, 114. For an isothermal reversible expansion process, the value, of q can be calculated by the expression, (a), , q, , 2.303nRT log, , (c), , q, , Pexp nRT log, , V2, V1 (b), , q, , 2.303nRT log, , V1, V2 (d) None of these, , V2, V1
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THERMODYNAMICS, , 103, , 115. A piston filled with 0.04 mol of an ideal gas expands, reversibly from 50.0 mL to 375 mL at a constant temperature, of 37.0ºC. As it does so, it absorbs 208 J of heat. The values, of q and w for the process will be:, (R = 8.314 J/mol K) (ln 7.5 = 2.01), (a) q = + 208 J, w = – 208 J, (b) q = – 208 J, w = – 208 J, (c) q = – 208 J, w = + 208 J, (d) q = + 208 J, w = + 208 J, 116. According to the first law of thermodynamics, U = q + W., In special cases the statement can be expressed in different, ways. Which of the following is not a correct expression ?, (a) At constant temperature q = –W, (b) When no work is done U = q, (c) In gaseous system U = q + P V, (d) When work is done by the system : U = q + W, 117. The internal energy change when a system goes from state A, to B is 40 kJ/mole. If the system goes from A to B by a reversible, path and returns to state A by an irreversible path what would, be the net change in internal energy ?, (a) > 40 kJ, (b) < 40 kJ, (c) Zero, (d) 40 kJ, 118. Under isothermal condition for one mole of ideal gas what, is the ratio of work done under reversible to irreversible, process, initially held at 20 atm undergoes expansion from, 1L to 2L, at 298K, under external pressure of 10 atm?, (a) 1.7, (b) 2.0, (c) 1.4, (d) 1.0, 119. Processes A to B, B to C and C to D shown in the figure, below respectively are ?, P1/ V1/ T1, , A, , P2/ V2/ T1, , B, , P2/ V3/ T2, , C, , P3/ V3/ T3, , D, , (a) Isothermal, isobaric and isochoric, (b) Isobaric, isothermal and isochoric, (c) Isothermal, isothermal and isobaric, (d) Isobaric, isobaric and isothermal, 120. What is the internal energy (kJ) change occurs when 36g, of H2O(l) converted to H2O(g)? H°(vapourisation) =, 40.79kJ/mol, (a), (b), (c), (d), 121. Consider the reaction :, 4NO2 ( g ) O 2 ( g ) 2N 2O5 ( g ),, rH = – 111 kJ., If N2O5(s) is formed instead of N2O5(g) in the above reaction,, the rH value will be :, (given, H of sublimation for N2O5 is –54 kJ mol–1), (a) + 54 kJ, (b) + 219 kJ, (c) – 219 J, (d) – 165 kJ, , 122. An ideal gas is allowed to expand both reversibly and, irreversibly in an isolated system. If T i is the initial, temperature and Tf is the final temperature, which of the, following statements is correct?, (a) (Tf)rev = (Tf)irrev, (b) Tf = Ti for both reversible and irreversible processes, (c) (Tf)irrev > (Tf)rev, (d) Tf > Ti for reversible process but Tf = Ti for irreversible, process, 123. Given, Reaction, Energy Change, (in kJ), Li(s), Li(g), 161, +, Li(g), Li (g), 520, 1, F (g), 2 2, F(g) + e–, , F(g), , 77, , F–(g), , Li+ (g) + F–(g), , (Electron gain, enthalpy), – 1047, , Li F(s), , 1, F (g), Li F(s), – 617, 2 2, Based on data provided, the value of electron gain enthalpy, of fluorine would be :, (a) – 300 kJ mol–1, (b) – 350 kJ mol–1, –1, (c) – 328 kJ mol, (d) – 228 kJ mol–1, 124. The standard enthalpy of formation ( f H° 298) for methane,, CH4 is – 74.9 kJ mol–1. In order to calculate the average, energy given out in the formation of a C – H bond from this, it is necessary to know which one of the following?, (a) The dissociation energy of the hydrogen molecule,, H2., (b) The first four ionisation energies of carbon., (c) The dissociation energy of H2 and enthalpy and, sublimation of carbon (graphite)., (d) The first four ionisation energies of carbon and, electron affinity of hydrogen., 125. For complete combustion of ethanol,, , Li (s) +, , C2 H5 OH l, , 3O2 g, , 2CO2 g, , 3H 2 O l ,, , the amount of heat produced as measured in bomb, calorimeter, is 1364.47 kJ mol–1 at 25ºC. Assuming ideality, the enthalpy of combustion, cH, for the reaction will be:, (R = 8.314 kJ mol–1), (a), , 1366.95 kJ mol, , 1, , (b), , 1361.95 kJ mol, , 1, , (c), , 1460.95 kJ mol, , 1, , (d), , 1350.50 kJ mol, , 1, , 126. Standard enthalpy of vapourisation vap H° for water at, 100°C is 40.66 kJ mol–1. The internal energy of vaporisation, of water at 100°C (in kJ mol–1) is :, (a) + 37.56, (b) – 43.76, (c) + 43.76, (d) + 40.66, (Assume water vapour to behave like an ideal gas).
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EBD_7207, THERMODYNAMICS, , 104, , 127. Consider the following reactions:, (i) H +(aq) + OH–(aq) = H2O(l),, H = – X1 kJ mol–1, 1, = H2O(l),, O, 2 2(g), H = – X2 kJ mol–1, (iii) CO2(g) + H2(g) = CO(g) + H2O,, H = – X3 kJ mol–1, , (ii) H2(g) +, , 5, O2(g) = 2CO2(g) + H2O(l)’, 2, H = + 4X4 kJ mol–1, Enthalpy of formation of H2O (l) is, (a) + X3 kJ mol– 1, (b) – X4 kJ mol– 1, –, 1, (c) + X1 kJ mol, (d) – X2 kJ mol– 1, 128. Diborane is a potential rocket fuel which undergoes, combustion according to the equation, , (iv) C 2 H 2(g), , B2 H 6 ( g ) 3O 2 ( s ), B 2 O3 ( s ) 3H 2O( g ), Calculate the enthalpy change for the combustion of, diborane. Given, , (i), , 2B( s), , (ii), , H2 ( g ), , (iii) H 2O(l ), , 3, O (g), 2 2, , B2 O3 ( s); H = –1273 kJ per mol, , 1, O2 (g ), 2, H 2O( g );, , H 2O(l ); H = –286 kJ per mol, , H = 44 kJ per mol, , B2H6 (g); H = 36 kJ per mol, (iv) 2B (s) + 3H2 (g), (a) + 2035 kJ per mol, (b) – 2035 kJ per mol, (c) + 2167 kJ per mol, (d) – 2167 kJ per mol, 129. How many molecules of ATP, undergo hydrolysis to raise, the temperature of 180 kg of water which was originally at, room temperature by 1°C ? C{P,m} water = 75.32J/mol/K,, H{P} for ATP hydrolysis= 7 kcal/mol, (a) 1.5 × 1025, (b) 2.00 × 1023, 25, (c) 3.4 × 10, (d) 4.0 × 1024, 130. What is the amount of heat (in Joules) absorbed by 18 g of, water initially at room temperature heated to 100°C ? If 10g, of Cu is added to this water , than decrease in temperature, (in Kelvin) of water was found to be? C (p,m) for water, 75.32J/mol K ; C (p,m) for Cu = 24.47J/mol K., (a) 5649, 369, (b) 5544, 324, (c) 5278, 342, (d) 3425, 425, , 131. The enthalpy changes for the following processes are listed, below :, Cl2(g) 2Cl(g),, 242.3 kJ mol–1, I2(g) 2I(g),, 151.0 kJ mol–1, ICl(g) I(g) + Cl(g),, 211.3 kJ mol–1, I2(s) I2(g),, 62.76 kJ mol–1, Given that the standard states for iodine and chlorine are, I2(s) and Cl2(g), the standard enthalpy of formation for ICl(g), is :, (a) +16.8 kJ mol–1, (b) +244.8 kJ mol–1, (c) –14.6 kJ mol–1, (d) –16.8 kJ mol–1, 132. What is the equilibrium constant if ATP hydrolysis by water, produce standard free energy of –50 kJ/mole under normal, body conditions ?, (a) 2.66 × 108, (b) 5.81 × 108, (c) 1.18 × 107, (d) 1.98 × 108, 133. A reaction with H = 0, is found to be spontaneous. This is, due to, (a), S is negative, (b) S is positive, (c) T S is positive, (d) Both (b) and (c), 134. In an irreversible process taking place at constant T and P and, in which only pressure-volume work is being done, the change, in Gibbs free energy (dG) and change in entropy (dS), satisfy, the criteria, (a) (dS)V, E > 0, (dG)T, P < 0, (b) (dS)V, E = 0, (dG)T, P = 0, (c) (dS)V, E = 0, (dG)T, P > 0, (d) (dS)V, E < 0, (dG)T, P < 0, 135. In conversion of lime-stone to lime,, , CaCO3(s), , CaO(s), , CO2(g) the values of H and S, -1, are + 179.1 kJ mol and 160.2 J/K respectively at 298 K and, 1 bar. Assuming that H and S do not change with, temperature, temperature above which conversion of, limestone to lime will be spontaneous is, (a) 1118 K, (b) 1008 K, (c) 1200 K, (d) 845 K., 136. For vaporization of water at 1 atmospheric pressure, the, values of H and S are 40.63 kJmol–1 and 108.8 JK–1 mol–1,, respectively. The temperature when Gibbs energy change, ( G) for this transformation will be zero, is:, (a) 293.4 K, (b) 273.4 K, (c) 393.4 K, (d) 373.4 K.
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THERMODYNAMICS, , FACT / DEFINITION TYPE QUESTIONS, 1., 2., , 3., 4., 5., 6., 7., 8., 9., , 10., 11., 12., , 13., 15., 16., 17., , 18., 19., 20., , 21., , (c), (b) The laws of thermodynamics deal with energy changes, of macroscopic systems involving a large number of, molecules rather than microscopic systems containing, a few molecules., (b), (c) The universe = The system + The surroundings, (c), (c) Closed system can exchange energy and not matter, with surroundings. Pressure cooker provides closed, system., (c) Isolated system can not exchange mass or energy., (b) We can describe the state of a gas by quoting its, pressure (P), volume (V), temperature (T ), amount (n), etc., (d) We know that q (heat) and work (w) are not state, functions but (q + w) is a state function. H – TS (i.e. G), is also a state functions. Thus II and III are not state, functions so the correct answer is option (d)., (c) Internal energy and molar enthalpy are state functions., Work (reversible or irreversible) is a path function., (c), (a) Internal energy is a quantity which represents the total, energy of the system. It may be chemical, electrical,, and mechanical or any other type of energy you may, think of, the sum of all these is the internal energy of, the system., (d), 14. (b), (c) In accordance with Hess’s law., (a), (a), E = Q–W, For adiabatic expansion, Q = 0, E = –W, The negative sign shows decrease in Internal energy,, which is equal to the work done on the system by the, surroundings., (d) q is a path dependent function, H is a state function, because it depends on U, p and V, all of which are, state functions., (a) The shaded area shows work done on an ideal gas in a, cylinder when it is compressed by a constant external, pressure, (b) As volume is constant hence work done in this proces, is zero hence heat supplied is equal to change in, internal energy., (b) W = – p V, 3(6 4), 6 litre atmosphere, 6 101.32, 608 J, , 105, , 22. (c) For isothermal reversible expansion., V, w = –nRT ln 2, V, 23. (c), , W, , 105 (1 10, , P V, , 2, , 1 10 3 ), , 900J, , 24. (d) The difference between H and U is not usually, significant for systems consisting of only solids or, liquids. Solids and liquids do not suffer any significant, volume changes upon heating. The difference,, however, becomes significant when gases are, involved., 25. (b), H = E + P V, for solid and liquid,, V = or H = E + n RT, for solids and liquids n = 0., 26. (c) During isothermal expansion of an ideal gas,, T = 0. Now H = E + PV, H, E, ( PV ), H = E + (nRT);, T);, Thus if T = 0., H = E, i.e., remain unaffected, 27. (d) We know that, H= E + P V, In the reactions, H2 + Br 2, change in volume or V = 0, So, H = E for this reaction, 28. (b), , n, , 1, ; H, 2, , E, , 1, RT ;, 2, , 2HBr there is no, , E, , 29. (c) As all reactant and product are liquid, , H, , E, , nRT, , H, , E, , (, , 30. (a), , H, , E P V, , 31. (c), , H, E nRT, n = 3 – (1 + 5), = 3 – 6 = –3, H, , 32. (b), , E, , n, , n (g), , 0, , 0), , ( 3RT ), , H, U, nRT for N 2 3H 2, ng = 2 – 4 = – 2, H=, , H, , U 2 RT or, , U=, , H+ 2R T, , 2 NH 3, U>, , H, , 33. (a) Mass independent properties (molar conductivity and, electromotive force) are intensive properties., Resistance and heat capacity are mass dependent,, hence extensive properties., 34. (a) Volume depends upon mass. Hence it is extensive, property.
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EBD_7207, THERMODYNAMICS, , 106, , 35., , 36., 37., , 38., 39., , (c) An extensive property is a property whose value, depends on the quantity or size of matter present in, the system. For example, mass, volume, internal energy,, enthalpy, heat capacity, etc. are extensive properties, (d ) The magnitude of the heat capacity depends on the, size, composition and nature of the system., (b) The heat required to raise the temperature of body by, 1C° is called thermal capacity or heat capacity., , 48., , 5, R, 5, 2, 1.67, (d), 3, 3, R, 2, (c) Given Cp = 75 JK–1 mol–1, 100, n, mole , Q = 1000 J, 18, , 49., , CP, CV, , Q = nCp T, 40., 41., , 42., 43., , T, , ( H), , 44., 46., , H products, , 2H 2O., , (b), , 1, H2, 2, , 1, Cl 2, 2, , H HCl, , 47., , 52., , H), , 2FeO(s)+ CO2(g), , H = –26.8 + 33.0 = + 6.2 kJ, (b) Given H, 1, A, B, + 150, 2, 2C D –125, 3B, +350, E A, 2D, To calculate H operate, 2 × eq. (1) + eq. (2) – eq. (3), H = 300 – 125 – 350 = – 175, (b), , H, , H f products, , ...(1), ...(2), ...(3), , H f reactants, H f (H 2 O)(g)] –, , [ H f (CO)(g), , (b), , C 2 H 4 3O 2, 2CO 2, Change in enthalpy,, , H, , H products, , 2H 2O., , H reactants, , 53., , 2 ( 394) 2 ( 286) (52 0), = –1412 kJ/ mol., (a) Hess’s law is used for calculating enthalpy of reaction., , 54., , (c), , 55., 57., , 1, 240 B.E. of HCl, 2, , B.E. of HCl = 215 + 120 + 90, = 425 kJ mol–1, (c) The reaction for formation of HCl can be written as, H2 + Cl2 2HCI, H – H + Cl – Cl 2 (H – Cl), Substituting the given values, we get enthalpy of, formation of, 2HCl = – ( 862 – 676) = –186 kJ., Enthalpy of formation of, 186, HCl =, kJ = –93 kJ., 2, , H), , H f (H 2 )(g)], [ H f (CO 2 )(g), = [– 110.5 + ( – 241.8)]– [– 393.5 + 0] = 41.2, , B.E. of reactant, , 1, 1, 430, 2, 2, , B.E.(H, , 6 B.E.(C, , C), , (a) Fe2O3(s)+ CO(g), , H°, , HCl, , B.E. of products, 90, , 51., , H reactants, , 2 ( 394) 2 ( 286) (52 0), = – 1412 kJ/ mol., (b), 45. (c), , 4 B.E.(C–H), , = [606.1 + (4 × 410.5) + 431.37)] – [336.49 + (6 × 410.5)], = –120.0 kJ mol–1, , 2.4 K, , (b) Enthalpy of formation of C 2 H 4 , CO2 and H 2 O are, 52, – 394 and – 286 kJ/ mol respectively. (Given), The reaction is, , C), , B.E.(C, , T=?, , (b) 1 calorie = 4.184 joule, (b) The coefficients in a balanced thermo-chemical, equation refer to the number of moles (not to molecules), of reactants and products involved in the reaction., (c), , C 2 H 4 3O 2, 2CO 2, change in enthalpy,, , = B.E (C, , 50., , 1000 18, 100 75, , (b) Enthalpy of reaction, = B.E(Reactant)– B.E(Product), , 58., 59., , X, , H, , Y, , X, , H1, , P, , H2, , Q, , H3, , Y, , H, H1, H2, H3, (a), 56. (b), (b) Heat of combustion of a substance is always negative, as it is the amount of heat evolved (i.e. decrease in, enthalpy) when one mole of the substance is, completely burnt in air or oxygen., (c), (a) Conc. of HCl = 0.25 mole, Conc. of NaOH = 0.25 mole, Heat of neutralization of strong acid by strong base, = – 57.1 kJ, HCl + NaOH, NaCl + H2O – 57.1kJ, 1 mole of HCl neutralise 1 mole of NaOH, heat evolved, = 57.1 kJ, 0.25 mole of HCl neutralise 0.25 mole of NaOH, Heat evolved = 57.1 × 0.25 = 14.275 kJ
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THERMODYNAMICS, , 60., 61., 62., , 63., , 64., 65., , 66., , 67., , (c), (d) A process is spontaneous only when there is decrease in, the value of free energy, i.e., G is –ve., (a) If Gsystem = 0 the system has attained equilibrium is, right choice., In it alternative (d) is most confusing as when G > 0,, the process may be spontaneous when it is coupled, with a reaction which has G < 0 and total G is, negative, so right answer is (a)., (b) Spontaneity of reaction depends on tendency to, acquire minimum energy state and maximum, randomness. For a spontaneous process in an isolated, system the change in entropy is positive., (d), G is negative for a spontaneous process., (a) Crystallization of sucrose solution. Entropy is a, measure of randomness during the crystallisation of, sucrose solution liquid state is changing into solid, state hence entropy decreases., (d), , G = H – T S; G is positive for a reaction to be, non-spontaneous when H is positive and S is, negative., (b) For the reaction, PCl5 (g), , 68., , 69., 70., , 107, , PCl3 (g) Cl2 (g), , The reaction given is an example of decomposition, reaction and we know that decomposition reactions, are endothermic in nature, i.e, H > 0., Further, n = (1 + 1) – 1= + 1, Hence more number of molecules are present in, products which shows more randomness i.e. S > 0, ( S is positive), (b) For the reaction, 2ZnS 2Zn + S2 ; G1º = 293 kJ, ..........(1), 2Zn + O2 2ZnO ; G2º = – 480 kJ, ..........(2), S2 + 2 O2 2SO2 ; G3º = – 544 kJ, .........(3), Gº for the reaction, 2ZnS + 3 O2 2ZnO + 2SO2, can be obtained by adding eqn. (1), (2) and (3), Gº = 293 – 480 – 544 = – 731 kJ, (a) Third law of Thermodynamics., q, S, (a), T, q, required heat per mole, T, constant absolute temperatur e, Unit of entropy is JK–1 mol–1, S total is always positive., , 71., , (a) For a spontaneous process,, , 72., , (d) For an exothermic reaction all three enthalpy, entropy, and Gibb's free energy change have negative values., (d) Gibb’s-Helmholtz equation is, G= H– T S, , 73., , 74. (d), , 75. (d), 76. (c), , 77. (d), , For a reaction to be non-spontaneous at all, temperatures, H should be +ve and S should be –ve, G = +ve – T × (–ve); the value of G is always, positive for such a reaction and hence it will be nonspontaneous at all temperatures., We know that G = H – T S, When H < 0 and S < 0 then G will be negative at, low temperatures (positive at high temperature) and, the reaction will be spontaneous., Since the process is at equilibrium G = 0 for G 0,, they should be H > 0, S > 0., S has negative value if number of gaseous moles, decreases during a reaction, ng = –ve, For the reaction, 2SO2 + O2 2SO3, ng = 2 – 3 = –1, G= H– T S, For a reaction to be spontaneous,, H = –ve, S = +ve, at all temperatures., but at high temperature,, , G, , H T S, ve, , ve, , Thus the second term will have high positive value, and reaction will be non-spontaneous., 78. (b) At equilibrium G = 0, Hence, G = H – Te S = 0, H = Te S or Te = DH, DS, , For a spontaneous reaction, G must be negative which is possible only if, H– T S<0, H < T S or T >, , DH, ; Te < T, DS, , 79. (a) Measure of disorder of a system is nothing but, Entropy. For a spontaneous reaction, G < 0. As per, Gibbs Helmholtz equation,, G= H– T S, Thus G is –ve only, When H = –ve (exothermic), and S = +ve (increasing disorder), 80. (a) Since, in the first reaction gaseous products are, forming from solid carbon hence entropy will increase, i.e. S = +ve., 1, O2(g) CO(g); S° = + ve, 2, Since, G° = H° – T S hence the value of G decrease, on increasing temperature., 81. (b) This is combustion reaction, which is always exothermic, hence, H = –ve, As the no. of gaseous molecules are increasing hence, entropy increases, now G, H T S, , C (gr.) +
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EBD_7207, THERMODYNAMICS, , 108, , For a spontaneous reaction, , 82., , G, ve, Which is possible in this case as H = –ve and S, = +ve., (c) For a spontaneous reaction, G(–ve), which is possible if S = +ve,, H = +ve, and T S > H [As G = H – T S], , 83., , (c), , S, , H, T, , ASSERTION-REASON TYPE QUESTIONS, 98., 99., 100., , 1.435 103, 273, , 5.260 cal / (mol K), , 101., , STATEMENT TYPE QUESTIONS, 84., , 85., , 86., 89., 90., , (c) Variables like P, V and T which describes the state of, system are called state variables or state functions, because their values depend only on the state of the, system and not on how it is reached., (d) The positive sign expresses when work is done on the, system. Similarly, negative sign expresses when work, is done by the system, (b), 87. (a), 88. (a), (c) All the statements regarding spontaneity of a reaction, are correct., (c) The standard enthalpy of reaction is the enthalpy, change for a reaction when all the participating, substances are in their standard states. The standard, state of a substance at a specified temperature is its, pure form at 1 bar. For example, the standard state of, liquid ethanol at 298 K is pure liquid ethanol at 1 bar., Standard state of solid ion at 500 K is pure iron at 1 bar., The standard conditions are denoted by adding the, superscript to the symbol H e.g., H ., , 102., , 103., , 104., , 105., , 106., , 107., , MATCHING TYPE QUESTIONS, 91., 93., , 94., , 95., , (b), 92. (c), (a) A – (p), B – (s), C – (r), D – (q), Expansion of a gas in vacuum (pext = 0) is called free, expansion., For isothermal irreversible change, q = –W = pext(Vf – Vi), for isothermal reversible change, q = – W = nRT ln (Vf/Vi), = 2.303 nRT log Vf/Vi, For adiabatic change, q = 0, U = Wad, (b) (A) ng = 2 – 2 = 0 hence H = U, (B) ng = 2 – 1 = 1 hence H = U + RT, (C) ng = 2 – 4 = –2 hence H = U – 2RT, (D) ng = 5 – 2 = 3 hence H = U + 3RT, (c), 96. (d), 97. (b), , 108., 109., 110., , (c) Values of state functions depend only on the state of, the system and not on how it is reached., (a) Q = –W if E = 0, (b) In an isothermal process change in internal energy, ( E) is zero (as it is a function of temperature)., According to first law of thermodynamics, Q + W = E. Hence Q = –W (if E = 0), If a system undergoes a change in which internal, energy of the system remains constant (i.e. E = 0), then –W = Q. This means that work done by the system, equals the heat absorbed by the system., (a) It is fact that absolute values of internal energy of, substances cannot be determined. It is also true that it, is not possible to determine exact values of constitutent, energies of a substance., (c) It may involve increase or decrease in temperature of, the system. Systems in which such processes occur,, are thermally insulated from the surroundings., (a) As internal energy is a state function so its value, depends on intial and final states of the system. In, case of cyclic system initial and final states are same., So E = 0, and similarly H = 0., (a) The properties whose magnitude depends upon the, quantity of matter present in the system are called, extensive properties eg, internal energy., (b) The mass and volume depend upon the quantity of, matter so these are extensive properties while ratio of, mass to its volume does not depend upon the quantity, of matter so this ratio is an extensive property., (a) In case of electric fan electrical energy is converted, into mechanical energy and in case of heater, electrical, energy is converted into heat energy. Therefore, these, follow the first law of thermodynamics., (c) The value of enthalpy of neutralisation of weak acid, by strong base is less than 57.1 kJ. This is due to the, reason that the part of energy liberated during, combination of H+ and OH+ ions is utilised in the, ionisation of weak acid., (d) When a solid melts, increase in enthalpy is observed., (b) The factor T S increases with increase in temperature., (b) Both assertion and reason are true but reason is not, the correct explanation of assertion., For a process to be spontaneous G must be negative., G= H–T S, Exothermic process ( H is negative) is nonspontaneous if S is negative and temperature is high, because in such condition, H., T S, ( G, , H T S, , tive ). When temperature is, , decreased, T S, H T S, H ( G, and so the reaction becomes spontaneous., , tive)
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THERMODYNAMICS, , 109, , CRITICAL THINKING TYPE QUESTIONS, , Uo = 40.79 kJ/mol, , 111. (c) Justification : free expansion w = 0, adiabatic process q = 0, U = q + w = 0, this means that internal energy remains, constant. Therefore,, T = 0., 112. (d) Mathematical expression of first law of, thermodynamics, E = q + w, E is a state function., 113. (d) As H = E + ngRT, if np < nr;, Hence, 114. (a) q = –W, , ng = np – n r = – ve., , H<, , 2.303nRTlog, , A, , kJ, 36g, mol 18g / mol, , U = 75.98 kJ, 4NO 2 (g) O 2 (g), , 2N 2 O5 (g),, , rH, , = – 111 kJ, , – 54 kJ, 2N 2 O5 (s), , B, , 118. (a) –Wirreversible = Pext (V2 – V1), = 10 atm (2L – 1L), = 10 atm – L, V2, , Pex dv, V1, , V2, = 2.303 nRT log V, 1, , = 1 2.303 0.0821 atm–L /K /mol × log, , – 111 – 54 =, , H', , H' = – 165 kJ, 122. (c) In a reversible process the work done is greater than in, irreversible process. Hence the heat absorbed in, reversible process would be greater than in the latter, case. So, Tf (rev.) < Tf (irr.), 123. (c) Applying Hess’s Law, 1, fH, sub H, diss H I.E. E.A, lattice H, 2, 617 161 520 77 E.A. (–1047), E.A. = –617 + 289 = –328 kJ mol–1, electron affinity of fluorine, = –328 kJ mol–1, 124. (a) To calculate average enthalpy of C – H bond in, methane following informations are needed, (i) dissociation energy of H2 i.e., 1, H (g), H( g ); H x suppose, 2 2, (ii) Sublimation energy of C(graphite) to C(g), C( graphite), C( g ); H y Suppose, Given, , irreversible path, , 2, 1, , = 16.96 atm–L, =, , 37.69, , H', , reversible path, , Wreversible, Wirreversible, , kJ, mol, Internal energy change for 36 g of water, , = 37.69, , V2, V1, , 115. (a) Process is isothermal reversible expansion, hence, U = 0, therefore q = – W., Since q = + 208 J, W = – 208 J, 116. (d) When work is done by the system, U = q – W, 117. (c) For a cyclic process the net change in the internal, energy is zero because the change in internal energy, does not depend on the path., , –Wreversible =, , = (40.79 – 3.10) kJ/mol, , 121. (d), , E., , 8.314 373, kJ / mol, 1000, , 16.96, 1.69 1.7, 10.00, , 119. (a), 120. (a) H2O (l ) H2O (g), Hvap = 40.79 kJ/mol, H = U + ngRT, 40.79 kJ/mol = U + (1) (8.314 JK–1 mol–1) (373 K), , C( graphite ), , 2H 2 ( g ), , CH 4 ( g ); H, , 75 kJ mol, , 125. (a) C 2 H 5 OH( ) 3O 2 ( g ), 2CO2 ( g ) 3H2O( ), Bomb calorimeter gives U of the reaction, Given, U = –1364.47 kJ mol–1, ng = – 1, H = U + ngRT = 1364.47, 126. (a), , = – 1366.93 kJ mol–1, H 2O ( ), H 2O (g) + Q, E = 37558 J / mol, E = 37.56 kJ mol–1, , 1 8.314 298, 1000, , 1
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EBD_7207, THERMODYNAMICS, , 110, , If now 10g of copper is added Cp, m = 24.47 J/ mol K, Amount of heat gained by Cu, , 127. (d) This reaction shows the formation of H2O, and the X2, represents the enthalpy of formation of H2O because, as the definition suggests that the enthalpy of formation, is the heat evolved or absorbed when one mole of, substance is formed from its constituent atoms., 128. (b) For the equation, B 2 H 6 ( g ) 3O 2 ( g ), , qp, , H, , 75.32, , J, K mol, , qp, , 75.32, , J, K mol, , J, T2 373K, K, –3.947 K = T2 –373 K, T2 = 369.05 K, , – 291.30 J = 75.32, , Cp dT, , qp, , 131. (a), , I 2 ( s ) Cl2 ( g ), , [ H(I2(s) I2(g)) + HI–I + HCl–Cl] – [ HI –, ], Cl, = 151.0 + 242.3 + 62.76 –2 × 211.3 = 33.46, Df H°(ICl) =, , 132. (a), , 50, , J, 10 4 moles, mol, = 753.2 × 103 J = 753.2 kJ, H for ATP = 7 kcal / mol, = 7 × 4.184 kJ/mol, = 29.2 kJ/mol, 6.022 × 1023 molecules of ATP produce = 29.2 kJ, 29.2 kJ produced from 6.022 × 1023 molecules, , or H° – T S° < 0, T>, , o, , qp = Cp, m dT, J, 18g, = 75.32, 373 298 K, K mol 18g / mol, , J, × 75 K, K, = 5.649 × 103 J, , = 75.32, , 8.314, , J, 310 lnK eq, K mol, , K eq 2.6 108, 133. (b), G= H– T S, G = – T S (when H = 0 and S = +ve), G = –ve, 134. (a) For spontaneous reaction, dS > 0 and dG should be, negative i.e. < 0., 135. (a), G° = H° – T S°, For a spontaneous reaction G° < 0, , = 1.5 × 1025 molecules, 130. (a) 18gm of water at 100oC, 10gm of Cu at 25oC is added., , 100 C, , kJ, mol, , 19.39 = lnKeq, , 75.32, , 75.8, 29.2, , 33.46, = 16.73 kJ / mol, 2, , RT ln K eq : Normal body temperature = 37oC, , G, , 180 103 g, 104 g moles, 18g / mol, , 753.2 kJ produced from 6.022 × 1023 ×, , 2ICl( g ), , rH =, , 299 298 K, , For 180 kg of water, no. of moles of water, , qp, , 10g, (373 – 298) K, 63g / mol, , = 291.3 J, Heat lost by water = 291.30 J, , B 2 O 3 ( g ) 3H 2 O( g ), , Eqs. (i) + 3 (ii) + 3 (iii) – (iv), H = – 1273 + 3(–286) + 3(44) – 36, = – 1273 – 858 + 132 – 36, = – 2035 kJ/mol, 129. (a), , J, K mol, , = 24.47, , 136. (d), , H 2O, , 179.3 103, 160.2, 1atm, , T, , 1117.9K, , 1118K, , H2O(g), , H = 40630 J mol –1, S = 108.8 JK–1 mol –1, G = H T S When, H T S =0, T =, , H, S, , G = 0,, , 40630 J mol 1, H, = 373.4 K., =, S, 108.8 J mol 1
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7, EQUILIBRIUM, FACT / DEFINITION TYPE QUESTIONS, 1., , 2., , 3., , 4., , 5., , Which of the following is not a general characteristic of, equilibria involving physical processes ?, (a) Equilibrium is possible only in a closed system at a, given temperature., (b) All measurable properties of the system remain, constant., (c) All the physical processes stop at equilibrium., (d) The opposing processes occur at the same rate and, there is dynamic but stable condition., The liquid which has a ………..vapour pressure is more, volatile and has a ………….boiling point., (a) Higher , higher, (b) Lower, lower, (c) Higher, lower, (d) Lower, higher, Boiling point of the liquid depends on the atmospheric, pressure. It depends on the altitude of the place; at high, altitude the boiling point………….., (a) increases, (b) decreases, (c) either decreases or increases, (d) remains same, In an experiment three watch glasses containing separately, 1mL each of acetone, ethyl alcohol, and water are exposed, to atmosphere and the experiment with different volumes of, the liquids in a warmer room is repeated, it is observed that, in all such cases the liquid eventually disappears and the, time taken for complete evaporation in each case was, different. The possible reason is/are, (a) the nature of the liquids is different, (b) the amount of the liquids is different, (c) the temperature is different, (d) All of the above, A small amount of acetone is taken in a watch glass and it is, kept open in atmosphere. Which statement is correct for, the given experiment?, (a) The rate of condensation from vapour to liquid state is, higher than the rate of evaporation., , 6., , 7., , 8., , 9., , (b) The rate of condensation from vapour to liquid state is, equal to the rate of evaporation., (c) The rate of condensation from vapour to liquid state is, much less than the rate of evaporation., (d) The rate of condensation from vapour to liquid state is, equal or less than the rate of evaporation., When pressure is applied to the equilibrium system, Ice, Water, Which of the following phenomenon will happen?, (a) More ice will be formed, (b) Water will evaporate, (c) More water will be formed, (d) Equilibrium will not be formed, A reaction is said to be in equilibrium when, (a) the rate of transformation of reactant to products is, equal to the rate of transformation of products to the, reactants., (b) 50% of the reactants are converted to products., (c) the reaction is near completion and all the reactants, are converted to products., (d) the volume of reactants is just equal to the volume of, the products., Which of the following is not true about a reversible, reaction?, (a) The reaction does not proceed to completion, (b) It cannot be influenced by a catalyst, (c) Number of moles of reactants and products is always, equal, (d) It can be attained only in a closed container, If the synthesis of ammonia from Haber's process is carried, out with exactly the same starting conditions (of partial, pressure and temperature) but using D2 (deuterium) in place, of H2. Then, (a) the equilibrium will be disturbed, (b) the composition of reaction mixture will remain same at, equilibrium., (c) Use of isotope in reaction will not produce ammonia., (d) At equilibrium rate of forward reaction will be greater, than the rate of reverse reaction
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EBD_7207, EQUILIBRIUM, , 112, , 10., , Consider the following graph and mark the correct statement., , (a), , HI, , H2+I2, , Time, , 11., , 12., , 13., , (c) K1 =, 14., , H 2 + I2, , 1, K2, , 1, K2, , 2, , 1, N2, 2, , 3, H2, 2, , A reaction is A + B, C + D. Initially we start with, equal concentrations of A and B. At equilibrium we find, that the moles of C is two times of A. What is the equilibrium, constant of the reaction?, 1, 1, (a), (b), 4, 2, (d) 2, (c) 4, , 18., , In A B, C . The unit of equilibrium constant is :, –1, (a) Litre mole, (b) Mole litre, (c) Mole litre–1, (d) No unit, For the reaction C(s) CO 2 (g), 2CO(g) , the partial, pressures of CO2 and CO are 2.0 and 4.0 atm respectively at, equilibrium. The Kp for the reaction is., (a) 0.5, (b) 4.0, (c) 8.0, (d) 32.0, In which of the following equilibrium Kc and Kp are not, equal?, , 19., , 20., , 2NH3 is K and for reaction, , K', (c) K ', (d) K, K, In the following equilibrium reaction, B + C,, 2A, the equilibrium concentrations of A, B and C are 1 × 10–3 M,, 2 × 10–3 M and 3 × 10–3 M respectively at 300 K. The value, of Kc for this equilibrium at the same temperature is, , (b), , SO 2 (g)+NO2 ( g ), , (c), , H 2 (g)+I 2 ( g ), , SO3 (g)+NO( g ), , 2HI(g), , (d), , 21., , NH3, the equilibrium constant is K ', K= K', , N 2 (g)+O 2 ( g ), , (a) 2 NO( g ), , (d) K1 = (K2)0, , The K and K ' will be related as:, (a) K × K ' = 1, (b), 15., , 1, (d) 36, 36, Given the reaction between 2 gases represented by A2 and, B2 to give the compound AB(g)., A2(g) + B2(g), 2 AB(g)., At equilibrium, the concentration, of A2 = 3.0 × 10–3 M, of B2= 4.2 × 10–3 M, of AB = 2.8 × 10–3 M, lf the reaction takes place in a sealed vessel at 527°C, then, the value of KC will be :, (a) 2.0, (b) 1.9, (c) 0.62, (d) 4.5, , 17., , (b) K1 = K22, , The equilibrium constant for the reversible reaction, N2 + 3H2, , 16., , Equal Time, , (a) Chemical equilibrium in the reaction, H 2 I2, 2HI, can be attained from other directions., (b) Equilibrium can be detained when H2 and I2 are mixed, in an open vessel., (c) The concentrations of H2 and I2 keep decreasing while, concentration of HI keeps increasing with time., (d) We can find out equilibrium concentration of H2 and, I2 from the given graph., What are the product formed when Deuterium is added, equilibrium reaction of H2 and I2 ?, (i) HD, (ii) DI, (iii) D2, (iv) HI, (a) (i), (ii) and (iv), (b) (i) and (ii), (c) (ii) and (iv), (d) All of these, If a system is at equilibrium, the rate of forward to the reverse, reaction is :, (a) less, (b) equal, (c) high, (d) at equilibrium, K1 and K2 are equilibrium constant for reactions (1) and (2), N2(g) + O2 (g), 2 NO (g) .............(1), 1, 1, N 2 (g ), O 2 (g ) ......(2), NO(g), 2, 2, Then,, (a) K1 =, , (b) 6, , (c), , HI, , Conc., , 1, 6, , 22., , 2CO2(g), 2C(s)+O 2 ( g ), For the following reaction in gaseous phase, 1, O 2 ( g ) CO 2 ( g ), K p / K c is, CO( g ), 2, (a) (RT)1/2, (b) (RT)–1/2, (c) (RT), (d) (RT)–1, The KP/KC ratio will be highest in case of, (a) CO (g), , 1, O (g), 2 2, , (b) H2 (g) + I2 (g), (c) PCl5 (g), , CO2 (g), 2HI (g), , PCl3(g) + Cl2 (g), , (d) 7H2 (g) + 2NO2 (g), , 2NH3 (g) + 4H2O(g)
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EQUILIBRIUM, , 23., , 24., , For a chemical reaction ;, A (g) + B ( ), D (g) + E (g), Hypothetically at what temperature, Kp = Kc, (when, R = 0.08 -atm/mole-K), (a) T = 0 K, (b) T = 1K, (c) T = 12.5 K, (d) T = 273 K, Steam reacts with iron at high temperature to give hydrogen, gas and Fe3O4 (s). The correct expression for the equilibrium, constant is, (a), , (c), 25., , 26., , 27., , 29., , PH2, , 2, , (b), , PH2 O, 2, (PH 2 ) 4 [Fe 3O 4 ], , (d), , 4, , (PH 2O ) [Fe], , (PH 2 ) 4, (PH 2O ) 4, , [Fe 3O 4 ], [Fe], , For the reaction C(s) + CO2(g) 2CO(g), Kp = 63 atm at 1000, K. If at equilibrium : Pco = 10 Pco2, then the total pressure, of the gases at equilibrium is, (a) 6.3 atm, (b) 6.93 atm, (c) 0.63 atm, (d) 0.693 atm, The rate constant for forward and backward reaction of, hydrolysis of ester are 1.1 10 2 and 1 .5 10 3 per minute, respectively. Equilibrium constant for the reaction, CH 3COOH C 2 H 5OH is, CH3COOC2 H 5 H, (a) 4.33, (b) 5.33, (c) 6.33, (d) 7.33, Value of KP in the reaction, , MgCO3(s), , MgO(s) CO 2(g ) is, , (a) KP = PCO, , 2, , (b), , KP, , (c), , KP, , (d), 28., , 113, , KP, , PCO 2, , PCO 2, , PMgO, , PMgCO 3, , PCO 2 PMgO, PMgCO 3, PMgO, , Which of the following is an example of homogeneous, equilibrium ?, (a), , 2SO2 (g) O2 (g), , (b), , C(s) H 2 O(g), , (c), , CaCO3 (s), , (d), , NH 4 HS(s), , CaCO 3 (s), , 2SO3 (g), CO(g) H 2 (g), , CaO(s) CO 2 (g), , NH3 (g) H 2S(g), Unit of equilibrium constant for the given reaction is, Ni(s) + 4CO(g), Ni (CO)4(g), –3, (a) (mol/l), (b) (mol/l)3, –4, (c) (mol/l), (d) (mol/l)4, , CaO(s) CO 2 (g), , For this reactions which of the following is/are true, (i) Kć = [CO2(g)], (ii) Kp = pCO2, (iii) [CaCO 3 (s)] and [CaO(s)] are both con stant, (iv) [ CO2(g)] is constant, (a) (i), (ii) and (iv), (b) (i), (ii) and (iii), (c) (ii) and (iv), (d) (i), (iii) and (iv), 31. In a reversible chemical reaction having two reactants in, equilibrium, if the concentration of the reactants are doubled, then the equilibrium constant will, (a) Also be doubled, (b) Be halved, (c) Become one-fourth, (d) Remain the same, 32. On doubling P and V with constant temperature the, equilibrium constant will, (a) remain constant, (b) become double, (c) become one-fourth, (d) None of these, 33. If for the reaction, N2 + 3H2, 2NH3, H = –92.38KJ/mole than what, happens if the temperature is increased?, (a) Reaction proceed forward, (b) Reaction proceed backward, (c) No effect on the formation of product, (d) None of these, 34. If Kc is in the range of ………………… appreciable, concentrations of both reactants and products are present., (a) 10–4 to 104, (b) 10 – 3 to 103, (c) 10+3 to 10 – 3, (d) 10 – 5 to 103, 35. The reaction quotient (Q) for the reaction, N 2 ( g ) 3H 2 ( g ), , is given by Q, , PMgCO 3, PCO 2, , 30. The thermal dissociation of calcium carbonate showing, heterogeneous equilibrium is, , 2NH 3 ( g ), , [ NH 3 ]2, [ N 2 ][ H 2 ] 3, , . The reaction will proceed from, , right to left if, (a) Q = 0, (b) Q = Kc, (c) Q < Kc, (d) Q > Kc, where Kc is the equilibrium constant, 36. The reaction quotient Q is used to, (a) predict the extent of a reaction on the basis of its, magnitude, (b) predict the direction of the reaction, (c) calculate equilibrium concentrations, (d) calculate equilibrium constant, 37. The correct relationship between free energy change in a, reaction and the corresponding equilibrium constant, Kc is, (a), G = RT ln Kc, (b) – G = RT ln Kc, (c), G° = RT ln Kc, (d) – G° = RT ln Kc
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EBD_7207, EQUILIBRIUM, , 114, , 38., , Usin g the equation ( K e G /RT ), the reaction, spontaneity can be interpreted in terms of the value of G°, is/are, (a) If G > 0, then – G /RT is positive, and e G /RT> 1, making K > 1, which implies a spontaneous reaction or, the reaction which proceeds in the forward direction, to such an extent that the products are present, predominantly., , 39., , 40., , 41., , 42., , (b) If G > 0, then – G /RT is negative, and e G /RT, < 1 making K < 1, which implies a non-spontaneous, reaction or a reaction which proceeds in the forward, direction to such a small degree that only a very minute, quantity of product is formed., (c) Both (a) and (b), (d) None of the above, Which of the following relation represents correct relation, between standard electrode potential and equilibrium, constant?, nFE, I., log K, 2.303 RT, nFE, e RT, , II., , K, , III., , log K, , 43., , N2 3H2, 2NH3 2 kcal, the favourable conditions are, (a) Low temperature, low pressure and catalyst, (b) Low temperature, high pressure and catalyst, (c) High temperature, low pressure and catalyst, (d) High temperature, high pressure and catalyst, , (b), , N 2 3H 2, , (c), , PCl5, , 2NH3, PCl3 Cl 2, , (d), , 44., , N2 O2, 2NO, The equilibrium which remains unaffected by pressure, change is, , (a), , N 2 (g) O2 (g), , (b), , 2SO2 (g) O2 (g), , (c), , 2O3 (g), , 2NO(g), , 2SO3 (g), , 3O2 (g), , (d), , 45., , 46., , nFE, 2.303 RT, , nFE, IV. log K 0.4342, RT, Choose the correct statement(s)., (a) I, II and III are correct, (b) II and III are correct, (c) I, II and IV are correct, (d) I and IV are correct, According to Le-chatelier’s principle, adding heat to a solid, liquid equilibrium will cause the, (a) temperature to increase, (b) temperature to decrease, (c) amount of liquid to decrease, (d) amount of solid to decrease, Which one of the following information can be obtained on, the basis of Le Chatelier principle?, (a) Dissociation constant of a weak acid, (b) Entropy change in a reaction, (c) Equilibrium constant of a chemical reaction, (d) Shift in equilibrium position on changing value of a, constraint, For the manufacture of ammonia by the reaction, , Which of the following reaction will be favoured at low, pressure ?, (a) H 2 I 2, 2HI, , N2 O4 (g), 2NO 2 (g), Suitable conditions for melting of ice :, (a) high temperature and high pressure, (b) high temperature and low pressure, (c) low temperature and low pressure, (d) low temperature and high pressure, In which of the following reactions, the equilibrium remains, unaffected on addition of small amount of argon at constant, volume?, (a) H2 (g) + I2 (g), 2HI (g), (b) PCl5 (g), , 47., , 48., , 49., , PCl3 (g) + Cl2 (g), , (c) N2 (g) + 3H2 (g), 2NH3 (g), (d) The equilibrium will remain unaffected in all, the three cases., Le-Chatelier principle is not applicable to, (a) H 2 ( g ) I 2 ( g ), 2HI( g ), (b), , Fe(s) S(s), , (c), , N 2 ( g ) 3H 2 ( g ), , (d), , N 2 ( g ) O2 ( g ), , FeS(s), , 2NH 3 ( g ), , 2NO( g ), , In an equilibrium reaction if temperature increases, (a) equilibrium constant increases, (b) equilibrium constant decreases, (c) any of the above, (d) no effect, In a two-step exothermic reaction, 3C(g), D(g), A2(g) + B2(g), Step 1, , Step 2, , Steps 1 and 2 are favoured respectively by, (a) high pressure, high temperature and low pressure, low, temperature, (b) high pressure, low temperature and low pressure, high, temperature, (c) low pressure, high temperature and high pressure, high, temperature, (d) low pressure, low temperature and high pressure, low, temperature
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EQUILIBRIUM, , 50., , 51., , 52., , 53., , 54., , 55., , 56., , 57., , 58., , What happens when an inert gas is added to an equilibrium, keeping volume unchanged?, (a) More product will form, (b) Less product will form, (c) More reactant will form, (d) Equilibrium will remain unchanged, In a vessel N2, H2 and NH3 are at equilibrium. Some helium, gas is introduced into the vessel so that total pressure, increases while temperature and volume remain constant., According to Le Chatelier’s principle, the dissociation of, NH3, (a) increases, (b) decreases, (c) remains unchanged, (d) equilibrium is disturbed, Effect of a catalyst on a equilibrium reaction., (i) A catalyst increases the rate of the chemical reaction, by making available a new low energy pathway for the, conversion of reactants to products., (ii) It increases the rate of forward and reverse reactions, that pass through the same transition state and does, not affect equilibrium., (iii) It lowers the activation energy for the forward and, reverse reactions by exactly the same amount., Which of the above statement(s) is/are correct ?, (a) Only (i), (b) (i) and (ii), (c) (i), (ii) and (iii), (d) (ii) and (iii), Which of the following is/are electrolytes?, (i) Sugar solution, (ii) Sodium chloride, (iii) Acetic acid, (iv) Starch solution, (a) (i) and (iv), (b) (ii) and (iv), (c) (ii) and (iii), (d) (i) and (iii), The geometry of hydronium ion is, (a) tetrahedral, (b) linear, (c) trigonal pyramidal, (d) trigonal planer, Which of the following statements are correct regarding, Arrhenius theory of acid and base?, (a) This theory was applicable to only aqueous solutions, (b) This theory was applicable to all solutions, (c) This theory could not explain the basicity of, substances like ammonia which do not possess a, hydroxyl group, (d) Both (a) and (c), Would gaseous HCl be considered as an Arrhenius acid ?, (a) Yes, (b) No, (c) Not known, (d) Gaseous HCl does not exist, A base, as defined by Bronsted theory, is a substance which, can, (a) lose a pair of electrons, (b) donate protons, (c) gain a pair of electrons, (d) accept protons, BF3 is an acid according to, (a) Arrhenius concept, (b) Bronsted-Lowry concept, (c) Lewis Concept, (d) Both (b) and (c), , 115, , 59. Which of the following can act as both Bronsted acid and, Bronsted base?, (a) Na2CO3, (b) OH–, –, (d) NH3, (c) HCO3, 60. Conjugate acid of NH 2 is :, , 61., , 62., , 63., , 64., , 65., , 66., , (b) NH3, (a) NH4+, (c) NH2, (d) NH, Among boron trifluoride, stannic chloride and stannous, chloride, Lewis acid is represented by, (a) only stannic chloride, (b) boron trifluoride and stannic chloride, (c) boron trifluoride and stannous chloride, (d) only boron trifluoride, Which of the following molecules acts as a Lewis acid ?, (a) (CH3)2 O, (b) (CH3)3 P, (c) (CH3)3 N, (d) (CH3)3 B, Which one of the following molecular hydrides acts as a, Lewis acid?, (a) NH3, (b) H2O, (c) B2H6, (d) CH4, Which of these is least likely to act as Lewis base?, (a) F–, (b) BF3, (c) PF3, (d) CO, Which one of the following is the correct statement ?, (a) HCO3– is the conjugate base of CO32–., (b) NH2– is the conjugate acid of NH3., (c) H2SO4 is the conjugate acid of HSO4–., (d) NH3 is the conjugate base of NH2–., Water is well known amphoprotic solvent. In which chemical, reaction water is behaving as a base?, (a) H 2SO 4 H 2 O, H 3 O + HSO 4 –, (b), (c), , H 2O, , H2O, , H2O NH2 –, , H3O +, , OH –, , NH3 OH –, , H 2 O NH 3, NH 4, OH –, An acid/ base dissociation equilibrium is dynamic involving, a transfer of proton in forward and reverse directions. Now, with, passage of time in which direction equilibrium is favoured ?, (a) in the direction of stronger base and stronger acid, (b) in the direction of formation of stronger base and, weaker acid, (c) in the direction of formation of weaker base and weaker, acid, (d) in the direction of formation of weaker base and, stronger acid, Three reactions involving H2PO4– are given below:, (i) H3PO4 + H2, H3O+ + H2PO4–, –, (ii) H2PO4 + H2O HPO42– + H3O+, (iii) H2PO4– + OH– H3PO4 + O2–, In which of the above does H 2 PO 4 act as an acid ?, , (d), , 67., , 68., , (a) (ii) only, (c) (iii) only, , (b) (i) and (ii), (d) (i) only
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EBD_7207, EQUILIBRIUM, , 116, , 69., , 70., , 71., , 72., , The value of the ionic product of water, (a) depends on volume of water, (b) depends on temperature, (c) changes by adding acid or alkali, (d) always remains constant, A base when dissolved in water yields a solution with a, hydroxyl ion concentration of 0.05 mol litre–1. The solution, is, (a) basic, (b) acidic, (c) neutral, (d) either (b) or (c), pH scale was introduced by :, (a) Arrhenius, (b) Sorensen, (c) Lewis, (d) Lowry, pH of solution is defined by expression, (a), , (c), 73., , 74., , 75., , 76., , 77., , 78., , 79., , 80., , 81., , log [H ], 1, log [H ], , (b), , (d), , log, , 1, H, , 1, log [H ], , The pH of a 10–3 M HCl solution at 25°C if it is diluted 1000, times, will be –, (a) 3, (b) zero, (c) 5.98, (d) 6.02, How many litres of water must be added to 1 litre an aqueous, solution of HCl with a pH of 1 to create an aqueous solution, with pH of 2 ?, (a) 0.1 L, (b) 0.9 L, (c) 2.0 L, (d) 9.0 L, What is the approximate pH of a 1 × 10–3 M NaOH solution?, (a) 3, (b) 11, (c) 7, (d) 1 × 10–11, Calculate the pOH of a solution at 25°C that contains, 1× 10– 10 M of hydronium ions, i.e. H3O+., (a) 4.000, (b) 9.0000, (c) 1.000, (d) 7.000, The pH value of a 10 M solution of HCl is, (a) less than 0, (b) equal to 0, (c) equal to 1, (d) equal to 2, +, What is the H ion concentration of a solution prepared by, dissolving 4 g of NaOH (Atomic weight of Na = 23 amu) in, 1000 ml?, (a) 10–10 M, (b) 10–4 M, –1, (c) 10 M, (d) 10–13 M, Calculate the pOH of a solution at 25°C that contains 1× 10– 10, M of hydronium ions, i.e. H3O+., (a) 4.000, (b) 9.0000, (c) 1.000, (d) 7.000, The pH of 0.005 molar solution of H 2 SO 4 is approximately:, (a) 0.010, (b) 1, (c) 2, (d) 0.005, Which solution has pH equal to 10 ?, (a) 10–4 M KOH, (b) 10–10 M KOH, –10, (c) 10 M HCl, (d) 10–4 M HCl, , 82., , Which of the following has highest pH ?, (a), , M, KOH, 4, , (b), , M, NaOH, 4, , M, M, Ca (OH) 2, (d), NH 4 OH, 4, 4, 83. A weak acid, HA, has a Ka of 1.00 × 10–5. If 0.100 mole of, this acid dissolved in one litre of water, the percentage of, acid dissociated at equilbrium is closest to, (a) 1.00%, (b) 99.9%, (c) 0.100%, (d) 99.0%, 84. A monobasic weak acid solution has a molarity of 0.005 and, pH of 5. What is the percentage ionization in this solution?, (a) 2.0, (b) 0.2, (c) 0.5, (d) 0.25, 85. Calculate the pH of a solution obtained by diluting 1 mL of, 0.10 M weak monoacidic base to 100 mL at constant, temperature if Kb of the base is 1 × 10–5 ?, (a) 8, (b) 9, (c) 10, (d) 11, 86. The ionisation constant of an acid, Ka, is the measure of, strength of an acid. The K a values of acetic acid,, hypochlorous acid and formic acid are 1.74 × 10–5, 3.0 × 10–, 8 and 1.8 × 10–4 respectively. Which of the following orders, of pH of 0.1 mol dm–3 solutions of these acids is correct?, (a) acetic acid > hypochlorous acid > formic acid, (b) hypochlorous acid > acetic acid > formic acid, (c) formic acid > hypochlorous acid > acetic acid, (d) formic acid > acetic acid > hypochlorous acid, 87. The first and second dissociation constants of an acid H2 A, are 1.0 × 10–5 and 5.0 × 10–10 respectively. The overall, dissociation constant of the acid will be, (a) 0.2 × 105, (b) 5.0 × 10–5, 15, (c) 5.0 × 10, (d) 5.0 × 10–15., 88. Equimolar solutions of HF, HCOOH and HCN at 298 K have, the values of Ka as 6.8 × 10–4 and 4.8 × 10–9 respectively., What is the observed trend of dissociation constants in, successive stages ?, (a) HF > HCN > HCOOH (b) HF > HCOOH > HCN, (c) HCN > HF > HCOOH (d) HCOOH > HCN > HF, 89. At 25 C, the dissociation constant of a base, BOH, is, 1.0 10 12 . The concentration of hydroxyl ions in, 0.01 M aqueous solution of the base would be, (a) 1.0 10 5 mol L 1 (b) 1.0 10 6 mol L 1, (c) 2.0 10 6 mol L 1, (d) 1.0 10 7 mol L 1, 90. Which of the following pKa value represents the strongest, acid ?, (a) 10–4, (b) 10–8, –5, (c) 10, (d) 10–2, 91. The dissociation constant of two acids HA1 and HA2 are, 3.14 × 10– 4 and 1.96 × 10– 5 respectively. The relative, strength of the acids will be approximately, (a) 1 : 4, (b) 4 : 1, (c) 1 : 16, (d) 16 : 1, , (c)
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EQUILIBRIUM, , 92., , 117, , Given, HF H 2 O, , Ka, , H 3O, , F, , Kb, , HF OH, , H 2O, , F, , 100., , Which of the following reaction is correct, 1, (a) Kb = Kw, (b) Kb, Kw, Ka, Kw, Kb, At 298K a 0.1 M CH3COOH solution is 1.34% ionized. The, ionization constant Ka for acetic acid will be, (a) 1.82 × 10– 5, (b) 18.2 × 10 –5, (c) 0.182 × 10 –5, (d) None of these, For dibasic acid correct order is, (a) K a1 K a 2, (b) K a1 K a 2, (c) Ka × Kb = Kw, , 93., , 94., , 95., , (d), , (c) K a1 K a 2, (d) not certain, For a polybasic acid, the dissociation constants have a, different values for each step, e.g.,, H 3A, , H, , H2A ; K, , 101., , 102., , K a1, , H 2A, , H, , HA 2 ; K, , HA 2, , H, , A3 ; K, , Ka2, Ka3, , What is the observed trend of dissociation constants in, successive stages ?, (a) K a1 K a 2 K a 3, (b) K a1 K a 2 K a 3, (c), 96., , K a1, , Ka2, , K a3, , (d), , K a1, , Ka2 Ka3, , 103., , K a1 , Ka2 and K a3 are the respective ionisation constants, , for the following reactions., , H 2S, , H, , HS, , HS, , H, , S2, , H 2S, , 2H, , S2, , 104., , The correct relationship between K a1 , Ka2 and Ka3 is, , 97., , 98., , 99., , (a), , K a3, , K a1, , K a2, , (b), , K a3, , K a1, , K a2, , (c), , K a3, , K a1, , K a2, , (d), , K a3, , K a1 / K a2, , Cationic hydrolysis gives the following solution:, (a) acidic, (b) basic, (c) neutral, (d) amphoteric, In qualitative analysis, in III group NH4Cl is added before, NH4OH because, (a) to increase the concentration of NH4+ions, (b) to increase concentration of Cl– ions, (c) to reduce the concentration of OH– ions, (d) to increase concentration of OH– ions, The solubility of AgI in NaI solution is less than that in pure, water because :, (a) the temperature of the solution decreases, , 105., , 106., , 107., , (b) solubility product to AgI is less than that of NaI, (c) of common ion effect, (d) AgI forms complex with NaI, When sodium acetate is added to an aqueous solution of, acetic acid :, (a) The pH of the solution decreases, (b) The pH of the solution increases, (c) The pH of the solution remains unchanged, (d) An acid salt is produced, Which of the following statements about pH and H+ ion, concentration is incorrect?, (a) Addition of one drop of concentrated HCl in NH4OH, solution decreases pH of the solution., (b) A solution of the mixture of one equivalent of each of, CH3COOH and NaOH has a pH of 7, (c) pH of pure neutral water is not zero, (d) A cold and concentrated H2SO4 has lower H+ ion, concentration than a dilute solution of H2SO4, H2 S gas when passed through a solution of cations, containing HCl precipitates the cations of second group, of qualitative analysis but not those belonging to the, fourth group. It is because, (a) presence of HCl decreases th e sulph ide ion, concentration., (b) solubility product of group II sulphides is more than, that of group IV sulphides., (c) presence of HCl increases th e sulphide ion, concentration., (d) sulphides of group IV cations are unstable in HCl., A salt ‘X’ is dissolved in water of pH = 7. The salt is made, resulting solution becomes alkaline in nature. The salt is, made, (a) A strong acid and strong base, (b) A strong acid and weak base, (c) A weak acid and weak base, (d) A weak acid and strong base, Aqueous solution of ferric chloride is acidic due to, (a) ionization, (b) polarization, (c) dissociation, (d) hydrolysis, The pKa of a weak acid, HA, is 4.80. The pKb of a weak base,, BOH, is 4.78. The pH of an aqueous solution of the, correspondng salt, BA, will be, (a) 9.58, (b) 4.79, (c) 7.01, (d) 9.22, The pKa of a weak acid (HA) is 4.5. The pOH of an aqueous, buffer solution of HA in which 50% of the acid is ionized is, (a) 7.0, (b) 4.5, (c) 2.5, (d) 9.5, A buffer solution is prepared in which the concentration of, NH3 is 0.30M and the concentration of NH4+ is 0.20 M. If, the equilibrium constant, Kb for NH3 equals 1.8 × 10–5,, what is the pH of this solution ? (log 2.7 = 0.433)., (a) 9.08, (b) 9.43, (c) 11.72, (d) 8.73
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EBD_7207, EQUILIBRIUM, , 118, , [H+], , in mol/L of a solution that is 0.20 M in, 108. What is, CH3COONa and 0.10 M in CH3COOH ? Ka for CH3COOH, = 1.8 × 10–5., (a) 3.5 × 10, , 4, , (b) 1.1 × 10, , 5, , (c) 1.8 × 10 5, (d) 9.0 × 10 6, 109. Which of the following pairs constitutes a buffer?, (a) NaOH and NaCl, (b) HNO3 and NH4NO3, (c) HCl and KCl, (d) HNO2 and NaNO2, 110. Buffer solutions have constant acidity and alkalinity, because, (a) these give unionised acid or base on reaction with, added acid or alkali., (b) acids and alkalies in these solutions are shielded from, attack by other ions., (c) they have large excess of H+ or OH– ions, (d) they have fixed value of pH, 111. The buffering action of an acidic buffer is maximum when, its pH is equal, (a) 5, (b) 7, (c) 1, (d) pKa, 112. When a buffer solution, sodium acetate and acetic acid is, diluted with water :, (a) Acetate ion concentration increases, (b) H+ ion concentration increases, (c) OH– ion conc. increases, (d) H+ ion concentration remains unaltered, 113. The product of ionic concentration in a saturated solution, of an electrolyte at a given temperature is constant and is, known as, (a) Ionic product of the electrolyte, (b) Solubility product, (c) Ionization constant, (d) Dissociation constant, 114. The Ksp for Cr(OH)3 is 1.6 × 10–30. The solubility of this, compound in water is :, (a), , 4, , 1.6 10, , 30, , (b), , 4, , 1.6 10, , 30, , / 27, , (c) 1.6 10 30/ 27, (d), 1.6 10 30, 115. At 25°C, the solubility product of Mg(OH)2 is 1.0 × 10–11., At which pH, will Mg2+ ions start precipitating in the form, of Mg(OH)2 from a solution of 0.001 M Mg2+ ions?, (a) 9, (b) 10, (c) 11, (d) 8, 116. pH of a saturated solution of Ba(OH)2 is 12. The value of, solubility product (Ksp) of Ba(OH)2 is :, (a) 3.3 × 10– 7, (b) 5.0 × 10–7, –6, (c) 4.0 × 10, (d) 5.0 × 10–6, 117. If s and S are respectively solubility and solubility product, of a sparingly soluble binary electrolyte then :, (a) s = S, (b) s = S2, 1, (c) s S1/ 2, (d) s, S, 2, 118. Why only As+3 gets precipitated as As2S3 and not Zn+2 as, ZnS when H2S is passed through an acidic solution, containing As+3 and Zn+2?, , (a) Solubility product of As2S3 is less than that of ZnS, (b) Enough As+3 are present in acidic medium, (c) Zinc salt does not ionise in acidic medium, (d) Solubility product changes in presence of an acid, 119. Solid Ba(NO 3 ) 2 is gradually dissolved in a, 1.0 × 10–4 M Na2CO3 solution. At which concentration of, Ba2+, precipitate of BaCO3 begins to form ? (Ksp for BaCO3, = 5.1 × 10–9), (a) 5.1 × 10–5 M, (b) 7.1 × 10–8 M, –5, (c) 4.1 × 10 M, (d) 8.1 × 1–7 M, 120. Solubility product of silver bromide is 5.0 × 10–13. The, quantity of potassium bromide (molar mass taken as, 120 g mol–1) to be added to 1 litre of 0.05 M solution of, silver nitrate to start the precipitation of AgBr is, (a) 1.2 × 10–10 g, (b) 1.2 × 10–9 g, (c) 6.2 × 10–5 g, (d) 5.0 × 10–8 g, 121. At 25°C, the solubility product of Mg(OH) 2 is, 1.0 × 10–11. At which pH, will Mg2+ ions start precipitating, in the form of Mg(OH) 2 from a solution of, 0.001 M Mg2+ ions?, (a) 9, (b) 10, (c) 11, (d) 8, , STATEMENT TYPE QUESTIONS, 122. Read the following statements carefully and choose the, correct answer, (i) Water and water vapour remain in equilibrium position, at atmospheric pressure (1.013 bar) and at 100°C in a, closed vessel., (ii) The boiling point of water is 100°C at 1.013 bar pressure, (iii) Boiling point of the liquid depends on the atmospheric, pressure., (iv) Boiling point depends on the altitude of the place; at, high altitude the boiling point increases., (a) (i), (ii) and (iv) are correct, (b) (i), (iii) and (iv), (c) (i), (ii) and (iii) are correct, (d) only (iii) is correct, 123. You must have seen that when a soda water bottle is opened,, some of the carbon dioxide gas dissolved in it fizzes out, rapidly. There is equilibrium between the molecules in the, gaseous state and the molecules dissolved in the liquid, under pressure i.e.,, CO2(gas), CO2(in solution), Which of the following statements is/are correct regarding, this?, (i) The phenomenon arises due to difference in solubility, of carbon dioxide at different pressures., (ii) This equilibrium is governed by Henry's law., (iii) The amount of CO2 gas dissolved in liquid increases, with decrease of temperature., (iv) The amount of CO2 gas dissolved in liquid decreases, with increase of temperature., (a) (i), (ii) and (iv) are correct, (b) (i) , (iii) and (iv), (c) (i), (ii) and (iii) are correct, (d) only (iii) is correct
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EQUILIBRIUM, , 124. Identify the CORRECT statements below regarding chemical, equilibrium:, (i) All chemical reactions which are in equilibrium are, irreversible., (ii) Equilibrium is achieved when the forward reaction rate, equals the reverse reaction rate., (iii) Equilibrium is achieved when the concentrations of, reactants and product remain constant., (iv) Equilibrium is dynamic in nature, (a) (i), (ii) and (iv) are correct, (b) (i), (ii), (iii) and (iv) are correct, (c) (i), (ii) and (iii) are correct, (d) only (ii) is correct, 125. Nobel gas is added to a reaction at equilibrium involving, gaseous reactant and gaseous product., Which of the following statement is true for above reaction?, Statement 1 : Reaction will proceed forward, as total, pressure has increased due to addition of Nobel gas., Statement 2 : Reaction will proceed backward, if Nobel gas, react with reactant., (a) Statement 1 and 2 are both correct., (b) Statement 1 is correct but statement 2 is incorrect., (c) Statement 1 is incorrect but statement 2 is correct., (d) Statement 1 and 2 both are incorrect., 126 Read the following statements and choose the correct, option., (i) The value of equilibrium constant is independent of, initial concentrations of the reactants and products., (ii) Equilibrium constant is temperature dependent, (iii) The equilibrium constant for the reverse reaction is, equal to the inverse of the equilibrium constant for the, forward reaction., (iv) The equilibrium constant for the reverse reaction is, equal to the equilibrium constant for the forward, reaction., (a) (i), (ii) and (iv) are correct, (b) (i), (iii) and (iv), (c) (i), (ii) and (iii) are correct, (d) only (iii) is correct, 127. Read the following statements and choose the correct option, (i) The numerical value of the equilibrium constant for a, reaction indicates the extent of the reaction., (ii) An equilibrium constant give information about the, rate at which the equilibrium is reached., (iii) If Kc > 103, products predominate over reactants, i.e.,, if Kc is very large, the reaction proceeds nearly to, completion., (iv) If Kc < 10–3, reactants predominate over products,, i.e., if Kc is very small, the reaction proceeds rarely., (a) (i), (ii) and (iv) are correct, (b) (i) , (iii) and (iv), (c) (i), (ii) and (iii) are correct, (d) only (iii) is correct, , 119, , 128. Which of the following statement(s) is/are correct ?, (i), G is negative, then the reaction is spontaneous and, proceeds in the forward direction., (ii), G is positive, then reaction is non-spontaneous, (iii) G is 0, then reaction is at equilibrium, (a) (i), (ii) and (iii) are correct, (b) (i) and (ii), (c) (ii) and (iii) are correct, (d) only (iii) is correct, 129. Read the following statements and choose the correct option, (i) Most of the acids taste sour, (ii) Acids turns blue litmus paper into red, (iii) Bases turns red litmus paper blue, (iv) Bases taste bitter and feel soapy, (a) (i), (ii) and (iv) are correct, (b) (i), (iii) and (iv), (c) (i), (ii) and (iii) are correct, (d) All statements are correct, 130. Which of the following statements are correct ?, (i) Strong acids have very weak conjugate bases, (ii) Weak acids have very strong conjugate bases, (iii) Strong bases have strong conjugate acids, (iv) Weak bases have weak conjugate acids, (a) (i) and (ii), (b) (i) and (iii), (c) (ii) and (iv), (d) (iii) and (iv), 131. Which of the following statement(s) is/are correct ?, (i) Water has ability to act both as an acid and a base, (ii) In pure water one H2O molecule donate proton and, acts an acid and another water molecule accepts a, proton and acts as a base., (a) Both (i) and (ii), (b) Neither (i) nor (ii), (c) Only (i), (d) Only (ii), 132. Which of the following statements are correct ?, (i) Ionic product of water (Kw) = [H+] [OH–] = 10–14M2, (ii) At 298K [H+] = [OH–] = 10–7, (iii) Kw does not depends upon temperature, (iv) Molarity of pure water = 55.55M, (a) (i), (ii) and (iii), (b) (i), (ii) and (iv), (c) (i) and (iv), (d) (ii) and (iii), 133. Read the following statements and choose the correct option, (i) Ka (ionization constant) is a measure of the strength, of the acids, (ii) Smaller the value of Ka, the stronger is the acid, (iii) Ka is a dimensionless quantity, (a) Statements (i) and (ii) are correct, (b) Statements (ii) and (iii) are correct, (c) Statements (i), (ii) and (iii) are correct, (d) Statements (i) and (iii) are correct, 134. Which of the following statement(s) is/are correct ?, (i) In a tribasic acid 2nd and 3rd (Ka , Ka ) ionization, 3, 2, constants are smaller than the first ionisation (Ka ), 1, (ii) It is difficult to remove a positively charged proton, from a negative ion due to electrostatic force., (a) Both (i) and (ii), (b) Neither (i) nor (ii), (c) Only (i), (d) Only (ii)
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EBD_7207, EQUILIBRIUM, , 120, , 135. Which of the following statements are correct ?, (i) The extent of dissociation of an acid depends on the, strength and polarity of the H –– A bond (where A is, an electronegative element.), (ii) As the strength of H–A bond increases, the energy, required to break the bond decreases., (iii) As the electronegativity difference between the atoms, H and A increases, acidity increases, (a) (i) and (ii), (b) (ii) and (iii), (c) (i) and (iii), (d) (i), (ii) and (iii), , MATCHING TYPE QUESTIONS, 136. Match the columns, Column-I, Column-II, (A) H2O (l), H2O (vap) (p) rate of melting = rate of, freezing, (B) I2 (solid), (q) rate of evaporation, I2 (vapour), = rate of condensation, (C) Ice, (r) rate of sublimation= rate, water, of condensation, (a) A – (p), B – (q), C – (r), (b) A – (r), B – (q), C – (p), (c) A – (p), B – (r), C – (q), (d) A – (q), B – (r), C – (p), 137. Match the Column-I with Column-II and mark the appropriate, choice., Column-I, Column-II, (A) Liquid, Vapour, (p) Saturated solution, (B) Solid, (q) Boiling point, Liquid, (C) Solid, Vapour, (r) Sublimation point, Solute, (s) Melting point, (D) Solute (s), (solution), (a) A – (p) ; B – (r) ; C – (q) ; D – (s), (b) A – (q) ; B – (s) ; C – (r) ; D – (p), (c) A – (s) ; B – (q) ; C – (p) ; D – (r), (d) A – (r) ; B – (s) ; C – (q) ; D – (p), 138. Match the columns., Column-I, Column-II, (Reactions), (Effect of increase in, pressure), 2HI(g) (p) Reaction proceed, (A) H2(g) + I2(g), backward., 1, O (g), 2 2, CO2(g), , (B) CO(g) +, , (C) N2O4(g), (a), (b), (c), (d), , (q) No effect on, , 2NO2(g) (r), , A – (q), B – (r), C – (p), A – (r), B – (q), C – (p), A – (p), B – (r), C – (q), A – (q), B – (p), C – (r), , reaction., Reaction proceed, forward, , 139. Match the columns :, Column-I, Column-II, (A) N2(g) + 3H2(g), (p), n>0, 2NH3(g) (t = 300ºC), (B) PCl5(g), PCl3(g), (q) KP < KC, + Cl2(g) (t = 50ºC), (C) C(s) + H2O(g), (r) KP not defined, CO(g) + H2(g), (D) CH3COOH(l) + C2H5OH(l) (s), n=1, CH3COOC2H5(l), + H2O(l), (a) A – (q), B – (p), C – (s), D – (r), (b) A – (p), B – (q), C – (r), D – (s), (c) A – (r), B – (p), C – (s), D – (q), (d) A – (s), B – (q), C – (p), D – (r), 140. Match the columns :, Column-I, Column-II, (A) For the equilibrium NH4I(s) (p) Forward shift, NH3(g) + HI(g),, if pressure is increased, at equilibrium, (B) For the equilibrium, (q) No change, N2 + 3H2, 2NH3, If volume is increased, at equilibrium, (C) For the equilibrium, (r) Backward shift, H2O(g) + CO(g), H2(g) + CO2 (g) inert gas is, added at constant pressure, at equilibrium, (D) For the equilibrium, (s) More N2 and H2 is, PCl5, formed., PCl3 + Cl2, what happens if more, PCl5 is added, (a) A – (p), B – (q), C – (r), D – (s), (b) A – (r), B – (s), C – (q), D – (p), (c) A – (s), B – (p), C – (q), D – (r), (d) A – (q), B – (s), C – (r), D – (p), 141. Match the columns, Column-I, Column-II, (A) Qc < Kc,, (p) Net reaction goes from, right to left., (B) Qc > Kc,, (q) Net reaction goes from, left to right., (C) Qc = Kc,, (r) No net reaction occurs., (a) A – (p), B – (q), C – (r), (b) A – (r), B – (q), C – (p), (c) A – (p), B – (r), C – (q), (d) A – (q), B – (p), C – (r)
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EQUILIBRIUM, , 142. Match the columns, Column-I, (A) Hydrochloric acid, (B) Acetic acid, (C) Citric and ascorbic, acids, (D) Tartaric acid, , 121, , Column-II, (p) Lemon and orange, (q) Tamarind paste., (r) Digestive juice, , (s) Constituent of, vinegar, (a) A – (q), B – (r), C – (p), D – (s), (b) A – (r), B – (s), C – (p), D – (q), (c) A – (s), B – (p), C – (q), D – (r), (d) A – (r), B – (p), C – (s), D – (q), 143. Match the columns, Column-I, Column-II, (A) HClO4, (p) Strong base, (B) HNO2, (q) Strong acid, (r) Weak base, (C) NH2–, –, (s) Weak acid, (D) HSO4, (a) A – (s), B – (q), C – (p), D – (r), (b) A – (q), B – (s), C – (p), D – (r), (c) A – (r), B – (p), C – (q), D – (s), (d) A – (s), B – (q), C – (p), D – (r), , ASSERTION-REASON TYPE QUESTIONS, Directions : Each of these questions contain two statements,, Assertion and Reason. Each of these questions also has four, alternative choices, only one of which is the correct answer. You, have to select one of the codes (a), (b), (c) and (d) given below., (a) Assertion is correct, reason is correct; reason is a correct, explanation for assertion., (b) Assertion is correct, reason is correct; reason is not a, correct explanation for assertion, (c) Assertion is correct, reason is incorrect, (d) Assertion is incorrect, reason is correct., 144. Assertion : Kp can be less than, greater than or equal to Kc., Reason : Relation between Kp and Kc depends on the, change in number of moles of gaseous reactants and, products ( n)., 145. Assertion : If a volume is kept constant and an inert gas, such as argon is added which does not take part in the, reaction, the equilibrium remains undisturbed., Reason : It is because the addition of an inert gas at constant, volume does not change the partial pressure or the molar, concentrations of the substance involved in the reaction., 146. Assertion : Buffer system of carbonic acid and sodium, bicarbonate is used for the precipitation of hydroxides of, third group elements., Reason : It maintains the pH to a constant value, about 7.4., 147. Assertion : Addition of silver ions to a mixture of aqueous, sodium chloride and sodium bromide solution will first, precipitate AgBr rather than AgCl., Reason : Ksp of AgCl > Ksp of AgBr., , CRITICAL THINKING TYPE QUESTIONS, 148. N2(g) + 3H2(g), N2(g) + O2(g), H2(g) +, , 2NH3(g), K1, 2NO(g), K2, , 1, O (g), 2 2, , (1), (2), , H2O(g), K3, , (3), , The equation for the equilibrium constant of the reaction, 5, O2 (g), 2, of K1, K2 and K3 is :, , 2NH3(g) +, , (a), , 2NO(g) + 3H2O(g), (K4) in terms, , K1.K 2, K3, , (c) K1 K2 K3, , (b), , K1.K 23, K2, , (d), , K 2 .K 33, K1, , A, B and, 149. Two equilibria, AB, are, simultaneously, maintained in a, AB B, AB2, solution with equilibrium constants, K1 and K2 respectively., The ratio of [A+] to [AB2–] in the solution is, (a) directly proportional to [B–], (b) inversely proportional to [B–], (c) directly proportional to the square of [B–], (d) inversely proportional to the square of [B–], 150. Equilibrium constant (K) for the reaction, Ni(s) + 4CO(g), Ni(CO)4(g) can be written in terms of, (1) Ni(s) + 2CO2 (g) + 2C(s), Ni(CO)4(g);, equilibrium constant = K1., 2CO(g);, (2) CO2(g) + C(s), equilibrium constant = K2., What is the relation between K, K2 and K2 ?, (a) K = (K1)/(K2)2, (b) K = (K1 . K2), 2, (c) K = (K1) (K2), (d) K = K1/K2, 151. K1, K2 and K3 are the equilibrium constants of the following, reactions (I), (II) and (III) respectively:, (I), , N2 + 2O2, , 2NO2, , (II) 2NO2, , N2 + 2O2, , (III) NO2, , 1, N 2 + O2, 2, , The correct relation from the following is, (a), , K1, , (c), , K1, , 1, K2, K2, , 1, K3, K3, , (b), , K1, , 1, K2, , (d), , K1, , 1, K2, , 1, K3, , K3, , 2
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EBD_7207, EQUILIBRIUM, , 122, , 152. For the following three reactions a, b and c, equilibrium, constants are given:, (i) CO(g) + H 2O(g), , CO 2 (g) + H 2 (g); K1, , (ii) CH4 (g) + H2O(g), , CO(g) + 3H2 (g); K 2, , (iii) CH4 (g) 2H2O(g), , CO2 (g) 4H2 (g);K3, , (a), , K1 K 2, , (b), , K3, , K 2 K3, , K1, , (d) K3 .K 23 K12, (c) K3 = K1 K2, 153. The value of equilibrium constant of the reaction, 1, 1, H 2 (g), I 2 is 8.0, HI g, 2, 2, The equilibrium constant of the reaction, , H2 g, , I2 (g), , 2HI(g) will be:, , 1, 16, , (b), , (c) 16, , (d), , (a), , 1, 64, 1, 8, , 154. For the reversible reaction,, N2(g) + 3H2(g), 2NH3(g) at 500°C, the value of Kp is, 1.44 10 5 when partial pressure is measured in, atmospheres. The corresponding value of K C , with, concentration in mole litre–1, is, (a), , (c), , 5, , 1.44 10, , 0.082 500, 1.44 10, , 2, , (b), , 5, , 0.082 773 2, , (d), , 1.44 10, , 5, , 8.314 773, 1.44 10, , 2, , 2, , 155. Two moles of PCl5 were heated in a closed vessel of 2L. At, equilibrium 40% of PCl5 is dissociated into PCl3 and Cl2., The value of equilibrium constant is, (a) 0.53, (b) 0.267, (c) 2.63, (d) 5.3, 156. PCl5 is dissociating 50% at 250°C at a total pressure of, P atm. If equilibrium constant is Kp, then which of the, following relation is numerically correct ?, (a) Kp = 3P, (b) P = 3Kp, 2P, 2K P, (d) Kp =, (c) P =, 3, 3, 157. For the decomposition of the compound, represented as, NH 2 COONH 4 ( s ), 2NH 3 ( g ) CO 2 ( g ), the Kp = 2.9 × 10–5 atm3., If the reaction is started with 1 mol of the compound, the, total pressure at equilibrium would be :, (a) 1.94 × 10–2 atm, (b) 5.82 × 10–2 atm, –2, (c) 7.66 × 10 atm, (d) 38.8 × 10–2 atm, 158. The values of Kp1 and Kp2 for the reactions, , X, and A, , Y Z, 2B, , ...(1), ...(2), , 2AB2 (g), , 2AB(g) B2 (g), , The degree of dissociation is ‘x’ and is small compared to 1., The expression relating the degree of dissociation (x) with, equilibrium constant Kp and total pressure P is :, (a) (2Kp/P), (b) (2Kp/P) 1/3, 1/2, (c) (2Kp/P), (d) (Kp/P), 160. On increasing the pressure, the gas phase reaction proceed, forward to re-establish equilibrium, as predicted by applying, the Le Chatelier’s principle. Consider the reaction., N 2 (g) 3H 2 (g), 2NH3 (g), Which of the following is correct, if the total pressure at, which the equilibrium is established, is increased without, changing the temperature?, (a) K will remain same, (b) K will decrease, (c) K will increase, (d) K will increase initially and decrease when, pressure is very high, 161. The exothermic formation of ClF3 is represented by the, equation :, Cl 2 (g), , 5, , 0.082 773, , are in the ratio of 9 : 1. If degree of dissociation of X and A, be equal, then total pressure at equilibrium (1) and (2) are in, the ratio :, (a) 3 : 1, (b) 1 : 9, (c) 36 : 1, (d) 1 : 1, 159. The dissociation equilibrium of a gas AB2 can be represented, as :, , 3F2 (g), , 2ClF3 (g) ;, , H = – 329 kJ, Which of the following will increase the quantity of ClF3 in, an equilibrium mixture of Cl 2 , F2 and ClF3 ?, (a) Adding F2, (b) Increasing the volume of the container, (c) Removing Cl2, (d) Increasing the temperature, 162. When hydrochloric acid is added to cobalt nitrate solution, at room temperature, the following reaction takes place out, the reaction mixture becomes blue. On cooling the mixture it, becomes pink. On the basis of this information mark the, correct answer., CO H 2 O 6, , 3, , (aq) 4Cl (aq), , (pink), , COCl4, , (blue), , 2, , (aq) 6H 2 O(l ), , H > 0 for the reaction, (a), (b) H < 0 for the reaction, (c), H = 0 for the reaction, (d) The sign of H cannot be predicted on the, basis of this information.
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EQUILIBRIUM, , 123, , HS–, I–,, , 163. In, RNH2 and NH3, order of proton accepting, tendency will be, (a) I– > NH3 > RNH2 > HS–, (b) HS– > RNH2 > NH3 > I–, (c) RNH2 > NH3 > HS– > I–, (d) NH3 > RNH2 > HS– > I–, 164. Which equilibrium can be described as an acid-base reaction, using the Lewis acid-base definition but not using the, Bronsted-Lowry definition?, 2NH4+ + SO42–, , (a) 2NH3 + H2SO4, (b) NH3 + CH3COOH, , NH4+ + CH3COO–, , (c) H2O + CH3COOH, , H3O+ + CH3COO–, , (d) [Cu(H2O)4]2– + 4 NH3, , [Cu(NH3)4]2+ + 4H2O, , 165. Equal volumes of three acid solutions of pH 3, 4 and 5 are, mixed in a vessel. What will be the H+ ion concentration in, the mixture ?, (a) 1.11 × 10–4 M, (b) 3.7 × 10–4 M, (c) 3.7 × 10– 3 M, (d) 1.11× 10–3 M, –10, 166. The pH of 10 M NaOH solution is nearest to:, (a) 6, (b) – 10, (c) 4, (d) 7, 167. 100 mL of 0.04 N HCl aqueous solution is mixed with 100 mL, of 0.02 N NaOH solution. The pH of the resulting solution, is:, (a) 1.0, (b) 1.7, (c) 2.0, (d) 2.3, 168. Equal volumes of three acid solutions of pH 3, 4 and 5 are, mixed in a vessel. What will be the H+ ion concentration in, the mixture ?, (a) 1.11 × 10–4 M, (b) 3.7 × 10–4 M, (c) 3.7 × 10– 3 M, (d) 1.11× 10–3 M, 169. At 100°C the Kw of water is 55 times its value at 25°C. What, will be the pH of neutral solution? (log 55 = 1.74), (a) 6.13, (b) 7.00, (c) 7.87, (d) 5.13, 170. Ionisation constant of CH3 COOH is 1.7 × 10 –5 if, concentration of H+ ions is 3.4 × 10–4M, then find out initial, concentration of CH3COOH molecules, (a) 3.4 × 10–4M, (b) 3.4 × 10–3M, –3, (c) 6.8 × 10 M, (d) 6.8 × 10–4M, 171. Values of dissociation constant, Ka are given as follows :, Acid, Ka, HCN, 6.2 × 10–10, HF, 7.2 × 10–4, HNO2, 4.0 × 10–4, Correct order of increasing base strength of the base CN–,, _, F– and NO 2 will be :, (a), (c), , F, , F, , CN, , _, , _, , NO2, , NO2, , (b), , CN, , (d), , _, , NO2, _, NO2, , CN, F, F, CN, , 172. The dissociation constants for acetic acid and HCN at 25°C, are 1.5 × 10–5 and 4.5 × 10–10 respectively. The equilibrium, constant for the equilibrium, CN– + CH3COOH, HCN + CH3COO– would be:, (a) 3.0 × 10– 5, (b) 3.0 × 10– 4, 4, (c) 3.0 × 10, (d) 3.0 × 105, 173. If degree of dissociation of pure water at 100°C is, 1.8 × 10–8, then the dissociation constant of water will be, (density of H2O = 1 gm/cc), (a) 1 × 10–12, (b) 1 × 10–14, –12, (c) 1.8 × 10, (d) 1.8 × 10–14, 174. Ionisation of weak acid can be calculated by the formula, (a), , 100, , Ka, c, , (b), , 100, ( pK a pH ), , 1 10, (c) Both (a) and (b), (d) None of these, 175. Equimolar solutions of the following were prepared in water, separately. Which one of the solutions will record the highest, pH ?, (a) SrCl2, (b) BaCl2, (c) MgCl2, (d) CaCl2, 176. Solubility product constant (Ksp) of salts of types MX, MX2, and M3X at temperature T are 4.0 × 10–8, 3.2 × 10–14 and, 2.7 × 10–15, respectively. Solubilities (mol dm–3) of the salts, at temperature 'T' are in the order –, (a) MX > MX2 > M3X, (b) M3X > MX2 > MX, (c) MX2 > M3X > MX, (d) MX > M3X > MX2, 177. In qualitative analysis, the metals of Group I can be separated, from other ions by precipitating them as chloride salts. A, solution initially contains Ag+ and Pb2+ at a concentration, of 0.10 M. Aqueous HCl is added to this solution until the, Cl– concentration is 0.10 M. What will the concentrations, of Ag+ and Pb2+ be at equilibrium?, (Ksp for AgCl = 1.8 × 10–10, Ksp for PbCl2 = 1.7 × 10–5), (a) [Ag+] = 1.8 × 10–7 M ; [Pb2+] = 1.7 × 10–6 M, (b) [Ag+] = 1.8 × 10–11 M ; [Pb2+] = 8.5 × 10–5 M, (c) [Ag+] = 1.8 × 10–9 M ; [Pb2+] = 1.7 × 10–3 M, (d) [Ag+] = 1.8 × 10–11 M ; [Pb2+] = 8.5 × 10–4 M, 178. The solubility product (Ksp) of the following compounds, are given at 25°C., Compound, Ksp, AgCl, 1.1 × 10–10, AgI, 1.0 × 10–16, PbCrO4, 4.0 × 10–14, Ag2CO3, 8.0 × 10–12, The most soluble and least soluble compounds are, respectively., (a) AgCl and PbCrO4, (b) AgI and Ag2CO3, (c) AgCl and Ag2CO3, (d) Ag2CO3 and AgI, 179. What is the molar solubility of Fe(OH)3 if, Ksp = 1.0 ×10–38 ?, (a) 3.16×10–10, (b) 1.386×10–10, –9, (c) 1.45×10, (d) 1.12×10–11
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EBD_7207, EQUILIBRIUM, , 124, , FACT / DEFINITION TYPE QUESTIONS, 1., 4., 5., , 6., 7., , 8., 9., 10., 11., 12., 13., , (c), 2. (c), 3. (b), (d) The time taken for complete evaporation depends on, (i) the nature of the liquid, (ii) the amount of the liquid, and (iii) the temperature., (c) When the watch glass is open to the atmosphere, the, rate of evaporation remains constant but the molecules, are dispersed into large volume of the room. As a, consequence the rate of condensation from vapour to, liquid state is much less than the rate of evaporation., (c) Ice melts with a reduction in volume. So Increase in, pressure shifts equilibrium to water side, result in, melting of ice according to Lechatelier's principle., (a) A reaction is said to be in equilibrium when rate of, forward reaction is equal to the rate of backward, reaction., (c), (b) The reaction mixtures starting either with H2 or D2, reach equilibrium with the same composition, except, that D2 and ND3 are present instead of H2 and NH3., (a) Equilibrium can be attained by either side of the, reactions of equilibrium., (d) According to the idea of dynamic equilibrium there is, possibility of formation of all product., (b) At equilibrium, the rate of forward and backward, reactions is equal., (a) For reation (1), [NO]2, K1 =, [N 2 ][O2 ], , 16., , 17., , 14., , 18., , K, (b) Given reaction, 2A, , Kc, , B C, , [B][C], 3, , 3 10, , (10 3 ) 2, , 4.2 10, , 2x.2x, x.x, , (2.8) 2, 3 4.2, , 3, , 0.62, , C +D, 2x 2x, , 4, , (a) For A B, , C,, , Unit of K c, , mol, litre, , n 1 2, n, , mol, litre, , 1, 1, , = Litre mole–1, P 2 CO, ;K p, PCO 2, , 4 4, 2, , 19., , (c) Kp =, , 20., , The concentration of solids and liquids are taken as, unity., 2 CO2 ( g ), (d) 2C(s) +O 2 ( g ), , 21., , n =2–1=+1, Kc and Kp are not equal., (b) For a gaseous phase reaction Kp and Kc are related as, K c ( RT ), , 8; C(s) 1;, , ng, , For the given reaction,, 1, CO(g) + O2 ( g ) CO2(g), 2, ng = 1– (1 + 0.5) = – 0.5 or, , .... (i), , Kp = Kc ( RT ), or, 22., , Kp, Kc, , = ( RT ), , 1, 2, , 1, 2, 1, 2, , (c) Using the relation KP = KC. (RT) n, we get, KP, KC, , ( RT ) n, , KP, Thus K will be highest for the reaction having, C, , [A]2, 2 10, , 3, , A+ B, At eqb. x, x, Kc =, , 1, [N 2 ]½ [O 2 ]½, therefore K1 =, [NO], K 22, , (c) N2 + 3H2, 2NH3, K = [NH3]2 / [N2] [H2]3, 1, 3, N2, H2, NH3, 2, 2, , Kc, , 3 10, , Kp, , .... (ii), K ' [NH3 ]/[N 2 ]1/ 2 [H 2 ]3/ 2, Dividing equation (i) by equation (ii), we get K' =, 15., , (c), , [AB]2, [A 2 ][B2 ], , Kc, , 2AB, (2.8 10 3 ) 2, , Kc, , and for reaction (2), K2 =, , (c) A2 + B2, , 3, , =6, , highest value of n.
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EQUILIBRIUM, , 125, , The n values for various reactions are, , 1, 1, –, 2, 2, (b) n = 2 – (1 + 1) = 0, (c) n = (1 +1) – 1= 1, (d) n = (2 + 4) – (7 + 2) = – 3, Thus maximum value of n = 1, (a) n 1– 1, , 23., , 24., , n, , (c) As Kp = Kc RT g, Here ng = 1, So, Kp = Kc when RT = 1, Thus T = 12.5 K, (b) 3Fe(s) + 4H2O (steam), (p H 2 ) 4, , Kp =, 25., , (p H 2 O ), , 4, , Fe3O4 (s) + 4 H2 (g), , only gaseous products and reactants., , (b) C(s) + CO2(g), 2CO(g), Apply law of mass action,, (10PCO2 )2, , (PCO )2, or 63, PCO2, , KP, , PCO2, , (Given KP = 65) and PCO = 10PCO2, or 63, , 100(PCO2 ) 2, , PCO 2, , 63, 100, , PCO, , PCO2, , 0.63 atm, , PCO = 0.63 + 6.3 = 6.93 atm., , 0.4342, , (d) Rate constant of forward reaction (Kf) = 1.1 × 10–2 and, rate constant of backward reaction (Kb) = 1.5 × 10–3, per minute., Kf, Kb, , Equilibrium constant (Kc), 27., , (a), , 1.1 10, , 2, , 1.5 10, , 3, , Therefore K P, , PCO2, , 28., , (a) All the reactants and products are in same physical, state., , 29., , (a), , 30., , [Ni(CO) 4 ], , mol l, , 1, , ln K, , (mol l 1) 3, [CO]4, (mol l 1 )4, (b) For the reaction, CaCO3 (s), CaO (s) + CO2 (g), On the basis of the stoichiometric equation, we can, write,, Kc =[CaO(s)] [CO2(g)/[CaCO3(s)], , K, , nFE, RT, , ........ (i), , nFE, RT, nFE, , 7.33, , MgCO3 (s) MgO(s) CO 2 (g), MgO & MgCO3 are solid and they donot exert any, pressure and hence only pressure exerted is by CO2., , K, , Q K c i.e the reaction will be fast in backward, direction i.e r b > rf., 36. (b) The equilibrium constant helps in predicting the, direction in which a given reaction will proceed at any, stage. For this purpose, we calculate the reaction, quotient Q. The reaction quotient Q (Qc with molar, concentration and Qp with partial pressures) is defined, in the same way as the equilibrium constant Kc except, that the concentrations in Qc are not necessary, equilibrium values., 37. (d), 38. (c) Both (a) and (b) are correct for the equation,, K = e G /RT, 39. (c), G = –2.303 RT log K, –nFE° = –2.303 RT log K, nFE (I), log K, 2.303 RT, , 10PCO2 = 10 × 0.63 = 6.3 atm, , Ptotal = PCO2, 26., , or 63 100 PCO2, , Since [CaCO3 (s)] and [CaO(s)] are both constant,, therefore modified equilibrium constant for the thermal, decomposition of calcium carbonate will be, Kć = [CO2(g)], K p = [p CO2(g)], 31. (d) Equilibrium constant (K) is independent of, concentrations of reactions and products., 32. (a) Equilibrium constant is not effected by change in, conditions like P and V. These changes can change, only the time required to attain equilibrium., 33. (b) Reaction proceed forward according to Le-chatelier’s, principle., 34. (b), 35. (d) For reaction to proceed from right to left, , e RT, , ....... (ii), , 40. (d) Solid, Liquid, It is an endothermic process. So when temperature is, raised, more liquid is formed. Hence adding heat will, shift the equilbrium in the forward direction., 41. (d) According to Le-chatelier's principle" whenever a, constraint is applied to a system in equilibrium, the, system tends to readjust so as to nullify the effect of, the constraint., 42. (b) The most favourable conditions are :, (i) High pressure ( n < 0), (ii) Low temperature (Exothermic reaction), (iii) Catalyst Fe is presence of Mo., 43. (c) As in this no. of moles are increasing hence low, pressure will favour the forward direction., n = (1 + 1) – 1 = 1, 44. (a) For n = 0, no effect of pressure.
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EBD_7207, EQUILIBRIUM, , 126, , 45., , 46., 47., 48., , 49., , 50., , 51., 52., , 53., , 54., , 55., 56., 57., 58., 59., , (a) Melting of ice involve absorption of heat i.e, Endothermic hence high temperature favour the, process., Further for a given mass volume of water is less than, ice thus high pressure favour the process. High, pressure and high temperature convert ice into liquid., (d), (b) Le chatelier principle is not applicable to solid-solid, equilibrium., (c) Effect of increase of temperature on equilibrium, constant depends on the fact that whether the reaction, is exothermic, or endothermic. If the reaction is, exothermic, it is favoured by low temperature and if, the reaction is endothermic, it is favoured by high, temperature., (d), , (a), , (c), (c), , (c), , (c), , A2 (g) B2(g), , D(g), 3C(g), step-1 step-2, since the steps 1 and 2 are exothermic hence low, temprature will favour both the reactions. In step - 1, moles are increasing hence low pressure will favour it., In step 2 moles are decreasing, hence high pressure, will favour it., On adding inert gas at constant volume the total, pressure of the system is increased, but the partial, pressure of each reactant and product remains the, same. Hence no effect on the state of equilibrium., The backward reaction is not favoured at high pressure., A catalyst increases the rate of the chemical reaction, by making available a new low energy pathway for the, conversion of reactants to products. It increases the, rate of forward and reverse reaction that pass through, the same transition state and does not affect, equilibrium., Catalyst lowers the activation energy for the forward, and reverse reactions by exactly the same amount., An aqueous solution of sodium chloride is comprised, entirely of sodium ions and chloride ions, while that of, acetic acid mainly contains unionized acetic acid, molecules and only some acetate ions and hydronium, ions. This is because there is almost 100% ionization, in case of sodium chloride which is a strong electrolyte, as compared to less than 5% ionization of acetic acid, which is a weak electrolyte., The hydronium ion has a trigonal pyramidal geometry, and is composed of three hydrogen atoms and one, oxygen atom. There is a lone pair of electrons on the, oxygen giving it this shape. The bond angle between, the atoms is 113°., , (d), (b) According to Arrhenius, acids are those substances, which give proton in aqueous solution, hence gaseous, HCl is not an Arrhenius acid., (d) Base accepts protons and acid donates protons., (c) Lewis concept., (c) H2CO3, H+ + HCO3–, H+ + CO3–., HCO 3 can donate and accept H+., , 60., , (b) Because NH3 after losing a proton, , 61., , (c), , 62., , (d), , 63., 64., 65., , (c), (b), (c), , NH3 + H2O, NH 2– H 3O, (Conjugate acid-base pair differ only by a proton), Lewis acid is that compound which have electron, deficiency. eg. BF3, SnCl2., (CH3)3 B - is an electron deficient, thus behave as a, lewis acid., Boron in B2H6 is electron deficient, BF3 is Lewis acid (e– pair acceptor), HSO4– accepts a proton to form H2SO4., Thus H2SO4 is the conjugate acid of HSO4–., H, , HSO4, , H 2SO 4, conjugate acid, of HSO 4, , base, , 66., 67., 68, , (a) Bronsted base is a substance which accepts proton., In option (a), H2O is accepting proton, i.e., acting as a, base., (c), (a) (i) H 3PO 4 H 2O, H 3O H 2PO 4, acid1, , base 2, , (ii) H 2PO4 H2O, acid1, , base 2, , acid 2, , base1, , HPO 4, , H 3O, , base1, , 2 PO 4 OH, (iii) Hbase, acid 2, , 70., , acid 2, , H 3 PO 4 O, acid1, , 1, , 69., , (H+) gives NH2–, , base 2, , Hence only in (ii) reaction H2PO4– is acting as an acid., (b) The value of ionic product of water changes with the, temperature., (a) Given : Hydroxyl ion concentration, [OH–] = 0.05 mol L–1. We know that, 14, , [H ][OH ] 1 10, , or [H ] 1 10, , 14, , 0.05, , 2 10 13 mol L–1, , We also know that, pH, , =, , log[H ], , log [2 10, , log 2 log 10, , 13, , 13, , ], , log 2 ( 13) log 10, , 0.3010 13.0000 12.6990., Since the value of pH > 7, therefore the solution is, basic., , 71., , (b), , 72., , (b) pH = – log [H+] = log, , 73., , 74., , 1, , H, , [H+] = 10–6 M, , (c) On dilution, = 10–6 mol, Now dissociation of water cannot be neglected,, Total [H+] = 10–6 + 10–7 = 11 × 10–7, pH = –log [H+], = –log (11× 10–7) = 5.98, (d), pH = 1 ; H+ = 10–1 = 0.1 M, pH = 2 ; H+ = 10–2 = 0.01 M, M1 = 0.1 V1 = 1, M2 = 0.01 V2 = ?
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EQUILIBRIUM, , 75., , 76., , 127, , From, M1V1 = M2V2, 0.1 × 1 = 0.01 × V2, V2 = 10 litres, Volume of water added = 10 – 1 = 9 litres, (b) Given [OH–] = 10–3, pOH = 3, pH + pOH = 14, pH = 14 – 3 = 11, (a) Given [H3O+] = 1 × 10–10 M, at 25º [H3O+] [OH–] = 10–14, [OH ], , 10, , 14, , 10, , 10, , 10, , 78., , 83. (a) Given Ka, C= 0.100 mol, for a weak electrolyte,, degree of dissociation, , 84. (b) HA, , H+ + A–, , Ka, , [H ][A ], ,, [HA], , 0.1, , Concentration of OH– = 1 litre, , 14, , ( OH –, , 10, , 10, , 10, OH, , 80., , 81., , 82., , 2 10, , 8, , 4 10, , 6, , 2 10, , 3, , Percentage ionization = 0.2, 85. (c) M1V1 = M2V2, 1 × 0.10 = M2 × 100, M2 = 0.001 = 10–3, BOH, C, C(1 ), , B, 0, C, , OH, 0, C, , C C, C(1 ), , Kb, , 2, , Kb = C, , ( 1, , 1), , Kb / C, , 14, , 10 –1 ), , Kb, C, , [OH ] C, , (a) Given [H3, at 25º [H3O+] [OH–] = 10–14, 10, , 5, , 2 10 8, .005, , O+] = 1 × 10–10 M, , [OH ], , [H+] = 10–pH, , 10 5 10, 0.0015, , Ka, C, , 0.1Mole / L, , As we know that, [ H ] [OH ] 10, 79., , 1%, , OH, , No. of moles of OH– = 0.1, , 13, , 10 2, , [H+] = 10–5 ; and at equilibrium [H+] = [A– ], , Ka, , 10 4 10 p, Now, [OH ] 10 p, pOH = 4, (a) Molarity (M) = 10M. HCl is a strong acid and it is, completely dissociated in aqueous solutions as : HCl, (10), H+(10) + Cl–., So, for every moles of HCl, there is one H+. Therefore, [H+] = [HCl] or [H+] = 10., pH = – log[H+] = – log [10] = – 1., 4, 0.1, (d) No. of moles of NaOH =, 40, [Molecular weight of NaOH = 40], , [H ] 10, , 1 10 –5, 0.100, , Ka, C, , ( ), , 4, , OH, , 77., , = 1.00×10–5,, , 10, , 5, , 10, , 3, , 10, , C, , K bC, , 4, , pH + pOH = 14, pH = 14 – 4 = 10, , 4, OH, , 10 4 10 p, Now, [OH ] 10 p, pOH = 4, (c) H2SO4, 2H+ + SO42–, Given concentration of H2SO4 = 0.005 M, [H+] = 0.005 × 2 = 0.01 = 10–2, pH = – log [H+] = – log 10–2 = 2, (a) pOH = – log [OH–], pH + pOH = 14, For 10–4 KOH, [OH–] = 10–4, pOH = – log [OH–] = – log 10–4 = 4, pH = 14 – pOH = 10, (d) Among M/4 KOH, M/4 NaOH, M/4 NH4OH, and, M/4 Ca(OH)2, Ca(OH)2 furnishes highest number of, OH– ions ( Ca(OH)2, Ca2+ + 2OH–)., So pH of M/4 Ca(OH)2 is highest., , 86. (d), 87. (d), , H 2A, , H+ + HA-, , K1 = 1.0 × 10–5 =, HA, , H, , K2, , K, , [H ][HA ], (Given), [H 2 A], , A, , 5.0 10, , [H ]2 [A 2 ], [H 2 A], , 10, , [H ][A, , ] (Given), , [HA ], , K1 K2, , = (1.0 × 10–5) × (5 × 10–10) = 5 × 10–15, 88. (b) Acidic strength, , Ka
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EBD_7207, EQUILIBRIUM, , 128, , 89., , = 1.0 × 10–12, , (d) Given Kb, [BOH] = 0.01 M, BOH, , B, , 0, , t o, c, teq c(1 x), , OH, , cx, , c 2 x2, c (1 x), , Kb, , [OH] = ?, 0, , cx, , cx 2, (1 x ), , 1.0 × 10–12 =, , 0.01x 2, (1 x ), , 90., , On calculation, we get, x = 1.0 × 10–5, Now, [OH–] = cx = 0.01 × 10–5 = 1 × 10–7mol L–1, (b) pKa = –log Ka, Smaller the value of pKa, stronger will be acid, Acid having pKa value of 10–8 is strongest acid., (b), , 1, , Ka1, , 3.14 10, , 4, , 91., , 2, , K a2, , 1.96 10, , 5, , 92., , (c), , Ka, , [H3O ][F ], [HF], , ...(ii), , [F ], , From (i) and (ii), KaKb = [H3O+][OH–]=Kw, (ionic product of water), 93., 94., , K, , c 2, , Base, , 2, , H2 X, , HX-, , H+ + HX- (K a1 ), , K a1, , K a2, , 106. (d) For acidic buffer pH = pKa + log, , H+ + X2- (K a 2 ), , (a) The values of dissociation constants for successive, stages decrease., 96. (a), 97. (a) NH 4 Cl H 2 O NH 4OH HCl, or, Cl, H 2O, HCl H i.e., acidic, 98. (c) Due to common ion effect addition of NH4Cl in group, (III) suppresses the ionisation of NH4OH with the, result concentration of OH– decreases., 99. (c) Solubility of weak electrolyte decreases in solvent, having common ion. So solubility of AgI in NaI solution, is less than in pure water because of common ion effect., 100. (b) Dissociation of CH 3COOH is suppressed by the, addition of sodium acetate (CH 3COONa ) due to, common ion (CH3COO ) effect. The [ H ] decreases, raising the pH of the acid solution., Note : After the addition of CH3COONa to CH3COOH,, a buffer solution is formed which has reserved pH, value., , acid, , Now pH is given by, 1, 1, 1, pK w, pKa, pK b, pH, 2, 2, 2, substituting given values, we get, 1, pH = (14 4.80 4.78) 7.01, 2, , 0.1, , by at least a factor of 10–3 i.e.,, , 95., , Due to this, there is predominance of H+ ions in, solution, hence the solution is acidic., 105. (c) In aqueous solution BA(salt) hydrolyses to give, BA + H2O, BOH + HA, , 1.34, 1.8 10 5, 100, (b) In polyprotic acids the loss of second proton occurs, much less readily than the first. Usually the Ka values, for successive loss of protons from these acids differ, , (a), , strong acid, , weak base, , ...(i), , [HF][OH ], , Kb, , 4 :1, , 101. (b) CH3COOH is weak acid while NaOH is strong base, so, one equivalent of NaOH can not be neutralized with, one equivalent of CH3COOH. Hence the solution of one, equivalent of each does not have pH value as 7. Its pH, will be towards basic side as NaOH is a strong base, hence conc. of OH– will be more than the conc. of H+., 102. (a) IVth group needs higher S2– ion concentration. In, presence of HCl, the dissociation of H2 S decreases, hence produces less amount of sulphide ions due, to common ion effect, thus HCl decreases the, solubility of H2 S which is sufficient to precipitate, IInd group radicals., 103. (d) A salt of strong base with weak acid undergoes anionic, hydrolysis to give basic solution., 104. (d) Ferric chloride is the salt of a strong acid and a weak, base, hence on hydrolysis it yields a mixture of weak, base and strong acid, FeCl3 3H2O Fe OH 3, 3HCl, , or pH, , pK a, , log, , salt, acid, , A, HA, , Given pKa = 4.5 and acid is 50% ionised., [HA] = [A–] (when acid is 50% ionised), pH = pKa + log 1, pH = pKa = 4.5, pOH = 14 – pH = 14 – 4.5 = 9.5, 107. (b) Given [NH3] = 0.3 M, [NH4+] = 0.2 M, Kb = 1.8 × 10–5 ., pOH = pK b, , log, , pKb = 4.74, , [salt], [base], , [pKb = –log Kb;, , pKb = –log 1.8 × 10–5], , 0.2, = 4.74 + 0.3010 – 0.4771 = 4.56, 0.3, pH = 14 – 4.56 = 9.436, , 4.74 log, , 108. (d), , pH, , p Ka+ log, , log H, , Salt, Acid, , log Ka – log, , Salt, Acid
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EQUILIBRIUM, , 129, , log H, , H, , = log Ka + log, , 110. (a), , 111. (d), , 112. (d), 113. (b), , 114. (b), , s, , 10 2, 2, Ksp = 4s3, , 5, , Cr 3 (aq.) 3OH (aq.), , Cr(OH)3 (s), , s, , (3s)3 = Ksp, , 27 S 4, , 2s = 10–2, , s, , 0.1, = 9 × 10 6 M, 0.2, HNO2 is a weak acid and NaNO2 is salt of that weak, acid and strong base (NaOH)., Lets take an example of an acidic buffer CH3COOH, and CH3COONa., CH3COOH, CH3COO – + H + ;, CH3COONa, CH3COO– + Na+, when few drops of HCl are added to this buffer, the H+, of HCl immediatly combine with CH3COO– ions to, form undissociated acetic acid molecules. Thus there, will be no appreciable change in its pH value. Like, wise if few drops of NaOH are added, the OH – ions, will combine with H+ ions to form unionised water, molecule. Thus pH of solution will remain constant., Buffering action is maximum when, [Salt] = [Acid], i.e., pH = pKa, pH or [H+] of a buffer does not change with dilution., Solubility product is the product of ionic concentration, in a saturated solution of an electrolyte at a given, temperature., , (s), , 115. (b), , Acid, Salt, , Acid, Salt, , = Ka, , = 1.8 × 10, , 109. (d), , [OH–] = 10–2, , = 4, , 1/ 4, , 1.6 10, 27, , 27, , AB, , 10-11, 10-3, , [Ba++] =, 120. (b), , [OH–] =, , 10, , 14, , 10, , 12, , Ba(OH)2, , s, , Ksp, [CO3, , ], , 5.1 10, , 9, , 1.0 10, , 4, , = 5.1 × 10–5 M, , Ag+ + Br-, , AgBr, , Ksp = [Ag+] [Br–], For precipitation to occur, Ionic product > Solubility product, [Br- ] =, , 30 1/ 4, , K sp, , [Ag + ], , =, , 5´10-13, = 10-11, 0.05, , i.e., precipitation just starts when 10–11 moles of KBr, is added to 1 AgNO3 solution, Number of moles of Br– needed from, KBr = 10–11, Mass of KBr = 10–11 × 120 = 1.2 × 10–9 g, 121. (b), , pOH = 4, pH + pOH = 14, pH = 10, 116. (b) Given pH = 12, or [H+] = 10–12, Since, [H+] [OH–] = 10–14, , B, , s, , Hence, solubility product of AB, Ksp = [A+] [B–], S = [s.] [s] s = S½, 118. (a) Ksp of As2 S3 is less than ZnS. In acid medium, ionisation of H2S is suppresed (common ion effect), and Ksp of ZnS does not exceed., 119. (a) Given Na2CO3 = 1.0 × 10–4 M, [CO3– – ] = 1.0 × 10–4 M, i.e., S = 1.0 × 10–4 M, At equilibrium, [Ba++] [CO3– –] = Ksp of BaCO3, , 3s, , = 10-4, , A, , s, , Mg(OH) 2, Mg, 2 OH, Ksp = [Mg++][OH–]2, 1.0 × 10–11 = 10–3 × [OH–]2, , [OH- ] =, , 3, , = 5 × 10–7, 117. (c) Let binary electrolyte be AB, , K sp, , K sp, , 10 2, 2, , Mg(OH)2, Mg, 2OH, ++, –, 2, Ksp = [Mg ][OH ], 1.0 × 10–11 = 10–3 × [OH–]2, [OH- ] =, , 10-11, 10-3, , = 10-4, , pOH = 4, pH + pOH = 14, , pH = 10, , STATEMENT TYPE QUESTIONS, = 10–2, , Ba 2+ + 2OH, s, , 2s, , 122. (c) Boiling point depends on the altitude of the place; at, high altitude the boiling point decreases., 123. (a)
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EBD_7207, EQUILIBRIUM, , 130, , 124. (b) Chemical reactions which are in equilibrium are, reversible, 125. (c) Since equilibrium constant is related to the partial, pressure of reactant and product therefore if nobel, gas is added, no change is observed, Statement 1 is, incorrect., If Nobel gas react with reactant, concentraction of, reactant will decrease and therefore reaction will, proceed backword according to Le– chatelier ’s, principle., 126. (c) Equilibrium constant is temperature dependent having, one unique value for a particular reaction represented, by a balanced equation at a given temperature., 127. (b) An equilibrium constant does not give any information, about the rate at which the equilibrium is reached., 128. (a), 129. (d), 130. (a) As strong acid dissociate completely in water hence, resulting base formed would be very weak. On the, other hand a weak acid is only partially dissociated in, aqueous solution, hence resulting base formed would, be strong., 131. (a) H2O(l) + H2O(l), H3O+(aq) + OH–(aq), acid, , base, , conjugate acid, , conjugate base, , 132. (c) Kw depends upon temperature as it is an equilibrium, constant., 133. (d) Larger the value of Ka, the stronger is the acid., 134. (a), 135. (c) Bond energy being directly related to bond strength, increases with increase in bond strength, , MATCHING TYPE QUESTIONS, 136. (d), 137. (b) (A) Liquid, Vapour equilibrium exists at the, boiling point., Liquid equilibrium exists at the, (B) Solid, melting point., (C) Solid, Vapour equilibrium exists at the, sublimation point., (D) Solute, Solute (solution) equilibrium exists, in a saturated solution., 138. (a) In case of A no. of moles of product and reactant are, same, in case of B no. of moles of reactant are greater, so reaction go forward, in case of C the no. of moles of, product are greater than no. of moles of reactant., 139. (a), (A) KP = KC (RT) n, KP, = (RT) n as n = –ve, KP < KC, KC, (B) n > 0, (C) n = 2 – 1 = 1, (D) As the reaction is not containing any gaseous, component therefore KP is not defined for this., 140. (b), (A) As n > 0 therefore if P , reaction will go in the, backward direction., , (B) As n < 0 therefore if V , P reaction will go in, the direction in which more number of gaseous, moles are formed i.e. backward direction., (C) As n = 0 hence no effect., (D) If concentration of reactant is increased reaction, will go in the forward direction., 141. (d), 142. (b), 143. (b) HClO4 is a strong acid, HNO2 is a weak acid., NH2– is a very good proton acceptor and thus, it is a, base., H2SO4 is a strong acid hence its conjugate base (HSO4–), will be a weak base., , ASSERTION-REASON TYPE QUESTIONS, 144. (a) Kp = Kc(RT) n, 145. (a) If the volume is kept constant and an inert gas such as, argon is added which does not take part in the reaction,, the equilibrium remains undisturbed. It is because the, addition of an inert gas at constant volume does not, change the partial pressure or the molar concentration, of the substance involved in the reaction. The reaction, quotient changes only if the added gas is reactant or, product involved in the reaction., 146. (d) In biological systems buffer system of carbonic acid, and sodium bicarbonate is found in our blood. It, maintains the pH of blood to a constant value of about, 7.4., 147. (a) Ionic product of AgBr is greater than that of AgCl in, comparison with there solubility product AgBr will, precipitate first rather than that of AgCl., , CRITICAL THINKING TYPE QUESTIONS, 148. (d) To calculate the value of K4 in the given equation we, should apply :, eqn. (2) + eqn.(3) × 3 – eqn. (1), hence K4 =, , K 2 K33, K1, , 149. (d) Given,, AB, , K1, , A, , B, , K1, , [A ] [B ], [AB], , AB, , B, , K2, , K2, , 1, , AB 2, , [AB2 ], [AB] [B ], , Dividing K1 and K2, we get, K, , K1, K2, , [A ] [B ]2, , [A ], , K, , [AB2 ], , [B ]2, , [AB2 ]
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EQUILIBRIUM, , 131, , 150. (a) K1 =, K=, , K=, , K, , Ni(CO)4, 2, , [CO 2 ], , ; K2 =, , [CO]2, [CO2 ], , 154. (d), , (R in L. atm. K–1 mole–1)., (0.082 773) 2, , [CO]4, , [Ni(CO)4 ], , [CO 2 ], , 2, , 155. (b), , ×, [CO2 ]2, [CO]2, , K 22, , K1, , K1, , ...(i), , [N 2 ][O 2 ]2, K2, , N2, , 2O 2, , 2, , K2, (III), , [N 2 ] [O2 ], , K3, , ...(ii), , [NO 2 ]2, 1, N2, 2, [N 2 ]1/2 [O2 ], [NO 2 ], , NO2, , K3, , 2, , K3, , ax, , ax, , PCl3, , 2NO 2, , [NO 2 ]2, , (II) 2NO 2, , PCl3 + Cl2, , PCl5, a (1-x), , a = 2, x = 0.4, V = 2 L, 2 1 0.4, PCl5, 2, , K1, , 2O 2, , (RT) n, 1.44 10 5, , [Ni(CO) 4 ], , 151. (b) (I) N 2, , Kp, , Kc, , O2, , [N 2 ][O 2 ]2, [NO2 ]2, , ...(iii), , from equations (i), (ii) and (iii), 1, 1, K1, K 2 ( K3 )2, 152. (c) Reaction (c) can be obtained by adding reactions (a), and (b) therefore K3 = K1. K2, Hence (c) is the correct answer., 153. (b) Given: Equilibrium constant (K1) for the reaction, K1, 1, 1, .........(i), HI(g), H 2 (g), I 2 (g); K1 = 8;, 2, 2, To find equilibrium constant for the following reaction, H 2 (g) I2 (g), 2HI(g); K2 = ?, For this multiply (i) by 2, we get, , .....(ii), , 2HI(g), H 2 (g) I2 (g); K1 = 82 = 64 .....(iii), [Note: When the equation for an equilibrium is, multiplied by a factor, the equilibrium constant must, be raised to the power equal to the factor], Now reverse equation (iii), we get, 1, .....(iv), H 2 (g) I 2 (g), 2HI(g); K, 64, [Note: For a reversible reaction, the equilibrium, constant of the backward reaction is inverse of the, equilibrium constant for the forward reaction.], Equation (iv) is the same as the required equation (ii),, 1, thus K2 for equation (ii) is, i.e. option (b) is correct., 64, , 0.6 mol L 1, , 2 0.4, 2, , Cl2, , 0.4 mol L, , 1, , 0.4 0.4, 0.267, 0.6, 156. (b) PCl5, PCl3 + Cl2, Moles at equilibrium, 1, 1, 1, 2, 2, 2, Mole fraction at equilibrium, 1, 1, 1, 3, 3, 3, Partial pressure at equilibrium, P, P, P, 3, 3, 3, Kc, , Kp, , 157. (b), , P P, 3 3, P/3, , P, 3, , NH 2 COONH 4 ( s ), , KP, , PNH3, , 2, , 2NH 3 ( g ) CO 2 ( g ), , PCO2, , PNH 2COONH 4 ( s), , = PNH3, , 2, , PCO2, , As evident by the reaction, NH3 and CO2 are formed, in molar ratio of 2 : 1. Thus if P is the total pressure of, the system at equilibrium, then, 1 P, 2 P, PCO2, PNH3, 3, 3, KP, , 2P, 3, , 2, , P, 4 P3, =, 3, 27, , Given KP = 2.9 × 10–5, 4 P3, 27, , 2.9 10 5, P3, , 2.9 10, 4, , 5, , 27, , 5, , P, , 2.9 10, 4, , 27, , 1, , 3, , = 5.82 × 10–2 atm
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EBD_7207, EQUILIBRIUM, , 132, , 158. (c) Given reaction are, X, Y Z, , Total number of moles, = 2 (1–x) + 2x + x = (2 + x), 2(1 x), PAB =, P where P is the total pressure., 2, (2 x), x, 2x, PAB =, P, P , PB, 2, (2 x), (2 x), Since x is very small so can be neglected in denominator, Thus, we get, PAB = x × P, PAB = (1 – x) × P, 2, x, PB, P, 2, 2, , ..... (i), , and A, ......(ii), 2B, Let the total pressure for reaction (i) and (ii) be P1 and, P2 respectively, then, K P1, , 9, 1, , KP2, , (given), , After dissociation,, X, Y Z, At equilibrium (1– ), [Let 1 mole of X dissociate with, dissociation ], Total number of moles = 1– + +, = (1+ ), , 1, 1, , Thus PX =, PZ =, K P1, , . P1 ; PY =, , 1, .P1, , 1, 1, , ., , 1, , ... (i), , 2, , 1, 1, , /, , ...(ii), , P2, , Dividing (i) by (ii), we get, 2, , K P1, K P2, , 4, , .P2, , or, , P1, P2, , K P1, K P2, K P1, , 1 P1, ., 4 P2, , or 9, , or, , .P1, , 2, , K P2, , 1 P1, ., 4 P2, 9, 1, , 36, or P1 : P2 = 36 : 1, 1, , Kc, , 2AB2 (g), , 2AB(g) B2 (g), 2x, , 2(1 x), , AB, , 2, , AB2, , B2, 2, , or K c, , PB2, 2, , 1], , x, , (2x) 2, , 2.K p, 2K p 3, x 3 .P, or x3 =, or x =, =, P, 2, P, 160. (a) Justification : According to Le-Chatelier’s principle, at, constant temperature, the equilibrium composition will, change but K will remain same., 161. (a) The reaction given is an exothermic reaction thus, according to Lechatalier’s principle lowering of, temperature, addition of F2 and / or Cl2 favour the for, ward direction and hence the production of ClF3 ., 162. (a), 163. (c) Strong base has higher tendency to accept the proton., Increasing order of base and hence the order of, accepting tendency of proton is, I, HS, NH3 RNH 2, , [Cu(H2O)4]2+ + 4NH3, [Cu(NH3)4]2+ + 4H2O, involves lose and gain of electrons. H2O is coordinated, to Cu by donating electrons (LHS). It is then removed, by withdrawing electrons., 165. (b) [H3O]+ for a solution having pH = 3 is given by, [H3O]+ = 1×10–3 moles/litre, [ [H3O]+ = 10–pH], Similarly for solution having pH = 4,, [H3O]+ = 1 × 10–4 moles/ litre and for pH = 5, [H3O+] = 1×10–5 moles/ litre, Let the volume of each solution in mixture be IL, then, total volume of mixture solution L = (1 + 1 + 1) L = 3L, Total [H3 O] + ion present in mixture solution, = (10–3 + 10–4 + 10–5) moles, Then [H3O]+ ion concentration of mixture solution, 164. (d), , 159. (b) For the reaction, at equi, , 2, , 1, , .P1, , 2 P2, 1, , PAB, , PAB2, x, (x)2 P 2 .P, 2, =, (1 x) 2 P 2, x 3 .P3, =, [ 1–x, 2 1 P2, , P1 ;, , 1, , A, 2B, Similarly for, At equilibrium (1– ) 2, We have,, K P2, , Now, K P, , .P1, , 1, , P1 /, , as degree of, , x, , {2(1 x)}2, , = x3 [(1–x) can be neglected in denominator (1– x) 1], The partial pressure at equilibrium are calculated on, the basis of total number of moles at equilibrium., , 0.00111, 10 4 10 5, M=, M, 3, 3, = 0.00037 M = 3.7 ×10–4 M., , =, , 10, , 3
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EQUILIBRIUM, , 133, , 166. (d) Given concentration of NaOH = 10, NaOH, Na, OH, 10 M, 10, 10, 10, , 10, , On taking log on both side, – log [H+] = –log (55 × 10–14)1/2, 1, pH = – log 55 7 log10, 2, pH = – 0.87 + 7, = 6.13, , M, , 10, , [OH ] from NaOH = 10, We have to consider dissociation of H2O, [OH ] from H 2 O = 10, Total [OH ] = 10, , 7, , 7, , 10, , = 10 7 (0.001 1) = 10, , 7, , 170. (c), 10, , 1001, 1000, , = 10, , 10, , 1001, , = (log1001 10 10 ) = – 3.004 + 10 = 6.9996, pH = 14 – pOH = 14 – 6.996 = 7.004, pH of 10–10 M NaOH solution is nearest to 7., 167. (c) Number of meq. of the acid = 0.04 × 100 = 4, Number of meq. of the base = 0.02 × 100 = 2, Number of meq. of the acid left on mixing = 4 – 2 = 2, Total volume of the solution = 200 mL, No. of meq of the acid present in 1000 mL of the, solution = 10, or No. of eq. of the acid in 1000 mL of the solution, 10, =, = 0.01, 1000, Since the acid is monobasic and completely ionises in, solution, 0.01 N HCl = 0.01 M HCl, Thus [H ] = 0.01, pH = – log (0.01) = – (– 2) = 2, 168. (b) [H3O]+ for a solution having pH = 3 is given by, [H3O]+ = 1×10–3 moles/litre, [ [H3O]+ = 10–pH], Similarly for solution having pH = 4,, [H3O]+ = 1 × 10–4 moles/ litre and for pH=5, [H3O+] = 1×10–5 moles/ litre, Let the volume of each solution in mixture be IL, then, total volume of mixture solution L = (1 + 1 + 1) L =3L, Total [H3O]+ ion present in mixture solution, = (10–3 + 10–4 + 10–5) moles, Then [H3O]+ ion concentration of mixture solution, 10 4 10 5, 0.00111, M=, M, 3, 3, = 0.00037 M = 3.7 ×10–4 M., 169. (a) Kw at 25°C = 1 × 10–14, At 25ºC, Kw = [H+] [OH–] = 10–14, At 100°C (given), Kw = [H+] [OH–] = 55 × 10–14, for a neutral solution, [H+], = [OH–], +, 2, [H ] = 55 × 10–14, or [H+] = (55 × 10–14)1/2, pH = – log [H+], 10, , 3, , +, , CH3COO + H, , [ CH 3COO ][ H ], [ CH 3COOH ], Given that,, Ka, , pOH = – log [OH ], , =, , –, , CH3COOH, , [CH 3 COO ] [H ] 3.4 10 4 M, Ka for CH3COOH = 1.7 × 10–5, CH3COOH is weak acid, so in it [CH3COOH] is equal, to initial concentration. Hence, (3.4 10 4 )(3.4 10 4 ), [CH 3 COOH ], , 1.7 10 5, , 3.4 10, , [ CH 3 COOH ], , 4, , 3.4 10, , 4, , 1.7 10 5, = 6.8 × 10–3M, 171. (c) Higher the value of Ka lower will be the value of pKa, i.e. higher will be the acidic nature. Further since, CN–, F– and NO2– are conjugate base of the acids, HCN, HF and HNO2 respectively hence the correct, order of base strength will be, F– < NO2– < CN–, ( stronger the acid weaker will be its conjugate base), , CH3COO– + H+ ;, , 172. (c) Given, CH3COOH, , Ka1 , = 1.5 × 10– 5, , ....(i), , H++ CN–; K a 2 = 4.5 × 10–10, , HCN, , or H+ + CN–, 1, K 'a 2, Ka 2, , HCN;, 1, , ...(ii), 4.5 10 –10, From (i) and (ii), we find that the equilibrium constant, (Ka) for the reaction,, CN– + CH3COOH, CH3COO– + HCN, is, Ka, , K a1, =, , K a' 2, , 1.5×10 –5, 4.5×10, , 173. (d) As, molarity,, , –10, , 1, 105, 3, , 3.33 10 4, , wt. of solute per litre of solution, Mol. wt. of solute, 1000, mole/litre, 18, H+ + OH–, c, c, , Molarity of H2O =, H2O, c (1 –, Thus, K a, , c 2, 1, , c 2 = 1.8 × 10–14
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EBD_7207, EQUILIBRIUM, , 134, , 174. (c) For weak acid dissociation equilibria, degree of, dissociation is given as :, Ka, c, , %, , 100, , log K a, , log H, , or pK a, , pH log, , pK a, , pH, , 1, , 10pK a, , or,, , 1, , log, , log, , Ka, c, , 1, , [H ]c, c (1 ), , [H ][A ], [HA], , Also, K a, , or, , 2´10-5, 1, , 1, , = éê10-16 ùú 4 or 10–4, ë, û, Thus the solubilities are in the order MX > M3 X> MX2, i.e the correct answer is (d)., 177. (c) Ksp = [Ag+] [Cl–], 1.8 × 10–10 = [Ag+] [0.1], [Ag+] = 1.8 × 10–9 M, Ksp = [Pb+2] [Cl–]2, 1.7 × 10–5 = [Pb+2] [0.1]2, [Pb+2] = 1.7 × 10–3 M, 178. (d) The solubility equilibrium for AgI is, , 1, , 1, , 1, [1 10, , 1, , é 2.7 ´10-15 ù 4, ú, Solubility of M 3 X = êê, ú, 27, ëê, ûú, , 1, , 1, , pH, , = éê8´10-15 ùú 3, ë, û, , [H ], (1 ), , pH, , 10pK a, , 1, , é 3.2 ´10-14 ù 3, ú or éê 32 ´10-15 ùú 3, Solubility of MX 2 = êê, ú, êë 4, úû, 4, êë, úû, , pKa pH, , ], , AgI (aq), , aq + I, , Ag, , aq ;, , 175. (b) The highest pH will be recorded by the most basic, solution. The basic nature of hydroxides of alkaline, earth metals increase as we move from Mg to Ba and, thus the solution of BaCl2 in water will be most basic, and so it will have highest pH., , Ksp = [Ag+][I–], Let solubility of AgI be S moles per litre,, [Ag+] = S, [I–] = S, Ksp = [Ag+][I–], 1 × 10–16 = (S) × (S) = S2, , 176. (d), , S = 1 10 16 2 1 10 8, On calculating solubility of all given compounds, , MX, , M, s, , Then Ksp, , 1, , X (Where s is the solubility), s, , s 2 or, , s, , K sp, , Compound Solubility, , Similarly for MX 2 ¾¾, ® M 2+ + 2X s, K sp, , s (2 s ) 2, , and for M 3 X, , Ksp, , (3s )3 s, , 2s, 4 s3 or, 3M, , 3s, , 1, K sp 3, , s, X, , 4, , 179. (b), , 3, , s, , 27 s 4 or s, , Ksp, , 1, 4, , 27, , From the given values of Ksp for MX, MX2 and, M3 X, we can find the solubilities of those salts at, temperature, T., -8, , Solubility of MX = 4´10, , -4, , = 2´10, , AgCl, , 1 10, , 5, , AgI, , 1 10, , 8, , PbCrO4, , 2 10, , 7, , Ag 2 CO3, , 1.26 10, , 4, , Ag2CO3 is most soluble and AgI is least soluble., Ksp = [Fe3+].[3OH–], So molar solubility of Fe3+ = S and [3OH–] = 3S, Fe(OH)3, 3OH, Fe3, [S] [3S], 1.0 ×10–38 = [S] [3S]3, 1.0 ×10–38 = S4 × 27, 1.0 10 38, S4 =, 27, S4 = 3.703×10–40, S = 3.703 10, , 40 1/ 4, , = 1.386 × 10–10
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8, , REDOX REACTIONS, , FACT / DEFINITION TYPE QUESTIONS, 1., , 2., , 3., , 4., , 5., , Which of the following process takes place in oxidation, process?, (a) Addition of oxygen, (b) Addition of hydrogen, (c) Removal of oxygen, (d) Addition of chlorine, Given reaction,, 2K4[Fe(CN)6] (aq) + H2O2 (aq), 2K3[Fe(CN)6](aq) + 2KOH(aq), The above given reaction is oxidation reaction due to, (a) removal of a hydrogen from H2O2, (b) addition of electropositive potassium to H2O2, (c) removal of electropositive element potassium from, potassium ferrocyanide (K4[Fe(CN)6]), (d) All of the above are the correct reasons., In the reaction given below, identify the species undergoing, redox reaction, 2Na(s) + H2(g), 2NaH(s), (a) Na is reduced and hydrogen is oxidised, (b) Na is oxidised and hydrogen is reduced, (c) Na undergoes oxidation and hydrogen undergoes, reduction, (d) Both (b) and (c), The loss of electron is termed as, (a) oxidation, (b) reduction, (c) combustion, (d) neutralization, Which of the following is correct code for x and y in the, following reaction., , 6., , 7., , 8., , Na +2 S2– (s), y, , (i), (ii), (iii), (iv), (a), (c), , x = oxidation reaction, y = reduction reaction, x = gain of two electrons, y = loss of two electrons,, x = reduction reaction, y = oxidation reaction, x = loss of two electrons, y = gain of two electrons, (i) and (ii), (b) (i) and (iv), (ii) and (iii), (d) (iii) and (iv), , (a), , MnO, , (c), , MnO24, , 9., , Mn, MnO 2, , (b), , CrO, , Cr, , (d), , Cr O, , Cr, , Which reaction involves neither oxidation nor reduction?, (a) CrO24- ¾¾, ® Cr2 O 72- (b) Cr ¾¾, ® CrCl3, , ® Na+, (c) Na ¾¾, In the following reaction, 4P 3KOH, , 3H 2 O, , (d), , 2S2 O 32- ¾¾, ® S4 O 62-, , 3KH 2 PO 2, , PH 3, , (a) phosphorus is both oxidised and reduced., (b) only phosphorus is reduced., (c) phosphorus is not oxidised, (d) None of these, Which one of the following reaction involves oxidationreduction ?, (a) H2 Br2, 2HBr, (b) NaBr HCl, NaCl HBr, (c) HBr AgNO 3, AgBr HNO 3, , (d) 2 NaOH H 2 SO 4, 10. In reaction, 4Na + O2, (a) oxidising agent, (c) Both (a) and (b), 11., , Zn 2 (aq.) 2e, (a) oxidation, (c) redox reaction, , 12., , Co(s) Cu 2 (aq), , x, 2Na(s) + S(s), , Which of the following involves transfer of five, electrons ?, , Na 2SO 4, , 2H 2 O, , 2Na2O, sodium behaves as, (b) reducing agent, (d) None of these, Zn(s) . This is, (b) reduction, (d) None of the above, Co 2 (aq) Cu(s), , The above reaction is, (a) oxidation reaction, (b) reduction reaction, (c) redox reaction, (d) None of these, 13. One mole of N2H4 loses 10 moles of electrons to form a new, compound, y. Assuming that all nitrogen appear in the new, compound, what is the oxidation state of nitrogen in y (There, is no change in the oxidation state of hydrogen ), (a) –1, (b) –3, (c) +3, (d) +5
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EBD_7207, REDOX REACTIONS, , 136, , 14., , 15., , 16., , 17., , 18., , 19., , 20., , 21., , 22., , 23., , 24., , 25., , 26., , When a strip of metallic zinc is placed in an aqueous, solution of copper nitrate the blue colour of the solution, disappear due to formation of, (a) Cu2+, (b) Zn 2+, (c) ZnS, (d) Cus, The correct order of electron releasing tendency of the, metals Cu, Zn and Ag is in the order:, (a) Cu > Zn > Ag, (b) Zn > Ag > Cu, (c) Ag > Zn > Cu, (d) Zn > Cu > Ag, What is the oxidation number of elements in the free or, in the uncombined state ?, (a) +1, (b) 0, (c) +2, (d) –1, In which of the following compounds oxygen has highest, oxidation state and in which it has lowest oxidation state?, OF2, H2O2, KO2, O2F2, (a) Highest = KO2, lowest = H2O2, (b) Highest = OF2, lowest = K2O2, (c) Highest = OF2, lowest = KO2, (d) Highest = KO2, lowest = H2O2, ‘Oxidation number of H in NaH, CaH2 and LiH, respectively, is, (a) +1, +1, –1, (b) –1, +1, + 1, (c) +1, + 1, + 1, (d) –1,–1, –1, Which of the following is the correct representative of, stock notation for auric chloride?, (a) Au(III)Cl3, (b) Au(II)Cl2, (c) Au(I)Cl2, (d) None of these, Oxidation number of N in HNO3 is, (a) – 3.5, (b) + 3.5, (c) –5, (d) + 5, In which of the following reactions, there is no change in, valency ?, (a) 4 KClO3, 3KClO4 + KCl, 2H2O + 3S, (b) SO2 + 2H2S, (c) BaO2 + H2SO4, BaSO4 + H2O2, (d) 3 BaO + O2, 2 BaO2., The oxidation number of chromium in potassium dichromate is, (a) + 6, (b) – 5, (c) – 2, (d) + 2, The oxidation number of sulphur in S 8 , S2 F 2 , H2 S, respectively, are, (a) 0, +1 and –2, (b) +2, +1 and –2, (c) 0, +1 and +2, (d) –2, +1 and –2, Oxidation number of cobalt in K[Co(CO)4] is, (a) +1, (b) +3, (c) –1, (d) –3, Oxidation number of nitrogen in (NH4)2SO4 is, (a) –1/3, (b) –1, (c) +1, (d) –3, Oxidation number of carbon in CH2Cl2 is, (a) –4, (b) +4, (c) 0, (d) –2, , 27., , 28., , 29., , 30., , 31., , 32., , 33., , 34., , 35., , 36., , 37., , 38., , 39., , In which of the following compounds, iron has lowest, oxidation state?, (a), , K 3[Fe(CN ) 6 ], , (b), , K 4 [Fe(CN) 6 ], , (c), , FeSO 4 .( NH 4 ) 2 SO 4 .6H 2 O, , (d), , Fe(CO)5, , The oxidation state of osmium (Os) in OsO4 is, (a) +7, (b) +6, (c) +4, (d) +8, Which of the following transition metal has zero oxidation, state ?, (a) [Fe(CO)5], (b) NH2.NH2, (c) NOClO4, (d) CrO5, In which of the compounds does 'manganese' exhibit highest, oxidation number ?, (a) MnO2, (b) Mn 3O4, (c) K2MnO4, (d) MnSO4, Among the following, identify the species with an atom in, +6 oxidation state, (a), , MnO–, , (c), , NiF, , –, , (b) Cr ( CN ), (d), , –, , CrO Cl, , In which of the following compounds the oxidation number, of carbon is not zero?, (a) HCHO, (b) CH3COOH, (c) C12 H22 O11, (d) CH3CHO, In which of the following compounds, the oxidation number, of iodine is fractional ?, (a) IF7, (b) I, (d) IF3, (c) IF5, A metal ion M3+ loses 3 electrons, its oxidation number will, be, (a) +3, (b) +6, (c) 0, (d) –3, The correct name for NO2 using stock notation is, (a) nitrogen dioxide, (b) nitrogen (iv) oxide, (c) nitrogen per oxide, (d) All of these, The oxide, which cannot act as a reducing agent, is, (a) NO2, (b) SO2, (c) CO2, (d) ClO2, The oxidation state of Fe in Fe3O4 is, (a) + 3, (b) 8/3, (c) + 6, (d) + 2, In oxygen difluoride, the oxidation number of oxygen is, (a) – 2, (b) – 1, (c) + 2, (d) +1, – 2, Oxygen has an oxidation state of +2 in the compound, (a), , H 2 O2, , (b), , CO2, , (c), , H2 O, , (d), , F2 O
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REDOX REACTIONS, , 40., , 41., , 42., , 43., , 44., , The number of electrons involved in the reduction of one, nitrate ion to hydrazine is, (a) 8, (b) 5, (c) 3, (d) 7, The average oxidation state of sulphur in Na2S4O6 is, (a) +2.5, (b) +2, (c) +3.0, (d) +3.5, Which of the following species can function both as, oxidizing as well as reducing agent ?, (a) Cl–, (b) ClO4–, –, (c) ClO, (d) MnO4 –, The oxidation number of an element in a compound is, evaluated on the basis of certian rules. Which of the, following rules is not correct in this respect?, (a) The oxidation number of hydrogen is always + 1., (b) The algebraic sum of all the oxidation numbers in a, compound is zero., (c) An element in the free or the uncombined state bears, oxidation number zero., (d) In all its compounds, the oxidation number of fluorine, is – 1., Nitric oxide acts as a reducing agent in the reaction, (a) 4 NH 3 5 O 2, 4 NO 6 H 2O, (b), (c), , 45., , 46., , 47., , 48., , 137, , NO, , I, , H O, , 2 NO H 2SO 3, , NO –, , l–, , H, , N 2 O H 2SO 4, , (d), NO H S N O S H O, In the compounds KMnO4 and K2 Cr 2 O7 the highest, oxidation state is of the element, (a) potassium, (b) manganese, (c) chromium, (d) oxygen, Atomic number of an element is 22. The highest O.S., exhibited by it in its compounds is, (a) 1, (b) 2, (c) 3, (d) 4, Why the displacement reactions of chlorine, bromine and, iodine using fluorine are not generally carried out in, aqueous solution?, (a) chlorine, bromine and iodine reacts with water and, displace oxygen of water, (b) Fluorine being very reactive attacks water and, displaces oxygen of water, (c) Fluorine does not react with chlorine, bromine and, iodine in aqueous media, (d) None of these, Which of the following statement is not true ?, (a) Displacement reaction of chlorine with Br– and I–, form the basis of identifying Br– and I– in laboratory, using layer test, (b) F2, Cl2, Br2 and I2 can be recovered by halogen, displacement reactions by using their respective, halides, (c) F2 can be recovered from F– by oxidising it, electrolytically., (d) None of these., , 49. Which of the following do not show disproportionation, reaction?, ClO –4 , F2, Cl2, ClO –2 , P4, S8, and ClO–, , (a) ClO –2 , ClO –4 , and ClO–, (b) F2 only, (c) F2 and ClO –4, (d), , ClO –4 only, , 50. Which one of the following reactions involves, disproportionation?, (a) 2H2SO4 + Cu CuSO4 + 2H2O + SO2, (b) As2O3 + 3H2S As2S3 + 3H2O, (c) 2KOH + Cl2 KCl + KOCl + H2O, (d) Ca3P2 + 6H2O 3Ca(OH)2 + 2PH3, 51. The following species will not exhibit disproportionation, reaction, (a) ClO–, , (b), , ClO 2, , (c), , (d), , ClO 4, , ClO3, , 52. In the reaction, 3Br2, , 6CO 32, , 3H 2 O, , 5Br, , BrO 3–, , 6HCO 3, , (a) Bromine is oxidised and carbonate is reduced., (b) Bromine is reduced and water is oxidised, (c) Bromine is neither reduced nor oxidised, (d) Bromine is both reduced and oxidised, 53. Which of the following elements does not show, disproportionation tendency?, (a) Cl, (b) Br, (c) F, (d) I, 54. Phosphorus, sulphur and chlorine undergo disproportion, in the ...A... medium., Here, A refers to, (a) acidic, (b) alkaline, (c) neutral, (d) Both (a) and (b), +1 -2, , 0, , 0, , D, 55. The reaction, 2H 2 O(l ) ¾¾, ® 2H 2 ( g ) + O2 ( g ) is an, , example of, (a) addition reaction, (b) decomposition reaction, (c) displacement reaction (d) None of these, 56. How will you balance the total ionic charge of reactant, and products if reaction is carried out in acidic solution?, (a) By using H+ ions, (b) By using OH– ions, (c) Adding H2O molecules to the reactant or product, (d) Multiplying by suitable coefficients., 57. Consider the following reaction occuring in basic medium, 2MnO –4 (aq)+Br – (aq), , 2MnO2 (s) + BrO3– (aq), How the above reaction can be balanced further?, (a) By adding 2 OH– ions on right side, (b) By adding one H2O molecule to left side, (c) By adding 2H+ ions on right side, (d) Both (a) and (b)
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EBD_7207, REDOX REACTIONS, , 138, , 58., , 59., , 60., , 61., , 62., , 63., , 64., , 65., , 66., , For the reaction : NH 3 OCl, N 2 H 4 Cl in basic, medium, the coefficients of NH3, OCl– and N2H4 for the, balanced equation are respectively, (a) 2, 2, 2, (b) 2, 2, 1, (c) 2, 1, 1, (d) 4, 4, 2, C2H6 (g) + nO2(g) CO2(g) + H2O(l), In this equation, the ratio of the coefficients of CO2 and, H2O is, (a) 1 : 1, (b) 2 : 3, (c) 3 : 2, (d) 1 : 3, 2MnO 4 5H 2 O 2 6H, 2 Z 5O 2 8H 2 O. In this, reaction Z is, (a) Mn +2, (b) Mn +4, (c) MnO2, (d) Mn, In the redox reaction,, xKMnO4 + NH3, yKNO3 + MnO2 + KOH + H2O, (a) x = 4, y = 6, (b) x = 3, y = 8, (c) x = 8, y = 6, (d) x = 8, y = 3, What is ‘A’ in the following reaction, 2Fe3+(aq) +Sn2+(aq) 2Fe2+(aq) + A, (a) Sn3+ (aq), (b) Sn4+ (aq), 2+, (c) Sn (aq), (d) Sn, Given :, NaBr, X Na2HAsO3 + Y NaBrO3 + Z HCl, + H3AsO4 + NaCl, The values of X, Y and Z in the above redox reaction are, respectively, (a) 2, 1, 2, (b) 2, 1, 3, (c) 3, 1, 6, (d) 3, 1, 4, The values of x and y in the following redox reaction, x Cl 2 + 6OH, ClO3 + y Cl + 3H 2O are, (a) x = 5, y = 3, (b) x = 2, y = 4, (c) x = 3, y = 5, (d) x = 4, y = 2, A negative E means that redox couple is a ___A___, than the H+/H2 couple, A positive E means that the redox couple is a ____B___, than H+/H2 couple, (a) A = stronger reducing agent, B = weaker reducing agent, (b) A = stronger oxidising agent, B = weaker oxidising agent, (c) A = weaker oxidising agent, B = stronger oxidising agent, (d) Both (a) and (c), Given E, (i), , Mg2+/Mg(s), E, , (ii), , Ag+/Ag(s), E, , = –2.36, = 0.80, , (iii) Al3+/Al(s), E, , = –1.66, , (iv) Cu2+/Cu(s), E, , = 0.52, , 67., , 68., , 69., , 70., , 71., , Out of the above given elements which is the strongest, oxidising agent and which is the weakest oxidising agent ?, (a) (iv) is the strong whereas (ii) is the weakest oxidising, agent, (b) (ii) is the strongest whereas (i) is the weakest, oxidising agent, (c) (i) is the strongest whereas (ii) is the weakest, oxidising agent, (d) (ii) is the strongest whereas (iii) is the weakest, oxidising agent, Stronger is oxidising agent, more is, (a) standard reduction potential of that species, (b) the tendency to get it self oxidised, (c) the tendency to lose electrons by that species, (d) standard oxidation potential of that species, Standard reduction potentials of the half reactions are, given below :, F2(g) + 2e– 2F– (aq); E° = + 2.85 V, Cl2(g) + 2e– 2Cl–(aq); E° = + 1.36 V, Br2(l) + 2e– 2Br–(aq); E° = + 1.06 V, I2(s) + 2e– 2I–(aq);, E° = + 0.53 V, The strongest oxidising and reducing agents respectively, are :, (a) F2 and I–, (b) Br2 and Cl–, (c) Cl2 and Br–, (d) Cl2 and I2, Standard electrode potentials of redox couples, A2+/A, B2+/B, C/C2+ and D2+/D are 0.3V, – 0.5V, – 0.75V, and 0.9V respectively. Which of these is best oxidising agent, and reducing agent respectively –, (a) D2+/D and B2+/B, (b) B2+/B and D2+/D, 2+, 2+, (c) D /D and C /C, (d) C2+/C and D2+/D, The standard reduction potentials at 298K for the following, half reactions are given against each, Zn2+ (aq) + 2e, Zn(s) ; –0.762 V, Cr3+ (aq) + 3e Cr (s); –0.740 V, 2H+ (aq) + 2e, H2 (g) ; 0.00 V, Fe3+ (aq) + e, Fe2+ (aq) ; 0.770 V, Which is the strongest reducing agent?, (a) Zn (s), (b) Cr (s), (c) H2(g), (d) Fe3+ (aq), Electrode potential data are given below :, 3, Fe (aq, ), , e, , Al3(aq) 3e, , 2, Fe (aq, );, , Al (s) ;, , E° = +0.77 V, E° = – 1.66 V, , Br2 (aq) 2e, 2Br (aq) ; E° = + 1.08V, Based on the data, the reducing power of Fe2+, Al and Br –, will increase in the order, (a) Br– < Fe2+ < Al, (b) Fe2+ < Al < Br –, –, 2+, (c) Al < Br < Fe, (d) Al < Fe2+ < Br–, 72. The standard reduction potentials for Cu2+/Cu; Zn2+/Zn;, Li +/Li; Ag+ /Ag and H+/H2 are + 0.34 V, – 0.762 V,, – 3.05 V, + 0.80 V and 0.00 V respectively. Choose the, strongest reducing agent among the following, (a) Zn, (b) H2, (c) Ag, (d) Li
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REDOX REACTIONS, , 73., , 139, , Given :, o, , E 1, Cl2 / Cl, 2, , E, , o, , Cr2O 72 / Cr3, , 1.36 V, E, , o, Cr3 / Cr, , 1.33V, E o, , 0.74 V,, , MnO 4 / Mn 2, , 1.51V, , The correct order of reducing power of the species, (Cr, Cr3+, Mn2+ and Cl–) will be, (a) Mn2+ < Cl– < Cr3+ < Cr, (b) Mn2+ < Cl3+ < Cl– < Cr, (c) Cr3+ < Cl– < Mn2+ < Cr, (d) Cr3+ < Cl– < Cr < Mn2+, 74., , –, , E Values of some redox couples are given below. On the, basis of these values choose the correct option., –, , 75., , 76., , E values : Br2/Br – = + 1.90; Ag+ /Ag(s) = + 0.80, –, Cu2+ /Cu(s) = + 0.34; I2(s) /I = 0.54, (a) Cu will reduce Br –, (b) Cu will reduce Ag, –, (c) Cu will reduce I, (d) Cu will reduce Br2, Arrange the following in the order of their decreasing, electrode potentials : Mg, K, Ba and Ca, (a) K, Ca, Ba, Mg, (b) Ba, Ca, K, Mg, (c) Ca, Mg, K, Ba, (d) Mg, Ca, Ba, K, The standard electrode potentials of four elements A, B, C, and D are –3.05, –1.66, –0.40 and +0.80. The highest chemical, reactivity will be exhibited by, (a) A, (b) B, (c) C, (d) D, , STATEMENT TYPE QUESTIONS, 77., , 78., , 79., , Which of the following statement(s) is/are correct for the, given reaction?, 2HgCl2 (aq) + SnCl2(aq), Hg2Cl2(s) + SnCl4(aq), (i) Mercuric chloride is reduced to Hg2Cl2, (ii) Stannous chloride is oxidised to stannic chloride, (iii) HgCl2 is oxidised to Hg2Cl2, (iv) It is an example of redox reaction, (a) (i), (ii) and (iv), (b) (i) and (ii), (c) (iii) and (iv), (d) (iii) only, Which of the following sequences of T and F is correct, for given statements. Here T stands for true and F stands, for false statements, (i) Reducing agents lower the oxidation number of an, element in a given substance. These reagents are, also called as reductants, (ii) Reducing agents are acceptor of electrons, (iii) Loss of electron(s) by any species is called oxidation, reaction, (iv) Oxidation and reduction always occur simultaneously., (a) TTTT, (b) TFTT, (c) TFFT, (d) FTTT, If aqueous solution of H2O2 is made acidic. For this which, of the following statement(s) is/are correct ?, (i) This aqueous solution oxidizes I–, (ii) This aqueous solution oxidizes F–, , (a) Both statements (i) and (ii) are correct., (b) Statement (i) is correct and (ii) is incorrect., (c) Statement (ii) is correct and (i) is incorrect., (d) Both statements (i) and (ii) are incorrect., 80. Which of the following statement(s) is/are correct ?, (i) All alkali metals and some alkaline earth metals (Ca,, Sr and Ba) displace hydrogen from cold water., (ii) Magnesium and iron react with steam as well as, acids to produce hydrogen gas., (iii) Cadmium and tin do not react with steam but, displace hydrogen from acids., (a) (i) and (ii), (b) (ii) only, (c) (i) and (iii), (d) (i), (ii) and (iii), 81. Which of the following statements are correct concerning, redox properties?, (i) A metal M for which E° for the half life reaction, Mn+ + ne–, M is very negative will be a good, reducing agent., (ii) The oxidizing power of the halogens decreases from, chlorine to iodine., (iii) The reducing power of hydrogen halides increases, from hydrogen chloride to hydrogen iodide, (a) (i), (ii) and (iii), (b) (i) and (ii), (c) (i) only, (d) (ii) and (iii), 82 Which of the following statement(s) is/are correct ?, (i) A negative value of E– means that the redox couple is, a weaker reducing agent than the H+/H2 couple., (ii) A positive E– means that the redox couple is weaker, reducing agent than the H+/H2., Which of the following code is incorrect regarding above, statements?, (a) Only (i), (b) only (ii), (c) Both (i) and (ii), (d) Neither (i) nor (ii), 83. Which of the following statement(s) is/are correct ?, (i), , Oxidation state of carbon in C3H 4 is –(4/3)., , (ii) Electrons are never shared in fraction., (a) (i) and (ii), (b) Only (i), (c) Only (ii), (d) Neither (i) nor (ii), , MATCHING TYPE QUESTIONS, 84. Match the columns, Column-I, (A) Addition of, electronegative element, (B) Removal of hydrogen, (C) Addition of, electropositive element, (D) Removal of oxygen, (a) (A) – (p), (B) – (q), (C), (b) (A) – (p), (B) – (p), (C), (c) (A) – (p), (B) – (q), (C), (d) (A) – (q), (B) – (q), (C), , Column-II, (p) Oxidation reaction, (q) Reduction reaction, , –, –, –, –, , (q),, (q),, (p),, (p),, , (D), (D), (D), (D), , –, –, –, –, , (p), (q), (q), (p)
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EBD_7207, REDOX REACTIONS, , 140, , 85., , Match the columns, Column-I, (A) 2Mg + O2, 2MgO, (B) Mg + Cl2, MgCl2, (C) 2H2S + O2, , 86., , 87., , 88., Column-II, (p) Removal of hydrogen, (q) Removal of, electropositive element, (r) Addition of oxygen, , 2S + 2H2O, (D) 2KI + H2O + O3, (s) Addition of, 2KOH + I2 + O2, electronegative, element, chlorine, (a) A – (s), B – (q), C – (p), D – (r), (b) A – (r), B – (s), C – (p), D – (q), (c) A – (s), B – (r), C – (q), D – (p), (d) A – (r), B – (p), C – (s), D – (q), Match Column-I (compound) with Column-II (oxidation, state of underlined element) and choose the correct, option., Column - I, Column - II, (A) CuO, (p) 4, (B) MnO2, (q) 3, (r) 2, (C) HAuCl4, (D) Tl2O, (s) 1, (a) A – (r), B – (p), C – (q), D – (s), (b) A – (s), B – (r), C – (p), D – (q), (c) A – (r), B – (s), C – (p), D – (q), (d) A – (s), B – (q), C – (p), D – (r), Match the columns, Column-I, Column-II, (A) V2O5(s) + 5 Ca(s), (p) Disproportionation, 2V(s) + 5 CaO(s), reaction, , 89., , 90., , 91., , (D) 2 KClO3(s), 2KCl(s) + 3O2(g), (a) A – (s), B – (q), C – (r), D, (b) A – (s), B – (r), C – (p), D, (c) A – (r), B – (s), C – (q), D, (d) A – (r), B – (s), C – (p), D, , 92., , 93., , reaction, (s) Displacement, reaction, – (p), – (q), – (p), – (q), , ASSERTION-REASON TYPE QUESTIONS, Directions : Each of these questions contain two statements,, Assertion and Reason. Each of these questions also has four, alternative choices, only one of which is the correct answer. You, have to select one of the codes (a), (b), (c) and (d) given below., (a) Assertion is correct, reason is correct; reason is a correct, explanation for assertion., (b) Assertion is correct, reason is correct; reason is not a, correct explanation for assertion, (c) Assertion is correct, reason is incorrect, (d) Assertion is incorrect, reason is correct., , CaCO3 (s), CaO(s) CO 2 (g) is an example of, decomposition reaction, Reason : Above reaction is not a redox reaction., Assertion : In a reaction, Zn(s) + CuSO4 (aq), ZnSO4(aq) + Cu(s), Zn is a reductant but itself get oxidized., Reason : In a redox reaction, oxidant is reduced by accepting, electrons and reductant is oxidized by losing electrons., , CRITICAL THINKING TYPE QUESTIONS, , (q) Decomposition, (B) CH4(g) + 2O2(g), CO2(g) + 2 H2O(l), reaction, (C) P4(s) + 3OH–(aq) + 3H2O(l) (r) Combination, PH3(g) + 3H 2 PO 2– ( aq ), , Assertion : In the reaction 2Na(s) + Cl2(g), 2NaCl(s), sodium is oxidised., Reason : Sodium acts as an oxidising agent in given, reaction., Assertion : HClO4 is a stronger acid than HClO3., Reason : Oxidation state of Cl in HClO4 is +VII and in HClO3, +V., Assertion : The reaction :, , 94., , 95., , 96., , Among NH3 , HNO3, NaN3 and Mg3N2 the number of, molecules having nitrogen in negative oxidation state is, (a) 1, (b) 2, (c) 3, (d) 4, Fill up the table from the given choice., Element, Oxidation number, in H2O2, Oxygen, –2 in most compounds (i), and (ii) in OF2, Halogen, –1 for (iii) in all its compounds, Hydrogen, (iv) in most of its compounds (v) in, binary metallic hydrides, Sulphur, (vi) in all sulphides, (i) (ii) (iii) (iv) (v) (vi), (a) +1 +1 Cl +1 –1 +2, (b) –1 +2 F, +1 –1 –2, (c) –1 +1 F, +1 +2 +2, (d) +1 +2 Cl +1 +1 +6, The correct decreasing order of oxidation number of oxygen, in compounds BaF2, O3, KO2 and OF2 is, (a) BaO2 > KO2 > O3 > OF2, (b) OF2 > O3 > KO2 > BaO2, (c) KO2 > OF2 > O3 > BaO2, (d) BaO2 > O3 > OF2 > KO2, 2–, Oxidation numbers of P in PO3–, 4 , of S in SO 4 and that of, Cr in Cr2 O 72– are respectively, (a) + 3, + 6 and + 5, (b) + 5, + 3 and + 6, (c) – 3, + 6 and + 6, (d) + 5, + 6 and + 6, When Cl2 gas reacts with hot and concentrated sodium, hydroxide solution, the oxidation number of chlorine, changes from, (a) zero to +1 and zero to –5, (b) zero to –1 and zero to +5, (c) zero to –1 and zero to +3, (d) zero to +1 and zero to –3
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REDOX REACTIONS, , 97., , 98., , Which of the following arrangements represent increasing, oxidation number of the central atom?, (a), , CrO2 , ClO3 , CrO24 , MnO 4, , (b), , ClO3 , CrO42, , (c), , CrO2 , ClO3 , MnO4 , CrO42, , (d), , CrO24 , MnO4 , CrO2 ,ClO3, , 103., , , MnO 4 , CrO 2, , Which of the following act as reducing agents ?, (i), , 99., , 141, , PO34, , 104., , (ii) SO3, , (iii) PO32, (iv) NH3, (a) (i), (ii) and (iii), (b) Only (iii), (c) (i), (iii) and (iv), (d) (iii) and (iv), In the reaction shown below, oxidation state of the carbon, in reactant and product are (i) and (ii) respectively? Is the, given reaction a redox reaction?, Na2CO3(aq) + HCl (aq), Na, , aq, , Cl, , aq, , H 2O, , 105., , 106., , CO 2 g, , (a) (i) 6, (ii) 4, yes, (b) (i) 6, (ii) 6, No, (c) (i) 4, (ii) 4, No, (d) (i) 4, (ii) 4, yes, 100. What products are expected from the disproportionation, reaction of hypochlorous acid?, (a) HCl and Cl2O, (b) HCl and HClO3, (c) HClO3 and Cl2O, (d) HClO2 and HClO4, 101. In the disproportionation reaction, 3 HClO3 HClO4 + Cl2 + 2O2 + H2O, the equivalent mass, of the oxidizing agent is (molar mass of HClO3 = 84.45), (a) 16.89, (b) 32.22, (c) 84.45, (d) 28.15, 102. Consider the following reaction :, z, xMNO4, yC2 O42, zH, xMn 2 2yCO 2, H 2O, 2, , 107., , The value’s of x, y and z in the reaction are, respectively :, (a) 5, 2 and 16, (b) 2, 5 and 8, (c) 2, 5 and 16, (d) 5, 2 and 8, In the balanced chemical reaction, IO3 aI, bH, cH 2 O dI 2, a, b, c and d respectively corresponds to, (a) 5, 6, 3, 3, (b) 5, 3, 6, 3, (c) 3, 5, 3, 6, (d) 5, 6, 5, 5, If equal volume of reactants are used, than no. moles of, KMnO4 (moles per liter) used in acidic medium required to, completely oxidises the 0.5 M FeSO3?, (a) 0.3, (b) 0.1, (c) 0.2, (d) 0.4, Acidic medium used in KMnO4 can be made from which of, the following acids?, (a) HCl, (b) H2SO4, (c) HI, (d) HBr, If rod of a metal (x) is put in a metal ion solution which is, blue in colour , solution turn colorless. The metal rod and, solution respectively are?, (a) Zinc and Cu(II), (b) Zinc and Ni(II), (c) Aluminium and Cu(II) (d) Both (a) and (c), What could be the X–in the system, Where X signifies, halogen ; formation of shown below X2 takes place, when, F2 is purge into aqueous solution of X–?, X2(g), , Aqueous solution of X, , (a) Br –, (c) I –, , (b) Cl –, (d) All of these
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EBD_7207, REDOX REACTIONS, , 142, , FACT / DEFINITION TYPE QUESTIONS, 1., 2., , (a) Addition of oxygen takes place in oxidation., (c) Given reaction is oxidation reaction due to removal, of electropositive element potassium from potassium, ferrocyanide., , 15., 16., 17., , D, , 2Na(s) + H 2 (g) ¾¾® 2NaH(s), , 3., , (d), , 4., 5., , With the careful application of the concept of, electronegativity only S we can find that sodium is, oxidised and hydrogen is reduced., (a) Losing of electron is called oxidation., (b) Oxidation reaction (loss of 2e–), +, , 6., , (a) O.N. of Mn in MnO – is +7 and in Mn, , 7., , difference is of 5 electrons., (a) Ox. no. of Cr on both side is + 6., , 8., , (a), , Br 2, , (d) Let the oxidation no. of N in HNO3 = x, , 21., , it is +2. The, , 4P 3KOH 3H 2 O KH 2 PO 2 PH 3, O.N of P = 0, In KH2PO2 it is + 1, In PH3 it is –3., Hence P is oxidised and reduced., (a) In a redox reaction, one molecule is oxidised and other, molecule is reduced i.e. oxidation number of reactants, are changed., H2, , 20., , 18., , 1 + x + (3 × – 2) = 0, x= +5, , Reduction (gain of 2e ), reaction, , 0, , 19., , 1, 2, (d) Oxidation number of hydrogen when it is bonded to, metals in binary compounds is –1, (a) Auric Chloride = Au(III)Cl3, , In KO2 =, , –, , –, , 0, , (b) Blue colour of the solution disappear due to, formation of Zn 2+., (d) Correct order is Zn > Cu > Ag., (b) For elements, in the free or the uncombined state,, each atom bears an oxidation number of zero., (c) Oxidation number of oxygen in OF2 = + 2., , (Na )2S (s), , 2Na(s) + S(s), , 9., , 14., , (c), , (b), , 12., , (c), , (c), , 1 × ( 1), , x, , for K, , for Co, , N 24 H 44, , loss of 10e, N, , 25., , O.N.of N changes from – 2 to +3, , 0, , (d) (NH4)2 SO4 is split into ions. NH 4 . Let O.N. of N be x, then, 1 × (x) + 4 × (+1) = 1, H, , 26., , (c), , 1, , x = –3, , 1, , |, , H – C 22 – Cl 1 O.N. of C is zero, |, , Cl, , Co 2 (aq) Cu(s), , N 2 6 Y;, , 4× (0), for CO, , O.N. of Co is = –1, , +, , 4Na + O 2 ¾¾, ® 2Na 2 O, , This reaction is a redox reaction as Co undergoes, oxidation whereas Cu+2 undergoes reduction., 13., , K[Co(CO) ], Let O.N. of Co be x then, , 2 HBr, , Co(s) Cu 2 (aq), , +1 -1, , 24., , 1 1, , In this reaction, Na converts into ion (Na+) and Na, donates electrons to oxygen atoms, So, Na behaves, as reducing agent., (b) Zn2+ + 2e– Zn(s), Here electrons are reducing from Zn 2+ to Zn., (c), , +2 +6 -2, , 23., , 22., , Loss of electrons (oxidation), , 11., , +1 +6 -2, , In this reaction, none of the elements undergoes a change, in oxidation number or valency., (a) Let x = oxidation no. of Cr in K2Cr2O7., (2 × 1) + (2 × x) + 7 (– 2) = 0, or 2 + 2x – 14 = 0 or x = + 6., (a) (i) Oxidation state of element in its free state is zero., (ii) Sum of oxidation states of all atoms in compound is, zero., O.N. of S in S8 = 0; O.N. of S in S2F2 = + 1;, O.N. of S in H2S = –2;, , Here H2 is oxidised and Br2 is reduced, thus it is, oxidation-reduction reaction., 10., , +2 -1, , ® Ba S O4 + H 2 O2, Ba O2 + H 2 S O4 ¾¾, , 27., 28., , 1, , (d) O.N. of Fe in (a), (b), (c) and (d) respectively are : +3,, +2, +2 and 0., (d) OsO4, Let O.N. of Os be x then 1 × (x) + 4(–2) = 0, x=8
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REDOX REACTIONS, , 143, , 29., 30., , (a) Fe(CO)5 is metal carbonyl, hence O.N. of Fe is zero., (c) O.N. of Mn in K2MnO4 is +6, , 31., , (d), , 47. (b) Fluorine is so reactive that it attacks water and, displaces the oxygen of water :, , MnO – (O.S. of Mn +7); Cr (CN) 36– (O.S. of Cr +3),, , NiF62 – (O.S. of Ni +4) and CrO2Cl2 (O.S. of Cr +6), 32., , (d) O.N. of carbon in CH3CHO is –1; in other cases it is, zero., , 33., , (b) O.N. of iodine in I3 is –1/3, , 34., 35., , (b) M3+ on losing 3 elections will become M+6 and O.N. = + 6., (b) The method of representing oxidation number by a, Roman numeral within the paranthesis represents Stock, notation., (c) Carbon has the maximum oxidation state of + 4,, therefore carbon dioxide (CO2 ) cannot act as a, reducing agent., , 36., , 37., , 38., , 39., , 40., , 41., 42., , 43., 44., 45., , 46., , (b) Let the oxidation no. of Fe in Fe3O 4, , x, , 3x + (– 2 × 4) = 0 or, 3x = 8, 8, x, 3, (c) Let oxidation state of oxygen in OF2 = x, x + ( – 1 × 2) = 0, x= +2, (d) In H2O2 : 2 × (+1) + 2 × x = 0, x = –1, In CO2 :, 4 + 2x = 0 x = –2, In H2O :, 2 × (+1) + x = 0 x = –2, In F2O :, 2 × (–1) + x = 0 x = +2, (d), , NO3, 5, , 1 2, 0, 2H 2O( I ) 2F2 ( g ), , 1 1, 0, 4HF( aq ) O 2 ( g ), , 48. (b) As fluorine is the strongest oxidising agent; there is, no way to convert F– ions to F2 by chemical means., The only way to achieve F2 from F– is to oxidise it, electrolytically., 49. (c) F2 being most electronegative element cannot exhibit, any positive oxidation state., In ClO –4 chlorine is present in its highest oxidation, state i.e + 7. Therefore it does not show, disproportionation reaction., 50. (c) A reaction, in which a substance undergoes, simultaneous oxidation and reduction, is called, disproportionation reaction. In these reactions, the, same substance simultaneously acts as an oxidising, agent and as a reducing agent. Here Cl undergoes, simultaneous oxidation and reduction., , 2KOH Cl 2, 0, , KCl KOCl H 2O., 1, , 1, , 51. (d) In disproportionation reaction, one element of a, compound will simultaneously get reduced and, oxidised. In ClO4 , oxidation number of Cl is + 7 and it, can not increase it further. So, ClO4 will not get, oxidised and so will not undergo disporportionation, reaction., , –, , N 2 H 4 So, for reduction of 1 mole of NO, 2, , 3 number of electrons required is 7., (a) Let the oxidation state of S be x., S4O62– 4x – 12 = – 2 4x = 10 x = 10/4 = 2.5, (c) Species, O.N., Cl –, –1, +7, ClO4–, ClO–, +1, +7, MnO4 –, In ClO– chlorine is in +1 oxidation state which can be, increased or decreased thus it acts as an oxidising or, reducing agent., In other given species the underlined elements are either, in their minimum or maximum oxidation state., (a), (b) O.N. of N changes from + 2 to + 5 hence NO is reducing., (b) In KMnO4 : Let O.N. of Mn be x, +1 + x + 4(–2) = 0 x = +7, In K2Cr2O7 : Let O.N. of Cr be x, 2(1) + 2x + 7(–2) = 0 x = +6, (d) The element is Ti (At. no. 22). Electronic configuration, is 1s2, 2s2p6, 3s2p6d2, 4s2. the energy level of 3d and, 4s is very close. It can have Ti4+ O.S., , 52. (d), , 3Br2, , 6CO32 –, , 3H 2 O, , 5Br –, , BrO3–, , 6HCO 3–, , O.N. of Br 2 changes from 0 to –1 and +5, hence it is reduced as well as oxidised., 53. (c), 54. (b) Phosphorus, sulphur and chlorine disproportionate in, the alkaline medium., 55. (b), , D, , 2H 2 O ¾¾® 2H 2 + O 2, , There is decomposition of H2O molecule into H2 and, O2., 56. (a) H+ ions are added to the expression on the, appropriate side so that the total ionic charges of, reactants and products become equal., 57. (d) Since reaction is occuring in basic medium therefore, 2OH– are added on right side., 2MnO 4 (aq) + Br– (aq), , 2MnO2(s) + Br O3– (aq) + 2OH–(aq), Now, hydrogen atoms balanced by adding one H2O, molecule to the left side, 2MnO 4 (aq) Br (aq) H 2 O( ), 2MnO2(s) + BrO 3 (aq) 2 OH (aq)
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EBD_7207, REDOX REACTIONS, , 144, , 58., , 2 NH 3, , 59., , 73., , (c) The balanced equation :, OCl, , N 2H 4, , Cl, , H 2O, , (b) The balanced equation is, 2C2 H6 7O2, 4CO 2 6H 2O., Ratio of the coefficients of CO2 and H2O is 4 : 6 or 2 : 3., , 60., , (a), , 2MnO 4–, , 61., , (d), , 8KMnO 4, , 5H 2 O 2, , 2Mn 2, , 6H, , 3NH 3, , 8MnO 2, , 5O 2, , 8H 2 O., , 3KNO 3, , 5KOH, , 74., 75., , 2H 2O, , 62., , 64., , E, , Reduction, , (b), 2Fe3, , 63., , Sn 2, , 2Fe2, , E, , Sn 4, , E, , Oxidation, (c) On balancing the given reaction, we find, 3Na2HAsO3 + NaBrO3 + 6HCl, 6NaCl + 3H3AsO4 + NaBr, , (c), o, , x Cl 2 6 H –, , 5, , –1, , ClO 3 yCl, –1, , 3H 2 O, , E, , 3Cl2, , 6OH, , (d) Negative E, , ClO 3, , 5Cl, , 76., , 3H 2 O, , Stronger reducing agent or weaker, oxidising agent, , Positive E, , 66., 67., 68., , 69., , 70., 71., , 72., , Weaker reducing agent or stronger, oxidising agent., (b) Strongest oxidising agent = Ag+/Ag(s), Weakest oxidising agent = Mg2+/Mg(s), (a) More is E°RP , more is the tendency to get itself reduced, or more is oxidising power., (a) Higher the value of reduction potential higher will be, the oxidising power whereas the lower the value of, reduction potential higher will be the reducing power., (c) The redox couple with maximum reduction potential, will be best oxidising agent and with minimum reduction, potential will be best reducing agent., (a) Since oxidation potential of Zn is highest hence, strongest reducing agent., (a) Fe, Al, Br, 0.77, –1.66 1.08 E°Red, –0.77 1.66, –1.08 E°Oxi, Hence, reducing power Al > Fe2+ > Br–, (d) More the negative reduction potential, more is the, tendency to lose electron. The reducing power, increases as the standard reduction potential becomes, more and more negative., Thus, Li is the strongest reducing agent as the, standard reduction potential of Li+/ Li is most negative,, –3.05 V., , 2.925, , K |K, , Ba 2 |Ba, Ca 2 |Ca, , Mg 2 |Mg, , 2.90, 2.87, , 2.37, , Highly negative value of E red shows the least value, , on balancing the eq we get, 65., , (a) Lower the value of reduction potential higher will be, reducing power hence the correct order will be, Mn2+ < Cl– < Cr3+ < Cr, (d), (d) Order of decreasing electrode potentials of Mg, K, Ba, and Ca is, Mg > Ca > Ba > K, It can be explained by their standard reduction, potentials., , of electrode potential., (a) Standard electrode potential i.e. reduction potential of, A is minimum (–3.05V) i.e. its oxidation potential is, maximum which implies ‘A’ is most reactive chemically., , STATEMENT TYPE QUESTIONS, 77., 78., 79., 80., , (a) For statement (iii), HgCl2 is reduced to Hg2Cl2, (b) For statement (ii) reducing agents are donor of electrons., (b) H2O2 is strong oxidizing than I2, reduction potential, of H2O2 is greater than that of I2., (d) All the given statements are correct., , 81., , (a) (i) Mn+ + ne–, , M, for this reaction, high, , negative value of E° indicates lower reduction, potential, that means M will be a good reducing, agent., Stronger reducing agent, Lower reduction potential, , Easy to oxidise, , higher oxidation potential, , (ii) Element, F, Cl, Br, I, Reduction potential +2.87 +1.36 +1.06 +0.54, (E° volt), As reduction potential decreases from fluorine to, iodine, oxidising nature also decreases from fluorine, to iodine., (iii) The size of halide ions increases from F – to I –., The bigger ion can loose electron easily. Hence the, reducing nature increases from HF to HI., 82., 83., , (a), (a) –(4/3) is the average oxidation state of C in C3H4.
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REDOX REACTIONS, , 145, , Oxidation state of N in NaN3 is, , MATCHING TYPE QUESTIONS, 84., , 85., 86., , 87., , (b) Oxidation is addition of electronegative or removal, of electroposition element to a substance or removal, of hydrogen from a substance., Reduction is addition of electropositive or removal, of electropositive element or removal of oxygen from, a substance., (b), (a) CuO, +2, MnO2, +4, HAuCl4, +3, Tl2O, +1, (b), , ASSERTION-REASON TYPE QUESTIONS, 88., , 89., , 90., , (c) In reaction 2Na(s) + Cl2(g), 2NaCl(s) sodium is, oxidised by loss of electrons and acts as a reducing, agent (donor of electrons)., (b) Both Assertion and Reason are true but reason is not, the correct explanation of assertion. Greater the number, of negative atoms present in the oxy-acid make the, acid stronger. In general, the strengths of acids that, have general formula (HO)m ZOn can be related to the, value of n. As the value of n increases, acidic character, also increases. The negative atoms draw electrons, away from the Z-atom and make it more positive. The, Z-atom, therefore, becomes more effective in, withdrawing electron density away from the oxygen, atom that bonded to hydrogen. In turn, the electrons, of H – O bond are drawn more strongly away from the, H-atom. The net effect makes it easier from the proton, release and increases the acid a strength., (b) Decomposition of calcium carbonate is not a redox, reaction., , 1, 3, Oxidation state of N in Mg3N2 is, 3 × 2 + 2x = 0 or x = –3, Thus 3 molecules (i.e. NH3, NaN3 and Mg3N2 have, nitrogen in negative oxidation state., + 1 + 3x = 0 or x = –, , 93. (b), 94. (b) Oxidation no. of O are + 2, 0, – 1/2 and – 1 respectively, 95. (d), , PO3–, 4 = x + 4 (– 2) = – 3; x – 8 = – 3; x = + 5, SO 2–, 4 = x + 4 (– 2) = – 2; x – 8 = – 2; x = + 6, Cr2 O72– = 2x + 7 (– 2) = – 2; 2x – 14 = – 2;, , 2x =12; x = + 6, 96. (b) On reaction with hot and concentrated alkali a mixture, of chloride and chlorate is formed, 3Cl2 + 3 NaOH(excess), 1, , 5NaCl, , Zn2+ (aq) + Cu(s), , CRITICAL THINKING TYPE QUESTIONS, 92., , 1, , 3HO Cl, , Reduction gain of 2e–, , (c) Calculating the oxidation state of nitrogen in given, molecules;, Oxidation state of N in NH3 is, x + 3 × (+ 1) = 0 or x = – 3, Oxidation state on N in NaNO3 is, 1 + x + 3 × (– 2) = 0 or x = + 5, , 3H 2O, , Oxidation, , (a), Zn(s) + Cu2+(aq), , 5, , NaClO3, , 97. (a), 98. (d) In (i) and (ii) both P and S are in highest oxidation, state. In (iii) and (iv) ; P has oxidation state of +4, which can be oxidized to +5 state, while in case of, NH3 nitrogen has oxidation state of –3 which can be, oxidised., 99. (c) The redox reaction involve loss or gain of electron(s), i.e. change in oxidation state. Given reaction is not a, redox reaction as this reaction involves no change in, oxidation state of reactant or product., 100. (b) During disproportionation same compound undergo, simultaneous oxidation reduction., , Oxidation loss of 2e–, 91., , Hot, , 5, , 2HCl HClO 3, , Reduction, 101. (a), , Cl20, , ClO3, x 6, x, , 1, 5, , x, , 0, , x, , 0 x = oxidation number, , Equivalent mass =, , 84.45, Molecular mass, =, 5, Oxidation number, , 16.89, , 102. (c) On balancing the given equations, we get, 2MnO 4, , 5C 2 O 4 2, , 16H, , 2Mn, , 10CO 2 8H 2O, , So, x = 2, y = 5 & z = 16
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9, , HYDROGEN, , FACT / DEFINITION TYPE QUESTIONS, 1., , 2., , 3., , 4., , 5., , Following are some properties of hydrogen which of the, following properties resemble with alkali metals and which, with halogens, (i) Hydrogen lose one electron to form unipositive ions, (ii) Hydrogen gain one electron to form uninegative ions, (iii) Hydrogen forms oxides, halides and sulphides, (iv) Hydrogen has a very high ionization enthalpy, (v) Hydrogen forms a diatomic molecule, combines with, elements to form hydrides and covalent compounds., (a) Alkali metals resemble (i), (iii) and (iv), Halogens resemble (ii) and (v), (b) Alkali metals resemble (i) and (iii), Halogens resemble (ii), (iii) and (v), (c) Alkali metals resemble (i) and (iii), Halogens resemble (ii), (iv) and (v), (d) Alkali metals resemble (i) only, Halogens resemble (iv) and (v), Hydrogen molecules differs from chlorine molecule in the, following respect, (a) Hydrogen molecule is non-polar but chlorine molecule, is polar, (b) Hydrogen molecule is polar while chlorine molecule is, non-polar, (c) Hydrogen molecule can form intermolecular hydrogen, bonds but chlorine molecule does not, (d) Hydrogen molecule cannot participate in coordination, bond formation but chlorine molecule can, Hydrogen can behave as a metal, (a) at very high temperature, (b) at very low temperature, (c) at very high pressure, (d) at very low pressure, The property of hydrogen which distinguishes it from alkali, metals is, (a) its electropositive character, (b) its affinity for non metal, (c) its reducing character, (d) its non-metallic character, Hydrogen accepts an electron to form inert gas, configuration. In this it resembles, , (a) halogen, (b) alkali metals, (c) chalcogens, (d) alkaline earth metals, 6., Which of the following statements is correct ?, (a) Hydrogen has same IP as alkali metals, (b) Hydrogen has same electronegativity as halogens, (c) It has oxidation number of –1 and +1, (d) It will not be liberated at anode, 7., Why does H+ ion always get associated with other atoms, or molecules?, (a) Ionisation enthalpy of hydrogen resembles that of, alkali metals., (b) Its reactivity is similar to halogens., (c) It resembles both alkali metals and halogens., (d) Loss of an electron from hydrogen atom results in a, nucleus of very small size as compared to other atoms, or ions. Due to small size it cannot exist free., 8., Which one of the following is not an isotope of hydrogen ?, (a) Deuterium, (b) Tritium, (c) Ortho hydrogen, (d) None of these, 9., Number of neutrons in three isotopes of hydrogen,, protium, deuterium and tritium respectively is, (a) 0, 1, 2, (b) 1, 1,1, (c) 2, 1, 0, (d) 2, 0, 1, 10. Which isotope(s) of hydrogen is/are radioactive and, emits low energy – particles?, (i) Protium, (ii) Tritium, (iii) Deuterium, (a) (i) and (ii), (b) (iii) only, (c) (ii) only, (d) (ii) and (iii), 11. Hydrogen bond energy is equal to :, (a) 3-7 cals, (b) 30-70 cals, (c) 3-10 kcals, (d) 30-70 kcals, 12. Which of the following reaction(s) represents commercial, method for production of dihydrogen?, (i), , CO(g) + H2O(g), , (ii) 2H2O(l), (iii), (iv), (a), (c), , 673K, catalyst, , electrolysis, traces of acid/base, , CO2(g) + H2(g), 2H2(g) + O2(g), , Zn + 2H +, Zn2+ + H2, CH4(g) + H2O(g) 1270, CO(g) + 3H2(g), Ni, (i), (ii) and (iii), (b) (iii) only, (i), (ii) and (iv), (d) (ii), (iii) and (iv)
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EBD_7207, HYDROGEN, , 148, , 13., , 14., , Which of the following is formed when zinc reacts with, sodium hydroxide?, (a) Hydrogen gas, (b) Sodium zincate, (c) Zinc oxide, (d) Both (a) and (b), Identify x and y in following reaction. What is the mixture, of x and y called?, CH4(g) + H2O(g), , 15., , 16., , 1270 K, Ni, , (a) x = CO2, y = H2O, water gas, (b) x = CO, y = H2O, syn gas, (c) x = CO, y = H2, water gas, (d) x = CO2, y = H2, syn gas, Why is water gas (mixture of CO and H2) also called ‘syn, gas’?, (a) Because it is synthesised from sewage, saw – dust,, scrap wood etc., (b) Because it is synthesised from methane gas, (c) Because it is used in the synthesis of methanol and, a number of hydrocarbons., (d) None of these, Which of the following statements is correct?, (a) Production of syngas from coal is called coal, gasification., (b) CO(g) + H2O (g), , 17., , 18., , 19., , 20., , 21., , 22., , 24., , x+y, , 673K, catalyst, , 23., , 25., , 26., , 27., , CO2(g) + H2 (g), , represents water gas shift reaction., (c) CO2 formed in water gas shift reaction is removed by, scrubbing with sodium zincate solution., (d) Both (a) and (b), Which one of the following pairs of substances on reaction, will not evolve H2 gas?, (a) Iron and H2SO4 (aqueous), (b) Iron and steam, (c) Copper and HCl (aqueous), (d) Sodium and ethyl alcohol, Which of the following metal evolves hydrogen on reacting, with cold dilute HNO3 ?, (a) Mg, (b) Al, (c) Fe, (d) Cu, Hydrogen is evolved by the action of cold dil. HNO3 on, (a) Fe, (b) Mn, (c) Cu, (d) Al, In Bosch’s process which gas is utilised for the production, of hydrogen gas ?, (a) Producer gas, (b) Water gas, (c) Coal gas, (d) None of these, Hydrogen is not obtained when zinc reacts with, (a) Cold water, (b) dil. HCl, (c) dil. H2SO4, (d) Hot NaOH (20%), Which one of the following pairs of substances will not, produce hydrogen when reacted together?, (a) Copper and conc. nitric acid, (b) Ethanol and metallic sodium, (c) Magnesium and steam, (d) Phenol and metallic sodium, , 28., , 29., , 30., , 31., , Very pure hydrogen (99.9) can be made by which of the, following processes ?, (a) Reaction of methane with steam, (b) Mixing natural hydrocarbons of high molecular weight, (c) Electrolysis of water, (d) Reaction of salts like hydrides with water, Which of the following is formed on reaction of carbon, monoxide gas with dihydrogen in presence of cobalt as, a catalyst?, (a) Methanal, (b) Methanol, (c) Methane, (d) Formic acid, Which of the following is not a use of dihydrogen ?, (a) It used in fuel cells for generating electrical energy., (b) Atomic hydrogen and oxy-hydrogen torches are, used for cutting and welding purposes., (c) It used in the synthesis of hydroquinone and tartaric, acid., (d) Both (b) and (c), Elements of which of the following group do not form, hydrides?, (a) Alkali metals, (b) Halogens, (c) Alkaline earth metals (d) Noble gases, Which of the following statements is incorrect?, (a) Ionic hydrides are stoichiometric compounds of, dihydrogen formed with most of s-block elements, (b) Ionic hydrides are crystalline, non-volatile and nonconducting in solid state., (c) Melts of ionic hydrides conduct electricity and, liberate dihydrogen gas at cathode., (d) Both (a) and (c), Saline hydrides react explosively with water, such fires can, be extinguished by, (a) water, (b) carbon dioxide, (c) sand, (d) None of these, Choose the correct option for following hydrides., B2H6, CH4, NH3 and HF, (a) Electron deficient hydride = B2H6 and HF, Electron precise hydride = CH4, Electron rich hydride = NH3, (b) Electron deficient hydride = B2H6, Electron precise hydride = CH4, Electron rich hydride = NH3 and HF, (c) Electron deficient hydride = CH4, Electron precise hydride = B2H6, Electron rich hydride = NH3 and HF, (d) Electron deficient hydride = CH4 and HF, Electron precise = B2H6, Electron rich hydride = NH3,, Elements of which of the following group(s) of periodic, table do not form hydrides., (a) Groups 7, 8, 9, (b) Group 13, (c) Groups 15, 16, 17, (d) Group 14, Which hydride is an ionic hydride ?, (a) H2S, (b) TiH1.73, (c) NH3, (d) NaH
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HYDROGEN, , 32., 33., 34., 35., , 36., , 37., , 38., , 39., , 40., 41., , 149, , Metal hydride on treatment with water gives, (a) H2O2, (b) H2O, (c) Acid, (d) Hydrogen, The polymeric hydride is, (a) CaH2, (b) NaH, (c) BaH2, (d) MgH2, Ionic hydrides reacts with water to give, (a) acidic solutions, (b) hydride ions, (c) basic solutions, (d) electorns, Which of the following in incorrect statement?, (a) s-block elements, except Be and Mg, form ionic hydride, (b) BeH4 , MgH2, CuH2, ZnH2 , CaH2 and HgH2 are, intermediate hydride, (c) p-block elements form covalent hydride, (d) d-and f-block elements form ionic hydride, Metal hydrides are ionic, covalent or molecular in nature., Among LiH, NaH, KH, RbH, CsH, the correct order of, increasing ionic character is, (a) LiH > NaH > CsH > KH > RbH, (b) LiH < NaH < KH < RbH < CsH, (c) RbH > CsH > NaH > KH > LiH, (d) NaH > CsH > RbH > LiH > KH, LiAlH4 is used as :, (a) An oxidizing agent, (b) A reducing agent, (c) A mordant, (d) A water softener, Water is :, (a) more polar than H2S, (b) more or less identical in polarity with H2S, (c) less polar than H2S, (d) None of these, In gas phase water is A molecule with a bond angle, of, B, and O–H bond length of, C d, (a) A = Bent, B = 100.5°, C = 95.7 pm, (b) A = Bent, B = 104.5°, C = 95.7 pm, (c) A = Bent, B = 109.5°,C = 99.7 pm, (d) A = Bent, B = 104.5°, C = 99.7 pm, The H–O–H angle in water molecule is about, (a) 90º, (b) 180º, (c) 102.5°, (d) 104.5º, Identify the structuer of water in the gaseous phase., (a), , +, , (b) H – O – H, H, , H–O–H, 2 –, , (c), , O, , +, , H, , 42., , 95., 7p, m, 104.5°, , +, , (d) None of these, , H, , The unusual properties of water in the condensed phase, (liquid and solid states) are due to the, (a) presence of hydrogen and covalent bonding between, the water molecules, (b) presence of covalent bonding between the water, molecules, (c) presence of extensive hydrogen bonding between, water molecules, (d) presence of ionic bonding, , 43. The boiling point of water is exceptionally high because, (a) there is covalent bond between H and O, (b) water molecule is linear, (c) water molecules associate due to hydrogen bonding, (d) water molecule is not linear, 44. Water possesses a high dielectric constant, therefore :, (a) it always contains ions, (b) it is a universal solvent, (c) can dissolve covalent compounds, (d) can conduct electricity, 45. At its melting point ice is lighter than water because, (a) H2O molecules are more closely packed in solid state, (b) ice crystals have hollow hexagonal arrangement of H2O, molecules., (c) on melting of ice the H2O molecule shrinks in size, (d) ice froms mostly heavy water on first melting., 46. The low density of ice compared to water is due to, (a) hydrogen-bonding interactions, (b) dipole-dipole interactions, (c) dipole-induced dipole interactions, (d) induced dipole-induced dipole interactions, 47. When two ice cubes are pressed over each other, they unite, to form one cube. Which of the following forces is, responsible to hold them together ?, (a) Hydrogen bond formation, (b) Van der Waals forces, (c) Covalent attraction, (d) Ionic interaction, 48. Which of the following reactions is an example of use of, water gas in the synthesis of other compounds?, 1270K, Ni, , (a) CH4(g) + H2O(g), (b) CO(g) + H2O(g), , 673K, Catalyst, , (c) CnH2n+2 + nH2O (g), (d) CO(g) + 2H2(g), , CO(g) + H2(g), CO2(g) + H2(g), , 1270K, Ni, , Cobalt, Catalyst, , nCO + (2n + 1)H2, , CH3OH(l), , 49. Which of the following metals reacts with H 2 O at room, temp?, (a) Ag, (b) Fe, (c) Al, (d) Na, 50. Which of the following statements do not define the, characteristic property of water “Water is a universal, solvent”, (a) It can dissolve maximum number of compounds, (b) It has very low dielectric constant, (c) It has high liquid range, (d) None of these, 51. Which of the following groups of ions makes the water, hard?, (a) Sodium and bicarbonate, (b) Magnesium and chloride, (c) Potassium and sulphate, (d) Ammonium and chloride.
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EBD_7207, HYDROGEN, , 150, , 52., 53., , The process used for the removal of hardness of water is, (a) Calgon, (b) Baeyer, (c) Serpeck, (d) Hoope, When zeolite (hydrated sodium aluminium silicate) is treated, with hard water the sodium ions are exchanged with, (a) H+ ions, (b) Ca2+ ions, 2 ions, (c), (d) OH– ions, SO 4, , 54., , 55., , 56., , 57., , 58., , 59., 60., , 61., , 62., , 63., , 64., , 65., 66., , 67., , Calgon used as a water softener is, (a) Na 2 [Na 4 (PO 3 )6 ], (b) Na 4 [ Na 2 (PO 3 ) 6 ], (d) Na 4 [ Na 2 ( PO 4 ) 6 ], (c) Na 4 [Na 4 (PO 4 )5 ], Polyphosphates are used as water softening agents because, they, (a) form soluble complexes with anionic specise, (b) precipitate anionic species, (c) forms soluble complexes with cationic species, (d) precipitate cationic species, Permanent hardness of water can be removed by adding, Calgon (NaPO3)n. This is an example of, (a) Adsorption, (b) Exchange of ion, (c) Precipitation, (d) None, Which one the following removes temporary hardness of, water ?, (a) Slaked lime, (b) Plaster of Paris, (c) Epsom, (d) Hydrolith, Permanent hardness of water is due to the presence of, (a) bicarbonates of sodium and potassium, (b) chlorides and sulphates of sodium and potassium, (c) chlorides and sulphates of calcium and magnesium, (d) bicarbonates of calcium and magnesium, In lab H2O2 is prepared by, (a) Cold H2SO4 + BaO2 (b) HCl + BaO2, (c) Conc. H2SO4 + Na2O2 (d) H2 + O2, HCl is added to following oxides. Which one would give, H2O2, (a) MnO2, (b) PbO2, (c) BaO, (d) None, The oxide that gives H2O2 on treatment with dilute H2SO4, is–, (a) PbO2, (b) BaO2, (c) MnO2, (d) TiO2, 30 volume hydrogen peroxide means, (a) 30% of H 2 O 2 solution, (b) 30 cm3 solution contains 1g of H 2 O 2, (c) 1 cm3 of solution liberates 30 cm3 of O2 at STP, (d) 30 cm3 of solution contains 1 mole of H2O2, The volume strength of 1.5 N H 2 O 2 solution is :, (a) 8.4, (b) 8.0, (c) 4.8, (d) 3.0, Commercial 10 volume H2O2 is a solution with a strength of, approximately, (a) 15%, (b) 3%, (c) 1%, (d) 10%, , The structure of H2O2 is, (a) planar, (b) non planar, (c) spherical, (d) linear, The O – O – H bond angle in H2O2 is, (a) 106°, (b) 109 28', (c) 120°, (d) 94.8°, Which of the following is the true structure of H2O2 ?, (a) H– O – O – H, , (c), , 68., , 69., 70., 71., , 72., , 73., , 74., , 75., , H, , (b), , H, , O O., H, H, In the hydrogen peroxide molecule :, (a) O – H bonds are polar but molecule is, non-polar., (b) The four atoms are arranged in a non-linear and nonplanar manner., (c) All the four atoms are in same plane., (d) Two hydrogen atoms are connected to one of the, oxygen., H2O2 is a, (a) Weak acid, (b) Weak base, (c) Neutral, (d) None of these, When H2O2 is oxidised the product is, (a) OH–, (b) O2, (c) O2–, (d) HO2 –, Which of the following is false about H2O2, (a) Act as both oxidising and reducing agent, (b) Two OH bonds lies in the same plane, (c) Pale blue liquid, (d) Can be oxidised by ozone, In which of the following reactions, H2O2 acts as a reducing, agent, (a) PbO2(s) + H2O2 (aq) PbO(s) + H2O( ) + O2 (g), (b) Na2SO3 (aq) + H2O2(aq) Na2SO4(aq) + H2O( ), (c) 2KI(aq) + H2O2 (aq) 2KOH (aq) + I2(s), (d) KNO2(aq) + H2O2 (aq) KNO3 (aq) + H2O( ), H2O2, 2H+ + O2 + 2e– ; E° = –0.68 V. This equation, represents which of the following behaviour of H2O2., (a) Reducing, (b) Oxidising, (c) Acidic, (d) Catalytic, The reaction, H 2S H 2 O 2, S 2H 2 O manifests, (a) Acidic nature of H2O2, (b) Alkaline nature of H2O2, (c) Oxidising action of H2O2, (d) Reducing action of H2O2., Which of the following statements is incorrect ?, (a) H2O2 can act as an oxidising agent, (b) H2O2 can act as a reducing agent, (c) H2O2 has acidic properties, (d) H2O2 has basic properties, O=O, , (d), , H, |, O O, |, H
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HYDROGEN, , 76., , 77., , Consider the reactions, (A) H2O2 + 2HI I2 + 2H2O, (B) HOCl + H2O2 H3O+ + Cl– + O2, Which of the following statements is correct about H2O2, with reference to these reactions? Hydrogen peroxide is, ______ ., (a) an oxidising agent in both (A) and (B), (b) an oxidising agent in (A) and reducing agent in (B), (c) a reducing agent in (A) and oxidising agent in (B), (d) a reducing agent in both (A) and (B), Which of the following equations depict the oxidising nature, of H2O2?, , 2Mn2+ + 8H2O + 5O2, 2MnO4 6H, 5H 2 O2, (b) 2Fe3+ + 2H+ + H2O2 2Fe2+ + 2H2O + O2, (c) 2I– + 2H+ + H2O2 I2 + 2H2O, (d) KIO4 + H2O2 KIO3 + H2O + O2, 78. Which one of the following undergoes reduction with, hydrogen peroxide in an alkaline medium ?, (a) Mn 2+, (b) HOCl, (c) PbS, (d) I2, 79. Which of the following does not represent reducing, action of H2O2?, (a) PbS(s) + 4H2O2(aq), PbSO4(s) + 4H2O(l), (b) HOCl + H2O2, H3O + + Cl¯ + O2, 2MnO2 + 3O2 +2H2O+2OH¯, (c) 2MnO4– + 3H2O2, (d) I2 + H2O2 + 2OH¯, 2I¯ + 2H2O + O2, 80. Which of the following is not true for hydrogen peroxide?, (a) H2O2 decomposes slowly on exposure to light., (b) It is kept away from dust because dust can induce, explosive decomposition of the compound., (c) H2O2 is used as bleaching agent for textiles, paper, pulp etc., (d) It is used as a moderator in nuclear reactor., 81. The decomposition of H2O2 is accelerated by –, (a) glycerine, (b) alcohol, (c) phosphoric acid, (d) Pt powder, 82. H2O2 is always stored in black bottles because, (a) It is highly unstable, (b) Its enthalpy of decomposition is high, (c) It undergo auto-oxidation on prolonged standing, (d) None of these, 83. H2O2 is, (a) Poor polar solvent than water, (b) Better polar solvent than H2O, (c) Both have equal polarity, (d) Better polar solvent but its strong auto-oxidising ability, limits its use as such, 84. Which of the following is wrong about H2O2? It is used, (a) As aerating agent in production of spong rubber, (b) As an antichlor, (c) For restoring white colour of blackened lead painting, (d) None of these, 85.. Heavy water is represented as, (b) D2O, (a) H2 18O, (d) H2O at 4°C, (c) D2 18 O, (a), , 151, , 86. What is formed when calcium carbide reacts with heavy, water?, (a) C2D2, (b) CaD2, (c) Ca2D2O, (d) CD2, 87. D2O is used in, (a) motor vehicles, (b) nuclear reactor, (c) medicine, (d) insecticide, 88. Complete the following reaction., Al4C3 + D2O, x+y, (a) x = C2D2 and y = Al(OD)3, (b) x = CD4 and y = Al(OD)3, (c) x = CO2 and y = Al2D3, (d) x = CD4 and y = Al2D3, 89. Which of the following is correct about heavy water ?, (a) Water at 4°C having maximum density is known as, heavy water, (b) It is heavier than water (H2O), (c) It is formed by the combination of heavier isotope of, hydrogen with oxygen, (d) None of these, 90. D2O is preferred to H2O, as a moderator, in nuclear reactors, because, (a) D2O slows down fast neutrons better, (b) D2O has high specific heat, (c) D2O is cheaper, (d) None of these, 91. The numbers of protons, electrons and neutrons in a, molecule of heavy water are respectively :, (a) 8, 10, 11, (b) 10, 10, 10, (c) 10, 11, 10, (d) 11, 10, 10, 92. Choose the incorrect statement, (a) Dihydrogen can release more energy than petrol., (b) The only pollutant in combustion of dihydrogen is, carbon dioxide., (c) Hydrogen economy is based on the principle of, transportation and storage of energy in the form of, liquid or gaseous dihydrogen, (d) Hydrogen economy has advantage that energy is, transmitted in the form of dihydrogen and not as, electric power., 93. Which of the following fuel is used for runnning the, automobiles first time in the history of India during October, 2005?, (a) D2O, (b) H2O2, (c) D2, (d) H2, , STATEMENT TYPE QUESTIONS, 94. The storage tanks used for H2 are made up of which metal, alloy(s), (i) NaNi5, (ii) B2H6, (iii) Ti–TiH2, (iv) Mg–MgH2, (a) (iii) and (iv), (b) (i) and (ii), (c) (i), (iii) and (iv), (d) (ii), (iii) and (iv)
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EBD_7207, HYDROGEN, , 152, , 95., , 96., , 97., , 98., , 99., , Which of the following sequence of T and F is correct for, given statements? Here T stands for true and F stands for, false statement, (i) The H–H bond dissociation enthalpy is highest for, a single bond between two atoms of any element, (ii) H2 is relatively inert at room temperature., (iii) Hydrogen combines with almost all the elements due, to its incomplete orbital, (iv) The atomic hydrogen is produced at high temperature, in an electric arc or under UV radiations., (a) TTTT, (b) FTFT, (c) FTTT, (d) FTTF, Which of the following statement(s) is/are incorrect?, (i) Dihydrogen reduces copper (II) oxide to copper, (ii) Reaction of dihydrogen with sodium gives sodium, hydride., (iii) Hydroformylation of olefins yields aldehydes which, further undergo reduction to give alcohols., (iv) Hydrogenation of vegetable oils using iron as, catalyst gives edible fats., (a) (i), (ii) and (iii), (b) (i) and (iv), (c) (iv) only, (d) (iii) and (iv), Choose the correct sequence of T and F for following, statements. Here T stands for true and F stands for false, statement., (i) At atmospheric pressure ice crystallises in the, hexagonal form, but at very low temperatures it, condenses to cubic form., (ii) Density of ice is less than that of water. Therefore,, an ice cube floats on water., (iii) In winter season ice formed on the surface of a lake, makes survival of the aquatic life difficult., (iv) Hydrogen bonding gives ice a open type structure, with wide holes., (a) TTFT, (b) FTFT, (c) FTTT, (d) TFTT, Which of the following statements are correct ?, (i) Hydrogen peroxide is industrially prepared by the, auto-oxidation of 2-alkylanthraquinols, (ii) One millilitre of 30% H2O2 means that solution will, give 100 V of oxygen at STP, (iii) Dihedral angle of H2O2 in gas phase is 90.2° and in, solid phase dihedral angle is 111.5°, (a) (i), (ii) and (iii), (b) (i) and (iii), (c) (ii) and (iii), (d) (i) and (ii), Some statements about heavy water are given below:, (i) Heavy water is used as a moderator in nuclear reactors., (ii) Heavy water is more associated than ordinary water., (iii) Heavy water is more effective solvent than ordinary, water., Which of the above statements are correct?, (a) (i) and (iii), (b) (i) and (ii), (c) (i), (ii) and (iii), (d) (ii) and (iii), , MATCHING TYPE QUESTIONS, 100. Match the columns, Column-I, Column-II, (A) Ionic hydrides, (p) NiH0.6-0.7, (B) Molecular hydrides (q) LiH, (C) Metallic hydrides, (r) HF, (a) A – (q), B – (r), C – (p), (b) A – (r), B – (q), C – (p), (c) A – (q), B – (p), C – (r), (d) A – (r), B – (p), C – (q), 101. Match the columns, Column - I, Column - II, (Chemical property, (Chemical equation), of water), (A) Basic nature, (p) 2H2O(l) + 2Na(s), 2NaOH(aq) + H2(g), (B) Auto-protolysis, (q) H2O(l) + H2O(l), H 3O (aq)+ OH–(aq), , (C) Oxidising nature, , (r) 2F2(g) + 2H2O(l), 4H+(aq) + 4F–(aq) + O2(g), (D) Reducing nature, (s) H2O(l) + H2S(aq), H3O+(aq) + HS–(aq), (a) A – (s), B – (q), C – (r), D – (p), (b) A – (s), B – (q), C – (p), D – (r), (c) A – (r), B – (q), C – (s), D – (p), (d) A – (p), B – (q), C – (s), D – (r), 102. Match the columns, Column-I, Column-II, (A) Clark’s method, (p) Mg(HCO3)2 + 2Ca(OH)2, 2CaCO3 + Mg(OH)2, + 2H2O, (B) Calgon’s method (q) 2NaZ(s) + M2 +(aq), MZ2(s)+ 2Na+(aq), (C) Boiling, (r) Ca(HCO3)2, CaCO3 + H2O + CO2, 2–, (D) Ion exchange, (s) M2+ + Na4P6O18, method, [Na2MP6O18]2– + 2Na+, (a) A – (s), B – (q), C – (r), D – (p), (b) A – (q), B – (p), C – (r), D – (s), (c) A – (p), B – (s), C – (r), D – (q), (d) A – (r), B – (q), C – (p), D – (s), 103. Match the columns, Column-I, Column-II, (A) Coordinated water (p) [Cu(H2O)4]4+SO2–, 4. H2O, (B) Interstitial water, (q) C17H35COONa, (C) Hydrogen-bonded (r) BaCl2.2H2O, water, (s) [Cr(H2O)6]3+ 3Cl–, (a) A – (r), B – (s), C – (q), (b) A – (q), B – (r), C – (s), (c) A – (r), B – (q), C – (p), (d) A – (s), B – (r), C – (p)
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HYDROGEN, , 104. Match the columns, Column-I, (A) Heavy water, , Column-II, (p) Bicarbonates of, Mg and Ca in water, (B) Temporary, (q) No foreign ions, hard water, in water, (C) Soft water, (r) D2O, (D) Permanent hard, (s) Sulphates & chlorides of, water, Mg & Ca in water, (a) A – (r), B – (s), C – (q), D – (p), (b) A – (q), B – (r), C – (s), D – (s), (c) A – (q), B – (s), C – (r), D – (p), (d) A – (r), B – (p), C – (q), D – (s), 105. Match the Column-I with Column-II and mark the appropriate, choice., Column-I, Column-II, (A) Syngas, (p) Na6P6O18, (B) Calgon, (q) NaAlSiO4, (C) Permutit, (r) CO + H2, (D) Producer gas, (s) CO + N2, (a) (A) – (p), (B) – (q), (C) – (r), (D) – (s), (b) (A) – (r), (B) – (p), (C) – (q), (D) – (s), (c) (A) – (r), (B) – (q), (C) – (s), (D) – (p), (d) (A) – (r), (B) – (q), (C) – (p), (D) – (s), , ASSERTION-REASON TYPE QUESTIONS, Directions : Each of these questions contain two statements,, Assertion and Reason. Each of these questions also has four, alternative choices, only one of which is the correct answer. You, have to select one of the codes (a), (b), (c) and (d) given below., (a) Assertion is correct, reason is correct; reason is a correct, explanation for assertion., (b) Assertion is correct, reason is correct; reason is not a, correct explanation for assertion, (c) Assertion is correct, reason is incorrect, (d) Assertion is incorrect, reason is correct., 106. Assertion : H+ does not exist freely and is always, associated with other atoms or molecules., Reason : Loss of the electron from hydrogen atom, results in nucleus (H+) of ~ 1.5 × 10–3 pm size. This is, extremely small as compared to normal atomic and ionic, sizes of 50 to 200 pm., 107. Assertion : Hydrogen combines with other elements by, losing, gaining or sharing of electrons., Reason : Hydrogen forms electrovalent and covalent bonds, with other elements., 108. Assertion : Temporary hardness can be removed by boiling., Reason : On boiling the soluble bicarbonates change to, carbonates which being insoluble, get precipitated., 109. Assertion : Calgon is used for removing permanent, hardness of water., Reason : Calgon forms precipitates with Ca2+ and Mg2+., 110. Assertion : Hard water is not suitable for laundary., Reason : Soap containing sodium stearate reacts with, hard water to precipitate out as calcium or magnesium, stearate., , 153, , 111. Assertion : Decomposition of H2O2 is a disproportionation, reaction., Reason : H2 O2 molecule simultaneously undergoes, oxidation and reduction., 112. Assertion : H2O2 is not stored in glass bottles., Reason : Alkali oxides present in glass catalyse the, decomposition of H2O2, , CRITICAL THINKING TYPE QUESTIONS, 113. Hydrogen will not reduce, (a) heated cupric oxide, (d) heated ferric oxide, (c) heated stannic oxide (d) heated aluminium oxide, 114. Which of the following terms is not correct for hydrogen ?, (a) Its molecule is diatomic, (b) It exists both as H+ and H– in different chemical, compounds, (c) It is the only species which has no neutrons in the, nucleus, (d) Heavy water is unstable because hydrogen is, substituted by its isotope deuterium, 115. The sum of the number of neutrons and protons in all the, three isotopes of hydrogen is, (a) 6, (b) 5, (c) 4, (d) 3, 116. The hydride ion, H–, is a stronger base than the hydroxide, ion, OH–. Which one of the following reactions will occur if, sodium hydride (NaH) is dissolved in water?, (a) H (aq) H 2 O(l) H 3O (aq), (b) H (aq) H 2O(l) OH (aq) H 2 (g), (c) H (aq ) H 2 O( l) OH (aq ) 2 H ( aq ) 2e, (d) H ( aq ) H 2 O ( l), No reaction, 117. The reaction of H2S + H2O2 S + 2H2O manifests, (a) Acidic nature of H2O2, (b) Alkaline nature of H2O2, (c) Oxidising nature of H2O2, (d) Reducing action of H2O2, 118. Which of the following is not true?, (a) D2O freezes at lower temperature than H2O, (b) Reaction between H2 and Cl2 is much faster than D2, and Cl2, (c) Ordinary water gets electrolysed more rapidly than D2O, (d) Bond dissociation energy of D2 is greater than H2, 119. Heavy water reacts respectively with CO2, SO3, P2O5 and, N2O5 to give the compounds :, (a) D2CO3, D2SO4, D3PO2, DNO2, (b) D2CO3, D2SO4, D3PO4, DNO2, (c) D2CO3, D2SO3, D3PO4, DNO2, (d) D2CO3, D2SO4, D3PO4, DNO3, 120. Identify x and y in following reaction:, , 2HSO4 (aq), , electrolysis, , x, , hydrolysis, , y + 2H+(aq) + H2O2(aq), (a) x = H2SO4 (aq), y = 2HSO 4 (aq), (b) x = HO3SOOSO3H(aq), y = 2HSO 4 (aq), (c) x = HO3SOOSO3H (aq), y = H2SO4(aq), (d) x = H2SO4(aq) , y = HO3SOOSO3H(aq)
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EBD_7207, HYDROGEN, , 154, , FACT / DEFINITION TYPE QUESTIONS, 1., 2., 3., 4., 5., 6., 7., 8., , 9., 10., 11., 12., 13., , (c) (i) and (iii) are properties of hydrogen which shows, its resemblance with alkali metals whereas (ii) , (iv), and (v) shows resemblance with halogens., (d) Chlorine has lone pair which it can donate to form, coordinate bond while hydrogen cannot., (c) Hydrogen behaves as a metal at very high pressure., (d) Hydrogen is a non-metal while all other members of, group 1 (alkali metals) are metals., (a) H + e– (1s1) H– (1s2 or [He]), F + e–([He]2s22p5), F– ( [He] 2s2 2p6 or [Ne]), (c) In metal hydrides the O.S. of hydrogen –1 otherwise it, is +1., (d), (c) (i) Protium, deuterium and tritium are isotopes of, hydrogen., (ii) Ortho and para hydrogens are allotropes of, hydrogen. In ortho hydrogen, protons are, spinning in same direction (parallel spin), while in, para hydrogen, protons spin in opposite direction, (antiparallel)., (a) Number of neutrons in protium, deuterium and, tritium respectively is = 0, 1 and 2, (c) Tritium is radioactive and emits low energy – particles., (c) Hydrogen bond is weak force of attraction existing, between molecules. Its energy is equal to 3-10 k cal, (c) Except method given in statement (iii) all other are, commercial methods for production of dihydrogen., Na2ZnO2 + H2, , (d) Zn+ 2NaOH, , 19., 20., , 15., 16., 17., , (c) CH4(g) + H2O(g), , 1270 K, Ni, , CO + 3H2, , Mixture of CO and H2 is called water gas., (c) Mixture of CO and H2 is used in synthesis of, methanol and a number of hydrocarbons due to this, reason it is also called syn gas., (d) Carbon dioxide formed in water gas shift reaction is, removed by scrubbing with sodium arsenite solution., (c), , Fe dil. H 2SO 4, , 3Fe 4H 2 O, Steam, , FeSO 4, , Fe 3O 4, , H2, , 21., 22., , 2Na + 2C 2 H 5OH ® 2C 2H 5ONa + H 2 -, , 18., , (a), , Mg dil. HNO 3, Mg( NO 3 ) 2, give H2 with dil HNO3), , H 2 (Mg and Mn, , CO + H 2+ H2O, , CO 2+ 2H2, , (a) Zinc has no action on cold water., (a) Cu + 4HNO3(conc.), Cu(NO3)2 + 2NO2 + 2H2O, C2 H 5OH, , Na, , C2 H 5O Na, , Mg 2H 2 O(steam), C6 H 5OH Na, NaH H 2 O, , 23., , Mg(OH) 2, C6 H 5O Na, , 1 / 2H 2, H2, 1 / 2H 2, , NaOH H 2, , (d) Very pure hydrogen can be prepared by the action of, water on sodium hydride., NaH + H2O, NaOH + H2, (very pure Hydrogen), , cobalt, catalyst, , CH3OH(l), , 24., , (b) CO(g) + 2H2(g), , 25., , (c) Hydrogen is not used in the synthesis of, hydroquinone and tartaric acid., (d) Almost all elements except noble gases, forms, hydrides., (c) Melts of ionic hydrides conduct electricity and, liberate dihydrogen gas at anode., (c) Fire due to action of water on saline hydrides cannot, be extinguished with water or CO2. These hydrides, can reduce CO2 at high temperature to produce O2., (b) Electron deficient hydride = B2H6, Electron precise = CH4, Electron rich = NH3 and HF, (a), (d) All metal hydrides are ionic in nature., (d) Metal hydride +H2O Metal hydroxide + H2, (d) Due to its covalent nature MgH2 is Polymeric in nature., (c) Ionic hydrides give the basic solution when it reacts, with water, e.g.,, LiH + H2O, LiOH + H2, (d) d- and f-block elements form metallic hydride., While p-block elements form covalent hydrides, s-block, elements except Be and Mg form ionic hydrides., Hydrides of Be, Mg, Cu, Zn, Ca and Hg are intermediate, hydride., (b), (b) LiH + AlCl3, (AlH3)n excess Li[AlH 4 ], LiH, Lithuim aluminium hydride is a most useful organic, reducing agent. It reduces functional groups but does, not attack double bonds., , 26., 27., 28., 29., , 30., 31., 32., 33., 34., , 35., , 4H 2, , Cu + dil. HCl, No reaction, Copper does not evolve H2 from acid as it is below, hydrogen in electrochemical series., , (b), , Mn(NO3)2 + H2, , catalyst, , water gas, , (Sodium zincate), , 14., , (b) Mn + 2HNO3 (dil.), , 36., 37.
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HYDROGEN, , 38., , 155, , 40., , (a) Polarity of bond depends on difference in, electronegativity of the two concerned atoms. H2O is, more polar than H2S because oxygen (in O–H) is more, electronegative than sulphur (in S–H)., (b) In gas phase water is a bent molecule with a bond, angle of 104.5° and O–H bond length of 95.7 pm., (d) The hybridisation in water is sp3 and bond angle 104.5º, , 41., , (c), , 39., , O, , 104.5°, , H, , 2 –, , O, , 95.7 pm, H, , +, , +, , H, , H, , (i), , (ii), , 52. (a) Calgon process is used to remove permanent hardness, of water, 53. (b) Na zeolite + CaCl 2, Ca zeolite 2NaCl, 54. (a) The complex salt of metaphosphoric acid sodium, hexametaphosphate (NaPO3)6, is known as calgon. It, is represented as Na2[Na4(PO3)6], 55. (c) Polyphosphates (sodium hexametaphosphates,, sodium tripolyphosphate or STPP) form soluble, complexes with Ca+2, Mg+2 present in hard water., 56. (b), 57. (a) This method is known as Clark's process. In this, method temporary hardness is removed by adding lime, water or milk of lime., Ca(OH)2 Ca(HCO3 )2, , O, +, , +, , H, , H, , H, H, (iii), , 42., , 43., 44., 45., , 46., , 47., 48., 49., , (c) The unusual properties of water in the condensed, phase (liquid an solid states) are due to the presence, of extensive hydrogen bonding between the water, molecules., (c) The high boiling point of water is due to H-bonding., (b) Due to high dielectric constant, water acts as a good, solvent therefore it is also called a universal solvent., (b) In the structure of ice each molecule of H2O is, surrounded by three H2O molecules in hexagonal, honey comb manner which results an open cage like, structure. As a result there are a number of 'hole' or, open spaces. In such a structure lesser number of, molecules are packed per ml. When ice melts a large, no. of hydrogen bonds are broken. The molecules, therefore move into the holes or open spaces and come, closer to each other than they were in solid state. This, result sharp increase in the density. Therefore ice has, lower density than water., (a) We know that due to polar nature, water molecules are, held together by intermolecular hydrogen bonds. The, structure of ice is open with large number of vacant, spaces, therfore the density of ice is less than water., (a) Two ice cubes stick to each other due to H-bonding., (d), (d) Sodium is most electropositive element among those, given., 2Na 2H 2O, 2NaOH H 2, Room temp., , 50., , 51., , (b) Water has high dielectric constant i.e., 78.39 C2/Nm2,, high liquid range and can dissolve maximum number, of compounds. That is why it is used as universal, solvent., (b) Temporary hardness is due to presence of bicarbonates, of calcium and magnesium and permanent hardness is, due to the sulphates and chlorides of both of calcium, and magnesium., , 2CaCO3, ppt., , 2H 2O, , 58. (c) Permanent hardness of water is due to chlorides and, sulphates of calcium and magnesium., 59. (a) H2SO4 + BaO2 BaSO4 + H2O2, 60. (d) MnO2 , PbO2 and BaO will not give H2O2 with HCl., MnO2 and PbO2 will give Cl2 and BaO will react with, HCl to give BaCl2 and water., 61. (b), 62. (c) 30 vol of H2O2 means one volume of H2O2 on, decomposition will give 30 volume of oxygen., , vol. strength, 5.6, Volume of normal (1N) H2O2 solution = 5.6 volumes., Volume of strength of 1.5 N H2O2, = 1.5 × 5.6 = 8.4 volumes., 68 10, 64. (b) Strength of 10V H2O2 =, 100 3.035%, 22400, 65. (b) Structure of H2O2 is nonplanar, 66. (d) O – O – H bond angle in H2O2 is 94.8°., , 63. (a) Normality of H2O2 =, , 67. (b), , H, , O–O, H, , 68. (b), 69. (a), , H 2O 2, weak acid, , is the true structure of H2O2., , H 2 O [O], , 70. (b) H 2 O2 [O] Oxidation H 2 O O 2, 71. (b) The value of dipole moment of H2O2 is 2.1 D, which, suggest the structure of H2O2 cannot be planar., An open-book structure is suggested for H2 O2 in, which O – H bonds lie in different plane., 72. (a) In the following reaction H2O2 acts as a reducing agent., PbO2(s) + H2O2 (aq) PbO(s) + H2O( ) + O2 (g), 73. (a) As H2O2 is loosing electrons so it is acting as reducing, agent., 74. (c) H2S is oxidised to S by H2O2., 75. (d) H2O2 does not have basic properties., 76. (b), 77. (c)
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EBD_7207, HYDROGEN, , 156, , 78., , (b, d) HOCl(aq) H 2 O 2 (aq), , MATCHING TYPE QUESTIONS, H 3O (aq) Cl (aq) O 2 (g), , I 2 (s) H 2 O 2 (aq) 2OH (aq), , 79., 80., 81., 82., 83., , 84., 85., 86., 87., 88., 89., , 2I (aq) 2H 2 O(l ) O 2 (g), (a) Option (a) represents oxidising action of H2O2 in, acidic medium., (d) H2O2 is not used as a moderator in nuclear reactors, (d) Decomposition of H2O2 can be accelerated by finely, divided metals such as Ag, Au, Pt, Co, Fe etc., (c) H2O2 is unstable liquid and decomposes into water, and oxygen either on standing or on heating., (d) Although H2O2 is a better polar solvent than H2O., However it cannot be used as such because of the, strong auto-oxidation ability., (d) H2O2 show all these properties., (b) The formula of heavy water (deuterium oxide) is D2O., (a) CaC2 2D 2 O C 2 D 2 Ca(OD) 2, (b) D2O is used in nuclear reactors as moderator., (b) Al4C3 + D2O, 3CD4 + 4Al(OD)3, (c) Heavy water is formed by the combination of heavier, isotope (1H2 or D) with oxygen., 2D 2 O, 2D 2 O 2, Heavy water, , 90., 91., , 92., 93., , (d) H2 O absorbs neutrons more than D2 O and this, decreases the number of neutrons for the fission, process., (b) Heavy water is D2O hence, number of electrons = 2 + 8 = 10, number of protons = 10, Atomic mass of D2O = 4 + 16 = 20, hence number of neutron, = Atomic mass – number of protons, = 20 – 10 = 10, (b) The only pollutant in combustion of dihydrogen is, oxides of dinitrogen (due to the presence of, dinitrogen as impurity with dihydrogen)., (d) It is for the first time in the history of India that a pilot, project using dihydrogen as fuel was launched in, October 2005 for running automobiles. Initially 5% H2, has been mixed in CNG for use in four wheeler vehicles., , STATEMENT TYPE QUESTIONS, 94., 95., 96., 97., , 98., 99., , (c) Tanks of metal alloy like NaNi5, Ti – TiH2, Mg – MgH2, etc are used for storage of dihydrogen in small, quantities., (a), (c) Hydrogenation occurs in presence of nickel as a, catalyst., (a) In winter seasons ice formed on the surface of a lake, provides thermal insulation which ensures the, survival of the aquatic life., (d) Dihedral angle of H2O2 in gas phase is 111.5° and, in solid phase it is 90.2°, (b), , 100. (a), 101. (b), 102. (c), 103. (d) Many salts can be crystallised as hydrated salts from, an aqueous solutions such an association of water is, of different types viz.,, (i) Coordinated water e.g., [Cr(H2O)6]3+ 3Cl–, (ii) Interstitial water e.g., BaCl2 . 2H2O, (iii) Hydrogen-bonded water, e.g., [Cu(H2O)4]4+ SO2–, 4. H2O in CuSO4.5H2O, 104. (d) Heavy water is D2O (1 – C); Temporary hard water, contains the bi-carbonates of Mg and Ca (2 – A); Soft, water contains no foreign ions (3 – B); Permanent, hard water contains the sulphates and chlorides of, Mg and Ca (4 – D) therefore the answer is D., 105. (b), , ASSERTION-REASON TYPE QUESTIONS, 106. (a) Due to extremely small size of H+ as compared to, normal atomic and ionic size H+ does not exist freely., 107. (a), 108. (a), 109. (c) Both assertion is correct reason is not true., Correct reason : Calgon forms soluble complexes with, Ca2+ and Mg2+ in which properties of these ions are, masked., 110. (a) 2C17H35COONa(aq) + M2+ (aq), (C17H35COO)2M (M = Ca or Mg) + 2Na+(aq), 111. (a) Both assertion and reason are true and reason is the, correct explanation of assertion., 112. (a), , CRITICAL THINKING TYPE QUESTIONS, 113. (d) H2 will not reduce heated Al2O3. As Al is more electropositive than hydrogen. therefore, its oxide will not be, reduced by hydrogen., 114. (d) Heavy water is stable., 115. (a) 1H1 1D2 1T3, no. of neutrons respectively are 0 , 1 , 2, no. of protons respectively are 1 , 1 , 1, Hence the sum of protons + neutrons = 1 + 2 + 3 = 6, 116. (b), , H (aq) H 2 O(l), base 1, , OH (aq) H 2 (g), base 2, , acid 1, , acid2, , H–, , In this reaction, acts as bronsted base as it accepts, one proton (H+) from H2O and for H2., 117. (c), , 2, , H 2 S H 2 O2, , 0, , S 2H 2O, , In this reaction H2O2 shows oxidising nature., 118. (a) D2O actually has higher freezing point (3.8°C) than, water H2O (0°C), 119. (d), 120. (b), , 2HSO4– (aq), Hydrolysis, , electrolysis, , HO3SOOSO3H(aq), , 2HSO 4 (aq) + 2H+(aq) + H2O2(aq)
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10, , THE s-BLOCK ELEMENTS, , FACT / DEFINITION TYPE QUESTIONS, 1., , 2., , 3., , 4., , 5., , 6., , 7., , Group 2 elements are called alkaline earth metals why?, Choose the correct reason(s)., (i) Hydroxides formed by group 2 elements are alkaline, in nature., (ii) Their metal oxides are found in the earth’s crust., (iii) Their oxides are alkaline in nature, (iv) Group 2 elements react with alkalies., (a) (i) and (ii), (b) (ii) and (iv), (c) (i), (ii) and (iii), (d) (ii) and (iii), Which of the following alkali metal is highly radioactive?, (a) Rubidium, (b) Caesium, (c) Francium, (d) Both (a) and (c), Which of the following are found in biological fluids Na+,, Mg2+, Ca2+, K+, Sr2+, Li+ and Ba2+, (a) Mg2+, Ca2+, and Sr2+, (b) Na2 +and K+, (c) Na+, K+, Mg2+and Ca2+, (d) Sr+, Li and Ba2+, Which of the following statements is not correct for alkali, metals?, (a) Alkali metals are the most electropositive metals., (b) Alkali metals exist in free state in nature., (c) These metals have the largest size in a particular, period of the periodic table., (d) Both (b) and (c), Which of the following has largest size ?, (a) Na, (b) Na+, –, (c) Na, (d) Can’t be predicted, Ionization potential of Na would be numerically the same as, (a) electron affinity of Na+, (b) electronegativity of Na+, (c) electron affinity of He, (d) ionization potential of Mg, Which one of the following properties of alkali metals, increases in magnitude as the atomic number rises ?, (a) Ionic radius, (b) Melting point, (c) Electronegativity, (d) First ionization energy., , 8., , Which of the following has density greater than water?, (a) Li, (b) Na, (c) K, (d) Rb, 9., The elements of group 1 provide a colour to the flame of, Bunsen burner due to, (a) low ionization potential, (b) low melting point, (c) softness, (d) presence of one electron in the outermost orbit, 10. The metal that produces red-violet colour in the nonluminous flame is, (a) Ba, (b) Ag, (c) Rb, (d) Pb, 11. The alkali metals have low melting point. Which of the, following alkali metal is expected to melt if the room, temperature rises to 30°C?, (a) Na, (b) K, (c) Rb, (d) Cs, 12. In the case of the alkali metals, (a) the cation is less stable than the atom, (b) the cation is smaller than the atom, (c) the cation and the atom have about the same size, (d) the cation is larger than the atom, 13. Which of the following is not correct ?, (a), , 2Li 2 O, , heat, 673K, , Li 2 O 2, , (b), , 2K 2O, , heat, 673K, , K 2O2, , (c), , 2Na 2 O, , heat, 673K, , Na 2O 2, , 2Na, , (d), , 2Rb 2O, , heat, 673K, , Rb 2 O 2, , 2Rb, , 2Li, 2K, , 14. The element which on burning in air gives peroxide is, (a) lithium, (b) sodium, (c) rubidium, (d) caesium
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EBD_7207, THE s-BLOCK ELEMENTS, , 158, , 15., , 16., , 17., , 18., , 19., , 20., , 21., , 22., , 23., , 24., , 25., , Which one of the alkali metals, forms only, the normal oxide,, M2O on heating in air ?, (a) Rb, (b) K, (c) Li, (d) Na, Which of the following is used as a source of oxygen in, space capsules, submarines and breathing masks ?, (a), , Li 2O, , (c), , KO2, , (b), , 26., , (a) C6H6, 27., , Na 2O 2, , (d) K 2O 2, The ionic mobility of alkali metal ions in aqueous solution is, maximum for, (a) Li+, (b) Na+, +, (c) K, (d) Rb+, For an aqueous solution under an electric field which of, the following have lowest mobility ?, (a) Li+, (b) Na+, +, (c) K, (d) Rb+, Which of the following pairs of substances would give, same gaseous product on reaction with water?, (a) Na and Na2O2, (b) Ca and CaH2, (c) Ca and CaO, (d) Ba and BaO2, Which is the most basic of the following?, (a), , Na 2O, , (b), , BaO, , (c), , As2O3, , (d), , Al2O3, , Which hydride is most stable, (a) NaH, (b) KH, (c) CsH, (d) LiH, The most stable compound is, (a) LiF, (b) LiCl, (c) LiBr, (d) LiI, Which of the following represents a correct sequence of, reducing power of the following elements?, (a) Li > Cs > Rb, (b) Rb > Cs > Li, (c) Cs > Li > Rb, (d) Li > Rb > Cs, What is the colour of solution of alkali metals in liquid, ammonia?, (a) Bronze, (b) Blue, (c) Green, (d) Violet, The alkali metals dissolve in liquid ammonia giving deep, blue solution. The solution is __x, ___. In concentrated, solution, the blue colour changes to ___y__ and becomes, __z___, (a) x = paramagnetic, y = colourless, z = diamagnetic, (b) x = diamagnetic, y = colourless, z = paramagnetic, (c) x = paramagnetic, y = bronze, z = diamagnetic, (d) x = paramagnetic, y = black, z = diamagnetic, , Na metal is stored in, , 28., , 29., , 30., , 31., , 32., , 33., , 34., , 35., , 36., , (b) kerosene, , (c) alcohol, (d) toluene, Which of the following metal is used along with lithium, to make the alloy named ‘white metal’ ?, (a) Nickel, (b) Aluminium, (c) Silver, (d) Lead, Which of the following metal is used as a coolant in, breeder nuclear reactors?, (a) Potassium, (b) Sodium, (c) Caesium, (d) Rubidium, Which is most basic in character ?, (a) CsOH, (b) KOH, (c) NaOH, (d) LiOH, Which compound will show the highest lattice energy ?, (a) RbF, (b) CsF, (c) NaF, (d) KF, In crystals which one of the following ionic compounds, would you expect maximum distance between centres of, cations and anions?, (a) LiF, (b) CsF, (c) CsI, (d) LiI, Among LiI, NaI, KI, the one which is more ionic and more, soluble in water is, (a) KI, (b) NaI, (c) LiI, (d) None of these, The products obtained on heating LiNO2 will be, (a), , Li 2 O NO2 O2, , (b), , Li 3 N, , (c), , Li 2 O NO O 2, , (d), , LiNO3, , O2, O2, , On heating anhydrous Na2CO3,.......is evolved, (a) CO2, (b) water vapour, (c) CO, (d) no gas, Complete the following two reactions., (i) 4LiNO3, x + O2, (ii) 2NaNO3, y + O2, (a) x = LiNO2, y = NaNO2, (b) x = Li2O + NO2, y = Na2O + NO2, (c) x = Li2O + NO2, y = NaNO2, (d) x = LiNO2, y = Na2O + NO2, Which of the following does not illustrate the anomalous, properties of lithium?, (a) The melting point and boiling point of Li are, comparatively high, (b) Li is much softer than the other group I metals, (c) Li forms a nitride Li3N unlike group I metals, (d) The ion of Li and its compounds are more heavily, hydrated than those of the rest of the group
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THE s-BLOCK ELEMENTS, , 37., , 38., , 39., , 40., , 41., , 159, , Why lithium react less vigorously with water than other, alkali metals?, (a) Lithium has most negative E value, (b) Lithium has small size and very high hydration, energy., (c) Lithium has least negative E value, (d) Both (a) and (b), Identify the correct statement, (a) Elemental sodium can be prepared and isolated by, electrolysing an aqueous solution of sodium chloride, (b) Elemental sodium is a strong oxidising agent, (c) Elemental sodium is insoluble in ammonia, (d) Elemental sodium is easily oxidised, Washing soda has formula, (a) Na2CO3.7H2O, (b) Na2CO3.10H2O, (c) Na2CO3.3H2O, (d) Na2CO3, The process associated with sodium carbonate manufacture, is known as, (a) Chamber, , (b) Haber, , (c) LeBlanc, , (d) Castner, , In Solvay ammonia process, sodium bicarbonate is, precipitate due to, (a) presence of NH3, (b) reaction with CO2, , 45. Which of the following is/are present as impurity in crude, sodium chloride, obtained by crystallisation of brine, solution?, (i) Sodium sulphate, (ii) Calcium chloride, (iii) Magnesium chloride (iv) Potassium chloride, (a) (i), (ii) and (iv), (b) (ii) and (iii), (c) (iii) and (iv), (d) (i), (ii) and (iii), 46. Which is manufactured by electrolysis of fused NaCl ?, (a) NaOH, 47., , 48., , 49., , 50., , 51., , (c) reaction with brine solution, 42., , 43., , 44., , (d) reaction with NaOH, Sodium carbonate is manufactured by Solvay process. The, products which can be recycled are, (a) CO2 and NH3, (b) CO2 and NH4Cl, (c) NaCl and CaO, (d) CaCl2 and CaO., How NH3 is recovered in Solvay process?, (a) By reaction of NH4Cl and Ca(OH)2, (b) By reaction of NH4HCO3 and NaCl, (c) By reaction of (NH4)2CO3 with H2O, (d) By any of the above, Why Solvay process cannot be extended to the, manufacture of potassium carbonate?, (a) Ammonium hydrogen carbonate does not react with, potassium chloride., (b) Potassium hydrogen carbonate is too soluble to be, precipitated by the addition of ammonium, hydrogencarbonate to a saturated solution of, potassium chloride., (c) Ammonium carbonate is precipitated out instead of, potassium hydrogen carbonate on reaction of, ammonium hydrogen carbonate with potassium, chloride, (d) None of the above, , 52., , 53., , 54., , 55., , (b) Na, , (c) NaClO, (d) NaClO3., Baking soda is, (a) NaHCO3, (b) K2CO3, (c) Na2CO3, (d) NaOH, Baking powder contains :, (a) NaHCO3, Ca(H2PO2)2 and starch, (b) NaHCO3, Ca(H2PO2)2, (c) NaHCO3, starch, (d) NaHCO3, Which of the following is the most abundant ion within, cell fluids?, (a) Sodium ions, (b) Potassium ions, (c) Calcium ions, (d) None of these, Which of the following is non-metallic?, (a) B, (b) Be, (c) Mg, (d) Al, Electronic configuration of calcium atom may be written as, (a) [Ne], 4p2, (b) [Ar], 4s2, 2, (c) [Ne], 4s, (d) [Ar], 4p2, The outer electronic configuration of alkaline earth metal is, (a) ns 2, (b) ns 1, 6, (c) np, (d) nd10, Which of the following atoms will have the smallest size ?, (a) Mg, (b) Na, (c) Be, (d) Li, The first ionization energy of magnesium is lower than the, first ionization energy of, (a) Lithium, (b) Sodium, (c) Calcium, (d) Beryllium, Which of the following relations is correct with respect to, first (I) and second (II) ionization potentials of sodium and, magnesium?, (a), , I Mg = II Na, , (b), , I Mg, , II Na, , (c) I Na I Mg, (d) II Na IIMg, 56. The first ionization energies of alkaline earth metals are, higher than those of alkali metals. This is because, (a) there is no change in the nuclear charge, (b) there is decrease in the nuclear charge of alkaline earth, metals, (c) there is increase in the nuclear charge of alkaline earth, metals, (d) none of these
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EBD_7207, THE s-BLOCK ELEMENTS, , 160, , 57., , Which of the following has maximum ionization energy, (b) Be, (a) Ba, Ba, e, Be, e, (c), , 58., , 59., , 60., , 61., , 62., , 63., , 64., , 65., , 66., , Ca, , Ca 2, , 2e, , (d), , Mg, , Mg 2, , 2e, , The most electropositive amongst the alkaline earth metals, is, (a) beryllium, (b) magnesium, (c) calcium, (d) barium, Alkaline earth metals are not found free in nature because, of, (a) their thermal instability, (b) their low melting points, (c) their high boiling points, (d) their greater chemical reactivity, A firework gives out crimson coloured light. It contains a, salt of, (a) Ca, (b) Na, (c) Sr, (d) Ba, Following are colours shown by some alkaline earth, metals in flame test. Which of the following are not, correctly matched?, Metal, Colour, (i) Calcium, Apple green, (ii) Strontium, Crimson, (iii) Barium, Brick red, (a) (i) and (iii), (b) (i) only, (c) (ii) only, (d) (ii) and (iii), Which one of the following properties of alkali metals, increases in magnitude as the atomic number rises ?, (a) Ionic radius, (b) Melting point, (c) Electronegativity, (d) First ionization energy, Out of the following elements which one do you expect to, be most reactive, chemically ?, (a) Mg, (b) Ca, (c) Sr, (d) Ba, In the reaction Mg, , H2O, (steam), , 67., , X H 2 ; X is, , (a) MgO, (b) Mg(OH)2, (c) MgH2, (d) None of these, The metals A and B form oxide but B also forms nitride when, both burn in air. The A and B are, (a) Cs, K, (b) Mg, Ca, (c) Li, Na, (d) K, Mg, Which of the following is the best method for preparation, of BeF2?, (a) Reaction of Be with F2, (b) Thermal decomposition of (NH4)2BeF4, (c) Reaction of Be with HF, (d) All of the above are equally effective, , 68., , 69., , 70., , Arrange the following compounds in order of increasing, solubility, (i) MgF2, (ii) CaF2,, (iii) BaF2, (a) (i) < (ii) < (iii), (b) (ii) < (i) < (iii), (c) (ii) < (iii) < (ii), (d) (iii) < (ii) < (i), Alkaline earth metals are, (a) reducing agent, (b) amphoteric, (c) dehydrating agent, (d) oxidizing agent, The oxidation state shown by alkaline earth metals is, (a) +2, (b) +1, +2, (c) –2, (d) –1, –2, Which one of the following is the most soluble in water?, (a), , 71., , 72., , 73., , 74., , 75., , 76., , 77., , Mg(OH)2, , (b), , Sr(OH)2, , (c) Ca(OH) 2, (d) Ba(OH) 2, Which of the following alkaline earth metal hydroxides is, amphoteric in character, (a) Be(OH)2, (b) Ca(OH)2, (c) Sr(OH)2, (d) Ba(OH)2, Of the metals Be, Mg, Ca and Sr of group 2 A. In the periodic, table the least ionic chloride would be formed by, (a) Be, (b) Mg, (c) Ca, (d) Sr, The order of solubility of sulphates of alkaline earth metals, in water is, (a) Be > Mg > Ca > Sr > Ba, (b) Mg > Be >> Ba > Ca > Sr, (c) Be > Ca > Mg > Ba >> Sr, (d) Mg > Ca > Ba >> Be > Sr, The solubilities of carbonates decrease down the magnesium, group due to a decrease in, (a) hydration energies of cations, (b) inter-ionic attraction, (c) entropy of solution formation, (d) lattice energies of solids, The correct order of increasing thermal stability of K2CO3,, MgCO3, CaCO3 and BeCO3 is, (a) BeCO3< MgCO3 < CaCO3 < K2CO3, (b) MgCO3 < BeCO3 < CaCO3 < K2CO3, (c) K2CO3 < MgCO3 < CaCO3 < BeCO3, (d) BeCO3 < MgCO3 < K2CO3 < CaCO3, In which of the following the hydration energy is higher, than the lattice energy?, (a) MgSO4, (b) RaSO4, (c) SrSO4, (d) BaSO4, Which of the following alkaline earth metal sulphates has, hydration enthalpy higher than the lattice enthalpy?, (a) CaSO 4, (b) BeSO 4, (c), , BaSO 4, , (d), , SrSO 4
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THE s-BLOCK ELEMENTS, , 78., , 161, , Beryllium shows diagonal relationship with aluminium., Which of the following similarity is incorrect ?, (a) Be forms beryllates and Al forms aluminates, (b) Be(OH)2 like Al(OH)3 is basic., (c) Be like Al is rendered passive by HNO 3 ., , 79., , 80., , 81., , 82., , 83., , 84., , 85., , 86., , 87., , 88., , 89., , (d) Be2C like Al4C3 yields methane on hydrolysis., The substance not likely to contain CaCO3 is, (a) gypsum, (b) sea shells, (c) dolomite, (d) a marble statue, Plaster of Paris is, (a), , CaSO 4 .2H 2 O, , (b), , CaSO 4 .H 2 O, , (c), , 1, CaSO 4 . H 2 O, 2, , (d), , CaSO 4 .4H 2 O, , Gypsum on heating at 120 – 130°C gives, (a) anhydrous salt, (b) hemihydrate, (c) monohydrate, (d) dihydrate, Plaster of Paris on making paste with little water sets to, hard mass due to formation of, (a) CaSO4, (b) CaSO4.1/2H2O, (c) CaSO4.H2O, (d) CaSO4.2H2O, The chemical which is used for plastering the broken bones, is, (a) (CaSO4)2H2O, (b) MgSO4.7H2O, (c) FeSO4. 7H2O, (d) CuSO4. 5H2O, Dead burn plaster is, (a) CaSO4.2H2O, (b) MgSO4. 7H2O, (c) CaSO4.½ H2O, (d) CaSO4, The formula for calcium chlorite is, (a) Ca(ClO4)2, (b) Ca(ClO3)2, (c) CaClO2, (d) Ca(ClO2)2, Bone ash contains, (a) CaO, (b) CaSO4, (c) Ca3 (PO4)2, (d) Ca(H2PO4)2, Mortar is a mixture of, (a) CaCO3, sand and water, (b) slaked lime and water, (c) slaked lime, sand and water, (d) CaCO3 and CaO, Which gas is released when CaCO3 reacts with dilute, HCl?, (a) H2, (b) CO2, (c) O2, (d) Cl2, Setting of cement is an, (a) exothermic reaction, (b) endothermic reaction, (c) neither endothermic nor exothermic, (d) example of neutralisation reaction, , 90. For a good quality cement what should be the ratio of, following :, I. Silica to alumina, II. CaO to the total of oxides of SiO2, Al2O3 and Fe2O3, (a) I = 2.5 to 4, II = Greater than 2, (b) I = Nearly 4, II = Less than 2, (c) I = 2.5, II = Closer to 2, (d) I = 2.5 to 4, II = Closer to 2, 91. Calcitonin and parathyroid hormone regulate concentration, of which of the following element in plasma?, (a) Calcium, (b) Magnesium, (c) Sodium, (d) Potassium, 92. Which of the following metal is found in green colouring, pigment chlorophyll of plants?, (a) Fe, (b) Mg, (c) Na, (d) Al, , STATEMENT TYPE QUESTIONS, 93. Select the correct statements, (i) Cs+ is more highly hydrated that the other alkali metal, ions, (ii) Among the alkali metals Li, Na, K and Rb, lithium has, the highest melting point, (iii) Among the alkali metals only lithium forms a stable, nitride by direct combination with nitrogen, (a) (i), (ii) and (iii), (b) (i) and (ii), (c) (i) and (iii), (d) (ii) and (iii), 94. Which of the following sequence of T and F is correct for, alkali metals ? Here T represents True and F represents, False statement., (i) Alkali metal hydrides are ionic solids with high, melting point., (ii) All alkali halides are ionic in nature., (iii) Li is the least powerful reducing agent and Na is the, most powerful reducing agent., (a) TTT, (b) TFT, (c) FTF, (d) TFF, 95. Which of the following statement(s) is/are correct regarding, Li2CO3 and Na2CO3 ?, (i) Sodium salt evolve CO2 at higher temperature., (ii) Polarization of Na+ is lesser than that of Li+., (a) Both statements (i) and (ii) are correct, (b) Both statements (i) and (ii) are incorrect, (c) Statement (ii) is correct explanation for statement (i), (d) Statement (i) is correct explanation for statement (ii)
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EBD_7207, THE s-BLOCK ELEMENTS, , 162, , 96., , Which of the following sequence of T and F is correct for, given statements?, (i) The alkali metal hydroxides are the strongest of all, bases., (ii) All alkali metal halides have high negative enthalpies, of formation., (iii) The stability of the carbonates and hydrogen, carbonates of alkali metals decrease with increase in, electropositive character down the group., (iv) Only LiHCO3 exist as solid., (a) TTFF, (b) TTTT, (c) FTFT, (d) TFFT, 97. Which of the following statement(s) is/are correct?, (i) The atomic and ionic radii of alkaline earth metals are, smaller than those of the corresponding alkali metals, in the same periods., (ii) Second ionisation enthalpies of the alkaline earth, metals are smaller than those of the corresponding, alkali metals., (iii) Compounds of alkaline earth metals are more, extensively hydrated than those of alkali metals, (a) (i) and (ii), (b) (ii) and (iii), (c) (i) and (iii), (d) (i), (ii) and (iii), 98. Which of the following statements are correct ?, (i) Copper - beryllium alloys are used in the preparation, of high strength springs, (ii) Metallic beryllium is used for making window X-ray, tubes., (iii) Magnesium powder is used in incendiary bombs and, singnals., (iv) Barium is used in treatment of cancer., (a) (i), (ii) and (iv), (b) (i) and (iii), (c) (i), (ii) and (iii), (d) (i), (ii), (iii) and (iv), 99. Which of the following is/are not characteristic, property(ies) of alkaline earth metals ?, (i) All alkaline earth metal oxides are basic in nature and, forms sparingly soluble hydroxides with water., (ii) The hydrated chlorides, bromides and iodies of Ca,, Sr and Ba on heating undergoes hydrolysis while, corresponding hydrated halides of Be and Mg on, heating undergo dehydration., (iii) Nitrates of alkaline earth metals decompose on, heating as below, 2M(NO3)2, 2MO + 4NO2 + O2, (a) (i) only, (b) (ii) only, (c) (i) and (iii), (d) (i) and (ii), 100. Which of the following statement(s) is/are correct regarding, Al and Be ?, (i) Both of these react with alkali., (ii) There is diagonal relationship among these elements., (a) Both (i) and (ii), (b) Only (i), (c) Only (ii), (d) Neither (i) nor (ii), , MATCHING TYPE QUESTIONS, 101. Match the columns, Column-I, (Alkali metal), , Column-II, (Colour imparted to an, oxidizing flame), (A) Cs, (p) Yellow, (B) Rb, (q) Blue, (C) K, (r) Violet, (D) Na, (s) Red violet, (E) Li, (t) Crimson red, (a) A – (q), B – (s), C – (r), D – (p), E – (t), (b) A – (s), B – (q), C – (r), D – (p), E – (t), (c) A – (t), B – (r), C – (s), D – (p), E – (q), (d) A – (q), B – (p), C – (r), D – (p), E – (t), 102. Match the columns, Column-I, Column-II, (Metal), (Oxide formed on burning), (A) Caesium, (p) Superoxide, (B) Lithium, (q) Peroxide, (C) Sodium, (r) Monoxide, (a) A – (q), B – (p), C – (r), (b) A – (r), B – (q), C – (p), (c) A – (p), B – (r), C – (q), (d) A – (q), B – (r), C – (p), 103. Match the columns. Here Column-I shows the names of, the metals used with lithium to make useful alloys and, Column-II shows the uses of these alloys, Column-I, Column-II, (A) Aluminium, (p) Armour plates, (B) Magnesium, (q) Aircraft parts, (C) Lead, (r) Bearings for motor, engines., (a) A – (q), B – (p), C – (r), (b) A – (q), B – (r), C – (p), (c) A – (p), B – (q), C – (r), (d) A – (p), B – (r), C – (q), 104. Match the columns., Column-I, Column-II, (Sodium compound), (Uses), (A) Sodium carbonate, (p) In fire extinguisher, (B) Sodium chloride, (q) In manufacture of, glass, soap, borax, and caustic soda., (C) Sodium hydroxide, (r) In preparation of Na2O2,, NaOH and Na2CO3, (D) Sodium hydrogen, (s) In petroleum refining, carbonate, (a) A – (q), B – (r), C – (s), D – (p), (b) A – (s), B – (q), C – (r), D – (p), (c) A – (p), B – (s), C – (r), D – (q), (d) A – (s), B – (r), C – (p), D – (q)
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THE s-BLOCK ELEMENTS, , 105. Match the columns, Column-I, Column-II, (A) Quick lime, (p) Ca(OH)2, (B) Slaked lime, (q) CaO, (C) Bleaching powder (r) Ca(OCl)2, (D) Plaster of Paris, (s) CaSO4. H2O, (a), A – (p), B – (q), C – (r), D – (s), (b), A – (s), B – (r), C – (q), D – (p), (c), A – (q), B – (p), C – (r), D – (s), (d), A – (q), B – (p), C – (s), D – (r), 106. Match the columns, (A) Quick lime, (p) Setting fractured bones, (B) Plaster of Paris, (q) A constituent of chewing gum, (C) Slaked lime, (r) Manufacture of bleaching, powder, (D) Limestone, (s) Manufacture of dyestuffs, (a) A – (p), B – (s), C – (q), D – (r), (b) A – (s), B – (p), C – (r), D – (q), (c) A – (q), B – (r), C – (p), D – (s), (d) A – (r), B – (q), C – (s), D – (p), , ASSERTION-REASON TYPE QUESTIONS, Directions : Each of these questions contain two statements,, Assertion and Reason. Each of these questions also has four, alternative choices, only one of which is the correct answer. You, have to select one of the codes (a), (b), (c) and (d) given below., (a) Assertion is correct, reason is correct; reason is a correct, explanation for assertion., (b) Assertion is correct, reason is correct; reason is not a, correct explanation for assertion, (c) Assertion is correct, reason is incorrect, (d) Assertion is incorrect, reason is correct., 107. Assertion: Lithium salts are mostly hydrated., Reason : The hydration enthalpy of alkali metal ions, decreases with increase in ionic sizes., 108. Assertion : Lithium carbonate is not so stable to heat., Reason : Lithium being very small in size polarizes large, CO32, , ion leading to the formation of more stable Li2O, , and CO2, 109. Assertion : Compounds of beryllium is largely covalent, and get hydrolysed easily., Reason : This is due to high value of ionisation potential, and small size of Be., 110. Assertion : Radium is most abundant s-block element., Reason : s-block elements are non-radioactive in nature., , 163, , CRITICAL THINKING TYPE QUESTIONS, 111. The melting point of lithium (181°C) is just double the melting, point of sodium (98°C) because, (a) down the group, the hydration energy decreases, (b) down the group, the ionization energy decreases, (c) down the group the cohesive energy decreases, (d) None of these, 112. Li has the maximum value of ionisation potential among, alkali metals i.e. lithium has the minimum tendency to ionise, to give LI+ ion. Thus, in aq. solution lithium is, (a) strongest reducing agent, (b) poorest reducing agent, (c) strongest oxidising agent, (d) poorest oxidising agent, 113. Lithium is strongest reducing agent among alkali metals, due to which of the following factor?, (a) Ionization energy, (b) Electron affinity, (c) Hydration energy, (d) Lattice energy, 114. Which of the following statements is incorrect?, (a) Alkali metal hydroxide are hygroscopic, (b) Dissolution of alkali metal hydroxide is endothermic, (c) Aqueous solution of alkali metal hydroxides are, strongly basic, (d) Alkali metal hydroxides form ionic crystals, 115. Which of the following on thermal decomposition yields a, basic as well as acidic oxide ?, (a) NaNO3, (b) KClO3, (c) CaCO3, (d) NH4NO3, 116. Which one of the following on hydrolysis, gives the, corresponding metallic hydroxide, H2O2 and O2?, (a) Li2O, (b) Na2O2, (c) NaO2, (d) Na2O, 117. Which of the following oxides of potassium is not known ?, (a), , K 2O, , (b), , K 2O 4, , (c), , KO3, , (d), , K 2O 3, , 118. Suppose an element is kept in air chamber, than air content, was evaluated after sometime , oxygen and nitrogen content, was found to be low comparitively. The given element will, be, (a) Li, (b) Rb, (c) Na, (d) K, 119. Suppose metal react with the oxygen to form oxide, than, aqueous solution of this oxide when added to a solution of, HI, solution turn yellowish brown in colour. This compound, is, (a) Na2O, (b) Li2O, (c) NaOH, (d) Na2O2, 120. Which of the following salt of lithium is most soluble in, organic solvent ?, (a) LiF, (b) LiCl, (c) LiBr, (d) LiI
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EBD_7207, THE s-BLOCK ELEMENTS, , 164, , 121. Arrange the following in increasing order of their melting, point ?, (A) LiCl, (B) NaCl, (C) KCl, (a) A < B < C, (b) B < A < C, (c) C < A < B, (d) A < B C, 122. The raw materials in Solvay Process are, (a), , Na 2 CO 3 , CaCO 3 and NH 3, , (b), , Na 2SO 4 , CaCO3 and NH 3, , 128., , 129., , (c) NaCl , NH3 and CaCO3., (d) NaOH, CaO and NH 3 ., 123. Compared with the alkaline earth metals, the alkali metals, exhibit, (a) smaller ionic radii, (b) highest boiling points, (c) greater hardness, (d) lower ionization energies., 124. Property of the alkaline earth metals that increases with, their atomic number is, (a) solubility of their hydroxides in water, (b) solubility of their sulphates in water, (c) ionization energy, (d) electronegativity, 125. Which one of the following does not react with water even, under red hot condition?, (a) Na, (b) Be, (c) Ca, (d) K, 126. Magnesium burns in CO2 to form, (a) MgO + C, (b) MgO + CO, (c) MgCO3, (d) MgO., 127. Arrange the following in increasing order of their solubility?, MgCO3(A), CaCO3(B), SrCO3(C), Na2CO3(D), , 130., , 131., , 132., , (a) A < B < C < D, (b) A < C < B < D, (c) C < A < B < D, (d) C < B < A < D, Which of the following will precipitate first when aqueous, solution containing sulphate ions are added?, (a) Mg2+, (b) Ca2+, 2+, (c) Sr, (d) Ba2+, If the fluoride salts of group 2 metals are dissolved in, water,than which of the following will show high solubility?, (a) BaF2, (b) RbF2, (c) CaF2, (d) BeF2, Aqueous solution of group 2 is precipitated by adding, Na2CO3, then this precipitate is tested on flame, no light in, visible region is observed , this element can be, (a) Ba, (b) Mg, (c) Ca, (d) Sr, Which of the following statement is false ?, (a) Strontium decomposes water readily than beryllium, (b) Barium carbonate melts at a higher temperature than, calcium carbonate, (c) Barium hydroxide is more soluble in water than, magnesium hydroxide, (d) Beryllium hydroxide is more basic than barium, hydroxide., Bleaching powder is obtained by the interaction of chlorine, with, (a) dil. solution of Ca(OH) 2, (b) dry CaO, (c) conc. solution of Ca(OH) 2, (d) dry slaked lime
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THE s-BLOCK ELEMENTS, , FACT / DEFINITION TYPE QUESTIONS, 1., , 2., 3., , 4., 5., , 6., , 7., , 8., 9., 10., , 11., 12., , 13., 14., , (c) Group 2 elements are called alkaline earth metals as, their oxides and hydroxides are alkaline in nature and, these metal oxides are found in the earth’s crust., (c) Francium is highly radioactive., (c) Monovalent sodium and potassium ions and divalent, magnesium and calcium ions are found in large, proportions in biological fluids., (b) Alkali metals readily lose electron to give monovalent, M+ ion. Hence they are never found in free state in nature., (c) A cation is always much smaller than the, corresponding atom, whereas an anion is always larger, than the corresponding atom., Hence, correct order of the size is, Na– > Na > Na+, (a) Na Na+ + e– ; IE of Na = + ve, Na+ + e– Na ; E.A. of Na+ = – ve, Both are equal but opposite in nature, (a) Within a group, ionic radius increases with increase in, atomic number. The melting points decrease down the, group due to weakening of metallic bond. The, electronegativity and the 1st ionization energy also, decreases down the group., (d) Li, Na, K are lighter than water but Rb is heavier than, water., (a), (c) Alkali metals have large size. When they are heated in, the flame of Bunsen burner, the electrons present in, the valence shell move from lower energy level to, higher energy level by absorption of heat from the, flame. When they come back to the ground state, they, emit the extra energy in the form of visible light to, provide colour to the flame., (d), (b) Gp 1 metals form cations M+ by loss of electron from, outermost shell. Electronic configuration of Gp 1 metals, is ns1. When the outer electron is removed to give a, positive ion, the size decreases because the outermost, shell is completely removed. After removal of an, electron, the positive charge of the nucleus is greater, on the remaining electrons so that each of the remaining, electrons are attracted more strongly towards the, nucleus. This further reduces the size., (a) Lithium does not form peroxide., (b) Sodium metal on burning in air gives sodium peroxide., , 165, , 15. (c) All the alkali metals when heated with oxygen form, different types of oxides for example lithium forms, lithium oxide (Li2O), sodium forms sodium peroxide, (Na2O2), while K, Rb and Cs form their respective, superoxides., 1, 2Li, O, Li 2O, 2 2, 16. (c) Because KO2 not only provides O2 but also removes, 1CO2 as follows, 4KO2, , 2CO2, , 4KO 2, , 4CO 2, , 2K 2 CO3, 2H 2 O, , 3O 2, , KHCO3 3O 2, , 17. (d) Smaller the size of cation higher is its hydration energy, and lesser is its ionic mobility hence the correct order, is Li+ < Na+ < K+ < Rb+, 18. (a) In aqueous solution because of high charge density, of Li+ it is heavily hydrated, therefore due to its, extensive hydration which increases its size to highest, the mobility of Li+ ion will be lowest., 19. (b) Both Ca and CaH2 produce H2 gas with water., Ca, , 2H 2 O, , CaH 2, , 2H 2 O, , Ca(OH) 2, , H2, , Ca(OH) 2, , 2H 2, , 20. (a), 21. (d) The basic character and stability of hydrides decrease, down the group., 22. (a) For a given metal, order of stability of halides is, MF > MCl > MBr > MI, 23. (a) A reducing agent is a substance which can loose, electron and hence a reducing agent should have low, ionisation energy. Now since ionisation energy, decreases from Li to Cs, the reducing property should, increase from Li to Cs. The only exception to this is, lithium. This is because the net process of converting, an atom to an ion takes place in 3 steps., (i) M(s) M(g), H = Sublimation energy, +, (ii) M(g) M (g) + e–, H = Ionisation energy, (iii) M+(g)+H2O M+ (aq) H = Hydration energy, The large amount of energy liberated in hydration of, Li (because of its small size) makes the overall H, negative. This accounts for the higher oxidation, potential of lithium i.e., its high reducing power., 24. (b) The alkali metals dissolve in liquid ammonia giving, deep blue solution., 25. (c) x = paramagnetic y = bronze, z = diamagnetic, 26. (b) Na reacts violently and may catch fire on exposure to, moisture (air + water). So it is always stored in, kerosene. Na reacts with alcohol to produce H2.
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EBD_7207, THE s-BLOCK ELEMENTS, , 166, , 27., 28., , 39., 40., , (d) Lithium with lead is used to make white metal., (b) Liquid sodium is used as a coolant in fast breeder, nuclear reactors., (a) Since the ionization energies of alkali metals decrease, down the group, the ionic character and consequently, basic property of their hydroxides increases in the same, order, i.e. from LiOH to CsOH., (c) With the same anion, smaller the size of the cation,, higher is the lattice energy. The correct order of size of, cations is –, Na+ < K+ < Rb+ < Cs+, Hence, the lattice energy of NaF will be maximum., i.e., NaF., (c) As Cs+ ion has larger size than Li+ and I– has larger, size than F–, therefore maximum distance between, centres of cations and anions is in CsI., (a) Larger cation (K+) develops less polarisation in anion, and thus KI has more ionic nature and more soluble in, water., (a) 4LiNO3, 2Li 2 O 4NO 2 O 2, (d) Anhydrous form of Na2CO3 does not decompose on, heating even to redness. It is a amorphous powder, called soda ash., (c) 4LiNO3, 2Li2O + 4NO2 + O2, 2 NaNO3, 2NaNO2 + O2, (b) Li is much softer than the other group I metals. Actually, Li is harder then other alkali metals., (b) Lithium although has most negative E value reacts, less vigorously with water than other alkali metals, due to its small size and very high hydration energy., (d) Elemental sodium is easily oxidised ( has low I.P.) and, acts as reductant., (b) Washing soda is Na2CO3.10 H2O., (c), , 41., , (c), , 42., 43., , (a) CO2 and NH3 formed are reused (See Solvay process), (a) NH3 is recovered when the solution containing, NH4Cl is treated with Ca(OH)2, 2NH4Cl + Ca(OH)2, 2NH3 + CaCl2 + H2O, (b), (d) Sodium sulphate, calcium chloride and magnesium, chloride are present as impurities in crude sodium, chloride., (b) Na metal is manufactured by electrolysis of fused NaCl, by two methods., (i) Castner's process, (ii) Down's process, In both the above processes electrolysis of fused, sodium chloride produces Na at cathode., , 29., , 30., , 31., 32., , 33., 34., , 35., 36., 37., , 38., , 44., 45., , 46., , NH 4 HCO3, , NaCl, , NaHCO3, , NH 4 Cl, Sod. bicarbonat e ppt ., , Brine, , 2Na, , 2NaOH, , at cathode 2Na, at anode 4OH, , 2e, , OH, , 48., 49., 50., , 51., 52., 53., , 54., 55., , 4e, , (a) NaHCO3 (baking soda) is one of the major constituents, of baking powder., (a) Baking powder has starch, NaHCO3 and Ca(H2PO2)2., (b) Potassium ions are the most abundant cations within, cell fluids., (a) Metallic character decreases, as we go to the right, side in a period and increases when we move, downwards in a group., (b) Ca (20) = 1s2 2s2 2p6 3s2 3p6 4s2 = [Ar], 4s2., (a), (c) Within a period, the size decreases from left to right,, i.e., Na > Mg > Li > Be. Atomic size increases down the, group., (d), (d) The IInd ionisation potential of Na is higher than Mg, because it requires more energy to remove an electron, from a saturated shell or stable (fully filled) orbital., 11 Na, , 1s 2 , 2s 2 2p 6 , 3s1, , I, , 1s 2 , 2s 2 2 p 6 , 3s 0, , II, , 1s 2 , 2s 2 2p5 , 3s 0, 12 Mg, , 1s 2 , 2s 2 2p 6 , 3s 2, , I, , II, , 56., , (c), , 57., 61., , (d), (a), , 62., , (a), , 63., , (d), , 64., , (a), , 65., , (d), , 1s 2 , 2s 2 2p 6 , 3s 0, 1s 2 , 2s 2 2p 6 , 3s1, Here Na-I < Mg-I and Na-II > Mg-II., As we go from grp I element to grp II element in a, period, an extra electron is added in same shell which, results in increase in nuclear charge due to which, force of attraction by the nucleus increases and hence, ionic radii decreases and consequently I.E. increases., 58. (d), 59. (d), 60. (c), Calcium gives brick red colour and barium gives, apple green colour in flame test., Within a group, ionic radius increases with increase in, atomic number. The melting points decrease down the, group due to weakning of metallic bond. The, electronegativity and the 1st ionization energy also, decreases down the group., Barium is most electropostive element among those, given. Hence it is most reactive., Mg(OH)2 is not formed because of poor solubility of, MgO in H2O., K and Mg, both form oxides, , K O2, , 66., 67., 68., , 2Na, 2H 2 O O 2, , 47., , KO 2 ;2Mg O 2, , 2MgO, , Mg form nitride also 3Mg N 2, Mg 3 N 2, K does not form nitride., (b) Thermal decomposition of (NH4)2BeF4 is the best, method for preparation of BeF2., (b) BaF2 > MgF2 > CaF2, (a) Alkaline earth metals have a fairly strong tendency to, lose their outermost electrons due to which they act, as reducing agent.
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THE s-BLOCK ELEMENTS, , 69., 70., , 71., 72., 73., 74., , 75., , 76., , (a), (d) For a compoud to be soluble, the hydration energy, must exceed lattice energy. For Gp.II hydroxides, (Mg(OH)2, Sr(OH)2, Ca(OH)2, Ba(OH)2, lattice energy, decrease more rapidly than the hydration energy & so, their solubility increases down the group. Ba(OH)2>, Sr(OH)2 > Ca(OH)2 > Mg(OH)2, (a) Be(OH)2 is amphoteric while Ca(OH)2, Sr(OH)2 and, Ba(OH)2 are all basic., (a) Because of small atomic size and high I.E. Be forms, covalent chloride., (a), (a) As we move down the group, the lattice energies of, carbonates remain approximately the same. However, the hydration energies of the metal cation decreases, from Be++ to Ba++, hence the solubilities of carbonates, of the alkaline earth metal decrease down the group, mainly due to decreasing hydration energies of the, cations from Be++ to Ba++., (a) As the basicity of metal hydroxides increases down, the group from Be to Ba, the thermal stability of their, carbonates also increases in the same order. Further, group 1 compounds are more thermally stable than, group 2 because their hydroxide are much basic than, group 2 hydroxides therefore, the order of thermal, stability is, BeCO3 < MgCO3< CaCO3< K2CO3., (a) In alkaline earth metals ionic size increases down the, group. The lattice energy remains constant because, sulphate ion is so large, so that small change in cationic, sizes do not make any difference. On moving down, the group the degree of hydration of metal ions, decreases very much leading to decrease in solubility, BeSO 4, , 77., 78., 79., 80., 81., , 167, , MgSO 4, , CaSO 4, , SrSO 4, , BaSO 4, , Be 2+ is very small, hence its hydration enthalpy is, greater than its lattice enthalpy, (b) The Be(OH)2 and Al(OH)3 are amphoteric in nature., (a) Gypsum is CaSO4.2H2O, (c) Chemically plaster of Paris is CaSO4.1/2H2O., , (b), , (b), , CaSO 4 2H 2 O, , 120 C, , CaSO4, , 1, H 2O, 2, , Plaster of Paris is hemihydrate., 82., , 1, H O) on making paste with, 2 2, little water sets to a hard mass due to formation of, gypsum (CaSO4.2H2O)., , (d) Plaster of Paris (CaSO4 ., , CaSO4 ., , 1, 3, H2O + H2O, 2, 2, , Plaster of Paris, , 83., , CaSO4.2H2O + Heat, Gypsum, , (a) (CaSO4)2.H2O – Plaster of paris is used for plastering, the broken bones., , 84. (d), 88. (b), , 85. (d), CaCO 3 2HCl, , 86. (c), , 87. (c), , 2CaCl 2, , H 2 O CO 2, , 89. (a) During setting of cement, silicates and aluminates of, calcium are hydrated. Hydration is an exothermic, process. Therefore setting of cement is exothermic, process., 90. (d) For a good quality cement, the ratio of silica (SiO2), to alumina (Al2O3) should be between 2.5 and 4 and, the ratio of lime (CaO) to the total of the oxides of, silicon (SiO2), aluminium (Al2O3) and iron (Fe2O3), should be as close as possible to 2., 91. (a) The calcium concentration in plasma is regulated at, about 100 mgL–1. It is maintained by two hormones :, calcitonin and parathyroid hormone., 92. (b), , STATEMENT TYPE QUESTIONS, 93. (d) Amongst alkali metal Li ions are highly hydrated., 94. (d) Lithium halides are some what covalent in nature., Li is the most powerful reducing agent and Na is the, least powerful reducing agent., 95. (c), 96. (a) For statement (iii), stability of the carbonates and, hydrogen carbonates of alkali metals increases with, increase in electropositive character down the group., Hydrogen carbonate of lithium does not exist as a, solid., 97. (d) All the given statements are correct., 98. (c) Radium is used in treatment of cancer., 99. (d) All alkaline earth metal oxides except BeO are basic, in nature. BeO is amphoteric in nature., Hydrated halides of Ca, Sr and Ba on heating, undergo dehydration while corresponding hydrated, halides of Be and Mg on heating suffer hydrolysis., 100. (a), , MATCHING TYPE QUESTIONS, 101. (a), 102. (c) Cs + O2, CsO2 (Superoxide), 4Li + O2, 2Li2O (Oxide), 2Na + O2, Na2O2 (Peroxide), 103. (a) Lithium metal is used to make useful alloys, for, example with lead to make ‘white metal’ bearings for, motor engines, with aluminium to make aircraft parts,, and with magnesium to make armour plates., 104. (a), 105. (c), 106. (b) Quick lime is used for the manufacture of dyestuffs., Plaster of Paris is used for setting of fractured bones., Slaked lime is used for the manufacture of bleaching, powder., Limestone is a constituent of chewing gum.
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EBD_7207, THE s-BLOCK ELEMENTS, , 168, , ASSERTION-REASON TYPE QUESTIONS, 107. (a) Li+ has maximum degree of hydration among other, alkali metals., 108. (a) Lithium carbonate is unstable to heat; lithium being, very small in size polarises a large CO32 ion leading, to the formation of more stable Li2O and CO2., 109. (a) Because of high value of ionisation enthalpy and small, size it forms compound which are highly covalent in, nature, hence, it get hydrolysed easily., 110. (d) Both assertion and reason are false., Radium is the rarest of all s-block elements comprising, only 10–10 percent of igneous rocks. Francium (s-block, member) is radioactive; its long lived isotope 223Fr, ahs a half life of only 21 minutes., , 120. (d) LiI is more soluble as the degree of covalent character, is high due to larger size of anion i.e., iodide ion by, greater polarization of lithium cation., 121. (a), 122. (c) NaCl (brine), NH 3 and CO 2 are raw materials. CaCO3, is source of CO 2 ., 123. (d) Because of larger size and smaller nuclear charge, alkali, metals have low ionization potential relative to alkaline, earth metals., 124. (a) Lattice energy decreases more rapidly than hydration, energy for alkaline earth metal hydroxides. On moving, down a group, solubility of their hydroxides, increases., 125. (b) 2Na 2H 2 O, 2NaOH H 2, 2K 2H 2O, 2KOH H 2, All alkali metals decompose water with the evolution, of hydrogen., , CRITICAL THINKING TYPE QUESTIONS, 111. (c) The atom becomes larger on descending the group, so, the bonds becomes weaker (metallic bond), the, cohesive force/energy decreases and accordingly, melting point also decreases., 112. (a) The ionisation potential value of lithium is maximum, among alkali metals i.e., its tendency to ionise to give, Li+ ions should be the minimum i.e. Li should be the, poorest reducing agent. But, lithium is the strongest, reducing agent in aq. solution. This is due to the largest, value of hydration energy of Li+ ions., 113. (c) Li due to highest hydration energy among the alkali, metals is the strongest reducing agent., 114. (b) During the dissolution of alkali metal hydrides energy, is released in large amount, i.e., it is exothermic in nature., 115. (c) Calcium carbonate on thermal decomposition gives, CaO (Basic oxide) and CO2 (Acidic oxide), CaCO 3, , Basic oxide Acidic oxide, , O 24 ion is not possible and K 2O4 is unknown ., 118. (a) All the given elements react with oxygen to form, oxides but only Li also react with nitrogen to form, Li3N., 119. (d) (a) and (b) forms corresponding hydroxides (NaOH, and LiOH) in aqueous solution, , 117. (b), , M 2 O H 2O, 2M, 2OH (M Na or Li), Therefore reaction of HI with (a), (b) and (c) is simply, a neutralization reaction, while aqueous solution of, (d) form H2O2 which act as oxidizing agent, hence, convert Iodide to Iodine( I2)., Na 2O 2, , 2H 2 O, , 2Na, , 126. (a), 127. (d), , 129. (a), , 2NaOH + H2O2 + O, , 2OH, , H 2O 2, , 2H 2 O, , Sr 2H 2O, , 128. (d), , CaO + CO 2, , 116. (c) 2NaO2 + 2H2O, , Ca, , 130. (b), 131. (d), , 132. (d), , C a(OH) 2, , Sr(OH) 2, , H2, , H2, , Be + 2H2O, No reaction, Ca, Sr, Ba and Ra decompose cold water readily with, evolution of hydrogen. Mg decomposes boiling water, but Be is not attacked by water even at high, temperatures as its oxidation potential is lower than, the other members., Mg burns in CO2 to give MgO and C., Group1 carbonates are more soluble than group 2, which are sparingly soluble, and also in case of group, 2, down the group the solubility of carbonates, decreases., Down the group solubility of sulphate decreases. Thus, Ba2+ ions will precipitate out most easily., BeF2 is highly soluble in water due to the high, hydration enthalpy of the small Be2+ ion., Electrons in Mg due to its small size are tightly bound, so they cannot be excited by the flame., Be(OH)2 is amphoteric, but the hydroxides of other, alkaline earth metals are basic. The basic strength, increases gradually., When cold calcium hydroxide reacts with chlorine, then, bleaching powder is obtained., 3Ca(OH) 2 2Cl 2, slaked lim e, , Ca(OCl)2 .Ca(OH) 2 .CaCl 2 .2H 2 O, Bleaching powder
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11, , THE p-BLOCK ELEMENTS, (GROUP 13 AND 14), FACT / DEFINITION TYPE QUESTIONS, 1., , 2., , The non-metal oxides are ___x, _____ whereas metal oxides, y, are _______, in nature., (a) x = acidic or neutral, y = basic, (b) x = acidic, y = neutral, (c) x = basic, y = acidic, (d) x = neutral, y = basic, Which of the following is most abundant in the earth crust ?, (a) Boron, , 3., , 4., , 5., , 6., , 7., , 8., , (b) Aluminium, , (c) Gallium, (d) Thallium, Ionisation enthalpy ( iH1kJ mol–1) for the elements of Group, 13 follows the order., (a) B > Al > Ga > In > Tl (b) B < Al < Ga < In < Tl, (c) B < Al > Ga < In > Tl (d) B > Al < Ga > In < Tl, The relationship between first, second and third ionisation, enthalpies of each group-13 element is, (a), (b) iH1 < iH2 < iH3, iH1 > iH2 > iH3, (c), (d) iH3 > iH1 > iH2, iH1 = iH2 > iH3, Which of the following properties of aluminium makes it, useful for food packaging ?, (a) Good electrical conductivity, (b) Good thermal conductivity, (c) Low density, (d) Non toxicity, Which of the following is/are true regarding gallium?, (i) It has unusually low melting point (303 K)., (ii) It exist in liquid state during summer., (iii) It has a high boiling point (2676 K)., The correct option is, (a) (i) and (ii), (b) (i) and (iii), (c) (i), (ii) and (iii), (d) (ii) and (iii), The element which shows least metallic character is, (a) Indium, (b) Boron, (c) Aluminium, (d) Gallium, Which one of the following has the lowest m.p.?, (a) B, , (b) Al, , (c) Ga, , (d) Tl, , Which of the following does not form M3+ ion?, (a) Boron, (b) Aluminium, (c) Indium, (d) Gallium, 10. The group 13 element that is liquid during summer and used, for measuring high temperature is, (a) Boron, (b) Aluminium, (c) Gallium, (d) Indium, 11. Thallium shows different oxidation states because, (a) it is transition element, (b) of inert pair effect, (c) of its amphoteric character, (d) of its higher reactivity, 12. The exhibition of highest co-ordination number depends, on the availability of vacant orbitals in the central atom., Which of the following elements is not likely to act as central, 9., , 13., 14., 15., 16., 17., 18., , atom in MF63 ?, (a) B, (b) Al, (c) Ga, (d) In, Which out of the following compounds does not exist?, (a) BF3, (b) TlCl3, (c) TlCl5, (d) Both (b) and (c), Aluminium chloride is a/an, (a) Bronsted - Lowery acid (b) Arhenius acid, (c) Lewis acid, (d) Lewis base, The strongest Lewis acid is, (a) BF3, (b) BCl3, (c) BBr3, (d) BI3, AlCl3 on hydrolysis gives, (a) Al2O3. H2O, (b) Al(OH)3, (c) Al2O3, (d) AlCl3.6H2O, Which metal is protected by a layer of its own oxide?, (a) Al, (b) Ag, (c) Au, (d) Fe, Aluminium vessels should not be washed with materials, containing washing soda because, (a) washing soda is expensive, (b) washing soda is easily decomposed, (c) washing soda reacts with aluminium to form soluble, aluminate, (d) washing soda reacts with aluminium to form insoluble, aluminium oxide
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EBD_7207, THE p-BLOCK ELEMENTS (GROUP 13 AND 14), , 170, , 19., , 20., , 21., , 22., , 23., , 24., , 25., , 26., , 27., , 28., , 29., , 30., , 31., , 32., , When Al is added to KOH solution, (a) no action takes place (b) oxygen is evolved, (c) water is produced, (d) hydrogen is evolved, Which of the following does not react with aqueous, NaOH ?, (a) B, (b) Al, (c) Ga, (d) Tl, Amphoteric oxide among the following is, (a) B2O3, (b) Ga2O3, (c) In2O3, (d) Tl2O3, Boron forms covalent compound due to, (a) higher ionization energy, (b) lower ionization energy, (c) small size, (d) Both (a) and (c), NH3 and BF3 form an adduct readily because they form, (a) a coordinate bond, (b) a hydrogen bond, (c) an ionic bond, (d) a covalent bond, The factor responsible for weak acidic nature of B–F bonds, in BF3 is, (a) large electronegativity of fluorine, (b) three centred two electron bonds in BF3, (c) p - d back bonding, (d) p - p back bonding, In borax bead test which compound is formed?, (a) Ortho-borate, (b) Meta-borate, (c) Double oxide, (d) Tetra-borate, The formula of mineral borax is, (a), , Na 2 B 4 O 7, , (b), , Na 2 B4 O 7 .4H 2 O, , (c), , Na 2 B4 O 7 .5H 2O, , (d), , Na 2 B 4O 7 .10H 2O, , Which of the following hydroxide is acidic ?, (a) Al(OH)3, (b) Ca(OH)3, (c) Tl(OH)3, (d) B(OH)3, Orthoboric acid, (a) donate proton to form H2BO3–, (b) accept proton of form H4BO3+, (c) donate OH– to form H2BO2+, (d) accept OH– to form [B(OH)4]–, H3BO3 on heating up to 373 K yields:, (a) boric anhydride, (b) orthoboric acid, (c) metaboric acid, (d) tetraboric acid, Boric acid is polymeric due to, (a) its acidic nature, (b) the presence of hydrogen bonds, (c) its monobasic nature, (d) its geometry, B(OH)3 is, (a) monobasic acid, (b) dibasic acid, (c) tribasic acid, (d) triacidic base, Orthoboric acid when heated to red hot gives, (a) metaboric acid, (b) pyroboric acid, (c) boron and water, (d) boric anhydride, , 33., , 34., , 35., , 36., , 37., , 38., , 39., , 40., , 41., , 42., , 43., , 44., , Which is false in case of boric acid H3BO3?, (a) It acts as a tribasic acid., (b) It has a planar structure., (c) It acts as a monobasic acid., (d) It is soluble in hot water., BCl3 does not exist as dimer but BH3 exists as dimer (B2H6), because, (a) chlorine is more electronegative than hydrogen, (b) there is p -p back bonding in BCl3 but BH3 does not, contain such multiple bonding, (c) large sized chlorine atoms do not fit in between the, small boron atoms where as small sized hydrogen, atoms get fitted in between boron atoms, (d) None of the above, In reaction, BF3 + 3LiBH4 3LiF + X ; X is, (a) B4 H10, (b) B2H6, (c) BH3, (d) B3H8, Inorganic benzene is, (a), , B3 H 3 N 3, , (b), , BH 3 NH 3, , (c), , B3 H 6 N 3, , (d), , H 3B3 N 6, , The structure of diborane ( B 2 H 6 ) contains, (a) four 2c-2e bonds and four 3c-2e bonds, (b) two 2c-2e bonds and two 3c-3e bonds, (c) two 2c-2e bonds and four 3c-2e bonds, (d) four 2c-2e bonds and two 3c-2e bonds, In diborane, (a) 4–bridged hydrogens and two terminal hydrogens are, present, (b) 2– bridged hydrogens and four terminal hydrogens, are present, (c) 3–bridged and three terminal hydrogens are present, (d) None of these, Diborane upon hydrolysis gives, (a) boric anhydride, (b) metaboric acid, (c) orthoboric acid, (d) boron oxide, Borazole is known as, (a) organic benzene, (b) organic xylene, (c) inorganic benzene, (d) inorganic xylene, The compounds of boron and hydrogen are collectively, called, (a) diboranes, (b) borazoles, (c) boracits, (d) boranes, The bonds present in borazole or inorganic benzene are, (a) 9 , 6, (b) 12 , 3, (c) 6 , 9, (d) 15 only, The two type of bonds present in B2H6 are covalent and, (a) ionic, (b) co-ordinate, (c) hydrogen bridge bond (d) None of these, Reaction of diborane with ammonia gives initially, (b) Borazole, (a) B2H6 . NH3, (d) [BH2(NH3)2]+[BH4]–, (c) B2H6 . 3NH3
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THE p-BLOCK ELEMENTS (GROUP 13 AND 14), , 45., , Which of the following compounds is not matched correctly, with its structure?, , H, , (a), , H–N, H–B, , B, , N, , (c), , H, H, , – Borazine, , B–H, , H, B, , Cl, Cl, , (d) Cl, Cl, 46., , 47., , 48., , 49., , 50., , 51., , 52., , 53., , B, H, , Al, , Cl, Cl, Al, , B – Cl, , 54. Red lead is, (a) Pb3O4, (b) Pb2O3, (c) Pb2 O, (d) PbO, 55. The oxide of lead used in lead accumulators is, (a) PbO, , N–H, , H, (b), , 171, , H, H, , – Diborane, , Cl, Al, , – Aluminium chloride, , – Boron trichloride, , The electronic configuration of four different elements is, given below. Identify the group 14 element among these, (a) [He] 2s1, , (b) [Ne] 3s 2, , (c) [Ne] 3s 2 3 p 2, , (d) [Ne] 3s 2 3 p 5, , Which of the following is most electronegative?, (a) Pb, (b) Si, (c) C, (d) Sn, Which of the following isotope of carbon is radioactive?, (a) 12 C, (b) 13 C, 14, (c), C, (d) All of these, Carbon and silicon belong to group 14. The maximum, coordination number of carbon in commonly occurring, compounds is 4, whereas that of silicon is 6. This is due to, (a) large size of silicon, (b) more electropositive nature of silicon, (c) availability of d-orbitals in silicon, (d) Both (a) and (b), The inert pair effect is most prominent in, (a) C, (b) Pb, (c) Ge, (d) Si, The most stable +2 oxidation state is exhibited by, (a) Fe, (b) Sn, (c) Pb, (d) Si, Which of the following lead oxides is present in ‘Sindhur’?, (a) PbO, (b) PbO 2, (c) Pb2 O3, (d) Pb3O4, Mark the oxide which is amphoteric in character, (a), , CO2, , (b), , SiO 2, , (c), , SnO2, , (d) CaO, , (b), , Pb 2 O3, , (c) Pb3O 4, (d) PbO 2, 56. Which of the following is/are not correctly matched ?, (i) GeO2 – Acidic, (ii) PbO2– Amphoteric, (iii) CO – Neutral, (iv) SiO2 – Amphoteric, (a) (i) and (iv), (b) (iv) only, (c) (ii) only, (d) (iii) only, 57. Least thermally stable is, (a) CCl4, (b) SiCl4, (c) GeCl4, (d) GeBr4, 58. Unlike the other elements of its group carbon and silicon, does not form MX2 type molecules because, (a) energetically this is not possible, (b) carbon undergoes catenation, (c) it is non-metallic, (d) carbon does not contain d-orbital, 59. Which of the following halides is the most stable?, (a) CF4, (b) CI4, (c) CBr4, (d) CCl4, 60. The stability of dihalides of Si, Ge, Sn and Pb increases, steadily in the sequence, (a), , PbX 2, , SnX 2, , GeX 2, , SiX 2, , (b) GeX2 << SiX2 << SnX2 << PbX2, (c) SiX2 << GeX2 << PbX2 << SnX2, (d) SiX2 << GeX2 << SnX2 << PbX2., 61. Which of the following is not correct?, (a) Ge(OH)2 is amphoteric, (b) GeCl2 is more stable than GeCl4, (c) GeO2 is weakly acidic, (d) GeCl4 in HCl forms [ GeCl2]2– ion, 62. The main reason that SiCl4 is easily hydrolysed as compared, to CCl4 is that, (a) Si-Si bond is weaker, (b) SiCl4 can form hydrogen bonds, (c) SiCl4 is covalent, (d) Si can extend its coordination number beyond four, 63. Which halide is least stable and has doubtful existence, (a) CI4, (b) GeI4, (c) SnI4, (d) PbI4, 64. PbF4, PbCl4 exist but PbBr4 and PbI4 do not exist because, of, (a) large size of Br – and I–, (b) strong oxidising character of Pb4+, (c) strong reducing character of Pb4+, (d) low electronegativity of Br – and I–.
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EBD_7207, THE p-BLOCK ELEMENTS (GROUP 13 AND 14), , 172, , 66., , Catenation i.e., linking of similar atoms depends on size and, electronic configuration of atoms. The tendency of, catenation in Group 14 elements follows the order :, (a) C > Si > Ge > Sn, (b) C >> Si > Ge Sn, (c) Si > C > Sn > Ge, (d) Ge > Sn > Si > C, The catenation tendency of C,Si and Ge is in the order, , 67., , Ge < Si < C.The bond energies (in kJ mol ) of C-C,Si-Si, and Ge-Ge bonds, respectively are, (a) 167, 180, 348, (b) 180, 167, 348, (c) 348, 167, 180, (d) 348, 180, 167, Lead pipes are readily corroded by, , 65., , 1, , 68., , 69., , 70., , 71., , 72., , 73., , 74., , 75., , 76., , (a), , H 2SO 4, , (b) HCl, , (c), , CH 3COOH, , (d) pure water, , Lead pipes are not suitable for drinking water because, (a) lead forms basic lead carbonate, (b) lead reacts with water containing air to form, Pb(OH)2, (c) a layer of lead dioxide is deposited over pipes, (d) lead reacts with air to form litharge, The reducing power of divalent species decreases in the, order, (a) Ge > Sn > Pb, (b) Sn > Ge > Pb, (c) Pb > Sn > Ge, (d) None of these, The element that does not show catenation among the, following p-block elements is, (a) carbon, (b) silicon, (c) germanium, (d) lead, How many six membered and five membered rings are, present in fullerene?, (a) Six membered = 20, five membered = 10, (b) Six membered = 20, five membered = 12, (c) Six membered = 25, five membered = 10, (d) Six membered = 12, five membered = 25, Which of the following is the pure form of carbon ?, (a) Diamond, (b) Fullerene, (c) Graphite, (d) All three forms are equally pure, Which one of the following is not an allotrope of carbon ?, (a) Carborundum, (b) Diamond, (c) Soot, (d) Graphite, Which of the following types of forces bind together the, carbon atoms in diamond ?, (a) Ionic, (b) Covalent, (c) Dipolar, (d) van der Waal’s, Carborundum is, (a) SiC, (b) CaC2, (c) Mg2C3, (d) None of these, Buckminster fullerene is, (a) pure graphite, (b) C-60, (c) diamond, (d) C-90, , 77., , 78., , 79., , 80., , 81., , 82., , 83., , 84., , 85., , 86., , The hybridisation state of carbon in fullerene is, (a) sp, (b) sp 2, 3, (c) sp, (d) sp 3 d, The number of carbon atoms in Buckminsterfullerene is, (a) 50, (b) 350, (c) 60, (d) 70, Graphite is a soft solid lubricant extremely difficult to melt., The reason for this anomalous behaviour is that graphite, (a) is an allotropic form of diamond, (b) has molecules of variable molecular masses like, polymers, (c) has carbon atoms arranged in large plates of rings of, strongly bound carbon atoms with weak inter plate, bonds, (d) is a non-crystalline substance, In graphite, electrons are, (a) localised on every third C-atom, (b) present in anti-bonding orbital, (c) localised on each C-atom, (d) spread out between the structure, The elements commonly used for making transistors are, (a) C and Si, (b) Ga and In, (c) P and As, (d) Si and Ge, The element which is exclusively applied as semi-conductor, (a) Au, (b) Ge, (c) Pt, (d) Si, Glass is a, (a) liquid, (b) solid, (c) supercooled liquid, (d) transparent organic polymer, Glass reacts with HF to produce, (a) SiF4, (b) H2SiF6, (c) H2SiO3, (d) Na3AlF6, Producer gas is the mixture of, (a) CO + N2, (b) CO + H2, (c) CO + water vapours (d) N2 + CH4, Coal gas is a mixture of, (a) H 2 O and CO, (b) H 2 ,CO, N 2 and CH 4, (c), , 87., , 88., , 89., , H 2 and CO, , (d), , CH 4 and CO, , Crystalline form of silica is called, (a) crystalline silicon, (b) quartz, (c) rock, (d) talc, Dry ice is, (a) solid SO2, (b) solid NH3, (c) solid O2, (d) solid CO2, In silica (SiO2), each silicon atom is bonded to, (a) two oxygen atoms, (b) four oxygen atoms, (c) one silicon and two oxygen atoms, (d) one silicon and three oxygen atoms
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THE p-BLOCK ELEMENTS (GROUP 13 AND 14), , 90., , 91., , 92., , 93., , 94., , 95., , 96., , 97., , 98., , R3SiCl on hydrolysis forms, (a) R3SiOH, (b), , (c) R 2Si O, (d) None of these, Which of the following statements is false?, (a) Water gas is a mixture of hydrogen and carbon, monoxide, (b) Producer gas is a mixture of CO and nitrogen, (c) Water gas is a mixture of water vapour and hydrogen, (d) Natural gas consists of methane, ethane and gaseous, hydrocarbons., Which gas is essential constituent of almost all fuel gases ?, (a) CO2, (b) N2, (c) CO, (d) H2O, CO2 is used for extinguishing fire because, (a) it has a relatively high critical temperature, (b) in solid state, it is called dry ice, (c) it is neither combustible nor a supporter of combustion, (d) it is a colourless gas, The correct statement with respect to CO is, (a) it combines with H2O to give carbonic acid, (b) it reacts with haemoglobin in RBC, (c) it is powerful oxidising agent, (d) it is used to prepare aerated drinks, Producer gas, a fuel and also a source of nitrogen is obtained, by, (a) passing a mixture of steam and air over incandescent, coke., (b) spraying oil into hot retorts., (c) restricted supply of air through a bed of incandescent, coke., (d) passing steam over incandescent coke., Which of the following shows bond in silicone :, (a) Si – Si – Si – Si, (b) – Si – O – Si – O – Si, (c) Si – C – Si – C – Si, (d) Si – C – Si – O – Si, Which of the following is formed on dehydration of, formic acid with concentrated H2SO4 ?, (a) CO, (b) CO2, (c) CH4, (d) H2, _____ helps to maintain pH of blood between 7.26 to 7.42, (a) CO2, (b) H2CO3, (c), , 99., , R 3Si O SiR 3, , CO32, , (d) H2CO3/HCO3–, , Which of the following is not the crystalline form of, silica?, (a) Quartz, (b) Cristobalite, (c) Tridymite, (d) All are crystalline form of silica., 100. Which of the following is used in surgical and cosmetic, plants?, (a) Silicones, (b) Silicates, (c) Silica, (d) None of these, , 173, , 101. Which of the following is not a man-made silicate ?, (a) Glass, (b) Cement, (c) Zeolites, (d) All are man-made silicates, 102. Which type of zeolite is used to convert alcohols directly, into gasoline ?, (a) ZSM – 3, (b) ZSM – 5, (c) ZSM – 2, (d) All of these, , STATEMENT TYPE QUESTIONS, 103. Which of the following statement(s) is/are not correct ?, (i) Valence shell electronic configuration of p-block, elements is ns2 np1-6, (ii) Non metals and metalloids exist only in the p-block, of the periodic table., (iii) In boron, carbon and nitrogen families the group, oxidation state is the most stable state for the lighter, elements in the group., (iv) For heavier elements in each group oxidation state, two unit less than the group oxidation state becomes, more stable due to inert pair effect, (a) (ii) only, (b) (ii), (iii) and (iv), (c) (iii) and (iv), (d) All given statements are correct, 104. Which of the following sequence of T and F is correct for, given statements. Here T represents ‘True’ and F, represents ‘False’ statement., (i) Aluminium forms [AlF6]3– ion while boron forms, only [BF4]– ion due to presence of d-orbitals in, aluminium., (ii) The first member of a group differs from the heavier, members in its ability to form p -p multiple bonds, to itself and to other second row elements. While, heavier member forms d -p bonds., (iii) d-orbitals contribute more to the overall stability of, molecules than p -p bonding of second row, elements., (a) TTT, (b) FTF, (c) TTF, (d) FTT, 105. Which of the following statement(s) is/are incorrect ?, (i) Trichlorides on hydrolysis in water form tetrahedral, [M(OH)4]– species., (ii) Hybridisation state of metal in tetrahedral species is, sp3 ., (iii) Aluminium chloride in acidified aqueous solution, forms [Al(OH)4]– ion., (a) (i) and (ii), (b) (ii) only, (c) (iii) only, (d) (i) and (iii), 106. Which of the following statement(s) regarding BCl3 and, AlCl3 is/are correct ?, (i) BCl3 possess lower melting point than AlCl3., (ii) BCl3 is more covalent in character than AlCl3., (a) Statement (i) is correct explanation for statement (ii)., (b) Statement (i) and (ii) both are incorrect, (c) Statement (i) and (ii) both are correct, (d) Statement (ii) is correct explanation for statement (i)
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EBD_7207, THE p-BLOCK ELEMENTS (GROUP 13 AND 14), , 174, , 107. Which of the following statement(s) is/are incorrect ?, (i) Higher boranes are not flammable., (ii) Boranes are hydrolysed by water to give orthoboric, acid., (iii) Boranes undergoes cleavage reactions with Lewis, bases to give borane adducts., (a) (i) only, (b) (ii) and (iii), (c) (iii) only, (d) (i) and (ii), 108. Select the correct statements for diborane :, (i) Boron is approximately sp3 hybridized, (ii) B – H – B angle is 180°, (iii) There are two terminal B – H bonds for each boron, atom, (iv) There are only 12 bonding electrons, (a) (i), (ii) and (iv), (b) (i), (ii) and (iii), (c) (ii), (iii) and (iv), (d) (i), (iii) and (iv), 109. Which of the following sequence of T and F is correct for, given statements. Here T stands for the true and F stands, for false statement., (i) The tendency to show +2 oxidation state increase in, the sequence Ge < Sn < Pb., (ii) Tin in +2 state is a reducing agent., (iii) Lead compounds in +2 state are strong oxidising, agents., (iv) In tetravalent state molecules of group 13 elements, act as electrons donor species., (a) TTTT, (b) TTFF, (c) TTFT, (d) TFFT, 110. Which of the following statement(s) is / are incorrect for, CO2?, (i) In laboratory CO2 is prepared by the action of dilute, HCl on calcium carbonate, (ii) Carbon dioxide is a poisonous gas, (iii) Increase in carbon dioxide content in atmosphere, lead to increase in green house effect., (iv) CO2 as dry ice is used as a refrigerant for ice cream, and frozen food., (a) (i) and (ii), (b) Only (ii), (c) (i), (ii) and (iii), (d) (ii) and (iii), 111. Which of the following sequence of T and F is correct for, given statements. Here T stands for true and F stands for, false statement., (i) Quartz is extensively used as a piezoelectric material., (ii) Kieselghur is an amorphous form of silica which is, used in filteration plants., (iii) Silica does not react with halogens, dihydrogen and, most of the acids and metals even at elevated, temperature., (a) TTT, (b) TFF, (b) TFT, (d) FFT, , MATCHING TYPE QUESTIONS, 112. Match the columns, Column-I, Column-II, (A) Borax-bead, (p) Alum, (B) Inorganic benzene, (q) Diborane, (C) Antiseptic, (r) Metaborate, (D) Bridged hydrogens, (s) Borazole, (a) A – (p), B – (r), C – (q), D – (s), (b) A – (r), B – (s), C – (p), D – (q), (c) A – (s), B – (r), C – (p), D – (q), (d) A – (q), B – (r), C – (s), D – (p), 113. Identify (i) to (v) in reactions (1) and (2) on the basis of, your identification choose the correct code for matching, Column-I with Column-II., 1., 2., , Na2B4O7.10H2O, , (i), , (ii) + (iii), , Na2B4O7.7H2O, (iv) + (v), Column-I, Column-II, (A) (i), (p) H3BO3, (B) (ii), (q) B2O3, (C) (iii), (r) NaBO2, (D) (iv), (s) NaOH, (E) (v), (t) Na2B4O7, (a) A – (t), B – (s), C – (p), D – (q), E – (r), (b) A – (r), B – (q), C – (s), D – (p), E – (t), (c) A – (t), B – (r), C – (q), D – (p), E – (s), (d) A – (t), B – (r), C – (s), D – (q), E – (p), 114. Match Column-I (Compound of boron) with Column-II, (Use) and choose the correct option., Column-I, Column-II, (A) Metal borides, (p) Flux for soldering metals, (B) Boron fibres, (q) Bullet-proof vest, (C) Borax, (r) As a mild antiseptic, (D) Boric acid, (s) As control rods in, nuclear industry, (a) A – (q), B – (s), C – (r), D – (p), (b) A – (q), B – (s), C – (p), D – (r), (c) A – (s), B – (q), C – (r), D – (p), (d) A – (s), B – (q), C – (p), D – (r), 115. Match the columns, Column-I, Column-II, (A) Carbon, (p) Metal, (B) Silicon, (q) Non-metal, (C) Germanium, (r) Metalloid, (D) Tin, (E) Lead, (a) A – (q), B – (q), C – (r), D – (p), E – (p), (b) A – (q), B – (r), C – (r), D – (p), E – (p), (c) A – (q), B – (r), C – (r), D – (p), E – (q), (d) A – (q), B – (q), C – (q), D – (r), E – (p)
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THE p-BLOCK ELEMENTS (GROUP 13 AND 14), , 116. Match columns, Column-I, (A) Graphite fibres, (B) Carbon black, (C) Charcoal, (D) Diamond, (a) A – (s), B – (q),, (b) A – (q), B – (s),, (c) A – (q), B – (r),, (d) A – (p), B – (r),, 117. Match the columns, Column-I, (A), (B), (C), (D), (E), (a), (b), (c), (d), , C, C, C, C, , –, –, –, –, , Column-II, (p) Abrasive for sharpening, hard tools, (q) Formation of light, weight composites., (r) Used in water filters to, remove organic, contaminators, (s) As filler in automobile, tyres, (r), D – (p), (r), D – (p), (s), D – (p), (s), D – (q), , Column-II, 1, Borazole, (p) CaSO4. H2O, 2, Plaster of Paris, (q) C60, Boric acid, (r) SiO2, Quartz, (s) B3N3H6, Buckminsterfullerene (t) H3BO3, A – (r); B – (p); C – (q); D – (t); E – (s), A – (p); B – (t); C – (r); D – (s); E – (q), A – (t); B – (q); C – (p); D – (r); E – (s), A – (s); B – (p); C – (t); D – (r); E – (q), , ASSERTION-REASON TYPE QUESTIONS, Directions : Each of these questions contain two statements,, Assertion and Reason. Each of these questions also has four, alternative choices, only one of which is the correct answer. You, have to select one of the codes (a), (b), (c) and (d) given below., (a) Assertion is correct, reason is correct; reason is a correct, explanation for assertion., (b) Assertion is correct, reason is correct; reason is not a, correct explanation for assertion, (c) Assertion is correct, reason is incorrect, (d) Assertion is incorrect, reason is correct., 118. Assertion : Atomic radius of gallium is higher than that, of aluminium, Reason : The presence of additional d-electron offer poor, screening effect for the outer electrons from increased, nuclear charge., 119. Assertion : Boron is metalloid., Reason : Boron shows metallic nature., 120. Assertion : The use of aluminium and its compounds for, domestic purposes is now reduced considerably., Reason : The highly toxic nature of aluminium is the, responsible factor., 121. Assertion : Pb4+ compounds are stronger oxidizing agents, than Sn4+ compounds., Reason : The higher oxidation states for the group 14, elements are more stable for the heavier members of the, group due to ‘inert pair effect’., , 175, , 122. Assertion : PbI4 of lead does not exist., Reason : Pb–I bond initially formed during the reaction, does not release enough energy to unpair 6s2 electrons., 123. Assertion : Graphite is thermodynamically most stable, allotrope of carbon., Reason :, , fH, , of graphite is taken as zero., , CRITICAL THINKING TYPE QUESTIONS, 124. The liquefied metal which expands on solidification is :, (a) Ga, (b) Al, (c) Zn, (d) In, 125. What is x in the following reaction?, Al(s) + NaOH (aq) + H2O (l), x + H2(g), –, (a) Na2[Al(OH)4], (b) Na+[Al(OH)4]–, –, (c) Na2[Al(OH)6], (d) Na+ [Al(OH)6]–, 126. Which among the following oxides react with alkali?, B2O3, Al2O3 and Tl2O, (a) B2O3 and Al2O3, (b) Al2O3 and Tl2O, (c) Only B2O3, (d) B2O3 and Tl2O, 127. White fumes appeared around the bottle of anhydrous, aluminium chloride is due to _____, (a) Cl2 gas, (b) moist HCl, (c) condensation of aluminium chloride vapours, (d) None of these, 128. What is the oxidation state and hybridisation of boron in, compound formed when BCl3 undergoes reaction with the, water?, (a) 3, sp2d, (b) 3, sp3, 3, (c) 4, sp, (d) 3, sp2d, 129. Which is not correct?, (a) Al acts as a reducing agent, (b) Al does not react with steam even at higher, temperature, (c) Al forms a number of alloys with other metals, (d) Al is ionic in all its compounds, 130. Which one of the following is the correct statement?, (a) Boric acid is a protonic acid, (b) Beryllium exhibits coordination number of six, (c) Chlorides of both beryllium and aluminium have, bridged structures in solid phase, (d) B2H6.2NH3 is known as ‘inorganic benzene’, 131. BF3 is used as a catalyst in several industrial processes, due to its, (a) strong reducing nature, (b) weak reducing action, (c) strong Lewis acid nature, (d) weak Lewis acid character, 132. What is the colour obtained when borax is heated in a, Bunsen burner flame with CoO?, (a) Blue, (b) Black, (c) Green, (d) Violet
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EBD_7207, THE p-BLOCK ELEMENTS (GROUP 13 AND 14), , 176, , 133. Which of the following statements about H3BO3 is not, correct?, (a) It is a strong tribasic acid, (b) It is prepared by acidifying an aqueous solution of, borax, (c) It has a layer structure in which planar BO3 units are, joined by hydrogen bonds, (d) It does not act as proton donor but acts as a Lewis, acid by accepting a lone pair of electrons, 134. The hybridisation of boron atom in orthoboric acid is, (a) sp, (b) sp 2, 3, (c) sp, (d) sp3d, 135. Which is not the use of orthoboric acid?, (a) As an antiseptic and eye wash., (b) In glass industry., (c) In glazes for pottery., (d) In borax - bead test., 136. Which of the following reaction shows production of, diborane on industrial scale ?, (a) 4BF3 + 3LiAlH4, (b) 2NaBH4 + I2, , 2B2H6 + 3LiF + 3AlF3, B2H6 + 2NaI + H2, 450K, , 137., , 138., , 139., , 140., , 141., , 142., , (c) 2BF3 + 6NaH, B2H6 + 6NaF, (d) Both (b) and (c), Identify the statement that is not correct as far as structure, of diborane is concerned, (a) There are two bridging hydrogen atoms and four, terminal hydrogen atoms in diborane, (b) Each boron atom forms four bonds in diborane, (c) The hydrogen atoms are not in the same plane in, diborane, (d) All, B – H bonds in diborane are similar, Which of the following structure is similar to graphite?, (a) B, (b) B4C, (c) B2H6, (d) BN, A compound X, of boron reacts with NH3 on heating to, give another compound Y which is called inorganic benzene., The compound X can be prepared by treating BF3 with, lithium aluminium hydride. The compounds X and Y are, represented by the formulas., (a) B2H6, B3N3H6, (b) B2O3, B3N3H6, (c) BF3, B3N3H6, (d) B3N3H6, B2H6, The product/s formed when diborane is hydrolysed is/are, (a) B2O3 and H3BO3, (b) B2O3 only, (c) H3BO3 and H2, (d) H3BO3 only, Which of the following species exists (A) [SiF 6 ]2– ,, (B) [GeCl6]2– and (C) [CCl6]2– ?, (a) (A) and (B), (b) (B) and (C), (c) Only (C), (d) (A) and (C), Ge(II)compounds are powerful reducing agents, whereas Pb(IV)compounds are strong oxidants .It is because, (a) Pb is more electropositive than Ge, (b) ionization potential of lead is less than that of Ge, , (c) ionic radii of Pb2+ and Pb4+ are larger than those of, Ge2+ and Ge4+, (d) of more pronounced inert pair effect in lead than, in Ge, 143. Which of the following statements is not correct ?, (a) Fullerene is formed by condensation of vapourised, Cn small molecules consists of mainly C60., (b) In fullerene a six membered ring can only fuse with, five membered ring and a five membered ring can, only fuse with six membered rings., (c) All carbon atoms are sp2 hybridised in fullerene, (d) All the above are correct., 144. The element that does not form a monoxide is, (a) lead, (b) tin, (c) germanium, (d) silicon, 145. A group 14 element is oxidised to form corresponding oxide, which is gaseous in nature, when dissolved in water pH of, the water decreases further addition of group 2 hydroxides, leads to precipitation. This oxide can be, (a) GeO2, (b) CO, (c) CO2, (d) SnO2, 146. Which among the following can act as reducing agent, (A)SnCl2, (B)CO and (C)PbCl2 ?, (a) (A) and (B), (b) (B) and (C), (c) (C) and (A), (d) Only (B), 147. Lead is not affected by dil. HCl in cold because, (a) Pb is less electronegative than H, (b) PbO film is formed which resists chemical attack by, acid, (c) PbCl2 protective coating gets formed on Pb surface, (d) PbO2 film is always present on Pb surface, which resist, chemical attack, 148. The percentage of s-character of the hybrid orbitals of, carbon in graphite and diamond are respectively, (a) 33, 25, (b) 50, 50, (c) 67, 25, (d) 33, 67, 149. What is the hybridisations of carbon atoms present in, diamond, graphite and fullerene respectively ?, (a) sp3,sp2 and sp2, (b) sp2,sp3 and sp2, 2, 2, 3, (c) sp ,sp and sp, (d) sp3,sp3 and sp2, 150. Which one of the following allotropic forms of carbon is, isomorphous with crystalline silicon?, (a) Graphite, (b) Coal, (c) Coke, (d) Diamond, 151. Which one of the following statements about the zeolites is, false ?, (a) They are used as cation exchangers, (b) They have open structure which enables them to take, up small molecules, (c) Zeolites are aluminosilicates having three dimensional, network, (d) None of the above, 152. Which of the following attacks glass, (a) HCl, (b) HF, (c) HI, (d) HBr
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THE p-BLOCK ELEMENTS (GROUP 13 AND 14), , 2., 3., 4., 5., 6., 8., 9., 10., 11., , Na 2 B4O7, anhydrous, , CuO B2 O3, 26., , (d), , 10H 2O, , Na2B4O7, , 2 NaBO2, B2 O3, sod.metaborate Boric anhydride, Cu(BO 2 )2, , cupric meta borate(Blue beed), , H 2O, , 30. (b) In Boric acid each B atom is sp2 hybridized and, contains BO 33, , units which are held together by, , hydrogen bonds., 31. (a), 32. (b), , H 3 BO 3, , 100 C, , HBO 2, , H 2 B4 O 7, , 160 C, , H, , H 2O, , 2B2 O3, , H 2O, , 33. (a) H3BO3 is monobasic acid., 34. (c), 35. (b), H, 36. (c), |, B, N H, H N, |, ||, H B, B H, N, |, H, , |, , (b) Na2B4O7. 10 H2O, , HBO2, metaboricacid, (orthorombic form), , |, , 25., , 373K, , |, , 22., 24., , (d) 2KOH 2Al 2H 2 O 2KAlO 2 3H 2, (a), (b) Down the group basic character of oxides increases., B2O3 - Acidc, Al2O3 - Amphoteric, Ga2O3 - Amphoteric, In2O3 - Basic, Tl2O3 - Basic, (d), 23. (a), (d) It is p p back bonding involving B and F. The, smaller atoms show more back bonding., , H3BO3, , ||, , 19., 20., 21., , B(OH) 4, , 28. (d) H3BO3 acts as a Lewis acid and accepts OH– ions to, form [B(OH)4]–, 29. (c) H3BO3 on heating at 373K yields metaboric acid, (HBO2), , |, , 16., , H3BO3 or B(OH)3 + OH–, , |, , 12., 13., 14., 15., , (a) The non – metal oxides are acidic or neutral whereas, metal oxides are basic in nature., (b) Aluminium does not occur in the free state in nature, but is most abundant metal in the earth’s crust., (d), (b) The order of ionisation enthalpies, as expected, is, iH1 < iH2 < iH3., (c) Due to the low density of aluminium it is useful for, food packaging., (c), 7. (b), (c) The m.p decreases from B to Ga , hence gallium (Ga), has least m.p. (303 K) among group of 13 element., (a) Due to its small size and high ionization energy boron, does not form B3+ ion., (c) Gallium is remarkable for its unusually low M.P., (29.7°C)., (b) Thallium shows different (+1 and +3) oxidation states, because of inert pair effect., (a), (c) Because Tl +5 does not exist, (c), (d) The order of strength of Lewis acid character for boron, halides is, BF3 < BCl3 < BBr3 < BI3 (due to back, bonding), (b), 17. (a), 18. (c), , |, , 1., , 27. (d) B(OH)3 is acid because it can take OH– ions., , ||, , FACT / DEFINITION TYPE QUESTIONS, , 177, , Inorganic benzene, B 3N 3H6, , It is isoelectronic with benzene., 37. (d) In diborane (B2H6) structure there are four 2c-2e bonds, and two 3c–2e bonds (see structure of diborane)., Structure of B 2H6 :, , Hb, Ht, , Ht, , ••, B, , B, , Ht, , Ht, ••, Hb, 38. (b), 39. (c), 40. (c), 41. (d), 42. (b), 43. (c) B2H6 contains hydrogen bridge bonds. These are one, electron bonds also known as banana bonds., 44. (d) B2H6 + NH3, , excess NH3, low temperature, , B2H6.2NH3, , Diborane with ammonia gives B2H6 .2NH3 that is, formulated as [BH2(NH3)]+[BH4]– which when heated, to 473K decomposes to give borazole.
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EBD_7207, THE p-BLOCK ELEMENTS (GROUP 13 AND 14), , 178, , 45., 46., , Cl, Cl, Cl, AlCl3 (dimer), Al, Cl, Cl, Al, Al, (c) Valence shell electronic configuration of group 14, , (c), , 2 2., , elements is ns p, 47., 48., 49., , 50., , 51., , (c) Electronegativity decreases down the group., (c) 14C is a radioactive isotope with half life of 5770, years and used for radiocarbon dating., (c) Due to non-availability of vacant d-orbitals, it cannot, exceed its coordination number more than four. Thus, carbon never forms complexes e.g., [CCl6]2– deos not, exist but [SiCl6]2– exists., (b) The inert pair effect is most prominent in the heavier, members of the group. Inert pair effect increases as we, move the group down the group., (c) Inert pair effect increases down the group. Hence for, , 64., , Pb in (IV) O.S. but Br2 and I 2 can not achieve Pb in, (IV) O.S. secondly Pb 4 is strong in oxidising nature, and in its presence, Br and I can not exist., 65., 66., 67., 68., 69., , 53., , (c) CO2, SiO2 are acidic, CaO is basic and SnO 2 is, amphoteric., (a) Red lead is Pb3O4. It is a mixed oxide of Pb (II) and, Pb (IV). It acts as a powerful oxidising agent., (d) PbO2 is a strong oxidising agent and is produced in, situ in lead storage batteries. The anode is oxidized to, PbO2 and cathode is reduced to spongy Pb., (b) SiO2 is acidic oxide., (d) The thermal stability of tetrahalides decreases in order, CX4 > SiX4 > GeX4 > SnX4 and in terms of same metal, with different halides is in order of, MF4 > MCl4 > MBr4 > MI4., (a) The stability of dihalides (MX2) increases down the, group. Except C and Si, the other members form, dihalides., (a) Since bond energy of C-F >C-Cl > C-Br > C-I, , 54., 55., , 56., 57., , 58., , 59., , Pb3O4 is also known as Sindhur.., , Hence CF4 is most stable., 60., , 61., 62., , 63., , (d) Reluctance of valence shell electrons to participate in, bonding is called inert pair effect. The stability of lower, oxidation state (+2 for group 14 element) increases on, going down the group. So the correct order is, SiX2 < GeX2 < SnX2 < PbX2, (b), , Ge4 is more stable than Ge2+ . Hence GeCl 4 is more, , stable than GeCl 2, (d) Carbon halides are not hydrolysed due to absence of, d-orbitals. On the other hand SiCl4 is easily hydrolysed, due to the availability of d-orbitals in Si., SiX4 + 2H2O, SiO2 + 4HX, (d) In nature Pb 4, , is strong oxidant and I, , reductant. Hence PbI 4 cannot exist., , is strong, , (b) 2Pb 2H 2O O 2, 2Pb(OH) 2, (a) The stability of +2 O.S. follows the order, Sn 2, , Ge 2, , 83., , Hence reducing power Ge Sn Pb, (d) The order of tendency of catenation for elements of C, family is, C >> Si > Ge Sn > Pb, (b) Fullerene contains twenty six membered rings and, twelve five membered rings., (b) Fullerenes are the only pure form of carbon because, they have smooth structure without having dangling, bonds., (a) Carborundum is chemically SiC. It is not an allotrope, of carbon., (b) In diamond each carbon atom is sp3 hybridized and, thus forms covalent bonds with four other carbon atoms, lying at the corners of a regular tetrahedron., (a), (b) Buckminster fullerene is C60. The molecule has shape, of soccer ball., (b) In fullerene each carbon atom is bonded to three other, carbon atoms and is sp2 hybridised., (c) Buckminster fullerene has the formula C60 and is made, from interlocking hexagonal and pentagonal rings of, C-atoms., (c), (d) In graphite, each carbon is sp2 -hybridized and the, single occupied unhybridized p-orbitals of C-atoms, overlap side wise to give -electron cloud which is, delocalized and thus the electrons are spread out, between the structure., (d) Si and Ge are semiconductors and are used in making, transistors., (b) Both Ge and Si are extensively used as semiconductors., Semiconductors are solids where there is only a small, difference in energy, called band gap, between the filled, valency band of electrons and a conduction band since, the band gap of Ge is less than Si, it is a better element, to be used as semiconductor., (c) Glass is a super cooled liquid., , 84., , (b), , 70., , 71., , (d), , (b), (d) The more the bond energy, the more is the, catenation., (c) Lead pipes are readily corroded by water containing, organic acids., , Pb 2, , Pb2 ,O.S. is most stable., 52., , (b) F and Cl are more oxidising in nature and can achieve, , 72., , 73., 74., , 75., 76., 77., 78., , 79., 80., , 81., 82., , 6 HF SiO 2, , H 2SiF6, , 2 H2O
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THE p-BLOCK ELEMENTS (GROUP 13 AND 14), , Producer gas is a fuel gas and is mixture of CO and N2., Coal gas is a mixture of H2 + CO + N2 + CH4, Quartz is crystalline form of silica., , — —, , — —, , In silica (SiO2); each Si atom is surrounded by four, oxygen atom., , — —, , (a), (b), (b), (d), (b), , — —, , 85., 86., 87., 88., 89., , — Si — O —Si — O — Si — O —Si —, —, , —, , —, , —, , —, , —, , — —, , — —, , — —, , O, , —, , O, , —, , O, , — —, , O, , — Si — O —Si — O — Si — O —Si —, O, , O, , O, , O, , — Si — O —Si — O — Si — O —Si —, Structure of SiO2, Only Si – O bonds exist and no Si = O., 90., , (b), , R 3SiCl HOH, , R 3SiOH HCl, , R 3SiOH HOSiR 3, 91., 92., 93., 94., 95., , R 3Si O SiR 3 H 2O, , (a) Water gas is CO H 2, (c) CO is essential constituent of almost all fuel gases., (c) CO 2 is incombustible and non supporter of, combustion., (b) CO react with haemoglobin, forms carboxy, haemoglobin and stopes the supply of O2, , 96., , (c) Producer gas is mixture of CO + N 2 . It is prepared by, incomplete combustion of coal in limited supply of air., (b) Silicone has Si – O – Si – O – Si linkage., , 97., , (a) HCOOH, , 373K, conc. H2SO4, , H2O + CO, , (d) H2CO3/HCO3– buffer system help to maintain pH of, blood between 7.26 to 7.42., 99. (d) Quartz, Cristobalite and Tridymite are crystalline, form of silica., 100. (a) Being biocompatible silicones are used in surgical, and cosmetic plants., 101. (c) Zeolite is not a man-made silicate., 102. (b) ZSM – 5 type of zeolite is used to convert alcohols, directly into gasoline., 98., , STATEMENT TYPE QUESTIONS, 103. (d) All the given statements are correct., 104. (c) d-orbitals are of higher energy than the p-orbitals,, they contribute less to the overall stability of, molecules than p -p bonding of the second row, elements., 105. (c) Aluminium chloride in acidified aqueous solution, forms octahedral [Al(H2O)6]3+ ion., , 179, , 106. (d) BCl3 is a covalent compound hence lower melting, point., 107. (a) Higher boranes are also spontaneously flammable in, air., 108. (d), , H, , H, , H, B, , 197°, , B, , H, H, H, 3, B is sp hybridized, Only 12 bonding electrons available, BHB angle is 97° not 180°., 109. (b) Lead compounds in +4 state are strong oxidising, agents. In tetravalent state the number of electrons, around the central atom in a molecule is eight. Being, electron precise molecules, they are normally not, expected to act as electron acceptor or electron, donor species., 110. (b) Carbon dioxide is not a poisonous gas., 111. (a) All the given statements are true., , MATCHING TYPE QUESTIONS, 112. (b), 113. (c) Na2B4O7.10H2O, , Na2B4O7, (i), , 2NaBO2 + B2O3, (ii), , Na2B4O7 + 7H2O, , (iii), , 4H3BO3 + 2NaOH, (iv), (v), , 114. (d), 115. (a) Carbon and silicon are non-metals. Germanium is a, metalloid. Tin and lead are metals., 116. (b), 117. (d), , ASSERTION-REASON TYPE QUESTIONS, 118. (c) Atomic radius of gallium is less than that of, aluminium., 119. (c) Boron is metalloid. Thus assertion is correct. Metalloids, possess, metallic as well as non-metallic nature. Hence,, reason is false., 120. (a) The use of aluminium and its compounds for domestic, purposes is now reduced considerably because of their, toxic nature., 121. (c) Assertion is true because lower oxidation state, becomes more & more stable for heavier elements in pblock due to inert pair effect. Hence Reason is false., 122. (a) PbI4 does not exist because Pb–I bond initially, formed during the reaction does not release enough, energy to unpair 6s2 electrons and excite one of, them to higher orbital to have four unpaired electrons, around lead atom., 123. (a)
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EBD_7207, THE p-BLOCK ELEMENTS (GROUP 13 AND 14), , 180, , CRITICAL THINKING TYPE QUESTIONS, 124. (a) Gallium (Ga) is soft , silvery metal. Its melting point is, 30ºC. This metal expands by 3.1% when it solidifies, and hence, it should not be stored in glass or metal, containers., 125. (b) 2Al(s) + 2NaOH(aq) + 6H2O(l), 2Na+[Al(OH)4]–(aq) + 3H2 (g), 126. (a) B2O3 is acidic and Al2O3 is amphoteric., 127. (b) Anhydrous aluminium chloride gets partially, hydrolysed with atmospheric moisture to liberate, HCl gas. Moist HCl appears white in colour., 128. (b) BCl3 forms [B(OH)4]– in which B is sp3 hybridized, and have +3 oxidation state., 129. (d) Al in its compounds forms covalent bonds., 130. (c) The correct formula of inorganic benzene is B3N3H6, so (d) is incorrect statement, OH, |, , Boric acid (H3BO3 or B, , OH ) is a lewis acid so (a), , |, , OH, is incorrect statement., The coordination number exhibited by beryllium is 4, and not 6 so statement (b) is incorrect., Both BeCl2 and AlCl3 exhibit bridged structures in solid, state so (c) is correct statement., , Cl, , m Cl, 2p, 20, Be 98°, 82°, , Cl, , Cl, , Be, , Cl, , 263 pm, , Be, , Be, , Cl, , Cl, , Cl, , Cl, , Cl, , Al, , Cl, , Cl, , Cl, , Al, , Cl, Cl, Cl, 131. (d), 132. (a) When borax is heated in a Bunsen burner flame with, CoO on a loop of platinum wire a blue coloured, Co(BO2)2 bead is formed., 133. (a) H3BO3 is a weak monobasic acid., 134. (b) The hybridizations of B in H3BO3 is sp2, 135. (d) Borax on heating gives B2O3 and NaBO2 which is, glassy mass and used for borax-bead test., 136. (c) Diborane is produced on industrial scale by the, reaction of BF3 with sodium hydride., 137. (d), , 138. (d) Boron nitride (BN) is known as inorganic graphite. The, most stable form is hexagonal one. It has layered, structure similar to graphite., , B–, N+, , –B, +N, B–, N+, , N, +, , –, B, N, +, , –, B, , N+, –B, +N, B–, , –B, , –, B, N, +, , –B, , N+ – + N, B, , +N, B–, , N, +, , 139. (a), 140. (c) When diborane is hydrolysed one can get both, orthoboric acid and H2., B2H6 + 6H2O 2H3BO3 + 6H2, 141. (a) Carbon does not contain d-orbital hence it cannot, expand its octet., 142. (d) Ge(II) tends to acquire Ge (IV) state by loss of electrons., Hence it is reducing in nature. Pb (IV) tends to acquire, Pb (II) O.S. by gain of electrons. Hence it is oxidising, in nature. This is due to inert pair effect., 143. (b) In fullerene a six membered ring can fuse with five, as well as with six membered ring while a five, membered ring can only fuse with a six membered, ring., 144. (d) Silicon does not form mono oxide., 145. (c) CO2 forms carbonic acid H2CO3, when dissolved in, water, CO is neutral, whereas other two GeO2 and, SnO2 are solids., 146. (a) Lead in +2 oxidation is stable while Sn and C are both, stable in +4 oxidation., 147. (c) Pb with dil HCl forms protective coating of PbCl 2, 148. (a) Graphite sp3, % s character = 33%, Diamond sp3, % s character = 25%, 149. (a), 150. (d) Diamond and crystalline silicon are isomorphous., 151. (d), 152. (b) Glass being a mixture of sodium and calcium silicates, reacts with hydrofluoric acid forming sodium and, calcium fluorosilicates respectively., Na 2SiO 3 6HF, , Na 2SiF6 3H 2 O, , CaSiO3 6HF CaSiF6 3H 2 O, The etching of glass is based on these reactions.
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12, ORGANIC CHEMISTRY–SOME BASIC, PRINCIPLES & TECHNIQUES, FACT / DEFINITION TYPE QUESTIONS, 1., , 2., , 3., , 4., , Which of the following scientist proposed that a ‘vital, force’ was responsible for the formation of organic, compounds ?, (a) Berzilius, , (b) Wohler, , (c) Berthelot, , (d) Kolbe, , First organic compound to be synthesised was, (a) methane, , (b) cane sugar, , (c) acetic acid, , (d) urea, , Which of the following organic compound was, synthesised by F. Wohler from an inorganic compound?, (a) Methane, , (b) Urea, , (c) Acetic acid, , (d) Chloroform, , The discovery that shook the belief in the vital force theory, was, , The state of hybrization of carbons 1, 3 and 5 are in the, following sequence, (a) sp2, sp, sp3, (b) sp, sp3, sp2, (c) sp, sp2, sp3, (d) sp3, sp2, sp, 9., The percentage of s- character of the hybrid orbitals in, ethane, ethene and ethyne are respectively., (a) 50, 75, 100, (b) 10, 20, 40, (c) 25, 33, 50, (d) 25, 50, 75, 10. Select the molecule which has only one -bond, (a) CH CH, (b) CH2 = CHCHO, (c) CH3CH = CH2, (d) CH3CH=CHCOOH, 11. 2- Pentene contains, (a) 15 - and one - bond (b) 14 -and one - bond, (c) 15 - and two - bonds (d) 14 - and two - bonds, 12. Which of the following does not represent the 2 – bromo, pentane ?, , (a) Stereoisomerism, (b) Synthesis of indigo, , (i), , (c) Wholer’s synthesis of urea from ammonium cyanate, , H, , H H H Br H, C C C C C, H H H H, , H, , (d) Fermentation of sugars, 5., , 6., , 7., , 8., , Br, , In laboratory, first organic compound was synthesised by, (a) Kekule, , (b) Hennel, , (c) Wohler, , (d) Liebig, , (ii), , Who is known as the “Father of Chemistry”?, (a) Faraday, , (b) Priestley, , (c) Rutherford, , (d) Lavoisier, , Br, (iii), , The hybridisation of carbon atom in C — C single bond of, H2C = CH — CH = CH2 is, (a), , sp3, , — sp, , (c), , sp2, , — sp2, , (b), , sp2, , — sp, , (b), , sp3, , — sp3, , In the hydrocarbon, CH3 – CH = CH – CH2 – C, 6, 5, 4, 3, 2, , CH, 1, , Br, (iv), (v) CH3CH2 CH2CHBrCH3, (a) (ii), (iii) and (v), (b) Only (ii), (c) (ii) and (iii), (d) (iii) and (v)
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EBD_7207, ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES & TECHNIQUES, , 182, , 13., , Which of the following correctly represents the expanded, form of following organic compound ?, , 1, , 2, , 3, , 4, , 5, , |, CH3, , H, , H, , H, C, , (a), , H, , H, , 1, , H, , 8, , H, , H, , H, , C, , C, C, , C, H, , H, , (c), , H, , H, , H, C, , H, H, , C, , C, , 18., H, H, , C, , C, C, H, , H, , H H, H, , (d), , H, , H, , 19., , H H, , C, , H, , C, , C, , C, , C, , C, H, , H, , H, , H, , 14., , Structural formula of benzene is, , 15., , H, H, C, H, H, C, H C C H, H C C H, H, H, (a) H, (b) H C C H, C C H, C, H C, H H, H, H, C H, C, H C C H, H C C H, (c) H C C H, (d) H C C H, H, C, C, H, H, The successive members in a homologues series differ, from each other by ________, (a) – CH2CH2– unit, (b) – CH2 unit, (c) – OCH3 unit, (d) – CH3 unit, Which of the following have incorrect molecular formula?, A. Icosane, –, C10H22, B. Triacontane, –, C30H62, C. Nonane, –, C9H20, D. Heptane, –, C7H14, (a) (A) and (D), (b) Only (D), (c) (B) and (D), (d) Only (B), Which of the following are incorrect methods of selecting, parent chain ?, , 16., , 17., , (i), , 7, , |, CH3, , H, , H, , 4, , 5, , 6, , C H3, 8, , 7, , 8, , 9, , |, CH 2 CH3, , 6, , 5, , 4, , 3, , (iv) CH3 CH CH 2 CH 2 CH 2 CH CH 2 CH 2 CH3, C, , C, , 3, , |, CH3, , C, , H, , 2, , |, , 7CH 2, , (iii) CH3 CH CH 2 CH 2 CH 2 CH CH 2 CH2 CH3, , C, , C, , C, , H, , (b), , H, , C, , H, , 6, , (ii) CH3 CH CH 2 CH 2 CH 2 CH CH 2 CH 2 CH 3, , 1, 2, 3, 4, 5, 6, CH3 CH CH 2 CH 2 CH 2 CH CH 2 CH 2 CH3, |, |, 7CH 2 CH 3, CH3, 8, , 20., 21., 22., , 23., 24., , 25., , 26., , 27., , |, CH 2 CH3, 2, 1, , (a) (i) and (ii), (b) (iv) only, (c) (i), (ii) and (iv), (d) (ii) only, The correct decreasing order of priority of functional, groups is, (a) – SO3H , – OH, – COCl,, C=C, (b) – COOH, – SO3H, – COOR, – OH, C=O, (c) – C C, – NH2, – OH,, (d) – CN, – CONH2,, C = O, – OH, Which of the following is incorrectly matched –, (a) vinegar carboxylic acid, (b) C2H6 alkane, (c) ethanol alcohol, (d) methanol ketone, The functional group present in organic, acid is –, (a) – OH, (b) – CHO, (c) –COOH, (d) > C = O, Which of these contains the carbonyl group?, (a) ketones, (b) aldehydes, (c) esters, (d) all of these, Butanone is a four-carbon compound with the functional, group –, (a) carboxylic acid, (b) aldehyde., (c) ketone, (d) alcohol., The functional group present in CH3COOC2H5 is –, (a) ketonic, (b) aldehydic, (c) ester, (d) carboxylic, Which of the following compounds contains 1°, 2°, 3° as, well as 4° carbon atoms ?, (a) Neopentane, (b) 2-methyl pentane, (c) 2,3-dimethyl butane (d) 2,2,3-trimethyl pentane, The number of secondary hydrogens in 2, 2-dimethylbutane, is, (a) 8, (b) 6, (c) 4, (d) 2, The compound which has one isopropyl group is, (a) 2, 2, 3, 3 - Tetramethylpentane, (b) 2, 2 - Dimethylpentane, (c) 2, 2, 3- Trimethylpentane, (d) 2- Methypentane, Which of the following statements is false for isopentane ?, (a) It has three CH3 groups, (b) It has one CH2 group, (c) It has one CH group, (d) It has a carbon which is not bonded to hydrogen
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ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES & TECHNIQUES, , 59., , 60., , 61., , 62., , 63., , 64., , 65., , 66., , 67., , 68., , Which of the following compounds is isomeric with 2, 2, 4,, 4- tetramethylhexane?, (a) 3-ethyl -2, 2- dimethylpentane, (b) 4-isopropylheptane, (c) 4-ethyl-3-methyl-4-n propyloctane, (d) 4, 4-diethyl-3-methylheptane, Which are isomers ?, (a) ethyl alcohol and dimethyl ether, (b) acetone and acetaldehyde, (c) propionic acid and propanone, (d) methyl alcohol and dimethyl ether, Methoxyethane and propanol are the examples of isomerism, of the type, (a) structural, (b) position, (c) functional, (d) tautomerism, Isomers of propionic acid are, (a) HCOOC2H5 and CH3COOCH3, (b) HCOOC2H5 and C3H7COOH, (c) CH3COOCH3 and C3H7OH, (d) C3H7OH and CH3COCH3, C6H5C N and C6H5N C are which type of isomers?, (a) Position, (b) Functional, (c) Tautomerism, (d) Linkage, A functional isomer of 1-butyne is, (a) 2-butyne, (b) 1-butene, (c) 2-butene, (d) 1, 3-butadiene, In which of the following, functional group isomerism is not, possible?, (a) Alcohols, (b) Aldehydes, (c) Alkyl halides, (d) Cyanides, The compounds CH3CH == CHCH3 and, CH3CH2CH == CH2, (a) are tautomers, (b) are position isomers, (c) contain same number of sp3– sp3, sp3– sp2 and sp2– sp2, carbon-carbon bonds, (d) exist together in dynamic equilibrium, Heterolytic fission of a covalent bond in organic molecules, gives, (a) free radicals, (b) cations and anions, (c) only cations, (d) only anions, Which of the following statements is not correct ?, (a) Carbocation posses sextet of electrons., (b) The order of carbocation stability is :, , CH3, , 69., , 70., , (CH3 )2 CH, , (CH3 )3 C, , (c) Carbocations have trigonal planar shape, (d) Carbocations are formed by heterolytic cleavage, Heterolytic fission of C – Br bond results in the formation of, (a) free radical, (b) carbanion, (c) carbocation, (d) Both (b) and (c), Which of the following carbocations is least stable?, (a) tert-Alkyl, (b) sec-Alkyl, (c) pri-Alkyl, (d) Methyl, , 185, , 71. Which of the following ions is most stable ?, (a), , CH 3 C CH 3, |, CH 3, , (b), , CH 3 CH 2 CH 2, , (c) CH 3CHCH 2 CH 3, (d) None of these, 72. The order of stability of the following carbocations :, CH 2, CH 2, , CH C H 2 ; CH3 CH 2 CH 2 ;, I, II, , is :, , III, (a) III > II > I, (b) II > III > I, (c) I > II > III, (d) III > I > II, 73. Select the most stable carbocation amongst the following, +, (a), (b), , +, , (c), , +, , (d), , +, , 74. What is the correct order of decreasing stability of the, following cations., I., , CH3 — CH— CH3, , II., , CH3 — CH— OCH 3, , III. CH3 — CH— CH 2 — OCH3, (a) II > I > III, (b) II > III > I, (c) III > I > II, (d) I > II > III, 75. The most stable carbonium ion among the following is, (a), , +, , C6 H5 CHC6 H5, , (b), , C6 H5CH 2, , (c) CH3 CH 2, (d) C6 H5CH 2CH 2, 76. The organic reaction which proceed through heterolytic, bond cleavage are called ________, (a) ionic, (b) polar, (c) nonpolar, (d) Both (a) and (b), 77. Among the following, the true property about, CH3, , +, C – CH3 is, , CH3, , (a), (b), (c), (d), , it is non-planar, its C+ is sp2-hybridized, an electrophile can attack on its C+, it does not undergo hydrolysis
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EBD_7207, ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES & TECHNIQUES, , 186, , 78., 79., , The shape of methyl carbanion is similar to that of –, (a) BF3, (b) NH3, (c) methyl free radical, (d) methyl carbocation, Arrange the carbanions,, , (CH3 )3 C , C Cl3 , (CH3 ) 2 C H , C6 H5 CH 2, in order of their decreasing stability :, , 90., , 91., 92., , (a) (CH3 ) 2 C H > C Cl3 > C6 H5 C H2 > (CH3 )3 C, (b) C Cl3 > C6 H5 CH 2 > (CH 3 )2 CH > (CH3 )3 C, (c), 80., 81., , 82., 83., , 84., , 85., , (CH3 )3 C > (CH3 )2 CH > C6 H 5 CH 2 > C Cl3, , (d) C6 H5 CH 2 > CCl3 > (CH3 )3 C > (CH3 ) 2 CH, The homolytic fission of a covalent bond liberates, (a) Carbonium ions, (b) Carbanions, (c) Free radicals, (d) Carbenes, Homolytic fission of C–C bond in ethane gives an, intermediate in which carbon is, (b) sp2-hybridised, (a) sp3-hybridised, (c) sp-hybridised, (d) sp2d-hybridised, Geometry of methyl free-radical is, (a) pyramidal, (b) planar, (c) tetrahedral, (d) linear, In which of the following homolytic bond fission takes place ?, (a) Alkaline hydrolysis of ethyl chloride, (b) Addition of HBr to double bond, (c) Photochlorination of methane, (d) Nitration of benzene, On exciting, Cl2 molecules by UV light, we get, (a) Cl, (b) Cl+, –, (c) Cl, (d) all of these, The increasing order of stability of the following free radicals, is, •, , •, , •, , 93., , 94., 95., , 96., , •, , (a) (C6H5)2 C H < (C6H5)3 C < (CH3)3 C < (CH3)2 C H, •, , •, , •, , •, , •, , •, , (b) (CH3)2 C H < (CH3)3 C < (C6H5)2 C H < (C6H5)3C, •, , •, , (c) (CH3)2 C H < (CH3)3 C < (C6H5)2 C H < (C6H5)3 C, •, , 86., , 87., , •, , •, , •, , (d) (C6H5)3C < (C6H5)2 C H < (CH3)3 C < (CH3)2 C H, Which of the following orders regarding relative stability of, free radicals is correct?, (a) 3° < 2° < 1°, (b) 3° > 2° > 1°, (c) 1° < 2° > 3°, (d) 3° > 2° < 1°, The most stable free radical among the following is, (a), , C 6 H5 CH 2CH 2, , (b), , 97., , 98., , (d) CH 3 CHCH 3, CH 3CH 2, For the reaction of phenol with CHCl3 in presence of KOH,, the electrophile is, (a), (b) : CCl2, CHCl 2, (d) CCl4, CHCl 2, The least stable free radical is, , 89., , (a) CH3 CH 2, , (b) CH3CH2 CH 2, , (c) (CH3)2 CH, , (d), , CH 3, , Electrophile, Nucleophile, A. HS, Cl, (CH3 )3N, B. BF3, C. H 2 N, C O, C2 H5O, D. R 3C X, (X H alogen), (a) B, C and D, (b) C and D, (c) C only, (d) B and D, Arrangement of (CH3)3 – C –, (CH3)2 – CH –, CH3 – CH2 –, when attached to benzyl or an unsaturated group in, increasing order of inductive effect is, (a) (CH3)3 –C – < (CH3)2 – CH – < CH3 – CH2–, (b) CH3 –CH2– < (CH3)2– CH – < (CH3)3 –C –, (c) (CH3)2 – CH– < (CH3)3 –C – < CH3—CH2–, (d) (CH3)3 – C– < CH3 –CH2 – < (CH3)2 –CH –, Polarization of electrons in acrolein may be written as, , CH 2, , CH CH, , O (b), , CH 2, , CH CH, , (c), , C6 H 5 CHCH 3, , (c), , (c) : CN, (d) C2 H 5O :, Which of the following represents a set of nucleophiles?, (b) AlCl3, BF3, NH3, (a) BF3, H2O, NH2–, (d) All of these, (c) CN–, RCH2–, ROH, Which of the following species does not acts as a, nucleophile?, (a) ROH, (b) ROR, (c) PCl3, (d) BF3, Which of the following is an electrophile ?, (a) Lewis acid, (b) Lewis base, (c) Negatively charged species, (d) None of the above, Which of the following pairs represent electrophiles?, (a) AlCl3, H2O, (b) SO3, NO2+, (c) BF3, H2O, (d) NH3, SO3, Which out of A, B, C and D is/are not correctly, categorised., , (a), , (c), , 88., , Which of the following is strongest nucleophile, (a) Br–, (b) : OH–, , 99., , O, , CH 2 CH CH O (d) CH 2 CH CH O, Point out the incorrect statement about resonance?, (a) Resonance structures should have equal energy, (b) In resonating structures, the constituent atoms must, be in the same position, (c) In resonating structures, there should not be same, number of electron pairs, (d) Resonating structures should differ only in the location, of electrons around the constituent atoms, , : CH 2, , C CH 3 and CH 2 C CH 3 are, |, ||, O, : .O. :, . .:, (a) resonating structures (b) tautomers, (c) geometrical isomers (d) optical isomers
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ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES & TECHNIQUES, , 100. In which of the following, resonance will be possible?, (a), , CH3, , CH 2, , (b), , CH 2, , CH CH, , (c), , CH 3COCH 3, , CH 2, , CHO, O, , (d) CH 2 CH CH 2 CH CH 2, 101. Which of the following statements regarding the resonance, energy of benzene is correct?, (a) Resonance energy is the energy required to break the, C–H bond in benzene, (b) Resonance energy is the energy required to break the, C–C bond in benzene, (c) Resonance energy is a measure of stability of benzene, (d) Resonance energy is the energy required to convert, , 102. Which of the following is not correctly matched ?, Group showing + R effect Group showing – R effect, (a) – NHCOR, – COOH, (b), , – OH, , C=O, , (c) – OR, – CHO, (d) – OCOR, – NO2, 103. The polarity is produced in the molecule by the interaction, of two – bonds or between a – bond and lone pair, of electrons present on an adjacent atom., The above statement is true for which of the following ?, (a) Inductive effect, (b) Electromeric effect, (c) Resonance effect, (d) Hyperconjugation, 104. Electromeric effect is a, (a) permanent effect, (b) temporary effect, (c) resonance effect, (d) inductive effect, 105. The kind of delocalization involving sigma bond orbitals is, called, (a) inductive effect, (b) hyperconjugation effect, (c) electromeric effect, (d) mesomeric effect, 106. Hyperconjugation involves overlap of the following orbitals, (a) (b), (c) p- p, (d) 107. Choose the correct order of stability of carbocation using, concept of hyperconjugation., CH3, |, C, |, CH3, , CH3, I, , CH3, , CH3, |, C, |, CH3, , CH 3CH 2 CH 3, III, , IV, , II, , (a) I < II < III < IV, (b) IV < III < II < I, (c) III < IV < II < I, (d) None of these, 108. Hyperconjugation is most useful for stabilizing which of, the following carbocations ?, (a) neo-Pentyl, (b) tert-Butyl, (c) iso-Propyl, (d) Ethyl, , 187, , 109. Which of the following is an example of elimination reaction?, (a) Chlorination of methane, (b) Dehydration of ethanol, (c) Nitration of benzene, (d) Hydroxylation of ethylene, 110. CH3 – Br + NH3, CH 3 – NH 2 HBr, The above reaction is classified as, (a) substitution, (b) addition, (c) elimination, (d) rearrangement, 111. Which of the following method is not used for determining, purity of a compound ?, (a) Chromatographic techniques, (b) Spectroscopic techniques, (c) Melting point, (d) All of the above parameters are used for determining, the purity of a compound., 112. Which of the following is not the criteria of purity of a, substance?, (a) solubility, (b) melting point, (c) boiling point, (d) density, 113. In crystallisation process impurities which impart colour, to the solution are removed by which of the following ?, (a) Repeated crystallisation, (b) Activated charcoal, (c) Bleaching agent, (d) Both (a) and (b), 114. Aniline is purified by, (a) steam distillation, (b) simple distillation, (c) vacuum distillation, (d) extraction with a solvent, 115. Which is purified by steam distillation, (a) Aniline, (b) Benzoic acid, (c) Petroleum, (d) Naphthalene, 116. The best method for the separation of naphthalene and, benzoic acid from their mixture is:, (a) distillation, (b) sublimation, (c) chromatography, (d) crystallisation, 117. In steam distillation the vapour pressure of volatile organic, compound is, (a) equal to atmospheric pressure, (b) double the atmospheric pressure, (c) less than atmospheric pressure, (d) more than atmospheric pressure, 118. Fractional distillation is used when, (a) there is a large difference in the boiling point of liquids, (b) there is a small difference in the boiling points of liquids, (c) boiling points of liquids are same, (d) liquids form a constant boiling mixture, 119. Purification of petroleum is carried out by, (a) fractional distillation (b) steam distillation, (c) vacuum distillation, (d) simple distillation, 120. A liquid decomposes at its boiling point. It can be purified, by, (a) steam distillation, (b) fractional distillation, (c) vacuum distillation, (d) sublimation
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EBD_7207, 188, , ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES & TECHNIQUES, , 121. Distillation under reduced pressure is employed for, (a) C6H6, (b) petrol, (c) CH2OHCHOHCH2OH, (d) organic compounds used in medicine, 122. Impure glycerine is purified by, (a) steam distillation, (b) simple distillation, (c) vacuum distillation, (d) None of these, 123. Glycerol can be separated from spent lye in soap industry, by which of the following method ?, (a) Steam distillation, (b) Fractional distillation, (c) Distillation under reduced pressure, (d) Differential extraction, 124. The latest technique for the purification of organic, compounds is, (a) chromatography, (b) fractional distillation, (c) crystallization, (d) vacuum distillation, 125. Which of the following is used as an adsorbent in, adsorption chromatography ?, (a) Silica gel, (b) Alumina, (c) Zeolite, (d) Both (a) and (b), 126. Which of the following acts as the stationary phase in, paper chromatography ?, (a) Water, (b) Alumina, (c) Silica gel, (d) None of these, 127. The most satisfactory method to separate mixture of sugars, is, (a) fractional crystallisation, (b) sublimation, (c) chromatography, (d) benedict reagent, 128. Chromatography is a valuable method for the separation,, isolation, purification and identification of the constituents, of a mixture and it is based on general principle of, (a) phase rule, (b) phase distribution, (c) interphase separation (d) phase operation, 129. In paper chromatography, (a) moving phase is liquid and stationary phase in solid, (b) moving phase is liquid and stationary phase is liquid, (c) moving phase is solid and stationary phase is solid, (d) moving phase is solid and stationary phase is liquid, 130. Which of the following is used for detection of carbon, and hydrogen ?, (a) Ca(OH)2, (b) CuO, (c) CaCl2, (d) KOH, 131. In sodium fusion test of organic compounds, the nitrogen, of the organic compound is converted into, (a) sodamide, (b) sodium cyanide, (c) sodium nitrite, (d) sodium nitrate, 132. Which of the following compounds does not show, Lassaigne’s test for nitrogen ?, (a) Urea, (b) Hydrazine, (c) Phenylhydrazine, (d) Azobenzene, 133. The compound formed in the positive test for nitrogen with, the Lassaigne solution of an organic compound is, (a) Fe4[Fe(CN)6]3, (b) Na3[Fe(CN)6], (c) Fe(CN)3, (d) Na4[Fe(CN)5NOS], , 134. In quantitative analysis of carbon and hydrogen, the, mass of water produced is determined by passing the, mixture through a weighed U – tube containing ___X___, and carbon dioxide is absorbed in concentrated solution, of ___Y___, (a) X = CaCl2 , Y = NaOH, (b) X = Ca(OH)2 , Y = CuSO4, (c) X = CuSO4 , Y = Ca(OH)2, (d) X = CaCl2 , Y = KOH, 135. Kjeldahl method is not applicable to which of the, following ?, (a) Nitro compounds, (b) Azo compounds, (c) Pyridine, (d) All of these., 136. Nitrogen in an organic compound can be estimated by, (a) Kjeldahl’s method only(b) Duma’s method only, (c) Both (a) and (b), (d) Neither (a) nor (b), 137. Duma's method involves the determination of nitrogen, content in the organic compound in form of, (a) NH3, (b) N2, (c) NaCN, (d) (NH4)2SO4, 138. In Kjeldahl’s method nitrogen present is estimated as, (a) N2, (b) NH3, (c) NO2, (d) None of these, 139. In kjeldahl's method of estimation of nitrogen, K2SO4 acts, as, (a) oxidising agent, (b) catalytic agent, (c) hydrolysing agent, (d) boiling point elevator, 140. 0.5g of an organic compound containing nitrogen on, Kjeldahlising required 29 mL of N/5 H2SO4 for complete, neutralization of ammonia. The percentage of nitrogen in, the compound is, (a) 34.3, (b) 16.2, (c) 21.6, (d) 14.8, 141. The percentage of sulphur in an organic compound whose, 0.32 g produces 0.233 g of BaSO4 [At. wt. Ba = 137, S = 32], is, (a) 1.0, (b) 10.0, (c) 23.5, (d) 32.1, 142. An organic compound contains C = 40%, H = 13.33% and, N = 46.67%. Its empirical formula would be, (a) CHN, (b) C2H2N, (c) CH4N, (d) C3H7N, 143. 2.79 g of an organic compound when heated in Carius tube, with conc. HNO 3 and H 3PO 4 formed converted into, MgNH 4 .PO 4 ppt. The ppt. on heating gave 1.332 g of, Mg 2 P2O 7 . The percentage of P in the compound is, (a) 23.33%, (b) 13.33%, (c) 33.33%, (d) 26.66%, 144. A compound contains 38.8% C, 16% H and 45.2% N. The, formula of compound would be :, (a), , CH 3 NH 2, , (b), , CH 3CN, , (c), , C 2 H 5CN, , (d) C H2(NH2)2
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ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES & TECHNIQUES, , 145. In estimation of percentage of oxygen. The mixture of, gaseous products containing oxygen is passed over red, hot coke. All oxygen is converted to A . This mixture, is passed through, B, when, A, is converted to, _ C, ., What is A, B and C in above statement., (a) A = CO2 , B = KOH , C = pure O2, (b) A = CO , B = I2O5 , C = CO2, (c) A = CO , B = I2 , C = CO2, (d) A = CO2 , B = Ca(OH)2 , C = CaCO3, , STATEMENT TYPE QUESTIONS, 146. Which of the following statement(s) is/are correct ?, (i) A carbon atom having an sp hybrid orbital is less, electronegative than carbon atoms possessing sp2, or sp3 hybridised orbitals., (ii) -bonds provide the most reactive centres in the, molecules containing multiple bonds, (iii) The number of, and, bonds in compound, CH2= C= CHCH3 are 7 and 2 respectively., (a) (i) and (iii), (b) (ii) and (iii), (c) (ii) only, (d) (i) only, 147. Which of the following sequence of T and F is correct., Here T stands for true statement and F stands for false, statement, (i) In heterolytic cleavage the bond breaks in such a, fashion that the shared pair of electrons remain with, one of the fragment., (ii) In homolytic cleavage, one of the electrons of the, shared pair in a covalent bond goes with each of the, bonded atoms., (iii) R – X, , heat or, light, , R +X, , This equation represents heterolytic cleavage, (a) TTF, (b) FFT, (c) FFF, (d) TTT, 148. Which of the following is/are correct for inductive effect ?, (i) In inductive effect polarisation of sigma bond is, caused by the adjacent bond., (ii) Halogens, –NO2 , –CN, and –CH3 are electron withdrawing, groups., (iii) –CH2CH3and –OC6H5 are electron donating groups., (a) (i) only, (b) (ii) only, (c) (i) and (iii), (d) (i), (ii) and (iii), 149. Which of the following sequence of T and F is correct for, given statements. Here T stands for correct and F stands, for false statement, (i) The more the number of contributing structures, the, more is the resonance energy., (ii) The resonance structures have different positions of, nuclei but same number of unpaired electrons, (iii) The energy of actual structure of the molecule (the, resonance hybrid) is lower than that of any of the, canonical structures., (a) TTT, (b) TFT, (c) FFF, (d) TFF, , 189, , 150. Which of the following statements are correct for, fractional distillation ?, (i) Fractional distillation method is used if the two, liquids have sufficiently large difference in their, boiling points., (ii) A fractionating column provides many surfaces for, heat exchange between the ascending vapours and, the descending condensed liquid., (iii) Each successive condensation and vaporisation unit, in the fractionating column is called a theoretical, plate., (iv) Fractional distillation method is used to separate, different fractions of crude oil in petroleum industry., (a) (i), (ii) and (iv), (b) (ii), (iii) and (iv), (c) (i), (ii) and (iii), (d) (i), (ii), (iii) and (iv), 151. Which of the following sequence of T and F is currect for, given statements. Here ‘T’ stands for True and ‘F’ stands, for False statement., (i) The relative adsorption of each component of, mixture is expressed in terms of its retardation factor, (RF), (ii) Retardation factor is given as :, RF =, , Distance moved by the solvent from base line, Distance moved by the substance from baseline, , (iii) In TLC the spots of colourless compounds can be, detected by ultraviolet light., (iv) Spots of amino acids may be detected by iodine., (a) TTTF, (b) TFFF, (c) TTTT, (d) TFTF, 152. In Kjeldahl’s method for the estimation of N 2 , potassium, sulphate and copper sulphate are used. On the basis of, their functions which of the following statement(s) is/are, correct?, (i) Potassium sulphate raises the bpt. and ensures, complete reaction., (ii) Copper sulphate acts as catalyst., (iii) Potassium sulphate acts as catalyst and copper, sulphate raises the bpt., (a) Only (iii) is correct, (b) (i) and (ii) are correct, (c) Only (ii) is correct, (d) None is correct, 153. In the estimation of carbon and hydrogen by combustion, method which of the following is/are correct ?, (i) A spiral of copper is introduced at the right extreme of, combustion tube if the organic compound contains, nitrogen., (ii) A spiral of silver is introduced if the organic compound, contains halogens., (iii) The copper oxide in the combustion tube is replaced, by lead chromate if the organic compound contains, sulphur., (a) (i) and (ii) are correct (b) (i) and (iii) are correct, (c) (ii) and (iii) are correct (d) All are correct
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EBD_7207, ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES & TECHNIQUES, , 190, , MATCHING TYPE QUESTIONS, 154. Match the columns, Column-I, , Column-II, , (A) Non – benzenoid compound, , (p), O, , (B) Alicyclic compound, , (q), , (C) Benzenoid compound, , (r), S, , (D) Heterocyclic aromatic compound (s), O, (a) A – (r), B – (p), C – (s), D – (q), (b) A – (s), B – (p), C – (q), D – (r), (c) A – (p), B – (r), C – (s), D – (q), (d) A – (r), B – (p), C – (q), D – (s), 155. Match Column-I (organic compound) with Column-II, (common name of the compound) and choose the correct, option., Column-I, Column-II, (Organic compound), (Common name of compound), (A) C6H5OCH3, (p) Neopentane, (B) H3CCH2CH2 OH, (q) Anisole, (C) (H3C)4C, (r) Acetophenone, (D) C6H5COCH3, (s) n – propyl alcohol., (a) A – (r), B – (s), C – (p), D – (q), (b) A – (r), B – (p), C – (s), D – (q), (c) A – (q), B – (s), C – (p), D – (r), (d) A – (q), B – (s), C – (r), D – (p), 156. Match the columns, Column-I, Column-II, (A) Aldehyde, (p) Chloro, (B) Ketone, (q) ol, (C) Alcohol, (r) one, (D) Halogen, (s) al, (a) A – (s), B – (r), C – (q), D – (p), (b) A – (s), B – (q), C – (p), D – (r), (c) A – (p), B – (q), C – (r), D – (s), (d) A – (r), B – (s), C – (q), D – (p), 157. Identify (i), (ii), (iii) and (iv) in the structure of given, organic compound. On the basis of your identification, match the columns., (iii), , OH, , (i), Column-I, A (i), B (ii), C (iii), D (iv), , Br, (ii), (p), (q), (r), (s), , CH 3, , Column-II, Functional group, Branch chain, Parent chain, Homologues unit, , (C) CH3 CH CH 2, |, CH3, , (r), , Neopentyl, , CH3, |, (s) tert – Butyl, (D) CH3 C, |, CH3, (a) A – (r), B – (q), C – (p), D –(s), (b) A – (s), B – (p), C – (r), D –(q), (c) A – (s), B – (p), C – (q), D –(r), (d) A – (s), B – (q), C – (p), D –(r), 159. Column-II give formula for compounds given in Column-I,, match them correctly., Column-I, Column-II, (A) Propane, (p) C2H5OH, (B) ethyl alcohol, (q) C3H8, (C) carboxylic acid, (r) CH3COOH, (D) ethyl ethanoate, (s) CH3COOC2H5, (a) A – (q), B – (p), C – (r), D – (s), (b) A – (p), B – (q), C – (s), D – (r), (c) A – (q), B – (s), C – (p), D – (r), (d) A – (q), B – (p), C – (s), D – (r), 160. Match the columns, Column-I, Column-II, (Organic compounds), (Type of isomerism), (A) CH3CH2CH2CH2CH3, (p) Functional group, isomerism, CH3, |, & CH3 CH CH 2CH3, (B) CH3CH2CH2OH, (q) Chain isomerism, OH, |, & CH 3 CH CH 3, , O, |, C CH 3, (D), (a), (b), (c), (d), , O, ||, C CH3, , H, , & CH3 (r), CH2Metamerism, C O, H, |, & CH 3 CH 2 C O, CH3OC3H7 & C2H5OC2H5 (s) Position isomerism, A – (s), B – (q), C – (r), D – (p), A – (s), B – (q), C – (p), D – (r), A – (q), B – (s), C – (r), D – (p), A – (q), B – (s), C – (p), D – (r), , (C) CH3, , CH3 – CH – CH2 – CH – CH – CH3 (iv), H3C, , (a) A – (q), B – (p), C – (s), D – (r), (b) A – (p), B – (s), C – (q), D – (r), (c) A – (q), B – (p), C – (p), D – (r), (d) A – (q), B – (p), C – (q), D – (r), 158. Match the columns., Column-I, Column--II, CH3, |, (A) CH3 — C— CH 2, (p) Isobutyl, |, CH3, (q) sec – Butyl, (B) CH3 CH 2 CH, |, CH3
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ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES & TECHNIQUES, , 161. Match the columns, Column-I, Column-II, (A) CH3COOH &, (p) Functional isomers, HCOOCH3, (B) 1 butene &, (q) Metamers, 2-butene, (C) diethyl ether &, (r) Position isomers, methyl propyl ether, (D) dimethyl ether, (s) Chain isomers and ethanol, and ethanol, (a) A – (p), B – (r), C – (q), D – (s), (b) A – (q), B – (r), C – (s), D – (p), (c) A – (q), B – (s), C – (p), D – (r), (d) A – (q), B – (p), C – (s), D – (r), 162. Match the columns, Column-I, Column-II, (A) Free radical, (p) Trigonal planar, (B) Carbocation, (q Pyramidal, (C) Carbanion, (r) Linear, (a) A – (p), B – (q), C – (r), (b) A – (p), B – (p), C – (q), (c) A – (r), B – (p), C – (q), (d) A – (p), B – (p), C – (r), 163. Match the columns, Column - I, Column - II, (A) Separation of sublimable (p) Steam distillation, compounds from non, sublimable, (B) Method based on the, (q) Sublimation, difference in the solubilities, of the compound and the, impurities in a suitable, solvent, (C) Separation of liquids, (r) Distillation, having sufficient difference, in their boiling points., (D) Separation of substances (s) Crystallisation, which are steam volatile, and are immiscible with water., (a) A – (q), B – (s), C – (r), D – (p), (b) A – (q), B – (r), C – (p), D – (s), (c) A – (s), B – (q), C – (r), D – (p), (d) A – (q), B – (s), C – (p), D – (r), 164. Match the columns, Column - I, Column - II, (Elements), (Colour of precipitate formed, in Lassaigne’s test), (A) Nitrogen, (p) Yellow, (B) Sulphur, (q) Prussian blue, (C) Chlorine, (r) Violet, (D) Phosphorus, (s) White, (a) A – (q), B – (r), C – (p), D – (s), (b) A – (r), B – (q), C – (p), D – (s), (c) A – (q), B – (r), C – (s), D – (p), (d) A – (r), B – (q), C – (s), D – (p), , 191, , 165. Match the columns, Column - I, , Column - II, , (A) Duma’s method, , (p), , 80 m1 100, 188 m, , (B) Kjeldahl’s method, , (q), , 31 m1 100, %, 1877 m, , (C) Carius method, , (r), , 1.4 M 2 v, , v1, 2, , m, , for bromine, , %, , (D) Percentage of phosphorus, (s), (a), (b), (c), (d), , A, A, A, A, , –, –, –, –, , 28 V 100, %, 22400 m, , (s), B – (r), C – (p), D – (q), (r), B – (s), C – (q), D – (p), (s), B – (p), C – (q), D – (r), (p), B – (r), C – (q), D – (s), , ASSERTION-REASON TYPE QUESTIONS, Directions : Each of these questions contain two statements,, Assertion and Reason. Each of these questions also has four, alternative choices, only one of which is the correct answer. You, have to select one of the codes (a), (b), (c) and (d) given below., (a) Assertion is correct, reason is correct; reason is a correct, explanation for assertion., (b) Assertion is correct, reason is correct; reason is not a, correct explanation for assertion, (c) Assertion is correct, reason is incorrect, (d) Assertion is incorrect, reason is correct., 166. Assertion : A primary suffix indicates the type of linkage in, the carbon atom., Reason : CN is a Primary suffix, 167. Assertion : The general formula for a dihydric alcohol is, CnH2n(OH)2, Reason : Ethylene glycol is a dihydric alcohol., 168. Assertion : IUPAC name of the following organic, compound is 3, 4, 7 – trimethyloctane, CH3 CH CH 2 CH 2 CH CH CH 2 CH3, |, |, |, CH3, CH3 CH 3, Reason : The numbering is done in such a way that the, branched carbon atoms get the lowest possible numbers., 169. Assertion : Chain isomerism is observed in compounds, containining four or more than four carbon atoms, Reason : Only alkanes show chain isomerism, 170. Assertion : But-1-ene and 2-methylprop-1-ene are position, isomers., Reason : Position isomers have same molecular formula but, differ in position of functional group or C = C.
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EBD_7207, ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES & TECHNIQUES, , 192, , 171. Assertion : Benzene exhibit two different bond lengths,, due to C – C single and C = C double bonds., Reason : Actual structure of benzene is a hybrid of, following two structures., , OH, 179 IUPAC name of, , is :, , OH, (a) But – 2 ene – 2, 3– diol, (b) Pent – 2 –ene – 2, 3 – diol, 172. Assertion : Aniline is better nucleophile than anilium ion., Reason : Anilium ion have +ve charge., 173. Assertion : Different number of electron pairs are present, in resonance structures., , (c) 2 – methylbut – 2 – ene – 2, 3 – diol, (d) Hex – 2 – ene – 2, 3 – diol, 180. The state of hybridization of C2, C3, C5 and C6 of the, hydrocarbon,, , Reason : Resonance structures differ in the location of, electrons around the constituent atoms., CH3, , 174. Assertion : Energy of resonance hybrid is equal to the, average of energies of all canonical forms., Reason : Resonance hybrid cannot be presented by a single, structure., 175. Assertion : Simple distillation can help in separating a, mixture of propan-1-ol (boiling point 97°C) and propanone, (boiling point 56°C)., Reason : Liquids with a difference of more thatn 20°C in, their boiling points can be separated by simple distillation., 176. Assertion : Components of a mixture of red and blue inks, can be separated by distributing the components between, stationary and mobile phases in paper chromatography., Reason : The coloured components of inks migrate at, different rates because paper selectively retains different, components according to the difference in their partition, between the two phases., 177. Assertion : Sulphur present in an organic compound can, be estimated quantitatively by Carius method., Reason : Sulphur is separated easily from other atoms in, the molecule and gets precipitated as light yellow solid., , CRITICAL THINKING TYPE QUESTIONS, 178. The IUPAC name of the following compounds is, N, |, H, , (a) N – phenyl ethanamide, (b) N – phenyl ethanone, , O, , 7, , CH3, |, C, CH = CH, 6|, 5, 4, CH3, , CH3, |, CH, , 2, , 1, , is in the following sequence, (a) sp3, sp2, sp2 and sp, , (b) sp, sp2, sp2 and sp3, , (c) sp, sp2, sp3 and sp2, , (d) sp, sp3, sp2 and sp3, , 181. Which of the following numberings is correct ?, FF 1, , 2, , A., 4, , 5, , F 5, , C., , 1, , FF 3, 3, , 4, 2, , (a) A, , 3, , Br, , 2, , B., 4, , 5, , FF 5, , 1, , Br, , D., , (b) B, , (c) C, , 1, , Br, , 2, , 4, , 3, , Br, , (d) D, , 182. The ratio of - to - bonds in benzene is, (a) 1 : 4, , (b) 1 : 2, , (c) 3 : 1, , (d) 1 : 6, , 183. In which of the compounds given below there is more than, one kind of hybridization (sp, sp2, sp3) for carbon ?, (i), , CH3CH2CH2CH3, , (ii) CH3CH = CHCH3, , (iii) CH2=CH–CH=CH2, , (iv) H –C C – H, , (a) (ii), , (b) (iii) and (iv), , (c) (i) and (iv), , (d) (ii) and (iii), , 184. Which of the following represents the given mode of, hybridisation sp2–sp2 – sp – sp from left to right?, (a) H2C = CH – C N, , (b) CH C – C CH, CH2, , (c) N – phenyl methanamide, (d) None of these, , C CH, , 3, , (c) H2C = C = C = CH2, , (d) CH2
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EBD_7207, ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES & TECHNIQUES, , 194, , 197. Which of the following is least reactive in a nucleophilic, substitution reaction., (a), , (CH 3 )3 C Cl, , (b), , CH 2, , (c), , CH 3CH 2Cl, , (d), , CH 2, , CHCl, , CHCH 2 Cl, , 198. Which of the following does not represent formation of, reactive intermediate correctly ?, (i), , CH3 CN, , C H3 CN, , (ii), , CH 3 C u, , CH3 Cu, , (iii) CH 3 B r, , CH3 Br, , (iv) CH 3 Cl, , CH3 Cl, , (a) (ii) only, (b) (ii) and (iii), (c) (ii) and (iv), (d) (iii) and (iv), 199. In Lassaigne’s test, the organic compound is fused with a, piece of sodium metal in order to, (a) increase the ionisation of the compound, (b) decrease the melting point of the compound, (c) increase the reactivity of the compound, (d) convert the covalent compound into a mixture of ionic, compounds, , 200. The most suitable method for separtion of a 1 : 1 mixture of ortho, and para nitrophenols is, (a) Sublimation, (b) Chromatography, (c) Crystallization, (d) Steam distillation, 201. The Lassaigne’s extract is boiled with dil. HNO3 before, testing for halogens because, (a) silver halides are soluble in HNO3, (b) Na2S and NaCN are decomposed by HNO3, (c) Ag2S is soluble in HNO3, (d) AgCN is soluble is HNO3, 202. The molecular mass of an organic compound which contains, only one nitrogen atom can be, (a) 152, (b) 146, (c) 76, (d) 73, 203. 0.25 g of an organic compound on Kjeldahl's analysis gave, enough ammonia to just neutralize 10cm3 of 0.5 M H2SO4., The percentage of nitrogen in the compound is, (a) 28, (b) 56, (c) 14, (d) 112, 204. During hearing of a court case, the judge suspected that, some changes in the documents had been carried out. He, asked the forensic department to check the ink used at two, different places. According to you which technique can, give the best results?, (a) Column chromatography, (b) Solvent extraction, (c) Distillation, (d) Thin layer chromatography
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ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES & TECHNIQUES, , FACT / DEFINITION TYPE QUESTIONS, 1., 2., , 3., , (a) Berzilius, a Swedish chemist proposed the concept, of ‘vital force’., (d) Urea was first discovered in human urine by H.M., Rouelle in 1773. It was synthesised in 1828 by Friedrich, Wohler and was the first organic compound to be, synthesised from inorganic starting materials. It was, found when Wohler attempted to synthesize ammonium cyanate, to continue a study of cyanates which, he had be carrying out for several years. On treating, silver cyanate with ammonium chloride solution he, obtained a white crystalline material which proved identical to urea obtained from urine., (b) F. Wohler synthesised urea from an inorganic, compound ammonium cyanate, Heat, , ® NH2CONH2, NH4CNO ¾¾¾, , Ammonium, cyanate, , 4., , 5., , 6., , 7., , Urea, , (c) According to vital force theory, organic compounds, could only be produced by living matter by a vital, force. It was in 1828, Friedrich Wholer heated NH4CNO, (derived from inorganic substance) and obtained urea, (an organic compound)., (c) Wholer synthesized urea from ammonium cyanate in, 1828. Kekule proposed catenation and structure of, benzene. Liebig is a history maker in sports science, (energy metabolism)., (d) Antoine-Laurent de Lavoisier (August 26, 1743 – May, 8, 1794) is known as the “father of modern chemistry.”, He was a French nobleman prominent in the histories, of chemistry, finance, biology, and economics. He, stated the first version of the Law of conservation of, mass, co-discovered, recognized and named oxygen, (1778) as well as hydrogen, disproved the phlogiston, theory, introduced the Metric system, invented the first, periodic table including 33 elements, and helped to, reform chemical nomenclature., (c) Hybridisation on the particular carbon can be, established by number of and bonds attached to it., Bond, Bond, Hybridisation, 4, –, sp3, 3, 1, sp2, 2, 2, sp, 1, , C H2, , 2, , 3, , CH CH, , 4, , C H2, , 3, 3, 3, 3, 1, 1, 1, 1, sp 2, sp 2 sp 2 sp 2, Both carbon atoms forming C—C single bond (C2 and, C3) are sp2 hybridised, , 8., , 9., , 195, , (b) C – 1 is sp hybridized C C, C – 3 is sp3 hybridized (C– C), C – 5 is sp2 hybridized (C = C), Thus the correct sequence is sp, sp3, sp2., (c), 10. (c), , H, , 11., , (b), , H, H H, |, | |, C C C C C H, | |, | | |, ;, H H H H H, 2 Pentene, , No. of bonds = 14, No. of bonds = 1, 12. (c) (ii) and (iii) do not represent 2-bromopentane, 13. (c) Correct expanded form of given structure is shown, in option (c)., 14. (c), 15. (b) The successive members of a homologous series, differ by a – CH2 unit., 16. (a) Correct molecular formula of icosane is C20H42, Correct molecular formula of heptane is C7H16, 17. (c) (iii) is the only correct method of selecting parent, chain., 18. (b) Correct order of decreasing priority is, – COOH, – SO3H, – COOR, – OH., 19. (d), 20. (c), 21. (d), 22. (c), 23. (c), 24. (d) 2, 2,3-trimethyl pentane, 1, CH3, 1, 2 |, H3C H 2 C CH, 3, , 1, , 25. (d), , 1, , CH 3, , 4, , 1, CH3, | 1, C CH 3, |4, CH3, 1, , CH 3, | 2, 1, C CH 2 CH 3, |, CH 3, , 1, , Thus number of secondary hydrogens is two., , 26. (d), , CH 3, , CH3, |, C, |, CH3, (a ), , CH3, |, C CH 2 CH 3, |, CH3, , CH3, |, , CH 3 – C – CH 2 CH 3, |, , CH3, ( b)
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EBD_7207, ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES & TECHNIQUES, , 196, , CH3 CH3, |, |, C HCH 2 CH 3, CH 3 C, |, CH3, (c ), , 43., , (c), , 44., , (a), , 45., , (d), , Cl, |, 3, CH3 — C — CH3, |, CH3, 1, , 2, , 2-chloro-2-methyl propane, , CH 3, |, CH 3 C HCH 2 CH 2CH 3, , (CH 3 ) 2 CHCH 2CH 2, 3-methylbutyl group, , 46., , (a), , (d ), , 27., , 28., , 29., , (d) In isopentane, (CH3)2CH CH2 CH3, every carbon is, having hydrogen atom(s)., (a), , (c), , 47., , (d), , H H H H H, 1° | 2° | 2° | 3° |3° |2° 1°, H3C – C – C – C – C – C – CH3, | | 1°| 1°|, |, H H CH3 CH3 H, 3, 4-dimethylheptane, 48., , 4, , (a), , CH3, , 3, , 2, , CH, , CO, , 6, , |, , 9, , 49., , (a), , It has 1 quaternary and 4 primary carbons., , 34., , (a), , 35., , (c) The correct name is 3 - methylbutan - 2 - ol, , 32. (c), , 1, O CH 2, , 33. (d), , 2, 3, CH 2 CH 2, , O CH 2 CH 3, , 3-ethoxy-1-methoxypropane, , 1, , CH3, , CH3, |, 2, 3, H3C C CH3, |, CH3, Neopentane, or 2, 2- Dimethylpropane, , 50., 51., , Cl, 1, 6, , 2, , 5, 4, , 52., , (b), (b) The compound is a derivative of benzoic acid. The, positions of substituents attached to benzene nucleus, are represented by number of C-atoms and not by, ortho, meta and para., (a) C3H6 has 2 structural isomers., CH 3 CH, , 3 Br, , propene, , CH 2 and H 2 C, , 3 Methyl pentan, , 41., 42., , Cl, , CH3, |, , CH3, |, , CH 2 Cl, |, , Cl, Cl, o-Chlorotoluene m-Chlorotoluene p-Chlorotoluene benzyl chloride, |, , 39., 40., , C7 H7 Cl has 4 isomers, , CH3, |, , 3 ol, , (a) The IUPAC name of the given compound is, 5-chlorohex-2-ene., (b) The compound is a derivative of butane., (b) The compound contains longest chain of 6C atoms, and amino group. Hence it is an alkanamine., (d) The compound is an ester. Its IUPAC name is derived, from alkyl alkanoate., (d) The compound is an aldehyde containing longest, chain of 6 C-atoms and side chains., , (c), , |, , 38., , cyclopropane, , 53., , |, , OH, 2, 3|, CH 2 C CH 3, 5 |4, CH3 CH 2, , CH 2, , CH 2, , 3-bromo-1chlorocyclohexene, , (c), , 10, , 1, , (b), , 37., , CH3, , 3-methyl-2-butanone, , 30., , 1, CH3, , 8, , 7, , CH CH, , methyl group, , CH3, , CH3, , (a), , 5, , CH CH, , Ketonic group, , H3C — C— CH3, , 36., , 4, , CH2 — CH3, , CH 3, , CH 3, , 3, , CH3CH CH 2CH, , 2,8-Dimethyl-4,6-decadiene, , There are four 1° C-atoms, three 2° C-atoms and two, 3° C-atoms, The structure of neopentane is, , 31. (d), , CH3, , 2|, , 1, , 54., 55., , 56., , (b) Alcohols and ethers are functional isomers., (b) Structures (a), (c) and (d) have the same molecular, formula (C6H12O) while (b) has C6H10O as molecular, formula, (d)
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ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES & TECHNIQUES, , 57., 58., , (b), , CH3CH2CH2C CH,(CH3)2CHC CH, CH3CH2C CCH3, I, II, III, (b) CH3CH2CH=CH2 CH3CH=CHCH3 (CH3)2C=CH2, 1-butene (i), , 2- butene (ii), (iii), (cis,- trans), , 197, +, , CH2, , cyclobutane (v), , 60., 65., 66., , (b), , 67., 68., , (b), (b) The order of stability of carbocations is :, , The two isomers differ in the position of the double bond so, they are called position isomers., , (CH3 )3 C (CH 3 )2 CH CH 3, , 71., , (c), (d) Greater the number of alkyl groups attached to a, positively charged C atom, greater is the, hyperconjugation (no bond resonance) and stable is, the cation., Thus order of decreasing stability of carbocation is,, tert – Alkyl > Sec-Alky > Pri-Alkyl > Methyl., (a) Carbonium ions are electron deficient species. More, the number of alkyl groups attached to it, more will be, stability due to + I effect., , whereas in alkyl carbocations dispersal of positive, charge on different hydrogen atoms is due to, hyper conjugation hence the correct order of stability, will be, , CH2, > CH 2, Benzyl, , C, |, , CH3, , >, , 3°carbonium ion, (+ve charge dispersed to maximum extent), (9 hyperconjugative H's), , CH 3, , C– H, |, , 72., , C H2, 1° carbonium ion, (+ve charge least dispersed), (2 hyper. H's), , Alternatively, above order of stability order can be, explained in terms of hyperconjugation., (d) Higher stability of allyl and aryl substituted methyl, carbocation is due to dispersal of positive charge due, to resonance, , CH 2, , CH C H 2, , CH 2, , CH, , Resonating structures of allyl carbocation, , Propyl, , Cl, , –, , C, , –, –, Cl > C6H5CH2 > (CH3)2 CH, Cl, , –, , > (CH3)3C, , +I effect of CH 3 group, intensifies the –ve charge, , –M effect, delocalises, –ve charge, , 80. (c) In homolytic fission each of the atoms acquires one, of the bonding electrons producing free radicals, (species having one unpaired electron)., , > CH3CH 2, , CH 2CH3, 2° carbonium ion, (5 hyper. H's), , CH3 CH 2 CH 2, , stability; in C6 H5CHC6 H5 , +ve charge can delocalise, over two benzene rings., 76. (d) The organic reaction which proceed through, heterolytic bond cleavage are called ionic or, heteropolar or just polar reactions., 77. (b) In carbocations, carbon bearing positive charge is, always sp2-hybridised, 78. (b) Methyl carbanion is sp3 hybridised, with three bond, pairs and one lone pair same is the case with NH3., , –ve charge, highly dispersed, due to – I effect, , CH3, , CH C H 2, , Allyl, , 73. (b) Structure (b) is a 3º carbocation, while (a) is 2º and (c), and (d) are 1º carbocations; thus (b) is the most stable., 74. (a), 75. (a) Higher the possibility of delocalisation, greater is its, , 79. (b), CH3, , CH2, , Resonating structures of benzyl carbocation, , methylcyclopropa ne (vi), , (b) 2, 2, 4, 4 - Tetramethylhexane has 10 carbon atoms,, only 4-isopropylheptane has also 10 carbon atoms so, these two are isomers., (a), 61. (c), 62. (a), 63. (b), 64. (d), (c), , 69., 70., , CH2, , 2-methylpropene, (iv), , CH3, , 59., , CH2, , CH 2, , A + B, A, B, 81. (b) Homolytic fission of the C – C bond gives free radicals, in which carbon is sp2- hybridised., 82. (b) The carbon atom of alkyl free radicals which is bonded, to only three atoms or groups of atoms is sp 2 hybridized. Thus free radicals have a planar structure, with odd electrons situated in the unused p-orbital at, right angles to the plane of hybrid orbitals., R, R, , 120°, R
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EBD_7207, ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES & TECHNIQUES, , 198, , 83., 84., , (c), (a) On exposure to UV light, Cl2 molecule undergoes, homolytic fission, to form chlorine free radicals., ., U.V., Cl Cl, 2Cl, , 85., , (b) The order of stability of free radicals, , (Chlorine free radicals), •, , (C 6 H 5 ) 3 C, , 86., , 87., 88., , •, , (C 6 H 5 ) 2 CH, , •, , (CH 3 ) 3 C, , •, , (CH 3 ) 2 C H, , The stabilisation of first two is due to resonance and, last two is due to inductive effect., (d) Free radicals are stabilized by hyperconjugation, thus, 3° free radicals having maximum number of, hyperconjugative structures are the most stable, and, primary free radical the least., , ., , C6 H5CHCH3 is a 2º benzylic free radical, hence, stabilized most due to resonance., (b) Dichlorocarbene, : CCl2 (a carbene) is the electrophile, formed as an intermediate in Reimer-Tiemann reaction., (b), , 89., , (d), , 90., , (c) The strength of nucleophile depends upon the nature, of alkyl group R on which nucleophile has to attack, and also on the nature of solvent. The order of strength, of nucleophiles follows the order :, CN– > I– > C6H5O– > OH– > Br– > Cl–, (c), 92. (d), (a) Electrophile is positivly charged or electron deficient, species. Lewis acids are electron acceptors that is, electron deficient species., (b) Electrophiles are electron deficient or positively, charged species., (d) BF3 and R3C – X are electrophile while (CH3)3N and, C2H5O– are nucleophile, (b) –CH3 group has +I effect, as number of –CH3 group, increases, the inductive effect increases., (d) Due to – I effect of the – CHO group, oxygen acquires- charge and the terminal carbon acquires + charge., , Order of stability of free radicals is, 3° > 2° > 1° > C H3, , 91., 93., , 94., 95., 96., 97., , 103. (c) Resonance effect is the polarity produced in the, molecule by the interactions of two – bonds or, between a – bond and a lone pair of electrons, present on an adjacent atom., 104. (b) Electromeric effect is purely a temporary effect and is, brought into play only at the requirement of attacking, reagent, it vanishes out as soon as the attacking, reagent is removed from reaction mixture., 105. (b), 106. (b) Alkyl groups with at least one hydrogen atom on the, -carbon atom, attached to an unsaturated carbon, atom, are able to release electrons in the following, way., , H, (c) All resonating structures should have same number, of electron pairs., 99. (a) The two structures involve only movement of, electrons and not of atoms or groups, hence these are, resonating structures., 100. (b) Only structure (b) has a conjugated system, which is, necessary for resonance., 101. (c), , 98., , effect., , H H, | |, H C C, | |, H H, , C = O shows – R, , H, |, H C, |, H, , H, |, C, |, H, , H, |, HC, |, H, , H, |, C, |, H, , H, , H, |, C, , H, , H, |, C, |, H, , In general greater the number of alkyl groups attached, to a positively charged carbon atom, the greater is the, hyperconjugation interaction and stabilisation of the, cation. Thus, we have the following relative stability, of carbocation., , CH3, , CH2 = CH – C = O, , 102. (b) – OH shows + R effect while, , Note that the delocalisation involves and bond, orbitals (or p orbitals in case of free radicals) ; thus it, is also known as – conjugation. This type of, electron release due to the presence of the system, H—C—C = C is known as hyperconjugation, 107. (b) The stability of carbocation on the basis of hyperconjugation can be explained as hyperconjugation, stabilises the carbocation because electron density, from the adjacent -bond helps in dispersing the, positive charge., , CH3, |, C, |, CH3, , (CH3 ) 2 CH, , CH3 CH 2, , CH3, , Hence, stability of carbocation is directly proportional, to number of alkyl group directly attached to, carbocations., 108. (b) Stability order of different alkyl carbocations on the basis, of hyperconjugation is :, 3° > 2° > 1° > methyl, In t-butyl cation, the C-atom bearing the positive charge, is attached to three methyl groups therefore it possess, nine -hydrogens. It will give maximum nine, hyperconjugative structures leading to maximum, stability.
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ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES & TECHNIQUES, , 109. (b) In elimination reactions one or two molecules are lost, from the substrate to form a multiple bond. Dehydration, of ethanol is an example of elimination reaction., C2 H5 OH, , Conc, H 2SO4, , CH 2, , CH 2, , H2O ., , 110. (a), 111. (d), 112. (a), 113. (b) Coloured impurities are removed by adsorbing over, activated charcoal., 114. (a) This method is applied for the purification of, substances which (i) are insoluble in water, (ii) are, volatile in steam, (iii) are associated with non steam, volatile impurities, (iv) have high molecular weights, and (v) possess a fairly high vapour pressure at about, the boiling point of water e.g. Aniline., 115. (a) Aniline is purified by steam distillation., A mixture of water and aniline boils at 371 K and 760, mm pressure which is less than boiling point of water., 116. (b) Among the given compounds naphthelene is volatile, but benzoic acid is non-volatile (it forms a dimer). So,, the best method for their separation is sublimation,, which is applicable to compounds which can be, converted directly into the vapour phase from its solid, state on heating and back to the solid state on cooling., Hence it is the most appropriate method., 117. (d), 118. (b) If there is a small difference (10 or less) in the boiling, points of liquids fractional distillation is used e.g. acetone b.p. 333 K and methanol b.p. 338 K., 119. (a) Fractional distillation is used for the distillation of petroleum. This method is used for separating a mixture, of two or more miscible, volatile liquids having close, (less than 40 degrees) boiling points. (For example, a, mixture of acetone, b.p., 56°C and methanol, b.p. 65°C), 120. (c) If any liquid decomposes at its boiling point, it can be, purified by vacuum distillation., 121. (c) Glycerol decomposes at its boiling point, hence it, should be purified by distillation under reduced, pressure., 122. (c) Vaccum distillation means distillation under reduced, pressure., 123. (c), 124. (a) The latest technique for the purification of organic, compounds is chromatography. These are of various, types like column, paper and gas-chromatography., 125. (d) Both silica gel and alumina are used as adsorbents, in adsorption chromatography., 126. (a) Chromatography paper contains water trapped in it,, which acts as the stationary phase., 127. (c) The mixture of sugars is a homogenous one., Homogeneous mixtures of a solvent and one or more, solutes (dissolved substances) are often separated by, chromatography. Chromatography works to separate, a mixture because the components of a mixture, distribute themselves differently when they are in, , 199, , contact with a “two phase system”. One phase is, stationary and the other is moving or mobile. The, stationary phase may be a solid packed in a tube or a, piece of paper. The mobile phase may be liquid of, gaseous., 128. (b), 129. (b), 130. (b) Carbon and hydrogen are detected by heating the, compound with copper (II) oxide. Carbon present in, the compound is oxidised to CO2 and hydrogen to, H2O., 131. (b), 132. (b) Hydrazine (NH2NH2) does not contain carbon and, hence on fusion with Na metal, it cannot form NaCN;, consequently hydrazine does not show Lassaigne’s, test for nitrogen., 133. (a) Prussian blue Fe 4 [Fe(CN) 6 ]3 is formed in lassaigne, test for nitrogen., , 3Na 4 [Fe(CN)6 Fe 3, Fe 4 [Fe(CN) 4 ]3 12Na, Prussian blue, , 134. (d), 135. (d) Kjeldahl method is not applicable to any of the given, compounds. As nitrogen of these compounds does, not change to ammonium sulphate on heating with, conc. H2SO4., 136. (c), 137. (b), 138. (b) In Kjeldahl’s method nitrogen is converted into, (NH4)2 SO4, then to NH3, 139. (d) To increase the bpt of H2SO4,K2SO4 is added, 140. (b) N%, , 1.4 N V, wt.of organic compound, , 1.4 29 1/ 5, 16.24%, 0.5, 141. (b), , % of S, , 32 0.233, 100 10 %, 233 0.32, , 142. (c) As in above question,, 40, 13.33, 3.33; H, 13.33; N, 12, 1, Relative No. of atoms,, C, , C, , 3.33, 3.33, , 1; H, , 13.33, 3.33, , 46.67, 14, , 4; N, , 3.34, , 3.34, 3.33, , Empirical formula = CH4N, 143. (b) Percentage of P =, , =, , wt . of Mg 2 P2 O 7, 62, × 100, ×, 222, wt. of compound, , 62, 1.332, × 100 = 13.33%, ×, 222, 2.79, , 1
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EBD_7207, ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES & TECHNIQUES, , 200, , 144. (a), , % of element, , Relative, no. of atoms, , C, , 38.8, , 38.8, 12, , H, , 16.0, , 16, 1, , 45.28, , 45.28, 14, , N, , heat, , 145. (b) Compound, , 1373K, , 2C O 2, , 16.0, , 1, 5, , (A), , 3.2, , 1, , O2 + Other gaseous products, 2CO, , I 2 5CO 2, (C), , STATEMENT TYPE QUESTIONS, 146. (c) A carbon having an sp hybrid orbital with 50%, s-character is more electronegative than carbon, atoms having sp2 and sp3 hybrid orbitals with 33%, and 25% s-character respectively., In CH2 = C = CHCH3, Number of bonds :, C – C = 3,, C – H = 6, total = 6 + 3 = 9, Number of, bonds = 2, 147. (a), 148. (c), 149. (b), 150. (b), 151. (d), , R–X, , heat or, light, , R +X, , Above equation is an example of homolytic cleavge, – CH3 is an electron donating group., The resonance structures have same positions of, nuclei and same number of unpaired electrons., Fractional distillation method is used if the difference, in boiling points of two liquids is not much., For statement (ii),, Distance moved by the substance from baseline, RF = Distance moved by the solvent from base line, , 152. (b), 153. (d), , MATCHING TYPE QUESTIONS, 154. (b), , 155. (c), , 156. (a), , 157. (c), , 158. (a), , 159. (a), , 160. (d), , 161. (a), , 162. (b), , 163. (d), , 164. (c), , 165. (a), , ASSERTION-REASON TYPE QUESTIONS, , (A), , I 2O 5 5CO, (B), , 3.2, , Simple ratio, , 166. (c) – CN is a secondary suffix., 167. (b), 168. (d) The correct name of the given compound is, 2, 5, 6 -trimethyloctane, 169. (c), , 170. (d), , 171. (c) Benzene has a uniform C – C bond distance of 139, pm, a value intermediate between the C – C single., (154 pm) and C = C double (134 pm) bonds., 172. (a) It is fact that aniline is better nucleophile than anilium, ion. Anilium ion contain +ve charge, which reduces, the tendency to donate lone pair of electron, C6 H5 NH3+ ., Anilium ion, , 173. (d) Resonance structures contain the same number of, unpaired electrons. However, they differ in the way of, distribution of electrons., 174. (d), , 175. (a), , 178. (a) It is derivative of ethanamide having N-phenyl group., 179. (b) The compound contains longest chain of 5C - atoms, and e of ene is retained as the suffix name starts with, constant, CH3, , 180. (d) CH3, , K 2SO 4 raises bpt. and CuSO 4 acts as catalyst., When organic compound contains nitrogen, upon, combustion it will produce oxides of nitrogen soluble, in KOH solution. The copper will convert them into, , 181. (d), , 2NO + 2Cu, , 2CuO + N 2, , 2 NO 2 + 4Cu, 4CuO + N 2 etc., Halogens will be removed as AgX. In case of sulphur, SO 2 formed will be removed as PbSO 4 ., , 177. (c), , CRITICAL THINKING TYPE QUESTIONS, , For statement (iv), amino acids sports may be, detected by spraying the TLC plate with ninhydrin, solution., , N2, , 176. (a), , 7, , F, , 5, , **, 4, , sp, , 3|, , sp 2, C, CH, 6|, 5, CH3, , CH3, sp, sp3 |, CH, CH — C C H, 4, , 3, , 2, , 1, , *1, , 2, , Br, , 3, , **, *, The numbering of C-atom starts from C or C . But, *, numbering from C give minimum locant (2) to Br which, is correct.
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ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES & TECHNIQUES, , 201, , 191. (c) Metamerism shown among compounds of the same, functional group., , 182. (a), No. of, , 192. (b) – I group destablises carbocation and since inductive, effect decreases with increasing length of carbon, chain. Therefore (b) is the correct option., , bonds = 12 ; No. of bonds = 3, , Ratio of : bonds = 3 : 12 = 1 : 4, 183. (a) In compounds (i), (iii) and (iv), all carbon atoms are, sp3, sp2 and sp hybridised, respectivley. However,, compound (ii) has sp2 and sp3 hybridised carbon atoms;, sp, , 3, , sp, , 2, , sp, , CH 3 CH, , 2, , sp, , 3, , CH CH 3, , 193. (d), , –NO2 group, being strong electron-withdrawing,, disperses the –ve charge, hence stabilizes the, concerned carbanion., , 194. (b) In the presence of UV rays or energy, by boiling chlorine,, free radical is generated which attack the methyl carbon, atom of the toluene., , 184. (a), , CH3, , 185. (c) See the number of, , bonds formed by C in each case., x, , x, , x, , benzyl free, radical, , x, , In HCOOH , (H2 N)2CO and CH3C HO,C forms 3, bonds an d 1, bond, hybridisation is sp 2. In, , Cl 2, , (CH3 )3COH , C forms 4 bonds, hence hybridisation, is sp3, , 2Cl•, , h, , •CH, 2, , x, , x, , CH2, +H, , x, , CH2Cl, , •, + Cl, , 195. (c) Cl– is the best leaving group among the given option., 196. (a) Nucleophilicity increases down the periodic table., , 186. (a), , 1, , 2, , 3, , IUPAC name – 3, 3-Dimethyl -1, , HO, , I, , 197. (b), , 187. (d) The compound contains longest chain of 3 C - atoms, and three -COOH groups and one -OH group attached, to it (latest convention)., , CH3CH2CH2CH =CH2, 1-pentene (i), , CH3CH2CH=CHCH3, 2- pentene, (cis,- trans) (ii), (iii), , CH3, |, , CH 3 CH CH, , CH3, CH 2, , 3-methyl-1-butene, (iv), CH3, |, , CH 3 C CHCH 3, , 2-methyl-2-butene, (vi), 189. (b), , 190. (d), , |, , CH 3 CH 2 C CH 2, , 2-methyl-1-butene, (v), , Cl, , F, , CHCl is capable of showing resonance which, , develops a partial double bond character on C–Cl bond,, thereby making it less reactive toward nucleophilic, substitution., , cyclohexanol, , 188. (c) C5H10 has 1º degree of unsaturation since the isomers, are acyclic, all of these are alkenes. For writing the, isomers, first introduce the double bond at different, possible positions, and then consider the possibility, of branching in the alkyl group., , H2C, , Br, , H 2C, , 198. (d), , .., , _, , CH Cl, :, .., , CH3 Br, CH 3 Cl, , H 2 C CH, , Cl, , CH3 Br, CH 3, , Cl, , 199. (d) To convert covalent compounds into ionic compounds, such as NaCN, Na2S, NaX, etc., 200. (d) The boiling point of o-nitrophenol is less than paranitrophenol due to presence of intramolecular hydrogen, bonding. Since p-nitrophenol is less volatile in than onitrophenol due to presence of inter molecular hydrogen, bonding hence they can be separated by steam, distillation., 201. (b), , Na2S and NaCN, formed during fusion with metallic, sodium, must be removed before adding AgNO3,, otherwise black ppt. due to Na2S or white precipitate, due to AgCN will be formed and thus white precipitate, of AgCl will not be identified easily.
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EBD_7207, ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES & TECHNIQUES, , 202, , Na 2S 2AgNO3, , 2NaNO3 Ag 2S, , 203. (b), , Percentage of N in a compound, , Black, , NaCN AgNO 3, , NaNO 3, , =, , AgCN, , 1.4 Normality of acid Volume of acid used, Mass of the substance taken, , Given, 0.5 M H2SO4 is used., , White, , Normality = Molarity × n, , NaCl AgNO 3, , NaNO 3, , AgCl, , where n, , white, , 202. (d), , Na 2S 2HNO 3, , boil, , NaCN HNO 3, , boil, , 2 NaNO 3, NaNO 3, , Mol.mass, Eq.mass, , 98, 49, , 2, , Normality = 0.5 × 2 = 1 N H2SO4, , H 2S, , Volume of acid used to neutralise, NH3 = 10 cm3, , HCN, , Mass of organic compound taken = 0.25 g, , The compounds with odd number of N-atoms have, odd masses and with even number of N-atoms have, even masses. This is “nitrogen rule”., , %N, , 204. (d), , 1.4 1 10, 0.25, , 56.
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13, HYDROCARBONS, FACT / DEFINITION TYPE QUESTIONS, 1., 2., 3., 4., , 5., , 6., , 7., , 8., , 9., , Which of the following fuel cause the least pollution ?, (a) Petrol, (b) CNG, (c) Kerosene, (d) LPG, LPG mainly contains :, (a) ethyne, (b) butane, (c) methane, (d) ethane, Natural gas is a mixture of :, (a) CH4 + C2H6 + C3H8 (b) CO + H2 + CH4, (c) CO + H2, (d) H2O + CO2, Which of the following gas is find in coal mines and marshy, places?, (a) Methane, (b) Ethane, (c) Benzene, (d) Propane, Which of the following represents the correct general, formula of alkanes ?, (a) CnH2n, (b) CnH2n + 2, (c) CnH2n – 2, (d) CnHn, Two adjacent members of a homologous series have, (a) a difference of CH2 in their structure, (b) a different of 14 amu in molecular mass, (c) same general method of preparation, (d) All the above, Methane, ethane and propane are said to form a homologous, series because all are, (a) hydrocarbons, (b) saturated compounds, (c) aliphatic compounds, (d) differ from each other by a CH2 group, Which of the following does not belong to the same, homologous series?, (a) CH4, (b) C2H6, (c) C3H8, (d) C4H8, In which of the following compounds only primary carbon, atoms are present?, (a) CH3 CH CH2 CH3 (b) CH3 CH CH3, , CH3, CH3 C CH3, CH3, , H3C, , CH, , CH2, , CH, , CH2CH3, , CH3, , is, (a) 3-ethyl-5-methylheptane, (b) 5-ethyl-3-methylheptane, (c) 3,5-diethylhexane, (d) 1,1-diethyl-3-methylpentane, 11. The number of chain isomers possible for the hydrocarbon, C5H12 is, (a) 1, (b) 2, (c) 3, (d) 4, 12. The number of primary, secondary and tertiary carbons in 3,, 4-dimethylheptane are respectively, (a) 4, 3 and 2, (b) 2, 3 and 4, (c) 4, 2 and 3, (d) 3, 4 and 2, 13. Name of the given compound CH3, CH3, , CH3, , 14., , 15., , 16., 17., , (d) CH3 – CH3, , CH2, , CH2CH3, , CH3, , CH3, (c), , 10. The IUPAC n ame of the following compound, , CH3, (a) 2, 3-diethyl heptane, (b) 5-ethyl-6-methyl octane, (c) 4-ethyl-3-methyl octane (d) 3-methyl-4-ethyl octane, Which of the following statements is false for isopentane–, (a) It has three CH3 groups, (b) It has one CH2 group, (c) It has one CH group, (d) It has a carbon which is not bonded to hydrogen, Molecular formula of which of the following alkane can, exist in more than one structure ?, (a) CH4, (b) C2H6, (c) C3H8, (d) C4H10, How many isomers are possible for the C5H12 ?, (a) 2, (b) 3, (c) 4, (d) 5, The number of 4° carbon atoms in 2,2,4,4-tetramethyl, pentane is –, (a) 1, (b) 2, (c) 3, (d) 4
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EBD_7207, HYDROCARBONS, , 204, , 18., , 19., , 20., , 21., , 22., 23., 24., , Which one of the following cannot be prepared by Wurtz, reaction ?, (a) CH4, (b) C2H6, (c) C3H8, (d) C4H10, The reaction,, CH 3 Br + 2Na + Br – CH 3, the product, is called, (a) Wurtz reaction, (b) Perkin’s reaction, (c) Aldol condensation (d) Levit reaction, Pure methane can be produced by, (a) Wurtz reaction, (b) Kolbe’s electrolytic method, (c) Soda-lime decarboxylation, (d) Reduction with H2, Sodium salts of carboxylic acids on heating with soda lime, give alkanes containing _______ than the carboxylic acid., (a) one carbon more, (b) one carbon less, (c) two carbon less, (d) Either (a) or (b), Which one of the following has the least boiling point?, (a) 2, 2– dimethylpropane (b) n-butane, (c) 2-methylpropane, (d) n-pentane, Which one of the following has highest boiling point?, (a) n-Octane, (b) 2,2 dimethyl pentane, (c) Iso-octan, (d) All have equal values, Which of the following reactions of methane is incomplete, combustion ?, (a) 2CH4 + O2, (b) CH4 + O2, , 25., , 26., , 27., , 28., , 29., , Cu / 523K /100 atm, Mo 2O3, , 2CH3OH, , HCHO + H2O, , (c) CH4 + O2 C(s) + 2H2O(l), (d) CH4 + 2O2 CO2(g) + 2H2O(l), In the free radical chlorination of methane, the chain initiating, step involves the formation of, (a) chlorine free radical, (b) hydrogen chloride, (c) methyl radical, (d) chloromethyl radical., Which one of the following gives only one monochloro, derivative?, (a) n-hexane, (b) 2-methylpentane, (c) 2, 3-dimethylpentane, (d) neo-pentane, Photochemical halogenation of alkane is an example of, (a) electrophilic substitution, (b) electrophilic addition, (c) nucleophilic substitution, (d) free radical substitution, 2-Methylbutane on reacting with bromine in the presence, of sunlight gives mainly, (a) 1-bromo-3-methylbutane, (b) 2-bromo-3-methylbutane, (c) 2-bromo-2-methylbutane, (d) 1-bromo-2-methylbutane, Complete combustion of CH 4 gives :, (a) CO 2 H 2 O, (c) COCl2, , (b) CO 2 H 2, (d) CO + CO2 + H2O, , 30., , 31., , 32., , 33., , 34., , 35., , 36., , 37., , 38., , 39., , 40., , Aromatisation of n-hexane gives :, (a) cyclohexane, (b) benzene, (c) cycloheptane, (d) toluene, Liquid hydrocarbons can be converted to a mixture of, gaseous hydrocarbons by :, (a) oxidation, (b) cracking, (c) distillation under reduced pressure, (d) hydrolysis, n-Hexane isomerises in presence of anhydrous aluminium, chloride and hydrogen chloride gas to give, (a) 2-Methyl pentane, (b) 3-Methyl pentane, (c) Both (a) and (b), (d) Neither (a) nor (b), Which of the following represents the correct reaction ?, Ni, , (a), , CH 4, , 2H 2 O, , (b), , CH 4, , H 2O, , Ni, , CO 3H 2, , (c), , CH 4, , H 2O, , Ni, , CH 3OH H 2, , (d), , CH 4, , H 2O, , Ni, , HCHO 2H 2, , CO 2, , 4H 2, , How many conformations are possible for ethane ?, (a) 2, (b) 3, (c) infinite, (d) one, Spatial arrangements of atoms which can be converted into, one another by rotation around a C–C single bond are called, (a) Stereoisomers, (b) Tautomers, (c) Optical isomers, (d) Conformers, General formula of alkenes and alkyl radicals are, respectively:, (a) CnH2n and CnH2n+1 (b) CnH2n and CnH2n+2, (c) CnH2n–1 and CnH2n (d) CnH2n+1 and CnH2n+2, The restricted rotation about carbon-carbon double, bond in 2- butene is due to, (a) overlap of one s- and one sp2-hybridized orbitals, (b) overlap of two sp2-hybridized orbitals, (c) overlap of one p-and one sp2-hybridized orbitals, (d) sideways overlap of two p-orbitals, Bond angle in alkenes is equal to, (a) 120°, (b) 109°28', (c) 180°, (d) 60°, The molecular formula of a compound in which double bond, is present between C & C :, (a) CnH2n + 2, (b) CnHn, (c) CnH2n, (d) CnH2n–2, IUPAC name of the following compound is, H3C CH CH 2 CH CH CH3, |, Cl, (b) 4-chloropent-1-ene, (a) 5-chloroheptene, (c) 5-chloropent-3-ene, (d) 5-chlorohex-2-ene
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HYDROCARBONS, , 80., , 81., , 82., , Isopropyl alcohol is obtained by reacting which of the, following alkenes with conc. H2SO4 and H2O, (a) Ethylene, (b) Propylene, (c) 2-methyl propene, (d) Isoprene, Which one of the following is the strongest bond?, (a) > C = C <, (b) — C C —, |, |, |, (c) — C — C =, (d) — C — C —, |, |, |, |, An alkyne has general formula :, (a), , 83., , 84., , 85., , 86., , 87., , 88., , 89., , 90., , 91., , 207, , Cn H 2n, , (b), , Cn H 2n 1, , (c) Cn H 2n 2, (d) Cn H 2n 2, The IUPAC name of the compound, CH3CH = CHC CH is, (a) Pent-l-yn-3-ene, (b) Pent-4-yn-2-ene, (c) Pent-3-en-1-yne, (d) Pent-2-en-4-yne, Number of alkynes for formula C5H8 is –, (a) 2, (b) 3, (c) 4, (d) 5, The IUPAC name of the compound having the formula, CH C – CH = CH2 is :, (a) 1-butyn-3-ene, (b) but-1-yne-3-ene, (c) 1-butene-3-yne, (d) 3-butene-1-yne, The homologue of ethyne is, (a) C2H4, (b) C2H6, (c) C3H8, (d) C3H6, The C - H bond length is minimum in the bond formed by, (a) sp - s overlapping (as in alkynes), (b) sp2 – s overlapping (as in alkenes), (c) sp3 – s overlapping (as in alkanes), (d) None of these, Triple bond of ethyne is made of, (a) Three – bonds, (b) Three – bonds, (c) Two and one – bond, (d) Two and one – bond, Maximum carbon-carbon bond distance is found in –, (a) ethyne, (b) ethene, (c) ethane, (d) benzene, The acetylene molecule contains :, (a) 5 sigma bonds, (b) 4 sigma and 1 pi bonds, (c) 3 sigma and 2 pi bonds (d) 2 sigma and 3 pi bonds, Butyne-2 contains :, (a) sp hybridised carbon atoms only, (b), , sp3 hybridised carbon atoms only, , (c) both sp and sp 2 hybridised carbon atoms, , 92., , (d) both sp and sp3 hybridised carbon atoms, The correct order towards bond length is, (a) C C C C C C (b) C C C C C C, (c) C C C C C C (d) C C C C C C, , 93. Which C-atom is the most electronegative in this structure?, III, II, CH 3 CH 2, , (a), (b), (c), (d), , C, , I, CH, , I, II, III, all are equal electronegative, Reagent, , R CH 2 CCl 2 R, R C C R, The reagent is, (a) Na, (b) HCl in H2O, (c) KOH in C2H5OH, (d) Zn in alcohol., 95. Calcium carbide when treated with water gives :, (a) ethylene, (b) methane, (c) acetylene, (d) ethane, 96. Which one of the following has the minimum boiling, point ?, (a) 1-Butene, (b) 1-Butyne, (c) n- Butane, (d) Isobutane, 97. Ammonical silver nitrate forms a white precipitate easily, with, (a) CH3C CH, (b) CH3C CCH3, 94., , (d) CH2 = CH2, (c) CH3CH CH 2, 98. When acetylene is passed through dil. H2SO4 in presence, of HgSO4, the compound formed is, (a) ether, (b) acetaldehyde, (c) acetic acid, (d) ketone, 99. Which of the following will be the final product when C2H2, reacts with HCl, , CH, (a), , (b), , CHCl, , CH3, |, CHCl 2, , CHCl, , (d) None of these, CHCl, 100. The hydrocarbon which can react with sodium in liquid, ammonia is, (c), , (a), , CH 3CH 2 CH 2 C, , (b), , CH 3CH 2C, , (c), , CH 3CH, , CCH 2CH 2 CH3, , CH, , CHCH3, , (d) CH 3CH 2C CCH 2CH3, 101.Which of these will not react with acetylene?, (a) NaOH, (b) Ammonical AgNO3, (c) Na, (d) HCl., 102. When acetylene is passed over heated iron tube, the product, obtained is –, (a) C2H2, (b) C4H4, (c) C6H6, (d) C8H8, 103. But–2–yne on chlorination gives, (a) 1 –chlorobutane, (b) 1, 2 –dichlorobutane, (c) 1, 1, 2, 2 –tetrachlorobutane, (d) 2, 2, 3, 3 –tetrachlorobutane
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EBD_7207, HYDROCARBONS, , 208, , 104. When propyne reacts with aqueous H2SO4 in the presence, of HgSO4, the major product is, (a) Propanal, (b) Propyl hydrogen sulphate, (c) Acetone, (d) Propanol, 105. Propyne on polymerisation gives, (a) Mesitylene, (b) Benzene, (c) Ethyl benzene, (d) Propyl benzene, 106. What happens when a mixture of acetylene and hydrogen, is passed over heated Lindlar’s catalyst ?, (a) Ethane and water are formed, (b) Ethylene is formed, (c) Acetylene and ethane are formed, (d) None of these, 107. Which of the following reaction is shown by alkynes ?, (a) Addition, (b) Substitution, (c) Polymerization, (d) All of these, 108. Which of the following reactions will yield 2,, 2-dibromopropane ?, (a) HC CH 2HBr, (b), , CH3C CH 2HBr, , (c), , CH3CH CH 2, , HBr, , (d) CH3CH CHBr HBr, 109. In the given reactions, CH 3C, , CH, , A, , CH 3CBr2 CHBr2, , CH 3C, , CH, , B, , CH 3CBr2 CH 3, , CH3C CH, HC CH, , Hg 2 H, 333 K, Hg 2 H, 333 K, , C, D, , A, B, C and D are respectively, (a) HBr, Br2, CH3COCH3, CH3CHO, (b) Br2, HBr, CH3COCH3, CH3CHO, (c) HBr, HBr, CH3COCH3, CH3CHO, (d) Br2, HBr, CH3CH2CHO, CH3CHO, 110. Which of the following polymer can be used as electrodes, in batteries ?, (a) Polypropene, (b) Polyacetylene, (c) Polyethene, (d) Polyisoprene, 111. Which of the following catalyst is used for the following, conversion ?, 3CH, , CH, , 113. Benzene was discovered by, (a) Ramsay, (b) Dalton, (c) Faraday, (d) Priestley, 114.The ring structure of benzene was proposed by, (a) Faraday, (b) Davy, (c) Kekule, (d) Wohler, 115. Six carbon atoms f benzene are of, (a) one type, (b) two types, (c) three types, (d) six types, 116. Select the true statement about benzene amongst the, following, (a) Because of unsaturation benzene easily undergoes, addition, (b) There are two types of C – C bonds in benzene molecule, (c) There is cyclic delocalisation of pi-electrons in benzene, (d) Monosubstitution of benzene gives three isomeric, products., 117. The benzene molecule contains, (a) 6 sp2 hybrid carbons (b) 3 sp2 hybrid carbons, (c) 6 sp3 hybrid carbons (d) 3 sp3 hybrid carbons, 118. Aromatic compounds burn with sooty flame because, (a) they have a ring structure of carbon atoms, (b) they have a relatively high percentage of hydrogen, (c) they have a relatively high percentage of carbon, (d) they resist reaction with oxygen of air, 119. Carbon atom in benzene molecule is inclined at an angle of, (a) 120°, (b) 180°, (c) 109° 28', (d) 60°, 120. The conditions for aromaticity is :, (a) molecule must have cyclic clouds of delocalised, electrons, (b) molecule must contain (4n + 2) electrons, (c) Both (a) and (b), (d) None of the above, 121. The chemical system that is non-aromatic is, (a), +, , (c), , (d), , +, , –, , 122. Benzene can be directly obtained from, (a) Acetylene, (b) Phenol, (c) Chlorobenzene, (d) All the above, , COONa, , 873 K, , (a) Platinized Asbestos (b), (c) Platinized Nickel, (d), 112. Which one of the following is a, compound?, (a) Aniline, (b), (c) Naphthalene, (d), , (b), , Red hot iron tube, Iron-molybdenum, non-benzenoid aromatic, Benzoic acid, Tropolone, , 123., , NaOH + CaO, The product A is, (a) Benzene, (c) Toluene, , A, , (b) Benzaldehyde, (d) Benzoic acid
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HYDROCARBONS, , 209, , 124. In a reaction of C6H5Y, the major product (> 60%) is, m-isomer, so the group Y is, (a) –COOH, (b) –NH2, (c) –OH, (d) –Cl, HNO3, H 2SO4, , 125., , A, , Br2, FeBr2, , B. The compound B is, , NO2, , NO2, , Br, , STATEMENT TYPE QUESTIONS, , (b), , (a), , Br, NO2, (c), , NO2, (d), , Br, , Br, , Br, , 126. Chlorobenzene is o, p-directing in electrophilic substitution, reaction. The directing influence is explained by, (a) + M of Ph, (b) +I of Cl, (c) + M of Cl, (d) + I of Ph, 127. Catalytic hydrogenation of benzene gives, (a) xylene, (b) cyclohexane, (c) benzoic acid, (d) toluene, 128. The strongest ortho - para and strongest meta - directing, groups respectively are, (a) –NO2 and –NH2, (b) –CONH2 and –NH2, (c) –NH2 and –CONH2 (d) –NH2 and –NO2, 129. For the formation of toluene by Friedal Craft reaction,, reactants used in presence of anhydrous AlCl 3 are, (a) C2H2 and CCl4, (b) CH4 and CaCN2, (c) C6H6 and CH3Cl, (d) C2H5 OH and Zn, 130. Benzene can be obtained in the reaction, (a) Ethene + 1, 3-butadiene, (b) Trimerisation of ethyne, (c) Reduction of PhCHO, (d) All of these, 131. Nitration of benzene by nitric acid and sulphuric acid is, (a) Electrophilic substitution, (b) Electrophilic addition, (c) Nucleophilic substitution, (d) Free radical substitution, 132. C6 H6 CH3Cl, , BHC, anhydrous, AlCl3, , 134. Benzene on reaction with ozone forms __________., (a) 2 molecules of aldehyde and 1 molecule of ketone, (b) 2 molecules of ketone and 1 molecule of aldehyde, (c) triozonide, (d) hexaozonide, 135. AlCl3 acts as ________ in Friedel-Crafts reaction, (a) nucleophile, (b) electrophile, (c) free radical, (d) intermediate, , C6 H5CH3 HCl, , is an example of, (a) Friedel - Craft’s reaction, (b) Kolbe’s synthesis, (c) Wurtz reaction, (d) Grignard reaction, 133. Benzene reacts with CH3COCl + AlCl3 to give, (a) chlorobenzene, (b) toluene, (c) benzyl chloride, (d) acetophenone, , 136. The electrophilic substitutions reactions of benzene takes, place via, (i) generation of electrophile, (ii) generation of nucleophile, (iii) formation of carbocation intermediate, (iv) removal of proton from the carbocation intermediate, (a) (i), (iii) and (iv), (b) (ii), (iii) and (iv), (c) (i) and (iv), (d) (ii) and (iv), 137. During the nitration of benzene. In the process of generation, of nitronium ion sulphuric acid behaves as a/an ______, and nitric acid behave as a/an _________., (a) base, acid, (b) acid, base, (c) strong acid, weak acid (d) weak acid, strong acid, 138. Benzene is highly unsaturated but it does not undergo, addition reaction because, (a) -electrons of benzene are delocalised., (b) cyclic structures do not show addition reaction, (c) benzene is a non-reactive compound, (d) All of the above, 139. Which of the following statements are correct ?, (i) LNG is obtained by liquefaction of natural gas., (ii) Petrol is obtained by fractional distillation of petroleum., (iii) Coal gas is obtained by destructive distillation of coal., (iv) CNG is found in upper strata during drilling of oil wells., (a) (i), (ii) and (iv), (b) (i), (ii) and (iii), (c) (i) and (iii), (d) (ii) and (iv), 140. Which of the following statements are correct ?, (i) Saturated hydrocarbons contain only carbon-carbon, single bonds., (ii) Saturated hydrocarbons contain both carbon-carbon, and carbon-hydrogen single bond., (iii) Unsaturated hydrocarbons contain carbon-carbon, double bonds., (iv) Unsaturated hydrocarbons contain carbon-carbon, double and triple bonds both., (a) (i) and (iii), (b) (ii) and (iv), (c) (i) and (ii), (d) (i) and (iv), 141. Which of the following statements are correct regarding, structure of methane ?, (i) Methane has tetrahedral structure., (ii) The bond angle between all H – C – H bonds is 109.5°., (iii) The carbon atom is sp2 hybridized., (iv) C – C and C – H bond lengths are 154 pm and 112 pm, respectively., (a) (i), (ii) and (iii), (b) (i), (iii) and (iv), (c) (i), (ii) and (iv), (d) (i), (ii), (iii) and (iv)
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EBD_7207, HYDROCARBONS, , 210, , 142. In the preparation of alkanes from hydrogenation of alkenes, and alkynes. Finely divided catalysts are used which of the, following statement(s) is/are correct regarding these, catalysts, (i) Platinum and palladium catalyse the reaction at room, temperature., (ii) Nickel catalyse the reaction at relatively higher, temperature and pressure., (iii) Platinum and palladium catalyse the reaction at higher, temperature., (a) (i) and (iii), (b) (i) and (ii), (c) (ii) and (iii), (d) (i) only, 143. Which of the following statements are correct ?, (i) The rate of reactivity of alkanes with halogens is, F2 > Cl2 > Br2 > I2., (ii) Rate of replacement of hydrogens of alkanes is, 3° > 2º > 1°, (iii) Fluorination of alkanes is a very slow process., (iv) Iodination of alkanes is too violent to be controlled., (a) (i), (ii) and (iii), (b) (i) and (ii), (c) (ii) and (iii), (d) (i) and (iv), 144. Which of the following statements are correct ?, (i) Decomposition reaction of higher alkanes into smaller, fragments by the application of heat is called pyrolysis., (ii) Pyrolysis and cracking are different processes., (iii) Dodecane on pyrolysis gives a mixture of heptane and, pentene., (iv) Pyrolysis follow free radical mechanism., (a) (i), (ii) and (iii), (b) (i), (ii) and (iv), (c) (i), (iii) and (iv), (d) (ii) and (iv), 145. Which of the following statement(s) is/are correct ?, (i) Alkanes can have infinite number of conformations, by rotation around a C – C single bonds., (ii) Rotation around C – C single bond is completely free., (iii) Rotation is hindered by a small energy barrier of, 1-20 kJ mol–1 due to torsional strain., (a) (i) and (ii), (b) (i) and (iii), (c) (ii) and (iii), (d) Only (iii), 146. Which of the following statements are correct ?, (i) Stability of conformation is affected due to torsional, strain., (ii) Magnitude of torsional strain depends upon the angle, of rotation about C – C bond., (iii) Eclipsed form has least torsional strain., (iv) Staggered form has maximum torsional strain., (a) (i) and (iii), (b) (i) and (ii), (c) (iii) and (iv), (d) (i) and (iv), 147. Which of the following statements are correct ?, (i) The general formula of alkenes is CnH2n., (ii) Alkenes are also known as paraffins., (iii) Bond length of C–C double bond in alkene is shorter, than C–C single bond in alkane., (iv) Carbon–Carbon double bond in alkene consists of two, sigma bonds., (v) Alkenes are easily attacked by electrophilic reagent., (a) (i) and (iv), (b) (i), (iii) and (v), (c) (i) and (iii), (d) (i), (ii), (iv) and (v), , 148. Which of the following statements are correct ?, (i) Cis form of alkene is polar whereas trans form is, non-polar, (ii) Cis form of alkene is non-polar whereas trans form is, polar., (iii) In case of solid alkenes the trans isomer has higher, melting point than the cis isomer., (iv) Cis and trans both form have same properties., (a) (i) and (iii), (b) (ii) and (iii), (c) (i), (iii) and (iv), (d) (i) and (iv), 149. Which of the following statements are correct ?, (i) Alkynes on reduction with palladised charcoal form, cis alkenes., (ii) Alkynes on reduction with palladised charcoal form, trans alkenes., (iii) Alkynes on reduction with sodium in liquid ammonia, form trans alkenes., (iv) Propyne on reduction with palladised charcoal form a, mixture of cis and trans propene., (a) (i) and (iv), (b) (i) and (iii), (c) (ii) and (iv), (d) (i), (iii) and (iv), 150. Which of the following statements are correct ?, (i) Polynuclear hydrocarbons contain two or more, benzene rings fused together., (ii) Polynuclear hydrocarbons have carcinogenic property., (iii) Polynuclear hydrocarbons are formed on incomplete, combustion of organic materials like tobacco, coal and, petroleum., (iv) They are also produced in human body due to various, biochemical reactions., (a) (i), (ii) and (iv), (b) (i), (iii) and (iv), (c) (ii), (iii) and (iv), (d) (i), (ii) and (iii), , MATCHING TYPE QUESTIONS, 151. Match the columns, Column-I, , (A) Eclipsed, , Column-II, H, , (p), , H, , H, , H, , H, H, HH, , (B) Staggered, , (q) H, H, , H, HH, , (C) Skew, , (r), , H, , H, H, H, H, (a) A – (r), B – (p), C – (q) (b) A – (r), B – (q), C – (p), (c) A – (p), B – (q), C – (r) (d) A – (q), B – ( p), C – (r)
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HYDROCARBONS, , 211, , 152. Match the columns, Column-I, (A) CH2 = CH2 CH3 – CH3 (p), (B) CH3Cl CH4, (q), (C) CH3Br CH3CH3, (r), (D) CH3COONa CH4, (s), (a) A – (r), B – (p), C – (s), D – (q), (b) A – (p), B – (s), C – (r), D – (q), (c) A – (s), B – (q), C – (p), D – (r), (d) A – (q), B – (p), C – (s), D – (r), 153. Match the columns, Column-I, (A) CH4 + O2 Cu/523K/100 atm, (B) CH4 + O2, , (q) (CH3)3COH, , (CH3COO) 2 Mn, KMnO 4, oxidation, , (D) (CH3)3CH, , Column-II, (p) HCHO, , Mo 2O3, , (C) C2H6 + O2, , Column-II, H2, Zn, H+, NaOH, CaO, H2, Pt/Pd, Na, dry ether, , (r), , CH3OH, , (s), , CH3COOH, , (a) A – (s), B – (p), C – (r), D – (s), (b) A – (q), B – (p), C – (s), D – (r), (c) A – (r), B – (p), C – (s), D – (q), (d) A – (p), B – (q), C – (r), D – (s), 154. Match the columns, Column-I, Column-II, (A) CH CH H 2, (p) Zn, CH 2 CH 2, (B) CH 3CH 2 Br, (q) Conc. H2SO4, CH 2 CH2, (C) CH 2 BrCH 2 Br, , (r), , Pd/C, , CH 2 CH2, (s) Alc. KOH, (D) CH3CH2OH, CH2 = CH2, (a) A – (r), B – (s), C – (p), D – (q), (b) A – (s), B – (r), C – (q), D – (p), (c) A – (q), B – (p), C – (s), D – (r), (d) A – (r), B – (s), C – (q), D – (p), 155. Match the columns, Column-I, Column-II, Cl, , (A), , (p) Cl2, uv, 500 K, CH2Cl, , (q) anhy. AlCl3, , (B), Cl, , (C), , Cl, , Cl, , Cl, , Cl, , (r) CH2Cl2, anhy. AlCl3, , Cl, Cl, , (D), , Cl, , Cl, , Cl, , Cl, Cl, , (s) Cl2, anhy. AlCl3,, , (a) A – (s), B – (r), C – (q), D – (p), (b) A – (q), B – (r), C – (s), D – (p), (c) A – (r), B – (p), C – (q), D – (s), (d) A – (q), B – (p), C – (s), D – (r), 156. Match the following reactants in Column I with the, corresponding reaction products in Column II and choose, the correct option from the codes given below., Column - I, Column - II, AlCl3, , (A) Benzene + Cl2, (B) Benzene + CH3Cl, , (p) Benzoic acid, (q) Methyl phenyl, , AlCl3, , (C) Benzene + CH3COCl, , ketone, (r) Toluene, , AlCl3, , (D) Toluene, , KMnO4 / NaOH, , (s) Chlorobenzene, , (a) A – (s), B – (r), C – (q), D – (p), (b) A – (s), B – (r), C – (p), D – (q), (c) A – (r), B – (s), C – (p), D – (q), (d) A – (r), B – (s), C – (q), D – (p), 157. Match the columns, Column - I, Column - II, (A) Alkyl + Acid halide, (p) Sulphonation, in presence of dry ether, (B) Arene + Acid halide, (q) Wurtz reaction, in presence of AlCl3, (C) Arene + Fuming sulphuric (r) Catalytic, in presence of AlCl3, hydrogenation, (D) Arene + Hydrogen, (s) Friedel-Crafts, in presence of Ni, reaction, (a) A - (p), B - (r), C - (q); D - (s), (b) A - (s), B - (q), C - (r); D - (p), (c) A - (r), B - (p), C - (s); D - (q), (d) A - (q), B - (s), C - (p); D - (r), 158. Match the columns, Column - I, Column - II, (A) Aromatic, (p) Planar, (B) Antiaromatic, (q) Non-planar, (C) Huckel rule, (r) 4 n localised, electrons, (D) Cyclo-octatetraene, (s) (4 n + 2) delocalised, electrons, (a) A - (p, s), B - (p, r), C - (s), D - (q, r), (b) A - (p, r), B - (p, s), C - (s), D - (q, r), (c) A - (p, s), B - (s), C - (p, r), D - (q, r), (d) A - (q, r), B - (p, r), C - (s), D - (p, s)
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EBD_7207, HYDROCARBONS, , 212, , 159., , Match the columns, Column - I, (Reactants), , Column - II, (No. of chlorinated, products), , (A) Benzene, , Cl2 , light, , p., , Three compounds, , (B) Toluene, , Cl2 , light, , q., , Four compounds, , (C) Methane, , Cl2 , light, , r., , (a), (b), (c), (d), , Cl2 , AlCl3, , s., , 165. In cyclopropane, cyclobutane and cyclohexane, the, common group is, , Single monochloro, derivative, , (D) Benzene, , CRITICAL THINKING TYPE QUESTIONS, , Six isomeric, , compounds, A – (r), B – (p, r), C – (q, r), D – (s), A – (s), B – (p, r), C – (q, r), D – (r), A – (p, r), B – (s), C – (q, r), D – (r), A – (s), B – (p, r), C – (r), D – (q, r), , (a), , |, C, |, , (b), , |, CH, |, , (c), , CH3, , (d), , |, CH2, |, , 166. The number of primary, secondary, tertiary and quaternary, carbons in neopentane are respectively, (a) 4, 3, 2 and 1, (b) 5, 0, 0 and 1, (c) 4, 0, 0 and 1, (d) 4, 0, 1 and 1, , 167. The IUPAC name of CH3, , ASSERTION-REASON TYPE QUESTIONS, Directions : Each of these questions contain two statements,, Assertion and Reason. Each of these questions also has four, alternative choices, only one of which is the correct answer. You, have to select one of the codes (a), (b), (c) and (d) given below., (a) Assertion is correct, reason is correct; reason is a correct, explanation for assertion., (b) Assertion is correct, reason is correct; reason is not a, correct explanation for assertion, (c) Assertion is correct, reason is incorrect, (d) Assertion is incorrect, reason is correct., 160. Statement-1 : 1-Butene on reaction with HBr in the presence, of a peroxide produces 1-bromobutane., Statement-2 : It involves the free radical mechanism., 161. Statement-1 : CH4 does not react with Cl2 in dark., Statement-2 : Chlorination of CH4 takes place in sunlight., 162. Statement-1 : Iodination of alkanes is reversible., Statement-2 : Iodination is carried out in presence of iodic, acid., 163. Statement-1 : All the hydrogen atoms in CH2 = C = CH2 lie, in one plane., Statement-2 : Carbon atoms are sp2 and sp hybridized., 164. Statement-1 : Tropylium cation is aromatic in nature, , Statement-2 : The only property that determines its aromatic, behaviour is its planar structure., , 168., , 169., , 170., , 171., , CH 2, , H C 4 H9, |, |, C C CH3 : –, |, |, CH3CH3, , (a) 3, 4, 4–Trimethyl octane, (b) 3, 4, 4–Trimethyl heptane, (c) 2–Ethyl, 3,3–dimethyl heptane, (d) 2–Butyl, 2 methyl,3–ethyl butane, Which one of the following has the lowest boiling point?, (a) 2-methylbutane, (b) 2-methyl propane, (c) 2, 2-dimethyl propane (d) n-pentane, Arrange the following in decreasing order of their boiling, points., (A) n–butane, (B) 2-methylbutane, (C) n-pentane, (D) 2, 2–dimethylpropane, (a) A > B > C > D, (b) B > C > D > A, (c) D > C > B > A, (d) C > B > D > A, When neo-pentyl bromide is subjected to Wurtz reaction,, the product formed is, (a) 2,2,4,4-tetramethylhexane, (b) 2,2,4,4-tetramethylpentane, (c) 2,2,5,5-tetramethylhexane, (d) 2,2,3,3-tetramethylhexane, Which one of the following reactions is expected to readily, give a hydrocarbon product in good yields ?, Electrolyt ic, , (a), , RCOOK, , (b), , RCOO Ag, , (c), , CH 3 CH 3, , (d), , (CH 3 ) 3 CCl, , oxidation, Br2, , Cl2, h, C 2 H 5OH
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EBD_7207, HYDROCARBONS, , 214, , 186.Which of the following types of reaction occur when a, reactant has got a double bond ?, (i) Addition, (ii) Photolysis, (iii) Nucleophilic substitution, (iv) Polymerization, (a) (i) and (iv), (b) (i), (ii) and (iii), (c) (iii) and (iv), (d) (ii) and (iii), 187. The disappearance of the characteristic purple colour of, KMnO4 in its reaction with an alkene is the test for, unsaturation. It is known as, (a) Markownikoff test, (b) Baeyer test, (c) Wurtz test, (d) Grignard test, 188. CH2 = CHCl reacts with HCl to form, (a), , CH 2 Cl CH 2 Cl, , (b), , (c), , CH 2, , (d) None of these, , CHCl.HCl, , CH3 CHCl2, , 189. The only alcohol that can be prepared by the indirect, hydration of alkene is, (a) Ethyl alcohol, (b) Propyl alcohol, (c) Isobutyl alcohol, (d) Methyl alcohol, 190. Which reactions are most common in alkenes, (a) Electrophilic substitution reactions, (b) Nucleophillic substitution reactions, (c) Electrophilic addition reactions, (d) Nucleophilic addition reactions, 191. In the presence of peroxide, hydrogen chloride and hydrogen, iodide do not give anti-Markownikov’s addition to alkenes, because, (a) Both are highly ionic, (b) One is oxidising and the other is reducing, (c) One of the steps is endothermic in both the cases, (d) All the steps are exothermic in both the cases, 192. Which of the following statements is incorrect regarding, dehydrohalogenation of alkenes ?, (a) During the reaction hydrogen atom is eliminated from, the - carbon atom., (b) Rate of reaction for same alkyl group;, Iodine > Bromine > Chlorine, (c) Rate of reaction; (CH3)3C – > (CH3)2CH – > CH3CH2 –, (d) Only nature of halogen atom determine rate of the, reaction., 193. How many structural isomers are possible for the alkyne, C6H10 ?, (a) 7, (b) 6, (c) 8, (d) 5, 194. Which of the following will have least hindered rotation, around carbon - carbon bond ?, (a) Ethane, (b) Ethylene, (c) Acetylene, (d) Hexachloroethane, , 195. Acetylenic hydrogens are acidic because, (a) Sigma electron density of C – H bond in acetylene is, nearer to carbon, which has 50% s-character, (b) Acetylene has only open hydrogen in each carbon, (c) Acetylene contains least number of hydrogens among, the possible hydrocarbons having two carbons, (d) Acetylene belongs to the class of alkynes with, molecular formula, CnH2n – 2., 196. Propyne can be prepared by dehydrohalogenation of, (a) 1–chloropropane, (b) 1, 2–dichloropropane, (c) 1, 2–dichloroethane, (d) 1, 1, 2, 2–tetrachloroethane, 197. Which is the most suitable reagent among the following to, distinguish compound (3) from rest of the compounds ?, 1., , CH 3 C, , C CH 3, , 2., , CH 3 CH 2 CH 2 CH 3, , 3., , CH 3 CH 2C CH, , 4., , CH 3 CH, , CH 2., , (a) Bromine in carbon tetrachloride, (b) Bromine in acetic acid, (c) Alk KMnO4, (d) Ammonical silver nitrate., 198. Predict the product C obtained in the following reaction of, butyne-1., CH 3CH 2, , (a), , (b), , (c), , C, , CH, , HCl, , CH3 CH2 CH 2, , CH 3 CH 2, , CH3CH 2, , B, , HI, , C, , I, |, C H, |, Cl, , I, |, CH CH 2 Cl, , I, |, C CH3, |, Cl, , (d), , CH3 CH CH2CH 2 I, |, Cl, 199. The correct increasing order of acidity of the following, alkynes, (1), , CH3 C, , C CH3, , (2), , CH3 C, , CH, , (3) CH CH, (a) 1 < 2 < 3, (c) 3 < 2 < 1, , (b) 2 < 3 < 1, (d) 1 < 3 < 2
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HYDROCARBONS, , 215, , (a), , H3C C C CH3, , (b), , H3C CH 2 C CH, , (a) It forms only one monosubstitution product, (b) The C - C bond distance in benzene is uniformly 1.397Å, (c) It is a resonance hybrid of a number of canonical forms, (d) It has three delocalised -molecular orbitals, 206. The ratio of to bonds in benzene is :, (a) 2, (b) 3, (c) 4, (d) 8, , (c), , H 2C CH C CH, , 207. The radical,, , (d), , HC C CH 2 C CH, , 200. Identify the alkyne in the following sequence of reactions., Alkyne, , H2, Lindlar's catalyst, , A, , Ozonolysis, , B, , only, , Wacker, Process, , CH2 = CH2, , 201. Which of the following represent the correct order of acidic, strength ?, (i), , HC CH, , H 2C CH 2, , CH3 CH3, , (ii), , HC CH CH3 CH3, , H 2C CH 2, , (iii) CH3C CH, , HC CH CH3 C, , 208., C CH3, , (iv) HC CH CH3 C CH CH3 C C CH3, (a) (i) and (iii), (b) (ii) and (iv), (c) (i) and (iv), (d) (i) and (iv), 202. Which one of these is not compatible with arenes?, (a) Greater stability, (b) Delocalisation of -electrons, (c) Electrophilic additions, (d) Resonance, COOH, 203., , COOH, CHO, and, , 209., , 210., are, , (a) Position isomer, (b) Chain isomer, (c) Functional isomer, (d) Stereoisomer, 204. The carbon-carbon bond length in benzene is, (a) Same as in C2H4, (b) In between C2H6 and C2H2, (c) In between C2H4 and C2H2, (d) In between C2H6 and C2H4, 205. Point out the wrong statement in relation to the structure of, benzene, , 211., , ., CH2, , is aromatic because it has :, , (a) 7 p-orbitals and 6 unpaired electrons, (b) 7 p-orbitals and 7 unpaired electrons, (c) 6 p-orbitals and 7 unpaired electrons, (d) 6 p-orbitals and 6 unpaired electrons, (i) Chlorobenzene and (ii) benzene hexachloride are, obtained from benzene by the reaction of chlorine, in the, pesence of, (a) (i) Direct sunlight and (ii) anhydrous AlCl3, (b) (i) Sodium hydroxide and (ii) sulphuric acid, (c) (i) Ultraviolet light and (ii) anhydrous FeCl3, (d) (i) Anhydrous AlCl3 and (ii) direct sunlight, A group which deactivates the benzene ring towards, electrophilic substitution but which directs the incoming, group principally to the o- and p-positions is, (a) –NH2, (b) –Cl, (c) –NO2, (d) –C2H5, Benzene can be obtained by heating either benzoic acid, with X or phenol with Y. X and Y are respectively., (a) Zinc dust and soda lime, (b) Soda lime and zinc dust, (c) Zinc dust and sodium hydroxide, (d) Soda lime and copper, Which of the following chemical system is non aromatic?, (a), , (b), , (c), , (d), S
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EBD_7207, HYDROCARBONS, , 216, , H, H–C–H, –, , FACT / DEFINITION TYPE QUESTIONS, , CH3CH 2 CH 2CH 2 CH3, n -pentane, , CH3 CH CH2, |, CH3, , CH3, , 17., 18., 19., , R X 2Na X R ', , 2,3-dimethylpropane (neo pentane), , 21., , H H H H H, | 2° | 2° | 3° |3° |2° 1°, H3C – C – C – C – C – C – CH3, |, | | 1°| 1°|, H H CH3 CH3 H, 3, 4-dimethylheptane, 1°, , 12., , (a), , 13., 15., , There are four 1° C-atoms, three 2° C-atoms and two, 3° C-atoms, (c), 14. (d), (d) CH4, C2H6 and C3H8 can have only one structure but, C4H10 can have more than one structure. Possible, structures of C4H10 are following, , CH3COO Na, , NaOH, , CaO, , CH 4, , – –, , Na 2CO3, , Methane, , (a) Higher is the branching lesser will be the boiling point, further increase in molecular weight increases boiling, point in alkane. Hence 2, 2– dimethyl propane will have, least boiling point., , H H H H, 4, 3, 2, H C C C C H, , CH3, |, CH3 C CH 3, |, CH3, , H H H H, Butane (n- butane), (b.p. 273 K), , dry ether, , Sodium ethanoate, , 22., , R R ' 2 NaBr, , CH 3 CH 3 2 NaBr, (c) Other three methods can be used for the preparation, of alkane having at least two carbon atoms., (b) Sodium salts of carboxylic acids on heating with soda, lime (mixture of sodium hydroxide and calcium oxide), give alkanes containing one carbon atom less than the, carboxylic acid. This process of elimination of carbon, dioxide from a carboxylic acid is known as, decarboxylation, , 1, , C, C H, H C, HH C HH, , 24., , (a) n-octane has highest boiling point due to unbranched, chain and maximum carbon atoms. It has max. Van der, Waal forces., (c), , H, , 25., , (a), , Cl2, , (d), , CH3, |, Neo-pentane, H3C C CH3 , has only 1° hydrogen, |, CH3, , H, , 1, , H, , H, , 23., , 3, , 2, , 2-Methylpropane (isobutane), (b.p.261 K), , (b) Possible isomers of C5H12 are, – –, , – –, , – –, , – –, , H H H H H, H–C–C–C–C–C–H, H H H H H, , – –, , 16., , dry ether, , CH3–Br + 2Na+Br–CH3, 20., , –— —–, , (b), (a) CH4 has only one carbon atom, hence it can’t be, prepared by Wurtz reaction, which involves two, molecules of alkyl halide., (a) When alkyl halide is treated with sodium metal in, presence of ether, alkane is obtained, this reaction is, called as Wurtz reaction., , 2 methylbutane (iso pentane), , CH3, |, H3C C CH3, |, CH3, , –, , (a), (c) Pentane (C5H12) exists as three chain isomers, , 1°, , – –, , 10., 11., , 1°, , CH3, CH3, Both carbon atoms in ethane are primary., , – –, , (d), , – –, , 9., , H, H, H – C –— C –— C – H, H, H, H–C–H, H, , H H H H, H –C—C —C –C –H, H H– C –H H H, H, , – – –, , (d) LPG is used as a domestic fuel with the least pollution., (b) LPG mainly contains butane., (a) Natural gas is a mixture of CH4, C2H6 and C3H8., (a), 5. (b), 6. (d), 7. (d), 8. (d), , – –, , 1., 2., 3., 4., , 26., , hv, Chain initiation, , 2Cl•, , and hence gives only one monochloro derivative.
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HYDROCARBONS, , 27., , 217, , (d), , 44. (b), CH3, , 28., , (c), , CH3, , |, , CH3 CH CH2 CH3, , Br2, sunlight, , 30., , CH 3, , 4, , |, , CH3 C CH2CH3, , Ease of replacement of H-atom 3° > 2° > 1°., (a) Complete combustion of all organic compounds leads, to formation of CO2 + H2O., (b) CH 3 CH 2 CH 2 CH 2 CH 2 CH 3, , 3, , CH, |, OH, , CH, , 2, , C 1CHO, |, CH 3, , 4-Hydroxy-2-methylpent-2-en-1-al, , |, , Br, , 2 bromo 2 methyl butane, , 29., , 5, , 45. (b), , H3C, , CH3, , H3C, , C=C, H, , H, C=C, , H, , H, , CH3, , 46. (d) Alkenes having double bonds with two different, groups on each end of the double bond show, geometrical isomerism. A2b2c2, A2b2cd, A2bcde., , Aromatisation, , Br, Benzene, , 31., , (b), , 32., , (c), , 33., , (b), , 34., , (c), , Aromatisation is a process in which aromatic, compounds are formed from open chain compounds., During cracking higher hydrocarbons (liquid) are, converted to lower gaseous hydrocarbons., n-Alkanes on heating with anhydrous aluminium, chloride and hydrogen chloride gas isomerise to, branched chain alkanes., Methane reacts with steam at 1273 K in the presence, of nickel catalyst to form carbon monoxide and, dihydrogen. This method is used for industrial, preparation of dihydrogen gas., A conformation is defined as the relative arrangement, of atoms or groups around a central atom, obtained by, the free rotation of one part of the molecule with respect, to rest of the molecule. For a complete rotation of 360º,, one part may rotate through any degree say 0.1º, 0.5º,, 1º etc. giving rise to infinite number of relative, arrangements of group (atom) around a central atom,, keeping other part fixed., Spatial arrangements of atoms which can be converted, around a C – C single bond are called conformations, or conformers or rotamers., , 35., , (d), , 36., 37., 38., , (a), (d), (a) As predicted by the VSEPR model of electron pair, repulsion, the molecular geometry of alkenes, includes bond angles about each carbon in a double, bond of about 120°., (c) Double bond in between carbon-carbon is present in, alkenes whose general formula is CnH2n., , 39., , 40., , (d), , 6 5, 4, 3, H3C CH CH 2 CH, |, Cl, , (a), , 42., , (c), , 43., , 1, , 2, , 3, , Br, H3C, , C, , C, , C, , Br, CH3, Br, CH3, , 1,2-dibromopropene, , 2,3-dibromobut-2-ene, , 47. (b) The two isomers differ in the position of the double, bond so they are called position isomers., 48. (b) CH3CH2CH=CH2 CH3CH=CHCH3 (CH3)2C=CH2, 1-butene (i), , 2- butene (ii), (iii), (cis,- trans), , CH2= CH–CH2Cl, (a) Since b (from bromo) comes earlier in alphabetical order, than c (from chloro), the correct name should be 2, 3dibromo-1, 4-dichlorobutene-2 and not 1,4-dichloro2, 3-dibromobutene-2., , 2-methylpropene, (iv), , CH3, , cyclobutane (v), , methylcyclopropa ne (vi), , 49. (a) As sketched in the above question , C5H10 may be, monosubstituted (i) and (iv), disubstituted as in (ii),, (iii) and (v) and trisubstituted as in (vi), 50. (c) The condition for geometrical isomerism is, a, a, a, e, C=C, C, =, C, or, b, d, b, b, 51. (a), 52. (d) CH3 CH2 Br KOH, (alc), , CH 2 CH 2, , KBr H 2O, , 53. (c), , 54. (d), , Br, |, CH3– CH – CH2 – CH3 alc. KOH, 2-Bromobutane, +, , CH3–CH–CH2–CH3, +, , 2, 1, CH CH3, , IUPAC name : 5- chlorohex-2-ene, 41., , H, , C, , H3C, H, , –H, C=C, , H, CH3, , trans-2-butene, , 55. (c), , C2 H5 OH, , conc. H 2SO 4, KMnO4, , C2H 4, , H 2O, , Note : If ethyl alcohol is taken in excess and the reaction, is carried out at a temperature of 433-443 K diethyl, ether is formed., , 2C2 H5OH, (excess), , conc. H2SO4, 433 443 K, , C2H5OC2H5 H2O
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HYDROCARBONS, , n(CH 2, , 219, , CH 2 ), , 473K,1500atm, Trace of oxygen, , CH3CH CHCH3, , (–CH 2 – CH 2 –)n, , 2 butene, , The polythene manufactured in this way is called low, density of polythene., 69., , (a), , O, , H, , O3, , CH 2 = CH 2 + H - O - SO2 OH, , H, , C, , C, , O, , O, , Zn / H 2O, , 2CH3CHO, , H 2SO 4, , CH 3 CH 2, , OSO 2 OH, , CH 3CH 2 OSO 2 OH, , Ethyl hydrogen sulphate, , Addition of sulphuric acid takes place according to, Markownikoff’s rule. Alkanes do not absorb cold conc., H 2SO4 ., 70., , (c) According to Markownikoff’s rule, "in case of addition, of an unsymmetrical reagent (H– X), the positive part, get attached to the C which is least substituted or, which bears larger number of hydrogen atoms.", R, , H, |, , H, C=C, , +H X, , |, H, , Markownikoff’s rule is based on the stability of, carbocations (3° > 2° > 1° > methyl)., 71. (c) As per Markovnikoff’s law, the positive part (e.g. H, of HX) or the less negative part of the reagent adds, to that carbon atom of alkene which has more, number of hydrogen atoms (the rich gets richer). So, (c) is the correct option as the two carbons, containing the double bond have one H atom each, i.e. symmetric., 72. (d) Completing the sequence of given reactions,, CH 3 – CH, , CH CH 3, , 75. (d), , CH 2, , CH 2, , CH2 CH 2, |, |, + MnO 2, OH OH, , 76. (b) When unsymmetrical unsaturated hydrocarbon reacts, with unsymmetrical reagent, then negative part of, reagents attacks that carbon which has less H-atom., [Markownikoff's rule], , CH 3 CH, , CH 2, , 400 C, , HI, , Propene-2, , 77. (a), , CH 2, , CH 2, , Br2, , CH 3, , I, |, CH CH 3, , 2-Iodopropane, , CCl4, , CH 2 CH 2, |, |, Br, Br, 1, 2-dibromo ethane, , CH 2, , CH 2, , HOCl, , CH 2 CH 2, |, |, Cl, OH, , aq. NaHCO3, , O, CH3– CH, , CH – CH3, , KOH, , Glycol, , 78. (b), , O3, , OH, , KMnO4, , H, |, , R C C H, , H, , H, , 74. (a) The addition of HBr takes place according to antiMarkovnikoff’s rule in presence of peroxide for, unsymmetrical alkenes., The addition of HBr to symmetrical alkenes is not, affected by the presence or absence of peroxide., , CH 2OH, |, CH 2OH, , Zn / H 2O, , Glycol, , O, , O, , 79. (c) By adding bromine water to a solution, if the colour of, bromine water decolourise then the compound is, unsaturated. This is a confirmatory test for unsaturation., , ‘A’, (ozonide), , 2CH3CHO H 2 O ZnO, 'B ', , 73., , Thus ‘B’ is CH3CHO, Hence (d) is correct answer., (c) The given molecular formula suggests that the aldehyde, formed will be acetaldehyde hence the alkene will be, , 80. (b), , CH 3 CH, , CH 2, , H 2O, , Conc. H 2 SO4, Markonikoff's rule, , CH 3 CH CH3, |, OH, Isopropyl alcohol
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EBD_7207, HYDROCARBONS, , 220, , 81., , 82., 83., , (b) Greater the s-character of C, higher is its, electronegativity, shorter and stronger will be the bond, formed by it. Thus C C is the strongest bond., (b) General formula for alkynes is CnH2n–2, (c) When both double and triple bonds are present, then, triple bond is considered as the principal group., , 95., , (c), , 96., , (d), , CH3 CH CH C CH, 5, , 84., , (b), , 3, , 4, , 2, , 1, , Three alkynes are possible for the formula C5H8., CH3CH2CH2C CH, CH3, , 85., , 86., 87., 88., 89., 90., , CH3, CH3CH2C CCH3, (c) If both the double and triple bonds are present, the, compound is regarded as derivative of alkyne. Further, if double and triple bonds are at equidistance from, either side, the preference is given to double bond., (d) C2H2 and C3H4 are homologue because they differ by, – CH2 group. Both have triple bond in their molecule., (a), (d) In CH CH triple bond consists of one and two, bonds., (c) In C2H6, C – C bond length is 1.54 Å., (c) Acetylene molecule can be represented as,, , H, 91., , C, , C, , H, , H3 C C, , C, , C C, , (b), , CH, , dil H 2SO 4, HgSO 4 .60 C, , CH H 2 O, , 99., , (b), , CH HCl, , CH, , C, , CH3CH 2 C, , CH, , C C, , (a) As the number of bonds between carbon atoms, increases, electronegativity of that carbon also, increases due to increasing active power of electrons., Also sp hybrid is more electronegative than sp2 which, is more electronegative than sp 3 ( s character, decreases) Hence, option (a) is correct., (c) On heating ethylene chloride (1, 1 dichloro ethane), with alcoholic potash followed by sodamide alkyne is, obtained, alc.KOH, , R CH, , NaNH 2, , HCl, , Cl, Cl, , Na in, liquid NH3, , C – Na, , Na, liq. NH3, , CNa, , CHCl, , CH, , CH, , CCl R, , R C C R, , CH, , CH2, , CH +HCl, , CHCl, , CH3, CHCl2, , [AgNO3+NH4OH], , AgC CAg + NH4NO3, white ppt., , 102. (c), 104. (c), , 103. (d), CH3 C CH H 2O, , 1.54A, , R, , CH Cl, , 1, 1-dichloroethane, , +HCl, , 93., , R CH 2 CCl 2, , CH 2, , CH3CHO, , 100. (b) Alkynes having terminal –C, H react with Na in, liquid ammonia to yield H2 gas. CH3CH2C CH can, react with Na in liquid NH3 so the correct answer is (b)., , CH3, , (b), , 1.34A, , CHOH], , CH3 – CH, , (3) (4), , C, , [CH 2, , unstable, , CH, , 92., , 94., , Acetylene, , 1, H2 (g ), 2, 101. (a) Acetylene reacts with the other three as:, , (1) (2), (3) (4), No. of bonds : 4 2, 2 4, – 2, 2 –, Hybridisation : sp3 sp, sp sp3, Thus, butyne-2 has sp and sp3 hybridised carbon, atoms., 1.20A, , 98., , CH + Ca (OH)2, , HC, , Among isomeric alkanes, the straight chain isomer, has higher boiling point than the branched chain, isomer. The greater the branching of the chain, the, lower is the boiling point. Further due to the presence, of electrons, these molecules are slightly polar and, hence have higher boiling points than the, corrosponding alkanes., (a) Terminal alkyenes give a white precipitate easily on, reaction with ammonical silver nitrate solution., , CH3CH 2C, , So, it contains 3 and 2 bonds., (d) We know that carbon having, (i) 4 bonds correspond to sp3, (ii) 3 and 1 bond correspond to sp2, (iii) 2 and 2 bonds correspond to sp, (1) (2), , 97., , CaC2 + 2H2O, Calcium, carbide, , CH3, , C CH 2, |, OH, , 40% H 2SO4, 1% HgSO4, , Rearrangement, , CH3 C CH3, , O, Acetone, , CH3, , 105. (a) 3CH 3 C CH, , CH, , CH3, , C, , C, , CH, , CH, C, CH3, , Mesitylene or 1, 3, 5-trimethyl benzene
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HYDROCARBONS, , 106. (b), , CH, , 221, , CH H 2, , Acetyle, , 107. (d) Addition –, , Lindlar's, Catalyst, Pd. BaSO4, , CH, , CH 2, , Y, , CH 2, , Y, , Ethylene, , X, , 124. (a), Ni, , CH 3H 2, , +, , CH 3 CH 3, , m-isomer, (> 60%), , Substitution –, CH CH Na, , CH C Na, , 1, H2, 2, , Polymerization –, 3CH CH, , hot Cu tube, Polymerization, , C6 H 6, Benzene, , Y = –COOH because it is meta directing group while, –NH2, – OH and –Cl are o and p directing groups., 125. (a) – NO2 is a meta-directing group. As it is also a, deactivating group so no chance of introduction of, second – Br atom., 126. (c) Cl exhibits –I effect and +M effect., , Br, , 108. (b), , CH3C CH 2HBr, , |, , CH3 C CH3, , 127. (b), , Br, , 128. (d), , |, , 109. (b), 110. (b) Thin film of polyacetylene can be used as electrodes, in batteries. These films are good conductors, lighter, and cheaper than the metal conductors., 111. (b), 112. (d) Amongst all tropolone is a non-benzenoid aromatic, compound., 113. (c), 114. (c) Kekule in 1865 suggested a ring structure of benzene, in which the ring was composed of six carbon atoms,, each of which carries one atom of hydrogen. To satisfy, the fourth valency of the carbon atom, he suggested, three alternate double bonds., , Ni, Cyclohexane, , CH3, , 129. (c), , + CH3Cl, , anhydrous AlCl3, , HC, , CH, , + HCl, , 130. (b) Benzene can be obtained by polymerisation of, acetylene., Red hot tube, 500 C, , 3HC CH, , 131. (a) In electrophilic substitution reaction an electrophile, (in this case NO2 ) replaces another atom (in this case, H) from the substrate (benzene)., , NO 2, , H, CH, , HNO3 H 2SO 4, Nitrobenzene, , 132. (a) This is an example of Friedel - Craft alkylation., 133. (d), , C, H, , 115. (a), 116. (c) Benzene do not show addition reactions like other, unsaturated hydrocarbons. However it show, substitution reactions. Due to resonance all the C – C, bonds have the same nature, which is possible because, of the cyclic delocalisation of -electrons in benzene., Monosubstitution will give only a single product., 117. (a) In the benzene molecule all the six carbons are sp2, hybridised as each C has one double bond., 118. (c) They have a relatively high percentage of carbon., 119. (a), 120. (c), 121. (c), 122. (d) Benzene can be obtained by all the compounds given., 123. (a) This is an example of decarboxylation reaction., COONa, NaOH + CaO, , + 3H2, Benzene, , CH, HC, , X, , + Na2CO3, , 134. (c), , + O3, , O, , O, O, , O O, O, O, O O, , Benzene triozonide, , 135. (b) Friedel- Craft reaction occurs in presence of an, attacking reagent which is an electrophile (AlCl3)., , STATEMENT TYPE QUESTIONS, 136. (a) According to experimental evidences, electrophilic, substitution reactions are supposed to proceed via the, following three steps:, (1) Generation of the electrophile, (2) Formation of carbocation intermediate, (3) Removal of proton from the carbocation intermediate
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EBD_7207, HYDROCARBONS, , 222, , 137. (b), , H, , .., HO3SO - H + H - O, .. - NO 2, , +, , H-O, .. - NO2 + HSO 4, , H, +, , H O, .. NO 2, , H 2O, , NO 2, , In the process of generation of nitronium ion, sulphuric, acid serves as an acid and nitric acid as a base., 138. (a) -electrons of benzene rings are delocalised, throughout the molecule. This makes the molecule very, stable. The stability resists breaking of double bonds, for addition., 139. (b) Natural gas is found in upper strata during drilling of, oil wells. The gas after compression is known as, Compressed Natural Gas (CNG)., 140. (b), 141. (c) In methane carbon atom is sp3 hybridized., 142. (b) Dihydrogen gas adds to alkenes and alkynes in the, presence of finely divided catalysts like platinum,, palladium or nickel to form alkanes. These metals, adsorb dihydrogen gas on their surfaces and activate, the hydrogen-hydrogen bond. Platinum and palladium, catalyse the reaction at room temperature but relatively, higher temperature and pressure are required with, nickel catalyst., 143. (b) For statement (iii), Fluorination is too violent to be, controlled. For statement (iv), Iodination is very slow, and a irreversible reaction. It can be carried out in the, presence of oxidizing agents like HIO3 or HNO3, , 144. (c), 145. (b), , 146. (b), 147. (b), , I2, , CH3, , Nitronium ion, , Protonated, nitric acid, , CH 4, , 149. (b) For statement (ii), alkynes on reduction with Pd/C form, cis-alkenes. For statement (iv), Propyne on reduction, with Pd/C form propene. Propene does not show, geometrical isomerism. Only those compounds show, geometrical isomerism which have two different atoms, or groups attached to each carbon atom., , CH 3 I HI, , 5 HI HIO3 3I 2 3H 2O, Pyrolysis and cracking are same process., Rotation around a C-C single bond is not completely, free. It is hindered by a small energy barrier of 1-20 kJ, mol-1 due to weak repulsive interaction between the, adjacent bonds. Such a type of repulsive interaction is, called torsional strain., Eclipsed form has maximum torsional strain while, staggered form has the least., For statement (ii), Alkenes are also known as olefins., For statement (iv), Carbon–Carbon double bond in, alkene consists of one sigma and one pi bond., H, , C, , C, , H, , H, H, 148. (a) For statement (i), cis form of alkenes have significant, dipole moment whereas dipole moment of trans form is, almost zero. For statement (iv), due to different, arrangements of atoms or groups in space cis and trans, isomers differ in their properties like melting point,, boiling point, dipole moment, solubility etc., , C, , H, , H, , C, , H, , Propene, , 150. (d) Polynuclear hydrocarbons are not produced in human, body by any biochemical reaction as when they enter, into human body they undergo various biochemical, reactions which finally damage DNA and cause cancer., , MATCHING TYPE QUESTIONS, 151. (a) Among the infinite number of conformations in the, staggered conformation hydrogen atoms are as far as, apart as possible. While in eclipse conformation, hydrogen atoms are perfectly eclipsed., In skew conformation, hydrogen atoms are closer than, in staggered but away than in eclipsed conformation., 152. (a), , CH 2, , CH 2, , Pt/Pd/ Ni, , H2, , CH3 CH3, Ethane, , Ethene, , CH3 Cl H 2, , Zn, H, , CH 4, Methane, , Chloromethane, , CH3Br 2Na BrCH3, , dry ether, , CH3 CH3 2NaBr, , Bromomethane, , CH 3COO Na, , NaOH, , HCl, , Ethane, , CaO, , CH 4, , Na 2 CO3, , 153. (c), 154. (a), , CH H 2, , CH, , Pd/C, , Ethyne, , H, H C, H, , H, C H, , CH 2, , CH 2, , Ethene, , alc.KOH, , X, , H, , H, C =C, , H, , H, , (X Cl, Br, I), CH 2 Br CH 2 Br Zn, , H, , H C, H, , CH 2 CH 2, , ZnBr2, , H, , C H, OH, , Ethanol, , Conc.H 2SO 4, , CH 2, , CH 2 H2 O, , Ethene
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EBD_7207, HYDROCARBONS, , 224, , CH3CH2CH2CH =CH2, , 171. (a) Electrolysis of a concentrated aqueous solution of, either sodium or potassium salts of saturated moncarboxylic acids yields higher alkane at anode., Electrolyt ic, , 2RCOOK, , 2RCOO, , Oxidation, , 1-pentene (i), , CH3, CH 3 CH CH, , Cathode, , |, , 2-methyl-2-butene, (vi), , 172. (a) Given, alc. / KOH, , B, , C, , O3 / H 2 O, , CH 2 O, , Hydrocarbon, , Since hydrocarbon C give only CH2O, on ozonolysis,, C should be CH2 = CH2 hence going backward A, should be ethane. Thus the reactions are, Cl 2 / h, , CH 3CH 3, , CH 3CH 2Cl, , (A), , (B), , CH 2 CH 2, , O3 / H 2 O, , (C ), , 2-methyl-1-butene, (v), , CH 3 C CHCH 3, , (Kolbe's Method), Cl 2, , CH 3 CH 2 C CH 2, , CH3, , 2K, 2e –, 2K, 2K H2O 2KOH H2, , A, , |, , CH 2, , 3-methyl-1-butene, (iv), , 2RCOO 2e –, R — R 2CO 2, , At anode 2RCOO, At Cathode, , CH3, , |, , 2K, , Anode, , CH3CH2CH=CHCH3, 2- pentene, (cis,- trans) (ii), (iii), , alc., KOH, , ..., , HCHO, , 179. (b) Stability of an alkene depends upon the heat of, hydrogenation of an alkene. The heat of hydration is, the standard enthalpy change in hydrogenation of on, alkene. The lower the heat of hydrogenation of an, alkene higher will be stability., Order of stability Heat of hydrogenation, (kJ/mol), trans-2-butene, – 115.5, cis-2-butene, – 119.6 and, 1-butene, – 126.8 respectively., 180. (c) In compounds, CH3, , ( D), , CH3, , CH 3, , 173. (b), , CH 3 — C— CH 3, , first has more dipole moment than second., Therefore its boiling point will be higher. Melting point, depends on symmetry therefore I has higher melting, point than II. Steric crowding in I is more than II, therefore I is more stable than II., , CH 3, , 2,2-dimethyl Propane, , All hydrogen atoms are equivalent in 2,, 2-dimethylpropan e. So it forms only one, monochlorinated product., 174. (c) In the eclipsed conformation of ethane, the dihedral, angle between hydrogen atoms of adjacent methyl, groups is 0º., HH, , H H, , H, , H, H, , CHH— CH — CH — CH, , 175. (a) The bond is formed by the sideways overlapping of, two p-orbitals of the two carbon atoms., The molecular plane does not have any electron, density as the p-orbitals are perpendicular to the plane, containing the ethene molecule. The nodal plane in, the -bond of ethene is located in the molecular plane., 176. (c), 177. (a), , Cl, CH3, 1, , 178. (c), , 4, , C=C, 2, , 3, , CH3, , 182. (b), , CH3 — CH — CH — CH3, , 5, , CH2CH3, I, , Correct IUPAC name of above compound is trans-2chloro-3-iodo-2-pentene, C5H10 has 1º degree of unsaturation since the isomers, are acyclic, all of these are alkenes. For writing the, isomers, first introduce the double bond at different, possible positions, and then consider the possibility, of branching in the alkyl group., , H+/ Heat, , OH, , H, H, , H, H, , 181. (b), , CH3, , CH3, H / Heat, , CH3 — C = CH — CH3 + CH3 — CH — CH = CH2, 3 methyl, 2 methyl, butene-1 (20%), butene-2, (B), (80%), (A), , In this case dehydration is governed by Saytzeff’s, rule according to which hydrogen is preferentially, eliminated from the carbon atom with fewer number of, hydrogen atoms i.e., poor becomes poorer. Thus, 2methyl butene-2 is the major product., CH3, , CH3 — C = CH — CH3, (A), , HBr, dark, , in absence, of peroxide, , CH3, , (CH3)2 — CH — CH — CH3 + CH3 — C — CH2 – CH3, Br, (Minor), , Br, (Major)
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EBD_7207, HYDROCARBONS, , 226, , (vii), , CH3, |, HC C C CH 3, |, CH3, 3,3- Dimethylbut-1-yne, , 194. (a) Ethylene has restricted rotation, acetylene has no, rotation, hexachloroethane has more rotation than, ethylene but less than ethane because of greater size, of the substituent (chlorine) in hexachloroethane than, in ethane (substituent is hydrogen)., 195. (a) The acidity of acetylene or 1–alkyne can be explained, on the basis of molecular orbital concept according to, which formation of C—H bond in acetylene involves, sp-hybridised carbon atom. Now since s electrons are, closer to the nucleus than p electrons, the electrons, present in a bond having more s character will be, correspondingly more closer to the nucleus., Thus owing to high s character of the C—H bond in, alkynes (s = 50%), the electrons constituting this bond, are more strongly held by the carbon nucleus i.e., the, acetylenic carbon atom or the sp orbital acts as more, electronegative species than the sp2 and sp3 with the, result the hydrogen present on such a carbon atom, ( C—H) can be easily removed as a proton., 196. (b), 197. (d) Br2 in CCl4 (a), Br2 in CH3 COOH (b) and alk. KMnO4, (c) will react with all unsaturated compounds, i.e., 1, 3, and 4 while ammonical AgNO3 (d) reacts only with, terminal alkynes, i.e., 3 and hence 3 can be distinguished, from 1, 2 and 4 by. ammonical AgNO3 (d)., 198. (c) This reaction occurs according to Markownikoff’s rule, which states that when an unsymmetrical alkene, undergo hydrohalogenation, the negative part goes, to that C-atom which contain lesser no. of H-atom., CH3 CH 2 C, , CH HCl, CH3 CH 2, , C CH 2, |, Cl, , I, |, HI, CH3 CH 2 C CH3, |, Cl, 199. (a) Only terminal alkynes show acidic nature. Ethyne is, more acidic than propyne. But-2-yne is not acidic as it, does not contain any hydrogen attached to sp, hybridised carbon., 200. (a), , 201. (c) Due to the maximum percentage of s character (50%),, the sp hybridised orbitals of carbon atoms in ethyne, molecules have highest electronegativity; hence, these, attract the shared electron pair of the C-H bond of, ethyne to a greater extent than that of the sp 2, hybridised orbitals of carbon in ethene and the sp3, hybridised orbital of carbon in ethane. Thus in ethyne,, hydrogen atoms can be liberated as protons more easily, as compared to ethene and ethane., 202. (c) In arenes, electrons are delocalised, hence arenes do, not undergo addition reactions easily. Aromatic, compounds (Arenes) are highly stable and show, resonance. eg. Benzene is the simplest example., 203. (c) Compounds having same molecular formula but, different functional groups in their molecules are called, functional isomers., 204. (d) In benzene due to delocalisation of - electrons, all, the C – C bond lengths are equal as each C– C bond, has some double bond character and thus the bond, length is between single and double bond, i.e., between, C2H6 and C2H4., 205. (a), 206. (c) Benzene has 12 and 3 bonds., Ratio of bonds to bonds = 12/3 = 4, 207. (d) Presence of 6p orbitals, each containing one, unpaired electron, in a six membered cyclic structure, is in accordance with Huckel rule of aromaticity., 208. (a), Cl, , + HCl, , Subtitution, Cl2, anly AlCl3, Addition, 3Cl2, sunlight, , Cl, , Cl, Cl, , Benzene, Cl, , Cl, Cl, , 209. (b) –Cl group is o-, p-directing due to +R effect ; however, it is deactivating due to strong –I effect of Cl (difference, from other o-, p-directing groups which are activating)., The net result is that chlorobenzene undergoes o, psubstitution, but with difficulty, 210. (b) C6 H5COOH NaOH C6 H 5COONa H 2O, (X), , C6 H5OH Zn, , (Y), , distill, , C6 H 6, , ZnO, , 211. (a) Huckel rule is not obeyed. It has only four electrons., Further it does not have continous conjugation.
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14, ENVIRONMENTAL CHEMISTRY, FACT / DEFINITION TYPE QUESTIONS, 1., , 2., , 3., , 4., , 5., , 6., , 7., , 8., , The type of pollution caused by spraying of DDT is, (a) air and soil, (b) air and water, (c) air, (d) air, water and soil, What is DDT among the following ?, (a) Greenhouse gas, (b) A fertilizer, (c) Biodegradable pollutant, (d) Non-biodegradable pollutant, The uppermost region of the atmosphere is called, (a) Ionosphere, (b) Stratosphere, (c) Troposphere, (d) Exosphere, Which of the following is the coldest region of atmosphere ?, (a) Thermosphere, (b) Mesosphere, (c) Troposphere, (d) Stratosphere, The region which is greatly affected by air pollution is, (a) Thermosphere, (b) Stratosphere, (c) Troposphere, (d) Mesosphere, The region containing water vapour is, (a) thermosphere, (b) stratosphere, (c) troposphere, (d) mesosphere, High concentration of which of the following in atmosphere, leads to stiffness of flower buds which eventually fall off, from plants?, (a) NO2, (b) SO2, (c) CFC, (d) Smog, The irritant red haze in the traffic and congested places, is due to presence of which of the following ?, (i) Oxides of sulphur, (ii) Oxides of nitrogen, (iii) Carbon dioxide, (iv) Mists, smoke and dust, (v) Smog, (a) (i), (iv) and (v), (b) (iii) only, (c) (ii) only, (d) (ii) and (v), , 9., , 10., , 11., , 12., , 13., , 14., , 15., , 16., , The quantity of CO2 in atmosphere is, (a) 3.34%, (d) 6.5%, (c) 0.034%, (d) 0.34%, The substance which is not regarded as a pollutant?, (a) NO2, (b) CO2, (c) O3, (d) Hydrocarbons, Which of the following is/are the hazardous pollutant(s), present in automobile exhaust gases?, (i) N2, (ii) CO, (iii) CH4, (iv) Oxides of nitrogen, (a) (ii) and (iii), (b) (i) and (ii), (c) (ii) and (iv), (d) (i) and (iii), The gas emitted by supersonic jet planes that slowly, depletes the concentration of ozone layer is, (a) CO, (b) NO, (c) SO2, (d) O2, Carbon monoxide (CO) is harmful to man because, (a) it forms carbolic acid, (b) it generates excess CO2, (c) it is carcinogenic, (d) it competes with O2 for haemoglobin, Increase in global temperature increases the incidence of, which of the following infectious disease(s), (i) Sleeping sickness, (ii) Yellow fever, (iii) Malaria, (iv) Dengue, (a) (ii) only, (b) (i) and (ii), (c) (iii) and (iv), (d) (i), (ii), (iii) and (iv), The green house effect is caused by, (a) CO2, (b) NO2, (c) NO, (d) CO, Which is related to ‘Green House Effect’?, (a) Farming of Green plants, (b) Farming of Vegetables in Houses, (c) Global Warming, (d) Biodegradable pollutant
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EBD_7207, ENVIRONMENTAL CHEMISTRY, , 228, , 17., , 18., , 19., , 20., , 21., , 22., , 23., , 24., , Green house gases, (a) allow shorter wavelength to enter earth's atmosphere, while doesn't allow longer wavelength to leave the, earth's atmosphere., (b) allow longer wavelength to enter earth atmosphere, while doesn't allow shorter wavelength to leave the, surface, (c) don't have wavelength specific character., (d) show wavelength specific behaviour near the earth, while far from earth these have wavelength independent behaviour., Today the concentration of green house gases is very high, because of, (a) use of refrigerator, (b) increased combustion of oils and coal, (c) deforestation, (d) All of the above, The greenhouse effect is because of the, (a) presence of gases, which in general are strong infrared, absorbers, in the atmosphere, (b) presence of CO2 only in the atmosphere, (c) pressure of O3 and CH4 in the atmosphere, (d) N2O and chlorofluorohydrocarbons in the atmosphere, The greenhouse gas is, (a) CO2, (b) SO2, (c) N2, (d) H2S, Which of the following gases is not a green house gas?, (a) CO, (b) O3, (c) CH4, (d) H2O vapour, Which of the following strategy is not a correct approach, to reduce global warming ?, (a) Reducing the green house gas emission by limiting, the use of fossil fuels, (b) Increase the vegetation cover particularly the forest, for photosynthetic utilization of CO2, (c) Minimizing the use of nitrogen fertilizers in agriculture, for reducing N2O emission, (d) Increasing the use of air conditioners, refrigeration, unit and production of plastic foams and propellants, in aerosal spray cans, The substance having the largest concentration in acid, rain, (a) H2CO3, (b) HNO3, (c) HCl, (d) H2SO4, When rain is accompanied by a thunderstorm, the collected, rain water will have a pH value, (a) slightly lower than that of rain water without, thunderstorm, (b) slightly higher than that when the thunderstorm is not, there, (c) uninfluenced by occurrence of thunderstorm, (d) which depends upon the amount of dust in air, , 25., , 26., , 27., , 28., , 29., , 30., , 31., , 32., , 33., , 34., , 35., , Acid rain is due to, (a) CH3, (b) N2O5, (c) SO2 and NO2, (d) C2H5OH, The pH of normal rain water is, (a) 6.5, (b) 7.5, (c) 5.6, (d) 3.5, Which of the following statements is incorrect ?, (a) Smoke particulates consist of solid or mixture of, solid and liquid particles formed during combustion, of organic matter., (b) Herbicides and insecticides that miss their target and, travel through air form mists., (c) Organic solvents, metals and metallic oxides form, fume particles, (d) None of these, Which of the following green house gas is released in paddy, field ?, I. CFCs, II. CH4, III. SO2, (a) Only I, (b) Only II, (c) Only III, (d) I and II, Photochemical smog is due to the presence of, (a) oxides of sulphur, (b) oxides of nitrogen, (c) oxides of carbon, (d) lead, The secondary precursors of photochemical smog are, (a) SO2 and NO2, (b) SO2 and hydrocarbons, (c) NO2 and hydrocarbons, (d) O3 and PAN, The main element of smog is, (a) O3 and PAN, (b) O3, (c) PAN, (d) PPN and PBN, Classical smog occurs in places of, (a) excess SO2, (b) low temperature, (c) high temperature, (d) excess NH3, The smog is essentially caused by the presence of, (a) Oxides of sulphur and nitrogen, (b) O2 and N2, (c) O2 and O3, (d) O3 and N2, Air pollution causing photochemical oxidants production, include, (a) Carbon monoxide, sulphur dioxide, (b) Nitrous oxide, nitric acid fumes, nitric oxide, (c) Ozone, peroxyacetyl nitrate, aldehydes, (d) Oxygen, chlorine, fuming nitric acid, Photochemical smog formed in congested metropolitan cities, mainly consists of, (a) ozone, peroxyacetyl nitrate and NOx, (b) smoke, peroxyacetyl nitrate and SO2, (c) hydrocarbons, SO2 and CO2, (d) hydrocarbons, ozone and SOx
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ENVIRONMENTAL CHEMISTRY, , 36., , 37., , 38., , 39., , 40., , 41., , 42., , 43., , 44., , 45., , In almost all Indian metropolitan cities like Delhi, the major, atmospheric pollutant(s) is/are, (a) suspended particulate matter (SPM), (b) oxides of sulphur, (c) carbon dioxide and carbon monoxide, (d) oxides of nitrogen, The non-viable particulate among the following is, (a) Dust, (b) Bacteria, (c) Moulds, (d) Fungi, Photochemical smog occurs in warm, dry and sunny climate., One of the following is not amongst the components of, photochemical smog, identify it., (a) NO2, (b) O3, (c) SO2, (d) Unsaturated hydrocarbon, The pollutants which came directly in the air from sources, are called primary pollutants. Primary pollutants are, sometimes converted into secondary pollutants. Which of, the following belongs to secondary air pollutants?, (a) C O, (b) Hydrocarbon, (c) Peroxyacetyl nitrate (d) NO, The main element of smog is, (a) O3 and PAN, (b) O3, (c) PAN, (d) Both (a) and (b), Which of the following statements is not true about classical, smog?, (a) Its main components are produced by the action of, sunlight on emissions of automobiles and factories., (b) Produced in cold and humid climate., (c) It contains compounds of reducing nature., (d) It contains smoke fog and sulphur dioxide, Which of the following statements about photochemical, smog is wrong?, (a) It has high concentration of oxidising agents, (b) It has low concentration of oxidising agent, (c) It can be controlled by controlling the release of NO2,, hydrocarbons ozone, etc., (d) Plantation of some plants like pinus helps in controlling, photochemical smog., Select the process that does not add particulate materials to, air., (a) Use of air conditioner, (b) Burning of fosssil fuels, (c) Paper industry, (d) Incomplete combustion of coal, The biggest particulate matter is, (a) HNO3 droplets, (b) Soot, (c) H2SO4 droplets, (d) Fly ash, The viable particulate among the following is, (a) Fumes, (b) Algae, (c) Smoke, (d) Mist, , 229, , 46. The aromatic compounds present as particulates are, (a) Polycyclic aromatic hydrocarbons, (b) Benzene, (c) Toluene, (d) Nitrobenzene, 47. Which of the following can control the photochemical, smog ?, (A) Use of catalytic converters in automobiles., (B) Plantation of trees like pinus, pyrus vitis etc., (C) Using less sulphur containing fossil fuels., (a) A and C, (b) B, (c) A and B, (d) A, B and C, 48. The gas responsible for ozone depletion :, (a) NO and freons, (b) SO2, (c) CO2, (d) CO, 49. Identify the incorrect statement from the following :, (a) Ozone absorbs the intense ultraviolet radiation of the, sun., (b) Depletion of ozone layer is because of its chemical, reactions with chlorofluoro alkanes., (c) Ozone absorbs infrared radiation., (d) Oxides of nitrogen in the atmosphere can cause the, depletion of ozone layer., 50. Identify the wrong statement in the following:, (a) Chlorofluorocarbons are responsible for ozone layer, depletion., (b) Greenhouse effect is responsible for global warming., (c) Acid rain is mostly because of oxides of nitrogen and, sulphur., (d) Ozone layer does not permit infrared radiation from, the sun to reach the earth., 51. Which of the following chemical, harmful to ozone, is, released by chlorofluoro carbon?, (a) Sulphur dioxide, (b) Fluorine, (c) Chlorine, (d) Nitrogen dioxide, 52. In Antarctica ozone depletion is due to the formation of, following compound, (a) acrolein, (b) peroxyacetyl nitrate, (c) SO2 and SO3, (d) chlorine nitrate, 53. Depletion of ozone layer causes, (a) breast cancer, (b) blood cancer, (c) lung cancer, (d) skin cancer, 54. Select the one that has an adverse effect on ozone layer., (a) Carbon dioxide, (b) Chlorofluorocarbons, (c) Soil, (d) Dust particles, 55. Ozone hole refers to, (a) Increase in concentration of ozone, (b) Hole in ozone layer, (c) Reduction in thickness of ozone layer in troposphere, (d) Reduction in thickness of ozone layer in stratsophere
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EBD_7207, ENVIRONMENTAL CHEMISTRY, , 230, , 56., , 57., , 58., , 59., , 60., , 61., , 62., , 63., , 64., , Which of the following statements is wrong?, (a) Ozone is not responsible for green house effect., (b) Ozone can oxidise sulphur dioxide present in the, atmosphere to sulphur trioxide., (c) Ozone hole is thinning of ozone layer present in, stratosphere., (d) Ozone is produced in upper stratosphere by the action, of UV rays on oxygen., Which of the following statements is correct?, (a) Ozone hole is a hole formed in stratosphere from which, ozone oozes out., (b) Ozone hole is a hole formed in the troposphere from, which ozone oozes out., (c) Ozone hole is thinning of ozone layer of stratosphere, at some places., (d) Ozone hole means vanishing of ozone layer around, the earth completely., Ozone is an important constituent of stratosphere because, it, (a) Destroys bacteria which are harmful to human life, (b) Prevents the formation of smog over large cities, (c) Removes poisonous gases of the atmosphere by, reacting with them, (d) Absorbs ultraviolet radiation which is harmful to, human life, The gas(es) not responsible for ozone depletion :, (a) NO and freons, (b) SO2, (c) CO2, (d) Both (b) and (c), What is the concentration of dissolved oxygen in cold, water ?, (a) 5 ppm, (b) 10 ppm, (c) 200, 000 ppm, (d) 100 ppm, Water pollution is caused by, (a) pesticides, (b) SO2, (c) O2, (d) CO2, Minamata disease of Japan is due to pollution of, (a) Aresenic, (b) Lead, (c) Cynide, (d) Mercury, The high amount of E. coli in water is the indicator of, (a) hardness of water, (b) industrial pollution, (c) sewage pollution, (d) presence of chlorine in water, A lake with an inflow of domestic sewage rich in organic, waste may result in, (a) drying of the lake very soon due to algal bloom, (b) an increase production of fish due to lot of nutrients, (c) death of fish due to lack of oxygen, (d) increased population of aquatic food web organisms, , 65., , 66., , 67., , 68., , 69., , 70., , 71., , 72., , 73., , 74., , Which of the following does not occur when the sewage is, discharged into water ?, (a) Increase in O2, (b) Cyanophycean blooms occur, (c) Depletion of O2 layers, (d) Eutrophication, Which of the following metal is a water pollutant and causes, sterility in human being, (a) As, (b) Mn, (c) Mg, (d) Hg, Sewage mostly constitutes, (a) Non-biodegradable pollutants, (b) Biodegradable pollutants, (c) Effluents, (d) Air pollutants, Sewage containing organic waste should not be disposed, in water bodies because it causes major water pollution., Fishes in such a polluted water die because of, (a) large number of mosquitoes, (b) increase in the amount of dissolved oxygen, (c) decrease in the amount of dissolved oxygen in water, (d) clogging of gills by mud, Sewage water is purified by, (a) aquatic plants, (b) microoganisms, (c) light, (d) fishes, Water is often treated with chlorine to, (a) remove hardness, (b) increase oxygen content, (c) kill germs, (d) remove suspended particles, Which causes death of fish in water bodies polluted by, sewage?, (a) Foul smell, (b) Pathogens, (c) Herbicides, (d) Decrease in D.O., B.O.D. test or biochemical oxygen demand test is made for, measuring, (a) air pollution, (b) water pollution, (c) noise pollution, (d) soil pollution, Brewery and sugar factory waste alters the quality of a, water body by increasing, (a) temperature, (b) turbidity, (c) pH, (d) COD and BOD, Which one of the following statement is not true ?, (a) pH of drinking water should be between 5.5 – 9.5., (b) Concentration of DO below 6 ppm is good for the, growth of fish., (c) Clean water would have a BOD value of less than 5, ppm., (d) Oxides of sulphur, nitrogen and carbon are the most, widespread air pollutant.
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ENVIRONMENTAL CHEMISTRY, , 75., , 76., , 77., , 78., , 79., , 80., , 81., , 82., , 83., , 84., , 85., , Limit of BOD prescribed by Central pollution Control Board, for the discharge of industrial and municipal waste waters, into natural surface waters, is, (a) < 100 ppm, (b) < 30 ppm, (c) < 3.0 ppm, (d) < 10 ppm, Biochemical Oxygen Demand, (BOD) is a measure of organic, material present in water. BOD value less than 5 ppm, indicates a water sample to be __________., (a) rich in dissolved oxygen, (b) poor in dissolved oxygen, (c) highly polluted, (d) not suitable for aquatic life, Phosphate fertilizers when added to water leads to, (a) Increased growth of decomposers, (b) Reduced algal growth, (c) Increased algal growth, (d) Nutrient enrichment (eutrophication), BOD of pond is connected with, (a) Microbes & organic matter, (b) Organic matter, (c) Microbes, (d) None of these, The maximum prescribed concentration of cadmium in, drinking water in ppm is, (a) 0.05, (b) 3, (c) 2, (d) 0.005, Excess nitrate in drinking water can cause, (a) methemoglobinemia (b) kidney damage, (c) liver damage, (d) laxative effect, Eutrophication causes reduction in, (a) dissolved oxygen, (b) nutrients, (c) dissolved salts, (d) All of the above, Water pollution is caused by, (a) pesticides, (b) fly ash, (c) auto exhausts, (d) aeroplanes, Which causes death of fishes in water bodies polluted by, sewage?, (a) Foul smell, (b) Pathogens, (c) Clogging of gills by silt, (d) Decrease in D.O., Chief source of soil and water pollution is, (a) mining, (b) agro industry, (c) thermal power plant (d) All of the above, What is DDT among the following ?, (a) Greenhouse gas, (b) A fertilizer, (c) Biodegradable pollutant, (d) Non-biodegradable pollutant, , 231, , 86. The quantity of DDT in food chain, (a) decreases, (b) remains same, (c) increases, (d) changes, 87. The effect of polluted water on soil is, that, (a) it decreases fertility, (b) it contaminates ground water, (c) it renders soil acidic or basic, (d) all of the above, 88. Soil is polluted by, I. pesticides, II. synthetic fertilizers, III. green manure, Choose the correct option., (a) I and III, (b) I and II, (c) II and III, (d) I, II and III, 89. Which of the following trophic level has least concentration, of toxins deposition ?, (a) Aquatic plant, (b) Small fish, (c) Human being, (d) Largest fish, 90. Green chemistry means such reactions which :, (a) produce colour during reactions, (b) reduce the use and production of hazardous chemicals, (c) are related to the depletion of ozone layer, (d) study the reactions in plants, 91. Which of the following practices will not come under green, chemistry?, (a) If possible, making use of soap made of vegetable oils, instead of using synthetic detergents., (b) Using H2O2 for bleaching purpose instead of using, chlorine based bleaching agents., (c) Using bicycle for travelling small distances instead of, using petrol/ diesel based vehicles., (d) Using plastic cans for neatly storing substances., 92. “Reducing potentially hazardous waste through smarter, production”., This represents a great step forward for, (a) green revolution, (b) green chemistry, (c) industrial revolution (d) green biotechnology, 93. Use of which of the following solvent in dry cleaning will, result in less harm to ground water ?, (a) Cl2C = CCl2, (b) Liquid CO2, (c) H2O2, (d) None of these, 94. Synthesis of ethanal commercially from which of the, following reagent is the part of green chemistry ?, (a) CH3 CH2OH, (b) CH2 = CH2, (c) HC CH, (d) All of these
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EBD_7207, ENVIRONMENTAL CHEMISTRY, , 232, , STATEMENT TYPE QUESTIONS, 95., , 96., , 97., , 98., , 99., , Which of the following sequence of T and F is correct for, given statements. Here T stands for True statement and, F stands for False statement., (i) Troposphere is the lowest region of atmosphere in, which the human beings along with other organisms, live., (ii) Troposphere extends up to the height of 10 km from, sea level., (iii) Stratosphere lies above troposphere, between 10, and 20 km above sea level., (iv) Troposphere contains much little water vapour,, dinitrogen, dioxygen and ozone, (v) Stratosphere contains ozone, and cloud formation, also takes place in this region., (a) TTTTT, (b) TFTFF, (c) TTFFF, (d) TFTFT, Which of the following statement(s) is / are correct ?, (i) Sulphuric acid, nitric acid as well as ammonium salts, are components of acid rain., (ii) Formation of acid rain can be reduced by using less, sulphur content fossil fuels for power plants and, industries., (iii) Catalytic converters must be used in cars to reduce, the harmful effect of exhaust., (iv) Main component of catalytic converter is ceramic, honey comb coated with metals like – Au, Ag, Pt etc., (a) (i), (ii) and (iii), (b) (ii) and (iii), (c) (ii), (iii) and (iv), (d) (i), (ii), (iii) and (iv), Which of the following statement(s) is/are correct ?, (i) Classical smog is a mixture of smoke, fog and, sulphur dioxide., (ii) Classical smog is also called oxidising smog, (iii) Hydrocarbons, NO2 and PAN are components of, photochemical smog., (a) (i) and (iii), (b) (i) and (ii), (c) (iii) only, (d) (i), (ii) and (iii), Which of the following statements are not correct?, (i) F– ion concentration above 2ppm causes brown, mottling in teeth., (ii) Excessive F– (over 10 ppm) causes harmful effect to, bones and teeth., (iii) Excessive lead in drinking water causes disease, methemoglobinemia, (iv) Deficiency of sulphate in drinking water causes, laxative effect., (a) (ii) and (iv), (b) (ii) and (iii), (c) (ii), (iii) and (iv), (d) (iii) and (iv), Which of the following statement(s) is/are true about, waste recycling ?, (i) Clothes can be made from recycled plastic waste., (ii) Fuel that has high octane rating and contains no, lead can be obtained from plastic waste., , (iii) Technology has now been developed to produce, electricity from the garbage., (a) (ii) only, (b) (ii) and (iii), (c) (iii) only, (d) All of these, , MATCHING TYPE QUESTIONS, 100. Match the columns, Column - I, Column - II, (A) Concentration of, (p) 6 ppm, dissolved oxygen in, cold water, (B) Concentration of, (q) 17 ppm, dissolved oxygen, below which growth, of fish gets inhibited, (C) BOD value of clean (r) 5 ppm, water, (D) BOD value of, (s) 10 ppm, polluted water., (a) A – (s), B – (s), C – (q), D – (p), (b) A – (p), B – (q), C – (r), D – (s), (c) A – (s), B – (p), C – (r), D – (q), (d) A – (p), B – (s), C – (q), D – (r), 101. Match the columns, Column I, Column II, (A) Acid rain, (p) CHCl2 – CHF2, (B) Photochemical smog (q) CO, (C) Combination with, (r) CO2, haemoglobin, (D) Depletion of ozone, (s) SO2, layer, (t) Unsaturated, hydrocarbons, (a) A – (r, s), B – (t, s), C – (q), D – (p), (b) A – (r), B – (s), C – (q), D – (p), (c) A – (t,s), B – (s), C – (q), D – (r), (d) A – (r), B – (t), C – (q), D – (p), 102. Match the columns, Column-I, Column-II, (A) Oxides of sulphur, (p) Global warming, (B) Nitrogen dioxide, (q) Damage to kidney, (C) Carbon dioxide, (r) ‘Blue baby’ syndrome, (D) Nitrate in drinking, (s) Respiratory diseases, water, (E) Lead, (t) Red haze in traffic and, congested areas, (a) A – (t), B – (p), C – (r), D – (s), E – (q), (b) A – (s), B – (t), C – (p), D – (r), E – (q), (c) A – (s), B – (q), C – (p), D – (t), E – (r), (d) A – (q), B – (s), C – (t), D – (r), E – (p)
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ENVIRONMENTAL CHEMISTRY, , 103. Match the columns, Column-I, (A) Nitrous oxide, from car exhausts, (B) Chlorofluorocarbon, (CFCs), (C) Methane, (D) Ozone (O3), , 233, , Column-II, (p) Secondary pollutant, , (q) Combustion of fossil, fuels, wood, etc, (r) Denitrification, (s) Refrigerators, aerosol,, sprays, (E) Carbon dioxide, (t) Cattle, rice fields, toilets., (a) A – (r), B – (s), C – (t), D – (p), E – (q), (b) A – (t), B – (p), C – (r), D – (s), E – (q), (c) A – (s), B – (t), C – (p), D – (q), E – (r), (d) A – (p), B – (r), C – (s), D – (t), E – (q), 104. Match the columns, Column-I, Column-II, (A) Releasing gases to the (p) Water pollution, atmosphere after, burning waste material, containing sulphur, (B) Using carbamates as (q) Photochemical smog,, pesticides, damage to plant life,, corrosion to building, material, induce, breathing problems,, water pollution, (C) Using synthetic, (r) Damaging ozone layer, detergents for washing, clothes, (D) Releasing gases, (s) May cause nerve, produced by, diseases in human, automobiles and, factories in the, atmosphere., (E) Using chlorofluoro(t) Classical smog, acid rain,, carbon compounds, water pollution, induce, for cleaning computer, breathing problems,, parts, damage to buildings,, corrosion of metals., (a) A – (t), B – (s), C – (p), D – q, E – (r), (b) A – (s), B – (t), C – (q), D – (p), E – (r), (c) A – (q), B – (t), C – (r), D – (p), E – (s), (d) A – (r), B – (s), C – (p), D – (q), E – (t), 105. Match the columns, Column I, Column II, (A) Phosphate fertilisers (p) BOD level of water, in water, increases, (B) Methane in air, (q) Acid rain, (C) Synthetic detergents (r) Global warming, in water, (D) Nitrogen oxides in air (s) Eutrophication, , (a), (b), (c), (d), , A – (p,s), B – (r), C – (p), D – (q), A – (p), B – (s), C – (r), D – (q), A – (s), B – (r), C – (q), D – (p), A – (p), B – (q), C – (s), D – (r), , ASSERTION-REASON TYPE QUESTIONS, Directions : Each of these questions contain two statements,, Assertion and Reason. Each of these questions also has four, alternative choices, only one of which is the correct answer. You, have to select one of the codes (a), (b), (c) and (d) given below., (a) Assertion is correct, reason is correct; reason is a correct, explanation for assertion., (b) Assertion is correct, reason is correct; reason is not a, correct explanation for assertion, (c) Assertion is correct, reason is incorrect, (d) Assertion is incorrect, reason is correct., 106. Assertion : Uncatalysed oxidation of sulphur dioxide is a, slow process., Reason : Particulate matter in polluted air catalyses the, oxidation of sulphur dioxide., 107. Assertion : Dinitrogen and dioxygen do not react with, each other at a normal temperature., Reason : At high altitudes dinitrogen combines with, dioxygen to form oxides of nitrogen, 108. Assertion : CO2 causes green house effect., Reason : Other gases do not show such effect., 109. Assertion : Green house effect was observed in houses, used to grow plants and these are made of green glass., Reason : Green house name has been given because glass, houses are made of green glass., 110. Assertion : The pH of acid rain is less than 5.6., Reason : Carbon dioxide present in the atmosphere, dissolves in rain water and forms carbonic acid., 111. Assertion : Photochemical smog is oxidising in nature., Reason : Photochemical smog contains NO2 and O3, which, are formed during the sequence of reactions., 112. Assertion : Suspended particulate matter (SPM) is an, important pollutant released by diesel vehicles., Reason : Catalytic converters greatly reduce pollution, caused by automobiles., 113. Assertion : Carbon dioxide is one of the important, greenhouse gases., Reason : It is largely produced by respiratory function of, animals and plants., 114. Assertion : Ozone is destroyed by solar radiation in upper, stratosphere., Reason : Thinning of the ozone layer allows excessive UV, radiations to reach the surface of earth., 115. Assertion : Excessive use of chlorinated synthetic, pesticides causes soil and water pollution., Reason : Such pesticides are non-biodegradables.
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EBD_7207, 234, , 116. Assertion : If BOD level of water in a reservoir is less than, 5 ppm it is highly polluted., Reason : High biological oxygen demand means low activity, of bacteria in water., 117. Assertion : Eutrophication shows increase in productivity, in water., Reason : With increasing eutrophication, the diversity of, the phytoplankton increases., 118. Assertion : The F – ions make the enamel on teeth much, harder., Reason : F– ions converts hydroxyapatite, [3(Ca(PO4)2 Ca(OH)2] into fluorapatite [3(Ca3(PO4)2. CaF2]., , CRITICAL THINKING TYPE QUESTIONS, 119. In which of the following regions hydrogen and helium are, found ?, (a) Stratosphere, (b) Mesosphere, (c) Exosphere, (d) Troposphere, 120. Which one of the following pairs is mismatched ?, (a) Fossil fuel burning, release of CO2, (b) Nuclear power, radioactive wastes, (c) Solar energy, Greenhouse effect, (d) Biomass burning, release of CO2, 121. Which of the following acts as a sink for CO?, (a) Plants, (b) Haemoglobin, (c) Microorganisms present in the soil, (d) Oceans, 122. How many time oxyhaemoglobin is less stable than, carboxyhaemoglobin?, (a) 50, (b) 200, (c) 500, (d) 300, 123. Dinitrogen and dioxygen are main constituents of air but, these do not react with each other to form oxides of nitrogen, because __________., (a) the reaction is endothermic and requires very high, temperature., (b) the reaction can be initiated only in presence of a, catalyst., (c) oxides of nitrogen are unstable., (d) N2 and O2 are unreactive, 124. SO2 is one of the air pollutants. SO2, (a) is a lung irritant, (b) dissolves in water to form acid rain, (c) both (a) and (b), (d) none of the above, 125 The greatest affinity for haemoglobin is shown by which, of the following :, (a) NO, (b) CO, (c) O2, (d) CO2, , ENVIRONMENTAL CHEMISTRY, , 126. Which pollutant is harmful for ‘Taj Mahal’?, (a) Hydrogen, (b) O2, (c) SO2, (d) Chlorine, 127. The beauty of Taj Mahal is endangered due to, (a) degradation of marble due to high temperature, (b) discharge of industrial waste in Yamuna river, (c) air pollutants released from oil refinery, (d) riparian erosion, 128. Acid rain is caused by or recent reports of acid rain in some, industrial cities are due to the effect of atmospheric pollution, by, (a) excessive release of CO2 by burning of fuels like wood, and charcoal, cutting of forests and increased animal, population, (b) excessive release of NO2 and SO2 in atmosphere by, burning of fossil fuel, (c) excessive release of NH3 by industrial plants and coal, gas, (d) excessive release of CO in atmosphere by incomplete, combustion of coke, charcoal and other carbonaceous, fuel in paucity of oxygen., 129. Which of the following is the major cause of global, warming?, (a) re-radiation of U.V. rays by CO2 and H2O, (b) re-radiation of I.R. rays by CO2 and H2O, (c) re-radiation of I.R. rays by O2 and N2, (d) re-radiation of U.V. rays by O2 and N2, 130. Formation of London smog takes place in, (a) winter during day time, (b) summer during day time, (c) summer during morning time, (d) winter during morning time, 131. The false statement among the followings :, (a) The average residence time of NO is one month, (b) Limestone acts as a sink for SOx, (c) SOx can be removed from flue gases by passing, through a solution of citrate ions, (d) Ammonia acts as a sink for NOx, 132. Which of the following statements about polar stratosphere, clouds (PSCs) is not correct?, (a) PSCs do not react with chlorine nitrate and HCl, (b) Type I clouds are formed at about –77ºC and contain, solid HNO3 . 3H2O, (c) Type II clouds are formed at about –85ºC and contain, some ice, (d) A tight whirlpool of wind called Polar Vortex is formed, which surrounds Antarctica, 133. Which of the following is/are formed when ozone reacts, with the unburnt hydrocarbons in polluted air ?, (i) Formaldehyde, (ii) Acrolein, (iii) Peroxyacetyl nitrate (iv) Formic acid, (a) (i) and (iv), (b) (ii) only, (c) (iii) only, (d) (i), (ii) and (iii)
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ENVIRONMENTAL CHEMISTRY, , 134. Thermal pollution affects mainly, (a) vegetation, (b) aquatic creature, (c) rocks, (d) air, 135. A dental disease characterised by mottling of teeth is due, to presence of a certain chemical element in drinking water., Which is that element?, (a) Boron, (b) Chlorine, (c) Fluorine, (d) Mercury, 136. Frequent occurrence of water blooms in a lake indicates, (a) nutrient deficiency, (b) oxygen deficiency, (c) excessive nutrient availability, (d) absence of herbivores in the lake, 137. Which one of the following statements is correct ?, (a) Extensive use of chemical fertilizers may lead to, eutrophication of nearby water bodies, (b) Both Azotobacter and Rhizobium fix atmospheric, nitrogen in root nodules of plants, (c) Cyanobacteria such as Anabaena and Nostoc are, important mobilizers of phosphates and potassium for, plant nutrition in soil, (d) At present it is not possible to grow maize without, chemical fertilizers, , 235, , 138. Lichens do not like to grow in cities, (a) because of absence of the right type of algae and fungi, (b) because of lack of moisture, (c) because of SO2 pollution, (d) because natural habitat is missing, 139. BOD of pond is connected with, (a) microbes & organic matter, (b) organic matter, (c) microbes, (d) None of these, 140. Which is known as ‘Third poison of environment’ and also, creates ‘Blue baby syndrome’, (a) Nitrate present in water, (b) Phosphate and detergents found in water, (c) Cynide, (d) Pesticides, 141. Negative soil pollution is, (a) reduction in soil productivity due to erosion and over, use, (b) reduction in soil productivity due to addition of, pesticides and industrial wastes, (c) converting fertile land into barren land by dumping, ash, sludge and garbage, (d) None of the above
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EBD_7207, ENVIRONMENTAL CHEMISTRY, , 236, , FACT / DEFINITION TYPE QUESTIONS, 1., 2., 3., 4., , (d), (d), (d), (b), , 5., 6., 7., , (c), (c), (b), , 8., , (c), , 9., 10., 11., , (c), (b) CO2 is generally not regarded as pollutant., (c) CO and oxides of Nitrogen are poisnous gases present, in automobile exhaust gases., (b) Nitric oxide (NO) which may be produced at the ground, level due to human activity or natural sources or is, produced in large amounts in the exhaust gases by the, engine of supersonic transport planes and introduced, directly into the strateosphere., , 12., , DDT causes air, water and soil pollution., DDT is a non-biodegradable pollutant., The uppermost region of atmosphere is exosphere., The coldest region is mesosphere (temp. –27ºC to, – 92ºC), Air pollution greatly affect the troposphere., Troposphere contains water vapour., High concentration of SO2 leads to stiffness of, flower buds., The irritant red haze in the traffic and congested, places is due to presence of oxides of nitrogen., , NO O3, , 13., 14., 15., 16., 17., , 18., 19., , 20., 21., 23., 24., 25., , NO 2, , 26., 27., 28., 29., , 30., 33., 34., 39., 44., 46., 47., , O2, , (d) CO is highly toxic and impairs respiration. CO combine, with haemoglobin of blood and reduces its O2 carry, capacity., (d), (a) CO2 causes green house effect., (c), (a) Radiation coming from sun or outerspace have high, energy or short wavelength, which are allowed to enter, by green house gases. However, radiation emitted by, earth is in infrared region, having long wavelength,, are reflected back by the envelope of green house, gases., (d), (a) Green house gases such as CO2, ozone, methane, the, chlorofluorocarbon compounds and water vapour form, a thick cover around the earth which prevents the IR, rays emitted by the earth to escape. It gradually leads, to increase in temperature of atmosphere., (a) CO2 is a green house gas., (a), 22. (d), (d) Acid rain contains H2SO4 > HNO3> HCl., (a) Normal rain water has pH 5.6. Thunderstorm results in, the formation of NO and HNO3 which lowers the pH., (c) Acid rain is rain or any other form of precipitation that, is unusually acidic. It has harmful effects on plants,, aquatic animals, and infastructure. Acid rain is mostly, , 48., 49., , caused by human emissions of sulfur and nitrogen, compounds which react in the atmosphere to produce, acids. In recent years, many governments have, introduced laws to reduce these emissions., (c) pH of normal rain water is 5.6 as CO2 present in, atmosphere combines with moisture to form H2CO3., (d), (b) Large amounts of CH4 are released in paddy fields,, coal mines and by fossil fuels., (b) The oxidised hydrocarbons and ozone in presence of, humidity cause photochemical smog., Hydrocarbons + O2, NO2, NO, O, O3 Peroxides, (d), 31. (a), 32. (b), (a) Smog is caused by oxides of sulphur and nitrogen., (c), 35. (a), 36. (a), 37. (a), 38. (c), (c), 40. (a), 41. (a), 42. (b), 43. (a), (d), 45. (b), (a) PAH (Poly Aromatic Hydrocarbon), (c) Usually catalytic converters are used in the, automobiles, which prevent the release of nitrogen, oxide and hydrocarbons to the atmosphere. Certain, plants e.g., Pinus, Juniparus. Quercus, Pyrus and, Vitis can metabolise nitrogen oxide and therefore,, their plantation could help in this matter., (a) NO and freons are responsible for ozone depletion., (c) The ozone layer, existing between 20 to 35 km above, the earth’s surface, shield the earth from the harmful, U. V. radiations from the sun., Depletion of ozone is caused by oxides of nitrogen, NO + N, N 2O + h, reactive nitric oxide, , NO + O 3, , NO 2 + O2, , O3 + h, , O2 + O, , NO 2 + O, , NO + O 2, , 50., , 2 O3 + h, 3 O 2 (Net reaction), The presence of oxides of nitrogen increase the, decomposition of O3., (d) Ozone layer acts as a shield and does not allow, ultraviolet radiation from sun to reach earth. It does, not prevent infra-red radiation from sun to reach earth,, thus option (d) is wrong statement and so it is the, correct answer., , 51., , (c), , CF2 Cl 2, •, , Cl, , O3, , h, , CF2 Cl, •, , Cl O, , •, , Cl, , O2
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ENVIRONMENTAL CHEMISTRY, , 52., 53., 54., 55., 56., 58., 59., 60., , 61., 62., 63., 67., 68., 69., 70., 71., 72., , 73., 74., , 75., 76., 77., 78., 79., 80., , 81., 82., 83., 84., 85., 86., 88., 89., 90., , (a) In antarctica ozone depletion is due to formation of, acrolein., (d) Depletion of ozone layer causes skin cancer., (b) They create holes in ozone layer., (d) Ozone hole is reduction in ozone layer in stratosphere., (a), 57. (c), (d) Ozone absorbs U.V. radiations harmful to human life., (d), (b) In cold water, dissolved oxygen can reach a, concentration upto 10 ppm, whereas oxygen in air is, about 200, 000 ppm., (a) Pesticides cause water pollution., (d) Minamata is caused by Hg poisoning., (c), 64. (c), 65. (a), 66. (b), (b) Domestic sewage constitute biodegradable pollutants., (c), (b) Sewage water is purified by micro-organisms., (c) Water is often treated with Cl2 to kill germs., (d) Decrease in D.O causes death of fish., (b) Strength of sewage or degree of water pollution is, measured in terms of BOD (Biochemical oxygen, demand) value., (d), (b) The ideal value of D.O for growth of fishes is 8 mg/ ., 7mg / is desirable range, below this value fishes get, susceptible to disease. A value of 2 mg/ or below is, lethal for fishes., (b) Water pollution is mainly caused by industrial wastes,, sewage, insecticide, herbicides, etc., (a), (d) Addition of phosphate fertilizers to water leads to, nutrient enrichment (eutrophication)., (a) BOD of pond is connected with microbes and organic, matter., (d), (a) Excessive concentration of nitrate in drinking water is, harmful and can cause methemoglobinemia (blue baby, syndrome)., (a) Eutrophication causes reduction in D.O, (a) Pesticides cause water pollution., (d) Decrease in D.O causes death of fish, (d), (d) DDT is a non-biodegradable pollutant., (c), 87. (d), (b) Pesticides and synthetic fertilizers pollute the soil., (a) Lower trophic level has lower toxins deposition than, higher trophic level., (b) Green chemistry may be defined as the programme of, developing new chemical products and chemical, processes or making improvements in the already, existing compounds and processes so as to make less, harmful to human health and environment. This means, the same as to reduce the use and production of, hazardous chemicals., , 237, , 91. (d), 92. (b) This represents a great step forward for green, chemistry., 93. (b) Replacement of earlier used tetra-chloroethene as, solvent for dry cleaning by liquid CO2 results in less, harm to ground water., 94. (b) Ethanal is commerically prepared by one step, oxidation of ethene in the presence of ionic catalyst, in aqueous medium with an yield of 90%., CH2 = CH2 + O2, , Catalyst, , Pb(II)|Cu(II), (in water), , CH3CHO, , STATEMENT TYPE QUESTIONS, 95. (c) For statement (iii), Stratosphere lies above, troposphere between 10 and 50 km above sea level, cloud formation takes place in troposphere., For statement (iv), Troposphere is a turbulent, dusty, zone containing air, much water vapour and clouds., For statement (v), Stratosphere contains dinitrogen,, dioxygen, ozone and little water vapour., 96. (d), 97. (a) Classical smog is also called reducing smog., 98. (d) For statement (iii), Methemoglobinemia (blue baby, syndrome) is caused due to excess of nitrate in, drinking water., For statement (iv), Excessive sulphate (> 500 ppm) in, drinking water causes laxative effect, otherwise at, moderate levels it is harmless., 99. (d) All the given statements are true for about waste, recycling., , MATCHING TYPE QUESTIONS, 100. (c), 103. (a), , 101. (a), 104. (a), , 102. (b), 105. (a), , ASSERTION-REASON TYPE QUESTIONS, 106. (a) The presence of particulate matter in polluted air, catalyses the oxidation of SO2 to SO3, 107. (b) At high altitudes when lightening strikes dinitrogen, and dioxygen combine to form oxides of nitrogen., 108. (c) Other gases like CFCs, ozone, water vapour and nitrous, oxide also show green house effect., 109. (c), 110. (b), 111. (a), 112. (b) SPM (Suspended Particulate Matter) is defined as, particles floating in the air with a diameter below 10, m. Studies have shown that high SPM concentrations, in the air can have a detrimental impact on respiratory, organs. SPM is generated from natural sources (e.g.,, volcanoes or dust storms) and human activities, (vehicles, incinerators and industrial plants).
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EBD_7207, ENVIRONMENTAL CHEMISTRY, , 238, , SPM, , Other aerosols, , Less than 10 m, Tend to float longer in, , Less than 100 m, Tend to settle fairly, , air due to small size, , quickly due to comparative, heaviness, , Catalytic converters is a device designed to reduce, the amount of emissions from automobiles. The current, (so-called three-way) systems use a heated metal, catalyst to reduce the emissions of carbon monoxide, (CO), hydrocarbons, and nitric oxide (NO), all of which, contribute to the formation of photochemical smog. In, an automobile’s exhaust system, a catalytic converter, provides an environment for a chemical reaction where, unburned hydrocarbons completely combust., 113. (b), 114. (d), 115. (a), 116. (c), 117. (b) Eutrophication is a natural process which literally, means well nourished or enriched. It is a natural state, in many lakes and ponds which have a rich supply of, nutrients. Eutrophication become excessive, however, when abnormally high amount of nutrient from sewage,, fertilizers, animal wastage and detergent, enter streams, and lakes causes excessive growth or blooms of, microorganisms. With increasing eutrophication, the, diversity of the phytoplankton community of a lake, increases and the lake finally becomes dominated by, blue - green algae., 118. (a) The F– ions make the enamel on teeth much harder, by converting hydroxyapatite, [3(Ca 3 (PO4)] 2 ., Ca(OH)2]. the enamel on the surface of the teeth,, into much harder fluorapatite, [3(Ca3(PO4)2. CaF2]., , CRITICAL THINKING TYPE QUESTIONS, 119. (c) H2, He and ionic oxygen are present in exosphere., 120. (c) Solar energy is not responsible for green house effect, instead it is a source of energy for the plants and, animals., 121. (c) CO is converted into CO2 by microorganism present, in soil., , 122. (d) Carboxyhaemoglobin is 300 times more stable than, oxyhaemoglobin., 123. (a) 124. (c), 125. (a) Haemoglobin has great affinity for NO., 126. (c), 127. (c) The beauty of Taj Mahal is endangered due to air, pollutants like SO2 released from oil refinery., 128. (b) When SO2 pollution in air is much higher. Sometimes,, SO2 mixes in the air with small particles of metals near, the factories and gets oxidised into sulphur trioxide, SO3. These gases are harmful and they react with water, to form sulphuric acid (H2SO4) or sulphurous acid, (H2SO3) and come down to earth with rain water, it is, called acid rain or acid precipitation., 129. (b), 130. (d) London smog is formed in morning during winter., 131. (a) The average residence time of NO is 4 days., 132. (a) PSCs react with chlorine nitrate and HCl to give HOCl, and Cl2., 133. (d) 3CH4 + 2O3, 3CH2= O + 3H2O, Formaldehyde, , CH2 == CHCH == OCH3COONO2, ||, O, Acrolein Peroxyacetyl nitrate (PAN), , 134. (b) Thermal pollution is caused by power plants. Power, plant requires a larger quantity of water for cooling., The water after cooling is left in the water body. The, temperature of left water is generally very high and, affects aquatic life., 135. (c) The excess of fluorine in water causes fluorosis. The, symptoms of fluorosis are mottling of teeth (yellowish, streaks) and abnormal bones liable to fracture etc. It is, an example of endemic disease., 136. (b), 137. (a), 138. (c) Because they are very sensitive to sulphur dioxide, and in cities the amount of SO2 is high so lichen do, not grow in cities., 139. (a) BOD of pond is connected with microbes and organic, matter., 140. (b), 141. (a)
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15, THE SOLID STATE, FACT / DEFINITION TYPE QUESTIONS, 1., , 2., , 3., , 4., , 5., , 6., , Which of the following is not a characteristic property of, solids?, (a) Intermolecular distances are short., (b) Intermolecular forces are weak., (c) Constituent particles have fixed positions., (d) Solids oscillate about their mean positions., Most crystals show good cleavage because their atoms, ions, or molecules are, (a) weakly bonded together, (b) strongly bonded together, (c) spherically symmetrical, (d) arranged in planes, “Crystalline solids are anisotropic in nature. What is the, meaning of anisotropic in the given statement?, (a) A regular pattern of arrangement of particles which, repeats itself periodically over the entire crystal., (b) Different values of some of physical properties are, shown when measured along different directions in, the same crystals., (c) An irregular arrangement of particles over the entire, crystal., (d) Same values of some of physical properties are shown, when measured along different directions in the same, crystals., A crystalline solid, (a) changes abruptly from solid to liquid when heated, (b) has no definite melting point, (c) undergoes deformation of its geometry easily, (d) has an irregular 3-dimensional arrangements, Which of the following is not a characteristic of a crystalline, solid ?, (a) Definite and characteristic heat of fusion., (b) Isotropic nature., (c) A regular periodically repeated pattern of arrangement, of constituent particles in the entire crystal., (d) A true solid, Which of the following is not a crystalline solid?, (a) KCl, (b) CsCl, (c) Glass, (d) Rhombic S, , 7., , Which of the following statements about amorphous solids, is incorrect ?, (a) They melt over a range of temperature, (b) They are anisotropic, (c) There is no orderly arrangement of particles, (d) They are rigid and incompressible, 8., Which of the following is not a crystalline solid?, (a) KCl, (b) CsCl, (c) Glass, (d) Rhombic S, 9., Which of the following is an amorphous solid ?, (a) Graphite (C), (b) Quartz glass (SiO2), (c) Chrome alum, (d) Silicon carbide (SiC), 10. Which of the following statement is not true about, amorphous solids ?, (a) On heating they may become crystalline at certain, temperature., (b) They may become crystalline on keeping for long time., (c) Amorphous solids can be moulded by heating., (d) They are anisotropic in nature., 11. The sharp melting point of crystalline solids is due to _____., (a) a regular arrangement of constituent particles observed, over a short distance in the crystal lattice., (b) a regular arrangement of constituent particles observed, over a long distance in the crystal lattice., (c) same arrangement of constituent particles in different, directions., (d) different arrangement of constituent particles in, different directions., 12. Why some glass objects from ancient civilisations are, found to become milky in appearance?, (a) Glass is a crystalline solid, milky appearance is due to, its crystalline nature., (b) Glass is amorphous but on heating it become, crystalline at some temperature., (c) Because of reaction of glass with impurities present, in the atmosphere., (d) None of these., 13. Which of the following amorphous solid is used as, photovoltaic material for conversion of sunlight into, electricity?, (a) Quartz glass, (b) Quartz, (c) Silicon, (d) Both (a) and (b)
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EBD_7207, , THE SOLID STATE, , 240, , 14., 15., 16., 17., , 18., 19., , 20., , 21., 22., 23., , 24., , 25., , 26., 27., , Solid CH4 is, (a) ionic solid, (b) covalent solid, (c) molecular solid, (d) does not exist, An example of a covalent crystalline solid is:, (a) Si, (b) Al, (c) NaF, (d) Ar, Among solids, the highest melting point is exhibited by, (a) Covalent solids, (b) Ionic solids, (c) Pseudo solids, (d) Molecular solids, Which of the following exists as covalent crystals in the, solid state ?, (a) Iodine, (b) Silicon, (c) Sulphur, (d) Phosphorus, The major binding force of diamond, silicon and quartz is, (a) electrostatic force, (b) electrical attraction, (c) covalent bond force (d) non-covalent bond force, In graphite electrons are :, (a) localised on each carbon atom, (b) spread out between the sheets, (c) localised on every third carbon atom, (d) present in antibonding orbital., Which one of the following forms a molecular solid when, solidified?, (a) Silicon carbide, (b) Calcium fluoride, (c) Rock salt, (d) Methane, Which of the following is a network solid ?, (a) SO2 (solid), (b) I2, (c) Diamond, (d) H2O (Ice), Which of the following solids is not an electrical conductor?, (a) Mg (s), (b) TiO (s), (c) I2 (s), (d) H2O (s), Iodine molecules are held in the crystals lattice by ______., (a) london forces, (b) dipole-dipole interactions, (c) covalent bonds, (d) coulombic forces, Which of the following is not the characteristic of ionic, solids?, (a) Very low value of electrical conductivity in the molten, state., (b) Brittle nature., (c) Very strong forces of interactions., (d) Anisotropic nature., Graphite is a good conductor of electricity due to the, presence of ______., (a) lone pair of electrons (b) free valence electrons, (c) cations, (d) anions, Graphite cannot be classified as ______., (a) conducting solid, (b) network solid, (c) covalent solid, (d) ionic solid, Which of the following cannot be regarded as molecular, solid ?, (i) SiC (Silicon carbide) (ii) AlN, (iii) Diamond, (iv) I2, (a) (i), (ii) and (iii), (b) (ii) and (iii), (c) (iv), (d) (ii) and (iv), , 28., , Crystals can be classified into basic crystal units, equal to, (a) 7, (b) 4, (c) 14, (d) 2, 29. How many three dimensional crystal lattice are possible?, (a) 20, (b) 7, (c) 14, (d) 10, 30. Which of the following represents monoclinic crystal, system?, (a) a b c, 90°, (b) a = b = c ,, =, °, (c) a b c ,, = 90°, 90°, (d) a = b = c ,, = = 90°, 31. In face-centred cubic lattice, a unit cell is shared equally, by how many unit cells, (a) 2, (b) 4, (c) 6, (d) 8, 32., , For orthorhombic system axial ratios are a b c and the, axial angles are, (a), , 33., , 34., , 90, , (b), , 90, , 90, (c), 90 , 90 (d), The unit cell dimensions of a cubic lattice (edges a, b, c and, the angles between them, , and ) are, (a) a = b = c, = = = 900, (b) a = b c, = = = 900, (c) a = b = c, = = 900,, 900, 0, (d) a b c, = = 90 ,, 900, Tetragonal crystal system has the following unit cell, dimensions, (a), , a, , b, , c,, , 90, , (b), , a, , b, , c,, , 90, , (c), , a, , b, , c,, , 90, , (d) a b c,, 90, 120, The number of atoms contained in a fcc unit cell of a, monoatomic substance is, (a) 1, (b) 2, (c) 4, (d) 6, 36. In face-centred cubic lattice, a unit cell is shared equally, by how many unit cells, (a) 2, (b) 4, (c) 6, (d) 8, 37. The number of atoms per unit cell of bcc structure is, (a) 1, (b) 2, (c) 4, (d) 6, 38. When molten zinc is converted into solid state, it acquires, hcp structure. The number of nearest neighbours of Zn will, be, (a) 6, (b) 12, (c) 8, (d) 4, 39. Hexagonal close packed arrangement of ions is described, as, (a) ABC ABA, (b) ABC ABC, (c) ABABA, (d) ABBAB, , 35.
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THE SOLID STATE, 40. In which of the following crystals alternate tetrahedral voids, are occupied?, (a) NaCl, (b) ZnS, (c) CaF2, (d) Na2O, 41. Which of the following metal(s) show(s) hexagonal close, packed structure (hcp) and which show face centred, cubic (fcc) structure?, hcp, fcc, (a) Ag, Zn, Mg, Cu, (b) Mg, Zn, Ag, Cu, (c) Cu, Fe, Al, Sn, (d) Na, Li, Zn, Cu, 42. The number of octahedral voids present in a lattice is A ., The number of closed packed particles, the number of, tetrahedral voids generated is B the number of closed, packed particles, (a) A- equal, B- half, (b) A- twice, B- equal, (c) A- twice , B- half, (d) A- equal, B- twice, 43. In the hexagonal close packed structure of a metallic lattice,, the number of nearest neighbours of a metallic atom is, (a) twelve, (b) four, (c) eight, (d) six, 44. The Ca2+ and F– are located in CaF2 crystal, respectively at, face centred cubic lattice points and in, (a) tetrahedral voids, (b) half of tetrahedral voids, (c) octahedral voids, (d) half of octahedral voids, 45. If Germanium crystallises in the same way as diamond, then, which of the following statement is not correct?, (a) Every atom in the structure is tetrahedrally bonded to, 4 atoms., (b) Unit cell consists of 8 Ge atoms and co-ordination, number is 4., (c) All the octahedral voids are occupied., (d) All the octahedral voids and 50% tetrahedral voids, remain unoccupied., 46. The arrangement ABC ABC .......... is referred to as, (a) Octahedral close packing, (b) Hexagonal close packing, (c) Tetrahedral close packing, (d) Cubic close packing, 47. The total number of tetrahedral voids in the face centred, unit cell is ______., (a) 6, (b) 8, (c) 10, (d) 12, 48. What is the coordination number in a square close packed, structure in two dimensions ?, (a) 2, (b) 3, (c) 4, (d) 6, 49. In the cubic close packing, the unit cell has ______., (a) 4 tetrahedral voids each of which is shared by four, adjacent unit cells., (b) 4 tetrahedral voids within the unit cell., (c) 8 tetrahedral voids each of the which is shared by four, adjacent unit cells., (d) 8 tetrahedral voids within the unit cells., , 241, , 50. In which of the following arrangements octahedral voids, are formed ?, (i) hcp, (ii) bcc, (iii) simple cubic, (iv) fcc, (a) (i), (ii), (b) (i), (iv), (c) (iii), (d) (ii), (iv), 51. Which of the following is the correct increasing order of, packing efficiency for hcp, bcc and simple cubic lattice?, (a) hcp < bcc < simple cubic, (b) bcc < hcp < simple cubic, (c) simple cubic < bcc < hcp, (d) simple cubic < hcp < bcc, 52. Total volume of atoms present in bcc unit cell is., 16 3, 4 3, r, r, (a), (b), 3, 3, 12 3, r, 3, Total volume of atoms present in a face-centred cubic unit, cell of a metal is (r is atomic radius), 16 3, 12 3, r, r, (b), (a), 3, 3, 24 3, 20 3, r, r, (d), (c), 3, 3, The interionic distance for cesium chloride crystal will be, a, (a) a, (b), 2, , (c), , 53., , 54., , 8 3, r, 3, , (d), , 2a, 3a, (d), 3, 2, The fraction of total volume occupied by the atoms present, in a simple cube is, , (c), , 55., , (a), (c), , 56., , 57., , 58., , 59., , 3 2, , (b), , 4 2, , (d), 6, 4, Percentages of free space in cubic close packed structure, and in body centered packed structure are respectively, (a) 30% and 26%, (b) 26% and 32%, (c) 32% and 48%, (d) 48% and 26%, The empty space in the body centred cubic lattice is, (a) 68%, (b) 52.4%, (c) 47.6%, (d) 32%, (e) 74%, Which one of the following statements about packing in, solids is incorrect ?, (a) Coordination number in bcc mode of packing is 8., (b) Coordination number in hcp mode of packing is 12., (c) Void space in hcp mode of packing is 32%., (d) Void space is ccp mode of packing is 26%., A metallic crystal crystallizes into a lattice containing a, sequence of layers AB AB AB......Any packing of spheres, leaves out voids in the lattice. What percentage of volume, of this lattice is empty space?, (a) 74%, (b) 26%, (c) 50%, (d) none of these.
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EBD_7207, , THE SOLID STATE, , 242, , 60., , In NaCl, the centre-to-centre nearest-neighbour distance of, ions is, (a), , 1, a, 4, , (b), , 70., , 3, a, 2, , 1, 1, a 2, a, (d), 2, 2, The edge lengths of the unit cells in terms of the radius of, spheres constituting fcc, bcc and simple cubic unit cell are, respectively ________., , (c), 61., , (a), , 62., , 63., , 64., , 65., , 66., , 67., , 68., , 69., , 2 2r,, , 4r, , 2r, 3, , (b), , 71., , 4r, , 2 2r, 2r, 3, , 4r, 4r, , 2 2r, (c) 2r, 2 2r,, (d) 2r,, 3, 3, CsBr crystallises in a body centered cubic lattice. The unit, cell length is 436.6 pm. Given that the atomic mass of Cs =, 133 and that of Br = 80 amu and Avogadro number being, 6.02 × 1023 mol–1, the density of CsBr is, (a) 0.425 g/cm3, (b) 8.5 g/cm3, 3, (c) 4.25 g/cm, (d) 82.5 g/cm3, An element occuring in the bcc structure has 12.08 × 1023 unit, cells. The total number of atoms of the element in these, cells will be, (a) 24.16 × 1023, (b) 36.18 × 1023, 23, (c) 6.04 × 10, (d) 12.08 × 1023, Pottasium has a bcc structure with nearest neighbour, distance 4.52 Å. Its atomic weight is 39. Its density (in kg m –3), will be, (a) 454, (b) 804, (c) 852, (d) 910, The cubic unit cell of a metal (molar mass = 63.55g mol–1), has an edge length of 362 pm. Its density is 8.92g cm–3., The type of unit cell is, (a) primitive, (b) face centered, (c) body centered, (d) end centered, AB; crystallizes in a body centred cubic lattice with edge, length ‘a’ equal to 387 pm. The distance between two, oppositely charged ions in the lattice is :, (a) 335 pm, (b) 250 pm, (c) 200 pm, (d) 300 pm, Iron crystallizes in a b.c.c. system with a lattice parameter of, 2.861 Å. Calculate the density of iron in the b.c.c. system, (Atomic weight of Fe = 56, NA = 6.02 × 1023 mol–1), (a) 7.92 g ml–1, (b) 8.96 g ml–1, –1, (c) 2.78 g ml, (d) 6.72 g ml–1, An element (atomic mass = 100 g / mol) having bcc structure, has unit cell edge 400 pm. Then, density of the element is, (a) 10.376 g/cm3, (b) 5.188 g/cm3, 3, (c) 7.289 g/cm, (d) 2.144 g/cm36, A metal crystallizes in 2 cubic phases fcc and bcc whose, unit cell lengths are 3.5 Å and 3.0Å respectively. The ratio, of their densities is, (a) 0.72, (b) 2.04, (c) 1.46, (d) 3.12, , Which of the following statements is not correct ?, (a) Vacancy defect results in decrease in density of, substance., (b) Vacancy defect can develop when a substance is, heated., (c) Interstitial defect increases the density of the, substance., (d) Ionic solids show interstitial defects only., Which defect is shown in the given figure?, Na, , +, , –, , Cl, , Na, , +, , –, , Cl, , 72., , 73., , 74., , 75., , 76., , 77., , –, , Cl, , 2+, , Sr, , Cl, , Na, , +, , –, , Cl, , –, , Na, , +, , –, , Cl, Na, Cl, , –, , Cl, , +, , –, , +, , Na, , (a) Frenkel defect, (b) Impurity defect, (c) Schottky defect, (d) Vacancy defect, Each of the following solids show, the Frenkel defect except, (a) ZnS, (b) AgBr, (c) AgI, (d) KCl, Schottky defect defines imperfection in the lattice structure, of, (a) solid, (b) gas, (c) liquid, (d) plasma, When electrons are trapped into the crystal in anion, vacancy, the defect is known as :, (a) Schottky defect, (b) Frenkel defect, (c) Stoichiometric defect (d) F-centre, Schottky defect in crystals is observed when, (a) an ion leaves its normal site and occupies an interstitial, site, (b) unequal number of cations and anions are missing from, the lattice, (c) density of the crystal increases, (d) equal number of cations and anions are missing from, the lattice, The appearance of colour in solid alkali metal halides is, generally due to, (a) Schottky defect, (b) Frenkel defect, (c) Interstitial positions (d) F-centre, Crystal defect indicated in the diagram below is, Na Cl Na Cl Na Cl, , Na+, Cl–, Cl– Na+, +, –, –, +, Na Cl, Cl Na Cl–, –, +, –, Cl Na Cl Na+, Na+, (a) Interstitial defect, (b) Schottky defect, (c) Frenkel defect, (d) Frenkel and Schottky defects
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THE SOLID STATE, 78., , 79., 80., , 81., 82., , 83., , 84., , 85., , 86., , 87., , 88., , 89., , 90., 91., , Schottky defect generally appears in :, (a) NaCl, (b) KCl, (c) CsCl, (d) all of these, Which defect causes decrease in the density of crystal, (a) Frenkel, (b) Schottky, (c) Interstitial, (d) F – centre, Which statement does not make sense?, (a) Frenkel defect is not found in alkali metal halides, (b) Schottky defect is very common in alkali metal halides, (c) Schottky defect lowers the density of the crystal, (d) Frenkel defect lowers the density of the crystal., Frenkel and Schottky defects are :, (a) nucleus defects, (b) non-crystal defects, (c) crystal defects, (d) nuclear defects, Equal number of atoms or ion missing from normal lattice, point creating a vacancy due to, (a) Frenkel defect, (b) Mass defect, (c) Schottky defect, (d) Interstitial defect, In stoichiometric defects, the types of compound exhibit, Frenkel defects have/has, (a) Low co-ordination nos., (b) High co-ordination, (c) Small difference in the size of cations and anions, (d) None of these, The crystal with metal deficiency defect is, (a) NaCl, (b) FeO, (c) KCl, (d) ZnO, Which of the following has Frenkel defects?, (a) Sodium chloride, (b) Graphite, (c) Silver bromide, (d) Diamond, Which of the following crystals does not exhibit Frenkel, defect?, (a) AgBr, (b) AgCl, (c) KBr, (d) ZnS, Due to Frenkel defect, the density of ionic solids, (a) decreases, (b) increases, (c) neither (a) nor (b), (d) does not change, In a solid lattice the cation has left a lattice site and is located, at an interstitial position, the lattice defect is :, (a) Interstitial defect, (b) Valency defect, (c) Frenkel defect, (d) Schottky defect, Doping of AgCl crystals with CdCl 2 results in, (a) Frenkel defect, (b) Schottky defect, (c) Substitutional cation vacancy, (d) Formation of F - centres, Cations are present in the interstitial sites in ________., (a) Frenkel defect, (b) Schottky defect, (c) Vacancy defect, (d) Metal deficiency defect, What is the energy gap between valence band and, conduction band in crystal of insulators ?, (a) Both the bands are overlapped with each other, (b) Very small, (c) Infinite, (d) Very large, , 243, , 92. Which of the following is non stoichiometric defect?, (i) Metal excess defect (ii) Impurity defect, (iii) F-centre, (iv) Metal deficiency defect, (a) (i) and (iv), (b) (i), (iii) and (iv), (c) (iii) and (ii), (d) All of these, 93. Which kind of defects are introduced by doping ?, (a) Dislocation defect, (b) Schottky defect, (c) Frenkel defects, (d) Electronic defects, 94. Silicon doped with electron – rich impurity forms _______., (a) p-type semiconductor, (b) n-type semiconductor, (c) intrinsic semiconductor, (d) insulator, 95. Which of the following defects is also known as dislocation, defect ?, (a) Frenkel defect, (b) Schottky defect, (c) Non – stoichiometric defect, (d) Simple interstitial defect, 96. Doping of silicon (Si) with boron (B) leads to :, (a) n-type semiconductor (b) p-type semiconductor, (c) metal, (d) insulator, 97. On doping Ge metal with a little of In or Ga, one gets, (a) p-type semi conductor (b) n-type semi conductor, (c) insulator, (d) rectifier, 98. With which one of the following elements silicon should be, doped so as to give p-type of semiconductor ?, (a) Germanium, (b) Arsenic, (c) Selenium, (d) Boron, 99. To get a n- type semiconductor, the impurity to be added to, silicon should have which of the following number of, valence electrons, (a) 1, (b) 2, (c) 3, (d) 5, 100. If we mix a pentavalent impurity in a crystal lattice of, germanium, what type of semiconductor formation will, occur?, (a) p-type, (b) n-type, (c) both (a) and (b), (d) None of the two., 101. The addition of arsenic to germanium makes the latter a, (a) metallic conductor, (b) intrinsic semiconductor, (c) mixed conductor, (d) extrinsic semiconductor, 102. Pure silicon doped with phosphorus is a, (a) metallic conductor, (b) insulator, (c) n-type semiconductor (d) p-type semiconductor, 103. Which of the folliowing metal oxides is anti-ferromagnetic, in nature?, (a) MnO2, (b) TiO2, (c) VO2, (d) CrO2, 104. Which of the following compound is like metallic copper in, its conductivity and appearance?, (a) VO3, (b) TiO3, (c) ReO3, (d) CrO2, 105. Magnetic moment of electron is due to which of the, following reason?, (a) Due to its orbital motion around the nucleus., (b) Due to its spin around its own axis., (c) Due to negative charge on electron., (d) Both (a) and (b).
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EBD_7207, , THE SOLID STATE, , 244, , :, , :, , :, , :, , :, , :, :, , :, , :, , :, , :, , :, , :, , :, , :, , :, , :, :, , :, , :, , Silicon atom, , :, , 115., , :, , 114., , :, , 113., , :, , :, , :, , :, , :, :, , :, , : As, :, , :, :, , :, , :, , :, , :, , :, , :, , :, , :, , :, , :, , :, , :, , :, , :, :, , :, , :, , :, , :, , :, , :, , :, , :, , :, , :, , :, , :, , :, , :, , :, , :, , :, , :, , :, , :, , :, , :, , :, , :, :, , :, , :, , :, , :, :, , :, , :, , :, , :, , : Al, :, , :, , :, , :, , :, , :, , :, , :, , :, , :, :, , :, , P, , :, , :, , :, , :, , :, , :, , :, :, , :, , :, , :, , :, , :, , :, , :, , :, , :, , :, , :, , (a) (i) and (iii), (c) (ii) and (iv), , :, :, , :, , B, , :, , (iv), , :, , :, :, , :, , :, , (iii), , :, , :, , :, , :, , :, , (ii), , :, , :, , 112., , :, , 111., , :, , 110., , :, , 109., , (c), (d), Which of the following statement is not correct?, (a) Paramagnetic substances lose their magnetism in the, absence of magnetic field., (b) Diamagnetic substances are weakly magnetised in, magnetic field in opposite direction., (c) Ferromagnetic substances becomes paramagnetic on, heating., (d) In antiferromagnetism domains are oppositely oriented, and cancel out each other’s magnetic moment., An element containing an odd number of electrons is:, (a) Paramagnetic, (b) Diamagnetic, (c) Antiferromagnetic, (d) None of these, Which of the following oxides behaves as conductor or, insulator depending upon temperature ?, (a) TiO, (b) SiO2, (c) TiO3, (d) MgO, Which of the following oxides shows electrical properties, like metals ?, (a) SiO2, (b) MgO, (c) SO2 (s), (d) CrO2, Which of the following statements is not true ?, (a) Paramagnetic substances are weakly attracted by, magnetic field., (b) Ferromagnetic substances cannot be magnetised, permanently, (c) The domains in antiferromagnetic substances are, oppositely oriented with respect to each other., (d) Pairing of electrons cancels their magnetic moment in, the diamagnetic substances., A ferromagnetic substance becomes a permanent magnet, when it is placed in a magnetic field because ________., (a) all the domains get oriented in the direction of magnetic, field., (b) all the domains get oriented in the direction opposite, to the direction of magnetic field., (c) domains get oriented randomly., (d) domains are not affected by magnetic field., A perfect crystal of silicon interchange is doped with some, elements as given in the options. Which of these options, show n-type semiconductors ?, , (i), , :, , 106. What is the value of Bohr, (a) 9.27 × 10–24 Am2, (c) 9.00 × 10–30 Am2, (d) 9.27 × 10–25 Am2, 107. Which of the following type of substances can be, permanently magnetised?, (a) Diamagnetic, (b) Ferromagnetic, (c) Ferrimagnetic, (d) Antiferromagnetic, 108. Which of the following structure represents, ferrimagnetism?, (a), (b), , :, , magneton B in A m2?, (b) 9.27 × 10–20 Am2, , :, , (b) (i), (d) (iii), , STATEMENT TYPE QUESTIONS, 116. Which of the following statement(s) is/are correct?, (i) Crystallin e solids have definite characteristic, geometrical shape., (ii) Crystalline solids have long range order., (iii) Sodium chloride and quartz glass are examples of, crystalline solids., (iv) Crystalline solids are isotropic in nature., (a) (i), (ii) and (iii), (b) (i), (ii) and (iv), (c) (i) and (ii), (d) (i) only, 117. Which of the following sequence of T and F is correct for, given statements. Here T stands for true statement and F, stands for false statement?, (i) Ionic solids are electrical insulators in the solid state, but conduct electricity in molten state., (ii) Graphite is a covalent solid., (iii) Covalent solids are conductor of electricity., (iv) Non polar molecular solids are held by weak, dispersion forces or London forces while polar, molecular solids are held by stronger dipole - dipole, interactions., (a) TTTF, (b) FTTF, (c) TFTT, (d) TTFT
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THE SOLID STATE, 118. Which of the following statements(s) is/are incorrect?, (i) Only 1/8th portion of an atom located at corner of a, cubic unit cell is its neighbouring unit cell., (ii) Total number of atoms per unit cell for a face centered, cubic unit cell is 3 ., (iii) Atom located at the body center is shared between, two adjacent unit cells., (a) (iii) only, (b) (ii) only, (c) (i) and (ii), (d) (ii) and (iii), 119. Which of the following is/are not true about the voids formed, in 3 dimensional hexagonal close packed structure ?, (i) A tetrahedral void is formed when a sphere of the, second layer is present above triangular void in the, first layer., (ii) All the triangular voids are not covered by the spheres, of the second layer., (iii) Tetrahedral voids are fomed when the triangular voids, in the second layer lie above the triangular voids in, the first layer and the triangular shapes of these voids, do not overlap., (iv) Octahedral voids are formed when the triangular voids, in the second layer exactly overlap with similar voids, in the first layer., (a) (i) and (iv), (b) (iii) and (iv), (c) (ii) and (iii), (d) (iv) only, 120. Which of the following sequence of T and F is true for, given statements. Here T stands for true statement and F, stands for false statement?, (i) Frenkel defect results in increase in density of the, solid., (ii) ZnS, AgCl, AgBr and AgI shows Frenkel defect., (iii) Schottky defect results in decrease in density of the, solid., (iv) AgBr shows Schottky defect only., (v) For NaCl there is one Schottky defect per 1016 ions., (a) TTTFT, (b) TTTFF, (c) FTTFT, (d) FTTFF, 121. Which of the following statements is /are correct?, (i) LiCl crystals are pink due to metal excess defect due, to presence of extra L+i ion at interstitial sites., (ii) Zinc oxide on heating turns yellow because its anionic, sites are occupied by unpaired electrons., (iii) In FeO crystals some Fe2+ are missing and the loss of, positive charge is made up by the presence of required, number of Fe3+ ions., (a) (i) and (ii), (b) (i), (ii) and (iii), (c) (ii) and (iii), (d) (iii) only, 122. Which of the following sequence of T and F is correct?, Here ‘T’ stands for true and ‘F’ stands for false statement., (i) Solids have conductivities in the order of 10–20 to 107, ohm–1 m–1., (ii) In semiconductors the gap between filled valence band, and conduction band is small., (iii) Electrical conductivity of insulators increases with rise, in temperature., (iv) Insulators have conductivities ranging between 10–6, to 104 ohm–1 m–1 ., (a) TTFF, (b) TTFT, (c) FTFF, (d) FTTF, , 245, , 123. White crystal of zinc oxide is heated, (i) Metal excess defect is created., (ii) Crystal become p-type semiconductor, (iii) Crystal become yellow in color., (iv) Free electron are created., (a) All statement(s) are correct. (b) (i), (ii) and (iv), (c) (i), (ii) and (iii), (d) (ii) and (iv), 124. Consider the Oxygen and chromium dioxide, both are placed, in magnetic field:, (i) Oxygen is attracted strongly in a magnetic field., (ii) Magnetic field persist in chromium dioxide while in, Oxygen not., (a) Both Statements are correct., (b) Statements (i) is correct only., (c) Statements (ii) is correct only., (d) Both Statements are incorrect., , MATCHING TYPE QUESTIONS, 125. Match the columns, Column-I, Column-II, (Type of solid), (Example of solid), (A) Molecular solid, (p) Ag, (B) Ionic solid, (q) SiC, (C) Metallic solid, (r) CCl4, (D) Covalent solid, (s) MgO, (a) A – (s), B – (r), C – (p), D – (q), (b) A – (q), B – (s), C – (p), D – (r), (c) A – (r), B – (q), C – (p), D – (s), (d) A – (r), B – (s), C – (p), D – (q), 126. Match the columns, Column-I, Column-II, (Type of unit cell), (Characteristic feature), (A) Primitive cubic unit, (p) Each of the three, cell, perpendicular edges, compulsorily have the, different edge length i.e.;, a b c., (B) Body centered cubic (q) Number of atoms per unit, unit cell, cell is one, (C) Face centered cubic (r) Each of the three, unit cell, perpendicular edges, compulsorily have the, same edge length i.e.;, a = b = c., (D) End centered, (s) In addition to the, orthorhombic unit, contribution from the, cell, corner atoms the number, of atoms present in a unit, cell is one., (t) In addition to the, contribution from the, corner atoms the number, of atoms present in a unit, cell is three., (a) A – (q), B – (s), C – (r, t), D – (p), (b ) A – (q, r), B – (r, s), C – (r, t), D – (p,s), (c) A – (r, s), B – (q, r), C – (r), D – (p), (d) A – (t), B – (r, s), C – (p, s), D – (q)
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EBD_7207, 246, , 127. Match the columns, Column-I, Column-II, (Crystal system), (Compounds), (A) Rhombohedral, (p) KNO3, (B) Orthorhombic, (q) Zinc blende, (C) Cubic, (r) CdS, (D) Hexagonal, (s) Calcite, (a) A – (p), B – (q), C – (s), D – (r), (b) A – (r), B – (p), C – (q), D – (s), (c) A – (s), B – (p), C – (q), D – (r), (d) A – (q), B – (r), C – (s), D – (p), 128. Match the columns, Column-I, Column-II, (A) Square close packing (p) Triangular voids, in two dimensions, (B) Hexagonal close, (q) Pattern of spheres is, packing in two, repeated every fourth, dimensions, layer, (C) Hexagonal close, (q) Coordination number 4, packing in three, dimensions, (D) Cubic close packing (s) Pattern of sphere is, in three dimensions, repeated alternate layers, (a) A – (r), B – (p), C – (s), D – (q), (b) A – (p), B – (s), C – (q), D – (r), (c) A – (s), B – (p), C – (q), D – (s), (d) A – (r), B – (p), C – (s), D – (q), 129. Match Column-I (Type of close packed structure) with, Column-II (Coordination number) and choose the correct, option., Column-I, Column-II, (A) One dimensional close, (p) 12, packed arrangement., (B) Square close packing in, (q) 6, two dimensions., (C) Two dimensional, (r) 2, hexagonal close packing., (D) Cubic close packed, (s) 4, arrangement., (a) A – (r), B – (s), C – (q), D – (p), (b) A – (r), B – (s), C – (p), D – (q), (c) A – (s), B – (r), C – (q), D – (p), (d) A – (s), B – (q), C – (p), D – (r), 130. Match the columns, Column-I, Column-II, (A) Impurity defect, (p) NaCl with anionic sites, F-centres, (B) Metal excess defect (q) FeO with Fe3+, (C) Metal deficiency, (r) NaCl with Sr2+ and some, defect, cationic sites vacant, (a) A – (r), B – (p), C – (q), (b) A – (p), B – (q), C – (r), (c) A – (r), B – (q), C – (p), (d) A – (q), B – (p), C – (r), , THE SOLID STATE, 131. Match the columns, Column-I, Column-II, (A) Mg in solid state, (p) p-Type semiconductor, (B) MgCl2 in molten state (q) n-Type semiconductor, (C) Silicon with, (r) Electrolytic conductors, phosphorus, (D) Germanium with boron (s) Electronic conductors, (a) A – (q), B – (p), C – (r), D – (s), (b) A – (p), B – (q), C – (s), D – (r), (c) A – (s), B – (r), C – (q), D – (p), (d) A – (r), B – (s), C – (p), D – (q), 132. Match the columns, Column-I, Column-II, (Molecule/ion), (Magnetic property), (A) C6H6, (p) Antiferromagnetic, (B) CrO2, (q) Ferrimagnetic, (C) MnO, (r) Ferromagnetic, (D) Fe3O4, (s) Paramagnetic, (E) Fe3+, (t) Diamagnetic, (a) A – (t), B – (r), C – (q), D – (p), E – (s), (b) A – (r), B – (t), C – (p), D – (s), E – (q), (c) A – (t), B – (r), C – (p), D – (q), E – (s), (d) A – (t), B – (r), C – (p), D – (s), E – (q), 133. Match the columns, Column-I, Column-II, (Compound), (Magnetic Property), (A) NaCl, (p) Ferrimagnetic, (B) MnO, (q) Paramagnetic, (C) CrCl3, (r) Ferromagnetic, (D) CrO2, (s) Diamagnetic, (E) MgFe2O4, (t) Antiferromagnetic, (a) A – (p), B – (r), C – (q), D – (t), E – (s), (b) A – (t), B – (q), C – (r), D – (p), E – (s), (c) A – (r), B – (t), C – (q), D – (p), E – (s), (d) A – (s), B – (t), C – (q), D – (r), E – (p), , ASSERTION-REASON TYPE QUESTIONS, Directions : Each of these questions contain two statements,, Assertion and Reason. Each of these questions also has four, alternative choices, only one of which is the correct answer. You, have to select one of the codes (a), (b), (c) and (d) given below., (a) Assertion is correct, reason is correct; reason is a correct, explanation for assertion., (b) Assertion is correct, reason is correct; reason is not a, correct explanation for assertion, (c) Assertion is correct, reason is incorrect, (d) Assertion is incorrect, reason is correct., 134. Assertion : Crystalline solids have long range order., Reason : Amorphous solids have short range order., 135. Assertion: Glass panes fixed to windows or panes of old, buildings are found to be slightly thicker at the bottom., Reason: Amorphous solids have a tendency to flow., 136. Assertion : In crystal lattice, the size of the tetrahedral hole, is larger than an octahedral hole., Reason : The cations occupy less space than anions in, crystal packing.
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THE SOLID STATE, 137. Assertion : In close packing of spheres, a tetrahedral void is, surrounded by four spheres whereas an octahedral void is, surrounded by six spheres., Reason : A tetrahedral void has a tetrahedral shape whereas, an octahedral void has an octahedral shape., 138. Assertion : The packing efficiency is maximum for the fcc, structure., Reason : The cordination number is 12 in fcc structures., 139. Assertion : In any ionic solid (MX) with Schottky defects,, the number of positive and negative ions are same., Reason : Equal number of cation and anion vacancies are, present., 140. Assertion : Electrical conductivity of semiconductors, increases with increasing temperature., Reason : With increase in temperature, large number of, electrons from the valence band can jump to the conduction, band., 141. Assertion : On heating ferromagnetic or ferrimagnetic, substances, they become paramagnetic., Reason : The electrons change their spin on heating., , CRITICAL THINKING TYPE QUESTIONS, 142. Which of the following type of solid has high melting point, and do not conduct electricity but its aqueous solution, and melt conduct electricity ?, (a) Covalent, (b) Ionic, (c) Molecular, (d) Metallic, 143. A group 1 hydride crystal when heated in presence of its, constituent metal vapour shows pink color. This metal can, be, (a) Na, (b) K, (c) Rb, (d) Li, 144. A solid with high electrical and thermal conductivity is, (a) Si, (b) Li, (c) NaCl, (d) Ice, 145. Which of the following is true about the value of refractive, index of quartz glass ?, (a) Same in all directions, (b) Different in different directions, (c) Cannot be measured, (d) Always zero, 146. Which of the following features are not shown by quartz, glass ?, (i) This is a crystalline solid., (ii) Refractive index is same in all the directions., (iii) This has definite heat of fusion., (iv) This is also called super cooled liquid., (a) (i) and (iii), (b) (iii) and (iv), (c) (i), (ii) and (iv), (d) (iii) only, 147. The number of carbon atoms per unit cell of diamond unit, cell is :, (a) 8, (b) 6, (c) 1, (d) 4, 148. Na and Mg crystallize in bcc and fcc type crystals respectively,, then the number of atoms of Na and Mg present in the unit, cell of their respective crystal is, (a) 4 and 2, (b) 9 and 14, (c) 14 and 9, (d) 2 and 4, , 247, , 149. In a cubic lattice A atom occupy all the corners. If B atom, occupy one of the opposite face, and atom C occupy the, remaining faces. The simplest formulae of the compound is, (a) ABC3, (b) ABC2, (c) ABC, (d) AB2C, 150. A solid has a structure in which ‘W’ atoms are located at the, corners of a cubic lattice ‘O’ atoms at the centre of edges, and Na atoms at the centre of the cube. The formula for the, compound is, (a) Na2WO3, (b) Na2WO2, (c) NaWO2, (d) NaWO3, 151. Potassium crystallizes with a, (a) body-centred cubic lattice, (b) face-centred cubic lattice, (c) simple cubic lattice, (d) orthorhombic lattice, 152. In a compound, atoms of element Y form ccp lattice and, those of element X occupy 2/3rd of tetrahedral voids. The, formula of the compound will be, (a) X4Y3, (b) X2Y3 (c), X2Y, (d) X3Y4, 153. A compound MpXq has cubic close packing (ccp) arrangement, of X. Its unit cell structure is shown below. The empirical, formula of the compound is, , M, X, , (a) MX, (b) MX2, (c) M2X, (d) M5 X14, 154. A substance AxBy crystallizes in a face centred cubic (fcc), lattice in which atoms ‘A’ occupy each corner of the cube, and atoms ‘B’ occupy the centres of each face of the cube., Identify the correct composition of the substance AxBy, (a) AB3, (b) A4B3, (c) A3 B, (d) Composition can’t be specified, 155. In which of the following structures coordination number, for cations and anions in the packed structure will be same?, (a) Cl – ion form fcc lattice and Na + ions occupy all, octahedral voids of the unit cell., (b) Ca2+ ions form fcc lattice and F– ions occupy all the, eight tetrahedral voids of the unit cell., (c) O2– ions form fcc lattice and Na+ ions occupy all the, eight tetrahedral voids of the unit cell., (d) S2– ions form fcc lattice and Zn2+ ions go into alternate, tetrahedral voids of the unit cell.
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EBD_7207, , THE SOLID STATE, , 248, , 156. If ‘a’ stands for the edge length of the cubic systems : simple, cubic, body centred cubic and face centred cubic, then the, ratio of radii of the spheres in these systems will be, respectively,, (a), , 1, 3, 1, a:, a, a:, 2, 4, 2 2, , (b), , 1, 1, a, a : 3a :, 2, 2, , 3, 3, 1, a:, a:, a, (d) 1a : 3a : 2a, 2, 2, 2, 157. Packing efficiency by arrangement of atoms in two, dimensional hexagonal close packing is, (a) 60.43, (b) 65.78, (c) 59.78, (d) 68.76, 158. The packing efficiency of the two-dimensional square unit, cell shown below is :, , 163. A metal has a fcc lattice. The edge length of the unit cell is, 404 pm. The density of the metal is 2.72 g cm-3. The molar, mass of the metal is :, (NA Avogadro’s constant = 6.02 × 1023 mol–1), (a) 30 g mol–1, (b) 27 g mol–1, –1, (c) 20 g mol, (d) 40 g mol–1, 164. Al (at. wt 27) crystallizes in the cubic system with a cell, , (c), , 165., , 166., , L, , (a) 39.27%, (b) 68.02%, (c) 74.05%, (d) 78.54%, 159. Edge length of unit cell is 3.608 × 10–8 cm, which crystallizes, in fcc and is determined to have a density of 8.92 g/cm 3., The mass of four atoms is, (a) 4.18 × 10–22, (b) 1.67 × 10–21, –22, (c) 2.09 × 10, (d) 8.37 × 10–22, 160. The edge length of unit cell of a metal having molecular, weight 75 g/mol is 5Å which crystallizes in cubic lattice. If, the density is 2g/cc then find the radius of metal atom., (NA = 6 × 1023). Give the answer in pm., (a) 217 pm, (b) 210 pm, (c) 220 pm, (d) 205 pm, 161. The number of atoms in 100 g of an fcc crystal with density,, d = 10 g/cm3 and cell edge equal to 100 pm, is equal to, (a) 1 × 1025, (b) 2 × 1025, 25, (c) 3 × 10, (d) 4 × 1025, 162. A metallic element exists as cubic lattice. Each edge of the, unit cell is 2.88 Å. The density of the metal is 7.20 g cm–3., How many unit cell will be present in 100 g of the metal?, (a) 6.85 × 102, (b) 5.82 × 1023, 5, (c) 4.37 × 10, (d) 2.12 × 106, , 167., , 168., , 169., , 170., , 171., , edge of 4.05 Å . Its density is 2.7 g per cm 3 . Determine the, unit cell type calculate the radius of the Al atom, (a) fcc, 2.432 Å, (b) bcc, 2.432 Å, (c) bcc, 1.432 Å, (d) fcc, 1.432 Å, A compound is formed by elements A and B. The crystalline, cubic structure has the A atoms at the corners of the cube, and B atoms at the body centre. The simplest formula of the, compound is, (a) AB, (b) A6 B, (c) AB6, (d) A8B4, What type of semiconductors respectively are formed when, the group 14 are doped with the group 13 and group 15?, (a) p,n, (b) n,p, (c) p,p, (d) n,n, Which of the following is ferroelectric compound?, (a) BaTlO3, (b) K4[Fe(CN)6], (c) Pb2O3, (d) None of these, Substance which is weakly repelled by a magnetic field is, (a) O2, (b) H2O, (c) CrO2, (d) Fe3O4, (e) ZnFe2O4, Which one of the following statements is correct?, (a) NaCl is a paramagnetic salt, (b) CuSO4 is a diamagnetic salt, (c) MnO is an example of ferromagnetic substance, (d) Ferrimagnetic substance like ZnFe2 O4 becomes, paramagentic on heating, Which of the following is true about the charge acquired by, p-type semiconductors ?, (a) positive, (b) neutral, (c) negative, (d) depends on concentration of p impurity, Which of the following represents correct order of, conductivity in solids ?, (a) Kmetals > > Kinsulators < Ksemiconductors, (b) Kmetals < < Kinsulators < Ksemiconductors, (c) Kmetals ; Kinsulators > Ksemiconductors = zero, (d) Kmetals < Ksemiconductors > Kinsulators zero
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THE SOLID STATE, , FACT / DEFINITION TYPE QUESTIONS, 1., 2., 3., , 4., , 5., 6., 7., 8., 9., 12., , 13., , 14., 15., , 16., 17., 18., 19., 20., 25., 29., 30., , (b) Intermolecular forces are strong in solids., (d) Crystals show good cleavage because their constituent, particles are arranged in planes., (b) Crystalline solids are anisotropic in nature that is some, of their physical properties like electrical resistance, or refractive index show different values when, measured along different directions in the same, crystals., (a) In crystalline solid there is perfect arrangement of the, constituent particles only at 0 K. As the temperature, increases the chance that a lattice site may be, unoccupied by an ion increases. As the number of, defects increases with temperature solid changes into, liquid., (b), (c) Glass is amorphous solid., (b) Amorphous solids are isotropic, because these, substances show same properties in all directions., (c) Glass is amorphous solid., (b), 10. (d), 11. (b), (b) On heating, amorphous solids become crystalline at, some temperature. The milky appearance of glass is, because of some crystallisation., (c) Amorphous silicon is used as best photovoltaic, material available for conversion of sunlight into, electricity., (c) Solid CH4 is a molecular solid. In this, the constituent, molecules are held together by van der Waal’s forces., (a) Si is an example of covalent crystalline solid among, the given choices. Si atoms are covalently linked in, tetrahedral manner., (a) Covalent as in case of diamond., (b) Among the given crystals, only silicon exists as a, covalent solid. It has diamond like structure., (c) Covalent bond force, (b) In graphite, the electrons are spread out between the, sheets., (d), 21. (c), 22. (c), 23. (a), 24. (a), (b), 26. (d), 27. (a), 28. (a), (c) There are only 14 possible three dimensional lattice., These are called Bravais lattices., (b), , 249, , 31. (c), , An isolated fcc cell is shown here. Each face of the, cell is common to two adjacent cells. Therefore, each, face centre atom contributes only half of its volume, and mass to one cell. Arranging six cells each, sharing the remaining half of the face centred atoms,, constitutes fcc cubic lattice. e.g., Cu and Al., 90, 32. (b) For orthorhombic system, 33. (a) It is based on the definition of the cubic lattice., , 90, 34. (b) For tetragonal a b c ,, 35. (c) The no. of atoms is a unit cell may be calculated by the, the formula, nc nb n f ne, Z, 8 1, 2, 4, Where nc = no. of atom at the corner, nb = no. of atoms at body centre, nf = no. of atoms at face centre, ne = no. of atoms at edge centre., An fcc crystal contains, 8 6, = + = 4 atoms in a unit cell., 8 2, , H, , 36. (c), , G, , E, F, , a, C, , A, a, , a, , B, An isolated fcc cell is shown here. Each face of the, cell is common to two adjacent cells. Therefore, each, face centre atom contributes only half of its volume, and mass to one cell. Arranging six cells each, sharing the remaining half of the face centred atoms,, constitutes fcc cubic lattice. e.g., Cu and Al.
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EBD_7207, , THE SOLID STATE, , 250, , 37., , 38., 39., 40., , 41., , 42., 43., 44., 46., 47., 51., , 52., , (b) In bcc structure,, no. of atoms at corner = (1/8) × 8 = 1, no. of atoms at body centre = 1, Total no. of atoms per unit cell = 1 + 1 = 2., (b) hcp is a closed packed arrangement in which the unit, cell is hexagonal and coordination number is 12., (c) ABAB...... is hexagonal close packing., (b) In ZnS structure, sulphide ions occupy all fcc lattice, points while Zn 2+ ions are present in alternate, tetrahedral voids., (b) Metals such as copper and silver crystallise in fcc, structure while metals Mg and Zn crystallise in hcp, structure., (d), (a) Co-ordination number in hcp and ccp arrangement, is 12., (a), 45. (c), (d) It represents ccp arrangement., (b), 48. (c), 49. (d), 50. (b), (c) hcp and ccp structures have maximum packing, efficiency = 74%., For bcc = 68%, For simple cubic = 52.4%, Total volume of atoms present in bcc unit cell is., (a), , 16 3, r (b), 3, , 57., 58., 59., 60., , (d) Packing fraction of bcc = 68%, Empty space = 100 – 68 = 32%, (c) The hcp arrangement of atoms occupies 74% of the, available space and thus has 26% vacant space., (b) In AB AB packing spheres occupy 74%. 26% is empty., (d) In NaCl the Cl– and Na+ touch along edge of cube the, distance between ions is, , 61., 62., , (a), (b) For body centred cubic lattice Z = 2, Atomic mass of unit cell = 133 + 80 = 213 a.m.u, Volume of cell = (436.6 × 10–10)3 cm3, , 54., 55., , 2 213, , 63., , (436.6 10 10 )3 6.02 1023, = 8.50 g/cm3, (a) There are two atoms in a bcc unit cell., So, number of atoms in 12.08 × 1023 unit cells, = 2 × 12.08 × 1023 = 24.16 × 1023 atoms., , 64., , (d) For bcc, d, , a, , 2d, 3, , 2 4.52, 1.732, , 5.219Å, , 3, , N A 10, , 30, , 2 39, , 4 3, r, 3, , (522), , 3, , (6.023 1023 ) 10, , 0.91g / cm3, , 8 3, = r, 3, (b) The face centered cubic unit cell contains 4 atom, 4 3 16 3, Total volume of atoms 4, r, r, 3, 3, , 65., , 910 kg m, , 30, , 3, , ZM, NA V, , (b), , Z, , NA V, M, , 8.92 6.02 1023 (362)3 10, 63.55, , 30, , =4, It has fcc unit cell, , (c) As CsCl is body-centred, d, 3a / 2., (d) Number of atoms per unit cell = 1, a, 2, , 66., , ( r = radius of atom and a = edge length), , 56., , 3, a or a, 2, = 522 pm, , z M, , 8 3, 12 3, r (d), r, 3, 3, (c) For bcc structure total number of atoms = 2., , Atoms touch each other along edges. Hence r =, , =, , a3 NA, , 4 3, r, 3, , Total volume = 2, , 53., , ZM, , Density,, , (c), 52., , a, 2, , 4, r, 3, Therefore % fraction =, (2 r )3 6, (b) Packing fraction is defined as the ratio of the volume, of the unit cell that is occupied by the spheres to the, total volume of the unit cell., P.F. for cpp and bcc are 0.74 and 0.68 respectively., So, the free space in ccp and bcc are 26% & 32%, respectively., , 67., , (a) For bcc lattice body diagonal = a 3 ., The distance between the two oppositely charged ions, a, 3, =, 2, 387 1.732, 335pm, =, 2, ZM, (a) For b.c.c., Z = 2, Now, d =, and V = a3, NV, d=, , 2 56, 23, , (6.02 10 ) (2.861 10 8 )3, , = 7.92 g ml–1
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THE SOLID STATE, 68., , (b), , 251, , Z M, , 2 100, , NA a 3, , 6.023 10 23 (400 10 10 ) 3, , 5.188 g / cm 3, d1, d2, , (a 2 )3, , z1, z2, , 3, 3.5, , 3, , 4, 2, , 1.46, , 69., , (c), , 70., , (d) Ionic solids must always maintain electrical neutrality., Ionic solids show vacancy or interstitial defects as, Frenkel and Schottky defect., (b), (d) In KCl, co-ordination number of cation and anion is 6, and 6 respectively. KCl is highly ionic so Schottky, defect is common., Note : Schottky defect is common in compounds having, high coordination nummber while Frenkel defect is, common in compounds with low coordination number., (a) Schottky defects are found in solid., (d) When electrons are trapped in anion vacancies, these, are called F-centre., , 71., 72., , 73., 74., , (a1 ), , e–, , 3, , +ve –ve, ion ion, F- centre in crystal, , 75., , 76., , 77., , 78., , 79., 80., 81., , (d) If in an ionic crystal of the type A+, B–, equal number, of cations and anions are missing from their lattice, sites so that the electrical neutrality is maintained. The, defect is called Schottky defect., (d) The appearance of colour in solid alkali metal halide is, due to presence of F-centre found as defect in the, crystal structure., (b) When equal number of cations (Na+) and anions (Cl–), are missing from their regular lattice positions, we have, Schottky defect., (d) Schottky defect occurs in ionic crystals of type A+ B,, when equal number of cations and anions are missing, from their lattice sites so that the electrical neutrality is, maintained. This defect generally appears in highly, ionic compounds which have high coordination, number. NaCl, KCl and CsCl all have high coordination, numbers i.e., 6, 6 and 8 respectively. So, Schottky defect, appear in all of the given compounds., (b) More is the Schottky defect in crystal more is the, decrease in density., (d) Frenkel defect does not lower the density of the crystal, since the ions do not leave the crystal lattice., (c) Frenkel and Schottky defects are crystal defects. It, arises due to dislodgement of cation or anion from, their places in the crystal lattice., , 82. (c) The vacancy created due to missing of equal no. of, atoms or ions form normal lattice point is called, Schottky defect. In this type of defect electrical, neutrality of ionic crystal is maintained., 83. (a) In stoichiometric Frenkel defects occurs in those, compound which have, (i) Low C.N., (ii) Large difference in size of cations and anions, 84. (b) Transition metals exhibit this defect due to metal, deficiency, the compound obtained are non, stoichiometric e.g. It is difficult to prepare ferrous oxide, with the ideal composition of FeO what we actually, obtain is Fe0.95O or FexO with x = 0.93 to 0.96, 85. (c) AgBr exhibit Frenkel defect., 86. (c) KBr does not exhibit Frenkel defect., 87. (d) No change in density, 88. (c) Frenkel defect is due to dislocation of ion from its usual, lattice site to interstitial position., 89. (c), 90. (a), 91. (d) When insulators (non metal atoms) interact to form a, solid, their atomic orbitals mix to form two bunch of, orbitals, separated by a large band gap. Electrons, cannot therefore be promoted to an empty level, where, they could move freely., 92. (b), 93. (d), 94. (b), 95. (a), 96. (b), 97. (a) p-type of semiconductors are produced, (i) due to metal deficiency defects, (ii) by adding impurity containing less electrons (i.e.,, atoms of group 13), Ge belongs to Group 14 and In to Group 13. Hence on, doping, p-type semicondutor is obtained., 98. (d) The semiconductors formed by the introduction of, impurity atoms containing one elecron less than the, parent atoms of insulators are termed as p-type, semiconductors. Therefore silicon containing 14, electrons is to be doped with boron containing 13, electrons to give a p-type semi-conductor., 99. (d) For n-type, impurity added to silicon should have more, than 4 valence electrons., 100. (b) n-type, since electron is set free., 101. (d) Extrinsic semiconductor, 102. (c) Pure silicon doped with phosphorus is a n-type, semiconductor, as n-type extrinsic semiconductor (Si), is made by doping the semiconductor with pentavalent, element., 103. (a) MnO2, 104. (c) Rhenium oxide ReO3 is like metallic copper in, conductivity., 105. (d), 106. (a), 107. (b), 108. (d) Ferrimagnetism is observed when the magentic, moments of the domains in the substance are aligned, in parallel and antiparallel directions in unequal, numbers.
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EBD_7207, , THE SOLID STATE, , 252, , 109. (c) Ferrimagnetic substances lose ferrimagnetism on, heating and become paramagnetic., 110. (a) An element containing an odd number of electrons is, paramagnetic., 111. (c), 112. (d), 113. (b), 114. (a), 115. (a), , STATEMENT TYPE QUESTIONS, 116. (c) Quartz glass is an example of amorphous solid and, crystalline solids are anisotropic in nature., 117. (d) Covalent solids are insulator of electricity. Graphite is, a covalent solid but it is a conductor of electricity due, to its structure., 118. (d) Total number of atoms per unit cell for a face centered, cubic unit is 4., The atom at the body center completely belongs to, the unit cell in which it is present., 119. (b), 120. (c) Frenkel defect does not change the density of the, solid., AgBr shows both Frenkel and Schottky defects., 121. (d) LiCl crystals are pink because its anionic sites are, occupied by unpaired eonline . Zinc oxide shows metal, excess defect due to presence of extra cations at, interstitial sites., 122. (a) Electrical conductivity of semiconductors increases, with rise in temperature. Insulators have, conductivities ranging between 10–20 to 10–10 ohm–, 1 m–1, ., 123. (b) Due to release of electrons, the crystal can conducts, electricity but conductivity is not as high as that of, metals. Its conductivity is very low and because, conduction is due to electrons so it is n-type, semiconductor, also excitation of these electron give, rise to yellow color in crystal., 124. (c) CrO2 is a ferromagnetic whereas O2 is paramagnetic., , MATCHING TYPE QUESTIONS, 125. (d), 130. (a), , 126. (b), 131. (c), , 127. (c), 132. (c), , 128. (a), 133. (d), , 129. (a), , ASSERTION-REASON TYPE QUESTIONS, 134. (b) In crystalline solids constituents are arranged in, definite orderly arrangement. This regular arrangement, of constituents extends throughout the three, dimensional network of crystal. Thus crystalline, substances said to have long range order. Whereas, amorphous solids have no regular arrangement., 135. (a), 136. (d) Tetrahedral holes are smaller in size than octahedral, holes. Cations usually occupy less space than anions., , 137. (c) Tetrahedral void is so called because it is surrounded, by four spheres tetrahedrally while octahedral void is, so called because it is surrounded by six spheres, octahedrally., 138. (b), 139. (a) Schottky defect is due to missing of equal number of, cations and anions., 140. (a) In case of semiconductors, the gap between valence, band and the conduction band is small and therefore, some of the electrons may jump from valence band to, conduction band and thus on increasing temperature, conductivity is also increased., 141. (a) All magnetically ordered solids (ferromagnetic,, ferrimagnetic and antiferromagnetic solids) transform, to the paramagnetic state at high temperature due to, the randomisation of spins., , CRITICAL THINKING TYPE QUESTIONS, 142. (b), 143. (d) Excess of lithium makes LiCl crystal pink., 144. (b) Out of the given substances, only Li has high electrical, and thermal conductivity as Li is a metallic solid., 145. (a), 146. (a), 147. (a) Diamond is like ZnS. In diamond cubic unit cell, there, are eight corner atoms, six face centered atoms and, four more atoms inside the structure., Number of atoms present in a diamond cubic cell, 1, 8, , 1, 2, , = 8´ + 6´ + 4 = 8, (corners) (face (inside, centered) body), 148. (d) The bcc cell consists of 8 atoms at the corners and one, atom at centre. Contribution of each atom at each corner, is equal to, , 1, ., 8, , 1, 1 2, 8, The fcc cell consists of 8 atoms at the eight corners, and one atom at each of the six faces. This atom at the, face is shared by two unit cells., n, , 8, , n 8, , 1, 8, , 6, , 1, 2, , 4, , 1, 1, 1, 2:, 4 =1:1:2, 8:, 8, 2, 2, 1, 8 1, 150. (d) In a unit cell, W atoms at the corner, 8, 1, 12 3, O-atoms at the centre of edges, 4, Na-atoms at the centre of the cube = 1, W : O : Na = 1 : 3 : 1, Hence, formula = NaWO3, 149. (b) A : B : C =
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THE SOLID STATE, , 253, , 151. (a) Potassium crystallises in bcc lattice., 152. (a) From the given data, we have, Number of Y atoms in a unit cell = 4, , Area of hexagonal, , 2 16, 3 3, From the above we get the formula of the compound, , Number of X atoms in a unit cell, , 8, , 1, 4, , 6 3 r2, , 4+1=1+1=2, , 1, 1, ×6+ ×8=3+1=4, 2, 8, So, formula = M2X4 = MX2, , 1, 6, 2, , 3, , 155. (a), 156. (a) Following generalization can be easily derived for, various types of lattice arrangements in cubic cells, between the edge length (a) of the cell and r the radius, of the sphere., a, For simple cubic : a = 2r or r, 2, For body centred cubic :, 4, r or r, 3, , 2 r2, L2, , 2( 2 r ) 2, , 100 =, , NA V 2 6 1023 (5 10 8 )3, M, 75, Z = 2, which represents bcc structure, Z, , r, , 1, , a, 2 2, Thus the ratio of radii of spheres for these will be, simple : bcc : fcc, , 2 r2, , 100, , 100 78.54%, 4, 159. (a) In fcc structure one unit cell consist 4 atoms, hence, density × a3 = The mass of four atoms, = 8.92 × (3.608 × 10–8)3, = 4.18 × 10–22, ZM, 160. (a), NA V, , =, , 3, a, 4, , 2 2r or r, , 60.43%, , 158. (d) Packing efficiency, , For face centred cubic :, a, , 100, , = Area occupied by circles within the square, Area of square, , Formula AB3, , a, , 3 r2, , 6, , 2 3.14, 100, 6 3, , 1, 8 1, 8, , Number of B atoms =, , 2 r2, , % occupied by, , No. of X atoms =, , 154. (a) Number of A atoms =, , 3, 4r 2, 4, , =6, , as X16 / 3Y4 or X 4 Y3, 153. (b) No. of M atoms =, , 3, (2r)2, 4, , 6, , 161. (d), , 3, 3, a, 5 2.165Å, 4, 4, 217 pm, a3, , M, , N A 10, Z, , 30, , 10 (100)3 6.02 1023 10, 4, , 1, a 3, = :, a:, a, 2, 2, 2 4, i.e. option (a) is correct answer., , Number of atoms in 100 g =, , 157. (a), , 216.5 pm, , 30, , 15.05, , 6.02 1023, 100, 15.05, , = 4 × 1025, 162. (b) The volume of the unit cell, = (2.88 Å)3 = 23.9 × 10–24 cm3., The volume of 100 g of the metal, Let radius of the sphere = r, Area occupied by sphere in hexagonal close packing, , r2 6, , 1, 6, , r2, , 2 r2, , m, , 100, = 13.9 cm3, 7.20, Number of unit cells in this volume, , =, , =, , =, , 13.9 cm3, 23.9 10, , 24, , cm3, , = 5.82 × 1023
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EBD_7207, , THE SOLID STATE, , 254, , 163. (b) Density is given by, Z M, , d, , NA a 3, , ; where Z = number of formula units present, , 165. (a) Given: Atoms are present in the corners of cube = A, and atom present at body centre = B. We know that a, cubic unit cell has 8 corners. Therefore contribution of, 1, . Since number of atoms, 8, per unit cell is 8, therefore total contribution =, , in unit cell, which is 4 for fcc, a = edge length of unit cell. M = Molecular mass, 4 M, , 2.72 =, 6.02 10, , M=, 164. (d), , =, , 2.7 =, , 23, , 404 10, , 2.72 ´ 6.02 ´ (404)3, , 4 ´107, , Z M, No a3, , 10 3, , (, , 1pm 10, , 10, , each atom at the corner =, , cm), , = 26.99 = 27 gm mole–1, 166. (a), , ,, , 167. (a), 168. (b), Z 27, , 6.02 10, , 23, , ( 4.05) 3 10, , 24, , Hence it is face centred cubic unit lattice., Again 4r = a 2 = 5.727 Å, r = 1.432 Å, , 1, 1 . We also know that atoms in the body centre,, 8, therefore number of atoms per unit cell = 1. Thus, formula of the compound is AB., In first case conduction is due to hole,while in second, case it is due electron., BaTlO3, Substances which are weakly repelled by external, magnetic field are called diamagnetic substances, e.g.,, H2O., Ferrimagnetic substance become para-magnetic on, heating. This is due to randomisation of spins on, heating., 171. (a), 8, , Z=4, 169. (d), , 170. (b)
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16, SOLUTIONS, FACT / DEFINITION TYPE QUESTIONS, 1., , 2., , 3., , 4., , 5., , 6., , 7., , 8., , “The importance of many pure substance in life depends, on their composition.”, Which of the following statement justify the above fact?, (a) 1 ppm of fluoride ions in water prevents tooth decay., (b) 1.5 ppm of fluoride ions causes tooth decay., (c) Concentration above 1.5 ppm can be poisonous., (d) All of the above., Which of the following fluoride is used as rat poison?, (a) CaF2, (b) KF, (c) NaF, (d) MgF2, Most of the processes in our body occur in, (a) solid solution, (b) liquid solution, (c) gaseous solution, (d) colloidal solution, The term homogenous mixtures signifies that, (a) its composition is uniform throughout the mixture., (b) its properties are uniform throughout the mixture., (c) both composition an d properties are un iform, throughout the mixture., (d) neither composition nor properties are uniform, throughout the mixture., Which of the following mixture is(are) called solution?, (i) water + ammonia, (ii) water + acetone, (iii) acetone + alcohol, (iv) hexane + water, (a) (i), (ii) and (iii), (b) (i), (iii) and (iv), (c) (i) and (iv), (d) (ii) and (iii), Which of the following is a quantitative description of the, solution?, (a) Dilute, (b) Concentrated, (c) Saturated, (d) Molar, When a solute is present in trace quantities the following, expression is used, (a) Gram per million, (b) Milligram percent, (c) Microgram percent, (d) Parts per million, Molarity of liquid HCl will be, if density of solution is, 1.17 gm/cc, (a) 36.5, (b) 32.05, (c) 18.25, (d) 42.10, , 9., , 10., , 11., , 12., , 13., , 14., , 15., , 1 M, 2.5 litre NaOH solution is mixed with another 0.5 M, 3, litre NaOH solution. Then find out the molarity of resultant, solution, (a) 0.80 M, (b) 1.0 M, (c) 0.73 M, (d) 0.50 M, An X molal solution of a compound in benzene has mole, fraction of solute equal to 0.2. The value of X is, (a) 14, (b) 3.2, (c) 1.4, (d) 2, The molarity of the solution containing 7.1 g of Na2SO4 in, 100 ml of aqueous solution is, (a) 2 M, (b) 0.5 M, (c) 1 M, (d) 0.05 M, The vapour pressure of pure benzene at 25°C is 640 mm Hg, and that of solution of solute A is 630 mm Hg. The molality, of solution is, (a) 0.2 m, (b) 0.4 m, (c) 0.5 m, (d) 0.1 m, 4.0 g of NaOH is dissolved in 100 ml solution. The normality, of the solution is, (a) 0.1 N, (b) 0.5 N, (c) 4.0 N, (d) 1.0 N, The molarity of pure water is, (a) 50 M, (b) 18 M, (c) 55.6 M, (d) 100 M, An aqueous solution of glucose is 10% in strength. The, volume in which 1 g mole of it is dissolved, will be, (a) 9 litre, (b) 1.8 litre, (c) 8 litre, (d) 0.9 litre, , 16. 10 g of NaCl is dissolved in 10 6 g of the solution. Its, concentration is, (a) 100 ppm, (b) 0.1 ppm, (c) 1 ppm, (d) 10 ppm, 17. On adding a solute to a solvent having vapour pressure, 0.80 atm, vapour pressure reduces to 0.60 atm. Mole fraction, of solute is, (a) 0.25, (b) 0.75, (c) 0.50, (d) 0.33
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EBD_7207, SOLUTIONS, , 256, , 18., , 2.5 litres of NaCl solution contain 5 moles of the solute., What is the molarity?, (a) 5 molar, (b) 2 molar, (c) 2.5 molar, (d) 12.5 molar, 19. The mole fraction of the solute in one molal aqueous, solution i, (a) 0.009, (b) 0.018, (c) 0.027, (d) 0.036, 20. 5 ml of N HCl, 20 ml of N/2 H2SO4 and 30 ml of N/3 HNO3, are mixed together and volume made to one litre. The, normality of the resulting solution is, N, 10, N, N, (c), (d), 20, 40, 25ml of a solution of barium hydroxide on titration with a 0.1, molar solution of hydrochloric acid gave a titre value of 35, ml. The molarity of barium hydroxide solution was, (a) 0.07, (b) 0.14, (c) 0.28, (d) 0.35, Mole fraction of the solute in a 1.00 molal aqueous solution, is, (a) 0.1770, (b) 0.0177, (c) 0.0344, (d) 1.7700, What is the normality of a 1 M solution of H3PO4 ?, (a) 0.5 N, (b) 1.0 N, (c) 2.0 N, (d) 3.0 N, The volume of 4 N HCl and 10 N HCl required to make 1 litre, of 6 N HCl are, (a) 0.75 litre of 10 N HCl and 0.25 litre of 4 N HCl, (b) 0.50 litre of 4 N HCl and 0.50 litre of 10 N HCl, (c) 0.67 litre of 4 N HCl and 0.33 litre of 10 N HCl, (d) 0.80 litre of 4 N HCl and 0.20 litre of 10 N HCl, Molarity of H2SO4 is 18 M. Its density is 1.8 g/ml. Hence, molality is, (a) 36, (b) 200, (c) 500, (d) 18, 200 ml of water is added to 500 ml of 0.2 M solution. What is, the molarity of this diluted solution ?, (a) 0.5010 M, (b) 0.2897 M, (c) 0.7093 M, (d) 0.1428 M, How many grams of concentrated nitric acid solution should, be used to prepare 250 mL of 2.0M HNO3 ? The concentrated, acid is 70% HNO3, (a) 90.0 g conc. HNO3, (b) 70.0 g conc. HNO3, (c) 54.0 g conc. HNO3, (d) 45.0 g conc. HNO3, For preparing 0.1 N solution of a compound from its impure, sample of which the percentage purity is known, the weight, of the substance required will be, (a) Less than the theoretical weight, (b) More than the theoretical weight, (c) Same as the theoretical weight, (d) None of these, , (a), , 21., , 22., , 23., , 24., , 25., , 26., , 27., , 28., , N, 5, , 29., , N, N, N, 50 ml H2SO4,, 30 ml HNO3,, 10 ml HCl is mixed, 10, 3, 2, and solution is made to 1L. Then normality of resultant, solution is, , If, , (a), , N, 20, , (b), , N, 40, , N, (d) N, 50, A solution made by dissolving 40 g NaOH in 1000 g of water, is, (a) 1 molar, (b) 1 normal, (c) 1 molal, (d) None of these, Which of the following concentration terms is/are, independent of temperature?, (a) Molality only, (b) Molality and mole fraction, (c) Molarity and mole fraction, (d) Molality and normality, A solution is prepared by dissolving 10 g NaOH in 1250 mL, of a solvent of density 0.8 mL/g. The molality of the solution, in mol kg–1 is, (a) 0.25, (b) 0.2, (c) 0.008, (d) 0.0064, Which of the following units is useful in relating, concentration of solution with its vapour pressure?, (a) mole fraction, (b) parts per million, (c) mass percentage, (d) molality, For mixture containing “four” components which of the, following is correct in term of mole fraction?, (a) x1+ x2+ x3+ x4, n3, x3, (b), n1 n 2 n 3, , (c), , 30., , (b), , 31., , 32., , 33., , 34., , n1, n1, n1 n 2 n 3 n 4, n, (d) n1 + n2 + n3 + n4 = 1, Which of the following concentration unit is independent, of temperature ?, (a) Normality, (b) Molarity, (c) Formality, (d) Molality, Which of the following factor do not affect solubility of, solid solute in liquid ?, (a) Temperature, (b) Pressure, (c) Nature of solute, (d) All of these, When a solid solute is added to the solvent, some solute, dissolves and its concentration increases in solution. This, process is known as ______. Some solute particles in, solution collide with the solid solute particles and get, separated out of solution. This process is known as, ______., (a) Crystallization, dissolution., (b) Dissolution, saturation., (c) Saturation, crystallization., (d) Dissolution, crystallization., , (c, , 35., , 36., , 37., , x1
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SOLUTIONS, , 38., , 39., , 40., , 41., , 42., , 257, , At the state of dynamic equilibrium, for, solute + solvent, solution., (a) Rate of dissolution = Rate of unsaturation., (b) Rate of dissolution = Rate of unsaturation., (c) Rate of dissolution = Rate of saturation, (d) Rate of crystallization = Rate of saturation., Which of the following statements is incorrect?, (a) A solution in which no more solute can be dissolved, at the same temperature and pressure is called a, saturated solution., (b) An unsaturated solution is one in which more solute, can be dissolved at the same temperature., (c) The solution which is in dynamic equilibrium with, undissolved solute is the saturated solution., (d) The minimum amount of solute dissolved in a given, amount of solvent is its solubility., On dissolving sugar in water at room temperature solution, feels cool to touch. Under which of the following cases, dissolution of sugar will be most rapid ?, (a) Sugar crystals in cold water., (b) Sugar crystals in hot water., (c) Powdered sugar in cold water., (d) Powdered sugar in hot water., The solubility of a solid in a liquid is significantly affected, by temperature changes., Solute + Solvent, Solution., The system being in a dynamic equilibrium must follow, Le-chatelier’s principle. Considering the Le-chatelier’s, principle which of the following is correct?, (a), , sol > 0; solubility, , ; temperature, , (b), , sol < 0; solubility, , ; temperature, , (c), , sol > 0; solubility, , ; temperature, , (d), , sol < 0;, , ;, , solubility, , temperature, , W1, Piston, , (a), , W1 W2 W3, , (b), , On the basis of the figure given above which of the, following is not true?, (a) In figure (a) assuming the state of dynamic equilibrium, rate of gaseous particles entering and leaving the, solution phase is same., (b) In figure (b) on compressing the gas number of, gaseous particles per unit volume over the solution, increases., (c) Rate at which gaseous particles are striking the, solution to enter it, decreases., (d) Rate at which gaseous particles are striking the, solution to enter it, increases., , 43. The statement “If 0.003 moles of a gas are dissolved in 900, g of water under a pressure of 1 atmosphere, 0.006 moles, will be dissolved under a pressure of 2 atmospheres”,, illustrates, (a) Dalton’s law of partial pressure, (b) Graham’s law, (c) Raoult’s law, (d) Henry’s law, 44. According to Henry’s law, the amount of gas that will, dissolve in blood plasma or any other liquid is determined, by which of these factor?, (a) Solubility of the gas in the liquid., (b) The total pressure of the gas mixture ., (c) pH of the liquid., (d) The osmotic pressure of the gas mixture., 45. Henry’s law constant of oxygen is 1.4 × 10–3 mol. lit–1. atm–1, at 298 K. How much of oxygen is dissolved in 100 ml at 298, K when the partial pressure of oxygen is 0.5 atm?, (a) 1.4 g, (b) 3.2 g, (c) 22.4 mg, (d) 2.24 mg, 46. At equillibrium the rate of dissolution of a solid solute in a, volatile liquid solvent is ______., (a) less than the rate of crystallisation., (b) greater than the rate of crystallisation., (c) equal to the rate of crystallisation., (d) zero, 47. A beaker contains a solution of substance ‘A’. Precipitation, of substance ‘A’ takes place when small amount of ‘A’ is, added to the solution. The solution is ______., (a) saturated, (b) supersaturated, (c) unsaturated, (d) concentrated, 48. Maximum amount of a solid solute that can be dissolved in, a specified amount of a given liquid solvent does not depend, upon __________., (a) Temperature, (b) Nature of solute, (c) Pressure, (d) Nature of solvent, 49. Low concentration of oxygen in the blood and tissues of, people living at high altitude is due to _________., (a) low temperature, (b) low atmospheric pressure, (c) high atmospheric pressure, (d) both low temperature and high atmospheric pressure, 50. Value of Henry's constant KH _______., (a) increases with increase in temperature., (b) decreases with increase in temperature., (c) remains constant., (d) first increases then decreases., 51. The value of Henry's constant KH is _______., (a) greater for gases with higher solubility., (b) greater for gases with lower solubility., (c) constant for all gases., (d) not related to the solubility of gases.
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EBD_7207, SOLUTIONS, , 258, , 55., , 56., , 57., , 58., , (d), , 63., , Iodine and sulphur dissolve in, (a) water, (b) benzene, (c) carbon disulphide, (d) ethanol, The liquids at a given temperature vapourise and under, equilibrium conditions the pressure exerted by the vapours, of the liquid over the liquid phase is called, (a) osmotic pressure, (b) atmospheric pressure, (c) hydrostatic pressure (d) vapour pressure, The vapour pressure of the solution at a given temperature, is found to be …………… than the vapour pressure of the, pure solvent at the same temperature., (a) higher, (b) lower, (c) equal, (d) can't calculate, The decrease in the vapour pressure of solvent depends, on the, (a) quantity of non-volatile solute present in the solution, (b) nature of non-volatile solute present in the solution, (c) molar mass of non-volatile solute present in the, solution, (d) physical state of non-volatile solute present in the, solution, A plot of p1 or p2 vs the mole fractions x1 and x2 is given, as., , Mole fraction of gas, in solution, , 1, , K H < p1°, , 61., , 62., , Which is an application of Henry’s law?, (a) Spray paint, (b) Bottled water, (c) Filling up atire, (d) Soft drinks (soda), Scuba divers may experience a condition called ______., To avoids this, the tanks used by scuba divers are filled, with air diluted with _____ ., (a) Migrains, Hydrogen (b) Cramps, Nitrogen, (c) Nausea, Oxygen, (d) Bends, Helium, People living at high attitudes often reported with a problem, of feeling weak and inability to think clearly. The reason for, this is., (a) at high altitudes the partial pressure of oxygen is less, than at the ground level., (b) at high altitudes the partial pressure of oxygen is more, than at the ground level., (c) at high altitudes the partial pressure of oxygen is equal, to at the ground level., (d) None of these., ____ a contemporary of Henry concluded independently, that solubility of a gas in a liquid solution is a function of, ____ of the gas., (a) Mosley, temperature, (b) Dalton, temperature, (c) Dalton, partial pressure, (d) Mosley, partial pressure, Raoult’s law becomes a special case of Henry’s law when, (a) K H = p °, (b) K H > p °, (c), , 60., , Mole fraction of gas, in solution, , Mole fraction of gas, in solution, , 54., , Partial pressure of gas, , Mole fraction of gas, in solution, , 59., , 1, , (d), , K H ³ p1°, , Vapour pressure, , (c), , (b), , Partial pressure of gas, , (a), , Partial pressure of gas, , 53., , Which of the followingfactor(s) affect the solubility of a, gaseous solute in the fixed volume of liquid solvent ?, (i) Nature of solute (ii) Temperature (iii) Pressure, (a) (i) and (iii) at constant T, (b) (i) and (ii) at constant P, (c) (ii) and (iii) only, (d) (iii) only, Which of the following graph is a correct representation of, Henry’s law?, , Partial pressure of gas, , 52., , p Total, , p1°, , = p1 +, , p2, , III p2°, II, , p2, p, , 1, , I, , x1 = 1 Mole fraction x1 = 0, x2 = 0, x2 = 1, x2, In this figure, lines I and II pass through the point for, which., (a) x1 1; x2 = 1, (b) x1 = x2 1, (c) x1 = 1; x2, d) x1 = x2 = 1, 64. The vapour pressure of two liquids ‘P’ and ‘Q’ are 80, and 60 torr, respectively. The total vapour pressure of, solution obtained by mixing 3 mole of P and 2 mole of Q, would be, (a) 72 torr, (b) 140 torr, (c) 68 torr, (d) 20 torr, 65. 18 g of glucose (C6H12O6) is added to 178.2 g of water. The, vapour pressure of water for this aqueous solution is, (a) 76.00 torr, (b) 752.40 torr, (c) 759.00 torr, (d) 7.60 torr, 66. PA and PB are the vapour pressure of pure liquid components,, A and B, respectively of an ideal binary solution. If XA, represents the mole fraction of component A, the total, pressure of the solution will be., (a) PA + XA (PB – PA), (b) PA + XA (PA – PB), (c) PB + XA (PB – PA), (d) PB + XA (PA – PB)
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SOLUTIONS, , 67., , 68., , 69., , 70., , 71., , 72., , 73., , 74., , 75., , A mixture of ethyl alcohol and propyl alcohol has a vapour, pressure of 290 mm Hg at 300 K. The vapour pressure of, propyl alcohol is 200 mm Hg. If the mole fraction of ethyl, alcohol is 0.6, its vapour pressure (in mm Hg) at the same, temperature will be, (a) 360, (b) 350, (c) 300, (d) 700, Two liquids X and Y form an ideal solution. At 300 K, vapour, pressure of the solution containing 1 mol of X and 3 mol of, Y is 550 mm Hg. At the same temperature, if 1 mol of Y is, further added to this solution, vapour pressure of the, solution increases by 10 mm Hg. Vapour pressure ( in mm, Hg) of X and Y in their pure states will be, respectively, (a) 300 and 400, (b) 400 and 600, (c) 500 and 600, (d) 200 and 300, The vapour pressure of two liquids X and Y are 80 and 60, torr respectively. The total vapour pressure of the ideal, solution obtained by mixing 3 moles of X and 2 moles of Y, would be, (a) 68 Torr, (b) 140 Torr, (c) 48 Torr, (d) 72 Torr, The vapour pressure of pure benzene and toluene at a, particular temperature are 100 mm and 50 mm respectively., Then the mole fraction of benzene in vapour phase in, contact with equimolar solution of benzene and toluene is, (a) 0.67, (b) 0.75, (c) 0.33, (d) 0.50, A solution containing components A and B follows Raoult's, law when, (a) A – B attraction force is greater than A – A and B – B, (b) A – B attraction force is less than A – A and B – B, (c) A – B attraction force remains same as A–A and B –B, (d) volume of solution is different from sum of volume of, solute and solvent, Relation between partial pressure and mole fraction is stated, by, (a) Graham’s law, (b) Raoult’s law, (c) Le-Chatelier, (d) Avogadro law, Which one of the following is not correct for an ideal, solution?, (a) It must obey Raoult’s law, (b), H 0, (c), H, V 0, (d) All are correct, An ideal solution is formed when its components, (a) have no volume change on mixing, (b) have no enthalpy change on mixing, (c) Both (a) and (b) are correct, (d) Neither (a) nor (b) is correct, All form ideal solution except, (a) C6H6 and C6H5CH3 (b) C2H6 and C2H5I, (c) C6H5Cl and C6H5Br (d) C2H5 I and C2H5OH., , 259, , 76. Which one of the following is non-ideal solution, (a) Benzene + toluene, (b) n-hexane + n-heptane, (c) Ethyl bromide + ethyl iodide, (d) CCl4 + CHCl3, 77. Mixtures of ethanol and acetone show positive deviation., The reason is, (a) In pure ethanol, molecules are hydrogen bonded., (b) In pure acetone, molecules are hydrogen bonded, (c) In both molecules are hydrogen bonded, (d) None of these, 78. A mixture of components A and B will show –ve deviation, when, (a), Vmix > 0, (b) Hmix < 0, (c) A – B interaction is weaker than A – A and B – B, interactions, (d) A–B interaction is stronger than A–A and B–B, interactions., 79. Which of the following liquid pairs shows a positive, deviation from Raoult’s law ?, (a) Water - Nitric acid, (b) Benzene - Methanol, (c) Water - Hydrochloric acid, (d) Acetone - Chloroform, 80. A solution of acetone in ethanol, (a) shows a positive deviation from Raoult’s law, (b) behaves like a non ideal solution, (c) obeys Raoult’s law, (d) shows a negative deviation from Raoult’s law, 81. Negative deviation from Raoult’s law is observed in, which one of the following binary liquid mixtures?, (a) Ethanol and acetone, (b) Benzene and toluene, (c) Acetone and chloroform, (d) Chloroethane and bromoethane, 82. Which one of the following binary liquid systems shows, positive deviation from Raoult’s law?, (a) Benzene-toluene, (b) Carbon disulphide-acetone, (c) Phenol-aniline, (d) Chloroform-acetone, 83. A binary liquid solution is prepared by mixing n-heptane, and ethanol. Which one of the following statements is, correct regarding the behaviour of the solution?, (a) The solution is non-ideal, showing negative deviation, from Raoult’s Law., (b) The solution is non-ideal, showing positive deviation, from Raoult’s Law., (c) n-heptane shows positive deviation while ethanol, shows negative deviation from Raoult’s Law., (d) The solution formed is an ideal solution., 84. Which one is not equal to zero for an ideal solution:, (a), Smix, (b) Vmix, (c), P = Pobserved - PRaoult, (d), Hmix
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EBD_7207, SOLUTIONS, , 260, , 85., , 86., , 87., , 88., , 89., , 90., , 91., , 92., , A mixture of two completely miscible non-ideal liquids which, distill as such without change in its composition at a, constant temperature as though it were a pure liquid. This, mixture is known as, (a) binary liquid mixture (b) azeotropic mixture, (c) eutectic mixture, (d) ideal mixture, The azeotropic mixture of water (b.p.100°C) and HCl, (b.p.85°C) boils at 108.5°C. When this mixture is distilled it, is possible to obtain, (a) pure HCl, (b) pure water, (c) pure water as well as pure HCl, (d) neither HCl nor H2O in their pure states, The system that forms maximum boiling azeotrope is, (a) carbondisulphide – acetone, (b) benzene – toluene, (c) acetone – chloroform, (d) n-hexane – n-heptane, Which one of the following binary mixtures forms an, azeotrope with minimum boiling point type?, (a) acetone-ethanol, (b) H2O-HNO3, (c) benzene-toluene, (d) n-hexane-n-heptane, On the basis of information given below mark the correct, option., Information: On adding acetone to methanol some of the, hydrogen bonds between methanol molecules break., (a) At specific composition methanol-acetone mixture will, form boiling azeotrope and will show positive deviation, from Raoult’s law., (b) At specific composition methanol-acetone mixture, forms boiling azeotrope and will show positive, deviation from Raoult’s law., (c) At specific composition methanol-acetone mixture will, form minimum boiling azeotrope and will show negative, deviation from Raoult’s law., (d) At specific composition methanol-acetone mixture will, form boiling azeotrope and will show negative, deviation from Raoult’s law., According to Raoult's law, relative lowering of vapour, pressure for a solution is equal to, (a) moles of solute, (b) moles of solvent, (c) mole fraction of solute (d) mole fraction of solvent, The relative lowering of the vapour pressure is equal to the, ratio between the number of, (a) solute molecules to the solvent molecules, (b) solute molecules to the total molecules in the solution, (c) solvent molecules to the total molecules in the solution, (d) solvent molecules to the total number of ions of the, solute., Vapour pressure of benzene at 30°C is 121.8 mm Hg. When, 15 g of a non volatile solute is dissolved in 250 g of benzene, its vapour pressure decreased to 120.2 mm Hg. The, molecular weight of the solute (Mo. wt. of solvent = 78), (a) 356.2, (b) 456.8, (c) 530.1, (d) 656.7, , 93., , The value of P° for benzene is 640 mm of Hg. The vapour, pressure of solution containing 2.5gm substance in 39gm., benzene is 600mm of Hg the molecular mass of X is, (a) 65.25, (b) 130, (c) 40, (d) 75, 94. The vapour pressure at a given temperature of an ideal, solution containing 0.2 mol of a non-volatile solute and 0.8, mol of solvent is 60 mm of Hg. The vapour pressure of the, pure solvent at the same temperature is, (a) 150 mm of Hg, (b) 60 mm of Hg, (c) 75 mm of Hg, (d) 120 mm of Hg, 95. 12 g of a nonvolatile solute dissolved in 108 g of water, produces the relative lowering of vapour pressure of 0.1., The molecular mass of the solute is, (a) 80, (b) 60, (c) 20, (d) 40, 96. The amount of solute (molar mass 60 g.mol –1) that must be, added to 180 g of water so that the vapour pressure of water, is lowered by 10% is, (a) 30 g, (b) 60 g, (c) 120 g, (d) 12 g, 97. The vapour pressure of water at 20°C is 17.5 mm Hg. If 18 g, of glucose (C6H12O6) is added to 178.2 g of water at 20°C,, the vapour pressure of the resulting solution will be, (a) 17.325 mm Hg, (b) 15.750 mm Hg, (c) 16.500 mm Hg, (d) 17.500 mm Hg, 98. Which one of the following is a colligative property ?, (a) Boiling point, (b) Vapour pressure, (c) Osmotic pressure, (d) Freezing point, 99. Which one of the following aqueous solutions will exihibit, highest boiling point ?, (a) 0.015 M urea, (b) 0.01 M KNO3, (c) 0.01 M Na2SO4, (d) 0.015 M glucose, 100. The rise in the boiling point of a solution containing 1.8 g of, glucose in 100 g of solvent is 0.1°C. The molal elevation, constant of the liquid is, (a) 0.01 K/m, (b) 0.1 K/m, (c) 1 K/m, (d) 10 K/m, 101. For an electrolyte, elevation of B.P. is directly proportional to, (a) molarity, (b) molality, (c) mole fraction, (d) All of these, 102. Which of the following aqueous solution has minimum, freezing point ?, (a) 0.01 m NaCl, (b) 0.005 m C2H5OH, (c) 0.005 m MgI2, (d) 0.005 m MgSO4., 103. 1.00 g of a non-electrolyte solute (molar mass 250 g mol–1), was dissolved in 51.2 g of benzene. If the freezing point, depression constant, Kf of benzene is 5.12 K kg mol –1, the, freezing point of benzene will be lowered by, (a) 0.3 K, (b) 0.5 K, (c) 0.4 K, (d) 0.2, 104. In a 0.2 molal aqueous solution of a weak acid HX the degree, of ionization is 0.3. Taking kf for water as 1.85, the freezing, point of the solution will be nearest to, (a) – 0.360ºC, (b) – 0.260ºC, (c) + 0.481ºC, (d) – 0.481ºC
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SOLUTIONS, , 105. The molecular weight of benzoic acid in benzene as, determined by depression in freezing point method, corresponds to, (a) ionization of benzoic acid., (b) dimerization of benzoic acid., (c) trimerization of benzoic acid., (d) solvation of benzoic acid., 106. A 0.5 molal solution of ethylene glycol in water is used as, coolant in a car. If the freezing point constant of water be, 1.86°C per mole, the mixture shall freeze at, (a) 0.93°C, (b) –0.93°C, (c) 1.86°C, (d) –1.86°C, 107. A solution of urea (mol. mass 56 g mol 1) boils at 100.18 C, at the atmospheric pressure. If Kf and Kb for water are, 1.86 and 0.512 K kg mol 1 respectively, the above solution, will freeze at, (a) 0.654 C, (b), 0.654 C, (c) 6.54 C, (d), 6.54 C, 108. The freezing point of 1% solution of lead nitrate in water will, be, (a) 2°C, (b) 1°C, (c) 0°C, (d) below 0°C, 109. A solution of sucrose (molar mass = 342 g mol–1) has been, prepared by dissolving 68.5 g of sucrose in 1000 g of water., The freezing point of the solution obtained will be, (Kf for water = 1.86 K kg mol–1)., (a) – 0.372°C, (b) – 0.520°C, (d) + 0.372°C, (d) – 0.570°C, 110. A solution containing 1.8 g of a compound (empirical formula, CH2O) in 40 g of water is observed to freeze at –0.465° C., The molecular formula of the compound is, (Kf of water = 1.86 kg K mol–1), (a) C2H4O2, (b) C3H6O3, (c) C4H8O4, (d) C6H12O6, 111. Blood cells retain their normal shape in solution which are, (a) hypotonic to blood, (b) isotonic to blood, (c) hypertonic to blood (d) equinormal to blood., 112. Isotonic solutions have same, (a) molar concentration (b) molality, (c) normality, (d) None of these, 113. As a result of osmosis, the volume of more concentrated, solution, (a) gradually decreases (b) gradually increases, (c) is not affected, (d) suddenly increases, 114. Which of the following pairs of solution are isotonic at the, same temperature ?, (a) 0.1 M Ca(NO3)2 and 0.1 M Na2SO4, (b) 0.1 M NaCl and 0.1 M Na2SO4, (c) 0.1 M urea and 0.1 M MgCl2, (d) 0.2 M urea and 0.1 M NaCl, 115. Osmotic pressure of 0.4% urea solution is 1.64 atm and that, of 3.42% cane sugar is 2.46 atm. When the above two, solutions are mixed, the osmotic pressure of the resulting, solution is :, (a) 0.82 atm, (b) 2.46 atm, (c) 1.64 atm, (d) 4.10 atm, , 261, , 116. Osmotic pressure of a solution at a given temperature, (a) increases with concentration, (b) decreases with concentration, (c) remains same, (d) initially increases and then decreases, 117. At 25°C, at 5 % aqueous solution of glucose (molecular, weight = 180 g mol–1) is isotonic with a 2% aqueous solution, containing an unknown solute. What is the molecular weight, of the unknown solute?, (a) 60, (b) 80, (c) 72, (d) 63, 118. Which one of the following statements is false?, (a) Raoult’s law states that the vapour pressure of a, component over a binary solution of volatile liquids is, directly proportional to its mole fraction, (b) Two sucrose solutions of the same molality prepared, in different solvents will have the same depression of, freezing point, (c) The correct order of osmotic pressures of 0.01 M, solution of each compound is, BaCl2 > KCl > CH3COOH > glucose, (d) In the equation osmotic pressure = MRT, M is the, molarity of the solution, 119. Which of the following statements is false?, (a) Units of atmospheric pressure and osmotic pressure, are the same., (b) In reverse osmosis, solvent molecules move through, a semipermeable membrane from a region of lower, concentration of solute to a region of higher, concentration., (c) The value of molal depression constant depends on, nature of solvent., (d) Relative lowering of vapour pressure, is a, dimensionless quantity., 120. During osmosis, flow of water through a semipermeable, membrane is, (a) from both sides of semipermeable membrane with equal, flow rates, (b) from both sides of semipermeable membrane with, unequal flow rates, (c) from solution having lower concentration only, (d) from solution having higher concentration only, 121. If 0.1 M solution of glucose and 0.1 M solution of urea are, placed on two sides of the semipermeable membrane to, equal heights, then it will be correct to say that, (a) there will be no net movement across the membrane, (b) glucose will flow towards urea solution, (c) urea will flow towards glucose solution, (d) water will flow from urea solution to glucose, 122. The van’t Hoff factor i for a compound which undergoes, dissociation in one solvent and association in other solvent, is respectively, (a) less than one and greater than one., (b) less than one and less than one., (c) greater than one and less than one., (d) greater than one and greater than one.
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EBD_7207, SOLUTIONS, , 262, , 123. If the various terms in the given below expressions have, usual meanings, the van’t Hoff factor (i) cannot be calculated, by which one of the following expressions, (a) V, inRT, (b) T f iK f .m, (c), (d), , Tb, , iK b .m, , Psolvent, , Psolution, , Psolvent, , i, , n, N n, , 124. Van’t Hoff factor is given by the expression ________., , i, , Normal molar mass, Abnormal molar mass, , (b) i, , Abnormal molar mass, Normal molar mass, , (a), , (c), , i, , Observed colligative property, Calculated colligative property, , (d) Both (a) and (c), 125. We have three aqueous solutions of NaCl labelled as 'A', 'B', and 'C' with concentrations 0.1M, 0.01M and 0.001M,, respectively. The value of van't Hoff factor for these, solutions will be in the order ________., (a) iA < iB < iC, (b) iA > iB > iC, (c) iA = iB = iC, (d) iA < iB > iC, 126. If is the degree of dissociation of Na 2SO4, the Vant Hoff’s, factor (i ) used for calculating the molecular mass is, (a) 1 +, (b) 1 –, (c) 1 + 2, (d) 1 – 2, 127. The freezing point of equimolal aqueous solutions will be, highest for, (a) C6H5NH3+Cl– (aniline hydrochloride), (b) Ca(NO3)2, (c) La(NO3)3, (d) C6H12O6 (glucose), 128. The correct relationship between the boiling points of very, dilute solutions of AlCl3 (t1) and CaCl2 (t2), having the, same molar concentration is, (a) t1 = t2, (b) t1 > t2, (c) t2 > t1, (d) t2 t1, 129. At 25°C, the highest osmotic pressure is exhibited by 0.1 M, solution of, (a) CaCl2, (b) KCl, (c) glucose, (d) urea, 130. Which one of the following salts will have the same value of, van’t Hoff factor (i) as that of K4[Fe(CN)6]., (a) Al2(SO4)3, (b) NaCl (c), Al(NO3)3, (d) Na2SO4., 131. Which of the following 0.10 m aqueous solutions will have, the lowest freezing point ?, (a) Al2(SO4)3, (b) C6H12O6, (c) KCl, (d) C12 H22 O11, , 132. The elevation in boiling point of a solution of 13.44 g of, CuCl2 in 1 kg of water using the following information will, be (Molecular weight of CuCl2 = 134.4 g and Kb= 0.52 K, kg mol 1), (a) 0.16, (b) 0.05, (c) 0.1, (d) 0.2, 133. Freezing point of an aqueous solution is, – 0.186°C. If the values of Kb and Kf of water are respectively, 0.52 K kg mol–1 and 1.86 K kg mol–1, then the elevation of, boiling point of the solution in K is, (a) 0.52, (b) 1.04, (c) 1.34, (d) 0.052, , STATEMENT TYPE QUESTIONS, 134. Study the given statements and choose the correct option., (i) 3.62 mass percentage of sodium hypochlorite in water, is used as commercial bleaching solution., (ii) 35% volume percentage of ethylene glycol is used as, an antifreeze (as coolent in car engines)., (iii) Concentration of dissolved oxygen in a litre of sea, water is 5.8 ppm., (a) Statements (i) and (ii) are correct, (b) Statements (i) and (iii) are correct, (c) Statements (ii) and (iii) are correct, (d) Statements (i),(ii) and (iii) are correct, 135. Molarity and molality of a solution of NaOH is calculated., If now temperature of the solution is increased then which, of the following statement(s) is/are correct ?, (i) Molarity of solution decreases, (ii) Molality of the solution increases, (a) Both statements are correct, (b) Statement (i) is correct only, (c) Statement (ii) is correct only, (d) Both statements are incorrect., 136. “If temperature increases solubility of gas decreases”. For, this situation which of the following statement(s) is/are, correct ?, (i) Reaction is endothermic, (ii) Le-chatelier’s principle can be applied, (a) Statement (i) and (ii) both are correct, (b) Statement (i) is correct only, (c) Statement (ii) is correct only, (d) Both statement(s) (i) and (ii) are incorrect, 137. Read the following statements carefully and choose the, correct option., (i) Different gases have different KH values at the same, temperature., (ii) Higher the value of KH at a given temperature, lower, is the solubility of the nature of gas in the liquid., (iii) KH is a function of the nature of the gas., (iv) Solubility of gases increases with increase of, temperature., (a) (i), (ii) and (iv) are correct., (b) (ii) and (iv) are correct., (c) (i), (ii) and (iii) are correct., (d) (i) and (iv) are correct.
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SOLUTIONS, , 138. Read the following statements and choose the correct, option., (i) Polar solutes dissolve in a polar solvent., (ii) Polar solutes dissolve in a non-polar solvent., (iii) Non-polar solutes dissolve in a non-polar solvent., (iv) Non-polar solutes dissolve in a polar solvent., (a) (i) and (ii) are correct., (b) (i), (ii) and (iii) are correct., (c) (i) and (iii) are correct., (d) (ii) and (iv) are correct., 139. Read the following statements carefully and choose the, correct option, (i) The vapour pressure of a liquid decreases with, increase of temperature., (ii) The liquid boils at the temperature at which its vapour, pressure is equal to the atmospheric pressure., (iii) Vapour pressure of the solvent decreases in the, presence of non-volatile solute., (iv) Vapour pressure of the pure solvent and solution is a, function of temperature., (a) (i), (ii) and (iv) are correct, (b) (i), (iii), and (iv) are correct, (c) (ii), (iii), and (iv) are correct, (d) (i), (ii) and (iii) are correct, 140. On the basis of information given below mark the correct, option., (i) In bromoethane and chloroethane mixture, intermolecular interactions of A-A and B-B type are, nearly same as A-B type interactions., (ii) In ethanol and acetone mixture A-A or B-B type, intermolecular interactions are stronger than A-B type, interactions., (iii) In chloroform and acetone mixture A-A or B-B type, intermolecular interactions are weaker than A-B type, interactions., (a) Solution (ii) and (iii) will follow Raoult's law., (b) Solution (i) will follow Raoult's law., (c) Solution (ii) will show negative deviation from Raoult's, law., (d) Solution (iii) will show positive deviation from Raoult's, law., 141. Which observation(s) reflect(s) colligative properties?, (i) A 0.5 m NaBr solution has a higher vapour pressure, than a 0.5 m BaCl2 solution at the same temperature, (ii) Pure water freezes at the higher temperature than pure, methanol, (iii) a 0.1 m NaOH solution freezes at a lower temperature, than pure water, Choose the correct answer from the codes given below, (a) (i), (ii) and (iii), (b) (i) and (ii), (c) (ii) and (iii), (d) (i) and (iii), , 263, , 142. Read the following statements carefully and choose the, correct option, (i) Osmotic pressure is not a colligative property., (ii) For dilute solutions, osmotic pressure is proportional, to the molarity, C of the solution at a given temperature, T., (iii) During osmosis ,solvent molecules always flow from, higher concentration to lower concentration of, solution., (iv) The osmotic pressure has been found to depend on, the concentration of the solution, (a) (i), (ii) and (iv) are correct, (b) (ii) and (iv) are correct, (c) (iii), and (iv) are correct, (d) (i), (ii) and (iii) are correct, , MATCHING TYPE QUESTIONS, 143. Match the columns, Column -I, (A) Mass percentage, (B) Mass by volume, , Column-II, (p) Medicine and pharmacy, (q) Concentration of, pollutants in water, (C) ppm, (r) Industrial chemical, application, (D) Volume percentage, (s) Liquid solutions, (a) A – (q), B – (p), C – (s), D – (r), (b) A – (s), B – (r), C – (p), D – (q), (c) A – (r), B – (q), C – (s), D – (p), (d) A – (r), B – (p), C – (q), D – (s), 144. Match the columns, Column-I, Column-II, (A) Na-Hg Amalgam, (p) gas - solid, (B) H2 in Pd, (q) gas - liquid, (C) Camphor in nitrogen gas, (r) liquid - solid, (D) Oxygen dissolved in water (s) solid - gas, (a) A – (q), B – (s), C – (r), D – (p), (b) A – (t), B – (p), C – (q), D – (s), (c) A – (r), B – (p), C – (s), D – (q), (d) A – (s), B – (q), C – (p), D – (p), 145. Match the Column I, II & III and choose the correct option., Column-I, Column-II, Column-III, (A) Gaseous (p) Solid-liquid (h) Copper dissolved, solutions, in gold, (B) Liquid, (q) Solid-solid (i) Chloroform mixed, solutions, with nitrogen, (C) Solid, (r) Liquid-gas (j) Common salt, solutions, dissolved in water, (a) (A) – (r) – (h), (B) – (r) – (i), (C) – (p) – (j), (b) (A) – (r) – (i), (B) – (p) – (j), (C) – (q) – (h), (c) (A) – (r) – (j), (B) – (p) – (h), (C) – (q) – (i), (d) (A) – (r) – (j), (B) – (q) – (i), (C) – (p) – (h)
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EBD_7207, 264, , 146. Match the columns, Column-I, (A) Saturated solution, , Column-II, (p) Solution having same, osmotic pressure at a given, temperature as that of given, solution., (B) Binary solution, (q) A solution whose osmotic, pressure is less than that, of another., (C) Isotonic solution, (r) Solution with two, components, (D) Hypotonic solution (s) A solution which contains, maximum amount of solute, that can be dissolved in a, given amount of solvent at, a given temperature., (a) A – (s), B – (r), C – (p), D – (q), (b) A – (r), B – (p), C – (s), D – (q), (c) A – (r), B – (s), C – (q), D – (p), (d) A – (q), B – (p), C – (r), D – (s), 147. Match the laws given in the Column-I with expression given, in Column-II., Column-I, Column-II, (A) Raoult’s law, (p) Tf = Kfm, (B) Henry’s law, (q) = CRT, o, o, (C) Elevation of boiling point, (r) p x1p1 x 2 p 2, (D) Depression in freezing point (s) Tb = Kbm, (E) Osmotic pressure, (t) p = KH.x, (a) A – (r), B – (t), C – (s), D – (p), E – (q), (b) A – (t), B – (r), C – (q), D – (s), E – (p), (c) A – (p), B – (t), C – (r), D – (q), E – (s), (d) A – (s), B – (p), C – (q), D – (r), E – (t), , ASSERTION-REASON TYPE QUESTIONS, Directions : Each of these questions contain two statements,, Assertion and Reason. Each of these questions also has four, alternative choices, only one of which is the correct answer. You, have to select one of the codes (a), (b), (c) and (d) given below., (a) Assertion is correct, reason is correct; reason is a correct, explanation for assertion., (b) Assertion is correct, reason is correct; reason is not a, correct explanation for assertion, (c) Assertion is correct, reason is incorrect, (d) Assertion is incorrect, reason is correct., 148. Assertion : Molarity of a solution in liquid state changes, with temperature., Reason : The volume of a solution changes with change in, temperature., 149. Assertion : If a liquid solute more volatile than the solvent, is added to the solvent, the vapour pressure of the solution, may increase i.e., ps > po., Reason : In the presence of a more volatile liquid solute,, only the solute will form the vapours and solvent will not., 150. Assertion : If one component of a solution obeys Raoult’s, law over a certain range of composition, the other component, will not obey Henry’s law in that range., Reason : Raoult’s law is a special case of Henry’s law., , SOLUTIONS, , 151. Assertion : Azeotropic mixtures are formed only by, non-ideal solutions and they may have boiling points either, greater than both the components or less than both the, components., Reason : The composition of the vapour phase is same as, that of the liquid phase of an azeotropic mixture., 152. Assertion : When methyl alcohol is added to water, boiling, point of water increases., Reason : When a volatile solute is added to a volatile solvent, elevation in boiling point is observed., 153. Assertion : When NaCl is added to water a depression in, freezing point is observed., Reason : The lowering of vapour pressure of a solution, causes depression in the freezing point., 154. Assertion : When a solution is separated from the pure, solvent by a semi- permeable membrane, the solvent, molecules pass through it from pure solvent side to the, solution side, Reason : Diffusion of solvent occurs from a region of high, concentration solution to a region of low concentration, solution., , CRITICAL THINKING TYPE QUESTIONS, 155. The normality of orthophosphoric acid having purity of 70, % by weight and specific gravity 1.54 is, (a) 11 N, (b) 22 N, (c) 33 N, (d) 44 N, 156. Which of the following statements, regarding the mole, fraction (x) of a component in solution, is incorrect?, (a) 0 x 1, (b) x 1, (c) x is always non-negative, (d) None of these, 157. Which one of the following gases has the lowest value of, Henry’s law constant?, (a) N2, (b) He, (c) H2, (d) CO2, 158. Equal masses of methane and oxygen are mixed in an empty, container at 25°C. The fraction of the total pressure exerted, by oxygen is, (a) 1/2, (b) 2/3, 1 273, (c), (d) 1/3, 3 298, 159. When a gas is bubbled through water at 298 K, a very dilute, solution of the gas is obtained. Henry’s law constant for, the gas at 298 K is 100 kbar. If the gas exerts a partial pressure, of 1 bar, the number of millimoles of the gas dissolved in, one litre of water is, (a) 0.555, (b) 5.55, (c) 0.0555, (d) 55.5, 160. KH value for Ar(g), CO2(g), HCHO (g) and CH4(g) are 40.39,, 1.67, 1.83 × 10–5 and 0.413 respectively., Arrange these gases in the order of their increasing, solubility., (a) HCHO < CH4 < CO2 < Ar, (b) HCHO < CO2 < CH4 < Ar, (c) Ar < CO2 < CH4 < HCHO, (d) Ar < CH4 < CO2 < HCHO
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SOLUTIONS, , 161. What is the ratio of no. of moles of nitrogen to that of, oxygen in a container of 5 litre at atmospheric pressure?, (a) 1 : 1.71, (b) 1 : 2, (c) 2 : 1, (d) 1 : 24, 162. Consider a and b are two components of a liquid mixture,, their corresponding vapour pressure (mmHg) are, respectively 450 and 700 in pure states and total pressure, given is 600. Then corresponding composition in liquid, phase will be, (a) 0.4, 0.6, (b) 0.5, 0.5, (a) 0.6, 0.4, (d) 0.3, 0.7, 163. Which will form maximum boiling point azeotrope, (a) HNO3 + H2O solution (b) C2H5OH + H2O solution, (c) C6H6 + C6H5CH3 solution (d) None of these, 164. If two liquids A and B form minimum boiling azeotrope at, some specific composition then _______., (a) A – B interactions are stronger than those between, A – A or B – B, (b) vapour pressure of solution increases because more, number of molecules of liquids A and B can escape, from the solution., (c) vapour pressure of solution decreases because less, number of molecules of only one of the liquids escape, from the solution, (d) A – B interactions are weaker than those between A –, A or B – B, 165. Chloroform and acetone are added to each other, Raoult’s, law shows negative deviation.what does this suggests ?, (a) Exothermic reaction, (b) Endothermic reaction, (c) Zero change in enthalpy, (d) None of these, 166. At 300 K the vapour pressure of an ideal solution containing, 1 mole of liquid A and 2 moles of liquid B is 500 mm of Hg., The vapour pressure of the solution increases by 25 mm of, Hg, if one more mole of B is added to the above ideal solution, at 300 K. Then the vapour pressure of A in its pure state is, (a) 300 mm of Hg, (b) 400 mm of Hg, (c) 500 mm of Hg, (d) 600 mm of Hg, 167. Someone has added a non electrolyte solid to the pure, liquid but forgot that among which of the two beakers he, has added that solid. This problem can be solved by, checking, (a) relative lower in vapour pressure, (b) elevation in boiling point, (c) depression in Freezing point, (d) all above, 168. The vapour pressure of a solvent decreases by 10 mm of Hg, when a non-volatile solute was added to the solvent. The, mole fraction of the solute in the solution is 0.2. What should, be the mole fraction of the solvent if the decrease in the, vapour pressure is to be 20 mm of Hg ?, (a) 0.8, (b) 0.6, (c) 0.4, (d) 0.2, 169. Vapour pressure of benzene at 30°C is 121.8 mm. When 15g, of a non-volatile solute is dissolved in 250 g of benzene, its, vapour pressure is decreased to 120.2 mm. The molecular, weight of the solute is, , 265, , (a) 35.67 g, (b) 356.7 g, (c) 432.8 g, (d) 502.7 g, 170. For a dilute solution containing 2.5 g of a non-volatile nonelectrolyte solute in 100 g of water, the elevation, in boiling point at 1 atm pressure is 2°C. Assuming, concentration of solute is much lower than the concentration, of solvent, the vapour pressure (mm of Hg) of the solution, is (take Kb = 0.76 K kg mol–1), (a) 724, (b) 740, (c) 736, (d) 718, 171. The difference between the boiling point and freezing point, of an aqueous solution containing sucrose (molecular wt =, 342 g mole–1 ) in 100 g of water is 105°C. If Kf and, Kb of water are 1.86 and 0.51 K kg mol –1 respectively, the, weight of sucrose in the solution is about, (a) 34.2 g, (b) 342 g, (c) 7.2 g, (d) 72 g, 172. If the elevation in boiling point of a solution of non-volatile,, non-electrolytic and non-associating solute in a solvent, (Kb = x K kg mol–1) is y K, then the depression in freezing, point of solution of same concentration would be (Kf of the, solvent = z K kg mol–1), (a), , 174., , 175., , 176., , (b), , xz, y, , (d), , yz, x, , yz, 2x, 1 g of a non-volatile, non-electrolyte solute of molar mass, 250 g/mol was dissolved in 51.2 g of benzene. If the, freezing point depression constant Kf of benzene is 5.12, kg K mol–1. The freezing point of benzene is lowered by, (a) 0.3 K, (b) 0.5 K, (c) 0.2 K, (d) 0.4 K, The boiling point of 0.2 mol kg–1 solution of X in water is, greater than equimolal solution of Y in water. Which one of, the following statements is true in this case ?, (a) Molecular mass of X is greater than the molecular mass, of Y., (b) Molecular mass of X is less than the molecular mass of, Y., (c) Y is undergoing dissociation in water while X, undergoes no change., (d) X is undergoing dissociation in water., What is the freezing point of a solution containing 8.1 g, HBr in 100 g water assuming the acid to be 90% ionised ?, (Kf for water = 1.86 K kg mol–1) :, (a) 0.85°K, (b) – 3.53°K, (c) 0°K, (d) – 0.35°K, An 1% solution of KCl (I), NaCl (II), BaCl2 (III) and urea (IV), have their osmotic pressure at the same temperature in the, ascending order (molar masses of NaCl, KCl,, BaCl2 and urea are respectively 58.5, 74.5, 208.4 and, 60 g mole–1). Assume 100% ionization of the electrolytes at, this temperature, (a) I < III < II < IV, (b) III < I < II < IV, (c) I < II < III < IV, (d) III < IV < I < II, , (c), , 173., , 2xz, y
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EBD_7207, SOLUTIONS, , 266, , FACT / DEFINITION TYPE QUESTIONS, 1., 2., 3., 4., 5., 6., , 7., 8., , 14., , (d), (c) Sodium fluoride is used as rat poison., (b) Almost all the processes in our body occur in liquid, solution., (c) In homogeneous mixtures composition and properties, both are uniform throughout the mixture., (a) Hexane is not water soluble, hence solution is not, formed., (d) Dilute, concentrated and saturated terms are, qualitative methods of description of concentration, of solution whereas molar or molarity is quantitative, method., (d), (b) Density = 1.17 gm/cc, (Given), Mass, As d =, Volume, volume = 1cc, mass = d = 1.17g, No. of moles, 1.17 ´1000, Molarity =, =, Volume in litre, 36.5 ´1, , 10., , Wt 1000, ; M, M.Wt . V, , (b), , M, , 12., , (a), , p, p, , n2, n1 n2, , or, , 7.1 1000, 142 100, , n2, n1 n2, , 13., , (d) Normality, , 15., , Normality, , 4 / 40, 100 /1000, , 1N, , (b), , =, 16., , 100, 180 = 1800 ml = 1.8 L, 10, Mass of solute, 106, Mass of solution, , (d) ppm, ppm, , 17., , (a), , p, , p, p, , 10, 106, , 106, , 10 ppm, , xsolute, , 0.2, 0.8 0.6, xsolute or, 0.8, 0.8, or xsolute = 0.25, , 18., 19., , 0.0156, 1, , 20., , 5, No. of moles of solute, 2M, 2.5, Volume in litres, (b) One molal solution means one mole of solute is present, in 1 kg (1000 g) solvent, i.e., mole of solute = 1, , 21., , 22., , 1000g 1000, =, 18, 18g, 1, 18, =, = 0.018., 1000, 1008, 1, 18, , (d) NV = N1V1 + N2V2 + N3V3, or, 1000N 1 5, , 0.2 m, , xsolute ,, , (b) Molarity, , Mole of solvent (H2O) =, , Moles of solute, Volume of solution in litre, , Given mass of solute = 4.0 g, No. of moles of NaOH = 4/40, Volume of solution = 100 ml = 100/1000 L, , 55.6, = 55, 1, 10 g glucose is dissolved in = 100 ml solution., 180 g (g mole) is dissolved in, , Mole fraction of solute =, , or 1 mole i.e., 78g benzene contains solute = 0.0156 mol, 0.0156 103, Molality of solution =, 78, , 55.6, , Molarity =, , 0.5M, , 640 630, 640, , 1000, 18, , Given, p° = 0.8 atm, p = 0.6 atm, xsolute = ?, , 1170, =32.05M, 36.5, (c) From molarity equation, M1V1 + M2V2 = M3(V1 + V2), 1× 2.5 + 0.5 × 3 = M3 × 5.5, 4, = 0.73M, M3 =, 5.5, (b) Relation between molality and mole fraction is, 1000 x2 1000 .2, m, 3.2, x1M1, 0.8 78, Thus, X (m) = 3.2, , 11., , Mass, Molar mass, , Moles of water, , =, , 9., , Number of moles, Volume of solution (L), , (c) Molarity, , 1, 1, 20, 30 or N, 2, 3, , 1, ., 40, , (a) For HCl, M = N = 0.1, N1V1 = N2V2 ; 25 × N1 = 0.1 × 35, 0.1 35, 0.1 35, ; M, 0.07 ., N1, 25, 25 2, (b) 1 molal solution means 1 mole of solute dissolved in, 1000 gm solvent., nsolute = 1, wsolvent = 1000 gm
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SOLUTIONS, , 267, , nsolvent =, , 1000, = 55.56, 18, , 1, = 0.0177, 1 55.56, (d) H3PO4 is tribasic so N = 3M = 3 × 1N = 3N., (c) N1V1 + N2V2 = NV, 4x + 10 (1 – x) = 6 × 1; –6x = –4 ; x = 0.67, Thus 0.67 litre of 4N HCl, 1 – x = 1 – 0.67 = 0.33 litre of 10 N HCl, , xsolute =, , 23., 24., , 25., , (c), , Normality =, Molarity =, , 36., , wt 1000, (d) Molarity (M) =, mol. wt. vol (ml), 2, , wt. 1000, 63 250, , wt. =, , 37., 38., , 63, gm, 2, , 100, 31.5 = 45 gm, 70, (b) More than theoretical weight since impurity will not, contribute., (c) Applying the law of equivalence,, N1V1 N 2V2 N3V3 N RVR, , wt. of 70% acid =, , 28., 29., , N, 50, 10, , N, 3, , 5 N 10 N, , 30, , N, 10, 2, , N R 1000, , 30., , 39., 40., 41., , 5 N = 1000 N R, , N, 50, (c) 40 g NaOH = 1 mole, , 42., , NR, , moles of the solute, 1, 1 molal, =, mass of the solvent in kg 1, (b) Both molality and mole fraction are not related to the, volume of solution, thus they are both independent of, temperature., (a) Given w = 10 g, Mol. mass = 40, Weight of solvent = 1250 × 0.8 g = 10000 g = 1 kg, molality =, , 31., 32., , molality, , 10, 40 1, , 0.25, , n3, n3 n 4, , x3, , 35. (d) Volume is temperature dependent, hence expression, involving volume term (normality, molarity and, formality) varies with temperature, , 18, 5000, 18 98, 1.8, 1000, (d) No. of millimoles = 500 × 0.2 = 100, Thus, molarity of diluted solution, , 100, 700, (Molarity = No. of moles L–1 = No. of millimoles mL–1), 0.1428 M, , 27., , (b) n n, 1, 2, , Molarity, Molarity Molecular mass, 1000, , Molality (m), Density, , 26., , 33. (a), 34. (c) In a given solution sum of all the mole fraction is unity, i.e.,, (a) x1 + x2 + x3 + x4 = 1, , 43., 44., , geq of solute, vol. of solution in L, , moles of solute, vol. of solution in L, , Formality =, , formula mass, vol. of solution in L, , Molality =, , moles of solute, mass of solvent in kg, , Since molality does not include the volume term, it is, independent of temperature., (b) An increase in temperature of the solution increases, the solubility of a solid solute., The amount of solute that dissolve depends on what, type of solute it is., For solids and liquid solutes, changes in pressure have, practically no effect on solubility., (d), (b) Number of solute particles going into solution will be, equal to the solute particles separating out and a state, of dynamic equilibrium is reached., solute + solvent, solution., i.e., rate of dissolution = rate of crystallization., (d) The maximum amount of solute dissolved in a, given amount of solvent is its solubility., (d), (b) According to Le-chateliers principle, for an exother, mic reaction ( < 0) increase in temperature decreases, the solubility., (c) On increasing the pressure over the solution phase, by compressing the gas to a smaller volume (in fig b), increase the number of gaseous particles per unit, volume over the solution and also the rate at which, the gaseous particles are striking the surface of, solution to enter it. The solubility of the gas will, increase until a new equilibrium is reached resulting, in an increase in the pressure of a gas above the, solution and thus its solubility increases., (d), (a) According to Henry’s law at a constant temperature,, the solubility of a gas in a liquid is directly proportional, to the pressure of the gas.
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EBD_7207, SOLUTIONS, , 268, , 45., , (d) According to Henry’s law,, m=k×p, given KH = 1.4 × 10–3, pO2 = 0.5 or, KH, , pO2, x O2, , x O2, 3, , m, M, , No. of moles; n, 0.7 10 4, , 48., 49., 50., 53., , 54., 55., , 56., , 57., 58., 59., 61., , m, 32, , 22.4 10, , (a) For example, decrease in the vapour pressure of water, by adding 1.0 mol of sucrose to one kg of water is, nearly similar to that produced by adding 1.0 mol of, urea to the same quantity of water at the same, temperature., (d), (a) Given V.PP = 80 torr, V.PQ = 60 torr, Ptotal = V·PP × xp + V·Pq × xq, = 80, , 4, , g 2.24 mg, (c), (b) [ Hint : If added substance dissolves, the solution is, unsaturated. If it does not dissolve solution is, saturated. If precipitation occurs solution is, supersaturated.], (c), (b) Body temperature of human beings remains constant., (a), 51. (b), 52. (a), (c) The partial pressure of the gas in vapour phase (p) is, proportional to the mole fraction of the gas (x)., p = KH .x, Where KH is Henry’s constant., (d) To increase the solubility of CO2 in soft drinks and, soda water, the bottle is sealed under high pressure., (d) Scuba divers must cope with high concentrations of, dissolved gases while breathing air at high pressure, underwater. Increased pressure increases the solubility, of atmospheric gases in blood. When the divers come, towards surface, the pressure gradually decreases., This releases the dissolved gases and leads to the, formation of bubbles of nitrogen in the blood. This, blocks capillaries and creates a medical condition, known as bends, which are painful and dangerous to, life. To avoid bends, as well as, the toxic effects of, high concentrations of nitrogen in the blood, the tanks, used by scuba divers are filled with air diluted with, helium (11.7% helium, 56.2% nitrogen and 32.1%, oxygen)., (a) At high altitudes the partial pressure of oxygen is, less than that at the ground level. This leads to low, concentrations of oxygen in the blood and tissues of, people living at high altitudes or climbers. Low blood, oxygen causes climbers to become weak and unable, to think clearly, symptoms of a condition known as, anoxia., (c), (a) Raoult’s law becomes special case of Henry’s law when, KH become equal to p °., 1, (c), 60. (d), (b) In the solution, the surface has both solute and solvent, molecules; thereby the fraction of the surface covered, by the solvent molecules gets reduced. Consequently,, the number of solvent molecules escaping from the, surface is correspondingly reduced, thus, the vapour, pressure is also reduced., m, , 46., 47., , 63., 64., , 0.5, 1.4 10, , 62., , 3, 2, 60, = 16 × 3 + 12 × 2, 5, 5, , Ptotal = 48 + 24 = 72 torr, 65., , (b) Moles of glucose, , 18, 180, , 0. 1, , 78.2, 9.9, 18, Total moles = 0.1 + 9.9 = 10, , Moles of water, , pH2O = Mole fraction × Total pressure, , 66., , 67., , 9.9, 760, 10, , = 752.4 Torr, (d) p = pAxA + pBxB, = pAxA + pB (1 – xA), pAxA + pB – pBxA, pB+ xA (pA – pB), (b), , ? , Given p B, , pA, , 200 mm of Hg, xA = 0.6,, , xB = 1 – 0.6 = 0.4, P = 290 of Hg, P = PA + PB = PA x A, , PB xB, , 290 = PA × 0.6 + 200 × 0.4, 68., , (b), , ptotal, , pA xA, , p A = 350 mm of Hg., , pB x B, , 1, 3, p, 4 B 4, pA 3 pB 550 4, In second case, 1, 4, ptotal pA, pB, 5, 5, pA 4 pB 560 5, Subtract (i) from (ii), , 550, , 69., , pA, , ...(i), , ...(ii), , pB, , 560 5 550 4 600, , pA, , 400, , (d) According to given information, pX = 80 Torr, pY = 60 Torr, nX = 3 moles, nY = 2 moles, mole fraction of X(xX), , nX, nX nY, , 3, 3 2, , 3, 5
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SOLUTIONS, , 269, , mole fraction of Y(xY), , nY, 2 2, nX nY 3 2 5, , Total Pressure, P = pXxX + pYxY, , 70., , 3, 2, 80, 60 = 48 + 24 = 72 Torr.., 5, 5, (a) Total vapour pressure = vapour pressure of pure, benzene + vapour pressure of toluene, = 100 + 50 = 150 mm, We know,, PC6H6, , P xC6 H6, , 100 150 xC6 H 6, , 100, 0.67, 150, (c) These two components A and B follows the condition, of Raoult’s law if the force of attraction between A and, B is equal to the force of attraction between A and A or, B and B., (b) According to Raoult’s law “The partial pressure of a, volatile component of a solution is directly proportional, to its mole fraction in solution at any temperature”., p = Px, xC6 H 6, , 71., , 72., , 73., 74., , where, p = Partial pressure of component, P° = Vapour pressure of component in pure form, x = mole fraction of component in solution., (c) For an ideal solution, H 0 , V 0, Hence, option (c) is incorrect., (c) For ideal solution,, , Vmixing, 75., 76., 77., , 78., , 79., , 0 and, , H mixing, , 0., , (d) C2H5I and C2H5OH form non-ideal solution., (d), (a) On adding acetone, its molecules get in between the, host molecules and break some of the hydrogen bonds, between them. Due to weakening of interactions, the, solution shows positive deviation from Raoult's law., (d) A solution containing A and B components shows, negative deviation when A–A and B–B interactions, are weaker than that of A–B interactions. For such, solutions., H = –ve and V = –ve, (b) Positive deviations are shown by such solutions in, which solvent-solvent and solute-solute interactions, are stronger than the solute-solvent interactions. In, such solution, the interactions among molecules, becomes weaker. Therefore their escaping tendency, increases which results in the increase in their partial, vapour pressures., In pure methanol there exists intermolecular, H–bonding., ---O – H--- O—H --- O—H--|, |, |, CH3, CH3, CH3, , On adding benzene, its molecules come between, ethanol molecules there by breaking H-bonds which, weaken intermolecular forces. This results in increase, in vapour pressure., 80. (a) A solution of acetone in ethanol shows positive, deviation from Raoult's law. It is because ethanol, molecules are strongly hydrogen bonded. When, acetone is added, these molecules break the hydrogen, bonds and ethanol becomes more volatile. Therefore, its vapour pressure is increased., 81. (a) Acetone and chloroform shows negative deviation, from Raoult's law when these are mixed, the hydrogen, bonding takes place between the two molecular species, due to which escaping tendency of either liquid, molecules becomes less and boiling point of solution, increases., 82. (b), 83. (b) For this solution intermolecular interactions between, n-heptane and ethanol are weaker than n-heptane-nheptane & ethanol-ethanol interactions hence the, solution of n-heptane and ethanol is non-ideal and, shows positive deviation from Raoult’s law., 84. (a) For an ideal solution Smix > 0, 85. (b), 86. (d) Azeotropic mixture is constant boiling mixture, it is not, possible to separate the components of azeotropic, mixture by boiling., 87. (c), 88. (a) Minimum boiling azeotrope is formed by solution, showing positive deviation. e.g. acetone – ethanol., 89. (b), 90. (c) Relative lowering of vapour pressure depends upon, the mole fraction of solute., P P, i.e.,, mole fraction of solute, P, 91. (b) According to Raoult's law, the relative lowering in, vapour pressure of a dilute solution is equal to the mole, fraction of the solute present in the solution., p, , p, , n, n N, 92. (a) Given vapour pressure of pure solvent, (P°) = 121.8 mm Hg; Weight of solute (w) = 15 g, Weight of solvent (W) = 250 g; Vapour pressure of, solution (P) = 120.2 mm Hg and Molecular weight of, solvent (M) = 78, From Raoult’s law, p, , Mole fraction of solute, , Po, P, , P, o, , w M, m W, , 121.8 120.2 15 78, 121.8, m 250, 15 78 121.8, or, m, 356.2, 250, 1.6
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EBD_7207, SOLUTIONS, , 270, , 93., , (d), , P, , P, , n2, n1 n2, , P, , 640 600, 640, , 0.1 180 100, 1K/m, 1.8 1000, 101. (b), Tb = Kb × i × m, Where Tb = Elevation in boiling point, Kb = molal elevation constant, i = vant Hoff factor, , 100. (c), , 2.5 x, 39 78, , 640 78 2.5, 80, 39 40, (c) According to Raoult's law, p p, xB, p, x, , 94., , Mole fraction of solute, , xB, p, , p, , .2, .2 .8, , 1, 5, , 60 mm of Hg, , T, , 1, or 4 p ( p) 5, p, 5, 60´5, = 75mm of Hg, Þ p° =, 4, , 95., , (c), , P, , Ps, , Po, , 0.1 =, m=, 96., , 97., , (b), , =, , 100, , P, , ps, ps, ps, , [, , p, , 17.5 ], , 0.1 ps, or ps = 17.325 mm Hg., 9.9, Hence (a) is correct answer., (c) Osmotic pressure is a colligative property., (c), Tb Kb m, Elevation in boiling point is a colligative property,, which depends upon the no. of particles, (concentration of solution). Thus greater the number, of particles, greater is the elevation in boiling point, and hence greater will be its boiling point., Na2SO4, 2Na+ + SO42–, Since Na2SO4 has maximum number of particles (c), hence has maximum boiling point., , or 17.5 – ps, , 98., 99., , 18/180, 178.2/18, , 0.4K, , X , i = 1.3); Tf = Kf × m × i, , 0.3, , 0. 3, , Tf, , Hence, we have m, , M1, 100 10, w1, , Moles of glucose in solution, Moles of water in solution, , 17.5 ps, , 1 1000, 250 51.2, , Tf = T f, T f = 0 – 0.481ºC = – 0.481ºC, 105. (b) Benzoic acid exists as dimer in benzene., 0.93 C, T f K f m 1.86 0.5 0.93 C; T f, 106. (b), 107. (b) As Tf = Kf. m, Tb = Kb. m, , w 2 18, 100 10 or w2 = 60 g, 60 180, Thus, 60 g of the solute must be added to 180 g of, water so that the vapour pressure of water is lowered, by 10%., (a) The vapour pressure of a solution of glucose in water, can be calculated using the relation, , p, , M 2W1, , 5.12, , Tf = 1.85 × 0.2 × 1.3 = 0.481º C, , 20, , w2, M2, , K f W2 1000, , H, , 1 0.3, , 12 18, m 108, , P, , Kfm, , 104. (d) ( HX, , w M, m W, , n, N, , 12 18, 0.1 108, , P, , or, , Tb molality.., 102. (a), Tf = i × Kf × m, Van't Hoff factor, i = 2 for NaCl, m = 0.01, hence Tf = 0.02 Kf which is maximum in the present, case., Hence Tf is maximum or freezing point is minimum., 103. (c), , p, , o, , Kb, , or, , Tf, , Tb, , [ Tb, , Kf, , Tb, Kb, , Kf, Kb, , 100.18 100 0.18 C], , 1.86, = 0.654°C, 0.512, As the Freezing Point of pure water is 0°C,, Tf = 0 –Tf, 0.654 = 0 – Tf, Tf = – 0.654, Thus the freezing point of solution will be – 0.654°C., 108. (d) Addition of solute to water decreases the freezing point, of water (pure solvent)., When 1% lead nitrate (solute) is added to water, the, freezing point of water will be below 0°C., , = 0.18 ×, , 109. (a), , Tf, , Kf, , 1000W2 1.86×1000×68.5, = 0.372, =, 342×1000, M 2W1, , Tf = T °f – Tf, Tf = – 0.372°C, 110. (d), Tf = Kf × m, M, , 1000 K f, , Tf, , w 2 solute, , w1 solvent, , 1000 1.86 1.8, 0.465 40, , M = 180
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SOLUTIONS, , 271, , Molecular formula = (empirical formula)n, n, , Molecular mass, Empirical formula mass, , 180, 30, , 6, , Molecular formula = (CH2O)6 = C6H12O6., 111. (b) Blood cells neither swell nor shrink in isotonic solution., As isotonic solutions have equal concentration, therefore there is no flow of solvent occurs and hence, solvent neither enters nor flow out of the blood cells., 112. (a) Isotonic solutions have same molar concentration at, given temperature provided the Van't Hoff factor (i) is, same., 113. (b), 114. (a) The solution which provide same number of ions are, isotonic., Ca (NO3)2, Ca2+ + 2 NO3–, Total ions produced = 3, Na2SO4, 2 Na+ + SO42–, Total ions produced = 3, 0.1 M Ca (NO3)2 and 0.1 M Na2SO4 are isotonic., 115. (d) Osmotic pressure is a colligative property. Hence, resulting osmotic pressure of the solution is given by, T = 1 + 2 + 3 .........., T = 1.64 + 2.46 = 4.10 atm., 116. (a) According to Boyle-van't Hoff law,, C (at constant temp), 117. (c) Isotonic solutions have same osmotic pressure, glucose = unknown solute, m1, M1, , 118. (b ), , m2, 5, or, M2, 180, , 2, M2, , M2, , 119. (b), 120. (c) During osmosis water flows through semipermeable, membrane from lower concentration to higher, concentration., 121. (a) As both the solutions are isotonic hence there is no, net movement of the solvent occurs through the, semipermeable membrane between two solutions., 122. (c) If compound dissociates in solvent i > 1 and on, association i < 1., 123. (a) Van’t Hoff equation is, V = inRT, For depression in freezing point., Tf = i × Kf × m, For elevation in boiling point., Tb = i × Kb × m, For lowering of vapour pressure,, , Psolution, , Psolvent, , i, , n, N n, , ., , 125. (c), , 126. (c) Na2SO4, 2Na+ + SO 24, Mol. before dissociation 1, Mol. after dissociation 1 –, , 0, 2, , 0, 1, , i 1, 2, 1 2, 127. (d) The salt that ionises to least extent will have highest, freezing point. [ i.e., minimum Tf ], 128. (b) AlCl3 furnishes more ions than CaCl2 and thus possess, higher boiling point i.e., t1 > t2., 129. (a) Concentration of particles in CaCl2 solution will be, maximum as i = 3 for CaCl2 and i = 2 for KCl., Glucose and Urea do not dissociate into ions, as they, are nonelectrolytes., , 130. (a), , K 4 [Fe(CN) 6 ] and Al 2(SO 4 )3 both dissociates to, give 5 ions or i = 5, K4 [Fe(CN)6], 4K+ + [Fe(CN)6]4–, 2Al3+ + 3SO 4--, , and Al2 (SO4 )3, , 131. (a) Depression in freezing point No. of particles., (when concentration of different solutions is equal), Al 2 (SO4 )3 provides five ions on ionisation, Al 2 (SO 4 ) 3, 2 Al 3, 3SO 42 –, while KCl provides two ions, , KCl ¾¾, ® K + + ClC6 H12 O6 and C12 H22 O11 are not ionised so they, have single particle in solution., Hence, Al2(SO4)3 have maximum value of depression, in freezing point or lowest freezing point., , 72, , Tf = Kf × m, Kf is a characteristic of a particular solvent i.e., it will, be different for different solvents., , Psolvent, , 124. (d), , No. of particles after ionisation, No. of particles before ionisation, , 132. (a) (i) i, (ii), , Tb = i × Kb × m, , Cu2+ + 2Cl, 0, 0, 2, , CuCl2, 1, (1 ), , 1 2, , i=1+2, 1, Assuming 100% ionization, So, i = 1 + 2 = 3, i, , Tb = 3 0.52 0.1 = 0.156 0.16 [m, 133. (d), , Tf, , i.k f .m ;, , Tf, Tb, , kf, kb, , Tf, , 0 ( 0.186 C), , 0.186, Tb, , 1.86, 0.52, , Tb, , Tb, , 13.44, 134.4, , i.k b .m, , 0.186 C, 0.52 0.186, 1.86, , 0.052, , 0.1]
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EBD_7207, SOLUTIONS, , 272, , 156. (a) Mole fraction of any component A in solution, , STATEMENT TYPE QUESTIONS, 134. (d), 135. (b) Molarity include volume thus with increase, intemperature increases’ volume increases, hence, molarity decreases while in case of molality mass of, solvent is taken, which is not effected by temperature., 136. (c) As temperature increases solubility of gas decreases,, so dissolution of gas can be considered as exothermic, process., 137. (c) Solubility of gases increase with decrease of, temperature., 138. (c) A solute dissolves in a solvent if intermolecular, interactions are similar in the two or we may say like, dissolves like., 139. (c) The vapour pressure of a liquid increases with, increase of temperature., 140. (b), 141. (d) Colligative properties depends upon the no. of, particles. Since methanol is non electrolyte hence, cannot be considered., 142. (b) Osmotic pressure is a colligative property as it, depends on the number of solute molecules and not, on their identity. For dilute solutions, during osmosis, solvent molecules always flow from lower, concentration to higher concentration of solution., , MATCHING TYPE QUESTIONS, 143. (d), , 144. (c), , 145. (b), , 146. (a), , 147. (a), , No. of moles of A, Total No. of moles of solution, , x, , As total no. of moles of solution > No. of moles of A, Thus x can never be equal to one or zero., 157. (d) According to Henry’s law the mass of a gas dissolved, per unit volume of solvent is proportional to the, pressure of the gas at constant temperature m = K p, i.e. as the solubility increases, value of Henry’s law, constant decreases. Since CO2 is most soluble in water, among the given set of gases. Therefore CO2 has the, lowest value of Henry’s law constant., 158. (d) Let the mass of methane and oxygen = m gm., Mole fraction of O2, =, , Moles of O2, Moles of O 2 + Moles of CH 4, , =, , m / 32, 1, m / 32, =, =, 3, m / 32 m /16 3m / 32, , Partial pressure of O2 = Total pressure × mole fraction, of O2 , PO2 = P ×, , 159. (a) kH = 100 kbar = 105 bar, p = 1 bar, p = k H × xA, p, kH, , xA =, , ASSERTION-REASON TYPE QUESTIONS, 148. (a), 149. (c) Both the solute and solvent will form the vapours but, vapour phase will become richer in the more volatile, component., 150. (b), 151. (b), 152. (d), 153. (a), 154. (b), , CRITICAL THINKING TYPE QUESTIONS, 155. (c) Equivalent weight of orthophosphoric acid, (H3PO4) =, , 3 31 64, 3, , 98, 3, , 100, 70, litre of solution contains, gm, 1000 1.54, 98 / 3, equivalent of H3PO4, Normality of solution, , 70 3, 10 1.54, 98, , 1, 100 103, , Moles of water =, , 1000, 18, , 10, , 5, , 55.5, , Weight of water = 1000 g (, Mole fraction = 10–5 =, , 1000 mL = 1000 g), , x, 55.5 x, , As 55.5 >>> x, thus neglecting x from denominator, 10, , 5, , x, 55.5, , 55.5 10 5 moles, , x, , or 0.555 millimoles., , Now 100 gm solution contains 70 gm H3PO4, , 70 3, 98, =, 1, 10 1.54, , 1 1, = P, 3 3, , 33 N, , 160. (c), 161. (a) % of N2 in atmosphere = 78.9%, % of O2 in atmosphere = 20.95%, Partial pressure of N2 = 0.789 atm = 0.799 bar, Partial pressure of O2 = 0.2095 atm = 0.212 bar, According to Henry’s law,, P, , (K H )O2 x, , 0.212, (34.86 1000), , x O2, , 6.08 10, , 6
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SOLUTIONS, , P, , 273, , 0.799, (76.48 1000), , (K H ) N 2 x, , n O2, n O2, , n H 2O, , n N2, , :, , x N2, , 1.0447 10, , 5, , n N2, n O2, , n H 2O, , n N2, , x O2 : x N 2, , 6.08 10 6 :1.04 10 5, = 1 : 1.71, 162. (a) According to idea of Raoult’s law “partial pressure of, one of the component is proportional to mole fraction, of that component in the solution.”, P = P1° x1 + P2° x 2, , 163., , 164., 165., , 166., , 600 = 450 x1 + 700 x2, 4.5 x1 + 7x2 = 6, x1 + x2 = 1, x1 = 0.6 x2 = 0.4, x1 = 0.4, x2 = 0.6, (a) The solutions (liquid mixture) which boils at constant, temperature and can distil as such without any change, in composition are called azeotropes., Solution of HNO3 and H2O will form maximum boiling, point azeotrope. Maximum boiling azeotropes show, negative deviation from Raoult’s law., Composition (%), Boiling Point, HNO3, 68.0, 359 K, H2O, 32.0, 373 K, Boiling point of the azeotrope of these two solutions, is 393.5 K., (a), (a) Since formation of hydrogen bonding takes place, due, to this bond energy is supposed to be released and, hence change in enthalpy is negative,so reaction is, exothermic., (a) According to Raoult’s law,, PT = xAp°A + xBp°B, Given, = PT1 500 mm Hg, nA = 1 and nB = 2, 500, , 1, pA, 3, , xA = 1/3 and xB = 2/3, , 2, pB, 3, , PT2 = 500 + 25 = 525 mm Hg, 525, , 1, pA, 4, , xA = 1/4 and xB = 3/4, 3, pB, 4, , p, p, , n, (mole fraction of solute), n N, , 10, p, , 0.2, , ...(ii), , Subtract (i) and (ii),, 2100 PA 3PB, p°B = 600 mm Hg, p°A + 2p°B = 1500 p°A = 300 mm Hg., , p° = 50 mm of Hg, , For other solution of same solvent, 20, p, , n, (Mole fraction of solute), n N, , 20, Mole fraction of solute, 50, Mole fraction of solute = 0.4, As mole fraction of solute + mole fraction of solvent = 1, Hence, mole fraction of solvent = 1 – 0.4 = 0.6, 169 (b) Relative lowering of vapour pressure is given by :, P – Ps, P, , w/m, w/ m W / M, , where, P° = vapour pressure of pure solvent, Ps = vapour pressure of solution, w = mass of solute, m = molecular mass of solute, W = mass of solvent, M = molecular mass of solvent, For dilute solution, 121.8 120.2, 121.8, , 15, 15/ m, =, 250 / 78 m, , 78, 1.3 10, 250, , 2, , m = 356.265, 170. (a) From Raoult law, p –p, p, , No.of moles of solute, No. of moles of solvent+ No. of moles of solute, , =, , When the concentration of solute is much lower than, the concentration of solvent,, p, , ...(i), 1500 pA 2pB, Also given Qn that one more mole of B is added to the, solution, the pressure of the ideal solution increases, by 25 mm Hg., Also, nB = 3, , 167. (d), 168. (b) According to Raoult's law, , p, , No. of moles of solute, No. of moles of solvent, , p, , Tb = Kb × m, Number of moles of the solute, 1000, Mass of solvent in grams, , m, Tb, , Kb, , Number of moles of the solute, × 1000, Mass of solvent in grams, , Number of moles of solute, Tb Mass of solvent in grams, K b 1000, , 2 100, 0.76 1000, , 0.26,
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EBD_7207, SOLUTIONS, , 274, , 100, 18, 760, p, From equation (i) we get,, 760, On solving, p = 724.46 724, 171. (d) (100 + Tb) – (0 – Tf) = 105, Tb + Tf = 5, m (kb + kf) = 5, , Number of moles of solvent, , m, , (or), , 5, 2.37, , i.e.,, , moles of solute, × 1.9, mass of solvent in kg, , 8.1/ 81, 1.9, 100 /1000, = 1.86 × 1 × 1.9 = 3.534K, , Tf, , Tf, , or, , Tf = 0 – 3.534 K, Tf = – 3.534K, 176. (d) 1% solution contains 1 g of the solute in 100 g of, solution., Osmotic pressure, = CRT, The value of R and T is same for all the solute however,, all of them undergo 100% dissociation, i×C, iKCl = 2, iNaCl = 2, iBaCl = 3 and iurea = 1., 2, nKCl = 1/74.5, , We know, Tf = kf × m, On substituting value of m,, yz, Tt =, x, 173. (d) Mass of non-volatile solute = 1g, Molar mass of solute = 250g mol–1, Mass of benzene = 51.2g, Kf = 5.12 K kg mol–1, 1000 w 2, M 2 w1, , 1/ 74.5, 1000, 100, , CKCl, , or, , 0.13, , KCl = 2 × 0.13 = 0.26, , nNaCl=, , 1, 58.5, , C NaCl, , 1/ 58.5, 1000, 100, , 0.17, , NaCl = 2 × 0.17 = 0.34, , where, w2= mass of the solute, M2 = molar mass of solute, w1 = mass of the solvent, , 174. (d), , 0, , = 1.86 ×, , Tf, , 5, 342 = 72g, 2.37 10, , y, x, , Tf, , ), , = 90% = 0.9, i = 1 + = 1 + 0.9 = 1.9, , 5, moles in 1000 g water, 2.37, , On substituting given values,, , Br, , 0, , = 1.86 ×, , 172. (b) Given kb, Tb = kb × m, y= x × m, , Tf, , (1, , 1, , T f = Kf × m × i, , = x K kg mol–1, , Kf, , H, , HBr, , 0.26, 5.56, , 5, moles in 100 g water, 2.37 10, , Wt. of sucrose, , m, , For HBr, , 5.56, , n, BaCl2, , Tf, , 5.12 1000 1, 51.2 250, , 0.4K, , Tb = iKb m, Given, ( Tb)x > ( Tb)y, ix Kb m > iyKb m, (Kb is same for same solvent), ix > iy, So, x is undergoing dissociation in water., 175. (b) Given mass of solute = 8.1 g, Mass of solvent = 100 g, , CBaCl2, , or, , =, , 1, 208.4, 1/ 208.4, 1000, 100, , BaCl2, , n urea, Curea, , 0.048, , 3 0.048 0.14, , 1, 60, 1/ 60, 1000, 100, , 0.16, , urea = 1 × 0.16 = 0.16, , BaCl2, , urea, , or III < IV < I < II, , KCl, , NaCl
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17, ELECTROCHEMISTRY, FACT / DEFINITION TYPE QUESTIONS, 1., , 2., , 3., , 4., , 5., , 6., , Which of the following statements is incorrect regarding, electrochemistry?, (a) It is the study of production of electricity from energy, released during spontaneous chemical reactions., (b) NaOH, Cl2, alkali and alkaline earth metals are prepared, by electrochemical methods., (c) The demerit associated with electrochemical methods, is that they are more polluting. Thus they are ecodestructive., (d) Electrochemical reactions are more energy efficient, and less polluting., What flows in the internal circuit of a galvanic cell?, (a) Ions, (b) Electrons, (c) Electricity, (d) Atoms, Which of the following statements about galvanic cell is, incorrect, (a) anode is positive, (b) oxidation occurs at the electrode with lower reduction, potential, (c) cathode is positive, (d) reduction occurs at cathode, Reaction that takes place at graphite anode in dry cell is, (a) Zn 2 2e, Zn(s), 2, (b) Zn(s), Zn, 2e, (c) Mn 2, 2e, Mn(s), (d) Mn(s), Mn, e 1.5V, In which of the following conditions salt bridge is not, required in a galvanic cell?, (a) When galvanic cell is used in geyser., (b) When distance between oxidation half cell and, reduction half cell is negligible., (c) Electrolytic solutions used in both the half cells are of, same concentration., (d) When both the electrodes are dipped in the same, electrolytic solution., Which device converts chemical energy of a spontaneous, redox reaction into electrical energy?, (a) Galvanic cell, (b) Electrolytic cell, (c) Daniell cell, (d) Both (a) and (c), , 7., , 8., , Which one is not called a anode reaction from the following?, (a), , Cl, , 1, Cl2, 2, , (c), , Hg, , Hg, , e, e, , (b), , Cu, , (d), , Zn2, , The cell reaction Cu 2Ag, represented by, , Cu, , 2e, , 2e, , Zn, , Cu, , (a), , Cu(s ) | Cu 2 (aq) | | Ag (aq) | Ag( s ), , (b), , Pt | Cu, , (c), , Cu, , 2, , 2, , 2, , Ag is best, , | | Ag ( aq ) | Ag( s), , | Cu | | Pt | Ag, , (d) None of the above representations, 9., , Zn (s ) | Zn 2 ( aq) | | Cu 2 ( aq) |Cu( s) is, anode, , cathode, , (a) Weston cell, (b) Daniel cell, (c) Calomel cell, (d) Faraday cell, 10. The tendency of an electrode to lose electrons is known as, (a) electrode potential, (b) reduction potential, (c) oxidation potential, (d) e.m.f., 11. The chemical reaction,, 2AgCl(s) H 2 (g), , 2HCl(aq) 2Ag(s), , taking place in a galvanic cell is represented by the notation, (a) Pt(s) | H 2 (g),1 bar |1M KCl(aq) | AgCl(s) | Ag(s), (b) Pt(s) | H 2 (g),1 bar |1M HCl(aq) |1MAg (aq) | Ag(s), (c), , Pt(s) | H 2 (g),1 bar |1M HCl(aq) | AgCl(s) | Ag(s), , (d), , Pt(s) | H 2 (g),1 bar |1M HCl(aq) | Ag(s) | AgCl(s), , 12. Given that the standard reduction potentials for M+/M and, N+/N electrodes at 298 K are 0.52 V and 0.25 V respectively., Which of the following is correct in respect of the following, electrochemical cell ?, M/M+ | | N+/N, (a) The overall cell reaction is a spontaneous reaction., (b) The standard EMF of the cell is – 0.27 V., (c) The standard EMF of the cell is 0.77 V., (d) The standard EMF of the cell is – 0.77 V.
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EBD_7207, ELECTROCHEMISTRY, , 276, , 13., , 14., , The difference between the electrode potentials of two, electrodes when no current is drawn through the cell is, called _________., (a) Cell potentials, (b) Cell emf, (c) Potential difference, (d) Cell voltage, For the given Nernst equation, E cell, , 15., , 16., , 17., , 18., , 19., , 20., , 21., , E cell, , 22., , Mg 2, RT, ln, 2, 2F, Ag, , Which of the following representation is correct?, (a) Ag+|Ag||Mg2+|Mg, (b) Mg2+|Mg||Ag|Ag+, (c) Mg|Mg2+||Ag+|Ag, (d) Mg|Mg2+||Ag|Ag+, For cell representation:, Cu(s)|Cu2+(aq)||Ag+(aq)|Ag(s), Which of the following is correct?, (i) Cu is reducing agent., (ii) Overall cell reaction is, , Cu s 2Ag aq, Cu 2 aq 2Ag s, (iii) Cu is cathode, (iv) Ag is anode, (a) (ii), (iii) and (iv), (b) (ii), (iii) and (iv), (c) (iii) and (iv), (d) (i) and (ii), The reference electrode is made by using, (a) ZnCl2, (b) CuSO4, (c) HgCl2, (d) Hg2Cl2, The standard hydrogen electrode potential is zero, because, (a) hydrogen oxidized easily, (b) electrode potential is considered as zero, (c) hydrogen atom has only one electron, (d) hydrogen is a very light element, Without losing its concentration ZnCl2 solution cannot be, kept in contact with, (a) Au, (b) Al, (c) Pb, (d) Ag, On the basis of the following E° values, the strongest, oxidizing agent is :, [Fe(CN)6]4– [Fe(CN)6]3– + e– ; E° = – 0.35 V, Fe2+ Fe3+ + e–;, E° = – 0.77 V, (a) [Fe(CN)6]4–, (b) Fe2+, (c) Fe3+, (d) [Fe(CN)6]3–, Standard electrode potential of three metals X, Y and Z are, – 1.2 V, + 0.5 V and – 3.0 V, respectively. The reducing power, of these metals will be :, (a) Y > Z > X, (b) X > Y > Z, (c) Z > X > Y, (d) X > Y > Z, Standard electrode potential for Sn4+ / Sn2+ couple is + 0.15 V, and that for the Cr3+ / Cr couple is – 0.74 V. These two couples, in their standard state are connected to make a cell. The cell, potential will be, (a) + 1.19 V, (b) + 0.89 V, (c) + 0.18 V, (d) + 1.83 V, , 23., , 24., , 25., , 26., , 27., , 28., , Standard reduction potentials of the half reactions are given, below :, F2(g) + 2e– 2F– (aq); E° = + 2.85 V, Cl2(g) + 2e– 2Cl–(aq); E° = + 1.36 V, Br2(l) + 2e– 2Br–(aq); E° = + 1.06 V, I2(s) + 2e– 2I–(aq); E° = + 0.53 V, The strongest oxidising and reducing agents respectively, are, (a) F2 and I–, (b) Br2 and Cl–, –, (c) Cl2 and Br, (d) Cl2 and I2, A button cell used in watches functions as following, Zn(s) + Ag2O(s) + H2O(l), 2Ag(s) + Zn2+(aq) + 2OH–(aq), If half cell potentials are :, Zn(s); Eo = – 0.76 V, Zn2+(aq) + 2e–, Ag2O(s) + H2O (l) + 2e–, 2Ag(s) + 2OH–(aq); Eo = 0.34 V, The cell potential will be :, (a) 0.42 V, (b) 0.84 V, (c) 1.34 V, (d) 1.10 V, The oxidation potentials of A and B are +2.37 and +1.66 V, respectively. In chemical reactions, (a) A will be replaced by B, (b) A will replace B, (c) A will not replace B, (d) A and B will not replace each other, A smuggler could not carry gold by depositing iron on the, gold surface since, (a) gold is denser, (b) iron rusts, (c) gold has higher reduction potential than iron, (d) gold has lower reduction potential than iron, Which cell will measure standard electrode potential of, copper electrode ?, (a) Pt (s) |H2 (g, 0.1 bar) |H+ (aq., 1 M) ||Cu2+ (aq., 1 M) |, Cu, (b) Pt (s) |H2 (g, 1 bar) |H+ (aq., 1 M) ||Cu2+ (aq., 2 M) |, Cu, (c) Pt (s) |H2 (g, 1 bar) |H+ (aq., 1 M) ||Cu2+ (aq., 1 M) |, Cu (d), Pt (s) |H2 (g, 1 bar) |H+ (aq., 0.1, 2+, M) ||Cu (aq., 1 M) | Cu, Which of the following statement is not correct about an, inert electrode in a cell ?, (a) It does not participate in the cell reaction., (b) It provides surface either for oxidation or for reduction, reaction., (c) It provides surface for conduction of electrons., (d) It provides surface for redox reaction., In the electrochemical reaction, 2Fe 3 Zn, Zn 2, 2Fe 2 ,, on increasing the concentration of Fe2+, (a) increases cell emf, (b) increases the current flow, (c) decreases the cell emf, (d) alters the pH of the solution
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ELECTROCHEMISTRY, 29., , 30., , 31., , 32., , The standard e.m.f. of a galvanic cell involving cell reaction, with n = 2 is found to be 0.295 V at 25°C. The equilibrium, constant of the reaction would be, (Given F = 96500 C mol–1; R = 8.314JK–1mol–1), (a) 2.0 1011, (b) 4.0 1012, , 37. The electrode potential E, , (c), , (a) + 0.73, (c) – 0.82, 38. In the cell reaction, , (d) 1 .0 1010, What will be the emf for the given cell, Pt | H2 (P1) | H+ (aq) | | H2 (P2) | Pt, 1 .0 10 2, , P, RT, log e 1, F, P2, , (b), , (c), , P, RT, log e 2, F, P1, , (d) None of these, , The value of electrode potential (10–4 M) H+ | H2(1 atm) | Pt, at 298 K would be, (a) – 0.236 V, (b) + 0.404 V, (c) + 0.236 V, (d) – 0.476 V, According to Nernst equation, which is not correct if, Q = Kc :, , (c), , 34., , 35., , 36., , P, RT, log e 1, 2F, P2, , (a), , (a) Ecell = 0, , 33., , 277, , Kc, , nFEcell, e RT, , (b), , RT, ln Q, nF, , E cell, , (Zn 2, , Zn), , of a zinc electrode at, , 25°C with an aqueous solution of 0.1 M ZnSO 4 is, [E, , (Zn 2, , Zn), , 2.303RT, = 0.06 at 298 K]., F, (b) – 0.79, (d) – 0.70, , = – 0.76 V. Assume, , Cu(s) 2Ag (aq), Cu 2 (aq) 2Ag(s) ,, o, E cell = 0.46 V. By doubling the concentration of Cu2+, Eocell, will become, (a) doubled, (b) halved, (c) increases but less than double, (d) decreases by a small fraction, , 39. E° of a cell aA bB, [a ]A [b]B, , cC dD is, , (a), , E, , RT ln, , (b), , E, , RT [C]c [D]d, ln, nF [A]a [B]b, , (c), , E, , RT, [C]c [d]D, (d), ln, nF [A]A [B]B, , E, , RT [a ]A [B]B, ln, nF [C]C [d]D, , [c]C [d]D, , 40. E° for the cell,, (d), , E cell, , E cell, , The standard emf of a cell, involving one electron change is, found to be 0.591 V at 25°C. The equilibrium constant of the, reaction is (F = 96500 C mol–1), (a) 1.0 × 101, (b) 1.0 × 105, (c) 1.0 × 1010, (d) 1.0 ×1030, For the galvanic cell, Zn | Zn2+ (0.1M) || Cu2+ (1.0M)|Cu the cell potential increase, if:, (a) [Zn2+] is increased, (b) [Cu2+] is increased, (c) [Cu2+] is decreased, (d) surface area of anode is increased, Consider the following cell reaction:, , Zn | Zn 2 (aq) | | Cu 2 (aq)| Cu is 1.10 V at 25°C. The, equilibrium constant for the cell reaction, Zn, , Cu 2 (aq), , Cu Zn 2 (aq), , is of the order of, (a) 10–37, (b) 1037, –17, (c) 10, (d) 1017, 41. What is the standard cell potential E° for an electrochemical, cell in which the following reaction takes place, spontaneously ?, , Cl2 (g) 2Br, Br2 (aq) 2Cl G, 50.6 kJ, (a) 1.2 V, (b) 0.53 V, (c) 0.26 V, (d) –0.53 V, 42. If 0.01 M solution of an electrolyte has a resistance of 40, ohms in a cell having a cell constant of 0.4 cm–1, then its, 2Fe( s) O2 ( g) 4H (aq) 2Fe2 (aq) 2H2O(l);E 1.67V, 2+, –3, molar, conductance in ohm–1 cm2 mol–1 is, At [Fe ] = 10 M, p(O2) = 0.1 atm and pH = 3, the cell, 2, (a), 10, (b) 104, potential at 25ºC is, (c), 10, (d), 103, (a) 1.47 V, (b) 1.77 V, 43. Specific conductance of a 0.1 N KCl solution at 23ºC is, (c) 1.87 V, (d) 1.57 V, 0.012 ohm–1 cm–1. Resistance of cell containing the solution, The e.m.f. of a Daniell cell at 298 K is E1., at same temperature was found to be 55 ohm. The cell, constant is, ZnSO4 CuSO4, Zn, Cu, (a) 0.0616 cm–1, (b) 0.66 cm–1, (0.01 M) (1.0 M), (c) 6.60 cm–1, (d) 660 cm–1, When the concentration of ZnSO4 is 1.0 M and that of 44. The unit of equivalent conductivity is, CuSO4 is 0.01 M, the e.m.f. changed to E2. What is the, (a) ohm cm, relationship between E1 and E2?, (b) ohm–1 cm2 (g equivalent)–1, (a) E 2 0 E1, (b) E1 E 2, (c) ohm cm2 (g equivalent), (d) S cm–2, (c) E1 E 2, (d) E1 E 2
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EBD_7207, 48., 49., , 50., , 51., , 52., , 53., , (a), , K, C, , (b), , m, , –1, 2, , c / (mol, , 1, L) 2, , 2, , 2, , –1, , –1, , m/(S cm mol ), , 1, , 1, , c / (mol L) 2, , c / (mol L) 2, , 2, , –1, , m/(S cm mol ), , (v), , 1, , c / (mol L) 2, , (a), (b), (c), (d), , Weak acid, (iv), (ii), (i), (iii), , Strong acid, (v), (iv), (ii), (ii), KCl KNO 3 HCl NaOAc NaCl, , Electrolyte:, , 55., , 2, , –1, , (S cm mol ) : 149.9, , Calculate, , 56., , KA, l, , (c), (d) All of these, m KV, Which of the following represents variation of molar, conductance of electrolyte with (concentration) ½, respectively for weak and strong electrolyte ?, (ii), (i), m/(S cm mol ), , 54., , m, , (iv), , 57., , HOAc, , 145, , 426.2, , 91, , 126.5, , using appropriate molar conductances, , of the electrolytes listed above at infinite dilution in H 2 O, at 25°C, (a) 217.5, (b) 390.7, (c) 552.7, (d) 517.2, Kohlrausch’s law states that at, (a) finite dilution, each ion makes definite contribution to, equivalent conductance of an electrolyte, whatever, be the nature of the other ion of the electrolyte., (b) infinite dilution each ion makes definite contribution, to equivalent conductance of an electrolyte depending, on the nature of the other ion of the electrolyte., (c) infinite dilution, each ion makes definite contribution, to conductance of an electrolyte whatever be the nature, of the other ion of the electrolyte., (d) infinite dilution, each ion makes definite contribution, to equivalent conductance of an electrolyte, whatever, be the nature of the other ion of the electrolyte., Which of the following expressions correctly represents, the equivalent conductance at infinite dilution of Al2(SO4)3,, , –1, , 47., , (iii), , Given that, , 2, , 46., , The resistance of 0.01 N solution of an electrolyte was found, to be 220 ohm at 298 K using a conductivity cell with a cell, constant of 0.88cm–1. The value of equivalent conductance, of solution is –, (a) 400 mho cm2 g eq–1, (b) 295 mho cm2 g eq–1, 2, –1, (c) 419 mho cm g eq, (d) 425 mho cm2 g eq–1, Specific conductance of 0.1 M HNO3 is 6.3×10–2 ohm–1 cm–1., The molar conductance of the solution is, (a) 100 ohm–1 cm2, (b) 515 ohm–1 cm2, –1, 2, (c) 630 ohm cm, (d) 6300 ohm–1 cm2, The specific conductance of a 0.1 N KCl solution at 23°C is, 0.012 ohm–1cm–1. The resistance of cell containing the, solution at the same temperature was found to be 55 ohm., The cell constant will be, (a) 0.142 cm–1, (b) 0.66 cm–1, –1, (c) 0.918 cm, (d) 1.12 cm–1, The unit of specific conductivity is, (a) ohm cm–1, (b) ohm cm–2, –1, (c) ohm cm, (d) ohm–1 cm–1, Which of the following solutions of KCl will have the highest, value of specific conductance?, (a) 1.0 N, (b) 0.1 N, (c) 1.0 ×10–2N, (d) 1.0 ×10–3N, The cell constant of a conductivity cell ___________., (a) changes with change of electrolyte., (b) changes with change of concentration of electrolyte., (c) changes with temperature of electrolyte., (d) remains constant for a cell., Which of the following pair(s) is/are incorrectly matched?, (i) R (resistance) – ohm ( ), (ii), (resistivity) – ohm metre ( m), (iii) G (conductance) – seimens or ohm (S), (iv) (conductivity) – seimens metre–1 (Sm–1), (a) (i), (ii) and (iii), (b) (ii) and (iii), (c) (i), (ii) and (iv), (d) (iii) only, On which of the following magnitude of conductivity does, not depends?, (a) Nature of material, (b) Temperature, (c) Pressure, (d) Mass of the material, Which of the following expression correctly represents, molar conductivity?, , conductances at infinite dilution of the respective ions?, , m/(S cm mol ), , 45., , m/(S cm mol ), , ELECTROCHEMISTRY, , 278, , (a), 1, , c / (mol L) 2, , (c), , 1, 3, , Al3, , Al3, , Al3, , 1, 2, SO42, , and, , SO42, , are the equivalent, , SO24, , (b), (d), , 2, , Al3, Al3, , 3, , SO24, SO24, , 6
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ELECTROCHEMISTRY, 58., , 279, , Limiting molar conductivity of NH4OH, i.e., m(NH 4OH is equal to :, , (a), , 59., , 60., , m NH 4Cl, , (b), , m NaOH, , (c), , m NH 4OH, , (d), , m NH 4Cl, , m NaCl, , m NaOH, , m NH 4Cl, , m NaCl, , m NH 4Cl, m NaOH, , m HCl, m NaCl, , Molar conductivities ( m ) at infinite dilution of NaCl, HCl, and CH3COONa are 126.4, 425.9 and 91.0 S cm2 mol–1, respectively. m for CH3COOH will be, (a) 425.5 S cm2 mol–1, (b) 180.5 S cm2 mol–1, (c) 290.8 S cm2 mol–1, (d) 390.5 S cm2 mol–1, At 25°C, the molar conductance at infinite dilution for the, strong electrolytes NaOH, NaCl and BaCl2 are 248 × 10–4,, 126 × 10 –4 and 280 × 10 – 4 Sm 2 mol –1 respectively., , 65. Molar ionic conductivities of a two-bivalent electrolytes, x 2 and y2 are 57 and 73 respectively. The molar, conductivity of the solution formed by them will be, (a) 130 S cm2 mol–1, (b) 65 S cm2 mol–1, 2, –1, (c) 260 S cm mol, (d) 187 S cm2 mol–1, 66. On electrolysis of dilute sulphuric acid using platinum, electrodes, the product obtained at the anode will be, (a) hydrogen, (b) oxygen, (c) hydrogen sulphide, (d) Sulphur dioxide, 67. If 0.5 amp current is passed through acidified silver nitrate, solution for 100 minutes. The mass of silver deposited on, cathode, is (eq.wt.of silver nitrate = 108), (a) 2.3523 g, (b) 3.3575 g, (c) 5.3578 g, (d) 6.3575 g, 68. Aluminium oxide may be electrolysed at 1000°C to furnish, aluminium metal (At. Mass = 27 amu; 1 Faraday = 96,500, , Coulombs). The cathode reaction is– Al, , Ba(OH)2 in S m2 mol –1 is, (a) 52.4 × 10–4, (b) 524 × 10–4, –4, (c) 402 × 10, (d) 262 × 10–4, CICH2 COONa, NaCl, HCl, , 224 ohm 1cm 2gm eq 1,, , 38.2 ohm 1cm 2gm eq 1,, 203 ohm 1cm 2 gm eq 1,, , What is the value of, , 62., , 63., , 64., , (a) 5.49 × 101 C, 69., , 70., , CICH2COOH, , (a) 288.5 ohm–1cm2gmeq–1, (b) 289.5 ohm–1cm2gmeq–1, (c) 388.8 ohm–1cm2gmeq–1, (d) 59.5 ohm–1cm2gmeq–1, At 25°C molar conductance of 0.1 molar aqueous solution, of ammonium hydroxide is 9.54 ohm-1 cm2mol-1 and at, infinite dilution its molar conductance is 238 ohm-1 cm2 mol-1., The degree or ionisation of ammonium hydroxide at the, same concentration and temperature is:, (a) 20.800%, (b) 4.008%, (c) 40.800%, (d) 2.080%, The electrical properties and their r espective, SI units are given below. Identify the wrongly matched pair., Electrical property, SI unit, (a) Specific conductance, S m–1, (b) Conductance, S, (c) Equivalent conductance, S m2 gequiv–1, (d) Cell constant, m, The ion of least limiting molar conductivity among the, following is, (a), , SO 24, , (b) H+, , (c), , Ca 2, , (d), , CH 3COO, , + 3e, , Al, , To prepare 5.12 kg of aluminium metal by this method we, require electricity of, , 0, m, , 61., , 3+, , 71., , 72., , 73., , 74., , 75., , (b) 5.49 × 10 4 C, , (c) 1.83 × 10 7 C, (d) 5.49 × 10 7 C, Which of the following is the use of electrolysis?, (a) Electrorefining, (b) Electroplating, (c) Both (a) & (b), (d) None of these, An electrolytic cell contains a solution of Ag2SO4 and has, platinum electrodes. A current is passed until 1.6 gm of O2, has been liberated at anode. The amount of silver deposited, at cathode would be, (a) 107.88 gm, (b) 1.6 gm, (c) 0.8 gm, (d) 21.60 gm, When 9650 coulombs of electricity is passed through a, solution of copper sulphate, the amount of copper deposited, is (given at. wt. of Cu = 63.6), (a) 0318g, (b) 3.18 g, (c) 31.8g, (d) 63.6g, Find the charge in coulombs required to convert 0.2 mole, VO3–2 into VO4–3 –, (a) 1.93 × 104, (b) 9.65 × 104, (c) 1.93 × 105, (d) 9.65 × 105, A silver cup is plated with silver by passing 965 coulombs, of electricity. The amount of Ag deposited is :, (a) 107.89 g, (b) 9.89 g, (c) 1.0002 g, (d) 1.08 g, The number of coulombs required to reduce 12.3 g of, nitrobenzene to aniline is :, (a) 115800 C, (b) 5790 C, (c) 28950 C, (d) 57900 C, The amount of electricity that can deposit 108 g of Ag from, AgNO3 solution is:, (a) 1 F, (b) 2 A, (c) 1 C, (d) 1 A
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EBD_7207, ELECTROCHEMISTRY, , 280, , 76., , 77., , 78., , 79., , To deposit one equivalent weight of silver at cathode, the, charge required will be, (a) 9.65 × 104 C, (b) 9.65 × 103 C, 5, (c) 9.65 × 10 C, (d) 9.65 × 107 C, The volume of oxygen gas liberated at NTP by passing a, current of 9650 coulombs through acidified water is, (a) 1.12 litre, (b) 2.24 litre, (c) 11.2 litre, (d) 22.4 litre, Three faradays electricity was passed through an aqueous, solution of iron (II) bromide. The weight of iron metal, (at. wt = 65) deposited at the cathode (in gm) is, (a) 56, (b) 84, (c) 112, (d) 168, On passing C ampere of electricity through a electrolyte, solution for t second. m gram metal deposits on cathode., The equivalent weight E of the metal is, (a), , 81., , 82., , 83., , 84., , 85., , 86., , C t, m 96500, , (b), , E, , 88., , 89., , C m, t 96500, , 96500 m, C t 96500, (d) E, m, C t, The number of electrons passing per second through a, cross-section of copper wire carrying 10–6 amperes of, current per second is found to be, (a) 1.6 × 10–19, (b) 6 × 10–35, –16, (c) 6 × 10, (d) 6 × 1012, Faraday’s laws of electrolysis will fail when, (a) temperature is increased, (b) inert electrodes are used, (c) a mixture of electrolytes is used, (d) None of these cases, The electric charge for electrode decomposition of one gram, equivalent of a substance is, (a) one ampere per second, (b) 96500 coulombs per second, (c) one ampere for one hour, (d) charge on one mole of electrons, In electrolysis of dilute H2SO4 using platinum electrodes, (a) H2 is evolved at cathode, (b) NH2 is produced at anode, (c) Cl2 is obtained at cathode, (d) O2 is produced, In electrolysis of NaCl when Pt electrode is taken then H2 is, liberated at cathode while with Hg cathode it forms sodium, amalgam. This is because, (a) Hg is more inert than Pt, (b) more voltage is required to reduce H+ at Hg than at Pt, (c) Na is dissolved in Hg while it does not dissolve in Pt, (d) conc. of H+ ions is larger when Pt electrode is taken, Electrolysis of fused NaCl will give, (a) Na, (b) NaOH, (c) NaClO, (d) None of these, How many moles of Pt may be deposited on the cathode, when 0.80 F of electricity is passed through a 1.0 M, solution of Pt4+?, , (c), , 80., , E, , 87., , 90., , E, , 91., , 92., , 93., , 94., , (a) 1.0 mol, (b) 0.20 mol, (c) 0.40 mol, (d) 0.80 mol, A current strength of 9.65 amperes is passed through excess, fused AlCl3 for 5 hours. How many litres of chlorine will be, liberated at STP? (F = 96500 C), (a) 2.016, (b) 1.008, (c) 11.2, (d) 20.16, A solution of copper sulphate (CuSO4) is electrolysed for, 10 minutes with a current of 1.5 amperes. The mass of copper, deposited at th e cathode (at. mass of Cu = 63u), is, (a) 0.3892g, (b) 0.2938g, (c) 0.2398g, (d) 0.3928g, When 0.1 mol MnO42– is oxidised the quantity of electricity, required to completely oxidise MnO42– to MnO4– is, (a) 96500 C, (b) 2 × 96500 C, (c) 9650 C, (d) 96.50 C, The weight of silver (at wt. = 108) displaced by a quantity of, electricity which displaces 5600 mL of O2 at STP will be, (a) 5.4 g, (b) 10.8 g, (c) 54.9 g, (d) 108.0 g, Electrolysis of a salt solution was carried out, after some, time solution turned yellow than salt can be, (i) NaCl, (ii) KCl, (iii) RbCl, (iv) KBr, (a) (i), (ii) and (iii), (b) (ii), (ii) and (iv), (c) (i), (ii) and (iv), (d) (i), (iii) and (iv), Which of the following statements is incorrect?, (a) Both electronic and electrolytic conductance depends, on the nature of conducting material., (b) Both electronic and electrolytic conductance varies, similarly with temperature., (c) Electronic conductance is independent but electrolytic, conductance depends on the amount of the, conducting substance., (d) All the above statements are incorrect., Which of the following statements is incorrect?, (a) Electrodes made up of gold participates in the chemical, reaction., (b) Electrolytic products of NaCl are Na and Cl2 whereas, of aqueous NaCl are NaOH, Cl2 and H2., (c) During electrolysis at cathode, reaction with higher, value of E is preferred., (d) All of the above statements are incorrect., During electrolysis of sulphuric acid, which of the following, processes is possible at anode?, A., , 2H 2 O(l), , B., , 2SO 24 (a q), , O 2 (g) 4H (aq) 4e, , Ecell, = + 1.23 V, , S2O82 (aq) 2e, , Ecell, = 1.96 V
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ELECTROCHEMISTRY, , 95., , 96., , 97., , 98., , 99., , 100., , 101., , 102., , 103., , Choose the correct option based on following statements., (i) Process A is preferred at higher concentration of, sulphuric acid., (ii) Process B is preferred at higher concentration of, sulphuric acid., (iii) Process A is preferred for dilute sulphuric acid., (iv) Process B is preferred for dilute sulphuric acid., (v) Both A and B are equally possible at higher, concentration., (a) (v) and (iii), (b) (iii) and (ii), (c) (i) and (iv), (d) (v) and (iv), Which of the following metals is not produced by, electrochemical reduction?, (a) Na, (b) Fe, (c) Mg, (d) Al, As lead storage battery is charged, (a) lead dioxide dissolves, (b) sulphuric acid is regenerated, (c) lead electrode becomes coated with lead sulphate, (d) the concentration of sulphuric acid decreases, During the charging of lead storage battery, the reaction at, anode is represented by, (a) Pb2 SO24, PbSO 4, (b) PbSO 4 2H 2 O, PbO2 SO42, 4H, 2e, 2, (c) Pb, Pb, 2e, (d) Pb2, 2e, Pb, Which colourless gas evolves, when NH4Cl reacts with, zinc in a dry cell battery, (a) NH4, (b) N2, (c) H2, (d) Cl2, In a hydrogen-oxygen fuel cell, combustion of hydrogen, occurs to, (a) produce high purity water, (b) create potential difference between two electrodes, (c) generate heat, (d) remove adsorbed oxygen from elctrode surfaces, Among the following cells:, Leclanche cell (i), Nickel-Cadmium cell (ii), Lead storage battery (iii), Mercury cell (iv), primary cells are, (a) (i) and (ii), (b) (i) and (iii), (c) (ii) and (iii), (d) (i) and (iv), The electrolyte used in Leclanche cell is, (a) paste of KOH and ZnO, (b) 38% solution of H2SO4, (c) moist paste of NH4Cl and ZnCl2, (d) moist sodium hydroxide, A device that converts energy of combustion of fuels like, hydrogen and methane, directly into electrical energy is, known as :, (a) Electrolytic cell, (b) Dynamo, (c) Ni-Cd cell, (d) Fuel Cell, Which one of the following cells can convert chemical, energy of H2 and O2 directly into electrical energy?, (a) Mercury cell, (b) Daniell cell, (c) Fuel cell, (d) Lead storage cell, , 281, , 104. Hydrogen-Oxygen fuel cells are used in space craft to supply, (a) power for heat and light, (b) power for pressure, (c) oxygen, (d) water, 105. Prevention of corrosion of iron by zinc coating is called, (a) electrolysis, (b) photoelectrolysis, (c) cathodic protection, (d) galvanization, 106. The best way to prevent rusting of iron is, (a) making it cathode, (b) putting in saline water, (c) Both of these, (d) None of these, 107. Several blocks of magnesium are fixed to the bottom of a ship, to, (a) make the ship lighter, (b) prevent action of water and salt, (c) prevent puncturing by under-sea rocks, (d) keep away the sharks, 108. Which of the following batteries cannot be reused?, (a) Lead storage battery (b) Ni-Cd cell, (c) Mercury cell, (d) Both (b) and (c), 109. Which of the following is a merit of Ni–Cd cell over lead, storage battery?, (a) Ni–Cd cell can be re-used., (b) Ni–Cd cell is comparatively economical to manufacture, (c) Ni–Cd cell has comparatively longer life, (d) All the above are the merits of Ni–Cd cell over lead, storage battery., 110. Which of the following statements regarding fuel cell is, incorrect?, (a) These cells are eco-friendly., (b) These cells convert energy of combustion of fuels, like H2, CH4, CH3OH etc., directly into electrical energy., (c) H2 – O2 fuel cell is used in Apollo space programme., (d) Fuel cells produce electricity with an efficiency of, about 100%., , STATEMENT TYPE QUESTIONS, =1.1 V, , <1.1 V, , 111., , e–, , anode, Zn, –ve, , current, salt, bridge, , cathode, Cu, +ve, , I=0, , Zn, , ZnSO4, CuSO4, Figure -1, , CuSO4, ZnSO4, Figure -2, >1.1 V, e–, , cathode, +ve, Zn, , current anode, –ve, Cu, , ZnSO4, , CuSO4, , Figure -3, , Cu
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EBD_7207, ELECTROCHEMISTRY, , 282, , (i), , Figure 1 represents electrochemical and Figure 3, represents electrolytic cell., (ii) Figure 2 represents electrolytic an d Figure 3, represents electrochemical cell., (iii) Figure 2 represents a cell which is not working i.e. no, current flows through the cell., (iv) Energy conversion shown in Figure 1 is chemical to, electrical whereas energy conversion shown in Figure, 2 is electrical to chemical., Which of the following is the correct coding for the, statements above., (a) TFTT, (b) TTTT, (c) TFFT, (d) FTFF, 112. Which of the following statements regarding given cell, representation is/are correct?, Cd(s)/Cd2+(aq)//Ag+(aq)/Ag(s), (i) In the given cell Cd electrode act as an anode whereas, Ag electrode acts as a cathode., (ii) In the given cell Cd electrode acts as a cathode whereas, Ag electrode acts as a annode., (iii) E cell, , E, , Ag /Ag, , E, , Cd 2 /Cd, , (a) (i) and (ii), (b) Only (ii), (c) Only (i), (d) (i) and (iii), 113. Which of the following is/are correct statement(s) for the, addition of Li, K, Rb to the aqueous solution of Na+., (i) The correct order of metals in which they reduce the, Na+ ion is Rb < K < Li., (ii) Reduction of metal ions would not take place., (a) Statement (i) and (ii) are correct., (b) Statement (i) is correct only., (c) Statement (ii) is correct only., (d) Neither (i) nor (ii) is correct., 114. Read the following statements carefully., (i) According to a convention cell potential of hydrogen, electrode (S.H.E.) is considered to be zero at all, temperatures., (ii) e.m.f. of the cell Pt(s)/H2(g, 1 bar)/H+(aq, 1 M), || Zn 2 (aq,1M) / Z n is – 0.76. This negative value, indicates that Zn 2+ ion reduces less easily then H+, ions., (iii) Copper does not dissolve in HCl but dissolves in HNO3, as in nitric acid it gets oxidised by nitrate ion., (iv) Inert metals like Pt or Au are used in certain electrodes, i.e., these metals does not participate in reaction but, provide surface for oxidation and reduction reactions., , (v) Fluorine has the highest electrode potential thereby, making it strongest oxidising agent whereas lithium, with lowest electrode potential is the weakest oxidising, and strongest reducing agent., , Which of the following is the correct coding for the, statements above., (a) TTTTT, (b) TTFTF, (c) FFTTT, (d) FFFTT, 115. Which of the following statement(s) is/are correct?, (i) Molar conductivity for strong electrolytes increases, gradually and of weak electrolytes increases rapidly, on dilution., (ii) If is the degree of dissociation of weak electrolytes., m, , Then,, , m, , (iii) Molar conductivity of CaX2 increases rapidly on, dilution., (a) (i) and (iii), (b) (ii) only, (c) (i) only, (d) (i) and (ii), 116. Read the following statements., (i), , According to Faraday’s second law amounts of, different substances liberated by same quantity of, electricity passing through the electrolytic solution, are proportional to their chemical equivalent weights., , (ii) 1 F = 96487 Cmol–1, calculation)., , 96500 Cmol–1 (for more accurate, , (iii) As per electrode reactions, , K, , e, , Al3, , K, 3e, , Al, , one mole of K+ and Al3+ require 1(1F) and 3(3F) mol of, electrons respectively., Which of the following is the correct coding for the above, statements?, (a) TTT, , (b) FFT, , (c) TFT, , (d) FTF, , 117. Which of the following statement(s) is/ are incorrect for, corrosion of iron?, (i), , Reaction occurring at anode is, O2 g, , 4H, , aq, , 4e, , 2H 2 O l, , (ii) Reaction occurring at cathode is, 2Fe s, , 2Fe2, , 4e, , (iii) Rust is Fe2O3.xH2O, (iv) H+ involved in corrosion reaction is provided from, H2CO3 which is formed due to dissolution of carbon, dioxide from air in to water., (a) (iv) only, , (b) (i) only, , (c) (i) and (ii), , (d) (i), (ii) and (iv)
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ELECTROCHEMISTRY, , 283, , MATCHING TYPE QUESTIONS, 118. Match the Column-I (functioning of Daniel cell) with, Column-II (value of Eext) and choose the correct option., Column-I, Column-II, (A) Flow of electrons from (p) E = 1.1 V, Cu to Zn and current, flows from Zn to Cu, (B) No flow of electrons, (q) E < 1.1 V, or current, (C) Zn dissolves at anode, (r) E > 1.1 V, and copper deposits at, cathode, (a) (A) – (q), (B) – (p), (C) – (r), (b) (A) – (r), (B) – (p), (C) – (q), (c) (A) – (p), (B) – (r), (C) – (q), (d) (A) – (r), (B) – (q), (C) – (p), 119. Match the items of Column I and Column II on the basis of, data given below :, , EO, , F2 /F, , O, = 2.87 V, E Li, , EO, , Br2 /Br, , /Li, , O, = –3.5 V, E Au 3, , /Au, , =1.4 V,, , =1.09 V, , Column-I, (A) F2, , Column-II, (p) metal is the strongest, reducing agent., (B) Li, (q) anion that can be oxidised, by Au3+, (C) Au3+, (r) non metal which is the, best oxidising agent, (D) Br–, (s) metal ion which is an, oxidising agent, (a) (A) – (r), (B) – (p), (C) – (s), (D) – (q), (b) (A) – (p), (B) – (r), (C) – (s), (D) – (q), (c) (A) – (q), (B) – (p), (C) – (s), (D) – (r), (d) (A) – (r), (B) – (s), (C) – (p), (D) – (q), 120. Match the columns., Column-I, Column-II, (A), , m, , (B), , EO, cell, , (C), (D), (a), (b), (c), (d), , ions/ volume, (r) Extensive property, G, (s), Increases with dilution, r cell, (A) – (p), (B) – (s), (C) – (q), (D) – (r), (A) – (s), (B) – (p), (C) – (q), (D) – (r), (A) – (s), (B) – (q), (C) – (p), (D) – (r), (A) – (s), (B) – (p), (C) – (r), (D) – (q), , (p) intensive property, (q) Depends on number of, , 121. Match the columns., Column-I, (A), (B), , m, , Column-II, (p) I × t, (q), , (C), , (r), , (D) Q, , (s), , m, , /, , o, m, , c, G*, R, , (a) (A) – (p), (B) – (r), (C) – (q), (D) – (s), (b) (A) – (s), (B) – (q), (C) – (r), (D) – (p), (c) (A) – (r), (B) – (s), (C) – (q), (D) – (p), (d) (A) – (s), (B) – (r), (C) – (q), (D) – (p), 122. Match the columns, Column-I, Column-II, (A) Cell in which electrolyte (p) H2 – O2 fuel cell, is a paste of KOH and, ZnO. This cell is used in, low current devices like, hearing aids, watches, etc., (B) Cell in which 38% H2SO4 (q) Mercury cell, solution is used as an, electrolyte., (C) Cell in which vapours, (r) Lead storage battery, produced during electrochemical reaction were, condensed and added, to drinking water, (D) Cell having longer life, (s) Nickel – Cadmium cell, than lead storage cell, and is expensive to, manufacture, (a) (A) – (r), (B) – (q), (C) – (p), (D) – (s), (b) (A) – (q), (B) – (r), (C) – (p), (D) – (s), (c) (A) – (q), (B) – (r), (C) – (p), D) – (s), (d) (A) – (q), (B) – (r), (C) – (s), (D) – (p), , ASSERTION-REASON TYPE QUESTIONS, Directions : Each of these questions contain two statements,, Assertion and Reason. Each of these questions also has four, alternative choices, only one of which is the correct answer. You, have to select one of the codes (a), (b), (c) and (d) given below., (a) Assertion is correct, reason is correct; reason is a correct, explanation for assertion., (b) Assertion is correct, reason is correct; reason is not a, correct explanation for assertion, (c) Assertion is correct, reason is incorrect, (d) Assertion is incorrect, reason is correct., 123. Assertion : The resistivity for a substance is its resistance, when it is one meter long and its area of cross section is one, square meter., Reason : The SI units of resistivity is ohm metre ( m).
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EBD_7207, ELECTROCHEMISTRY, , 284, , 124. Assertion : On increasing dilution, the specific conductance, keep on increasing., Reason : On increasing dilution, degree of ionisation of, weak electrolyte increases and molality of ions also, increases., 125. Assertion : Galvanised iron does not rust., Reason : Zinc has a more negative electrode potential than, iron., , CRITICAL THINKING TYPE QUESTIONS, 126. If salt bridge is removed from two half-cells the voltage, (a) drops to zero, (b) does not change, (c) increases gradually, (d) increases rapidly, 127. In the electrolytic cell, flow of electrons is from, (a) cathode to anode in solution, (b) cathode to anode through external supply, (c) cathode to anode through internal supply, (d) anode to cathode through internal supply, 128. Standard potentials (Eº) for some half-reactions are given, below :, (i), , Sn 4, , (ii), , 2 Hg 2, , 2e, 2e, , Sn 2 ; E º, Hg 2 2 ; E º, , 0.15 V, 0.92 V, , (iii) PbO2 4H, 2e, Pb 2, 2H 2O ; E º 1.45 V, Based on the above, which one of the following statements, is correct ?, (a) Sn4+ is a stronger oxidising agent than Pb4+, (b) Sn2+ is a stronger reducing agent than Hg22+, (c) Hg2+ is a stronger oxidising agent than Pb4+, (d) Pb2+ is a stronger reducing agent than Sn 2+, 129. Consider the following relations for emf of a electrochemical, cell:, (i) emf of cell = (Oxidation potential of anode), – (Reduction potential of cathode), (ii) emf of cell = (Oxidation potential of anode), + (Reduction potential of cathode), (iii) emf of cell = (Reduction potential of anode), + (Reduction potential of cathode), (iv) emf of cell = (Oxidation potential of anode), – (Oxidation potential of cathode), Which of the above relations are correct?, (a) (ii) and (iv), (b) (iii) and (i), (c) (i) and (ii), (d) (iii) and (iv), 130. The correct order of E 2, values with negative sign, M, , /M, , for the four successive elements Cr, Mn, Fe and Co is, (a) Mn > Cr > Fe > Co, (b) Cr < Fe > Mn > Co, (c) Fe > Mn > Cr > Co, (d) Cr > Mn > Fe > Co, 131. Consider the following four electrodes:, P = Cu2+ (0.0001 M)/Cu(s), Q = Cu2+ (0.1 M)/Cu(s), R = Cu2+ (0.01 M)/Cu(s), S = Cu2+ (0.001 M)/Cu(s), , If the standard reduction potential of Cu2+/Cu is +0.34 V,, the reduction potentials in volts of the above electrodes, follow the order., (a) P > S > R > Q, (b) S > R > Q > P, (c) R > S > Q > P, (d) Q > R > S > P, 132. At 298K the standard free energy of formation of H2O ( ) is, –237.20 kJ/mole while that of its ionisation into H+ ion, and hydroxyl ions is 80 kJ/mole, then the emf of the following, cell at 298 K will be, [Take Faraday constant F = 96500C], H2 (g , 1 bar) | H+ (1M) | | OH– (1M) | O2(g, 1 bar), (a) 0.40V, (b) 0.81V, (c) 1.23 V, (d) – 0.40 V, 133. If the following half cells have E° values as, A3+ + e– –––– A2+, E° = y2V, A2+ + 2e– –––– A, E° = –y1V, The E° of the half cell A3+ + 3e –––– A will be, (a), , 2 y1, , y2, , 3, (c) 2y1 – 3y2, , y2 2 y1, 3, (d) y2 – 2y1, , (b), , 134. Cu + (aq ) is unstable in solution and undergoes, simultaneous oxidation and reduction according to the, reaction :, 2Cu (aq), Cu 2 (aq) Cu( s ), Choose correct Eº for given reaction if Eº Cu2+/Cu = 0.34 V, and Eº Cu2+/Cu+ = 0.15 V, (a) –0.38 V, (b) +0.49 V, (c) +0.38 V, (d) –0.19 V, 135. In a cell that utilises the reaction, Zn( s) 2H (aq) Zn 2 (aq) H 2 ( g ) addition of H2SO4, to cathode compartment, will, (a) increase the E and shift equilibrium to the right, (b) lower the E and shift equilibrium to the right, (c) lower the E and shift equlibrium to the left, (d) increase the E and shift equilibrium to the left, 136. For a cell reaction involving two electron change, the, standard EMF of the cell is 0.295 V at 2°C. The equilibrium, constant of the reaction at 25°C will be:, (a) 29.5 × 10–2, (b) 10, (c) 1 × 1010, (d) 2.95 × 10–10, 137. Standard cell voltage for the cell Pb | Pb2+ || Sn2+ | Sn is, – 0.01 V. If the cell is to exhibit Ecell = 0, the value of, [Sn 2+] / [Pb2+] should be antilog of –, (a) + 0.3, (b) 0.5, (c) 1.5, (d) – 0.5, , 138. The cell, Zn | Zn 2 (1 M) || Cu 2 (1 M) | Cu ( E cell 1.10 V), was allowed to be completely discharged at 298 K. The, relative concentration of Zn2+ to Cu2+, (a) 9.65 × 104, (c) 37.3, , [Zn 2 ], [Cu 2 ], , (b) antilog (24.08), (d) 1037.3., , is
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ELECTROCHEMISTRY, , 285, , 139. What is the potential of half-cell consisting of, zinc electrode in 0.01 M ZnSO 4 solution at 25°C, , Eox, , 0.763 V, , (a) 0.8221 V, (b) 8.221 V, (c) 0.5282 V, (d) 9.282 V, 140. The oxidation potential of 0.05 M H2SO4 is, (a) –2 × 0.0591, (b) –0.01 × 0.0591, (c) –2.321 × 0.0591, (d) +1 × 0.0591, 141. For a relation, nFE cell, rG, , (i), , E cell E cell in which of the following condition?, (a) Concentration of any one of the reacting species, should be unity, (b) Concentration of all the product species should be, unity., (c) Concentration of all the reacting species should be, unity., (d) Concentration of all reacting and product species, should be unity., 142. A 0.5 M NaOH solution offers a resistance of 31.6 ohm in a, conductivity cell at room temperature. What shall be the, approximate molar conductance of this NaOH solution if, cell constant of the cell is 0.367 cm–1 ., (a) 234 S cm2 mole–1, (b) 23.2 S cm2 mole–1, 2, –1, (c) 4645 S cm mole, (d) 5464 S cm2 mole–1, 143. The equivalent conductances of two strong electrolytes at, infinite dilution in H2O (where ions move freely through a, solution) at 25°C are given below :, CH3COONa, , 146. A weak electrolyte having the limiting equivalent, conductance of 400 S cm2. equivalent–1 at 298 K is 2%, ionized in its 0.1 N solution. The resistance of this solution, (in ohms) in an electrolytic cell of cell constant 0.4 cm–1 at, this temperature is, (a) 200, (b) 300, (c) 400, (d) 500, 147. Conductivity , is equal to _________., , 91.0 S cm 2 / equiv., , 426.2 S cm 2 / equiv., What additional information/ quantity one needs to calcuof an aqueous solution of acetic acid?, late, , )., , 144. Resistance of 0.2 M solution of an electrolyte is 50 . The, specific conductance of the solution is 1.3 S m –1 . If, resistance of the 0.4 M solution of the same electrolyte is, 260 , its molar conductivity is :, (a) 6.25 × 10–4 S m2 mol–1 (b) 625 × 10–4 S m2 mol–1, (c) 62.5 S m2 mol–1, (d) 6250 S m2 mol–1, 145. The limiting molar conductivities of HCl, CH3COONa and, NaCl are respectively 425, 90 and 125 mho cm2 mol–1 at, 25°C. The molar conductivity of 0.1 M CH3COOH solutions, is 7.8 mho cm2 mol–1 at the same temperature. The degree of, dissociation of 0.1 M acetic acid solution at the same, temperature is, (a) 0.10, (b) 0.02, (c) 0.15, (d) 0.03, , G, R, , 152., , 153., , 154., , (iv), , m, , RT, log a M n, nF, (d) The efficiency of a hydrogen-oxygen fuel cell is 23%, When electric current is passed through acidified water,, 112 ml of hydrogen gas at STP collected at the cathode in, 965 seconds. The current passed in amperes is, (a) 1.0, (b) 0.5, (c) 0.1, (d) 2.0, On passing current through two cells, connected in series, containing solution of AgNO3 and CuSO4, 0.18 g of Ag is, deposited. The amount of the Cu deposited is:, (a) 0.529 g, (b) 10.623 g, (c) 0.0529 g, (d) 1.2708 g, In the electrolysis of water, one faraday of electrical energy, would liberate, (a) one mole of oxygen, (b) one gram atom of oxygen, (c) 8 g oxygen, (d) 22.4 lit. of oxygen, Electrolysis of dilute aqueous NaCl solution was carried, out by passing 10 milli ampere current. The time required to, liberate 0.01 mol of H2 gas at the cathode is, (1 Faraday = 96500 C mol–1), (a) 9.65 × 104 sec, (b) 19.3 × 104 sec, 4, (c) 28.95 × 10 sec, (d) 38.6 × 104 sec, E, , 151., , H, , (ii), , l, A, (a) (i) and (iii), (b) (i) and (ii), (c) (i), (ii) and (iii), (d) (ii), (iii) and (iv), 148. Arrange the following in increasing order of their, conductivity Na+ (A), K+ (B), Ca2+ (C), Mg2+(D), (a) A, B, C, D, (b) B, A, C, D, (c) C, A, D, B, (d) A, B, D, C, 149. The conductivity of electrolytic solutions depends upon, which of the following?, (i) Size of ions produced, (ii) Viscosity of the solvent, (iii) Concentration of electrolyte, (iv) Solvation of ions produced, (a) (i) and (iii), (b) (i), (ii) and (iii), (c) (i), (iii) and (iv), (d) All of these, 150. Mark the false statement?, (a) A salt bridge is used to eliminate liquid junction, potential, (b) The Gibbs free energy change, G is related with, electromotive force E as G = –nFE, (c) Nernst equation for single electrode potential is, , (iii), , HCl, , (a), of chloroacetic acid (ClCH2COOH), (b), of NaCl, (c), of CH3COOK, (d) the limiting equivalent coductance of H (, , 1 l, RA, , Eo
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EBD_7207, ELECTROCHEMISTRY, , 286, , 155. What is the amount of chlorine evolved when 2 amperes of, current is passed for 30 minutes in an aqueous solution of, NaCl ?, (a) 66 g, (b) 1.32 g, (c) 33 g, (d) 99 g, 156. On passing a current of 1.0 ampere for 16 min and 5 sec, through one litre solution of CuCl2, all copper of the solution, was deposited at cathode. The strength of CuCl2 solution, was (Molar mass of Cu= 63.5; Faraday constant = 96,500, Cmol–1), (a) 0.01 N, (b) 0.01 M, (c) 0.02 M, (d) 0.2 N, 157. 0.2964 g of copper was deposited on passage of a current of, 0.5 amp for 30 mins through a solution of copper sulphate., Calculate the oxidation state of Cu (At. mass 63.56)., (a) +1, (b) +2, (c) +3, (d) +4, 158. One Faraday of electricity is passed through molten Al 2O3,, aqueous solution of CuSO4 and molten NaCl taken in three, different electrolytic cells connected in series. The mole, ratio of Al, Cu and Na deposited at the respective cathode, is, (a) 2 : 3 : 6, (b) 6 : 2 : 3, (c) 6 : 3 : 2, (d) 1 : 2 : 3, 159. What will happen during the electrolysis of aqueous, solution of CuSO4 by using platinum electrodes ?, , (i), (ii), (iii), (iv), (a), (c), , Copper will deposit at cathode., Copper will deposit at anode., Oxygen will be released at anode., Copper will dissolve at anode., (i) and (iii), (b) (ii) and (iv), (i) and (ii), (d) (ii) and (iii), , 160. How much charge is required, when 1 mole of Cr2O72, reduce to form 1 mole of Cr 3+ ?, (a) 6F, (b) 3F, (c) 1F, (d) 2F, 161. When a lead storage battery is discharged, (a) SO2 is evolved, (b) Lead sulphate is consumed, (c) Lead is formed, (d) Sulphuric acid is consumed, 162. The most durable metal plating on iron to protect against, corrosion is, (a) nickel plating, (b) copper plating, (c) tin plating, (d) zinc plating, 163. Which of the following statements is incorrect regarding, dry (Leclanche) cell?, (a) Cathode used in the cell is coated by powdered, manganese dioxide and carbon., (b) Most common application of this cell is in our, transistors and clocks., (c) At cathode, Mn is oxidised from + 3 to + 4., (d) At anode Zn is oxidised from 0 to + 2.
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ELECTROCHEMISTRY, , 287, , FACT / DEFINITION TYPE QUESTIONS, 1., , 2., , 3., 4., 5., 6., , (c) Electrochemical reactions are more energy efficient, and less polluting. Thus study of electrochemistry is, important to create new eco-friendly technologies., (a) We know that in the internal circuit of a galvanic cell, ions flow whereas in the external circuit, the electrons, flow from one electrode to another., (a) Anode has negative polarity., (b), (d) When both the electrodes are kept in the same solution, there will be no requirement of salt bridge., (d), , Reduction, 7., , (d), , 8., 9., , (a), (b) The cell in which Cu and Zn rods are dipped in its, solution is called Daniel cell., (c) The magnitude of the electrode potential of a metal is, a measure of its relative tendency to loose or gain, electrons. i.e., it is a measure of the relative tendency, to undergo oxidation (loss of electrons) or reduction, (gain of electrons)., , 10., , Zn 2+, , M, M, 11., , (b), , 2e-, , M, n, , n, , Zn, , It shows reduction reaction., , 13., 15., 16., 17., 18., , 19., , (b), (b), (d), (d), (b), (b), , Z, , ne, , M (reduction potential), , 2AgCl(s) H 2 (g) 2HCl(aq) 2Ag(s), The activities of solids and liquids are taken as unity, and at low concentrations, the activity of a solute is, approximated to its molarity., The cell reaction will be, , Eocell, , o, Ecathode, , o, Eanode, , Eocell, , 0.25 0.52, , Eoright, , 21. (b) Given ESn, E Cr, , 3, , Cr, , E cell, , 4, , Sn, , ( 1.2), 2, , Y, , ( 0.5), , = + 0.15 V, , = – 0.74 V, , E ox, , E red, , = 0.74 + 0.15, = 0.89 V, 22. (a) Higher the value of reduction potential higher will be, the oxidising power whereas the lower the value of, reduction potential higher will be the reducing power., 23. (d) E°Cell = E°OP + E°RP, = 0.76 + 0.314 = 1.10 V, 24. (b) Follow ECS, A will replace B., 25. (c) Gold having higher E oRe d and oxidises Fe, 26. (c), 27. (d), 28. (c) Nernst equation Ecell = E ocell, , Fe, , RT ln[Fe 2 ]2, Zn 2, 3, 2, nF [Fe ], , increasing [Fe2+] will decrease the Ecell., 29. (d), , 0.0591, log K, n, , E, , Here, n, , 2, E, , 0.295, , 2 0.295, = 9.98 10 or K 1010, 0.0591, 30. (b) RHS : 2H+ + 2e–, H2 (P2), LHS : H2(P1), 2H+ + 2e–, H2(P2), overall reaction : H2 (P1), log K, , o, E left, , E = E°-, , RT P2, RT P2 RT P1, n = 0n =, n, nF, P1, nF, P1, nF P2, , E, , E, , 0.059, 1, log, n, [H ], , =0, , 0.059, 1, log, 1, 10 4, , 0.27V, , 14. (c), Cu is anode and Ag+ is cathode., Calomel electrode is used as reference electrode., Electrode potential is considered as zero., Without losing its concentration ZnCl 2 solution, cannot kept in contact with Al because Al is more, reactive than Zn due to its highly negative electrode, reduction potential., (c) From the given data we find Fe3+ is strongest oxidising, agent. More the positive value of E°, more is the, tendency to get oxidized. Thus correct option is (c)., , X, , ( 3.0), , ne (oxidation potential), , Pt(s) | H 2 (g),1bar | H (aq)1M | AgCl(aq)1M | Ag(s), , 12., , 20. (c) As the value of standard reduction potential decreases, the reducing power increases i.e.,, , 31. (a), , 32. (d), , Ecell, , Ecell, , RT, ln Q, nF, , At equilibrium,, Ecell = 0 and Q = Kc, Ecell, , Ecell, , 0.236V, , .
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EBD_7207, ELECTROCHEMISTRY, , 288, , 33., , (c) The E°cell is given by, 0.0591, log Keq, n, , E°cell, , 0.0591, log Keq, 1, , 0.591, , or log Keq =, 34., , 40., , 41., , 0.591, = 10, 0.0591, , or Keq = 1 × 1010, (b) For the given cell, , 42., , 0.059V, [Zn 2 ( aq )], log, 2, [Cu 2 ( aq)], The cell potential will decreases with increase in, [Zn 2+ (aq)] and will increases with increase in, [Cu2+(aq)]., (d) Here n = 4, and [H+] = 10– 3 (as pH = 3), Applying Nernst equation, , E = Eº –, , 0.059, [Fe 2 ]2, log, n, [H ]4 ( pO2 ), , 36., , Ecell, , Greater the factor, , 37., , Cu 2, , Zn 2, , Cu, , RT [Zn 2 ], n, nF [Cu 2 ], , °, Ecell, , ( Zn, , 2, , Zn 2 /Zn, , E, , ), , (Cu, , 2, , ), , 0.76, , E, , Zn 2 /Zn, , (d) emf will decrease., , 39., , (b), , E cell, , o, E cell, , Hence, , E ocell, , m), , 1000, Molarity, , 1000 0.01, 103 ohm 1 cm 2 mol 1, .01, (b) Specific conductance of the solution ( ) = 0.012 ohm–1, cm–1 and resistance (R) = 55 ohm., Cell constant = Specific conductance × Observed, , resistance, 44., , (b), , ohm–1, , 45., , (a), , eq, , , less is the EMF, 46., , cm2, , = 0.012 × 55= 0.66 cm–1., (geq)–1, , 1000, 1, =, N, R, , l 1000, a, N, , 1, 1, 1000, 1000, cell constant, 0.88, =, R, N, 220, 0.01, = 400 mho cm2 g eq–1, (c) Molar conductance of solution is related to specific, conductance as follows :, , 1000, C, where C is molar concentration., Putting = 6.3 × 10–2 ohm–1 cm–1 and, C = 0.1M, , 2.303RT, log, nF, Zn 2, , 1, 0.06, log, 2, 0.1, , RT [C]c [D]d, ., ln, nF [A]a [B]b, E cell, , G, ;, nF, , ( 50.61kJ), , Molar conductance (, , 0.76 0.03, , 0.79V, , 38., , nFE ; E, , cell constant 0.4, = 0.01 ohm 1 cm 1, =, 40, resistance, , Zn, , Zn 2 /Zn, , G, , =, , Hence E1 > E2, (b) For Zn2+ Zn, , E, , 1.9 1037, , 0.26V, 2 96500 10 3, (d) Molarity = 0.01 M ; Resistance = 40 ohm;, l, -1, Cell constant = 0.4cm ., A, , 0.059, log107 = 1.67 – 0.103 = 1.567, 4, , (b) Cell reaction is, Zn, , KC, , log K C, , Specific conductivity ( ), , 43., , 0.059, (10 3 )2, 1.67, log, 4, (10 3 ) 4 0.1, , = 1.67 -, , (c), , 0.059, 1.10 2, log K C or, 0.059, 2, , o, E cell, , E, , Ecell = E°cell, , 35., , (b), , RT [C]c [D]d, ln, nF [A]a [B]b, , 47., 48., 49., , m, , =, , m, , = (6.3 × 10–2 ohm–1 cm–1) ×, , ....(a), , 1000, , (0.1mol / cm3 ), = 6.3 × 10–2 × 104 ohm–1cm2 mol–1, = 630 ohm–1 cm2 mol–1, 1, Cell constant, (b), R, Cell constant = × R; 0.012 × 55 = 0.66 cm–1., (d) ohm–1 cm–1, (a) The specific conductance increases with concentration., The number of ions per cm–3 increase with increase of, concentration.
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ELECTROCHEMISTRY, 50., 51., 52., 53., 54., 55., , 289, , (d), (c) Correct matching for pair (iii) will be, [G (conductance) – siemens or ohm–1(S).], (d) Conductivity does not depend upon mass or weight, of material., (d), (c) (i) represents weak electrolyte, (ii) represents strong electrolyte., (b), , NaCl, AcOH, , 57., , 58., , 61. (c), , m NH 4Cl, , m NH, , m NaOH, , m Na, , m NaCl, , m NH, 4, , 280 10, , 2Cl, , Ba(OH) 2, , 4, , 1, , S m 2 mol, , 2 NaOH, , BaCl2, , 1, , 2 NaCl, , 280 10 4 2 248 10 4 2 126 10 4, 524 10 4 Sm 2 mol 1., , ClCH 2COONa HCl, , ClCH 2COOH NaCl, , LClCH 2COOH = 427 - 38.2, = 388.8 ohm-1cm 2gm eq-1, , 62. (b), , =, , M, , =, , M, , 9.54, = 0.04008 = 4.008 %., 238, , 63. (d) Cell constant = l/a, Unit = m/m2 = m–1., 64. (d) Larger the size, lower the speed., 2, 1, 65. (a), m 57 73 130 Scm mol, 66. (b) At anode :, 1, H 2O + O2, 2HO–, 2, 67. (b) Given current (i) = 0.5 amp;, Time (t) = 100 minutes × 60 = 6000 sec, Equivalent weight of silver nitrate (E) = 108., According to Faraday's first law of electrolysis, , mOH, , m Na, , 126 10 4 Sm 2 mol, , 1, , 224 + 203 = LClCH2COOH + 38.2, , m Cl, , 4, , 248 10 4 Sm 2 mol, , LClCH 2COONa + LHCl = LClCH 2COOH + L NaCl, , (i) (ii) (iii), , (d) Kohlrausch’s Law states that at infinite dilution, each, ion migrates independent of its co-ion and contributes, to the total equivalent conductance of an electrolyte a, definite share which depends only on its own nature., From this definition we can see that option (d) is the, correct answer., (c) Conductivity of an electrolyte depends on the mobility, of ions and concentration of ions. The motion of an, ionic species in an electric field is retarded by the, oppositely charged ions due to their interionic, attraction. On dilution, concentration of electrolyte, decreases and the retarding influence of oppositely, charged ions decreases. Therefore mobility of ions, increases., (d), , Cl, , Ba(OH) 2, , [426.2 91.0 126.5] 390.7, 56., , Na, , Ba(OH)2, , (iii), , 126.5, , OH, , Now,, , (ii), , 91.0, , AcONa, , Na, , Ba 2, , (i), , 426.2, , HCl, , 60. (b), , Eit, 108 0.5 6000, 3.3575 g., 96500, 96500, (d) 1 mole of e– = 1F = 96500 C, 27g of Al is deposited by 3 × 96500 C, 5120 g of Al will be deposited by, W, , m Cl, , 68., , m OH, , 3 96500 5120, 5.49 10 7 C, 27, (c) Electrorefining and electroplating are done by, electrolysis., , =, , m NH, 4, , m Cl, , m Na, , m OH, , m NH 4OH, , 59., , (d), , CH 3COOH, , m NH 4Cl, CH3COONa, , 69., m Na, , m NaOH, , HCl, , m Cl, , m NaCl, NaCl, , = 91 + 425.9 – 126.4 = 390.5 S cm2mol–1, , 70. (d), , WA, EA, , WB 1.6, ;, EB, 8, , Wt . of Ag, , 71. (b), , Wt. of Ag, 108, , 21.6 g, , Cu+2 +, , 2e–, 2 × 96500 C, , 9650 C will deposit =, , Cu (s), 63.6g, 63.6, 9650 = 3.18 g, 2 96500
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EBD_7207, ELECTROCHEMISTRY, , 290, , 72., , (a) Charge = 0.2 × 1 Faraday, = 0.2 × 96500 coulombs, = 19300 = 1.93 × 104 coulombs, , 73., , (d), , e, Ag, Ag, 96500 coulombs deposit = 108 g of Ag, , 965 coulombs deposit =, 74., , (d), , 86., , 108, 965 1.08 g Ag, 96500, , C 6 H 5 NO 2 + 6H + + 6e- ¾ ¾, ® C 6 H 5 NH 2 + 2H 2 O, , E C6 H 5 NO 2 (eq.wt) =, , Na+ (l) + Cl– (l), , 87., , 123, = 20.5, 6, , Number of coulombs required =, , NaCl (s), Na+ + e–, Na (l), (at cathode), Cl– (l), Cl (g) + e–, (at anode), Cl (g) + Cl(g), Cl2 (g), (b) Pt4+ + 4e–— Pt, 4F electricity is required to deposit 1 mole of Pt., 0.80 F of electricity will deposit, = 1/4 × 0.80 moles of Pt = 0.20 mol., (d) 1F 11.2 L Cl2 at STP, No. of Faradays =, , w ´96500, Eq. wt, , 12.3´ 96500, = 57900 C, 20.5, (a) According to Faraday law's of electrolysis, amount of, electricity required to deposit 1 mole of metal, = 96500 C = 1 F, i.e., for deposition of 108g Ag electricity required = 1 F, (a) For deposition of silver, reaction is, Ag+ + e–, Ag, 1 mol of Ag will be deposited by, = 1 F = 96500 C = 9.65 × 104 C, Since 1 equivalent weight of Ag is also equal to the, weight of its 1 mol, hence 1 equivalent weight of Ag, will be deposited by = 9.65 × 104 C, (b) A current of 96500 coulombs liberate 1 mole of O2., 96500 C liberates = 22.4 L of O2 at NTP, , 88., , 76., , 77., , 22.4, ´ 9650, 96500, = 2.24 L of O2 at NTP, , Eq. wt. of copper =, Z=, , 78., , 79., 80., , 2, , 81., 82., 83., 84., , 85., , 91., , (a), , 92., , (b), , 93., , (a), , 94., , (b), , 95., , (b), , 96., 97., , (b), (b), , 2, , 10, , 6, , 1.602 10, , 19, , 6.24 1012, , (d), (d) Charge on one mole of electrons = 96500 C., (a) When platinum electrodes are dipped in dilute solution, H2SO4 than H2 is evolved at cathode., (b) In electrolysis of NaCl when Pt electrode is taken then, H2 liberated at cathode while with Hg cathode it forms, sodium amalgam because more voltage is required to, reduce H+ at Hg than Pt., (a) When molten or fused NaCl is electrolysed, it yields, metallic sodium and gaseous chlorine. Reactions, involved are as follows:, , MnO4, , (c), , 2, , 5600, 32 8g, wO =, 2, 22400, = 1 equivalent of O2, = 1 equivalent of Ag = 108, Electrolysis of these (i), (ii) and (iii) salt release Chlorine, which is yellowish in colour while Br2 is reddish brown, in colour, Electronic conductance decreases with increase in, temperature whereas electrolytic conductance, increases with increase in temperature as no. of ions, or charge carriers increases with increase in, temperature., Gold is an inert metal. Electrodes made up of inert, metals does not participate in chemical reaction., Reaction A is preferred for electrolysis of dilute, sulphuric acid and B is preferred for electrolysis of, concentrated sulphuric acid., Iron can be easily reduced with carbon it does not, require electrochemical reduction., H2SO4 is regenerated., During charging, the lead storage battery behaves like, an electrolytic cell. So, at anode the reaction is, PbSO 4, , 98., , e, , Quantity of electricity required, = 0.1F = 0.1 × 96500 = 9650 C, (d) wO = nO × 32, , 9650 C liberates =, , 56, 2e, Fe; E Fe, 28, (b) Fe, 2, 1 Faraday liberates = 28 g of Fe, 3 Faraday liberates = 3 × 28 = 84 gm, (c), (d) Charge (Coulombs) pass per second = 10–6, number of electrons passed per second, , 0.2938g, , ( 7), , MnO 24, , 0.1 mole, , 90., , 31.5, , 31.5, 1.5 10 60, 96500, , ( 6), , (c), , 63, 2, , 31.5, 96500, , W = Zit =, , 89., , 1.8, , Vol. of Cl2 = 1.8 × 11.2 L = 20.16, (b) W = Zit, where Z = Electrochemical equivalent, , =, , 75., , 9.65 5 60 60, 96500, , 2H 2 O, , 2NH 4 Cl Zn, , PbO 2, , 2NH 3, , 4H, , ZnCl 2, , SO 42, , H2, , ., , 2e
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ELECTROCHEMISTRY, 99., , 100., , 101., 102., 103., 104., 105., 106., , 107., 108., 109., 110., , 291, , (b) In H 2 O 2 fuel cell, the combustion of H2 occurs to, create potential difference between the two, electrodes., (d) Primary cells are those cells, in which the reaction, occurs only once and after use over a period of time, it, becomes dead and cannot be reused again. e.g.,, Leclanche cell and mercury cell., (c) The electrolyte used in Leclanche cell is moist paste of, NH4Cl and ZnCl2., (d) A device that converts energy of combustion of fuels,, directly into electrical energy is known as fuel cell., (c), (b) H2 – O2 fuel cell supply power for pressure., (d) Prevention of corrosion by zinc coating is called, galvanization., (a) Cathodic protection is best method to prevent iron, from rusting. In this method iron is made cathode by, application of external current., Saline water is highly conducting and hence, accelerates the formation of rust., (b) Magnesium provides cathodic protection and prevent, rusting or corrosion., (c) Mercury cell being primary in nature can be used only, once., (c), (d) Fuel cells produce electricity with an efficiency of, about 70% compared to thermal plants whose, efficiency is about 40%., , STATEMENT TYPE QUESTIONS, 111. (a) Statement (ii) is false as in Figure 2 equal potential is, applied from either side thereby making his cell, non-functional. Figure 3 represents electrolytic cell, as current flows from anode to cathode., 112. (d) According to an accepted convention anode is written, on the left side and cathode on the right while, representing the galvanic cell., 113. (c) Because reduction potential of water is higher than, that of Na+ so H2 will be evolved and no reduction of, metal ions occurs., 114. (a) All statements given above are correct., 115. (d) CaX2 is a strong electrolyte thus its molar conductivity, increases slowly on dilution., 116. (c) We consider,, 96487Cmol 1 96500Cmol 1 for approximate, calculations not for accurate calculations., 117. (c) Reaction occurring at anode, , 2Fe2, , 2Fe s, , 4e, , Reaction occuring at cathode, O2 g, , 4H, , aq, , 4e, , 2H 2 O, , MATCHING TYPE QUESTIONS, 118. (b), , 119. (a), , 120. (b), , 121. (d), , 122. (b), , ASSERTION-REASON TYPE QUESTIONS, 123. (b) We know, R, , A, , or R, , A, , , where proportionality, , is called resistivity. If = 1m and A = 1m2,, i.e., Resistance = Resistivity.., , constant, then R =, , 124. (d) The specific conductivity decreases while equivalent, and molar conductivities increase with dilution., 125. (a) Zinc metal which has a more negative electrode, potential than iron will provide electrons in preference, of the iron, and therefore corrodes first. Only when all, the zinc has been oxidised, the iron start to rust., , CRITICAL THINKING TYPE QUESTIONS, 126. (a) Salt bridge allows the flow of current by completing, circuit. No current will flow and voltage will drop to, zero, if salt bridge is removed., 127. (d) In electrolytic cell the flow of electrons is from anode, to cathode through internal supply., 128. (b), , E°, , Sn 2+ / Sn 4+, , = -0.15V > E°, , Hg 22+ / Hg 2+, , = -0.92, , Hence, Sn 2+ is stronger reducing agent than Hg 22 ., 129. (a) Option (ii) and (iv) are correct., 130. (a) The value of E °, , M 2+ M, , E°, , Mn 2+ Mn, , for given metal ions are, , = -1.18 V,, , E°, , = -0.9 V,, , E°, , = -0.44 V and, , E°, , = -0.28 V., , Cr 2+ Cr, , Fe2+ Fe, Co 2+ Co, , Th e correct order of E ° 2+, M, , M, , values without, , considering negative sign would be, Mn2+ > Cr2+ > Fe2+ > Co2+., 131. (d), , E red, , E ored, , 0.591, log[M n ], n, , Lower the concentration of Mn+, lower is the reduction, potential., Hence order of reduction potential is :, Q> R>S >P
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ELECTROCHEMISTRY, , 293, , 141. (c) When the concentration of all reacting species kept, unity, then Ecell, become, , 146. (d), , R, , 1, 1, ohm 1 = 0.0316 ohm–1, R 31.6, Specific conductance = conductance × cell constant., = 0.0316 ohm–1 × 0.367 cm–1, = 0.0116 ohm–1 cm–1, Now, molar concentration = 0.5M, (given), = 0.5 × 10–3 mole cm–3, , Conductance =, , Molar conductance =, , 0.0116, , 0.5 10 3, = 23.2 S cm2 mol–1, 143. (b) According to Kohlrausch’s law, molar conductivity of, weak electrolyte acetic acid (CH3COOH) can be, calculated as follows:, CH3COOH, , CH3COONa, NaCl, , calculating value of, 1, R, , 144. (a), 1.3, , A, , C, , HCl, , 65m, , Eo, , EH (Eq. wt) =, , 2, 22400, = 11200 ml (STP), =1 g =, 2, 2, , Total charge passed =, , 96500´112, = 965, 11200, , Q = It = 965, 965, = 1amp., 965, 152. (c) Using Faraday’s second law of electrolysis,, Weight of Cu deposited, Weight of Ag deposited, , A, , w Cu, 0.18, , 1, , Equ. wt. of Cu, Equ. wt. of Ag, , 63.5 1, 2 108, , 63.5 18, = 0.0529 g., 2 108 100, 153. (c) According to the definition 1 F or 96500 C is the charge, carried by 1 mol of electrons when water is electrolysed, 2H2O, 4H+ + O2 + 4e–, So, 4 Faraday of electricity liberate = 32 g of O2., Thus 1 Faraday of electricity liberate, 32, g of O2 = 8 g of O2, =, 4, , wCu =, , [molarity is in moles/litre but 1000 is used to convert, liter into cm3], 1, 65 m 1 1000 cm 3, 260, 0.4 moles, , 650 m 1, 1, m3, 260 4 mol 1000, , 154. (b), , = 6.25 × 10–4 S m2 mol–1, 145. (b), for CH3COOH, , =, , 0.4 1000, = 500 Ohms, 8 0.1, , I=, , 1000, molarity, , =, , 1000, N, , A, , 2.303RT, log a M n ., nF, ® H2, 151. (a) 2H+ + 2e– ¾ ¾, E, , should also be known for, , A, , 1, 50, , 1, R, , 147. ( b), 148. (d) Charge on Mg and Ca ion is greater than that of Na, and K, so Mg and Ca ions possess higher conductivity,, also solvation of metal ion decreases as we move down, the group, hence conductivity increases), 149. (d) The conductivity of electrolytic solution depends, upon all of the given factors., 150. (c) Correct Nernst equation is, , NaCl, , CH 3COOH ., , 8, , 0, , C, , nFE cell, rG, 142. (b) Here, R = 31.6 ohm, , Value of, , C, , E cell and the given relation will, , CH3COO, , Na, , CH3COO, , H, , H, , Cl, , Na, , Cl, , H+ + OH-, , 1, ® H2, H+ + e- ¾¾, 2, 0.5 mole of H2 is liberated by 1 F = 96500 C, 0.01 mole of H2 will be liberated by, , 96500, ´0.01 = 1930 C, 0.5, Q=I ×t, =, , = 90 + 425 – 125 = 390 mho cm2 mol–1., Degree of dissociation ( ) =, , H 2O, , c, m, m, , 7.8, 390, , 0.02, , t=, , Q, 1930 C, =, = 19.3´10 4 sec, I 10 ´10-3 A
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EBD_7207, ELECTROCHEMISTRY, , 294, , 155. (b), , 1, Cl, 2 2, , At Anode, Cl, , Equivalent wt. of chlorine (ECl2 ), WCl2, , ECl2, , I t, , 96500, , 158. (a) The charge carried by 1 mole of electrons is one faraday., Thus for a reaction, , e, , 35.5 2, 2, , 35.5 2 30 60, 96500, , Mn, ne, M, nF = 1 mole of M, , 35.5, , Al3, 3F, 1F, , 1.32 gm., , W, Q, 156. (a) By Faraday's 1st Law of electrolysis,, =, E 96500, (where Q it = charge of ion ), We know that no. of gram equivalent, , it, 1, W, 1 965, E 96500 96500 100, (where i= 1 A, t = 16×60+5 = 965 sec.), Since, we know that, , Normality, , No. of gram equivalent, Volume (in litre), , 0.2964, 96500 = 31.75 g of copper, 900, , Thus, 31.75 is the eq. mass of copper, At. mass = Eq. mass × Valency, 63.56 = 31.75 × x, x = + 2., , Cu 2, , 2e, , Na, , e, , 2F, 1F, , 1F, , Al, , 1 mol, 1/3 mol, , Cu, , 1 mol, 1/2 mol, , Na, , 1 mol, , The mole ratio of Al, Cu and Na deposited at the, 1 1, 3 2, , respective cathode is : :1 or 2 : 3 : 6., , 1, 100 = 0.01 N, 1, , 157. (b) Quantity of charge passed = 0.5 × 30 × 60 = 900 coulomb, 900 coulomb will deposit = 0.2964g of copper, 96500 coulomb will deposit, =, , 3e, , 159. (a), 160. (b) Total of 6 electrons are required to form 2 moles of, Cr3+ therefore to form 1 mole 3F of charge is required., 161. (d), , Pb PbO 2, , 2H 2SO 4, , Discharge, Recharge, , 2PbSO 4, , 2H 2 O., , Sulphuric acid is consumed on discharging., 162. (d) This is because zinc has higher oxidation potential, than Ni, Cu and Sn. The process of coating of iron, surface with zinc is known as galvanization. Galvanized, iron sheets maintain their lustre due to the formation, of protective layer of basic zinc carbonate., 163. (c) At cathode reduction occurs according to following, reaction., 4, , MnO 2, , NH 4, , e, , 3, , MnO OH, , NH3
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18, CHEMICAL KINETICS, FACT / DEFINITION TYPE QUESTIONS, 1., , 2., , 3., , 4., , 5., , 6., , 7., , The term – dc/dt in a rate equation refers to :, (a) the conc. of a reactant, (b) the decrease in conc. of the reactant with time, (c) the velocity constant of reaction, (d) None of these, The rate law for the single- step reaction 2A B, 2C,, is given by:, (a) rate = k [A].[B], (b) rate = k [A]2.[B], (c) rate = k [2A].[B], (d) rate = k [A]2.[B]o, Rate of which reaction increases with temperature :, (a) of any type of reactions, (b) of exothemic reactions, (c) of endothemic reactions, (d) of none, In a slow reaction, rate of reaction generally .......... with, time:, (a) decreases, (b) increases, (c) sometimes increases and sometimes decreases., (d) remains constant, The rate of a chemical reaction tells us about,, (a) the reactants taking part in reaction, (b) the products formed in the reaction, (c) how slow or fast the reaction is taking place, (d) None of the above, For the reaction 2 A + B 3C + D, which of the following does not express the reaction rate ?, d[D], d[ B], (a), (b), dt, dt, 1 d[A], 1 d[C], (c) –, (d) 3 dt, 2 dt, Consider the reaction, N2 (g) + 3H2 (g) 2 NH3 (g), d[ NH 3 ], d[H 2 ], and, The equality relationship between, dt, dt, is, (a), , d[ NH 3 ], dt, , 2 d[ H 2 ], 3 dt, , d[ NH 3 ], dt, , (b), (c), , d[ NH 3 ], dt, , 3 d[H 2 ], 2 dt, , d[H 2 ], dt, , d[ NH 3 ], 1 d[ H 2 ], dt, 3 dt, For the reaction 2 A + B 3C + D, which of the following does not express the reaction rate ?, d[ B], d[ D], (a), (b), dt, dt, , (d), 8., , 1 d[A], 1 d[C], (d) 2 dt, 3 dt, Which of the following reaction does not occur fastly ?, (a) Precipitation of AgCl by mixing aqueous solutions of, AgNO3 and NaCl., (b) Burning of gasoline, (c) Rusting of iron, (d) Burning of LPG for cooking, Chemical kinetics is a study to find out, (a) the feasibility of a chemical reaction, (b) extent to which a reaction will proceed, (c) speed of a reaction, (d) All of the above, Rate of a reaction can be defined as, (a) the rate of decrease in concentration of any one of the, reactants, (b) the rate of increase in concentration of any one of the, products, (c) the rate of decrease in concentration of any one of the, reactants or the rate of increase in concentration of, any one of the products, (d) the sum of rate of decrease in concentration of all the, reactants or the rate of increase in concentration of all, the products, The rate of reaction, (a) increases as the reaction proceeds, (b) decreases as the reaction proceeds, (c) remains the same as the reaction proceeds, (d) may decrease or increase as the reaction proceeds, , (c), , 9., , 10., , 11., , 12., , –
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EBD_7207, CHEMICAL KINETICS, , 296, , 13., , 14., , 15., , 16., , 17., , 18., , 19., , 20., , 21., , 22., , The unit of rate of reaction is, (a) mole/dm3, (b) mole/pound, (c) mole/dm3 sec, (d) mole/cm3, In the rate equation, when the conc. of reactants is unity, then rate is equal to, (a) specific rate constant, (b) average rate constant, (c) instantaneous rate constant, (d) None of above, The rate of reaction between two specific time intervals is, called, (a) instantaneous rate, (b) average rate, (c) specific rate, (d) ordinary rate, Instantaneous rate of a chemical reaction is, (a) rate of reaction in the beginning, (b) rate of reaction at the end, (c) rate of reaction at a given instant, (d) rate of reaction between two specific time intervals, At the beginning the decrease in the conc. of reactants is, (a) slow, (b) moderate, (c) rapid, (d) None of above, The average rate and instantaneous rate of a reaction are, equal, (a) at the start, (b) at the end, (c) in the middle, (d) when two rate have time interval equal to zero, The rate of reaction depends upon the, (a) volume, (b) force, (c) pressure, (d) conc. of reactants, For the following reaction: NO2(g) + CO(g), NO(g) +, CO2(g), the rate law is: Rate = k [NO2]2. If 0.1 mole of, gaseous carbon monoxide is added at constant temperature, to the reaction mixture which of the following statements is, true?, (a) Both k and the reaction rate remain the same, (b) Both k and the reaction rate increase, (c) Both k and the reaction rate decrease, (d) Only k increases, the reaction rate remain the same, Which one of the following statements for the order of a, reaction is incorrect ?, (a) Order can be determined only experimentally., (b) Order is not influenced by stoichiometric coefficient, of the reactants., (c) Order of reaction is sum of power to the concentration, terms of reactants to express the rate of reaction., (d) Order of reaction is always whole number., The rate of the reaction 2NO Cl2, 2NOCl is, given by the rate equation rate = k [NO]2 [Cl2], The value of the rate constant can be increased by:, (a) increasing the concentration of NO., (b) increasing the temperature., (c) increasing the concentration of the Cl 2, (d) doing all of the above, , 23., 24., , 25., , 26., , 27., , Order of reaction can be, (a) 0, (b) fraction, (c) whole number, (d) integer, fraction, zero, Units of rate constant of first and zero order reactions in, terms of molarity M unit are respectively, (a) sec–1 , Msec–1, (b) sec–1, M, –1, –1, (c) Msec , sec, (d) M, sec–1., A reaction involving two different reactants can never, be, (a) bimolecular reaction (b) second order reaction, (c) first order reaction, (d) unimolecular reaction, ,, it, would, be, a, zero, order reaction when, 3A B C, (a) the rate of reaction is proportional to square of, concentration of A, (b) the rate of reaction remains same at any concentration, of A, (c) the rate remains unchanged at any concentration of B, and C, (d) the rate of reaction doubles if concentration of B is, increased to double, For the following homogeneous reaction,, k, , 28., , 29., , 30., , 31., , 32., , C, A B, the unit of rate constant is, (a) sec –1, (b) sec–1 mol L–1, (c) sec–1 mol–1 L, (d) sec–1 mol–2 L2, Order of reaction is decided by, (a) temperature, (b) mechanism of reaction as well as relative concentration, of reactants, (c) molecularity, (d) pressure, Velocity constant k of a reaction is affected by, (a) change in the concentration of the reactant, (b) change of temperature, (c) change in the concentration of the product, (d) None of the above, The, rate, constant, for, the, reaction, 2N 2O5, 4 NO 2 O 2 is 3.10 × 10–5 sec–1. If the rate is, 2.4 × 10–5 mol litre–1 sec–1 then the concentration of N 2 O 5, (in mol litre–1) is :, (a) 0.04, (b) 0.8, (c) 0.07, (d) 1.4, A zero order reaction is one whose rate is independent of, (a) the concentration of the reactants, (b) the temperature of reaction, (c) the concentration of the product, (d) the material of the vessel in which reaction is carried, out, The rate law for a reaction between the substances A and B, is given by Rate = k [A]n [B]m, On doubling the concentration of A and halving the, concentration of B, the ratio of the new rate to the earlier, rate of the reaction will be as, (a) (m + n), (b) (n – m), (c) 2(n –, , m), , (d), , 1, 2, , (m n )
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CHEMICAL KINETICS, , 33., , 34., , 35., , 36., , 37., , 38., , 39., , 40., , 41., , 297, , In the reaction 2A B A 2 B, if the concentration of A is, doubled and that of B is halved, then the rate of the reaction will:, (a) increase 2 times, (b) increase 4 times, (c) decrease 2 times, (d) remain the same, The order of a reaction, with respect to one of the reacting, component Y, is zero. It implies that:, (a) the reaction is going on at a constant rate, (b) the rate of reaction does not vary with temperature, (c) the reaction rate is independent of the concentration, of Y, (d) the rate of formation of the activated complex is zero, If the rate of a gaseous reaction is independent of pressure,, the order of reaction is:, (a) 0, (b) 1, (c) 2, (d) 3, If the rate of the reaction is equal to the rate constant, the, order of the reaction is, (a) 3, (b) 0, (c) 1, (d) 2, In a reaction, when the concentration of reactant is increased, two times, the increase in rate of reaction was four times., Order of reaction is :, (a) zero, (b) 1, (c) 2, (d) 3, For the reaction A + 2B, C, rate is given by R = [A] [B]2, then the order of the reaction is, (a) 3, (b) 6, (c) 5, (d) 7, The unit of rate constant for a zero order reaction is, (a) mol L–1 s–1, (b) L mol–1 s–1, 2, –2, –1, (c) L mol s, (d) s – 1, Which one of the following reactions is a true first order, reaction?, (a) Alkaline hydrolysis of ethyl acetate, (b) Acid catalyst hydrolysis of ethyl acetate, (c) Decomposition of N2O, (d) Decomposition of gaseous ammonia on a hot platinum, surface, C + 2D, experimental results, For a reaction A + B, were collected for three trials and the data obtained are, given below:, , Trial [A], M [B], M Initial Rate, M s–1, , 43. What is order with respect to A, B, C, respectively, [A], [B], [C], rate (M/sec.), 0.2, 0.1, 0.02, 8.08 × 10–3, 0.1, 0.2, 0.02, 2.01 × 10–3, 0.1, 1.8, 0.18, 6.03 × 10–3, 0.2, 0.1, 0.08, 6.464 × 10–2, (a) –1, 1, 3/2, (b) –1, 1, 1/2, (c) 1, 3/2, –1, (d) 1, –1, 3/2, 44. Select the rate law that corresponds to the data shown for, the following reaction :, A B, C, Expt. No., 1, 2, 3, 4, , (A), 0.012, 0.024, 0.024, 0.012, , (B), 0.035, 0.070, 0.035, 0.070, , (a) Rate = k[B]3, (b) Rate = k [B]4, 3, (c) Rate = k [A] [B], (d) Rate = k [A]2 [B]2, 45. The order of a reaction with rate equal to k[A]3/2 [B]–1/2 is, :, 1, (a) 1, (b), 2, 3, 2, For the reaction,, , (c), , 46., , (d) 2, , H, , products, CH3COCH3 + I2, The rate is governed by expression, dx, k[acetone][H ], dt, The order w.r.t. I2 is:, (a) 1, (b) 0, (c) 3, (d) 2, 47. The rate constant of a reaction is 3.00 × 103 L mol–1 sec–1., The order of this reaction will be:, (a) 0, (b) 1, (c) 2, (d) 3, 48. During the kinetic study of the reaction, 2A + B, C + D,, following results were obtained:, Run [A ](mol L–1 ) [B](mol L– 1 ), , –4, , 42., , 1, 0.40, 0.20, 5.5 × 10, 2, 0.80, 0.20, 5.5 × 10–4, 3, 0.40, 0.40, 2.2 × 10–3, The correct rate law of the reaction is, (a) rate = k[A]0 [B]2, (b) rate = k[A] [B]2, (c) rate = k[A] [B], (d) rate = k[A] [B]0, The rate law for the reaction, xA + yB, mP + nQ is Rate = k [A]c [B]d. What is the, total order of the reaction?, (a) (x + y), (b) (m + n), (c) (c + d), (d) x/y, , Initial Rate, 0.10, 0.80, 0.10, 0.80, , Initial rate of, formation of, –1, , –1, , D (mol L min ), I, , 0.1, , 0.1, , 6.0 × 10, , –3, , II, , 0.3, , 0.2, , 7.2 × 10, , –2, , III, , 0.3, , 0.4, , 2.88 × 10, , –1, , 2.40 × 10, , –2, , IV, , 0.4, , 0.1, , Based on the above data which one of the following is, correct?, (a) rate = k [A]2 [B], 2, , 2, , (c) rate = k [A] [B], , (b) rate = k[A] [B], (d) rate = k [A] [B]2
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EBD_7207, CHEMICAL KINETICS, , 298, , 49., , 50., , 51., , 52., , 53., , For the reaction H2(g) + Br2 (g) 2HBr (g), the experimental, data suggest, rate = k[H2][Br2]1/2. The molecularity and, order of the reaction are respectively, 3 3, 3, ,, (a) 2,, (b), 2, 2 2, 1, (c) 1, 1, (d) 1,, 2, The chemical reaction 2O 3, 3O 2 proceeds as follows:, Slow, Fast, 2 O 2 the rate law, O3, O2 O ; O O 3, expression should be, (a) r = k[O3]2, (b) r = k [O3]2[O2]–1, 3, 2, (c) r = k [O3][O2], (d) r = [O3][O2]2, Nitrogen monoxide, NO, reacts with hydrogen, H2, according, to the following equation:, 2NO(g) + 2H2(g) N2(g) + 2H2O(g), If the mechanism for this reaction were,, 2NO(g) + H2(g) N2(g) + H2O2(g) ; slow, H2O2(g) + H2(g) 2H2O(g) ; fast, , Which of the following rate laws would we expect to obtain, experimentally?, (a) Rate = k[H2O2][H2], (b) Rate = k[NO]2[H2], 2, 2, (c) Rate = k[NO] [H2], (d) Rate = k[NO][H2], Which of the following is not a first order reaction ?, (a) Hydrogenation of ethene, (b) Natural radioactive decay of unstable nuclei, (c) Decomposition of HI on gold surface, (d) Decomposition of N2O, The plot that represents the zero order reaction is :, (a), , [R], , (b), , [R], , 54., , 55., , 56., , 59., , 60., , 61., , 62., , t, , [R], , (d) In[R], t, t, The plot of concentration of the reactant vs time for a, reaction is a straight line with a negative slope. The reaction, follows a rate equation, (a) zero order, (b) first order, (c) second order, (d) third order, The half-life of a reaction is inversely proportional to the, square of the initial concentration of the reactant. Then the, order of the reaction is, (a) 0, (b) 1, (c) 2, (d) 3, The rate equation for a reaction,, N2O, N2 + 1/2O2, is Rate = k[N2O]0 = k. If the initial concentration of the, reactant is a mol Lit–1, the half-life period of the reaction is, a, t 1 ka, (a) t 1, (b), 2, k, 2, 2, (c), , 58., , 63., , t, (c), , 57., , t1, 2, , a, k, , (d), , t1, 2, , k, a, , 64., , Half life of a first order reaction is 4s and the initial, concentration of the reactant is 0.12 M. The concentration, of the reactant left after 16 s is, (a) 0.0075 M, (b) 0.06 M, (c) 0.03 M, (d) 0.015 M, The reaction A B follows first order kinetics. The time, taken for 0.8 mole of A to produce 0.6 mole of B is 1 hour., What is the time taken for conversion of 0.9 mole of A to, produce 0.675 mole of B?, (a) 2 hours, (b) 1 hour, (c) 0.5 hour, (d) 0.25 hour, The rate of a first order reaction is 1.5 × 10–2 mol L–1 min–1 at, 0.5 M concentration of the reactant. The half life of the reaction, is, (a) 0.383 min, (b) 23.1 min, (c) 8.73 min, (d) 7.53 min, The rate constant for a first order reaction whose half-life, is, 480 seconds is :, (a) 2.88 × 10–3 sec–1, (b) 2.72 × 10–3 sec–1, –3, –1, (c) 1.44 × 10 sec, (d) 1.44 sec–1, The rate constant of a first order reaction is 6.9 10 3 s 1 ., How much time will it take to reduce the initial concentration, to its 1/8th value?, (a) 100 s, (b) 200 s, (c) 300 s, (d) 400 s, A reaction proceeds by first order, 75% of this reaction, was completed in 32 min. The time required for 50%, completion is, (a) 8 min, (b) 16 min, (c) 20 min, (d) 24 min, Point out the wrong statement:, For a first order reaction, (a) time for half-change (t 1/2) is independent of initial, concentration, (b) change in the concentration unit does not change the, rate constant (k), (c) time for half-change × rate constant = 0.693, (d) the unit of k is mole–1 min–1, t 1 can be taken as the time taken for the concentration of a, 4, , 3, of its initial value. If the rate constant, 4, t, for a first order reaction is k, the 1 can be written as, , reactant to drop to, , 4, , 65., , (a) 0.75/k, (b) 0.69/k, (c) 0.29/k, (d) 0.10/k, In a first-order reaction A, B, if k is rate constant and, inital concentration of the reactant A is 0.5 M, then the halflife is, (a), , log 2, k, , (b), , log 2, k 0.5, , (c), , ln 2, k, , (d), , 0.693, 0.5k
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CHEMICAL KINETICS, , 66., , 67., , 68., , 299, , Consider the reaction, 2A + B, products. When, concentration of B alone was doubled, the half-life did not, change. When the concentration of A alone was doubled,, the rate increased by two times. The unit of rate constant, for this reaction is, (a) s – 1, (b) L mol–1 s–1, (c) no unit, (d) mol L–1 s–1., The decomposition of N2O5 occurs as, 2N2O5, 4NO2 + O2 and follows Ist order kinetics,, hence:, (a) the reaction is unimolecular, (b) the reaction is bimolecular, (c) t1/2 a0, (d) None of these, B, if k is rate constant and, In a first-order reaction A, inital concentration of the reactant A is 0.5 M, then the halflife is, (a), , log 2, k, , (b), , log 2, k 0.5, , ln 2, 0.693, (d), k, 0.5k, For a first order reaction, a plot of log (a – x) against time is, a straight line with a negative slope equal to, k, (a), (b) – 2.303 k, 2.303, , (c), 69., , Ea, 2.303, (d), 2, ., 303, R, k, In a reaction A, Products, when start is made from, 8.0 × 10–2 M of A, half-life is found to be 120 minute. For the, initial concentration 4.0 × 10–2 M, the half-life of the reaction, becomes 240 minute. The order of the reaction is :, (a) zero, (b) one, (c) two, (d) 0.5, The value of rate constant of a pseudo first order reaction, _________., (a) depends on the concentration of reactants present in, small amount., (b) depends on the concentration of reactants present in, excess., (c) is independent on the concentration of reactants., (d) depends only on temperature., In the Haber process for the manufacture of ammonia the, following catalyst is used, (a) Platinized asbestos, (b) Iron with molybdenum as promoter, (c) Copper oxide, (d) Alumina, What is the activation energy for a reaction if its rate doubles, when the temperature is raised from 20°C to 35°C?, (R = 8.314 J mol–1 K–1), (a) 269 kJ mol–1, (b) 34.7 kJ mol–1, –1, (c) 15.1 kJ mol, (d) 342 kJ mol–1, , 74. A reaction having equal energies of activation for forward, and reverse reaction has :, (a), G= 0, (b) H = 0, (c), H= G= S= 0, (d) S = 0, 75. In an exothermic reaction if H is the enthalpy then activation, energy is, (a) more than H, (b) less than H, (c) equal to H, (d) none of the above, 1, 76. In the Arrhenius plot of ln k vs , a linear plot is obtained, T, with a slope of –2 × 104 K. The energy of activation of the, reaction (in kJ mole–1) is (R value is 8.3 J K–1 mol–1), (a) 83, (b) 166, (c) 249, (d) 332, 77. The rate of reaction is doubled for every 10°C rise in, temperature. The increase in reaction rate as a result of, temperature rise from 10°C to 100°C is, (a) 112, (b) 512, (c) 400, (d) 614, 78. Plots showing the variation of the rate constant (k) with, temperature (T) are given below. The plot that follows, Arrhenius equation is, , (a), , (b), , (c), , (d), , (c), , 70., , 71., , 72., , 73., , 79. If the activation energy for the forward reaction is 150 kJ, mol–1 and that of the reverse reaction is 260 kJ mol–1, what, is the enthalpy change for the reaction ?, (a) 410 kJ mol–1, (b) –110 kJ mol–1, (c) 110 kJ mol–1, (d) – 410 kJ mol–1, 80. Activation energy of a chemical reaction can be determined, by, (a) evaluating rate constant at standard temperature, (b) evaluating velocities of reaction at two different, temperatures, (c) evaluating rate constants at two different temperatures, (d) changing concentration of reactants, 81. In respect of the equation k Ae Ea / RT in chemical, kinetics, which one of the following statements is correct ?, (a) A is adsorption factor, (b) Ea is energy of activation, (c) R is Rydberg’s constant, (d) k is equilibrium constant
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EBD_7207, CHEMICAL KINETICS, , 300, , 82., , 83., , 84., , 85., , 86., , 87., , 88., , 89., , Rate of a reaction can be expressed by Arrhenius equation, as : k Ae Ea / RT, In this equation, Ea represents, (a) the total energy of the reacting molecules at a, temperature, T, (b) the fraction of molecules with energy greater than the, activation energy of the reaction, (c) the energy below which all the colliding molecules will, react, (d) the energy below which colliding molecules will not, react, The minimum energy required for the reacting molecules to, undergo reaction is :, (a) potential energy, (b) kinetic energy, (c) thermal energy, (d) activation energy, The reason for almost doubling the rate of reaction on, increasing the temperature of the reaction system by 10°C, is, (a) the value of threshold energy increases, (b) collision frequency increases, (c) the fraction of the molecule having energy equal to, threshold energy or more increases, (d) activation energy decreases, The slope in Arrhenius plot, is equal to:, , 90., , 91., , 92., , Ea, R, , (a), , Ea, 2.303 R, , (b), , (c), , R, 2.303 Ea, , (d) None of these, , The activation energy for a reaction is 9.0 kcal/mol. The, increase in the rate constant when its temperature is, increased from 298K to 308K is, (a) 63%, (b) 50%, (c) 100%, (d) 10%, In a reversible reaction the energy of activation of the, forward reaction is 50 kcal. The energy of activation for the, reverse reaction will be, (a) < 50 kcal, (b) either greater than or less than 50 kcal, (c) 50 kcal, (d) > 50 kcal, A catalyst, (a) increases the rate of reaction by decreasing G of a, reaction., (b) increases the rate of reaction by increasing G of a, reaction., (c) increases the rate of reaction by decreasing activation, energy of the forward reaction., (d) increases the rate of reaction by providing an, alternative pathway via an intermediate with lower, activation energy., Which of the following statements best describes how a, catalyst works?, , 93., , 94., , (a) A catalyst changes the potential energies of the, reactants and products., (b) A catalyst decreases the temperature of the reaction, which leads to a faster rate., (c) A catalyst lowers the activation energy for the reaction, by providing a different reaction mechanism., (d) A catalyst destroys some of the reactants, which lowers, the concentration of the reactants., In terms of the ‘Collision Theory of Chemical Kinetics’, the, rate of a chemical reaction is proportional to, (a) the change in free energy per second, (b) the change in temperature per second, (c) the number of collisions per second, (d) the number of products molecules, According to collision theory, which of the following is, NOT a true statement concerning a catalyst?, (a) A catalyst changes the temperature of reaction., (b) The mechanism of a reaction will change when a, catalyst is added., (c) A catalyst provides a different activation energy for a, reaction., (d) A catalyst changes the speed of a reaction, but not the, equilibrium constant., Which of the following influences the rate of a chemical, reaction performed in solution?, (a) Temperature, (b) Activation energy, (c) Presence of a catalyst, (d) All of the above influence the rate, How can be activation energy for a reaction be determined, graphically?, (a) Plot k versus T, the slope of the line will be equal to Ea, (b) Plot 1/[A]t versus t, the slope of the line will be equal, to Ea, (c) Plot ln [A]t versus t, the slope of the line will be equal, to – Ea, (d) Plot ln k versus 1/T, the slope of the line will be equal, to – Ea / R, The Arrhenius equation expressing the effect of temperature, on the rate constant of the reaction is, (a), , E a / RT, , (b), , k, , Ea, RT, , k, , log e, , E a / RT, , Ea, (b) ln A, R, (c) ln k, (b) log10 a, A chemical reaction was carried out at 300 K and 280 K. The, rate constants were found to be k1 and k2 respectively., then, (a) k1 = 4k1, (b) k2 = 2k1, (c) k2 = 0.25 k1, (d) k2 = 0.5 k1, , (a), , 96., , e, , Ea, (d) k Ae, RT, In Arrhenius plot, intercept is equal to, , (c), 95., , k
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CHEMICAL KINETICS, , 98., , 99., , 100., , 101., , 102., , 103., , 104., , For a first order reaction, the plot of log K against 1/T is a, straight line. The slope of the line is equal to, (a), , Ea, R, , (b), , 2.303, Ea R, , (c), , Ea, 2.303, , (d), , Ea, 2.303 R, , Collision theory is applicable to, (a) first order reactions, (b) zero order reactions, (c) bimolecular reactions (d) intra-molecular reactions, According to the collision theory of reaction rates, the rate, of reaction increases with temperature due to, (a) greater number of collision, (b) higher velocity of reacting molecules, (c) greater number of molecules having the activation, energy, (d) decrease in the activation energy, Which of the following has been used to explain the subject, of chemical kinetics, (a) Collision theory of bimolecular reactions, (b) The activated complex theory, (c) Arrhenius equation, (d) All of these, A catalyst increases rate of reaction by, (a) decreasing enthalpy, (b) decreasing internal energy, (c) decreasing activation energy, (d) increasing activation energy, Activation energy of the reaction is, (a) the energy released during the reaction, (b) the energy evolved when activated complex is formed, (c) minimum amount of energy needed to overcome the, potential barrier, (d) the energy needed to form one mole of the product, In a reaction, the threshold energy is equal to, (a) activation energy + normal energy of reactants, (b) activation energy - normal energy of reactants, (c) normal energy of reactants - activation energy, (d) average kinetic energy of molecules of reactants, The temperature dependence of rate constant (k) of a, chemical reaction is written in terms of Arrhenius equation,, k = A . e E a Activation energy (Ea) of the reaction can be, calculated by plotting, (a) k vs., , 1, log T, , (c) log k vs., , 1, log T, , (b), , log k vs, , 1, T, , (d) k vs. T, , 105. Consider an endothermic reaction X, , Y with the, , activation energies E b and E f for the backward and, forward reactions, respectively. In general, (a) there is no definite relation between E b and E f, , (b), , Eb, , Ef, , (c), , Eb, , Ef, , (d), , Eb, , Ef, , 106. The energies of activation for forward and reverse reactions, for A2 + B2, 2AB are 180 kJ mol–1 and 200 kJ mol–1, respectively. The presence of a catalyst lowers the activation, energy of both (forward and reverse) reactions by 100 kJ, mol–1. The enthalpy change of the reaction (A2 + B2 ®, 2AB) in the presence of a catalyst will be (in kJ mol–1), (a) 20, (b) 300, (c) 120, (d) 280, 107. For the exothermic reaction A B C D, H is the heat, of reaction and Ea is the energy of activation. The energy of, activation for the formation of A + B will be, (a) Ea, (b) H, (c) Ea + H, (d) H – Ea, 108. In most cases, for a rise of 10K temperature the rate constant, is doubled to tribled. This is due to the reason that, (a) collision frequency increases by a factor of 2 to 3., (b) fraction of molecules possessing threshold energy, increases by a factor of 2 to 3, (c) Activation energy is lowered by a factor of 2 to 3., (d) none of these, 109. Consider Fig. and mark the correct option., Activated complex, , E1, Products, , Energy, , 97., , 301, , E2, Reactants, Reaction coordinate, , (a) Activation energy of forward reaction is E1 + E2 and, product is less stable than reactant., (b) Activation energy of forward reaction is E 1 + E2 and, product is more stable than reactant., (c) Activation energy of both forward and backward, reaction is E1 + E2 and reactant is more stable than, product., (d) Activation energy of backward reaction is E 1 and, product is more stable than reactant., 110. Activation energy of a chemical reaction can be determined, by ___________, (a) determining the rate constant at standard temperature., (b) determining the rate constants at two temperatures., (c) determining probability of collision., (d) using catalyst.
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EBD_7207, CHEMICAL KINETICS, , 302, , 111. According to which theory activation energy and proper, orientation of the molecules together determine the criteria, for an effective collision ?, (a) Arrhenius theory, (b) Activated complex theory, (c) Collision theory, (d) Both (a) and (c), , C, , (ii) in fig. b, D represents average rate and C represents, instantaneous rate, (iii) fig. a, A represents instantaneous rate and B represents, average rate, (iv) fig. b, C represents average rate and D represents, instantaneous rate, (a) (i) and (ii) are correct (b) (ii) and (iv) are correct, (c) (i) and (iv) are correct (d) (ii) and (iii) are correct, Choose correct option based on following statements. Here, T stands for true statement and F for false statement., (i) Molecularity is defined as the number of reacting, species taking part in a complex reaction,, (ii) Molecularity helps in understanding the mechanism, of reaction., (iii) Reactions with the molecularity three are very rare and, slow to proceed., (iv) Complex reactions involving more than three molecules, take place in more than one step., (a) TTTF, (b) TFTF, (c) FTTF, (d) FTTT, Read the following statements, (i) Order of reaction can be fractional or zero., (ii) Molecularity of a reaction can be fractional but cannot, be zero., (iii) Slowest step in the complex reaction is considered as, a rate determining step., (iv) Units of rate constant for second order reaction are, mol L s–1., (v) Order is applicable to elementary as well as complex, reactions whereas molecularity is applicable only for, elementary reactions., Which of the following is the correct code for the statements, above ?, (a) TTFFT, (b) TFTFT, (c) FFFTT, (d) FTTFF, Consider the following statements with respect of zero order, reaction, (i) The rate of the reaction is independent of reactant, concentration, (ii) The rate of the reaction is independent of temperature, (iii) The rate constant of the reaction is independent of, temperature, (iv) The rate constant of the reaction is independent of, reactant concentration, Choose the correct statement(s)., (a) (i) only, (b) (i) and (ii) only, (c) (iii) and (iv) only, (d) (i) and (iv)only, Which of the following statement(s) is/are correct ?, (i) For a zero order reaction concentration [R] vs time (t), gives a straight line plot, , D, , (ii) For a first order reaction log, , 115., , STATEMENT TYPE QUESTIONS, 112. Consider the following reaction :, 4NH 3 (g) 5O 2 (g), , (i), , Pt(s), , 4NO(g) 6H 2O(g), , Rate of reaction with respect to NH 3 will be, 1 [NH3 ], 4, t, , (ii) For the given reaction, (iii) For the given reaction, , 1 [O2 ], t, 5, 1 [NH3 ], t, 4, , 1 [H 2O], 6, t, 1 [NO], 4, t, , (iv) For the given reaction,, 1 [NH3 ], 1 [O2 ], Rate =, t, 5, t, 4, 4 [NO] 6 [H 2O], =, t, t, Which of the following is the correct code for the statements, above., (a) TTTT, (b) TFTF, (c) FTFT, (d) TFFT, 113. Which of the following statement(s) is/are correct?, (i) Rate of reaction decreases with passage of time as the, concentration of reactants decrease., (ii) For a reaction, pP qQ, rR sS, x, , 116., , 117., , y, , Rate = k[P] [Q] where x = p and y = q, (iii) Rate law is the expression in which reaction rate is, given in terms of molar concentration of reactants with, each term raised to some power, which may or may not, be same as the stichiometric coefficient of the reacting, species in a balanced chemical equation., (a) (i) and (iii), (b) (i) and (ii), (c) (ii) and (iii), (d) (i) only, 114. Study the following graphs and choose the correct option, , c1, c2, , A, B, , t4 t2 t, (i), , (P), Concentration, , Concentration, , [R]0, , c2, c1, , 118., , [P], t, , t4 t2 t, Time, Time, (a), (b), in fig. a, A represents average rate and B represents, instantaneous rate, , [R]0, does not vary linearly, [R], , with time., (iii) Inversion of cane sugar is a pseudo first order reaction., (a) (i) and (iii), (b) (i) only, (c) (ii) and (iii), (d) (iii) only
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CHEMICAL KINETICS, , 303, , 119. At high pressure the following reaction is of zero order., 2NH3 g, , 1130 K, Platinum catalyst, , N2 g, , 3H 2 g, , Which of the following statements are correct for above, reaction?, (i) Rate of reaction = Rate constant, (ii) Rate of reaction depends on concentration of ammonia., (iii) Rate of decomposition of ammonia will remain constant, until ammonia disappears completely., (iv) Further increase in pressure will change the rate of, reaction., (a) (i), (iii) and (iv), (b) (i), (ii) and (iii), (c) (ii) and (iv), (d) (i), (ii) and (iv), 120. Consider the following statements:, (i) Increase in concentration of reactant increases the, rate of a zero order reaction., (ii) Rate constant k is equal to collision frequency A if, Ea = 0., (iii) Rate constant k is equal to collision frequency A if, Ea = ., (iv) lnk vs T is a straight line., (v) lnk vs 1/T is a straight line., Correct statements are, (a) (i) and (iv), (b) (ii) and (v), (c) (iii) and (iv), (d) (ii) and (iii), 121. According to collision theory, not all collisions between, molecules lead to reaction. Which of the following, statements provide reasons for the same ?, (i) The total energy of the two colliding molecules is less, than some minimum amount of energy., (ii) Molecules cannot react with each other unless a, catalyst is present., (iii) Molecules that are improperly oriented during collision, will not react., (iv) Molecules in different states of matter cannot react, with each other., (a) (i) and (ii), (b) (i) and (iii), (c) (ii) and (iii), (d) (i) and (iv), 122. Consider the following statements, (i) Rate constant for every physical and chemical change, gets doubled with 10°C rise in temperature, (ii) On taking log both side Arrhenius equation will, become, Ea, log A, RT, (iii) The energy required to form activated complex is known, as activation energy, Which of the following is the correct code for statements, above?, (a) TTT, (b) FTT, (c) FTF, (d) TFT, log k, , 123. Read the following statements., (i) e E a /RT corresponds to the fraction of molecules that, have kinetic energy greater than Ea., (ii) Ea can be calculated as follows, k, log 1, k2, , T2 T1, Ea, 2.303R T1T2, , (iii) Catalyst can alter a reaction both ways means it can, either decrease on increase rate of reaction, (iv) A catalyst always decreases the activation energy of, the reaction but does not alter Gibb’s energy., (v) A catalyst does not alter equilibrium constant rather, it, helps in attaining the equilibrium faster., Which of the following is the correct codes for above, statements ?, (a) TTTFF, (b) TFFTT, (c) TFFTF, (d) FTFTT, 124. The following statement(s) is (are) correct :, (i) A plot of log kp versus 1/T is linear, (ii) A plot of log [X] versus time is linear for a first order, reaction, X, P, (iii) A plot of log p versus 1/T is linear at constant volume, (iv) A plot of p versus 1/V is linear at constant temperature, (a) (i) only, (b) (ii) only, (c) (i) and (iv), (d) (i), (ii) and (iv), , MATCHING TYPE QUESTIONS, 125. Match the columns, Column-I, (A) Mathematical expression for rate, of reaction, (B) Rate of reaction for zero order, reaction is equal to, (C) Units of rate constant for zero, order reaction is same as that of, (D) Order of a complex reaction is, determined by, (a) A – (q), B – (p), C – (s), D – (r), (b) A – (r), B – (p), C – (s), D – (q), (c) A – (q), B – (s), C – (p), D – (r), (d) A – (p), B – (q), C – (s), D – (r), 126. Match the columns, Column-1, (A) Zero order reaction, (B) First order reaction, (C) Second order reaction, (a) A – (q), B – (r) ,C – (p), (b) A – (q), B – (p) ,C – (r), (c) A – (p), B – (q) ,C – (r), (d) A – (p), B – (r) ,C – (q), , Column-II, (p) rate constant, (q) rate law, (r) order of slowest, step, (s) rate of reaction, , Column-II, (p) L mole–1 sec–1, (q) mole L–1 sec–1, (r) sec –1
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EBD_7207, CHEMICAL KINETICS, , 304, , [R ], 2.303, log 0, (B) k =, [R], t, , 12, , (q) Rate constant for zero, order reaction, , (C) Value of k for first, , (r), , k = [R]0, 2t1 2, , order reaction when, [R]0, t = t 1 and[R], 2, , (a), (b), (c), (d), , (s) Rate constant for first, order reaction, , [R]0, 2, A – (s), B – (q), C – (p), D – (r), A – (q), B – (s), C – (p), D – (r), A – (q), B – (p), C – (s), D – (r), A – (q), B – (s), C – (p), D – (t), , t = t1 2 and [R], , Column - II, (p) Effective collisions., , (q) Collision frequency, , (C) Molecules for which, Rate = Z ABe, , (r) e, , E a /RT, , Ea /RT, , shows significant deviations, (D) Collision in which molecules (s) Complex molecules, collide with sufficient K.E., and proper orientation., (a) A – (q), B – (r), C – (s), D – (p), (b) A – (r), B – (q), C – (s), D – (p), (c) A – (q), B – (s), C – (r), D – (p), (d) A – (q), B – (r), C – (p), D – (s), 131. Consider the energy diagram of a reaction : B A, on the, basis of given diagram select the correct code for matching, Column-I and Column-II., X, , B, , A, Progress of reaction, Column-II, (p) Enthalpy of reaction, (q) Energy of transition state, (r) Activation energy of, forward reaction, X, (s) Activation energy of, backward reaction, A – (s), B – (r), C – (q), D – (p), A – (q), B – (r), C – (p), D – (s), A – (r), B – (s), C – (p), D – (q), A – (s), B – (r), C – (p), D – (q), , Column-I, (A) X – A, (B) X – B, (C) A – B, (D), (a), (b), (c), (d), , CRITICAL THINKING TYPE QUESTIONS, , 2, , (D) Value of k for zero, order reaction when, , 130. Match the columns, Column - I, (A) Number of collisions per, second per unit volume, of the reaction mixture., (B) Fraction of molecules, with energies equal to, or greater than Ea, , Energy, , 127. Match the columns, Column-I, Column-II, (A) The decomposition, (p) Zero order reaction, of gaseous ammonia, on a hot platinum, surface, (B) The thermal, (q) Pseudo first order, decomposition of HI, reaction., on gold surface, (C) All natural and, (r) Zero order reaction at, artificial radioactive, high pressure, decay of unstable, nuclei, (D) Inversion of cane sugar (s) First order reaction., (a) A – (r), B – (p), C – (s), D – (q), (b) A – (r), B – (s), C – (q), D – (p), (c) A – (q), B – (s), C – (p), D – (r), (d) A – (q), B – (p), C – (s), D – (p), 128. Match the columns., Column-I, Column-II, (A) Catalyst alters the rate (p) cannot be fraction or zero, of reaction, (B) Molecularity, (q) proper orientation is not, there always., (C) Second half life of first (r) by lowering the activation, order reaction, enrgy, (D) Energetically favourable (s) is same as the first, reactions are sometimes, slow, (a) A – (q), B – (r), C – (s), D – (p), (b) A – (r), B – (s), C – (p), D – (q), (c) A – (r), B – (p), C – (s), D – (q), (d) A – (p), B – (r), C – (s), D – (q), 129. Match the columns, Column - I, Column - II, [R ] [R], (p) k = 2.303 log 2, (A) k = 0, t, t, , 132. In the following reaction, how is the rate of appearance of, underlined product related to the rate of disappearance of, the underlined reactant ?, BrO 3 (aq), , 5Br (aq), , 6H, , 3Br 2 (l), , 3H 2 O(l), , (a), , d[Br2 ], dt, , d[Br ], dt, , (b), , d[Br2 ], dt, , 3 d[Br ], 5 dt, , (c), , d[Br2 ], dt, , 3 d[Br ], 5 dt, , (d), , d[Br2 ], dt, , 5 d[Br ], 3 dt
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CHEMICAL KINETICS, , 305, , 133. For the reaction A + B, C + D. The variation of the, concentration of the products is given by the curve, , Y, Z, , the following mechanism A 2, , Conc, W, Time, , X, , (a) Z, (b) Y, (c) W, (d) X, 134. The rate of the reaction 2N2O5 4NO2 + O2 can be written, in three ways :, d[N 2O5 ], dt, d[NO 2 ], dt, d[O 2 ], dt, , k [N 2O5 ], k [N 2O5 ], , k [N 2O 5 ], , The relationship between k and k' and between k and k, are:, (a) k = 2k ; k = k, (b) k = 2k ; k = k / 2, (c) k = 2k ; k = 2k, (d) k = k ; k = k, 135. CHCl3 Cl2, , CCl 4, , HCl, , Rate law for above reaction will be, Rate = k[CHCl3 ][Cl2, , 1, ]2, , On the basis of information provided which of the following, option will be correct ?, (a) Rate law for any chemical reaction can be predicted, accurately by looking at balanced chemical equation., (b) Rate law for a chemical reaction has to determine, experimentally., (c) Either determined experimentally or obtained from, balanced chemical reaction, rate law will be same., (d) None of the above is correct., 136. The reaction of hydrogen and iodine monochloride is given, as:, H 2 (g) 2ICl(g), , 2HCl(g) I 2 (g), , The reaction is of first order with respect to H2(g) and ICI(g),, following mechanisms were proposed., Mechanism A:, H 2 (g) 2ICl(g), , Which of the above mechanism(s) can be consistent with, the given information about the reaction?, (a) A and B both, (b) Neither A nor B, (c) A only, (d) B only, 137. The hypothetical reaction A 2 B 2, 2 AB ; follows, , 2HCl(g) I 2 (g), , Mechanism B:, H 2 (g) ICl(g), , HI(g);slow, , HI(g) ICl(g), , HCl(g) I 2 (g);fast, , A, , B2, , Slow, , AB, , Fast, , B, A B, , A A,, Fast, , AB ., , The order of the overall reaction is, (a) 0, (b) 1, (c) 2, (d) 3/2, 138. The initial rates of reaction, Products, at different initial, 3A + 2B + C, concentrations are given below:, Initial rate,, [A]0, M, [B]0, M, [C]0, M, Ms–1, 5.0 × 10–3, 0.010, 0.005, 0.010, 5.0 × 10–3, 0.010, 0.005, 0.015, 1.0 × 10–2, 0.010, 0.010, 0.010, 1.25 × 10–3, 0.005, 0.005, 0.010, The order with respect to the reactants, A, B and C are, respectively, (a) 3, 2, 0, (b) 3, 2, 1, (c) 2, 2, 0, (d) 2, 1, 0, 139. The rate law for the reaction 2X + Y, Z is Rate =, k[X][Y]. The correct statement with regard to this relation, is, (a) the rate of the reaction is independent of [X] and [Y], (b) for this reaction t1/2 is independent of initial, concentrations of reactant, (c) the rate of formation of Z is twice the rate of, disappearance of X, (d) the rate of disappearance of X is equal to rate of, disappearance of Y, 140. The bromination of acetone that occurs in acid solution is, represented by this equation., CH3COCH3 (aq) + Br2 (aq) CH3COCH2Br (aq) + H+ (aq), + Br– (aq), These kinetic data were obtained for given reaction, concentrations., Initial, Concentrations, M, [CH3 COCH3], 0.30, 0.30, 0.30, 0.40, , [Br2], 0.05, 0.10, 0.10, 0.05, , [H+], 0.05, 0.05, 0.10, 0.20, , Initial rate,, disappearance of, Br2, Ms–1, 5.7×10–5, 5.7 × 10–5, 1.2 × 10–4, 3.1 × 10–4, , Based on given data, the rate equations is:, (a) Rate = k[CH3COCH3][H+], (b) Rate = k [CH3COCH3][Br2], (c) Rate = k [CH3COCH3] [Br2] [H+]2, (d) Rate = k [CH3COCH3][Br2] [H+]
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EBD_7207, CHEMICAL KINETICS, , 306, , 145., , 146., , (i), , Reaction rate, , 144., , C 2 H 4 O(g) CH 4 (g) CO(g) ,the initial pressure of, C2H4O(g) is 80 torr and total pressure at the end of 20, minutes is 120 torr. The time needed for 75% decomposition, of C2H4O would be :, (a) 20 minutes, (b) 40 minutes, (c) 80 minutes, (d) 120 minutes, 150. Which of the following graph(s) is/are correct for a zero, order reaction?, , Time, , (iii), , Concentration, of reactant rate, , 143., , C6 H5Cl N 2, , Time, , (iii), , Reaction rate, , C6 H 5 N 2 Cl, , At 0°C, the evolution of N2 becomes two times faster when, the initial concentration of the salt is doubled. Therefore, it, is, (a) a first order reaction, (b) a second order reaction, (c) independent of the initial concentration of the salt, (d) a zero order reaction, Consider the following reaction at 25°C:, (CH3)3COH(l) + HCl(aq) (CH3)3CCl(l) + H2O(l), The experimentally determined rate law for this reaction, indicates that the reaction is of first order in (CH3)3COH, and that the reaction is of first order overall. Which of the, following would produce an increase in the rate of this, reaction?, (a) Increasing the concentration of (CH3)3COH, (b) Increasing the concentration of HCl, (c) Decreasing the concentration of HCl, (d) Decreasing the concentration of (CH3)3CCl, The following data pertains to reaction between A and B :, S. No. [A] mol L–1 [B] mol L–1 Rate (mol L–1 time–1), 1, 1.0 × 10–2, 2.0 × 10–2, 2.0 × 10–4, –2, –2, 2, 2.0 × 10, 2.0 × 10, 4.0 × 10–4, –2, –2, 3, 2.0 × 10, 4.0 × 10, 8.0 × 10–4, Which of the following inference(s) can be drawn from the, above data ?, (i) Rate constant of the reaction is 1.0 × 10–4., (ii) Rate law of the reaction is : rate = k[A][B], (iii) Rate of reaction increases four times on doubling the, concentration of both the reactants., Select the correct answer using the codes given below :, (a) (i), (ii) and (iii), (b) (i) and (ii), (c) (ii) and (iii), (d) (iii) only, The decomposition of ammonia on tungsten surface at 500 K, follows zero order kinetics. The half-life period of this, reaction is 45 minutes when the initial pressure is 4 bar. The, half-life period (minutes) of the reaction when the initial, pressure is 16 bar at the same temperature is, (a) 120, (b) 60, (c) 240, (d) 180, A substance 'A' decomposes by a first order reaction starting, initially with [A] = 2.00 M and after 200 min, [A] becomes, 0.15 M. For this reaction t1/2 is, (a) 53.72 min, (b) 50.49 min, (c) 48.45 min, (d) 46.45 min, , 147. If half-life of a substance is 5 yrs, then the total amount of, substance left after 15 years, when initial amount is 64 grams, is, (a) 16 grams, (b) 2 grams, (c) 32 grams, (d) 8 grams., 148. In a 1st order reaction, reactant concentration C varies with, time t as :, 1, (a), increases linearly with t, C, (b) log C decreases linearly with t, 1, (c) C decreases with, t, 1, (d) log C decreases with, t, 149. For the first order reaction, , Time, , (iv), , Concentration, of reactant rate, , 141. Consider a reaction aG + bH, Products. When, concentration of both the reactants G and H is doubled, the, rate increases by eight times. However, when concentration, of G is doubled keeping the concentration of H fixed, the, rate is doubled. The overall order of the reaction is, (a) 0, (b) 1, (c) 2, (d) 3, 142. Diazonium salt decomposes as, , slope = –k, , Time, , (a) (ii) and (iii), (c) (ii), (iii) and (iv), , (b) (i), (ii) and (iii), (d) (i) and (iv)
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CHEMICAL KINETICS, , 307, , 1, 6.0 . The pre-exponential factor, T, A and the activation energy Ea, respectively, are, (a) 1.0 × 106 s–1 and 9.2 kJ mol–1, (b) 6.0 s–1 and 16.6 kJ mol–1, (c) 1.0 × 106 s–1 and 16.6 kJ mol–1, (d) 1.0 × 106 s–1 and 38.3 kJ mol–1, 154. The activation energies of two reactions are E1 and, E2 (E1 > E2). If the temperature of the system is increased, from T1 to T2, the rate constant of the reactions changes, from k1 to k1' in the first reaction and k2 to k2' in the second, reaction. Predict which of the following expression is, correct?, , equation log k = – (2000), , (a), , k1', k1, , k 2', k2, , (c), , k1', k1, , k 2', k2, , (b), , k1', k1, , k '2, k2, , (d), , k1', k1, , k '2, k2, , (a) The temperature of the system, (b) The geometry or orientation of the collision, (c) The velocity of the reactants at the point of collision, (d) All of the above influence the rate, 158. The activation energy for a hypothetical, Product, is 12.49 kcal/mole. If temperature is, reaction, A, raised from 295 to 305, the rate of reaction increased by, (a) 60%, (b) 100%, (c) 50%, (d) 20%, 159. A reactant (A) froms two products :, A, , k1, , B, Activation Energy Ea, , 1, , k2, , A, C, Activation Energy Ea, 2, If Ea = 2 Ea , then k1 and k2 are related as :, 2, , 1, , (a), , k2, , k1e Ea1 / RT, , (b), , k2, , k1e Ea2 / RT, , (c), , k1, , Ak 2 e Ea1 / RT, , (d), , k1, , 2k 2 e Ea2 / RT, , 160. Which of the following graph(s) represents exothermic, reaction?, (A), , Activated complex, , Energy, , 151. The integrated rate equations can be determined for, (a) zero order reactions, (b) first order reactions, (c) second order reactions, (d) Both (a) and (b), 152. In a zero-order reaction for every 10° rise of temperature,, the rate is doubled. If the temperature is increased from, 10°C to 100°C, the rate of the reaction will become :, (a) 256 times, (b) 512 times, (c) 64 times, (d) 128 times, 153. For a first order reaction A P, the temperature (T), dependent rate constant (k) was found to follow the, , Reactants, Products, Reaction coordinate, , 1, , (B), , Energy, , 155. A graph plotted between log k vs 1/T for calculating, activation energy is shown by, , Activated complex, , Products, Reactants, , log k, 1/T, , (c), , Reaction coordinate, , (b) log k, , log k, , 1/T, , (d) log k, 1/T, , 1/T, 156. The rate constant, the activation energy and the arrhenius, parameter of a chemical reaction at 25°C are 3.0 × 10–4s–1,, 104.4 kJ mol–1 and 6.0 × 1014 s–1 respectively. The value of, the rate constant as T, is, (a) 2.0 × 1018 s–1, (b) 6.0 × 1014 s–1, (c) Infinity, (d) 3.6 × 1030 s–1, 157. Collision theory is used to explain how chemical species, undergo a reaction. Using this theory and the kinetic, molecular model, which of the following does NOT influence, the rate of a chemcial reaction?, , Activated complex, , (C), Energy, , (a), , Reactants, , Products, , Reaction coordinate, , (a) (A) only, (b) (B) only, (c) (C) only, (d) (A) and (B), 161. Which of the following statements is not correct for the, catalyst?, (a) It catalyses the forward and backward reaction to the, same extent., (b) It alters G of the reaction., (c) It is a substance that does not change the equilibrium, constant of a reaction., (d) It provides an alternate mechanism by reducing, activation energy between reactants and products.
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EBD_7207, CHEMICAL KINETICS, , 308, , (a), , Concentration, , [A], , (d), , Concentration, , 162. Consider the reaction A f B. The concentration of both the, reactants and the products varies exponentially with time., Which of the following figures correctly describes the, change in concentration of reactants and products with, time?, [B], , [B], Time, , [A], Time, , [B], , 163. During decomposition of an activated complex., (i) energy is always released, (ii) energy is always absorbed, (iii) energy does not change, , Concentration, , (b), , (iv) reactants may be formed, (a) (i), (ii) and (iii), (b) (i) and (iv), [A], Time, , (c), , Concentration, , [B], , [A], Time, , (c) (ii) and (iii), , (d) (ii), (iii) and (iv), , 164. Which of the following statements is incorrect ?, (a) Energy is always released when activated complex, decomposes to form products., (b) Peak of the energy distribution curve corresponds to, the most probable potential energy., (c) Peak of the energy distribution curve corresponds to, the most probable kinectic energy., (d) When the temperature is raised maximum of energy, distribution curve moves to higher energy value and, broadens out.
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CHEMICAL KINETICS, , 309, , FACT / DEFINITION TYPE QUESTIONS, 1., , (b), , 2. (b), , 3. (a), , 6., , (d) In the given options, , 4. (a), , 5. (c), , d[C], will not represent the, 3dt, reaction rate. It should not have –ve sign as it is product., , 1 dC, show the rate of formation of product C, 3 dt, which will be positive., (a) If we write rate of reaction in terms of concentration of, NH3 and H2,then, , since, , 7., , Rate of reaction, So,, 8., , d[ NH 3 ], dt, , 1 d[ NH 3 ], 2, dt, , 1 d[H 2 ], 3 dt, , 2 d[H 2 ], 3 dt, , d[C], will not represent the, 3.dt, reaction rate. It should not have –ve sign as it is product., , (d) In the given options, , 1 dC, show the rate of formation of product C, 3 dt, which will be positive., Rusting of iron is a slow change., The feasibility of a chemical reaction can be predicted, by thermodynamics. Extent to which a reaction will, proceed can be determined from chemical equilibrium., Speed of a reaction i.e. time taken by a reaction to, reach equilibrium, can be predicted by chemical kinetics, 12. (d), 13. (c), 14. (a), 15. (b), 17. (c), 18. (d), The rate of a reaction is the speed at which the, reactants are converted into products. It depends, upon the concentration of reactants. e.g for the, reaction, , since, 9., 10., , 11., 16., 19., , (c), (c), , (c), (c), (d), , 20., , (a), , 21., , (d), , 22., , 23., , (b), , (d), , A + B ¾¾, ® Product ;, r µ [A][B], k remains constant at constant temperature and CO, does not effect the rate of reaction., order of reaction may be zero, whole number or, fractional., 2 NO (g) + Cl2(g), 2 NOCl(g), Rate = k [NO]2 [Cl], The value of rate constant can be increased by, increasing the temperature., Order of reaction is equal to the number of molecules, whose concentration is changing with time. It can be, zero or in fractions or an integer., , 24. (a) For a zero order reaction., rate =k[A]º i.e. rate = k, hence unit of k = M.sec–1, For a first order reaction., rate = k [A], k = M.sec–1/M = sec–1, 25. (d) The reaction involving two different reactant can never, be unimolecular., 26. (b) For reaction 3A, B C, If it is zero order reaction r = k [A]0, i.e the rate remains, same at any concentration of 'A'. i.e independent upon, concentration of A., 27. (c) k = (mol lit –1 ) 1–n time –1 . For given reaction, n = 2. k = mol–1 lit sec–1, 28. (b) The order of a chemical reaction is given by, concentration of reactants appearing in the lowest, step., 29. (b) Velocity constant ‘k’ is characteristic constant of a, reaction and depends only on temperature and catalyst., 30. (b) Given dx/dt = 2.400 × 10–5 mol litre–1 sec–1, k = 3.10 × 10–5 sec–1, For first order reaction, 2N 2 O 5, , 2NO 2, , O2, , dx, k [N 2 O 5 ], dt, or 2.4 × 10–5 = 3.0 × 10–5 [N2O5], , or [N2O5] =, , 2.4 10, , 5, , 3.0 10, , 5, , = 0.8 mol. litre–1, , dx, k [reactant]0, dt, Thus the rate of zero order reaction is independent of, concentration of reactants., 32. (c) Rate1 = k [A]n [B]m; Rate2 = k [2A]n [½B]m, , 31. (a) For zero order reaction,, , Rate 2, Rate1, , k [2A]n [½B]m, k [A]n [B]m, , = [2]n [½]m = 2n.2–m = 2n–m, , 33. (a) 2A + B, A2B, r1 = k[A]2 [B], , ...(i), , B, 2, , When, [A] = [2A], [B] =, r2, , = k 2A, , = k 2[A]2[B] = 2r1, , 2, , B, 2 [B], = k 4[A], 2, 2, , r1 k[A]2[B], , Rate of reaction is increased two times.
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EBD_7207, CHEMICAL KINETICS, , 310, , 34., , 35., , (c) Let us consider a reaction,, xX+yY, aA+bB, rate = [X]x [Y]y, It is given that order of reaction w.r.t. component Y is, zero., Hence, rate = [X]x, i.e., rate becomes independent of the concentration of Y., (a), , rate, , (b), , 38., , 39., , 40., , 41., , 42., 43., , k [A]0, , or r = k thus for zero order reactions rate is equal to, t, h, e, rate constant., (c) Since rate of reaction becomes four times on doubling, concentration of reactant, it is second order, reaction., (b) Order is the sum of the power of the concentration, terms in rate law expression., Hence the order or reaction is = 1 + 2 = 3, (a) Rate = k[A]°, Unit of k = mol L–1 sec–1, (c), , N 2O, , N2, , Divide (2) by (3), , 45., , (a), , 46., , (b), , 47., , (c), , 48., , (d), , 1, O2, 2, , dx, [N 2 O]1, dt, i.e. order of reaction = 1, (a) From the first set of data (i) and (ii) it is observed that, on keeping concentration of [B] constant and on, doubling the concentration of [A] rate does not, changes hence order of reaction with respect to A is, zero., From the second set of data (i) and (iii) it is observed, that rate becomes 4 times on doubling the concentration, of [B] keeping [A] constant hence order with respect to, [B] will be 2, rate = k[A]0 [B]2, (c) Order is the sum of the powers to which the, concentration terms are raised in the rate equation., (d) If rate = k[A]x [B]y [C]z, From first two given data, 8.08 × 10–3 = k [0.2]x [0.1]y [0.02]z, .... (1), 2.01 × 10–3 = k [0.1]x [0.2]y [0.02]z, .... (2), Divide (1) by (2) we get, 4 = 2x (1/2)y, Similarly, from second and third data, (9)y (9)z = 3, 2y + 2z = 1 ., From first and fourth data 4z = 8 = 23, 2z = 3. So z = 3/2, y = – 1, x = 1, , 0.10, 0.10, , [0.024]x [0.035] y, [0.012]x [0.035] y, , 1 = [2]x, x = 0, , [preactant ], , if n = 0, , 37., , Divide (3) by (1), , k[A]n, , r, r, , (a) Let the rate law be r, , 0, , i.e., rate = k, So, the order of reaction will be zero., 36., , 44., , = k [A]x[B]y, , 0.80, 0.10, , [0.024] x [0.070] y, [0.024] x [0.035] y, , 8 = (2)y , y = 3, Hence, rate equation, R = k[A]0[B]3 =k[B]3, Given r = k [A]3/2 [B]–1/2, 3 1 2, =1, Order = 3/2 – 1/2 =, 2, 2, The order w.r.t. I2 is zero because the rate is not, dependent on the concentration of I2., As we know that, units of rate constant., = (unit of conc.)1–n (unit of time)–1, = (mol L–1)1–n (sec)–1, On comparing these units with the given units of rate, constant, we get, (mol L–1)1–n (sec)–1 = L mol–1 sec–1, Þ Ln–1 mol1–n sec–1 = L mol–1 sec–1, On comparing the powers, we get, n–1=1 Þ n=2, So, reaction is of second order., In case of (II) and (III), keeping concentration of [A], constant, when the concentration of [B] is doubled,, the rate quadruples. Hence it is second order with, respect to B. In case of I & IV Keeping the, concentration of [B] constant. when the concentration, of [A] is increased four times, rate also increases four, times. Hence, the order with respect to A is one. hence, Rate = k [A] [B]2, , 49., , (a) The order of reaction is, , 50., , (b), , O3, , Fast, , O2, , 3, and molecularity is 2., 2, , O; O O 3, , Slow, , 2O 2, , k, , [O 2 ][O], (I) Rate = k' [O3][O] put [O] from (I), [O 3 ], , r, , k '[O 3 ]K[O 3 ], [O 2 ], , k[O 3 ]2 [O 2 ], , 1, , Note intermediates are never represented in rate law, equation., 51., 52., 53., , (c), (c) Thermal decomposition of HI on gold surface is an, example of zero order reaction., (c) For zero order reaction,, rate, r = k[R]°, dR, k, dt, R = kt + R0, where R0 is the concentration of reactant at time t = 0., Thus [R] increases with time
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CHEMICAL KINETICS, , 54., , (a) Plots of conc. [A] Vs time, t, , [A], t, , (d), , 61. (c), , First, order, , log [A], , Zero, order, , 55., , 311, , 1, , 1, [ A], , t, , t, , Second, order, , [ A]2, , 56., , 57., , a, 2k, (a) t1/2 = 4s, , Thus, k =, =, , T = 16s, T, , t1/ 2, , 16, 4, , 4, , (, , T = n × t½), , Where [A] o = initial concentration an d [A], = concentration left after time t, (b) A B For a first order reaction, Given a = 0.8 mol, (a – x) = 0.8 – 0.6 = 0.2, , t=, , 2.303, 0.8, log, or k = 2.303 log 4, 1, 0.2, again a = 0.9, a – x = 0.9 – 0.675 = 0.225, 2.303, 0.9, log, t, 0.225, , 2.303, log 4, t, Hence t = 1 hour, (b) For a first order reaction, A, 2.303log 4, , 59., , r, , k[A] or k, , k, , 1.5 10, 0.5, , Further, t1/ 2, 60., , products, , =, , 2.303, log 2 = 16 min, 0.0433, , 63. (d) Unit of k for Ist order reaction is (Time)–1, 64. (c), , 2.303, 1, log, 3/ 4, k, , t1/ 4, , 3 10, , 2, , 0.693, t1/ 2, , where k = rate constant, t1/2 = half life period = 480 sec., k, , 2.303, (2log 2 log 3), k, , 2.303, (2 0.301 0.4771), k, 65. (c) For a first order reaction, , 0.29, k, , 2.303, a, log, t, a x, when t = t½, 2.303, a, log, t½, a a/2, , 2.303, ln 2, log 2, k, k, 66. (b) Since doubling the concentration of B does not change, half life, the reaction is of 1st order w.r.t. B., Order of reaction with respect to A = 1 because rate of, reaction doubles when concentration of A is doubled, keeping concentration of B constant., Order of reaction = 1 + 1 = 2 and units of second, order reaction are L mol–1 sec–1., 67. (c) Half life time (t1/2) for nth order reaction is given by,, t1/2 [a]1– n, where n is the order of reaction and a is concentration, of reactant., or t½, , 0.693, , (c) For first order reaction, k, , 2.303, 4, log, 3, k, , 2.303, (log 4 log 3), k, , k, , = 3 × 10–2, , 0.693, k, , 2.303, log 4 = 0.0433 min–1, 32, , k, , r, [A], 2, , 2.303, 100, 2.303, a, log, log, =, t, (a x ), 32, (100 75), , 2.303, a, 2.303, 100, log, =, log, k, (a x ), 0.0433, 50, , k, , k, , = 100 sec, , Now we can use this value of k to get the value of time, required for 50% completion of reaction, , æ 1 ön, æ 1 ö4 0.12, [ A] = [ A]o çç ÷÷÷ = 0.12´çç ÷÷÷ =, = 0.0075 M, çè 2 ø, èç 2 ø, 16, , 58., , 3, , T n t1/ 2 = 3 × 100 = 300 sec., 62. (b) Given: 75% reaction gets completed in 32 min, , t1/ 2, , n, , 6.9 10, , 1, n 1, , i.e. n = 3, Thus reaction is of 3rd order., (a) For a zero order reaction, , ;n=3, , 0.693, , t1/ 2, a, , n, , n, , 1, 1, 1, 8, 2, , a2, , We know that t1/ 2, , 1, 2, , where [A]0 = initial concentration, , Third, order, , 1, , t1/2, , [A] [A]0, , 0.693, = 1.44 × 10–3 sec–1, 480, , 23.1
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EBD_7207, CHEMICAL KINETICS, , 312, , 68., , As decomposition of N2O5 follows 1st order kinetic., So,, t1/2 [a]1 – 1, t1/2 a0, (c) For a first order reaction, a, 2.303, log10, t, a x, when t = t½, , k, , 69., , (a), , t, , (a) As per Arrhenius equation (k Ae Ea / RT ) , the rate, constant increases exponentially with temperature., (b) For a reversible reaction,, H = Ea (forward) – Ea (backward), H = 150 – 260 = –110 kJ mol–1, (c) We know that the activation energy of chemical, r, e, a, c, t, i, o, n, , 80., , Ea, T2 T1, k2, , where, k1 2.303R T1T2, k1 is the rate constant at temperature T1 and k2 is the, rate constant at temperature T2 and Ea is the activation, energy. Therefore activation energy of chemical, reaction is determined by evaluating rate constant at, two different temperatures., , is given by formula =, , a, 2.303, log, k, a x, , or t, , 2.303, log a, k, , (t1/ 2 )1, (t1/ 2 )2, , a2, a1, , n 1, , 2.303, log(a x), k, n 1, 4 10 2, , 120, ;, 240, , ;n, , 2, , 70., , (c), , 71., 72., , (b), (b) In Haber’s process, ammonia is manufactured from N2, and H2 using iron as catalyst with molybdenum as, promoter at high temperature and pressure, N 2 3H 2, , 73., , 78., , ln 2, k, , 2.303, log10 2, k, , or t ½, , (b) As the rate of reaction get doubled for every 10°C rise, in temperature. Hence the increase in reaction rate as a, result of temperature rise from 10°C to 100°C is equal, to = 29 = 512, , 79., , 2.303, a, log10, a a/2, t½, , k, , 77., , Fe 2O3 (catalyst), Mo (catalytic promoter), , Ea, k, (b) log 2 =, 2.303R, k1, , log 2 =, , 8 10 2, , 1, T1, , Ea, 2.303 8.314, , (b) In equation k = Ae Ea / RT ; A = Frequency factor, k = velocity constant, R = gas constant and Ea = energy, of activation, , 82., , (d) In Arrhenius equation k = Ae Ea / RT , Ea is the energy, of activation, which is required by the colliding, molecules to react resulting in the formation of, products., (d), (b) When the temperature is increased, energy in form of, heat is supplied which increases the kinetic energy of, the reacting molecules. This will increase the number, of collisions and ultimately the rate of reaction will be, enhanced., (a) Arrhenius equation is given by, , 2NH3, , 83., 84., , 1, T2, 1, 293, , 81., , 1, 308, , Ea, 15, ×, 0.3 =, 2.303 8.314, 293 308, , 85., , k = Ae Ea /(2.303 RT ), Taking log on both sides, we get, , 0.3 2.303 8.314 293 308, ., 15, = 34673 J mole–1, = 34.7 J mole–1, , Ea =, , H = Ea f, , (b), , 75., , (d) The activation energy depends upon the nature of, chemical bonds undergoing rupture during chemical, reaction and is independent of enthalpies of reactants, and products., E /RT, (b) k = Ae– a, lnk = ln A – Ea/RT, For ln k vs 1/T, ln A = intercept, – Ea/R = slope = –2 × 104 K, Ea = 8.3 × 2 × 104 J mol–1, = 16.6 × 104 J mol–1 or 166 kJ mol–1, , Ea, 2.303 RT, , Arrhenius plot a graph between log k and, , Ea b = 0, , 74., , 76., , log k = log A –, , slope is, 86., , (a), , Ea, ., 2.303 R, , 2.303log, , log, k2, k1, , k2, k1, , k2, k1, , E a T2 T1, R T1T2, , 9.0 103 308 298, 2.303 2 308 298, , 1.63; k 2, , 1.63k1;, , 1, whose, T
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CHEMICAL KINETICS, , 313, , k1, , =, 87., , H, , (b), , k1, , Ea ( f ), , 108. (b) For a 10 K rise in temperature, collision frequency, increases merely by 1 to 2% but the number of, effective collisions increases by 100 to 200%., 109. (a), 110. (b), 111. (c), , 100, , 1.63k1 k1, 100 63.0%, k1, , STATEMENT TYPE QUESTIONS, , Ea (b), , Thus energy of activation for reverse reaction depend, upon whether reaction is exothermic or, endothermic., ve , Ea(b), , If reaction is exothermic, H, If reaction is endothermic,, 88., 93., , (d), (d), , 89. (c), , 94., , (d), , 95., , (b) ln k = ln A –, , k, , Ae, , 90. (c), , ve Ea(b), , H, , 91. (a), , Ea( f ), Ea ( f ), , 92. (d), , E a / RT, , 96., , Ea, , intercept is ln A., RT, (c) The rate constant doubles for 10º C rise in temperature., For 20º C rise, the rate constant will be 4 times, k1 = 4k2 or k2 = 0.25 K1, , 97., , (d), , k, , Ae, , Ea, 1, ., 2.303R T, , log k = log A –, , E a / RT, , Ea, ., 2.303R, (c) Applicable to bimolecular reactions., (a), (d) All the statements are correct (see text)., (c) Activation energy is lowered in presence of +ve, catalyst., (c), (a) Threshold Energy = Energy of activation + Internal, energy, , Equation of straight line slope =, , 104. (b), , k, , Ae, , E a / RT, , log k = log A, , Plot of log k Vs., , Ea, 1, ., 2.303R T, , 1, T, , Ea, 2.303R, 105. (d) Enthalpy of reaction ( H) = Ea, , 1, 4, , Rate =, , NH 3, t, , 1, 5, , pt(s), , 4NO(g) 6H 2O(g), , O2, , 1, 4, , t, , NO, t, , =, , 1, 6, , H 2O, t, , 113. (a) For given reaction x and y may or may not be equal to, p and q respectively., 114. (a) Average rate depends upon the chan ge in, concentration of reactants or products and the time, taken for that change to occur. Average rate cannot, be used to predict the rate of a reaction at a particular, instant as it would be constant for the time interval for, which it is calculated. So, to express the rate at a, particular moment of time we determine the, instantaneous rate. It is obtained when we consider, the average rate at the smallest time interval say dt (, i.e. when it approaches zero)., 115. (d) Molecularity is defined as the number of reacting, species taking part in an elementary reaction,, 116. (b) Molecularity cannot be fractional or zero. Correct units, of rate constant for second order reaction are mol–1 Ls–1., 117. (d) For zero order reaction, rate of reaction as well as rate, constant are independent of reactant concentration., 118. (a) For a first order reaction log, , [R]0, varies linearly with, [R], , time as shown below., , [R]0, , 102., 103., , 4NH 3 (g) 5O 2 (g), , slop = k/2.303, , log, , 98., 99., 100., 101., , 112. (b) For the given reaction, , [R], , k2, , Increase in k1, , Straight line Slope =, , 0, (f), , – Ea, , (b), , for an endothermic reaction H = +ve hence for H to, be positive, Ea < Ea, (f), (b), 106. (a) Presence of catalyst does not affect enthalpy change, of reaction H R E f E b = 180 – 200 = – 20 kJ/mol, 107. (c) For the exothermic reaction the energy of products is, always less than the reactants. If Ea is the energy of, activation for the forward reaction, the energy of, activation for backward reaction is Ea + H, , Time, , 119. (a), 120. (b) According to Arrhenius equation, k = Ae–Ea/RT, when Ea = 0, k = A, Also ln k s 1/T is a straight line with slope = –Ea/R., Statements (ii) and (v) are correct., 121. (b), 122. (b) Rate constant gets doubled with every 10°C in, temperature for chemical change only not for physical, change.
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EBD_7207, CHEMICAL KINETICS, , 314, , 123. (b) Correct formula for calculation of Ea is, k, log 2, k1, , 124. (d), , 137. (d), , Ea, T2 T1, 2.303R T1T2, , A2, A2, , The word catalyst should not be used when the added, substance reduces the rate of reaction. The substance, is then called inhibiter., The relevant expressions are as follows., , A, , 127. (a), , 1 d[Br ], 5 dt, d[Br2 ], dt, , 128. (c), , 129. (b), , 1 d[Br2 ], 3 dt, 3 d[Br ], 5 dt, , 133. (b) The curve Y shows the increase in concentration of, products with time., 134. (b) Rate of disappearance of reactants = Rate of appearance, of products, 1 d(N 2 O5 ), 2, dt, 1, k(N 2 O5 ), 2, , k, 2, , k, 4, , 1 d(NO2 ), 4, dt, 1, k (N 2 O5 ), 4, , d(O 2 ), dt, k (N 2 O5 ), , k, , k, 2, 135. (b) Rate law has to be determined experimentally as Cl2 is, , k = 2k, k, , 1, in rate law whereas its stichiometric, 2, coefficient in balanced chemical equation is 1., 136. (d) As the slowest step is the rate determining step thus, the mechanism B will be more consistent with the given, information also because it involve one molecule of, H2 and one molecule of ICl it can expressed as, r = k [H2][ICl], Which shows that the reaction is first order w.r.t. both, H2 & ICl., raised to power, , AB, , B (Slow ), , k[A 2 ], , since A is intermediate, , 1, 1, 2, , Order, , CRITICAL THINKING TYPE QUESTIONS, 132. (c), , B2, , A ( Fast );, , Rate law equation, , MATCHING TYPE QUESTIONS, 126. (b), 131. (d), , A, , Rate law = k[A][B 2 ] put value of [A] from Ist reaction, , H 1, I, log Kp =, R T, log [X] = log [X]0 + kt, P/T = constant (V constant), PV = constant (T constant), , 125. (a), 130. (a), , 2AB ;, , B2, , A, , K k[A 2 ][B2 ], , 3, 2, , 138. (d) From 1st and 2nd sets of data - no change in rate is, observed with the change in concentration of ‘C’. So, the order with respect to ‘C’ is zero., From 1st and 4th sets of data, Dividing eq. (4) by eq. (1), 1.25 10, , 3, , 5.0 10, , 3, , 0.005, 0.010, , x, , or 0.25 = (0.5)x or (0.5)2 = (0.5)x, x=2, The order with respect to ‘A’ is 2 from the 1st and 3rd, sets of data dividing eq. (1) by eq. (3), , 5.0 10, , 3, , 1.0 10, , 2, , 0.005, 0.010, , y, , or (0.5)1 = (0.5)y, y= 1, The order with respect to ‘B’ is 1, So the order with respective the reactants A, B and C, is 2, 1 and 0., 139. (N) None of the given options is correct., The given reaction is : 2X + Y — Z, –, , d[X], 2dt, , d[Z], dt, , Rate of formation of Z is half of the rate of, disappearance of X., d[X], 2dt, , d[Y], dt, , Rate of disappearrance of X is not equal to rate of, disappearance of Y., 140. (a) Rewriting the given data for the reaction, CH 3COCH 3 ( aq ) Br2 ( aq ), , H, , CH3COCH 2 Br(aq) H ( aq) Br ( aq)
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CHEMICAL KINETICS, , 315, , S. Initial concent Initialconcentr Initialconcentr, No., -ration of, -ation of Br2, -ation of H, CH3COCH3, in M, in M, in M, , Rate of, disappearance, , d [ A], k [ A]x [ B ] y, dt, Doubling [A], rate is doubled. Hence 2x = 2, x =1, d [ A], Similarly y = 1;, k [ A][ B ], dt, , 144. (c) Rate law :, , of Br2 in Ms 1, d, dx, i.e., [Br2 ]or, dt, dt, , 1, , 0.30, , 0.05, , 0.05, , 5.7 10, , 5, , 2, , 0.30, , 0.10, , 0.05, , 5.7 10, , 5, , 3, , 0.30, , 0.10, , 0.10, , 1.2 10, , 4, , 4, , 0.40, , 0.05, , 0.20, , 3.1 10, , 4, , Actually this reaction is autocatalyzed and involves, complex calculation for concentration terms., We can look at the above results in a simple way to, find the dependence of reaction rate (i.e., rate of, disappearance of Br2)., From data (1) and (2) in which concentration of, CH3COCH3 and H+ remain unchanged and only the, concentration of Br 2 is doubled, there is no change in, rate of reaction. It means the rate of reaction is, independent of concentration of Br2., Again from (2) and (3) in which (CH3CO CH3) and, (Br2) remain constant but H+ increases from 0.05 M to, 0.10 i.e. doubled, the rate of reaction changes from, 5.7×10–5 to 1.2 × 10–4 (or 12 × 10–5), thus it also, becomes almost doubled. It shows that rate of reaction, is directly proportional to [H+]. From (3) and (4), the, rate should have doubled due to increase in conc of, [H+] from 0.10 M to 0.20 M but the rate has changed, from 1.2× 10–4 to 3.1×10–4. This is due to change in, concentration of CH3COCH3 from 0.30 M to 0.40 M., Thus the rate is directly proportional to [CH3 COCH3]., We now get, rate = k [CH3COCH3]1[Br2]0[H+]1, = k [CH3COCH3][H+]., 141. (d) Overall order = sum of orders w.r.t each reactant., Let the order be x and y for G and H respectively, , k, , rate, [ A][ B ], , (rate)2, (rate)1, , 2.0 10 4, 1 10 2, , k (2[ A])(2[b]), k[ A][ B ], , t1/2, , 2, , t1/2, , 1, , P2 t1/2 2, ,, P1, 45, , 1, , 1, , litre, b, , 1, , 2, , 2a, , 2b, , 8r, , 2a, , b, , 2r, , For (1) and (3), the rate is doubled when conc. of, G is doubled keeping that of H constant i.e.,, , rate, , [G], , x=1, , From (2) and (3), y = 2, Overall order is 3., 142. (a) As doubling the initial conc. doubles the rate of, reaction, order =1, 143. (a), , 16, 4, , 16, 45 180 min, 4, 146. (a) Given initial concentration (a) = 2.00 M; Time taken, (t) = 200 min and final concentration (a – x) = 0.15 M., For a first order reaction rate constant,, k, , a, 2.303, log, t, a x, , 2.303, 2.00, log, 200, 0.15, , 2.303, (0.301 0.824), 200, Further, , 1.29 10 2 min 1 ., , 0.693, 0.693, 53.72 min ., k, 1.29 10 2, 147. (d) t1/2 = 5 years, T = 15 years hence total number of half, (t1/ 2 ), , life periods, , litre time 1 ), r, , 3, , 4, , (t1/2)2 =, , 15, 5, , 3., , 64, ( 2) 3, , [G]mole [H]mole rate(mole, litre, a, , 1, , 145. (d) For a zero order reaction,, t1/2 a0 (initial concentration or initial pressure), (t1/2)1 P1, (t1/2)2 P2, , Amount left, Exp.No., , 2 10 2, , Slope, , 148. (b), , 8g, , k / 2.303, , log C, t, , 149. (b) Let x torr of C2H4O decompose after 20 min. Then,, 80 – x + 2x =120 ; x = 40 torr = 50% of initial pressure., Hence t1/2 = 20 min. For 75% reaction, fraction left, =, , 25, 100, , 1, 4, , 1, 2, , 2, , No. of half lives = 2. Time needed for 75% reaction.,, 2 × 20 = 40 min
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EBD_7207, CHEMICAL KINETICS, , 316, , 150. (d), 151. (d) The integrated rate equations are different for the, reactions of different reaction orders. We shall, determine these equations only for zero and first order, chemical reactions., 152. (b), , r, , 100 C, , r, , 10 C, , 2, , T2 T1, 10, , 2, , 100 10, 10, , 29, , 156. (b) T2 = T (say), T1 = 25°C = 298K,, Ea = 104.4 kJ mol–1 = 104.4 × 103 J mol–1, k1=3 × 10 – 4, k2 = ?,, log, , 512 (where 2 is, log, , temperature coefficient of reaction), 153. (d), , log k, , log A, , Ea, 2.303RT, , Also given log k, , 6.0 (2000), , 1, T, , Ea, 2.303 R, , Ea, 1, 2.303 R T1, , 3 10, , …(2), , log, , k2, , 2000 ;, , Ea = 2000 × 2.303 × 8.314 = 38.29 kJ mol–1, 154. (b) We know more will be activation energy lesser will be, rate constant. Thus if E 1 > E 2 then k 1 < k2 . As, temperature increases, it will effect both rates in similar, way., 1, 155. (b) A graph plotted between log k vs for calculating, T, activation energy is shown as, , k2, , 1, 2.303 (8.314 J K mol ) 298, , 1, T, , ,, , 3 10, , 1, , 1, , 4, , As T, , log, , 1, T2, , 104.4 103 J mol, , k2, , …(1), , On comparing equations, (1) and (2), log A = 6.0, A = 106 s–1, and, , k2, k1, , 1, , 1, T, , 0, 104.4 103 J mol 1, 2.303 8.314 298, , 4, , k2, 3 10 4, , 18.297,, , k2, 3 10 4, , (1.98 1018 ) (3 10 4 ), , 1.98 1018, , 6 1014 s, , 1, , 157. (d), 158. (b) For 10°C rise of temperature the rate is almost doubled., 159. (c), , k1, , A1e, , k2, , A2e, , Ea / RT, 1, , Ea / RT, 2, , .........(i), ........(ii), , On dividing eqn (i) from eqn. (ii), k1, k2, , log k, , A1, ( Ea2, A2, , Ea1 ) / RT, , ........(iii), , Given Ea2 = 2Ea1, On substituting this value in eqn. (iii), , 1/T, , from Arrhenius equation, Ea, log k log A, 2.303 RT, , k1, 160. (a), , k2 A e, 161. (b), , Ea / RT, 1, , 162. (b), , 163. (b), , 164. (b)
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19, SURFACE CHEMISTRY, FACT / DEFINITION TYPE QUESTIONS, 1., , 2., , 3., , 4., , 5., , 6., , 7., , 8., , 9., , Adsorbed acetic acid on activated charcoal is:, (a) adsorber, (b) absorber, (c) adsorbent, (d) adsorbate, Adsorption is always, (a) endothermic, (b) exothermic, (c) exothermic in case of physical and endothermic in case, of chemical, (d) Either (a) or (b), Which is not correct regarding the physical adsorption of a, gas on surface of solid ?, (a) On increasing temperature, adsorption increases, continuously, (b) Enthalpy and entropy changes are negative, (c) Adsorption is more for some specific substance, (d) Reversible, How many layers are adsorbed in chemical adsorption ?, (a) One, (b) Two, (c) Many, (d) Zero, Adsorption due to strong chemical forces is called, (a) Chemisorption, (b) Physisorption, (c) Reversible adsorption (d) Both (b) and (c), In physical adsorption, gas molecules are bound on the, solid surface by, (a) chemical forces, (b) electrostatic forces, (c) gravitational forces, (d) van der Waal’s forces, Which of the following statements is not correct ?, (a) Physical adsorption is due to van der Waal’s forces, (b) Chemical adsorption first decreases with increase in, temperature., (c) Physical adsorption is reversible, (d) Adsorption energy for a chemical adsorption is, generally greater than that of physical adsorption., Adsorption of gases on solid surface is exothermic reaction, because, (a) free energy increases (b) enthalpy is positive, (c) entropy increases, (d) enthalpy is negative, The gas which is least adsorbed on charcoal (under identical, conditions) is, , 10., , 11., , 12., , 13., , 14., , 15., , 16., , (a) HCl, (b) O2, (c) CO2, (d) NH3, Adsorption is accompanied by, (a) decrease in enthalpy and increase in entropy, (b) increase in enthalpy and increase in entropy, (c) decrease in enthalpy and decrease in entropy, (d) increase in enthalpy and decrease in entropy, Choose the incorrect statement in respect of physisorption?, (a) It is not specific in nature, (b) It arises because of van der Waal’s force, (c) It is reversible in nature, (d) Enthalpy of adsorption is in the range 80-240 kJ mol–1, The term ‘sorption’ stands for ___________., (a) absorption, (b) adsorption, (c) both absorption and adsorption, (d) desorption, Extent of physisorption of a gas increases with _______., (a) increase in temperature., (b) decrease in temperature., (c) decrease in surface area of adsorbent., (d) decrease in strength of van der Waal’s forces., Extent of adsorption of adsorbate from solution phase, increases with _________., (a) increase in amount of adsorbate in solution., (b) decrease in surface area of adsorbent., (c) increase in temperature of solution., (d) decrease in amount of adsorbate in solution., Which of the following is not a favourable condition for, physical adsorption ?, (a) High pressure, (b) Negative H, (c) Higher critical temperature of adsorbate, (d) High temperature, Physical adsorption of a gaseous species may change to, chemical adsorption with ________., (a) decrease in temperature, (b) increase in temperature, (c) increase in surface area of adsorbent, (d) decrease in surface area of adsorbent
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EBD_7207, SURFACE CHEMISTRY, , 318, , 1, n, The adsorption of a gas on a solid surface varies with, pressure of the gas in which of the following manner, (a) Fast, slow independent of the pressure, (b) Slow fast independent of the pressure, (c) Independent of the pressure fast, slow, (d) Independent of the pressure slow, fast, If x is amount of adsorbate and m is amount of adsorbent,, which of the following relations is not related to adsorption, process ?, (a) x / m = f (p) at constant T., (b) x / m = f (T) at constant p., (c) p = f (T) at constant (x / m)., x, p T, (d), m, In Freundlich adsorption isotherm, the value of 1/n is :, (a) between 0 and 1 in all cases, (b) between 2 and 4 in all cases, (c) 1 in case of physical adsorption, (d) 1 in case of chemisorption, Which is adsorbed in maximum amount by activated charcoal, ?, (a) N2, (b) CO2, (c) Cl2, (d) O2, Freundlich equation for adsorption of gases (in amount of x g), on a solid (in amount of m g) at constant temperature can be, expressed as, , (c) n, , 20., , 21., , 22., , 23., , 24., , (a) log, , 27., , (a), , x, m, , log p, , 1, x, log K (b) log, n, m, , log K, , 1, log p, n, , x, x, 1, pn, log p, log K, (d), m, m, n, According to Freundlich adsorption isotherm, the amount, of gas adsorbed at very high pressure, (a) reaches a constant limiting value, (b) goes on increasing with pressure, (c) goes on decreasing with pressure, (d) increase first and decreases later with pressure, , log x/m, , (b), , p, , p, , (d), , (d), , (c), , log p, , 28., , 29., , 30., , 31., , 32., , (c), 25., , Which is not correct regarding the adsorption of a gas on, surface of solid?, (a) On increasing temperature, adsorption increases, continuously, (b) Enthalpy and entropy changes are –ve, (c) Adsorption is more for some specific substance, (d) This Phenomenon is reversible, Which of the following curves is in accordance with, Freundlich adsorption isotherm ?, , log x/m, , 19., , 26., , log x/m, , 18., , In physisorption adsorbent does not show specificity for, any particular gas because _________., (a) involved van der Waal’s forces are universal., (b) gases involved behave like ideal gases., (c) enthalpy of adsorption is low., (d) it is a reversible process., Which of the following is an example of absorption ?, (a) Water on silica gel, ., (b) Water on calcium chloride., (c) Hydrogen on finely divided nickel., (d) Oxygen on metal surface., For adsorption of a gas on a solid, the plot of log x/m vs, log P is linear with slope equal to (n being whole number), (a) k, (b) log k, , log x/m, , 17., , 33., , log p, , Which of the following is related to adsorption?, (i), H = – ve, (ii), S = – ve, (iii) –T S = – ve, (iv) G = – ve, (a) (i), (ii) and (iv), (b) (ii) and (iii), (c) (iii) only, (d) (i), (iii) and (iv), The role of a catalyst in a reversible reaction is to, (a) increase the rate of forward reaction, (b) decrease the rate of backward reaction, (c) alter the equilibrium constant of the reaction, (d) allow the equilibrium to be achieved quickly, Catalytic poisons act by :, (a) making the products chemically inactive., (b) increasing the rate of the backward reaction., (c) chemical combination with any one of the reactants., (d) preferential adsorption on the catalyst surface., A catalyst :, (a) lowers the activation energy, (b) changes the rate constant, (c) changes the product, (d) itself destroyed in the reaction, Active charcoal is a good catalyst because it, (a) is made up of carbon atoms., (b) is very reactive., (c) has more adsorption power., (d) has inert nature toward reagents., Which of the following kind of catalysis can be explained, by the adsorption theory ?, (a) Homogeneous catalysis, (b) Acid - base catalysis, (c) Heterogeneous catalysis, (d) Enzyme catalysis
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SURFACE CHEMISTRY, , 34., , 35., , 36., , 37., , 38., , 39., , 40., , 41., , 42., , According to the adsorption theory of catalysis, the speed, of the reaction increases because(a) Adsorption lowers the activation energy of the reaction, (b) The concentration of reactant molecules at the active, centres of the catalyst becomes high due to strong, adsorption, (c) In the process of adsorption, the activation energy of, the molecules becomes large, (d) Adsorption produces heat which increases the speed, of the reaction, Catalyst increases the rate of reaction by, (a) decreasing threshold energy, (b) decreasing activation energy, (c) increasing activation energy, (d) decreasing equilibrium constant, A catalyst can affect reversible reaction by, (a) changing equilibrium constant, (b) slowing forward reaction, (c) attaining equilibria in both directions, (d) None of these, Which one of the following is an example of homogeneous, catalysis ?, (a) Haber’s process of synthesis of ammonia, (b) Catalytic conversion of SO2 to SO3 in contact process, (c) Catalytic hydrogenation of oils, (d) Acid hydrolysis of methyl acetate., Identify the correct statement regarding enzymes, (a) Enzymes are specific biological catalysts that cannot, be poisoned., (b) Enzymes are normally heterogeneous catalysts that, are very specific in their action., (c) Enzymes are specific biological catalysts that can, normally function at very high temperatures (T, 1000K)., (d) Enzymes are specific biological catalysts that possess, well-defined active sites., A biological catalyst is, (a) an enzyme, (b) a carbohydrate, (c) an amino acid, (d) a nitrogenous base, The action of enzymes in living system is to :, (a) supply energy to tissues, (b) enhance immunity, (c) circulate oxygen, (d) enhance the rate of biochemical reactions., Hydrolysis of urea is an example of, (a) homogenous catalysis (b) heterogenous catalysis, (c) biochemical catalysis (d) zeolite catalysis, The efficiency of an enzyme in catalysing a reaction is due, to its capacity, (a) to form a strong enzyme-substrate complex, (b) to decrease the bond energies of substrate molecule, (c) to change the shape of the substrate molecule, (d) to lower the activation energy of the reaction, , 319, , 43. What is the role of molybdenum in Haber’s process for, manufacture of ammonia?, (a) As catalytic poison (b) As a catalytic promoter, (c) As a catalyst, (d) As a reactant, 44. Which of the following step(s) is/are not involved in the, mechanism of adsorption theory of heterogeneous, catalyst?, (i) Diffusion of reactants to the surface of the catalyst., (ii) Sorption of reactant molecules on the surface of the, catalyst., (iii) Occurrence of chemical reaction on the catalyst’s, surface through formation of an intermediate., (iv) Desorption of reaction products from the catalyst’s, surface., (v) Diffusion of reaction products away from the catalyst’s, surface., (a) (i) only, (b) (ii) and (iv), (c) (ii) only, (d) (i), (ii) and (v), 45. Which of the following equation does not represent, homogeneous catalysis?, (a), , CH3 COOCH3 l, , H2O l, , H 2SO 4 l, , CH3COOH aq, (b), (c), , 4NH3 g, 2SO2 g, , 5O 2 g, O2 g, , Pt s, NO g, , 4NO g, , CH3OH aq, , 6H 2O g, , 2SO 3 g, , (d) Both (a) and (b), 46. Milk is a colloid in which a, (a) liquid is dispersed in a liquid, (b) solid is dispersed in a liquid, (c) gas is dispersed in a liquid, (d) sugar is dispersed in a liquid, 47. Butter is a colloid formed when, (a) Fat is dispersed in water, (b) Fat globules are dispersed in water, (c) Water is dispersed in fat, (d) None of the above, 48. The size of colloidal particles is between, (a) 10–7 – 10–9 cm, (b) 10–9 – 10–11 cm, –5, –7, (c) 10 – 10 cm, (d) 10–2 – 10–3 cm, 49. An aerosol is a :, (a) dispersion of a solid or liquid in a gas, (b) dispersion of a solid in a liquid, (c) dispersion of a liquid in a liquid, (d) solid solution, 50. An example of dispersion of a liquid in a gas is :, (a) milk, (b) vegetable oil, (c) foam, (d) mist, 51. Alloy is an example of, (a) gel, (b) solidified emulsion, (c) solid solution, (d) sol
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EBD_7207, SURFACE CHEMISTRY, , 320, , 52., , 53., , 54., , 55., , 56., , 57., , 58., , 59., , 60., , 61., , 62., , 63., , 64., , 65., , If dispersed phase is a liquid and the dispersion medium is, a solid, the colloid is known as, (a) a sol, (b) a gel, (c) an emulsion, (d) a foam, Hair cream is an example of, (a) gel, (b) sol, (c) aerosol, (d) foam, Which one of the following is correctly matched?, (a) Emulsion-smoke, (b) Gel-butter, (c) Aerosol-hair cream, (d) Sol-whipped cream, Cheese is an example of, (a) solid sol, (b) emulsion, (c) gel, (d) foam, Which one of the following in not a colloidal solution?, (a) Smoke, (b) Ink, (c) Blood, (d) Air, Small liquid droplets dispersed in another liquid is called, (a) gel, (b) suspension, (c) emulsion, (d) true solution, When dispersed phase is liquid and dispersion medium is, gas then the colloidal system is called, (a) Smoke, (b) Clouds, (c) Jellies, (d) Emulsions, Which one is a colloid?, (a) Sodium chloride, (b) Urea, (c) Cane sugar, (d) Blood, Suspensions are, (a) Visible to naked eye, (b) Not visible by any means, (c) Invisible under electron microscope, (d) Invisible through microscope, Cloud or fog is an example of colloidal system of, (a) Liquid dispersed in gas, (b) Gas dispersed in gas, (c) Solid dispersed in gas, (d) Solid dispersed in liquid, A colloid always :, (a) Contains two phases, (b) Is a true solution, (c) Contains three phases, (d) Contains only water soluble particles, Which one of the following is correctly matched ?, (a) Emulsion - curd, (b) Foam - mist, (c) Aerosol - smoke, (d) Solid sol - cake, At the critical micelle concentration (CMC) the surfactant, molecules, (a) decompose, (b) dissociate, (c) associate, (d) become completely soluble, How non-polar and polar part in micelle are arranged ?, (a) Polar at outer surface and non-polar at inner surface, (b) Polar at inner surface and non-polar at outer surface, (c) Both polar and non-polar at inner surface, (d) Distributed all over the surface, , 66., , 67., , 68., , 69., , 70., , 71., , 72., , 73., , 74., , 75., , 76., , 77., , 78., , Surface tension of lyophilic sols is, (a) lower than that of H2O, (b) more than that of H2O, (c) equal to that of H2O, (d) either less or more than H2O depending upon the, nature of disperse phase, Which of the following is a lyophilic colloid ?, (a) Milk, (b) Gum, (c) Fog, (d) Blood, Lyophobic colloids are :, (a) gun proteins, (b) protective colloids, (c) irreversible colloids, (d) reversible colloids, Which one is an example of multimolecular colloid system, (a) Soap dispersed in water, (b) Protein dispersed in water, (c) Gold dispersed in water, (d) Gum dispersed in water, Example of intrinsic colloid is, (a) glue, (b) sulphur, (c) Fe, (d) As 2 S 3, Associated colloid among the following is, (a) enzymes, (b) proteins, (c) cellulose, (d) sodium stearate, The formation of micelles takes place only above, (a) inversion temperature, (b) Boyle temperature, (c) critical temperature, (d) Kraft temperature, A precipitate is changed to colloidal solution by the, following process :, (a) dialysis, (b) ultrafiltration, (c) peptization, (d) electrophoresis, Which of the following is used for neutralising charge on, colloidal solution?, (a) Electrons, (b) Electrolytes, (c) Positively charged ions, (d) Compounds, Pure water can be obtained from sea water by, (a) Centrifugation, (b) Plasmolysis, (c) Reverse osmosis, (d) Sedimentation, Blood may be purified by, (a) Dialysis, (b) Electro-osmosis, (c) Coagulation, (d) Filtration, During dialysis, (a) only solvent molecules can diffuse, (b) solvent molecules, ions and colloidal particles can, diffuse, (c) all kinds of particles can diffuse through the semipermeable membrane, (d) solvent molecules and ions can diffuse, The electrolytic impurities of a sol can most easily be, separated by, (a) dialysis, (b) electrosmosis, (c) electrophoresis, (d) electrodialysis
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SURFACE CHEMISTRY, , 79., , 80., , 81., , 82., , 83., , 84., , 85., , 86., , 87., , 88., , 89., , 90., , 91., , The formation of colloid from suspension is, (a) Peptisation, (b) Condensation, (c) Sedimentation, (d) Fragmentation, The separation of colloidal particles from particles of, molecular dimensions is known as, (a) sedimentation, (b) dispersion, (c) pyrolysis, (d) dialysis, Which one of the following impurities present in colloidal, solution cannot be removed by electrodialysis?, (a) Sodium chloride, (b) Potassium sulphate, (c) Urea, (d) Calcium chloride, The migration of dispersion medium under the influence of, an electric potential is called :, (a) Cataphoresis, (b) Electroosmosis, (c) Electrophoresis, (d) Sedimentation, The movement of colloidal particles towards their respective, electrodes in the presence of an electric field is known as :, (a) electrolysis, (b) Brownian movement, (c) electrodialysis, (d) electrophoresis, Peptization denotes, (a) Digestion of food, (b) Hydrolysis of proteins, (c) Breaking and dispersion into the colloidal state, (d) Precipitation of solid from colloidal dispersion, Colloidal gold is prepared by, (a) Mechanical dispersion (b) Peptisation, (c) Bredig’s Arc method (d) Hydrolysis, Peptization involves, (a) precipitation of colloidal particles, (b) disintegration of colloidal aggregates, (c) evaporation of dispersion medium, (d) impact of molecules of the dispersion medium on the, colloidal particles, Hardy-Schulze rule explains the effect of electrolytes on the, coagulation of colloidal solution. According to this rule,, coagulation power of cations follow the order, (a) Ba+2 > Na+ > Al+3, (b) Al+3 > Na+ > Ba+2, +3, +2, +, (c) Al > Ba > Na, (d) Ba+2 > Al+3 > Na+, Tyndall effect is shown by, (a) sol, (b) solution, (c) plasma, (d) precipitate, The cause of Brownian movement is, (a) heat changes in liquid state, (b) convectional currents, (c) the impact of molecules of the dispersion medium on, the colloidal particles., (d) attractive forces between the colloidal particles and, molecules of dispersion medium., When a strong beam of light is passed through a colloidal, solution, the light will, (a) be reflected, (b) be scattered, (c) be refracted, (d) give a rainbow, The simplest way to check whether a system is colloidal or, not is by, , 321, , 92., , 93., , 94., , 95., , 96., , 97., , 98., , 99., , 100., , 101., , 102., , 103., , 104., , (a) Tyndall effect, (b) Brownian movement, (c) Electrodialysis, (d) Measuring particle size, Which of the following is most effective in causing the, coagulation of ferric hydroxide sol?, (a) KCl, (b) KNO3, (c) K2SO4, (d) K3[Fe(CN)6], The ability of an ion to bring about coagulation of a given, colloid depends upon, (a) its size, (b) the magnitude of its charge, (c) the sign of its charge, (d) both magnitude and sign of its charge, Which of the following electrolytes is least effective in, coagulating ferric hydroxide solution?, (a) KBr, (b) K2SO4, (c) K2CrO4, (d) K4 [Fe(CN)6], Which of the following acts as protective colloid?, (a) Silica gel, (b) Gelatin, (c) Sodium acetate, (d) None of these, Tyndall effect shown by colloids is due to :, (a) scattering of light by the particles, (b) movement of particles, (c) reflection of light by the particles, (d) coagulation of particles, Which of the following is not a property of colloidal, solution?, (a) Heterogenity, (b) Particle size > 100 nm, (c) Tyndall effect, (d) Brownian movement, Which of the following is most powerful to coagulate the, negative colloid?, (a) ZnSO4, (b) Na3PO4, (c) AlCl3, (d) K4[Fe(CN)6], The charge on colloidal particles is due to, (a) presence of electrolyte, (b) very small size of particles, (c) adsorption of ions from the solution, (d) None of these, The ion that is more effective for the coagulation of As2S3, sol is, (a) Ba2+, (b) Na+, 3–, (d) AI3+, (c) PO4, Which one of the following impurities present in colloidal, solution cannot be removed by electrodialysis?, (a) Sodium chloride, (b) Potassium sulphate, (c) Urea, (d) Calcium chloride, Brownian movement is found in, (a) Colloidal solution, (b) Suspension, (c) Saturated solution, (d) Unsaturated solution, Random motion of colloidal particles is known as, (a) Dialysis, (b) Brownian movement, (c) Electroosmosis, (d) Tyndall effect, In which of the following Tyndall effect is not observed ?, (a) Suspensions, (b) Emulsions, (c) Sugar solution, (d) Gold sol
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EBD_7207, SURFACE CHEMISTRY, , 322, , 105. Which of the following is not true about the emulsion?, (a) Emulsion can be broken into constituent liquids by, heat, freezing, (b) Emulsion of oil in water is generally unstable, (c) Emulsion do not show the Tyndall effect, (d) They show brownian motion, 106. Which of the following process is responsible for the, formation of delta at a place where rivers meet the sea?, (a) Emulsification, (b) Colloid formation, (c) Coagulation, (d) Peptisation, 107. Which of the following colloid does not contain liquid as a, dispersion medium?, (a) Sol, (b) Gel, (c) Emulsion, (d) Foam, 108. Which of the following method is used for coagulation of, the sol?, (a) By mixing two oppositely charged sols., (b) By electrophoresis., (c) By addition of electrolytes., (d) All of the above., 109. Which of the following phenomenon occurs when the, precipitate of Mg(OH)2 attains blue colour in presence of, magneson reagent?, (i) Absorption of solvent, (ii) Adsorption of coloured substance, (iii) Absorption and adsorption both of solvent, (iv) Adsorption of solvent, (a) (i) and (ii), (b) (ii) only, (c) (ii) and (iv), (d) (iii) only, 110. Which of the following is not the condition for Tyndall, effect?, (a) The refractive indices of dispersed phase and, dispersion medium should differ greatly in magnitude., (b) The diameter of the dispersed particles is not much, smaller than the wavelength of light used., (c) Tyndall effect is observed only when viewed from the, direction of the passage of light., (d) All of these are required conditions for Tyndall effect., 111. Which of the following is not emulsifying agent for W/O, emulsion?, (a) Lampblack, (b) Long chain alcohol, (c) Proteins, (d) Heavy metal salts of fatty acids, 112. Emulsions can be broken into constituent liquid by ______., (a) heating, (b) freezing, (c) centrifuging, (d) All of these, , STATEMENT TYPE QUESTIONS, 113. Read the following statements regarding adsorption and, choose the correct option., (i) It is a surface phenomenon., (ii) The material which is adsorbed is termed as adsorbate., (iii) The material on the surface of which the adsorption, takes place is called adsorbent., (iv) Adsorption is a bulk phenomenon., , 114., , 115., , 116., , 117., , 118., , (a) Only (iv) is correct, (b) (i) and (ii) are correct, (c) (i), (ii) and (iv) are correct, (d) (i), (ii) and (iii) are correct, Read the following statements related to physisorption., (i) Adsorbent shows preference for gases with high, molecular weight., (ii) Easily liquefiable gases gets readily adsorbed., (iii) Adsorption varies with change in temperature and, pressure., (iv) Finely divided and solid metals adsorb gases equally., (v) It is exothermic with low value of enthalpy of, adsorption., Which of the following is the correct code for the statements, above ?, (a) TFFTF, (b) FTTFT, (c) TFFTT, (d) FTTTF, Read the following statements related to chemisorption, (i) It is highly specific., (ii) It increases with increase in temperature and pressure., (iii) It is reversible., (iv) It increases with increase in surface area of adsorbent., Which of the following is correct code for the statements, above?, (a) TTFT, (b) TFFT, (c) FTFT, (d) FFTF, Which of the following statement(s) is/are correct about, solid catalyst?, (i) Same reactants may give different product by using, different catalysts., (ii) Catalyst is required in large quantities to catalyse, reactions., (iii) Catalyst does not change H of reaction., (iv) Catalytic activity of a solid catalyst does not depend, upon the strength of chemisorption., (a) (i) and (iii), (b) (i) only, (c) (ii), (iii) and (iv), (d) (iii) and (iv), Which of the following statement(s) is/are correct?, (i) Zeolites are good shape selective catalysts because, of their honeycomb-like structures., (ii) All zeolites are naturally occurring substance., (iii) An important zeolite catalyst used in the petroleum, industry in ZSM-5., (a) (i) only, (b) (ii) only, (c) (i) and (iii), (d) (ii) and (iii), Read the following statements regarding enzyme catalysis, (i) Enzyme catalysis is highly specific in nature., (ii) Enzyme catalysis to work effectively requires optimum, temperature (298-310 K) and optimum pH (3-5), (iii) Metal ions like Na+, Mn2+, Co2+, Cu2+ etc. increases, the activity of enzymes., (iv) Catalyst used in Ostwald’s process is platinised, asbestos at 673 K., (v) Catalyst used in contact process is platinised asbestos, or V2O5 at 673-723 K., Which of the following is the correct coding for the above, statements?, (a) FTFTF, (b) TFTFT, (c) TTFFF, (d) FTTFT
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SURFACE CHEMISTRY, , 323, , 119. Which of the following statements are correct?, (i) Gelatine sol if evaporated off it can be reobtained, simply by mixing gelatine obtained on evaporation, with suitable dispersion medium., (ii) Metal sulphide sols need stabilising agents for their, preservation, (iii) S8 being a macromolecule forms macromolecular, colloid., (iv) Starch and proteins are natural whereas polythene, and polystyrene are man-made macromolecules., (v) Micelles are formed above kraft temperature at any, concentration, (a) (i), (ii) and (iii), (b) (i), (ii) and (iv), (c) (iii), (iv) and (v), (d) (ii), (iv) and (v), 120. Read the following statements, (i) Tyndall effect is used to distinguish between a colloidal, and true solution., (ii) Values of colligative properties are same for true and, colloidal solutions., (iii) Random bombardment of the colloidal particles by, the molecules of the dispersion medium does not allow, colloids to settle thereby providing stability to them., (iv) Most acceptable phenomena to account for the charge, of sol particles is electrodispersion., Which of the following is the correct code for statements, above?, (a) TFTF, (b) TTFF, (c) FTFT, (d) TFFT, MATCHING TYPE QUESTIONS, 121. Match the columns, Column-I, x, kc1 n, (A), m, (B) log, 1, n, , (C) log, , x, m, , log k, , (p) Adsorption varies, directly with pressure, 1, log p, n, , x, m, , (q) Adsorption from, solution phase, , 0, log k, , 1, 1, n, x, kp1 n, (D), m, (high pressure), , (a), (b), (c), (d), , Column-II, , 1, log p, n, , (r) Freudlich isotherm, cannot be explained, , (s) Adsorption is, independent of pressure, , A – (q), B – (s), C – (p), D – (r), A – (q), B – (p), C – (s), D – (r), A – (r), B – (p), C – (s), D – (q), A – (r), B – (s), C – (p), D – (q), , 122. Match the columns, Column-I, Column-II, (Biochemical reactions), (Enzymes), (A) C6H12O6(aq), (p) Zymase, 2C2H5OH(aq) + 2CO2(g), (B) NH2CONH2(aq) + H2O(l), (q) Pepsin, 2NH3(g) + CO2(g), (C) Proteins, Peptides, (r) Urease, (D) C12H22O11(aq) + H2O(l), (s) Invertase, C6H12O6(aq) + C6H12O6(aq), (a) A – (p), B – (r), C – (q), D – (s), (b) A – (p), B – (q), C – (r), D – (s), (c) A – (r), B – (p), C – (q), D – (s), (d) A – (p), B – (r), C – (s), D – (q), 123. Match the columns, Column-I, Column-II, (Catalyst), (Industrial product), (A) V2O5, (p) High density, poly-ethylene, (B) Ziegler-Natta, (q) Polyacrylonitrile, (C) Peroxide, (r) NH3, (D) Finely divided Fe, (s) H2SO4, (a) A – (s), B – (p), C – (q), D – (r), (b) A – (s), B – (r), C – (q), D – (p), (c) A – (r), B – (p), C – (q), D – (s), (d) A – (s), B – (q), C – (p), D – (r), 124. Match the columns, Column-I, Column-II, (A) Oil in water emulsion, (p) Clouds, (B) Aerosols containing, (q) Vanishing cream, small droplets of water, suspended in air, (C) When river water meets (r) Smoke, the sea water, (D) Colloidal solution of, (s) Formation of delta, carbon, arsenic, compounds, dust etc., in air, (a) A – (q), B – (p), C – (s), D – (r), (b) A – (p), B – (q), C – (s), D – (r), (c) A – (q), B – (s), C – (p), D – (r), (d) A – (q), B – (p), C – (r), D – (s), 125. Match the columns, Column-I, Column-II, (A) As2S3 sol, (p) Bredig’s Arc method, (B) Fe(OH)3 sol, (q) Double decomposition, (C) Colloidal sols of metals (r) Peptization, like Au, Ag, Pt, etc., (D) Conversion of freshly (s) Hydrolysis, prepared precipitate, into a colloidal sol, (a) A – (q), B – (s), C – (r), D – (p), (b) A – (q), B – (p), C – (s), D – (r), (c) A – (s), B – (q), C – (p), D – (r), (d) A – (q), B – (s), C – (p), D – (r)
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EBD_7207, SURFACE CHEMISTRY, , 324, , 126. Match the columns, Column-I, Column-II, (A) In this process molecules, (p) Dialysis, and ions diffuse through, membrane outside and, pure colloidal solution, is left behind., (B) This process is used, (q) Ultrafilteration, if the dissolved substance, in the impure colloidal, solution is only an, electrolyte, (C) In this process ordinary, (r) Electro-dialysis, filter paper is soaked into, collodion (4% solution of, nitrocellulose in a mixture, of alcohol and ether), (a) A – (p), B – (r), C – (q), (b) A – (r), B – (p), C – (q), (c) A – (p), B – (q), C – (r), (d) A – (q), B – (r), C – (p), 127. Match the columns, Column-I, Column-II, (A) Sulphur vapours passed (p) Normal electrolyte, through cold water, solution, (B) Soap mixed with water, (q) Molecular colloids, above critical micelle, concentration, (C) White of egg whipped, (r) Associated colloid, with water, (D) Soap mixed with water, (s) Macro molecular, below critical micelle, colloids, concentration, (a) A – (q), B – (r), C – (s), D – (p), (b) A – (r), B – (q), C – (s), D – (p), (c) A – (p), B – (r), C – (s), D – (q), (d) A – (q), B – (s), C – (r), D – (p), 128. Match the columns, Column-I, Column-II, (A) Protective colloid, (p) FeCl3 + NaOH, (B) Liquid - liquid colloid, (q) Lyophilic colloids, (C) Positively charged colloid (r) Emulsion, (D) Negatively charged colloid (s) FeCl3 + hot water, (a) A – (q), B – (r), C – (p), D – (s), (b) A – (p), B – (r), C – (s), D – (q), (c) A – (q), B – (r), C – (s), D – (p), (d) A – (r), B – (q), C – (s), D – (p), 129. Match the columns, Column-I, Column-II, (A) Dialysis, (p) Cleansing action of soap, (B) Peptisation, (q) Coagulation, (C) Emulsification, (r) Colloidal sol formatioin, (D) Electrophoresis, (s) Purification, , (a) A – (s), B – (r), C – (p), D – (q), (b) A – (q), B – (r), C – (p), D – (s), (c) A – (s), B – (p), C – (r), D – (q), (d) A – (s), B – (r), C – (q), D – (p), 130. Match the columns, Column-I, Column-II, (A) Butter, (p) dispersion of liquid in liquid, (B) Pumice stone (q) dispersion of solid in liquid, (C) Milk, (r) dispersion of gas in solid, (D) Paints, (s) dispersion of liquid in solid, (a) A – (r), B – (s), C – (p), D – (q), (b) A – (s), B – (r), C – (p), D – (q), (c) A – (q), B – (r), C – (p), D – (s), (d) A – (s), B – (r), C – (q), D – (p), 131. Match the columns, Column-I, Column-II, (A) Argyrol, (p) Kalazar, (B) Antimony, (q) Intramuscular injection, (C) Colloidal gold, (r) Stomach disorders, (D) Milk of magnesia (s) Eye lotion, (a) A – (r), B – (p), C – (s), D – (q), (b) A – (r), B – (p), C – (q), D – (s), (c) A – (s), B – (q), C – (p), D – (s), (d) A – (s), B – (p), C – (q), D – (r), , ASSERTION-REASON TYPE QUESTIONS, Directions : Each of these questions contains two statements,, Assertion and Reason. Each of these questions also has four, alternative choices, only one of which is the correct answer. You, have to select one of the codes (a), (b), (c) and (d) given below., (a) Assertion is correct, reason is correct; reason is a correct, explanation for assertion., (b) Assertion is correct, reason is correct; reason is not a, correct explanation for assertion, (c) Assertion is correct, reason is incorrect, (d) Assertion is incorrect, reason is correct., x, k . p1/ n is known as Freundlich, m, adsorption isotherm, where x is the mass of gas adsorbed, by m grams of adsorbate, p is the equilibrium pressure, k, and n are constants for given system and temperature., , 132. Assertion : The relation, , 1, ,, n, the lines by which their adsorption isotherms can be, represented will meet at a point., 133. Assertion : The enthalpy of physisorption is greater than, chemisorption., Reason : Molecules of adsorbate and adsorbent are held, by van der Waal’s forces in physisorption and by chemical, bonds in chemisorption., , Reason : When several substances have same value of
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SURFACE CHEMISTRY, , 134., , 135., , 136., , 137., , x, 1/ n, Assertion : According to Freundlich: = K . p ., m, Reason : The isotherm shows variation of the amount of, gas adsorbed by the adsorbent with temperature., Assertion: Detergents with low CMC are more economical, to use., Reason: Cleansing action of detergents involves the, formation of micelles. These are formed when the, concentration of detergents becomes equal to CMC., Assertion: An ordinary filter paper impregnated with, collodion solution stops the flow of colloidal particles., Reason: Pore size of the filter paper becomes more than the, size of colloidal particle., Assertion : The value of colligative properties are of small, order for colloids as compared to true solution., Reason : Number of particles in colloidal solution is, comparatively smaller than true solutions., , CRITICAL THINKING TYPE QUESTIONS, 138. Which of the following statements is incorrect regarding, physisorptions?, (a) More easily liquefiable gases are adsorbed readily., (b) Under high pressure it results into multimolecular layer, on adsorbent surface., (c) Enthalpy of adsorption ( H adsorption) is low and, positive., (d) It occurs because of van der Waal’s forces., 139. Which is correct about physical adsorption?, (a) High temperature and high pressure favour adsorption, (b) High temperature and low pressure favour adsorption, (c) Low temperature and high pressure favour adsorption, (d) Low temperature and low pressure favour adsorption, 140. Which one of the following is not applicable to the, phenomenon of adsorption ?, (a), H> 0, (b) G < 0, (c), S< 0, (d) H < 0, 141. Methylene blue, from its aqueous solution, is adsorbed on, activated charcoal at 25°C. For this process, which of the, following statement is correct ?, (a) The adsorption requires activation at 25°C, (b) The adsorption is accompanied by a decrease in, enthalpy, (c) The adsorption increases with increase of temperature, (d) The adsorption is irreversible, 142. In the adsorption of a gas on solid, Freundlich isotherm is, obeyed. The slope of the plot is zero. Then the extent of, adsorption is, (a) directly proportional to the pressure of the gas, (b) inversely proportional to the pressure of the gas, (c) directly proportional to the square root of the pressure, of the gas, (d) independent of the pressure of the gas, , 325, , 143. On the basis of data given below predict which of the, following gases shows least adsorption on a definite amount, of charcoal?, Gas, CO2 SO2 CH4 H2, Critical temp./K 304 630 190 33, (a) CO2, (b) SO2, (c) CH4, (d) H2, 144. Which of the following statements regarding difference, between adsorption and absorption is incorrect?, (a) Adsorption is a surface whereas absorption is a bulk, phenomena., (b) Water vapours are absorbed by anhydrous CaCl2 but, adsorbed by silica gel., (c) Adsorption and absorption take place individually., They can not occur simultaneously., (d) All of the above statements are correct., 145. Which of the following is not an application of adsorption?, (a) In metallurgy for concentration of sulphide ores., (b) In heterogeneous catalysis involving solid catalyst., (c) In homogeneous catalysis., (d) Separation of inert gas., 146. Which of the following statements regarding catalyst is not, true ?, (a) A catalyst remains unchanged in composition and, quantity at the end of the reaction, (b) A catalyst can initiate a reaction, (c) A catalyst does not alter the equilibrium in a reversible, reaction, (d) Catalysts are sometimes very specific in respect of, reaction, 147. Which of the following statements about a catalyst is true ?, (a) A catalyst accelerates the reaction by bringing down, the free energy of activation, (b) A catalyst also takes part in the reaction mechanism, (c) A catalyst makes the reaction more feasible by making, the Gº more negative, (d) A catalyst makes the equilibrium constant of the, reaction more favourable for the forward reaction, 148. Which one of the following, statements is incorrect about, enzyme catalysis?, (a) Enzymes are mostly protenious in nature., (b) Enzyme action is specific., (c) Enzymes are denaturated by ultraviolet rays and at, high temperature., (d) Enzymes are least reactive at optimum temperature., 149. Given below, catalyst and corresponding process/reaction, are matched. The one with mismatch is, (a) [RhCl(PPh3)2] : Hydrogenation, (b) TiCl4 + Al (C2H5)3 : Polymerization, (c) V2O5 : Haber-Bosch process, (d) Nickel : Hydrogenation
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EBD_7207, SURFACE CHEMISTRY, , 326, , 150. Which one of the following statements is incorrect in the, case of heterogeneous catalysis ?, (a) The catalyst lowers the energy of activation, (b) The catalyst actually forms a compound with the, reactant, (c) The surface of the catalyst plays a very important role, (d) There is no change in the energy of activation., 151. In petrochemical industry alcohols are directly converted, to gasoline by passing over heated, (a) Platinum, (b) ZSM-5, (c) Iron, (d) Nickel, 152. Which of the following feature of catalysts is described in, reactions given below?, (i), (ii), , 153., , 154., , 155., , 156., , 157., , 158., , 159., , 160., , CO g, CO g, , 2H2 g, H2 g, , Cu / ZnO Cr2O3, Cu, , CH3OH g, , HCHO g, , Ni, (iii) CO g 3H2 g, CH4 g H2O g, (a) Activity, (b) Selectivity, (c) Catalytic promoter, (d) Catalytic poison, The dispersed phase and dispersion medium in soap lather, are respectively, (a) gas and liquid, (b) liquid and gas, (c) solid and gas, (d) solid and liquid, Which of the following is not a colloid ?, (a) Chlorophyll, (b) Smoke, (c) Ruby glass, (d) Milk, Which of the following forms a colloidal solution in water ?, (a) NaCl, (b) Glucose, (c) Starch, (d) Barium nitrate, Which of the following forms cationic micelles above certain, concentration?, (a) Sodium dodecyl sulphate, (b) Sodium acetate, (c) Urea, (d) Cetyl trimethyl ammonium bromide, Which of the following does not contain a hydrophobic, structure ?, (a) Linseed oil, (b) Lanolin, (c) Glycogen, (d) Rubber, Which one of the following is an example for multimolecular, colloid?, (a) Aqueous starch sol, (b) Aqueous enzyme sol, (c) Alcoholic polystyrene sol, (d) Aqueous sol of sodium laurylsulphate, Bredig arc method cannot be used to prepare colloidal, solution of which of the following, (a) Pt, (b) Fe, (c) Ag, (d) Au, Colloidal solutions are not purified by, (a) Dialysis, (b) Electrodialysis, (c) Ultrafiltration, (d) Electrophoresis, , 161. Colloid of which one of the following can be prepared by, electrical dispersion method as well as reduction method ?, (a) Sulphur, (b) Ferric hydroxide, (c) Arsenious sulphide (d) Gold, 162. Which of the following ions can cause coagulation of, proteins ?, (a) Ag+, (b) Na+, 2+, (c) Mg, (d) Ca2+, 163. Which of the following will be most effective in the, coagulation of Al(OH)3 sol ?, (a) KCN, (b) BaCl2, (c) NaCl, (d) Mg3(PO4)2, 164. Point out the false statement :, (a) The colloidal solution of a liquid in liquid is called gel, (b) Hardy Schulze rule is related with coagulation, (c) Brownian movement and Tyndall effect are shown by, colloidal system, (d) Gold number is a measure of the protective power of, lyophilic colloid, 165. The disperse phase in colloidal iron (III) hydroxide and, colloidal gold is positively and negatively charged,, respectively. Which of the following statements is NOT, correct?, (a) Coagulation in both sols can be brought about by, electrophoresis, (b) Mixing the sols has no effect, (c) Sodium sulphate solution causes coagulation in both, sols, (d) Magnesium chloride solution coagulates, the gold sol, more readily than the iron (III) hydroxide sol., 166. A colloidal solution is subjected to an electric field. The, particles move towards anode. The coagulation of same sol, is studied using NaCl, BaCl2 and AlCl3 solutions. The order, of their coagulation power should be –, (a) NaCl > BaCl2 > AlCl3 (b) BaCl2 > AlCl3 > NaCl, (c) AlCl3 > BaCl2 > NaCl (d) BaCl2 > NaCl > AlCl3, 167. Flocculation value of BaCl2 is much less than that of KCl, for sol A and flocculation value of Na2SO4 is much less, than that of NaBr for sol B. The correct statement among, the following is :, (a) Both the sols A and B are negatively charged., (b) Sol A is positively charged arid Sol B is negatively, charged., (c) Both the sols A and B are positively charged., (d) Sol A is negatively charged and sol B is positively, charged., 168. In Brownian movement or motion, the paths of the particles, are, (a) Linear, (b) Zig-zag, (c) Uncertain, (d) Curved
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SURFACE CHEMISTRY, , 169. How does a delta form at the meeting place of sea and river, water?, (a) The electrolyte present in sea water coagulate the clay, (b) the electrolyte present in sea water has no role, (c) the electrolyte present in river water coagulate the clay, (d) Both (a) and (c) are correct, 170. .......... is a silver sol used as an eye lotion. Fill in the blank, with an appropriate answer., (a) Amytol, (b) Argyrol, (c) Ciprofloxacin, (d) Both (a) and (b), 171. Which of the following will show Tyndall effect?, (a) Aqueous solution of soap below critical micelle, concentration., (b) Aqueous solution of soap above critical micelle, concentration., (c) Aqueous solution of sodium chloride., (d) Aqueous solution of sugar., 172. Which of the following combinations of dispersed phase, and dispersion medium will not form a colloid ?, (a) Dispersed phase – Solid, Dispersion medium – Solid., (b) Dispersed phase – Solid, Dispersion medium – Gas., , 327, , (c) Dispersed phase – Gas, Dispersion medium – Gas., (d) Dispersed phase – Liquid, Dispersion medium – Gas., 173. Which of the following statements is incorrect?, (a) Colloidal gold is used for intramuscular injection., (b) Colloidal solution of latex is used in preparation of, rubber., (c) Photographic films are prepared by coating an, emulsion of AgBr in gelatin over glass plate., (d) Tannin used in leather industry contains positively, charged colloidal particles., 174. Which of the following is not correctly matched ?, (a) Sulphur sol – Oxidation, (b) Gold sol – Double decomposition, (c) Fe(OH)3 sol – Hydrolysis, (d) Both (b) and (c), 175. How many of the following are negatively charged sols?, Eosin dye, sol of charcoal, haemoglobin, Al2O3.xH2O,, As2S3, TiO2.sol, copper sol, (a) 3, (b) 4, (c) 6, (d) All of these
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EBD_7207, SURFACE CHEMISTRY, , 328, , 20., , FACT / DEFINITION TYPE QUESTIONS, , 7., 8., , 9., , 10., , 11., 12., 17., 19., , S, , Gas (Adsorbate) + Solid (Adsorbent), Gas adsorbed on solid + Heat, Initially adsorption increases with increase in pressure, at a particular temperature then got slow. After attaining, equilibrium adsorption become independent of, pressure., 21., 22., , x, m, , 23., , 24., , m, , 1, n, , 1, =1, n, , P1, , at high pressure, , 1, =0, n, , x, P, m, i.e., the value of n varies between 0 to 1., (b) The gases having higher values for critical temperature, are easily liquified and are adsorbed to the greater, extent. CO2 has highest critical temperature of 304K., (b) According to Freundlich equation., x, x, Kp1/ n, p1/ n or, m, m, x, x, 1, log Kp1/ n or log, log K, log p, m, n, m, (a) According to Freundlich adsorption isotherm, or log, , 25., , 29., , plot of log x vs log P is linear with slope =, , 1, kP n, , at low pressure, , 27., 28., , Intercept = log k, log P, , (d), (a) According to Freundlich adsorption isotherm, x, m, , 26., , 1, = n, e, lop, , x, m, , 4., 5., 6., , (d) The substance which is adsorbed is called adsorbate., (b) Adsorption is an exothermic process., (a) On increasing temperature physical adsorption of a, gas on surface of solid decreases. Solid adsorb greater, amount of gas at lower temperature., (a) Chemical adsorption involves formation of monolayer., (a) Chemisorption involves strong chemical forces., (d) In physisorption, gas molecules are held on the solid, surface by weak van der Waal’s forces., (b), (d) Adsorption is accompanied by evolution of heat as, the residual forces acting along the surface of, adsorbent decrease i.e., adsorption is accompanied by, decrease in enthalpy., (b) The more readily soluble and easily liquefiable gases, such as NH3, HCl and SO2 are adsorbed more than the, so called permanent gases like O2. This is due to the, van der Waals or intermolecular forces which are, involved in adsorption., (c) Adsorption is an exothermic process i.e. DH of, adsorption is always negative. When a gas is adsorbed,, the freedom of movement of its molecules becomes, restricted i.e. DS is negative. Hence adsorption, accompanied by decrease in enthalpy as well as, decrease in entropy of the system., (d) The heat evolved in physisorption is quite low varying, generally between 20-40 kJ mol–1., (c), 13. (b), 14. (a), 15. (d), 16. (b), (a), 18. (b), (d) According to Freundlich adsorption isotherm., At intermediate pressure, extent of adsorption, x, 1, x, log k, log P;, kP1/ n or log, m, n, m, , log, , 1., 2., 3., , (a) Adsorption of a gas on solid is represented by, following equilibria,, , x, kp1/n (wheren n > 1), m, At very high pressure x/m = kp0, (a) On increasing temperature adsorption of a gas on, surface of solid decreases. Solid adsorb greater amount, of substances at lower temperature., (c), (a) For adsorption to occur G = –ve as in adsorption, H = –ve, S = –ve. –T S is positive for adsorption., (d) For a reaction in equilibrium, the increase in rate of, reaction in forward direction by catalyst increases the, concentration of product(s) and thus the rate of, backward reaction also increases with the same, magnitude and allow the equilibrium to be achieved, quickly.
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SURFACE CHEMISTRY, , 30., , 31., 32., 33., 34., , 35., , 36., 37., 38., 39., 40., 41., , 329, , (d) The catalytic poisons decrease the activity of the, catalyst because they are preferentially adsorbed on, the surface of catalyst., (a) A catalyst increases the rate of reaction by decreasing, the activation energy., (c) Active charcoal has more adsorption power due to, greater surface area., (c) Adsorption theory is applied to heterogeneous, catalysis., (a) According to the adsorption theory of catalysis, the, activity of catalysis is due to the presence of free, valencies on its surface due to which surface of catalyst, has chemical force of altraction. When a gas comes in, contact with this surface molecules get attached, attached through these valencies. Further the rate of, reaction is always increases by decreases in activation, energy. When any of the reactants is strongly, adsorbed on the surface of catalyst, the rate becomes, inversely proportional to the concentration of that, reaction. The reaction is then said to be inhibited by, such reactant. The lowering of activation energy always, leads to the increase in speed of reaction., (b) Catalyst decreases the activation energy of the, reaction by forming an intermediate product. So no of, molecules having activation energy increases hence, rate of reaction increases., (c) A catalyst can affect reversible reaction by attaining, equilibria in both directions., (d) In acid hydrolysis of methyl acetate all are present in, one phase (liquid)., (d) Enzymes are specific biological catalysts possessing, well - defined active sites., (a) Enzymes are biological catalysts., (d) Enzymes are biological catalysts and enhance the rate, of biochemical reactions., (c) Hydrolysis of urea can be represented as follows, Urease, , 42., 43., 44., 45., 46., 47., 48., 49., 50., 51., , H 2 N C NH 2 H 2O, 2NH3 CO 2, (enzyme), ||, O, Since it involves biological catalyst (enzyme) so it is, an example of biochemical catalysis., (d) Efficiency of catalysing property of a catalyst is, inversely proportional to activation energy., (b) Molybdenum acts as a promoter for iron which is used, as a catalyst in Haber’s process., (c) Second step involves adsorption of reactant, molecules on the surface of the catalyst., (b), (a) Milk is a emulsion in which liquid is dispersed in liquid., (c) In butter (liquid – solid) water is dispersed in fat., (c) Size of colloidal particles is 10–5 - 10–7 cm., (a), (d) Mist is a colloid (aerosol) in which liquid is dispersed, in gas., (c) Alloy is an example of solid solution., , 52. (b) Colloid, Dispersed phase Dispersion medium, Sol, Solid, Liquid, Gel, Liquid, Solid, Emulsion, Liquid, Liquid, Foam, Gas, Liquid, 53. (d) Emulsions are liquid-liquid colloidal systems, Generally, one of the two liquids is water., 54. (b) Butter is an example of gel., 55. (c) Cheese is a liquid dispersed in solid phase., 56. (d) Air is a homogeneous mixture of gases, mainly nitrogen, and oxygen., 57. (c) When the dispersed phase and dispersion medium both, are liquid, the colloidal system is called as an emulsion, like milk, vasnishing cream etc., 58. (b) Cloud consists of fine droplets of water suspended in, air., 59. (d) Blood is a –vely charged colloidal system. Rest of the, compounds, i.e., NaCl, urea & cane sugar form true, solution in water., 60. (a) Suspension particles are visible under a microscope, and sometimes even to a naked eye., 61. (a) Fog is a colloidal system consisting water droplets, dispersed in air., 62. (a) A collidal solution is biphasic and heterogeneous. It is, composed of two phases :, 1. Dispersed phase; 2. Dispersion medium., 63. (c) Smoke is solid gas system, solid (D. P) and gas (D.M.), Note : D.P. : Dispersed phase, D.M. : Dispersion medium, 64. (c) The critical micelle concentration is the lowest, concentration at which micelle formation appears when, surfactants are present above that CMC, they can act, as emulsifiers that will solubilise a compound which is, normally insoluble in the solvent being used., 65. (a), , Polarhead, Non-polar tail, (micelle), , 66. (a) Surface tension of lyophilic sols is lower than water, (dispersion medium)., 67. (b) Gum is lyophilic colloid., 68. (c) Lyophobic colloids are irreversible colloids. They are, protected by lyophilic colloids., 69. (c) Example of multimolecular colloid system is a gold, dispersed in water., 70. (a) On shaking with the dispersion medium, colloids, directly form the colloidal sol. Hence they are called, intrinsic colloids. i.e., glue., 71. (d) Sodium stearate is a soap. Soaps and detergents are, surface agents which when dissolved in a medium,, forms aggregated particles, called associated colloids., 72. (d) The formation of micelles takes place only above a, particular temperature called kraft temperature (TK).
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EBD_7207, SURFACE CHEMISTRY, , 330, , 73., 74., 75., , 76., 77., , 78., 79., 80., , 81., , 82., , 83., 84., 85., 86., 87., , 88., 89., 90., 91., , 92., , 93., , (c), (b) Electrolytes are used for neutralising charge on, colloidal particles., (c) The osmotic pressure of sea water is 25 atm at 15°C., When pressure greater than 26 atm is applied on sea, water separated by a rigid emipermeable membrane,, pure water is obtained., (a) Blood is purified by dialysis., (d) The use of membrane for separating colloidal particles, is termed as dialysis. Hence it is clear that colloidal, particle cannot pass through animal membrane. Hence, only solvent molecules and ions (in case of, electrodialysis) can diffuse., (d) Electrolytic (Ionic) impurities can be most easily, removed on application of electric field., (a) Formation of colloid from suspension is known as, peptization., (d) The separation of colloidal particles of molecular, dimension is known as dialysis. It is a purification, method of colloid., (c) Electrodialysis involves movement of ions towards, oppositely charged electrodes., Urea being a covalent compound does not dissociate, to give ions and hence it cannot be removed by, electrodialysis.However all the other given compounds, are ionic which can undergo dissociation to give, oppositely charged ions and thus can be separated., (b) The motion of a liquid through a membrane under the, influence of an applied electric field is known as electroosmosis., (d), (c) Peptization comes under dispersion methods of, preparation of colloids., (c) Colloidal gold is prepared by Bredig's arc method., (b) Peptisation is disintegration of colloidal aggregate., (c) According to this law the coagulating effect of an ion, on dispersed phase of opposite charge increases with, the increase in valency of the ion. The precipitating, power of Al3+ , Ba++, Na+ ions is in order Al3+ > Ba2+, > Na+., (a) Tyndall effect is shown by sols., (c) It is due to impact of molecules of dispersion medium, on the colloidal particles., (b) It is due to Tyndall effect., (a) Tyndall effect is the simplest way to check colloidal, system since path of light beam becomes visible due, to scattering of light., (d) Fe(OH)3 is positive sol. K3[Fe(CN)6 ] will provide, [Fe(CN)6]3– for coagulation having highest magnitude, of –ve charge among given options., (d) According to the Hardy schulze rule the coagulating, effect of an ion on dispersed phase of opposite charge, increases with the valency of the ion. Therefore more, the charge on oppositely charged ion higher is the, coagulation value., , 94., , 95., 97., , 98., , 99., 100., , 101., , 102., 103., 104., 105., 106., 107., 108., 111., 112., , (a) Smaller the charge on anion, lesser will be its, coagulating power., KBr have Br– with least charge of – 1 on Br thus, KBr is least effective in coagulating Fe(OH)3., (b), 96. (a), (b) The size of colloidal particles is between 1 nm and, 1000 nm i.e., it is not always greater than 100 nm. So (b), is not a property of colloidal solution. All others are, the properties of colloidal solution., (c) According to Hardy-Schulze rule "The amount of, electrolyte required to coagulate a fixed amount of a, sol depends upon the sign of charge and valency of, the flocculating ion.", Thus, the coagulating power vary in the order., Al3+ > Zn++ > Na+, (c), (d) A negative ion causes the precipitation of positively, charged sol and vice-versa. Since As2S3 is a negative, sol so more will be the positive charge on cation more, effective it will be in causing coagulation of As2S3 sol., Among the given ions, Al3+ has the greatest valency, and thus is the most effective coagulating agent., (c) Electrodialysis involves movement of ions towards, oppositely charged electrodes., Urea being a covalent compound does not dissociate, to give ions and hence it cannot be removed by, electrodialysis.However all the other given compounds, are ionic which can undergo dissociation to give, oppositely charged ions and thus can be separated., (a) Brownian movement is exhibited by colloidal system., (b) Brownian movement is random motion., (c) Sugar forms homogeneous solution hence no Tyndall, effect is exhibited., (c) Emulsion show the tyndal! effect. Refers to ans 280., (c), (b) For gel dispersed phase is liquid and dispersion, medium is solid., (d), 109. (c), 110. (c), (c) Protein is an emulsifying agent for O/W emulsion., (d), , STATEMENT TYPE QUESTIONS, 113. (d) Adsorption is a surface phenomenon., 114. (b) Statements (i) and (iv) are incorrect. A given surface, of an adsorbent does not show any preference for a, particular gas as the van der Waal’s forces are, universal., Finely divided metals are better adsorbent as compared, to solid metals because they have large surface area, and extent of adsorption increases with surface area., 115. (a) As chemisorption involves compound formation, it is, usually irreversible in nature., 116. (a), 117. (c) Zeolites are found in nature as well as synthesised, for catalytic selectivity.
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SURFACE CHEMISTRY, , 118. (b) Optimum pH for the enzyme catalysis is 5-7. Catalyst, used in Ostwald’s process is platinised asbestos at, 573 K., 119. (b) S8 forms a multimolecular colloid. Micelles are formed, above kraft temperature and above a particular, concentration called Critical Micelle Concentration, (CMC)., 120. (a) Values of colligative properties for colloids are of small, order as compared to values shown by true solution., Most acceptable phenomena to account for the charge, of sol particles in preferential adsorption., MATCHING TYPE QUESTIONS, 121. (a), 122. (a), 123. (a) (A) V2O5 is used as a catalyst during the preparation, of H2SO4, (B) Ziegler-Natta is used as a catalyst during the, preparation of HDPE., (C) Peroxide is used as a catalyst durin g the, preparation of polyacrylonitrile., (D) Finely divided Fe is used as a catalyst during the, preparation of ammonia., 124. (a), 125. (d), 126. (a), 127. (a), 128. (c), 129. (a), 130. (b), 131. (d) Argyrol is used as an eye lotion., Antimony is used in Kalazar., Collidal gold is used in intramuscular injection., Milk of magnesia is used in the stomach disorder., , ASSERTION-REASON TYPE QUESTIONS, 132. (c) Assertion is true, reason is false. When several lines, 1, , then the lines by which, n, their adsorption isotherms can be represented will be, parallel and will not meet at a point., Assertion is false but Reason is true. The enthalpy, of chemisorption is of the order of 40 - 400 kJmol –1, while for physical adsorption it is of the order of, 20 - 40 kJmol–1., Assertion is true but Reason is false., Freundlich adsorption isotherm gives an empirical, relationship between the quantity of gas adsorbed by, unit mass of solid adsorbent and pressure at a particular, temperature., 136. (c), Colligative properties depend upon number of, particles., , have the same value of, , 133. (d), , 134. (c), , 135. (a), 137. (a), , CRITICAL THINKING TYPE QUESTIONS, 138. (c) Adsorption is an exothermic process, hence H will, always be negative., , 331, , 139. (c) Physical adsorption involves weak forces, physical in, nature with small heat of adsorption. Thus low, temperature and high pressure favours physical, adsorption., 140. (a), 141. (b) The adsorption of methylene blue on activated, charcoal is an example of physiosorption which is, exothermic, multilayer and does not have energy barrier., 142. (d) Freundlich’s isothermal adsorption equation can be, given as, x, m, log, , kp1/n, x, m, , Thus,, , log k, x, m, , 1, log p; slope, n, , 1, n, , 0, , kp 0, , 143. (d), 144. (c) Both adsorption and absorption can take place, simultaneously. The term sorption is used to describe, both the processes., 145. (c) Homogenous catalysis does not involves adsorption., 146. (b) A catalyst can not initiate a reaction., 147. (a), 148. (d) Enzymes are most reactive at optimum temperature., The optimum temperature for enzyme activity lies, between 40°C to 60°C., 149. (c) V2 O 5 is used as catalyst in contact process of, manufacturing H2SO4., 150. (c) The theory of heterogeneous catalysis is based upon, the phenomenon of adsorption. The activity of catalyst, is due to the presence of free valencies on its sufrace, due to which surface of catalyst has force of attraction., 151. (b) ZSM-5 is a shape selective catalyst. Zeolites are good, shape selective catalysts because of the honey comb, like structure., 152. (b) Given reactions shows that the selectivity of different, catalysts for same reactants is different., 153. (a) Soap lather is a colloid containing gas as a dispersed, phase and liquid as a dispersion medium., 154. (a) Chlorophyll. Smoke is an example of solid-gas colloid, system Ruby glass is an example of solid-solid colloid, system. Milk is an liquid -liquid colloid system., 155. (c) Starch molecules have colloidal dimensions whereas, NaCl, glucose and Ba(NO3)2 are crystalloids and, soluble in water., 156. (d) Cetyl trimethyl ammonium bromide,, , [C16 H33 (CH3 )3 N Br ] is a cationic micelle., 157. (d) Linseed oil, lanolin and glycogen attract water hence, contain a hydrophilic structure but rubber does not, attract water and thus does not contain a hydrophobic, structure.
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EBD_7207, SURFACE CHEMISTRY, , 332, , 158. (a) Multimolecular colloids consist of aggregates of atoms, or small molecules. Sulphur sol is an example of, multimolecular colloids, 159. (b) Bredig’s arc method is suitable for the preparation of, colloidal solution of metals like gold, silver, platinum, etc. An arc is struck between the metal electrode under, the surface of water containing some stabilzing agent, such as a trace of KOH. However, Fe does not react, with alkalies that is why it is not obtained by Bredig’sarc method., 160. (d) Colloidal solutions are not purified by electrophoresis., Movement of colloidal particles under the influence of, electric field is called electrophoresis. So, it can make, easier. Electrophoresis is the property of colloids not, the purification method., 161. (d) Gold by Bredig’s method (Dispersion method) and by, reduction method, AuCl3 + Tannic acid Gold sol, 162. (a) Proteins are coagulated by some heavy metal ions like, Ag+, Hg2+ and Pb2+., 163. (d) Al(OH)3 is a positive sol so salt having anion with, maximum negative charged (i.e. phosphate ion) will be, most effective in coagulation., 164. (a) Colloid of liquid in liquid is called emulsion. Colloid of, liquid in solid is gel., 165. (b) When oppositely charged sols are mixed their charges, are neutralised. Both sols may be partially or, completely precipitated., , 166. (c) As colloidal particles move towards anode so these, particles are negatively charged and coagulated by, cations of electrolyte., According to Hardy Schulze rule,, Coagulation power, charge of ion, Order of coagulation power is Al3+ > Ba2+ > Na+, 167. (b) In first case the given compounds have same anion, but different cations having different charge hence, they will precipitate negatively charged sol i.e. ‘A’., In second case the given compounds have similar, cation but different anion with different charge. Hence, they will precipitate positively charged sol. i.e. ‘B’., 168. (b) Brownian movement is zig zag motion of sol particles., 169. (a), 170. (a), 171. (b), 172. (c) A gas mixed with another gas forms a homogeneous, mixture and hence is not a colloidal system., 173. (d) Tannin used in leather industry contains negatively, charged colloidal particles., 174. (b) Gold sol is prepared by reduction as, 2AuCl3 3HCHO 3H 2 O, , Reduction, , 2Au sol, , 3HCOOH 6HCl, , 175. (b) Eosin dye, sol of charcoal, As2S3 and copper sol are, example of negatively charged sol.
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20, GENERAL PRINCIPLES AND PROCESSES, OF ISOLATION OF ELEMENTS, FACT / DEFINITION TYPE QUESTIONS, 1., , 2., , 3., , 4., , 5., , 6., , 7., , 8., , 9., , 10., , 11., , Which one of the following is an ore of silver ?, (a) Argentite, (b) Stibnite, (c) Haematite, (d) Bauxite, Cinnabar is an ore of, (a) Hg, (b) Cu, (c) Pb, (d) Zn, An example of an oxide ore is, (a) Bauxite, (b) Malachite, (c) Zinc blende, (d) Feldspar, The natural materials from which an element can be extracted, economically are called, (a) ores, (b) minerals, (c) gangue, (d) None of these, The most abundant metal on the surface of the earth is, (a) Fe, (b) Al, (c) Ca, (d) Na, Which of the following is an ore of tin ?, (a) Carborundum, (b) Epsomite, (c) Cassiterite, (d) Spodumene, Which of the following is chalcopyrite?, (a) CuFeS2, (b) FeS2, (c) KMgCl3.6H2O, (d) Al2O3.2H2O, Haematite is the ore of, (a) Pb, (b) Cu, (c) Fe, (d) Au, Composition of azurite mineral is, (a) CuCO3CuO, (b) Cu(HCO3)2. Cu(OH)2, (c) 2CuCO3.Cu(OH)2, (d) CuCO3. 2Cu(OH)2, Which one of the following is a mineral of iron ?, (a) Malachite, (b) Cassiterite, (c) Pyrolusite, (d) Magnetite, All ores are minerals, while all minerals are not ores because, (a) the metal can’t be extracted economically from all the, minerals, (b) minerals are complex compounds, (c) the minerals are obtained from mines, (d) all of these are correct, , 12. Which one of the following is not a sulphide ore?, (a) Magnetite, (b) Iron pyrites, (c) Copper glance, (d) Sphalerite, 13. The impurities associated with mineral used in, metallurgy are called collectively?, (a) Slag, (b) Flux, (c) Gangue, (d) Ore, 14. The most abundant element in the earth’s crust (by weight), is, (a) Si, (b) Al, (c) O, (d) Fe, 15. Malachite is an ore of, (a) iron, (b) copper, (c) mercury, (d) zinc, 16. Cassiterite is an ore of, (a) Mn, (b) Ni, (c) Sb, (d) Sn, 17. Galena is an ore of, (a) Pb, (b) Hg, (c) Zn, (d) None of these, 18. The metal always found in the free states is, (a) Au, (b) Ag, (c) Cu, (d) Na, 19. Matrix is defined as –, (a) the unwanted foreign material present in the ore, (b) the flux added to remove the unwanted impurities from, ore, (c) the slag formed as a result of the reaction of flux with, gangue, (d) the material used in the reduction of metal oxide to, metal, 20. Which of the following pair is incorrectly matched ?, (a) Magnetite – Fe3O4, (b) Copper glance – Cu2S, (c) Calamine – ZnCO3, (d) Zincite – ZnS, 21. Which one of the following ores is best concentrated by, froth-flotation method ?, (a) Galena, (b) Cassiterite, (c) Magnetite, (d) Malachite, 22. Froth floatation process is used for the metallurgy of, (a) chloride ores, (b) amalgams, (c) oxide ores, (d) sulphide ores
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EBD_7207, GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF ELEMENTS, , 334, , 23., , 24., , 25., , 26., , 27., , 28., , 29., , 30., , 31., , 32., , Cassiterite is concentrated by, (a) levigation, (b) electromagnetic separation, (c) floatation, (d) liquefaction, While extracting an element from its ore, the ore is grounded, and leached with dil. potassium cyanide solution to form, the soluble product potassium argento cyanide. The element, is, (a) Lead, (b) Chromium, (c) Manganese, (d) Silver, The method of concentrating the ore which makes use of, the difference in density between ore and impurities is called, (a) levigation, (b) leaching, (c) magnetic separation (d) liquifaction, Leaching is a process of, (a) reduction, (b) concentration, (c) refining, (d) oxidation, Which one of the following ores is concentrated by chemical, leaching method?, (a) Galena, (b) Copper pyrite, (c) Cinnabar, (d) Argentite, Electromagnetic separation is used in the concentration of, (a) copper pyrites, (b) bauxite, (c) cassiterite, (d) cinnabar, For which ore of the metal, froth floatation method is used, for concentration?, (a) Horn silver, (b) Bauxite, (c) Cinnabar, (d) Heamatite, Which of the following metal is leached by cyanide process ?, (a) Ag, (b) Na, (c) Al, (d) Cu, Which one of the following ores is not concentrated by, froth floatation process?, (a) Copper pyrites, (b) Pyrargyrite, (c) Pyrolusite, (d) Zinc blende, In froth flotation process many chemicals (frother , collector,, activator, and depressant) are used. Which of the following, is a frother ?, (a), , 33., , 34., , CuSO 4, , 35., , 36., , 37., , 38., , 39., , 40., , 41., , 42., , (b) NaCN+ alkali, , (c) Pine oil, (d) Potassium xanthate, Froth flotation process is based on, (a) wetting properties of ore particle, (b) specific gravity of ore particles, (c) magnetic properties of ore particles, (d) electrical properties of ore particles, In the froth flotation process of concentration of ores, the, ore particles float because they:, (a) are light, (b) are insoluble, (c) have the surface which is not wetted easily, (d) have a constant electrical charge, , 43., , 44., , Main function of roasting is, (a) to remove volatile substances, (b) oxidation, (c) reduction, (d) slag formation, Roasting is generally done in case of the, (a) oxide ores, (b) silicate ores, (c) sulphide ores, (d) carbonate ores, Heating of pyrites in air for oxidation of sulphur is called, (a) roasting, (b) calcination, (c) smelting, (d) slagging, The role of calcination in metallurgical operations is, (a) to remove moisture, (b) to decompose carbonates, (c) to drive off organic matter, (d) to decompose carbonates and drive off moisture and, organic matter, General method for the extraction of metal from oxide ore is, (a) carbon reduction, (b) reduction by aluminium, (c) reduction by hydrogen, (d) electrolytic reduction, Function of the flux added during smelting is, (a) to make ore porous, (b) to remove gangue, (c) to make reduction easier, (d) to precipitate slag, Process followed before reduction of carbonate ore is, (a) calcination, (b) roasting, (c) liquation, (d) polling, Calcination is the process in which :, (a) ore is heated above its melting point to expel H2O or, CO2 or SO2, (b) ore is heated below its melting point to expel volatile, impurities, (c) ore is heated above its melting point to remove S, As, and Sb as SO2 ,As2O3 and Sb2O3 respectively, (d) ore is heated below its melting point to expel H2O or, CO2, When a metal is to be extracted from its ore and the gangue, associated with the ore is silica, then, (a) an acidic flux is needed, (b) a basic flux is needed, (c) both acidic and basic fluxes are needed, (d) Neither of them is needed, Which of the following fluxes is used to remove acidic, impurities in metallurgical process?, (a) Silica, (b) Lime stone, (c) Sodium chloride, (d) Sodium carbonate
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GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF ELEMENTS, , 45., , 46., , 47., , 48., , 49., , 50., , Which of the following reactions is an example for calcination, process ?, (a), , 2Ag 2HCl, , (b), , 2Zn O 2, , (c), , 2ZnS 3O 2, , (d), , MgCO3, , O, , 2AgCl H 2 O, , 2ZnO, 2ZnO 2SO 2, MgO CO 2, , After partial roasting the sulphide of copper is reduced by, (a) cyanide process, (b) electrolysis, (c) reduction with carbon (d) self reduction, Hydro-metallurgical process of extraction of metals is based, on, (a) complex formation, (b) hydrolysis, (c) dehydration, (d) dehydrogenation, 2CuFeS2 O 2, Cu 2 S 2FeS SO 2, Which process of metallurgy of copper is represented by, above equation?, (a) Concentration, (b) Roasting, (c) Reduction, (d) Purification, Which of the following is not used as a collector ?, (a) Pine oil, (b) Xanthates, (c) Cresols, (d) Fatty acids, Which of the following reaction represents calcination, process ?, (a) 2PbS 3O 2, 2PbO 2SO 2, (b) CaCO3.MgCO3 (s), CaO(s) MgO(s) 2CO2(g), coke, 1673K, , 51., , 52., , 53., , 54., , 55., , (c) ZnO C, Zn CO, 2FeO CO 2, (d) Fe 2 O3 CO, According to Ellingham diagram, the oxidation reaction of, carbon to carbon monoxide may be used to reduce which, one of the following oxides at the lowest temperature ?, (a) Al2O3, (b) Cu2O, (c) MgO, (d) ZnO, Which of the following condition favours the reduction, of a metal oxide to metal?, (a), H = +ve, T S = + ve at low temperature, (b) H = +ve, T S = – ve at any temperature, (c), H = –ve, T S = – ve at high temperature, (d) H = –ve, T S = + ve at any temperature, Ellingham diagram normally consists of plots of, (a), Sº vs T, (b), fGº vs Sº, (c), Gº vs T, (d) Hº vs T, G° vs T plot in the Ellingham’s diagram slopes downward, for the reaction, 1, 1, O2, MgO (b) 2Ag, O2, Ag 2O, (a) Mg, 2, 2, 1, 1, O2, CO, O2, CO 2, (d) CO, (c) C, 2, 2, In the blast furnace iron oxide is reduced by, (a) silica, (b) CO, (c) carbon, (d) limestone, , 335, , 56. Furnaces are lined with calcium oxide because, (a) it gives off oxygen on heating, (b) it gives strong light on heating, (c) it is refractory and basic, (d) it is not affected by acids, 57. The following reactions take place in the blast furnace in, the preparation of impure iron. Identify the reaction, pertaining to the formation of the slag., (a) Fe2O3(s) + 3 CO(g) 2 Fe (l) + 3 CO2 (g), (b) CaCO3 (s) CaO (s) + CO2 (g), (c) CaO (s) + SiO2(s) CaSiO3(s), (d) 2C(s) + O2 (g) 2 CO(g), 58. Refractory materials are generally used in furnaces because, (a) they possess great structural strength, (b) they can withstand high temperature, (c) they are chemically inert, (d) they do not require replacement, 59. Which of the following reactions taking place in the blast, furnace during extraction of iron is endothermic?, (a) CaCO3, CaO CO2, (b) 2C O 2, 2CO, (c), , C, , O2, , CO 2, , (d) Fe 2 O 3 3CO, 2Fe 3CO 2, 60. Cast iron is, (a) made by melting pig iron with scrap iron and coke, using hot air blast, (b) having slightly lower carbon content (about 3%) as, compared to pig iron, (c) extremely hard and brittle, (d) All of the above statements are true, 61. In the extraction of copper from its sulphide ore, the metal is, finally obtained by the reduction of cuprous oxide with :, (a) Copper (I) sulphide (Cu2S), (b) Sulphur dioxide (SO2), (c) Iron sulphide (FeS), (d) Carbon monoxide (CO), 62. Extraction of zinc from zinc blende is achieved by, (a) electrolytic reduction, (b) roasting followed by reduction with carbon, (c) roasting followed by reduction with another metal, (d) roasting followed by self-reduction, 63. In the extraction of Cu, the metal is formed in the bessemer, converter due to the reaction :, 6Cu + SO2, (a) Cu2S + 2Cu2O, (b) Cu2S, 2Cu + S, 2Cu + FeO, (c) Fe + Cu2O, (d) 2Cu2O, 4Cu + O2, 64. Aluminothermic process is used for the extraction of metals,, whose oxides are, (a) fusible, (b) not easily reduced by carbon, (c) not easily reduced by hydrogen, (d) strongly basic
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EBD_7207, GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF ELEMENTS, , 336, , 65., 66., , 67, , 68., , 69., , 70., , 71., 72., , 73., 74., , Electrometallurgical process is used to extract, (a) Fe, (b) Pb, (c) Na, (d) Ag, The electrolytic method of reduction is employed for the, preparation of metals that, (a) are weakly electropositive, (b) are moderately electropositive, (c) are strongly electropositive, (d) form oxides, Aluminium is extracted from alumina (Al2O3 ) by electrolysis, of a molten mixture of, (a) Al2O3 + HF + NaAlF4, (b) Al2O3 + CaF2 + NaAlF4, (c) Al2O3 + Na3AlF6 + CaF2, (d) Al2O3 + KF + Na3AlF6, In the extraction of aluminium by Hall-Heroult process,, purified Al2O3 is mixed with CaF2 to, (i) lower the melting point of Al2O3., (ii) increase the conductivity of molten mixture., (iii) reduce Al3+ into Al(s)., (iv) acts as catalyst., (a) (i) and (ii), (b) (i), (ii) and (iii), (c) (iii) and (iv), (d) (ii), (iii) and (iv), In the extraction of chlorine by electrolysis of brine______., (a) oxidation of Cl– ion to chlorine gas occurs., (b) reduction of Cl– ion to chlorine gas occurs., (c) For overall reaction G has negative value., (d) a displacement reaction takes place., Brine is electrolysed by using inert electrodes. The reaction, at anode is ________., (a), , Cl – (aq.), , 1, Cl 2 (g) e – ;, 2, , (b), , 2H 2 O(1), , (c), (d), , ECell, , 1.36V, , O2 (g) 4 H, , 4e – ; ECell, , 1.23V, , Na (aq.) e –, , Na(s);, , ECell, , 2.71V, , H (aq.) e –, , 1, H 2 (g);, 2, , ECell, , 0.00V, , Blister copper is, (a) Impure Cu, (b) Cu alloy, (c) Pure Cu, (d) Cu having 1% impurity, The furnace used to prepare commercial iron is lined with, which of the following ?, (a) Haematite, (b) Magnetite, (c) Ironpyrites, (d) Both (a) and (b), Which form of the iron contains 4% carbon ?, (a) Cast iron, (b) Pig iron, (c) Wrought iron, (d) Both (a) and (b), Which of the following reaction takes place in blast furnace, during extraction of copper ?, (a) 2Cu 2S 3O 2, 2Cu 2O 2SO 2, 2FeO 2SO 2, (b) 2FeS 3O 2, 6Cu SO 2, (c) 2Cu 2O Cu 2S, (d) All of these, , 75., , 76., , 77., , 78., , 79., , 80., , 81., , 82., , 83., , The main reactions occurring in blast furnace during, extraction of iron from haematite are________., (i) Fe2O3 + 3CO — 2Fe + 3CO2, (ii) FeO + SiO2 — FeSiO3, (iii) Fe2O3 + 3C — 2Fe + 3CO, (iv) CaO + SiO2 — CaSiO3, (a) (i) and (iii), (b) (ii) and (iv), (c) (i) and (iv), (d) (i), (ii) and (iii), The process of zone refining is used in the purification of, (a) Si, (b) Al, (c) Ag, (d) Cu, Van Arkel method of purification of metals involves, converting the metal to a, (a) volatile stable compound, (b) volatile unstable compound, (c) non volatile stable compound, (d) None of the above, The method not used in metallurgy to refine the impure, metal is, (a) Mond’s process, (b) Van–Arkel process, (c) Amalgamation process, (d) Liquation, Which of the following pairs of metals is purified by van, Arkel method ?, (a) Ga and In, (b) Zr and Ti, (c) Ag and Au, (d) Ni and Fe, The method of zone refining of metals is based on the principle, of, (a) greater solubility of the impurities in the molten state, than in the solid, (b) greater mobility of the pure metal than that of the, impurite, (c) higher melting point of the impurities than that of the, pure metal, (d) greater noble character of the solid metal than that of, the impurities, Method used for obtaining highly pure silicon which is, used as a semiconductor material, is, (a) oxidation, (b) electrochemical, (c) crystallization, (d) zone refining, What is anode mud?, (a) Fan of anode, (b) Metal of anode, (c) Impurities collected at anode in electrolysis during, purification of metals, (d) All of these, The process of zone refining is used in the purification of, (a) Si, (b) Al, (c) Ag, (d) Cu
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GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF ELEMENTS, , 84., , 85., , Which of the following statements regarding, chromatography is incorrect ?, (a) It is based on the principle that different components, of mixture gets adsorbed differently on an adsorbent, (b) Column chromatography involves column of Al2O3 in, a glass tube as a stationary phase., (c) The mobile phase may be a gas, a liquid or a solid., (d) Component which is more soluble is stationary phase, takes longer time to travel., Which of the following metal is used in the manufacture of, dye-stuffs and paints ?, (a) Copper, (b) Zinc, (c) Aluminium, (d) Magnesium, , 337, , 89. Read the following statements, (i) The principle that the impurities are more soluble in, the melt than in the solid state is used in the, manufacture of high purity semiconductors., (ii) Van Arkel method of refining Zr involves heating of, crude metal with Cl2 to form corresponding halide., (iii) Mond process for refining of nickel involves formation, of metal carbonyls as an intermediate., Which of the following is the correct code for the statements, above ?, (a) TTT, (b) FFT, (c) TFT, (d) FTF, , MATCHING TYPE QUESTIONS, STATEMENT TYPE QUESTIONS, 86., , 87., , 88., , Read the following statements, (i) Magnetic separation method is employed when one, component either ore or gangue is magnetic in nature., (ii) Depressant NaCN used in case of ore containing, mixture of ZnS and PbS allows ZnS to come with froth, and prevents PbS from coming to the froth., (iii) For concentration powdered bauxite ore is digested, with conc.NaOH at 473–523K and 35–36 bar pressure., Which of the following is the correct code for the statements, above ?, (a) TFT, (b) TTF, (c) FTF, (d) FFT, Which of the following statements related to Ellingham, diagrams are correct ?, (i) It provides a sound basis for the choice of reducing, agent in the reduction of oxides., (ii) Each Ellingham plot is represented by a straight line, untill unless there is some change in phase i.e., solid liquid, liquid gas and gas liquid occurs., (iii) Diagrams similar to Ellingham can be constructed for, sulphides and halides which clearly indicates why, reduction of MxS is difficult in comparison to MxO., (iv) Ellingham diagrams predicts the tendency of reduction, with a reducing agent and kinetics of the reduction, process., (a) (i), (ii) and (iii), (b) (i) and (iii), (c) (i), (ii) and (iv), (d) (ii) and (iv), Which of the following statement(s) is/are correct ?, (i) Cast iron is used in the manufacture of railway sleepers, (ii) Wrought iron is used in the manufacture of anchors,, bolts, chains etc., (iii) Nickel steel is used in making pendulums., (a) Only (i), (b) (i) and (ii), (c) (i), (ii) and (iii), (d) Only (iii), , 90., , Match the columns, Column - I, , Column - II, , (A) Fe 2 O3 .xH 2O(s), , (p) Slag formation, , Fe2O3 (s) xH 2O(g), (B) FeO SiO2, , FeSiO3, , (q) Reduction of, iron oxide, (r) Calcination, , (C) Discharge gas produced, during this process is utilised, in manufacture of H2SO4., (D) Fe2 O3 3C, 2Fe 3CO (s) Roasting, (a) A– (r), B – (p), C – (s), D – (q), (b) A– (p), B – (r), C – (s), D – (q), (c) A– (r), B – (s), C – (p), D – (q), (d) A– (r), B – (p), C – (q), D – (s), 91. Match the columns, Column - I, Column - II, (A) According to r G vsT, graph, oxide of this metal, can be easily reduced to, corresponding metal by, heating with coke, (B) Substance responsible for, the blistered appearence, of the copper obtained as, result of extraction of, copper from cuprous oxide, (C) Metal which during, purification is distilled, off and collected by, rapid chilling, (D) On addition to Al2O3 its, melting point gets reduced, and conductivity gets, enhanced, (a) A – (p), B – (q), C – (s), D – (r), (b) A – (q), B – (s), C – (p), D – (r), (c) A – (q), B – (p), C – (s), D – (r), (d) A – (q), B – (p), C – (r), D – (s), , (p) Sulphur oxide, , (q) Copper, , (r) Na3AlF6 or CaF2, , (s) Zinc
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EBD_7207, GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF ELEMENTS, , 338, , 92., , 93., , 94., , 95., , 96., , Match the columns., Column-I, (A) Blisterred Cu, (B) Blast furnace, (C) Reverberatory, furnace, (D) Hall-Heroult, process, , Column-II, (p) Aluminium, (q) 2Cu2O + Cu2S, (r) Iron, (s) FeO + SiO2, , 6Cu + SO2, FeSiO3, , (t) 2Cu2S + 3O2 2Cu2O + 2SO2, (a) A – (q), B – (r), C – (s), D – (p), (b) A – (p), B – (q), C – (r), D – (t), (c) A – (t), B – (s), C – (r), D – (q), (d) A – (s), B – (t), C – (r), D – (q), Match the columns., Column-I, Column-II, (A) Coloured bands, (p) Zone refining, (B) Impure metal to volatile, (q) Fractional, complex, distillation, (C) Purification of Ge and Si, (r) Mond Process, (D) Purification of mercury, (s) Chromatography, (t) Liquation, (a) A – (p), B – (q), C – (s), D – (t), (b) A – (s), B – (r), C – (p), D – (q), (c) A – (r), B – (s), C – (p), D – (q), (d) A – (t), B – (s), C – (r), D – (q), Match the columns., Column-I, Column-II, (A) Cyanide process, (p) Ultrapure Ge, (B) Froth Floatation Process, (q) Dressing of ZnS, (C) Electrolytic reduction, (r) Extraction of Al, (D) Zone refining, (s) Extraction of Au, (t) Purification of Ni, (a) A – (s), B – (q), C – (r), D – (p), (b) A – (q), B – (r), C – (p), D – (t), (c) A – (p), B – (q), C – (r), D – (s), (d) A – (r), B – (s), C – (t), D – (p), Match the columns, Column-I, Column-II, (A) Cyanide process, (p) Ultrapure Ge, (B) Floatation process, (q) Pine oil, (C) Electrolytic reduction, (r) Extraction of Al, (D) Zone refining, (s) Extraction of Au, (a) A – (r), B – (p), C – (s), D – (q), (b) A – (s), B – (q), C – (r), D – (p), (c) A – (r), B – (q), C – (s), D – (p), (d) A – (s), B – (p), C – (r), D – (q), Match the columns, Column-I, Column-II, (A) Distillation, (p) Zr, (B) Electrolytic refining, (q) Ga, (C) Liquation, (r) Cu, (D) Zone refining, (s) Hg, (E) Vapour phase refining, (t) Sn, , 97., , (a) A – (r), B – (s), C – (t), D – (q), E – (p), (b) A – (s), B – (r), C – (t), D – (q), E – (p), (c) A – (s), B – (t), C – (r), D – (q), E – (p), (d) A – (s), B – (r), C – (p), D – (q), E – (t), Match the columns, Column-I, Column-II, (A) This metal is used in, (p) Zinc, extraction of chromium, and manganese., (B) Common metal in brass, (q) Aluminium, and bronze., (C) Common metal in brass, (r) Copper, and german silver., (D) Substance used in making (s) Stainless steel, cycles, automobiles,, utensils, etc., (a) A – (q), B – (r), C – (p), D – (s), (b) A – (r), B – (q), C – (p), D – (s), (c) A – (q), B – (p), C – (r), D – (s), (d) A – (q), B – (r), C – (s), D – (p), , ASSERTION-REASON TYPE QUESTIONS, Directions : Each of these questions contain two statements,, Assertion and Reason. Each of these questions also has four, alternative choices, only one of which is the correct answer. You, have to select one of the codes (a), (b), (c) and (d) given below., (a) Assertion is correct, reason is correct; reason is a correct, explanation for assertion., (b) Assertion is correct, reason is correct; reason is not a, correct explanation for assertion, (c) Assertion is correct, reason is incorrect, (d) Assertion is incorrect, reason is correct., 98. Assertion : Levigation is used for the separation of oxide, ores from impurities., Reason : Ore particles are removed by washing in a current, of water., 99. Assertion : Zinc can be used while copper cannot be used, in the recovery of Ag from the complex [Ag(CN)2]–., Reason : Zinc is a powerful reducing agent than copper., 100. Assertion : Leaching is a process of reduction., Reason : Leaching involves treatment of the ore with a, suitable reagent so as to make it soluble while impurities, remains insoluble., 101. Assertion : Coke and flux are used in smelting., Reason : The phenomenon in which ore is mixed with, suitable flux and coke is heated to fusion is known as, smelting., 102. Assertion : Copper obtained after bessemerization is known, as blister copper., Reason : Blisters are produced on the surface of the metal, due to escaping of dissolved SO2., 103. Assertion : Lead, tin and bismuth are purified by liquation, method., Reason : Lead, tin and bismuth have low m.p. as compared, to impurities.
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GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF ELEMENTS, , CRITICAL THINKING TYPE QUESTIONS, 104. Copper can be extracted from, (a) Kupfernical, (b) Dolomite, (c) Malachite, (d) Galena, 105. Which of the following metal is correctly matched with its, ore?, Metal, Ore, (a) Zinc, Calamine, (b) Silver, Ilmenite, (c) Magnesium, Cassiterite, (d) Tin, Azurite, 106. Which ore contains both iron and copper?, (a) Cuprite, (b) Chalcocite, (c) Chalcopyrite, (d) Malachite, 107. Sulfide ores are common for the metals, (a) Ag, Cu and Pb, (b) Ag, Mg and Pb, (c) Ag, Cu and Sn, (d) Al, Cu and Pb, 108. Which one of the following does not occur as sulphide, ore?, (a) Zn, (b) Cr, (c) Ag, (d) Fe, 109. Pyrolusite is a/an, (a) oxide ore, (b) sulphide ore, (c) carbide ore, (d) Not an ore, 110. Sulphide ores of metals are usually concentrated by froth, flotation process. Which one of the following sulphide ores, offer an exception and is concentrated by chemical leaching?, (a) Galena, (b) Copper pyrite, (c) Sphalerite, (d) Argentite, 111. Which of the following statements is correct ?, (a) Gangues are carefully chosen to combine with the slag, present in the ore to produce easily fusible flux to carry, away the impurities, (b) Slags are carefully chosen to combine with the flux, present in the ore to produce easily fusible gangue to, carry away the impurities, (c) Gangues are carefully chosen to combine with the flux, present in the ore to produce easily fusible slag to, carry away the impurities, (d) Fluxes are carefully chosen to combine with the gangue, present in the ore to produce easily fusible slag to, carry away the impurities, 112. Carbon and CO gas are used to reduce which of the following, pairs of metal oxides for extraction of metals ?, (a) FeO, SnO, (b) SnO, ZnO, (c) BaO, Na2O2, (d) FeO, ZnO, 113. In the cyanide extraction process of silver from argentite, ore, the oxidising and reducing agents used are, (a) O2 and CO respectively, (b) O2 and Zn dust respectively, (c) HNO3 and Zn dust respectively, (d) HNO3 and CO respectively, , 339, , 114. Consider the following reactions at 1000°C, 1, A. Zn(s), 360kJ mol 1, O 2 (g), ZnO(s); G, 2, 1, O 2 (g), CO(g); G, B. C(gr), 460kJ mol –1, 2, Choose the correct statement at 1000°C, (a) zinc can be oxidised by carbon monoxide., (b) zinc oxide can be reduced by graphite, (c) carbon monoxide can be reduced by zinc., (d) both statements (a) and (b) are true, 115. Which of the following statements, about the advantage of, roasting of sulphide ore before reduction is not true?, (a) The G of of the sulphide is greater than those for CS2, and H2S., (b) The Gof is negative for roasting of sulphide ore to, oxide., (c) Roasting of the sulphide to the oxide is thermodynamically feasible., (d) Carbon and hydrogen are suitable reducing agents for, reduction of metal sulphides., 116. Which of the following statement is not correct about, Ellingham diagram?, (a), G increases with an increase in temperature, (b) It consists of plots of fGº vs T for formation of oxides, (c) a coupling reaction can be well expressed by this, diagram, (d) It express the kinetics of the reduction process, 117. A coupled reaction takes place as follow–, A + B –––– C + D,, Gº = + x kj, D + E –––– F, Gº = – y kj, for the spontaneity of reaction A + B + E ––– C+F,, which of the following is correct?, (a) 2x = y, (b) x < y, (c) x > y, (d) x = (y)× T S, 118. The value of f Gº for formation of Cr2O3 is – 540 kJmol–1, and that of Al2O3 is – 827 kJ mol–1 What is the value of, °, rG for the reaction?, 4, 3, , Al(s) +, , 2, 3, , Cr2 O 3 (s), , – 574 kJ mol–1, , 2, 3, , Al 2 O 3 (s), , 4, Cr(s)., 3, , (a), (b) –287 kJ mol–1, –1, (c) + 574 kJ mol, (d) +287 kJ mol–1, 119. Before introducing FeO in blast furnace , it is converted to, Fe2O3 by roasting so that, (a) it may not be removed as slag with silica, (b) it may not evaporate in the furnace, (c) presence of it may increase the m.pt. of charge, (d) None of these., 120. The temperature in °C at which Fe2O3 is finally reduced to, Fe in the blast furnace is, (a) 993, (b) 797, (c) 897, (d) 1597
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EBD_7207, GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF ELEMENTS, , 340, , 121. When copper ore is mixed with silica, in a reverberatory, furnace copper matte is produced. The copper matte, contains________., (a) sulphides of copper (II) and iron (II), (b) sulphides of copper (II) and iron (III), (c) sulphides of copper (I) and iron (II), (d) sulphides of copper (I) and iron (III), 122. In the metallurgy of aluminium _________., (a) Al3+ is oxidised to Al (s)., (b) graphide anode is oxidised to carbon monoxide and, carbon dioxide., (c) oxidation state of oxygen changes in the reaction at, anode., (d) oxidation state of oxygen changes in the overall, reaction involved in the process., 123. In the extraction of chlorine from brine_________., (i), (ii), (iii), (iv), (a), (c), , G for the overall reaction is negative., G for the overall reaction is positive., , for overall reaction has negative value., E for overall reaction has positive value., (i) and (ii), (b) (ii) and (iii), (i) and (iv), (d) (iii) and (iv), E, , 6Cu SO2, 124. Cu 2S 2Cu 2O, In which process of metallurgy of copper, above equation, is involved?, (a) Roasting, (b) Self reduction, (c) Refining, (d) Purification, 125. Which of the following statements regarding metallurgy of, iron is incorrect ?, (a) Reaction Fe3O 4 4CO, 3Fe 4CO 2 belongs to, lower temperature range (500 – 800K) of the blast, furnace., Fe CO 2 belongs to, (b) Reaction FeO CO, higher temperature range (900 – 1500K) of the blast, furnace., (c) The iron obtained from blast furnace is cast iron with, 3% carbon., (d) For reduction of iron oxide to occur G of the couple, of following reactions should be negative, 1, FeO(s), Fe(s), O 2 (g), 2, 1, C(s), O2 (g), CO(g), 2, 126. Extraction of which of the following is based on oxidation ?, (a) Highly reactive metals, (b) Moderately reactive metals, (c) Non-metals, (d) Both (a) and (c), , 127. Which of the following reaction(s) occur in temperature, range 500 – 800 K in blast furnace., 2FeO CO2, (i) Fe2O3 CO, 3Fe 4CO2, (ii) Fe3O4 4CO, (iii) FeO CO, Fe CO 2, 2CO, (iv) C CO 2, (b) (i), (ii) and (iii), (a) (i) and (ii), (c) (iii) and (iv), (d) (iv) only, 128. In Hall-Heroult process how much carbon anode is burnt, away to produce each 1kg of aluminium ?, (a) 0.3 kg, (b) 0.5 kg, (c) 1 kg, (d) 0.1 kg, 129. In electro-refining of metal the impure metal is used to make, the anode and a strip of pure metal as the cathode, during, the electrolysis of an aqueous solution of a complex metal, salt. This method cannot be used for refining of, (a) Silver, (b) Copper, (c) Aluminium, (d) Sodium, 130. During the process of electrolytic refining of copper, some, metals present as impurity settle as ‘anode mud’. These are, (a) Fe and Ni, (b) Ag and Au, (c) Pb and Zn, (d) Sn and Ag, 131. If the impurities in a metal has a greater affinity for oxygen, and is more easily oxidised than the metal, then the, purification of metal may be carried out by, (a) distillation, (b) zone refining, (c) electrolytic refining, (d) cupellation, 132. Germanium of very high purity is obtained by, (a) liquation, (b) vapour phase refining, (c) distillation, (d) zone refining, 133. Which of the following statements regarding electrolytic, refining of copper is incorrect ?, (a) In this process anode is made up of impure copper and, pure copper strips are taken as cathode., (b) Acidic or basic solution of copper sulphate is used as, electrolyte, (c) Antimony, tellurium, silver and gold are some of the, metals deposits as anode mud during this process, (d) Zinc can be also refined by electrolytic refining method., 134. Which of the following is incorrectly matched ?, Metal, Uses, (a) Wrought iron, Casting stoves, gutter, pipes, toys etc., (b) Copper, Coinage alloy, (c) Aluminium, Extraction of chromium and, manganese, (d) Nickel steel, Measuring tapes
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GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF ELEMENTS, , FACT / DEFINITION TYPE QUESTIONS, 1., 2., 3., 4., 5., 6., , (a), (a), (a), (a), (b), (c), , 7., , (a), , 8., 9., , (c), (c), , 10., , (d), , 11., 12., 13., , (a), (a) The formula of magnetite is Fe3O4., (c) Impurities associated with minerals are called gangue, or matrix., (c), , 14., 15., 16., 17., 18., 19., 20., 21., , 22., 23., 24., , Argentite or silver glance (Ag 2S) is an ore of Ag., Cinnabar (HgS) is an ore of Hg., Bauxite ore of aluminium is Al 2 O3.2H 2O ., Al is most abundant metal on the surface of the earth., Carborundum - SiC, Epsomite or Epsom salt - MgSO4.7H2O, Cassiterite - SnO2, Spodumene - Ore of lithium, Chalcopyrite : CuFeS2, Fool's gold : FeS2, Carnalite : KMgCl3.6H2O, Bauxite : Al2O3.2H2O, Haematite is Fe2O3. Thus it is the ore of iron (Fe)., Azurite is a basic carbonate ore of copper., 2CuCO3. Cu(OH)2, Fe3O4 – Magnetite, CuCO3·Cu(OH)2 – Malachite, Pyrolusite – MnO2 and Cassiterite – SnO2., , (b) Malachite is an ore of copper Cu OH 2 .CuCO3 ., (d) Cassiterite is an ore of Sn also known as tin stone, SnO2., (a) Galena is an ore of lead. It is PbS., (a) Gold being least reactive found native., (a), (c) Zincite is ZnO., (c) Galena is PbS and thus purified by froth floatation, method., Froth flotation method is used to concentrate sulphide, ores. This method is based on the fact that the surface, of sulphide ores is preferentially wetted by oils while, that of gangue is preferentially wetted by water., (d) Froth floatation process is used for the concentration, of sulphide ores., (b) Cassiterite contains the magnetic impurities of FeSO 4, and thus concentrated by electromagnetic separation., (d) Cyanide process is used in the metallurgy of Ag, 2Ag 2S 8 NaCN O 2 2 H 2 O, 4 Na[ Ag (CN ) 2 ] 4 NaOH 2S, 2 Na [ Ag ( CN ) 2 ] Zn, Na 2 [ Zn (CN ) 4 ] 2 Ag, , 341, , 25. (a), 26. (b) Leaching is a process used for concentration of ore. In, this process, a powdered ore is treated with a suitable, reagent (such as acids, bases or other chemicals), which can selectively dissolve the ore, but not the, impurities., 27. (d) Au and Ag can be extracted from their native ores by, leaching (Mac-Arthur Forrest cyanide process)., 28. (c), 29. (c) Cinnabar is sulphide ore (HgS). Hence purified by froth, floatation process., 30. (a) Ag is leached by cyanide process., 31. (c) Pyrolusite is MnO2. Hence not concentrated by froth, floatation process., 32. (c) Froth reduces the surface tension of water and the, solution forms froth., 33. (a) Froth flotation process is based on wetting properties, of ore particles., 34. (c) The surface of particles not wetted hence they float at, the surface, 35. (a) To remove moisture and non-metallic impurities like S,, P and As are oxidised and are removed as volatile, substances., S8 8O 2, 8SO 2 ; P4 5O2, P4 O10, , 4As 3O2 2As2O3, 36. (c) In this process sulphide ores are converted into oxide, ores., 2ZnS + 3O2 2ZnO + 2SO2, 37. (a), 38. (d), 39. (a) Carbon reduction, Fe2O3 + 3 C 2Fe + 3CO, 40. (b) Flux + Gangue Slag, 41. (a) Calcination is heating ore in absence of air to remove, moisture and volatile impurities. Carbonate ores, decomposed to corresponding oxides as a result of, calcination., 42. (d) Calcination is a process of heating a substance to a, high temperature but below the melting or fusion point,, causing loss of moisture, reduction or oxidation and, dissociation into simpler substances., 43. (b) Since silica is acidic impurity the flux must be basic., CaO SiO2, CaSiO3, 44. (b) To remove acidic impurities basic flux is added which, is CaCO3 ., 45. (d) Decomposition of carbonates and hydrated oxides., 46. (d) 2CuO CuS 3Cu SO 2 (Self - reduction), 47. (a) For example, Ag2S is converted into Na[Ag(CN)2]., When Zn is added, Ag is displaced., 48. (b)
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EBD_7207, 342, , 49., 50., 51., , 52., 53., 54., 55., 56., 57., , 58., 59., 60., , 61., , 62., , 63., 64., , 65., , 66., 67., , GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF ELEMENTS, , (c) Cresol is used as froth stabiliser., (b) Calcination involves heating when the volatile matter, escapes leaving behind the metal oxide., (b) In the graph of r G° vs T for formation of oxides, the, Cu2O line is almost at the top. So, it is quite easy to, reduce oxide ores of copper directly to the metal by, heating with coke both the lines of C, CO and C, CO 2, are at much lower temperature (500 - 600 K)., 2Cu CO, Cu 2 O C, (d), (c) Ellingham diagram normally consists of plots of fGº, Vs T for the formation of oxides of elements., (c), (b) Fe2 O3 3CO 2Fe 3CO 2, (c), (c) In blast furnace at about 1270 K, calcium carbonate is, almost completely decomposed to give CaO which acts, as a flux and combines with SiO2 present as impurity, (gangue) in the ore to form calcium silicate (fusible, slag), CaO(s) (basic flux) + SiO2 (s) (acidic flux), CaSiO3 (s) (slag), (b) These are the substances which can withstand very, high temperature without melting or becoming soft., (a), (d) Cast iron is different from pig iron and is made by, melting pig iron with scrap iron and coke using hot air, blast. It has slightly lower carbon content (about 3%), and is extremely hard and brittle., (a) Cuprous oxide formed during roasting of cuprous, sulphide is mixed with few amount of cuprous sulphide, and heated in a reverberatory furnace to get metallic, copper., 2Cu 2O Cu 2S 6Cu SO2, (b) Extraction of Zn from ZnS (Zinc blende) is achieved, by roasting followed by reduction with carbon., 2ZnS + 3O2, 2ZnO + 2SO2, ZnO + C, Zn + CO, (d) Decomposition of carbonates and hydrated oxides., (b) Aluminothermite process involves reduction of oxides, which are not satisfactorily reduced by carbon such, as Fe 2 O 3 , Mn 3 O 4 , Cr 2 O 3 , etc. to metals with, aluminium., Cr2 O3 2Al Al 2 O 3 2Cr H, ve, (c) Because Na is very reactive and cannot be extracted, by means of the reduction by C, CO etc. So it is extracted, by electrolysis., (c), (c) Fused alumina (Al2O3) is a bad conductor of electricity., Therefore, cryolite (Na3AlF6) and fluorspar (CaF2) are, added to purified alumina which not only make alumina, a good conductor of electricity but also reduce the, melting point of the mixture to around 1140 K., , 68., 71., , 74., , (a), 69. (c), 70. (a), (d) Blister-Copper contains 1 – 2 % impurities. It is, obtained after Bessemerisation of crude copper., (a), (b) Pig iron contains 4% carbon and many impurities in, smaller amount., (d), 75. (c), 76. (a), , 77., , (a), , 72., 73., , 78., , Ti 2I2, , 1700K, , TiI4, , Ti 2I 2, Pure metal, , Volatile, Stable compound, , (c) Liquation process, Mond’s process and, van Arkel, process these are the refining processes that are applied, depending upon the nature of the metal under treatment, and nature of the impurities whereas amalgamation, process is used for the extraction of noble metals like, gold, silver, etc, from native ores. The metal is, recovered from the amalgam by subjecting it to, distillation, where the mercury distils over leaving, behind the metal., Ore + Hg, , 79., , 523K, , Amalgam Distilled, , Hg-vapours, , Metal, (b) Zr and Ti are purified by van Arkel method., 870K, , Zr(s ) 2I2 ( g ), , ZrI4 ( g ), , ZrI4 ( g ), , 2075K, Tugsten filament, , Ti(s) + 2I2(s), , 523K, , Zr(s ) 2I 2 ( g ), , TiI4(g), , 1700K, , Ti(s) + 2I2(g), Pure titanium, , 80., , 81., , 82., 84., 85., , (a) Zone refining is based on the difference in solubility, of impurities in molten and solid state of the metal., This method is used for obtaining metals of very high, purity., (d) Si obtained by reduction of SiCl4 with H2 is further, purified by zone refining method to get Si of very high, purity. Silicon is purified by zone-refining process, because the impurities present in it are more soluble in, the liquid phase than in the solid phase., (c), 83. (a), (c) Mobile phase cannot be solid., (b) Zinc dust is used as a reducing agent in the manufacture, of dye-stuffs, paints etc., , STATEMENT TYPE QUESTIONS, 86., , 87., , (a) For ore containing mixture of ZnS and PbS, depressant, NaCN allows PbS to come with froth and prevents, ZnS from coming to the froth., (b) Ellingham diagram represents plot between G and T, therefore with increase in temperature phase change, Gas Liquid is not possible. Ellingham diagram does, not give any information about kinetics of the reduction, reaction.
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GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF ELEMENTS, , 88., 89., , (c), (c) Van Arkel method involves heating crude Zr with, iodine to form corresponding iodide. The metal iodide, being more covalent volatilises., , MATCHING TYPE QUESTIONS, 90., 95., , 96., , (a), 91. (c), 92. (a), 93. (b), 94. (a), (b) Cyanide process is for gold (A - s); floatation process, - pine oil (B - q); Electrolytic reduction - Al (C - r);, Zone refining -Ge (D - p)., (b), 97. (a), , ASSERTION-REASON TYPE QUESTIONS, 98., , 99., 100., 101., , 102., 103., , (c) Assertion is true but reason is false., Oxide ores being heavier than the earthy or rocky, gangue particles, settle down while lighter impurities, are washed away., (a), (d) Assertion is false but reason is true. Leaching is a, process of concentration., (b) Both assertion and reason are true but reason is not, the correct explanation of assertion. Non fusible mass, present in ore in mixing with suitable flux are fused, which are then reduced by coke to give free metal., (a) Both assertion and reason are correct and reason is, the correct explanation of assertion., (a), , CRITICAL THINKING TYPE QUESTIONS, 104. (c) Malachite is CuCO3 . Cu(OH)2 it is ore of copper., 105. (a) (a), Zinc, Calamine is ZnCO3, (b), Silver, Ilmenite is FeTiO3, (c), Magnesium, Cassiterite is SnO2, (d), Tin, Azurite is, [2CuCO3.Cu (OH)2], 106. (c) Cuprite : Cu2O; Chalcocite : Cu2S; Chalcopyrite : CuFeS2;, Malachite: Cu(OH)2 .CuCO3. We see that CuFeS2, contains both Cu and Fe., 107. (a) Silver, copper and lead are commonly found in earth's, crust as Ag2S (silver glance), CuFeS2 (copper pyrites), and PbS (galena), 108. (b) Except chromium all the given metals exists as their, sulphides., Zn exists as zinc blende ZnS., Silver exists as silver glance Ag2S., Iron exists as iron pyrites FeS2., Mercury exists as mercuric sulphide HgS., 109. (a) MnO2 is pyrolusite (oxide ore)., , 343, , 110. (d) Leaching is the selective dissolution of the desired, mineral leaving behind the impurities in a suitable, dissolving agent e.g.,, Argentitie or Silver glance, Ag2S is an ore of silver., Silver is extracted from argentite by the mac-Arthur, and Forest process (leaching process)., Ag 2S 4NaCN, , 2Na[Ag CN 2 ] Na 2S, , 4Au 8KCN 2H 2O O 2, , 4K[Au CN 2 ] 4KOH, , 111. (d), 112. (d), 113. (b) The reactions involved in cyanide extraction process, are :, Ag 2 S, (argentite), , + 4NaCN, , 2Na [Ag(CN)2] + Na2S, , 4Na2S + 5O 2 + 2H2O, , 2Na2SO4 + 4NaOH + 2S, , Oxiding, agent, , 2Na[Ag(CN)2] +, , Zn, , (reducing, agent), , Na2 [Zn(CN)4] + 2, , Ag, 114. (b), 115. (d) The sulphide ore is roasted to oxide before reduction, because the Gof of most of the sulphides are greater, than those of CS2 and H2S, therefore neither C nor H, can reduce metal sulphide to metal. Further, the, standard free energies of formation of oxide are much, less than those of SO2. Hence oxidation of metal, sulphides to metal oxide is thermodynamically, favourable., 116. (d) Ellingham diagrams are based on thermodynamic, concepts. It does not tell anything about the kinetics, of the reduction process., 117. (d) For a spontaneous reaction , Gº must be negative, and it can be possible only in this case when x < y, 118. (b) The two equation are:, 4, Al(s) O 2 (g), 3, , 2, Al2 O3 (s),, 3, , f Gº, , 827kJ mol 1, , … (1), 4, Cr(s) O 2 (g), 3, , 2, Cr2 O3 (s),, 3, , f Gº, , 540kJ mol 1, , … (2), Subtracting equation (ii) from equation (i) we have,, 4, 2, Al(s), Cr2O3 (s), 3, 3, , 2, Al 2O 3 (s), 3, rG, , 4, Cr(s),, 3, , 287kJ mol, , 1
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EBD_7207, GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF ELEMENTS, , 344, , 119. (a) FeO is capable forming slag with SiO2, SiO 2, , FeO, , FeSiO 3, , 120. (a) In blast furnace Fe2O3 is finally reduced to Fe at 993°C, 121. (c), , 122. (b), , 123. (b), , 124. (b) This process is also called autoreduction process or, air reduction process. The sulphide ores of less, electropositive metals are heated in air to convert part, of the ore into oxide or sulphate which then react with, the remaining sulphide ore to give the metal and, sulphur dioxide., , 2Cu 2S 3O2, Cu 2S 2Cu 2O, , 2Cu 2O 2SO2, 6Cu SO2, , 125. (c) The iron obtained from blast furnace is pig iron with, 4% carbon and impurities like S, P, Mn etc., in small, amount., , 126. (c) Extraction of non-metals are based on oxidation. For, example extraction of chlorine from brine., 2Cl–(aq) + 2H2O(l) 2OH–(aq) + H2(g) + Cl2(g), 127. (a) (iii) and (iv) reactions occur in the temperature range of, 900 – 1500K in blast furnace., 128. (b) For each kg of Al produced, about 0.5 kg of carbon, anode is burnt away., 129. (d) Na reacts vigorously with water (exothermic process ), 130. (b) During the process of electrolytic refining of copper, Ag and Au are obtained as anode mud., 131. (d), 132. (d) Metals of high purity are obtained by zone refining, e.g., silicon, germanium, boron, gallium, indium., 133. (b) During electrolytic refining of copper electrolyte used, is acidified solution of copper sulphate., 134. (a) Wrought iron is used in making anchors, wires, bolts, chains and agricultural implements.
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21, THE p-BLOCK ELEMENTS, (GROUP 15, 16, 17 AND 18), FACT / DEFINITION TYPE QUESTIONS, 1., , 2., , 3., 4., 5., , 6., , 7., 8., , 9., , Ionic radii (in Å) of As3+, Sb3+ and Bi3+ follow the order, (a) As3+ > Sb3+ > Bi3+, (b) Sb3+ > Bi3+ >As3+, 3+, 3+, 3+, (c) Bi > As > Sb, (d) Bi3+ > Sb3+ > As3+, Which of the following statements is not correct for, nitrogen?, (a) Its electronegativity is very high, (b) d-orbitals are available for bonding, (c) It is a typical non-metal, (d) Its molecular size is small, Collectively the elements of group 15 are called –, (a) pnicogens, (b) pnicopens, (c) nicopen, (d) None of these, Which one of the following elements is most metallic ?, (a) P, (b) As, (c) Sb, (d) Bi, Which of the following statement is incorrect for group 15, elements ?, (a) Order of ionization enthalpies is, H, H, H, i 1, i 2, i 3, (b) The boiling point and melting point increases from top, to bottom in the group, (c) Dinitrogen is a gas while all others are solids, (d) All statements are correct, Which of the follow group 15 element forms metallic bonds, in elemental state ?, (a) As, (b) P, (c) Sb, (d) Bi, The three important oxidation states of phosphorus are, (a) –3, +3 and +5, (b) –3, +3 and –5, (c) –3, +3 and +2, (d) –3, +3 and +4, Nitrogen is relatively inactive element because, (a) its atom has a stable electronic configuration, (b) it has low atomic radius, (c) its electronegativity is fairly high, (d) dissociation energy of its molecule is fairly high, Which of the following has the highest p – p bonding, tendency ?, (a) N, (b) P, (c) As, (d) Sb, , 10. Pick out the wrong statement., (a) Nitrogen has the ability to form p -p bonds with itself., (b) Bismuth forms metallic bonds in elemental state., (c) Catenation tendency is higher in nitrogen when, compared with other elements of the same group., (d) Nitrogen has higher first ionisation enthalpy when, compared with other elements of the same group., 11. Nitrogen forms N2, but phosphorus is converted into P4, from P, the reason is, (a) Triple bond is present between phosphorus atom, (b) p – p bonding is strong, (c) p – p bonding is weak, (d) Multiple bond is formed easily, 12. What causes nitrogen to be chemically inert ?, (a) Multiple bond formation in the molecule, (b) Absence of bond polarity, (c) Short internuclear distance, (d) High bond energy, 13. Among the 15th group elements, as we move from nitrogen, to bismuth, the pentavalency becomes less pronounced, and trivalency becomes more pronounced due to, (a) Non metallic character (b) Inert pair effect, (c) High electronegativity (d) Large ionization energy, 14. Pentavalence in phosphorus is more stable when compared, to that of nitrogen even though they belong to same group., This is due to, (a) dissimilar electronic configuration, (b) due to presence of vacant d-orbitals, (c) reactivity of phosphorus, (d) inert nature of nitrogen, 15. Which one has the lowest boiling point ?, (a) NH3, (b) PH3, (c) AsH3, (d) SbH3, 16. Most acidic oxide among the following is –, (a) N2O5, (b) P2O5, (c) N2O4, (d) As 2 O 3, 17. Which of the following species has the highest dipole moment?, (a) NH3, (b) PH3, (c) AsH3, (d) SbH3
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EBD_7207, THE p-BLOCK ELEMENTS (GROUP 15, 16, 17 AND 18), , 346, , 18., , 19., , 20., , 21., , 22., , 23., , 24., , 25., , 26., , 27., , 28., , The correct decreasing order of basic strength is:, (a) AsH3 SbH3 PH3 NH3, (b) SbH 3, , AsH 3, , (c), , PH 3, , NH 3, , PH 3, , NH 3, , AsH 3, , SbH 3, , (d) PH 3 AsH 3 SbH 3 NH 3, Which of the following fluorides does not exist?, (a) NF5, (b) PF5, (c) AsF 5, (d) SbF5, The p-block element of group 15 that forms predominantly, basic oxide is, (a) N, (b) P, (c) As, (d) Bi, With respect to protonic acids, which of the following, statements is correct ?, (a) PH3 is more basic than NH3, (b) PH3 is less basic than NH3, (c) PH3 is equally basic as NH3, (d) PH3 is amphoteric while NH3 is basic, PCl5 is possible but NCl5 does not exist :, (a) in N, d-sub-shell is absent, (b) ionization energy of N is very high, (c) it does not like Cl, (d) None of these, Maximum covalency of nitrogen is __________., (a) 3, (b) 5, (c) 4, (d) 6, Elements of group-15 form compounds in +5 oxidation state., However, bismuth forms only one well characterised, compound in +5 oxidation state. The compound is, (a) Bi2O5, (b) BiF5, (c) BiCl5, (d) Bi2S5, Pure nitrogen is prepared in the laboratory by heating a mixture, of, (a) NH4OH + NaCl, (b) NH4NO3 + NaCl, (c) NH4Cl + NaOH, (d) NH4Cl + NaNO2., On heating ammonium dichromate and barium azide, separately we get, (a) N2 in both cases, (b) N2 with ammonium dichromate and NO with barium, azide, (c) N2O with ammonium dichromate and N2 with barium, azide, (d) N2O with ammonium dichromate and NO2 with barium, azide, In Haber’s process for the manufacture of NH 3 :, (a) finely divided nickel is used as a catalyst, (b) finely divided iron is used as a catalyst, (c) finely divided molybdenum is used as a catalyst, (d) no catalyst is necessary, Ammonia on reaction with hypochlorite anion can form :, (a) NO, (b) N2H4, (c) NH4Cl, (d) Both (b) and (c), , 29., , 30., , 31., , 32., , 33., , 34., , 35., , 36., , 37., , 38., , 39., , 40., , NH3 gas is dried over :, (a) CaO, (b) HNO3, (c) P2O5, (d) CuSO4, The shape of ammonia molecule is, (a) tetrahedral, (b) pyramidal, (c) planar triangle, (d) octahedral, When ammonia is heated with cupric oxide, a molecule of, ammonia will, (a) gain 3 electrons, (b) lose 3 electrons, (c) gain 2 electrons, (d) lose 2 electrons, In which the NH3 is not used ?, (a) Cold storage, (b) Anaesthetic, (c) Manufacture of rayon and plastic, (d) None of these, Liquid ammonia bottles are opened after cooling them in ice, for sometime. It is because liquid NH3, (a) Brings tears to the eyes, (b) Has a high vapour pressure, (c) Is a corrosive liquid, (d) Is a mild explosive, Ammonia is generally manufactured for fertilizers by the, reaction, (a) 2 NH 4 Cl Ca (OH ) 2 CaCl 2 2H 2 O 2 NH 3, (b) By passing an electric discharge in a mixture of N2 and, H2, (c) By passing a mixture of N2 and H2 under high pressure, and moderate temperature over a catalyst, (d) None of these, Nitrogen dioxide cannot be obtained by heating :, (a) KNO3, (b) Pb(NO3)2, (c) Cu(NO3)2, (d) AgNO3, Which of the following oxides is neutral ?, (a) N2O3, (b) N2O4, (c) N2O5, (d) N2O, The bonds present in N2O5 are :, (a) only ionic, (b) covalent and coordinate, (c) only covalent, (d) covalent and ionic, Which of the following oxides of nitrogen is a coloured, gas?, (a) N2O, (b) NO, (c) N2O5, (d) NO2, Which of the following shows nitrogen with its increasing, order of oxidation number?, (a) NO < N2O < NO2 < NO3– < NH4+, (b) NH4+ < N2O < NO2 < NO3– < NO, (c) NH4+ < N2O < NO < NO2 < NO3–, (d) NH4+ < NO < N2O < NO2 < NO3–, In which one of the following oxides of nitrogen, one, nitrogen atom is not directly linked to oxygen?, (a) NO, (b) N2O4, (c) N2O, (d) N2O3
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p-BLOCK ELEMENTS (GROUP 15, 16, 17 AND 18), , 41., , 42., , 43., , 44., , 45., , 46., , 47., , 48., , 49., , Which of the following oxides of nitrogen reacts with FeSO4, to form a dark brown compound, (a) N2O, (b) NO, (c) NO2, (d) N2O3, Which oxide of nitrogen is obtained on heating ammonium, nitrate at 250ºC ?, (a) Nitric oxide, (b) Nitrous oxide, (c) Nitrogen dioxide, (d) Dinitrogen tetraoxide, Which of the following can be used as an anaesthesia ?, (a) N2O, (b) NO, (c) NCl3, (d) NO2, A deep brown gas is formed by mixing two colourless gases, which are, (a) NO2 and O2, (b) N2O and NO, (c) NO and O2, (d) NH3 and HCl, Which of the following elements does not form stable, diatomic molecules ?, (a) Iodine, (b) Phosphorus, (c) Nitrogen, (d) Oxygen, The catalyst used in the manufacture of HNO 3 by, Ostwald’s process is :, (a) platinum gauze, (b) vanadium pentoxide, (c) finely divided nickel (d) platinum black ., Concentrated nitric acid, upon long standing, turns yellow, brown due to the formation of, (a) NO, (b) NO2, (c) N2O, (d) N2O4, Which of the following trihalide is unstable?, (a) NF3`, (b) AsCl3, (c) SbBr3, (d) NCl3, What will be the A and B in the following equations., 8NH3 3Cl2, A, , NH3 3Cl 2, B, , 50., , 51., , 52., , 6NH 4Cl N 2, NCl3 3HCl, , (a) A = Excess, B = Excess, (b) A = Limited, B = Excess, (c) A = Excess, B = Limited, (d) A = Limited, B = Limited, Which of the following is the strongest reducing agent ?, (a) NH3, (b) PH3, (c) BiH3, (d) SbH3, Which of the following element will form acidic oxides of, type E2O3?, (a) As, (b) Sb, (c) Bi, (d) P, Which one of the following is not an use of ammonia ?, (a) To produce various nitrogenous fertilizers., (b) In manufacture of nitric acid, (c) As a refrigerate, (d) In the pickling of stainless steel, , 347, , 53. The nitrogen oxides that contain(s) N–N bond(s) is /are, (i) N2O, (ii) N2O3, (iii) N2O4, (iv) N2O5, (a) (i) , (ii), (b) (ii) , (iii) , (iv), (c) (iii) , (iv), (d) (i), (ii) and (iii), 54. Zinc on reaction with dilute HNO3 gives x and zinc on, reaction with concentrated HNO3 gives y. Identify x and y., (a) x = NO2 , y = N2O, (b) x = N2O , y = NO, (c) x = NO , y = NO2, (d) x = N2O , y = NO2, 55. Which of the following is incorrect for white and red, phosphorus ?, (a) They are both soluble in CS2, (b) They can be oxidised by heating in air, (c) They consist of the same kind of atoms, (d) They can be converted into one another, 56. Which of the following phosphorus is most reactive ?, (a) Red phosphorus, (b) White phosphorus, (c) Scarlet phosphorus, (d) Violet phosphorus, 57. White phosphorus is, (a) a monoatomic gas, (b) P4, a tetrahedral solid, (c) P8, a crown, (d) a linear diatomic molecule, 58. Which property of white phosphorus is common to, red phosphorous ?, (a) It burns when heated in air., (b) It reacts with hot caustic soda solution to give, phosphine., (c) It shows chemiluminescence., (d) It is soluble in carbon disulphide., 59. Which of the following statements regarding allotropic, forms of phosphorus is incorrect?, (a) White phosphorus is more reactive than red and black, due to high angular strain., (b) Red phosphorus on heating catches fire and give, dense red fumes of P4O10., (c) Red phosphorus is polymeric in nature consisting of, chains of P4 tetrahedral., (d) Black phosphorus has two forms -black and -black, phosphorus, 60. Which of the following is incorrect?, (a) M.p of monoclinic sulphur > m.p. of rhombic sulphur., (b) Specific gravity of rhombic sulphur > specific gravity, of monoclinic sulphur., (c) Monoclinic sulphur is stable below 369 K., (d) Both rhombic sulphur and monoclinic sulphur have, S8 molecules., 61. One mole of calcium phosphide on reaction with excess, water gives, (a) one mole of phosphine, (b) two moles of phosphoric acid, (c) two moles of phosphine, (d) one mole of phosphorus pentoxide, 62. PH3, the hydride of phosphorus is, (a) metallic, (b) ionic, (c) non-metallic, (d) covalent
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EBD_7207, THE p-BLOCK ELEMENTS (GROUP 15, 16, 17 AND 18), , 348, , 63., , 64., , 65., , Phosphine is not obtained by which of the following reaction, (a) White P is heated with NaOH, (b) Red P is heated with NaOH, (c) Ca3P2 reacts with water, (d) Phosphorus trioxide is boiled with water, Phosphine is not evolved when, (a) white phosphorus is boiled with a strong solution of, Ba(OH)2, (b) phosphorus acid is heated, (c) calcium hypophosphite is heated, (d) metaphosphoric acid is heated., Pure phosphine is not combustible while impure phosphine, is combustible, this combustibility is due to presence of, (a), , 66., , 67., , 68., , 69., , 70., , 71., , O, , O, , H, , P, H, , OH, , (b) H, , 72., , 73., , 74., , HO, , 76., , 77., , OH, , (d) H, , 79., , 80., , 81., , 82., , 83., , OH, , O, , P, , 78., , P, OH, , O, (c), , 75., , (b) N2, , P2 H 4, , (c) PH 5, (d) P2 O5, When orthophosphoric acid is heated to 600°C, the product, formed is, (a) PH3, (b) P2O5, (c) H3PO3, (d) HPO3, P2O5 is heated with water to give, (a) hypophosphorous acid(b) phosphorous acid, (c) hypophosphoric acid (d) orthophosphoric acid, Basicity of orthophosphoric acid is, (a) 2, (b) 3, (c) 4, (d) 5, PCl3 reacts with water to form, (a) PH3, (b) H3PO4 and HCl, (c) POCl3, (d) H3PO4, H3PO2 is the molecular formula of an acid of phosphorus., Its name and basicity respectively are, (a) phosphorus acid and two, (b) hypophosphorous acid and two, (c) hypophosphorous acid and one, (d) hypophosphoric acid and two, The structural formula of hypophosphorous acid is, , (a), , (a) zero, (c) three, , 84., , P, , OOH, OH, OH, Number of sigma bonds in P4O10 is, (a) 6, (b) 7, (c) 17, (d) 16., The number of hydrogen atom(s) attached to phosphorus, atom in hypophosphorous acid is, (a) three, (b) one, (c) two, (d) zero, The number of P – O – P bonds in cyclic metaphosphoric, acid is, , 85., , 86., , 87., , (b) two, (d) four, , Oxidation states of P in H4 P2O5 , H4 P2O6 , and H4 P2O7 ,, are respectively:, (a) + 3, + 5, + 4, (b) + 5, + 3, + 4, (c) + 5, + 4, + 3, (d) + 3, + 4, + 5, How many bridging oxygen atoms are present in P4O10?, (a) 5, (b) 6, (c) 4, (d) 2, Which of the following statements is not valid for oxoacids, of phosphorus?, (a) Orthophosphoric acid is used in the manufacture of, triple superphosphate., (b) Hypophosphorous acid is a diprotic acid., (c) All oxoacids contain tetrahedral four coordinated, phosphorus., (d) All oxoacids contain atleast one P = O and one P —, OH group., What is hybridization of P in PCl5 ?, (a) sp3, (b) sp3 d 2, 3, (c) sp d, (d) sp2, Which of the following is a cyclic phosphate ?, (a) H3 P3 O10, (b) H6 P4 O13, (c) H5 P5 O15, (d) H7 P5 O16, P—O—P bond is present in, (a) H4P2O6, (b) H4P2O5, (c) Both (a) and (b), (d) Neither (a) nor (b), Orthophosphoric acid is, (a) monobasic, (b) dibasic, (c) tribasic, (d) tetrabasic, The oxyacid of phosphorous in which phosphorous has, the lowest oxidation state is, (a) hypophosphorous acid, (b) orthophosphoric acid, (c) pyrophosphoric acid, (d) metaphosphoric acid, The number of P—O—P bonds in cyclic metaphosphoric, acid is, (a) zero, (b) two, (c) three, (d) four, Among the oxyacids of phosphorus, the dibasic acid is, (a) H4P2O7, (b) H3PO2, (c) HPO3, (d) H3PO3, The basicity of pyrophosphorus acid is, (a) 2, (b) 4, (c) 1, (d) 5, The, oxidation, state, of, phosphorus, in, cyclotrimetaphosphoric acid is, (a) +3, (b) +5, (c) –3, (d) +2, Which acid has P – P linkage ?, (a) Hypophosphoric acid (b) Pyrophosphoric acid, (c) Metaphosphoric acid (d) Orthophosphoric acid
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p-BLOCK ELEMENTS (GROUP 15, 16, 17 AND 18), , 88., , 89., , 90., , 91., , 92., , 93., , 94., , 95., , 96., , 97., , 98., , 99., , In a cyclotrimetaphosphoric acid molecule, how many single, and double bonds are present?, (a) 3 double bonds; 9 single bonds, (b) 6 double bonds; 6 single bonds, (c) 3 double bonds; 12 single bonds, (d) Zero double bonds; 12 single bonds, Strong reducing behaviour of H3PO2 is due to, (a) Low oxidation state of phosphorus, (b) Presence of two –OH groups and one P–H bond, (c) Presence of one –OH group and two P–H bonds, (d) High electron gain enthalpy of phosphorus, In solid state PCl5 is a ________., (a) covalent solid, (b) octahedral structure, (c) ionic solid with [PCl6 ]+ octahedral and [PCl 4] –, tetrahedra, (d) ionic solid with [PCl4 ]+ tetrahedral and [PCl 6 ]–, octahedra, Electron affinity of sulphur is, (a) more than O and Se, (b) more than O but less than Se, (c) less than O but more than Se, (d) equal to O and Se, All the elements of oxygen family are, (a) non metals, (b) metalloids, (c) radioactive, (d) polymorphic, Which shows maximum catenation property ?, (a) S, (b) Se, (c) Te, (d) O, Oxygen and sulphur both are the members of the same group, in periodic table but H2O is liquid while H2S is gas because, (a) molecular weight of water is more, (b) electronegativity of sulphur is more, (c) H2S is weak acid, (d) water molecules are having weak hydrogen bonds, between them, Which of the following hydrides has the lowest boiling, point?, (a) H2O, (b) H2S, (c) H2Se, (d) H2Te, Which of the following hydrides is most acidic ?, (a) H2Te, (b) H2Se, (c) H2O, (d) H2S, Which of the following hydrides shows the highest boiling, point ?, (a) H2O, (b) H2S, (c) H2Se, (d) H2Te, Which is the best oxidising agent among the following ?, (a) S, (b) O, (c) Se, (d) Te, Which of the following oxide is amphoteric ?, (a) SnO2, (b) CaO, (c) SiO2, (d) CO2, , 349, , 100. Which of the following is not correctly matched ?, (a) SF4 – gas, (b) SeF4 – liquid, (c) TeF4 – solid, (d) SF6 – solid, 101. The compound which gives off oxygen on moderate heating, is :, (a) cupric oxide, (b) mercuric oxide, (c) zinc oxide, (d) aluminium oxide, 102. Oxygen molecule is, (a) diamagnetic with no-unpaired electron(s), (b) diamagnetic with two unpaired electrons, (c) paramagnetic with two unpaired electrons, (d) paramagnetic with no unpaired electron(s), 103. The number of electrons that are paired in oxygen molecule, are, (a) 16, (b) 12, (c) 14, (d) 7, 104. On heating KClO3 we get, (a) KClO2 + O2, (b) KCl + O2, (c) KCl + O3, (d) KCl + O2 + O3, 105. Which of the following is not oxidized by O3 ?, (a) KI, (b) FeSO4, (c) KMnO4, (d) K2MnO4, 106. About 20 km above the earth, there is an ozone layer. Which, one of the following statements about ozone and ozone, layer is true?, (a) Ozone has a triatomic linear molecule, (b) It is harmful as it stops useful radiation, (c) It is beneficial to us as it stops U.V radiation, (d) Conversion of O3 to O2 is an endothermic reaction, 107. Oxygen gas can be prepared from solid KMnO4 by :, (a) treating the solid with H 2 gas, (b) strongly heating the solid, , 108., , 109., , 110., , 111., , 112., , (c) dissolving the solid in dil. H 2SO 4, (d) dissolving solid in dil. HCl, Which of the following statements is correct :, (a) Ozone is a resonance hybrid of oxygen, (b) Ozone is an isomer of oxygen, (c) Ozone has no relationship with oxygen, (d) Ozone is an allotropic modification of oxygen, Which of the following on thermal decomposition gives, oxygen gas ?, (a) Ag2 O, (b) Pb3O4, (c) PbO2, (d) All of these, Which of the following is an acidic oxide?, (a) Mn 2O7, (b) Na2O, (a) N2O, (b) BaO, Atomicity of sulphur in rhombic sulphur is, (a) 1, (b) 2, (c) 8, (d) 6, Which of the following form of the sulphur shows, paramagnetic behaviour ?, (a) S8, (b) S6, (c) S2, (d) All of these
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EBD_7207, THE p-BLOCK ELEMENTS (GROUP 15, 16, 17 AND 18), , 350, , 113. What is X in the following reaction ?, X, , 114., , 115., , 116., , 117., , 118., , 119., , 120., , 121., , 122., , 2SO2(g) + O2(g), 2SO3(g), (a) V2O5, (b) CuO, (c) CuCl2, (d) MnO2, Which of the following oxo acid of sulphur has O–O bond ?, (a) H2S2O7, (b) H2S2O8, (c) H2S2O6, (d) H2S2O5, Carbohydrates on reaction with conc. H2SO4 becomes, charred due to, (a) hydrolysis, (b) dehydration, (c) hydration, (d) oxidation, Which of the following is the key step in the manufacture of, sulphuric acid ?, (a) Burning of sulphur or sulphide ores in air to generate, SO2, (b) Conversion of SO2 to SO3 by the reaction with oxygen, in presence of catalyst., (c) Absorption of SO3 in H2SO4 to give oleum., (d) Both (b) and (c), Hybridization of S in SO3 is, (a) sp 2, (b) sp 3, 2, (c) sp d, (d) sp 3 d 2, By which of the following SO2 is formed ?, (a) Reaction of dil. H2SO4 with O2, (b) Hydrolysis of dil. H2SO4, (c) Reaction of conc. H2SO4 with Cu, (d) None of these, Number of bonds in SO2 are, (a) two and two, (b) two and one, (c) two and three, (d) None of these, Bleaching action of SO2 is due to its, (a) oxidising property, (b) acidic property, (c) reducing property, (d) basic property, The acid which has a peroxy linkage is, (a) Sulphurous acid, (b) Pyrosulphuric acid, (c) Dithionic acid, (d) Caro’s acid, S – S bond is not present in, (a), , S 2 O 24, , (b), , S2 O 52, , (c), , S2 O 32, , (d), , S2 O 27, , 123. Oleum is, (a) castor Oil, (b) oil of vitriol, (c) fuming H 2SO 4, (d) None of them, 124. On addition of conc. H2SO4 to a chloride salt, colourless, fumes are evolved but in case of iodide salt, violet fumes, come out. This is because, (a) H2SO4reduces HI to I2, (b) HI is of violet colour, (c) HI gets oxidised to I2, (d) HI changes to HIO3, 125. Which of the following are peroxoacids of sulphur?, (a) H2SO5 and H2S2O8 (b) H2SO5 and H2S2O7, (c) H2S2O7 and H2S2O8 (d) H2S2O6 and H2S2O7, , 126. Hot conc. H2SO4 acts as moderately strong oxidising agent., It oxidises both metals and nonmetals. Which of the, following element is oxidised by conc. H2SO4 into two, gaseous products?, (a) Cu, (b) S, (c) C, (d) Zn, 127. Caro’s acid is, (a) H2SO3, (b) H3S2O5, (c) H2SO5, (d) H2S2O8, 128. Sulphuric acid reacts with PCl5 to give, (a) thionyl chloride, (b) sulphur monochloride, (c) sulphuryl chloride, (d) sulphur tetrachloride, 129. Which one of the following reacts with conc. H2SO4?, (a) Au, (b) Ag, (c) Pt, (d) Pb, 130. The number of dative bonds in sulphuric acid molecule is, (a) 0, (b) 1, (c) 2, (d) 4, 131. What is the number of sigma ( ) and pi ( ) bonds present, in sulphuric acid molecule ?, (a) 6 , 2, (b) 6 , 0, (c) 2 , 4, (d) 2 , 2, 132. Which characteristic is not correct about H2SO4 ?, (a) Reducing agent, (b) Oxidising agent, (c) Sulphonating agent (d) Highly viscous, 133. Among F, Cl, Br and I the lowest ionization potential will be, of, (a) fluorine, (b) chlorine, (c) bromine, (d) iodine, 134. The electronegativity follows the order, (a) F > O > Cl > Br, (b) F > Cl > Br > O, (c) O > F > Cl > Br, (d) Cl > F > O > Br, 135. The bond energies of F2 , Cl2, Br2 and I2 are 155, 244, 193, and 151 kJ mol–1 respectively. The weakest bond will be in, (a) Br2, (b) Cl2, (c) F2, (d) I2, 136. The outer electronic structure of 3s2 3p5 is possessed by, (a) O, (b) Cl, (c) Br, (d) Ar, 137. Electron gain enthalpy with negative sign of fluorine is less, than that of chlorine due to :, (a) High ionization enthalpy of fluorine, (b) Smaller size of chlorine atom, (c) Smaller size of fluorine atom, (d) Bigger size of 2p orbital of fluorine, 138. Which one of the following order is correct for the bond, energies of halogen molecules ?, (a) I2 > Cl2 > Br2, (b) Br2 > Cl2 > I2, (c) I2 > Br2 > Cl2, (d) Cl2 > Br2 > I2, 139. The correct order of reactivity of halogens with alkalies is, (a) F > Cl > Br > I, (b) F < Cl > Br < I, (c) F < Cl < Br < I, (d) F < Cl < Br > I
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p-BLOCK ELEMENTS (GROUP 15, 16, 17 AND 18), , 140. The correct order of increasing oxidising power is, (a) F2 > Br2 > Cl2 > I2, (b) F2 < Cl2 < Br2 < I2, (c) Cl2> Br2 > F2 > I2, (d) I2 < Br 2 < Cl2 < F2, 141. Fluorine is a stronger oxidising agent than chlorine in, aqueous solution. This is attributed to many factors except, (a) heat of dissociation (b) ionisation potential, (c) heat of hydration, (d) electron affinity, 142. Fluorine exhibits an oxidation state of only –1 because, (a) it can readily accept an electron, (b) it is very strongly electronegative, (c) it is a non-metal, (d) it belongs to halogen family, 143. Which of the following halogen does not exhibit positive, oxidation state in its compounds?, (a) Cl, (b) Br, (c) I, (d) F, 144. The halogen that is most easily reduced is, (a) F2, (b) Cl2, (c) Br2, (d) I2, 145. Which one of the following elements shows more than one, oxidation states ?, (a) Sodium, (b) Fluorine, (c) Chlorine, (d) Potassium, 146. Which of the following halogens exhibit only one oxidation, state in its compounds ?, (a) Bromine, (b) Chlorine, (c) Fluorine, (d) Iodine, 147. Which of the following is the best description for the, behaviour of bromine in the reaction given below ?, H 2 O Br2, HOBr HBr, (a) Proton acceptor only, (b) Both oxidized and reduced, (c) Oxidized only, (d) Reduced only, 148. Among the following which is the strongest oxidising agent?, (a) Br2, (b) I2, (c) Cl2, (d) F2, 149. The correct order of heat of formation of halogen acids is, (a) HI > HBr > HCl > HF (b) HF > HCl > HBr > HI, (c) HCl > HF > HBr > HI (d) HCl > HBr > HF > HI, 150. Which is the weakest out of HF, HCl, HBr and HI?, (a) HF, (b) HCl, (c) HBr, (d) HI, 151. Which of the following is most volatile ?, (a) HI, (b) HBr, (c) HCl, (d) HF, 152. At room temperature, HCl is a gas while HF is a low boiling, liquid. This is because, (a) H- F bond is covalent (b) H- F bond is ionic, (c) HF has metallic bond (d) HF has hydrogen bond, 153. The bleaching action of chlorine is due to, (a) reduction, (b) hydrogenation, (c) chlorination, (d) oxidation, , 351, , 154. Cl2 reacts with hot and conc. NaOH to give –, (a) NaClO, (b) NaClO3, (c) NaClO2, (d) NaClO4, 155. When chlorine reacts with cold and dilute solution of sodium, hydroxide, the products obtained are, (a) Cl– + ClO–, , 156., , 157., , 158., , 159., , (b) Cl– + ClO 2, , (c) Cl– + ClO3, (d) Cl– + ClO 4, Chlorine is liberated when we heat, (a) KMnO4 + NaCl, (b) K2Cr2O7 + MnO2, (c) Pb(NO3)2 + MnO2, (d) K2Cr2O7 + HCl, Which of the following is used in the preparation of chlorine?, (a) Only MnO2, (b) Only KMnO4, (c) Both MnO2 and KMnO4, (d) Either MnO2 or KMnO4, The reaction of KMnO4 and HCl results in, (a) oxidation of Mn in KMnO4 and production of Cl2, (b) reduction of Mn in KMnO4 and production of H2, (c) oxidation of Mn in KMnO4 and production of H2, (d) reduction of Mn in KMnO4 and production of Cl2, Bleaching powder on standing forms mixture of :, (a), , CaO Cl 2, , (b), , CaO CaCl 2, , (c) HOCl Cl 2, (d) CaCl 2 Ca (ClO3 ) 2, 160. Hydrochloric acid at 25ºC is, (a) ionic and liquid, (b) covalent and liquid, (c) ionic and gas, (d) None of these, 161. Gaseous HCl is a poor conductor of electricity while its, aqueous solution is a good conductor this is because, , 162., , 163., , 164., , 165., , (a) H 2 O is a good conductor of electricity, (b) a gas cannot conduct electricity but a liquid can, (c) HCl gas does not obey Ohm’s law, whereas the solution, does, (d) HCl ionises in aqueous solution, Which one is most stable to heat –, (a) HClO, (b) HClO2, (c) HClO3, (d) HClO4, Interhalogen compounds are more reactive than the, individual halogen because, (a) two halogens are present in place of one, (b) they are more ionic, (c) their bond energy is less than the bond energy of the, halogen molecule, (d) they carry more energy, Which of the following is not the characteristic of, interhalogen compounds ?, (a) They are more reactive than halogens, (b) They are quite unstable but none of them is explosive, (c) They are covalent in nature, (d) They have low boiling points and are highly volatile., The hybridization in ICl7 is, (a) sp3 d 3, (b) d2 sp 3, 3, (c) sp d, (d) sp3
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EBD_7207, THE p-BLOCK ELEMENTS (GROUP 15, 16, 17 AND 18), , 352, , 166. In which of the following reactions chlorine is both reduced, and oxidized?, (a) 2KMnO4 + 16HCl, 2KCl + 2MnCl2 + 8H2O + 5Cl2, (b) 6NaOH + 3Cl2, 5NaCl + NaClO3 + 3H2O, (c) NH3 + 3Cl2, NCl3 + 3HCl, 2HIO3 + 10HCl, (d) I2 + 6H2O + 5Cl2, 167. Which of the following is observed when Cl2 reacts with, hot and concentrated NaOH?, (a) NaCl , NaOCl, (b) NaCl , NaClO2, (c) NaCl , NaClO3, (d) NaOCl , NaClO3, 168. Which one of the following noble gases is not found in the, atmosphere, (a) Rn, (b) Kr, (c) Ne, (d) Ar, 169. The last member of the family of inert gases is, (a) argon, (b) radon, (c) xenon, (d) neon, 170. Which of the following is the correct sequence of the noble, gases in their group in the periodic table?, (a) Ar, He, Kr, Ne, Rn, Xe (b) He, Ar, Ne, Kr, Xe, Rn, (c) He, Ne, Kr, Ar, Xe, Rn (d) He, Ne, Ar, Kr, Xe, Rn, 171. Which of the following noble gases do not have an octet, of electrons in its outermost shell?, (a) Neon, (b) Radon, (c) Argon, (d) Helium, 172. Number of unpaired electrons in inert gas is, (a) zero, (b) 8, (c) 4, (d) 18, 173. In the following four elements, the ionisation potential of, which one is the highest ?, (a) Oxygen, (b) Argon, (c) Barium, (d) Cesium, 174. Gradual addition of electronic shells in the noble gases, causes a decrease in their, (a) ionisation energy, (b) atomic radius, (c) boiling point, (d) density, 175. Which of the following noble gas is least polarisable?, (a) He, (b) Xe, (c) Ar, (d) Ne, 176. In which of the following groups, when He is placed, its all, the properties are satisfied, (a) with alkali metals, (b) with halogens, (c) with inert gases, (d) None of these, 177. The most abundant inert gas in the atmosphere is, (a) He, (b) Ne, (c) Ar, (d) Kr, 178. The lowest boiling point of helium is due to its, (a) inertness, (b) gaseous nature, (c) high polarisability, (d) weak van der Waal’s forces between atoms, 179. Which of the noble gas has highest polarisability?, (a) He, (b) Ar, (c) Kr, (d) Xe, , 180. The noble gas which was discovered first in the sun and, then on the earth, (a) argon, (b) xenon, (c) neon, (d) helium, 181. A radioactive element X decays to give two inert gases X is, (a), 182., , 183., , 184., , 185., , 186., , 187., , 188., , 189., , 190., , 191., , 238, 92 U, , (b), , 226, 88 Ra, , (c) Both (a) and (b), (d) Neither (a) nor (b), Which of the following noble gases has the highest positive, electron gain enthalpy value?, (a) Helium, (b) Krypton, (c) Argon, (d) Neon, Which inert gas show abnormal behaviour on liquefaction, (a) Xe, (b) He, (c) Ar, (d) Kr, The ease of liquefaction of noble gases increases in the, order, (a) He < Ne < Ar < Kr < Xe, (b) Xe < Kr < Ne < Ar < He, (c) Kr < Xe < He < Ne < Ar, (d) Ar < Kr < Xe < Ne < He, The correct order of solubility in water for He, Ne, Ar, Kr,, Xe is, (a) He > Ne > Ar > Kr > Xe, (b) Ne > Ar > Kr > He > Xe, (c) Xe > Kr > Ar > Ne > He, (d) Ar > Ne > He > Kr > Xe, Which one of the following elements is most reactive ?, (a) He, (b) Ne, (c) Ar, (d) Xe, Noble gases are group of elements which exhibit very, (a) high chemical activity, (b) low chemical activity, (c) minimum electronegativity, (d) much paramagnetic properties, In XeF2, XeF4, XeF6 the number of lone pairs on Xe are, respectively, (a) 2, 3, 1, (b) 1, 2, 3, (c) 4, 1, 2, (d) 3, 2, 1., Total number of lone pair of electrons in XeOF4 is, (a) 0, (b) 1, (c) 2, (d) 3, Noble gases do not react with other elements because, (a) they have completely filled valence shell (ns2np6), (b) the sizes of their atoms are very small, (c) they are not found in abundance, (d) they are monoatomic, Which one of the following reactions of xenon compounds, is not feasible?, (a) 3XeF4 + 6H 2 O, 2Xe + XeO 3 +12HF +1.5O 2, (b) 2XeF2 + 2H 2 O, 2Xe + 4HF + O 2, (c), (d), , XeF6 + RbF, XeO3 + 6HF, , Rb[XeF7 ], XeF6 + 3H 2 O
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p-BLOCK ELEMENTS (GROUP 15, 16, 17 AND 18), , 192. Which of the following has maximum number of lone pairs, associated with Xe ?, (a) XeF4, (b) XeF6, (c) XeF2, (d) XeO3, 193. The shape of XeO2F2 molecule is, (a) trigonal bipyramidal (b) square planar, (c) tetrahedral, (d) see-saw, 194. XeF4 on partial hydrolysis produces, (a) XeF4, (b) XeOF2, (c) XeOF4, (d) XeO3, 195. Which element out of He, Ar, Kr and Xe forms least number, of compounds ?, (a) He, (b) Ar, (c) Kr, (d) Xe, 196. The element which has not yet been reacted with F2 is, (a) Ar, (b) Xe, (c) Kr, (d) Rn, 197. XeF6 on complete hydrolysis gives, (a) Xe, (b) XeO2, (c) XeO3, (d) XeO4, 198. XeF4 involves which hybridization, (a) sp, (b) sp 2, 2, (c) sp d, (d) sp 3 d 2, 199. Shape of XeOF4 is, (a) octahedral, (b) square pyramidal, (c) pyramidal, (d) T-shaped, 200. The hybridization of Xe in XeF2 is, (a) sp 3, (b) sp 2, 3, (c) sp d, (d) sp 2 d, 201. Which is a planar molecule ?, (a) XeO4, (b) XeF4, (c) XeOF4, (d) XeO2F2, 202. Which of the following has sp3 hybridization ?, (a) XeO3, (b) BCl3, (c) XeF4, (d) BBr3, 203. The number of lone pair of electrons present on Xe in XeF 2, is, (a) 3, (b) 4, (c) 2, (d) 1, 204. Hybridization and structure of XeF4 is, (a) sp3d, trigonal bipyramidal, (b) sp3, tetrahedral, (c) sp3d2, square planar, (d) sp3d2, hexagonal, 205. Number of lon e pairs of electrons on Xe atoms, XeF2, XeF4 and XeF6 molecules are respectively, (a) 3, 2 and 1, (b) 4, 3 and 2, (c) 2, 3 and 1, (d) 3, 2 and 0, 206. Which one of the following is correct pair with respect to, molecular formula of xenon compound and hybridization, state of Xenon in it?, (a) XeF4, sp3, (b) XeF2, sp, (c) XeF2, sp3d, (d) XeF4, sp2, , 353, , 207. Which statement about noble gases is not correct?, (a) Xe forms XeF6, (b) Ar is used in electric bulbs, (c) Kr is obtained during radioactive disintegration, (d) He has the lowest b.pt among all the noble gases, 208. The geometry of XeF6 is, (a) planar hexagon, (b) regular octahedron, (c) distorted octahedron (d) square bipyramid, 209. Trigonal bipyramidal geometry is shown by :, (a) XeO3F2, (b) XeO3F2, (c) FXeOSO2F, (d) [XeF8]2–, 210. Which has trigonal bipyramidal shape ?, (a) XeOF4, (b) XeO3, (c) XeO3F2, (d) XeOF2, 211. Argon is used, (a) in filling airships, (b) to obtain low temperature, (c) in high temperature welding, (d) in radiotherapy for treatment of cancer, 212. Noble gases are used in discharge tubes to gives different, colours. Reddish orange glow is due to, (a) Ar, (b) Ne, (c) Xe, (d) Kr, 213. Which one of the following statements regarding helium is, incorrect ?, (a) It is used to produce an d sustain powerful, superconducting magnets., (b) It is used as a cryogenic agent for carrying out, experiments at low temperatures., (c) It is used to fill gas balloons instead of hydrogen, because it is lighter and non-inflammable., (d) It is used in gas-cooled nuclear reactors., 214. The coloured discharge tubes for advertisement mainly, contain, (a) xenon, (b) helium, (c) neon, (d) argon, 215. Sea divers go deep in the sea water with a mixture of which, of the following gases, (a) O2 and He, (b) O2 and Ar, (c) O2 and CO2, (d) CO2 and Ar, 216. Which of the following is the life saving mixture for an, asthma patient ?, (a) Mixture of helium and oxygen, (b) Mixture of neon and oxygen, (c) Mixture of xenon and nitrogen, (d) Mixture of argon and oxygen, 217. Which of the following statements are true?, (i) Only type of interactions between particles of noble, gases are due to weak dispersion forces., (ii) Ionisation enthalpy of molecular oxygen is very close, to that of xenon., (iii) Hydrolysis of XeF6 is redox reaction., (iv) Xenon fluorides are not reactive., (a) (i) and (iii), (b) (i) and (ii), (c) (ii) and (iii), (d) (iii) and (iv)
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EBD_7207, THE p-BLOCK ELEMENTS (GROUP 15, 16, 17 AND 18), , 354, , 218. Which of the following element has the property of diffusing, through most commonly used laboratory materials such as, rubber, glass or plastics., (a) Xe, (b) Rn, (c) He, (d) Ar, 219. Which of the following is used to produce and sustain, powerful superconducting magnets to form an essential, part of NMR spectrometer ?, (a) Ar, (b) Ne, (c) Rn, (d) He, , STATEMENT TYPE QUESTIONS, 220. Which of the following statements are correct?, (i) Arsenic and antimony are metalloids., (ii) Phosphorus, arsenic and antimony are found mainly, as sulphide minerals., (iii) Covalent redii increases equally from N to Bi., (iv) Elements of group 15 have extra stability and higher, ionisation energy due to exactly half filled ns2np3, electronic configuration., (v) In group 15 elements only nitrogen is gas whereas all, others are solids., (a) (i), (iv) and (v), (b) (ii), (iii) and (iv), (c) (i), (ii) and (iii), (d) (ii), (iii) and (v), 221. Read the following statements regarding chemical reactivity, of group 15 elements., (i) Only compound of Bi with +5 oxidation state is BiF5., (ii) Intermediate oxidation states for both nitrogen and, phosphorus disproportionate in both acid and alkali., (iii) Nitrogen due to absence of d-orbitals in its valence, shell cannot form d -p bond as the heavier elements, thus R3P = O exists but R3N = O does not exists., (iv) BiH3 is the strongest reducing agent amongst the, hydrides of nitrogen family., (v) P2O3 is more acidic than P2O5., Which of the following is the correct code for the statements, above?, (a) FTFFT, (b) FFTTF, (c) TFTTF, (d) TFTFT, 222. Which of the following statements are correct?, (i) All the three N—O bond lengths in HNO3 are equal., (ii) All P—Cl bond lengths in PCl5 molecule in gaseous, state are equal., (iii) P4 molecule in white phohsphorus have angular strain, therefore white phosphorus is very reactive., (iv) PCl5 is ionic in solid state in which cation is tetrahedral, and anion is octahedral., (a) (i) and (iv), (b) (iii) and (iv), (c) (ii) and (iii), (d) (ii) only, 223. Which of the following is the correct code for statements, below ?, (i) Due to small size oxygen has less negative electron, gain enthalpy than sulphur., (ii) Oxygen shows only –2 oxidation state whereas S, Se, and Te shows +4 O.S in their compounds with oxygen, and +6 with fluorine., , (iii) All hydrides of oxygen family possess reducing, property which increases from H2S to H2Te., (iv) Among hexahalides of group 16 hexafluorides are the, onlys table halides., (v) Dimeric monohalides of group 16 un dergo, disproportionation., (a) TFFTT, (b) FTTFF, (c) FTFTF, (d) TFTFT, 224. The correct statement(s) about O3 is(are), (i) O—O bond lengths are equal, (ii) Thermal decomposition of O3 is endothermic, (iii) O3 is diamagnetic in nature, (iv) O3 has a bent structure, (a) (i) and (iii), (b) (ii) and (iii), (c) (i), (ii) and (iv), (d) (i) and (iv), 225. Consider the following statements, (i), , Reaction 2Fe3+ + SO2 + 2H2O, , 2Fe2+ + SO24 + 4H+, , shows reducing character of sulphur dioxide, (ii) H2S2O8 contains four S = O, two S – OH and one O–, O bond, (iii) NH3 gas can be dried effectively by using conc., H2SO4., (iv) One of the major use of H2SO4 is in the manufacture, of fertilizers., Which of the following is the correct code for the statements, above?, (a) TTFF, (b) TTFT, (c) FTFT, (d) TFFT, 226. Which of the following statements regarding properties of, halogens are correct?, (i) Due to small size electron gain enthalpy of fluorine is, less than that of chlorine., (ii) Iodine has same physical state but different colour as, compare to other members of the group., (iii) Fluorine shows no positive oxidation state., (iv) In X2(g) + H2O(l), HX(aq) + HOX(aq), (where X2 = Cl or Br), (a) (i), (ii) and (iv), (b) (i), (iii) and (iv), (c) (ii), (iii) and (iv), (d) (iii) and (iv), 227. Consider the following statements regarding interhalogen, compounds, (i) For all types of interhalogen compounds, XX1 , XX13 , XX15 and XX17, , (ii), (iii), (iv), (a), (c), , X is the halogen of, , lesser electronegativity in comparison to X1., At 298 K all interhalogen compounds are either, volatile solids or liquids., ClF undergoes hydrolysis as below,, ClF + H2O, HF + HOCl, Fluorine containing interhalogen compounds are very, useful as fluorinating agents., TTFF, (b) TFTT, FTFT, (d) TFFT
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p-BLOCK ELEMENTS (GROUP 15, 16, 17 AND 18), , 228. Which of the following statements are correct?, (i) Among halogens, radius ratio between iodine and, fluorine is maximum., (ii) Leaving F—F bond, all halogens have weaker X—X, bond than X—X bond in interhalogens., (iii) Among interhalogen compounds maximum number of, atoms are present in iodine fluoride., (iv) Interhalogen compounds are more reactive than, halogen compounds., (a) (i) and (ii), (b) (i), (ii) and (iii), (c) (ii) and (iii), (d) (i), (iii) and (iv), 229. Which of the following statements are correct?, (i) Natural abundance of noble gases is ~ 1% by volume, of which Ar is the major constituent., (ii) Noble gases have high positive values of electron, gain enthalpy., (iii) Preparation of XeF2 requires F2 in excess amount., (iv) Complete hydrolysis of all three XeF2, XeF4 and XeF6, gives Xe as one of product., (a) (i) and (iii), (b) (ii) and (iv), (c) (i) and (ii), (d) (ii) and (iii), , MATCHING TYPE QUESTIONS, 230. Match the columns, Column-I, (A) 2Pb(NO3)2, , Column-II, 673K, , (p) High pressure favours, , 4NO2 + 2PbO + O2, , the formation of, product, (q) Product formed is acidic, (B) N2(g) + O2(g), 2NO(g), brown gas, (C) NH4NO3, , (r) This reaction occurs at, N2O + 2H2O, , a high temperature, about 2000 K, , (D) N2(g) + 3H2(g), (s) Product formed is a, 2NH3(g), neutral colourless gas, (a) A – (r, s), B – (q), C – (s), D – (p), (b) A – (q), B – (r,s), C – (s), D – (p), (c) A – (q), B – (s), C – (r, s), D – (p), (d) A – (q), B – (r, s), C – (p), D – (s), 231. Match the columns, Column - I, Column - II, (A) Used in manufacture, (p) Ammonia, of calcium cyanamide, (B) Used in manufacture, (q) Nitric acid, of nitric acid, (C) Used in pickling of, (r) Dinitrogen, stainless steel, (a) A – (r), B – (p), C – (q) (b) A – (p), B – (r), C – (q), (c) A – (r), B – (q), C – (p) (d) A – (q), B – (p), C – (r), , 355, , 232. Match the columns., Column-I, (Oxyacid), , Column-II, (Materials for, preparation), (A) H3PO2, (p) Red P + alkali, (B) H3PO3, (q) P4O10 + H2O, (C) H3PO4, (r) P2O3 + H2O, (D) H4P2O6, (s) White P + alkali, (a) (A) – (s), (B) – (r), (C) – (q), (D) – (p), (b) (A) – (p), (B) – (r), (C) – (q), (D) – (s), (c) (A) – (s), (B) – (r), (C) – (p), (D) – (q), (d) (A) – (q), (B) – (r), (C) – (p), (D) – (s), 233. Match the columns, Column - I, Column - II, (A) POCl3, (p) Contains four P – OH, two P = O and one, P–O–P, (B) H4P2O5, (q) Yellowish white, chloride of phosphorus, reacts with moist air, (C) H4P2O6, (r) Contains four P – OH,, two P = O and one P – P, bond, (D) H4P2O7, (s) Colourless oily chloride, of phosphorus reacts, with orthophosphoric, acid, (a) A – (q), B – (s), C – (p), D – (r), (b) A – (s), B – (q), C – (r), D – (p), (c) A – (q), B – (s), C – (r), D – (p), (d) A – (q), B – (r), C – (s), D – (p), 234. Match the columns, Column - I, Column - II, (A) Metal that shows no, (p) Platinum, reaction with dioxygen, (B) Metal forms strong, (q) Nitrogen, acidic oxide with oxygen, (C) A non-metal discharge (r) Manganese, of whose oxide might, be slowly depleting the, concentration of the, ozone layer, (D) Metal which forms, (s) Aluminium, amphoteric oxide, (a) A – (p), B – (r), C – (q), D – (s), (b) A – (r), B – (p), C – (q), D – (s), (c) A – (p), B – (q), C – (r), D – (s), (d) A – (p), B – (r), C – (s), D – (q)
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EBD_7207, 356, , 235. Match the columns., Column-I, Column-II, (A) Pb3O4, (p) Neutral oxide, (B) N2O, (q) Acidic oxide, (C) Mn 2O7, (r) Basic oxide, (D) Bi2O3, (s) Mixed oxide, (a) A – (p), B – (q), C – (r), D – (s), (b) A – (s), B – (p), C – (q), D – (r), (c) A – (r), B – (q), C – (s), D – (p), (d) A – (s), B – (r), C – (p), D – (q), 236. Match the columns., Column-I, Column-II, (A) SF4, (p) Tetrahedral, (B) BrF3, (q) Pyramidal, (r) Sea-saw shaped, (C) BrO3–, (s) Bent T-shaped, (D) NH4+, (a) A – (r), B – (q), C – (p), D – (s), (b) A – (r), B – (s), C – (q), D – (p), (c) A – (p), B – (q), C – (r), D – (s), (d) A – (p), B – (s), C – (r), D – (q), 237. Match the columns, Column - I, Column - II, (A) HClO2, (p) Contains all different, bonds, (B) HClO3, (q) Contains maximum, Cl = O bond, (C) HClO, (r) Contains Cl with lowest, O.S., (D) HClO4, (s) Contains three types of, bonds, (a) A – (s), B – (p, s), C – (p, r), D – (q, s), (b) A – (p, s), B – (s), C – (p, r), D – (q, s), (c) A – (s), B – (p, r), C – (p, s), D – (q, s), (d) A – (p, s), B – (s), C – (q, s), D – (p, r), 238. Match the columns., Column - I, Column - II, (Oxides of halogens), (Uses), (A) O2F2, (p) in water treatment, (B) ClO2, (q) in estimation of CO, (C) I2O5, (r) for removing plutonium, from spent nuclear fuel., (a) A – (q) , B – (p), C – (r), (b) A – (r), B – (p), C – (q), (c) A – (p), B – (r), C – (q), (d) A – (r), B – (q), C – (p), 239. Match the columns, Column - I, Column - II, (A) XeF4, (p) Contains similar types, of bonds, (B) XeOF4, (q) Contains maximum lone, pair, (C) XeF2, (r) Square pyramidal, geometry, (D) XeO3, (s) Contains one lone pair, , THE p-BLOCK ELEMENTS (GROUP 15, 16, 17 AND 18), , (a) A – (p), B – (r, s), C – (p, q), D – (p, s), (b) A – (r, s), B – (p), C – (r, s), D – (p, s), (c) A – (p), B – (p, q), C – (r, s), D – (p, s), (d) A – (p), B – (r, s), C – (p, s), D – (p, q), 240. Match the columns., Column-I, Column-II, (A) Partial hydrolysis of the, (p) He, compound does not change, oxidation state of central atom, (B) It is used in modern diving, (q) XeF6, apparatus, (C) It is used to provide inert, (r) XeF4, atmosphere for filling electrical, bulbs, (D) Its central atom is in sp3d2, (s) Ar, hybridisation, (a) A – (p), B – (s), C – (p), D – (r), (b) A – (p), B – (q), C – (r), D – (s), (c) A – (q), B – (p), C – (s), D – (r), (d) A – (p), B – (r), C – (q), D – (s), 241. Match the columns., Column-I, Column-II, (A) XeF6, (p) sp3d3– distorted octahedral, (B) XeO3, (q) sp3d2 – square planar, (C) XeOF4, (r) sp3 – pyramidal, (D) XeF4, (s) sp3d2 – square pyramidal, (a) A – (p), B – (r), C – (s), D – (q), (b) A – (p), B – (q), C – (s), D – (r), (c) A – (s), B – (r), C – (p), D – (q), (d) A – (s), B – (p), C – (q), D – (s), , ASSERTION-REASON TYPE QUESTIONS, Directions : Each of these questions contain two statements,, Assertion and Reason. Each of these questions also has four, alternative choices, only one of which is the correct answer. You, have to select one of the codes (a), (b), (c) and (d) given below., (a) Assertion is correct, reason is correct; reason is a correct, explanation for assertion., (b) Assertion is correct, reason is correct; reason is not a, correct explanation for assertion, (c) Assertion is correct, reason is incorrect, (d) Assertion is incorrect, reason is correct., 242. Assertion : Dinitrogen is inert at room temperature., Reason : Dinitrogen directly combines with lithium to form, ionic nitrides., 243. Assertion : N2 is less reactive than P4., Reason : Nitrogen has more electron gain enthalpy than, phosphorus., 244. Assertion : When a metal is treated with conc. HNO3 it, generally yields a nitrate, NO2 and H2O., Reason : Conc. HNO3 reacts with metal and first produces, a metal nitrate and nascent hydrogen. The nascent hydrogen, then further reduces HNO3 to NO2.
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p-BLOCK ELEMENTS (GROUP 15, 16, 17 AND 18), , 245. Assertion : White phosphorus is more reactive than red, phosphorus., Reason : Red phosphorus consists of P4 tetrahedral units, linked to one another to form linear chains., 246. Assertion : Bond angle of H2S is smaller than H2O., Reason : Electronegativity of the central atom increases,, bond angle decreases., 247. Assertion : Both rhombic and monoclinic sulphur exist as, S8 but oxygen exists as O2., Reason : Oxfygen forms p – p multiple bond due to small, size and small bond length but p – p bonding is not, possible in sulphur., 248. Assertion : SF6 cannot be hydrolysed but SF4 can be., Reason : Six F atoms in SF6 prevent the attack of H2O on, sulphur atom of SF6., , 357, , 256. What is the change observed when AgCl reacts with NH3?, (a) White ppt is formed, (b) Solution become colourless, (c) Yellow ppt is formed, (d) No change is observed, 257. In which of the following equations the product formed, has similar oxidation state for nitrogen?, (i), , (a), , Cu 2, , (c), , Cu ( NH 3 ) 26, , (b), , Cu ( NH 3 ) 24, , (d), , Cu( NH3 ) 22, , 252. Blue solid which is obtained on reacting equimolar amounts, of two gases at, is?, (a) N2O, (b) N2O3, (c) N2O4, (d) N2O5, 253. Concentrated nitric acid, upon long standing, turns yellow, brown due to the formation of, (a) NO, (b) NO2, (c) N2O, (d) N2O4, 254. In the reaction, 4HNO 3 P4 O10, 4HPO 3 X , the product X is, (a) N2O5, (b) N2O3, (c) NO2, (d) H2O, 255. Ammonia on catalytic oxidation gives an oxide from which, nitric acid is obtained. The oxide is :, (a), , N 2 O3, , (b) NO, , (c), , NO 2, , (d), , N 2 O5, , N2O + 2H2O, , (ii) 2Pb(NO3)2, , 673K, , (iii) 4HNO3 + P4O10, Cool, , (iv) 2NO2, , CRITICAL THINKING TYPE QUESTIONS, 249. In nitrogen family, the H-M-H bond angle in the hydrides, gradually becomes closer to 90º on going from N to Sb. This, shows that gradually, (a) The basic strength of the hydrides increases, (b) Almost pure p-orbitals are used for M-H bonding, (c) The bond energies of M-H bonds increase, (d) The bond pairs of electrons become nearer to the, central atom, 250. Bond dissociation enthalpy of E—H (E = element) bonds is, given below. Which of the compounds will act as strongest, reducing agent?, Compound, NH3, PH3, AsH3 SbH3, –1 389, (E—H)/kJ, mol, 322, 297, 255, diss, (a) NH3, (b) PH3, (c) AsH3, (d) SbH3, 251. The deep blue colour produced on adding excess of, ammonia to copper sulphate is due to presence of, , NH4NO3, , 258., , 259., , 260., , 261., , Heat, , 4NO2 + 2PbO + O2, 4HPO3 + 2N2O5, , N2O4, , (a) (i) and (iii), (b) (ii) and (iv), (c) (i) and (v), (d) (iii) and (iv), What is Z in following reaction, CuSO4 + Z Cu3P2 +H2SO4, HgCl2 + Z Hg3P2 +HCl, (a) White phosphorus, (b) Red phosphorus, (c) Phosphine, (d) Orthophosphoric acid, Electronegativity of oxygen is more than sulphur yet H2S is, acidic while water is neutral. This is because, (a) water is highly associated compound, (b) molecular mass of H2S is more than H2O, (c) H2S is gas while H2O is a liquid, (d) H–S bond is weaker than H–O bond, It is possible to obtain oxygen from air by fractional, distillation because, (a) oxygen is in a different group of the periodic table, from nitrogen, (b) oxygen is more reactive than nitrogen, (c) oxygen has higher b.p. than nitrogen, (d) oxygen has a lower density than nitrogen., Which of the following structures is the most preferred and, hence of lowest energy for SO3 ?, , O, (a), , S, , S, O, , O, , (b), , O, , O, (c), , S, O, , O O, , O, , O, (d), , S, O, , O, , 262. Which one of the following arrangements does not give, the correct picture of the trends indicated against it ?, (i) F2 > Cl2 > Br2 > I2 : Oxidizing power, (ii) F2 > Cl2 > Br2 > I2 : Electron gain enthalpy, (iii) F2 > Cl2 > Br2 > I2 : Bond dissociation energy, (iv) F2 > Cl2 > Br2 > I2 : Electronegativity., (a) (ii) and (iv), (b) (i) and (iii), (c) (ii) and (iii), (d) (ii), (iii) and (iv)
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p-BLOCK ELEMENTS (GROUP 15, 16, 17 AND 18), , FACT / DEFINITION TYPE QUESTIONS, 1., 2., 3., 4., 5., 6., 7., 8., 9., 10., 11., 12., , (d) Ionic radii increases down the group, (b) In case of nitrogen, d-orbitals are not available., (a) Collectively these elements are called pnicogens and, their compound pniconides., (d) Metallic character increases down the group, Bi is, most metallic, (b) The melting point in group 15 increases upto arsenic, and then decreases upto bismuth., (d) Bismuth forms metallic bonds in elemental state., (a) –3, +3, +5, (d) N2 molecule contains triple bond between N atoms, having very high dissociation energy (946 kJ mol–1), due to which it is relatively inactive., (a) Nitrogen due to small size is able to show p -p lateral, overlap forming N N, rest elements due to bigger size, are not able to show p -p lateral overlap., (c) Catenation tendency is higher in phosphorus when, compared with other elements of same group., (c) Nitrogen form N2 (i.e. N N) but phosphorus form, P4, because in P2, p — p bonding is present which, is a weaker bonding., (d) The cause of inert nature of N2 is the presence of, .., , triple bond N, 13., 14., 15., 16., , 17., , 18., , 19., , .., , N, , (b), (b) Phosphorous can achieve coordination number 5 due, to vacant d atomic orbitals in valence shell which is, not possible in nitrogen, (b) The order of boiling points of the group 15 hydrides, is : BiH3 > SbH3 > NH3 > AsH3 > PH3, (a) Oxide in which central atom has higher charge and, more electronegativity is more acidic, i.e., N2O5 > N2O4 > P2O5 > As2O3., (a) Order of dipole moment, NH3 > PH3 > AsH3 > SbH3, (Based upon electronegativity), (c) As the size of central atom increases the lone pair of, electrons occupies a larger volume. In other words, electron density on the central atom decreases and, consequently its tendency to donate a pair of electrons, decreases along with basic character from NH3 to, BiH3., (a) NF 5 does not exist because N does not form, pentahalides due to the absence of d-orbital in its, valence shell. While P, As and Sb form pentahalides of, the general formula MX5 (where, M = P, As and Sb), due to the presence of vacant d-orbitals in their, respective valence shell., , 359, , 20. (d) Bi forms basic oxides whereas N and P form acidic and, As and Sb form amphoteric oxides., 21. (b) The basic character decreases from NH3 to BiH3. The, basic nature is due to the presence of lone pair of, electrons on the central atom. NH3 is the strongest, electron pair donor due to its small size as the electron, density of the electron pair is concentrated over a small, region. As the size increases the electron density gets, diffused over a large region and hence the ability to, donate the electron pair (basic nature) decreases., 22. (a) NCl5 in not possible because N does not contain, d-orbitals., Only nitrogen has a tendency to form p – p multiple, bonds. Other forms d – p multiple bonds easily.., 23. (c), 25. (d), , 24. (b), NH 4 Cl NaNO 2, , Heat, NaCl, , NH 4 NO 2, Heat, , N2, , 2H 2O., , 26. (a), 27. (b) In Haber’s process for manufacture of NH3, finely, divided iron is used as catalyst and molybdenum is, used as catalytic promoter, , N2 ( g ) 3H2 ( g ), , Fe Mo, 800K, High P, , 2NH 3 ( g ), , 28. (d) N2H4 and NH4Cl are obtained by reaction of ammonia, with hypochlorite anion., 29., , 30., , 3NH 3 NaOCl, N 2 H 4 NH 4 Cl NaOH, (a) HNO3 and CuSO4 are not drying agents, while P2O5, reacts with NH3. The moisture present in NH3 is, removed by passing it through a tower packed with, quicklime (CaO)., (b) Ammonia has pyramidal shape with sp3 hybridisation., , 31. (b), , 32. (b), 33. (b), , 34. (c), 35. (a), 36. (d), , 3CuO 2 NH 3, , 3Cu 3H 2 O N 2 ,, O.S. of N in NH3 is –3 and in N2 is zero. Hence loss of, 3 electrons, NH3 is not used as anaesthetic, Liquid ammonia has high vapour pressure which is, lowered down by cooling, otherwise the liquid will, bump., By Haber’s process, Only nitrates of heavy metals and lithium decompose, on heating to produce NO2., N2O3, N2O4 and N2O5 are acidic oxides. Only N2O is, neutral oxide.
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EBD_7207, THE p-BLOCK ELEMENTS (GROUP 15, 16, 17 AND 18), , 360, , 37., , 38., 39., , (b), , 56., 57., 58., , O, N–O–N, , O, O, The structure clearly shows the presence of covalent, and co-ordinate bonds., (d) NO2 is reddish brown coloured gas. Rest of the oxides, are colourless., (c) Compound, O.S. of N, N2O, +1, NO, +2, NO2, +4, +5, NO3–, –3, NH4+, Therefore increasing order of oxidation state of N is:, NH 4 < N2O < NO < NO2 < NO3 ., , 40., , 41., , (c) In N2O (nitrous oxide) two N atoms are covalently, bonded through triple bond, [N N, O], (b) FeSO 4 NO FeSO 4 .NO, , 42., 43., , (b), (a) N2O is used as anaesthetic, , 44., 45., , (c) 2NO O 2, 2 NO 2 brown, (b) Phosphorus from stable P4 molecule., , 46., 47., , (a) 4NH 3 5O 2, 4NO 6H 2 O, (b) The slow decomposition of HNO3 is represented by, the eqn., 4HNO3, 4NO2 + 2H2O + O2, , P4 + 5O2, 60., 61., , 62., 63., 64., 65., , (c) Monoclinic sulphur is stable above 369 K., (c) Ca3P2 + 6H2O, 3Ca(OH)2 + 2PH3 ; i.e 2 moles of, phosphine are produced from one mole of calcium, phosphide., (d) PH3 is covalent hydride, (b) Red P does not react with NaOH to give PH3., (d) PH3 is not obtained when metaphosphoric acid is, heated., (a) The combustibility of PH3 is due to presence of P2H4., The pure PH3 is not combustible., , (d) For nitrogen, only NF3 is known to be stable., 6NH 4Cl N 2, (a) 8NH3 3Cl2, , (d), , 2H 3 PO 4, , 52., , 3Cl 2, excess, , 67., , (d), , P2 O 5, , 68., , (b) Orthophosphoric acid, H 3PO 4 contains three P – OH, bonds and is therefore, tribasic., , 3H 2 O, , orthophosphoric acid, , 69., , (b) PCl3 + H2O, POCl3 + 3H2O, , 70., , (c), , O, ||, P, H | OH, H, (a) We know that empirical formula of hypophosphorus, acid is H3PO2. In this only one ionisable hydrogen, atom is present i.e. it is monobasic. Therefore option, (a) is correct structural formula of it., , 71., , =, , =, , =, , 72., , O, , O, , (d), , O, , O, , P, , P, O, , P, ||, O, , O, , ||, , O, , ||, , 55., , O, ||, P, , =, , =, , 54., , POCl3 + 2HCl, H3PO4 + 3HCl, , H 3PO 2 is named as hypophosphorous acid. It is, monobasic as it contains only one P – OH bond, its, basicity is one., , =, , 53., , 2 H 3 PO 4, , O, ||, P, | OH, OH, , NCl3 3HCl, , (c) BiH3 is the strongest reducing agent while NH3 is the, weakest reducing agent., (d) The oxides of the type E 2 O 3 of nitrogen and, phosphorus are purely acidic., (d) NH3 is not used in the pickling of stainless steel., O, O, N–N =, (d) N, O, O, (ii), (i), O, O, O, O, N–N, N–O–N, O, O, O, O, (iii), (iv), (d) 4 Zn + 10 HNO3 (dil.) 4 Zn(NO3)2 + 5H2O + N2O, Zn + 4 HNO3 (conc.), Zn(NO3)2 + 2H2O + 2NO2, (a) Both white and red phosphorus are not soluble in CS2, only white phosphorus is soluble in CS2., , 2 HPO 3, , 2H 2O, , |, , NH3, , 51., , 600 C, , 66., , (excess), , 50., , P4O10., , Pt. gauge, , (yellow-brown), , 48., 49., , 59., , (b) White phosphorous is most reactive, (b) White phosphorous is P4 and tetrahedral, (a) Except (a) all other properties are shown by white, phosphorous., (b) White phosphorus on heating readily catches fire in, air to give dense white fumes of P4O10., , |, , O, , O
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p-BLOCK ELEMENTS (GROUP 15, 16, 17 AND 18), , 73., , 361, , (c) Structure of hypophosphorous acid, , O, , H, |, , H O P, , (b), , O, , |, , P, HO, , 75., , Two H-atoms are attached to P atom., (c) In cyclic metaphosporic acid number of P–O–P bonds, is three., O OH, P, O, O, O, O, P, P, O, OH, HO, (d), , 76., , (b), , O, bridging, , bridging, , P, , O, P, , (d), , (e), , bridging, , OH, , H, , P, O, , P, , bridging, oxygen, , H3PO3, , The H–atom of the –OH group is ionisable whereas, H–atom which is directly linked to P–atom is nonionisable. Thus H3PO3 is dibasic acid., 85. (a) Pyrophosphorous acid (H4P2O5) is a dibasic acid as it, contains two P—OH bonds., O, , OH, , OH, , 86. (b) Formula of cyclotrimetaphosphoric acid is (HPO3)3, Oxidation state of ‘P’ is 3(+ 1 + x + 3 (– 2)) = 0, x+ –6+1 =0, x= +5, 87. (a), , O, , O, , HO — P — P — OH, , H Hypophosphorous acid (H3PO2) is a, , HO OH, Hypophosphoric acid, OH, , H, , monobasic acid. i.e., it has only one ionisable hydrogen, atom or one OH is present., 1, Hybridisation in PCl5 =, 5 5 0 0 5 sp3d, 2, H5P5O15 (HPO3)5. It is metaphosphoric acid which is, a cyclic phosphate., H4P2O5 is pyrophosphorous acid it contains P–O–P, bond, H3PO4 is tribasic, Hypophosphorous acid is H3PO2 in which O.S. of P, is +1, , 78., , (c), , 79., , (c), , 80., , (b), , 81., 82., , (c), (a), , 83., 84., , (c), (d) Structures of given oxyacids are following, O, , O, , P, , (a) HO, , OH, , H 3PO 4, , O, , O, (b), , OH, , H, , O, , O, O, , O, , 77., , P, , OH, , HO, , H—P—O—P—H, , P, O, , O, , O, , O, P, , O, , P, HPO3, , H3PO2, , O, O, , (c) HO, , H, , H, , H, , 74., , O, , OH, , P, OH, , O, OH, H4P2O7, , O, , P, , OH, O, , P, , OH, , O, , OH, , Pyrophosphoric acid, , HO, , O, , P, , P, , OH, , O, O, O, Metaphosphoric acid, O, HO, , P, , OH, , OH, Orthophosphoric acid, , 88. (a), 89. (c), 90. (d), 91. (a) Electron affinity increases from left to right in period, and decreases from top to bottom in a group but, electron affinity of O is less than S due to small size., 92. (d) All exhibit polymorphism, 93. (a)
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EBD_7207, THE p-BLOCK ELEMENTS (GROUP 15, 16, 17 AND 18), , 362, , 94., , (d) H2O is liquid but H2S is a gas. This can be attributed, to the presence of intermolecular hydrogen bonding, in case of H2O., 95. (b), 96. (a), 97. (a) H2O (due to intermolecular H - bonding), 98. (b) Oxygen being more electronegative, 99. (a) SnO2 is an amphoteric oxide because it reacts with, acids as well as with bases to form corresponding salts., Sn(SO4)2 + 2H2O, SnO2 + 2H2SO4(conc), SnO2 + 2NaOH, Na2SnO3 + H2O, 100. (d) All hexafluorides of group 16 elements are gaseous in, nature., 101. (b) Oxygen can be prepared by heating oxides of Hg, Pb,, Ag, Mn and Ba., 2Hg + O2, 2HgO, 102. (c) It is paramagnetic with two unpaired electrons, 103. (c) Total number of electrons in O2 is 16. It has 2 unpaired, electrons, the rest 14 are paired., , 120. (c), , SO 2 2H 2 O H 2SO 4, due to reduction., , 2H . Bleaching action is, , 121. (d) Caro’s acid is H 2SO 5 which contains one, S – O, – O – H peroxy linkage. It is also known as, permonosulphuric acids., O, ||, H – O – O – S – OH, ||, O, Caro's acid, , 122. (d), 123. (c) Oleum is H 2S2 O 7 ( H 2SO4 SO3 ) which is obtained, by dissolving SO3 in H2SO4 and is called fuming, sulphuric acid., 124. (c), 125. (a), 126. (c), 127. (c) It is H2SO5., 128. (c), , HO.SO 2OH 2PCl5 ClSO2Cl 2POCl3 2HCl, Sulphuryl chloride, , 105. (c) In KMnO4 manganese is already present in its highest, possible oxidation state i.e. +7.So no further oxidation, is possible., 106. (c) Ozone layer is beneficial to us, because it stops, harmful ultraviolet radiations from reaching the earth., , 129. (b), , 2Ag 2H 2SO 4, 2H 2 O SO 2 Ag 2SO 4 ., Au, Pt does not react. Pb forms insoluble PbSO4, , K 2 MnO4 4MnO 2 O 2, 107. (b) 2KMnO 4, 108. (d) Ozone is an allotrope of oxygen., 109. (d) 2Ag2O (s) 4Ag (s) + O2 (g), 2Pb3O4 (s) 6PbO (s) + O2 (g), 2PbO2 (s) 2PbO (s) + O2 (g), 110. (a) Mn2O7 is an acidic oxide. BaO and Na2O are basic, oxides while N2O is a neutral oxides., 111. (c) It is 8, 112. (c) S2 is paramagnetic. It contains two unpaired electrons, in the antibonding * orbital, , 131. (a), , 104. (b), , 2KClO 3, , 2 KCl 3O 2, , 113. (a) 2SO2(g) + O2(g), , 2SO3(g), , O, , OH, , Peroxodisulphuric acid, (H2S2O8), , 115. (b) Conc. H2SO4 is a strong dehydrating agent due to, which carbohydrates becomes charred on reaction with, conc. H2SO4 acid., 116. (b) The key step in the manufacture of H2SO4 is catalystic, oxidation of SO2 with O2 to give SO3 in presence of, V2O5 ., 117. (a) In SO3, sp2 hybridisation, 118. (c) Cu 2H 2SO 4 (conc), CuSO 4 SO 2 2H 2O, 119. (b), , 133., 134., 135., 136., 137., , O = S–O–O–S=O, OH, , 2 , one, , see structure, , O, ||, , H O S O H;, ||, O, , 6 &2, , 132. (a) In H 2SO 4 , the S atom is present in its highest, , =, , 114. (b), , =, , O, , V2O5, , 130. (c), , 138., 139., 140., 141., , oxidation state of +6. Hence H 2SO 4 can act an oxidant, only by gain of electrons, (d) Ionisation potential decreases down the group., (a), (d) The lesser the bond energy, the weaker is the bond, (b) 3s2 3p5 is electronic configuration of Cl, (c) The electron gain enthalpy order for halogens is, Cl > F > Br > I, Due to small size of fluorine the extra electron to be, added feels more electron-electron repulsion., Therefore fluorine has less value for electron affinity, than chlorine., (d), (a) Reactivity follows the order F > Cl > Br > I, (d), (b) Except ionisation potential other factors are true to, explain the oxidising (strong) behaviour of F2 ., , 142., 143., 144., 145., , (b), (d) Fluorine exhibit -ve oxidation state, (a) Since F2 is most oxidising, it is easily reduced, (c) Chlorine shows O.S. from –1,+1 to +7, whereas others, show O.S. as Na +1 ; K +1 ; F –1
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p-BLOCK ELEMENTS (GROUP 15, 16, 17 AND 18), , 363, , 146. (c) Fluorine always exhibit –1 oxidation state., 147. (b), , 148. (d), , 149. (b), , 150. (a), 151. (c), 152. (d), 153. (d), , 154. (b), , 0, , H 2O Br2, , 1, , 166. (b), , 1, , HOBr HBr, , Thus here oxidation number of Br increases from 0 to, +1 and also decreases from 0 to –1. Thus it is oxidised, as well as reduced., Since all the halogens have a strong tendency to, accept electrons. Therefore halogens act as strong, oxidising agents and their oxidising power decreases, from fluorine to iodine., On moving from top to bottom of halogen group the, bond dissociation energy of hydrogen h alides, decreases and so the heat of formation of halogen, acids also decreases., HF, due to intermolecular H-bonding is weakest among, HX acids, Volatile character HCl > HBr >HI > HF, Due to hydrogen bonding HF is a liquid, Bleaching action of chlorine is due to oxidation in, presence of moisture., Cl 2 H 2O HCl HClO, HClO, HCl O, Colouring matter + | O |, Colourless matter, Cl2 + 2NaOH, NaCl + NaClO + H2O, (cold & dil), , 3Cl2 + 6NaOH, , 5NaCl + NaClO3 + 3H2O, , (hot & conc.), , 155. (a), , 2NaOH Cl2, , NaCl NaOCl H 2 O, , 156. (d) K 2 Cr2O 7 conc.HCl Cl2, 157. (d) MnO2 or KMnO4 with conc HCl give Cl2., 7, , 158. (d) 2 KMnO 4 16HCl, , 2, , 2 MnCl 2 2 KCl 8H 2 O 5Cl 2, , O.S of Mn changes from +7 to +2 hence reduction, occurs and Cl2 is formed., 159. (d), 160. (d), 161. (d), 162. (d), , 163. (c), 164. (d), , 6CaOCl 2, Ca (ClO3 ) 2 5CaCl 2, It is autooxidation., HCl acid at 25º C is a gas and polar in nature, In gaseous state the HCl is covalent in nature while in, aqueous solution it ionises to give H+ and C l ions, As the oxidation state of the central halogen atom, increases, the halogen-oxygen bond becomes more, and more covalent. As a result the thermal stability of, the oxoacid increases. Thus, HClO4 is most stable to, heat, whereas HClO is least stable to heat., The bond energy of interhalogen compounds is less, than the bond energy of halogens., Interhalogen compounds are not highly volatile, , 165. (a) ICl7. The hybridisation is, , 1, 7 7 0 0, 2, , 167. (c) 6NaOH + 3Cl2, , 7 (sp3d3 ), , 1, , +5, , 5NaCl + NaCl O3 + 3H 2O, , 5NaCl + NaClO3 + 3H2O, , (hot and conc.), , 168. (a) Rn because it is radioactive element obtained by the, disintegration of radium, 169., 170., 171., 172., 173., 174., , 175., 176., , 177., 178., 179., 180., , Rn 222 2 He4, (b) Radon is the last member of family, (d), (d) Electronic configuration of He is 1s2, (a) Inert gases do not contain unpaired electrons, (b) Ionization potential of inert gases is highest in periodic, table due to stable electronic configuration., (a) Ionisation energy decreases as we move away from, nucleus due to less electrostatic attraction between, electrons and nucleus, (a) The smaller the size the least is the polarisability, (c) The differentiating electron enter in s subshell in case, of He, hence it is s- block element. Its electronic, configuration l s2 makes it inert in nature hence it is, placed with inert gases., (c) Ar is the most abundant in atmosphere, (d) Due to weak van der Waal’s forces, He has lowest, boiling point, (d) The larger the size the more is the polarisiability, (d) He was observed in the spectrum of the sun, , 181. (b), , hence Cl and OCl, , 0, , 6NaOH + 3Cl2, , 88 Ra, , 226, , 88 Ra, , 226, , 86, , 86, , Rn 222, , 2, , He 4 . Both are inert gases, , 182. (d) Electron gain enthalpy for noble gases is positive and, it becomes less positive with increase in size of atom., Value of electron gain enthalpy, He – 48 kJ mol–1, Ne – 116 kJ mol–1, Ar, Kr – 96 kJ mol–1, Xe – 77 kJ mol–1, Hence, Ne has highest positive electron gain enthalpy., 183. (b), 184. (a) As size increases, van der Waal's forces of attraction, between n oble gas atoms also increases., Consequently, ease of their liquefaction increases., 185. (c) Solubility increases from He to Rn, 186. (d) Xe forms maximum compounds hence it is most, reactive, 187. (b) Noble gases exhibit low chemical activity, 188. (d), 189. (b) In XeOF4, Xenon is sp3d 2 hybridised and has one, lone pair of electrons., 190. (a), 191. (d) The products of the concerned reaction react each, other forming back the reactants., XeF6 3H 2 O, XeO3 6HF .
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EBD_7207, THE p-BLOCK ELEMENTS (GROUP 15, 16, 17 AND 18), , 364, , 198. (d) Hybridisation in, 1, (8 4 0 0) 6 sp3d 2, XeF4, 2, 199. (b) XeOF4 square pyramidal, 200. (c) Hybridisation of XeF2 is sp3d, 201. (b) XeF4 is planar, 202. (a) In XeO3 the hybridisation is sp3, , F, , Xe, , 192. (c) XeF2 :, , F, F, 3lp, , 203. (a) XeF2 has, , F, , F, , XeF4 :, , 205. (a), , Xe, F, , 204. (c), , F, , 2lp, , 206. (c), 207. (c), 208. (c), , F, , F, XeF6 : F, , Xe, , F, , Xe, , F, pair of electrons 3, Hybridisation of XeF4 is sp3d2 and structure is square, planar, XeF2 XeF4 XeF6, Valence electrons of Xe 8, 8, 8, Electrons involved, 2, 4, 6, in bond formation, Lone pairs left, 3, 2, 1, Hybridisation in each case is XeF4sp3d2, XeF2sp3d,, He is obtained during radioactive decay, The geometry of XeF6 is distorted octahedral in which, all the six positions are occupied by fluorine atoms, and the lone pair of electrons of Xe atom is present at, the corner of one of the triangular faces., , Xe, , O, , F, , Xe, F, , XeO3 :, , F, , F, , F, , F, , structure hence number of lone, , O, , O, 1 lp, , 209. (b) The hybridization of XeO3F2 is sp3d and its structure, is trigonal bipyramidal in which oxygen atoms are, situated on the plane and the fluoride atoms are on the, top and bottom., F, , Hence XeF2 has maximum no. of lone pairs of electrons., 193. (d) XeO2F2 has trigonal bipyramidal geometry, but due to, presence of lone pair of electrons on equitorial position,, its actual shape is see-saw., , O, , F, , F, , F, , O, , O, , Xe, , F, , O, , 210. (b) The shape of XeO3 is Trigonal Pyramidal., , Xe, O, , Xe, , F, , 194. (b) XeF4 H 2O 2HF XeOF2, 195. (a) No compound of He as yet been reported, 196. (a) No compound of Ar as yet been reported with F2, 197. (c), , XeF6 3H 2 O, , 6HF XeO 3, , O, , O, , O, , (Trigonal Pyramidal Structure), , 211. (c) Argon is used in high temperature welding and other, operations which require a non-oxidising atmosphere, and the absence of nitrogen., 212. (b) Neon gives a distinct reddish glow when used in either, low-voltage neon glow lamps or in high voltage, discharge tube.
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p-BLOCK ELEMENTS (GROUP 15, 16, 17 AND 18), , 213. (c) Helium is twice as heavy as hydrogen it is inflammable, but not lighter than hydrogen. Helium has the lowest, melting and boiling point of any element which makes, liquid helium an ideal coolant for many extremely low, temperature application such as super conducting, magnet and cryogenic research where temperature, close to absolute zero are needed. He is used in gas, cooled atomic reactors as a heat transfer agent., 214. (c) Coloured discharge tubes mainly contain Neon, 215. (a) Breathing mixture is (O2 + He), 216. (a) Mixture of (He + O2) is used for asthma patient, 217. ( b), 218. (c), 219. (d), , STATEMENT TYPE QUESTIONS, 220. (a) Phosphorus occurs in minerals of the apatite family,, Ca9(PO4)6, CaX2 (X = F, Cl or OH) which are main, components of phosphate rocks whereas arsenic and, antimony are found as sulphide minerals. The increase, in covalent radii from N to P is greater in comparison, to increase from As to Bi., 221. (c) For nitrogen oxidation states from +1 to +4, disproportionate in acidic solution only. Oxidation, state of phosphorous in P2O5 is + 5 whereas in P2O5, is +3 thus P2O5 is more acidic than P2O3., 222. (b), 223. (a) Oxygen shows oxidation state of +2 in OF2. H2O which, is a hydride of oxygen element of group 16 is neutral, in nature., 224. (c), , O, O, , O, O–, , –O, , excess, , F2 g, , 673K,1 bar, , For statements (iv), 2XeF2(s) + 2H2O(l), 6XeF4 + 12H2O, XeF6 + 3H2O, , MATCHING TYPE QUESTIONS, 230. (b), 235. (b), 240. (c), , 231. (a), 236. (b), 241. (a), , 232. (a), 237. (b), , 233. (c), 238. (b), , 234. (a), 239. (a), , ASSERTION-REASON TYPE QUESTIONS, 242. (c) At higher temperatures, dinitrogen combines with, metals to form ionic nitrides., 243. (c), 244. (a) Both assertion and reason are true and reason is the, correct explanation of assertion., M, , (metal), , HNO3, , 2HNO3, , MNO3, H, (metal nitrate) (nascent hydrogen), , (conc.), , 2H, , 2NO2, , (nascent hydrogen), , 2H 2 O, , 245. (b) White phosphorus exists as P4 tetrahedral molecule, having P-P-P bond angle 60º. Hence the molecule is, under strain and more reactive. On the other hand red, phosphorus exists as P4 tetrahedra which are joined, together through covalent bonds giving polymeric, structure., 246. (c) Bond angle of H2 S (92°) < H2O (104°31). As the, electronegativity of the central atom decreases, bond, angle decreases. In the present case, S is less, electronegative than oxygen. Thus bond pairs in H2S, are more away from the central atom than in H2O and, thus repulsive forces between bond pairs are smaller, producing smaller bond angle., 247. (a), 248. (a), , CRITICAL THINKING TYPE QUESTIONS, , O, , Ozone is diamagnetic in nature (due to presence of, paired electron) and both the O – O bond length are, equal. It has a bent structure., 225. (b) NH3 being basic reacts with acidic H2SO4 thus H2SO4, cannot be used for drying NH3., 226. (b) Physical state of iodine is different from other halogens, as iodine is solid, bromine is a liquid whereas fluorine, and chlorine are gases., 227. (b) At 298K, ClF exits as a gas., 228. (d), 229. (c) For statement (iii) preparation of XeF2 requires Xe in, excess amount, Xe g, , 365, , XeF2 s, , 249. (b) With the decrease in the electronegativity of central, atom the bond angle decreases, 250. (d), 251. (b) CuSO 4 4NH 3, [Cu ( NH 3 ) 4 ] SO 4, Blue complex due to Cu(NH 3 ) 2, 4, 252. (b) 2NO + N2O4, , 250K, , 253. (b) The slow decomposition of HNO3 is represented by, the eqn., 4HNO3, 4NO2 + 2H2O + O2, (yellow-brown), , 254. (a), 255. (c) [Fe(H2O)5NO]2+ ion is formed, 256. (b) Ag+ (aq) + Cl– (aq) AgCl(s), Colourless, , White ppt, , AgCl (s) + 2NH3 (aq), 2Xe(g) + 4HF(aq) + O2(g), 4Xe + XeO3 + 24HF + 3O2, XeO3 + 6HF, , 2N2O3, , White ppt, , [Ag(NH3 )2] Cl (aq), , Colourless, , 257. (b) NO2 and N2O4 has + 4 oxidation state for nitrogen., 258. (c) 3CuSO4 + 2PH3 Cu3P2 + 3H2SO4, 3HgCl2 + 2PH3 Hg3P2 + 6HCl
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EBD_7207, THE p-BLOCK ELEMENTS (GROUP 15, 16, 17 AND 18), , 366, , 259. (d) SH–bond is weaker than, O–H bond. Hence H2S will, furnish more H+ ions, 260. (c) Air is liquified by making use of the joule-Thompson, effect (cooling by expansion of the gas) Water vapour, and CO2 are removed by solidification. The remaining, major constituents of liquid air i.e., liquid oxygen and, liquid nitrogen are separated by means of fractional, distillation (b.p. of O2 = –183°C : b. P. of N2 = – 195.8°C), 261. (d) Formal charges help in the selection of the lowest, energy structure from a number of possible Lewis, structures for a given compound. The lowest energy, structure means the structure with the smallest formal, charge on each atom of the compound. A Lewis dot, structure is preferable when all formal charges are zero., 262. (c) From the given options we find option (a) is correct. The, oxidising power of halogens follow the order, F2 > Cl2 > Br2 > I2. Option (b) is incorrect because it in, not the correct order of electron gain enthalpy of, halogens., The correct order is Cl2 > F2 > Br 2 > I2. The low value, of F2 than Cl2 is due to its small size., Option (c) is incorrect. The correct order of bond, dissociation energies of halogens is, Cl2 > Br2 > F2 > I2., Option (d) is correct. It is the correct order of, electronegativity values of halogens. Thus option (b), and (c) are incorrect., 263. (c) The H–X bond strength decreases from HF to HI. i.e., HF > HCl > HBr > HI. Thus HF is most stable while HI, is least stable. The decreasing stability of the hydrogen, halide is also reflected in the values of dissociation, energy of the H–X bond, H, , F, , 135kcal mol, , H Cl, 103kcal mol, , H, , Br, , 87 kcal mol, , 266. (c) 2F2 (g) + 2H2O (l), Cl2 (g) + 2H2O (l), 267. (d), , 1, , 4H+ (aq) + 4F– (aq) + O2 (g), HCl (aq) + HOCl, 5, , 1, , NaClO3 2NaCl All statements are, NaClO, correct as evident from the reaction, , 268. (b) 2KMnO 4 16HCl, , 2KCl 2MnCl 2 8H 2 O 5Cl 2, , 269. (d), 270. (c) XeF6 + 3H2O, XeO3 + 6HF, Complete hydrolysis of XeF6 gives XeO3 (an, explosive)., 271. (d) (i) The first ionization energy of xenon (1, 170 kJ mol–, 1) is quite close to that of dioxygen (1,180 kJ mol–1)., (ii) The molecular diameters of xenon and dioxygen are, almost identical., Based on the above similarities Barlett (who prepared, O2+[PtF6]– compound) suggested that since oxygen, combines with PtF6, so xenon should also form similar, compound with PtF6., 272. (c), , 2XeF6 SiO 2, SiF4 2XeOF4, 273. (c) Xenon undergo sp3 hybridization., , 5p, , 5s, , 5s, , (ground, state), , 5p, , 5d, , 5d, , (third excited, state), , In the fourth excited state xenon atom, has 8 unpaired, electrons, , H I, 71kcal mol, , 264. (c) MI > MBr > MCl > MF. As the size of the anion, decreases covalent character also decreases., 265. (a) Metal halides with higher oxidation state are more, covalent than the one in lower oxidation state., , 5p, 5d, 5s, 3, One s and three p orbital undergo sp hybridization., Four sp3 hybrid orbitals form four bonds with oxygen, atoms. They are sp3 – p. Four p – d bonds are also, formed with oxygen atoms by the unpaired electrons.
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22, THE d-AND f-BLOCK ELEMENTS, FACT / DEFINITION TYPE QUESTIONS, 1., , 2., , 3., , The transition elements have a general electronic, configuration, (a), , ns 2, np 6, nd 1, , (b), , ( n 1) d 1, , (c), , (n 1) d1 10, ns1 2, , 10, , 10, , , ns 0, , 2, , , np 0, , 6, , (d) n d1 10, ns1 2, Correct electronic configuration of Cr (Z = 24) is, (a) 1s2 2s2 2p6 3s2 3p6 3d 7 4s1, (b) 1s2 2s2 2p6 3s2 3p6 3d 5 4s1, (c) 1s2 2s2 2p6 3s2 3p6 3d 7 4s2, (d) 1s2 2s2 2p6 3s2 3p6 3d 6 4s2, Which of the following configuration is correct for iron ?, (a), , 1s2 ,2 s2 2 p6 ,3s2 3 p6 3d 4, 2, , 2, , 6, , 2, , 6, , 6, , (b) 1s , 2s 2 p ,3s 3 p 3d 4s, (c), , 5., , 8., , 12., , 1s2 ,2s2 2 p6 ,3s2 3 p6 3d 2, , (a) Ni3+, (b) Mn 3+, (c) Fe3+, (d) Co3+, (At. Nos. Mn = 25, Fe = 26, Co = 27, Ni = 28), Which of the following element does not belong to first, transition series?, (a) Fe, (b) V, (c) Ag, (d) Cu, , 7., , 11., , 13., , 4., , n 1 d 10 ns 2 is the general electronic configuration of, (a) Fe, Co, Ni, (b) Cu, Ag, Au, (c) Zn, Cd, Hg, (d) Se, Y, La, The last electron in d-block elements goes to, (a) (n-1) d, (b) nd, (c) np, (d) (n-1) s, The elements which exhibit both vertical and horizontal, similarites are, (a) inert gas elements, (b) representative elements, (c) rare elements, (d) transition elements, , An atom has electronic configuration, , 1s 2 2s 2 2 p 6 3s 2 3 p6 3d 3 4s 2 in which group would it be, placed?, (a) Fifth, (b) Fifteenth, (c) Second, (d) Third, 10. In 3d-series atomic number (Z) varies from, (a) Z 21 30, (b) Z 22 30, , 2, , 2, 2 6, 2 6 2 2, (d) 1s , 2s 2 p ,3s 3 p 3d 4s, Which one of the following ions has electronic, configuration [Ar] 3d 6 ?, , 6., , 9., , 14., , 15., , 16., , (c) Z 20 30, (d) Z 31 40, The valence shell of transition elements consists of, (a) nd orbitals, (b) (n-1) d orbitals, (c) ns np nd orbitals, (d) (n-1) d ns orbitals, Number of unpaired electrons in Ni2+(Z=28) is, (a) 4, (b) 2, (c) 6, (d) 8, Which of the following element is not a member of transition, elements ?, (a) Zn, (b) Pt, (c) Ce, (d) Mo, The number of unpaired electrons in gaseous species of, Mn3+, Cr 3+ and V3+ respectively are., (a) 4, 3 and 2, (b) 3, 3 and 2, (c) 4, 3 and 2, (d) 3, 3 and 3, The first element in the 3d-transition series is, (a) Sc, (b) Ti, (c) V, (d) Ca, Which of the following has more unpaired d-electrons?, (a), , (b), , Zn, , Fe 2, , (c) Ni, (d) Cu, 17. The number of unpaired electrons in a nickel atom in ground, state are At. No. of Ni, , 28, , (a) 2, (b) 5, (c) 3, (d) 7, 18. Which one of the following is an example of non-typical, transition elements ?, (a) Li, K, Na, (b) Be, Al, Pb, (c) Zn, Cd, Hg, (d) Ba, Ga, Sr.
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EBD_7207, THE d-AND f-BLOCK ELEMENTS, , 368, , 19., , 20., , 21., , 22., , 23., , 24., , 25., , 26., , 27., , 28., , 29., , 30., , Which of the following has the maximum number of unpaired, electrons?, (a) Ti2+, (b) Fe2+, (c) Cr+, (d) Cu+, The outer electronic configuration of Ag is 4d 10 5s1, it, belongs to, (a) 5th period, group 4, (b) 4th period, group 5, th, (c) 5 period, group 11 (d) 6th period, group 9, Manganese belongs to, (a) 1st transition series, (b) 2nd transition series, rd, (c) 3 transition series, (d) 4th transition series, The no. of unpaired electrons in Mn7+ ions, (At. no. of Mn = 25) is, (a) 0, (b) 1, (c) 2, (d) 3, Which one of the following species is paramagnetic?, (a) N2, (b) Co, (d) Zn, (c) Cu+, Which of the following species is/are paramagnetic?, Fe2+, Zn0, Hg2+, Ti4+, (a) Fe2+ only, (b) Zn 0 and Ti4+, 2+, 2+, (c) Fe and Hg, (d) Zn 0 and Hg2+, In first transition series, the melting point of Mn is low, because, (a) due to d10 configuration, metallic bonds are strong, (b) due to d7 configuration, metallic bonds are weak, (c) due to d5 configuration, metallic bonds are weak, (d) None of these, The transition metals have a less tendency to form, ions due to, (a) high ionisation energy, (b) low heat of hydration of ions, (c) high heat of sublimation, (d) All of these, The common oxidation states of Ti are, (a) + 2 and + 3, (b) + 3 and + 4, (c) – 3 and – 4, (d) + 2, + 3 and + 4, Maximum oxidation state is shown by, (a) Os, (b) Mn, (c) Co, (d) Cr, Which one of the elements with the following outer orbital, configurations may exhibit the largest number of oxidation, states?, (a) 3d 54s1, (b) 3d 54s2, 2, 2, (c) 3d 4s, (d) 3d 34s2, Which of the following pairs has the same size?, (a), , 31., , Fe 2+ , Ni 2+, , (b), , Zr 4+ , Ti 4+, , (c) Zr 4+ , Hf 4+, (d) Zn 2+ , Hf 4+, For the four successive transition elements (Cr, Mn, Fe and, Co), the stability of +2 oxidation state will be there in which, of the following order?, (a) Mn > Fe > Cr > Co, (b) Fe > Mn > Co > Cr, (c) Co > Mn > Fe > Cr, (d) Cr > Mn > Co > Fe, , 32., , Iron exhibits +2 and + 3 oxidation states. Which of the, following statements about iron is incorrect ?, (a) Ferrous oxide is more basic in nature than the ferric, oxide., (b) Ferrous compounds are relatively more ionic than the, corresponding ferric compounds., (c) Ferrous compounds are less volatile than the, corresponding ferric compounds., (d) Ferrous compounds are more easily hydrolysed than, the corresponding ferric compounds., 33. Four successive members of the first row transition, elements are listed below with their atomic numbers., Which one of them is expected to have the highest third, ionization enthalpy?, (a) Vanadium (Z = 23), (b) Chromium (Z = 24), (c) Manganese (Z = 25) (d) Iron (Z = 26), 34. Of the following outer electronic configurations of atoms,, the highest oxidation state is achieved by which one of, them ?, (a) (n – 1)d 3 ns2, (b) (n – 1)d 5 ns1, 8, 2, (c) (n – 1)d ns, (d) (n – 1)d 5 ns2, 35. For d block elements the first ionization potential is of the, order, (a) Zn > Fe > Cu > Cr, (b) Sc = Ti < V = Cr, (c) Zn < Cu < Ni < Co, (d) V > Cr > Mn > Fe, 36. Which of the following does not represent the correct order, of the properties indicated ?, (a) Ni2+ > Cr2+ > Fe2+ > Mn2+ (size), (b) Sc > Ti > Cr > Mn (size), (c) Mn2+ > Ni2+ < Co2+ <Fe2+ (unpaired electron), (d) Fe2+ > Co2+ > Ni2+ > Cu2+ (unpaired electron), 37. Zinc and mercury do not show variable valency like d-block, elements because, (a) they are soft, (b) their d-shells are complete, (c) they have only two electrons in the outermost subshell, (d) their d-shells are incomplete, 38. Which of the following transition element shows the, highest oxidation state ?, (a) Mn, (b) Fe, (c) V, (d) Cr, 39. Which of the following elements does not show vari able, oxidation states?, (a) Copper, (b) Iron, (c) Zinc, (d) Titanium, 40. Which one of the following transition elements does not, exhibit variable oxidation state?, (a) Ni, (b) Cu, (c) Fe, (d) Sc, 41. Electronic configuration of a transition element X in +3, oxidation state is [Ar]3d5. What is its atomic number ?, (a) 25, (b) 26, (c) 27, (d) 24
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THE d-AND f-BLOCK ELEMENTS, , 42., , 43., , 44., , 45., , 46., , 47., , 48., , 49., , 50., , Metallic radii of some transition elements are given below., Which of these elements will have highest density ?, Element, Fe, Co Ni, Cu, Metallic radii/pm, 126 125 125 128, (a) Fe, (b) Ni, (c) Co, (d) Cu, Transition metals mostly are, (a) diamagnetic, (b) paramagnetic, (c) neither diamagnetic nor paramagnetic, (d) both diamagnetic and paramagnetic, Transition metals usually exhibit highest oxidation states in, their, (a) chlorides, (b) fluorides, (c) bromides, (d) iodides, Which of the following statements is incorrect?, (a) Zn,Cd and Hg due to presence of completely filled, d-orbitals [(n–1)d10ns2] are not studied along with, other transition metals., (b) Zn, Cd and Hg have low m.p and are comparitively, softer than other transition metals., (c) Metallic bond made by elements with d 5 configuration, is stronger as compared to metalic bond made by, elements with d 3 configuration., (d) Metals of 5d series forms strong metallic bonds as, compared with metals of 3d series., Which of the following is incorrect?, (a) Mn shows oxidation state of +7 in MnF7, (b) Fe and Co shows +3 oxidation state in FeX3 and CoF3., (c) V shows oxidation state of + 5 in VF5., (d) Cu does not shows +2 oxidation state with I–., Which of the following is not correct about transition, metals?, (a) Their melting and boiling points are high, (b) Their compounds are generally coloured, (c) They can form ionic or covalent compounds, (d) They do not exhibit variable valency, Transition elements, (a) have low melting point, (b) exhibit variable oxidation states, (c) do not form coloured ions, (d) show inert pair effect, Which one of the following ions is the most stable in, aqueous solution?, (a) V3+, (b) Ti3+, 3+, (c) Mn, (d) Cr3+, (At.No. Ti = 22, V = 23, Cr = 24, Mn = 25), Which one of the following does not correctly represent, the correct order of the property indicated against it?, (a) Ti < V < Cr < Mn : increasing number of oxidation, states, (b) Ti3+ < V3+ < Cr3+ < Mn3+ : increasing magnetic moment, (c) Ti < V < Cr < Mn : increasing melting points, (d) Ti < V < Mn < Cr : increasing 2nd ionization enthalpy, , 369, , 51. What is wrong about transition metals?, (a) Diamagnetic, (b) Paramagnetic, (c) Form complexes, (d) Shows variable oxidation state, 52. Which of the following ions has the maximum magnetic, moment?, (a) Mn +2, (b) Fe+2, 3+, (c) Ti, (d) Cr+2., 53. Four successive members of the first row transition elements, are listed below with atomic numbers. Which one of them is, expected to have the highest E, , M3 / M 2, , value ?, , (a) Cr(Z = 24), (b) Mn(Z = 25), (c) Fe(Z = 26), (d) Co(Z = 27), 54. Which one of the following ions exhibit highest magnetic, moment?, (a) Cu 2, (b) Ti3, (c), , Ni 2, , (d), , 55. A compound of a metal ion M x, , Mn 2, Z, , 24 has a spin, , only magnetic moment of 15 Bohr Magnetons. The, number of unpaired electrons in the compound are, (a) 2, (b) 4, (c) 5, (d) 3, 56. Titanium shows magnetic moment of 1.73 B.M. in its, compound. What is the oxidation number of Ti in the, compound?, (a) +1, (b) +4, (c) +3, (d) +2, 57. Which of the following ions having following electronic, structure would have maximum magnetic moment?, (a) 1s 2 2 s 2 2 p 6 3s 2 3 p 6 3d 3, (b) 1s 2 2 s 2 2 p 6 3s 2 3 p 6 3d 5, (c), , 1s 2 2s 2 2 p 6 3s 2 3 p 6 3d 7, , (d) 1s 2 2 s 2 2 p 6 3s 2 3 p 6 3d 9, 58. If n is the number of unpaired electrons, the magnetic, moment (in BM) of transition metal/ion is given by, (a), , n n 2, , (b), , 2n n 1, , (c), , n n 2, , (d), , 2n n 1, , 59. Which one of the following ions has the maximum, magnetic moment?, (a) Sc3+, (b) Ti3+, (c) Cr3+, (d) Fe3+, 60. The magnetic nature of elements depend on the presence, of unpaired electrons. Identify the configuration of, transition element, which shows highest magnetic moment., (a) 3d 7, (b) 3d 5, 8, (c) 3d, (d) 3d 2
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EBD_7207, THE d-AND f-BLOCK ELEMENTS, , 370, , 61., , Transition elements show magnetic moment due to spin, and orbital motion of electrons. Which of the following, metallic ions have almost same spin only magnetic moment ?, (i) Co2+, (ii) Cr2+, (iii) Mn 2+, (iv) Cr3+, (a) (i) and (iii), (b) (i) and (iv), (c) (ii) and (iii), (d) (ii) and (iv), 62. The aqueous solution containing which one of the, following ions will be colourless? (Atomic number:, Sc = 21, Fe = 26, Ti = 22, Mn = 25), (a) Sc3+, (b) Fe2+, 3+, (c) Ti, (d) Mn 2+, 63. Transition elements form coloured ions due to, (a) d-d transition, (b) fully filled d-orbitals, (c) smaller atomic radii, (d) availability of s-electrons, 64. The catalytic activity of transition metals and their, compounds is mainly due to, (a) their magnetic behaviour, (b) their unfilled d-orbitals, (c) their ability to adopt variable oxidation state, (d) their chemical reactivity, 65. Which of the following is colourless in water?, (a), 66., , Ti3, , (b), , 73., , 74., , 75., , VO2 , VO2 , TiO 2 , CrO 24, , 76., , V3, , (c) Cu 3, (d) Sc3, Which group contains coloured ions out of, (i), , Cu, , 2, , (ii), , Ti, , 2, , 67., , 68., , 69., , 70., , 71., , 72., , 77., , 78., , 79., , 80., , TiO 2, , CrO 24, , (a), , VO2, , VO2, , (b), , VO2, , TiO 2, , VO 2, , CrO 42, , (c), , CrO 24, , TiO 2, , VO2, , VO2, , (d), , TiO 2, , VO 2, , VO 2, , CrO 42, , Which of the following ion(s) is/are oxidising in nature?, (i), , V2+ EM 2 / M, , (ii), , Mn3, , (iii) Cr 2, , 4, , (iii) Co, (iv) Fe 2, (b) (i), (iii), (iv), (a) (i), (ii), (iii), (iv), (c) (ii), (iii), (d) (i), (ii), Which of the following statements about the interstitial, compounds is incorrect ?, (a) They are chemically reactive., (b) They are much harder then the pure metal., (c) They have higher melting points than the pure metal., (d) They retain metallic conductivity., Formation of interstitial compound makes the transition, metal, (a) more soft, (b) more ductile, (c) more metallic, (d) more hard, If a non metal is added to the interstital sites of a metal, then, the metal becomes, (a) softer, (b) less tensile, (c) less malleable, (d) more ductile, Gun metal is an alloy of, (a) Cu and Al, (b) Cu and Sn, (c) Cu, Zn and Sn, (d) Cu, Zn and Ni, Brass is an alloy of, (a) Zn and Sn, (b) Zn and Cu, (c) Cu, Zn and Sn, (d) Cu and Sn, Which one of the following is coinage metal ?, (a) Zn, (b) Cu, (c) Sn, (d) Pb., , Bronze is an alloy of, (a) Pb + Sn + Zn, (b) Cu + Sn, (c) Pb + Zn, (d) Cu + Zn, An alloy of transition metal containing a non transition, metal as a constituent is, (a) invar, (b) bronze, (c) chrome steel, (d) stainless steel, Choose the correct increasing order of the oxidation state, of the central metal atom in the following oxoanions., , E, E, , 1.18, 1.57, , M3 / M 2, , M2 / M, , 0.91, , (a) (i) and (iii), (b) only (ii), (c) (ii) and (iii), (d) only (iii), Which of the following transition metal ion is colourless in, aqueous solution?, (a) Ti4+, (b) Zn 2+, 4+, (c) V, (d) Both (a) and (b), Transition metals show catalytic activity, (a) Due to their ability to form complexes., (b) Due to their ability to show multiple oxidation state., (c) Due to availabiltiy of d orbitals for bond formation., (d) Both (a) and (b)., Which of the following transition metal on catalysis the, reaction between iodide and persulphate ion?, (a) Fe2+, (b) Fe3+, 2+, (c) Ni, (d) Both (a) and (c), Which of the following reactions are disproportionation, reactions ?, (i), , Cu, , (ii), , 3MnO 4, , Cu 2, , 4H, , (iii) 2KMnO 4, (iv) 2MnO 4, , 3Mn 2, , (a) (i) and (ii), (c) (ii), (iii) and (iv), , Cu, , 2MnO4, , MnO 2, , K 2 MnO4, , MnO 2 O 2, , 2H 2O, , 5MnO 2, , 2H 2 O, , (b) (i), (ii) and (iii), (d) (i) and (iv), , 4H
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THE d-AND f-BLOCK ELEMENTS, , 81., , 82., , 83., , 84., , 85., , 86., , 87., , 88., , 89., , 371, , In the form of dichromate, Cr (VI) is a strong oxidising agent, in acidic medium but Mo (VI) in MoO3 and W (VI) in WO3, are not because ____________ ., (i) Cr (VI) is more stable than Mo(VI) and W (VI)., (ii) Mo (VI) and W(VI) are more stable than Cr(VI)., (iii) Higher oxidation states of heavier members of group6 of transition series are more stable., (iv) Lower oxidation states of heavier members of group-6, of transition series are more stable., (a) (i) and (ii), (b) (ii) and (iii), (c) (i) and (iv), (d) (ii) and (iv), K2Cr2O7 on heating with aqueous NaOH gives, (a), , CrO 24, , (b) Cr(OH)3, , (c), , Cr2O 72, , (d) Cr(OH)2, , CrO3 dissolves in aqueous NaOH to give, (b) CrO42–, (a) Cr2O72–, (c) Cr(OH)3, (d) Cr(OH)2, The oxidation state of chromium in the final product formed, by the reaction between KI and acidified potassium, dichromate solution is, (a) + 3, (b) + 2, (c) + 6, (d) + 4, The bonds present in the structure of dichromate ion are, (a) four equivalent Cr – O bonds only, (b) six equivalent Cr – O bonds and one O – O bond, (c) six equivalent Cr – O bonds and one Cr – Cr bond, (d) six equivalent Cr – O bonds and one Cr – O – Cr bond, Potassium dichromate when heated with concentrated, sulphuric acid and a soluble chloride, gives brown-red, vapours of, (a) CrO3, (b) CrCl3, (c) CrO2Cl2, (d) Cr2O3, The acidic, basic or amphoteric nature of Mn2O7, V2O5 and, CrO are respectively, (a) acidic, acidic and basic, (b) basic, amphoteric and acidic, (c) acidic, amphoteric and basic, (d) acidic, basic and amphoteric, Which of the following oxides of Cr is amphoteric, (a) CrO2, (b) Cr2O3, (c) CrO5, (d) CrO3, Which of the following is amphoteric oxide ?, Mn 2 O7 , CrO3 , Cr2O3 ,CrO, V2 O5 , V2 O4, , 90., , (a), , V2 O5 , Cr2 O3, , (b), , Mn 2 O7 , CrO3, , (c), , CrO, V2 O5, , (d), , V2O5 , V2 O 4, , When acidified K 2Cr2 O7 solution is added to Sn2+ salts, then Sn2+ changes to, (a) Sn, (b) Sn3+, (c) Sn4+, (d) Sn+, , 91. In neutral or faintly alkaline medium, thiosulphate is, quantitatively oxidized by KMnO4 to, (b) SO42–, (a) SO32–, (c) SO2, (d) SO52–, 92. KMnO4 can be prepared from K2MnO4 as per the reaction:, 3MnO 24, , 2MnO24, , 2H 2 O, , MnO2 + 4OH, , The reaction can go to completion by removing OH– ions, by adding., (a) KOH, (b) CO2, (c) SO2, (d) HCl, 93. In the laboratory, manganese (II) salt is oxidised to, permanganate ion in aqueous solution by, (a) hydrogen peroxide, (b) conc. nitric acid, (c) peroxy disulphate, (d) dichromate, 94. The starting material for the manufacture of KMnO4 is, (a) pyrolusite, (b) manganite, (c) magnatite, (d) haematite, 95. An explosion take place when conc. H2SO4 is added to, KMnO4. Which of the following is formed?, (a), , Mn 2O7, , (b), , MnO 2, , (c) MnSO4, (d) M 2O3, 96. If KMnO4 is reduced by oxalic acid in an acidic medium, then oxidation number of Mn changes from, (a) 4 to 2, (b) 6 to 4, (c) +7 to +2, (d) 7 to 4, 97. KMnO4 acts as an oxidising agent in alkaline medium. When, alkaline KMnO4 is treated with KI, iodide ion is oxidised to, ________ ., (a), , I2, , (b), , IO, , (c), , IO3, , (d), , IO4, , 98. On the basis of data given below,, E, , E, , Sc3 /Sc2, , Cr 2 /Cr, , 0.37 , E, , 0.90 , E, , Mn 3 /Mn 2, , Cu 2 /Cu, , 1.57, , 0.34, , Which of the following statements is incorrect?, (a) Sc 3+ has good stability due of [Ar]3d 0 4s 0, configuration., (b) Mn3+ is more stable than Mn 2+., (c) Cr2+ is reducing in nature., (d) Copper does not give H2 on reaction with dil. H2SO4., 99. Which of the following is most acidic?, (a) Mn 2O7, (b) V2O5, (c) Fe2O3, (d) Cr2O3, 100. Which of the following is the use of potassium, permanganate?, (a) Bleaching of wool, cotton and silk fibers., (b) decolourisation of oils., (c) In analytical chemistry., (d) All of these.
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EBD_7207, THE d-AND f-BLOCK ELEMENTS, , 372, , 101. Which of the following is not correctly matched?, Compound of, Use, transition metal, (a) TiO, Pigment industry, (b) MnO2, Dry battery cell, (c) V2O5, Manufacture of H2SO4, (d) PdCl2, Manufacture of polyethylene, 102. A series1 metal ion , M(II) aqueous solution react with the, KI to form iodine and a precipitate is formed, this M(II) can, be:, (a) Zn2+, (b) Mn 2+, (c) Cu2+, (d) Ni2+, 103. Total number of inner transition elements in the periodic, table is, (a) 10, (b) 14, (c) 28, (d) 30, 104. Which of the following ions will exhibit colour in aqueous, solutions?, (a), 105., , 106., , 107., , 108., , 109., , La 3+ (Z = 57), , (b), , Ti3+ (Z = 22), , (c) Lu 3+ (Z = 71), (d) Sc3+ (Z = 21), The lanthanoide contraction is responsible for the fact that, (a) Zr and Y have about the same radius, (b) Zr and Nb have similar oxidation state, (c) Zr and Hf have about the same radius, (d) Zr and Zn have the same oxidation state, (Atomic numbers : Zr = 40, Y = 39, Nb = 41, Hf = 72, Zn = 30), Which one of the following elements shows maximum, number of different oxidation states in its compounds?, (a) Eu, (b) La, (c) Gd, (d) Am, Lanthanoids are, (a) 14 elements in the sixth period (atomic no. = 90 to 103), that are filling 4f sublevel, (b) 14 elements in the seventh period (atomic no. = 90 to, 103) that are filling 5f sublevel, (c) 14 elements in the sixth period (atomic no. = 58 to 71), that are filling 4f sublevel, (d) 14 elements in the seventh period (atomic no. = 58 to, 71) that are filling 4f sublevel, Which of the following factors may be regarded as the main, cause of lanthanoide contraction?, (a) Greater shielding of 5d electrons by 4f electrons, (b) Poorer shielding of 5d electrons by 4f electrons, (c) Effective shielding of one of 4f electrons by another in, the subshell, (d) Poor shielding of one of 4f electron by another in the, subshell, Lanthanoid which has the smallest size in +3 state is, (a) Tb, (b) Er, (c) Ce, (d) Lu, , 110. Lanthanum is grouped with f-block elements because, (a) it has partially filled f-orbitals, (b) it is just before Ce in the periodic table, (c) it has both partially filled f and d-orbitals, (d) properties of lanthanum are very similar to the elements, of f-block, 111. A reduction in atomic size with increase in atomic number is, a characteristic of elements of, (a) high atomic masses, (b) d-block, (c) f-block, (d) radioactive series, 112. Which of the following oxidation states is the most common, among the lanthanoids?, (a) 3, (b) 4, (c) 2, (d) 5, 113. Identify the incorrect statement among the following:, (a) 4f and 5f orbitals are equally shielded., (b) d-Block elements show irregular and erratic chemical, properties among themselves., (c) La and Lu have partially filled d-orbitals and no other, partially filled orbitals., (d) The chemistry of various lanthanoids is very similar., 114. In context of the lanthanoids, which of the following, statements is not correct?, (a) There is a gradual decrease in the radii of the members, with increasing atomic number in the series., (b) All the members exhibit +3 oxidation state., (c) Because of similar properties the separation of, lanthanoids is not easy., (d) Availability of 4f electrons results in the formation of, compounds in +4 state for all the members of the series., 115. The outer electronic configuration of Gd (Atomic No. : 64) is, (a) 4f 3 5d5 6s2, (b) 4f 8 5d 0 6s2, 4, 4, 2, (c) 4f 5d 6s, (d) 4f 7 5d 1 6s2, 116. The correct order of ionic radii of Y3+, La3+, Eu3+ and Lu3+, is, (a), , La 3, , (b), , Y3, , La 3, , Eu 3, , Lu 3, , (c), , Y3, , Lu 3, , Eu 3, , La 3, , 118., , 119., , 120., , Lu 3, , Y3, , Lu 3, Eu 3, La 3, Y3, (Atomic nos. Y =39, La = 57, Eu = 63, Lu = 71), Which of the following lanthanoid ions is diamagnetic ?, (At nos. Ce = 58, Sm = 62, Eu = 63, Yb = 70), (a) Sm2+, (b) Eu2+, 2+, (c) Yb, (d) Ce2+, Lanthanide contraction can be observed in, (a) At, (b) Gd, (c) Ac, (d) Lw, The approximate percentage of iron in mischmetal is, (a) 10, (b) 20, (c) 50, (d) 5, The most common lanthanide is, (a) lanthanum, (b) cerium, (c) samarium, (d) plutonium, , (d), 117., , Eu 3
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THE d-AND f-BLOCK ELEMENTS, , 121. Non-lanthanide atom is, (a) La, (b) Lu, (c) Pr, (d) Pm, 122. In which of the following lanthanides oxidation state +2 is, most stable?, (a) Ce, (b) Eu, (c) Tb, (d) Dy, 123. Actinoides, (a) are all synthetic elements, (b) include element 104, (c) have any short lived isotopes, (d) have variable valency, 124. Which of the following exhibit only + 3 oxidation state ?, (a) U, (b) Th, (c) Ac, (d) Pa, 125. Larger number of oxidation states are exhibited by the, actinoids than those by the lanthanoids, the main reason, being, (a) 4f orbitals more diffused than the 5f orbitals, (b) lesser energy difference between 5f and 6d than, between 4f and 5d orbitals, (c) more energy difference between 5f and 6d than between, 4f and 5d orbitals, (d) more reactive nature of the actionids than the, lanthanoids, 126. The maximum oxidation state exhibited by actinide ions is, (a) +5, (b) +4, (c) +7, (d) +8, 127. There are 14 elements in actinoid series. Which of the, following elements does not belong to this series ?, (a) U, (b) Np, (c) Tm, (d) Fm, 128. Which of the following actinoids show oxiation states upto, +7 ?, (i) Am, (ii) Pu, (iii) U, (iv) Np, (a) (i) and (ii), (b) (ii) and (iv), (c) (iii) and (iv), (d) (i) and (iii), 129. Which of the following lanthanoid element is steel hard in, nature?, (a) Eu, (b) Pm, (c) Sm, (d) Ce, 130. What is the percentage of lanthanoid metal in mischmetall?, (a) 90%, (b) 20%, (c) 5%, (d) 95%, 131. Which of the following is the use of mischmetall ?, (a) In bullets, (b) In lighter flint, (c) As catalyst in petroleum cracking, (d) Both (a) and (b), 132. Which of the following actinoid element has 5f 7 6d1 7s2, configuration?, (a) Bk, (b) Cm, (c) Pa, (d) No, , 373, , 133. The increasing order of the shielding of electrons by the, orbitals ns,np,nd,nf is, (a) ns,np,nd,nf, (b) np,ns,nd,nf, (c) nd,nf,np,ns, (d) nf,nd.np,ns, 134. Which of the following in its oxidation state shows the, paramagnetism ?, (a) Tb(IV), (b) Lu(III), (c) Ce(IV), (d) La(III), , STATEMENT TYPE QUESTIONS, 135. Mark the correct statement(s)., (i) Manganese exhibits +7 oxidation state, (ii) Zinc forms coloured ions, (iii) [CoF6]3– is diamagnetic, (iv) Sc forms +4 oxidation state, (v) Zn exhibits only +2 oxidation state, (a) (i) and (ii), (b) (i) and (v), (c) (ii) and (iv), (d) (iii) and (iv), 136. Which of the following statements are correct ?, (i) The maximum oxidation state of Mn with the oxygen is, +VII while with fluorine is +IV., (ii) Fluorine is more oxidizing in nature than oxygen., (iii) Fluorine exhibit an oxidation state of –1., (iv) Seven fluorine cannot be accommodated around Mn., (a) (i), (ii) and (iii), (b) (ii), (iii) and (iv), (c) (i) and (iv), (d) (i), (ii), (iii) and (iv), 137. Which of the following statements are correct ?, (i) Chromium has the highest melting point among the, series 1 metals., (ii) Number of unpaired electrons is greater in Cr than other, elements of series 1., (iii) In any row the melting point of transition metal, increases as the atomic number increases., (a) (i) and (iii), (b) (i) and (ii), (c) (ii) and (iii), (d) (i), (ii) and (iii), 138. Read the following statements?, (i) Aqueous solutions formed by all ions of Ti are, colourless., (ii) Aqueous solution of ferrous ions is green in colour., (iii) Small size and presence of vacant d-orbitals make, transition metal ions suitable for formation of complex, compounds., (iv) Catalytic action of transition metals involves the, increase of reactant concentration at catalyst surface, and weakening of the bonds in the reacting molecules., Which of the following is the correct code for above, statements?, (a) FTTT, (b) TFFT, (c) TFTT, (d) FFTT
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EBD_7207, 374, , 139. Which of the following statements are correct?, (i) Interstitial compounds contain non-metal atoms, trapped inside the metal crystal whereas alloys are, homogeneous blend of metals., (ii) Steel and bronze are alloys of transition and nontransition metals., (iii) Some boride containing interstitial compounds are very, hard comparable to that of diamond., (iv) Interstitial compounds are chemically more reactive, than parent metal., (a) (i) and (iii), (b) (ii) and (iv), (c) (ii) and (iii), (d) (i), (ii) and (iii), 140. Which of the following statements are correct?, (i) As a result of lanthanoid contraction members of 4d, and 5d series exhibit similar radii., (ii) IE2 is high for Cr and Cu whereas IE3 is very high for, Zn., (iii) Heavier members of d-block elements like p-block, elements favours lower oxidation states., (iv) In any transition series maximum number of oxidation, states is shown by middle elements or elements near, middle elements., (a) (i) and (ii), (b) (i), (ii) and (iv), (c) (i), (ii) and (iii), (d) (ii) and (iv), 141. Consider the following statements, (i) La(OH)3 is the least basic among hydroxides of, lanthanides., (ii) Zr4+ and Hf4+ posses almost the same ionic radii., (iii) Ce4+ can as an oxidizing agent., Which of the above is/are true ?, (a) (i) and (iii), (b) (ii) and (iii), (c) (ii) only, (d) (i) and (ii), 142. Read the following statements., (i) Chemistry of actinoids is complex in comparsion to, chemistry of lanthanoids., (ii) Ce4+ is very good reducing agent., (iii) Eu2+ is a strong reducing agent., (iv) Out of all lanthanides Ce,Pr,Nd,Dy and Ho shows +4, oxidation state., Which of the following is the correct code for the statements, above?, (a) TTFF, (b) TFTF, (c) FTFT, (d) FTTF, 143. Read the following statements?, (i) Only Pu show maximum oxidation state of +7 in, actinoids., (ii) M4+ ion of Th is the only diamagnetic M4+ ion of, actinoid series., (iii) Electrons present in the 5f orbitals of actinides can, participate in bonding to a firm greater extent as, compared to electrons present in 4f orbitals of, lanthanides., (iv) Magnetic properties of actinoids are more complex, than lanthanoids, , THE d-AND f-BLOCK ELEMENTS, , Which of the following is the correct code for the statements, above?, (a) FTTT, (b) TFTT, (c) TFFT, (d) FFTT, 144. Which of the following statement(s) regarding Hf and Zr, is/are correct ?, (i) Hf has greater density than Zr., (ii) Lanthanoid contraction is responsible for such radii., (a) Both (i) and (ii) are correct., (b) Both (i) and (ii) are incorrect, (c) Statement (i) is correct only, (d) Statement (ii) is correct only., , MATCHING TYPE QUESTIONS, 145. Match the columns, Column-I, Column -II, (A) Metal of the 3d-series which (p) Manganese, does not form MO type oxide., (B) Metal of the 3d-series which (q) Vanadium, forms most covalent oxide., (C) Metal of the 3d-series which (r) Scandium, forms the amphoteric oxide., (a) A – (p), B – (r), C – (q), (b) A – (r), B – (p), C – (q), (c) A – (r), B – (q), C – (p), (d) A – (q), B – (p), C – (r), 146. Match the columns, Column-I, Column-II, (Ion), (M calculated), (A) Ti2+, (p) 2.84, (B) Zn2+, (q) 5.92, (C) Mn 2+, (r) 0, (D) Sc3+, (s) 4.90, (a) A – (s), B – (p), C – (q), D – (r)., (b) A – (r), B – (p), C – (q), D – (s)., (c) A – (p), B – (r), C – (q), D – (s)., (d) A – (p), B – (s), C – (q), D – (r)., 147. Match the columns, Column-I, Column-II, (A) Compound formed when, (p) acidified, yellow CrO24 is acidified., , MnO4, , (B) reagent oxidises Fe2+ to Fe3+ (q) Cr2 O72, (C) Compound produced when, (r) K2MnO4, MnO2 is fused with KNO3, (D) Compound having dark, (s) KMnO4, purple crystals isostructural, with KClO4, (a) A – (q),B – (p), C – (r), D – (s), (a) A – (p),B – (q), C – (r), D – (s), (a) A – (q),B – (r), C – (p), D – (s), (a) A – (q),B – (p), C – (s), D – (r)
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THE d-AND f-BLOCK ELEMENTS, , 148. Match the columns, Column-I, (A) Lanthanide hard as steel., (B) Lanthanide with maximum, paramagnetic character in, Ln4+ state., (C) Lanthanide with maximum, value of E° for reaction, Ln(s)., Ln3+(aq)+3e–, (D) Lanthanide whose Ln 3+ ion is, diamagnetic in nature, (a) A – (r), B – (s), C – (p), D – (q), (b) A – (r), B – (q), C – (s), D – (p), (c) A – (s), B – (r), C – (q), D – (p), (d) A – (r), B – (s), C – (q), D – (p), , 375, , 154., , Column-II, (p) Lu, (q) Tb, , (r) Sm, , (s) Eu, , whereas that, of Cu(I) is 3d10. Which of the following is correct ?, (a) Cu (II) is more stable, (b) Cu (II) is less stable, (c) Cu (I) and (II) are equally stable, (d) Stability of Cu (I) and Cu (II) depends on nature of, copper salts, 155. Highest oxidation state of manganese in fluoride is +4, (MnF4) but highest oxidation state in oxides is +7 (Mn2O7), because _______., (a) fluorine is more electronegative than oxygen., (b) fluorine does not possess d-orbitals., (c) fluorine stabilises lower oxidation state., (d) in covalent compounds fluorine can form single bond, only while oxygen forms double bond., 156. Four successive members of the first series of the transition, metals are listed below. For which one of them the standard, , ASSERTION-REASON TYPE QUESTIONS, Directions : Each of these questions contain two statements,, Assertion and Reason. Each of these questions also has four, alternative choices, only one of which is the correct answer. You, have to select one of the codes (a), (b), (c) and (d) given below., (a) Assertion is correct, reason is correct; reason is a correct, explanation for assertion., (b) Assertion is correct, reason is correct; reason is not a, correct explanation for assertion, (c) Assertion is correct, reason is incorrect, (d) Assertion is incorrect, reason is correct., 149. Assertion : Cuprous ion (Cu+) has unpaired electrons while, cupric ion (Cu++) does not., Reason : Cuprous ion (Cu+) is colourless whereas cupric, ion (Cu++) is blue in the aqueous solution, 150. Assertion : Transition metals show variable valency., Reason : Transition metals have a large energy difference, between the ns2 and (n – 1)d electrons., 151. Assertion : Transition metals are good catalysts., Reason : V2O5 or Pt is used in the preparation of H2SO4 by, contact process., 152. Assertion : Magnetic moment values of actinides are lesser, than the theoretically predicted values., Reason : Actinide elements are strongly paramagnetic., , º, potential E M 2, , 157., , 158., , 159., , 160., , 161., , CRITICAL THINKING TYPE QUESTIONS, 153. Among the following series of transition metal ions, the, one where all metal ions have 3d2 electronic configuration, is (At. nos. Ti = 22; V = 23; Cr = 24; Mn = 25), 3, , 2, , 3, , 4, , (a), , Ti , V, , (b), , Ti , V 4 , Cr 6 , Mn 7, , (c), , Ti 4 , V 3 , Cr 2 , Mn 3, , (d), , Ti 2 , V 3 , Cr 4 , Mn 5, , , Cr, , , Mn, , The electronic configuration of Cu(II) is 3d9, , 162., , /M, , value has a positive sign?, , (a) Co (Z = 27), (b) Ni (Z = 28), (c) Cu (Z = 29), (d) Fe (Z = 26), The standard redox potentials for the reactions, Mn2+ + 2e– Mn and Mn3+ + e– Mn2+ are –1.18 V and, 1.51 V respectively. What is the redox potential for the, reaction Mn3+ + 3e– Mn?, (a) 0.33 V, (b) 1.69 V, (c) – 0.28 V, (d) – 0.85 V, Which one of the following transition metal ions shows, magnetic moment of 5.92 BM?, (a) Mn 2+, (b) Ti3+, 3+, (c) Cr, (d) Cu2+, In the following salts the lowest value of magnetic moment, is observed in, (a) MnSO4. 4H2O, (b) CuSO4.5H2O, (c) FeSO4.6H2O, (d) ZnSO4.7H2O, In which of the following pairs both the ions are coloured in, aqueous solutions ?, (a) Sc3+, Ti3+, (b) Sc3+, Co2+, 2+, +, (c) Ni , Cu, (d) Ni2+, Ti3+, (At. no. : Sc = 21, Ti = 22, Ni = 28, Cu = 29, Co = 27), For the ions Zn2+, Ni2+ and Cr3+ which among the following, statements is correct?, (atomic number of Zn = 30, Ni = 28 and Cr = 24), (a) All these are colourless, (b) All these are coloured, (c) Only Ni2+ is coloured and Zn2+ and Cr3+ are colourless, (d) Only Zn2+ is colourless and Ni2+ and Cr3+ are coloured, Cuprous ion is colourless while cupric ion is coloured, because, (a) both have half filled p-and d-orbitals, (b) cuprous ion has incomplete d-orbital and cupric ion, has a complete d-orbital, (c) both have unpaired electrons in the d-orbitals, (d) cuprous ion has complete d-orbital and cupric ion has, an imcomplete d-orbital.
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EBD_7207, THE d-AND f-BLOCK ELEMENTS, , 376, , of the following ions V2+, V3+, V4+, Fe2+, Fe3+, , 163. The colour, are respectively, (a) green, violet, blue, green, yellow, (b) yellow, green, violet, green, blue, (c) violet, green, yellow, green, blue, (d) yellow, green, blue, green, violet, 164. Which of the following arrangements does not represent, the correct order of the property stated against it ?, (a) V2+ < Cr2+ < Mn2+ < Fe2+ : Paramagnetic behaviour, (b) Ni2+ < Co2+ < Fe2+ < Mn2+ : Ionic size, (c) Co3+ < Fe3+ < Cr 3+ < Sc3+ : Stability in aqueous, solution, (d) Sc < Ti < Cr < Mn : Number of oxidation states, 165. Acidified K2Cr2O7 solution turns green when Na2SO3 is, added to it. This is due to the formation of :, (a) Cr2(SO4)3, (b) CrO42–, (c) Cr2(SO3)3, (d) CrSO4, 166. Which of the statements is not true?, (a) On passing H2S through acidified K2Cr2O7 solution,, a milky colour is observed., (b) Na2Cr2O7 is preferred over K2Cr2O7 in volumetric, analysis., (c) K2Cr2O7 solution in acidic medium is orange., (d) K2Cr2O7 solution becomes yellow on increasing the, pH beyond 7., 167. Which one of the following is an amphoteric oxide ?, (i) Mn 2O7, (ii) CrO, (iii) V2O4, (iv) Cr2O3, (a) (i) and (ii), (b) (ii), (iii) and (iv), (c) (iii) and (iv), (d) (ii) and (iv), 168. Among the oxides, Mn 2O7 (I), V2O3 (II), V2O5 (III), CrO, (IV) and Cr 2O3 (V) the basic oxides are, (a) I and II, (b) II and III, (c) III and IV, (d) II and IV, 169. When a small amount of KMnO4 is added to concentrated, H2SO4, a green oily compound is obtained which is highly, explosive in nature. Compound may be, (a) MnSO4, (b) Mn 2O7, (c) MnO2, (d) Mn 2O3, 170. Identify the product and its colour when MnO2 is fused, with solid KOH in the presence of O2., (a) KMnO4, purple, (b) K2MnO4, dark green, (c) MnO, colourless, (d) Mn2O3, brown, 171. When KMnO4 solution is added to oxalic acid solution, the, decolourisation is slow in the beginning but becomes, instantaneous after some time because, (a) CO2 is formed as the product., (b) reaction is exothermic., (c), (d), , MnO4 catalyses the reaction., , Mn, , 2, , acts as autocatalyst., , 172. Which of the following oxidising reaction of KMnO4 occurs, in acidic medium?, (i) Fe2+ (green) is converted to Fe3+ (yellow)., (ii) Iodide is converted to iodate., (iii) Thiosulphate oxidised to sulphate., (iv) Nitrite is oxidised to nitrate., (a) (i) and (iii), (b) (i) and (iv), (c) (iv) only, (d) (ii) and (iv), 173. Arrange the following increasing order of acidic character?, Mn2O7(A), Mn2O3(B), MnO(C)?, (a) C, A, B, (b) A, C, B, (c) B, A, C, (d) C, B, A, 174. Solution of oxalate is colourless. It is made acidic by adding, excess of H+, then titrated with KMnO4. Now at a moment, if someone has added large amount of KMnO4, in it then, no. of possible products are, (a) CO2, Mn2+, H2O, (b) CO2, MnO2, H2O, (c) MnO2, H2O, CO2, (d) CO2, MnO2, H2O, Mn2+, 175. Knowing that the chemistry of lanthanoids(Ln) is dominated, by its + 3 oxidation state, which of the following statements, is incorrect?, (a) The ionic size of Ln (III) decrease in general with, increasing atomic number, (b) Ln (III) compounds are generally colourless., (c) Ln (III) hydroxide are mainly basic in character., (d) Because of the large size of the Ln (III) ions the bonding, in its compounds is predominantly ionic in character., 176. The +3 ion of which one of the following has half filled 4f, subshell?, (a) La, (b) Lu, (c) Gd, (d) Ac, 177. Although + 3 is the characteristic oxidation state for, lanthanoids but cerium also shows + 4 oxidation state, because _________ ., (i) it has variable ionisation enthalpy, (ii) it has a tendency to attain noble gas configuration, (iii) it has a tendency to attain f 0 configuration, (iv) it resembles Pb4+, (a) (ii) and (iii), (b) (i) and (iv), (c) (ii) and (iv), (d) (i), (ii) and (iii), 178. Dichromate [Cr(VI) ] is a strong oxidizing agent whereas, Mo(VI) and W(VI) are found to be not. This is due to, (a) Lanthanoid contraction, (b) Down the group metallic character increases, (c) Down the group metallic character decreases, (d) Both (a) and (b), 179. Which of the following conversions can be carried out by, both acidified K2Cr2O4 and acidified KMnO4?, (i) Fe2+, (ii) I–, Fe3+ + e–, (iii) I–, I2, (a) (i) and (iii), (c) (i), (iii) and (iv), , (iv) H2S, S, (b) (ii) and (iv), (d) (i), (ii) and (iii)
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THE d-AND f-BLOCK ELEMENTS, , 377, , FACT / DEFINITION TYPE QUESTIONS, 1., , (c) General electronic configuration of transition elements, is ( n 1) d, , 2., 3., 4., , 1 10, , ns, , (b) Cr (24) = 1s , 2 s2 2 p 6 , 3s 2 3 p 6 , 3d 5 , 4 s1 ,, (b) Configuration of Fe ( Z = 26), 1s2, 2s2, 2p6, 3s2, 3p6, 3d 6, 4s2, (d), , Ni, , : [Ar] 3d, , 7, , Fe3+ : [Ar] 3d 5, , 5., 6., 8., , Co3+ : [Ar] 3d 6, (c) Ag belongs to second transition series., (c), 7. (a), (d) Transition elements due to similar (almost) sizes exhibit, both vertical and horizontal similarities., , 9., , (a) Group number is given by ns, , 2 3, , 11., 12., , 14., , (a) 3d series starts from Sc Z 21 and ends with, Zn(Z –30)., (d) Since transition metals can lose electrons from, n 1 d ns orbitals hence they are valence orbitals., (b) Atomic no. of Ni = 28, Ni (Ground state) = 1s2, 2s2, 2p6, 3s2, 3p6, 3d 8, 4s2,, , It has 2 unpaired electrons, (c) Cerium (Ce) belongs to lanthanide series and is member, of inner-transition metals., (c) Mn3+ = [Ar]3d4, Number of unpaired electrons = 4, Cr3+ = [Ar]3d3, = [Ar], No. of unpaired electrons = 3, V3+ = [Ar]3d2, = [Ar], No. of unpaired electrons = 2, (a), , Ni 28 Ni Ar 3d8 4s 2 contain 2 unpaired electrons., , 18. (c) Zn, Cd, Hg do not show properties of transition, elements hence they are known as non typical, transition elements., 19. (c) The outer electronic configuration of the given ions is, as, d, , s, , d, , s, , d, , s, , d, , s, , Ti2+, Fe2+, Cr+, , 5, , = [Ar], , 15., , 17. (a), , n 1 d electrons., , 2, 2, 6, 2, 6, 8, 0, Ni 2 = 1s , 2s , 2p , 3s , 3p , 3d , 4s, 3d, , 13., , Fe 2 contain maximum number of unpaired electrons., , 2, , 3+, , Zn [Ar]3d10 4s1,Fe 2 [Ar]3d 6 4s0 , Ni [Ar]3d 8 4s1,, Cu [Ar]3d10 4s 0 ;, , 1 2, , Mn3+ : [Ar] 3d 4, , 10., , 16. (b), , Cu+, , 20. (c), 21. (a), 22. (a) Mn7+ = 25 – 7 = 18e– = [Ar], 0 unpaired electrons., 23. (b), , Co, , [Ar]3d 7 4s 2, , Since it contains three unpaired electrons. Hence it is, paramagnetic., 24. (a) The outermost electronic configuration of Fe is, Fe = [Ar] 3d6 4s2, Fe2+ = [Ar] 3d6 4s0, , Since Fe2+ has 4 unpaired electrons it is paramagnetic, in nature., Zn = [Ar] 3d10 4s2 —— no unpaired e–, Hg2+ = [Ar] 4f 14 5d10 —— no unpaired e–, Ti4+ = [Ar] 3d0 4s0 —— no unpaired e–, 25. (c) Due to d 5 configuration, Mn has exactly half filled, d-orbitals. As a result the electronic configuration is, stable means 3d electrons are more tightly held by the, nucleus and this reduces the delocalization of electrons, resulting in weaker metallic bonding., 26. (d) All statements are correct.
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EBD_7207, THE d-AND f-BLOCK ELEMENTS, , 378, , 27., , 37., , (b), , 38., 39., , (a), (c), , 40., 41., 43., , (d), (b), (b), , 28., , (d) The minimum oxidation state in transition metal is equal, to the number of electrons in 4s shell and the maximum, oxidation state is equal to the sum of the 4s and 3d, electrons., Ti = [Ar] 3d24s2, Hence minimum oxidation state is +2 and maximum, oxidation state is +4. Thus the common oxidation states, of Ti are +2, +3 and +4, (a) Os shows maximum oxidation state of +8., , 29., , (b) Mn - 3d 5 4s2, , 44., , (b), , 45., , (a), , 46., , (a), , 47., 48., , (d), (b), , 49., 50., , (d), (c), , 51., 52., , (a), (a), , 53., , (d), , 30., 31., 32., 33., , The no. of various oxidation states possible are + 2,, + 3, + 4, + 5, + 6 and + 7., (c) Due to lanthanide contraction, the size of Zr and Hf, (atom and ions) become nearly similar., (a), (d) Fe3+ is easily hydrolysed than Fe2+ due to more, positive charge., (c) Electronic configuration, 4s, 3d, V2+ –3d 3 4s0, Cr – 3d 4 4s0, – 3d 5 4s0, , Mn, , 34., 35., , Fe – 3d 6 4s0, For third ionization enthalpy Mn has stable, configuration due to half filled d-orbital., (d) (n – 1)d 5ns2 attains the maximum O.S. of + 7., (a) The ionisation energies increase with increase in, atomic number. However, the trend is some irregular, among d-block elements. On the basis of electronic, configuration, the, Zn : 1s 2 2s 2 p 6 3s 2 p6 d 10 4s 2, Fe : 1s 2 2s 2 p 6 3s 2 p 6 d 6 4s 2, Cu : 1s 2 2s 2 p6 3s 2 p 6 d 10 4s1, 2, , 2 6, , 2 6 5, , 54., , Mn 2 (Five), , 55., , 56., , Fe (Four) Co (Three) >, , Ni2 (Two) Cu 2 (One), , / Fe, , (d), , E 3, Mn3 / Mn 2 = + 1.57 V,,, Co / Co 2 = + 1.97 V, 2+, Since Mn contains maximum number of unparied, , (d) Magetic moment, , (c) Magnetic moment, 1.73, , Ni 2 (Two), , 2, , Fe, , / Cr, , n n 2 where n = number of, , unpaired electrons 15, , n n 2, , nn 2, , n=3, , n n 2 BM, n 1 , it has one unpaired electron, , hence electronic configuration is, , Ar 3d 1 and, , electronic configuration for Z = 22 is Ar 3d 2 4s2 ., , Co 2 (Three) Fe2 (Four), (d) For unpaired electrons >, 2, , Cr, , electrons hence it has maximum magnetic moment, , 1, , IE1 follows the order : Zn > Fe > Cu > Cr, (a) In a period on moving from left to right, ionic radii, decreases., (a) So order of cationic radii is, Cr2+ > Mn2+ > Fe2+ > Ni2+ and, (b) Sc > Ti > Cr > Mn (correct order of atomic radii), (c) For unpaired electrons, , and 80Hg have their d orbitals completely filled, so they do not show any variable valency., Highest O.S. by Mn (+7), Zinc does not show variable oxidation state due to, completely filled d-orbitals., Sc does not show variable valency., 42. (d), Transition metals are generally paramagnetic since they, contain unpaired electrons., Since reduction potential of fluorine is highest, transition metals exhibit highest oxidation state with, fluorine., Zn, Cd and Hg due to presence of completely filled, d-orbitals in ground state as well as in their common, oxidation states are not regarded as a transition metals, but they are studied along with the transition metals., The +7 oxidation state of Mn is not represented in, simple halides but MnO3F is known, Transition metals exhibit variable valency, In transition metals d electrons also take part in, bonding, so they show variable oxidation states., For chromium ion + 3 oxidation state is most stable., The melting points of the transition element first rise, to a maximum and then fall as the atomic number, increases manganese have abnormally low melting, point., They may or may not be diamagnetic, Mn++ –5 unpaired electrons, Fe++ – 4 unpaired electrons, Ti++ – 2 unpaired electrons, Cr++ – 4 unpaired electrons, Hence maximum no. of unpaired electron is present in, Mn++., Magnetic moment µ number of unpaired electrons, E 3, E 3, 2 = + 0.77 V, 2 = – 0.41 V, E, , Cr : 1s 2s p 3s p d 4s, 36., , 30Zn, , 57., 58., , Hence charge on Ti is +3, (b) The more the number of unpaired electrons, the more, is magnetic moment.Therefore the answer is (b)., (a)
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THE d-AND f-BLOCK ELEMENTS, , 379, , Fe3+(d5) has 5 unpaired electrons therefore magnetic, , 59., , (d), , 60., , moment, n(n 2), 5(5 2) 5.91 which is, maximum among given options. As Sc3+, Ti3+, Cr 3+,, V3+ contains 0, 1, 3, and 2 number of unpaired electrons, respectively., (b), 61. (b), , 62., , 63., , 64., , 65., 66., 67., , 68., 69., , 70., 71., 72., 73., 74., , (a), , Sc3+, , 3d 04s 0, , Fe2+, , 3d 6, , Ti3+, , 3d 14s 0, , Mn2+, , 3d 5, , Cu 2, , 4s 0, , Ar 3d 9 , Ti 4, , Ar 3d 0 , Co 2, , Ar 3d 7 , Fe 2, , Ar 3d 6, , 1, 3, 4 are coloured ions hence the answer is b., (a) In interstitial compounds small atoms like H, B and C, enter into the void sites between the packed atoms of, crystalline metal. They retain metallic conductivity and, are chemically inert., (d) A covalent bond is formed between small interstial, non-metal and transition metal which make it hard, (c) If non metal is added to the interstital site the metal, becomes less malleable due to formation of covalent, bond between metal and non metal, (c) Gun metal is an alloy of Cu, Zn and Sn. It contains 88%, Cu, 10% Sn and 2% Zn., (b) Brass is an alloy of Cu and Zn, (b) Cu, Ag and Au are called coinage metals., (b) Bronze is an alloy of Cu and Sn., (b) Bronze - 10% Sn, 90% Cu, (Sn is a non transition element), 4, , VO2, , 4, , Ti O2, , 2Fe 2, , S2 O82, , 2Fe3, , 2SO 24, , 80. (a), 81. (b), 82. (a), , Cr2 O 72–, , 2OH –, , 2CrO 42, , H 2O, , Hence CrO 24 ion is obtained., , 4s 0, , In Sc3+ there is/are no unpaired electrons. So the, aqueous solution of Sc3+ will be colourless., (a) Transition elements form coloured ions due to d-d, transitions. In the presence of ligands, there is splitting, of energy levels of d-orbitals. They no longer remain, degenerated. So, electronic transition may occur, between two d-orbitals. The required amount of energy, to do this is obtained by absorption of light of a, particular wavelength in the region of visible light., (c) The transition metals and their compounds are used, as catalysts. Because of the variable oxidation states, they may form intermediate compound with one of the, readtants. These intermediate provides a new path with, lowe activation energy. V2O5 + SO2, V2O4 + SO3, 2V2O4+ O2 2V2O5, (d) Since Sc3+ does not contain any unpaired electron it, is colourless in water., (b), , 2Fe2+ + I2, , 79. (b) 2Fe3+ + 2I–, , 5, , 6, , Cr O42, , 75., , (b), , 76., 77., 78., , (b), (d) Ti4+ (3d0) and Zn2+ (3d10) are colourless., (d), , VO2, , 83. (b), , CrO3, , 2NaOH, , Na 2CrO 4, , 84. (a), , Cr2O 7 + 6I + 14H, , 2, , +, , H2O, , 3I 2 + 7H 2 O + 2Cr, , 3+, , oxidation state of Cr is +3., 85. (d), O, , 2–, , O, , O, , Cr, , Cr, , O, , O, , O, , O, , Dichromate ion, , There are six equivalent Cr — O bonds and one, Cr — O — Cr bond., 86. (c) Solid potassium dichromate when heated with, concentrated sulphuric acid and a soluble chloride, gives orange red vapours of a volatile oily liquid, CrO2Cl2, K2Cr2O7 + 4NaCl + 6H2SO4, 2KHSO4 + 4NaHSO4 + 2CrO2Cl2, chromyl chloride, , 87. (c) Mn2O7 is acidic, V2O5 is amphoteric acid and CrO is, basic., 88. (a) CrO2 is amphoteric in nature, 89. (a), 90. (c), 91. (b) In neutral or faintly alkaline medium thiosulphate is, quantitatively oxidized by KMnO4 to SO42–, 8KMnO4 + 3Na2S2O3 + H2O, 3K2SO4 + 8MnO2 + 3Na2SO4 + 2KOH, 92. (b) HCl and SO2 are reducing agents and can reduce, MnO4–. CO2 which is neither oxidising and nor, reducing will provide only acidic medium. It can shift, reaction in forward direction and reaction can go to, completion., 93. (c) In laboratory, manganese (II) ion salt is oxidised to, permagnate ion in aqueous solution by, peroxodisulphate., 2Mn 2, , S2O82, , 8H 2O, , peroxodisulphate ion, , 2MnO4 10SO42, , 16H, , 94. (a) Pyrolusite (It is MnO 2 ), 95. (a), , 2KMnO4 H 2SO4 Conc, K 2SO 4 Mn O, 2 7, , H 2O, , Explosive, , 96. (c) In acid medium MnO4 8H 5e, Mn 2, (O.S. of Mn changes form +7 to +2), , 4H 2O
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EBD_7207, THE d-AND f-BLOCK ELEMENTS, , 380, , 97., 98., , (c), (b) Mn2+ (d5)is more stable than Mn 3+ (d4), thus, , La 3+ : 54 e– = [Xe], , Ti3+ : 19 e– = [Ar] 3 d 1 (Coloured), , 122. (b), , Mn 3 /Mn 2, , ve, , 99., , (a) As the oxidation state of metal associated with oxygen, increases, the acidic character of oxide increases., 100. (d), 101. (d) PdCl2 is used as a catalyst in Wacker’s process., 102. (c) Only Cu in its +2 oxidation state is able to oxidizes the, I– to I2, 103. (c) The number is 28(14 lanthanide +14 Actinides), , Lu 3+ : 68 e– = [Xe] 4 f 14, , 105. (c), , 106. (d), , 107. (c), , 108. (b), , 109. (d), , 6s2 – 7 unpaired e–, , Yb2+(Z = 70), [Xe]4f14 6s2 – 0 unpaired e–, Ce2+(Z = 58), 1, 1, 2, [Xe]4f 5d 6s – 2 unpaired e–, Only Yb2+ is diamagnetic., 118. (b) Amongst the given elements, only Gd is a lanthanide ., 119. (d) Mischmetal is an alloy which contains rare earth, elements (94-95%), iron (5%) and traces of sulphur,, carbon, silicon, calcium and aluminium. It is used in, gas lighters, tracer bullets and shells., 120. (b) Cerium is the most common lanthanide, 121. (a) La ( lanthanum ) is non lanthanide atom, , E, , 104. (b), , [Xe]4f7, , Sc3+ : 18 e– = [Ar], A regular decrease in the size of the atoms and ions in, lanthanoid series from La3+ to Lu3+ is called lanthanide, contraction. The similarity in size of the atoms of Zr, and Hf is due to the lanthanide contraction., We know that lanthanides La, Gd shows +3, oxidation, state, while Eu shows oxidation state of +2 and + 3. Am, shows +3, +4, +5 and +6 oxidation states. Therefore, Americium (Am) has maximum number of oxidation, states., Lanthanides are 4 f-series elements starting from cerium, (Z= 58) to lutetium (Z = 71). These are placed in the, sixth period and in third group., In lanthanides, there is poorer shielding of 5d electrons, by 4 f electrons resulting in greater attraction of the, nucleus over 5 d electrons and contraction of the atomic, radii., On going from left to right in lanthanoid series ionic,, size decreases i.e., Ce+3 > Tb+3 > Er+3 > Lu+3., , 110. (d), 111. (c) Lanthanide contraction results into decrease in atomic, and ionic radii., 112. (a), 113. (a) 4f orbital is nearer to nucleus as compared to 5 f orbital, therefore, shielding of 4 f is more than 5 f., 114. (d), 115. (d) The configuration of Gd is [xe] 4f 7 5d1 6s2., 116. (c) In lanthanide series there is a regular decrease in the, atomic as well as ionic radii of trivalent ions (M3+) as, the atomic number increases. Although the atomic radii, do show some irregularities but ionic radii decreases, from La(103 pm) to Lu (86pm). Y3+ belong to second, transition series there fore have greater ionic radii then, other ions of third transition series., 117. (c) Sm2+(Z = 62), [Xe]4f6 6s2 – 6 unpaired e–, Eu2+(Z = 63), , Eu 2, , has electronic configuration Xe 4f 7 hence, , stable due to half filled atomic orbitals., 123. (d) Actinides have variable valency due to very small, difference in energies of 5f, 6d and 7s orbitals. Actinides, are the elements from atomic number 89 to 103., 124. (c) Ac (89) = [Rn] [6d1] [7s2], 125. (b) The main reason for exhibiting larger number of, oxidation states by actinoids as compared to, lanthanoids is lesser energy difference between 5 f and, 6d orbitals as compared to that between 4f and 5d, orbitals., In case of actinoids we can remove electrons from 5f, as well as from d and due to this actinoids exhibit larger, number of oxidation state than lanthanoids., 126. (c) Actinoids exhibit variable oxidation states, which vary, from +3 to +7., 127. (c), 128. (b ), 129. (c), 130. (d) Mischmetall consists of a lanthanoid metal, (~95%) and iron (~ 5%) and traces of S,C,Ca and Al., 131. (d), 132. (b) Curium (Cm) has configuration 5f 7 6d1 7s2., 133. (d), 134. (a) Tb4+ = 4f 7, — 3 unpaired e–, 3+, 14, Lu = 4f, — 0 unpaired e–, 4+, 0, Ce = 4f, — 0 unpaired e–, 3+, 0, La = 4f, — 0 unpaired e–, , STATEMENT TYPE QUESTIONS, 135. (b) (i) Outer electronic configuration of Mn is 3d54s2 and, hence exhibits +7 oxidation state., (ii) Zinc does not form coloured ions as it has, completely filled 3d104s7 configuration., (iii) In [CoF6]3–, Co3+ is a d7 system. Fluoride is a weak, field ligand and hence does not cause pairing of, electrons., Co3+, , ; Paramagnetic, (iv)Sc can form a maximum of +3 oxidation state as it, has an outer electronic configuration of 3d14s2., (v) Zn exhibits only +2 oxidation state as this O.S. is, the most stable one.
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THE d-AND f-BLOCK ELEMENTS, , 381, , 136. (d), 137. (b) In any row the melting points of transition metals rise, to a maximum at d5 except for anomalous values of, Mn and Tc and falls regularly as the atomic number, increases., 138. (a) Aqueous solution formed by Ti3+ ions has purple, colour., 139. (a) Steel is an alloy of Fe and C (non-metal). Interstitial, compounds are chemically inert., 140. (b) Heavier members of d-block elements unlike p-block, elements shows higher oxidation states. For example, W(VI) is more stable than Cr(VI)., 141. (b) As a result of lanthanide contraction Zr 4+ and Hf4+, possess almost the same ionic radii. Ce4+ is an, oxidising agent. Ce4+ gains electron to acquire more, stable Ce3+state. La(OH)3 is the most basic among, lanthanide hydroxides., 142. (b) Ce4+ is a strong oxidant reverting to the common +3, state., Ho does not show oxidation state of +4. Lanthanoids, showing +4 oxidation state are Ce, Pr, Nd, Dy and Tb., 143. (a) Both Np and Pu shows oxidation state of +7., 144. (a) Atomic mass of Hf is greater than that of Zr, Hf is a, series 3 metal, so for almost similar radius Hf has greater, density, Lanthanoid contraction is responsible for, almost similar radii., , Thus options (a) and (c) are discarded; now let us, observe the second point of difference., 23 V, , 4, , :1s 2 2s 2 p6 3s 2 p 6 d1, , Thus option (b) is discarded, 23 V, , 3, , 24 Cr, , 4, , 25 Mn, , 154. (a), 156. (c), , :1s 2 2 s 2 p 6 3s 2 p 6 d 2, , :1s 2 2s 2 p6 3s 2 p 6 d 2, , 5, , :1s 2 2 s 2 p 6 3s 2 p 6 d 2, , 155. (d), Eo, , = 0.34 V, , Cu 2 / Cu, , other has – ve E oR.P., Eo, , Co, , Eo, , Ni, , Eo, , Fe, , / Co =, , / Ni, , = – 0.25V, , / Fe, , = – 0.44V, , 157. (c), , E, Mn, , 2, , 2e, , 145. (b), , 146. (c), , 147. (a), , 148. (d), , ASSERTION-REASON TYPE QUESTIONS, 149. (d), 150. (c) The assertion is correct but the reason is false. Actually, transition metal show variable valency due to very, small difference between the ns2 and (n – 1)d electrons., 151. (b) Due to larger surface area and variable valencies to, form intermediate absorbed complex easily, transition, metals are used as catalysts., 152. (b) The magnetic moments are lesser than the fact that 5f, electrons of actinides are less effectively shielded, which results in quenching of orbital contribution., , CRITICAL THINKING TYPE QUESTIONS, 153. (d) The electronic configuration of different species given, in the question are, (a) 22 Ti3 :1s2 2s2 p6 3s2 p6 d1, (b) 22 Ti :1s 2 2s 2 p6 3s 2 . p6 d 2 4s1, (c) 22 Ti4 :1s2 2s2 p6 3s 2 p6, (d), , 22 Ti, , 2, , :1s 2 2 s 2 p 6 3s 2 p 6 d 2, , Mn, , Mn 3, , e, , Mn 2, , Mn 3, , 3e, , Mn, , 1.18, 1.51, 0.28, , nE, 2.36 V, 1.51 V, 0.85 V, , 158. (a) Given magnetic moment of transition metal, = n n 2, , MATCHING TYPE QUESTIONS, , – 0.28 V, , 5.92, , i.e., n = 5, Number of unpaired electrons in Mn2+ = 5, Number of unpaired electrons in Ti3+ = 1, Number of unpaired electrons in Cr3+ = 3, Number of unpaired electrons in Cu2+ = 1, Number of unpaired electrons in Co2+ = 3, Thus Mn2+ have magnetic moment = 5.92 BM, 159. (a) Mn++ = 3d5 i.e. no. of unpaired e– = 5, Cu++ = 3d9 i.e. no. of unpaired e– = 1, Fe++ = 3d6 i.e. no. of unpaired e– = 4, Zn++ = 3d10 i.e. no. of unpaired e– = 0, Ni++ = 3d8 i.e. no. of unpaired e– = 3, Higher the number of unpaired electrons higher will be, the magnetic moment. Hence Mn ++ having maximum, unpaired electrons will have the maximum magnetic, moment., 160. (d) Sc3+ : 1s2, 2s2p6, 3s2p6d0, 4s0; no unpaired electron., Cu+ : 1s2, 2s2p6, 3s2p6d 10, 4s0; no unpaired electron., Ni2+: 1s2, 2s2p6, 3s2p6d 8, 4s0;, unpaired electrons are present., Ti3+ : 1s2, 2s2p6, 3s2p6d 1, 4s 0;, unpaired electron is present, Co2+ : 1s2, 2s2p6, 3s2p6d 7, 4s0;, unpaired electrons are present, So from the given options the only correct combination, is Ni2+ and Ti3+.
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23, COORDINATION COMPOUNDS, (a), (b), (c), (d), , FACT / DEFINITION TYPE QUESTIONS, 1., , 2., , 3., 4., , 5., , 6., , According to the postulates of Werner for coordination, compounds, (a) primary valency is ionizable, (b) secondary valency is ionizable, (c) primary and secondary valencies are non-ionizable, (d) only primary valency is non-ionizable., Which of the following postulates of Werner’s theory is, incorrect?, (a) Primary valencies are satisfied by negative ions., (b) Secondary valencies are satisfied by neutral molecules, or negative ions., (c) Secondary valence is equal to the coordination, number and it depends upon the nature of ligand, attached to metal., (d) The ions/ groups bound by the secondary linkages, to the metal have charecteristic spatial arrangements., CrCl3 has primary valence of, (a) 3, (b) 4, (c) 2, (d) 1, One mole of the complex compound Co(NH3)5Cl3, gives 3, moles of ions on dissolution in water. One mole of the same, complex reacts with two moles of AgNO3 solution to yield, two moles of AgCl (s). The structure of the complex is, (a) [Co(NH3)3Cl3]. 2 NH3, (b) [Co(NH3)4Cl2] Cl . NH3, (c) [Co(NH3)4Cl] Cl2. NH3, (d) [Co(NH3)5Cl] Cl2, When AgNO3 is added to a solution of Co(NH3)5Cl3, the, precipitate of AgCl shows two ionisable chloride ions. This, means :, (a) Two chlorine atoms satisfy primary valency and one, secondary valency, (b) One chlorine atom satisfies primary as well as, secondary valency, (c) Three chlorine atoms satisfy primary valency, (d) Three chlorine atoms satisfy secondary valency, Which one is the most likely structure of CrCl3. 6H2O if 1/, 3 of total chlorine of the compound is precipitated by adding, AgNO3, , CrCl3. 6H2O, [ Cr (H2O)3 Cl3]. (H2O)3, [ CrCl2 (H2O)4 ] Cl . 2H2O, [ CrCl (H2O)5 ] Cl2 . H2O, , 7., , K 4 [Fe(CN) 6 ] is a :, (a) double salt, (b) complex compound, (c) acid, (d) base, 8., The number of ions formed on dissolving one molecule of, FeSO4(NH4)2SO4.6H2O in water is:, (a) 4, (b) 5, (c) 3, (d) 6, 9., The solution of K4[Fe(CN)6] in water will, (a) give a test K+, (b) give a test Fe2+, –, (c) give a test of CN, (d) give a test of [Fe(CN)6]4–, 10. In the coordination compound, K4[Ni(CN)4], the oxidation, state of nickel is, (a) 0, (b) +1, (c) +2, (d) –1, 11. The coordination number of a central metal atom in a, complex is determined by, (a) the number of ligands around a metal ion bonded by, sigma and pi-bonds both, (b) the number of ligands around a metal ion bonded by, pi-bonds, (c) the number of ligands around a metal ion bonded by, sigma bonds, (d) the number of only anionic ligands bonded to the metal, ion., 12. The oxidation state of Cr in [Cr ( NH 3 ) 4 Cl 2 ] is, (a) 0, (b) + 1, (c) + 2, (d) + 3, 13. In Ni(CO)4– , oxidation number of Ni is :, (a) 4, (b) – 4, (c) 0, (d) + 2, 14., , [EDTA]4 is a :, (a) monodentate ligand, (c) quadridentate ligand, , (b) bidentate ligand, (d) hexadentate ligand
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EBD_7207, COORDINATION COMPOUNDS, , 384, , 15., , The compound having the lowest oxidation state of iron is:, (a), (c), , 16., , 17., , 18., , 19., , 20., , 21., , 22., , 23., , 24., , 25., , K 4 Fe(CN ) 6, , (b), , 26., , K 2 FeO 4, , (d) Fe(CO)5, Fe 2O 3, The coordination number and the oxidation state of the, element ‘E’ in the complex, [E (en)2 (C2O4)]NO2 (where (en) is ethylene diamine) are,, respectively,, (a) 6 and 2, (b) 4 and 2, (c) 4 and 3, (d) 6 and 3, Some salts although containing two different metallic, elements give test for only one of them in solution. Such, salts are, (a) complex, (b) double salts, (c) normal salts, (d) None of these, Coordination number of Ni in [Ni(C2O4)3]4– is, (a) 3, (b) 6, (c) 4, (d) 5, According to Lewis, the ligands are, (a) acidic in nature, (b) basic in nature, (c) some are acidic and others are basic, (d) neither acidic nor basic, Ligand in a complex salt are, (a) anions linked by coordinate bonds to a central metal, atom or ion, (b) cations linked by coordinate bonds to a central metal, or ion, (c) molecules linked by coordinate bonds to a central, metal or ion, (d) ions or molecules linked by coordinate bonds to a, central atom or ion, The ligand N(CH2CH2NH2)3 is, (a) tridentate, (b) pentadentate, (c) tetradentate, (d) bidentate, An example of ambidentate ligand is, (a) Ammine, (b) Aquo, (c) Chloro, (d) Thiocyanato, Which of the following does not form a chelate ?, (a) EDTA, (b) Oxalate, (c) Pyridine, (d) Ethylenediamine, A bidenate ligand always, (a) has bonds formed to two metals ions, (b) has a charge of +2 or – 2, (c) forms complex ions with a charge of +2 or –2, (d) has two donor atoms forming simultaneously two, sigma ( ) bonds., An ambident ligand is one which, (a) is linked to the metal atom through two donor atoms, (b) has two donor atoms, but only one of them has the, capacity to form a coordinate bond [or a sigma ( ), bond], (c) has two donor atoms, but either of two can form a, coordinate bond, (d) forms chelate rings., , 27., , 28., , NH2-NH2 serves as, (a) Monodentate ligand (b) Chelating ligand, (c) Bridging ligand, (d) Both (a) and (c), Which one of the following is NOT a ligand ?, (a) PH3, (b) NO+, +, (c) Na, (d) F–, Glycinato ligand is:, NH2, , CH 2, –, COO, (b) bidentate ligant, (c) two donor sites N and O–, (d) All of the above, Which one does not belong to ligand?, (a) PH3, (b) NO+, (c) BF3, (d) Cl–, Which ligand is expected to be bidentate?, , (a), , 29., , 30., , (a), , C2 O42, , (b) CH3C N, , Br –, , 31., , 32., , 33., , 34., , 35., , 36., , (c), (d) CH3NH2, Which one of the following ligands forms a chelate, (a) Acetate, (b) Oxalate, (c) Ammonia, (d) Cyanide, Choose the correct statement., (a) Coordination number has nothing to do with the, number of groups or molecules attached to the central, atom, (b) Coordination number is the number of coordinating, sites of all the ligands connected to the central atom, or the number of coordinate bonds formed by the metal, atom with ligands, (c) Werner’s coordination theory postulates only one type, of valency, (d) All the above are correct, O2 is a, (a) Monodentate ligand (b) Bidenate ligand, (c) Tridentate ligand, (d) Hexadenate ligand, The stabilisation of cooordination compounds due to, chelation is called the chelate effect. Which of the following, is the most stable complex species ?, (a), , [Fe(CO)5 ], , (b) [Fe(CN)6 ]3, , (c), , [Fe(C2 O 4 )3 ]3, , (d) [Fe(H 2 O)6 ]3, , A chelating agent has two or more than two donor atoms to, bind to a single metal ion. Which of the following is not a, chelating agent ?, (a) thiosulphato, (b) oxalato, (c) glycinato, (d) ethane - 1, 2-diamine, Which of the following species is not expected to be a, ligand?, (a) NO, (c), , NH 2 CH 2 CH 2 NH 2, , (b), , NH 4, , (d) Both (a) and (b)
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EBD_7207, COORDINATION COMPOUNDS, , 386, , 58., , 59., , The type of isomerism present in Pentaminenitrochromium, (III) chloride is, (a) optical, (b) linkage, (c) ionisation, (d) polymerisation., Which of the following compounds shows optical, isomerism?, (a), , 61., , 62., , 63., , 64., , 65., , 66., , 67., , 68., , 69., , 71., , (b) [ Cr ( C 2 O 4 ) 3 ]3, , (d) [Cu ( NH 3 ) 4 ]2, [ ZnCl4 ]2, Which of the following ions can exhibit optical isomerism–, (a) [Co(NH3)4Cl2]+, (b) [Co(NH3)2Cl4]–, +, (c) Cis-[Co(en)2Cl2], (d) trans-[Co(en)2Cl2]+, Which would exhibit co-ordination isomerism, (a) [Cr(NH3)6][Co(CN)6] (b) [Co(en)2Cl2], (c) [Cr(NH3)6]Cl3, (d) [Cr(en)2Cl2]+, [Co(NH3)5NO2]Cl2 and [Co(NH3)5(ONO)]Cl2 are related to, each other as, (a) geometrical isomers (b) optical isomers, (c) linkage isomers, (d) coordination isomers, Coordination isomerism is caused by the interchange of, ligands between the, (a) cis and trans structure, (b) complex cation and complex anion, (c) inner sphere and outer sphere, (d) low oxidation and higher oxidation states, Change in composition of co-ordination sphere yields which, type of isomers, (a) optical, (b) geometrical, (c) ionisation, (d) None of these, Which of the following does not show optical isomerism?, (a) [Co(NH3)3Cl3]0, (b) [Co (en) Cl2 (NH3)2]+, (c) [Co (en)3]3+, (d) [Co (en)2Cl2]+, (en = ethylenediamine), The complexes [Co(NH3)6] [Cr(CN)6] and [Cr(NH3)6 ], [Co(CN)6] are the examples of which type of isomerism?, (a) Linkage isomerism, (b) Ionization isomerism, (c) Coordination isomerism, (d) Geometrical isomerism, The complex, [Pt(py)(NH3)BrCl] will have how many, geometrical isomers ?, (a) 3, (b) 4, (c) 0, (d) 2, Which of the following has a square planar geometry?, (a) [PtCl4]2–, (b) [CoCl4]2–, 2–, (c) [FeCl4], (d) [NiCl4]2–, (At. nos.: Fe = 26, Co = 27, Ni = 28, Pt = 78), Which of the following pairs represent linkage isomers?, (a) [Pd(PPh3)2(NCS)2] and [Pd ( PPh3)2 (SCN)2], (b) [Co(NH3)5NO3] SO4 and [Co(NH3)5SO4]NO3, (c) [PtCl2(NH3)4] Br2 and [Pt Br 2 (NH3)4] Cl2, (d) [Cu(NH3)4][Pt Cl4] and [Pt(NH3)4][CuCl4], (c), , 60., , [ Co (CN ) 6 ]3, , 70., , 72., , 73., , 74., , 75., , 76., , 77., , 78., , 79., , 80., , 81., , 82., , Which one of the following has an optical isomer?, (a) [Zn(en) (NH3)2]2+, (b) [Co(en)3]3+, (c) [Co(H2O)4(en)]3+, (d) [Zn(en)2]2+, (en = ethylenediamine), Which one of the following complex ions has geometrical, isomers ?, (a) [Ni(NH3)5Br]+, (b) [Co(NH3)2(en)2]3+, 3+, (c) [Cr(NH3)4(en)2], (d) [Co(en)3]3+, (en = ethylenediamine), The ionisation isomer of [Cr(H2O)4Cl(NO2)]Cl is, (a) [Cr(H2O)4(O2N)]Cl2, (b) [Cr(H2O)4Cl2](NO2), (c) [Cr(H2O)4Cl(ONO)]Cl, (d) [Cr(H2O)4Cl2(NO2)].H2O, Isomerism exhibited by [Cr(NH3)2(H2O)2Cl2]+ are –, (a) ionization, optical, (b) hydrate, optical, (c) geometrical, optical, (d) coordinate, geometrical, Type of isomerism which exists between [Pd(C6H5)2(SCN)2], and [Pd(C6H5)2(NCS)2] is :, (a) Linkage isomerism, (b) Coordination isomerism, (c) Ionisation isomerism (d) Solvate isomerism, Which of the following complex will show geometrical as, well as optical isomerism (en=ethylenediammine), (a) Pt(NH3)2Cl2, (b) [Pt(NH3)Cl4], (c) [Pt(en)3]4+, (d) [Pt(en)2Cl2], The number of geometrical isomers from [Co(NH3)3(NO2)3], is, (a) 2, (b) 3, (c) 4, (d) 0, The number of isomers exhibited by [Cr(NH3)3Cl3] is, (a) 2, (b) 3, (c) 4, (d) 5, For the square planar complex [M (a) (b) (c) (d)] (where M, = central metal and a, b, c and d are monodentate ligands),, the number of possible geometrical isomers are, (a) 1, (b) 2, (c) 3, (d) 4, Which of the following will exhibit optical isomerism ?, (a) [Cr(en) (H2O)4]3+, (b) [Cr(en)3]3+, (c) trans-[Cr(en)(Cl2)(NH3)2]+, (d) [Cr(NH3)6]3+, Which of the following will give maximum number of, isomers?, (a) [Co(NH3)4Cl2], (b) [Ni(en)(NH3)4]2+, (c) [Ni(C2O4)(en)2], (d) [Cr(SCN)2(NH3)4]2+, The compounds [PtCl2(NH3)4]Br2 and [PtBr2(NH3)4]Cl2, constitutes a pair of, (a) coordination isomers (b) linkage isomers, (c) ionization isomers, (d) optical isomers, Which one of the following will not show geometrical, isomerism ?, (a) [Cr(NH3)4Cl2]Cl, (b) [Co(en)2Cl2]Cl, (c) [Co(NH3)5NO2]Cl2, (d) [Pt(NH3)2Cl2 ]
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COORDINATION COMPOUNDS, , 83., , 84., , 85., , 387, , A similarity between optical and geometrical isomerism is, that, (a) each gives equal number of isomers for a given, compound, (b) if in a compound one is present then so is the other, (c) both are included in stereoisomerism, (d) they have no similarity, The type of isomerism present in n itropentaamminechromium (III) chloride is, (a) optical, (b) linkage, (c) ionization, (d) polymerization, What kind of isomerism exists between [Cr(H 2 O)6 ]Cl3, (violet) and [Cr(H 2O)5Cl]Cl 2 H 2O (greyish-green) ?, , 86., , (a) linkage isomerism, (b) solvate isomerism, (c) ionisation isomerism (d) coordination isomerism, Which of the following type of isomerism is shown by, given complex compound?, H3N, , NH3, , NO2, , Co, O2N, , 87., , 88., , 89., , 90., , 91., , NH3, , NO2, , (a) Facial, (b) Meridional, (c) Cis, (d) Both b and c, Due to the presence of ambidentate ligands coordination, compounds show isomerism. Palladium complexes of the, type [Pd(C6H5)2(SCN)2] and [Pd(C6H5)2(NCS)2] are, (a) linkage isomers, (b) coordination isomers, (c) ionisation isomers, (d) geometrical isomers, The geometry of Ni(CO)4 and Ni(PPh 3)2Cl2 are, (a) both square planar, (b) tetrahedral and square planar, (c) both tetrahedral, (d) None of these, The number of unpaired electrons in the complex, [Cr(NH3)6]Br3 is (Atomic number Cr = 24), (a) 4, (b) 1, (c) 2, (d) 3, Which of the following facts about the complex, [Cr(NH3)6]Cl3 is wrong?, (a) The complex involves d 2 sp3 hybridisation and is, octahedral in shape., (b) The complex is paramagnetic., (c) The complex is an outer orbital complex, (d) The complex gives white precipitate with silver nitrate, solution., Which of the following statements is correct ?, (Atomic number of Ni = 28), (a) Ni(CO)4 is diamagnetic and [NiCl4]2– and [Ni(CN)4]2–, are paramagnetic, (b) Ni(CO)4and [Ni(CN)4]2– are diamagnetic and [NiCl4]2–, is paramagnetic, , (c), , Ni(CO)4and [NiCl4]2–are diamagnetic and [Ni(CN)4]2–, , is paramagnetic, (d) [NiCl4]2– and [Ni(CN)4]2– are diamagnetic and Ni(CO)4, is paramagnetic, , 92. Which of the following species represent the example of, dsp2 - hybridisation ?, (a) [Fe(CN)6]3–, (b) [Ni(CN)4]2–, –, (c) [Ag(CN)2], (d) [Co(CN)6]3–, 93. Which one of the following will show paramagnetism, corresponding to 2 unpaired electrons?, (Atomic numbers : Ni = 28, Fe = 26), (a) [FeF6]3–, (b) [NiCl4]2–, 3–, (c) [Fe(CN)6], (d) [Ni(CN)4]2–, 94. Atomic number of Cr and Fe are respectively 25 and 26,, which of the following is paramagnetic?, (a) [Cr(CO)6], (b) [Fe(CO)5], (c) [Fe(CN)6]–4, (d) [Cr(NH3)6]+3, 95. CN– is a strong field ligand. This is due to the fact that, (a) it carries negative charge, (b) it is a pseudohalide, (c) it can accept electrons from metal species, (d) it forms high spin complexes with metal species, 96. Which one of the following cyano complexes would exhibit, the lowest value of paramagnetic behaviour ?, (a), , [Co(CN) 6 ]3, , (b) [Fe(CN) 6 ]3, , (c) [Mn (CN) 6 ]3, (d) [Cr (CN) 6 ]3, (At. Nos : Cr = 24, Mn = 25, Fe = 26, Co = 27), 97. [Sc(H2O)6]3+ ion is :, (a) colourless and diamagnetic, (b) coloured and octahedral, (c) colourless and paramagnetic, (d) coloured and paramagnetic, 98. Which has maximum paramagnetic nature ?, (a), , [Mn (H 2 O) 6 ]2, , 2, (b) [Cu( NH 3 ) 4 ], , (c) [Fe(CN) 6 ]4, (d) [Cu(H 2 O) 4 ]2, 99. The compound which is not coloured is, (a), , K 4 [Fe(CN)6 ], , (b), , K3[Fe(CN)6 ], , (c), , Na 2 CdCl 4, , (d), , Na 2 CuCl 4, , 100. Which of the following complexes exhibits the highest, paramagnetic behaviour ?, (a) [V(gly)2(OH)2(NH3)2]+ (b) [Fe(en)(bpy)(NH3)2]2+, (c) [Co(ox)2(OH)2]2–, (d) [Ti(NH3)6]3+, where gly = glycine, en = ethylenediamine and bpy =, bipyridyl moities), (At.nosTi = 22, V = 23, Fe = 26, Co = 27), 101. Which of the following complex ion is not expected to absorb, visible light ?, (a), , Ni(CN)4, , (c), , Fe(H 2 O)6, , 2, 2, , (b), , Cr(NH3 )6, , 3, , (d), , Ni(H 2O)6, , 2
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EBD_7207, COORDINATION COMPOUNDS, , 388, , 102. Of the following complex ions, which is diamagnetic in nature?, (a) [NiCl4]2–, (b) [Ni(CN)4]2–, (c) [CuCl4]2–, (d) [CoF6]3–, 103. Which one of the following is an outer orbital complex and, exhibits paramagnetic behaviour ?, (a) [Ni(NH3)6]2+, (b) [Zn(NH3)6)]2+, 3+, (c) [Cr(NH3)6], (d) [Co(NH3)6]3+, 104. Atomic number of Mn, Fe, Co and Ni are 25, 26, 27 and 28, respectively. Which of the following outer orbital octahedral, complexes have same number of unpaired electrons ?, (i), , [MnCl6 ]3, , (ii), , [FeF6 ]3, , (iii) [CoF6 ]3, , (iv) [Ni(NH 3 )6 ]2, , (a) (ii) and (iii), (c) (i) and (ii), , (b) (i) and (iv), (d) (i) and (iii), , 105. Which of the following options are correct for [Fe(CN)6 ]3, complex ?, (i) Possess d2sp3 hybridisation, (ii) Possess sp3 d2 hybridisation, (iii) It is paramagnetic, (iv) It is diamagnetic, (a) (i) and (iii), (b) (ii) and (iii), (c) (i) and (iv), (d) (ii) and (iv), 106. Which of these statements about [Co(CN)6]3– is true ?, (a) [Co(CN)6]3– has four unpaired electrons and will be in, a low-spin configuration., (b) [Co(CN)6]3– has four unpaired electrons and will be in, a high spin configuration., (c) [Co(CN)6]3– has no unpaired electrons and will be in a, high-spin configurtion., (d) [Co(CN)6]3– has no unpaired electrons and will be in a, low-spin configuration., 107. The complex ion which has highest magnetic moment among, the following is, (a) [CoF6]3–, (b) [Co(NH3)6]3+, 2+, (c) [Ni(NH3)4], (d) [Ni(CN)4]2–, 108. Which of the following complex ions is diamagnetic?, (a) [FeF6]3–, (b) [CoF6]3–, 3–, (c) [Co(C2O4)3], (d) [Fe(CN)6]3–, 109. Which of the following has square planar structure?, (a) [Ni(CO)4], (b) [NiCl4]2–, 2, (c) [Ni(CN)4]2–, (d) Ni H 2O, 6, 110. Which of the following correctly explains the fact that, [Co(NH3)6]3+ is an inner orbital complex whereas [CoF6]3–, is an outer orbital complex?, (a) NH3 being a strong ligand results into pairing of 3d, orbital electrons in Co3+., (b) F– being a strong ligand results into pairing of 3d, orbital electrons in Co3+., (c) F– being a weak ligand cannot cause the pairing of, electrons present in 3d orbital of Co3+., (d) Both (a) and (c)., , 111. Which of the following statements is incorrect?, (a) [MnCl6]3– is more paramagnetic than [Mn(CN)6]3–, (b) Both [Co(C2O4)3]3– and [CoF6]3– are paramagnetie., (c) [Fe(CN)6]3– forms inner orbital complex whereas, [FeF6]3– forms outer orbital complex., (d) Both (a) and (b)., 112. Which of the following is not correctly matched?, Coordination polyhedron, Shape, (A) [Co(NH3)6]3+, Octahedral, (B) [Ni(CO)4], Square planar, (C) [PtCl4]2–, Tetrahedral, (a) C, (b) B and C, (c) A and C, (d) B, 113. Which of the following are inner orbital complex (i.e.,, involving d2sp3 hybridisation) and is paramagnetic in, nature?, (a) [Mn(CN)6]3– , [Fe(CN)6]3–, [Co(C2O4)3]3–, (b) [MnCl6]3–, [FeF6]3–, [CoF6]3–, (c) [Mn(CN)6]3–, [Fe(CN)6]3–, (d) [MnCl6]3– , [Fe(CN)6]3–, [Co(C2O4)3]3–, 114. Which of the following is the limitation of valence bond, theory?, (a) It does not distinguish between weak and strong, ligands., (b) It does not give quantitative interpretation of magnetic, data., (c) It does not explain the colour exhibited by, coordination compounds, (d) All of these, 115. Which complex of Co2+ will have the weakest crystal field, splitting –, (a) [CoCl6]4–, (b) [Co(CN)6]4–, (c) [Co(NH3)6]2+, (d) [Co(en)3]2+, 116. The crystal field stabilization energy (CFSE) is the highest, for, (a) [CoF4]2–, (b) [Co(NCS)4]2–, 3+, (c) [Co(NH3)6], (d) [CoCl4]2–, 117. In which of the following complexes of the Co (at. no. 27),, will the magnitude of o be the highest?, (a) [Co(CN)6]3–, (b) [Co(C2O4)3]3–, 3+, (c) [Co(H2O)6], (d) [Co(NH3)6]3+, 118. Among the ligands NH3, en, CN– and CO the correct order, of their increasing field strength, is :, (a), , NH3, , (b) CN, (c), , en, , (d), , CO, , en, NH3, CN, NH3, , CN, CO, NH3, en, , CO, en, CO, CN, , 119. Among the following complexes the one which shows zero, crystal field stabilization energy (CFSE):, (a) [Mn(H2O)6]3+, (b) [Fe(H2O)6]3+, 2+, (c) [Co(H2O)6], (d) [Co(H2O)6]3+
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COORDINATION COMPOUNDS, , 389, , 120. The crystal field splitting energy for octahedral (, tetrahedral ( t) complexes is related as, (a), , t, , –, , 1, 2, , (b), , 0, , t, , –, , 4, 9, , 0), , and, , 0, , 3, 2, (d), 0, t –, 0, 5, 5, 121. Which one of the following is the correct order of field, strength of ligands in spectrochemical series?, (a) I – < Cl– < F – < H2O < CN–, (b) F – < H2O < I – < CN – < Cl–, (c) CN – < I – < F – < Cl– < H2O, (d) H2O < F – < CN – < Cl– < I–, 122. The colour of the coordination compounds depends on the, crystal field splitting. What will be the correct order of, absorption of wavelength of light in the visible region, for, the complexes,, , (c), , t, , –, , 3, , 3, , (iii) It does not explain how colour of coordination, compounds depends on ligand attached to central, metal atom/ion., (a) (i) and (ii), (b) (ii) and (iii), (c) (ii) only, (d) (i), (ii) and (iii), 128. If magnetic moment of [MnBr4]2– is 5.9 BM. Predict the, number of electrons?, (a) 2, (b) 3, (c) 6, (d) 5, 129. Arrange the following complexes in increasing order toward, the wavelength of light they absorb? Where M is metal ion., [M(NH3)]3+ = a, [M(CN)6]3– = b, [M(C2O4)3]3– = c,, [MF6]3– = d,, (a) d, c, a, b, (b) d, a, c, b, (c) b, a, c, d, (d) a, b, c, d, 130. Which of the following does not have a metal- carbon bond?, (a), , 3, , [Co(NH3 )6 ] ,[Co(CN)6 ] ,[Co(H 2 O)6 ], , (a), , [Co(NH)6 ]3, , [Co(NH3 )6 ]3, , [Co(H 2O)6 ]3, , (b) [Co(NH3 )6 ]3, , [Co(H 2O)6 ]3, , [Co(CN)6 ]3, , 3, , 3, , 3, , (c), , [Co(H 2 O)6 ], , [Co(NH3 ) 6 ], , (d) [Co(CN)6 ]3, , [Co(NH3 )6 ]3, , [Co(H 2 O)6 ]3, , electrons for d4 configuration will be t 32 g , eg1,, (d) In the presence of CN– as a ligand o < P., 124. Which of the following is incorrect regarding, spectrochemical series?, (b), , F, , 132., , [Co(CN)6 ], , 123. Which of the following statements related to crystal field, splitting in octahedral coordination entities is incorrect?, (a) The dx2 –y2 and dz2 orbitals has more energy as, compared to dxy, dyz and dxz orbitals., (b) Crystal field spitting energy ( o) depends directly on, the charge of the metal ion and on the field produced, by the ligand., (c) In the presence of Br– as a ligand the distribution of, , (a) NH3 > H2O, , 131., , C 2O 24, , (c) NCS– > SCN–, (d) en > edta4–, 125. For which of the following ligands magnitude of the crystal, field splitting ( o)will be greater than pairing energy (P)?, (a) Cl–, (b) SCN–, (c) CO, (d) S2–, 126. Violet colour of [Ti(H2O)6]Cl3 on heating changes to___., (a) Green, (b) Colourless, (c) White, (d) Red, 127. Which of the following is the limitation of crystal field, theory?, (i) Ligands are assumed as point charges., (ii) It does not accounts for the covalent character of, bonding between the ligand and the central atom., , 133., , 134., , Al(OC 2 H 5 ) 3, , (b), , C 2 H 5MgBr, , (c) K[Pt (C 2 H 4 )Cl 3 ], (d) Ni(CO) 4, In Fe(CO)5, the Fe – C bond possesses, (a) ionic character, (b) -character only, (c) -character, (d) both and characters, The charge on the central metal ion in the complex [Ni(CO)4], is, (a) + 2, (b) + 4, (c) 0, (d) + 3, The unpaired electrons in Ni(CO)4 are, (a) zero, (b) one, (c) three, (d) four, The correct structure of Fe(CO)5 is (Z=26 for Fe), (a) octahedral, (b) tetrahedral, (c) square pyramidal, (d) trigonal pyramidal, , 135. For the reaction of the type M 4L, , ML 4, , (a) larger the stability constant, lower the proportion of, ML4 that exists in solution, (b) larger the stability constant, higher the proportion of, ML4 that exists in solution, (c) smaller the stability constant, higher the proportion of, ML4 that exists in solution, (d) None of the above, 136. Coordination compounds have great importance in, biological systems. In this context which of the following, statements is incorrect ?, (a) Cyanocobalamin is B12 and contains cobalt, (b) Haemoglobin is the red pigment of blood and contains, iron, (c) Chlorophylls are green pigments in plants and contain, calcium, (d) Carboxypeptidase - A is an exzyme and contains zinc., 137. Which one of the following coordination compounds is, used to inhibit the growth of tumours?, (a) Trans-platin, (b) EDTA complex of calcium, (c) [(Ph3P)3RhCl], (d) Cis-platin
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EBD_7207, COORDINATION COMPOUNDS, , 390, , 138. For [Co2(CO)8], what is the total number of metal – carbon, bonds and number of metal–metal bonds., (a) 10 ,1, (b) 8, 2, (c) 8, 1, (d) 10, 0, 139. Consider the following reactions., X Y, , XY, , XY Y, , XY2, , Y, , XY2, XY3, , K1=, , [XY], [X][Y], , K2 =, , [XY2 ], [XY][Y], , XY3, K3 = XY Y, 2, , On the basis of reactions above which of the following is, incorrect?, (a) Overall stability constant = K1K2K3, (b) K1K2K3 =, , [XY3 ], , [X][Y]3, , (c) Dissociation constant =, , 1, ., Formation constant, , (d) All of the above are correct., 140. Calculate the value of log K3 when log values of K2, K1, K4, and 4 respectively are 4.0, 3.20, 4.0 and 11.9 ?, (a) 2.0, (b) 2.7, (c) 3.0, (d) 2.5, , STATEMENT TYPE QUESTIONS, 141. Identify the correct statements for the behaviour of, ethane- 1, 2-diamine as a ligand., (i) It is a neutral ligand., (ii) It is a didentate ligand., (iii) It is a chelating ligand., (iv) It is a unidentate ligand., (a) (i), (ii) and (iii), (b) (ii) and (iii), (c) (iii) and (iv), (d) (i), (iii) and (iv), 142. Read the following statements, (i) Macromolecules cannot behave as a ligand., (ii) [EDTA]4– can bind through two oxygen and four, nitrogen atom., (iii) Chelate complexes are more stable than similar, complexes containing unidentate ligands., (iv) Coordination number of the central atom/ion is, determined only by the number of sigma bonds formed, by the ligand with central atom/ion, Which of the following is the correct code for statements, above?, (a) FFTT, (b) FTFT, (c) TFTF, (d) FFFT, 143. Which of the following statements regarding formulas and, naming of coordination compounds are correct?, , (i), , During nomenclature names of neutral ligands are kept, same except for H2O, NH3 and CO., (ii) If the complex is anion, the name of the metal ends, with the suffix–ate., (iii) While writing formula of coordination compounds, polydentate ligands are listed alphabeticaly., (iv) The cation is named first in both positively and, negatively charged coordination entities., (a) (i), (ii) and (iii), (b) (ii), (iii) and (iv), (c) (i), (iii) and (iv), (d) (i), (ii), (iii) and (iv), 144. Which of the following statements are correct?, (i) Square planar complexes of MABXL type show three, isomers-two cis and one trans., (ii) Complexes of Ma3B3 type show three isomers-two, cis and one trans., (iii) Optical isomerism is common in octahedral complexes, involving bidentate ligands., (iv) [Co(NH3)4Cl (NO2)]Cl show linkage isomerism., (v) Hydrate isomerism is another name of solvate, isomerism., (a) (i), (ii) and (iii), (b) (i), (iii) and (iv), (c) (ii), (iii) and (v), (d) (iii), (iv) and (v), 145. Which of the following statements are correct?, (i) When light of wavelength 600nm is absorbed by, complex [Ti(H2O)6]3+ its configuration changes from, t12g e0g, , t 02g e1g and it appears violet in colour.., , (ii) Anhydrous CuSO4 is white but CuSO4 .5H2O is blue, in colour as presence of H2O as a ligand causes crystal, field spitting., (iii) Ruby is aluminum oxide containing 0.5 – 1% Cr 3+, ions with d3 configuration., (iv) Crystal field theory predict correctly that anionic, ligands should exert the greater splitting effect., (a) (i), (ii) and (iii), (b) (ii) and (iii), (c) (i), and (ii), (d) (ii), (iii) and (iv), 146. Which of the following statement(s) is/are incorrect?, (i) In metal carbonyls M–C bond is formed by the, donation of lone pair of electrons on the carbonyl, carbon into a vacant orbital of metal., (ii) M—C bond is formed by the donation of a pair of, electrons from a filled d orbital of metal into the vacant, antibonding * orbital of CO., (iii) Bonding in metal carbonyls is called synergic, bonding., (a) (i) and (ii), (b) (iii) only, (c) (ii) only, (d) None of these
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COORDINATION COMPOUNDS, , MATCHING TYPE QUESTIONS, 147. Match the columns., Column-I, Column-II, (Ligand), (Type of ligand), (A) Triphenylphosphine, (p) Unidenate, (B) BF3, (q) Didentate, (C) Ethylenediamine, (r) Not a ligand, (D) Ethylenediaminetetracetateion (s) Hexadenate, (a) A – (p), B – (r), C – (q), D – (s), (b) A – (p), B – (q), C – (r), D – (s), (c) A – (p), B – (r), C – (q), D – (s), (d) A – (p), B – (q), C – (s), D – (p), 148. Match the complex species given in Column-I with the, isomerism exhibited in Column-II and assign the correct, code:, Column-I, Column-II, (Complex species), (Isomerism), (A) [Co[NH3)4Cl2]+, (p) optical, (B) cis-[Co(en)2Cl2]+, (q) ionisation, (C) [Co(NH3)5(NO2)]Cl2, (r) coordination, (D) [Co(NH3)6][Cr(CN)6], (s) geometrical, (a) A – (s), B – (p), C – (q), D – (r), (b) A – (p), B – (r), C – (q), D – (s), (c) A – (q), B – (s), C – (p), D – (r), (d) A – (p), B – (r), C – (s), D – (q), 149. Match the columns., Column-I, Column-II, (A) [Ni(CN)4]2–, (p) Ti4+, (B) Chlorophyll, (q) sp3; paramagnetic, (C) Ziegler - Natta, (r) Non-planar, catalyst, (D) [NiCl4]2–, (s) Mg2+, (E) Deoxyhaemoglobin, (t) Planar, (u) dsp2; diamagnetic, (a) A – (u), B – (s), C – (p), D – (q), E – (r), (b) A – (q), B – (s), C – (p), D – (u), E – (r), (c) A – (q), B – (s), C – (p), D – (u), E – (t), (d) A – (u), B – (s), C – (p), D – (q), E – (t), 150. Match the columns., Column-I, Column-II, (Complexes), (Absorbed Light), (A) [Ni(H2O)4(en)]2+(aq), (p) Yellow Orange, (q) Blue–Green, (B) [Ni(H2O)4(en)2]2+(aq), (C) [Ni(en)3]2+(aq), (r) Red, (a) A – (r), B – (q), C – (p), (b) A – (p), B – (r), C – (q), (c) A – (q), B – (r), C – (p), (d) A – (r), B – (p), C – (q), , 391, , 151. Match the columns., Column-I, Column-I, (A) Estimation of, (p) [Ag(CN)2]–, water hardness., (B) Extraction of silver., (q) [Ni(CO)4], (C) Hydrogenation of, (r) Na2EDTA, alkenes., (D) Photography, (s) [(Ph3P)3RhCl], (E) Purification of, (t) [Ag(S2O3)2]3–, Nickel., (a) A – (r) , B – (p), C – (s), D – (t), E– (q), (b) A – (p) , B – (r), C – (s), D – (t), E– (q), (c) A – (r) , B – (s), C – (p), D – (t), E– (q), (d) A – (r) , B – (p), C – (s), D – (q), E– (t), 152. Match the columns., Column-I, Column-II, (Coordination compound), (Central metal atom), (A) Chlorophyll, (p) Rhodium, (B) Blood pigment, (q) Cobalt, (C) Wilkinson catalyst, (r) Calcium, (D) Vitamin B12, (s) Iron, (t) Magnesium, (a) A – (t), B – (s), C – (p), D – (q), (b) A – (s), B – (q), C – (p), D – (r), (c) A – (p), B – (q), C – (r), D – (s), (d) A – (r), B – (t), C – (p), D – (q), , ASSERTION-REASON TYPE QUESTIONS, Directions : Each of these questions contain two statements,, Assertion and Reason. Each of these questions also has four, alternative choices, only one of which is the correct answer. You, have to select one of the codes (a), (b), (c) and (d) given below., (a) Assertion is correct, reason is correct; reason is a correct, explanation for assertion., (b) Assertion is correct, reason is correct; reason is not a, correct explanation for assertion, (c) Assertion is correct, reason is incorrect, (d) Assertion is incorrect, reason is correct., 153. Assertion : NF3 is a weaker ligand than N(CH3)3., Reason : NF3 ionizes to give F– ions in aqueous solution., 154. Assertion : [Fe(CN)6]3– is weakly paramagnetic while, [Fe(CN)6]4– is diamagnetic., Reason : [Fe(CN)6 ]3– has +3 oxidation state while, [Fe(CN)6]4– has +2 oxidation state., 155. Assertion : [Ti(H2O)6]3+ is coloured while [Sc(H2O)6]3+ is, colourless., Reason : d-d transition is not possible in [Sc(H2O)6]3+.
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EBD_7207, COORDINATION COMPOUNDS, , 392, , CRITICAL THINKING TYPE QUESTIONS, 156. A co-ordination complex compound of cobalt has the, molecular formula containing five ammonia molecules, one, nitro group and two chlorine atoms for one cobalt atom., One mole of this compound produces three mole ions in an, aqueous solution. On reacting this solution with excess of, AgNO3 solution, we get two moles of AgCl precipitate. The, ionic formula for this complex would be, (a) [Co(NH3)4 (NO2) Cl] [(NH3) Cl], (b) [Co (NH3)5 Cl] [Cl (NO2)], (c) [Co (NH3)5 (NO2)] Cl2, (d) [Co (NH3)5] [(NO2)2Cl2], 157. What is the secondary valence of following compounds, PtCl2.2NH3, CoCl3.4NH3 and NiCl2.6H2O, if moles of AgCl precipitated per mole of the given, compounds with excess AgNO3 respectively are: 0, 1 and 2, (a) 6, 4, 4, (b) 4, 6, 6, (c) 4, 4, 6, (d) 2, 4, 6, 158. C63H88CoN14O14P is the formulae of the Cyanocobalamine,, (vitamin B12) it contain CN– and CN– is very poisonous,, than why this compound does not prove to be fatal for, us? (it inhibit the electron transport chain ?, (a) CN– forms covalent bond, (b) CN– is coordinating to the cobalt as the ligand, (c) CN– hydrolysis immediately, (d) All of these, 159. Suppose someone made aqueous solution of NiCl2 and, recrystallized its aqueous solution in excess of water and, if two moles of precipitate AgCl was formed on treatment, with AgNO3, what is the most probable structure of the, compound ?, (a) [Ni(Cl)2(H2O)4], (b) [Ni (H2O)6]Cl2, (c) [Ni(H2O)5Cl], (d) [Ni (H2O)4Cl2].2H2O, 160. Total number of electron count in Ni(CO)4 and Fe(CO)5, respectively are., (a) 36, 36, (b) 34, 36, (c) 36, 34, (d) 34, 34, 161. The formula for the complex, dichlorobis (urea) copper (II), is, (a) [Cu{O = C (NH2)2}] Cl2, (b) [Cu{O = C (NH2)2}Cl]Cl, (c) [CuCl2 {O = C(NH2)2}2], (d) [CuCl2] [{O = C (NH2)2}]2, 162. According to IUPAC nomenclature sodium nitroprusside is, named as, (a) Sodium pentacyanonitrosylferrate (III), (b) Sodium nitroferrocyanide, (c) Sodium nitroferricyanide, (d) Sodium pentacyanonitrosylferrate (II), , 163. The total number of possible isomers for the complex, compound [CuII (NH3)4] [PtII Cl4], (a) 3, (b) 6, (c) 5, (d) 4, 164. Which of the following will give a pair of enantiomorphs?, (a) [Cr(NH3)6][Co(CN)6] (b) [Co(en)2Cl2]Cl, (c) [Pt(NH3)4] [PtCl6], (d) [Co(NH3)4Cl2]NO2, (en =NH2CH2CH2NH2), 165. The existence of two different coloured complexes with the, composition of [Co(NH3 )4 Cl2 ]+ is due to :, (a) linkage isomerism, (b) geometrical isomerism, (c) coordination isomerism(d) ionization isomerism, 166. Which of the following has an optical isomer, (a) [Co(en) (NH3)2]2+, (b) [Co(H2O)4(en)]3+, 3+, (c) [Co(en)2 (NH3)2], (d) [Co(NH3)3Cl] +, 167. Among the following complexes, optical activity is possible, in, (a), , [ Co ( NH 3 ) 6 ]3, , (b) [ Co ( H 2 O) 2 ( NH 3 ) 2 Cl 2 ], (c), , [ Cr ( H 2 O) 2 Cl 2 ], , (d) [ Co ( CN ) 5 NC ], 168. [Co(NH3)4 (NO2)2] Cl exhibits, (a) linkage isomerism, ionization isomerism and, geometrical isomerism, (b) ionization isomerism, geometrical isomerism and optical, isomerism, (c) linkage isomerism, geometrical isomerism and optical, isomerism, (d) linkage isomerism, ionization isomerism and optical, isomerism, 169. Which of the following compounds exhibits linkage, isomerism ?, (a) [Co(en)2]Cl3, (b) [Co(NH3)6][Cr(en)3], (c) [Co(en)2NO2Cl]Br, (d) [Co(NH3)5Cl]Br2, 170. Among the following coordination compounds/ions, 3, , (i), , Fe CN 6, , (iii), , Co NH 3 6, , 3, , (ii), , Pt NH3 2 Cl 2, , (iv), , Cr H 2 O 6 Cl3, , Which species exhibit geometrical isomerism?, (a) (ii) only, (b) (i) and (ii), (c) (ii) and (iv), (d) (i) and (iii), 171. Identify the optically active compounds from the following :, (i), , [Co(en)3 ]3, , (ii), , trans [Co(en)2 Cl 2 ], , (iii) cis [Co(en)2 Cl2 ], , (iv) [Cr(NH3 )5 Cl], , (a) (i) and (iii), (c) (iii) and (iv), , (b) (ii) and (iii), (d) (i), (iii) and (iv)
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COORDINATION COMPOUNDS, , 172. For which value of the x, and y, the following square, planar compound shows geometrical isomers [Pt (Cl)x, (Br)y]2–, (a) 1, 3, (b) 3, 1, (c) 2, 2, (d) 1, 1, 173. How many geometrical isomers are possible for following, square planar compound [M (Cl) (Br) (I) (F)] (where M is, a metal ion), (a) 2, (b) 3, (c) 9, (d) 8, 174. The terahedral complex [M(A)(B)(X)(Y)], where A,B,X and, Y are different ligands and M is a metal ion is, (a) optically inactive, (b) rotate plane polarized light, (c) incomplete information, (d) can’t be said, 175. The complex given is, , 393, , sp3,, , 182., , Co, , 176., , 177., , 178., , 179., , en, (i) non-superimposable on its mirror images, (ii) optically inactive, (iii) rotate plane polarised light, (iv) planar, (a) (i) and (ii), (b) (i) and (iv), (c) (i), (ii) and (iii), (d) (ii) only, The correct statement with respect to the complexes Ni(CO)4, and [Ni(CN)4]2– is, (a) nickel is in the same oxidation state in both, (b) both have tetrahedral geometry, (c) both have square planar geometry, (d) have tetrahedral and square planar geometry, respectively, Which one of the following complexes is an outer orbital, complex ?, (a) [Co(NH3)6]3+, (b) [Mn(CN)6]4–, 4–, (c) [Fe(CN)6], (d) [Ni(NH3)6]2+, (Atomic nos. : Mn = 25; Fe = 26; Co = 27, Ni = 28), The d-electron configurations of Cr2+, Mn2+, Fe2+ and Co2+, are d 4 , d 5, d 6 and d 7, respectively. Which one of the, following will exhibit minimum paramagnetic behaviour?, (a) [Mn(H2O)6]2+, (b) [Fe(H2O)6]2+, 2+, (c) [Co(H2O)6], (d) [Cr(H2O)6]2+, (At, nos. Cr = 24, Mn = 25, Fe = 26, Co = 27), Both [Ni(CO)4 ] and [Ni(CN)4]2– are diamagnetic. The, hybridisations of nickel in these complexes, respectively,, are, , sp3,, , dsp2, , (a), (b), 2, 3, (c) dsp , sp, (d) dsp2, sp2, 180. Which one of the following is an inner orbital complex, as well as diamagnetic in behaviour? (Atomic number:, Zn = 30, Cr = 24, Co = 27, Ni = 28), (a) [Zn(NH3 )6]2+, (b) [Cr(NH3)6]3+, 3+, (c) [Co(NH3)6], (d) [Ni(NH3)6]2+, 181. The value of the ‘spin only’ magnetic moment for one of the, following configurations is 2.84 BM. The correct one is, , en 3+, , en, , sp3, , 183., , 184., , 185., , 186., , (a), , d 5 (in strong ligand field), , (b), , d 3 (in weak as well as in strong fields), , (c), , d 4 (in weak ligand fields), , (d) d 4 (in strong ligand fields), The d electron configurations of Cr2+, Mn2+, Fe2+ and Ni2+, are 3d 4, 3d 5, 3d 6 and 3d 8 respectively. Which one of the, following aqua complexes will exhibit the minimum, paramagnetic behaviour?, (a) [Fe(H2O)6]2+, (b) [Ni(H2O)6]2+, 2+, (c) [Cr(H2O)6], (d) [Mn(H2O)6]2+, (At. No. Cr = 24, Mn = 25, Fe = 26, Ni = 28), According to valence bond theory which of the following, statement is correct about the complexes Ni(CO)4 and, [Ni(CN)4]2– if both are diamagnetic in nature, (a) both are tetrahedral, (b) both are square planar, (c) one is square planar and other is tetrahedral, (d) one is tetrahedral and other is square planar, Correct statements about the following complexes, [MnCl6]3– and [Mn(CN)6]3– respectively are., (a) Magnetic moment is 4.8 and 2.8, (b) inner sphere and outer sphere complexes., (c) sp3d2 and d2sp3 complexes., (d) Both (a) and (c)., In which of the following coordination entities the magnitude, 0 (CFSE in octahedral field) will be maximum?, (a) [Co(H2O)6]3+, (b) [Co(NH3)6]3+, 3–, (c) [Co(CN)6], (d) [Co (C2O4)3]3–, (At. No. Co = 27), Which of the following complex ions is expected to absorb, visible light?, (a) [Ti (en)2(NH3)2]4 +, (b) [Cr (NH3)6]3 +, 2, +, (c) [Zn (NH3)6], (d) [Sc (H2O)3 (NH3)3]3+, (At. no. Zn = 30, Sc = 21, Ti = 22, Cr = 24), , 187. Crystal field stabilization energy for high spin d 4 octahedral, complex is:, (a) – 1.8, , 0, , (b) – 1.6 0 + P, , (c) – 1.2, , 0, , (d) – 0.6, , 0
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EBD_7207, COORDINATION COMPOUNDS, , 394, , of d 6 -cation, , 188. Low spin complex, have the following energy :, (a), , 12, 5, , 0, , P, , (b), , in an octahedral field will, 12, 5, , 0, , 3P, , 2, 2, (d), (c), 0 2P, 0 P, 5, 5, ( 0= Crystal Field Splitting Energy in an octahedral field,, P = Electron pairing energy), 189. Which of the following carbonyls will have the strongest, C – O bond ?, (a) [Mn (CO)6]+, (b) [Cr (CO)6], (c) [V (CO)6]–, (d) [Fe (CO)5], , 190. Which of the following complexes formed by Cu 2 ions is, most stable ?, , (a), , Cu 2, , 4NH3, , [Cu(NH3 ) 4 ]2 , log K = 11.6, , (b), , Cu 2, , 4NH, , [Cu(NH) 4 ]2 , log K = 27.3, , (c), , Cu 2, , 2en, , (d), , Cu 2, , 4H 2O, , [Cu(en)2 ]2 ,, , log K = 15.4, , [Cu(H 2O)4 ]2 , Log K = 8.9, , 191. Atomic number of Mn, Fe and Co are 25, 26 and 27, respectively. Which of the following inner orbital octahedral, complex ions are diamagnetic ?, (i), , [Co(NH3 )6 ]3, , (ii), , [Mn(CN)6 ]3, , (iii) [Fe(CN)6 ]4, , (iv) [Fe(CN)6 ]3, , (a) (i) and (ii), (c) (iii) and (iv), , (b) (i) and (iii), (d) (ii) and (iv)
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COORDINATION COMPOUNDS, , 395, , FACT / DEFINITION TYPE QUESTIONS, 1., 2., 3., 4., , (a) The primary valencies are ionisable and represented, by dotted line., (c), (a) CrCl3 has primary valence of 3., (d) Co (NH3)5 Cl3, [Co(NH3)5Cl]+2 + 2ClStructure is [Co (NH3)5 Cl] Cl2., Now [Co(NH3 )5 Cl]Cl2, , 2AgNO3, , [Co(NH3 )5 Cl](NO3 )2, 5., , 2AgCl, , (a) Since the precipitate of AgCl shows two ionisable, chloride ion the complex must have the structure., [Co( NH 3 ) 5 Cl]Cl 2, , 2AgNO 3, , [Co( NH 3 )5 Cl]( NO 3 ) 2, , 6., , 2AgCl, , Hence two chlorine atoms satisfy the primary valency, and one, secondary valency, (c) The ions present in the ionisation sphere are precipited, Hence [CrCl 2 (H 2O) 4 ]Cl.2H 2 O contains 1/3 Cl in, , 7., , 8., , ionisation sphere to be precipited by AgNO3 as AgCl, (b) Complex compounds do not dissociate into constituent, ions., K4[Fe(CN)6], 4K+ + [Fe(CN6)]4–, It is a complex because no CN– is formed on, dissociation., (b) It is a double salt:, FeSO 4 . NH 4 2 SO 4 .6H 2 O, Fe2, , 9., 10., , 11., , 2SO24, , 2NH 4, , (a), (a) Let the o.s. of Ni in K4[Ni(CN)4] be = x then, 4 (+ 1) + x + (–1) × 4 = 0, 4+x–4=0, x=0, (c) The coordination number of central metal atom in a, complex is equal to number of monovalent ligands,, twice the number of bidentate ligands and so on,, around the metal ion bonded by coordinate bonds., Hence coordination number = no. of bonds formed, by metals with ligands, , 12., , (d) Oxidation state of Cr in [Cr ( NH 3 ) 4 Cl 2 ] ., , 13., , Let it be x, 1 × x + 4 × 0 + 2 × (–1) = 1 Therefore x =3., (c) The oxidation state of metal in metal carbonyls is always, zero., , 14. (d) [EDTA]4– is a hexadentate ligand, because it has six, donor atoms and donate 6 pairs of electrons to central, metal atom in the complex., 15. (d) In Fe(CO)5, Fe is in minimum oxidation state (zero)., 16. (d) In the given complex we have two bidentate ligands, (i.e en and C2O4), so coordination number of E is 6, (2 × 2 + 1 × 2 = 6), Let the oxidation state of E in complex be x, then, [x + (–2) = 1] or x – 2 = 1, or x = + 3, so its oxidation state is + 3, Thus option (d) is correct., 17. (a) Salt may be complex salt. Metal atom present in the, coordination sphere appears in the form of complex, ion and not as simple cation, 18. (b) Coordinate number is = 2 (number of bindentate ligands, C.N. of Ni = (2 × 3) = 6, 19. (b) In the complex formation the ligands whether negative,, neutral or positive always donate electrons to the, central metal atom hence they act as Lewis bases., 20. (c) K[Co(CN)4] let the O. N. of Co be x then, 1 × (+1) + 1(+x) + 4(–1) = 0 x = + 3, 21. (c) Number of donor atoms (N) in N(CH2CH2NH2)3 are, four., So, N(CH2CH2NH2)3 is a tetradentate ligand., 22. (d) Ambidentate ligands are those unidentate ligands, which contain more than one coordinating atoms., Thiocyanate is an example of such a ligand., M SCN, M NCS, Thiocyanato, , Isothiocyanato, , 23. (c), 24. (d), 25. (c) Ambident ligand has two donor atoms, either of two, can form a coordinate bond., 26. (c), 27. (c) Na+ is not a ligand., 28. (d) These are facts about glycinato ligand., 29. (c) BF3 has incomplete octet and is Lewis acid; it cannot, donate electron pair., COO, 30. (a) It has two donor atoms, i.e., |, COO, 31. (b), 32. (b), 33. (b) O2 is a bidentate ligand., 34. (c), 35. (a), 36. (b), 37. (c), 38. (b), 39. (a), 40. (c) EDTA4– can bind through two nitrogen and four, oxygen atoms to a central metal ion.
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COORDINATION COMPOUNDS, , 67., , 397, , (a) Complexes of the type MABCD may exist in three, isomeric forms., A, B, A, C, en, M, , D, A, , C, , C, , 69., , 70., , B, , (III), , D, , Similarly, [Pt (py) (NH3) BrCl] may exist in three isomeric, form in which, M = Pt, A = Py, B = NH3, C = Br, D = Cl., (a) Complexes with dsp2 hybridisation are square planar., So, [PtCl4]2– is square planar in shape., (a) The SCN– ion can coordinate through S or N atom, giving rise to linkage isomerism, M SCN thiocyanato, M, NCS isothiocyanato., (b) Option (b) shows optical isomerism [Co(en)3]3+, 3+, , en, en, , 3+, , en, , Co, , Co, , en, , en, , en, Mirror, , d–form, , –form, , Complexes of Zn++ cannot show optical isomerism as, they are tetrahedral complexes with plane of symmetry., Co(H2 O)4 (en), , 3, , (b), , 3+, , en, NH3, Co, NH3, , en, cis, , 72. (b) Ionisation isomer of [Cr(H2 O) 4 Cl(NO 2 )]Cl is, [Cr(H2O)4Cl2]NO2., 73. (c) Ma 2 b2 c 2 can show both optical & geometrical, isomerism., 74. (a) The compound shows linkage isomerism because the, ligand in the compound is an ambidenate ligand that, can bond at more than one atomic site., i.e., NCS and SCN, 75. (d) [Pt(en)2Cl2] is a complex of the type M(AA)2 B2 which, is octahedral Such compounds exhibit optical and, geometrical isomerism both, 76. (a) [Co(NH3)3(NO2)3] is of the type MA3B3. They give, two geometrical Isomers, 77. (d) [Cr(NH3)3Cl3] is of the type MA3B3 and exists in two, isomeric forms., 78. (d) Square planar complex of the formula Mabcd give three, geometrical isomers, 79. (b), 80. (d) The complex ion [Cr(SCN)2(NH3)4]2+ can exhibit, geometrical and linkage isomerism, 81. (c) [PtCl2(NH3)4]Br2 and [PtBr2(NH3)4]Cl2 are ionisation, isomers, 82. (c) Octahedral complex of the type MA5B do not show, geometrical isomerism, 83. (c) Similarity between optical and geometrical isomerism, is that both are included in stereo isomerism, 84. (b) The compound is [Cr(NH3)5NO2]Cl2 and can exhibit, O, , linkage isomerism due to NO2 group (– N, , have two planes of symmetry, , hence it is also optically inactive., [Zn(en)2]2+ cannot show optical isomerism, 71., , NH 3, trans-, , B, , (II), , M, , B, , 68., , D, , A, , M, , D, , en, , Co, , M, , (I), C, , 3+, , NH 3, , 85., 86., 87., 88., , O – N= O), (b), (b) Given compound shows meridional isomerism., (a), (c), Ni(CO)4, Ni(PPh3)2 Cl2, O.S., , Ni0, , Ni2+, , E.C., , [Ar]3d84s2, , [Ar]3d84s0, , Pairing of e–, , No pairing of e–, , Hybridization, , sp3 (tetrahedral) sp3 (tetrahedral), , or, O
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EBD_7207, COORDINATION COMPOUNDS, , 398, , 89., , (d), , In [Cr (NH3)6]Br3, Cr is in +3 oxidation state, 3d, 4s, 4p, Cr, , 24, , 98., , (a) Paramagnetic species has unpaired electron. More the, no. of unpaired electrons, more will be paramagnetic, character., Complex, O. S. of metal, , d2sp3 hybridized, , Cr, , 3d, , +3, , 4p, , 4s, , 21, , 90., , 91., , Its ion is octahedral in nature. Due to the presence of, three unpaired electrons it is paramagnetic., (c) [Cr(NH3)6]Cl3 is an inner orbital complex, because in, this complex inner d-orbital is used for hybridisation, (d2 sp3 ), (b), , Atom/Ion, Complex, , Ni, , 2+, , 3d, , 8, , 4s, , 2, , 8 2, , 2, , Ni (d s ), [Ni(CO)4], , 92., 93., 94., , 95., , 96., , Co3+, , 3, , sp, , :, , [Co(CN)6]3– :, , CN–, , 97., , Fe2+, 2, , Cu2+, , 4, , However, CN– is a strong ligand, so pairing of electrons, will occur in the complex having CN– ions., Fe2+ in presence of CN–, , 0, , (b), (b) As in [NiCl4]–2 chloride ion being a weak ligand is not, able to pair the electrons in d orbital., (d) Cr 3+ has 4so 3d3 electronic configuration with 3, unpaired electrons, hence paramagnetic. In other cases, pairing of d-electrons take place in presence of strong, field ligands such as CO or CN–., In Cr(CO)6 molecule 12 electrons are contributed by, CO group and it contain no odd electron, (b) CN– is a strong field ligand as it is a pseudohalide ion., These ions are strong coordinating ligands and hence, have the tendency to form -bond (from the pseudo, halide to the metal) and -bond (from the metal to, pseudo halide), (a), , 4, , 6, , No. of unpaired electron = 1, Fe2+, , 0, , Rearrangement, , Cu2+, , No. of unpaired electrons = 5, Cu2+, , sp3, , 2–, , dsp, , 2, , Electronic configuration of the ion present in complex, 3d, 4s, 4p, 2+, Mn, , 2, , Rearrangement, , 4, , Cu H 2 O, , 2, , 2–, , [Ni(CN)4], , Cu NH3, , Mn 2+, , 4p, , (d ), , [NiCl4], , 6, , Fe CN, , No. of, Magnetic, unpaired, nature, electrons, , Configuration, , 2, , Mn H 2 O, , is a strong field ligand and it causes pairing of, electrons as a result number of unpaired electrons in, Co3+ becomes zero and hence it has lowest value of, paramagnetic behaviour., (a) Sc = [Ar] 3d 1, 4s2, Oxidation state of Sc in [Sc(H2O)6]3+ is Sc3+, Sc3+ = [Ar] 3d 0, 4s0., It does not have unpaired electron, Sc3+ is diamagnetic and colourless., , No. of unpaired electron = 0, Thus [Mn (H2O)6]2+ having maximum no. of unpaired, electrons has maximum paramagnetic nature., 99. (c) In Na2[CdCl4], Cd has oxidation state +2., So, its electronic configuration is 4d105s0, or all the 4d orbitals are fully filled., Hence, there will be no d-d transition. So, it is, colourless., 100. (c), 101. (a), , Ni(CN)4, , 2, , : Number of unpaired electrons = 0, , Cr(NH3 ) 6, , 3, , Fe(H 2 O) 6, , 2, , : Number of unpaired electrons = 4, , 2, , : Number of unpaired electrons = 2, , Ni(H 2O)6, , : Number of unpaired electrons = 3, , 102. (b) Ni++ = 3d 8 4s0, , Since, the coordination number of Ni in this complex is, 4, the configuration of Ni++ at first sight shows that, the complex is paramagnetic with two unpaired, electron. However, experiments show that the complex
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COORDINATION COMPOUNDS, , 399, , is diamagnetic. This is possible when the 3d electrons, rearrange against the Hund’s rule as shown below., This is in accordance with the fact that the ligand, involved here is strong i.e., CN– ion., Ni++ (after rearrangement), 4s, , 4p, , Hence, now dsp2 hybridization involving one 3d, one, 4s and two 4p orbitals, takes place leading to four dsp2, hybrid orbitals, each of which accepts four electron, pairsfrom CN– ion forming [Ni (CN)4]2– ion., [Ni (CN)4]2–, ××, , ××, , ××, , ×× ××, , 109. (c) Electronic configuration of Ni2+ is [Ar] 3d8 4s0 4p0., CN– is strong ligand and will do pairing of electrons, so will have one d orbital left empty. C.N. is 4 so dsp 2, hybridisation will take place which is square planar,, Ni(CO)4 and [Ni(CN)4]2– are tetrahedral., 110. (d), 111. (b) [Co(C2O4 )3]3– is dimagnetic as oxalate is a strong, ligand causing pairing of 3d electrons in Co3+thereby, leading to d2sp3 hybridisation., 3d, , 4s, , 4p, , 3d, , 4s, , 4p, , Orbitals of, 3+, Co ion, 2, , 3, , d sp hybridised, 3+, oribitals of Co, , 2, , four dsp hybrid bonds, Thus, the complex is diamagnetic as it has no unpaired, electron., 103. (a) [Ni(NH3)6]2+, Ni2+ = 3d 8, according to CFT = t 62g eg2 therefore,, hybridisation is sp3d 2 and complex is paramagnetic., 104. (d), 105. (a), 106. (d) In [Co(CN)6]3– O.N. of Co is +3, Co+3 = 3d6 4s0, CN– is a strong field ligand, Pairing of electrons occurs so in this complex no, unpaired electron is present and it is low spin, complex., 107. (a), Complex, Configuration, No. of unpaired, electrons, 3–, [CoF6], 4, [Co(NH3)6]3+, 0, [Ni(NH3)4]2+, [Ni(CN)4]2–, , 0, , 3d, , 4s 4p, : : : : :, , Rearrangement dsp, , 0, , 2, , Magnetic moment = n n 2 ,, where n = no. of unpaired electrons, [CoF6]3– has highest magnetic moment (4.8) due to, the presence of 4 unpaired electrons., 108. (c) [Co(C2O4)3]3– has Co3+ (d6 system), due to presence of stronger C2O42– chelating ligand, pairing of electrons occurs in this case., Co3+ :, ; Diamagnetic., , 3, , d sp hybrid, , 2, , 112. (b) [Ni(CO4)], tetrahedral, [PtCl4]2–, square planar, 113. (c) [Mn(CN)6 ]3– and [Fe(CN) 6 ] 3– are inner orbital, complexes and paramagnetic while [Co(C2O4)3]3– is, diamagnetic in nature., 114. (d), 115. (a) Cl– is a weak field ligand., 116. (c) Higher the oxidation state of the metal, greater the, crystal field splitting energy. In options (a), (b) and, (d), Co is present in + 2 oxidation state and in (c) it is, present in + 3 oxidation state and hence has a higher, value of CFSE., 117. (a) In octahedral complex the magnitude of o will be, highest in a complex having strongest ligand. Out of, the given ligands CN– is strongest. So, o will be, highest for [Co(CN)6]3–. Thus option (a) is correct., 118. (a) Ligands can be arranged in a series in the orders of, increasing field strength as given below :, Weak field ligands :, , I, , Br, , S2, , SCN, , Cl, , N3 , F, < Urea, OH– < oxalate, , Strong field ligands, , O, , H 2O, , NCS, , EDTA, , Py, NH3 <, , en = SO3– < bipy, Phen, , C6 H 5, , NO2, , CH3, , CN, CO, Such a series is termed as spectrochemical series. It is, an experimentally determined series based on the, absorption of light by complexes with different ligands., 119. (b) Due to d5 configuration CFSE is zero.
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EBD_7207, COORDINATION COMPOUNDS, , 400, , 120. (b) The crystal field splitting in tetrahedral complexes is, lower than that in octahedral complexes, and, t, , 137. (d), 138. (a) Structure of [Co2(CO)8], , 4, 0., 9, , O, , 122. (c), 121. (a), 123. (d) CN– is a strong field ligand and form low spin, complexes thus o > P., 124. (b) According to spectrochemical series C2 O24, , n(n 2)B.M. Where n = number of, , electrons and B.M. is Bohr magneton., 5.9 =, , Al, , 131. (d) Due to some backbonding by sidewise overlapping, between d-orbitals of metal and p-orbital of carbon,, the Fe–C bond in Fe(CO)5 has both and character., 132. (c) In case of [Ni(CO)]4, the ligand CO, is neutral thus the, charge on Ni is zero., 133. (a) Ni(CO) 4 . The O. S. of Ni is Zero. Electronic, configuration is [Ar] 3d8 4s2 4p0 . In presence of, strong ligand CO the paring of electrons take place, and electronic configuration will be [Ar] 3d10 4s0 4p0., Hence unpaired electrons is zero ., 134. (d) Fe(CO)5 (Z=26) O. S. of Fe is zero . Electronic, configuaration is [Ar]] 3d6, 4s24p0. After pairing of, electrons of d and s orbitals, we have one d atomic, orbital empty. C. N. is 5 so hybridisation is dsp3 which, is trigonal bipyramidal., 135. (b) For the reaction of the type M 4L, , ML4 , larger, , the stability constant, the higher the proportion of, ML4 that exists in solution., 136. (c) The chlorophyll molecule plays an important role in, photosynthesis, contain porphyrin ring and the metal, Mg not Ca., , CO, , C, , Total M – C bonds = 10, Total M – M bonds = 1, 1, [X][Y]3, =, 139. (a) Overall stability constant = K, [XY3 ], 1K 2 K 3, , 140. (b), , = K1K2K3K4, log, , O CH 2 CH 3, O CH 2 CH 3, O CH 2 CH 3, , CO, , Co, , O, , n(n 2), , n=5, 129. (c) Energy for excitation ( ) = hc /wavelength., Therefore lower the wavelength of light higher will be, the energy for excitation, ( ) i.e crystal field splitting, energy., correct order will be b < a < c < d, 130. (a) Triethoxyaluminium has no Al – C linkage, , Co, , CO, , F ., , 125. (c) CO is a strong field ligand and for strong field ligands, o > P., 126. (b) In the absence of ligand , crystal field splitting does, not occur and hence the substance is colourless., 127. (a), 128. (d) Magnetic moment can be calculated by using the, relation, , CO, , CO, , C, , CO, , = log (K1 K2 K3 K4), , log = logK1 + logK2 + logK3 + logK4, logK3 = 11.9– (3.20 + 2.0 + 4.0), logK3 = 2.7, , STATEMENT TYPE QUESTIONS, 141. (a), 142. (a) Macromolecules like proteins can acts as ligand., [EDTA]4– can bind through two nitrogen and four, oxygen atoms., 143. (d), 144. (b) Complexes of Ma3B3 type shows fac– meridional, isomerism., Solvate isomerism is refered to as a hydrate isomerism, when solvent is water., 145. (b), , [Ti(H 2O)6 ]3 gives, , violet, , colour, , if, , light, , corresponding to the energy of blue-green region of, wavelength 498 nm is absorbed by it., Irrespective of prediction of crystal field theory on, the basis of experimental observation shows that, anionic ligands are found at the low end of the, spectrochemical series., 146. (d), , MATCHING TYPE QUESTIONS, 147. (c), 148 (d) A – (p), B – (r), C – (s), D – (q), 149. (a) [NiCl4]2– is sp3 hybridised and paramagnetic in nature, [Ni(CN)4 ]2– is square planar and diamagnetic., Chlorophyll contains Mg2+ Ziegler – Natta catalyst, contains Ti4+ Deoxyhaemoglobin is nonplanar and, oxyhaemoglobin planar.
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COORDINATION COMPOUNDS, , 401, , 150. (d) Crystal field splitting energy increases with increase, in ligand field strength i.e., with increase in no. of ‘en’, groups and wavelength of absorbed light decrease, with increase in ligand field strength, [Ni(H2O)4(en)]2+(aq) will absorb light of higher, wavelength i.e., Red., [Ni(en)3]2+ will absorb light of lower wavelength i.e.,, blue-green and [Ni(H2O)4(en)2]2+ will absorb yellow, orange light., 151. (a), 152. (a), , 161. (c) [CuCl2{(O = C(NH2)2}2], 162. (a) IUPAC name of sodium nitroprusside Na2[Fe(CN)5NO], is sodium pentacyanonitrosylferrate (III) because in it, NO is neutral ligand. Hence, 2×O.N. of Na + O.N. of Fe + 5×O.N. of CN, 1×O.N. of NO = 0, 2×(+1) + O.N. of Fe + 5 ×(–1) +1×0 = 0, O.N. of Fe = 5 – 2 = +3, Hence ferrate (III), 163. (d) The total number of isomers for the complex compound, [Cu II ( NH 3 ) 4 ][Pt II Cl 4 ] is four.., , ASSERTION-REASON TYPE QUESTIONS, , These four isomers are, , 153. (c) It is correct statement that NF3 is a weaker ligand than, N(CH3 ) 3 , the reason is that fluorine is highly, electronegative therefore, it with draw electrons from, nitrogen atom. Hence, the lone pair of nitrogen atom, cannot be ligated. While N(CH3)3 is a strong ligand, because CH3 is electron releasing group., 154. (b) Both Assertion and Reason are true but Reason is not, the correct explanation of statement-1. [Fe(CN)6]3– is, weakly paramagnetic as it has unpaired electrons while, [Fe(CN)6]2– has no unpaired electron., It is diamagnetic., 155. (a) Both Assertion and Reason are true and Reason is the, correct explanation of statement-1. [Sc(H2O6]3+ has, no unpaired electron in its d subshell and thus d–d, transition is not possible whereas [Ti(H2O)6]3+ has, one unpaired electron in its d subshell which gives, rise to d–d transition to impart colour., , [Cu ( NH 3 )3 Cl] [Pt ( NH 3 )Cl 3 ],, [Cu ( NH 3 )Cl 3 ] [Pt ( NH 3 ) 3 Cl] ,, [CuCl 4 ][Pt ( NH 3 ) 4 ], and, , The isomer [Cu (NH3)2 Cl2][Pt (NH3)2 Cl2] does not, exist due to both parts being neutral., 164. (b) Non superimposable mirror images are called optical, isomers and may be described as “chiral’. They are, also called enantiomers and rotate plane polarised light, in opposite directions., Cl, , [Co(NH 3 )5 NO 2 ]Cl 2, , 2Cl, , 2AgNO3, , [Co(NH 3 )5 NO 2 ], , Cl, Cl, , en, , Cl, , Co, , CRITICAL THINKING TYPE QUESTIONS, 156. (c) As it forms two moles of silver chloride thus it has two, moles of ionisable Cl., , PtCl4 ., , Cu(NH3 )4, , Co, , en, , en, , 165. (b), , 2Cl, , 2AgCl 2NO 3, , 157. (b) Pt Cl2.2NH3= [Pt(NH3)2Cl2], CoCl3.4NH3= [Co(NH3)4Cl2] Cl, NiCl2.6H2O = [Ni(H2O)6]Cl2., 158. (b) CN– is coordinated to cobalt as the ligand and, coordinated compounds have different properties, than the individual species., 159. (b) Since complex compound gives 2 moles of AgCl on, treatment with AgNO3., most probable structure is (b)., 160. (a) Total number of electron count in Ni(CO)4, = Atomic number – oxidation state + 2 × no. of ligands, = 28 – 0 + 2 × 4 = 36, Similarly for Fe(CO)5,, = 26 – 0 + 10 = 36, , trans (green), HN, 3, , Cl, , Cl, , Co, HN, 3, , NH3, , NH3, , cis (violet), , +, , en
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EBD_7207, COORDINATION COMPOUNDS, , 402, , NH3, , 166. (c), , NH3, en, , NH3, , NH3, , Co, , F, , I, Br, 174. (b) Non –superimposable mirror images are optically, active, hence rotate plane polarized light., , en, , Enantiomers of cis- Co(en) 2 (NH 3 ) 2, , 3, , 167. (b) It is optically active., 168. (a) The given compound may have linkage isomerism due, to presence of NO2 group which may be in the form, –NO2 or –ONO., It may have ionisation isomerism due to presence of, two ionisable group –NO2 & –Cl. It may have, geometrical isomerism in the form of cis-trans form as, follows :, [Co(NH3)4Cl(NO2)]NO2 & [Co(NH3) (NO2)2]Cl, ––– Ionisation isomers., [Co(NH3)5(NO2)2]Cl & [Co(NH3)5(ONO)2]Cl, ––– Linkage isomers, NO2, , H3N, Co, NH3, , a, , M, , d, , NH3, , H3N, , NH3, , NH3, , Co, , b, c, whatever angle, 175. (c) Complex is not superimposable, onofitsmolecule, mirror image, hence optically active i.e., rotate plane polarized light., , b, , c, , Trans-form, , Cis-form, , Cl, , F, , M, , d, , 180°, 3+, , 3+, en, en, , en, Co, , Co, , NO2, NH3, , NH3, , M, , en, , en, cannot be, superimposed, , non-superimposable, , en, , en, 180°, 3+, , en, Co, en, , Geometrical isomers, 169. (c) The complex compound [Co(en)2NO2Cl] Br can have, NO2 group differently linked to central metal atom –, O, or O – N = O, N, O, 170. (a) Geometrical isomerism is possible only in square planar, complexes of the type MA2B2 and MA2BC and for, octahedral complexes of the type MA4B2 and MA4BC., Hence only (ii) will show geometrical isomerism., 171. (a), 172. (b) Geometrical isomers of following type of square, planar complexes is possible. Ma2b2 type, Ma2bc, type and Mabcd type., 173. (b) Three isomer are possible., , I, , a, , NO2, , NO2, , F, , M, , en, , Co, , en, , Cl, , I, , 176. (d) Nickel in Ni(CO)4 is sp3 hybridised therefore geometry, of Ni(CO)4 is tetrahedral whereas in [Ni(CN)4]2– nickel, is dsp2 hybridised therefore geometry of [Ni(CN)4]2–, is square planar., 177. (d) Hybridisation, [Fe(CN)6 ]4 ,[Mn(CN)6 ]4 ,, d 2 sp3, , d 2 sp 3, , [Co(NH 3 ] 3 ,[Ni(NH 3 ) 6 ] 2, d 2 sp 3, , sp 3d 2, , Hence [ Ni( NH 3 ) 6 ]2 is outer orbital complex., 178. (c) Cr2+ d4, , 4, , Mn2+ d5, , 5, , Fe2+ d6, , 4, , Co2+ d7, , 3, , M, Br, , Br, , Cl, , Minimum paramagnetic behaviour = [Co (H2O)6]2+
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COORDINATION COMPOUNDS, , 403, , 179. (b) In carbonyls O.S. of metal is zero, In [Ni(CO)4], the oxidation state of nickel is zero. Its, configuration in Ni(CO)4 is, 3d, , 4s, , = 3(5) = 15 = 3.87 B.M., d 4– in weak ligand field, , 4p, , [Ni(CO)4] ;, , t2g, sp3 hybridisation, , In [Ni(CN)4]2– the oxidation state of Ni is 2+ and its, configuration is, , 3d, , 2, , 4s, , 4p, , [Ni(CN)4], , dsp2 hybridisation, Thus the hybridisations of nickel in these compounds, are sp3 and dsp2 respectively., Hence (b) is the correct answer., 180. (c) [Co(NH3 )6 ]3+ , Co3+ (27 –3 = 24), , eg, , = 4(4+2) = 24 = 4.89, d 4– in strong ligand field, , t2g, , eg, , = 2(4) = 8 = 2.82., 182. (b) Lesser is the number of unpaired electrons smaller will, be the paramagnetic behaviour. As Cr ++, Mn++, Fe++, and Ni++ contains., Cr++ (3d 4) =, = 4 unpaired e–., Mn++ (3d 5) =, , 2, , d sp3, , (inner octahedral, complex & diamagnetic), , [Cr(NH3 )6]3+ , Cr 3+ (24 –3 = 21), 2, , d sp3, , (inner octahedral, complex & paramagnetic), , [Ni(NH3 )6]2+ , Ni2+ (28 – 2 = 26), sp3d, , 2, , (outer octahedral, complex & paramagnetic), , [Zn(NH3)6]2+ , Zn 2+ (30 – 2 = 28), sp3d 2, , (outer octahedral, complex &diamagnetic), , 181. (d) d 5 –––– strong ligand field, , t2g, , eg, , = n(n 2) = 3 = 1.73BM, d 3–– in weak as well as in strong field, , t2g, , eg, , = 5 unpaired e–., Fe++ (3d 6) =, = 4 unpaired e–., Ni++ (3d 8) =, = 2 unpaired e–., As Ni++ has minimum no. of unpaired e– thus this is, least paramagnetic., 183. (d) In case of diamagnetic complexes the electrons are, paired. In case of Ni (CO)4 3d orbital is fully filled give, rise to sp3 hybridisation while in case of [Ni(CN)4]2–, nickel is in +2 oxidation state, so one d-orbital is vacant, give rise to dsp2 hybridization which is square planar, in nature., 184. (d) Number of unpaired electrons in [MnCl6]3– and, [Mn(CN)6]3– respectively are 4 and 2, Magnetic moment will respectively be 4.8 and 2.8, [MnCl6]3– is sp3d2 hybridised and [Mn(CN)6]3– will, be d2sp3 hybridised., 185. (c) In octahedral field the crystal field splitting of d- orbitals, of a metal ion depends upon the field produced by the, ligands. In general ligands can be arranged in a series, in the order of increasing fields and splittings which, they produce around a central metal ion. A portion of, the series is given below., cyanide > ethylene - diamine > ammonia > pyridine >, thiocyanate > water > oxalate > hydroxide > fluoride >, chloride > bromide > iodide.
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EBD_7207, COORDINATION COMPOUNDS, , 404, , Out of the given ligands water, ammonia, cyanide and, oxalate, we can find from the above series of ligands that, the maximum splitting will occur in case of cyanide (CN–), i.e. the magnitude of 0 will be maximum in case of, [Co(CN)6]3+., 186. (b) Since Cr3+ in the complex has unpaired electrons in, the d orbital, hence it will absorb visible light and will, be coloured, Ti = [Ar]3d 2 4 s2 ; Ti4 + = 3d 0, Cr = [Ar] 3d 5 4s1; Cr3+ = 3d 3, Zn= [Ar] 3d 10 4s2; Zn2+= 3d 10, Sc = [Ar] 3d 1 4s2; Sc3+ = 3d 0, 187. (d) d 4 in high spin octahedral complex, , eg, , Where, x, y, , 6, , 188. (b) d – t2g, , 2, 2, 2, , eg, , 0,0, , C.F.S.E = – 0.4 × 6, =, 189. (a), , t 2g, 0, , electrons in eg orbital, , CFSE = [0.6 × 1] + [–0.4 × 3] = – 0.6, , —, , CFSE = (–0.4x + 0.6y), , electrons in t2g orbital, , 190. (b), , 12, 5, , 0, , (in low spin), 0 + 3P, , 0 + 3P, , As positive charge on the central metal atom increases,, the less readily the metal can donate electron density, into the * orbitals of CO ligand (donation of electron, density into * orbitals of CO result in weakening of, C – O bond). Hence, the C – O bond would be, strongest in [Mn(CO)6]+., 191. (b)
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EBD_7207, HALOALKANES AND HALOARENES, , 406, , 17., , 18., , 19., , 20., , 21., , 22., , 23., , 24., , 25., , 26., , Halogenation of alkanes is, (a) a reductive process, (b) an oxidative process, (c) an isothermal process (d) an endothermal process, Ethylene dichloride can be prepared by adding HCl to, (a) Ethane, (b) Ethylene, (c) Acetylene, (d) Ethylene glycol, In which of the following conversions, phosphorus, pentachloride is used as the reagent?, (a) H 2 C CH 2, CH 3CH 2 Cl, (b), , CH 3CH 2 OH, , (c), , H 3 C O CH 3, , CH 3 CH 2 Cl, , 27., , 28., , 29., , CH 3Cl, , CH2 = CHCl, (d) CH CH, The best method for the conversion of an alcohol into an, alkyl chloride is by treating the alcohol with, (a) PCl5, (b) dry HCl in the presence of anhydrous ZnCl2, (c) SOCl2 in presence of pyridine, (d) None of these, Which of the following is liquid at room temperature (b.p. is, shown against it) ?, (a) CH3I, 42ºC, (b) CH3Br, 3ºC, (c) C2H5Cl 12ºC, (d) CH3F, –78ºC, The catalyst used in the preparation of an alkyl chloride by, the action of dry HCl on an alcohol is, (a) anhydrous AlCl3, (b) FeCl3, (c) anhydrous ZnCl2, (d) Cu, Chlorobenzene is prepared commercially by, (a) Raschig process, (b) Wurtz Fittig reaction, (c) Friedel-Craft’s reaction, (d) Grignard reaction, In the preparation of chlorobenzene from aniline, the most, suitable reagent is, (a) Chlorine in the presence of ultraviolet light, (b) Chlorine in the presence of AlCl3, (c) Nitrous acid followed by heating with Cu2Cl2, (d) HCl and Cu2Cl2, Which of the following possesses highest melting point?, (a) Chlorobenzene, (b) m-dichlorobenzene, (c) o-dichlorobenzene, (d) p-dichlorobenzene, Conant Finkelstein reaction for the preparation of alkyl, iodide is based upon the fact that, (a) Sodium iodide is soluble in methanol, while sodium, chloride is insoluble in methanol, (b) Sodium iodide is soluble in methanol, while NaCl and, NaBr are insoluble in methanol, (c) Sodium iodide is insoluble in methanol, while NaCl, and NaBr are soluble, (d) The three halogens differ considerably in their, electronegativity, , 30., , 31., , 32., , 33., , 34., , 35., , 36., , CS, , 2, . The main product of this, Silver acetate Br2, reaction is, (a) CH3 Br, (b) CH 3COI, (c) CH 3COOH, (d) None of these, Which of the following reactions is an example of, nucleophilic substitution reaction?, (a) 2 RX + 2 Na R – R + 2 NaX, (b) RX + H2 RH + HX, (c) RX + Mg RMgX, (d) RX + KOH ROH + KX, Which one is most reactive towards SN 1 reaction ?, (a) C6 H5 CH(C6 H 5 )Br, (b) C6 H5 CH(CH3 )Br, (c) C6 H 5 C(CH 3 )(C6 H 5 )Br, , (d) C6 H 5 CH 2 Br, A Grignard reagent may be made by reacting magnesium, with, (a) Methyl amine, (b) Diethyl ether, (c) Ethyl iodide, (d) Ethyl alcohol, Which one of the following halogen compounds is difficult, to be hydrolysed by SN1 mechanism?, (a) Tertiary butyl chloride (b) Isopropyl chloride, (c) Benzyl chloride, (d) Chlorobenzene, The order of reactivity of the given haloalkanes towards, nucleophile is :, (a) RI > RBr > KCl, (b) RCl > RBr > RI, (c) RBr > RCl > RI, (d) RBr > RI > RCl, Most reactive halide towards SN1 reaction is, (a) n-Butyl chloride, (b) sec-Butyl chloride, (c) tert-Butyl chloride, (d) Allyl chloride, In SN1 reaction, the recemization takes place. It is due to, (a) inversion of configuration, (b) retention of configuration, (c) conversion of configuration, (d) Both (a) and (b), The order of reactivities of the following alkyl halides for a, SN2 reaction is, (a) RF > RCl > RBr > RI (b) RF > RBr > RCl > RI, (c) RCl > RBr > RF > RI (d) RI > RBr > RCl > RF, Which of the following is an example of SN2 reaction?, (a) CH 3 Br OH, CH 3OH Br, (b) CH 3 C H CH 3 OH, CH 3 C H CH 3, |, , |, , Br, , (c), , 37., , 38., , CH 3CH 2 OH, , OH, H 2O, , CH 2 CH 2, (d) (CH3 )3 C Br OH, (CH3 )3COH Br, SN2 mechanism proceeds through intervention of, (a) carbonium ion, (b) transition state, (c) free radical, (d) carbanion, Which among MeX, RCH2X, R2CHX and R3CX is most, reactive towards SN2 reaction?, (a) MeX, (b) RCH2X, (c) R2CHX, (d) R3CX
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HALOALKANES AND HALOARENES, , 78., , 79., , 80., , 81., , 82., , 83., , 84., , 409, , CFxCly [where x + y = 4]. These compounds are not used, because, (a) these are fluorocarbons, (b) these are difficult to synthesise, (c) they deplete ozone layer, (d) None of the these, Freon (dichlorodifluoro methane) is used, (a) as local anaesthetic, (b) for dissolving impurities in metallurgical process, (c) in refrigerator, (d) in printing industry, Use of chlorofluorocarbons is not encouraged because, (a) They are harmful to the eyes of people that use it, (b) They damage the refrigerators and air conditioners, (c) They eat away the ozone in the atmosphere, (d) They destroy the oxygen layer, Which of the following is used in fire extinguishers, (a) CH4, (b) CHCl3, (c) CH2Cl2, (d) CCl4, Solvent which is used in the synthesis of chlorofluorocarbons, (a) iodoform, (b) chloroform, (c) carbon tetrachloride (d) methylene chloride, Uses of dichloromethane is, (a) paint remover, (b) solvent in drugs manufacturing, (c) metal cleansing and finishing solvent, (d) All of the above, Trichloroacetaldehyde, CCl3CHO reacts with chlorobenzene, in presence of sulphuric acid and produces:, Cl, , (a), , Cl, , C, , 86. Which of the following is responsible for depletion of the, ozone layer in the upper strata of the atmosphere?, (a) Polyhalogens, (b) Ferrocene, (c) Fullerenes, (d) Freons, 87. Haloforms are trihalogen derivatives of, (a) Ethane, (b) Methane, (c) Propane, (d) Benzene, 88. Which of the following compounds is used as a refrigerant ?, (a) Acetone, (b) CCl4, (c) CF4, (d) CCl2F2, 89. Chloroform is used as :, (a) Fire extinguisher, (b) Industrial solvent, (c) Refrigerant, (d) Insecticide, 90. AgNO3 does not give precipitate with CHCl3 because, (a) CHCl3 does not ionise in water, (b) CHCl3 does not react with AgNO3, (c) CHCl3 is chemically inert, (d) None of these, 91. When chloroform is exposed to light and air, it forms, (a) chlorine gas, (b) methyl chloride, (c) phosgene gas, (d) carbon tetrachloride, 92., , C–CCl3, Cl, , 93., , Cl, , 94., , H, , OH, , (b) Cl, , C, , Cl, , Cl, , (c), , Cl, , CH, , Cl, , 95., , 96., , CCl3, , Cl, , (d) Cl, , C, , 97., Cl, , CH2Cl, , 85., , Chloroform on treatment with conc. HNO3 gives, (a) Chloropicrin, (b) Nitromethane, (c) Picric acid, (d) Acetylene, , Cl, , 98., , H, , The above structural formula refers to, (a) BHC, (b) DNA, (c) DDT, (d) RNA, If chloroform is left open in air in the presence of sunlight, it, gives, (a) carbon tetrachloride (b) carbonyl chloride, (c) mustard gas, (d) lewisite, Full name of DDT is, (a) 1, 1, 1-trichloro-2, 2-bis(p-chlorophenyl) ethane, (b) 1, 1-dichloro-2, 2-diphenyl trimethylethane, (c) 1, 1-dichloro-2, 2-diphenyl trichloroethane, (d) None of these, Freon(s) is/are :, (a) CClF3, (b) CFCl3, (c) CCl2F2, (d) All of these, Freon-12 is commonly used as, (a) insecticide, (b) refrigerant, (c) a solvent, (d) a fire extinguisher, Freon used as refrigerant is, (a) CF2 == CF2, (b) CH2F2, (c) CCl2F2, (d) CF4, Methylene chloride can be used as, (a) paint remover, (b) propellant in aerosols, (c) solvent in manufacturing of drugs, (d) All of these
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EBD_7207, HALOALKANES AND HALOARENES, , 410, , 99., , Which of the following are the harmful effects of methylene, chloride?, (a) Impaired hearing and vision, (b) Dizziness, nausea and tingling, (c) Skin burning, (d) All of these, , STATEMENT TYPE QUESTIONS, 100. Read the following statements and choose the correct, option., (i) The general formula of alkyl halides is CnH2n+1 X, (ii) The general formula of aryl halides is Cn Hn–1 X, (iii) In alkyl halides halogen atom(s) is attached to sp2, hybridised carbon atom, (iv) In aryl halides halogen atom(s) is attached to sp2, hybridised carbon atom., (a) (i), (ii) and (iii) are correct, (b) (i), (ii) and (iv) are correct, (c) (ii), (iii) and (iv) are correct, (d) (i), (ii), (iii) and (iv) are correct, 101. Following statements are given regarding the preparation, of aryl halides from toluene. Read the following statements, and choose the correct option., (i) Aryl chlorides and bromides can be easily prepared, by this method., (ii) The ortho and para isomers formed in the reaction can, not be separated easily due to small difference in their, melting point., (iii) Reactions with iodine are reversible in nature and, require the presence of an oxidising agent., (iv) Fluoro compounds are not prepared by this method, due to low reactivity of fluorine., (a) (i) and (iii) are correct, (b) (ii) and (iv) are correct, (c) (i), (ii), and (iii) are correct, (d) All statements are correct, 102. Read the following statements and choose the correct, option., (i) For the same alkyl group, the boiling points of alkyl, halides decreases in the order., RI > RBr > RCl > RF, (ii) With the increases in size and mass of halogen atom,, the magnitude of van der Waal’s forces increases., (a) Both statements (i) and (ii) are correct, (b) Statement (i) is correct and (ii) is incorrect, (c) Statement (ii) is correct and (i) is incorrect, (d) Both statement (i) and (ii) are incorrect, 103. Read the following statements and choose the correct, answer, (i) The boiling points of isomeric haloalkanes decrease, with increase in branching., , (ii) Among isomeric dihalobenzenes the para-isomers, have higher melting point than their ortho and metaisomers., (iii) The isomeric dihalobenzene have large difference in, their boiling and melting points, (iv) The isomeric dihalobenzene have nearly same boiling, point., (a) (i), (ii) and (iii) are correct, (b) (i) and (iii) are correct, (c) (ii) and (iv) are correct, (d) (i), (ii) and (iv) are correct, 104. Read the following statements and choose the correct code, (i) SN2 reactions follows a second order kinetics whereas, SN1 reactions follows the first order kinetics, (ii) SN1 reactions follows the second order kinetics, whereas SN2 follows the first order kinetics, (iii) SN2 reactions take place in a single step whereas, SN1 reactions take place in a two steps, (iv) Tertiary alkyl halides are least reactive towards SN2, reactions but we show high reactivity towards SN1, reaction., (a) (ii) and (iv) are correct, (b) (i) and (iii) are correct, (c) (i), (ii) and are correct, (d) (ii), (iii) and (iv) are correct, 105. Read the following statements and choose the correct, option., (i) SN1 reactions are carried out through formation of, carbocation as an intermediate., (ii) SN1 reactions are two step reactions in which step 1 is, fast and irreversible., (iii) Step 1 involves breaking of C–Br bond which obtain, energy through solvation of halide., (iv) SN1 reactions are two step reactions in which step 2 is, slow and reversible., (v) Allylic and benzylic halides show high reactivity, toward SN1 reactions., (a) (i), (iii) and (v) are correct, (b) (ii), (iii) and (v) are correct, (c) (i), (iii) and (iv) are correct, (d) (i), (ii) and (iv) are correct, 106. Read the following statements and choose the correct code, (i) SN2 reaction proceed with complete stereochemical, inversion., (ii) SN1 reaction proceed with recimisation., (iii) A dextrorotatory compound rotate the plane polarised, light to the left., (iv) A laevorotatory compound rotate the plane polarised, light to the right, (a) TFTT, (b) TTFF, (c) FFTT, (d) TFTF
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HALOALKANES AND HALOARENES, , C2H5, , C2H5, H, , Y, , 107. H, CH3, B, , Y, , 411, , X, , C2H5, H, , Y, Y, , CH3, , CH3, A, , Y, , For the reaction scheme given above some statements are, given. Read the statements and choose the correct option., If (A) is the only compound obtained, the process is, called retention of configuration., , (ii) If (B) is the only compound obtained, the process is, called inversion of configuration., (iii) If a 50 : 50 mixture of the above two is obtained then, the process is called racemisation, (iv) The product A + B is optically active, (a) TTTF, , (b) TFTF, , (c) TTFF, , (d) TFFT, , 108. Which of the statement(s) is/are true, regarding following, reaction?, R, R', R'', (i), , CBr, , Nu –, , R, R', R'', , –, , CNu + Br, , The reaction involves the formation of transition state., , (ii) Higher the nucleophilic character of the nucleophile,, faster will be the reaction., (iii) The product is always optically inactive., (a) (ii) only, , (b) (ii) and (iii), , (c) (i), (ii) and (iii), , (d) Neither (i), (ii) nor (iii), , MATCHING TYPE QUESTIONS, 109. Match the columns, Column - I, , (A) C2H6, , Column - II, , Cl2 / UV light, , (B) C6H5NH2, , C2H5Cl, , (p) Finkelstein reaction, (q) Free radical substitution, , NaNO 2 HCl/Cu 2 Cl 2, 273 278K, , A+B, , (i), , 110. Match the columns, Column - I, , Column - II, , (A) CH2 = CH – CH2Cl, , (p) Gem-dichloride, , (B) CH2 = CHX, , (q) Vinylic halide, , (C) CH3CHCl2, , (r) Dichloride, , (D) CH2Cl CH2Cl, , (s) Allylic halide, , (a) A – (r), B – (q), C – (p), D – (s), (b) A – (q), B – (p), C – (s), D – (r), (c) A – (s), B – (q), C – (p), D – (r), (d) A – (r), B – (p), C – (s), D – (q), , C6H5Cl, (C) CH3Cl + NaI, , (r) Swarts reaction, , CH3I + NaCl, (D) CH3 – Br + AgF, , (s) Sandmeyer’s reaction, , CH3F + AgBr, (a) A – (q), B – (s), C – (p), D – (r), (b) A – (q), B – (r), C – (p), D – (s), (c) A – (r), B – (p), C – (s), D – (q), (d) A – (s), B – (r), C – (p), D – (q), 111. Match the columns, Column - I, Column - II, (A) Chloroform, (p) Antiseptic, (B) Iodoform, (q) Insecticide, (C) Trichloromethane, (r) Anesthetic, (D) DDT, (s) Propellant, (a) A – (s), B – (p), C – (r), D – (q), (b) A – (r), B – (p), C – (s), D – (q), (c) A – (q), B – (s), C – (p), D – (q), (d) A – (r), B – (s), C – (p), D – (q), 112. Match the columns., Column - I, Column - II, (Haloalkane/arene), (Applications), (A) Iodoform, (p) CF4, (B) BHC, (q) Antiseptic, (C) Freon - 14, (r) Moth repellant, (D) Halothanes, (s) Inhalative anesthetic, (E) p-dichlorobenzene, (t) Termite pesticide, (a) A – (q), B – (s), C – (t), D – (r), E – (p), (b) A – (q), B – (t), C – (p), D – (s), E – (r), (c) A – (r), B – (s), C – (q), D – (p), E – (t), (d) A – (p), B – (r), C – (t), D – (q), E – (s), 113. Match the columns, Column-I, Column-II, (A) Chloramphenicol, (p) Goiter, (B) Thyroxine, (q) Surgery, (C) Chloroquine, (r) Typhoid, (D) Halothane, (s) Malaria, (a) A – (q), B – (p), C – (s), D – (r), (b) A – (r), B – (p), C – (s), D – (q), (c) A – (s), B – (r), C – (q), D – (p), (d) A – (p), B – (s), C – (q), D – (r)
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EBD_7207, HALOALKANES AND HALOARENES, , 412, , ASSERTION-REASON TYPE QUESTIONS, Directions : Each of these questions contain two statements,, Assertion and Reason. Each of these questions also has four, alternative choices, only one of which is the correct answer. You, have to select one of the codes (a), (b), (c) and (d) given below., (a) Assertion is correct, reason is correct; reason is a correct, explanation for assertion., (b) Assertion is correct, reason is correct; reason is not a, correct explanation for assertion, (c) Assertion is correct, reason is incorrect, (d) Assertion is incorrect, reason is correct., 114. Assertion : SN2 reaction of an optically active aryl halide, with an aqueous solution of KOH always gives an alcohol, with opposite sign of rotation., Reason : SN2 reactions always proceed with inversion of, configuration., 115. Assertion : Alkylbenzene is not prepared by Friedel-Crafts, alkylation of benzene., Reason : Alkyl halides are less reactive than acyl halides., 116. Assertion : Exposure of ultraviolet rays to human causes, the skin cancer, disorder and disrupt the immune system., Reason : Carbon tetrachloride is released into air it rises to, atmosphere and deplets the ozone layer., 117. Assertion : CHCl3 is stored in dark bottles., Reason : CHCl3 is oxidised in dark., 118. Assertion : CCl4 is not a fire extinguisher., Reason : CCl4 is insoluble in water., , CRITICAL THINKING TYPE QUESTIONS, , 122. The catalyst used in the preparation of an alkyl chloride by, the action of dry HCl on an alcohol is, (a) Anhydrous AlCl3, (b) FeCl3, (c) Anhydrous ZnCl2, (d) Cu, 123. CH 3 CH 2, , 124., , 125., , 126., , 127., , 128., , 119. The IUPAC name of the compound shown below is, , Cl, , 129., , Br, (a) 2-bromo-6-chlorocyclohex-1-ene, (b) 6-bromo-2-chlorocyclohexene, (c) 3-bromo-1-chlorocyclohexene, (d) 1-bromo-3-chlorocyclohexene, 120. A compound is formed by substitution of two chlorine for, two hydrogens in propane. The number of possible isomeric, compounds is, (a) 4, (b) 3, (c) 5, (d) 2, 121. Which one of the following is not an allylic halide?, (a) 4-Bromopent-2-ene, (b) 3-Bromo-2-methylbut-1-ene, (c) 1-Bromobut-2-ene, (d) 4-Bromobut-1-ene, , 130., , 131., , C H CH 3 obtained by chlorination of, |, , Cl, n-butane, will be, (a) l-form, (b) d-form, (c) Meso form, (d) Racemic mixture, The number of possible enantiomeric pairs that can be, produced during monochlorination of 2-methylbutane is, (a) 2, (b) 3, (c) 4, (d) 1, The number of structural and configurational isomers of a, bromo compound, C5H9Br, formed by the addition of HBr, to 2-pentyne respectively are, (a) 1 and 2, (b) 2 and 4, (c) 4 and 2, (d) 2 and 1, Which of the following reagent produces pure alkyl halides, when heated with alcohols?, (a) PCl5, (b) PCl3, (c) SOCl2, (d) dry HCl, If C5H12 undergoes reaction with chlorine in the presence, of sunlight, only one product is formed, than reactant is, (a) 3, 3-dimethylpropane (b) 2, 3-dimethylpropane, (c) 1, 3-dimethylpropane (d) 2, 2,-dimethylpropane, Hydrocarbon (CH3)3CH undergoes reaction with Br2 and, Cl2 in the presence of sunlight, if the reaction with Cl is, highly reactive and that with Br is highly selective so, no.of possible products respectively is (are), (a) 2, 2, (b) 2, 1, (c) 1, 2, (d) 1, 1, Possible major product formed in the reaction of, neopentylalcohol with HCl is, (a) 2 -chloro-2-methylbutane, (b) 2, 2 -dimethyl 1-chloropropane, (c) 2 -chloro -3-methylbutane, (d) 3, chloro -3-methylbutane, Fluorobenzene (C6H5F) can be synthesized in the laboratory, (a) by direct fluorination of benzene with F2 gas, (b) by reacting bromobenzene with NaF solution, (c) by heating phenol with HF and KF, (d) from aniline by diazotisation followed by heating the, diazonium salt with HBF4, Which chloride is least reactive with the hydrolysis point, of view?, (a) CH3Cl, (b) CH3CH2Cl, (c) (CH3)3CCl, (d) CH2 = CH – Cl
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HALOALKANES AND HALOARENES, , 413, , 132. In a SN2 substitution reaction of the type, R, , Br Cl, , DMF, , R, , Cl Br, , which one of the following has the highest relative rate ?, (a) CH3 – CH2 – CH2Br, , (c), , CH, | 3, CH3 C CH 2 Br, |, CH3, , (b), , CH3, , CH CH 2 Br, |, CH3, , (d) CH3CH2Br, , 140., , 133. Which will undergo SN2 reaction fastest among the following, halogen compounds?, (a) CH3CH2F, (b) CH3CH2Cl, (c) CH3CH2Br, (d) CH3CH2I, 134. Consider the following bromides :, Me, , Me, , Br, (A), , 139., , 141., , Me, , Me, , Br, (B), , Br, , The correct order of SN1 reactivity is, (a) B > C > A, (b) B > A > C, (c) C > B > A, (d) A > B > C, , 142., , 135. Which one is most reactive towards SN 1 reaction ?, (a), , C 6 H 5 CH(C 6 H 5 )Br, , (b) C6 H5CH(CH3 )Br, (c) C6 H5C(CH3 )(C6 H5 )Br, (d) C6 H5CH 2 Br, 136. Consider the reactions :, (i), , (CH 3 )2 CH CH 2Br, , C 2 H5OH, , 143., , (CH 3 )2 CH CH 2OC 2H 5, , (ii), , (CH 3 ) 2 CH CH 2 Br, , HBr, , C 2H 5O, , (CH 3 )2 CH, , CH 2OC 2 H 5, , Br, , The mechanisms of reactions (i) and (ii) are respectively :, (a) SN1 and SN2, (b) SN1 and SN1, (c) SN2 and SN2, (d) SN2 and SN1, 137. The organic chloro compound, which shows complete, stereochemical inversion during a SN2 reaction, is, (a) (C2H5)2CHCl, (b) (CH3)3CCl, (c) (CH3)2 CHCl, (d) CH3Cl, 138. Under certain conditions an alkyl halide reacts with base to, give an alkene and HCl [Elimination Reaction] for example, R – CH2 – CH2 – Cl R – CH = CH2 + HCl, The extent of these reactions depends on the structure of, alkyl halides (e.g. primary, secondary or tertiary). The relative, extent to which such reactions take place is in the order, (of haloalkanes) :, , 144., , (a) Primary < secondary < Tertiary, (b) Primary > Secondary > Tertiary, (c) Primary > Secondary < Tertiary, (d) Primary < Secondary > Tertiary, The correct order of reactivity of the halides, ethyl, chloride (I) iso-propyl chloride (II) and benzyl chloride, (III) in SN1 reaction is, (a) I > II > III, (b) III > II > I, (c) II > I > III, (d) I > III > II, Which of the following represents correct set of ambident, nucleophiles?, (a) CN– and NH3, (b) CN–, NO2, (c) OH–, RO–, (d) CN–, OH–, Which of the following statements is correct?, (a) S N 2 reactions of optically active halides are, accompanied by inversion of configuration., (b) S N 1 reactions of optically active halides are, accompanied by racemisation., (c) Carbocation formed in SN1 reaction is sp2 hybridized., (d) All of the above., The replacement of chlorine of chlorobenzene to give phenol, requires drastic conditions, but the chlorine of 2,, 4-dinitrochlorobenzene is readily replaced since,, (a) nitro groups make the aromatic ring electron rich at, ortho/para positions, (b) nitro groups withdraw electrons from the meta position, of the aromatic ring, (c) nitro groups donate electrons at meta position, (d) nitro groups withdraw electrons from ortho/para, positions of the aromatic ring, A set of compounds in which the reactivity of halogen atom, in the ascending order is, (a) chlorobenzene, vinyl chloride, chloroethane, (b) chloroethane, chlorobenzene, vinyl chloride, (c) vinyl chloride, chlorobenzene, chloroethane, (d) vinyl chloride, chloroethane, chlorobenzene, Aryl halides are extremely less reactive towards nucleophilic, substitution than alkylhalides. Which of the following, accounts for this ?, (i) Due to resonance in aryl halides., (ii) In alkyl halides carbon atom in C–X bond is sp2, hybridised whereas in aryl halides carbon atom in C–X, bond is sp3 hybridized., (iii) Due to stability of phenyl cation., (iv) Due to possible repulsion there are less chances of, nucleophile to approach electron rich arenes., (a) (i), (ii) and (iv), (b) (i), (ii) and (iii), (c) (i) and (iv), (d) (ii), (iii) and (iv)
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EBD_7207, HALOALKANES AND HALOARENES, , 414, , 145. The organic compound used as feedstock in the synthesis, of chlorofluorocarbons is, , 149. Exposure of CCl4 causes, (a) Liver cancer in human, , (a) CH2Cl2, , (b) CHCl3, , (b) Damage to nerve cells, , (c) CH3Cl, , (d) CCl4, , (c) Coma, unconsciousness, , 146. CCl4 is well known fire extinguisher. However after using it, to extinguish fire, the room should be well ventilated. This, is because, (a) It is flammable at higher temperatures, (b) It is toxic, (c) It produces phosgene by reaction with water vapour, at higher temperatures, (d) It is corrosive, 147. In which part of the atmosphere, does the freon remain, unchanged ?, , (d) All of these, 150. Chloroform cannot be prepared from which of the following?, (a) CH3OH, , (b) C2H5OH, , (c) CH3CHO, , (d) (CH3)2CO, , 151. Which one of the following has antiseptic property?, (a) Dichloromethane, , (b) Trifluoromethane, , (c) Triiodomethane, , (d) Tetrachloromethane, , 152. Chronic chloroform exposure may cause damage to liver, and kidney, due to the formation of, , (a) Stratosphere, , (b) Troposphere, , (a) phosgene, , (b) methylene chloride, , (c) Mesosphere, , (d) Thermosphere, , (c) methyl chloride, , (d) carbon tetrachloride, , 148. Natural ozone layer is unbalanced due to, (a) cloudiness of poisonous gases, (b) presence of rain in the atmosphere, , 153. The spatial arrangement of four groups around a central, carbon atom is tetrahedral and if all the substituents, attached to that a carbon are different. Such a carbon is, called ______., , (c) initiation of radical chain by freon, , (a) asymmetric carbon, , (b) stereocentre, , (d) All of the above, , (c) chiral, , (d) All of these
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HALOALKANES AND HALOARENES, , 415, , FACT / DEFINITION TYPE QUESTIONS, 1., , (d) Neohexyl chloride is a primary halide as in it Cl-atom is, attached to a primary carbon., , CH3, C, , CH3, , CH2, , CH2Cl, , CH3, 2., , 3., 4., , (b), , CH 2 Cl, |, CH 2 Cl, , (gem-dihalide), , (vic-dihalide), , (b) Gem-dihalides are those in which two halogen atoms, are attached on the same carbon atom., (a), , 5., , (b), , 1, , CH3, , CH3, |, CH, 3, , 1, , CH3, |, 2, 1, C CH 2 CH 3, |, 4, Cl, , 3-chloro-2-3-dimethylpentane, 1, , 6., , (b), , CH 3, |, 2, H 3C C Cl, |, 3 CH, 3, , IUPAC name : 2-Chloro-2-methylpropane., , 7., , (a), , Br, |, CH3 CH 2 C CH Cl, 4, , 3, , 2, , 2 - bromo - 1 - chloro but - 1 - ene, , 8., , (d), , CH 2, , 10., 12., 13., , Cl2, , UV light, , C2 H5Cl+HCl, , 19. (b) When ethyl alcohol is treated with PCl 5 , then ethyl, chloride is formed., CH 3 CH 2 OH + PCl 5, 20., , CH 3 CH 2 Cl + HCl + POCl 3, (c) The best method for the conversion of an alcohol into, an alkyl chloride is reaction of the alcohol with thionyl, chloride (SOCl2) in the presence of pyridine., , R – OH + SOCl2, , CH CH 2Cl, 2, , (a) Isopropyl chloride CH3 CH CH3 chlorine atom is, |, Cl, attached to 2° carbon atom., (a), 11. (d), (c) Because of the small size of F, the C–F bond is strongest, in CH3F., (b) CH3Cl has higher dipole moment than CH3F due to, much longer C–Cl bond length than the C–F bond., The much longer bond length of the C–C bond, outweighs the effect produced by lower, electronegativity of Cl than that of F., , Pyridine, , RCl + SO2 + HCl, , SO2 and HCl being gases escape leaving behind pure, alkyl halide., 21. (a) Boiling point of CH3I is 42°C which indicates that it is, liquid at room temperature. CH3I is larger molecule so, it has stronger vander Waal’s force of attraction than, others., 22. (c) In preparation of an alkyl chloride by the action of dry, HCl, the catalyst generally used is anhydrous ZnCl2., 23. (a), 24. (c), +, , N, , NH2, , (3 chloro 1 propene), , 9., , 16. (a) C2H6, , 17. (b), 18. (d) Ethylene dichloride can be prepared by adding HCl to, ethylene glycol (CH2OH. CH2OH)., , CHCl 2, |, CH3, , 1, , 14. (c) For the same alkyl group, the boiling points of alkyl, halides decrease in the order :, RI RBr RCl RF, This is because with the increase in size and mass of, halogen atom, the magnitude of van der Waal's forces, increases., 15. (a), , NCl, , –, , Cl, , HNO2, , Cu2Cl2, , HCl, (Diazotization), , Sandmeyer’s reaction, , 25. (d) Para-di-chlorobenzene has most symmetrical structure, than others. It is found as crystalline lattice form,, therefore, it has highest melting point (52°C) due to, symmetrical structure., Cl, , Cl
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EBD_7207, HALOALKANES AND HALOARENES, , 416, , 39., , Para-chlorobenzene, , 26., , (b), , R X, , acetone, , NaI, , R I, , NaX 40., , Soluble in, (CH 3OH, Me 2CO), , R, , X, , NaI, , R, , I, , NaX, Insoluble in, (CH 3OH, Me 2CO), , 41., , (where X = Cl or Br), CS2, , 27., , (a), , 28., , (d) In nucleophilic substitution, a nucleophile provides, an electron pair to the substrate and the leaving group, departs with an electron pair., , CH3COOAg Br2, , CH3Br AgBr CO 2, , 42., , H, , 43., , 29., , These are usually written as S N (S stands for, substitution and N for nucleophilic) and are common, in aliphatic compounds especially in alkyl halides and, acyl halides., (c) SN1 reactions involve the formation of carbocations,, hence higher the stability of carbocation, more will be, reactivity of the parent alkyl halide. Thus tertiary, carbocation formed from (c) is stabilized by two phenyl, groups and one methyl group, hence most stable., , 30., , (c), , 31., , 32., , CH3 CH 2 I Mg, , Dry, Ether, , 44., , CH 3 CH 2 MgI, , Ethyl magnesium, iodide, , (d) Chlorobenzene does not undergo hydrolysis by SN1, mechanism because in this halogen is present on sp2, hybridised carbon atom, such halogens are relatively, inert., (a) For a given alkyl group, the order of reactivity is, , 45., 46., , 33., , (c), , 34., , (d), , 35., 36., 37., 38., , (d), (a), (b), (a) 1° Alkyl halides (having least steric hindrance at, -carbon atom) are most reactive towards SN2 reaction., , (c), , H3C CH 2, , |, , C* CH3, |, Cl, , The compound containing a chiral carbon atom i.e., (a, carbon atom which is attached to four different atoms, is known as a chiral carbon atom) is optically active., A, s, 2-chlorobutane contains a chiral C* atom hence it is, optically active., (c) Diastereomers since they have different melting points,, boiling points, solubilities etc., (a), (b) Compound which are mirror image of each other and, are non superimposable are termed as enantiomers., CH3, , CH3, , OH, , H, , and, H, , HO, , R – I > R – Br > R – Cl > R – F, increasing bond energy, , decreasing halogen reactivity., This order depends on the carbon-halogen bond, energy; the carbon-fluorine bond energy is maximum, and thus fluorides are least reactive while carboniodine bond energy is minimum hence iodides are most, reactive., More stable the carbocation, more reactive will be the, parent alkyl halide towards SN1 reaction., 3° > Benzyl > Allyl > 2° > 1° > methyl, SN1 reaction involves carbocation which are planar, (sp2 hybridised) and thus can be attacked on either, face of the carbon., Weaker the C–X bond, greater is the reactivity., Only 1° alkyl halides (i.e. CH3Br) undergo SN2 reaction., , (c) Isopropyl chloride, being 2° alkyl halides, can undergo, SN1 as well as SN2 mechanism., (b) Due to steric hindrance tertiary alkyl halide do not, react, by S N2 mechanism they react by S N 1, mechanism. SN2 mechanisam is followed in case of, primary and secondary alkyl halides of, CH3 – X CH3 – CH2X (CH3)2 – CH.X (CH3)3– C–X, (c) In SN2 mechanism transition state is pentavalent. Thus, bulky alkyl group will be sterically hindered and smaller, alkyl group will favour the SN2 mechanism. So the, decreasing order of reactivity of alkyl halides is, RCH2X > R2CHX > R3CX, (c) Inversion in configuration occurs in SN2 reactions., , HO, , CH3, , 47., , H, OH, , H, CH3, , These are enantiomers, (b) The compound has two similar assymmetric, C-atoms. It has plane of symmetry and exist in meso, form., , Plane of symmetry, , Meso - 2, 3 dichlorobutane, , 48., , (a) Compounds having chiral carbon atom are optically, active., , CH3, , 49., , H, |, C* COOH (* is chiral carbon atom), |, Cl, , (a) A mixture of equal amounts of the two enantiomers is, called a racemic mixture.
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EBD_7207, HALOALKANES AND HALOARENES, , 418, , 69., 70., 71., , 72., , (c) Due to resonance in chlorobenzene., (a) N-Phenylacetanilide, C6H5N(C6H5)COCH3,, precipitates out to a complex with anhydrous AlCl3., (d) More the stability of the carbocation, higher will be, the reactivity of the parent chloride. Allyl chloride >, Vinyl chloride > Chlorobenzene, (c) SN1 reactions involve the formation of carbocations,, hence higher the stability of carbocation, more will be, reactivity of the parent alkyl halide. Thus tertiary, carbocation formed from (c) is stabilized by two phenyl, groups and one methyl group, hence most stable., HONO, HCl, , C6H5N2Cl, , CuCl, , (d) C6H5NH2, , 74., , (b) This method is not applicable for the preparation of, aryl halides because the C–O bond in phenol has a, partial double bond character and is difficult to break, being stronger than a single bond., (d) Due to resonance, the electron density increases more, at ortho- and para-positions than at meta-positions., Further, the halogen atom because of its – I effect has, some tendency to withdraw electrons from the benzene, ring. As a result, the ring gets somewhat deactivated, as compared to benzene and hence the electrophilic, substitution reactions in haloarenes occur slowly and, require more drastic conditions as compared to those, in benzene., (d) Chloropicrin is nitrochloroform. It is obtained by the, nitration of chloroform with HNO3., , 76., , HCCl 3, , Chloroform, , 77., 79., , 80., , 81., , (c), (c), , (c), , (d), , 82., , (c), , 83., , (d), , HNO 3, , O 2 NCCl 3, , Chloropicr in, , Chloropicrin is a liquids, poisonous and used as an, insecticide and a war gas, 78. (c), Freon (CCl2F2) is an odourless, non-corrosive, non, toxic gas which is stable even at high temperatures, and pressures. It has low b.p. low specific heat and, can be easily liquified by applying pressure at room, temperature. It is therefore, widely used as refrigerant, (cooling agent) in refrigerators and air conditioners., Chlorofluorocarbon is used in air-conditioners and in, domestic refrigerators for cooling purposes. Its main, drawback is this, it is responsible for ozone depletion., Its vapours are non-inflammable (i.e. do not catch fire)., Hence used as fire extinguishers under the name, pyrene., Tetrachloromethane (carbon tetrachloride) is also used, as feedstock in the synthesis of chlorofluorocarbon, and other pharmaceutical manufacturing and general, solvents etc., Dichloromethane is widely used as solvent as a paint, remover, as a propellant in aerosols and as a process, solvent in the manufacture of drugs. It is also used as, a metal cleaning and finishing solvent., , (c), , Cl 3C–C=O +, , Cl, , conc. H2SO4, , H, , Cl3C–CH, , Cl, , Cl, , Cl, , DDT, , 85., , (a), , 86., , (d) Chlorofluorocarbons, e.g. CF 2 Cl 2 , CHF 2 Cl 2 ,, HCF2CHCl2. These are non-inflammable colourless, and stable upto 550ºC. These are emitted as propellants, in aerosol spray, cans refrigerators, fire fighting reagents, etc. They are chemically inert and hence do not react, with any substance with which they come in contact, and therefore float through the atmosphere and as a, result enter the stratosphere. There they absorb UV-rays, and due to this they produce free atomic chlorine which, results decomposition of ozone which cause depletion, of ozone layer., (b) Haloform compounds with the formula CHX3, where, X is a halogen atom., Haloforms are trihalogen derivatives of methane., Example : Chloroform CHCl3., (d) Under ordinary conditions freon is a gas. Its boiling, point is –29.8°C. It can easily be liquified. It is chemically, intert. It is used in air-conditioning and in domestic, refrigeratiors for cooling purposes (as refrigerant), (b) Chloroform (CHCl3) is used as industrial solvent., (a) Chloroform is an organic compound which does not, ionise in water. Since it can not provide Cl–, therefore,, it is not precipitated with AgNO3., (c) When chloroform is exposed to light it is oxidised to a, poisonous gas known as phosgene., , C6H5Cl, , 73., , 75., , Cl, , 84., , 87., , 88., , 89., 90., , 91., , Cl3C, , H HO, , 2 CHCl 3, , 92., , (c), , NO 2, , 2O 2, , Cl3C NO 2, Chloropicrin (used, as an insecticide), , 2 COCl 2, , Cl 2, , 93. (b), Cl, , 94., , (a), , CCl3CH, Cl, (DDT), , 95., 96., 97., 98., , (d) Freons are chlorofluorocarbons., CClF3, CFCl3 and CCl2F2, all are freons., (b), (c) Freons are chlorofluorocarbon., (d), 99. (d), , STATEMENT TYPE QUESTIONS, 100. (b) In alkyl halides halogen atom(s) is attached to sp3, hybridised carbon atom.
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HALOALKANES AND HALOARENES, , 419, , 101. (a) Aryl chlorides and bromides can be easily prepared, by electrophilic substitution of arenes with chlorine, and bromine respectively in the presence of Lewis, acid catalysis like iron or iron (III) chloride., The ortho and para isomers can be easily separated, due to large difference in their melting points., Reactions with iodine are reversible in nature and, require the presence of an oxidising agent (HNO3,, HIO4) to oxidise the HI formed during iodination., Fluoro compounds are not prepared by this method, due to high reactivity of fluorine., 102. (a), 103. (d) The boiling points of isomeric haloalkanes decrease, with increase in branching. For example, 2-bromo-2methylpropane has the lowest boiling point among, the three isomers., CH3, |, CH3CH 2CH 2CH 2Br CH3CH 2CH CH3 H3C C CH 3, |, |, Br, Br, b.p. / K 375, 364, 346, Boiling points of isomeric dihalobenzenes are very, nearly the same. However, the para-isomers are high, melting as compared to their ortho and meta-isomers., It is due to symmetry of para-isomers that fits in crystal, lattice better as compared to ortho- and meta-isomers., , Cl, , Cl, , Cl, , Cl, Cl, Cl, b.p/K, 453, 446, 448, m.p.K, 256, 249, 323, 104. (b) SN2 reaction follow a 2nd order kinetic ie the rate, depends upon the concentration of both the reactants,, where in SN1 reactions rate depends only upon the, concentration of only one reactants., The order of reactivity order of alkyl halides for SN2, reaction 3° > 2° > 1° and for SN1 reactions 3° < 2° < 1°, 105. (a) In SN1 reactions step 1 is slow and reversible and the, slowest step is the rate determining step, 106. (b) If the compound rotates the plane polarised light to, the right, i.e., clockwise direction, it is called, dextrorotatory (Greek for right rotating) or the d-form, and is indicated by placing a positive(+) sign before, the degree of rotation. If the light is rotated towards, left (anticlockwise direction), the compound is said to, be laevorotatory or the l-form and a negative (–) sign, is placed before the degree of rotation., 107. (a) If a 50 : 50 mixture of the (A) and (B) is obtained then, the process is called racemisation and the product is, optically inactive, as one isomer will rotate light in the, direction opposite to another., , 108. (d) tert-Alkyl halides undergo SN1 reactions, hence they, involve the formation of quite stable carbocations, and, not the transition state. In S N1 reactions, the, nucleophile is not involved in rate determining (first), step, hence its stronger or weaker nature does not, influence the reaction rate. In SN1, the product has, more percentage of the inverted configuration than, the retained configuration, i.e. only partial racemization, takes place, hence the product will be having some, optical activity., , MATCHING TYPE QUESTIONS, 109. (c) In allylic halides hydrogen atom is bonded to sp3, hybridized carbon atom. Whereas in vinylic halide,, hydrogen atom is bonded to sp2 hybridized carbon, atom., CH 2 CH 2, CH3CHCl2, |, |, Ethylidene chloride, Cl, Cl, (gem-dihalide), , Ethylene dichloride (vic-dihalide), , 110. (a) Alkyl iodides are often prepared by the reaction of, alkyl chlorides/bromides with NaI in dry acetone. This, reaction is known as Finkelstein reaction., R X NaI, R I NaX, X = Cl, Br, NaCl or NaBr thus formed is precipitated in dry, acetone., It facilitates the forword reaction according to le, chatelier’s principle. The synthesis of alkyl fluorides, is best accomplished by heating an alkyl chloride/, bromide in the presence of a metallic fluoride such as, AgF, Hg2F2, CoF2 or SbF3. The reaction is termed as, Swarts reaction., H 3C Br AgF, H 3C F AgBr, 111. (b), 112. (b), 113. (b) Chloramphenicol, produced by soil microorganism is, very effective for the treatment of typhoid fever. Our, body produces iodine containing hormone thyroxine,, the deficiency of which causes a disease called goiter., Synthetic halogen compounds, viz chloroquine is used, for the treatment of malaria; halothane is used as an, anaesthetic during surgery. Certain fully fluorinated, compounds are being considered as potential blood, substitutes in surgery., , ASSERTION-REASON TYPE QUESTIONS, 114. (d) Assertion is false, because aryl halides do not undergo, nucleophilic substitution under ordinary conditions., This is due to resonance, because of which the carbon–, chlorine bond acquires partial double bond character,, hence it becomes shorter and stronger and thus cannot, be replaced by nucleophiles. However Reason is true., 115. (c) Alkyl halides give polyalkylation products.
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EBD_7207, HALOALKANES AND HALOARENES, , 420, , 116. (b) Carbon tetrachloride rises to atmosphere and deplete, the ozone layer. This depletion of ozone layer increases, exposure of UV rays to human being which lead to, increase of skin cancer, eye diseases and disorder with, discruption of the immune system., 117. (c) CHCl3 is stored in dark bottles to prevent oxidation of, CHCl3 in presence of sunlight., 118. (d) CCl4 is used as a fire extinguisher. The dense, non, combustible vapours cover the burning substance and, prevents the availability of oxygen around burning, material., , CRITICAL THINKING TYPE QUESTIONS, Cl, , 1, 2, , 3 Br, 120. (c) The compound is C3H 6Cl 2 and the number of, possible isomeric compunds is 5, , CH 3, CH 3 Cl, |, |*, |, (iii) CH 3 CH CH CH 3 (iv) CH 3 CH CH 2 CH2 Cl, Optically inactive, Optically active, Thus structures (i) and (iii) are optically active, each, has one chiral carbon; so each structure will give one, enantiomeric pair; thus total enantiomeric pairs will, be two., 125. (b) Addition of HBr of 2-pentyne gives two structural, isomers (I) and (II), CH3, , ROH + SOCl2, , 121. (d) 4-Bromobut-l-ene is not an allylic halide, BrH 2C —CH 2 —CH CH 2, 4 Bromobut 1 ene, , 122. (c) Primary and secondary alkyl chlorides are prepared, from the respective alcohols by using HCl gas and, anhydrous ZnCl2 (Groove’s process)., 123. (d) Chlorination of n-butane taken place via free radical, h, , Cl, , •, , •, , Cl, , CH3 — CH2 — CH2 — CH3, CH3, , Neopentyl alcohol, , Cl 2 / h, , –H2O, , CH3, , 1, 2 shift, , C2H5, , +, , CH3 – C – CH2.CH3, CH3, , l, , Racemic mixture, , Cl• may attack on either side and give a racemic mixture, of 2 chloro butane which contain 50% d form and 50%, l-form., , CH3 – C = CH – CH3, CH3, –, , 50% d form + 50% l form, , CH3, CH3 – C – CH2, CH3, , –, , d, , R–Cl + SO2 + HCl, , 127. (d) This compound have only one type of hydrogen, available., 128. (b) Chlorine atom is highly reactive so it will react with, all type of hydrogen available while the Br atom is, highly selective so it will react with that hydrogen, which give the highly stabilize tertiary alkyl radical, so only one product is formed., CH3, CH3, 129. (a) CH3 – C – CH2OH HCl CH3 – C – CH2 – OH2+, CH3, CH3, , H — C — Cl + Cl — C — H, , C2H5, , (II), , Each one of these will exist as a pair of geometrical, isomers. Thus, there are two structural and four, configurational isomers., 126. (c) Thionyl chloride is preferred because the other two, products formed in the reaction are escapable gases., Hence the reaction gives pure alkyl halides, , (d.l pair), , formation i.e., Cl2, , C(Br)CH 2CH3, , – –, , H Cl H, |, | |, H C C C H, |, | |, H Cl H, , CHCH 2 CH 3 CH3CH, , (I), , 2-methyl-2-butene, , Cl, CH3 – C – CH2 – CH3, CH3, – –, , H H H, |, | |, Cl C C C H, | |, |, Cl H H, , HBr, , – –, , H H H, |, | |, H C C* C H, |, | |, Cl Cl H, , C CH 2 CH 3, , CH3C(Br), , asymmetric carbon atom, , H H H, |, | |, H C C C H, | |, |, Cl H Cl, , C, , – –, , 119. (c), , 124. (d) First draw possible different structures obtained on, monochlorination of 2-methylbutane,, CH3CH(CH3)CH2CH3., CH3, CH3, |, |, (i) ClCH 2 * C HCH 2 CH3 (ii) CH 3 C CH 2 CH 3, |, Optically active, Cl, Optically inactive, , 2, chloro-2-methyl butane
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HALOALKANES AND HALOARENES, , 421, +, , N2 Cl–, , NH2, •, , Na NO 2 HCl, , 130. (d), , HBF4, , 0 5º diazotisation, , N2+BF4–, , F, + BF3+ N2, , Benzene diazonium, tetrafluoroborate, , 131. (d), , 135. (c), , (Balz-Schiemann, reaction), , CH 2 = CH - Cl, (Vinyl Chloride), , The halogen atom in vinyl chloride is not reactive as in, other alkyl halides. The non-reactivity of chlorine atom, is due to resonance stabilisation. The .p. on Cl-atom, can participate in delocalisation (Resonance) to give, two canonical structure., Q, , 136. (a), , +, , CH 2 = CH - Cl ¾¾, ® CH 2 - CH = C l, , 132. (d) The rate of SN2 substitution reaction is maximum in, case of CH3CH2Br because SN2 mechanism is followed, in case of primary and secondary halides i.e., SN2, reaction is favoured by small groups on the carbon, atom attached to halogens so, CH3 CH2 Br > CH3 CH2 CH2 Br >, , 137. (d), , CH3, |, CH3– CH – Br > CH3 – C – Br, |, |, CH3, CH3, , 138. (a), 139. (b), , 133. (d) Smaller the R group reactivity will be higher towards, SN2 reaction. For alkyl halides containing similar alkyl, group better will be the leaving group, more facile is, the nucleophilic substitution reaction., Amongst the halide ions, the order in which the leaving, groups depart follows the sequence :, I– > Br – > Cl– > F–, It is because of this reason that the order of reactivity, of haloalkanes follows the sequence :, iodoalkanes > bromoalkanes > chloroalkanes >, fluoroalkanes, 134. (a), , 140. (b), , +, , Br ionisation, , Me, , Me, , +Br, , –, , (A), Me, , ionisation, , Br, , Me, , Me, + Br –, , +, , (B), Me, Me, Br, (C), , ionisation, , Me, Me, , +, , + Br, , –, , Since S N 1 reactions involve the formation of, carbocation as intermediate in the rate determining, step, more is the stability of carbocation higher will, be the reactivity of alkyl halides towards SN1 route., Now we know that stability of carbocations follows, the order : 3° > 2° > 1°, so SN1 reactivity should also, follow the same order., 3° > 2° > 1° > Methyl (SN1 reactivity), SN1 reactions involve the formation of carbocations,, order of stability of carbocation is 3° > 2° > 1° hence, higher the stability of carbocation, more will be the, reactivity of the parent alkyl halide. Moreover the, tertiary carbocation formed from (c) is stabilized by, two phenyl groups., A strong nucleophile favours the SN2 reaction and a, weak nucleophile favours the SN1 reaction., First reaction is SN1 reaction because C2H5OH is used, as solvent which is a weak nucleophile., Second reaction is SN2 reaction because C2H5O– is, strong nucleophile., SN2 reaction is favoured by small groups on the carbon, atom attached to halogen., So, the order of reactivity is, CH 3 Cl (CH 3 ) 2 CHCl (CH 3 )3CCl, (C 2 H 5 )2CHCl, SN2 reaction is shown to maximum extent by primary, halides. The only primary halides given is CH3Cl so, the correct answer is (d)., Primary halide < Secondary halide < Tertiary halide., Since SN1 reaction involve the formation of carbocation, as intermediate in the rate determining step. More stable, the carbocation, more is the reactivity of the halide, toward SN1 route. As we know that the stability of the, carbocations decreases in the order :, Benzyl 2º > 1°., Hence the correct order of stability is, III > II > I, Cyanides and nitrites possess two nucleophilic, centres and are called ambident nucleophiles. Actually, cyanide group is a nucleophile in two different ways, [ C N : C N ] . Similarly nitrite ion also, represents an ambident nucleophile with two different, , points of linkage [ O N O]. The linkage through, oxygen results in alkyl nitrites while through nitrogen, atom, it leads to nitroalkanes., 141. (d) In case of optically active alkyl halides, the product, formed as a result of SN2 mechanism has the inverted, configuration as compared to the reactant. This is, because the nucleophile attaches itself on the side, opposite to the one where the halogen atom is present., In case of optically active alkyl halides, SN1 reactions, are accompanied by racemisation. The carbocation, formed in the slow step being sp2 hybridised is planar, (achiral).
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25, ALCOHOLS, PHENOLS AND, ETHERS, FACT / DEFINITION TYPE QUESTIONS, 1., , The characteristic grouping of secondary alcohols is, (a), , CH 2OH, , (c), , C OH, , (b), , CHOH, , (d), , C, , |, , 2., , 3., , |, , OH, OH, , The compound HOCH 2 CH 2 OH is, (a) ethane glycol, (b) ethylene glycol, (c) ethylidene alcohol, (d) dimethyl alcohol, The structural formula of cyclohexanol is, CH 2, , CH 2, (a), , H2 C, , CHOH, , H2 C, , CH 2, , (b), , H2 C, , CHOH, , H2 C, , CH 2, , CH 2 OH, , 4., , 5., , 6., , 7., , 8., , The IUPAC name of CH 3 CH CH 2, |, OH, , (a) 1, 1-dimethyl-1, 3-butanediol, (b) 2-methyl-2, 4-pentanediol, (c) 4-methyl-2, 4-pentanediol, (d) 1, 3, 3-trimethyl-1, 3-propanediol, 10. Number of metamers represented by molecular formula, C4H10O is, (a) 4, (b) 3, (c) 2, (d) 1, 11. Which of the following compounds is aromatic alcohol?, OH, , CH2OH, , CH2OH, , (A), , (B), , (C), , OH, , CH 2, , CH 2, , (c), , 9., , CH 3, |, C CH 3 is, |, OH, , (d), , Which of the following is dihydric alcohol ?, (a) Glycerol, (b) Ethylene glycol, (c) Catechol, (d) Resorcinol, An example of a compound with functional group – O – is :, (a) acetic acid, (b) methyl alcohol, (c) diethyl ether, (d) acetone, Butane-2-ol is, (a) primary alcohol, (b) secondary alcohol, (c) tertiary alcohol, (d) aldehyde, Cresol has, (a) Alcoholic – OH, (b) Phenolic – OH, (c) – COOH, (d) – CHO, How many isomers of C5H11OH will be primary alcohols ?, (a) 5, (b) 4, (c) 2, (d) 3, , CH3, , (D), , CH3, , (a) A, B, C, D, (b) A, D, (c) B, C, (d) A, 12. How many alcohol(s) with molecular formula C4H10O are, chiral in nature?, (a) 1, (b) 2, (c) 3, (d) 4, 13. Give IUPAC name of the compound given below, CH3 –CH – CH 2 – CH 2 –CH – CH 3, |, , Cl, , |, , OH, , (a) 2-Chloro-5-hydroxyhexane, (b) 2-Hydroxy-5-chlorohexane, (c) 5-Chlorohexane-2-ol, (d) 2-Chlorohexan-5-ol, 14. IUPAC name of m-cresol is ___________, (a) 2-methylphenol, (b) 3-chlorophenol, (c) 3-methoxyphenol, (d) benzene-1, 3-diol
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EBD_7207, ALCOHOLS, PHENOLS AND ETHERS, , 424, , 15., , IUPAC name of the compound CH3 – CH – OCH3 is, , 23., , |, , CH3, , 16., , _______________ ., (a) 1-methoxy-1-methylethane, (b) 2-methoxy-2-methylethane, (c) 2-methoxypropane, (d) isopropylmethyl ether, Which of the following are benzylic alcohols?, (i), , C6 H5 – CH 2 – CH 2OH, , (ii), , C6 H5 – CH 2 OH, , 24., , 25., , (iii) C 6 H5 – C H – OH, |, , CH3, , 26., , (iv) C6 H5 – CH 2 – C H – OH, |, , CH3, , 17., , (b) (ii) and (iii), (a) (i) and (ii), (c) (i), (ii) and (iv), (d) (i) and (iv), In which of the following structures hydroxyl group is, attached to sp2 carbon atom?, , CH2OH, (a), , CH, , 27., , OH, OH, , 28., , (b), , Ethyl alcohol is industrially prepared from ethylene by, (a) Permanganate oxidation, (b) Catalytic reduction, (c) Absorbing in H2SO4 followed by hydrolysis, (d) All the three, Sodium salt of benzene sulphonic acid on fusion with caustic, soda gives, (a) Benzene, (b) Phenol, (c) Thiophenol, (d) Benzoic acid, Acid catalyzed hydration of alkenes except ethene leads to, the formation of, (a) primary alcohol, (b) secondary or tertiary alcohol, (c) mixture of primary and secondary alcohols, (d) mixture of secondary and tertiary alcohols, Ethyl alcohol can be prepared from Grignard reagent by the, reaction of :, (a) HCHO, (b) R2CO, (c) RCN, (d) RCOCl, Isopropyl alcohol is obtained by reacting which of the, following alkenes with concentrated H2SO4 followed by, boiling with H2O?, (a) Ethylene, (b) Propylene, (c) 2-Methylpropene, (d) Isoprene, Alkenes convert into alcohols by, (a) hydrolysis by dil. H 2SO 4, (b) hydration of alkene by alkaline KMnO4, , OH, , CH(CH3)OH, CH3, , (c), 18., , 19., , 20., , 21., , Which of the following is an example of unsymmetrical, ether?, (a) C2H5OC2H5, (b) C6H5OC6H5, (c) C6H5OC2H5, (d) CH3OCH3, Which of the following will not form phenol or phenoxide ?, (a) C6H5N2Cl, (b) C6H5SO3Na, (c) C6H5Cl, (d) C6H5CO2H, Benzyl alcohol is obtained from benzaldehyde by, (a) Fittig’s reaction, (b) Cannizzaro’s reaction, (c) Kolbe’s reaction, (d) Wurtz’s reaction, In the reduction, R, , 22., , (d), , CHO, , H2, , (c) hydrolysis by water vapours and conc. HNO3, 29., , 30., , 31., , RCH 2 OH, , the catalyst used is :, (a) Ni, (b) Pd, (c) Pt, (d) Any of these, Ethylene reacts with Baeyer’s reagent to give, (a) ethane, (b) ethyl alcohol, (c) ethylene glycol, (d) None of these, , 32., , (d) hydration of alkene by aqueous KOH, Which of the following reacts with NaOH to give an alcohol?, (a) Propene, (b) Butene, (c) Ethanal, (d) Methanal, By which of the following methods alcohol can be prepared, in excellent yield?, (a) From alkenes, (b) By hydroboration-oxidation, (c) From carbonyl compounds, (d) From Grignard reagent, Which of the following are used to convert RCHO into, RCH2OH?, (i) H2/Pd, (ii) LiAlH4, (iii) NaBH4, (iv) Reaction with RMgX followed by hydrolysis, (a) (i) and (ii), (b) (i), (ii) and (iii), (c) (ii), (iii) and (iv), (d) (i) and (iii), Commercially carboxylic acids are reduced to alcohols by, converting them to the ______., (a) esters, (b) aldehydes, (c) ketones, (d) amines
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ALCOHOLS, PHENOLS AND ETHERS, , 33., , 425, , The hydrocarbon which produce phenol and acetone as a, by product in the large quantity is, CH2CH3, , CH3, (a), , (b), , CH3, , CH3, (c), , 34., , 35., , 36., , 37., , 38., , 39., , 41., , C, , CH3, , In the reaction, RNH 2, , (d) CH3, , HNO2, 273 278K, , CH, , ROH H 2 O C ;, , C is (where R = C6H5), (a) NH3, (b) N2, (c) O2, (b) CO2, The correct order of boiling points for primary (1º),, secondary (2º) and tertiary alcohol (3º) is, (a) 1º > 2º > 3º, (b) 3º > 2º > 1º, (c) 2º > 1º > 3º, (d) 2º > 3º > 1º, Alcohols of low molecular weight are, (a) soluble in water, (b) soluble in water on heating, (c) insoluble in water, (d) insoluble in all solvents, Which of the following has lowest boiling point ?, (a) p-Nitrophenol, (b) m-Nitrophenol, (c) o-Nitrophenol, (d) Phenol, Which statement is not correct about alcohol?, (a) Molecular weight of alcohol is higher than water, (b) Alcohol of less no. of carbon atoms is less soluble in, water than alcohol of more no. of carbon atoms, (c) Alcohol evaporates quickly, (d) All of the above, Which one of the following alcohols is least soluble in, water?, (a), , 40., , H3C, , CH 3 OH, , (b), , C 3 H 7 OH, , (c) C 4 H 9 OH, (d) C10 H 21OH, Methanol and ethanol are miscible in water due to, (a) covalent character, (b) hydrogen bonding character, (c) oxygen bonding character, (d) None of these, If ethanol dissolves in water, then which of the following, would be observed, (a) absorption of heat and contraction in volume, (b) emission of heat and contraction in volume, (c) absorption of heat and increase in volume, (d) emission of heat and increase in volume, , 42. Which of the following is correct ?, (a) On reduction of any aldehyde, secondary alcohol is, formed, (b) Reaction of vegetable oil with H2SO4 gives glycerine, (c) Sucrose on reaction with NaCl gives invert sugar, (d) Alcoholic iodine gives iodoform with NaOH, 43. Which of the following is not true in case of reaction with, heated copper at 300°C?, (a) Phenol, Benzyl alcohol, (b) Secondary alcohol, Ketone, (c) Primary alcohol, Aldehyde, (d) Tertiary alcohol, Olefin, 44. Phenol is more acidic than alcohol because, (a) phenol is more stable than water, (b) phenol is aromatic and alcohol is aliphatic, (c) phenoxide ion is resonance stabilised, (d) None of these, 45. Acidity of phenol is due to, (a) hydrogen bonding, (b) phenolic group, (c) benzene ring, (d) resonance stabilisation of its anion, 46. Which one of the following compounds has the most acidic, nature?, (a), , CH2OH, , (b), , OH, , OH, , (c), , OH, , (d), , CH, , 47. The ionization constant of phenol is higher than that of, ethanol because :, (a) phenoxide ion is bulkier than ethoxide, (b) phenoxide ion is stronger base than ethoxide, (c) phenoxide ion is stabilized through delocalization, (d) phenoxide ion is less stable than ethoxide, 48. Which one of the following on oxidation gives a ketone ?, (a) Primary alcohol, (b) Secondary alcohol, (c) Tertiary alcohol, (d) All of these, 49. Primary and secondary alcohols on action of reduced copper, give, (a) Aldehydes and ketones respectively, (b) Ketones and aldehydes respectively, (c) Only aldehydes, (d) Only ketones, 50. When ethyl alcohol reacts with acetic acid, the products, formed are, (a) Sodium ethoxide + hydrogen, (b) Ethyl acetate + water, (c) Ethyl acetate + soap, (d) Ethyl alcohol + water
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ALCOHOLS, PHENOLS AND ETHERS, , 76., , 77., , 78., , Phenol is less acidic than_______________ ., (a) ethanol, (b) o–nitrophenol, (c) o-methylphenol, (d) o-methoxyphenol, Which of the following is most acidic?, (a) Benzyl alcohol, (b) Cyclohexanol, (c) Phenol, (d) m-Chlorophenol, Mark the correct increasing order of reactivity of the, following compounds with HBr/HCl, CH2OH, , (A), , 79., , 80., , CH2OH, , CH2OH, , NO2, , Cl, , (B), , (C), , (a) A < B < C, (b) B < A < C, (c) B < C < A, (d) C < B < A, Arrange the following in increasing order of their acidity?, o–cresol(a), salicyclic acid(b), phenol(c), (a) c < a < b, (b) b < c < a, (c) a < b < a, (d) a < c < b, In the reaction, , OH, , ONa, + NaOH, , 81., , 427, , + H2O, , Phenol behaves as, (a) Bronsted base, (c) Lewis acid, In the given reaction, , (b) Bronsted acid, (d) Lewis base, , OH, Na 2 Cr2O 7, , A, , H 2SO 4, , A is, OH, (a), , O, , (b), OH, , O, , O, , CHO, , (c), , (d), CH2, , 82. Which enzyme converts glucose and fructose both into, ethanol ?, (a) Diastase, (b) Invertase, (c) Zymase, (d) Maltase, 83. An industrial method of preparation of methanol is :, (a) catalytic reduction of carbon monoxide in presence of, ZnO–Cr2O3, (b) by reacting methane with steam at 900ºC with a nickel, catalyst, (c) by reducing formaldehyde with lithium aluminium, hydride, (d) by reacting formaldehyde with aqueous sodium, hydroxide solution, 84. Ethyl alcohol is industrially prepared from ethylene by, (a) Permanganate oxidation, (b) Catalytic reduction, (c) Absorbing in H2SO4 followed by hydrolysis, (d) All the three, 85. ‘Drinking alcohol’ is very harmful and it ruins the health., ‘Drinking alcohol’ stands for, (a) drinking methyl alcohol, (b) drinking ethyl alcohol, (c) drinking propyl alcohol, (d) drinking isopropyl alcohol, 86. The fermentation reactions are carried out in temperature, range of, (a) 20-30°C, (b) 30-40°C, (c) 40-50°C, (d) 50-60°C, 87. Ethanol is prepared industrially by, (a) hydration of ethylene (b) fermentation of sugar, (c) Both the above, (d) None of these, 88. The fermentation of starch to give alcohol occurs mainly, with the help of, (a) O2, (b) air, (c) CO2, (d) enzymes, 89. In the commercial manufacture of ethyl alcohol from starchy, substances by fermentation method. Which enzymes, slipwise complete the fermentation reaction, (a) Diastase, maltase and zymase, (b) Maltase, zymase and invertase, (c) Diastase, zymase and lactase, (d) Diastase, invertase and zymase, 90. Methyl alcohol is toxic. The reason assigned is, (a) it stops respiratory track, (b) it reacts with nitrogen and forms CN– in the lungs, (c) it increses CO2 content in the blood, (d) it is a reduction product of formaldehyde, 91. In order to make alcohol undrinkable pyridine and methanol, are added to it. The resulting alcohol is called, (a) Power alcohol, (b) Proof spirit, (c) Denatured spirit, (d) Poison alcohol, 92. Wine (alcoholic beverages) contains, (a) CH3OH, (b) Glycerol, (c) C2H5OH, (d) 2-propanol
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EBD_7207, ALCOHOLS, PHENOLS AND ETHERS, , 428, , 93., , Tonics in general contain, (a) Ether, (b) Methanol, (c) Ethanol, (d) Rectified spirit, 94. Widespread deaths due to liquor poisoning occurs due to, (a) presence of carbonic acid in liquor, (b) presence of ethyl alcohol in liquor, (c) presence of methyl alcohol in liquor, (d) presence of lead compounds in liquor, 95. Select the incorrect statement about the fermentation., (a) When grapes are crushed, sugar and the enzyme come, in contact and fermentation starts, (b) Fermentation takes place in anaerobic conditions, (c) Carbon monoxide is released during fermentation, (d) If air gets into fermentation mixture, the oxygen of air, oxidises ethanol to ethanoic acid which in turn, destroys the taste of alcoholic drinks, 96. Denaturation of alcohol is the, (a) mixing of CuSO4 (a foul smelling solid) and pyridine, (to give the colour) to make the commercial alcohol, unfit for drinking, (b) mixing of CuSO4 (to give the colour) and pyridine (a, foul smelling solid) to make the commercial alcohol, unfit for drinking, (c) mixing of Cu(OAc)2 and ammonia to make the, commercial alcohol unfit for drinking, (d) mixing of Cu(OAc) 2 and pyridine to make the, commercial alcohol unfit for drinking, 97. Which one is formed when sodium phenoxide is heated, with ethyl iodide ?, (a) Phenetole, (b) Ethyl phenyl alcohol, (c) Phenol, (d) None of these, 98. Williamson’s synthesis is used to prepare, (a) acetone, (b) diethyl ether, (c) P.V.C., (d) bakelite, 99. The reaction of sodium ethoxide with ethyl iodide to form, diethyl ether is termed, (a) electrophilic substitution, (b) nucleophilic substitution, (c) electrophilic addition, (d) radical substitution, 100. Which of the following cannot be made by using, Williamson’s synthesis?, (a) Methoxybenzene, (b) Benzyl p-nitrophenyl ether, (c) Methyl tertiary butyl ether, (d) Di-tert-butyl ether, 101. The reaction given below is known as, C 2 H 5 ONa, , IC 2 H 5, , C 2 H 5 OC 2 H 5, , NaI, , (a) Kolbe’s synthesis, (b) Wurtz synthesis, (c) Williamson’s synthesis, (d) Grignard’s synthesis, 102. Ethanol and dimethyl ether form a pair of functional isomers., The boiling point of ethanol is higher than that of dimethyl, ether, due to the presence of, , 103., , 104., , 105., , 106., , 107., , 108., , 109., , 110., , 111., , 112., , 113., , (a) H-bonding in ethanol, (b) H-bonding in dimethyl ether, (c) CH3 group in ethanol, (d) CH3 group in dimethyl ether, Ether which is liquid at room temperature is, (a) C2H5OCH3, (b) CH3OCH3, (c) C2H5OC2H5, (d) None of these, Ether can be used, (a) as a general anaesthetic, (b) as a refrigerant, (c) in perfumery, (d) all of the above, Which of the following compound is soluble in ether?, (a) Oils and fats, (b) Water, (c) NaCl, (d) PCl5, An ether is more volatile than an alcohol having the same, molecular formula. This is due to, (a) dipolar character of ethers, (b) alcohols having resonance structures, (c) inter-molecular hydrogen bonding in ethers, (d) inter-molecular hydrogen bonding in alcohols, Which of the following has strongest hydrogen bonding?, (a) Ethyl amine, (b) Ethanal, (c) Ethyl alcohol, (d) Diethyl ether, Oxygen atom in ether is, (a) very active, (b) replaceable, (c) comparatively inert, (d) active, The ether that undergoes electrophilic substitution, reactions is, (a) CH3OC2H5, (b) C6H5OCH3, (c) CH3OCH3, (d) C2H5OC2H5, Diethyl ether on heating with conc. HI gives two moles of, (a) ethanol, (b) iodoform, (c) ethyl iodide, (d) methyl iodide, Methylphenyl ether can be obtained by reacting, (a) phenolate ions and methyl iodide, (b) methoxide ions and bromobenzene, (c) methanol and phenol, (d) bromo benzene and methyl bromide, Diethyl ether can be decomposed by heating with, (a) HI, (b) NaOH, (c) Water, (d) KMnO4, The major organic product in the reaction,, CH3 — O — CH(CH3)2 + HI Product is, (a) ICH2OCH(CH3)2, (b) CH 3O C( CH 3 ) 2, |, , I, (c) CH3I + (CH3)2CHOH (d) CH3OH + (CH3)2CHI, 114. An aromatic ether is not cleaved by HI even at 525 K. The, compound is, (a) C6H5OCH3, (b) C6H5OC6H5, (c) C6H5OC3H7, (d) Tetrahydrofuran
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ALCOHOLS, PHENOLS AND ETHERS, , 115. When 2-methoxypropane is heated with HI, in the mole, ratio 1 : 1, the major products formed are, (a) methanol and 2-iodopropane, (b) methyl iodide and 2-propanol, (c) methyl iodide and 2-iodopropane, (d) methanol and 2-propanol, 116. Formation of diethyl ether from ethanol is based on a, (a) dehydration reaction, (b) dehydrogenation reaction, (c) hydrogenation reaction, (d) heterolytic fission reaction, 117. The cleavage of an aryl-alkyl ether with cold HI gives :, (a) alkyl iodide and water, (b) aryl iodide and water, (c) alkyl iodide, aryl iodide and water, (d) phenol and alkyl iodide, 118. Which of the following compounds is resistant to, nucleophilic attack by hydroxyl ions?, (a) Methyl acetate, (b) Acetonitrile, (c) Acetamide, (d) Diethyl ether, , STATEMENT TYPE QUESTIONS, 119. When an alcohol is prepared by reaction of ethylmagnesiumbromide with 2–pentanone, product formed does not rotate, plane polarised light. For this reaction which of the following, statement(s) is/are correct ?, (i). Product formed is achiral., (ii) Racemic mixture is formed., (a) Both statements (i) and (ii) are correct., (b) Statement (i) is correct only., (c) Statement (ii) is correct only., (d) Both statements (i) and (ii) are incorrect., 120. Which of the following statements are correct ?, (i) Alcohols react as nucleophiles in the reactions, involving cleavage of O–H bond., (ii) Alcohols react as electrophiles in the reactions, involving cleavage of O–H bond., (iii) Alcohols react as nucleophile in the reaction involving, cleavage of C–O bond., (iv) Alcohols react as electrophiles in the reactions, involving C–O bond., (a) (i) only, (b) (i) and (iv), (c) (ii) and (iii), (d) (ii) only, 121. Which of the following are correct statement(s) ?, (i) Polar nature of O–H bond is responsible for acidic, character of alcohols., (ii) Acidic strength of alcohols follow the order 1° > 2° > 3°., (iii) Alcohols are stronger acids than water., (iv) Alcohols also react as Bronsted base., (a) (i), (ii) and (iii), (b) (i), (ii) and (iv), (c) (ii), (iii) and (iv), (d) (i), (iii) and (iv), 122. Read the following statements and choose the correct, option., (i) Ethanol on dehydration at 443 K gives ethene, (ii) Ethanol on dehydration at 413 K gives diethyl ether, , 429, , (iii) Only primary alcohols on dehydration give ethers., (iv) Secondary and tertiary alcohols on dehydration give, ethers having 2° and 3° carbon attached with O atom., (a) TTFF, (b) TFTF, (c) TTTF, (d) FTTF, 123. Which of the following statements are correct ?, (i) In phenols, the —OH group is attached to sp 2, hybridised carbon of an aromatic ring, (ii) The carbon – oxygen bond length (136 pm) in phenol, is slightly more than that in methanol, (iii) Partial double bond character is due to the, conjugation of unshared electron pair of oxygen with, the aromatic ring., (iv) sp2 hybridised state of carbon to which oxygen is, attached., (a) (i), (ii) and (v), (b) (i), (ii) and (iii), (c) (i), (iii) and (iv), (d) (i) and (iv), 124. Which of the following statements are correct ?, (i) Ethanol mixed with methanol is called denatured, alcohol., (ii) Excess of methanol in body may cause blindness., (iii) In the body methanol is oxidised to methanoic acid., (iv) A methanol poisoned patient is treated by giving, intravenous injections of ethanoic acid., (a) (i), (ii) and (iii), (b) (ii), (iii) and (iv), (c) (i) and (v), (d) (i), (iii) and (iv), , MATCHING TYPE QUESTIONS, 125. Match the columns, Column-I, OH, , Column-II, , (p) Quinol, , (A), , OH, OH, (B), , (q) Phenol, OH, (r) Catechol, , (C), OH, , (D), (a), (b), (c), (d), , OH, , (s) Resorcinol, OH, A – (q), B – (p), C – (s), D – (r), A – (r), B – (p), C – (s), D – (q), A – (s), B – (q), C – (p), D – (r), A – (q), B – (r), C – (s), D – (p)
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EBD_7207, ALCOHOLS, PHENOLS AND ETHERS, , 430, , 126. Match the columns, Column-I, (A) Methanol, , (p), , Column-II, Conversion of phenol, to o-hydroxysalicylic, acid, Wood spirit, Heated copper at 573 K, Reaction of alkyl halide, with sodium alkoxide, , (B) Kolbe’s reaction, (q), (C) Williamson’s synthesis (r), (D) Conversion of 2° alcohol (s), to ketone, (a) A – (s), B – (r), C – (q), D – (p), (b) A – (q), B – (s), C – (p), D – (r), (c) A – (q), B – (p), C – (s), D – (r), (d) A – (r), B – (q), C – (p), D – (s), 127. Match the columns, Column-I, Column-II, (A) Antifreeze used in, (p) Methanol, car engine, (B) Solvent used in, (q) Phenol, perfumes, (C) Starting material for, (r) Ethleneglycol, picric acid, (D) Wood spirit, (s) Ethanol, (a) A – (s), B – (q), C – (p), D – (r), (b) A – (r), B – (s), C – (q), D – (p), (c) A – (s), B – (q), C – (r), D – (p), (d) A – (p), B – (r), C – (q), D – (s), , ASSERTION-REASON TYPE QUESTIONS, Directions : Each of these questions contain two statements,, Assertion and Reason. Each of these questions also has four, alternative choices, only one of which is the correct answer. You, have to select one of the codes (a), (b), (c) and (d) given below., (a) Assertion is correct, reason is correct; reason is a correct, explanation for assertion., (b) Assertion is correct, reason is correct; reason is not a, correct explanation for assertion., (c) Assertion is correct, reason is incorrect., (d) Assertion is incorrect, reason is correct., 128. Assertion : The bond angle in alcohols is slightly less, than the tetrahedral angle., Reason : In alcohols, the oxygen of –OH group is attached, to sp3 hybridized carbon atom., 129. Assertion : In Lucas test, 3º alcohols react immediately., Reason : An equimolar mixture of anhyd. ZnCl2 and conc., HCl is called Lucas reagent., 130. Assertion : Reimer-Tiemann reaction of phenol with CCl4, in NaOH at 340 K gives salicyclic acid as the major product., Reason : The reaction occurs through intermediate, formation of dichlorocarbene., 131. Assertion : Phenol is more reactive than benzene towards, electrophilic substitution reaction., Reason : In the case of phenol, the intermediate carbocation, is more resonance stabilized., , 132. Assertion : In case of phenol, bromination takes place, even in absence of Lewis acid whereas bromination of, benzene takes place in presence of Lewis acid like FeBr 3., Reason : – OH group attached to benzene ring is highly, deactivating., 133. Assertion : ter - Butyl methyl ether is not prepared by the, reaction of ter-butyl bromide with sodium methoxide., Reason : Sodium methoxide is a strong nucleophile., 134. Assertion : Ethers behave as bases in the presence of, mineral acids., Reason : Due to the presence of lone pairs of electrons on, oxygen., 135. Assertion : With HI, anisole gives iodobenzene and methyl, alcohol., Reason : Iodide ion combines with smaller group to avoid, steric hindrance., 136. Assertion : With HI at 373 K, ter-butyl methyl ether gives, ter-butyl iodide and methanol., Reason : The reaction occurs by SN2 mechanism., 137. Assertion : Ethyl phenyl ether on reaction with HBr form, phenol and ethyl bromide., Reason : Cleavage of C–O bond takes place on ethyloxygen bond due to the more stable phenyl-oxygen bond., , CRITICAL THINKING TYPE QUESTIONS, 138. Vinyl carbinol is, (a), , HO CH 2 CH, , (b), , CH 3C(OH ), , (c), , CH 3 CH, , (d), , CH 3 C(CH 2 OH), , CH 2, , CH 2, CH OH, CH 2, , 139. Propene, CH 3CH CH 2 can be converted into 1-propanol, by oxidation. Indicate which set of reagents amongst the, following is ideal to effect the above conversion ?, (a) KMnO4 (alkaline), (b) Osmium tetraoxide (OsO4/CH2Cl2), (c) B6H6 and alk. H2O2, (d) O3/Zn, 140. The product of the following reaction is, (i ) BH 3 / THF, (ii ) H 2 O 2 , OH, , (a) 1-Pentanol, (c) Pentane, , (b) 2-Pentanol, (d) 1,2-Pentanediol, , 141. C6 H5 CH CHCHO X C6 H5 CH, In the above sequence X can be, , CHCH 2 OH ., , (a), , H 2 / Ni, , (b), , (c), , K 2 Cr2 O7 / H, , (d) Both (a) and (b), , NaBH 4
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ALCOHOLS, PHENOLS AND ETHERS, , 142. Which of the following reactions will yield phenol?, , Cl, (i) fusion with NaOH at 300 atm, , (i), , (ii)H 2O/H, , NH2, (i)NaNO2 /HCl, (ii)H 2O(Warming), , (ii), , (i)Oleum, (ii)NaOH,(Heating), (iii)H, , (iii), , Cl, (iv), , 143., , 144., , 145., , 146., , (i)NaOH(aq.).298K/1atm, (ii)HCl, , (a) (i), (ii) and (iii), (b) (i) and (iii), (c) (i), (iii) and (iv), (d) (ii), (iii) and (iv), Hydration of styrene is carried out in presence of acid as, catalyst. The major product is., (a) 1–hydroxy–2–phenylethane., (b) 1–hydroxy–1–phenylethane., (c) 2–hydroxy–1–phenylethane., (d) 2–hydroxy–2–phenylethane., Which of the following reagents can be used for preparation, of cumene ?, (i) C6H6, Cl2, hv; Mg.THF; acetone., (ii) C6H6, CH3CH2CH2Cl, AlCl3., (iii) C6H6, CH3CHClCH3, AlCl3., (iv) C6H6, CH3CH2Cl, AlCl3;, (a) (i) and (ii), (b) (ii) and (iii), (c) (i), (ii) and (iii), (d) (ii) and (iv), The hydroboration of an alkene is carried out, then on, oxidation with hydrogen peroxide, the alcohol so obtained, is achiral. Possible structure of alkene is (are) :, (i) 2, 3– dimethylbut–2–ene., (ii) 3, 4–dimethylbut –3–ene., (iii) 2–methyl–but–2–ene., (iv) 2–methylpropene., (a) (i) and (iv), (b) (ii) and (iii), (c) (iii) and (iv), (d) (i) and (iii), Which of the following shows structure of allylic alcohol?, (i), , CH 2 CH CH 2OH, , (ii), , CH 2 CH OH, , 431, , 147. Mechanism of acid catalysed hydration reaction involves, (i) Protonation of alkene to form carbocation by, electrophilic attack of H3O+, (ii) Nucleophilic attack of water on carbocation., (iii) Deprotonation to form alcohol., (a) (i) and (ii), (b) (i) and (iii), (c) (i), (ii) and (iii), (d) (ii) and (iii), 148. Phenol is less acidic than, (a) acetic acid, , (b) p-methoxyphenol, , (c) acetylene, (d) ethanol, 149. The correct order of acid strength of the following, compounds :, (A) Phenol, (B) p–Cresol, (C) m–Nitrophenol, (D) p–Nitrophenol, (a) D > C > A > B, (b) B > D > A > C, (c) A > B > D > C, (d) C > B > A > D, 150. Arrange the following compounds in order of decreasing, acidity :, , OH, , OH, ;, , ;, Cl, (I), , CH3, (II), , Zn dust, , (iii) CH 2 CH CH CH3 OH, , (a) (i), (iii) and (iv), (c) (ii), (iii) and (iv), , (b) (i), (ii) and (iv), (d) (i), (ii), (iii) and (iv), , OH, ;, , ;, NO2, (III), , OCH3, (IV), , (a) II > IV > I > III, (b) I > II > III > IV, (c) III > I > II > IV, (d) IV > III > I > II, 151. ClCH2CH2OH is stronger acid than CH3CH2OH because, of:, (a) – I effect of Cl increases negative charge on O atom of, alcohol, (b) – I effect of Cl disperses negative charge on O atom to, produce more stable cation, (c) – I effect of Cl disperses negative charge on O atom to, produce more stable anion, (d) None of these, 152. Which one of the following compounds will be most readily, attacked by an electrophile ?, (a) Chlorobenzene, (b) Benzene, (c) Phenol, (d) Toluene, 153. Consider the following reaction:, Phenol, , (iv) CH 2 CH C CH3 2OH, , OH, , X, , CH3Cl, Anhydrous AlCl3, , Y, Alkaline KMnO4, , The product Z is, (a) benzaldehyde, (c) benzene, , (b) benzoic acid, (d) toluene, , Z
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EBD_7207, ALCOHOLS, PHENOLS AND ETHERS, , 432, , 154. When phenol is reacted with CHCl3 and NaOH followed by, acidification, salicylaldehyde is obtained. Which of the, following species are involved in the above mentioned, reaction as intermediate?, OH, O, , CHCl2, , H, +, CCl2, , (a), , (b), , (c), , (d), , 155. The reagent used for dehydration of an alcohol is, (a) phosphorus pentachloride, (b) calcium chloride, (c) aluminium oxide, (d) sodium chloride, 156. The alcohol which does not give a stable compound on, dehydration is, (a) ethyl alcohol, (b) methyl alcohol, (c) n-Propyl alcohol, (d) n-Butyl alcohol, 157. A compound of the formula C4H10O reacts with sodium, and undergoes oxidation to give a carbonyl compound, which does not reduce Tollen’s reagent, the original, compound is, (a) Diethyl ether, (b) n-Butyl alcohol, (c) Isobutyl alcohol, (d) sec-Butyl alcohol, 158. Which of the following fact(s) explain as to why, p-nitrophenol is more acidic than phenol?, I. –I Effect of nitro group., II. Greater resonance effect of p-nitrophenoxy group, III. Steric effect of bulky nitro group, (a) I and II, (b) I and III, (c) II and III, (d) II alone, 159. In the following sequence of reactions,, CH 3CH 2 OH, , P I2, , A, , Mg, ether, , B, , H O, , OC6H5, , (d), , CH3, , O, NO2, , 161. Which of the following reagents can be used to oxidise, primary alcohols to aldehydes?, (i) CrO3 in anhydrous medium., (ii) KMnO4 in acidic medium., (iii) Pyridinium chlorochromate., (iv) Heat in the presence of Cu at 573K., (a) (i) and (iii), (b) (ii), (iii) and (iv), (c) (i), (iii) and (iv), (d) (i), (iii) and (iv), , CH3, |, CH CH 2, , O CH 2, , CH3, , Heated, , HI, , Which of the following compounds will be formed?, (a), , CH3 CH CH3 CH 3CH 2 OH, |, CH3, , (b), , CH3 CH CH 2 OH CH3CH3, |, CH3, , (c), , CH3, |, CH3 CH CH 2 OH CH3 CH 2, , (d), , CH3, |, CH3 CH CH 2, , HCHO, , 2, C, D, the compound D is, (a) propanal, (b) butanal, (c) n-butyl alcohol, (d) n-propyl alcohol., 160. Which of the following species can act as the strongest, base?, (a), (b), OH, OR, , (c), , 162. Which one of the following will show the highest pH value ?, (a) m–nitrophenol., (b) p–nitrophenol., (c) o–nitrophenol., (d) Both (b) and (c)., 163. Which of the following is most reactive towards aqueous, HBr ?, (a) 1-Phenyl-1-propanol, (b) 1-Phenyl-2-propanol, (c) 3-Phenyl-1-propanol, (d) All are equally reactive, 164. The major product of the reaction between tert-butyl chloride, and sodium ethoxide is, (a) 2-methylprop-1-ene, (b) 1-butene, (c) 2-butene, (d) ethene, 165. In Williamson synthesis if tertiary alkyl halide is used than, (a) ether is obtained in good yield, (b) ether is obtained in poor yield, (c) alkene is the only reaction product, (d) a mixture of alkene as a major product and ether as a, minor product forms., 166. In the reaction:, , 167. In the reaction, , I, , I CH3CH 2OH, OCH3, , (a), , OBr and CH4, , (b), , Br and CH3Br, , (c), , Br and CH3OH, , (d), , OH and CH3Br, , HBr, , the products are, , 168. An aromatic ether is not cleaved by HI even at 525 K. The, compound is, (a) C6H5OCH3, (b) C6H5OC6H5, (c) C6H5OC3H7, (d) Tetrahydrofuran
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ALCOHOLS, PHENOLS AND ETHERS, , 433, , FACT / DEFINITION TYPE QUESTIONS, 1., , (b), , 2. (b), CH 2, , 3., , (a), , H2 C, , CH OH, , H2 C, , CH 2, , OH, or, , Benzyl alcohol, , (b) Glycols are dihydric alcohols (having two hydroxyl, groups). Ethylene glycol is the first member of this, series., CH 2OH, |, CH 2 OH, , (c), (b), (b), (b), , Ethers contain the functional group – O –, CH3CH2CH(OH)CH3 – a secondary alcohol, Cresol has phenolic group – OH, Four primary alcohols of C5H11OH are possible. These, are:, (i) CH 3 CH 2 CH 2 CH 2 CH 2 OH, , 22. (c), , (iv) CH 3, , 5, , 4, , CH 3, |, C CH 2 OH, |, CH 3, 3, , CH3, |, 2, , 1, , 9., , (b), , 10., , (b) C4H10O : (i) C2H5OC2H5 (ii) CH3OC3H7, (iii) CH3OCH(CH3)2, (c), 12. (a), 13. (c), 14. (a), 15. (c), (b), (c) In this structure –OH group is directly attached to, double bonded carbon atom i.e. sp2 hybridized carbon, atom., (c) If two groups attached to the oxygen atom are, different then ethers are known as unsymmetrical or, mixed ethers., , CH3 CH CH 2 C CH3, |, |, OH, OH, 2-methyl- 2, 4-pentanediol., , 11., 16., 17., , 18., , CH 2, , KMnO4, , CH 2, , CH 2 CH 2, |, |, OH OH, , OH, , Glycol, , This reaction is known as Baeyer’s test for unsaturation., 23. (c) Ethylene is passed into concentrated sulphuric acid at, 75–80°C under pressure when a mixture of ethyl, hydrogen sulphate and diethyl sulphate is formed., H 2C, , CH 2 H 2SO 4, , 100 C, , Ethylene, , H 2O, , (ii) CH 3 CH 2 CH CH 2 OH, |, CH 3, (iii) CH 3 CH CH 2CH 2 OH, |, CH 3, , Sod. benzoate, , 21. (d) Any one of Ni, Pt or Pd can be used in the reduction of, aldehydes., , Baeyer’s, reagent, , (Ethylene glycol), , 5., 6., 7., 8., , Benzaldehyde, , C6 H 5 CH 2 OH C6 H 5 COONa, , CH 2, , 4., , 19. (d) Benzoic acid (C 6 H 5COOH) will not form phenol or, phenoxide., 20. (b) By heating benzaldehyde with conc. NaOH or KOH, (Cannizzaro reaction)., C6 H5 CHO NaOH, , SO 3 Na, , 24. (b), , C2 H5 HSO 4, , C 2 H 5 OH H 2SO 4, , O Na, , NaOH, 300°, , + Na2SO3.H2O, HCl, , OH, + NaCl, , 25. (b), 26. (a), , H, |, H C O, , CH3MgBr, , H, |, H C OMgBr, |, CH 3, , H, |, H3O, H C OH, |, CH 3, 27. (b) Since the compound is formed by hydration of an, alkene, to get the structure of alkene remove a molecule, of water from the alcohol., CH3 C H CH3, |, OH, Isopropyl alcohol, , H 2O, , CH 2 CHCH 3, Propylene
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EBD_7207, ALCOHOLS, PHENOLS AND ETHERS, , 434, , 28., , (b), , CH 2, , CH 2, , H 2 O [O], , alk. KMnO 4, , CH2–CH2, , 40., , –, , (b) Hydrogen bonding :, , Ethane, , NaOH, , Methanal, , R'OH, H, , Aniline, , 37., 38., , 39., , RCOOR, , H2, Catalyst, , N2 Cl, , CHI 3, , Iodoform, , 43., , Sod. formate, , +HCl, Warm, Benzene diazonium, chloride, , R CH 2 CH2 OH, 1° alcohol, , Alcohol, , HCOONa 5 NaI 5H 2O, , Cu, , R CH 2, , 300 C, , CHO + H 2, , Aldehyde, , H, , RCH 2OH R OH, , R, , C, , OH, , Cu, 300 C, , R, 2° alcohol, , R – C – R + H2, O, Ketone, , CH 3, , R, , C, , OH, , Cu, 300 C, , CH3, 3° alcohol, , + N2 + HCl, , (a) Among isomeric alcohols surface area decreases from, 1º to 2º to 3º alcohols and hence the boiling point., (a) The lower alcohols are readily soluble in water and the, solubility decreases with the increase in molecular, weight. The solubility of alcohols in water can be, explained due to the formation of hydrogen bond, between the highly polarised —OH groups present, both in alcohol and water., (c) o-Nitrophenol has intramolecular H-bonding., (b) The solubility of alcohols depend on number of, C-atoms of alcohols. The solubility of alcohols in water, is decreased by increasing number of C-atoms of, alcohol. As resulting molecular weight increases, the, polar nature of O – H bond decreases and hence, strength of hydrogen bond decreases., (d) Solubility of alcohol in water decreases with increase, in molecular mass due to increase in water repelling, alkyl part in alcohol., , Water, , |, , (a) When primary (1°) alcohols are treated with copper at, 300°C, then aldeh ydes are obtained by, dehydrogenation of alcohols. Similarly secondary (2°), alcohols form ketone and alkene is obtained by, dehydration of tertiary (3°) - alcohols. But phenol does, not respond to this test., , OH, H2O, , R, , |, , (b) When ethanol dissolves in water then there is emission, of heat and contraction in volume., (d) C 2 H 5 OH 4 I 2 6 NaOH, , (d), (b) A diazonium salt is formed by treating an aromatic, primary amine with nitrous acid (NaNO2 + HCl) at, 273-278 K. Diazonium salts are hydrolysed to phenols, by warming with water or by treating with dilute acids., , NaNO2, , 36., , 42., , CH3OH + HCOONa, Methyl, alcohol, , NH2, , 35., , 41., , (b), 31. (b), (a) Commercially, acids are reduced to alcohols by, converting them to the esters, followed by their, reduction using hydrogen in the presence of catalyst, (catalytic hydrogenation)., RCOOH, , 33., 34., , Ethanol, , (d) The aldehydes which do not have -hydrogen atom, react with NaOH when half of molecules are reduced, to alcohol and other half of molecules are oxidised to, acid (Cannizzaro reaction)., 2 HCHO, , 30., 32., , OH, , H, , Alcohol, , C 6 H 5 – OH, O–, , 44., , R C CH 2 + H 2 O, |, CH 3, Alkene, , Cu, 300 C, , No reaction, , O, , O, , –, , (c), , etc., . .–, , Resonance stabilisation of phenoxide ion, (conjugate base of phenol), , Conjugate base of ethyl alcohol, i.e., C2H5O– does, not show resonance., .., .. –, OH, O, .., , 29., , CH 3 CH 2, , R, , 45., , + H+, , (d), Phenol, (resonance not, significant), , 46., , .., .., , Conc. H 2SO 4, , O –H, , .., , CH 2 H 2 O, , –, , O–H, , |, , OH OH, Glycol, , CH 2, , –, , O–H, , Phenoxide ion, (resonance stabilised), , (b) Phenol is most acidic because its conjugate base is, stabilised due to resonance, while the rest three, compounds are alcohols, hence, their corrosponding, conjugate bases do not exhibit resonance.
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ALCOHOLS, PHENOLS AND ETHERS, , 47., , (c), , C6 H 5 OH H 2 O, , 435, , C6 H5O, , Phenoxide ion, , n propyl alcohal, , The phenoxide ion is stable due to resonance., , O, , –, , O, , –, ••, , O, , H3C, , O–, , O, , –, , ••, –, , –, , The negative charge is delocalized in the benzene ring, which is a stabilizing factor in the phenoxide ion and, because of this reason ionization constant of phenol, is higher whereas no resonance is possible in alkoxide, ions (RO–)derived from alcohol. The negative charge, is localized on oxygen atom in case of alcohols., (b) Secondary alcohols on oxidation give ketones., Note : – Primary alcohols from aldehydes., , R, , R, , Isopropyl, alcohol, , CH3, , CH3, |, C, |, CH3, , Br, , CH3, , CH3, |, C Br, |, CH3, , 56. (d), , OH, |, , OH, , Br, , Cu, 300, , CH, H3 C, , Br, 2, 4,6-Tribromophenol, , CH3CHO H 2, , Note : The –OH group in phenol, being activating, group, facilitates substitution in the o- and p-positions., , Aldehyde, , OH, , H3 C, OH, , Br, , 3Br2 (aq.), , (a) Alcohols are oxidized by removal of H2 in presence, of a heated metal catalyst (Cu), 1 alcohol, , Cu, 300, , 2° Alcohol, , CH3CCH3 H 2, ||, O, , 57. (b), , 50., , (b), , 51., , (c) Secondary alcohols oxidise to produce kenone., , C 2 H 5 OH CH 3 COOH, , CH 3 COO.C 2 H 5, , OH, , OH, , H 2O, 2KOH, , OH, , CH(OH)2, , CHO, , –H2O, , ( O), , CH3COCH2CH3, CH3CHOHCH2CH3, 2-Butanol, Ethyl methyl ketone, (b) Greater the stability of the intermediate carbocation,, more reactive is the alcohol. Since 2-methylpropan-2-ol, generates 3° carbocation, therefore, it reacts fastest, with HBr., (c) Primary alcohol on oxidation give aldehyde which on, further oxidation give carboxylic acid whereas, secondary alcohols give ketone., , CHCl2, , + CHCl3, , Ketone, , 53., , Ketone, , 2-Methyl-2-propanol, , Ketone, , CH3CH 2OH, , 52., , C=O, , C=O, , R, , 49., , [O] H3C, , CH 3 CH 2 COOH, , 54. (a) Lucas reagent is conc. HCl + anhyd. ZnCl2., 55. (d) The rates of reaction with lucas reagent follows the, order., 3° alcohol > 2° alcohol > 1° alcohol, since carbocations are formed as intermediate, more, stable the carbocation, higher will be the reactivity of, the parent compound (alcohol). 2-Methylpropan-2-ol, generates a 3º carbocation, so it will react fastest; other, three generates either 1º or 2º carbocations., CH3, |, H, CH3, C OH, |, CH3, , R, , [O], , CHOH, , CH – OH, , isopropyl alcohal, , OH, , 48., , [O], , CH 3 CH 2 CHO, , ••, –, , –, , [O], , CH 3CH 2 CH 2 OH, , H 3O, , Reimer-Tiemann reaction., 58. (c) When phenol reacts with Zinc dust, then benzene and, zinc oxide are formed, OH, , + Zn, Phenol, , + ZnO, Benzene
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ALCOHOLS, PHENOLS AND ETHERS, , 87., , 437, , (c) Hydration of alkenes, CH 2, , CH 2, , –, , H HSO 4, H 2O, Boil, , CH3 CH 2 HSO 4, , CH 3 – CH 2, , HSO 4, , CH 3 – CH 2 OH H 2SO 4, , Fermentation of sugar :, Invertase, , C12 H 22O11 H 2O, , Glucose, , Zymase, , C6H12O 6, , 88., , (d), , Starch, , 89., , (a), , 2(C6H10O5 )n nH 2O, , Fructose, , 2C2H 5OH 2CO 2, , Glucose or Fructose, Enzymes, , C2H12O6 C6H12O6, , Alcohol, , Starch, , Diastase, (from germinated barley), , n(C12H 22O11 ), Maltose, , C12 H 22 O11 H 2 O, C6 H12 O6, , 90., 91., 92., 93., 94., 95., , 96., , 97., 98., 99., , Maltase, (from yeast), , Zymase, (from yeast), , 2C6 H12 O6, , (b), , C 6 H 5 ONa C 2 H 5 I, , 111. (a), , C6 H5O CH3 I C6 H5 OCH3 I, 112. (a) Ethers are readily cleaved by HI as follows :, , 2C 2 H 5 OH 2CO 2, , C2 H 5 Br, , C 2 H 5ONa, , Sod. ethoxide, , C6 H 5OC 2 H 5, , NaI, , R O R, , C2 H5, , (b) Reaction of sodium ethoxide with ethyl iodide to, produce diethyl ether is known as Williamson, synthesis., It is a nucleophilic substitution reaction and proceeds, via SN2 mechanism., 100. (d) The two components should be (CH3 )3 CONa +, (CH3)3CBr. However, tert-alkyl halides tend to undergo, elimination reaction rather than substitution leading, to the formation of an alkene, Me2C = CH2, , R O R, |, H, , CH 3O CH(CH 3 ) 2, , I, , ROH + RI., , HI, , 373K, , CH 3I, Methyl, iodide, , (CH 3 )2CHOH, Isopropyl, alcohol, , The alkyl halide is always formed from the smaller alkyl, group., 114. (b) Due to greater electronegativity of sp2-hybridized, carbon atoms of the benzene ring, diaryl ethers are not, attacked by nucleophiles like I–., 115. (b), , CH3, CH3, , CH3, CHOCH3 + HI, , CH3I +, , CH3, , CHOH, , 2-propanol, , 116. (a) Dehydration of alcohols gives ethers, Ar – OH + RI, 117. (d) R – O – Ar + HI, Aryl-alkyl ether, , O C2H 5, , diethyl ether, , H, , 113. (c) In case of unsymmetrical ethers, the site of cleavage, depends on the nature of alkyl group e.g.,, , Phenetole, , NaBr, , 2 C2H5I + H2O, , 110. (c) C2H5OC2H5 + 2 HI, , Glucose, , (b), (c) Denaturing can also be done by adding 0.5% pyridine,, petroleum naptha, CuSO4 etc., (c), (c) Tonics contain ethyl alcohol., (c) Due to presence of methyl alcohol in liquor., (c) The quantity of sugar increases and yeast grows on, the outer skin as grapes ripen. When grapes are, crushed, sugar and the enzyme come in contact and, fermentation starts. Fermentation takes place in, anaerobic contidions i.e., in absence of air CO2 is, released during fermentation. If air gets into, fermentation mixture the oxygen of air oxidises ethanol, to ethanoic acid which in turn destroys the taste of, alcoholic drinks., (b) The commercial alcohol is made unfit for drinking by, mixing in it some copper sulphate (to give it colour), and pyridine (a foul smelling liquid). It is known as, denaturation of alcohol., (a), , 101. (c) Preparation of ethers by reacting sodium ethoxide with, alkyl halide is called Williamson synthesis., 102. (a) Due to H-bonding, the boiling point of ethanol is much, higher than that of the isomeric diethyl ether., 103. (c) CH3OCH3 and C2H5OCH3 are gases while C2H5OC2H5, (b. p. 308 K) is low boiling liquid., 104. (d) Ether is used (i) as a general anaesthetic, (ii) as, refrigerant since it produces cooling on evaporation,, (iii) as solvent for oils, fats, resins etc. (iv) for providing, inert medium in Wurtz reaction, (v) for preparing, Grignard reagent, (vi) in perfumery., 105. (a) Like dissolves like. Oils and fats, being covalent,, dissolve in ether, a non-polar solvent., 106. (d) Due to inter-molecular hydrogen bonding in alcohols, boiling point of alcohols is much higher than ether., 107. (c), 108. (c), 109. (b) Only alkyl aryl ethers e.g., C6 H5OCH3 undergoes, electrophilic substitution reactions., , 118. (d), , Phenol, , Alkyl iodide, , Due to steric hinderance, smaller alkyl group is always, attached to iodine., Diethyl ether, being a Lewis base, is not attacked by, nucleophiles, while all others contain electrophilic carbon,, hence attacked by nucleophiles like OH– ions., , CH3, , O, |, C OCH3, , CH3 C, , CH 3, , O, ||, C NH 2, , C2H5 O, . . C 2 H5, , .., , N
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EBD_7207, ALCOHOLS, PHENOLS AND ETHERS, , 438, , STATEMENT TYPE QUESTIONS, , ASSERTION-REASON TYPE QUESTIONS, , 119. (c) Product formed is 2–methyl–pentan–2–ol hence, carbon is attached to four different group therefore, the molecule is chiral but because the carbonyl group, is planar so attack of methyl group can take place either, ways above and below the plane of the of molecule, hence equal number of enantiomers are formed and, hence the racemic mixture is formed., 120. (b) Alcohols are versatile compounds. They react both, as nucleophiles and electrophiles. The bond between, O—H is broken when alcohols react as nucleophiles., Alcohols as nucleophiles, (i) R–O–H+, , +C –, , H, R –+O, .. – C –, , R CH 2 OH H, , 133., 134., 135., 137., , Br – CH2 + H2O, , R, , R, , 121. (b) Alcohols are, weaker acids than water. This can be, illustrated by the reaction of water with an alkoxide., –, , R – O – H + :OH, Conjugate Conjugate, acid, base, , : :, , : :, , : :, , R – O: + H – O – H, Base, Acid, , This reaction shows that water is a better proton donor, (i.e., stronger acid) than alcohol. Also in the above, reaction, we note that an alkoxide ion is a better proton, acceptor than hydroxide ion, which suggests that, alkoxide are stronger bases (sodiumethoxide is a, stronger base than sodium hydroxide)., 122. (c) The dehydration of secondary and tertiary alcohols, to give corresponding ethers is unsuccessful as, elimination competes over substitution and as a, consequence, alkenes are easily formed., 123. (c) The C — O bond length (136 pm) in phenol is slightly, less than that in methanol (142 pm)., 124. (a) A methanol poisoned patient is treated by giving, intravenous injection of ethanol., , MATCHING TYPE QUESTIONS, 125. (d), , 126. (c), , 132., , R CH2 OH 2, , Br– + CH2 – OH2+, , –, , .., , R–O– C– +H +, , (ii) The bond between C — O is broken when they, react as electrophiles. Protonated alcohols react, in this manner., Protonated alcohols as electrophiles, , 127. (b), , :O :, , in alcohols is slightly less, C H, than the tetrahedral angle (109°-28 ). It is due to the, repulsion between the unshared electron pairs of, oxygen., 129. (b) The correct explanation is : In Lucas test, tertiary, alcohols react immediately because of the formation, of the more stable tertiary carbocations., 130. (c) The correct reason is : Nucleophilic attack of phenolate, ion through the ortho-carbon atom occurs on CCl4 (a, neutral electrophile) to form an intermediate which on, hydrolysis gives salicylic acid (ArSE reaction)., 131. (a) R is the correct explanation of A. Due to +M effect of, , 128. (a) The bond angle, , O, .. H , its intermediate carbocation is more stable than, the one in benzene., (c) The usual halogenation of benzene takes place in the, presence of a Lewis acid, such as FeBr 3, which, polarises the halogen molecule. In case of phenol, the, polarisation of bromine molecule takes place even in, the absence of Lewis acid. It is due to the highly, activating effect of –OH group attached to the benzene, ring., (b), (a) R is the correct explanation of A., (d), 136. (c), (c) Alkyl aryl ethers are cleaved at the alkyl-oxygen bond, due to the more stable aryl-oxygen bond. The reaction, yields phenol and alkyl halide, O–R, OH, , +H–X, , +R–X, , Ethers with two different alkyl groups are also cleaved, in the same manner., R X R ' OH, R O R ' HX, , CRITICAL THINKING TYPE QUESTIONS, 138. (a) Methyl alcohol (CH3OH) is also known as carbinol., Hence vinyl carbinol is CH2 = CH – CH2OH., 139. (c) KMnO4 (alkaline) and OsO4 / CH2Cl2 are used for, hydroxylation of double bond while O3 /Zn is used for, ozonolysis. Therefore, the right option is (c), i.e.,, 3CH 3CH CH 2 BH3 in THF (CH 3CH 2 CH 2 )3 B, 3H 2O 2, NaOH, , 3CH 3CH 2CH 2OH + H3BO3, 1-propanol, , 140. (a) Hydroboration-oxidation leads to anti-Markownikoff’s, hydration, thus
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ALCOHOLS, PHENOLS AND ETHERS, , 439, –, , 141. (b) NaBH4 and LiAlH4 attacks only carbonyl group and, reduce it into alcohol group., , C6 H5, , CH, , CHCHO, , –, , CH3– C–O, , NaBH 4, , CH, , alcohols –OH group is attached to sp2 hybridized, carbon whereas in allylic alcohols – OH group is, attached to sp3 hybridized carbon., 147. (c) The mechanism of the reaction involves the following, three steps:, Step 1: Protonation of alkene to form carbocation by, electrophilic attack of H3O+, H2O + H+ H3O+, H, C=C, , >, OCH3, , OH, , CH3, +I, , NO 2, –I, – M, – I, (A), (B), (C), (D), Electron withdrawing substituents increases the, acidity of phenols; while electron releasing substituents, decreases acidity. Further the particular effect (acidity, increasing or decreasing) is more when a substituent, is present in o-ortho position to phenolic group. Thus, the correct order will be D > C > A > B., 150. (c) Electron withdrawing substituents like –NO2 , Cl, increase the acidity of phenol while electron releasing, substituents like – CH3, – OCH3 decreases acidity., hence the correct order of acidity will be, , H, , —C—C, , + H2O, , H, , III, (–M, –I), , —C—C—O —H, , 151. (c), , H, , Step 3: Deprotonation to form an alcohol., , H, , .., , +, — C — C — O — H + H2O, , H, , H, , Cl, , CH3, , I, (–I > +M), , II, (+I, + HC), , CH 2CH 2OH, Stronger acid, , CH 2CH 2O, ve charge on O, dispersed hence, conjugate base, stable, , .., , H, , 148. (a) More the stability of the conjugate base, higher is the, acidic character of the parent acid. Stability order of, the four conjugate bases is arranged below., , IV, (+ M), , Cl, , OH, +, , OCH3, , ClCH 2 CH 2 OH is stronger acid than CH3CH 2 OH due, to – I effect of Cl., , Cl, , — C — C — +H3 O, , OH, >, , .., , H, , OH, >, , NO2, , +, , H, , OH, >, , Step 2: Nucleophilic attack of water on carbocation., +, , OH, , NO2, , H, , H, , C, , 149. (a), , + H2O, , —C—C, , OH, , OH, , OH, +, , CH, , Least stable, (resonance not, possible), , –OCH3 group intensifies, –ve charge due to + M effect, , H, , +H—O, .. —H, , O–, , cinnamic alcohol, , CH OH represents vinylic alcohol. In vinylic, , +, , >, , CH.CH 2 OH, , 142. (a), 143. (b) Carbocation is formed as intermediate which is most, stabilized when protonation occurs on terminal carbon., 144. (b) Reaction of 1–chloropropane leads to the formation of, the primary carbocation which rearranges to more, stable secondary carbocation, hence (ii) and (iii) give, similar products., 145. (a) In case of (ii) and (iii), the alcohol so obtained contain, carbon which is attached to four different groups i.e.,, chiral carbon while in case of (i) and (iv) achiral alcohol, is obtained., 146. (a) CH 2, , >, , Most stable because, resonating structures, are equivalent, , cinnamic aldehyde, , C6 H 5, , O, , O, , CH3CH 2 OH, Weaker acid, , CH3, , CH 2O, ve charge intensified,, hence conjugate, base unstable, , H, , H, , 152. (c) Due to strong electron-donating effect of the OH, group, the electron density in phenol is much higher, than that in toluene, benzene and chlorobenzene and, hence phenol is readily attacked by the electrophile.
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26, ALDEHYDES, KETONES AND, CARBOXYLIC ACIDS, FACT / DEFINITION TYPE QUESTIONS, 1., , 8., , Choose the correct IUPAC name for, , CH3 C H CHO, |, , CH2 CH 3, , 2., , 3., , 4., , (a) Butan - 2- aldehyde, (b) 2- methylbutanal, (c) 3- methylisobutyraldehyde, (d) 2- ethylpropanal, The IUPAC name of the compound having the molecular, formula Cl3C –CH2CHO is, (a) 3, 3, 3- trichloropropanal, (b) 1, 1, 1- trichloropropanal, (c) 2, 2, 2- trichloropropanal, (d) Chloral, The IUPAC name of CH3COCH(CH3)2 is, (a) 2-methyl-3-butanone (b) 4-methylisopropyl ketone, (c) 3-methyl-2-butanone (d) Isopropylmethyl ketone, IUPAC name of following will be, CHO, CH3, , 5., , 6., , 7., , OH, (a) 4-formyl 3-methyl 1-hydroxy benzene, (b) 4-formyl 3-methyl phenol, (c) 4-hydroxy 2-methyl benzaldehyde, (d) 4-hydroxy 2-methyl carbaldehyde, IUPAC name of ethyl isopropyl ketone is, (a) 4-methyl pent-3-one (b) 2-methyl pent-3-one, (c) 4-methyl pent-2-one (d) 2-methyl pent-2-one, In > C = O group sigma bond is formed by, (a) sp2-p-overlapping, (b) sp3-p-overlapping, (c) sp-p-overlapping, (d) s-p-overlapping, The -bond in carbonyl group is formed by, (a) s-s-overlapping, (b) p-p-overlapping, (c) s-p-overlapping, (d) p-d-overlapping, , 9., 10., 11., , 12., 13., , 14., , 15., , Which of the following is correct for carbonyl compounds?, R, R, + –, – +, (b), (a), C=O, C=O, R, R, –, +, –, (c), (d) +, R–C=O, R–C=O, , R, R, Which of the following contain an aldehyde?, (a) Vanilla beans, (b) Meadow sweet, (c) Cinnamon, (d) All of these, Which of the following have pleasant smell?, (a) Methanal, (b) Propanal, (c) Ethanal, (d) Hexanal, Which one of the following can be oxidised to the, corresponding carbonyl compound?, (a) 2-hydroxy-propane, (b) Ortho-nitrophenol, (c) Phenol, (d) 2-methyl-2 hydroxy-propane, Which one of the following on oxidation gives a ketone ?, (a) Primary alcohol, (b) Secondary alcohol, (c) Tertiary alcohol, (d) All of these, What is formed when a primary alcohol undergoes catalytic, dehydrogenation ?, (a) Aldehyde, (b) Ketone, (c) Alkene, (d) Acid, Primary and secondary alcohols on action of reduced copper, give, (a) Aldehydes and ketones respectively, (b) Ketones and aldehydes respectively, (c) Only aldehydes, (d) Only ketones, Which alkene on ozonolysis gives CH3 CH2CHO and, CH3CCH3, ||, , O, (a), , CH3, , (b) CH3CH2CH = CHCH2CH3, CH3, (c) CH3CH2CH = CH CH3 (d) CH 3 C CHCH 3, CH3CH2CH = C, , |, , CH3
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EBD_7207, ALDEHYDES, KETONES AND CARBOXYLIC ACIDS, , 442, , 16., , 17., , 18., , 19., , 20., , 21., , The catalyst used in Rosenmund's reduction is, (a) HgSO4, (b) Pd/BaSO4, (c) anhydrous AlCl3, (d) anhydrous ZnCl2, C 6 H 5C N + [H], C 6 H 5CHO + NH 3 ., The above reaction is, (a) Mendius reaction, (b) Sandorn's reaction, (c) Rosenmund's reaction (d) Stephen’s reaction, Benzaldehyde can be prepared by oxidation of toluene by, (a) Acidic KMnO4, (b) K2Cr2O7 / H+, (c) CrO2Cl2, (d) All of these, The oxidation of toluene to benzaldehyde by chromyl, chloride is called, (a) Rosenmund reaction (b) Wurtz reaction, (c) Etard reaction, (d) Fittig reaction, An aldehyde group can be present, (a) in between carbon chain, (b) at any position in carbon atom, (c) only at the end of carbon chain, (d) at the second carbon atom of the carbon chain, Benzaldehyde is obtained from Rosenmund’s reduction of, , (a), , (b), , X, , 23., , 26., , 27., , 28., , 29., (d), , Cl, , OH, , Which of the following is not used in the preparation of, ketone?, (a) Oxidation of secondary alcohols, (b) Dehydrogenation of 2° alcohol, (c) Pyrolysis of calcium acetate, (d) Acid hydrolysis of alkyl cyanide, Product of the following reaction is, , 30., 31., , CN, SnCl 2 + HCl, , + 2(H), , +, , ?, , H3O, , COOH, (a), , (c), , 32., , CHO, (b), , O, , NH2, , (d), , C–CH3, , AlCl3, , 33., , A, , CrO3 in (CH3CO)2O, H3O, , +, , B, , (a) acetophenone, (b) benzaldehyde, (c) cyclohexyl carbaldehyde, (d) benzoic acid, The reaction, + CO + HCl, , O, , Cl, (c), , 25., , O, CH3, , Find out B in the given reactions, + CH3 – X, , SnCl 2 / HCl, , O, , 22., , 24., , AlCl3, , CHO, , (a) Rosenmund’s reaction (b) Stephen’s reaction, (c) Cannizzaro’s reaction (d) Gatterman-Koch reaction, Which aldehyde cannot be obtained by Rosenmund’s, reaction?, (a) CH3CHO, (b) HCHO, (c) CH3CH2CHO, (d) All of these, The conversion PhCN PhCOCH 3 , can be achieved, most conveniently by reaction with, (a) CH3MgBr followed by hydrolysis, (b) I2 – NaOH, CH3I, (c) Dil. H2SO4 followed by reaction with CH2N2, (d) LiAlH4 followed by reaction with CH3I, Which of the following is used to prepare ketone from acyl, chloride ?, (a) R-MgX, (b) R2Cd, (c) CO + HCl, (d) CrO3, Which of the following forces explain the boiling point of, aldehydes and ketones?, (a) Hydrogen bonding, (b) van der Waal’s forces, (c) Dipole-dipole attraction(d) None of these, Which is highly soluble in water?, (a) Methanal, (b) Propanal, (c) Propanone, (d) Butanone, Propanal and propanone, both have same molecular, formula(C3H6O), what do you expect about their boiling, points?, (a) Both have same boiling point, (b) Boiling point of propanal is higher than the boiling, point of propanone., (c) Boiling point of propanal is lower than the boiling point, of propanone, (d) Nothing can be predicted, Less reactivity of ketone is due to, (a) + I inductive effect decrease positive charge on, carbonyl carbon atom, (b) steric effect of two bulky alkyl groups, (c) sp2 hybridised carbon atom of carbonyl carbon atom, (d) Both (a) and (b), Acetaldehyde reacts with, (a) Electrophiles only, (b) Nucleophiles only, (c) Free radicals only, (d) Both electrophiles and nucleophiles
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ALDEHYDES, KETONES AND CARBOXYLIC ACIDS, , 34., , 35., , 36., , 37., , 38., , 39., , 40., , 41., , Carbonyl compounds undergo nucleophilic addition, because of, (a) electronegativity difference of carbon and oxygen, atoms, (b) electromeric effect, (c) more stable anion with negative charge on oxygen, atom and less stable carbonium ion, (d) None of the above, Which of the following statement is false ?, (a) Cannizzaro reaction is given by aldehydes in presence, of alkali, (b) Aldol condensation is given by aldehydes in presence, of alkali, (c) Aldol condensation is given by aldehydes and ketones, in presence of acids, (d) None of the above, If formaldehyde and KOH are heated, then we get, (a) methane, (b) methyl alcohol, (c) ethyl formate, (d) acetylene, The reagent which can be used to distinguish acetophenone, from benzophenone is, (a) 2,4- dinitrophenylhydrazine, (b) aqueous solution of NaHSO3, (c) benedict reagent, (d) I2and Na2CO3, Benzaldehyde reacts with ethanoic KCN to give, (a) C6H5CHOHCN, (b) C6H5CHOHCOC6H5, (c) C6H5CHOHCOOH, (d) C6H5CHOHCHOHC6H5, Acetone reacts with iodine (I2) to form iodoform in the, presence of, (a) CaCO3, (b) NaOH, (c) KOH, (d) MgCO3, (CH3)3C–CHO does not undergo aldol condensation due, to, (a) three electron donating methyl groups, (b) cleavage taking place between —C— CHO bond, (c) absence of alpha hydrogen atom in the molecule, (d) bulky (CH3)3C—group, Acetaldehyde reacts with semicarbazide and forms, semicarbazone. Its structure is, (a) CH3CH = NNHCON = CHCH3, (b) CH3CH = NNHCONH2, (c), , 42., , 43., , 44., , CH 3 CH, , 443, , (a) Cu, (b) CuO, (c) Cu2O, (d) Cu(OH)2, 45. Aldol condensation would not occur in :, (a) CH3COCH3, (b) CH3CH2CHO, (c) HCHO, (d) CH3CHO, 46. Cannizzaro reaction occurs with, (a), , |, , (b), , C 6 H 5 CHO, , (c) CH 3 CHO, (d) CH 3 CO CH 3, 47. Which of the following compound will show positive silver, mirror test ?, (a) HCOOH, 48., , 49., , 50., , 51., , 52., , 53., , 54., , N — N — CONH 2, , OH, (d) CH3CH = N—CONHNH2, Iodoform test is not given by, (a) 2-Pentanone, (b) Ethanol, (c) Ethanal, (d) 3-Pentanone, Phenylmethyl ketone can be converted into ethylbenzene, in one step by which of the following reagents?, (a) LiAlH4, (b) Zn-Hg/HCl, (c) NaBH4, (d) CH3MgI, When acetaldehyde is heated with Fehling’s solution it gives, a precipitate of, , CH 3 CH 2 OH, , 55., , 56., , 57., , (b), , CH 3 (CHOH ) 3 CHO, , (c) CH 3CO(CHOH )CH 3 (d) Both (a) and (b), Aldehydes and ketones are distinguished by which of the, following test ?, (a) Lucas test, (b) Tollen’s test, (c) KMnO4 solution (Baeyer’s test), (d) None of these, Aldehydes and ketones are generally reduced by :, (a) Clemmensen reduction (b) H2S, (c) H2/Ni, (d) None of these, In which reaction, > C = O can be reduced to > CH2?, (a) Wolf-Kishner reaction (b) Reimer-Tiemann reaction, (c) Wurtz reaction, (d) None of these, A compound does not react with 2, 4-dinitrophenylhydrazine,, the compound is :, (a) Acetone, (b) Acetaldehdye, (c) CH3OH, (d) CH3CH2COCH3, Which gives lactic acid on hydrolysis after reacting with, HCN ?, (a) HCHO, (b) CH3CHO, (c) C6H5CHO, (d) CH3COCH3, The most appropriate reagent to distinguish between, acetaldehyde and formaldehyde is :, (a) Fehling’s solution, (b) Tollen’s reagent, (c) Schiff’s reagent, (d) Iodine in presence of base, Aldehydes can be oxidised by :, (a) Tollen’s reagent, (b) Fehling solution, (c) Benedict solution, (d) All the above, 2-pentanone and 3-pentanone can be distinguished by :, (a) Cannizaro's reaction (b) Aldol condensation, (c) Iodoform reaction, (d) Clemmensen's reduction, Cross aldol condensation occurs between, (a) two same aldehydes, (b) two same ketones, (c) two different aldehydes and ketones, (d) None of these, Ketone upon treatment with Grignard Reagent gives, (a) primary alcohol, (b) secondary alcohol, (c) tertiary alcohol, (d) aldehyde
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EBD_7207, ALDEHYDES, KETONES AND CARBOXYLIC ACIDS, , 444, , 58., , 59., , 60., , 61., , 62., , When acetaldehyde reacts with alcohol then produce, (a) Acetal, (b) Ketal, (c) Acetone, (d) None, The product formed in Aldol condensation is, (a) a beta-hydroxy aldehyde or a beta-hydroxy ketone, (b) an alpha-hydroxy aldehyde or ketone, (c) an alpha, beta unsaturated ester, (d) a beta-hydroxy acid, Clemmensen reduction of a ketone is carried out in the, presence of which of the following ?, (a) Glycol with KOH, (b) Zn-Hg with HCl, (c) LiAlH4, (d) H2 and Pt as catalyst, Which of the following products is formed when, benzaldehyde is treated with CH3MgBr and the addition, product so obtained is subjected to acid hydrolysis ?, (a) A secondary alcohol (b) A primary alcohol, (c) Phenol, (d) tert-Butyl alcohol, During reduction of aldehydes with hydrazine and, potassium hydroxide, the first step is the formation of, (a) R — CH —, — N — NH 2 (b) R — C N, (c) R — C— NH, (d) R — CH —, — NH, O, A and B in the following reactions are, , R–C–R', , HCN, KCN, , A, , B, , O, , OH, , A = RR'C, , OH, , , B = LiAlH4, , OH, (b) A = RR'C, , 69., , 70., , 71., , COOH, , 72., , R– C, CH2NH2, R', , CN, (a), , 68., , 2, , ||, , 63., , 67., , , B = NH3, , 73., , 74., , CN, (c), , 64., , 65., , , B H 3O, OH, (d) A = RR'CH2CN, B = NaOH, The product obtained by the reaction of an aldehyde and, hydroxylamine is, (a) hydrazone, (b) aldoxime, (c) primary amine, (d) alcohol, Which one gives positive iodoform test ?, (a) (CH 3 ) 2 CHCH 2 OH, (b) C 6 H 5 OH, (c), , 66., , A = RR'C, , CH 3 CH 2, , H, |, C CH 2, |, OH, , CH 3, , (d) CH 3CH 2 OH, The compound that neither forms semicarbazone nor oxime, is, (a) HCHO, (b) CH3COCH2Cl, (c) CH3CHO, (d) CH3CONHCH3, , 75., , 76., , 77., , Schiff’s reagent gives pink colour with, (a) acetaldehyde, (b) acetone, (c) acetic acid, (d) methyl acetate, Benzophenone can be converted into benzene by using, (a) fused alkali, (b) anhydrous AlCl3, (c) sodium amalgam in water, (d) acidified dichromate, In the reaction of NaHSO3 with carbonyl compounds to, form bisulphite product, the nucleophile is, (b) SO3Na, (a) HSO3–, (d) None of the above, (c) SO3– –, Wolf-Kishner reduction is, (a) reduction of carbonyl compound into alcohol, (b) reduction of carbonyl compound into alkene, (c) reduction of carboxyl compound into alkane, (d) reduction of nitro compound into aniline, Tollen’s reagent is, (a) ammonical CuSO4, (b) ammonical AgNO3, (c) alkaline solution containing complex of copper nitrate, (d) none of these, R, OR, Compound of general formula, , C, , are called, , H, OR, (a) diester, (b ) acid anhydride, (c) hemiacetal, (d) acetal, Imine derivatives of aldehyde and ketone is called as, (a) Schiff’s reagent, (b) Fehling’s reagent, (c) Schiff’s base, (d) Schiff’s acid, Which reaction is used for detecting the presence of, carbonyl group?, (a) Reaction with hydrazine, (b) Reaction with phenyl hydrazine, (c) Reaction with hydroxylamine, (d) All of the above, The difference between aldol condensation and, Cannizzaro’s reaction is that:, (a) the former takes place in the presence of -H-atom., (b) the former takes place in the absence of -H-atom., (c) the former takes place in the presence of -H-atom., (d) none of the above, X, , C6H5CH=CHCHO, C6H5CH=CHCH2OH, In the above sequence X can be :, (a) H2/Ni, (b) NaBH4, +, (c) K2Cr2O7/H, (d) Both (a) and (b), Which of the following pairs of compounds will undergo, aldol and Cannizzaro reaction respectively ?, (i) acetone; benzaldehyde, (ii) acetaldehyde; butan–2–one, (iii) propanone; formaldehyde., (iv) cyclopentanone, benzaldehyde, (a) (i), (ii) and (iii), (b) (ii) and (iii), (c) (ii), (iii) and (iv), (d) (iii) and (iv)
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ALDEHYDES, KETONES AND CARBOXYLIC ACIDS, , 78., , 79., , 80., , 81., , Two compounds benzyl alcohol and benzoic acid are formed, from this compound, when this compound is heated in the, presence of conc.NaOH, this compound is., (a) Benzaldehyde, (b) Benzylalcohol, (c) Acetophenone, (d) Benzophenone, The reagent which does not react with both, acetone and, benzaldehyde., (a) Sodium hydrogensulphite, (b) Phenyl hydrazine, (c) Fehling’s solution, (d) Grignard reagent, Which of the following compounds will give butanone on, oxidation with alkaline KMnO4 solution?, (a) Butan-1-ol, (b) Butan-2-ol, (c) Both of these, (d) None of these, Which of the following compounds is most reactive towards, nucleophilic addition reactions?, O, O, (a), , ||, , CH3 – C– H, O, , C–H, , (c), 82., , 83., , 84., , 85., , 86., , 445, , (b), , ||, , CH3 – C– CH3, O, , C–CH3, , (d), , Which of the following does not represent the natural source, of the corresponding acids ?, (a) Formic acid : Red ant, (b) Acetic acid : Vinegar, (c) Butyric acid : Rancid butter, (d) Isobutyric acid : Automobile exhausts, Vinegar is a solution of acetic acid which is :, (a) 15 – 20%, (b) 20 –25%, (c) 6 – 8%, (d) 2 – 4%, Methyl cyanide can be converted into acetic acid by one of, the following reactions., (a) Reduction, (b) Hydrolysis, (c) Electrolysis, (d) Decarboxylation, Toluene can be oxidised to benzoic acid by, (a) KMnO4 (alk.), (b) K2Cr2O7 (alk.), (c) Both (a) and (b), (d) Neither (a) nor (b), Which of the following does the best represent the structure, of the carboxylate ion ?, –, , O, (b) R – C, , R–C, O+, , O–, , +, , H2O, , A, , 373K, , A is, CH2Cl, , (a), , CHCl2, , (b), CH3, , CCl3, , (c), , (d), Cl, , 89., , R, C = O +A, , R, , HCl gas, , R, , dil HCl, , R, , A is, (a) CH3OH, , C, , O, , CH2, , O, , CH2, , CH 2COOH, , (d), CH 2COOH, , CH 2OH, CH2OH, , 90. In the reaction, , CHO, HNO3 /H 2SO4, 273 283K, , A, , Benzaldehyde, , A is, , CHO, , CHO, NO2, , (a), , (b), , NO2, , CHO, (c), , (d) Both (a) and (b), , NO2, 91. Which of the following can not be oxidised to give carboxylic, acid?, CH2CH2CH3, CH3, (b), , (d) None of these, , R–C, , Select the acid(s) which cannot be prepared by Grignard, reagent., (a) Acetic acid, (b) Succinic acid, (c) Formic acid, (d) All of the above, , + H2O, , (b) CH3COOH, , (c), , O+, , 87., , CHO, , Cl2/h, , (a), , O, , (c), , CH3, , –, , O, (a), , 88. In the given reaction, , HC, , (c), , CH3, , CH3, , CH3, , CH3, , (d), , C, , CH3
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EBD_7207, ALDEHYDES, KETONES AND CARBOXYLIC ACIDS, , 446, , 92., 93., , 94., , 95., , Lower carboxylic acids are soluble in water due to, (a) low molecular weight (b) hydrogen bonding, (c) dissociation into ions (d) easy hydrolysis, Dimerisation of carboxylic acids is due to, (a) ionic bond, (b) covalent bond, (c) coordinate bond, (d) intermolecular hydrogen bond, Boiling points of carboxylic acids are, (a) lower than corresponding alcohols, (b) higher than corresponding alcohols, (c) equal to that of corresponding alcohols, (d) None of the above, MgBr, ( i ) CO 2, , P, , ( ii ) H 3O, , In the above reaction product 'P' is, CHO, COOH, (a), , (b), OH, O, , (c), 96., , 97., , 98., 99., , (d) C 6 H 5, , ||, , C C6H5, , In the anion HCOO– the two carbon-oxygen bonds are found, to be of equal length. What is the reason for it?, (a) Electronic orbitals of carbon atom are hybridised, (b) The C=O bond is weaker than the C–C bond, (c) The anion HCOO– has two reasonating structures, (d) The anion is obtained by removal of a proton from the, acid molecule, Carboxylic acids are more acidic than phenol and alcohol, because of, (a) intermolecular hydrogen bonding, (b) formation of dimers, (c) highly acidic hydrogen, (d) resonance stabilization of their conjugate base, Which of the following has the maximum acidic strength ?, (a) o- nitrobenzoic acid (b) m-nitrobenzoic acid, (c) p-nitrobenzoic acid, (d) p-nitrophenol, Which of the following is the weakest acid ?, OH, (a), , (b) CH3COOH, COOH, , (c) HCOOH, , (d), , 100. Which of the following acids has the smallest dissociation, constant ?, (a) CH3CHFCOOH, (b) FCH2CH2COOH, (c) BrCH2CH2COOH, (d) CH3CHBrCOOH, 101. Which one of the following esters is obtained by the, esterification of propan-2-ol with ethanoic acid ?, (a) (CH3)2CHCOOCH3, (b) CH3COOCH2CH3, (c) CH3COOCH(CH3)2, (d) (CH3)2CHCOOCH2CH3, 102. The major product of nitration of benzoic acid is, (a) 3- Nitrobenzoic acid (b) 4- Nitrobenzoic acid, (c) 2- Nitrobenzoic acid, (d) 2, 4- dinitrobenzoic acid, 103. Among the following acids which has the lowest pK a, value?, (a) CH 3 CH 2 COOH, (b) (CH3 ) 2 CH COOH, (d) CH 3COOH, (c) HCOOH, 104. The correct order of increasing acidic strength is, __________, (a) Phenol < Ethanol < Chloroacetic acid < Acetic acid, (b) Ethanol < Phenol < Chloroacetic acid < Acetic acid, (c) Ethanol < Phenol < Acetic acid < Chloroacetic acid, (d) Chloroacetic acid < Acetic acid < Phenol < Ethanol, 105. Which reagent can convert acetic acid into ethanol ?, (a) Na + alcohol, (b) LiAIH4 + ether, (c) H2 + Pt, (d) Sn + HCl, 106. Which is false in case of carboxylic acids?, (a) They are polar molecules, (b) They form H-bonds, (c) They are stronger than mineral acids, (d) They have higher b.p. than corresponding alcohols, 107. The elimination of CO2 from a carboxylic acid is known as, (a) hydration, (b) dehydration, (c) decarboxylation, (d) carboxylation, 108. The reaction of carboxylic acid gives effervescences of, CO2 with NaHCO3. The CO2 comes from., (a) R – COOH, (b) NaHCO3, (c) Both (a) and (b), (d) None of these, 109. Acetic anhydride is obtained by the reaction of, (a) sodium and acetic acid, (b) ammonia and acetic acid, (c) ethanol and acetic acid, (d) P2O5 and acetic acid, 110. Benzoic acid may be converted to ethyl benzoate by reaction, with, (a) sodium ethoxide, (b) ethyl chloride, (c) dry HCl—C2H5OH, (d) ethanol, 111. Propionic acid with Br2 /P yields a dibromo product. Its, structure would be:, Br, |, (a) H– C – CH COOH, (b) CH2Br – CH2 – COBr, 2, |, Br, Br, |, (c) CH3– C – COOH, (d) CH2 Br – CHBr – COOH, |, Br
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ALDEHYDES, KETONES AND CARBOXYLIC ACIDS, , 112. The product obtained when acetic acid is treated with, phosphorus trichloride is, (a) CH3COOPCl3, (b) ClCH2COCl, (c) CH3COCl, (d) ClCH2COOH, 113. The reaction, RCH 2 CH 2 COOH, , Red P, , R, , Br2, , CH 2 CH COOH, |, , Br, , is called as, (a) Reimer- Tiemann reaction, (b) Hell-Volhard Zelinsky reaction, (c) Cannizzaro reaction, (d) Sandmeyer reaction, 114. Benzoic acid reacts with conc. HNO3 and H2SO4 to give :, (a) 3-Nitrobenzoic acid, (b) 4-Benzene sulphonic acid, (c) 4-Nitrobenzoic acid, (d) 2-Nitrobenzoic acid, 115. In the following reaction, RCH 2 COOH, , Br2 / P, , X, , excess NH 3, , CH3COOCH2CH3 + H2O, , (a) removing water, (b) taking ethanol in excess, (c) taking acetic acid in excess, (d) all the above factors, 117. A carboxylic acid can best be converted into acid chloride, by using, (a) PCl5, (b) SOCl2, (c) HCl, (d) ClCOCOCl, 118. Arrange the following four acids in their decreasing order, of acidity, COOH, , COOH, , CH3, , I, , II, , COOH, , COOH, CMe3, , III, (a) I > II > III > IV, (c) II > IV > III > I, , H3C, , 119. The strongest acid among the following is –, (a) Salicylic acid, (b) m-hydroxybenzoic acid, (c) p-hydroxybenzoic acid (d) Benzoic acid, 120. Among the following, the most acidic is :, (a) CH3COOH, (b) ClCH2COOH, (c) Cl2CHCOOH, (d) Cl2CHCH2COOH, 121. Which of the following is the correct decreasing order of, acidic strength of, (i) Methanoic acid, (ii) Ethanoic acid, (iii) Propanoic acid, (iv) Butanoic acid, (a) (i) > (ii) > (iii) > (iv), (b) (ii) > (iii) > (iv) > (i), (c) (i) > (iv) > (iii) > (ii), (d) (iv) > (i) > (iii) > (ii), 122. Among the following the strongest acid is, (a), , CH 3COOH, , (b), , CH 2 ClCH 2COOH, , (c), , CH 2 ClCOOH, , (d), , CH3CH 2COOH, , 123. Arrange the following carboxylic acid in their decreasing, acidity., 1., , COOH, , 2., , COOH, HOOC – CH2 – COOH Malonic acid, , 3., , CH2– COOH, , Y, , The major compounds X and Y are, (a) RCHBrCONH2 ; RCH(NH2)COOH, (b) RCHBrCOOH ; RCH(NH2)COOH, (c) RCH2COBr ; RCH2COONH4, (d) RCHBrCOOH ; RCH2CONH2, 116. The yield of ester in esterification can be increased by, CH3CH2OH + CH3COOH, , 447, , CH3, , IV, (b) IV > III > II > I, (d) III > IV > II > I, , Oxalic acid, , Succinic acid, , CH2– COOH, (a) 3 > 2 > 1, (c) 2 > 3 > 1, , (b) 1 > 2 > 3, (d) 2 > 1 > 3, , STATEMENT TYPE QUESTIONS, 124. Read the following statements and choose the correct option, (i) The carbonyl carbon atom is sp2 -hybridised, (ii) The carbonyl carbon is an electrophilic (Lewis acid), centre, (iii) The carbonyl oxygen is a nucleophilic (Lewis base), centre, (iv) Carbonyl compounds are non- polar in nature., (a) (i), (ii) and (iv) are correct, (b) (i), (ii) and (iii) are correct, (c) (ii), (iii) and (iv) are correct, (d) (ii) and (iv) are correct, 125. Which of the following statement(s) is/are true regarding, preparation of aldehydes and ketones?, (i) Both can be prepared by the oxidation of the concerned, alcohol with copper at about 250ºC., (ii) Both can be prepared by the oxidation of the concerned, alcohol by Oppenauer oxidation., (iii) Both can be prepared by the oxidation of respective, alcohol with acidic dichromate., (a) (i) only, (b) (ii) and (iii), (c) (i) and (iii), (d) All the three
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EBD_7207, ALDEHYDES, KETONES AND CARBOXYLIC ACIDS, , 448, , 126. Which of the following statements are false?, (i) No aldehyde can be prepared by the oxidation of, primary alcohol with acidic KMnO4., (ii) Aldehydes having a boiling point less than 100°C can, be prepared by the oxidation of primary alcohol with, acidic dichromate., (iii) Secondary alcohols on oxidation with PCC in, dichloromethane give carboxylic acids having lesser, number of carbon atoms, (iv) Tertiary alcohols can't be oxidised at all, (a) (ii) and (iii), (b) (ii), (iii) and (iv), (c) (i), (iii) and (iv), (d) (i), (ii) and (iii), 127. Read the following statements and choose the correct option, (i) The boiling points of aldehydes and ketones are lower, than those of alcohols of similar molecular masses, (ii) Alcohols show intermolecular hydrogen bonding, whereas aldehydes and ketones do not show, intermolecular hydrogen bonding., (iii) The lower members of aldehydes and ketones are, miscible with water in all proportions, because they, form hydrogen bond with water., (iv) The solubility of aldehydes and ketones increases, rapidly on increasing the length of alkyl chain, (a) TTFF, (b) TFFT, (c) FTTT, (d) TTTF, 128. Aldehydes are generally more reactive than ketones in, nucleophilic addition reactions. Which of the following, statements accounts for this ?, (i) Sterically, the presence of two relatively large, substituents in ketones hinders the approach of, nucleophile to carbonyl carbon, (ii) Aldehydes show resonance whereas ketones do not, (iii) Electronically, the presence of two alkyl groups reduce, the electrophilicity of the carbonyl carbon more, effectively., (iv) Electronically carbonyl carbon atom in ketones is more, electrophilic than in aldehydes, (a) (i) and (iii), (b) (i) and (iv), (c) (ii) and (iii), (d) (ii) and (iv), 129. 2C6 H5CHO, , OH, , H 2O, , C6 H5CH 2OH, , C6 H5COO, , Which of the following statements are correct regarding, the above reduction of benzaldehyde to benzyl alcohol?, (i) One hydrogen is coming from H2O as H+ and another, from C6H5CHO as H–, (ii) One hydrogen is coming from H2O as H– and another, from C6H5CHO as H+, (iii) One hydrogen from H2O and another from C6H5CHO,, both in the form of H–, (iv) The reduction is an example of disproportionation, reaction, (a) (i), (ii) and (iii), (b) (i) and (iv), (c) (ii), (iii) and (iv), (d) (iii) and (iv), 130. Which of the following statement(s) is/are true regarding, esterification of a carboxylic acid with an alcohol ?, , (i), (ii), (iii), (iv), (a), (c), , It is carried out in presence of a strong acid which acts, as a catalyst., The strong acid makes the carbonyl carbon more, electrophilic, and hence causes the alcohol, a strong, nucleophile to attack on the carbonyl carbon., The strong acid makes the carbonyl group more, electrophilic which is thus attacked easily by an, alcohol, a weak nucleophile., Esterification can be done even in absence of a strong, acid., (i) and (ii), (b) (i) and (iii), (i) only, (d) (iv) only, , MATCHING TYPE QUESTIONS, 131. Match the columns, Column-I, Column-II, (Common names), (IUPAC names), (A) Cinnamaldehyde, (p) Pentanal, (B) Acetophenone, (q) Prop-2-enal, (C) Valeraldehyde, (r) 4-Methylpent-3-en-2-one, (D) Acrolein, (s) 3-Phenylprop-2-enal, (E) Mesityl oxide, (t) 1-Phenylethanone, (a) A – (s), B – (t), C – (p), D – (q), E – (r), (b) A – (p), B – (q), C – (s), D – (t), E – (r), (c) A – (t), B – (s), C – (p), D – (r), E – (q), (d) A – (q), B – (t), C – (r), D – (s), E – (p), 132. Match the columns, Column-I, Column-II, Zn Hg / HCl, , (A) R CO CH 3, , R CH 2, , (B) 2C6 H5 CHO, , CH 3, , NaOH, , (p) Friedel-Craft’s, reaction, (q) Kolbe’s reaction, , C6 H5 COONa C6 H5 CH 2 OH, , (C) C6 H 6, , CH 3 COCl, , Anhyd., , AlCl3, , C6 H5 COCH3, , (D) C 6 H 5 OH CO 2, , NaOH, , (r) Clemmensen’s, reaction, (s) Cannizzaro’s, reaction, , HOC6 H 4COONa, (a) A – (p), B – (q), C – (r), D – (s), (b) A – (q), B – (p), C – (r), D – (s), (c) A – (r), B – (s), C – (p), D – (q), (d) A – (s), B – (r), C – (p), D – (q), 133. Match the columns, Column-I, Column-II, (A) Etard reaction, (p) Alcoholic KOH, (B) Hydroxylation, (q) Anhydrous AlCl3, (C) Dehydrohalogenation (r) Chromyl chloride, (D) Friedel-Crafts reaction (s) Dilute alkaline KMnO4, (a) A – (p), B – (q), C – (r), D – (q), (b) A – (s), B – (r), C – (p), D – (q), (c) A – (r), B – (s), C – (p), D – (q), (d) A – (q), B – (p), C – (s), D – (r)
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ALDEHYDES, KETONES AND CARBOXYLIC ACIDS, , 134. Match the columns, Column-I, (Reactions), (A) Benzophenone, Diphenylmethane, (B) Benzaldehyde, 1-Phenylethanol, (C) Cyclohexanone, Cyclohexanol, (D) Phenyl benzoate, Benzaldehyde, , Column-II, (Reagents), (p) LiAlH4, , R, , (r) Zn(Hg)/Conc HCl, (s) CH3MgBr, , R, , Column-II, (p) Oxime, , R, , C = NOH, , (q) Semicarbazone, , C = N – NH2, , (r), , Imine, , O, , (D), , R, R, , C = N – NH – C – NH 2 (s) Hydrazone, , (a) A – (q), B – (s), C – (p), D – (r), (b) A – (r), B – (p), C– (s), D– (q), (c) A – (r), B – (s), C – (p), D– (q), (d) A – (s), B – (r), C – (q), D– (p), 136. Match the acids given in Column-I with their correct IUPAC, names given in Column-II., Column-I, Column-II, (Acids), (IUPAC names), (A) Phthalic acid, (p) Hexane-1, 6-dioic acid, (B) Oxalic acid, (q) Benzene-1, 2-dicarboxylic acid, (C) Succinic acid, (r) Pentane-1, 5-dioic acid, (D) Adipic acid, (s) Butane-1, 4-dioic acid, (E), (a), (b), (c), (d), , (p) Glutaric acid, , (B), (C), , CH2, , COOH, , CH2, , COOH, , CH2, , COOH, , CH2, , COOH, , (D) CH2, , COOH, , (q) Adipic acid, (r) Succinic acid, , Glutaric acid, (t) Ethane-1, 2-dioic acid, A – (t), B – (q), C – (r), D – (p), E – (s), A – (p), B – (s), C – (t), D – (q), E – (r), A – (q), B – (t), C – (s), D – (p), E – (r), A – (r), B – (t), C – (p), D – (s), E – (q), , (s) Malonic acid, , CH2, CH2, CH2, , R, (C), , (A) CH2, , CH2, , R, (B), , Column-II, , COOH, , (q) DIBAL–H, , R, C = NH, , 137. Match the columns, Column-I, , COOH, , (a) A – (p), B – (s), C – (r), D – (q), (b) A – (q), B – (s), C – (p), D – (r), (c) A – (s), B – (r), C – (q), D – (p), (d) A – (r), B – (s), C – (p), D – (q), 135. Match the columns, Column-I, (A), , 449, , COOH, , (a) A – (q), B – (p), C – (s), D – (r), (b) A – (r), B – (p), C – (s), D – (q), (c) A – (s), B – (r), C – (p), D – (q), (d) A – (r), B – (q), C – (s), D – (p), , ASSERTION-REASON TYPE QUESTIONS, Directions : Each of these questions contain two statements,, Assertion and Reason. Each of these questions also has four, alternative choices, only one of which is the correct answer. You, have to select one of the codes (a), (b), (c) and (d) given below., (a) Assertion is correct, reason is correct; reason is a correct, explanation for assertion., (b) Assertion is correct, reason is correct; reason is not a, correct explanation for assertion, (c) Assertion is correct, reason is incorrect, (d) Assertion is incorrect, reason is correct., 138. Assertion : The boiling points of aldehydes and ketones, are higher than hydrocarbons and ethers of comparable, molecular masses., Reason : There is a weak molecular association in aldehydes, and ketones arising out of the dipole-dipole interactions., 139. Assertion : Formaldehyde is a planar molecule., Reason : It contains sp2 hybridised carbon atom., 140. Assertion : Compounds containing –CHO group are easily, oxidised to corresponding carboxylic acids., Reason : Carboxylic acids can be reduced to alcohols by, treatment with LiAlH4.
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EBD_7207, ALDEHYDES, KETONES AND CARBOXYLIC ACIDS, , 450, , 141. Assertion : The molecular mass of acetic acid in benzene is, 120 instead of 60., Reason : The carboxylic acids exist as cyclic dimers in which, the two molecules of the acid are held together by two, strong hydrogen bonds., , O, , (a), , (b), , OH, , O, , CRITICAL THINKING TYPE QUESTIONS, 142. IUPAC name of the following compound is, , (c), , O, H, , (a) 2-(2-propenyl) butanal, (b) 2-(1-propenyl) butanal, (c) 4-formyl 4-ethyl but-2-ene, (d) 2-ethyl pent-3-en-l-al, 143. Observe the following structures and pick up the correct, statement., +, , C=O, , C = OH, , I, , II, , C = O group and lower alcohols, , have H-bonding., (d) alkanes are held together by weak van der Waal’s, forces (being non-polar), aldehydes and ketones, C = O group and held together by, , contain polar, , strong dipole-dipole attraction and lower alcohols, have H-bonding, which is stronger than dipole-dipole, attraction., 145. Product of the following reaction is, CN, +, , 146. Which of the following reagent reacts in different ways, with CH3CHO, HCHO and C6H5CHO ?, (a) Fehling solution, (b) C6H5NHNH2, (c) Ammonia, (d) HCl, 147. A new carbon - carbon bond is formed in, (i) Aldol condensation (ii) Kolbe’s reaction, (iii) Reimer-Tiemann reaction, (iv) Wurtz Fittig reaction, (a) (i) and (iii), (b) (ii) and (iii), (c) (i), (ii) and (iiv), (d) All the four, 148. Which of the following is an example of nucleophilic, addition ?, NNH 2, , O, , (a) Carbonyl carbon of I is more electrophilic than that, of II, (b) Carbonyl carbon of I is less electrophilic than that, of II, (c) Carbonyl carbon of both structures have equal, electrophilic character, (d) It depends upon the complete structure of the, compound, 144. The boiling points of aldehydes and ketones lie in between, alkanes and alcohols of comparable masses because, (a) alkanes are polar, (b) aldehydes and ketones are non-polar, (c) alkanes are non-polar and aldehydes and ketones, contain polar, , (d), , MgX, dry ether, +, , H3O, , (a), , ||, , C 6 H 5 C CH 3, O, , (b), , ||, , C6 H 5 CCH 3, , ||, , NH 2 NH 2 ,H, , C 6 H 5 C CH 3, OH, , LiAlH 4, , |, , C6 H5CHCH 3, , (c) Both (a) and (b), (d) None of the two, 149. Acetal formation is a reversible reaction, R, H, , C = O + R'OH, , H, , +, , R, H, , OH R'OH, H +, , C, , OR', R, , OR', , + H2O, H, OR', Under what conditions, the reaction can be forced to proceed, only in right (forward) direction ?, (a) Using excess of alcohol, (b) Using high temperature, (c) Using dilute acid and excess of alcohol, (d) Using dry acid and excess of alcohol, 150. In the crossed Cannizzaro reaction involving HCHO as one, of the components, (a) HCHO is always oxidised because of electronic effect, (b) HCHO is always oxidised because of steric effect, (c) both of the above statements are true, (d) none of the above statement is true, 151. Which of the following acts as a nucleophile in the aldol, condensation of ethanal?, , (i) OH–, (iii) –CH2CHO, (a) Only (i), (c) (i) and (iii), , (ii), , C, , H2O, , (b) (i) and (ii), (d) All the three
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ALDEHYDES, KETONES AND CARBOXYLIC ACIDS, , 451, , 152. Which of the following acts as a nucleophile in the, Cannizzaro reaction involving benzaldehyde ?, (i), , OH–, , (ii), , 154., , 155., , 156., , 157., , C6H4CHO, , (a), , (iv) H2O, (a) (i) and (iv), (b) (i) and (ii), (c) (i) and (iii), (d) Only (i), Which of the following undergoes haloform reaction ?, (i) CH3CH2COCH2Cl, (ii) C6H5COCH3, (iii) C6H5COCHCl2, (iv) CH3CH2COCCl3, (a) Only (ii), (b) (ii) and (iv), (c) (i), (ii) and (iv), (d) All the four, When ethanal reacts with propanal in the presence of a, base, the number of products formed is, (a) 2, (b) 3, (c) 4, (d) 5, Aldehydes and ketones will not form crystalline derivatives, with, (a) sodium bisulphite, (b) phenylhydrazine, (c) semicarbazide hydrochloride, (d) dihydrogen sodium phosphate., Which of the following compound will undergo self aldol, condensation in the presence of cold dilute alkali ?, (a) CH2 CH CHO, (b) CH C CHO, (c) C 6 H 5 CHO, (d) CH 3 CH 2 CHO ., Which of the following is an example of aldol condensation?, (a) 2CH3COCH3, (b) 2HCHO, , dil NaOH, , dil NaOH, , (c) C6H5CHO + HCHO, (d) None of the above, 158. Identify X,, H3C, H3C, (a), , C=O, , CH3MgI, dry ether, , CH3OH, , CH 3, |, CH 3 C(OH)CH 2 COCH 3, , CH3OH, dil NaOH, , C6H5CH2OH, , Intermediate, , H 2O, , X, , (b) Ethyl alcohol, , (c) Methyl cyanide, (d) tert-Butyl alcohol, 159. An organic compound of formula, C3H6O forms phenyl, hydrazone, but gives negative Tollen’s test. The compound is, (a) CH3CH2COCH3, (b) CH3CH2CHO, (c) CH3COCH3, (d) Both (a) and (c), 160., , R, , NH, , CHO, , –, , (iii) C6H5CH(OH)O–, , 153., , 161. Cannizaro’s reaction is not given by _____________, , Hydrolysis, , 3, C = O HCN (A), (C), (B), R, Compound (C) in above reaction is, (a), -hydroxy acid, (b) -amino acid, (c), -amino alkanol, (d) -amino -hydroxy acid, , CH3, , (b), , CHO, , (c) HCHO, (d) CH3CHO, 162. Benzophenone can be obtained by_____________., (i) Benzoyl chloride + Benzene + AlCl3, (ii) Benzoyl chloride + Diphenyl cadmium, (iii) Benzoyl chloride + Phenyl magnesium chloride, (iv) Benzene + Carbon monoxide + ZnCl2, (a) (i), (ii) and (iii), (b) (ii) and (iii), (c) (iii) and (iv), (d) (i), (ii) and (iv), 163. Which of the following conversions can be carried out by, Clemmensen Reduction ?, (i) Benzaldehyde into benzyl alcohol, (ii) Cyclohexanone into cyclohexane, (iii) Benzoyl chloride into benzaldehyde, (iv) Benzophenone into diphenyl methane, (a) (ii) and (iv), (b) (i) and (iv), (c) (i) and (iii), (d) (iii) and (iv), 164. Benzaldehyde is less reactive than propanal because, (i) the carbon atom of the carbonyl group of benzaldehyde, is less electrophilic as in propanal., (ii) the carbon atom of the carbonyl group of benzaldehyde, is more electrophilic as in propanal., (iii) carbonyl group in benzaldehyde is more polar due to, resonance, (iv) carbonyl group in benzaldehyde is less polar due to, resonance, (a) (i) and (iii), (b) (i) and (iv), (c) (i) only, (d) (iv) only, 165. Addition of hydrogen cyanide to aldehydes and ketones, occurs in presence of a base.The role of base is to, (i) catalyse the reaction, (ii) generate CN– ion, (iii) slow down the reaction, (iv) to stabilize the cyanohydrins, (a) (i) and (iii), (b) (i) and (ii), (c) (i) and (iv), (d) (ii) and (iv), 166. Addition of alcohols to aldehydes and ketones takes place, in presence of dry HCl gas because it, (i) Protonates the oxygen of the carbonyl compounds, (ii) Increases the electrophilicity of the carbonyl carbon, (iii) Removes the excess moisture from the reaction, (iv) Helps the reaction to move in the forward direction, (a) (i), (ii) and (iv), (b) (i), (ii), (iii) and (iv), (c) (ii) ,(iii), and (iv), (d) (i), (iii) and (iv), 167. When benzaldehyde and acetaldehyde undergoes reaction, with the 2, 4–DNP ?, (a) Benzaldehyde reacts slowly than acetaldehyde, (b) Acetaldehyde reacts slowly than benzaldehyde, (c) Both reacts equally, (d) Both do not react with 2, 4-DNP
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EBD_7207, ALDEHYDES, KETONES AND CARBOXYLIC ACIDS, , 452, , 168. Suppose the reaction of compound containing ketone as, functional group is carried in basic medium of NaOH. Which, of the following will one use to protect the unwanted, reaction due presence of carbonyl moiety., (a) NaHSO3, (b) HCN, (c) ethylene glycol and HCl, (d) None of these, 169. A compound C5H10O forms orange–red precipitate upon, reaction with 2,4–DNP, but does not give positive Tollen’s, test and iodoform test. Possible compound is, (a) 2, 2–dimethylpropanal (b) 3–methylbutan–2–one, (c) Pentan–3–one, (d) None of the above, 170. Nitration of the compound is carried out, this compound, gives red–orange ppt. with 2,4–DNP, this compound, undergoes Cannizzaro reaction but not aldol, than possible, product due to nitration is, (a) 3–nitroacetophenone, (b) (2–nitro)–2–phenylethanal, (c) (2–nitro)–1–phenylpropan–2–one, (d) 3–nitrobezaldehyde, 171. Structure of the compound whose IUPAC name is 3-ethyl2-hydroxy-4-methylhex-3-en-5-ynoic acid is :, OH, , OH, COOH, , (a), , COOH, , (b), OH, , COOH, , (c), , (d), , COOH, , OH, , 172. The end product B in the sequence of reactions,, R X, , CN, , A, , NaOH, , B is, , (a) an alkane, (b) a carboxylic acid, (c) sodium salt of carboxylic acid, (d) a ketone, 173. Which is the most suitable reagent for the following, conversion?, O, , CH3 – CH, , O, CH3 – CH, , ||, , CH – CH 2 – C– OH, , (b) Benzoyl peroxide, (a) Tollen’s reagent, (c) I2 and NaOH solution (d) Sn and NaOH solution, 174. In the given reaction,, (C 6 H 5 CO) 2 O, , C 6 H 5 COOCOCH 3, , I, H 2O, , (i), , C2 H5Cl, , (ii), , CH3Cl, , II, , (i) KCN, (ii) H3O, , (i) AgCN, (ii) H3O, , (iii) CH 3CH, , CH 2, , (iv) CH3Br, , (i) Mg, (ii) CO2, (iii) H3O, , KMnO 4 / OH, heat, , (b) (i) and (ii), (a) (iii) and (iv), (c) (ii) and (iii), (d) (i) and (iv), 176. Primary alcohols can be readily oxidised to carboxylic acids, by., (i) KMnO4 in neutral medium., (ii) KMnO4 in acidic or alkaline medium., (iii) K2Cr2O7 in alkaline medium., (iv) K2Cr2O7 in acidic medium., (a) (i), (ii) and (iv), (b) (i), (ii) and (iii), (c) (ii) and (iii), (d) (i) and (iii), 177. Which of the following is correct order of acidity?, (a) HCOOH > CH3COOH > ClCH2COOH > C2H5 COOH, (b) ClCH2COOH > HCOOH > CH3COOH > C2H5COOH, (c) CH3COOH > HCOOH > ClCH2COOH > C2H5COOH, (d) C2H5COOH > CH3COOH > HCOOH > ClCH2COOH, 178. An organic compound A upon reacting with NH3 gives B., On heating B gives C. C in presence of KOH reacts with Br2, to given CH3CH2NH2. A is :, (a) CH3COOH, (b) CH3CH2CH2COOH, (c), , ||, , CH – CH 2 – C– CH 3, , H 2O, , Identify the product(s) formed in the given reaction., I, II, (a) 2 molecules of benzoic acid 2 molecules of ethanoic acid, (b) 2 molecules of benzoic acid 1 molecules of benzoic acid, and 1 molecule of ethanoic acid, (c) 1 molecule of ethanoic acid 1 molecule of benzoic acid, (d) 1 molecule of benzoic acid 1 molecule of butanoic acid, 175. Ethanoic acid can’t be obtained by which of the following, reaction ?, , CH3 CH COOH, |, CH3, , (d) CH3CH2COOH, , 179. The correct order of increasing acid strength of the, compounds, (A) CH3CO2H, (B) MeOCH2CO2H, (C) CF3CO2H, (a) D < A < B < C, (c) B < D < A < C, , Me, , CO2H is, Me, (b) A < D < B < C, (d) D < A < C < B, , (D)
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EBD_7207, ALDEHYDES, KETONES AND CARBOXYLIC ACIDS, , 454, , FACT / DEFINITION TYPE QUESTIONS, , H, , 15., 1., , (b), , 2, , (a), , 1, , CH3 – CH – CHO, , (a), , 16., O CH3, , 1, , 2 ||, , 3|, , 4, , 3., , (c), , 4., 8., 9., , (c), 5. (b), 6. (a), 7. (b), (b) O is more electronegative than C., (d) Vanillin -vanilla beans, Salicylaldehyde - meadow sweet, Cinnamaldehyde -from cinnamon., (d) The lower aldehydes have sharp pungent odours. As, the size of the molecule increases, the odour becomes, less pungent and more fragrant., (a) Carbonyl compounds (aldehydes and ketones) are, obtained by the oxidation of 1° and 2° alcohols, respectively. Among the given options, only (a) is 2°, alcohol hence it can be oxidized to ketone., , 10., , 11., , CH 3, , C, , C H CH 3 ; 3- methyl-2-butanone, , OH, |, H3CCHCH3, , oxidation, , 2 hydroxypropane, , 12., , R, C=O, , R, , R, , Isopropyl, alcohol, , 13., , Ketone, , (a) 1° Alcohols on catalytic dehydrogenation give, aldehydes., RCH2OH, 1° alcohol, , 14., , Cu, 300°C, , RCHO + H2, Aldehyde, , (a) Alcohols are oxidized by removal of H2 in presence, of a heated metal catalyst (Cu), Cu, 300, , CH3CH 2OH, 1 alcohol, , H3 C, CH, H3 C, , OH, , 2° Alcohol, , 17., , O, , C 6 H 5C N, , C 6 H 5CH, 18., , Ketone, , (–H2O), , NH, , HCl, , H 2O / H, , C 6 H 5CHO NH 3, , (c), CH3, acidic K 2 Cr2 O7, or KMnO 4, , CHO, CrO 2 Cl2, , (Etard reaction), , 19., 24., 26., 27., , (c), (a), (b), (a), , 28., , (b), , 29., , (c), , 30., 31., 32., 33., , Aldehyde, , CH3CCH3 H2, ||, O, , CH3, , SnCl 2, , 2[H], , COOH, , CH3CHO H 2, , Cu, 300, , O, , CH3, , presence of SnCl2 / HCl reagent. This reaction is, known as Stephen’s reaction., , Acetone, , [O], , CHOH, , CH3– CH 2– C– C, , (b) Catalyst used in Rosenmund reduction is Pd/BaSO4., Rosenmund reduction is used for reduction of acid, chloride., O, O, Pd / BaSO4, R C Cl, R C H, (d) Phenyl cyanide is reduced into benzaldehyde in the, , O, ||, H3CCCH 3, , (b) Secondary alcohols on oxidation give ketones., Note : Primary alcohols form aldehydes., R, , O3, , CH3– C – CH 3+ CH3 – CH2 – CHO, , CH2 – CH3, , 2., , H O, , CH3, , C, , CH3, O, , O, , 4, , 3, , CH3– CH2– C, , 34., , Acidic KMnO4 and K2Cr 2O7 oxidise toluene to, benzoic acid but CrO2Cl2 oxidises it to benzaldehyde., 20. (c), 21. (b), 22. (d), 23. (b), 25. (d), Formyl chloride is unstable at room temperature., Alkanenitriles (other than methanenitrile) and, benzonitrile give ketones with Grignard reagents., C=O, , + –, C – O ; the polarity exists in, , carbonyl group due to resonance., (a) Solubility decreases with increase in mol. wt., (c) Propanone has symmetrical structure., (d), (b) Acetaldehyde reacts only with nucleophiles. Since the, mobile p electrons of carbon–oxygen double bond are, strongly pulled towards oxygen, carbonyl carbon is, electron-deficient and carbonyl oxygen is electron-rich., The electron deficient (acidic) carbonyl carbon is most, susceptible to attack by electron rich nucleophilic, reagents, that is, by base. Hence the typical reaction, of aldehydes and ketones is nucleophilic addition., (c)
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EBD_7207, ALDEHYDES, KETONES AND CARBOXYLIC ACIDS, , 458, , 111. (c) This reaction is an example of Hell - Volhard Zelinsky, reaction. In this reaction acids containing – H on, treatment with X2 /P give di-halo substituted acid., Br2 /P, , CH 3 – CH 2 COOH, , 112. (c), , CH 3, , CBr2, , COOH, , CH 3 COCl +, , 3CH 3COOH PCl3, , Acetyl, Chloride, , H3PO3, Phosphorous, acid, , 113. (b), 114. (a) –COOH group when attached to benzene ring, deactivates the ring and substitution occurs at, m-position. (HNO3 + H2SO4) is a source of + NO2, (electrophile) which attacks at m-position., , COOH, , COOH, , + HNO3, , H2SO4, , NO2, , 3- nitrobenzoic acid, , 115. (b), , RCH 2 COOH, , Br2 / P, HVZ reaction, , R CH COOH, |, , Br, 'X', , NH3, (Excess), , R CH COOH, |, NH 2, 'Y', , 116. (d) The yield of product in a reversible reaction can be, increased by (i) removing one of the products, (ii), taking either of the reactant in excess., 117. (d) Use of SOCl2 and ClCOCOCl forms gaseous byproducts which can be easily removed, giving better, yield of RCOCl. Further, oxalyl chaloride is particularly, easy to use becasue any excess of it can be easily, evaporated due to its low b.p. (62ºC), O O, O, ||, , ||, , ||, , R C OH Cl C C Cl, O, ||, , R C Cl HCl, , CO, CO 2, 118. (b), 119. (a) Salicylic acid, because it stabilizes the corresponding, salicylate ion by intramolecular H-bonding., , 120. (c), , 122. (c) Chlorine is electron withdrawing group. Further, inductive effect is stronger at, position than, -position. i.e.,, , Cl 2CHCOOH is most acidic because it has two, chlorine at -position., 121. (a) An electron releasing substituent (+I) intensify the, negative charge on the anion resulting in the decrease, of stability and thus decreases the acidity of the acid., Hence acid character decreases as the + I-effect of the, alkyl group increases as, CH3– < CH3CH2– < CH3CH2CH2– < CH3CH2CH2CH2–, Hence the order becomes : (i) > (ii) > (iii) > (iv), , CH 2ClCOOH CH 2 ClCH 2 COOH, 123. (b), , STATEMENT TYPE QUESTIONS, 124. (b) Carbonyl compounds have substantial dipole moments, and are polar in nature. The high polarity of the, carbonyl group is due to resonance., 125. (a) Primary alcohols on oxidation give carboxylic acids as, the final product, of course through aldehydes., Oppenauer oxidation involves oxidation of 2º alcohols, to ketones, and not for the oxidation of 1º alcohols., 126. (c) If the aldehyde has a boiling point less than 100°C, it, can be prepared by the oxidation of 1° alcohols with, regular oxidising agents like acidic permanganate or, dichromate. Since the aldehyde has a lower boiling, point than the alcohol, it is distilled off as soon as it is, formed ; so further oxidation to a carboxylic acid is, minimized., 127. (d) The solubility of aldehydes and ketones decreases, rapidly on increasing the length of alkyl chain., 128. (a) Aldehydes are generally more reactive than ketones, in nucleophilic addition reactions due to steric and, electronic reasons. Sterically, the presence of two, relatively large substituents in ketones hinders the, approach of nucleophile to carbonyl carbon than in, aldehydes having only one such substituent., Electronically, aldehydes are more reactive than, ketones because two alkyl groups reduce the, electrophilicity of the carbonyl carbon more effectively, than in former., 129. (b) The hydrogen atom that is added to the carbonyl, carbon of the aldehyde in the reduction is derived, directly from the other aldehyde molecule as a hydride, ion. The second hydrogen that is added to the, negatively charged oxygen is coming from the solvent, (consult mechanism of Cannizzaro reaction). Oxidation, of one molecule of the compound at the expense of, other molecule of the same compound is known as, disproportionation., 130. (b) First two steps of the esterification make the question, clear, +, , O, CH3 — C — OH, , OH, H, , +, , CH3 — C — OH, , R–O–H, , Protonated acid, (carbonyl C is more, electrophilic than, that of parent acid), , OH, CH — C — OH, Protonated acid, , R–O–H, , CH3 — C — OH, +, , HO R
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ALDEHYDES, KETONES AND CARBOXYLIC ACIDS, , 459, , MATCHING TYPE QUESTIONS, 131. (a), 136. (c), , 132. (c), 137. (c), , 133. (b), , 148. (c) (a) C = O + H2NNH2, , 134. (d), , 138. (a), 139. (a), 140. (b), 141. (a) The molecular mass of acetic acid in benzene is 120, instead of 60 because the carboxylic acids exists as, cyclic dimers in which two molecules of the acid are, held together by two strong hydrogen bond., , CRITICAL THINKING TYPE QUESTIONS, 142. (d), 143. (b) In structure II, presence of positive charge on oxygen, causes the displacement of electrons toward oxygen,, making carbon more electron deficient than that in, unprotonated carbonyl group., 144. (c) It is the reason for the given fact., 145. (d), 146. (c) With ammonia, HCHO forms hexamethylenetetramine,, CH3 CHO gives acetaldehydeammonia addition, product, while C6H5CHO gives hydrobenzamide., 147. (d) Aldol condensation :, , 2CH3CHO, , OH, , –, , 149. (d) Being reversible reaction, the backward reaction i.e., acetal -hemiacetal step can be restricted by minimizing, water content, i.e. by using dry HCl. The step hemiacetal, - aldehyde can be restricted by using excess of alcohol., 150. (c) First step in Cannizzaro reaction is the nucleophilic, addition of OH– on the carbonyl carbon., , H, , |, , H C, , O OH, , Easier because of electronic, and steric effects, , New C–C bond, , R–C–O, , –, , Ist, step, , H, , H, , CO, CO2,140ºC, ,140ºC, Pressure, Pressure, , H–C–O, , –, , H, –, , R–C=O, 2nd Step, (slow), , H–C=O +R–C –O, , OH, OH, H, I, 151. (c) OH– and –CH2CHO act as nucleophile in the first two, steps., , OH, CHCl3, , O, , 3KOH, , CH3 CHO, , OH, , OH, (fast), , OH, Higher the electron deficiency on cabonyl carbon, more, easier will be the attack of the nucleophile, (OH–) on its carbon. Futher, the attack of OH– on the, carbonyl carbon is more easy in case of HCHO because, its carbon is least hindered having two hydrogens, (steric effect). Thus the intermediate I is formed very, easily which donates hydride ion to another aldehyde, and thus itself oxidised., H, , Kolbe reaction :, , Reimer-Tiemann reaction :, , H, , –, , R–C=O, , H, , COOH, , OH, , C H 2 CHO, , CH 3CHO, , CH 2CHO, , OH, , CHO, , |, , H 2O, , + 3KCl + 2H2O, , 3KOH, , |, , CH 3 C H, |, , New C–C bond, , CHCl, , C = NNH2, , NHNH2, , New C–C bond, , OH, , +, , H, , C – OH, , (b) In the reduction of carbonyl group with LiAlH4 or, NaBH4, a hydride ion is transferred from the metal, to the carbonyl carbon (nucleophilic addition), H, –, –, C = O + H– AlH3, C – OAlH3, , H3C–C–CH2CHO, , ONa, ONa, , +, , 135. (b), , ASSERTION-REASON TYPE QUESTIONS, , OH, , H, , (as an acid), , Wurtz Fittig reaction :, , CH3 CH, |, , CH2CHO, , Na, , H3CCl + 2Na + Cl, , 152. (c), H3C, , H, , + 2NaCl, C6 H5CHO, New C–C bond, , –, , C6H5 – C – O + C6H5 – C = O, H, , OH
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EBD_7207, 462, , 186. (d) V is most stable because its anion is stabilized to a, greater extent through H – bonding with H atom of OH, present on both ortho-positions ; followed by II in, which one OH group is present. Compound IV comes, next to II because here –OCH3 group is present in ortho, position which although is not capable of forming, H–bonding yet more acidic than p-HOC6H4COOH (III), due to ortho effect. Compound III is less acidic than, benzoic acid because of electron-releasing group in, the para position. Thus, , ALDEHYDES, KETONES AND CARBOXYLIC ACIDS, , COOH, , OH, , HO, , COOH, , V, , >, , II, , COOH, OCH3, , IV, , OH, , >, , COOH, , >, , COOH, , >, I, , OH, III
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27, AMINES, FACT / DEFINITION TYPE QUESTIONS, 1., , 2., , 3., , 4., , A secondary amine is, (a) a compound with two carbon atoms and an –NH2, group., (b) a compound containing two –NH2 groups., (c) a compound in which hydrogens of NH3 have been, replaced by two alkyl groups., (d) a compound with an –NH2 group on carbon atom in, number two position., The general formula of quaternary ammonium compound, is, (a) R–NH2, (b) R3N, +, –, (c) R4N X, (d) NH4X, The total number of electrons around the nitrogen atom in, amines are, (a) 8, (b) 7, (c) 4, (d) 3, The IUPAC name of the compound having formula,, O, , C, , CH, , CH2 is, , OH, , NH2, , OH, , 7., , The correct IUPAC name for CH2 = CHCH2NHCH3 is, (a) Allylmethylamine, (b) 2-amino-4-pentene, (c) 4-aminopent-1-ene, (d) N-methylprop-2-en-1-amine, 8., Amines play an important role in the survival of life., Naturally they are found in, (a) proteins, (b) vitamins, (c) alkaloids, (d) All of these, 9., Intermediates formed during reaction of RCONH2 with Br 2, and KOH are, (a) RCONHBr and RNCO (b) RNHCOBr and RNCO, (c) RNHBr and RCONHBr (d) RCONBr2, 10. Which of the following reactions will not give a primary, amine?, , 11., , 5., , 6., , (a) 3-amino-hydroxy propine acid, (b) 2-amino-propan-3-oic acid, (c) amino hydroxy propanoic acid, (d) 2-amino-3-hydroxy propanoic acid, The number of primary amines of formula C4H11N is :, (a) 1, (b) 3, (c) 4, (d) 2, What is the IUPAC name of the following compound ?, NH 2, , (a), (b), (c), (d), , 2-methyl-4-hexanamine, 5-methyl-3-hexanamine, 2-methyl-4-amino hexane, 5-methyl-3-amino hexane, , Br2 / KOH, , (a), , CH 3 CONH 2, , (b), , CH 3CN, , LiAlH 4, , (c), , CH 3 NC, , LiAlH 4, , (d), , CH 3CONH 2, , LiAlH 4, , Propionamide on Hofmann degradation gives –, (a) methyl amine, (b) ethyl amine, (c) propyl amine, (d) ethyl cyanide, 12. Secondary amines could be prepared by, (a) reduction of nitriles, (b) Hofmann bromamide reaction, (c) reduction of amides, (d) reduction of isonitriles, 13. Gabriel’s phthalimide synthesis is used for the preparation, of, (a) Primary aromatic amines, (b) Secondary amines, (c) Primary aliphatic amines, (d) Tertiary amines, 14. Ethyl amine can be obtained by the, (a) Action of NH3 on ethyl iodide., (b) Action of NH3 on ethyl alcohol., (c) Both (a) and (b), (d) Neither (a) nor (b)
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EBD_7207, AMINES, , 464, , 15., , 16., , 17., , 18., , 19., , 20., , 21., , 22., , 23., , Treatment of ammonia with excess of ethyl iodide will yield, (a) diethylamine, (b) ethylamine, (c) triethylamine, (d) tetraethylammonium iodide, For alkylation of ammonia which of the following is not, used ?, (a) CH3–X, (b) CH3–CH2–X, (c) (CH3)2CH–X, (d) (CH3)3C–X, Which of the following amines can be prepared by Gabriel, method ?, (i) CH3CH2NH2, (ii) (CH3)2CHNH2, (iii) (CH3)3CNH2, (iv) C6H5NH2, (a) (i) and (iii), (b) (ii) and (iv), (c) (i), (ii) and (iii), (d) (i) and (ii), Amongst the given set of reactants, the most appropriate, for preparing 2° amine is ___________., (a) 2°R–Br + NH3, (b) 2°R–Br + NaCN followed by H2/Pt, (c) 1°R–NH2 + RCHO followed by H2/Pt, (d) 1°R–Br (2 mol) + Potassium phthalimide followed by, H3O+/heat, The best reagent for converting 2 – phenylpropanamide, into 2-phenylpropanamine is ___________., (a) excess H2, (b) Br2in aqueous NaOH, (c) iodine in the presence of red phosphorus, (d) LiAlH4in ether, Which of the following methods of preparation of amines, will give same number of carbon atoms in the chain of, amines as in the reactant?, (a) Reaction of nitrite with LiAlH4., (b) Reaction of amide with LiAlH4 followed by treatment, with water., (c) Heating alkylhalide with potassium salt of phthalimide, followed by hydrolysis., (d) Treatment of amide with bromine in aquesous solution, of sodium hydroxide., The reduction of nitro compounds is most preferred in the, presence of, (a) Pd/H2 in ethanol, (b) Sn + HCl, (c) finely divided Ni, (d) iron scrap and HCl., An alkyl or benzyl halide on reaction with an ethanolic, solution of ammonia undergoes, (a) electrophilic substitution reaction, (b) nucleophilic substitution reaction., (c) free radical mechanism., (d) nucleophilic addition reaction., In the ammonolysis of alkyl halides the halogen atom is, replaced by an amino(–NH2) group which of the following, represent the correct order of reactivity of halides with, amines., (a) RBr > RI > RCl, (b) RI > RCl > RBr, (c) RI > RBr > RCl, (d) RCl > RBr > RI, , 24., , 25., , 26., , 27., , Which of the following will give primary amine only ?, (i) ammonia + propylchloride, (ii) potassium pthalimide + ethylchloride, (iii) potassium pthalimide + chlorobenzene, (a) (i) and (ii), (b) (i) and (iii), (c) (ii) and (iii), (d) (i), (ii) and (iii), Amines have, (a) Garlic odour, (b) Fishy odour, (c) Jasmine odour, (d) Bitter almonds odour, Aniline is less soluble in water than ethyl amine due to, (a) resonance stablization of benzene ring, (b) resonance stabilization of anilium ion, (c) more hydrophobic nature of C6H5 group than C2H5, group, (d) more hydrophobic nature of C6H5 group than C2H5, group, Which of the following should be most volatile?, (I) CH3CH2CH2NH2, (II) (CH3)3N, (III), , 28., , 29., , 30., , 31., , 32., , 33., , 34., , CH3CH2, , CH3, , NH, , (IV) CH3CH2CH3, , (a) II, (b) IV, (c) I, (d) III, Amines behave as, (a) lewis acids, (b) lewis bases, (c) aprotic acids, (d) amphoteric compounds, The basic character of amines is due to, (a) presence of nitrogen atom, (b) lone pair of electrons on nitrogen atom, (c) tetrahedral structure, (d) high electronegativity of nitrogen, Aliphatic amines are.....basic than NH3 but aromatic amines, are......basic than NH3., (a) more, less, (b) less, more, (c) both (a) and (b), (d) None of these, Substitution of one alkyl group by replacing hydrogen of, primary amines, (a) increases the base strength, (b) decreases the base strength, (c) remains the same, (d) None of the above, Which of the following is not characteristic of amines?, (a) They smell like ammonia, (b) They are inflammable in air, (c) They show the property of hydrogen bonding, (d) They are amphoteric in nature, The correct order of basicity in amines, (i) C2H5NH2, (ii) CH3NH2, (iii) (CH3)2NH, (iv) (CH3)3N, (a) (i) < (iv) < (ii) < (iii), (b) (iv) < (ii) < (iii) < (i), (c) (i) < (ii) < (iii) < (iv), (d) (ii) < (iii) < (iv) < (i), The conjugate base of (CH3)2NH+2 is, (a) (CH3)2NH, (b) (CH3)2N+, +, (c) (CH3)3N, (d) (CH3)2N–
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AMINES, , 35., , 36., , 465, , High basicity of Me2 NH relative to Me3N is attributed to:, (a) effect of solvent, (b) inductive effect of Me, (c) shape of Me2NH, (d) shape of Me3N, The correct order of basicity of the following compounds, , NH 2, , NH 2, , B, , A, , 37., , 38., , NH 2, , C, , NO 2, (a) B > A > C, (b) A > B > C, (c) C > A > B, (d) C > B > A, Which of the following statement is correct ?, (a) Ammonia is more basic than methylamine., (b) Methylamine is more basic than ammonia., (c) Dimethylamine is less basic than methylamine., (d) Dimethylamine is less basic than trimethylamine., Which of the following compounds is most basic?, (a), , O2N, , NH2, , (b), , CH2NH2, , (c), , N – COCH3, H, , NH2, , (d), 39., , Which of the following compounds is the weakest Bro nsted, base?, NH2, NH, 2, , (a), , (b), , OH, (c), 40., , 41., , OH, (d), , The correct decreasing order of basic strength of the, following species is __________ H2O, NH3, OH–, NH2–, (a) NH2– > OH– > NH3 > H2O, (b) OH– > NH2– > H2O > NH3, (c) NH3 > H2O > NH2– > OH–, (d) H2O > NH3 > OH–> NH2–, Which of the following factors affect the basic strength of, amine?, (i) Inductive effect, (ii) Steric hinderance, (iii) Solvation effect, (iv) Solubility in organic solvents., (a) (i) and (iv), (b) (i), (ii) and (iii), (c) (ii) and (iii), (d) (ii) and (iv), , 42. Which statement is not true among the following?, (a) Amines are bases, (b) They turn red litmus blue, (c) Trimethyl amine is less basic than dimethyl amine, (d) Amines yield alcohols on aqueous hydrolysis., 43. Aniline is used, (a) in crimping of wool, (b) in dyeing industry, (c) in making of glue, (d) in fast drying vanish, 44. Which of the following statements about primary amines is, ‘False’ ?, (a) Alkyl amines are stronger bases than aryl amines, (b) Alkyl amines react with nitrous acid to produce, alcohols, (c) Aryl amines react with nitrous acid to produce phenols, (d) Alkyl amines are stronger bases than ammonia, 45. Mark the correct statement, (a) Methylamine is slightly acidic, (b) Methylamine is less basic than ammonia, (c) Methylamine is a stronger base than ammonia, (d) Methylamine forms salts with alkalies., 46. For carbylamine reaction, we need hot alcoholic KOH and, (a) any primary amine and chloroform, (b) chloroform and silver powder, (c) a primary amine and an alkyl halide, (d) a monoalkylamine and trichloromethane., 47. The compound obtained by heating a mixture of a primary, amine and chloroform with ethanolic potassium hydroxide, (KOH) is, (a) an alkyl cyanide, (b) a nitro compound, (c) an alkyl isocyanide, (d) an amide, 48., , R NH 2 CH3COCl, , A, , The product (A) will be –, (a) RNHCOCH3, , (b) RN(COCH3)2, , (excess), , (c) RN(COCH3 )3 Cl, (d) R – CONH2, 49. Carbylamine reaction is used for the detection of, (a) aliphatic 2° amines, (b) aliphatic 1° amines, (c) aromatic 1° amines, (d) Both (b) and (c), 50. In the reaction,, HNO, , 2, RNH 2, ROH H 2O C ; C is, (a) NH3, (b) N2, (c) O2, (d) CO2, 51. An organic amino compound reacts with aqueous nitrous, acid at low temperature to produce an oily nitrosoamine., The compound is, (a) CH3NH2, (b) CH3CH2NH2, (c) CH3CH2NHCH2CH3 (d) (CH3CH2)3N, 52. Ethylamine reacts with HNO2 giving :, (a) C2H5OH, (b) C2H5NO2, (c) NH3, (d) C2H6
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EBD_7207, AMINES, , 466, , 53., , 54., , 55., , 56., , 57., , 58., , 59., , 60., , 61., , 62., , 63., , 64., , 65., , Primary amines can be distinguished from secondary and, tertiary amines by reacting with, (a) Chloroform and alcoholic KOH, (b) Methyl iodide, (c) Chloroform alone, (d) Zinc dust, Which of the following is not correct ?, (a) Ethyl amine and aniline both have – NH2 group, (b) Ethyl amine and aniline dissolve in HCl, (c) Ethyl amine and aniline both react with CHCl3 and, KOH to form unpleasant smelling compound, (d) Ethyl amine and aniline both react with HNO2 in cold, to give hydroxy compounds, Hinsberg reagent is, (a) C6H5SO3H, (b) C6H5NO, (c) C6H5SO2Cl, (d) C6H5N2Cl, Reaction of aniline with benzaldehyde is, (a) substitution, (b) addition, (c) condensation, (d) polymerization, The amine that does not react with acetyl chloride is, (a) CH3NH2, (b) (CH3)2NH, (c) (CH3)3N, (d) None of these, Which of the following compounds cannot be identified, by carbylamine test?, (a) CH3CH2NH2, (b) CHCl3, (c) C6H5NH2, (d) C6H5–NH–C6H5, (CH3)2CHNH2 is reacted with excess acetic anhydride, the, compound formed is, (a) (CH3)2CHNCOCH3, (b) (CH3)2CN(COCH3)2, (c) (CH3)2CHOH, (d) (CH3)2CN(COOCH3)2, In order to distinguish between C2H5–NH2 and C6H5–NH2,, which of the following reagent is useful, (a) Hinsberg’s reagent, (b) HNO2, (c) CHCl3 + KOH, (d) NaOH, All three amines 1°, 2°, 3° react with, 1. H2O, 2. R–X, 3. HCl, 4. (CH3CO)2O, (a) 1, 2, (b) 4 only, (c) 1, 2, 4, (d) 1, 2, 3, –NH2 group in aniline is, (a) only o-directing, (b) only p-directing, (c) only m-directing, (d) o-and p-directing, Strong activating effect of –NH2 group is reduced by using, (a) CH3COCl, (b) CH3Cl, (c) CH3ONa, (d) CH3–CHO, When bromination of aniline is carried out by protecting, –NH2. The product is, (a) o-bromoaniline, (b) 2, 4, 6 tribromoaniline, (c) p-bromoaniline, (d) mixture of o-and p-bromoaniline, Hinsberg’s method to separate amines is based on the, use of, (a) benzene sulphonyl chloride, , 66., , 67., , 68., , 69., , (b) benzene sulphonic acid, (c) ethyl oxalate, (d) acetyl chloride, The reaction,, C6H5NH2+ ClCOC6H5, C6H5NHCOC6H5 is called:, (a) Friedel-crafts reaction, (b) Claisen condensation, (c) Benzoylation or Schotten Baumann reaction, (d) None of these, Which of the following statements is not correct regarding, aniline?, (a) It is less basic than ethylamine, (b) It can be steam-distilled, (c) It reacts with sodium to give hydrogen, (d) It is soluble in water, Benzylamine may be alkylated as shown in the following, equation:, C 6 H5CH 2 NHR, C6 H5 CH 2 NH 2 R – X, Which of the following alkylhalides is best suited for this, reaction through SN1 mechanism?, (a) CH3Br, (b) C6H5Br, (c) C6H5CH2Br, (d) C2H5Br, The product of the following reaction is _________., NHCOCH3, + Br2/CH3COOH, , NHCOCH3, (i), , NHCOCH3, Br, , (ii), , Br, NHCOCH3, (iii), , 70., , 71., , Br, , Br, , NHCOCH3, Br, , (iv), , Br, (b) (i) and (ii), (a) (i) and (iii), (c) (iii) and (iv), (d) (i), (ii) and (iii), Aniline and other arylamines are usually colourless but, get coloured on storage due to___________., (a) hydrolysis, (b) dehydration, (c) reduction, (d) atmospheric oxidation, The acylation reaction of amines is carried out in presence, of pyridine because, (i) pyridine is stronger base than amine., (ii) pyridine is weaker base than amine., (iii) pyridine removes HCl formed and shifts the equilibrium, to the right hand side., (iv) pyridine removes HCl formed and shifts the equilibrium, to the left hand side., (a) (i) and (iii), (b) (ii) and (iv), (c) (ii) and (iii), (d) (i) and (iv)
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EBD_7207, AMINES, , 468, , 85., , Which of the following are intermediates in Sandmeyer, reaction ?, (i), , 86., , 87., , 88., , 89., , 90., , 91., , 92., , C6 H 5 N, , NCl, , (ii), , C6 H5 N, , N, , (iii) C 6 H 5, (iv) C6H5Cl, (a) (ii) and (iii), (b) (i) and (iv), (c) (ii) and (iv), (d) (i) and (ii), In the diazotization of arylamines with sodium nitrite and, hydrochloric acid, an excess of hydrochloric acid is used, primarily to, (a) Supress the concentration of free aniline available for, coupling, (b) Supress hydrolysis of phenol, (c) Ensure a stoichiometric amount of nitrous acid, (d) Neutralise the base liberated, Which of the following reagent can be used to convert, benzenediazonium chloride into benzene?, (a) CH3OH, (b) H3PO2, (c) Br2–H2O, (d) LiAlH4, When benzenediazonium chloride in hydrochloric acid, reacts with cuprous chloride, then chlorobenzene is formed., The reaction is called, (a) Gattermann reaction (b) Perkin reaction, (c) Etard reaction, (d) Sandmeyer reaction, , 95, , 96., , Read the following statements and choose the correct, option., (i) Nitrogen atom in amines is sp3-hybridised., (ii) The geometry of amines is pyramidal., , NH2, , NH2, , Cu/HCl, , The reaction ArN 2Cl –, ArCl N 2 CuCl is, named as _________ ., (a) Sandmeyer reaction (b) Gatterman reaction, (c) Claisen reaction, (d) Carbylamine reaction, Which of the following compound will not undergo azo, coupling reaction with benzene diazonium chloride., (a) Aniline, (b) Phenol, (c) Anisole, (d) Nitrobenzene, Which of the following cannot be prepared by Sandmeyer’s, reaction?, (i) Chlorobenzene, (ii) Bromobenzene, (iii) Iodobenzene, (iv) Fluorobenzene, (a) (i) and (ii), (b) (ii) and (iii), (c) (iii) and (iv), (d) (i) and (iv), The reagents that can be used to convert benzenediazonium, chloride to benzene are _________., (i) SnCl2/HCl, (ii) CH3CH2OH, (iii) H3PO2, (iv) LiAlH4, (a) (i) and (ii), (b) (ii) and (iii), (c) (iii) and (iv), (d) (i) and (iii), , STATEMENT TYPE QUESTIONS, 93., , 94., , (iii) The angle C–N–C or C–N–H is slightly more than, 109.5°., (a) (i), (ii) and (iii), (b) (i) and (ii), (c) (i) and (iii), (d) (ii) and (iii), Which of the following statements are correct ?, (i) Lower aliphatic amines are soluble in water., (ii) Solubility increases with decrease in molar mass of, amines., (iii) Higher amines are insoluble in water., (iv) Amines are soluble in organic solvents., (a) (i), (ii) and (iii), (b) (i), (iii) and (iv), (c) (ii), (iii) and (iv), (d) (i) and (iv), Which of the following statements are correct ?, (i) Primary amines show more intermolecular association, than secondary amines., (ii) Tertiary amines do not show intermolecular, association., (iii) Boiling points of isomeric alkenes follow the order, 3° > 2° > 1°, (a) (i) and (iii), (b) (i) and (ii), (c) (i), (ii) and (iii), (d) (ii) and (iii), Which of the following is/are correct regarding nitration of, aniline with conc. HNO3 and conc. H2SO4 ?, , conc. HNO3, conc. H 2SO4, , (i), , NH2, , NO2, +, NO2, , +, , NH3, , +, , NH3, , (ii), , NO2, , 97., , (iii) The substitution can be explained on the basis of, inductive effect (– I), (iv) The substitution can be influenced by +M and +E, effects., (a) (i), (ii) and (iii), (b) (ii) and (iii), (c) (ii) and (iv), (d) (i) and (iii), Which of the following statements are correct ?, (i) In Sandmeyer reaction nucleophiles like Cl–, Br– and, CN– are indroduced in benzene ring in the presence of, Cu+ ion, (ii) In Gattermann reaction nucleophiles are introduced in, benzene ring in the presence of copper powder and, HCl., (iii) The yield in Gattermann reaction is found to be better, than Sandmayer reaction., (a) (i) and (ii), (b) (i), (ii) and (iii), (c) (ii) and (iii), (d) (i) and (iii)
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AMINES, , 469, , (a) A – (q), B – (p), C – (s), D – (r), (b) A – (s), B – (p), C – (q), D – (r), , MATCHING TYPE QUESTIONS, 98., , Match the columns, Column-I, Column-II, (A) Gabriel phthalimide, (p) C6H5CH2NH2, reaction, (B) Reduction with, (q) C6H5NH2, LiAlH4, (C) Reaction with, (r) C6H5CN, alc. KOH + CHCl3, (D) 1° Amide with, (s) CH3CH2NH2, Br2 + KOH, (a) A – (p, s); B – (p, s); C – (p, q, s); D – (p, q, s), (b) A – (s); B – (p); C – (q); D – (p, q), (c) A – (p, s); B – (r); C – (q); D – (s), (d) A – (p, q); B – (p); C – (p, q); D – (s), 99. Match the columns, Column-I, Column-II, (A) Ammonolysis, (p) Amine with lesser number, of carbon atoms, (B) Gabriel phthalimide, (q) Detection test for primary, synthesis, amines., (C) Hoffmann bromamide (r) Reaction of Phthalimide, reaction, with KOH and R–X, (D) Carbylamine reaction (s) Reaction of alkylhalides, with NH3, (a) A – (s), B – (r), C – (p), D – (q), (b) A – (r), B – (p), C – (r), D – (s), (c) A – (q), B – (r), C – (s), D – (p), (d) A – (s), B – (p), C – (q), D – (r), 100. Match the columns, Column-I, Column-II, (A) Benzene sulphonyl, (p) Zwitter ion, chloride, (B) Sulphanilic acid, (q) Hinsberg reagent, (C) Alkyl diazonium salts (r) Dyes, (D) Aryl diazonium salts (s) Conversion to alcohols, (a) A – (s), B – (q), C – (r), D – (p), (b) A – (q), B – (p), C – (s), D – (r), (c) A – (r), B – (s), C – (p), D – (q), (d) A – (s), B – (p), C – (r), D – (q), 101. Match the columns, Column-I, Column-II, (A) ArN 2 Cl, , ArOH (p) HBF4 / NaNO 2, , (B) ArN 2 Cl, , ArNO 2 (q) H2O, , (C) ArN2 Cl, , ArH (r) HBF4, , (D) ArN 2 Cl, , ArF, , (s) CH3CH2OH, , (c) A – (q), B – (s), C – (p), D – (r), (d) A – (q), B – (s), C – (r), D – (p), , ASSERTION-REASON TYPE QUESTIONS, Directions : Each of these questions contain two statements,, Assertion and Reason. Each of these questions also has four, alternative choices, only one of which is the correct answer. You, have to select one of the codes (a), (b), (c) and (d) given below., (a), , Assertion is correct, reason is correct; reason is a correct, explanation for assertion., , (b), (c), , Assertion is correct, reason is correct; reason is not a, correct explanation for assertion, Assertion is correct, reason is incorrect, , (d), , Assertion is incorrect, reason is correct., , 102. Assertion : Aromatic 1°amines can be prepared by Gabriel, phthalimide synthesis., Reason : Aryl halides undergo nucleophilic substitution, with anion formed by phthalimide., 103. Assertion : Only a small amount of HCl is required in the, reduction of nitro compounds with iron scrap and HCl in, the presence of steam., Reason : FeCl2 formed gets hydrolysed to release HCl, during the reaction., 104. Assertion : Amines are basic in nature., Reason : Amines have lone pair of electrons on nitrogen, atom., 105. Assertion : Acetanilide is less basic than aniline., Reason : Acetylation of aniline results in decrease of, electron density on nitrogen., 106. Assertion : Nitration of aniline can be conveniently done, by protecting the amino group by acetylation., Reason : Acetylation increases the electron-density in the, benzene ring., 107. Assertoin : Aniline does not undergo Friedel-Crafts, reaction., Reason : –NH2 group of aniline reacts with AlCl3 (Lewis, acid) to give acid-base reaction., 108. Assertion : Acylation of amines gives a monosubstituted, product whereas alkylation of amines gives, polysubstituted product., Reason : Acyl group sterically hinders the approach of, further acyl groups, 109. Assertion : Nitrating mixture used for carrying out nitration, of benzene consists of conc. HNO3 + conc. H2SO4., Reason : In presence of H2SO4, HNO3 acts as a base and, produces NO2+ ions.
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EBD_7207, AMINES, , 470, , CRITICAL THINKING TYPE QUESTIONS, 110. The IUPAC name of diethyl isopropyl amine is, (a) N, N-diethylpropan-2-amine, (b) N, N-diethylpropan-1-amine, (c) N, N-diethylisopropylamine, (d) N, N-diethylaminopropane, 111. IUPAC name of the following compound is, , 118. What is the decreasing order of basicity of primary,, secondary and tertiary ethylamines and NH3 ?, (a), , NH3, , (b), , (C2 H 5 )3 N, , (c), , (C2 H 5 )2 NH, , C2H5 NH 2, , (C2H5 )3N, , (C 2 H5 )2 NH, , (C2 H 5 )2 NH, , C2 H5 NH 2, , NH 3, , C2 H 5 NH 2, , (C 2 H 5 )3 N, , NH 3, , (d) (C2H5)2 NH > C2H5NH2 > (C2H5)3 NH > NH3, 119. The correct order of the increasing basicity of methyl amine,, ammonia and aniline is, (a) methyl amine < aniline < ammonia, , NH2, Cl, (a) 2-chloro pentanamine, (b) 4-chloro pentan-1-amine, (c) 4-chloro pent-2-en-1-amine, (d) 2-chloro pent-3-en-5-amine, 112. Which of the following is the correct IUPAC name of the, compound ?, , (b) aniline < ammonia < methyl amine, (c) aniline < methyl amine < ammonia, (d) ammonia < aniline < methyl amine, 120. Arrange the following amines in the decreasing order of, their basicity, NH2, , (1), , N (CH 3 ) 2, , NH2, , (2), , CH2 – CH2 – NH2, , Cl, , (3), , Cl, , (a) 1, 2-dichloro-4-(N, N-dimethyl) aniline, (b) Dimethyl – (3, 4-dichlorophenyl) amine, (c) 3, 4-dichloro - N, N-dimethyl aniline, (d) N, N-dimethylamino - 3, 4-dichlorobenzene, 113. Acetamide is treated with the following reagents separately., Which one of these would yield methylamine?, (a) NaOH – Br2, , (b) Sodalime, , (c) Hot conc. H 2SO4, , (d), , (a) 1 > 3 > 2, , (b) 3 > 2 > 1, , (c) 1 > 2 > 3, , (d) 2 > 1 > 3, , 121., , NH2, , (CH3CO) 2 O,Pyridine, , (I), , PCl5, , 114. Amine that cannot be prepared by Gabriel phthalimide, synthesis is, (a) aniline, (b) benzylamine, (c) methylamine, (d) iso-butylamine, 115. A primary amine is formed by an amide on treatment with, bromine and alkali. The primary amine has, (a) 1 carbon atom less than amide, (b) 1 carbon atom more than amide, (c) 1 hydrogen atom less than amide, (d) 1 hydrogen atom more than amide, , (II), , (i) LiAlH 4, (ii) H2O, , III, , The basicity order of I, II and III is –, (a) III > I > II, , (b) I > II > III, , (c) III > II > I, , (d) II > III > I, , 122. Which of the statement is true regarding the basicity of the, following two primary amines ?, CH2NH2, , CH2NH2, , 116. High basicity of Me 2 NH relative to Me3N is attributed to:, (a) effect of solvent, (b) inductive effect of Me, (c) shape of Me2NH, (d) shape of Me3N, 117. Which one of the following is the strongest base in aqueous, solution ?, (a) Methylamine, (b) Trimethylamine, (c) Aniline, (d) Dimethylamine, , I, , II, , (a) Both are equally basic because both are 1º amines, (b) I > II because it is an aromatic amine, (c) II > I because it is an aliphatic amine, (d) I < II because of difference in the nature of -carbon
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AMINES, , 473, , 14. (c), , FACT / DEFINITION TYPE QUESTIONS, , Ethyl, iodide, , 1., , (c), , 2., 5., , (c), 3. (a), 4. (d), (c) 1° amines have –NH2 group in their structure. 4 primary, amines are possible by C4H11N., CH 3 CH 2 CH 2 CH 2 NH 2, (i), CH 3 CH 2 CH CH 3, |, NH 2, (ii), CH3, CH3, |, |, CH3 C CH3, CH3 C CH 2 NH 2, |, |, NH 2, H, (iii), (iv), (b) The compound contains longest chain of 6C atoms, and amino group. Hence it is an alkanamine., (d), 8. (d), 9. (a), (c) CH3NC (methyl isocyanide) on reduction with LiAlH4, gives secondary amine, O, ||, (b) CH 3 CH 2 C NH 2 Br2 KOH, , 6., 7., 10., , 11., , RNH 2, Primary amine, , R 2 NH, Secondary amine, , R3N, Tertiary amine, , CH3 CH 2 NH 2, , 12., , (d), , R N, , C, , 4[H], , alkyl isocyanide, , RNH CH3, secondary amine, , O, , O, , 13., , (c), , NH, , KOH, , –+, , NK, O, , O, Phthalimide, , N-Potassium phthalimide, (a nucleophile), , O, RX / DMF, SN 2 rxn, , C 2 H 5 OH, , O, , 39 (c), 42. (d), , N-alkyl phthalimide, , O, OH, OH, O, Phthalic acid, , +, , H2N R, , Primary, amine, , NH 3, , C2 H5 NH 2 HI, C 2 H 5 NH 2, , H2O, , 15. (d), 16. (d), 17. (d) For the preparation of Me3CNH2, the required alkyl, halide is Me3CX which will react with potassium, phthalimide, a strong base, to form alkene rather than, substituted product. For preparing C6H5NH2, C6H5Cl, will be the starting halide in which Cl is non-reactive., 18. (c), 19. (d), 20. (c), 21. (d) Reduction with iron scrap and hydrochloric acid is, preferred because FeCl2 formed gets hydrolysed to, release hydrochloric acid during the reaction. Thus,, only a small amount of hydrochloric acid is required, to initiate the reaction., 22. (b), 23. (c), 24. (a), 25. (b) Amines possess fishy smell., 26. (c), 27. (b), 28. (b), 29. (b) Basic nature of amines arises due to presence of lone, pair of e–1s on the N-atom, which can be shared with, an electron deficient species., 30. (a), 31. (a) Secondary amine is more basic than primary amine., 32. (d) Amines are basic in nature, 33. (b) (CH3)3N < CH3NH2 < (CH3)2 NH < C2H5NH2, 34. (a), 35. (a) Secondary amines are more basic than tertiary amines, due to stabilisation of 2° amine by hydrogen bonding, with solvent molecule., 36. (c) Aliphatic amines are more basic than aromatic amines., Resonance decreases the basic character due to, delocalisation of shared pair of electrons on nitrogen, within benzene nucleus, 37. (b) Basic character of amines is, 2° > 1 ° > 3° > NH3, 38. (b), , RNH 2, , NR, , C2 H5I NH3, , 43. (b), 44. (c), , CH2–NH2 compound is most basic due, to localized lone pair of electron on nitrogen atom while, other compounds have delocalized lone pair of electron., 40. (a), 41. (b), Amines give alcohols only on reaction with HNO2, and not on hydrolysis., Aniline gives dyes on coupling reaction with phenols, and p-amines., Aryl amines do not produce phenol on treatment with, nitrous acid
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EBD_7207, AMINES, , 474, , 45., , (c) Methyl amine is a stronger base than ammonia due to, +I effect. The alkyl groups which are electron releasing, groups increase the electron density around the, nitrogen thereby increasing the availability of the lone, pair of electrons to proton or lewis acid and making, the amine more basic, NH3, CH3NH2, 10–5, , 46., , RNC, , alkyl, isocyanide, , 3KCl 3H 2 O, , 1 Amine, , .., R – NH – C – CH 3, , (d) R – NH2 + CHCl3, , R – NH – C – CH 3, alc. KOH, , R, , N, , +, , –, , 55., , +, N, , (b) RNH2, , 51., , (c) Since the organic amino compound on reaction with, nitrous acid at low temperature produces an oily, nitrosoamine so the organic amino compound is a, secondary aliphatic amines., , 58., 63., 65., , ROH + H2O + N2, , 52., , (a), , (A), , C 2 H 5 NH 2 +, Ethyl amine, , HNO 2, , (B), , (C), , (c), , Nitrous acid, , Ethanol, , 53., , 73., , 273 278K, , C2 H5OH + N2 + H2O, Note : This reaction is used as a test for aliphatic amines, since no other class of amines liberates N2 gas on, treatment with HNO2., (a) 1° amines (aliphatic and aromatic) react with, CHCl3/KOH to yield isocyanide (foul smelling) This is, known as carbylamine test which is not given by 2°, and 3° amines., , Benzene diazonium, chloride, , OHC, , (c) The compounds containing active H-atoms (H atoms, attached to N, O or S) react with CH3COCl to form, acetyl derivatives., (d), 59. (b), 60. (b), 61. (d), 62. (d), (a), 64. (d), (a) Hinsberg’s method is based on the use of benzene, sulphonyl chloride., (c) The inclusion of C6H5CO gp.in a molecule is called, benzoylation, (d) Aniline is insoluble in water, because its –NH2 group, can’t form H- bond with water due to bulky phenyl, group., (c), 69. (b), 70. (d), (a) Pyridine is a stronger base than the amine, pyridine, removes HCl formed in acylation reaction of amines, and shifts the equilibrium to the right hand side., (b) The hydrogen attached to nitrogen in sulphonamide, is strongly acidic due to the presence of strong, electron withdrawing sulphonyl group. Hence, it is, soluble in alkali., (d) In case of substituted aniline, electron releasing groups, like –OCH3, –CH3 increase basic strength whereas, electron withdrawing groups like –NO2 , –SO3H,, –COOH, –X decrease it., , –, C, , 50., , HCl), , 57., , 72., HNO2, , Cl, , (c), , 68., 71., , alc. KOH, , Cold, , 56., , 67., , NH2 + CHCl3, , HONO, (NaNO 2, , 66., , C, , Alcohol, , N N, , NH2, , isocyanide (CH 3CH 2 NC) is produced. This equation, is known as carbylamine reaction., (a) Acylation occurs in one step only because lone pair, of nitrogen is delocalized with acyl group., O, O–, , C2 H5OH N 2, , NH 2, , (c) We know that, , In this reaction, bad smelling compound ethyl, , 49., , C2 H5 NH 2 HONO, , Kb = 1.8 ×, (a) Any primary amine means both aliphatic as well as, aromatic but monoalkylamines means only 1° aliphatic, amines. Therefore, option (a) is correct while (d) is, wrong., , CH 3CH 2 NH 2 CHCl 3 3KOH, CH 3CH 2 NC 3KCl 3H 2O, , 48., , (d) Nitrous acid reacts differently with aliphatic and, aromatic amines in cold., , 44 × 10–5, , RNH2+ CHCl3+ 3KOH, 47., , 54., , +, , N = CH, , N, , NH2, , 74., , (a), , .. –, NCl, , NaNO2 /HCl, , 0C, Diazotisation, , Br, CuBr/HBr, Sandmeyer reaction
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AMINES, , 475, , NH2, 75., , ArNH2 + NaNO2 + 2HX, , NaNO2 ,HCl, , ¾¾¾¾¾®, , (b), , 1° Aromatic amine, , (0-5°C), , diazonium, chloride, , aniline, 76., , 81. (c) Primary aromatic amines react with nitrous acid to, yield arene diazonium salts., , N2 Cl, , –, , Arene diazonium salt, , OH, , Ar—N2+X–, , N=N, , +, Benzene, diazonium, chloride, , (– N = N –) group is called azo – group., , 77., , heat, , Ar—N2+BF4–, Ar—F + BF3 + N2, +, , N 2 Cl, , CuCN, , HCl, 278K, , +, , NaNO2 ,HCl, , Diazotization, , CuCN, , Benzene diazonium, chloride, (A), , 0°C, , C N, , (A), , (diazotisation), , Aniline, , –, , NaNO2, , 82. (a), , N NCl, , (d), , HBF4, , NH 2, , p–hydroxyazobenzene, , NH2, , Ar—N = N+X– + NaX + 2H2O, , The diazonium group can be replaced by fluorine by, treating the diazonium salt with fluoroboric acid, (HBF4). The precipitated diazonium fluoroborate is, isolated, dried and heated until decomposition occurs, to yield the aryl fluoride. This reaction is known as, Balz-Schiemann reaction., , (a) Azo dye is prepared by diazo coupling reaction of, phenol with diazonium salt., N+2Cl, , cold, , benzene, diazonium, chloride, , Benzonitrile, (B), Sandmeyer reaction, , 83. (d), 84. (c), 85. (a), 86. (a) Excess of HCl is used to convert free aniline to aniline, hydrochloride otherwise free aniline would undergo, coupling reaction with benzenediazonium chloride., 87. (b), 88. (d), 89. (b), 90. (d), 91. (b), 92. (b), 78., , (b) The given reaction is known as Sandmeyer’s reaction., , N2+Cl, 79., , –, , OH, H 2O, , (b), , + N2+HCl, , CH3, , 80., , (b), , CH3, , p- cresol, , +, , N, , –, , NCl +, , Benzene diazonium chloride, , OH, Phenol, , N= N, p-Hydroxyazobenzene, (orange dye), , OH, , STATEMENT TYPE QUESTIONS, 93. (b) The fourth orbital of nitrogen in all amines contains an, unshared pair of electrons. Due to the presence of, unshared pair of electrons, the angle C–N–E, (where E, is C or H) is less than 109.5°., 94. (b) Lower aliphatic amines are soluble in water solubility, decreases with increase in molar mass of amines., Higher amines are essentially insoluble in water., 95. (b) Primary and secondary amines are engaged in, intermolecular association due to hydrogen bonding, between nitrogen of one and hydrogen of another, molecule. This intermolecular association is more in, primary amines than in secondary amines as there are, two hydrogen atoms available for hydrogen bond, formation in it. Tertiary amines do not have, intermolecular association due to the absence of, hydrogen atom available for h ydrogen bond, formation. Therefore, the order of boiling points of, isomeric amines is as follows :, Primary > Secondary > Tertiary
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EBD_7207, AMINES, , 476, , 115. (a) The reaction is Hoffmann bromamide reaction, , 96., , (b), , 97., , (a) The yield in Sandmayer reaction is found to be better, than Gattermann reaction., , O, ||, R C NH 2 Br2, , MATCHING TYPE QUESTIONS, 98., , (a), , 99. (a), , 100. (b), , R NH 2, , 101. (c), , 103. (d), , 104. (a) Amines are basic due to the presence of a lone pair of, electrons on nitrogen atom. The lone pair can be easily, donated., 105. (d), 106. (c) Acetylation decreases the electron-density in the, benzene ring thereby preventing oxidation., 107. (a), 109. (a), , 108. (c), HNO3, , 2H 2SO4, , 2HSO4, , NO2, , H3O, , CRITICAL THINKING TYPE QUESTIONS, 110. (a), , CH3, , HC – N, , C2H5, C2H5, , CH3, , N, N- diethyl propan - 2-amine, , 111. (c), 112. (c) The compound is derivative of aniline. The positions, of groups are shown by numbering the nuclear Catoms., 113. (a), , CH3CONH2, , NaOH, Br2, , CH3 NH 2, , (Hofmann bromamide reaction), , 114. (a) Aniline cannot be prepared by this method because, aryl halides do not undergo nucleophilic substitution, with potassium phthalimide under ordinary conditions, to give N-phenyl phthalimide (i.e., cleavage of C–X, bond in haloarenes is quite difficult)., CO, , NH, , alc, KOH, , CO, , 2NaBr Na 2CO3 2H 2O, , R – NH2 contains one carbon less than, , ASSERTION-REASON TYPE QUESTIONS, 102. (a), , 4NaOH, , 116. (a) Secondary amines are more basic than tertiary amines, due to stabilisation of 2° amine by hydrogen bonding, with solvent molecule., 117. (d) Aromatic amines are less basic than aliphatic amines., Among aliphatic amines the order of basicity is, 2° > 1° > 3°. The electron density is decreased in 3°, amine due to crowding of alkyl group over N atom, which makes the approach and bonding by a proton, relatively difficult. Therefore the basicity decreases., Further Phenyl group show – I effect, thus decreases, the electron density on nitrogen atom and hence the, basicity., dimethylamine (2° aliphatic amine) is strongest base, among given choices., The correct order of basic strength is, Dimethylamine > Methyl amine > Trimethyl amine >, Aniline., 118. (d) All aliphatic amines are stronger bases than NH3 and, among different ethylamines order of basictity is, 2° > 1° > 3°. Thus, the correct order is (d) i.e.,, (C2H5)2 NH > C2H5NH2 > (C2H5)3N > NH3, This anomolous behaviour of tertiary ethyl amine is, due to steric factors i.e., crowding of alkyl groups cover, nitrogen atom from all sides and thus makes the, approach and bonding by a lewis acid relatively, difficult which results the maximum steric strain in, tertiary amines. The electrons are there but the path is, blocked resulting the reduction in its basicity., 119. (b) In aniline the lone pair on N is involved in delocalization, with benzene ring and is not available for protonation., Methyl amine is a stronger base than ammonia because, +I effect of methyl group increases electron density, on N making it more basic than NH3., NH 2, , CO, , NH 2, , N+K–, , CO, RX, , CO, N–R, , Cannot be protonated. least basic, , CO, , – NH 2, I Effect increases, basicity., , CH3, , H+/H2O, COOH, COOH, , + RNH2, , O, ||, R C NH 2, , 120. (a)
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28, BIOMOLECULES, FACT / DEFINITION TYPE QUESTIONS, 1., , 2., , 3., , 4., , 5., , 6., , 7., , 8., , 9., , 10., , Biomolecules are, (a) aldehydes and ketones, (b) acids and esters, (c) carbohydrates, proteins and fats, (d) alcohols and phenols, Which of the following is a disaccharide ?, (a) Lactose, (b) Starch, (c) Cellulose, (d) Fructose, The sugar that is characteristic of milk is, (a) maltose, (b) ribose, (c) lactose, (d) galactose, Which one is a disaccharide ?, (a) Glucose, (b) Fructose, (c) Xylose, (d) Sucrose, Which of the following monosaccharide is pentose ?, (a) Glucose, (b) Fructose, (c) Arabinose, (d) Galactose, The commonest disaccharide has the molecular formula, (a) C10 H18 O9, (b) C10 H20 O10, (c) C18 H22 O11, (d) C12 H22 O11, Monosaccharides usually contains ... carbon atoms., (a) C3 to C10, (b) C1 to C6, (c) C4 to C10, (d) C5 to C8, Which one of th e following compounds is found, abudnantly in nature?, (a) Fructose, (b) Starch, (c) Glucose, (d) Cellulose, A carbohydrate that cannot be hydrolysed into simpler, units is called, (a) polysaccharides, (b) trisaccharides, (c) disachharides, (d) monosaccharides, Which of the following statements is incorrect ?, (a) Maltose gives two molecules of glucose only., (b) Cellulose and sucrose are polysaccharide., (c) Polysaccharides are not sweet in taste., (d) Polysaccharides are also known as non-sugars., , 11., , 12., , 13., , 14., , 15., , 16., , 17., , 18., , Reducing sugars reduce., (a) only Fehling’s solution, (b) only Tollen’s solution., (c) both (a) & (b), (d) neither (a) nor (b), Which among the following is the simplest sugar?, (a) Glucose, (b) Starch, (c) Cellulose, (d) None of these, Glucose can’t be classified as, (a) hexose, (b) carbohydrate, (c) aldose, (d) oligosaccharide, Which of the following properties of glucose cannot be, explained by its open chain structure?, (i) Glucose does not form hydrogen sulphite with NaHSO3, (ii) On oxidation with HNO3 glucose gives saccharic acid., (iii) Glucose is found to exist in two different crystalline, forms which are named as and ., (a) (ii) only, (b) (i) and (iii), (c) (ii) and (iii), (d) (i) and (ii), Which of the following gives positive Fehling solution test?, (a) Protein, (b) Sucrose, (c) Glucose, (d) Fats, Which of the following statements is incorrect regarding, glucose?, (a) It is an aldohexose., (b) It is also known as dextrose, (c) It is monomer of cellulose., (d) It is the least abundant organic compound on earth., Glucose gives silver mirror test with Tollen’s reagent. It, shows the presence of, (a) acidic group, (b) alcoholic group, (c) ketonic group, (d) aldehyde group, The symbols D and L represents, (a) the optical activity of compounds., (b) the relative configuration of a particular stereoisomer., (c) the dextrorotatory nature of molecule., (d) the levorotatory nature of molecule
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EBD_7207, BIOMOLECULES, , 480, , 19., , 20., , 21., , 22., , 23., , 24., , 25., , 26., , 27., , 28., , 29., , 30., , Glucose is found to exist in two different and crystalline, forms. These forms can be obtained by., (i) The form of glucose is obtained by crystallisation, from concentrated solution of glucose at 303 K., (ii) The form of glucose is obtained by crystallisation, from concentrated solution of glucose at 303 K., (iii) The form is obtained by crystallisation from hot and, saturated aqueous solution at 371 K., (iv) The form is obtained by crystallisation from hot and, saturated aqueous solution at 371 K., (a) (i) and (iii), (b) (ii) and (iv), (c) (ii) and (iii), (d) (i) only, The function of glucose is to, (a) provides energy, (b) promote growth, (c) prevent diseases, (d) perform all above, Which one of the following compounds is different from, the rest?, (a) Sucrose, (b) Maltose, (c) Lactose, (d) Glucose, The two functional groups present in a typical carbohydrate, are:, (a) – CHO and – COOH (b) > C = O and – OH, (c) – OH and – CHO, (d) – OH and – COOH, When glucose reacts with bromine water, the main product, is, (a) gluconic acid, (b) glyceraldehyde, (c) saccharic acid, (d) acetic acid, Glucose does not react with, (a) Br2/H2O, (b) H2NOH, (c) HI, (d) NaHSO3, Glucose reacts with acetic anhydride to form, (a) monoacetate, (b) tetra-acetate, (c) penta-acetate, (d) hexa-acetate, Reduction of glucose by HI suggest that, (a) presence of OH groups, (b) presence of –CHO group, (c) cyclic structure of glucose, (d) six carbon atoms are arranged in straight chain, The reaction of glucose with red P + HI is called, (a) Sandmeyer’s reaction (b) Reformatsky reaction, (c) Gattermann’s reaction (d) Reduction, Which of the following reactions of glucose can be explained, only by its cyclic structure?, (a) Glucose forms pentaacetate, (b) Glucose reacts with hydroxylamine to form an oxime, (c) Pentaacetate of glucose does not react with, hydroxylamine, (d) Glucose is oxidised by nitric acid to gluconic acid, Which is the least stable form of glucose ?, (a), -D-Glucose, (b) -D-Glucose, (c) Open chain structure (d) All are equally stable, Isomerization of glucose produces, (a) galactose, (b) fructose, (c) mannose, (d) allose, , 31., , A solution of D-glucose in water rotates the plane polarised, light, (a) to the right, (b) to the left, (c) to either side, (d) None of these, 32. The number of chiral carbon atoms present in cyclic, structure -D(+) glucose, (a) 3, (b) 4, (c) 5, (d) 6, 33. The -D glucose and -D glucose differ from each other, due to difference in carbon atom with respect to its, (a) conformation, (b) configuration, (c) number of OH groups (d) size of hemiacetal ring, 34. The two forms of D glucopyranose obtained from the, solution of D glucose are called, (a) isomers, (b) anomers, (c) epimers, (d) enantiomers, 35. Which of the following carbohydrates are branched polymer, of glucose?, (i) Amylose, (ii) Amylopectin, (iii) Cellulose, (iv) Glycogen, (a) (i) and (ii), (b) (ii) and (iv), (c) (iii) and (iv), (d) (i), (ii) and (iii), 36. The number of chiral carbon atoms present in cyclic, structure -D(+) glucose, (a) 3, (b) 4, (c) 6, (d) 5, 37. Which of the following reagent cannot distinguish between, glucose and fructose?, (a) Fehling’s solution, (b) Tollen’s reagent, (c) Benedict’s solution, (d) All of these, 38. Maltose and glucose are, (a) oxidising sugar, (b) reducing sugar, (c) first is oxidising and second is reducing sugar, (d) both are non-reducing sugar, 39. Choose the correct relationship for glucose and fructose, (a) these are functional isomers, (b) these are chain isomers, (c) these are position isomers, (d) All of these, 40. The pair of compounds in which both the compounds give, positive test with Tollen’s reagent is, (a) Glucose and Sucrose, (b) Fructose and Sucrose, (c) Acetophenone and Hexanal, (d) Glucose and Fructose, 41. The letter D and L in carbohydrates represent, (a) its optical rotation, (b) its mutarotation, (c) its direct synthesis, (d) its configuration, 42. Which of the following statement is correct about fructose?, (a) It is dextrorotatory compound, (b) It exists in the two cyclic forms which is obtained by, the addition of OH at C-5 to the >C=O group, (c) It exists as six membered ring, (d) It is named as furanose as it contain one oxygen and, six carbon atom
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BIOMOLECULES, , 43., , 44., , 45., , 481, , Fructose is, (a) a hemiacetal, (b) an acetal, (c) a hemiketal, (d) a ketal, The sugar present in fruits is, (a) fructose, (b) glucose, (c) sucrose, (d) galactose, Three cyclic structures of monosaccharides are given below, which of these are anomers, , H, , OH, , HO, , H, , OH, , H, , HO, H, H, , H, OH, CH2OH, (I), , 46., , 47., , 48., , 49., , 50., , 51., , 52., , 53., , 54., , 55., , O HO, H, H, , H, OH, H, OH, CH2OH, (II), , O, , HO, , H, , HO, , H, , HO, , H, , H, HO, H, , OH, H, , 56., , 57., O, , 58., , CH2OH, (III), , (a) I and II, (b) II and III, (c) I and III, (d) III is anomer of I and II, The sugar present in honey is, (a) sucrose, (b) glucose, (c) fructose, (d) maltose, Which of the following is the sweetest sugar?, (a) Sucrose, (b) Glucose, (c) Fructose, (d) Maltose, Cellulose is a polymer of, (a) Glucose, (b) Fructose, (c) Ribose, (d) Sucrose, Sucrose on hydrolysis gives, (a) fructose+ribose, (b) glucose + fructose, (c) glucose+glucose, (d) fructose + fructose, The presence or absence of hydroxyl group on which carbon, atom of sugar differentiates RNA and DNA?, (a) 1st, (b) 2nd, rd, (c) 3, (d) 4 th, Carbohydrates are stored in the body as, (a) sugars, (b) starch, (c) glucose, (d) glycogen, A carbohydrate insoluble in water is, (a) glucose, (b) fructose, (c) cellulose, (d) sucrose, Which of the following carbohydrate does not correspond, to the general formula Cx(H2O)y ?, (a) Glucose, (b) 2-Deoxyribose, (c) Fructose, (d) Arabinose, Lactose is made of, (a), -D-glucose only, (b) -D-glucose and -D-glucose, (c), -D-galactose and -D-glucose, (d) -D-galactose and -D-glucose, Which of the following monosaccharides are present as, five membered cyclic structure (furanose structure)?, , 59., , (i) Ribose, (ii) Glucose, (iii) Fructose, (iv) Galactose, (a) (i) and (ii), (b) (i) and (iii), (c) (iii) and (iv), (d) (ii) and (iii), Invert sugar is, (a) chemically inactive form of sugar, (b) equimolecular mixture of glucose and fructose, (c) mixture of glucose and sucrose, (d) a variety of cane sugar, Which one of the followin g does not exhibit the, phenomenon of mutarotation ?, (a) (+) – Sucrose, (b) (+) – Lactose, (c) (+) – Maltose, (d) (–) – Fructose, Glycogen is a branched chain polymer of -D-glucose units, in which chain is formed by C1–C4 glycosidic linkage, whereas branching occurs by the formation of C1-C6, glycosidic linkage. Structure of glycogen is similar to, ______., (a) Amylose, (b) Amylopectin, (c) Cellulose, (d) Glucose, Which of the following correctly represents the cyclic, structure of –D–(–)– fructo furanose., (a), , 12, , HOH2C–C–OH, 3, HO——H, , H, , O, , 4, ——OH, , H, , 5, 6, , CH2OH, , (b), , 2, , 1, , HO–C–CH2OH, 3, HO——H, , O, , 4, H——OH, , H, , 5, 6, , CH2OH, , (c), , O, , 6, HOH2C, 5, H, , (d) HOH2C, , H, , 2, H, 4, , OH, 3, , OH, , H, , 1, CH2OH, OH, , O, , 6, , OH, 2, , 5, H, 4, , OH, 3, , OH, , H, , CH2OH, 1
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EBD_7207, BIOMOLECULES, , 482, , 60., , 61., , 62., , 63., , 64., 65., , 66., 67., 68., 69., , 70., 71., , Sucrose which is dextrorotatory in nature after hydrolysis, gives glucose and fructose, among which, (i) Glucose is laevorotatory and fructose is dextrorotatory., (ii) Glucose is dextrorotatory and fructose is laevorotatory, (iii) The mixture is laevorotatory., (iv) Both are dextrorotatory., (a) (i) and (iii), (b) (ii) and (iii), (b) (iii) and (iv), (d) (iii) only, Chemically amylose is a _________ with 200–1000, -D-(+)-glucose units held by ______ glycosidic linkage, (a) long unbranched chain, C1– C6., (b) branched chain, C1 – C4., (c) long unbranched chain, C1– C4., (d) branched chain, C1– C6., Amylopectin is a ________ polymer of -D-glucose units, in which chain is formed by _______ glycosidic linkage, whereas branching occurs by ________ glycosidic linkage., (a) branched chain, C1– C6, C1– C4., (b) branched chain, C1– C4, C1– C6., (c) unbranched chain, C1– C4, C1– C6., (d) unbranched chain, C1– C6, C1– C4., Which of the following is incorrect about cellulose?, (a) It is a major constituent of cell wall of plant cells., (b) It is a branched chain disaccharide, (c) It is composed of only -D-glucose units., (d) The glycosidic linkage between two units is found, between C1 of one unit and C4 of next unit., Which of the following is also known as animal starch?, (a) Glycine, (b) Glycogen, (c) Amylose, (d) Cellulose, Select the uses of carbohydrates., (a) Honey is used as instant source of energy by vaids in, ayurvedic system of medicine, (b) These are used as storage molecules, (c) They are used in furniture, cotton fibre, lacquers, (d) All of the above, The number of essential amino acids in man is, (a) 8, (b) 10, (c) 18, (d) 20, An acidic amino acid among the following is, (a) glycine, (b) valine, (c) proline, (d) leucine, Amino acids are the building blocks of, (a) fats, (b) proteins, (c) vitamins, (d) carbohydrates, Which one of the amino acids can be synthesised in the, body?, (a) Alanine, (b) Lysine, (c) Valine, (d) Histidine, One of essential -amino acid is, (a) lysin, (b) serine, (c) glycine, (d) proline, Two functional group that are present in all amino acids are, the, (a) hydroxy, amine, (b) hydroxy, amide, (c) carboxyl, amino, (d) carboxyl, amide, , 72., , 73., 74., , 75., , Which of the following is not an optically active amino, acid?, (a) Valine, (b) Glycine, (c) Leucine, (d) Arginine, In aqueous solution, an amino acid exists as, (a) cation, (b) anion, (c) dianion, (d) zwitter ion, Which one of the following statements is correct?, (a) All amino acids except lysine are optically active, (b) All amino acids are optically active, (c) All amino acids except glycine are optically active, (d) All amino acids except glutamic acids are optically, active, Amino acids generally exist in the form of Zwitter ions. This, means they contain, (a) basic—NH2 group and acidic —COOH group, (b) the basic— NH3 group and acidic —COO– group, (c) basic—NH2 and acidic —H+ group, , 76., , 77., , 78., , (d) basic–COO– group and acidic — N H 3 group, Which of the following molecules is capable of forming, Zwitter ion?, (a) NH2CH2COOH, (b) CH3CH2NH2, (c) CH3CH2COOH, (d) All of these, The structural feature which distinguishes proline from, natural -amino acids?, (a) Proline is optically inactive, (b) Proline contains aromatic group, (c) Proline is a dicarboxylic acid, (d) Proline is a secondary amine, The linkage present in proteins and peptides is, O, , O, , (a), , ||, , O, ||, , ||, , C O, , O, ||, , (d) – NH –, C O C, Which one of the following structures represents the, peptide chain?, (c), , 79., , (b), , C NH, , (a), , O, H, |, ||, |, N C N C NH C NH, || | |, O H, , (b), , H, H, |, | | | | | | |, N C C C C N C C C, || | | |, | | |, O, , (c), , H, O, H, H, | | ||, | |, | |, N C C N C C N C C, |, | ||, | ||, O, O, , (d), , H, O, H, | | | ||, | | |, | |, N C C C N C C N C C C, | |, | | |, || | |, H, O
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BIOMOLECULES, , 80., , 81., , 82., , 83., , 84., , 85., , 86., , 87., , 88., , 89., , 90., , 91., , 92., , Simplest proteins has one peptide linkage. It is, (a) tripeptide, (b) dipeptide, (c) tetrapeptide, (d) oligopeptide, A nanopeptide contain how many peptide bond, (a) 7, (b) 9, (c) 8, (d) 10, Proteins are polypeptide of, (a), -amino acid, (b) -hydroxy acid, (c) D- -amino acid, (d) L- -amino acid, Globular proteins are present in, (a) blood, (b) eggs, (c) milk, (d) all of these, In fibrous proteins, polypeptide chains are held together, by, (a) van der waals forces, (b) electrostatic forces of attraction, (c) hydrogen bonds, (d) covalent bonds, Which of the following is not a function of proteins?, (a) Formation of hair, wool, skin and nails, (b) As a biological catalysts in the form of enzymes., (c) As food in the form of meat, eggs, (d) As energy provider for metabolism, Which of the following is not a fibrous protein?, (a) Keratin, (b) Myosin, (c) Insulin, (d) Both (a) and (b), A polypeptide with more than hundred amino acid residues,, having molecular mass higher than 10,000 u is called_____., (a) nucleic acid, (b) hormone, (c) protein, (d) enzyme, An insulin is a ______ which contains _____ amino acids., (a) protein, 74, (b) protein, 51, (c) hormone, 51, (d) hormone, 74, Which of the following is an example of globular proteins?, (a) Glycine, (b) Albumin, (c) Alanine, (d) Both (a) and (b), Which of the following is not a characteristics of fibrous, proteins?, (a) In the fibrous proteins polypeptide chains are held, together by hydrogen and disulphide bonds., (b) These have fibre like structure., (c) These are generally soluble in water., (d) These have elongated shape., Which of the following statements is true about a peptide, bond (RCONHR)?, (a) It is non planar., (b) It is capable of forming a hydrogen bond., (c) The cis configuration is favoured over the trans, configuration., (d) Single bond rotation is permitted between nitrogen, and the carbonyl group., Proteins are condensation polymers of, (a), -amino acids, (b) -amino acids, (c), -hydroxy acids, (d) -hydroxy acids, , 483, , 93. Primary structure of a protein is, (a) sequence in which -amino acid are linked to one, another, (b) sequence in which amino acids of one polypeptide, chain are joined to other chain, (c) the folding patterns of polypeptide chains, (d) the pattern in which the polypeptide chain are arranged, 94. The protein that transport oxygen in the blood stream is, (a) haemoglobin, (b) insulin, (c) collagen, (d) albumin, 95. The helical structure of protein is stabilized by, (a) dipeptide bonds, (b) hydrogen bonds, (c) ether bonds, (d) peptide bonds, 96. Which of the following statements is incorrect?, (a) In -helix structure a polypeptide chain forms all, possible hydrogen bonds by twisting into a right, handed screw., (b) In -structure of proteins all peptide chains are, stretched out to nearly maximum extension., (c) During denaturation 1° and 2° structures are destroyed, but 3° structure remains intact., (d) All the above statements are incorrect., 97. Which of the following indicates the order in which amino, acids are linked together in a protein ?, (a) Primary structure, (b) Secondary structure, (c) Tertiary structure, (d) Quaternary structure, 98. Which of the following statement is not true about, secondary structure of protein ?, (a) The alpha helix, beta pleated sheet and beta turns are, examples of secondary structure of protein., (b) The ability of peptide bonds to form intramolecular, hydrogen bonds is important to secondary structure., (c) The steric influence of amino acid residues is important, to secondary structure., ( d) The hydrophilic/ hydrophobic character of amino acid, residues is important to secondary structure., 99. Which of the following terms indicates to the arrangement, of different protein subunits in a multiprotein complex ?, (a) primary structure, (b) secondary structure, (c) tertiary structure, (d) quaternary structure, 100. Secondary structure of protein is mainly governed by, (a) hydrogen bonds, (b) covalent bonds, (c) ionic bonds, (d) disulphide bonds, 101. The secondary structure of a protein refers to, (a) fixed configuration of the polypeptide backbone, (b), helical backbone, (c) hydrophobic interactions, (d) sequence of, amino acids, 102. Tertiary structure of protein arises due to, (a) folding of polypeptide chain, (b) folding, coiling and bonding of polypeptide chain, (c) linear sequence of amino acid in polypeptide chain, (d) denatured proteins
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EBD_7207, 484, , 103. Denaturation of proteins leads to loss of its biological, activity by, (a) Formation of amino acids, (b) Loss of primary structure, (c) Loss of both primary and secondary structures, (d) Loss of both secondary and tertiary structures, 104. Coagulation of protein is known as, (a) dehydration, (b) decay, (c) deamination, (d) denaturing, 105. Which of the following terms refers to the overall three, dimensional shape of a protein., (a) Primary structure, (b) Secondary structure, (c) Tertiary structure, (d) Quaternary structure, 106. Which of the following indicates to ‘regions of ordered, structure within a protein’., (a) Primary structure, (b) Secondary structure, (c) Tertiary structure, (d) Quaternary structure, 107. The strongest form of intermolecular bonding that could, be formed involving the residue of the amino acid serine is., (a) ionic bond, (b) hydrogen bond, (c) van der Waals interactions, (d) None of the above, 108. Which of the following protein destroy the antigen when it, enters in body cell?, (a) Antibodies, (b) Insulin, (c) Chromoprotein, (d) Phosphoprotein, 109. Which of the following is incorrect regarding enzymes?, (a) Most of them are globular proteins., (b) They are very specific for a particular reaction but not, for a particular substrate., (c) They are generally named after the compound or class, of compounds upon which they work., (d) All the above statements are incorrect., 110. Enzymes take part in a reaction and, (a) decrease the rate of a chemical reaction, (b) increase the rate of a chemical reaction, (c) both (a) and (b), (d) None of these, 111. Enzymes are made up of, (a) Edible proteins, (b) Proteins with specific structure, (c) Nitrogen containing carbohydrates, (d) Carbohydrates, 112. The enzyme which hydrolyses triglycerides to fatty acids, and glycerol is called, (a) Maltase, (b) Lipase, (c) Zymase, (d) Pepsin, 113. Which one of the following, statements is incorrect about, enzyme catalysis?, (a) Enzymes are mostly proteinous in nature., (b) Enzyme action is specific., (c) Enzymes are denaturated by ultraviolet rays and at, high temperature., (d) Enzymes are least reactive at optimum temperature., , BIOMOLECULES, , 114. Insulin production and its action in human body are, responsible for the level of diabetes. This compound belongs, to which of the following categories?, (a) An enzyme, (b) A hormone, (c) A co-enzyme, (d) An antibiotic, 115. Enzymes are essential as biocatalysts. They function in, (a) aqueous medium, temp = 30–35ºC; pH=7, (b) organic medium, (c) aqueous medium under extreme pH conditions, (d) None of these, 116. Which of the following statements is incorrect?, (a) Enzymes are organic catalysts, (b) Enzymes have a very large turnover number, (c) Enzymes action is specific, (d) Enzymes always require a coenzyme in their catalytic, action., 117. Among the following vitamins the one whose deficiency, causes rickets (bone deficiency) is :, (a) Vitamin A, (b) Vitamin B, (c) Vitamin D, (d) Vitamin C, 118. The vitamin that is not soluble in water is, (a) Vitamin B1, (b) Vitamin B2, (c) Vitamin B6, (d) Vitamin D, 119. Deficiency of vitamin B1 causes the disease, (a) Convulsions, (b) Beri-Beri, (c) Cheilosis, (d) Sterility, 120. Anaemia is caused by the deficiency of vitamin, (a) B6, (b) B1, (c) B2, (d) B12, 121. Vegetable oils like wheat gram oil, sunflower oil etc. are the, good source of, (a) vitamin K, (b) vitamin E, (c) vitamin D, (d) vitamin A, 122. Which is a fat soluble vitamin?, (a) Vitamin A, (b) Vitamin B6, (c) Vitamin C, (d) Vitamin B2, 123. Vitamin B2, a water soluble vitamin is also known as, (a) ascorbic acid, (b) riboflavin, (c) thiamine, (d) pyridoxine, 124. Which of the following statements about vitamin B12 is, incorrect ?, (a) It has a cobalt atom, (b) It also occurs in plants, (c) It is also present in rain water, (d) It is needed for human body in very small amounts, 125. The couplings between base units of DNA is through :, (a) Hydrogen bonding, (b) Electrostatic bonding, (c) Covalent bonding, (d) van der Waals forces, 126. Which of the following is correct about H-bonding in, nucleotide?, (a) A --- A and T --- T, (b) G --- T and A --- C, (c) A --- G and T --- C, (d) A --- T and G --- C
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BIOMOLECULES, , 127. In DNA, the complimentary bases are:, (a) Adenine and thymine; guanine and cytosine, (b) Adenine and thymine ; guanine and uracil, (c) Adenine and guanine; thymine and cytosine, (d) Uracil and adenine; cytosine and guanine, 128. The segment of DNA which acts as the instrumental manual, for the synthesis of the protein is:, (a) ribose, (b) gene, (c) nucleoside, (d) nucleotide, 129. In DNA the linkages between different nitrogenous bases, are :, (a) peptide linkage, (b) phosphate linkage, (c) H-bonding, (d) glycosidic linkage, 130. DNA multiplication is called as, (a) translation, (b) transduction, (c) transcription, (d) replication, 131. Chromosomes are made from, (a) proteins, (b) nucleic acids, (c) proteins and nucleic acids, (d) carbohydrates and nucleic acids, 132. The double helical structure of DNA was proposed by, (a) Watson and Crick, (b) Meichers, (c) Emil Fischer, (d) Khorana, 133. - Helix is found in, (a) DNA, (b) RNA, (c) lipid, (d) carbohydrates, 134. Which of the following compounds is responsible for the, transmission of heredity characters?, (a) RNA, (b) DNA, (c) Glucose, (d) Haemoglobin, 135. The latest discovery in cytology is that of, (a) respiration, (b) genetic code, (c) enzyme, (d) None of these, 136. Energy is stored in our body in the form of, (a) ATP, (b) ADP, (c) fats, (d) carbohydrates, 137. The chemical change in DNA molecule that could lead to, synthesis of protein with an altered amino acid sequence, is called, (a) replication, (b) lipid formation, (c) cellular membrane, (d) mutation, 138. DNA has deoxyribose, a base and the third component which, is, (a) phosphoric acid, (b) ribose, (c) adenine, (d) thymine, 139. The process by which synthesis of protein takes place based, on the genetic information present in m-RNA is called, (a) Translation, (b) Transcription, (c) Replication, (d) Messenger hypothesis, , 485, , 140. Which of the following structures represents thymine ?, , NH2, , O, , N, , HN, (a), , N, H, , O, , (b), , OH, , NH2, N, , (c), , N, H, , O, , (d), , N, , CH3, , N, , HO N, 141.When adenine is attached to ribose sugar, it is called, adenosine. To make a nucleotide from it, it would require, (a) oxygenation, (b) addition of a base, (c) addition of phosphate (d) hydrogenation, 142.Which of the following is not present in a nucleotide?, (a) Guanine, (b) Cytosine, (c) Adenine, (d) Tyrosine, 143.The function of DNA in an organism is, (a) to assist in the synthesis of RNA molecule, (b) to store information of heredity characteristics, (c) to assist in the synthesis of proteins and polypeptides, (d) All of these, 144. Which of the following statements regarding DNA, fingerprinting is incorrect?, (a) It is used in forensic laboratories for identification of, criminals., (b) It cannot be altered by surgery., (c) It is different for every cell and cannot be altered by, any known treatment., (d) It is used to determine paternity of an individual., HO, , STATEMENT TYPE QUESTIONS, 145. Read the following statements and choose the correct, answer?, (i) All monosaccharides are reducing sugars., (ii) All monosaccharides are not reducing sugars., (iii) In disaccharides if aldehydic or ketonic groups are, bonded, these are non– reducing sugars., (iv) In disaccharides if aldehydic or ketonic groups are, free, these are reducing sugars., (a) (i), (iii) and (iv), (b) (ii), (iii) and (iv), (c) (i) and (iv), (d) (ii) and (iv), 146. Which of the following statement(s) is/are correct?, (i) Glucose is reducing sugar, (ii) Sucrose is reducing sugar, (iii) Maltose is non reducing sugar, (iv) Lactose is reducing sugar, (a) (i) and (ii) only, (b) (i) and (iii) only, (c) (i) and (iv) only, (d) All of these
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EBD_7207, BIOMOLECULES, , 486, , 147. Which of the following statements regarding carbohydrates, are correct?, (i) Lactose is the carbohydrate found in milk., (ii) More than 25 monosaccharides occur naturally., (iii) Sucrose on hydrolysis gives one molecule each of, glucose and fructose., (iv) Maltose is a non-reducing sugar whereas sucrose is a, reducing disaccharide sugar., (a) (i), (ii) and (iii), (b) (i) and (iii), (c) (ii), (iii) and (iv), (d) (iii) and (iv), 148. Read the following statements., (i) Pyran is a cyclic organic compound with one oxygen, atom and five carbon atom., (ii) The cyclic structure of glucose is correctly represented, by Haworth strucure., (iii) Five membered cyclic structure of glucose is called, pyranose structure., Which of the following statement(s) is/are true?, (a) (i) and (iii), (b) (i) and (ii), (c) Only (iii), (d) (i), (ii) and (iii), 149. Consider the following statements., (i) Linkage between two monosaccharide units through, oxygen atom is called glycosidic linkage., (ii) Sucrose on hydrolysis gives an equimolar mixture of, fructose and glucose which is dextrorotatory., (iii) Lactose consists of linkage between C1 of galactose, and C4 of glucose., (iv) Out of two components of starch the component, present in greater proportion is insoluble in water., (v) Glycogen is also known as animal starch because it is, structurally similar to amylose a component of starch., Which of the following is the correct code for statements, above?, (a) FFFTT, (b) FTTTF, (c) TFTFT, (d) TFTTF, 150. Read the following statements., (i) Haworth structure of –D– glucose will be ., , 6, CH2OH, O, 5, H H, , 4, HO OH, 3, , 153., , 154., , OH, , 1, 2, , H, , H, , O, , 5, H, , 152., , 155., , OH, H, (ii) Fructose belongs to D–series and is a laevorotatory, compound., (iii) Haworth structure for –D–fructose will be., , 6, HOH2C, , 151., , 2, H, 4, , OH, 3, , OH, , H, , OH, CH2OH, 1, , 156., , (iv) Fructose contains a ketonic functional group at C–2, carbon atom., Which of the following is the correct code for the statements, above?, (a) FTTT, (b) FFTT, (c) TFFT, (d) FTFT, Read the following statements and choose the correct–, option?, (i) Starch is a polymer of – glucose., (ii) Starch consists of amylose and amylopectin., (iii) Amylose is insoluble in water., (iv) Amylopectin is soluble in water., (a) (i) (iii) and (iv), (b) (i), (ii) and (iii), (c) (i) and (ii), (d) (iii) and (iv), Which among the following statements are true for glycine?, (i) It exists in crystalline form, (ii) It is optically active, (iii) It is soluble in water, (iv) It can form Zwitter ions, (a) (i), (ii) and (iii), (b) (i), (ii) and (iv), (c) (i), (iii) and (iv), (d) (ii), (iii) and (iv), Which of the following statements are correct?, (i) Proteins on hydrolysis gives only -amino acids., (ii) Gln stands for glutamic acid., (iii) Amino acids with equal number of amino and carboxyl, groups are neutral., (iv) All naturally occuring -amino acids are optically, active., (a) (i) and (iii), (b) (i), (ii) and (iv), (c) (iii) and (iv), (d) (ii), (iii) and (iv), Which of the statements about "Denaturation" given below, are correct ?, (i) Denaturation of proteins causes loss of secondary and, tertiary structures of the protein., (ii) Denaturation leads to the conversion of double strand, of DNA into single strand, (iii) Denaturation affects primary strucrture which gets, distorted, (a) (ii) and (iii), (b) (i) and (iii), (c) (i) and (ii), (d) (i), (ii) and (iii), Of the following statements about enzymes which ones, are true?, (i) Enzymes lack in nucleophilic groups, (ii) Enzymes are highly specific both in binding chiral, substrates and in catalysing their reactions, (iii) Enzymes catalyse chemical reactions by lowering, the energy of activation, (iv) Pepsin is a proteolytic enzyme, (a) (i) and (iv), (b) (i) and (iii), (c) (ii), (iii) and (iv), (d) (i), Which of the following statements are correct ?, (i) Vitamins A, D, E and K are insoluble in water., (ii) Vitamins A, D, E and K are stored in liver and adipose, tissues., (iii) Vitamin B and vitamin C are water soluble., (iv) Water soluble vitamins should not be supplied, regularly in diet., (a) (i), (ii) and (iv), (b) (i), (ii) and (iii), (c) (i) and (iv), (d) (ii) and (iv)
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BIOMOLECULES, , 487, , 157. Which of the following statement(s) is/are correct?, (i) Information regarding the sequence of nucleotides in, the chain of a nucleic acid is called its primary structure., (ii) In secondary structure of DNA adenine forms, hydrogen bonds with guanine whereas cytosine forms, hydrogen bonds with thymine., (iii) RNA molecules are of three types m-RNA, r-RNA and, t-RNA and they all perform different functions., (a) (ii) only, (b) (i) and (iii), (c) (ii) and (iii), (d) (iii) only, 158. Consider the following statements., (i) Nucleic acids are long chain polymers of nucleotides., (ii) Sugar moiety in DNA molecules is -D-ribose whereas, in RNA molecules it is -D-2-deoxyribose., (iii) RNA contains four bases viz. adenine (A), guanine, (G), cytosine (C) and uracil (U), (iv) Nucleotide is a nucleoside linked to phosphoric acid, at 4 – position of sugar moiety., Which of the following is the correct code for the statements, above?, (a) TFFT, (b) TFTF, (c) FFTT, (d) FTFF, , MATCHING TYPE QUESTIONS, , (CHOH)4, , Column - II, Characteristic of, glucose molecule, CH3 (CH2) 4 CH3, , (p) Presence of, , CH2OH, , C = O group, , (B), , H–C, , CN, , OH, (i) NH2OH (CHOH)4, (CHOH)4, (ii) HCN, CH2OH, CH2OH, CHO, , CHO, , (C), , CHO, (CHOH)4, , (CH3COO)2O, , (CHOH)4, CH2OH, , (CH–O–C–CH3)4, , O, , CHO, Br2water, , (q) Presence of, aldehydic group, , (d) A – (r), B – (p), C – (q), D – (s), 160. Match the columns, Column - I, , Column - II, , (Enzymes), , (Reactions), , (A) Invertase, , (p) Decomposition of urea into, NH3 and CO2, , (B) Maltase, , (q) Conversion of glucose into, ethyl alcohol, , (C) Pepsin, , (r) Hydrolysis of maltose into, glucose, , (D) Urease, , (s) Hydrolysis of cane sugar, , (E) Zymase, , (t) Hydrolysis of proteins into, peptides, , (a) A – (s), B – (r), C – (t), D – (p), E – (q), (b) A – (r), B – (q), C – (s), D – (p), E – (t), , Column - I, , Column - II, , (A) Vitamin B6, , (p) Fat soluble, , (B) Vitamin K, , (q) Xerophthalmia, , (C) Vitamin D, , (r) Convulsions, , (D) Vitamin A, , (s) Delayed blood clotting, , (a) A – (p,q), B – (p,s), C – (p), D – (r), (b) A – (r), B – (p,s), C – (p), D – (p, q), (c) A – (p,s), B – (r), C – (p), D – (p,q), (d) A – (r), B – (p,s), C – (p,q), D – (p), Column - I, , (r) All six carbon, atoms are linked, in a straight, chain., , COOH, , (s) Presence of five, , (CHOH)4, , —OH groups, , CH2OH, , 161. Match the columns, , 162. Match the columns, , O, , CH2–O–C–CH3, , CH2OH, , (D), , (c) A – (r), B – (p), C – (s), D – (q), , (c) A – (q), B – (p), C – (r), D – (s), E – (t), , CHO, , (A), , (b) A – (r), B – (s), C – (p), D – (q), , (d) A – (s), B – (p), C – (t), D – (q), E – (r), , 159. Match the columns., Column - I, Reaction of glucose, , HI,, , (a) A – (p), B – (r), C – (s), D – (q), , Column - II, , (A) Vitamin A, , (p) Scurvy, , (B) Vitamin B12, , (q) Hemorrhagic condition, , (C) Vitamin C, , (r) Sterility, , (D) Vitamin E, , (s) Xerophthalmia, , (E) Vitamin K, , (t) Pernicious anaemia, , (a) A – (t), B – (s), C – (p), D – (r), E – (r), (b) A – (s), B – (t), C – (p), D – (q), E – (r), (c) A – (s), B – (t), C – (p), D – (r), E – (q), (d) A – (t), B – (s), C – (p), D – (r), E – (q)
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EBD_7207, BIOMOLECULES, , 488, , ASSERTION-REASON TYPE QUESTIONS, Directions : Each of these questions contain two statements,, Assertion and Reason. Each of these questions also has four, alternative choices, only one of which is the correct answer. You, have to select one of the codes (a), (b), (c) and (d) given below., (a) Assertion is correct, reason is correct; reason is a correct, explanation for assertion., (b) Assertion is correct, reason is correct; reason is not a, correct explanation for assertion, (c) Assertion is correct, reason is incorrect, (d) Assertion is incorrect, reason is correct., 163. Assertion : D(+)– Glucose is dextrorotatory in nature., Reason : ‘D’ represents its dextrorotatory nature., 164. Assertion : Sucrose is called an invert sugar., Reason : On hydrolysis, sucrose bring the change in the, sign of rotation from dextro (+) to laevo(–)., 165. Assertion : -glycosidic linkage is present in maltose,, , CH2OH, O, H, O, OH H, HO, H, H OH, , H, , H, , CH2OH, OH, H, OH H, OH, H OH, , Reason : Maltose is composed of two glucose units in which, C–1 of one glucose unit is linked to C–4 of another glucose, unit., 166. Assertion : At isoelectric point, the amino group does not, migrate under the influence of electric field., Reason : At isoelectric point, amino acid exists as a, zwitterion., 167. Assertion : Vitamin D cannot be stored in our body, Reason : Vitamin D is fat soluble vitamin and is excreted, from the body in urine., , CRITICAL THINKING TYPE QUESTIONS, 168. Which one of the following is the reagent used to identify, glucose?, (a) Neutral ferric chloride, (b) Chloroform and alcoholic KOH, (c) Ammoniacal silver nitrate, (d) Sodium ethoxide, 169. Glucose molecule reacts with 'X' number of molecules of, phenylhydrazine to yield osazone. The value of 'X' is, (a) four, (b) one, (c) two, (d) three, 170. In the acetylation of glucose, which group is involved in, the reaction, (a) CHO group, (b) >C = O group, (c) alcoholic OH group (d) all of these, , 171. Select the false statement about the cyclic glucose., (a) If the OH group is added to CHO group it will form, cyclic hemiacetal structure, (b) Glucose form six-membered ring in which –OH is at, C–5 position, (c) Melting point of -glucose is 423 K and of -glucose, is 419 K, , O, , H, H, HO, H, (d), H, , 1, C, 2, 3, 4, 5, 6, , OH, OH, H O, OH, , CH2OH, , H, H, HO, H, H, , 1, C, 2, 3, 4, 5, 6, , OH, H, OH, OH, CH2OH, , -D(+)-glucose, , 1, , HO C H, H 2 OH, HO 3 H O, H 4 OH, H 56, CH2OH, -D(+)-glucose, , 172.When -D-glucose and -D-glucose are dissolved in water, in two separate beakers I and II respectively and allowed to, stand, then –, (a) specific rotation in beaker I will decrease while in II will, increase upto a constant value, (b) the specific rotation of equilibrium mixture in two, beakers will be different, (c) the equilibrium mixture in both beakers will be, leavorotatory, (d) the equilibrium mixture in both beakers will contain, only cyclic form of glucose, 173. In disaccharides, if the reducing groups of monosaccharides, i.e., aldehydic or ketonic groups are bonded, these are nonreducing sugars. Which of the following disaccharide is a, non-reducing sugar?, , CH2OH, CH2OH, O, OH, H, H, H, H, H, O, OH H, OH H, (a), HO, OH, H OH, H OH, , CH2OH, O H HOH2C O, H, H, H, O, (b) HO OH H, H HO CH2OH, H OH, OH H, CH2OH, CH2OH, O, O OH, HO, HO, H, H, O, OH H, OH H, (c), H, H, H, H OH, H OH, CH2OH, CH2OH, O, O OH, H, H, H, H, O, OH H, OH H, (d), HO, H, H, H OH, H OH
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BIOMOLECULES, , 489, , 174. Which of the following pairs represents anomers?, , CHO, H, HO, (a), , OH, , HO, , H, , H, , HO, , H, , OH, , H, , OH, , H, , OH, CH2OH, , H, , OH, CH2OH, , CHO, , (b), , HO, , H, , H, HO, , H, , H, , OH, CH2OH, , HO, , H, CH2OH, , H, , OH, , HO, , H, , H, , OH, , H, O, , H, , OH, , H, , OH, , 179., , OH, , HO, , H, , HO, , OH, , O, , H, , H, H, , O, , CH2OH, , H, HO, , H, , H, H, , CH2OH, , 178., , OH, , HO, , OH, , 177., , OH, , OH, , H, H, , (d), , H, , H, , HO, , (c), , CHO, OH, , HO, , CH2OH, a, e, O H HOH2C O, H, H, H, b, a, d, e, O, OH H, H, HO, CH, HO c, b, 2OH, f, c, d, H OH, OH H, , CHO, , H, , H, , f, , OH O, , H, HO, H, , OH, CH2OH, , H, , 180., , CH2OH, , 175. Optical rotation of some compound along with their, structures are given below which of them have D, configuration., 181., , CHO, H, , CHO, H, , CH2OH, (+) rotation, , (I), , OH, , HO, OH, , CH2OH, , H, , C=O, HO, , H, , H, , OH, , H, , OH, , H, , OH, CH2OH, , H, , OH, CH2OH, , (+) rotation, , (–) rotation, , (II), , (III), , (a) I, II, III, (b) II, III, (c) I, II, (d) III, 176. Structure of a disaccharide formed by glucose and fructose, is given below. Identify anomeric carbon atoms in, monosaccharide units., , 182., , (a) ‘a’ carbon of glucose and ‘a’ carbon of fructose., (b) ‘a’ carbon of glucose and ‘e’ carbon of fructose., (c) ‘a’ carbon of glucose and ‘b’ carbon of fructose., (d) ‘f’ carbon of glucose and ‘f’ carbon of fructose., Cyclic structure of fructose resembles with, (a) pyran, (b) furan, (c) pyridine, (d) oxiran, Sucrose in water is dextro-rotatory, [ ]D= + 66.4º. When, boiled with dilute HCl, the solution becomes leavo-rotatory,, [ ]D= –20º. In this process the sucrose molecule breaks, into, (a) L-glucose + D-fructose, (b) L-glucose + L-fructose, (c) D-glucose + D-fructose, (d) D-glucose + L-fructose, Which one of the following statements is not true regarding, (+) Lactose ?, (a) On hydrolysis (+) Lactose gives equal amount of D(+), glucose and D(+) galactose., (b) (+) Lactose is a -glycoside formed by the union of a, molecule of D(+) glucose and a molecule of D(+), galactose., (c) (+) Lactose is a reducing sugar and does not exhibit, mutarotation., (d) (+) Lactose, C12H22O11 contains 8-OH groups., Which one of the following sets of monosaccharides forms, sucrose?, (a), –D-Galactopyranose and –D-Glucopyranose, (b) –D-Glucopyranose and –D-Fructofuranose, (c), –D-Glucopyranose and –D- Fructofuranose, (d) –D-Glucopyranose and –D-Fructopyranose, Which of the following statements is correct?, (a) Only the compounds following general formula, Cx(H2O)y are carbohydrates., (b) Acetic acid (CH3COOH) having general formula, C2(H2O)2 falls in this category., (c) Rhamnose having formula C6H12O5 is a carbohydrate., Though this is not according to general formula of, carbohydrates., (d) Chemically the carbohydrates may be defined as, optically inactive polyhydroxy aldehydes or ketones., The strongest form of intermolecular bonding that could, be formed involving the residue of the amino acid valine is, (a) ionic bond, (b) hydrogen bond, (c) van der Waals interactions, (d) none of the above
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EBD_7207, BIOMOLECULES, , 490, , 183. Which functional group participates in disulphide bond, formation in proteins?, (a) Thioester, , (b) Thioether, , (c) Thiol, , (d) Thiolactone, , 184. Glycosidic linkage is actually an, (a) Carbonyl bond, , (b) Ether bond, , (c) Ester bond, , (d) Amide bond, , O, , 185. For, , ||, , C N H (peptide bond ), , Which statement is incorrect about peptide bond?, (a) C — N bond length in proteins is longer than usual, bond length of the C — N bond, (b) Spectroscopic analysis shows planar structure of the, , 186. The function of enzymes in the living system is to, (a) transport oxygen, (b) provide energy, (c) provide immunity, (d) catalyse biochemical reactions, 187. Vitamin C must be supplied regularly in diet because, (a) it is water soluble hence excreted in urine and can’t be, stored in the body, (b) it is fat soluble hence stored in the body and cannot, be used on regular basis, (c) it is required in a large amount by the body hence, supplied regularly, (d) it is water soluble hence used by the body on daily, basis and is to be supplied regularly., 188. In both DNA and RNA, heterocylic base and phosphate, ester linkages are at –, (a), , C5' and C1' respectively of the sugar molecule, , (b), , C1' and C5' respectively of the sugar molecule, , (c) C — N bond length in proteins is smaller than usual, bond length of the C—N bond, , (c), , C '2 and C5' respectively of the sugar molecule, , (d) None of the above, , (d), , C5' and C '2 respectively of the sugar molecule, , — C — NH — group, ||, , O
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BIOMOLECULES, , 491, , FACT / DEFINITION TYPE QUESTIONS, 1., 2., 3., 4., , 5., , 6., 7., 9., 10., 11., , 12., , 13., , 14., , (c) Carbohydrates, proteins and fats are biomolecules., (a) Lactose is a disaccharide., (c) It is found in the milk of all animals and imparts, sweetness to milk (hence named milk sugar)., (d) The disaccharides are sugars which on hydrolysis give, two moles of the same or different monosaccharides., Sucrose, maltose and lactose (C12H22O11 ) are the, common examples., (c) Aldo-(keto) pentoses having 5 carbon, Aldo-(keto) hexoses having 6 carbon, is an example of Pentose Sugar, arabinose, (aldopentose) glucose, galactose and fructose are, important examples of hexose sugar., (d) The most common disaccharide, Lactose has the, molecular formula C12H22O11., (a), 8. (d), (d) Monosaccharides cannot be hydrolysed to simpler, molecules., (b) Sucrose is an oligosaccharide and cellulose is a, polysaccharide., (c) All those carbohydrates which reduce Fehling’s, solution and Tollens’ reagent are referred to as reducing, sugars., (a) Glucose (C 6 H12O 6 ) is the simplest molecule which, is monosaccharide while others are polysaccharides, which on hydrolysis give monosaccharides., Option (a) is correct., (d) Glucose is aldohexose. Glucose is a monosaccharide,, i.e. it can not be hydrolysed further to simple sugars., Oligosaccharides on hydrolysis give 2-10 molecules, of monosaccharides., (b) To explain the properties which can not be explained, by open chain structure of glucose it was proposed, that one of the –OH groups may add to the –CHO, group and form a cyclic hemiacetal structure as shown, below., O, , 1, H—C—OH, 2, H, OH, 3, HO, H, H 4, H 5, , OH, , 6CH2OH, , – D – (+) – Glucose, , 15., , 1, H —C, 2, H, OH, HO 3, H, H 4, OH, 5, H, OH, 6, CH, OH, 6, 2, , 1, HO —C—H, 2, H, OH, 3, HO, H, H 4, H 5, , OH, , 6CH2OH, , – D – (+) – Glucose, , (c) Glucose contain aldehyde group. Hence it give positive, Fehling solution test., , 16. (d) It is the most abundant organic compound on earth., 17. (d) Tollen’s reagent is reduced by glucose due to aldehydic, group and gives grey colour as silver metal., 18. (b) The letter ‘D’ or ‘L’ before the name of any compound, indicate the relative configuration of a particular, stereoisomer., 19. (a), 20. (a), 21. (d) Glucose is a monosaccharide, others are disaccharides., Sucrose is a combination of glucose and fructose., Maltose is a combination of two glucose units. Lactose, (or milk sugar) is a combination of glucose and, galactose (a hexose sugar)., 22. (c) Glucose is considered as a typical carbohydrate which, contains –CHO and –OH group., 23. (a) Glucose contains an aldehyde group. It is oxidised, into acidic group by bromine water and gluconic acid, is formed, CH 2 OH (CHOH) 4, , CHO, , (O), , CH 2OH (CHOH ) 4 COOH, Br2, , H 2O, , 2HBr, , O ob, , 24. (d) Weak reagent like NaHSO3 is unable to open the chain, and can’t react with glucose. This explains the inability, of glucose to form aldehyde bisulphite compound., 25. (c) Glucose has 5 hydroxyl groups, hence it reacts with, acetic anhydride to form a penta-acetate, 26. (d), 27. (d) Red P + HI is reducing agent., 28. (c) Pentaacetate of glucose does not react with, hydroxylamine, 29. (c) Open chain structure is unstable and converted to, cyclic., 30. (b), 31. (a) Natural glucose is dextrorotatory and thus, glucose is, also known as dextrose., 32. (c), H, H, HO, H, H, , 1, 2, 3, 4, , OH, OH, H, , O, , OH, , 5, CH2OH, -D-(+) - Glucose, (Fischer formula)
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EBD_7207, BIOMOLECULES, , 492, , 33., , (b), , -D glucose and –D glucose are the isomers which, differ in the orientation (configuration) of H and OH, groups around C1 atom., 1, , H —C2—OH, , H —C —OH, O, HO —C3—H, , HO —C3—H, , H —C4—OH, 5, , 6, , CH2OH, , – D - glucose, , – D - glucose, , –) of, 34. (b) T h e t wo i somer i c for m s ( – a n d, D-glucopyranose differ in configuration only at, C–1; hence these are called anomers., , 36., , (d), , 44., 45., 46., 47., 48., , H, , H, HO, , H, H, , 1, 2, 3, 4, , OH, OH, , H, , O, , 49., , OH, , 38., , (b) Maltose and glucose are reducing sugars., , 39., , (a) Glucose contains -CHO group and fructose contains, > C = O group,. Hence these are functional isomers., , 40., , (d) Glucose being an aldose responds to Tollen’s test, while fructose, although a ketose, undergoes, rearrangement in presence of basic medium (provided, by Tollen’s reagent) to form glucose, which then, responds to Tollen’s test., , – C – OR + H2O, , OR, A ketal, , A hemiketal, , (a) We know that cellulose (C 6 H12 O 6 ) n is the chief, constituent of cell walls of plants. It is the most, abundant organic substance found in nature. It is a, polymer of glucose with 3500 repeat units in a chain., (b) Sucrose is a disaccharide which on hydrolysis gives, one molecule of glucose (monosaccharide) and, fructose (monosaccharide)., H, , C6 H12 O 6 C6 H12 O 6, glucose, , fructose, , 51., 52., 53., , (d) Carbohydrates are stored in the body as glycogen., (c), (b) CH2OHCH2CHOHCHOHCH2OH does not correspond, to Cx(H2O)y., (d) Lactose (milk sugar) is a disaccharide, it is made of, -D-galactose and -D-glucose, , 54., , (d), (b) Fructose has the molecular formula C6 H12 O6. It, belongs to D-series and is laevorotatory compound. It, also exists in two cyclic forms which are obtained by, the addition of –OH at C-5 to the >C=O group. The, ring thus formed is a five membered ring and is named, as furanose with analogy to the compund Furan. Furan, is a five membered cyclic compound with one oxygen, and four carbon atoms., , ROH, , (b) RNA has D (–) – Ribose and the DNA has 2–Deoxy, D (–) – ribose as the carbohydrate unit., 5, O, O, OH, HOCH2, OH, HOCH2, 1, 4, H, H, H, H, H, H, H, H, 2, 3, OH OH, OH H, 2-deoxy ribose, ribose, From the structures it is clear that 2nd carbon in DNA, do not have OH group., , (Fischer formula), , 42., , – C – OH, , 50., , -D-(+) - Glucose, , (b) Glucose and fructose both are reduced by Fehling’s, solution, Tollen’s reagent and Bendict’s solution., Therefore, these three reagents can not be used to, distinguish between glucose and fructose., , OR, An acetal, , A hemiacetal, , sucrose, , 37., , – C – OR + H2O, , In cyclic structure of fructose, ketonic group has, reacted with an alcoholic group, it is said to be an, example of an intramolecular cyclic hemiketal., (a) Sweet taste of fruits is due to fructose., (a), (c) Honey is collected from flowers by honey bee which, contains fructose., (c) Fructose is the sweetest sugar., , C12 H 22 O11 H 2 O, , 5, CH2OH, , 41., , – C – OH, , OR, A ketone, , 5, , 6, , (b), , – C = O + ROH, , O, , H —C, , CH2OH, , – C = O + ROH, , H, ROH, , OR, , H —C4—OH, , H —C, , (c), , An aldehyde, , HO —C —H, , 2, , 35., , 43., , 1, , H —C —OH, , H, , H, , CH2OH, O, HO, H, OH H, H, H, H OH, , 55., 57., 58., , H, , CH2OH, O H, H, OH H, OH, H OH, , (b), 56. (b), (a) Sucrose does not have free — CHO or CO group, hence, it does not undergo mutarotation., (b), 59. (d)
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BIOMOLECULES, , 60., , 61., , 62., , 63., 64., , 65., , 66., , 67., 68., , 493, , (b) Sucrose is dextrorotatory but after hydrolysis gives, dextrorotatory glucose and laevorotatory fructose., Since the laevorotation of fructose (–92.4°) is more, than dextrorotation of glucose (+52.5°), the mixture is, laevororatory., (c) Chemically amylose is a long unbranched chain with, 200-1000 -D-(+ -glucose units held by C1-C4, glycosidic linkage., (b) It is a branched chain polymer of -D-glucose units in, which chain is formed by C1-C4 glycosidic linkage, whereas branching occurs by C1-C6 glycosidic linkage., (b) Cellulose is a straight chain polysaccharide., (b) The carbohydrates are stored in animal body as, glycogen. It is also known as animal starch because, its structure is similar to amylopectin and is rather more, highly branched., (d) Carbohydrates are essential for life in both plants and, animals. Honey has been used for a long time as an, instant source of energy by ‘Vaids’ in ayurvedic system, of medicine. Carbohydrates are used as storage, molecules as starch in plants and glycogen in animals., Cell wall of bacteria and plants is made up of cellulose., We build furniture etc., from cellulose in the form of, wood and cloth ourselves with cellulose in the form of, cotton fibre. They provide raw materials for many, important industries like textiles, paper lacquers and, breweries., (b) There are 20 amino acids in man out of which 10 amino, acids are essential amino acids. These essential amino, acids are supplied to our bodies by food which we, take because they cannot be synthesised in the body., These are (1) valine (2) leucine (3) Isoleucine (4) Phenyl, alanine (5) Threonine (6) Methionine (7) Lysine (8), Tryptho phone (9) Arginine (10) Histidine., (N) All the given options are example of neutral amino, acids., (b), - Amino acid is the building block unit of protein, which is formed by polymerisation of amino acid, through peptide linkage., NH C, ||, O, , 69., , 70., 71., , 72., , (a) Except alanine, all amino acids are essential amino acids, which cannot be synthesised in the body and must be, obtained through diet., (a), (c) Amino acids are the compounds having one or more, amino groups and one or more carboxyl groups in the, same molecule., (b) Except glycine, all other naturally occurring, - amino acids are optically active, since the -carbon, atom is asymmetric., , 73., , (d), , In neutral solution, amino acids exists as dipolar ion (also, known as zwitter ions or inner salts) where the proton of, –COOH group is transferred to the – NH2 group to form, inner salt, known as dipolar ion., , R, .. |, H2N–CH–COOH, -Amino acid, , R, .. |, H2N–CHOO–+H+, R, |, +, H3N–CH–COO–, H 2O, , Zwitter ion, , 74. (c) With the exception of glycine all the 19 other common, amino acids have a uniquely different functional group, on the central tetrahedral alpha carbon., H, |, H — C — COOH, |, NH 2, , glycine, 75. (d) Zwitter ion contains both +ve and –ve charge. Proton, of –COOH group is transferred to the –NH2 group.—, NH3+ group is acidic since it can donate a proton and, —COO– group is basic since it can accept a proton., 76. (a) Amino Acids are amphoteric in nature. So for it a special, term is coined called Zwitter ion., They have following structure in solution, , R, +, –, H3N–C–COO, , H, [Zwitter Ion], , 77. (d) Proline is a secondary amine, 78. (a) Proteins and peptides are linked by peptide linkages, O, ||, , ( C NH) ., , 79. (c) The bond formed between two amino acids by the, elimination of a water molecule is called a peptide, linkage or bond. The peptide bond is simply another, name for amide bond., C OH H — N—, |, , O, , |, , H, , Carboxyl group Amine group of, of one amino acid other amino acid, , — C– N— H 2 O, ||, |, O H, Peptide bond, , The product formed by linking amino acid molecules, through peptide linkages. —CO—NH—, is called a, peptide., 80. (b), 81. (c), 82. (d) Proteins are highly complex, natural compounds,, composed of a large number of different -amino-acids, joined together with peptide linkage, i.e., they are, naturally occuring polypeptides., 83. (d) All these are the examples of globular proteins. These, are soluble in water.
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EBD_7207, 494, , 84., , (c) Polypeptide chains in fibrous proteins are held together, by disulphide and hydrogen bonds., 85. (d) Proteins are building blocks of the body but they do, not provide energy for metabolism., 86. (c) Insulin is an example of globular protein., 87. (c), 88. (b), 89. (b), 90. (c) These are generally insoluble in water., 91. (b) The NH of the amide can act as a hydrogen bond donor, and the carbonyl group can act as a hydrogen bond, acceptor. Statements (a), (c) and (d) are false. The, peptide bond has double bond character due to the, interaction of the nitrogen lone pair with the carbonyl, group. This prevents bond rotation and makes the bond, planar. The trans isomer is favoured over the cis isomer., 92. (a), 93. (a) The sequence in which the -amino acids are linked to, one another in a protein molecule is called its primary, structure., 94. (a), 95. (b) The -helix structure is formed when the chain of, -amino acids coils as a right handed screw (called, -helix) because of the formation of hydrogen bonds, between amide groups of the same peptide chain, i.e.,, NH group in one unit is linked to carbonyl oxygen of, the third unit by hydrogen bonding. This hydrogen, bonding between different units is responsible for, holding helix in a position., 96. (c) During denaturation 2° and 3° structures are destroyed, but 1° structure remains intact., 97. (a) Primary structure refers to the order of the amino acids, in a protein., 98. (d) The hydrophilic/ hydrophobic character of amino acid, residues is important to tertiary structure of protein, rather than to secondary structure. In secondary, structure, it is the steric size of the residues that is, important and residues are positioned to minimise, interactions between each other and the peptide chain., 99. (d) Quaternary structure refers to the overall structure of, a multiprotein complex where as primary, secondary, and tertiary structure refer to the different structural, levels of a single protein., 100. (a) The arrangement of polypeptide chains formed as a, result of hydrogen bonding is called secondary, structure of proteins., -helix is formed by intramolecular H-bonding., -pleated sheet is formed by intermolecular H-bonding., 101. (b) The secondary structure of a protein refers to the, shape in which a long peptide chain can exist. There, are two different conformations of the peptide linkage, present in protein, these are -helix and -conformation., The -helix always has a right handed arrangement. In, -conformation all peptide chains are streched out to, nearly maximum extension and then laid side by side, and held together by intermolecular hydrogen bonds., The structure resembles the pleated folds of drapery, and therefore is known as -pleated sheet., , BIOMOLECULES, , 102. (b) In this structure of protein atoms are highly coiled, and form a spherical form., 103. (d), 104. (d) When a protein, in its native form, is subjected to a, physical change like change in temperature, or a, chemical change like change in pH, the native, conformation of the molecule is disrupted and proteins, so formed are called denaturated proteins., The denaturation may be reversible or irreversible. The, coagulation of egg on boiling is an example of, irreversible protein denaturation., However, it has been shown now that in some cases,, the process is actually reversible. The reverse process, is called renaturation., 105. (c) Tertiary structure indicates the overall structure of the, protein., 106. (b), 107. (b) Serine contains a hydroxyl functional group on its side, chain and so the strongest possible interaction will be, hydrogen bonding where the hydroxyl group could, act as a hydrogen bond donor or hydrogen bond, acceptor., 108. (a) When antigens enter in to the body cells and destroy, them, then antibodies being proteins are synthesised, in the body and combine with antigens and destroy, these antigens by forming inactive complexes., Therefore antibodies protein destroy antigens., 109. (b) Enzymes are highly specific for a particular reaction, and also for a particular substrate., 110. (b) Enzymes being biocatalyst can increase the rate of a, reaction upto 10 million times. Even very small amount, can accelerate a reaction., 111. (b) Enzymes are made up of protein with specific structure., 112. (b) Triglycerides are lipids, hence these are hydrolysed, by lipases to glycerol and fatty acids., 113. (d) Enzymes are most reactive at optimum temperature., The optimum temperature for enzyme activity lies, between 40°C to 60°C., 114. (b) Insulin is a biochemically active peptide harmone, secreted by pancreas., 115. (a), 116. (d) Enzymes may or may not require a coenzyme for their, catalytic action., 117. (c) Deficiency of vitamin D causes rickets., 118. (d) Vitamin D is a fat soluble vitamin., 119. (b) Beri-Beri., 120. (d) Vitamin, Disease caused by deficiency, B6, Dermatitis, B1, Beri-beri, B2, Photophobia, glossitis, B12, Pernicious anaemia, 121. (b) Vitamin E is mainly present in vegetable oils like wheat, gram oil, sunflower oil, etc., 122. (a) Vitamin A or retinol.
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BIOMOLECULES, , 495, , 123. (b), 124. (c) It is found in liver, egg, milk, meat, and fish. Minute, amounts are probably present in all animal cells., Peculiarly, unlike other vitamins, B12 is not found in, significant amounts in green plants., 125. (a) DNA consists of two polynucleotide chains, each, chain forms a right handed helical spiral with ten bases, in one turn of the spiral. The two chains coil to double, helix and run in opposite direction held together by, hydrogen bonding., 126. (d), , 127. (a), , 128. (b), , 129. (c), 130. (d), , 131. (c), , O, OH P, O, P, OH, O, OH P, , Deoxyribose-Adenine ... Thymine -Deoxyribose, Deoxyribose-Guanine ... Cytosine -Deoxyribose, Deoxyribose-Guanine ... Cytosine -Deoxyribose, Deoxyribose-Adenine ... Thymine -Deoxyribose, , O, P OH, O, P, OH, O, P OH, , The hydrogen bonds are formed between the base, (shown by dotted lines). Because of size and, geometrics of the bases, the only possible pairing in, DNA and between G(Guanine) and C(Cytosine), through three H-bonds and between A (Adenosine), and T (Thymine) through two H-bonds., In DNA the complimentary base are Adenine and, thymine., Guanine and cytosine, The genetic information for cell is contained in the, sequence of bases A, T, G and C in DNA molecule., The DNA sequence that codes for a specific protein is, called a Gene and thus every protein in a cell has a, corrosponding gene., The base pairs of the two strands of DNA are linked, together through H-bonds., DNA has the property of self - replication . It is therefore, a reproducing molecule. This unique property of DNA, is at the root of all reproduction. Through its, replication, DNA acts as the key to heredity. In the, replication of DNA, the two strands of a double helix, unwind and separate as a template for the formation of, a new complementary strand., Each chromosome is made up of DNA tightly coiled, many times around proteins called histones that, supports its structure., , 132. (a), 133. (a) DNA has double stranded -helical structure., 134. (b) DNA is responsible for transmission of heredity, character., 135. (b), 136. (a) Energy is stored in our body in the form of A.T.P., 137. (d), 138. (a) Phosphoric acid is the third component in DNA., 139. (a) Synthesis of polypeptide is known as translation. For, this process three type of RNA are essential., , 140. (d) The correct structure of thymine is, OH, N, HO, , CH3, N, , Thymine (T), , 141., 142., 143., 144., , (c), (d) Tyrosine is an -amino acid, and not a purine, (d), (c) DNA fingerprinting is same for every cell and cannot, be altered by any known treatment., , STATEMENT TYPE QUESTIONS, 145. (a), 146. (c) Sucrose is non-reducing in nature. It does not contain, a free aldehydic or ketonic group. Maltose is a, reducing sugar., 147. (b) Naturally occurring monosaccharides are 20 only., Sucrose is a non–reducing sugar whereas maltose is a, reducing sugar., 148. (b) The six membered cyclic structure of glucose is called, pyranose structure ( or ), in analogy with pyran., Pyran is a cyclic organic compound with one oxygen, atom and five carbon atoms in the ring. The cyclic, structure of glucose is correctly represented by, Haworth structure., 149. (d) Sucrose is a dextrorotatory but on hydrolysis gives, dextrorotatory glucose and laevorotatory fructose., Since the laevorotation of fructose (–92.4°) is more, than dextrorotation of glucose (+52.5°), thus the, resulting mixture is laevorotatory. Glycogen is, structurally similar to amylopectin not amylose., 150. (a) For statement (i) correct Haworth structure for –D, glucose will be., 6, CH2OH, O, 5, H, H, H, 1, 4, OH, OH OH, H, 3, 2, H, OH, 151. (c) Amylose is water soluble component which constitutes, about 15-20% of strach. Amylopectin is insoluble in, and constitutes about 80-85% of starch., 152. (c) Glycine is optically inactive., 153. (a) Gln stands for glutamine. Except glycine, all other, naturally occurring –amino acids are optically active., 154. (c) When the proteins are subjected to the action of heat,, mineral acids or alkali, the water soluble form of globular, protein changes to water insoluble fibrous protein. This, is called denaturation of proteins. During denaturation, secondary and tertiary structures of protein destroyed, but primary structures remains intact.
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EBD_7207, BIOMOLECULES, , 496, , group, the third molecule of phenylhydrazine condenses, to glucosazone. Therefore the value of X is 3, , 155. (c), 156. (b) Water soluble vitamins must be supplied regularly in, diet because they are readily excreted in urine and, cannot be stored (except vitamin B12) in our body., 157. (b) In secondary structure of DNA adenine forms, hydrogen bonds with thymine whereas cytosine forms, hydrogen bonds with guanine., 158. (b) The sugar moiety in DNA molecule is -D-2deoxyribose whereas in RNA molecule, it is -D-ribose., Nucleotide is a nucleoside linked to phosphoric acid, at 51– position of sugar moiety., , CHO + H2NNHC6H5, , CH = NNHC6H5, , CHOH, , CHOH, warm, , (CHOH)3, , CH(OH)3, , CH2OH, , CH2OH, , Glucose, , Glucose Phenylhydrozone, H2NNHC6 H5, , MATCHING TYPE QUESTIONS, 159. (c), 162. (c), , 160. (a), Vitamin A, Vitamin B12, Vitamin C, Vitamin E, Vitamin K, , CH = NNHC6H5, , 161. (b), -, , H2NNHC6 H5, – H2 O, , C=O, , C = NNHC6H5, , Xerophthalmia, Pernicious anaemia, Scurvy, Sterility, Haemorrhage, , CH = NNHC6H5, , (CHOH)3, , CH(OH)3, CH2OH, , CH2OH, , Glucosazone, , keto compound of glucose, phenyle hydra zone, , ASSERTION-REASON TYPE QUESTIONS, 163. (c), 164. (a) The hydrolysis of sucrose brings about a change in, the sign of rotation from dextro (+) to laevo (–) and the, product is named as invert sugar., 165. (d) Maltose is compound of two -D - glucose units in, which C1 of one glucose is linked to C4 of another, glucose unit., 6, , 6, , CH2OH, H 5 OH, H, 1, 4, OH, H, HO 3 2, H OH, , I, -D-glucose, , CH2OH, OH, 5, H, 4, 1, OH, H, 3, 2 OH, H OH, , H, , O, , 170. (c) OHC.(CHOH)4.CH2OH + 5Ac2O, Glucose, , or penta-acetyl Glucose, , 171. (c) Melting point of -glucose, 323 K., 172. (a), , 419 K and -glucose is, , -D-glucose or -D-glucose when dissolved in water, and allowed to stand, following equilibrium is, stablished, which is called mutarotation., -D-glucose, (+111°), , Open chain form, , -D-glucose, (+19°), , Specific rotation of -form falls until a constant value, of +52.5° is reached. On the other hand, specific, rotation of form increases. Specific rotation of, equilibrium mixture is 52.5°., , 166. (a) R is the correct explanation of A., 167. (d) Vitamin D is a fat soluble vitamin and can be stored in, the body since it is not excreted out of the body., , 168. (c) Glucose contains aldehyde group due to which it gives, positive test with ammoniacal silver nitrate., 169. (d) We know that glucose reacts with one molecule of, phenyl hydrazine to give phenyl hydrazone. When, warmed with excess of phenylhydrazine, the, secondary alcoholic group adjacent to the aldehyde, group is oxidised by another molecule of, phenylhydrazine to a ketonic group. With this ketonic, , Acetic anhydride, OHC. (CHOAc) 4 . CH2 OAc, Glucose penta-acetate, , II, -D-glucose, , CRITICAL THINKING TYPE QUESTIONS, , Au.ZnCl2, , 173. (b) The two monosoccharides are held together by a, glycosidic linkage between C1 of -glucose and C2 of, -fructose. Since the reducing groups and glucose and, fructose are involved in glycosidic bond formed., Sucrose is non-reducing sugar., 174. (c) Cyclic hemiacetal forms of monosaccharide which differ, only in the configuration of the hydroxyl group at C1, are anomers., 175. (a), , 176. (c)
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BIOMOLECULES, , 497, , O, , 6, , 177. (b), , HOH2C, 5, , 1, , CH2OH, 2, , OH, , H, , 3, , 4, , H, , Sucrose is a disaccharide of –D-Glucopyranose and, –D-Fructofuranose., 181. (c) Most of the carbohydrates have a general formula,, Cx(H2O)y, and were considered as hydrates of carbon., All the compounds which fit into this formula may not, be classified as carbohydrates. Acetic acid (CH3COOH), fits into this general formula, C2(H2O)2 but is not a, carbohydrate. Similarly, rhamnose, C6H12O5 is a, carbohydrate but does not fit in this definition. A large, number of their reactions have shown that they contain, specific functional groups. Chemically, the, carbohydrates may be defined as optically active, polyhydroxy aldehydes or ketones or the compounds, which produce such units on hydrolysis., 182. (c) Valine has no functional groups on its side chain. There, is only an alkyl group and so only van der Waals, interactions are possible., , OH, , H, , OH, , -D-(–)-Fructofuranose, , O, , 6, , HOH2C, 5, , OH, OH, , H, , 3, , 4, , H, , 2, , CH2OH, 1, , H, , OH, , -D-(–)-Fructofuranose, , 178. (c) The hydrolysis of sucrose by boiling with mineral acid, or by enzyme invertase or sucrase produces a mixture, of equal molecules of D(+) glucose and, D(–) Fructose., C12 H 22O11, , H 2O, , sucrose, , [ D], , 179. (c), , HO, H, , HCl, , C 6 H12O 6, , 66.5º, , CH2OH, O, H, OH, H, H, , C 6 H12O 6, , D glu cos e, , D Fructose, , [ D ] 52.5º [ D ] 92 º, Invert sugar ,[ D ] 20º, OH, , H, OH, H, , O, H, , H, , O, CH2OH, , H, , OH, , (b), , HO, O, OH, , CH2, OH, , OH, , HO, , OH, -D-Glucopyranose, , O, OH, , +, , H2C, OH, , OH, , -D-Fructofuranose, , HO, , OH, , CH2, O, OH, , CH2, O, , OH, OH, Fructose, , Disulphide, , Thiol, , Example :, 2HO 2C CHC H 2SH, |, NH2, Cystine, , HO 2CCHC H 2S SCH 2 CHC O 2 H, |, |, NH 2, NH2, , 184. (b) Glycosidic linkage is actually an ether bond as the, linkage forming the rings in an oligosaccharide or, polysaccharide is not just one bond, but the two bonds, sharing an oxygen atom e.g. sucrose, , OH, , OH, , CH2, , R S S R, , 2R S H, , Cystine, , H, , (Lactose), All reducing sugar shows mutarotation., 180., , 183. (c), , O, HO, , + H2O, CH2, , OH, , OH
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EBD_7207, BIOMOLECULES, , 498, , 185. (a) Due to resonance C — N bond in protein acquires, double bond character and is smaller than usual, C — N bond., O, , O, C, , C, , NH, , –, , NH, , 186. (d) The function of enzymes in the living system is to, catalyse biochemical reactions which occur in living, systems. e.g. invertase, pepsin, amylase., Sucrose, , Invertase, , glucose + fructose, , (polymer), , Starch, (polymer), , (monomer), , amylase, , glucose, (monomer), , 187. (a) Vitamin C is water soluble. Therefore, it is readily, excreted in urine and cannot be stored in our body and, is supplied regularly in diet., 188. (b) In DNA and RNA heterocyclic base and phosphate, ester are at C1' and C5' respectively of the sugar, molecule., HO, |, 5, HO – P – O – C H 2, ||, O, , H, , N, 4, C, , O, H, H, |, |, C3––– C2, |, |, OH OH, , N, C, , H, , N, N
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29, POLYMERS, FACT / DEFINITION TYPE QUESTIONS, 1., , 2., , 3., , 4., , 5., , 6., , 7., , 8., , 9., , 10., , 11., , Which is not true about polymers?, (a) Polymers do not carry any charge, (b) Polymers have high viscosity, (c) Polymers scatter light, (d) Polymers have low molecular weight, Which of the following belongs to the class of natural, polymers?, (a) Proteins, (b) Cellulose, (c) Rubber, (d) All of these, Which of the following natural products is not a polymer?, (a) DNA, (b) Cellulose, (c) ATP, (d) Urease, Among the following a natural polymer is, (a) cellulose, (b) PVC, (c) teflon, (d) polyethylene, Which of the following is not a biopolymer ?, (a) Proteins, (b) Rubber, (c) Cellulose, (d) RNA, Rayon is :, (a) synthetic plastic, (b) natural rubber, (c) natural silk, (d) artificial silk, Protein is a polymer of:, (a) glucose, (b) terephthalic acid, (c) amino acids, (d) None of these, Natural silk is a, (a) polyester, (b) polyamide, (c) polyacid, (d) polysaccharide, Polymers are:, (a) micromolecules, (b) macromolecules, (c) sub-micromolecules (d) None of the above, Which of the following is/are a semisynthetic polymers?, (a) Cellulose acetate, (b) Polyvinyl chloride, (c) Cellulose nitrate, (d) Both (a) and (c), Which of the following is not linear polymer ?, (a) Bakelite, (b) Polyester, (c) Cellulose, (d) High density polyethene, , 12. A polymer is formed when simple chemical units, (a) combine to form long chains, (b) combine to form helical chains, (c) break up, (d) become round, 13. Polymer formation from monomers starts by, (a) condensation reaction between monomers, (b) coordinate reaction between monomers, (c) conversion of monomer to monomer ions by protons, (d) hydrolysis of monomers., 14. On the basis of mode of formation, polymers can be, classified?, (a) as addition polymers only, (b) as condensation polymers only, (c) as copolymers, (d) both as addition and condensation polymers, 15. In addition polymer monomer used is, (a) unsaturated compounds, (b) saturated compounds, (c) bifunctional saturated compounds, (d) trifunctional saturated compounds, 16. Nylon 66 belongs to the class of, (a) Addition polymer, (b) Condensation polymer, (c) Addition homopolymer, (d) Condensation heteropolymer, 17. A polymer made from a polymerization reaction that, produces small molecules (such as water) as well as the, polymer is classified as a/an ..... polymer., (a) addition, (b) natural, (c) condensation, (d) elimination, 18. In elastomer, intermolecular forces are, (a) strong, (b) weak, (c) nil, (d) None of these, 19. A thermoplastic among the following is, (a) bakelite, (b) polystyrene, (c) terylene, (d) urea-formaldehyde resin, 20. Which is an example of thermosetting polymer?, (a) Polythene, (b) PVC, (c) Neoprene, (d) Bakelite
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EBD_7207, POLYMERS, , 500, , 21., , 22., , 23., , 24., , 25., , 26., , 27., , 28., , 29., , 30., , 31., , Which of the following is thermoplastic ?, (a) Bakelite, (b) Polyethylene, (c) Terylene, (d) All of these, Thermosets are:, (a) cross-linked polymers, (b) don’t melt or soften on heating, (c) cross-linking is usually developed at the time of, moulding where they harden reversibly, (d) all of the above, Which is/are true for elastomers?, (a) These are synthetic polymers possessing elasticity, (b) These possess very weak intramolecular forces or, attractions between polymer chains, (c) Vulcanised rubber is an example of elastomer, (d) All of the above, Among the following polymers the strongest molecular, forces are present in, (a) elastomers, (b) fibres, (c) thermoplastics, (d) thermosetting polymers, Three dimensional molecular structure with cross links are, formed in the case of a, (a) thermoplastic, (b) thermosetting plastic, (c) Both (a) and (b), (d) None of the above, Which of the following polymer is an example of fibre ?, (a) Silk, (b) Dacron, (c) Nylon-66, (d) All of these, Which of the following statements is not correct for fibres?, (a) Fibres possess high tensile strength and high, modulus., (b) Fibres impart crystalline nature., (c) Characteristic features of fibres are due to strong, intermolecular forces like hydrogen bonding., (d) All are correct., Which of the following is/are examples of fibres?, (a) Polyesters, (b) Polyamide, (c) Polythene, (d) Both (a) and (b), Which of the following can be repeatedly soften on, heating?, (i) Polystyrene, (ii) Melamine, (iii) Polyesters, (iv) Polyethylene, (v) Neoprene, (a) (i) and (iii), (b) (i) and (iv), (c) (iii), (iv) and (v), (d) (ii) and (iv), Which of the following does not undergo addition, polymerization?, (a) Vinylchloride, (b) Butadiene, (c) Styrene, (d) All of the above undergoes addition polymerizations, Which of the following is a cross linked polymer?, (a) PVC, (b) Bakelite, (c) Polyethylene, (d) Rubber, , 32., , 33., , 34., , 35., , 36., , 37., , 38., , 39., , 40., , 41., , 42., , 43., , Fibres that have good resistance to stains, chemicals,, insects and fungi is, (a) Acrylic, (b) Terylene, (c) Nylon, (d) All of these, Which of the following statements is not true about low, density polythene?, (a) Tough, (b) Hard, (c) Poor conductor of electricity, (d) Highly branched structure, Low density polythene is prepared by, (a) Free radical polymerisation, (b) Cationic polymerisation, (c) Anionic polymerisation, (d) Ziegler-Natta polymerisation, The monomer of teflon is, (a) CHF == CH2, (b) CF2 == CF2, (c) CHC1 == CHCl, (d) CHF == CHC1, The monomer(s) used in the preparation of Orlon, a, substitute for wool is/are, (a) caprolactam, (b) tetrafluoroethene, (c) styrene and 1, 3-butadiene, (d) acrylonitrile, Orlon is a polymer of, (a) styrene, (b) tetrafluoroethylene, (c) vinyl chloride, (d) acrylonitrile, Which of the following polymer is used for manufacturing, of buckets, dustbins, pipes etc ?, (a) Low density polythene, (b) High density polythene, (c) Teflon, (d) Polyacrylonitrile, Which of the following catalyst is used in preparation of, high density polythene ?, (a) Peroxide catalyst, (b) Ziegler - Natta catalyst, (c) Wilkinson’s catalyst, (d) Pd - catalyst, Which of the following statements is false?, (a) Artificial silk is derived from cellulose., (b) Nylon-66 is an example of elastomer., (c) The repeat unit in natural rubber is isoprene., (d) Both starch and cellulose are polymers of glucose., Melamine plastic crockery is a copolymer of:, (a) HCHO and melamine (b) HCHO and ethylene, (c) melamine and ethylene (d) None of these, Caprolactam polymerises to give, (a) terylene, (b) teflon, (c) glyptal, (d) nylon-6, Nylons, polysters and cotton, all posses strength due to:, (a) intermolecule H-bonding, (b) van der Waals’ attraction, (c) dipole-dipole interaction, (d) None of the above
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POLYMERS, , 501, , 44. Nylon 66 is a polyamide obtained by the reaction of, (a) COOH(CH2)4 COOH + NH2C6H4NH2, (b) COOH(CH2)4 COOH + NH2 (CH2)6 NH2, (c) COOH (CH2)6 COOH + NH2 (CH2)4 NH2, (d) COOHC6H4 COOH–(p) + NH2 (CH2)6 NH2, 45. Interparticle forces present in nylon-66 are, (a) van der waal’s, (b) hydrogen bonding, (c) dipole-dipole interactions, (d) None of these, 46. The plastic household crockery is prepared by using, (a) melamine and tetrafluoroethane, (b) malonic acid and hexamethyleneamine, (c) melamine and vinyl acetate, (d) melamine and formaldehyde, 47. Which of the following is currently used as a tyre cord ?, (a) Terylene, (b) Polyethylene, (c) Polypropylene, (d) Nylon - 6, 48. Of the following which one is classified as polyester, polymer ?, (a) Terylene, (b) Bakelite, (c) Melamine, (d) Nylon-66, 49. Which one of the following is not a condensation polymer ?, (a) Melamine, (b) Glyptal, (c) Dacron, (d) Neoprene, 50. Bakelite is obtained from phenol by reacting with, (a) (CH2OH)2, (b) CH3CHO, (c) CH3 COCH3, (d) HCHO, 51. The polymer containing strong intermolecular forces e.g., hydrogen bonding, is, (a) teflon, (b) nylon 6, 6, (c) polystyrene, (d) natural rubber, 52. Nylon threads are made of, (a) polyester polymer, (b) polyamide polymer, (c) polyethylene polymer (d) polyvinyl polymer, 53. Which compound/set of compounds is used in the, manufacture of nylon 6?, (a), , CH = CH2, , (b) HOOC(CH2)4COOH + NH2(CH2)6 NH2, , (c), , CH 2, , CH3, |, CH C CH 2, , O, (d), 54., , The repeating unit present in Nylon 6 is, (a) — [NH(CH2)6NHCO(CH2)4CO] —, (b) — [CO(CH2)5NH] —, (c) — [CO (CH2)6NH] —, (d) — [CO (CH2)4NH] —, , 55. Which of the following polymer is a polyamide ?, (a) Terylene, (b) Nylon, (b) Rubber, (d) Vulcanised rubber, 56. Which of the following compound is used for preparation, of melamine formaldehyde polymer ?, , NH2, (a), , N, (b), , H2N, , N, , NH2, , NH2, , H2N, , NH2, , N, N, , N, , (c), , N, , NH2, NH2, (d), H2N, , N, , 57. Acetic acid is added in the preparation of Nylon-6 due to, (a) initiate polymerisation, (b) avoid polymerisation at first step, (c) avoid oxidation, (d) removal of water, 58. The bakelite is made from phenol and formaldehyde. The, initial reaction between the two compounds is an example, of:, (a) aromatic electrophilic substitution, (b) aromatic nucleophilic substitution, (c) free radical reaction, (d) aldol reaction, 59. Melamine plastic crockery is a codensation polymer of, (a) HCHO and melamine, (b) HCHO and ethylene, (c) melamine and ethylene, (d) None of these, 60. Which of the following polymer is used for making, phonograph records ?, (a) Bakelite, (b) Dacron, (c) Teflon, (d) PVC, 61. Novolac is, (a) cross-linked polymer (b) linear polymer, (c) addition polymer, (d) synthetic rubber
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EBD_7207, POLYMERS, , 502, , 62., , 63., , Dacron is a –, (a) crease resistant, (b) polyamide, (c) addition polymer, (d) polymer of ethylene glycol and phthalic acid, The monomeric units of terylene are glycol and which of, the following, , 67., , 68., , OH, OH, , (a), , (b), OH, , OH, , 69., , OH, , (c), , COOH, , (d), , COOH, , OH, , 70., , OH, , 64., , Which of the following is novolac ?, CH3, , (a), , 71., , CH3, O–, , –O, , 72., OH, , –H2C, , CH3, , 73., , (b), , OH, CH2 –, , (c), , 74., , Which of the following is an use of butadiene – styrene, copolymer?, (a) Manufacture of autotyres, (b) Footwear components, (c) Cable insulation, (d) All of these, A homopolymer is obtained by polymerisation of:, (a) one type of monomer units, (b) two types of monomer units, (c) either of the above, (d) None of the above, A copolymer of isobutylene and isoprene is called:, (a) align butyl rubber, (b) buna-S, (c) buna-N, (d) thiokol, Which one is a homopolymer?, (a) Bakelite, (b) Nylon, (c) Terylene, (d) Neoprene, Polymerisation in which two or more chemically different, monomers take part is called:, (a) addition polymerisation, (b) copolymerisation, (c) chain polymerisation, (d) homo polymerisation, Natural rubber is a polymer of, (a) butadiene, (b) isoprene, (c) 2-methylbutadiene, (d) hexa-1, 3-diene, Which one of the following statement is not true?, (a) In vulcanization the formation of sulphur bridges, between different chains make rubber harder and, stronger., (b) Natural rubber has the trans -configuration at every, double bond, (c) Buna-S is a copolymer of butadiene and styrene, (d) Natural rubber is a 1, 4 - polymer of isoprene, Natural rubber is polymer of, (a) CH2 = CH – Cl, CH3, |, C CH, , CH2 –, , (b) cis CH 2, , OH, , OH, , CH2, , CH2 –, (c) trans CH 2, , (d), , 65., , 66., , Soft drinks and baby feeding bottles are generally made up, of, (a) Polystyrene, (b) Polyurethane, (c) Polyurea, (d) Polyamide, Which is not an example of copolymer ?, (a) SAN, (b) ABS, (c) Saran, (d) PVC, , (d) cis CH 2, 75., , CH3, |, C CH, , CH 2, , CH 2, , CH3 CH3, |, |, C, C, CH 2, , Which of the following is not the property of natural rubber, (a) Low tensile strength, (b) High water absorption capacity, (c) Soft and sticky, (d) High elasticity
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POLYMERS, , 76., , 77., , 78., , 79., , 80., , 81., , 82., , 503, , Natural rubber is, (a) all trans polyisoprene (b) all cis-polysioprene, (c) chloroprene, (d) Buna-N, The process involving heating of rubber with sulphur is, called, (a) Galvanisation, (b) Vulcanization, (c) Bessemerisaion, (d) Sulphonation, Isoprene is a valuable substance for making, (a) propene, (b) liquid fuel, (c) synthetic rubber, (d) petrol, Synthetic polymer which resembles natural rubber is :, (a) neoprene, (b) chloroprene, (c) glyptal, (d) nylon, Synthetic rubber is:, (a) polyester, (b) polyamide, (c) polysaccharide, (d) poly (halodiene), Which of the following are example of synthetic rubber?, (i) Polychloroprene, (ii) polyacrylonitrile, (iii) Buna-N, (iv) cis-polyisoprene, (a) (i) and (iii), (b) (i) and (ii), (c) (iii) and (iv), (d) (ii) and (iii), Buna-N synthetic rubber is a copolymer of :, (a) H2C = CH – CH = CH2 and H5C6 – CH = CH2, (b) H2C = CH – CN and H2C = CH – CH = CH2, (c) H2C = CH – CN and H 2 C = CH – C = CH 2, |, , CH3, Cl, |, , 83., , (d) H 2 C = CH – C = CH 2 and H 2C = CH – CH = CH 2, Which of the following structures represents neoprene, polymer?, (a), , (b), , CN, |, –(CH 2 – CH –) n, , (c), , Cl, |, –( CH 2 – CH –)n, , (d), , 84., , 85., , –(CH 2 – C = CH – CH 2 –) n, |, Cl, , –( CH – CH 2 –)n, |, C6 H5, , Which of the following is not a copolymer?, (a) Buna–S, (b) Baketite, (c) Neoprene, (d) Dacron, In the manufacture of tyre rubber, the percentage of sulphur, used as a crosslinking agent is, (a) 2%, (b) 5%, (c) 10%, (d) 0.5%, , 86. Which of the following polymer is formed on reaction of, 1,3–butadiene and acrylonitrile?, (a) Buna–S, (b) Buna – N, (c) Neoprene, (d) Dacron, 87. Which of the following monomers form biodegradable, polymers?, (a) 3-hydroxybutanoic acid + 3-hydroxypentanoic acid, (b) Glycine + amino caproic acid, (c) Ethylene glycol + phthalic acid, (d) both (a) and (b), 88. The polymer which undergoes environment degradation, by microoganism is known as, (a) chain-growth polymer, (b) chain step polymer, (c) biodegradable polymer, (d) non-biodegradable polymer, 89. Generally, molecular mass of a polymer is over, (a) 100, (b) 500, (c) 1,000, (d) 10,000, 90. For natural polymers PDI is generally, (a) 0, (b) 1, (c) 100, (d) 1000, , STATEMENT TYPE QUESTIONS, 91. Which of the following statement(s) is/are correct?, (i) Macromolecules have high molecular mass of order, 103 – 107u., (ii) Monomeric units are joined together by ionic or, covalent bond., (a) Only (i), (b) Only (ii), (c) Both (i) and (ii), (d) Neither (i) nor (ii), 92. Consider the following statements., (i) Polystyrene is a homopolymer whereas Buna–N is a, copolymer., (ii) Condensation polymers can be obtained by, condensation between two similar bi–functional, monomeric units., (iii) Elastomers are the polymers in which the polymeric, chains are held together by the weakest intermolecular, forces., (iv) Buna–S and Buna–N consist of close packing of chains, which impart them crystalline nature., Which of the following is the correct code for the statements, above?, (a) TTFF, (b) TFTF, (c) FTFT, (d) TFFT, 93. Two condensation polymers are made, (1) ethylene diamine + ethane–1, 2– dicarboxylic acid, (2) trimethylenediamine + ethane–1, 2– dicarboxylic acid, if both polymers of same molecular weight are obtained, then which of the following statements is/are correct ?, (i) Polymer (1) is found to melt at lower temperature., (ii) Polymer (2) is found to melt at lower temperature., (iii) H-bonding is major factor., (a) (i), (ii) and (iii), (b) Only (ii), (c) (i) and (iii), (d) (ii) and (iii)
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EBD_7207, POLYMERS, , 504, , 94., , 95., , 96., , Which of the following statements are correct?, (i) A polyamide nylon 6,6 prepared by the condensation, polymerisation of hexamethylene diamine with adipic, acid is used in the manufacture of tyre cords., (ii) Terylene is crease resistant and is blended with cotton, and wool fibres for various applications., (iii) Condensation reaction of phenol and formaldehyde, to form novolac can be catalysed either by acid or, base., (iv) Melamine formaldehyde polymer is mainly used in the, manufacture of electrical switches, (a) (i), (ii) and (iii), (b) (ii), (iii) and (iv), (c) (iii) and (iv), (d) (ii) and (iii), Read the following statements., (i) Rubber latex is a colloidal dispersion of rubber in water., (ii) Natural rubber is a cis –1, 4–polyisoprene having, elastic properties due to coiled structure and weak, van der Waal’s forces., (iii) Vulcanisation of natural rubber with sulphur and an, appropriate additive is carried out above 415K., (iv) In the manufacture of tyre rubber, 5% of sulphur is, used as a cross-linking agent., (v) Synthetic rubbers are homopolymers., Which of the following is the correct code for the, statements above?, (a) FFTFT, (b) TTFTF, (c) TTFFF, (d) FFTTF, Which of the following statements are correct?, (i) Buna–N being resistant to the action of petrol,, lubricating oil and organic solvents is used in making, oil seals., (ii) Biodegradable polymers are manufactured because, of low chemical resistance, strength and durability of, conventional polymers., (iii) PHBV is a copolymer used in the manufacture of, orthopaedic devices., (iv) Nylon 2–nylon 6 is a biodegradable polymer., (a) (i), (ii) and (iii), (b) (ii), (iii) and (iv), (c) (i), (iii) and (iv), (d) (i) and (iv), , MATCHING TYPE QUESTIONS, 97. Match the columns., Column-I, Column-II, (A) Linear polymer, (p) Melamine, (B) Semisynthetic polymer (q) Polyvinyl chloride, (C) Branched chain polymer (r) LDPE, (D) Network polymer, (s) Cellulose nitrate, (a) A– (s), B – (q), C – (r), D – (p), (b) A– (q), B – (s), C – (r), D – (p), (c) A– (q), B – (r), C – (s), D – (p), (d) A– (q), B – (s), C – (p), D – (r), , 98., , Match the columns., Column-I, Column-II, (A) Highly branched, (p) Teflon, chemically inert polymer, used in the insulation of, electric wires., (B) Linear polymer, (q) Polyacrylonitrile, prepared in presence of, Al(C2H5)3 and TiCl4., (C) Corrosion resistant, (r) HDPE, polymer used in, manufacture of non–stick, surface coated utensils., (D) Addition polymer, (s) LDPE, used as a substitute for, wool., (a) A–(s), B–(r), C–(q), D–(p), (b) A–(s), B–(p), C–(r), D–(q), (c) A–(s), B–(r), C–(p), D–(q), (d) A–(r), B–(s), C–(p), D–(q), 99. Match Column-I (Monomer) with Colum-II (Polymer) and, select the correct answer using the codes given below the, lists:, Column-I, Column-II, (A) Hexamethylenediamine, (p) Bakelite, (B) Phenol, (q) Dacron, (C) Phthalic acid, (r) Glyptal, (D) Terephthalic acid, (s) Melamine, (t) Nylon, (a) A – (t), B – (p), C – (q), D – (r), (b) A – (t), B – (p), C – (r), D – (q), (c) A – (s), B – (r), C – (p), D – (q), (d) A – (s), B – (r), C – (q), D – (p), 100. Match the columns, Column-I, Column-II, (A) Polyester of glycol and, (p) Novalac, phtalic acid, (B) Copolymer of 1, 3-butadiene (q) Glyptal, and styrene, (C) Phenol and formaldehyde, (r) Buna-S, resin, (E) Polyester of glycol and, (s) Buna-N, terephthalic acid, (F) Copolymer of 1, 3-butadiene (t) Dacron, and acrylonitrile, (a) A – (q), B – (s), C – (p), D – (t), E – (r), (b) A – (q), B – (r), C – (p), D – (t), E – (s), (c) A – (r), B – (p), C – (t), D – (s), E – (q), (d) A – (p), B – (s), C – (q), D – (t), E – (r)
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POLYMERS, , 101. Match the polymers given in Column-I with their chemical, names given in Column-II, Column-I, Column-II, (A) Nylon 6, (p) Polyvinyl chloride, (B) PVC, (q) Polyacrylonitrile, (C) Acralin, (r) Polycaprolactum, (D) Natural rubber, (s) Low density polythene, (E) LDP, (t) cis-polyisoprene, (a) A – (r), B – (p), C – (q), D – (t), E – (s), (b) A – (s), B – (q), C – (t), D – (r), E – (p), (c) A – (t), B – (s), C – (p), D – (q), E – (r), (d) A – (s), B – (t), C – (r), D – (q), E – (p), 102. Match the columns, Column-I (Polymers), Column-II (Uses), (A) Nylon 6,6, (p) Fabrics and ropes, (B) Nylon 6, (q) Electrical switches, (C) Dacron, (r) Bristles for brushes, (D) Bakelite, (s) As glass reinforcing, materials in safety, helmets., (a) A – (q), B – (s), C – (r), D – (p), (b) A – (r), B – (q), C – (s), D – (p), (c) A – (r), B – (p), C – (s), D – (q), (d) A – (r), B – (p), C – (q), D – (s), 103. Match the columns, Column-I, Column-II, (A) Polymer of styrene, (p) used in making handles, of utensils and computer, discs, (B) Polymer of ethylene, (q) used as an insulator, glycol and phtalic acid, (C) Polymer of phenol, (r) used in making paints, and formaldehyde, and lacquers., (D) Polymer of vinyl, (s) used in manufacture of, chloride, rain coats and flooring., (a) A – (q), B – (r), C– (p), D – (s), (b) A – (r), B – (q), C– (p), D – (s), (c) A – (q), B – (p), C– (r), D – (s), (d) A – (q), B – (r), C– (s), D – (p), , ASSERTION-REASON TYPE QUESTIONS, Directions : Each of these questions contain two statements,, Assertion and Reason. Each of these questions also has four, alternative choices, only one of which is the correct answer. You, have to select one of the codes (a), (b), (c) and (d) given below., (a) Assertion is correct, reason is correct; reason is a correct, explanation for assertion., (b) Assertion is correct, reason is correct; reason is not a, correct explanation for assertion, (c) Assertion is correct, reason is incorrect, (d) Assertion is incorrect, reason is correct., , 505, , 104. Assertion : Olefinic monomers undergo addition, polymerisation., Reason : Polymerisation of vinylchloride is initiated by, peroxides/ persulphates., 105. Assertion : Teflon has high thermal stability and chemical, inertness., Reason : Teflon is a thermoplastic., 106. Assertion : Bakelite is a thermosetting polymer., Reason : Bakelite can be melted again and again without, any change., 107. Assertion : In vulcanisation of rubber, sulphur cross links, are introduced., Reason : Vulcanisation is a free radical initiated chain, reaction., 108. Assertion : The time of vulcanisation and temperature is, increased by adding accelerators., Reason : By vulcanising, a material of high tensile strength, can be obtained., 109. Assertion : Most of the Synthetic polymers are not, biodegradable., Reason : Polymerisation process induces toxic character in, organic molecules., , CRITICAL THINKING TYPE QUESTIONS, 110. The polymer containing strong intermolecular forces e.g., hydrogen bonding, is, (a) teflon, , (b) nylon 6, 6, , (c) polystyrene, , (d) natural rubber, , 111. Among cellulose, poly (vinyl chloride), nylon and natural, rubber, the polymer in which the intermolecular force of, attraction is weakest is, (a) nylon, , (b) poly (vinyl chloride), , (c) cellulose, , (d) natural rubber, , 112. Which one of the following polymers is prepared by, condensation polymerisation?, (a) Teflon, , (b) Natural rubber, , (c) Styrene, , (d) Nylon-66, , 113. When condensation product of hexamethylenediamine and, adipic acid is heated to 525K in an atmosphere of nitrogen, for about 4-5 hours, the product obtained is, (a) solid polymer of nylon 66, (b) liquid polymer of nylon 66, (c) gaseous polymer of nylon 66, (d) liquid polymer of nylon 6
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EBD_7207, POLYMERS, , 506, , 114. Identify A, B and C in the following sequence of reactions, , O, , O, , B, , B + nCH2 = CH2, C+C, , 2A, , 2 C6H5–C–O, , A + CH2 = CH2, , O, , (c) Nylon-66 –NH–(CH2)6–NH–C–(CH2)4–C–O–n, , O, , C6H5–C–O–O–C–C6H5, , O, , CH3, (d) PMMA – CH2–C, COOCH3, , C, , C6H5–(CH2– CH2)n—CH2– CH2 – CH2, –(CH2–CH2)n– C6H5, , (a) A = C 6H5, B = C6H5 – CH2– CH 2 and, , n, , 118. In which of the following polymers ethylene glycol is one, of the monomer units?, (a), , (, —OCH, 2–CH2OOC, , CO)n, , C = C 6H5 –(CH2 – CH2)n—CH2 – CH 2 ., (, (b) —CH, n, 2–CH2 —, (, , (b) A = C6H5 CO, B = C6H5 – CH2– CH 2 and, C = C6H5 –(CH2–CH2)n—CH2– CH 2, , (c), , C = C6H5—(CH2 –CH2)n—CH2 CH 2, , (, (d) —O–CH–CH, 2–C–O–CH–CH2–C—, n, , CH3, , n, , (ethene)., , (a) Both are equally stable (b) Acrylonitrile., (c) Ethene, (d) Cannot say, 116. Which one of the following monomers gives the polymer, neoprene on polymerization ?, (a) CF2 = CF2, (b) CH2 = CHCl, (c) CCl2 = CCl2, , (, —CH, 2–C=CH–CH2—, n, Cl, , CN, , C — CH, , (c), , CH 2, , (, —O–CH–CH, 2–C–O–CH–CH2–C—, n, , CH3, , 117. Which of the following is not correctly matched?, (a) Terylene –OCH2–CH2– C–, , H, , O, , — CH 2 — C = CH — CH 2 —, , Cl, , H O, , O, , (, CH2)6–N–C–(CH2)4–C—n, (d) —N–(, , –C –, n, , (b) Neoprene, , CH2CH3 O, , (, , O, , O, , (, , CH 2, , (a), , (, , (d), , 119. Arrange the following in increasing order of their melting, point. Nylon 2,2 (1); Nylon 2,4 (2), Nylon 2,6 (3), Nylon, 2,10(4), (a) 1, 2, 3, 4, (b) 3, 4, 2, 1, (c) 2, 1, 3, 4, (d) 4, 3, 2, 1, 120. Which of the following rubber is not a polydiene?, (a) Polyisoprene, (b) Polychloroprene, (c) Thiokol rubber, (d) Nitrile rubber, 121. Which of the following polymer is biodegradable?, , (b) —CH, (, 2–CH=CH–CH2–CH 2–CH —, n, , Cl, |, , CH2CH3 O, , (, , C = C6H5 –(CH2– CH2)n—C6H5, 115. In which of the following the formation of radical was found, to be more stable; the formation of radical of (acrylonitrile), , O, , (, , (d) A= C 6H5, B=C6H5 –CH2– CH 2 and, , or CH 2 CH 2, , (, —CH, 2–CH=CH–CH2 –CH–CH2—, n, (, , (c) A = C 6H5, B = C6H5 – CH – CH3 and, , n, , 122. The mass average molecular mass & number average, molecular mass of a polymer are 40,000 and 30,000, respectively. The polydispersity index of polymer will be, (a) < 1, (b) > 1, (c) 1, (d) 0
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POLYMERS, , FACT / DEFINITION TYPE QUESTIONS, 1., 2., 3., 4., 5., , 6., , 7., 8., 9., , 10., 11., 13., , 14., 18., , 19., , 20., 21., 22., 23., 24., 25., , (d), (d), (c), (a), (b), , All these are natural polymers and exist in nature., ATP is a monomer molecule., It is present in the cell wall of plant cell., Since proteins, cellulose and RNA control various, activities of plants and animals, they are called, biopolymers., (d) Rayon is a manufactured from regenerated cellulosic, fiber. Rayon is produced from naturally occurring, polymers and therefore it is not a truly synthetic fiber,, nor is it a natural fiber. It is known by the names viscose, rayon and artificial silk in the textile industry. So,, option (d) is the correct choice., (c) Protein is a natural polymer of amino acids., (b), (b) Polymers are substances of high molecular weight, (usually more than a few thousand) formed by the union, of small molecular weight substances by covalent, bonds., (d) Cellulose acetate also known as rayon and cellulose, nitrate are semisynthetic polymers., (a), 12. (a), (a) Polymerisation starts either by condensation or addition, reactions between monomers. Condensation polymers, are formed by the combination of monomers with the, elimination of simple molecules.Whereas the addition, polymers are formed by the addition together of the, molecules of the monomer or monomers to form a large, molecule without elimination of any thing., (d), 15. (a), 16. (d), 17. (c), (b) Elastomers are the polymers having very weak, intermolecular forces of attraction between the polymer, chain. The weak forces permit the polymer to be, streched., (b) Those polymers in which process of heat softening, and cooling can be repeated as many times as desired., Example : polystyrene PVC, teflon, etc., (c) Bakelite is a thermosetting polymer. On heating it is, infusible and cannot be remoulded., (b), (d) These are characteristics of thermosets., (d) All these are characteristics of elastomers., (d) Thermosetting polymers have strongest molecular, forces. These are crosslinked polymers., (b) Thermosetting plastics have three dimensional cross, linked structure., , 507, , 26. (d) Silk is protein fibre. Dacron is polyester fibre and, Nylon-66 is polyamide fibre., 27. (d) All the given statements about fibres are correct., 28. (d) Both polyesters and polyamides are examples of fibres., 29. (b) Polystyrene and polyethylene belong to the category, of thermoplastic polymers which are capable of, repeatedly softening on heating and harden on cooling., 30. (d) Vinyl chloride, butadiene and styrene being, unsaturated undergoes addition polymerization., 31. (b), 32. (a), 33. (c), 34. (a) Ethene on free radical polymerisation gives low density, polythene., 35. (b) Monomer of teflon is to Tetrafluoro ethylene C2F4., 36. (d) Acrylonitrile is the monomer used in the preparation of, orlon., 37. (d) Orlon is a trade name of polyacrylonitrile, 38. (b) High density polythene is used for manufacturing of, buckets, dustbins, pipes etc., 39. (b) High density polythene is formed when addition, polymerisation of ethene takes place in a hydrocarbon, solvent in presence of catalyst such as ziegler-natta, catalyst., 40. (b) Nylon-66 is an example of first synthetic fibres produced, from the simple molecules. It is prepared by, condensation polymer-isation of adipic acid and, haxamethylene diamine., 41. (a) Melamine plastic crockery is a copolymer of HCHO, and Melamine., 42. (d), 43. (a), 44. (b), , nHOOC(CH 2 )4 COOH nH 2 N(CH 2 )6 NH 2, adipic acid, , Hexamethylene, diamine, , O, 525K, Polymerisation, , O, , [– C – (CH 2)4 – C –NH–(CH2) 6 – NH –]n, Nylon 6, 6, , 45. (b), 46. (d) The unbreakable plastic household crockery is made, from copolymer of formaldehyde (HCHO) and melamine., , N, , H2N, N, , NH2, N, , HCHO, formaldehyde, , Polymerisation, , Melamine, polymer, , NH2, , Melamine, , 47. (d) Nylon tyre cord is made from high tenacity continuous, filament yarn by twisting and plying.
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POLYMERS, , 70., 71., 72., , 73., , 74., 75., 78., 79., , 509, , (d) Neoprene is a homopolymer of 2-chloro-buta-1, 3-diene STATEMENT TYPE QUESTIONS, or chloroprene., 91. (a) Monomeric units in polymers are joined together by, (b) It is the definition of copolymerisation., covalent bonds only., Polymerisation, 92., (b), Condensation polymers are formed by repeated, (b) n CH2=CH–C=CH2, condensation reaction between two different biCH3, functional or tri–functional monomeric units., Isoprene, Buna–S and Buna–N being elastomeric consists of, –CH2–CH=C–CH2 –, n, polymeric chains held together by weak intermolecular, CH3, forces thus they are elastic in nature., Polyisoprene, 93. (d) Number of hydrogen bonds is greater in polymer (1), (Natural rubber), than in (2) as the density of amide bond is greater in (1), (b), therefore the chain links to each other strongly in (1), CH2 CH3, CH2, H, CH3, H, than in (2) hence (1) melts at higher temperature., C=C, C=C, C=C, 94., (d), Nylon–6 is used in the manufacturing of tyre cords, – CH 2, CH, CH, H, CH2, 3, 2, CH2–, not nylon–6,6 this is used in making sheets, bristles, Natural rubber, Natural rubber, for brushes and in textile industry. Melamine –, (All cis configuration), formaldehyde polymer is used in the manufacture of, unbreakable crockery., All statements except (b) are correct, (b) Natural rubber is a linear 1, 4-addition polymer of 95. (b) Vulcanisation of natural rubber with sulphur and an, appropriate additive is carried out within temperature, isoprene (2-methyl-1, 3-butadiene)., range of 373K to 415K. Synthetic rubbers are either, (d), 76. (b), 77. (b), homopolymersof 1, 3– butadiene derivatives or, (c) Rubber is a polymer of isoprene. Its chemical formula is, copolymers of 1, 3–butadiene or its derivatives with, (C5H8)n., another unsaturated monomer., (a) Neoprene is a synthetic polymer that resembles natural 96. (c) Biodegradable polymers are manufactured because, conventional polymers are quite resistant to the, rubber. Neoprene is a polymer of chloroprene which, environmental degradation which leads to accumulation, polymerises 700 times faster than the isoprene, of polymeric solid waste materials causing acute, (monomer of natural rubber) and no specific catalyst is, environmental problems., needed for this purpose., CH 2, , C CH, |, Cl, , MATCHING TYPE QUESTIONS, , CH 2, , Chloroprene, (2 chloro 1,3 butadiene), , ( CH 2, , C, |, Cl, , CH CH 2, , C CH CH 2 ), |, Cl, , n, , Neoprene, , 80., , (d) Synthetic rubber (neoprene) is a polymer of, CH 2, , 81., 82., 83., 84., 85., 86., 87., , Cl, |, , CH C, , CH 2 or chloroprene., , (a), (b) Buna – N is a copolymer of butadiene, (CH2=CH–CH=CH2) and acrylonitrile (CH2 = CHCN)., (a) Neoprene is a polymer of chloroprene, (2 – chloro – 1, 3 – butadiene)., (c) Neoprene is a homopolymer of chloroprene., (b) 5% of sulphur is used as a crosslinking agent in the, manufacture of tyre rubber., (b) Buna – N is obtained by copolymerisation of, 1, 3–butadiene and acrylonitrile., (d), 88. (c), 89. (d), 90. (b), , 97., 98., 99., 100., 101., 102., 103., , (b), (c), (b), (b), (a), (c), (a), (A), (B), , A – (q), B – (s), C – (r), D – (p), A – (s), B – (r), C – (p), D – (q), A – (t), B – (p), C – (r), D – (q), A – (q), B – (r), C – (p), D – (t), E – (s), A – (r), B – (p), C – (q), D – (t), E – (s), A – (r), B – (p), C – (s), D – (q), , Polystyrene is used as insulator., Glyptal a polymer of ethylene glycol and phthali acid, is used in manufacture of paints and lacquers., (C) Bakelite, a polymer of phenol & formal dehyde is used, for making electrical switches, handles of utensils and, computer disc’s., (D) PVC, a polymer of vinyl chloride is used in manufacture, of raincoat and flooring., , ASSERTION-REASON TYPE QUESTIONS, 104. (a), 105. (b) Due to the presence of strong C–F bonds, teflon has, high thermal stability and chemical inertness., 106. (c) Bakelite can be heated only once., 107. (b) Vulcanisation is a process of treating natural rubber, with sulphur or some compounds of sulphur under, heat so as to modify its properties. This cross-linking, give mechanical strength to the rubber.
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EBD_7207, POLYMERS, , 510, , 115. (b) Considering the resonance structure, the radical of, acrylonitrile is found to be more stable., H, , 108. (d) The time of vulcanisation is reduced by adding, accelerators and activators., 109. (d), , H, , CH = CH – C, , 110. (b) Nylon 6, 6 has amide linkage capable of forming, hydrogen bonding., 111. (d) Nylon and cellulose, both have intermolecular, hydrogen bonding, polyvinyl chloride has dipoledipole interactions, while natural rubber has van der, Waal forces which are weakest., 112. (d) Copolymer of adipic acid (6C) and hexamethylene, diamine (6C)., , C=C, , N, , H, , CRITICAL THINKING TYPE QUESTIONS, , H, , + C 6H 5, , + C6H5, , C6H5 – CH – CH – C, , N, , C6H5 – CH – CH, , N, , C 6H5 – CH 2 –CH 2, , n HOOC(CH 2 ) 4 COOH nH 2 N (CH 2 ) 6 NH 2, Adipic acid, , Hexamethyl ene diamine, , O, ||, , O, ||, , ( C (CH 2 ) 4 C NH (CH 2 ) 6, , NH ) n, , Nylon 66, , (Resonance stabilised), , It has high tenacity and elasticity. It is resistant to, abrasion and not affected by sea water. It is used for, reinforcement of rubber tyres, manufacture of, parachute, safety belts, carpets and fabrics., 113. (b) The condensation polymerisation of hexamethylene, diamine and adipic acid is done in solution form by, interface technique. In this liquid nylon polymer is, obtained., n.H 2 N (CH 2 )6, , 114. (a), , O, ||, , CH 2, , 6, , NHCO, Nylon, , O, ||, , Polymerisation, nH 2O, , CH 2, , 4, , CO ]n, O, ||, , C6 H5 C O O C C6 H5, , 2C6 H5 C O, , 2C6 H5, Phenyl radical, , (A), , CH 2, , 116. (d), , nCH 2, , CH, , |, , C, , Chloroprene, , CH 2, , O 2 or peroxides, , Cl, ( CH 2, , |, , CH, , C CH 2 )n, , Neoprene, , 117. (a) Terylene is prepared by condensing terephthalic acid, and ethylene glycol, nHOOC, , COOH + nHOCH 2CH 2OH, , Terephthalic acid, , O, ||, C, , Ethylene glycol, , O, ||, C – OCH2–CH2–O n, Terylene, , (A), , C6 H5 CH2, , Cl, , NH 2 +, , nHOOC – (CH 2 ) 4 – COOH, [ HN, , C, , C6 H5 CH2 CH 2 CH 2 C H 2, (B), , C6 H 5 CH 2 CH 2 CH 2 C H 2, (B), , nCH 2, , CH 2, , C6 H 5 (CH 2 CH 2 ) n CH 2 C H 2 ., (C), , C6 H 5— (CH 2 CH 2 )n—CH 2 C H 2 C6 H—(CH, 5, 2 CH 2 ) n—CH 2 C H 2, , C6 H5 (CH 2— CH 2 )n—CH 2, , CH 2 CH 2, , (CH 2 CH 2 )—C, n, 6 H5, , 118. (a), 119. (d) As the amide density along the chain increases the, melting point increases., 120. (c) Thiokol is polymer of CH2ClCH2Cl and sodium, polysulphide Na–S–S–Na and thus, not polydiene, rubber., 121. (d), 122. (b) Average number molecular weight M n, Average mass molecular weight M w, Polydispersity index (PDI) =, , Mw, Mn, , 30,000, 40,000, , 40,000, 1.33, 30,000
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30, CHEMISTRY IN, EVERYDAY LIFE, FACT / DEFINITION TYPE QUESTIONS, 1., , 2., , 3., , 4., , 5., , The use of chemicals for treatment of diseases is called as, (a) isothermotherapy, (b) angiotherapy, (c) physiotherapy, (d) chemotherapy, Which of the following statements is not true aboult enzyme, inhibitors?, (a) Inhibit the catalytic activity of the enzyme., (b) Prevent the binding of substrate., (c) Generally a strong covalent bond is formed between, an inhibitor and an enzyme, (d) Inhibitors can be competitive or non-competitive., Which of the following is not a target molecule for drug, function in body?, (a) Carbohydrates, (b) Lipids, (c) Vitamins, (d) Proteins, Which of the following comounds are administered as, antacids?, (i) Sodium carbonate, (ii) Sodium hydrogencarbonate, (iii) Aluminium carbonate, (iv) Magnesium hydroxide, (a) (i) and (ii), (b) (ii) and (iv), (c) (i), (ii) and (iii), (d) All of these, The drug H, is used as, N, , N, , 6., , 7., , CH2, , CH2 NH2, (a) Antacid, (b) Analgesic, (c) Antimicrobial, (d) Antiseptic, The function of enzymes in the living system is to, (a) transport oxygen, (b) provide energy, (c) provide immunity, (d) catalyse biochemical reactions, Which one of the following is employed as a tranquilizer?, (a) Naproxen, (b) Tetracycline, (c) Chlorpheninamine, (d) Equanil, , 8., , 9., 10., 11., , 12., , 13., 14., , 15., 16., 17., 18., , 19., , Which one of the following is employed as a tranquilizer, drug?, (a) Promethazine, (b) Valium, (c) Naproxen, (d) Mifepristone, Terfenadine is commonly used as a/an, (a) tranquilizer, (b) antihistamine, (c) antimicrobial, (d) antibiotic, Which one of the following is not a tranquilizer?, (a) Equanil, (b) Veronal, (c) Salvarsan, (d) Serotonin, Tranquillizers are substances used for the treatment of, (a) cancer, (b) AIDS, (c) mental diseases, (d) physical disorders, Which one of the following is employed as a tranquilizer, drug?, (a) Promethazine, (b) Valium, (c) Naproxen, (d) Mifepristone, Which of the following drugs is a tranquilizer and sedative, (a) Sulphadiazine, (b) Papaverine, (c) Equanil, (d) Mescaline, Drug which helps to reduce anxiety and brings about, calmness is, (a) tranquillizer, (b) diuretic, (c) analgesic, (d) antihistamine, The drug used as an antidepressant is, (a) Luminol, (b) Tofranil, (c) Mescaline, (d) Sulphadiazine, Barbituric acid and its derivatives are well known, (a) antipyretics, (b) analgesics, (c) antiseptics, (d) traquillizers, Which of the following is a hypnotic drug?, (a) luminal, (b) salol, (c) catechol, (d) chemisol, Which of the following is used for inducing sleep?, (a) Paracetamol, (b) Chloroquine, (c) Bithional, (d) Barbituric acid derivatives, Aspirin is, (a) antibiotic, (b) antipyretic, (c) sedative, (d) psychedelic
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EBD_7207, CHEMISTRY IN EVERYDAY LIFE, , 512, , 20., , 21., , An antipyretic is, (a) quinine, (b) paracetamol, (c) luminal, (d) piperazine, The following compound is used as, O, O – C – CH3, , 22., , 23., , 24., , 25., , 26., , 27., , 28., , 29., , 30., , 31., , 32., , COOH, (a) an anti-inflammatory compound, (b) analgesic, (c) hypnotic, (d) antiseptic, Barbituric acid and its derivatives are well known, (a) antipyretics, (b) analgesics, (c) antiseptics, (d) traquillizers, The drug used for prevention of heart attacks is, (a) aspirin, (b) valium, (c) chloramphenicol, (d) cephalsoprin, Sulpha drugs are used for, (a) precipitating bacteria, (b) removing bacteria, (c) decreasing the size of bacteria, (d) stopping the growth of bacteria, Aspirin falls under which class of drugs ?, (a) Analgesic, (b) Antibiotic, (c) Antifertility, (d) antacid, Which of the following term means pain killer, (a) Antibiotic, (b) Analgesic, (c) Antipyretic, (d) Penicillin, Which one of the following can possibly be used as analgesic, without causing addiction and mood modification ?, (a) Diazepam, (b) Morphine, (c) N-Acetyl-para-aminophenol, (d) Tetrahydrocannabinol, Aspirin is known as, (a) acetyl salicylic acid, (b) phenyl salicylate, (c) acetyl salicylate, (d) methyl salicylic acid, Which one among the following is not an analgesic?, (a) Ibuprofen, (b) Naproxen, (c) Aspirin, (d) Valium, Which of the following statements about aspirin is not true?, (a) It is effective in relieving pain., (b) It is a neurologically active drug., (c) It has antiblood clotting action., (d) It belongs to narcotic analgesics., Salol can be used as, (a) antiseptic, (b) antipyretic, (c) analgesic, (d) None of these, Various phenol derivatives, tincture of iodine (2 – 3%) I2, in (water / alcohol) and some dyes like methylene blue are, (a) antiseptics, (b) disinfectants, (c) analgesics, (d) antipyretics, , 33., , 34., , 35., , 36., , 37., , 38., , 39., , 40., , 41., , 42., , 43., , 44., , Sulpha drugs are used for, (a) precipitating bacteria, (b) removing bacteria, (c) decreasing the size of bacteria, (d) stopping the growth of bacteria, Streptomycin is effective in the treatment of, (a) tuberculosis, (b) malaria, (c) typhoid, (d) cholera, An antibiotic with a broad spectrum, (a) kills the antibodies, (b) acts on a specific antigen, (c) acts on different antigens, (d) acts on both the antigens and antibodies, Which of the following is not an antiseptic drug?, (a) Iodoform, (b) Dettol, (c) Gammexane, (d) Genation violet, Penicillin was first discovered by, (a) A. Fleming, (b) Tence and Salke, (c) S.A. Waksna, (d) Lewis Pasteur, Veronal, a barbiturate drug is used as, (a) anaesthetic, (b) sedative, (c) antiseptic, (d) None of these, A drug effective in the treatment of pneumonia, bronchitis,, etc, is, (a) streptomycin, (b) chloramphenicol, (c) penicillin, (d) sulphaguanidine, Commonly used antiseptic 'Dettol' is a mixture of, (a) o-chlorophenozylenol + terpeneol, (b) o-cresol + terpeneol, (c) phenol + terpeneol, (d) chloroxylenol + terpeneol, Chloroamphenicol is an :, (a) antifertility drug, (b) antihistaminic, (c) antiseptic and disinfectant, (d) antibiotic-broad spectrum, The drug which is effective in curing malaria is, (a) quinine, (b) aspirin, (c) analgin, (d) equanil, An antibiotic contains nitro group attached to aromatic, nucleus. It is, (a) penicillin, (b) streptomycin, (c) tetracycline, (d) chloramphenicol, The structure given below is known as, O, CH 2, , H, , S, , CH3, , C NH, O, , (a) Penicillin F, (c) Penicillin K, , H, , CH3, COOH, , N, H, , (b) Penicillin G, (d) Ampicillin
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CHEMISTRY IN EVERYDAY LIFE, , 45., , 46., , 47., , 48., , 49., , 50., , 51., , 52., , 53., , 54., , 55., , 56., , Arsenic drugs are mainly used in the treatment of, (a) Jaundice, (b) Typhoid, (c) Syphilis, (d) Cholera, Bithional is an example of, (a) disinfectant, (b) antiseptic, (c) antibiotic, (d) analgesic, Penicillin is an :, (a) antibiotic, (b) anaesthetic, (c) antiseptic, (d) antipyretic, Which of the following is a broad spectrum drug?, (a) Plasmoquine, (b) Chloroquine, (c) Chloramphenicol, (d) D.D.T., Bithional is added to soap as an additive to function as a/an, (a) softener, (b) hardener, (c) dryer, (d) antiseptic, Antiseptics and disinfectants either kill or prevent growth, of microorganisms. Identify which of the following, statements is not true:, (a) Chlorine and iodine are used as strong disinfectants., (b) Dilute solutions of boric acid and hydrogen Peroxide, are strong antiseptics., (c) Disinfectants harm the living tissues., (d) A 0.2% solution of phenol is an antiseptic while 1%, solution acts as a disinfectant., Arsenic containing medicine used for the treatment of, syphilis, is, (a) erythromycin, (b) ofloxacin, (c) tetracycline, (d) salvarsan, Novestrol is an, (a) antibiotic, (b) analgesic, (c) antacid, (d) antifertility drug, Which is the correct statement about birth control pills?, (a) Contain estrogen only, (b) Contain progesterone only, (c) Contain a mixture of estrogen and progesterone, derivatives., (d) Progesterone enhances ovulation., Compounds with antiseptic properties are ____________, (i) CHCl3, (ii) CHI3, (iii) Boric acid, (iv) 0.3 ppm aqueous solution of Cl2, (a) (i) and (ii), (b) (ii) and (iii), (c) (i) and (iv), (d) (i) and (iii), Which of the following is not a function of aspirin?, (a) Relief from arthritic pain, (b) Relief from postoperative pain., (c) Prevents platelet coagulation., (d) Prevention of heart attacks., Which of the following method of classification of drugs is, useful for doctors?, (a) On the basis of drug action., (b) On the basis of chemical structure., (c) On the basis of molecular targets., (d) On the basis of pharmacological effect., , 513, , 57. Which of the following method of classification of drugs is, useful for medicinal chemists?, (a) On the basis of molecular targets., (b) On the basis of chemical structure., (c) On the basis of drug action., (d) All of these., 58. Which of the following statements is true about the catalytic, activity of enzyme?, (a) Enzyme holds the substrate for a biochemical reaction., (b) En zyme binds substrate through a variety of, interactions such as ionic bonding, hydrogen bonding,, van der Waal’s interaction or dipole – dipole, interaction., (c) Enzyme provides functional group that will attack the, substrate and carry out biochemical reaction., (d) All of the above., 59. Drug tegamet is used as, (a) Antacid, (b) Antimalarial, (c) Analgesic, (d) Antiseptic, 60. Which of the following is an essential component of sleeping, pills?, (a) Analgesics, (b) Tranquilizers, (c) Antihistamines, (d) Both (b) and (c), 61. Which type of drugs inhibit the enzymes which catalyse, the degradation of noradrenaline?, (a) Narcotic analgesics, (b) Antacids, (c) Antidepressant, (d) Non–narcotic analgesic., 62. Which of the following is/are example(s) of narcotic, analgesics?, (a) Morphine, (b) Heroin, (c) Codeine, (d) All of these, 63. Which of the following was the first effective treatment, discovered for syphilis?, (a) Penicillin, (b) Arsphenamine, (c) Chloramphenicol, (d) Sulphanilamide, 64. Which of the following is an example of narrow spectrum, antibiotic?, (a) Chloramphenicol, (b) Penicillin G, (c) Ampicillin, (d) Ofloxacin, 65. Antibiotic that can be given orally in case of typhoid, acute, fever, dysentery, meningitis and pneumonia is_______., (a) vancomycin, (b) salvarsan, (c) ofloxacin, (d) chloramphenicol, 66. Veronal and luminal are derivatives of barbituric acid which, are______., (i) Tranquilizers, (ii) Non– narcotic analgesic., (iii) Antiallergic drug, (iv) Neurologically active drug., (a) (i) and (iv), (b) (i) and (iii), (c) (ii) and (iii), (d) (i) only
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EBD_7207, CHEMISTRY IN EVERYDAY LIFE, , 514, , 67., , Which is correct about saccharin?, O, , 68., 69., 70., 71., , 72., , 73., 74., 75., , 76., 77., , (c), 78., 79., , 80., , C, , NH, SO2, (b) It is 600 times sweeter than sugar, (c) It is used as sweetening agent, (d) All of these, Which of the following acts as an antioxidant in edible oils, (a) Vitamin B, (b) Vitamin C, (c) Vitamin D, (d) Vitamin E, Salts of sorbic acid and propionic acid are used as, (a) antioxidants, (b) flavouring agents, (c) food preservatives, (d) nutritional supplements, Which of the following add nutritive value to food?, (a) Sweeteners, (b) Antioxidants, (c) Fat emulsifiers, (d) None of these, Arrange the following artificial sweeteners in increasing, order of their sweetness value?, (a) Sucralose < Saccharin < Alitame < Aspartame, (b) Aspartame < Saccharin < Sucralose < Alitame, (c) Aspartame < Sucralose < Saccharine < Alitame, (d) Saccharine < Aspartame < Sucralose < Alitame, Which of the following artificial sweetener does not provide, calories?, (a) Alitame, (b) Aspartame, (c) Sucralose, (c) Both (b) and (c), Sodium benzoate is used as, (a) food preservative, (b) analgesic, (c) filler in detergents, (d) antiseptic, Structuraly biodegradable detergent should contain, (a) normal alkyl chain, (b) branched alkyl chain, (c) phenyl side chain, (d) cyclohexyl side chain, Detergents are prepared by the action of H2SO4 on which, of the following?, (a) Cholesterol, (b) Lauryl alcohol, (c) Cyclohexanol, (d) p-Nitrophenol, Sodium alkyl benzene sulphonate is used as, (a) soap, (b) fertilizers, (c) pesticides, (d) detergents, Which of the following represents a synthetic detergent?, (a) C15H31COOK, (b) CH3[CH2]16COONa, , (a) It is, , C12H25, , 81., , SO3Na, , (d) None of these, Which of the following represents soap, (a) C17H35COOK, (b) C17H35COOH, (c) C15H31COOH, (d) (C17H35COO)2Ca, Alkaline hydrolysis of esters is known as :, (a) Esterification, (b) Saponification, (c) dehydration, (d) alkalination, Commercial detergent contains mainly, (a) RCOONa, (b) RONa, (c) RSNa, (d) RSO3Na, , 82., 83., 84., 85., , 86., , 87., , 88., , 89., , 90., , Washing soap can be prepared by saponification with alkali, of which of the following oil, (a) Rose oil, (b) Paraffin oil, (c) Groundnut oil, (d) Kerosene oil, Palmitic acid and stearic acid are used as :, (a) medicine, (b) soap, (c) antiseptic cream, (d) pickle, Soaps can be classified as :, (a) esters, (b) salts of fatty acids, (c) alcohols, (d) phenols, Cetyltrimethyl ammonium bromide is a popular, (a) anionic detergent, (b) cationic detergent, (c) non-ionic detergent (d) sweetener, The cationic detergent that is used in hair conditioners is, (a) sodium dodecylbenzene sulphonate, (b) sodium lauryl sulphate, (c) tetramethyl ammonium chloride, (d) cetyltrimethyl ammonium bromide, Which of the following enhances leathering property of, soap?, (a) Sodium carbonate, (b) Sodium rosinate, (c) Sodium stearate, (d) Trisodium phosphate, Polyethyleneglycols are used in the preparation of which, type of detergents?, (a) Cationic detergents, (b) Anionic detergents, (c) Non-ionic detergents (d) Soaps, Which of the following are anionic detergents?, (i) Sodium salts of sulphonated long chain alcohol., (ii) Ester of stearic acid and polyethylene glycol., (iii) Quarternary ammonium salt of amine with acetate ion., (iv) Sodium salts of sulphonated long chain hydrocarbons., (a) (i) and (iv), (b) (ii) and (iii), (c) (iii) and (iv), (d) (i), (ii) and (iv), Which type of detergent is formed when stearic acid reacts, with polyethylene glycol?, (a) Cationic detergent, (b) Anionic detergent, (c) Non– ionic detergent (d) None of these, Glycerol is added to soap. It functions ___________, (a) as a filler., (b) to increase leathering., (c) to prevent rapid drying., (d) to make soap granules., , STATEMENT TYPE QUESTIONS, 91., , Which of the following statements are incorrect about, receptor proteins?, (i) Majority of receptor proteins are embedded in the cell, membranes., (ii) The active site of receptor proteins opens on the inside, region of the cell., (iii) Chemical messengers are received at the binding sites, of receptor proteins., (iv) Shape of receptor doesn’t change during attachment, of messenger., (a) (i), (ii) and (iii), (b) (ii) and (iv), (c) (ii), (iii) and (iv), (d) (i) and (iv)
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CHEMISTRY IN EVERYDAY LIFE, , 92., , 93., , 94., , 95., , 96., , Which of the following statements are incorrect about, penicillin?, (i) An antibacterial medicine., (ii) Ampicillin is its synthetic modification., (iii) It has bacteriostatic effect., (iv) It is a broad spectrum antibiotic., (a) (i) and (ii), (b) (ii) and (iv), (c) (iii) and (iv), (d) (i) and (ii), Which of the following statements are correct?, (i) Before 1970 for treatment of stomach acidity Al(OH)3, is a better antacid in comparison to NaHCO3 ., (ii) Discovery of cimetidine was a major break through in, the treatment of hyperacidity., (iii) Terfenadine is a drug which competes with histamine, for binding sites of receptor., (iv) Antidepressant drugs like equanil inhibit the enzymes, which catalyze the degradation of noradernaline., (v) Veronal and luminal belongs to the class of tranquilizers, called barbiturates., (a) (i), (ii) and (iii), (b) (i), (iii) and (v), (c) (ii), (iii) and (iv), (d) (ii), (iv) and (v), Consider the following statements., (i) Antiseptics are not ingested like antibiotics., (ii) 1% solution of iodine in alcohol–water mixture is known, as tincture of iodine., (iii) SO2 in low concentrations are used as antiseptics, whereas in higher concentration are used as, disinfectants., (iv) Birth control pills essentially contain a mixture of, synthetic estrogen and progesterone derivates., Which of the following is the correct code for the statements, above ?, (a) FTFT, (b) TTTF, (c) TFFT, (d) FFTT, Which of the following statements are correct?, (i) Cationic detergents have germicidal properties, (ii) Bacteria can degrade the detergents containing highly, branched chains., (iii) Some synthetic detergents can give foam even in ice, cold water., (iv) Synthetic detergents are not soaps., (a) (i), (ii) and (iii), (b) (i), (iii) and (iv), (c) (ii), (iii) and (iv), (d) (iii) and (iv), Consider the following statements., (i) Potassium soaps are soft to the skin than sodium, soaps., (ii) Shaving soaps contain ethanol to prevent rapid drying., (iii) Builders like Na2CO3 and Na3PO4 make soaps act more, rapidly., (iv) Hard water contains Ca2+ and Mg2+ ions which forms, insoluble Ca2+ and Mg 2+ soaps separates out as, scum., , 515, , Which of the following is the correct code for the statements, above ?, (a) TFTT, (b) FTTT, (c) FTFT, (d) TTTT, 97. Which of the following statements are correct?, (i) Detergents give foam easily even in hard water., (ii) Anionic detergents are used in toothpastes., (iii) Cationic detergents being inexpensive are extensively, used as germicide., (iv) Detergents with linear alkyl chains are more polluting, as compared to detergents having branched alkyl, chains., Which of the following is the correct code for the statements, above?, (a) TFTF, (b) TFFT, (c) FFFT, (d) TTFF, , MATCHING TYPE QUESTIONS, 98. Match the columns, Column-I, Column-II, (A) Sodium Perborate, (p) Disinfectant, (B) Chlorine, (q) Antiseptic, (C) Bithional, (r) Milk bleaching agent, (D) Potassium stearate, (s) Soap, (a) A – (p), B – (q), C – (r), D – (s), (b) A – (q), B – (r), C – (s), D – (p), (c) A – (r), B – (p), C – (q), D – (s), (d) A – (s), B – (p), C – (q), D – (r), 99. Match the columns, Column-I, Column-II, (A) Ranitidine, (p) Tranquilizer, (B) Furacine, (q) Antibiotic, (C) Phenelzine, (r) Antihistamine, (D) Chloramphenicol, (s) Antiseptic, (a) A – (r), B – (s), C – (p), D – (q), (b) A – (s), B – (p), C – (q), D – (r), (c) A – (p), B – (q), C – (r), D – (s), (d) A – (q), B – (r), C – (s), D – (p), 100. Match the columns, Column – I, Column – II, (A) First antibacterial, (p) Broad spectrum, drug, antibiotic, (B) Protosil, (q) Arsphenamine, (C) Chloramphenicol, (r) 1932, (D) Ofloxacin, (s) 1947, (a) A – (q), B – (r), C – (p, s), D – (p), (b) A – (r), B – (q), C – (p, s), D – (p), (c) A – (q), B – (p, s), C – (r), D – (p), (d) A – (p), B – (r), C – (p, s), D – (q)
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EBD_7207, CHEMISTRY IN EVERYDAY LIFE, , 516, , 101. Match the columns, Column-I, CH3, , Column-II, +, –, , (A) CH3(CH2)15–N–CH3 Br, , (p) Dishwashing, , CH3, , (B) CH3– (CH2)11, , powder, , –, , +, , SO3 Na, , (C) C17H35COO–Na+ + Na2CO3 + Rosin, , (q) Laundry soap, (r) Hair, conditioners, (s) Toothpaste, , (D) CH3(CH2)16COO(CH2CH2O)nCH2 CH2OH, (a) A – (p), B – (q), C – (r), D – (s), (b) A – (q), B – (r), C – (p), D – (s), (c) A – (r), B – (s), C – (q), D – (p), (d) A – (p), B – (r), C – (q), D – (s), 102. Match the columns, Column -I, Column -II, (A) Toilet soap, (p) Made by beating, tiny air bubbles before, their hardening., (B) Transparent soap, (q) Contain glycerol, to prevent rapid drying., (C) Shaving soaps, (r) Prepared by, using better grades of, fats and oils., (D) Soaps that float in, (s) Made by dissolving, water, the soap in ethanol, and then evaporating, excess alkali., (a) A – (s), B – (p), C – (q), D – (r), (b) A – (r), B – (s), C – (q), D – (p), (c) A – (r), B – (q), C – (p), D – (s), (d) A – (q), B – (s), C – (p), D – (r), , ASSERTION-REASON TYPE QUESTIONS, Directions : Each of these questions contain two statements,, Assertion and Reason. Each of these questions also has four, alternative choices, only one of which is the correct answer. You, have to select one of the codes (a), (b), (c) and (d) given below., (a) Assertion is correct, reason is correct; reason is a correct, explanation for assertion., (b) Assertion is correct, reason is correct; reason is not a, correct explanation for assertion, (c) Assertion is correct, reason is incorrect, (d) Assertion is incorrect, reason is correct., 103. Assertion : The drugs which act on the central nervous, system and help in reducing anxiety are called antibiotics., Reason : Pencillin is an antibiotic., 104. Assertion : Equanil is a tranquilizer., Reason : Equan il is used to cure depression and, hypertension., , 105. Assertion : Tetracycline is a broad spectrum antibiotic., Reason : Tetracyclin is effective against a number of types, of bacteria, large viruses and typhus fever., 106. Assertion : Antiseptics are applied to living tissues., Reason : Iodine is a powerful antiseptic., 107. Assertion : Sedatives are given to patients who are mentally, agitated and violent., Reason : Sedatives are used to suppress the activities of, central nervous system., 108. Assertion : Non-competitive inhibitor inhibits the catalyic, activity of enzyme by binding with its active site., Reason : Non-competitive inhibitor changes the shape of, the active site in such a way that substrate can’t recognise, it., 109. Assertion : Sodium chloride is added to precipitate soap, after saponification., Reason : Hydrolysis of esters of long chain fatty acids by, alkali produces soap in colloidal form., , CRITICAL THINKING TYPE QUESTIONS, 110. Which of the following are sulpha drugs?, (i) Sulphapyridine, (ii) Prontosil, (iii) Salvarsan, (iv) Nardil, (a) (i) and (ii), (b) (ii) and (iv), (c) (i), (ii) and (iv), (d) (ii), (iii) and (iv), 111. Among the following antihistamines, which are antacids, (i) Ranitidine, (ii) Brompheniramine, (iii) Terfenadine, (iv) Cimetidine, (a) (i) and (iii), (b) (i), (ii) and (iv), (c) (i) and (iv), (d) (ii) and (iii), 112. Which one of the following is an antihistamine?, (a) Iproniazid, (b) Salvarsan, (c) Zantac, (d) Chloramphenicol, 113. Morphine is, (a) an alkaloid, (b) an enzyme, (c) a carbohydrate, (d) a protein, 114. H1 – Receptor antagonists is a term associated with :, (a) Antiseptics, (b) Antihistamins, (c) Antacids, (d) Analgesics, 115. Amoxillin is semi-synthetic modification of, (a) penicillin, (b) streptomycin, (c) tetracycline, (d) chloroampheniol, 116. Which of the following is used as an antibiotic ?, (a) Ciprofloxacin, (b) Paracetamol, (c) Ibuprofen, (d) Tocopherol, 117. Select the incorrect statement., (a) Equanil is used to control depression and hypertension., (b) Mifepristone is a synthetic steroid used as “morning, after pill”., (c) 0.2 percent solution of phenol is an antiseptic while its, 1.0 percent solution is a disinfectant., (d) A drug which kills the organism in the body is called, bacteriostatic.
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CHEMISTRY IN EVERYDAY LIFE, , 118. A large number of antibiotics have been isolated from, (a) Bacteria actinomycetes, (b) Acids, (c) Alkanals, (d) Bacteria rhizobium, 119. Antiseptic chloroxylenol is, (a) 4-chloro-3, 5-dimethylphenol, (b) 3-chloro-4, 5-dimethylphenol, (c) 4-chloro-2, 5-dimethylphenol, (d) 5-chloro-3, 4-dimethylphenol, 120. Which of the following is not correctly matched?, (i) Proteins that are, – Receptors, crucial to body’s communication, process., (ii) Drugs that mimic, – Antagonists, the natural messenger by, switching on the receptor., (iii) Drugs that binds to, – Agonists, the receptor site and inhibit, its natural function., (a) (ii) only, (b) (iii) only, (c) (i) and (iii), (d) (ii) and (iii), 121. Which of the following drug inhibits the synthesis of, chemicals known as prostaglandins which stimulate, inflammation in tissue and cause pain?, (a) Barbiturates, (b) Aspirin, (c) Seldane, (d) Iproniazid, 122. Bactericidal antibiotics are those which, (a) have inhibitory effect on microbes., (b) have killing effect on microbes., (c) have both inhibitory and killing effect on microbes., (d) intervene in metabolic process of microorganism., 123. Which of the following antibiotics is not correctly classified?, Bactericidal, Bacteriostatic, (A) Penicillin, Erythromycin, (B) Aminoglycosides, Tetracycline, (C) Chloramphenicol, Ofloxacin, (a) A and B, (b) C only, (c) B and C, (d) B only, 124. Antibiotics that are effective mainly against Gram-positive, or Gram-negative bacteria X. Antibiotics that are effective, against a single organism or disease are Y, What is X and Y ?, (a) X = Broad spectrum antibiotics., Y = Narrow spectrum antibiotics., (b) X = Broad spectrum antibiotics., Y = Limited spectrum antibiotics., , 517, , 125., , 126., , 127., , 128., , 129., , 130., , (c) X = Narrow spectrum antibiotics., Y = Limited spectrum antibiotics., (d) X = Narrow spectrum antibiotics., Y = Broad spectrum antibiotics., Which of the following is an example of synthetic, progesterone derivative which is most widely used as, antifertility drug?, (a) Norethindrone, (b) Novestrol, (c) Ethynylestradiol, (d) All of these, Substance used for the preservation of coloured fruit juices, is, (a) benzene, (b) benzoic acid, (c) phenol, (d) sodium meta bisulphite, The artificial sweetener containing chlorine that has the, appearance and taste as that of sugar and is stable at cooking, temperature is, (a) Aspartame, (b) Saccharin, (c) Sucrolose, (d) Alitame, Benzalkonium chloride is a, (a) cationic surfactant and antiseptic, (b) anionic surfactant and soluble in most of organic, solvents, (c) cationic surfactant and insoluble in most of organic, solvents, (d) cationic surfactant and antimalarial, Which one of the following is not used as a filler in laundry, soaps?, (a) Sodium silicate, (b) Glycerol, (c) Sodium rosinate, (d) Borax, Which of the following is an example of liquid dishwashing, detergent?, (a) CH3(CH2)10–CH2OSO3–Na+, (b) C9H19 —, (c), , CH3—, , )5 –CH2CH2OH, (, —O—CH, 2–CH2–O—, –, , CH3, (d), , +, , —SO3 Na, +, –, , CH3(CH2)15–N–CH3 Br, CH3, , 131. Which of the following statements is incorrect?, (a) Saccharin is about 550 times as sweet as cane sugar., (b) Aspartame is used in the manufacture of baked sweets., (c) Alitame is more sweet than saccharin and aspartame., (d) Sodium benzoate is commonly used preservative.
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EBD_7207, CHEMISTRY IN EVERYDAY LIFE, , 518, , O–COCH3, , FACT / DEFINITION TYPE QUESTIONS, 1., 5., 6., , (d), 2. (c), 3. (c), 4. (b), (a) Given drug is used as Antacid., (d) The function of enzymes in the living system is to, catalyse biochemical reactions which occure in living, systems. e.g. invertase, pepsin, amylase., Sucrose, , Invertase, , (polymer), , Starch, (polymer), , 7., 8., 9., 10., , 11., 13., , 14., 17., , 18., 19., , 20., 21., 22., 23., 24., 25., 26., 27., , glucose + fructose, (monomer), , amylase, , glucose, , (a), , 29., , (d) Valium is a tranquilizer and not an analgesic. It is used, for treatment of stress, fatigue, mild and severe mental, diseases., (d) Aspirin is an non-narcotics analgesic., (a) Salol is phenyl salicylate used as antiseptic., (a) Antiseptic drugs cause destruction of micro-organism, that produce septic disease e.g. Dettol, Savlon, Boric, acid, Phenol, Iodoform, KMnO4 and some dye such, as methylene blue, genation violet., (d) Sulpha drugs (antibacterial and antibiotic) are group, of drugs which are derivative of sulphanilamide., (a) It is the very effective antibiotic for tuberculosis., (c) Broad spectrum antibiotics act on different antigens., (c) It is an insecticide., (a) A. Fleming discovered penicillin in 1929., (b), (c) Penicillin is an effective medicine for pneumonia, disease., (d) The mixture of chloroxylenol and terpenol is dettol, which is used as antiseptic., (d) Chloroamphenicol is a broad spectrum antibiotic., (a) Substances used for the treatment of malaria are, antimalarial e.g. Quinine, chloroquine., (d) Chloramphenicol is, , 30., 31., 32., , (monomer), , (d) Equanil is an important medicine used in depression, and hypertension., (b), (b) Terfenadine is commonly used as antihistamine., (c) Salvarsan is an organoarsenic compound, used in the, treatment of syphilis. It was the first modern, chemotherapeutic agent., (c), 12. (b), (c) Tranquilizers reduce anxiety and tension they are also, called psychototropic drugs. These are two type, (i) Sedative the drugs used for violent and mentaly, agitated patient e.g.., Equanil and diazepam., (ii) Antidepressant- The drug are used to patients who, are highly depressed and lose self confidence e.g., tofranil vitalin, amphetamine etc., (a), 15. (b), 16. (d), (a) These drugs induce sleep and are habit forming, common example of hypnotic drugs are Luminal and, Saconal., (d), (b) Aspirin is antipyretic i.e., a drug which is responsible, for lowering the temperature of feverish organism to, normal, other antipyretic drugs are Paracetamol,, Phenacetin., (b) Paracetamol is an antipyretic, (b) It is acetyl salicylic acid i.e., aspirin, analgesic and, antipyretic., (d), (a) Due to anti-blood clotting action of aspirin, it is used, to prevent heart attack., (d) Sulpha drugs (antibacterial and antibiotic) are group, of drugs which are derivative of sulphanilamide., (a) Analgesic are pain killers., (b) Analgesic means painkiller., (c) We know that N-acetyl-para-aminophenol (or, paracetamol) is an antipyretic which can also be used, as an analgesic to relieve pain., , COOH, Aspirin (Acetyl salicylic acid), , 28., , 33., 34., 35., 36., 37., 38., 39., 40., 41., 42., 43., , NO2, , 44., 45., 46., , O, ||, CH CH NH C – CHCl 2, |, |, OH CH 2OH, , (b) It is the known structure of Penicillin G, (c) Arsenic drugs are poisonous for syphilis., (b) Bithional is a well known antiseptic, added in soaps to, reduce odours produced by bacterial decomposition, of organic matter of skin., Cl, , OH HO, , Cl, , S, Cl, 47., 48., , Bithional, , Cl, , (a) Penicillin is an antibiotic., (c) Chloramphenicol is a broad spectrum drug., [Broad spectrum antibiotics are medicines effective, against gram positive as well as gram negative bacteria,, e.g., tetracycline, chloramphenicol, etc.]
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CHEMISTRY IN EVERYDAY LIFE, , 49., , 50., 51., 52., 53., 55., 56., , 57., 59., 60., 61., 62., 63., 64., 65., , 66., 67., 68., 69., 70., 71., , 72., 73., 74., 75., 76., 77., , 519, , (d) Bithional is another well known antiseptic which is, added to good quality soaps to reduce the odours, produced by bacterial decomposition of organic matter, on the skin., (b) Dilute solutions of boric acid and hydrogen peroxide, are weak antiseptics., (d), (d) Novestrol is an antifertility drug., (c), 54. (b), (b) Morphine narcotics are chiefly used for the relief of, post operative pain., (d) Classification of drugs on the basis of pharmacological, effect is useful for doctors because it provides them, the whole range of drugs available for the treatment of, a particular type of problem., (a), 58. (d), (a) Drug tegamet was designed to prevent the interaction, of histamine with the receptors present in the stomach, wall. This resulted in release of lesser amount of acid., (b), (c) Antidepressant drugs inhibit the enzymes which, catalyse the degradation of noradrenaline., (d), (b) Arsphenamine also known as salvarsan was the first, effective treatment discovered for syphilis., (b) Penicillin G has a narrow spectrum, while all other, options have broad spectrum., (d) Chloramphenicol is rapidly absorbed from the, gastrointestinal tract and hence can be given orally in, case of typhoid, acute fever, meningitis, pneumonia, etc., (a), (d) All are characteristics of Saccharin., (d) Vitamin E is an antioxidant present in edible oils., (c) Salts of sorbic acid and propionic acid are used as, food preservatives because these chemicals inhibit the, growth of yeast bacteria or moulds., (d) Neither any of the substances among given options, possess nutritive value., (b) Artificial sweetener, Sweetness value, Aspartame, 100, Saccharin, 550, Sucralose, 600, Alitame, 2000, (c) Sucralose does not provide calories., (a) Sodium benzoate is used as a food preservative., (b) Structurally biodegradable detergents should contain, branched alkyl chain., (b), (d) It is used as detergent., (c) The most widely used domestic detergent is the, sodium dodecyl benzene sulphonate (SDS)., , CH3 – (CH2)11, , SO3Na, , (Sodium dodecyl benzene sulphonate), , 78. (a) Soaps are the sodium or potassium salt of higher fatty, acids e.g., C17H37COOK (Potassium stearate). These, are obtained by alkaline hydrolysis of oils and fats., The reaction is called saponification., 79. (b) Alkaline hydr olysis of esters is known as, saponification., R 'OH RCOONa, R COOR ' NaOH, 80. (d) Commercial detergent are the sodium salts of long chain, (linear) alkyl substituted benzene sulphonic acids, (LAB) and are most widely used. The most common is, sodium dodecylbenzene sulphonate., –, , CH3(CH2)10 CH2, , +, , SO2O Na, , 81. (c) Any oils which are good for eating or cooking, can be, used in making soap. One of the best is said to be, coconut oil. Groundnut, Shea butter, Cocoa butter, Sun, flower and many other vegetable oils are also used., 82. (b) Sodium or potassium salts of palmitic acid, (CH 3 (CH 2 ) 14 COO – Na + ) and stearic acid, (CH3(CH2)16COO–Na+) are used as soaps., 83. (b) Soaps are actually salts of higher fatty acids., Example ; C17 H 35 COONa, (sodium stearate), , 84. (b), CH3 (CH 2 )15, , CH 3, |, N CH3 Br, |, CH3, , cetyl trimethyl ammonium bromide, , 85. (d) Cetyltrimethyl ammonium bromide possess germicidal, properties. Thus it is used as a cationic detergent in, hair conditioners., 86. (b), 87. (c), 88. (a), 89. (c) Non– ionic detergent is formed when stearic acid reacts, with polyethylene glycol., 90. (c), , STATEMENT TYPE QUESTIONS, 91. (b), 92. (c), 93. (b) For statement (ii), drug which brings major change in, the treatment of hyperacidity was histamine. For, statement (iv), antidepressant drugs like iproniazid and, phenelzine inhibit the enzymes which catalyse the, degradation of noradrenaline when the enzyme is, inhibited, this important neurotransmitter is slowly, metabolised and can activate its receptor for longer, periods of time, thus counteracting the effect of, depression., 94. (c) For statement (ii), 2–3% solution of iodine in alcohol, water mixture is known as tincture of iodine. For, statement (iii), SO2 in very low concentrations are used, as disinfectants.
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EBD_7207, CHEMISTRY IN EVERYDAY LIFE, , 520, , 95., 96., 97., , (b), (a) Shaving soaps contain glycerol to prevent rapid drying., (d) For statement (iii), cationic detergents are expensive, and thus have limited use. For statement (iv),, detergents having unbranched chains can be, biodegraded more easily thus are less polluting as, compared to detergents having branched chains., , 117. (d) Bacteriostatic drugs inhibit the growth of organism, while bactericidal drugs kill the microorganisms., 118. (a), OH, 119. (a), , H3C, , MATCHING TYPE QUESTIONS, 98., , (c), , 99. (a), , 100. (a), , 101. (c), , Cl, , CH3, , Chloroxylenol, , 102. (b), , ASSERTION-REASON TYPE QUESTIONS, 103. (d) The drugs which act on the central nervous system, and help in reducing anxiety are called tranquilizers., 104. (a) Tranquilizers are chemicals which are used to cure, mental diseases., 105. (a) Broad spectrum antibiotics are those medicines which, are effective against several different types of harmful, micro organisms., 106. (b) Antiseptics are those chemical which kill or prevent, the growth of micro organism. Antiseptics do not harm, the living tissues and can be applied on cuts and, wounds. They help to reduce odour resulting from the, bacterial decomposition in the mouth and on the body., 107. (a) A small quantity of sedative produces a feeling of, relaxation, calmness and drowsiness., 108. (d), 109. (b), , CRITICAL THINKING TYPE QUESTIONS, 110. (a), 111. (c), 112. (c) Iproniazid, Tranquilizer, Salvarsan, Antimicrobial, Zantac (ranitidine), Antihistamine, Chloramphenicol, Antibiotic, 113. (a) It is an alkaloid, a class of organic compound which is, basic in nature and of plant origin containing atleast, one nitrogen atom in a ring structure of molecule., 114. (b) The term “antihistamine” refers only to H1 antagonists,, which is also known as H1-receptor antagonists and, H1-antihistamine., 115. (a) Amoxillin is semisynthetic modification of Penicillin, 116. (a) Ciprofloxacin is used as antibiotic while paracetamol,, ibuprofen and tocopherol are respectively antipyretic,, pain killer and Vit. E., , (4-chloro-3, 5-dimethylphenol), , 120. (d) Drugs that mimic the natural messenger by switching, on the receptor are called agonists. While drugs that, binds to the receptor site and inhibit its natural function, are called antagonists., 121. (b), 122. (b) Bactericidal have killing effect on microbes while, bacteriostatic have inhibitory effect on microbes., 123. (b) Chloramphenicol is bacteriostatic antibiotic while, ofloxacin is bactericidal type antibiotic., 124. (c) Narrow spectrum antibiotics are effective against, Gram–positive or Gram-negative bacteria. Limited, spectrum antibiotics are effective against a single, organism or disease., 125. (a) Norethindrone is an example of synthetic progestrone, derivative most widely used as antifertility drug., 126. (b) Benzoic acid used as preservative as sodium benzoate., 127. (c), HO, OH, Cl, HO, , O, , Cl, , HO, O, , O, , OH Cl, Sucrolose, , 128. (a) Benzalkonium chloride, also known as, alkyldimethylbenzylammonium chloride is nitrogenous, cationic surface active agent belonging to the, quaternary ammonium group. It is used as antiseptic., –, Cl, +, , N, CnH2n+1, , n = 8, 10, 12, 14, 16, 18, , 129. (b) Laundry soaps contain fillers like sodium rosinate,, sodium silicate, borax and sodium carbonate., 130. (b), 131. (b) Aspartame cannot be used in baked food as it is, unstable at cooking temperature thus its use is limited, to cold foods and soft drinks.
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Mock Test-1, Time : 1 hr, 1., , 2., , 3., , 4., , 5., , 6., , Max. Marks -180, , An acidic solution of 'X' does not give precipitate on, passing H2S through it. 'X' gives white precipitate when, NH4OH is added to it. The white precipitate dissolves in, excess of NaOH solution. Pure 'X' fumes in air and dense, white fumes are obtained when a glass rod dipped in NH4OH, is put in the fumes. Compound 'X' can be, (a) ZnCl2, (b) FeCl3, (c) AlCl3, (d) SnCl2, CN– is a strong field ligand. This is due to the fact that, (a) it carries negative charge, (b) it is a pseudohalide, (c) it can accept electrons from metal species, (d) it forms high spin complexes with metal species, The weight of NaCl decomposed by 4.9g of H2SO4, if 6 g of, sodium hydrogen sulphate and 1.825 g of HCl, were, produced in the reaction is:, (a) 6.921 g, (b) 4.65 g, (c) 2.925 g, (d) 1.4 g, Which one of the following statement is not, true ?, (a) pH of dr inking water should be between, 5.5 – 9.5., (b) Concentration of DO below 6 ppm is good for the, growth of fish., (c) Clean water would have a BOD value of less than 5 ppm., (d) Oxides of sulphur, nitrogen and carbon are the most, widespread air pollutant., Which of the following statements is not correct for nitrogen ?, (a) Its electronegativity is very high, (b) d-orbitals are available for bonding, (c) It is a typical non-metal, (d) Its molecular size is small, Which of the following statement is false ?, (a) For 1 mole of an ideal gas, Cp – Cv = R, (b), , E, T, , 0 for an ideal gas, T, , q w p v, (c), (d) For reversible isothermal expansion of 1 mole of an, ideal gas from volume V1 to V2, work done is equal to, RT ln (V2/V1), 7., , 0.4 moles of HCl and 0.2 moles of CaCl 2 were dissolved in, water to have 500 mL of solution, the molarity of Cl– ion is:, (a) 0.8 M, (b) 1.6 M, (c) 1.2 M, (d) 10.0 M, , 8., , 9., , 10., , 11., , 12., , In sodium fusion test of organic compounds, the nitrogen, of the organic compound is converted into, (a) Sodamide, (b) Sodium cyanide, (c) Sodium nitrite, (d) Sodium nitrate, Specific volume of cylindrical virus particle is 6.02 × 10–2, cc/gm. whose radius and length 7 Å & 10 Å respectively. If, NA = 6.02 × 1023, find molecular weight of virus, (a) 3.08 × 103 kg/mol, (b) 3.08 × 104 kg/mol, 4, (c) 1.54 × 10 kg/mol, (d) 15.4 kg/mol, Inductive effect involves, (a) displacement of -electrons, (b) delocalisation of -electrons, (c) delocalisation of -electrons, (d) displacement of -electrons, The energy of a photon is 3 × 10–12 erg. What is its, wavelength in nm ?, (h = 6.62 × 10–27 erg-sec; c = 3 × 1010 cm/s), (a) 662, (b) 1324, (c) 66.2, (d) 6.62, Among the following compounds (I - III), the ease of their, reaction with electrophiles is,, , OCH3, , I, , 13., , 14., , NO2, , II, , III, , (a) II > III > I, (b) III > II > I, (c) II > I > III, (d) I > II > III, Aluminium vessels should not be washed with materials, containing washing soda since, (a) washing soda is expensive, (b) washing soda is easily decomposed, (c) washing soda reacts with Al to form soluble aluminate, (d) washing soda reacts with Al to form insoluble, aluminium oxide, The following data are for the decomposition of ammonium, nitrite in aqueous solution :, Vol. of N2 in cc, Time (min), 6.25, 10, 9.00, 15, 11.40, 20, 13.65, 25, 35.65, Infinity, The order of rection is :, (a) Zero, (b) One, (c) Two, (d) Three
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EBD_7207, MT-2, , 15., , 16., , 17., , CHEMISTRY, , Which of the following reagents convert propene to 1propanol?, (a) H2O, H2SO4, (b) aqueous KOH, (c) MgSO4, NaBH4/H2O (d) B2H6, H2O2, OH–, A closed container contains equal number of oxygen and, hydrogen molecules at a total pressure of 740 mm. If oxygen, is removed form the system then pressure will, (a) Become double of 740 mm, (b) Become half of 740 mm, (c) Become 1/9 of 740 mm, (d) Remains unchanged, Compound X of molecular formula C4H6 takes up one, equivalent of hydrogen in presence of Pt to form another, compound Y which on ozonolysis gives only ethanoic acid., The compound X can be, (a), , CH 2, , CH CH, , CH 2, , (b) CH2 = C = CHCH3, (c), , 18., , 19., , CH 3C CCH 3, (d) All the three, 1 M solution of CH3COOH should be diluted to ..............., times so that pH is doubled., (a) four times, (b) 5.55 × 104 times, 6, (c) 5.55 × 10 times, (d) 10–2 times, Which of the following statements, about the advantage, of roasting of sulphide ore before reduction is not true?, (a) The G of of the sulphide is greater than those for, CS2 and H2S., , (b) The G of is negative for roasting of sulphide ore to, oxide., (c) Roasting of the sulphide to the oxide is, thermodynamically feasible., (d) Carbon and hydrogen are suitable reducing agents, for metal sulphides., 20. Which one of the following is NOT a buffer solution?, (a) 0.8 M H2 S + 0.8 M KHS, (b) 2MC6H5NH2 + 2MC6H5 N H 3 Br –, (c) 3MH2CO3 + 3MKHCO3, (d) 0.05 M KClO4 + 0.05 M HClO4, 21. Which of the following statements is false ?, (a) Radon is obtained from the decay of radium, (b) Helium is inert gas, (c) Xenon is the most reactive among the rare gases, (d) The most abundant rare gas found in the atmosphere, is helium, 22. Which one of the following is expected to exhibit optical, isomerism?, (en = ethylenediamine), (a) cis-[Pt(NH3)2 Cl2], (b) trans-[Pt(NH3)2Cl 2], (c) cis-[Co(en)2Cl 2], (d) trans-[Co(en)2Cl 2], , 23., , The following equilibrium constants are given:, 2NH3 ; K1, N 2 3H 2, , N2 O2, 2NO; K 2, 1, O2, H 2 O; K 2, H2, 2, The equilibrium constant for the oxidation of NH3 by, oxygen to give NO is, (a), , K 2 K32, K1, , (b), , K 22 K3, K1, , K 2 K33, K1, 24. Four successive members of the first row transition, elements are listed below with their atomic numbers., Which one of them is expected to have the highest, third ionization enthalpy?, (a) Vanadium (Z = 23), (b) Chromium (Z = 24), (c) Manganese (Z = 25) (d) Iron (Z = 26), 25. Which of the following organometallic compound is and, bonded?, (a) [Fe ( 5 – C5H5)2], (b) Fe (CH3)3, (c) K [PtCl3( 2 – C2H4)] (d) [Co(CO)5 NH3]2+, 26. In the balanced chemical reaction, bH, cH 2 O dI 2, IO3 aI, a, b, c and d respectively corresponds to, (a) 5, 6, 3, 3, (b) 5, 3, 6, 3, (c) 3, 5, 3, 6, (d) 5, 6, 5, 5, 27. Which of the following statements is true?, (a) Silicon exhibits 4 coordination number in its, compound, (b) Bond energy of F2 is less than Cl2, (c) Mn(III) oxidation state is more stable than Mn(II) in, aqueous state, (d) Elements of 15th group shows only +3 and +5, oxidation states, 28. Which of the following compounds has the highest boiling, point?, (a) CH3CH2CH2 Cl, (b) CH 3CH 2 CH 2 CH 2 Cl, (c) CH 3CH (CH 3 )CH 2 Cl, (d) (CH3 ) 3 CCl, 29. Which one of the following statements is not correct ?, (a) Nickel forms Ni(CO)4, (b) All the transition metals form monometallic carbonyls, (c) Carbonyls are formed by transition metals, (d) Transition metals form complexes, 30. Hydrogen has an ionisation energy of 1311 kJ mol–1 and, for chlorine it is 1256 kJ mol–1. Hydrogen forms H+ (aq), ions but chlorine does not form Cl+ (aq) ions because, (a) H+ has lower hydration enthalpy, (b) Cl+ has lower hydration enthalpy, (c) Cl has high electron affinity, (d) Cl has high electronegativity, , (b), , K1 K 2, K3, , (d)
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MOCK TEST 1, 31., , 32., , MT-3, , The number of en antiomers of the compound, CH 3 CHBr CHBr COOH is :, (a) 2, (b) 3, (c) 4, (d) 6, Equivalent weighs of KMnO4 acidic medium, neutral, medium and concentrated alkaline medium respectively are, M M M, , , . Reduced products can be, 5 1 3, (a) MnO 2 , MnO 24 , Mn 2, (b), , MnO 2 , Mn 2 , MnO 24, , (c), , Mn 2 , MnO24 , MnO 2, , Mn 2 , MnO2 , MnO24, Which of these have no unit?, (a) Electronegativity, (b) Electron affinity, (c) Ionisation energy, (d) Excitation potential, Which of the following statements is not correct for sigma, and pi-bonds formed between two carbon atoms?, (a) Sigma-bond determines the direction between carbon, atoms but a pi-bond has no primary effect in this, regard, (b) Sigma-bond is stronger than a pi-bond, (c) Bond energies of sigma- and pi-bonds are of the order, of 264 kJ/mol and 347 kJ/mol, respectively, (d) Free rotation of atoms about a sigma-bond is allowed, but not in case of a pi-bond, The reactivity of metals with water is in the order of, (a) Na > Mg > Zn >Fe > Cu, (b) Cu > Fe > Zn > Mg > Na, (c) Mg > Zn > Na > Fe > Cu, (d) Zn > Na > Mg > Fe > Cu, The emf of Daniell cell at 298 K is E1, Zn | ZnSO4 (0.01 M) | | CuSO4 (1.0 M) | Cu, When the concentration of ZnSO4 is 1.0 M and that of, CuSO4 is 0.01 M, the emf changed to E2 What is the relation, between E1 and E2?, , (d), , 33., , 34., , 35., , 36., , (a) E1 = E2, , (b), , E2, , 0, , (c), , (d), , E1, , E2, , E1, , E2, , 42., , 43., , E2, , 44., , O, , The above shown polymer is obtained when a carbonyl, compound is allowed to stand. It is a white solid. The, polymer is, (a) Trioxane, (b) Formose, (c) Paraformaldehyde, (d) Metaldehyde., The correct order of atomic/ionic sizes is, (a) N < Li < B, (c), , 41., , CH 2, , O, CH 2, , 38., , 40., , O, , CH 2, 37., , 39., , Ca 2, , S2, , Cl, , (b), , F, , (d), , Na, , O2, , Mg 2, , N3, , Cl, , 45., , In qualitative analysis, the metals of Group I can be, separated from other ions by precipitating them as chloride, salts. A solution initially contains Ag + and Pb2+ at a, concentration of 0.10 M. Aqueous HCl is added to this, solution until the Cl– concentration is 0.10 M. What will, the concentrations of Ag+ and Pb2+ be at equilibrium?, (Ksp for AgCl = 1.8 × 10–10,, Ksp for PbCl2 = 1.7 × 10–5), (a) [Ag+] = 1.8 × 10–7 M ; [Pb2+] = 1.7 × 10–6 M, (b) [Ag+] = 1.8 × 10–11 M ; [Pb2+] = 8.5 ×10–5 M, (c) [Ag+] = 1.8 × 10–9 M ; [Pb2+] = 1.7 × 10–3 M, (d) [Ag+] = 1.8 × 10–11 M ; [Pb2+] = 8.5 ×10–4 M, In the diazotization of arylamines with sodium nitrite and, hydrochloric acid, an excess of hydrochloric acid is used, primarily to, (a) Supress the concentration of free aniline available, for coupling, (b) Supress hydrolysis of phenol, (c) Ensure a stoichiometric amount of nitrous acid, (d) Neutralise the base liberated, In lake test of Al 3 ion, there is formation of coloured, floating lake. It is due to, (a) adsorption of litmus by H2O, (b) adsorption of litmus by Al(OH)3, (c) adsorption of litmus by Al(OH)4–, (d) none of these, For the reaction, 2Cl(g) — Cl2(g), the signs of H and S, respectively, are:, (a) +, –, (b) +, +, (c) –, –, (d) –, +, Ethanol and dimethyl ether form a pair of functional isomers., The boiling point of ethanol is higher than that of dimethyl, ether, due to the presence of, (a) H-bonding in ethanol, (b) H-bonding in dimethyl ether, (c) CH3 group in ethanol, (d) CH3 group in dimethyl ether, Which of the following reactions will not result in the, formation of anisole?, (a) Phenol + dimethyl sulphate in presence of a base, (b) Sodium phenoxide is treated with methyl iodide, (c) Reaction of diazomethane with phenol, (d) Reaction of methylmagnesium iodide with phenol, What will be the heat of formation of methane, if the heat of, combustion of carbon is '–x' kJ, heat of formation of water, is '–y' kJ and heat of combustion of methane is 'z' kJ ?, (a) (–x – y + z) kJ, (c) (–x – 2y – z) kJ, , (b) (–z – x + 2y) kJ, (d) (–x – 2y + z) kJ
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EBD_7207, MT-4, , CHEMISTRY, ANSWER KEY, , 1. (a), , 2. (b), , 3. (c), , 4. (b), , 5. (b), , 6. (c), , 7. (b), , 8. (b), , 9. (d), , 10. (a), , 11. (a), , 12. (d), , 13. (d), , 14. (b), , 15. (d), , 16. (b), , 17. (d), , 18. (b), , 19. (d), , 20. (d), , 21. (d), , 22. (c), , 23. (d), , 24. (c), , 25. (d), , 26. (a), , 27. (b), , 28. (b), , 29. (b), , 30. (b), , 31. (c), , 32. (c), , 33. (a), , 34. (c), , 35. (a), , 36. (c), , 37. (a), , 38. (b), , 39. (c), , 40. (a), , 41. (b), , 42. (c), , 43. (a), , 44. (d), , 45. (d), , HINTS & SOLUTIONS, 1., , (a), , NH 4 OH, , X, , 8., , White ppt, excess, , Acidic solution (soluble ), , NaOH, , 9., , ( No. ppt with H 2S), , Given reactions (white precipitate with H2S in presence, of NH4OH) indicate that 'X' should be ZnCl2 which, explains all given reactions., ZnCl 2 2H 2 O, NH 4 OH, , HCl, , White fumes –H 2O, , ZnCl 2, Zn (OH ) 2, , 2., , 3., , 4., , 5., 6., 7., , Zn (OH ) 2, , HCl, , White fumes, , NH 4Cl, Dense white fumes, , 2 NaOH, , Zn(OH) 2, , 2NaCl, , 2 NaOH, , Na 2 ZnO 2, , 2H 2 O, , Excess, , (b) CN– is a strong field ligand as it is a psuedohalide ion., These ions are strong coordinating ligands and hence, have the tendency to form -bond (from the pseudo, halide to the metal) and -bond. (from the metal to, pseudo halide), (c) NaCl + H 2SO 4, NaHSO 4 HCl, xg, , 4.9g, , 6g, , CaCl2, , Wt. of one virus particle, , 10., 11., , Ca, , 0.2moles, , 2Cl, , 2 0.2 0.4moles, , Total Cl– moles = 0.4 + 0.4 = 0.8 moles, Moles, Molarity =, Vol.in L, , Molarity of Cl–, , 0.8, =, 0.5, , 154 10, , 6.62 10 27 erg-sec 3 1010 cm s 1, , 3 10 12 erg, , 12., 13., 14., , 15., , = 6.62 × 10–5 cm, = 6.62 × 10–5 × 107 nm [1 cm = 107 nm], = 662 nm, (d) –OCH3 activates the benzene ring. –NO2 deactivates, the ring. Hence the reaction of the given compounds, with electrophiles is in the order, I > II > III., (d), (b) NH 4 NO2, N 2 2H 2O, Volume of N2 formed in successive five minutes are, 2.75 cc, 2.40 cc and 2.25 cc which is in decreasing, order. So rate of reaction is dependen t on, concentration of NH4NO2. As decrease is not very, fast so it will be first order reaction., (d) We know that, 6 CH3 CH, , CH 2, , 1, Pr opene, , 2(CH3 CH 2 CH 2 )3, , 1.6 M., , 23, , 6.02 10 23, 6.02 10 2, = 15400 g/mol = 15.4 kg/mole, (a) Inductive effect involves displacement of -electrons., hc, (a) Using the relation, E, Substituting given values, we get, , =, , 0.4moles, , 2, , volume, specific volume, , Mol. wt. of virus = Wt. of NA particle, , 1.825g, , According to law of conservation of mass "mass is, neither created nor destroyed during a chemical, change", Mass of the reactants = Mass of products, x + 4.9 = 6 + 1.825, or, x = 2.925 g, (b) The ideal value of D.O for growth of fishes is 8 mg/ ., 7mg is desirable range, below this value fishes get, susceptible to desease. A value of 2 mg/ or below is, lethal for fishes., (b) In case of nitrogen, d-orbitals are not available., (c) C is incorrect ; The correct is E = q + w, (b), HCl, H, Cl, 0.4moles, , NaCN)., (b) Sodium cyanide (Na + C + N, (Lassaigne's test), (d) Specific volume (volume of 1 gm) of cylindrical virus, particle = 6.02 × 10–2 cc/gm, Radius of virus (r) = 7 Å = 7 × 10–8 cm, Length of virus = 10 × 10–8 cm, Volume of virus, 22, (7 10 8 ) 2 10 10 8, r 2l, 7, = 154 × 10–23 cc, , B 2H 6, ether, 0°C, H2O2, OH, , 6CH 3CH 2 CH 2 OH 2H 3 BO 3, Pr opanol
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MOCK TEST 1, 16., , (b), , MT-5, , p tot, , 740 mm, , p tot, , p O2, , 23., , pH 2, , (d) Given,, N 2 3H 2, , Number of moles of O 2 and H2 are equal, p O2 p H 2, pO 2, , pH 2, , 740, , pO2 370mm p H2, 17. (d) Formation of only CH3COOH by ozonolysis indicates, that the compound Y should be CH3CH = CHCH3, which can be formed by all of the three given, compounds, CH 2, , CH CH, , CH 3, , CH3C CCH3, , 1H 2 / Pt, , X, , CH 2, , C, X, , CH 3 CH, , 18., , 19., , 20., , 21., , Y, , CH, , CH CH 3, , Y, , CH3CH, , Y, , O3, , 22. (c), , 2CH 3 COOH, , en, , +, Cl, , Cl, , Co, , Co, Cl, , Cl, en, Cis-d-isomer, , en, Cis- -isomer, , Mirror, , 3, , [N 2 ] [H 2 ], , Now operate,, , +, , ....(i), , [NO]2 [H2O]3, [NH3 ]2 [O2 ]5 / 2, [NO]2, [N 2 ] [O 2 ], , , K2, , [H 2O], ½, , [H 2 ] [O 2 ], , or K 3, , [H 2O]3, [H 2 ]3 [O 2 ]3 / 2, , K 2 . K 33, K1, , [H 2 O]3, [N ] [H2 ]3, [NO]2, . 2, 3, 3, /, 2, [N 2 ] [O2 ] [H 2 ] [O2 ], [NH3 ]2, , pH, , en, , [NH3 ]2, , & K3, , CHCH3, , 1, pK a, [pK a log1], 2, 2, pH' (twice of pH) = pKa, 1, pK a, [pK a log C ], 2, – log C = pKa = – log Ka, C = Ka = 1.8 × 10–5 M, 1, dilution, 5.55 × 10 4 times, C, (d) The sulphide ore is roasted to oxide before reduction, because the Gof of most of the sulphides are greater, than those of CS2 and H2S, therefore neither C nor H, can reduce metal sulphide to metal. Further, the, standard free energies of formation of oxide are much, less than those of SO2. Hence oxidation of metal, sulphides to metal oxide is thermodynamically, favourable., (d) Buffer solution contains weak base + salt of weak, base with strong acid or weak acid + salt of weak acid, with strong base., In option (d) the acid used is HClO4 which is strong, acid and KClO4 is salt of this acid with strong base., So it is not an example of buffer solution., (d) The most abundant rare gas found in the atmosphere, is argon and not helium., , (b), , but K1, , 1H 2 / Pt, , CHCH3, CHCH3, , For this equation, K, , 1H 2 / Pt, , CH 2, , X, , or, , 2NH 3 ; K1, , N2 O2, 2NO; K 2, ....(ii), 1, H2, O2, H 2 O; K 3, ....(iii), 2, We have to calculate, 4NH 3 5O2, 4NO 6H 2O; K ?, 5, 2NH 3, O2, 2NO 3H 2 O, 2, , [NO]2 [H 2 O]3, [NH 3 ]2 [O 2 ]5 / 2, K, , 24., , K, , K 2 . K 33, K1, , (c) For third ionization enthalpy last configuration of, 4s, 3d, V – 4s0 3d3, Cr, , – 4s0 3d4, , Mn – 4s0 3d5, Fe – 4s0 3d6, For third Ionization enthalpy Mn has stable, configuration due to half filled d-orbital., 25., 26., , (d) [ Co (CO ) 5 NH 3 ]2 . In this complex. Co-atom attached, with NH3 through bonding with CO attached with, dative -bond., (a) Given reaction is, IO3 aI, bH, Ist half reaction, I, , I2, , –1, 0, IInd half reaction, , cH 2 O dI 2, , ...(i), (oxidation), , ...(ii), IO 3, I2, +5, 0, (reduction), On balancing equation (ii) we have, 10e, 2IO3 12H, I 2 6H 2 O ...(iii), Now, balance equation (i)
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EBD_7207, MT-6, , CHEMISTRY, , I2 2e ....(iv), 2I, Multiply eqn (iv) by 5 and add it to eqn (iii), we get, 2IO 3, , 27., , 10I, , 12H, , .., , 29., 30., 31., 32., , MnO 4, , H, , MnO 4, , H 2O, , Mn 2, 2, , 7, , 35., , 36., , MnO 42, 6, , M, 5, , 5, 1, , OH, , MnO2, , 3, , larger size than Cu due to (3d10 4s 2 ) configurations., Hence, the reactivity order is, Na > Mg > Zn > Fe > Cu., (c) Using the relation, E cell, , 0, E cell, , E0cell, , 0.0591, [anode], log, n, [cathode], 2+, , 0.0591 [Zn ], log, n, [Cu 2+ ], , Substituting the given values in two cases., 0.01, 0.0591, E1 E 0, log, 2, 1.0, 0.0591, log10 2, E0, 2, 0.0591, E0, 2 or (E 0 0.0591)V, 2, , E0, , On keeping, , O, , CH2, , O, , 37., , (a), , 38., , (b) Amongst the isoelectronic species, the anion having, more negative charge would have the larger size., (c) Ksp = [Ag+] [Cl–], 1.8 × 10–10 = [Ag+] [0.1], [Ag+] = 1.8 × 10–9 M, Ksp = [Pb+2] [Cl–]2, 1.7 × 10–5 = [Pb+2] [0.1]2, [Pb+2] = 1.7 × 10–3 M, (a) Excess of HCl is used to convert free aniline to aniline, hydrochloride otherwise free aniline would undergo, coupling reaction with benzenediazonium chloride., (b) In lake test of Al3+, there is formation of coloured, floating lake. It is due to the adsorption of litmus by, Al(OH)3., (c) 2Cl(g) — Cl2(g), Entropy is decreasing (–ve) in the reaction. Further, the reaction is exothermic since a bond is being, formed, i.e., H is also –ve., (a) Due to H-bonding, the boiling point of ethanol is much, higher than that of the isomeric diethyl ether., (d) Phenol has active (acidic) hydrogen so it reacts with, CH3MgI to give CH4, and not anisole, C 6 H 5 OH CH 3MgI, CH 4 C6 H5OMgI, (d) From given data, we have, CO2 – x kJ, … (i), C + O2, 1, O2, H 2 O y kJ, H2, …(ii), 2, …(iii), CH 4 2O 2, CO 2 2H 2 O z kJ, , 39., , 40., , M, , M, 3, 7, 4, (a) Electronegativity is the tendency of the atom to attract, electrons to itself when combined in a compound as, defined by Pauling. Electronegativity, is a relative term, so it does not have any unit., (c) As sigma bond is stronger than the (pi) bond, so it, must be having higher bond energy than (pi) bond., (a) The reactivity may be attributed to size factor, larger, is the size, higher is tendency to lose electron (low, I.E.). Zn is the last element of 3d series and it has, MnO 4, , 34., , .., , F, ..:, , (b) Molecules having higher molecular weight and less, branching have higher boiling point., (b) Always transition metals combines with more than, one carbonyl group., (b) Hydration energy of Cl+ is very less than H+ hence it, doesn’t form Cl+ (aq) ion., (c) No. of asymmetric carbon = 2, No. of enantiomers = 22 = 4., (c), Change Equiv. mass, 7, , 33., , 6H 2 O, , or, IO3 5I 6H, 3I 2 3H 2 O, Hence a = 5, b = 6, c = 3, d = 3, (b) This is because of inter-electronic replusions between, lone pairs., B.E. :, F – F Cl – Cl, (kJ mol–1) : 158.8 242.6, : .F., , 28., , 6I 2, , 1, 0.0591, log, 2, 0.01, 0.0591, log10 2, E0 –, 2, 2 0.0591, E0 –, or (E 0 – 0.0591)V, 2, Thus,, E1 E 2, , E2, , 41., 42., , 43., 44., , 45., , 3 HCHO, aq. solution, , CH2, , CH2, O, Trioxane, (meta formaldehyde), , The required equation is, C 2H 2, CH 4 Q, To get the required equation, operate, (i) + 2 × (ii) – (iii), Thus, we get, C, , 2H 2, , CH 4, , [( x ), , ( 2 y ) ( z )], , Thus,, heat of formation of methane is, (–x – 2y + z) kJ
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Mock Test-2, Time : 1 hr, 1., , The angular momentum of the electron in first excited, energy state of hydrogen atom is, (a), , 3., , 4., , 5., , 6., , 7., , 8., , h, , (b), 2(2 1), , (b), , CH 2, , (c), , CH 3COCH 3, , (d), , CH 2, , CH, , CH, , CH CH 2, , (c) n = 3, l = 2, m = +1, s =, , h, 2, , h, (d) None of these, 2, When NaCl is dopped with 1.0 × 10–3 mole of SrCl2, the, number of cation vacancy is, (a) 6.023 × 1018, (b) 6.023 × 1020, 20, (c) 2 × 6.023 × 10, (d) 3.011 × 1020, A 0.5 M NaOH solution offers a resistance of 31.6 ohm in a, conductivity cell at room temperature. What shall be the, approximate molar conductance of this NaOH solution if, cell constant of the cell is 0.367 cm–1 ., (a) 23.4 S cm2 mole–1, (b) 23.2 S cm2 mole–1, (c) 46.45 S cm2 mole–1, (d) 54.64 S cm2 mole–1, Ammonium dichromate on heating gives, (a) chromic acid & ammonia, (b) chromium sesquioxide & nitrogen, (c) chromium sesquioxide & ammonia, (d) chromic acid and N2, Predict the relative acidic strength among the following, (a) H2O, H2S, H2Se, H2Te, (b) H2O < H2S < H2Se < H2Te, (c) H2Te < H2Se < H2S < H2O, (d) H2O < H2Se < H2S < H2Te, The catalyst used in the preparation of an alkyl chloride by, the action of dry HCl on an alcohol is, (a) anhydrous AlCl3, (b) FeCl3, (c) anhydrous ZnCl2, (d) Cu, In which of the following, resonance will be possible?, (a) CH2 CH2 CH2 CHO, , (c), , 2., , Max. Marks -180, 1, 2, , 1, 2, 4+, –, 2+, Sn + 2e, Sn, E° = 0.13 V, –, –, 2Br, E° = 1.08 V, Br2 + 2e, Calculate Keq for the cell at 20°C formed by two electrodes, (a) 1041, (b) 1032, –32, (c) 10, (d) 10–42, Calculate the pH of a solution obtained by mixing 2 ml of, HCl of pH 2 and 3 ml of solution of KOH of pH = 12, (a) 10.30, (b) 3.70, (c) 11.30, (d) None of these, Which of the following represents a correct sequence of, reducing power of the following elements?, (a) Li > Cs > Rb, (b) Rb > Cs > Li, (c) Cs > Li > Rb, (d) Li > Rb > Cs, Paramagnetism of Cr (Z = 24), Mn2+ (Z = 25) and Fe3+, (Z = 26) are x, y and z respectively. They are in the order, (a) x = y = z, (b) x > y > z, (c) x = y > z, (d) x > y = z, Formaldehyde reacts with ammonia to give urotropine is, (a) (CH2)6N4, (b) (CH2)4N3, (c) (CH2)6N6, (d) (CH2)3N3, Indicate the wrongly named compound, , (d) n = 3, l = 1, m = –1, s =, , 9., , 10., , 11., , 12., , 13., , 14., , (a), , CH 3 C H CH 2 CH 2, |, CH 3, , CHO, , (4-methyl -1- pentanal), (b), , O, , CH 3 CH C, |, CH 3, , C COOH, , (4- methyl -2- pentyn -1- oic acid), CH, , CH 2, , The four quantum numbers that could identify the third 3p, electron in sulphur are, 1, (a) n = 3, l = 0, m = +1, s =, 2, (b) n = 2, l = 2, m = –1, s =, , 1, 2, , (c), , CH 3CH 2 CH 2 C H COOH, |, CH 3, , (2- methyl -1- pentanoic acid), , O, ||, (d) CH3 CH2 CH CH C CH3
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EBD_7207, MT-8, , 15., , (3- hexen -5- one), The favourable condition for a process to be spontaneous, is :, (a), , T S, , H,, , H, , ive,, , S, , ive, , (b) T S, , H,, , H, , ive,, , S, , ive, , T S, , H,, , H, , ive,, , S, , ive, , (d) T S, , H,, , H, , ive,, , S, , ive, , (c), 16., , 17., , CHEMISTRY, , (c), 18., , 19., , 20., , 21., , 2.303, 1, log, min, 24, 8, , 1, , (b), , 0.692, min, 24, , (d), , 2.303, 8, log, min, 24, 1, , (b), , 27., , 28., , C6 H 5 CHCHNO 2, , C 6 H 5 CH, , CHCOOH, , 29., , C6 H5 CH= CHCHO, , 30., , C 6 H 5CHO CH 2 (COOH) 2, , C 6 H 5 CH, , CHCO 2 H, , The product obtained on reaction of C2H5Cl with hydrogen, over palladium carbon is :, (a) C3H8, (b) C4H10, (c) C2H6, (d) C2H4, Which of the following can be predicted from, electronegativity values of elements ?, (a) Dipole moment of a molecule, (b) Valency of elements, (c) Polarity of bonds, (d) Position in electrochemical series, ClO 3( aq ) 2Cl ( aq ) is an, The reaction 3ClO ( aq ), example of, (a) oxidation reaction, (b) reduction reaction, (c) disproportionation reaction, (d) decomposition reaction, The concentration of a reactant X decreases from 0.1 M to, 0.005 M in 40 min. If the reaction follows first order kinetics,, the rate of the reaction when the concentration of X is 0.01, M will be, (a) 1.73 × 10–4 M min–1 (b) 3.47 × 10–4 M min–1, (c) 3.47 × 10–5 M min–1 (d) 7.5 × 10–4 M min–1, Mark the false statement?, (a) A salt bridge is used to eliminate liquid junction, potential, (b) The Gibbs free energy change, G is related with, electromotive force E as G = –nFE, (c) Nernst equation for single electrode potential is, E, , C6 H5CHO CH3CHO, NaOH, , (d), , 26., , C 6 H 5 CHO (CH 3CO ) 2 O, CH 3COONa, , (c), , 25., , 1, , C6 H5CHO CH3 NO 2, KOH, , 24., , 1, , Fluorine does not show highest oxidation state opposite, to other halogens, because, (a) it is most electronegative, (b) it has no d-orbital, (c) its atomic radius is very small, (d) F– ion is stable and isoelectronic with neon, Glucose contains in addition to aldehyde group., (a) one secondary –OH and four primary –OH groups, (b) one primary –OH and four secondary –OH groups, (c) two primary –OH and three secondary –OH groups, (d) three primary –OH and two secondary –OH groups, Which of the following product is obtained by treating 1butyne with HgSO4 and H2SO4?, (a) CH3CH2COCH3, (b) CH3CH2CH2CHO, (c) CH3CH2CH2COOH, (d) CH3CH2CH = CH2, An example of Perkin’s reaction is, (a), , 22., , 23., , Vapour pressure (in torr) of an ideal solution of two liquids, A and B is given by : P = 52XA + 114, where XA is the mole fraction of A in the mixture. The vapour, pressure (in torr) of equimolar mixture of the two liquids, will be :, (a) 166, (b) 83, (c) 140, (d) 280, The decomposition of a substance follows first order, kinetics. Its concentration is reduced to 1/8th of its initial, value in 24 minutes. The rate constant of the decomposition, process is, (a) 1/24 min–1, , Alc. NH 3, , Eo, , RT, log a M n, nF, , (d) The efficiency of a hydrogen-oxygen fuel cell is 23%, The paramagnetism of transition element compounds is, due to, (a) paired eletrons spining in opposite directions, (b) unpaired eletrons in d and f-orbitals, (c) shared valance electrons, (d) unpaired electrons in s or p-orbitals., Aniline, chloroform and alcoholic KOH react to produce a, bad smelling substance which is, (a) phenyl isocyanide, (b) phenyl cyanide, (c) chlorobenzene, (d) benzyl alcohol., The species with a radius less than that of Ne is, (a) Mg2+, (b) F–, 2–, (c) O, (d) K+, Vapour density of the equilibrium mixture of the reaction, SO2Cl2(g), SO2(g) + Cl2(g) is 50.0. Percent dissociation, of SO2Cl2 is :
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MOCK TEST 2, , 31., , 32., , 33., , 34., , 35., , 36., , 37., , (a) 33.33, (b) 35.0, (c) 30.0, (d) 66.67, What will be the emf for the given cell, Pt | H2 (P1) | H+ (aq) | H2 (P2) | Pt, (a), , RT P1, ln, F, P2, , (b), , (c), , RT P2, ln, F, P1, , (d) None of these, , RT P1, ln, 2F P2, , H3PO3 is, (a) neutral, (b) basic, (c) a tribasic acid, (d) a dibasic acid, Of the following which is diamagnetic in nature?, (a), , [ Co F6 ]3, , (b) [ Ni Cl 4 ]2, , (c), , [ Cu Cl 4 ]2, , (d) [Ni(CN) 4 ]2, , Which of the following products is formed when, benzaldehyde is treated with CH3MgBr and the addition, product so obtained is subjected to acid hydrolysis ?, (a) A secondary alcohol (b) A primary alcohol, (c) Phenol, (d) tert-Butyl alcohol, Mole fraction of methanol in its aqueous solution is 0.5., The concentration of solution in terms of percent by mass, of methanol is, (a) 36, (b) 50, (c) 64, (d) 72, The unit cell of an ionic compound is a cube in which, cations (A) occupy each of the corners and anions (B) are, at the centres of each face . The simplest formula of the, ionic compound is, (a) AB2, (b) A3B, (c) AB3, (d) A4 B 3, For orthorhombic system axial ratios are a, b c and, the axial angles are, (a), 90, (b), , 38., , MT-9, , 90, 90 ,, 90, , 90, (c), (d), Electrolytic reduction of alumina to aluminium by HallHeroult process is carried out, (a) in the presence of NaCl, (b) in the presence of fluorite, , 39., , 40., , 41., , 42., , 43., , 44., , (c) in the presence of cryolite which forms a melt with, lower melting temperature, (d) in the presence of cryolite which forms a melt with, higher melting temperature, Consider the following complex, [Co(NH3)5CO3]ClO4., The coordination number, oxidation number, number of delectrons and number of unpaired d-electrons on the meal, are respectively, (a) 6, 3, 6, 0, (b) 7, 2, 7, 1, (c) 7, 1, 6, 4, (d) 6, 2, 7, 3, Nylon is a :, (a) polysaccharide, (b) polyester, (c) polyamide, (d) all of the above, , OH, , Br2, H2O, , X,, , X is identified as, (a) 2, 4, 6-tribromophenol, (b) 2-bromo-4-hydroxylbenzene sulphonic acid, (c) 3, 5-dibromo-4-hydroxybenzene sulphonic acid, (d) 2-bromophenol, The non-polar molecule is :, (a) NF3, (b) SO3, (c) CHCl3, (d) ClO2, The hybridization of P in PO 34 is the same as of, (a) S in SO3, (b) N in NO–3, (d) I in ICl4–, (c) S in SO4– –, Solution of potash alum is acidic in nature. This is due to, hydrolysis of, (a), , 45., , SO3H, , SO 24, , (b) K+, , (c) Al2(SO4)3, (d) Al3+, When conc. HNO3 acts on our skin, the skin becomes, yellow, because, (a) HNO3 acts as an oxidising agent, (b) HNO3 acts as a dehydrating agent, (c) Nitro-cellulose is formed, (d) The proteins are converted into xanthoproteins
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EBD_7207, CHEMISTRY, , MT-10, , ANSWER KEY, 1. (a), , 2. (b), , 3. (b), , 4. (b), , 5. (a), , 6. (c), , 7. (b), , 8. (d), , 9. (b), , 10. (c), , 11. (a), , 12. (d), , 13. (a), , 14. (d), , 15. (b), , 16. (c), , 17. (d), , 18. (b), , 19. (b), , 20. (a), , 21. (b), , 22. (c), , 23. (c), , 24. (c), , 25. (d), , 26. (c), , 27. (b), , 28. (a), , 29. (a), , 30. (b), , 31. (b), , 32. (d), , 33. (d), , 34. (a), , 35. (c), , 36. (c), , 37. (b), , 38. (c), , 39. (a), , 40. (c), , 41. (c), , 42. (b), , 43. (c), , 44. (a), , 45. (d), , HINTS&SOLUTIONS, 1., 2., , h, 2, (n = 2 for first excited state), (b) Two Na+ ions are replaced by one Sr 2+ion to maintain, electrical neutrality., Hence, number of vacancies, = Number of Sr 2+ ions doped, , (a) Angular momentum, mvr, , 3, , 3., , = 1.0 10, 6.02 10, (b) Here, R = 31.6 ohm, , 23, , n, , 6.02 10, , 9., , (b) Sn 4, Br2, , 2e, , = E, , G, , 5., , (a) Assume that each has lost a proton . So we get : HO–, , HS–, HSe–, HTe–, It can be easily seen that the volume available for the, negative charge is increasing from HO– to HTe– ,, therefore, (i) volume available for the negative charge is, increasing from left to right, (ii) charge density is decreasing from left to right, (iii) basicity is decreasing from left to right, (iv) acidity of conjugate acids is increasing from left, to right, H2O < H2S < H2Se < H2Te, (c) In preparation of an alkyl chloride by the action of dry, HCl, the catalyst generally used is anhydrous ZnCl2., (b) Only structure (b) has a conjugated system, which is, necessary for resonance., (d), , 6., 7., 8., , (NH 4 ) 2 Cr2 O7, , G, , Also,, , log K eq, , E, , Sn 4 / Sn 2, , 2 96500 0.95 kJ/mol, , 2.303RT log K eq, G, 2.303 R, , T, , ( 2 96500 0.95), = 32.6820, 2.303 8.314 293, Keq = antilog 32.682, = 4.78 × 1032 1032, , 10., , (c), , [H ] in HCl solution (pH = 2) = 10, , 2, , M ; [OH ] in, , KOH solution (pOH = 14 – 12 = 2) = 10, , Cr2O3 4H 2O N 2, , (b), , Br2 / Br, , G nFE cell, n = 2, F = 96500., , Now, , 1, 1, ohm 1 = 0.0316 ohm–1, R 31.6, Specific conductance = conductance × cell, , 4., , 1.08 V, , = 1.08 – 0.13 = 0.95 V, , C=, , heat, , 2Br E, , 0.13 V, , E° values shows Br 2 has higher reduction potential., Hence Ecell = ER – EL, , 20, , constant, = 0.0316 ohm–1 × 0.367 cm–1, = 0.0116 ohm–1 cm–1, Now, molar concentration = 0.5 M, (given), = 0.5 × 10–3 mole cm–3, K, 0.0116, Molar conductance =, C 0.5 10 3, = 23.2 S cm2 mol–1, , Sn 2 E, , 2e, , 2, , M, , Excess m Mol of OH in 5 ml mixture, = 3 10, , 2, , 2 10, , 2, , [OH ] in mixture =, , 1.0 10, , 1.0 10, 5, , = 2 10, pOH = log 2 10, 11., , 3, , 2, , ;, , 2, , 3, , M;, , = 3 – log 2;, , pH = 14 – (3 – log 2) = 11.30, (a) A reducing agent is a substance which can loose, electron and hence a reducing agent should have low, ionisation energy. Now since ionisation energy, decreases from Li to Cs, the reducing property should, increase in the opposite manner. The only exception, to this is lithium. This is because the net process of, converting an atom to an ion takes place in 3 steps.
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MOCK TEST 2, , 12., 13., , MT-11, , (i) M(s) M(g) H = Sublimation energy, (ii) M(g) M+(g) + e– H = Ionisation energy, (iii) M+(g)+H2O M+ (aq) H = Hydration energy, The large amount of energy liberated in hydration of, Li (because of its small size) makes the overall H, negative. This accounts for the higher oxidation, potential of lithium i.e., its high reducing power., (d) Number of unpaired electrons in Cr, Mn 2+ and Fe3+, are 6, 5 and 5 respectively., (a), , urotropine, hexamethylene, tetramine, , 6H 2 O, , O, , 6, , 5, , 4, , 3, , (d), , 15., 16., , (hex 3-ene-5-one), (b), (c) Total V.P.,, PAº X A, , = (PAº, , PBº X B, , or, , PAº, , k, , 18., 19., , 23., 24., 25., , C2H 6, , N, 2.303, log 0, t, N, , 2.303, 0.1, log, 40 min, 0.005, 2.303, log 20, 40 min, , PBº ( 1 X A ), , PBº, , Pd, , which serves as catalyst. The accumulation of, hydrogen gas occurrs on the surface of active, palladium so that fast reduction may be achieved., (c) Greater the difference of electronegativities, more polar, is the bond., (c) O.N. of Cl in ClO– changes from +1 to +5 and –1., (d) For a first order reaction, we have, k, , 2.303, 1.3010, 40, , Now rate = k × [reactant], When [x] = 0.01 M, , 52, , 166 torr, , 2.303, × 1.3010 × 0.01 M min–1, 40, = 7.5 × 10–4 M min–1., (c) Correct Nernst equation is, , rate, , 1, 1, 114, 2, 2, , 140 torr, , 26., , 2.303, a, log, t, a x, , 2.303, 1, log, 1, 24, 8, , C2 H 5Cl H 2, , Acyloins, , Reduction of alkyl halide by H 2 in the presence of Pd, , PBº, , 114 torr ; PAº, , Hence P 166, , (c), , k, , PAº X A, , PBº )X A, , Thus, PBº, , (d), , 22., , ||, , Ar C H C Ar, , Aromatic aldehyde, , C H C C H3, , C H3 C H2 C H, , |, , CN, , ArCHO, , 2 || 1, , 14., , P, , (b) In general,, OH O, , 6HCHO 4NH 3, , (CH 2 )6 N 4, , 17., , 21., , 27., 28., , 2.303, log 8, 24, , 2.303 RT, log a M n ., nF, , Eo, , E, , (b), (a) This is isocyanide test, NH2, , (b), (b) Structural formula of glucose is, , + CHCl3 3KOH, Chloroform, , CHO, , Aniline, , |, , (CHOH)4, |, , N C, , CH 2OH, , In addition to – CHO group it contains one primary, and four secondary – OH groups., 20., , (a), , CH3 CH 2 C, , CH, , + 3KCl + 3H2O, , Hg 2, , Phenyl isocyanide, (Offensive smell), , H3O, , OH, |, CH3 CH 2 C CH 2, , tautomerization, , CH3CH 2COCH3, , 29., , (a), , 30., , (b) SO 2Cl 2(g ), , %, , SO 2(g ) Cl 2(g), , D d, 100, d ( y 1), , (y = 2)
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EBD_7207, CHEMISTRY, , MT-12, , molar mass of SO2 Cl 2, 2, d = 50.0 (given), , 67.5;, , 2H+ + 2e– Oxidation, , (b) LHS : H2 (P1), 2H+ + 2e–, , RHS :, Net reaction :, H2 (P1), , H2 (P2) Reduction, , 37., 38., 39., , H2 (P2), , RT, P2, E = E° – nF ln . P, 1, , 0, , RT P1, ln, 2F P2, , O, , 32., , (d), , H, , P, |, , OH , so dibasic acid, , OH, , 33., , 1, 6 3, 2, (An atom at the face centre is sh ared by, 2 unit cells), Hence, formula is : AB3, 90, (b) For orthorhombic, (c), (a) [Co(NH3)5CO3]ClO4. Six monodentate ligands are, allached to Co hence C. N. of Co = 6;, O. N. = x + 5 × (0) + 1× (–2) + 1× (–1) = 0, x = + 3 ; electronic configuration of Co3+[Ar] 3d64s0, hence number of d electrons is 6 : All d electrons are, paired due to strong ligand hence unpaired electrons, zero., , Number of B atoms per unit cell =, , 67.5 50.0, 100 = 35%, 50.0(2 1), , %, , 31., , (An atom at corner is shared by 8 unit cells), , 135, 2, , D, , 40., , (c) Nylon has, , 41., , (c), , O, ||, C NH – group linkage, , 3d, , (d) [Co F6 ]3 :, , .. .. 4p.. .. .. .., , 4s, , HO, , 4d, , SO3H, , Br2, H2 O, , Br, , 3 2, , sp d (Paramagnetic), [Ni Cl 4 ]2 :, , HO, , .. .. .. .., , Br, , sp 3(paramagnetic), , [CuCl 4 ]2 :, ., ., , dsp2 (paramagnetic), [Ni(CN)4]2– :, ., ., 34., 35., , 42., , .. .. .., , 43., , % by mass of methanol, , .. .. .., , 36., , 16 100, 16 9, , (c) Number of A atoms per unit cell =, , (c) Number of hybrid orbitals of P in PO 34 = ½ [5 + 0 +, , 1, [6 + 0 + 2 – 0] = 4 (sp3), 2, (a) Al2(SO4)3 is a salt of weak base and a strong acid, hence on hydrolysis it will produce Al(OH)3 and, H2SO4. Since H2SO4 is a strong acid and Al(OH)3 is, is a weak base hence the solution will be acidic due to, SO4– –., (d) It is the correct answer., , =, , 44., , 64, , 1, 8 1, 8, , – OH group is highly activating. This is a type of, electrophillic substitution reaction at ortho and para, position., (b) SO3 has trigonal planar geometry (sp2 hybridisation, of S) which is symmetrical., 3] = 4 (sp3), No. of hybrid orbitals of N in NO–3, = ½ [5 + 0 + 1] = 3 (sp2), No. of hybrid orbitals of S in SO4– –, , (diamagnetic), dsp2 :, (a) Aldehydes, other than formaldehyde, when treated, with RMgX give 2º alcohols., (c) Mass of methanol in 1 mol solution = 0.5 × 32 = 16 g, Mass of water in solution = 0.5 × 18 = 9 g, , SO3H, , 45.
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Mock Test-3, Time : 1 hr, 1., , 2., , Max. Marks -180, , Fluorine is more electronegative than either boron or, phosphorus. What conclusion can be drawn from the fact, that BF3 has no dipole moment but PF3 does ?, (a) BF3 is not spherically symmetrical but PF3 is spherically, symmetrical., (b) BF3 molecule must be linear, (c) The atomic radius of P is larger than that of B, (d) The BF3 molecule must be planar triangular, The volume-temperature graphs of a given mass of an ideal, gas at constant pressure are shown below., , p2 p, 3, p1, , V, , O, , 3., , 4., , 5., , 273, , CH 3, , CH 2, , 3., , CH 3, , CH 2 C, , CH 2, , CH 3, , CH, , 4. CH 3 CH CH 2., (a) Bromine in carbon tetrachloride, (b) Bromine in acetic acid, (c) Alk KMnO4, (d) Ammonical silver nitrate., In the extraction of Cu, the metal is formed in the bessemer, converter due to the reaction :, (a) Cu2S + 2Cu2O, 6Cu + SO2, (b) Cu2S, 2Cu + S, (c) Fe + Cu2O, 2Cu + FeO, (d) 2Cu2O, 4Cu + O2, An example of electrophilic substitution reaction is, (a) Chlorination of methane, (b) Conversion of methyl chloride to methyl alcohol, (c) Nitration of benzene, (d) Formation of ethylene from ethyl alcohol., , Which of the following molecules is most suitable to, disperse benzene in water ?, O, –, +, (a), O Na, , O, , + –, , (b) Na O, , –, , (c), , 7., , 8., , 9., , +, , O Na, , O, , T(K), , What is the correct order of pressures ?, (a) p1 > p3 > p2, (b) p1 > p2 > p3, (c) p2 > p3 > p1, (d) p2 > p1 > p3, Which is the most suitable reagent among the following to, distinguish compound (3) from rest of the compounds ?, 1. CH 3 C C CH 3, 2., , 6., , CH3, , (d) Cl, Ozone hole refers to, (a) Increase in concentration of ozone, (b) Hole in ozone layer, (c) Reduction in thickness of ozone layer in troposphere, (d) Reduction in thickness of ozone layer in stratsophere, Which of the following does not represent the correct order, of the properties indicated, (a) Ni2+ > Cr2+ > Fe2+ > Mn2+ (size), (b) Sc > Ti > Cr > Mn (size), (c) Mn2+ > Ni2+ < Co2+ < Fe2+, (unpaired electron), (d) Fe2+ > Co2+ > Ni2+ > Cu2+, (unpaired electron), The major product formed in the following reaction is :, CH 3, |, , CH 3O –, , |, , CH 3OH, , CH 3— C — CH 2 Br, H, , CH3, |, , (a), , CH 3—C — CH 2 OCH3, , (b), , H, CH 3— C H — CH 2 CH 3, , |, , |, , OCH3, CH 3, |, , (c), , CH 3 — C, CH 2, CH3, , (d), , CH3 — C — CH3, , |, , |, , OCH3
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EBD_7207, CHEMISTRY, , MT-14, , 10., , 12., , 13., , (d) 3Cl2 6OH, 14., , 5Cl, , ClO 3, , 3H 2 O, , 18., , O, ||, C, , 19., , CH 3O C( CH 3 ) 2, , 20., , 21., , For the reaction H 2(g), , 16., , is rate k[H 2 ][Br2 ]1/ 2 . Which of the following statement, is true about this reaction, (a) The reaction is of second order., (b) Molecularity of the reaction is 3/2, (c) The unit of k is s–1, (d) Molecularity of the reaction is 2, The correct order of solubility in water for He, Ne, Ar, Kr,, Xe is, (a) He > Ne > Ar > Kr > Xe, (b) Xe > Kr > Ar > Ne > He, (c) Ne > Ar > Kr > He > Xe, (d) Ar > Ne > He > Kr > Xe, If uncertainty in position and velocity are equal then, uncertainty in momentum will be, , 17., , 2HBr(g) , the rate law, , Br2(g), , (a), , 1 mh, 2, , (b), , 1, 2, , (c), , h, 4 m, , (d), , mh, 4, , h, m, , CH3, , NH 4 and NH 2, , (c), , SO 24 , PO 34 and [ BF4 ], , (d), , NH 4 and NH 3, , When dihydroxyacetone reacts with HIO4, the product is/, are :, (a) HCHO, (b) HCOOH, (c) HCHO and HCOOH (d) HCHO and CO2, Hard water when passed through ion exchange resin, containing R’COOH groups, becomes free from :, (a), , Cl, , (b), , SO 24, , (d) Ca 2, H3O, A certain compound (X) when treated with copper sulphate, solution yields a brown precipitate. On adding hypo, solution, the precipitate turns white. The compound is, (a) K2CO3, (b) KI, (c) KBr, (d) K3PO4, Consider the following reactions:, (i) H +(aq) + OH–(aq) = H2O(l),, H = – X1 kJ mol–1, (c), , I, (c) CH3I + (CH3)2CHOH, (d) CH3OH + (CH3)2CHI, , 15., , | CH3, , (b) CH 3 and CH 3, , 22., , |, , N, , (a) N, N-dimethylcyclopropanecarboxamide, (b) N-methylcyclopropanamide, (c) cyclopropionamide, (d) none of the above, Which of the following shows iso-structural species?, (a), , The major organic product in the reaction,, CH3 — O — CH(CH3)2 + HI Product is, (a) ICH2OCH(CH3)2, (b), , IUPAC name of the following compound :, , |, , 11., , (I) n = 3, l = 2, m1 = –2, (II) n = 3, l = 1, m1 = 0, (III) n = 3, l = 0, m1 = – 1, (IV) n = 3, l = 2, m1 = 0, (V) n = 3, l = 3, m1 = –2, Of these question state designation which does not, describe an allowed state for an electron in an atom ?, (a) I and IV, (b) III and V, (c) II and V, (d) IV and V, In an adiabatic process which of the following is true?, (a) q = + w, (b) q = 0, (c), E=q, (d) P V = 0, In which of the following cases, the stability of two, oxidation states is correctly represented, (a) Ti3+ > Ti4+, (b) Mn2+ > Mn3+, 2+, 3+, (c) Fe > Fe, (d) Cu+ > Cu2+, Which is not the disproportionation reaction ?, (a) 3H 3PO 2, 2H 3 PO 2 PH 3, (b) HCHO OH, HCOO, CH 3OH, (c) NH 4 NO 3, N 2 O 2H 2 O, , 23., , 1, O, = H2O(l),, 2 2(g), H = – X2 kJ mol–1, (iii) CO2(g) + H2(g) = CO(g) + H2O,, H = – X3 kJ mol–1, , (ii) H2(g) +, , 5, O 2(g) = 2CO2(g) + H2O(l)’, 2, H = + 4X4 kJ mol–1, Enthalpy of formation of H2O (l) is, (a) + X3 kJ mol– 1, (b) – X4 kJ mol– 1, (c) + X1 kJ mol– 1, (d) – X2 kJ mol– 1, The restricted rotation about carbon carbon double bond, in 2-butene is due to, (a) Overlap of one s- and sp2-hybridized orbitals, (b) Overlap of two sp2- hybridized orbitals, (c) Overlap of one p- and one sp2-hybridized orbitals, (d) Sideways overlap of two p- orbitals., , (iv) C 2 H 2(g), , 24.
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MOCK TEST 3, 25., , 26., , 27., , 28., , 29., , 30., , 31., , 32., , MT-15, , A balloon has maximum capacity of 20 L. At one atmospheric, pressure 10 L of air is filled in the balloon. It will burst when, pressure is (assuming isothermal condition), (a) > 0.5 atm, (b) < 0.5 atm, (c) = 0.5 atm, (d), 0.5 atm, Which one of the following complexes will have four, different isomers ?, (a), , [ Co (en ) 2 Cl 2 ]Cl, , (b) [ Co ( en )( NH 3 )Cl 2 ]Cl, , (c), , [Co(PPh 3 ) 2 Cl2 ]Cl, , (d) [Co (en )3 ]Cl 3, , Solubility product of a salt AB is 1 × 10–8 in a solution in, which the concentration of A+ ions is 10–3 M. The salt will, precipitate when the concentration of B– ions is kept, (a) between 10–8 M to 10–7 M, (b) between 10–7 M to 10–8 M, (c) > 10–5 M, (d) < 10–8 M, Sucrose in water is dextro-rotatory, [ ]D= + 66.4º. When, boiled with dilute HCl, the solution becomes leavo-rotatory,, [ ]D= –20º. In this process the sucrose molecule breaks, into, (a) L-glucose + D-fructose, (b) L-glucose + L-fructose, (c) D-glucose + D-fructose, (d) D-glucose + L-fructose, Select correct statement(s)., (a) Cyanamide ion (CN22–) is isoelectronic with CO2 and, has the same linear structure, (b) Mg2C3 reacts with water to form propyne, (c) CaC2 has NaCl type lattice, (d) All of the above, In DNA the complementary bases are, (a) adenine and thymine; guanine and cytosine, (b) uracil and adenine; cytosine and guanine, (c) adenine and guanine; thymine and cytosine, (d) adenine and thymine; guanine and uracil, Vapour pressure of benzene at 30°C is 121.8 mm. When 15, g of a non volatile solute is dissolved in 250 g of benzene, its vapour pressure decreased to 120.2 mm. The molecular, weight of the solute (Mo. wt. of solvent = 78), (a) 356.2, (b) 456.8, (c) 530.1, (d) 656.7, In nitrogen family, the H-M-H bond angle in the hydrides, gradually becomes closer to 90º on going from N to Sb., This shows that gradually, (a) The basic strength of the hydrides increases, (b) Almost pure p-orbitals are used for M-H bonding, (c) The bond energies of M-H bonds increase, (d) The bond pairs of electrons become nearer to the, central atom, , 33., , 34., , 35., , For reaction aA, xP , when [A] = 2.2 mM, the rate was, found to be 2.4 mMs–. On reducing concentration of A to, half, the rate changes to 0.6 mMs–1. The order of reaction, with respect to A is :, (a) 1.5, (b) 2.0, (c) 2.5, (d) 3.0, Which of the following is used for making optical, instruments?, (a) SiO2, (b) Si, (c) SiH4, (d) SiC, If ‘a’ stands for the edge length of the cubic systems :, simple cubic, body centred cubic and face centred cubic,, then the ratio of radii of the spheres in these systems will, be respectively,, (a), , 1, 3, 1, a:, a:, a, 2, 4, 2 2, , (b), , 1, 1, a : 3a :, a, 2, 2, , 1, 3, 3, a:, a:, a, (d) 1a : 3a : 2a, 2, 2, 2, Which of the following is correct order of acidity?, (a) HCOOH > CH3COOH > ClCH2COOH, > C2H5 COOH, (b) ClCH2COOH > HCOOH > CH3COOH, > C2H5 COOH, (c) CH3COOH > HCOOH > ClCH2COOH, > C2H5COOH, (d) C2H5COOH > CH3COOH > HCOOH, > ClCH2COOH, Penicillin is :, (a) analgesic, (b) antipyretic, (c) antimalarial, (d) antibiotic, The rate constant k, for the reaction, 1, N 2 O 5 (g ), 2 NO 2 (g ), O 2 (g), 2, , (c), , 36., , 37., , 38., , is 1.3 × 10–2s–1. Which equation given below describes the, change of [N2 O5 ] with time ? [N2 O5 ]0 and [N2O 5] t, corrospond to concentration of N2O5 initially and at time t., (a) [N2O5]t = [N2O5]0 + kt, (b) [N2O5]0 = [N2O5]t ekt, (c) log [N2O5]t = log [N2O5]0 + kt, [N 2 P5 ]0, (d) In [N P ], 2 5 t, , 39., , kt, , Which of the following reactions can produce aniline as, main product?, (a) C6H5NO2 + Zn/KOH, (b) C6H5NO2 + Zn/NH4Cl, (c) C6H5NO2 + LiAlH4, (d) C6H5NO2 + Zn/HCl
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MOCK TEST 3, , MT-17, , ANSWER KEY, 1. (d), , 2. (a), , 3. (d), , 4. (d), , 5. (c), , 6. (c), , 7. (d), , 8. (a), , 9. (d), , 10. (b), , 11. (b), , 12. (b), , 13. (c), , 14. (c), , 15. (d), , 16. (b), , 17. (a), , 18. (a), , 19. (c), , 20. (d), , 21. (d), , 22. (d), , 23. (d), , 24. (d), , 25. (b), , 26. (b), , 27. (c), , 28. (c), , 29. (d), , 30. (a), , 31. (a), , 32. (b), , 33. (b), , 34. (a), , 35. (a), , 36. (b), , 37. (d), , 38. (d), , 39. (d), , 40. (c), , 41. (c), , 42. (b), , 43. (d), , 44. (c), , 45. (b), , HINTS&SOLUTIONS, 1., 2., 3., , 4., 5., , 6., , 7., 8., , (d) BF3 is planar triangular while PF3 is pyramidal., (a) From the graph we can see the correct order of, pressures p1 > p3 > p2, (d) Br2 in CCl4 (a), Br2 in CH3 COOH (b) and alk. KMnO4, (c) will react with all unsaturated compounds, i.e., 1, 3, and 4 while ammonical AgNO3 (d) reacts only with, terminal alkynes, i.e., 3 and hence compund 3 can be, distinguished from 1, 2 and 4 by. ammonical AgNO3, (d)., (d) Decomposition of carbonates and hydrated oxides., (c) Chlorination of methane proceeds via free radical, mechanism. Conversion of methyl chloride to methyl, alcohol proceeds via nucleophilic substitution., Formation of ethylene from ethyl alcohol proceeds, via dehydration reaction. Nitration of benzene is, electrophilic substitution reaction., (c) Benzene is non-polar and hence dissolves non-polar, compounds (like dissolves like). Among the given, compounds, only (c) is non-polar hence it dissolves, in benzene., (d) Ozone hole is reduction in ozone layer in stratosphere., (a) In a period on moving from left to right ionic radii, decreases., (a), So order of cationic radii is, Cr2+ > Mn2+ > Fe2+ > Ni2+ and, (b), Sc > Ti > Cr > Mn, (correct order of atomic radii), (c), For unpaired electrons, , Mn 2 (Five), (d), , |, , –, , CH 3 — CH — CH 2 Br, , –Br, , –, , CH3, OCH3, , |, , CH3 — C — CH3, |, , OCH3, , 10., , 11., 12., 13., , 14., , (b) A set of question number is valid when, l < n and m1 lies between – l to + l. Thus sets I, II, IV, are valid, III, V invalid., (b) For adiabatic process, q = 0, (b) Mn2+ (3d5) is more stable than Mn3+ (3d4)., (c) In NH4NO3, there are two different N-atoms (NH4+,, NO3– ) with different oxidation numbers, thus reaction, is not disproportionation., (c) In case of unsymmetrical ethers, the site of cleavage, depends on the nature of alkyl group e.g.,, CH 3O CH(CH3 ) 2, , 15., , CH3 — C H — C H 2, 1 carbocation, , (CH3 )2 CHOH, Isopropyl, alcohol, , (d), , H 2(g ), , Br2(g ), , 2HBr(g ), , Rate law, R = k[H2] [Br2]½, Order of reaction = 1 + ½ = 3/2, Molecularity of reaction = 2, The unit of k, , R, [H 2 ] [Br2 ]½, 1, , [mole. lit 1 ] [mole. lit 1 ]½, , Cu 2 (One), , |, , 373K, , The alkyl halide is always formed from the smaller, alkyl group., , mole. lit 1s, , CH3, , HI, , CH3I, Methyl, iodide, , Fe2 (Four), , (d) The reaction is an example of SN1 reaction, CH3, , |, , CH3 — C — CH3, 3 carbocation, , Co 2 (Three), , Ni 2 (Two), , 9., , hydrideshift, , Ni2 (Two), , Co 2 (Three), For unpaired electrons, Fe2 (Four), , CH3, , 16., , = mole–½ . lit½ . s–1, (b) As the molecular weight of noble gas atoms increases, down the group its polarity increases due to which, van-der-waal’s force between them increases. Due to, increased polarity of heavier inert gas, its solubility in
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EBD_7207, CHEMISTRY, , MT-18, , water also increases. So, most soluble gas will be Xe, and least soluble will be He., So correct order is Xe > Kr > Ar > Ne > He, 17., , (a), , x, , v, , x. v, v, , 2, , h, 4 m, h, 4 m, , p, , 24., , h, 4 m, , v, , 18., 19., , 23., , m. v, , m, , h, 4 m, , 25., , 1 hm, 2, (a) It is N, N-dimethylcyclopropane –carboxamide., (c) Hybridisation can be calculated by calculating the no, of valence electron and dividing it by 8., , 26., , In SO 4 2 = Total no. of e–, = 6 + (6 × 4) + 2 = 32, So, no. of hybrid orbitals =, sp3 hybridization., , 32, 8, , 4, , en, , 20., , 21., , 22., , (d), , Co, , Cl, trans-form, , NH3, , +, , +, NH3, en, , NH3, , NH3, en, , Co, , 27., , (c), , 2CuI, I2, Cuprous iodide (Brown colour, (White ppt.), in solution), , Na 2S4 O6, , Sod. tetrathionate, (coloourless), , 2NaI, , Cl, , Co, , Cl, Cl, cis-form, , (d) An ion exchange resin containing R–COOH group, exchange cations like Ca2+, Mg2+, Na+, Fe2+ with H+, when hard water is passed through it. This resin is, called cation exchange resin., (b) KI reacts with CuSO4 solution to produce cuprous, iodide (white precipitate) and I2 (which gives brown, colour) Iodine reacts with hypo (Na 2S2O35H2 O), solution. Decolourisaiton of solution shows the, appearance of white precipitate., , 2Na 2S2 O3 I2, , NH3, , (ii) Optical isomers, , 2CH 2 O CO 2, , 2K 2SO4, , en, , Co, , NH3, cis-form, , Cl, trans-form, , NH3, , A B, , AB, , [A ][B ], [AB], , Ksp, , 2CuSO4 4KI, , Cl, , Cl, , Cl, , 5 24 3 32, 4, =, 8, 8, Hybridisation = sp3, Similarly, for BF4 , it is sp3., , HIO4, , +, , +, NH3, , Similarly, for PO 43 ; no. of hybrid orbitals, , CH 2 OH, |, CO, |, CH 2 OH, , (d) This reaction shows the formation of H2O, and the, X2 represents the enthalpy of formation of H2 O, because as the definition suggests that the enthalpy, of formation is the heat evolved or absorbed when, one mole of substance is formed from its constituent, atoms., (d) Rotation around bond is not possible. If any attempt, is made to rotate one of the carbon atoms, the lobes, of -orbital will no longer remain coplanar i.e no, parallel overlap will be possible and thus -bond will, break . This is known as concept of restricted rotation., In other words the presence of -bonds makes the, position of two carbon atom., (b) The balloon would burst when V > 20 L, P1V1 = P2V2, 1 × 10 = P2 × 20, P2 = 0.5 atm, (no bursting), Thus, a pressure below 0.5 atm, it would burst., (b) Complex Co(en)(NH3)2Cl2]Cl will have four different, isomers., (i) Geometrical isomers, , Salt will precipitate if ionic conc. > Ksp, [A+][B–] > 1 × 10–8, (1× 10–3)[B–] > 1 ×10–8, , 28., , 1 10, , 8, , or 1 × 10–5, 1 10 3, (c) The hydrolysis of sucrose by boiling with mineral, acid or by enzyme invertase or sucrase produces a, mixture of equal molecules of D(+) glucose and, D(–) Fructose., [B ], , C12 H 22O11, sucrose, , [ D], , 66.5º, , H 2O, , HCl, , C 6 H12O 6, D glu cos e, , C 6 H12O 6, , D Fructose, , [ D ] 52.5º [ D ] 92 º, Invert sugar ,[ D ] 20º
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MOCK TEST 3, 29., , MT-19, , (d) In CO2 we have 22 (6 + 8 + 8 = 22) electrons. In (CN22–, ), we have 22 (6 + 7 + 7 + 2 =22) electrons. Both CO2, and (CN22–) have linear structures. Thus, statement, (a) is correct., Mg 2C3, , 4H 2O, , 2Mg(OH) 2, , CH 3C, , Thus the ratio of radii of spheres for these will be, simple : bcc : fcc, =, , CH, , Propyne, , i.e., statement (b) is also correct ., , i.e. option (a) is correct answer., 36., , The structure of CaC2 is of NaCl type, i.e., statement (c) is also correct., 30., , (a) In DNA, adenine faces thymine and guanine faces, cytosine., , 31., , (a) Given vapour pressure of pure solute, , 1, a 3, :, a:, a, 2 2, 2 4, , (b) Recall that presence of electron-withdrawing group, increases, while presence of electron-releasing group, decreases the acidity of carboxylic acids., , ClCH 2COOH, (electron-withdrawing gp.), , O, ||, H C OH, , CH3, (Electron-releasing character, increasing from left to right), , (P 0 ) 121.8 mm; Weight of solute (w) = 15 g, , Weight of solvent (W) = 250 g; Vapour pressure of, pure solvent (P) = 120.2 mm and Molecular weight of, solvent (M) = 78, From Raoult’s law, Po, P, , P, , w M, m W, , o, , 15 78 121.8, 250, 1.6, , or m, , 121.8 120.2, 121.8, , 15 78, m 250, , 37., , (d) Penicillin is an antibiotic which was first obtained from, a fungus, penicillium notatum by the scientist,, Flemming., , 38., , (d) As the unit of rate constant is sec–1, so the reaction is, first order reaction. Hence, k, , (b) With the decrease in the electronegativity of central, atom the bond angle decreases, , 33., , (b) When the concentration of reactant is reduced to half, its initial value, the rate is reduced by 2.4, 0.6, It means, rate, , O, ||, C 2H 5 C OH, , 356.2, , 32., , 39., , NH4Cl), , (a) SiO2 is used for this purpose., , 35., , (a) Following generalization can be easily derived for, various types of lattice arrangements in cubic cells, between the edge length (a) of the cell and r the, radius of the sphere., , r or r, , a, , 2 2r or r, , 1, 2 2, , a, , ( m et a l /, , a, 2, , NHOH, , NH NH, Hydrazobenzene, , Thus, Aniline will be main product in case of (d)., 40., , (c) Reaction used in fuel cell is, 2H 2 (g) O 2 (g), , 3, a, 4, , For face centred cubic :, , Aniline (C6H5NH2), , In alkaline medium, , For body centred cubic :, , 3, , Main product, , In neutral medium Phenyl hydroxylamine,( Z n /, , 34., , 4, , (d) Various products are formed when nitroarenes are, reduced. These are given below for C6H5NO2., In acidic medium, HCl), , So, order of reaction = 2, , a, , 1, [N O ], a, = log, or kt = log 2 5 0, t, [N 2 O5 ]t, ( a x), , Medium, , 4 times, , [ reactant]2, , For simple cubic : a = 2r or r, , O, ||, C OH, , 41., , 2H 2 O( ), , At anode : [H 2, , 2H, , At cathode : O 2, , 2H 2 O 4e, , 2e] 2, , 4OH, , (c) Carbon cannot expand its coordination number beyond, four due to the absence of d-orbitals, hence it cannot, form [CCl6 ]2 ion
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Mock Test-4, Time : 1 hr, , Max. Marks -120, CH3, , 1., , IUPAC name of, , is, CH2CH3, , 2., , 3., , 4., , 5., , 6., , 8., , (a) 1-methyl-3 ethyl cyclohexane, (b) 1-ethyl-3 methyl benzene, (c) 1-ethyl-3 methyl cyclo hexane, (d) Cyclo hexane-1-ethyl-3-methyl, Which of the following structures does not contain any, chiral C atom but represent the chirality in the structure., (a) 2 – Ethyl – 3 – hexene (b) 2, 3-Pentadiene, (c) 1,3 – Butadiene, (d) Pent – 3 – en – 1 – yne, N2 and O2 are converted to mono cations N2+ and O2+, respectively, which of the following is wrong?, (a) In N2+, the N – N bond weakens, (b) In O2+, the O – O bond order increases, (c) In O2+, paramagnetism decreases, (d) N2+ becomes diamagnetic, In a compound AOH, electronegativity of ‘A’ is 2.1, the, compound would be, (a) Acidic, (b) Neutral towards acid & base, (c) Basic, (d) Amphoteric, Which of the following orders is wrong?, (a) Electron affinity– N < O < F < Cl, (b) Ist ionisation potential – Be < B < N < O, (c) Basic property– MgO < CaO < FeO < Fe2O3, (d) Reactivity–Be < Li < K < Cs, Cl, The dipole moment of chlorobenzene, , 9., , (a), (c), 10., , 7., , Cl, , 11., , is, , (a) 2.86 D, (b) 2.25 D, (c) 1.5 D, (d) 0 D, Following substances are in solid state :, (A) Methane, (B) Cesium chloride, (C) Ice, (D) Lithium, , Kb, , Ka, , (b), , 2–, , 1=, , 3, , (d), , 1+, , 2=, , 3, , H 3O, , F— ;, , HF OH — ., , (a) Kb = Kw, , (b), , Kb, , 1, Kw, , (c) Ka × Kb = Kw, , (d), , Ka, Kb, , Kw, , In an amino acid, the carboxyl group ionises at pK a1 =, 2.34 and ammonium ion at pK a 2 = 9.60. The isoelectric, , 12., , Cl, , H 2O, , 3), , Which relation is correct, , Cl, , dipole moment of, , 1, ( –, 2 1, 1– 2= 3, 3=, , Given, HF H 2 O, F—, , is 1.5 D. The, , Cl, , Which non-conductive solid when melts converts into, conductive liquid?, (a) C, D, (b) Only C, (c) Only B, (d) A, B and C, On applying pressure to the equilibrium, ice, water, which phenomenon will happen, (a) More ice will be formed, (b) More water will be formed, (c) Equilibrium will not be disturbed, (d) Water will evaporate, Let 1 be the frequency of the series limit of the Lyman series,, 2 be the frequency of the first line of the Lyman series, and 3, be the frequency of the series limit of the Balmer series, then –, , 13., , point of the amino acid is at pH, (a) 5.97, (b) 2.34, (c) 9.60, (d) 6.97, AB, A2 and B2 are diatomic molecules. If the bond enthalpies, of A2, AB and B2 are in the ratio 1:1 :0.5 and enthalpy of, formation of AB from A2 and B2 is –100 kJ mol–1 . What is, the bond energy of A2 :, (a) 200 kJ mol–1, (b) 100 kJ mol–1, –1, (c) 300 kJ mol, (d) 400 kJ mol–1, Equal volume of 0.1 M urea and 0.1 M glucose are mixed., The mixture will have, (a) Lower osmotic pressure, (b) Same osmotic pressure, (c) Higher osmotic pressure, (d) None of these
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EBD_7207, CHEMISTRY, , MT-22, , 14., , Fe, , 3, , –, , e, , Fe, , 2, , 2e, , 3, , 15., , 19., , If the following half cells have the E° values as, 2, , Fe, , ;, , E°, , =, , +, , 0.77V, , and, , Fe ; E° = – 0.44V. The E° of the half cell, –, , Fe, 3e, Fe will be, (a) 0.33 V, (b) 1.21 V, (c) 0.04 V, (d) 0.605 V, The oxidation states of sulphur in the anions, , 20., , SO 32 , S2 O 24 and S2 O 62 follow the order, , (a), , SO 32, , S 2 O 24, , (b) S 2O 24, , S2 O62, , SO 32, , S 2 O 62, , S2 O 24, , SO 32, , SO 32, , S2 O 62, , (c), , (d) S 2O 24, 16., , S2 O 62, , 21., , Which of the following statements is not correct?, (a) C– Cl bond in vinyl chloride is less polar than in CH3Cl, (b) C – Cl bond in vinyl chloride is stronger than in CH3Cl, (c) C – Cl bond in vinyl chloride is shorter than in CH3Cl, (d) Vinyle chloride undergo nucleophilic substitution, more readily than CH3Cl., , 22., , NO2, , 23., , 17., , Br2, Fe,, , X, , CH3ONa, CH3OH,, , Y, , Cl, , Product (Y) of this reaction is –, NO2, , NO2, , (a), , (b), , 24., Br, , OCH3, Cl, , OCH3, , NO2, , NO2, , (c), , 25., , (a) Ca, , (b) Ba, , (c) Al, , (d) Zn, , Choose the correct order of T (True) and F (False) –, (1) When the pH of rain water is below 3.6, it is called, acid rain., (2) Ozone hole occurs over Antarctica mainly during, September–October and it gets replenished in, November–December., (3) Methylcyclohexane is an ozone-depleting molecule., (4) COD is always larger than BOD., (a) TTTF, (b) FTFT, (c) FFFF, (d) FTTT, Which of the following compound can not used in, preparation of iodoform?, (a) CH3CHO, , (b) CH3COCH3, , (c) HCHO, , (d) 2-propanol, , The correct priorities for the substituents shown below,, according to the E-Z sequence rule is, I., , (d), OCH3, , OCH3, , 18., , A metal which is not affected by conc. H2SO4, HNO3 or, alkalis forms a compound X. This compound X can be, used to give a complex which finds its application for toning, in photography? The metal is, (a) Au, (b) Ag, (c) Hg, (d) Cu, If Cl2 gas is passed into aqueous solution of KI containing, some CCl4 and the mixture is shaken, then”., (a) Upper layer becomes violet, (b) Lower layer becomes violet, (c) Homogenous violet layer is formed, (d) None, In Lassaigne’s test, the organic compound is fused with a, piece of sodium metal in order to, (a) increase the ionisation of the compound, (b) decrease the melting point of the compound, (c) increase the reactivity of the compound, (d) convert the covalent compound into a mixture of ionic, compounds, An aqueous solution of colourless metal sulphate M gives, a white precipitate with NH4OH. This was soluble in excess, of NH4OH. On passing H2S through this solution a white, ppt. is formed. The metal M in the salt is, , Although Al has a high oxidation potential it resists, corrosion because of the formation of a tough, protective, coat of, (a) Al(NO3)2, (b) AlN, (c) Al2O3, (d) Al2(CO3)2, , – CN, , II., , III. – COOH, , V., , O, ||, , – C, , IV., , – CBr (CH 3 ) 2, , – CH 2, , O, ||, , C, , H, , (a) II, III, V, I, IV, , (b) V, II, I, IV, III, , (c) III, IV, I, II, V, , (d) II, V, I, IV, III, , OCH 3
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MOCK TEST 4, 26., , MT-23, , Identify X in the sequence given :, (b), , NH2, CHCl3, KOH, , (Y), , HCl, (300 K), , X + methanoic acid, , (c), , Cl, , (a), , NH2, , Cl, , (b) C N, , (d), , Cl, , 29., (c), 27., , N, , Cl, , C, , Cl, , N 2 (g) 3H 2 (g ), , Select the rate law that corresponds to the data shown for, the following reaction A B, C, Expt. No., , (1), , (2), , Initial Rate, , 1, , 0.012, , 0.035, , 0.10, , 2, , 0.024, , 0.070, , 0.80, , 3, , 0.024, , 0.035, , 0.10, , 4, , 0.012, , 0.070, , 0.80, , Rate = K[B]3, , 28., , (d) CH3–NH, , Rate = K [B]4, , (a), (b), 3, (c) Rate = K [A] [B], (d) Rate = K [A]2 [B]2, An alkene upon ozonolysis yield, CHO – CH2– CH2– CH2 – CHO only. The alkene is, (a) CH2= CH – CH2 –– CH2 –– CH2 –– CH2 –– CH3, , 1 mol of N 2 and 3 mol of H 2 are placed in a closed, container at a pressure of 4 atm. The pressure falls to 3 atm, at the same temperature when the following equilibrium is, attained, 2 NH 3 (g ) ., , The K p for the dissociation of NH 3 is, (a), , (c), 30., , 3 3, 0.5 (1.5)3, , atm, , 0.5 (1.5)3, atm 2, 3 3, , 2, , (b), , 0.5 (1.5) 3 atm 2, , (d), , (1.5) 3, atm, 0.5, , 2, , 0.5 g mixture of K2Cr2O7 and KMnO4 was treated with, excess of KI in acidic medium. I2 liberated required 100 cm3, of 0.15N. Na2S2O3 solution for titration. The percentage, amount of K2Cr2O7 in the mixture is, (a) 85.36 %, (b) 14.64 %, (c) 58.63 %, (d) 26.14 %
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EBD_7207, CHEMISTRY, , MT-24, , ANSWER KEY, 1. (d), , 2. (b), , 3. (d), , 4. (b), , 5. (b), , 6. (c), , 7. (c), , 8. (b), , 9. (c), , 10. (c), , 11. (a), , 12. (d), , 13. (b), , 14. (c), , 15. (d), , 16. (d), , 17. (b), , 18. (c), , 19. (a), , 20. (b), , 21. (d), , 22. (d), , 23. (b), , 24. (c), , 25. (a), , 26. (a), , 27. (a), , 28. (b), , 29. (b), , 30. (b), , HINTS&SOLUTIONS, , 1., , CH3, 3, 2, 1 CH CH Ethyl group comes first in alphab2, 3, , (d), , 2, , RcZ2, , 3, , RcZ2, , C2H5, 3, , atical order but IUPAC name of, , 2 Br , 2-bromo1, CH3, , 4-ethyl-1-methyl cyclohexane. It follows lowest sum, rule, (i.e., lowest set of locant is preferred.), , 10., , (c), , Lowest sum of locant is = 1 + 2+ 3 = 6, 2., , (b) The molecule 2,3 - pentadiene does not have any, chiral C but at the same time it does not have any, mirror plane which makes the molecule chiral., , there is one unpaired electron hence it is, (d) In, paramagnetic., , 4., , (b) In A — O — H, if EN of ‘A’ is 2.1 then it will be neutral,, as XA – X0 = X0 – XH. (where X is EN), , 5., , (b) Correct order is B < Be < O < N., , 6., , 6, , (c), , Cl, 1, , 11., , 2, 3, , Cl 5 4, , 12., , pKa1, , 2, 1, , 2, 2, , A2, , B2, , or 100, , 7., 8., , 9., , 13., , = 1.5 D, , (c) CsCl is ionic solid., (b) Volume of ice > volume of water & thus increase in, pressure favours forward reaction showing decrease, in volume., RcZ2, , (c), , 1, , RcZ2, , 1, , 1, , n12, , n 22, , 1, 2, , 1, , 1, 2, , pK a 2, , (b), , M total, , 1, A2, 2, , 2 AB;, , 5.97, , x, 2, , x, 4, , x, , 1, B2 AB;, 100KJ, 2, 2x x 4x, x 400 KJ, 4, , 100, , (n 1 n 2 ), st ,, v1 v 2, , 14., , (c), , E0, , 0.77 2( 0.44), 3, , M total, , 0.1 M, , 0.77 0.88, 3, , 0.11, 3, , 0.04, 15. (d) The chemical bond method gives the O.N., –, , O, , O O, –, , O O, –, , S = O: O – S – S – O :, –, O +4, +3, , 16., , RcZ2, , ., , [F – ][H 2 O], , 2.34 9.60, 2, , n1 = n2 = 0.1, V1 = V2 = 1litre, , 2 1.5 1.5 cos 120, , [HF][OH – ], , Enthalpy of formation of AB is – 100 KJ/mole:, , 2 1 2 cos, , = (1.5) 2 (1.5) 2, , RcZ2, 4, , 2, , 3, , 2, , 5Cl are vectorically cancelled., , =, , 1, , 2, , (d) Let bond energy of A2 be x then bond energy of AB, is also x and bond energy of B2 is x/2., , Cl, , It is due 1 Cl and 3 Cl, , 1, , [H 3 O ][F – ], and K b, [HF][H 2 O], , Ka, , =, , Dipole moments of 2Cl and, , 2, , 22, , (a) Isoelectric point (pH), , 120, Cl, , 12, , 3RcZ 2, 4, , Therefore, Ka × Kb = [H3O+] [OH–] = Kw., , N2+, , 3., , 1, , 2, , 2, , 1, , 1, , (d), , +, –, CH2 = CH– Cl, , Vinyl chloride, , –, , –, , O–S– S–O, , +5 O O, , +, –, CH2 – CH = Cl
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MOCK TEST 4, 17., , MT-25, , (b), , |, , OCH3, , AuCl3 HCl, , 25., , N, |, C, |, N, , (a) (i), , 2KCl I 2, , I2, , Violet Colour, , (iii), , OH, |, C O,, |, |, O C, , (v), , O C, |, |, C O, |, H, , I 2 to HIO 3, I 2 5Cl 2 6H 2 O, In case of Br 2 :, , 2HIO 3 10HCl, , Br2, , 22., , 2H 2 O Cl 2, 2HBrO 2HCl, (d) To convert covalent compounds into ionic, compounds such as NaCN, Na2S, NaX, etc., (d), , Zn, , 2, , C, |, N,, |, C, , CH 3, |, C CH 3 ,, |, Br, , (ii), , Complex, , Note: The excess of Cl2 should be avoided. The layer, may become color less due to conversion of, , 21., , ||, , potassium iodide and sod. hypochlorite yield, iodoform., , H[Au(Cl)4 ], , 2KI Cl 2, , CCl 4, , group on reaction with, , CH, O, , (c) Because the layer of Al2O3 (oxide) is inert, insoluble, and impervious., (a) Au, the gold is not attacked by acids and alkalis. It, forms AuCl3.AuCl3 further reacts with HCl to form, H[Au(Cl)4] which is used in photography., , (b), , group or CH3, , Br, , Cl, , 20., , those compound which contains either CH3 CH, , OH, , SNAr, , Br, , 19., , (c) Formaldehyde can not produce iodoform, as only, , CH3ONa, , X=, , 18., , 24., , NO2, , NO2, , 2NH 4 OH, , Zn(OH) 2, , O C, |, C OCH 3 ,, |, O, , Arrange (NNN), (BrCC), (OOO), (CHH), (OOH) in, increasing atomic number. The order is ii, iii, v, i, iv., 26., , 2NH 4, , (iv), , H, |, C, |, H, , (a), , CHCl3, , Cl, , KOH, , White ppt., , Zn(OH)2, , 2NH 4 OH, , (NH 4 ) 2 ZnO 2, , 2H 2 O, , Soluble, , (NH 4 )2 ZnO2, , H 2S, , HCl, 300 K, , ZnS 2NH 4 OH, White ppt., , 23., , NC [Y], , Cl, , (b). (1) When the pH of rain water is below 5.6, it is called, acid rain., (2) Ozone hole occurs over Antarctica mainly during, September–October and it gets replenished in, November–December., (3) Methylcyclohexane is not an ozone-depleting, molecule., (4) BOD (Biological oxygen demand) is a measure, of organic pollutant present in the sample of, water. Higher is the value of BOD, higher is the, level of organic pollution in water. The amount, of oxygen (in mg/L) consumed for oxidising all, organic and oxidisable inorganic material in a, sample of water is called chemical oxygen, demand (COD)., COD is always larger than BOD., , + HCOOH, , Cl, [X], 27., , (a) Let the rate law be r = [A]x[B]y, Divide (3) by (1), , 0.10, 0.10, , [0.024]x [0.035] y, [0.012]x [0.035] y, , 1 = [2]x, x = 0, Divide (2) by (3), , 0.80, 0.10, , [0.024]x [0.070] y, [0.024]x [0.035]y, , 8 = (2)y , y = 3, Hence rate equation, R = K[A]0[B]3 =K[B]3
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EBD_7207, CHEMISTRY, , MT-26, , CH2– C – H, , O, , 28., , (b) CH2, , Ozonolysis, , O, , CH2– CH, , =, , CH2, , 1 0 .5, 3, 3, , 2 0 .5, 3, , CH2– CH, , CH2– C – H, , 29., , (b), , p N2, , 3, , 3, , 3, , 2, , = 0.5 × (1.5) 3 atm 2, , N 2 (g) 3H 2 (g), 2 NH 3(g), 3 – 3x 2x at equilibrium, 1–x, Total moles,, 1 – x + 3 – 3x + 2x = 4 – 2x = 3 (given), (Since, 4 moles = 4 atm given), x = 0.5, , K p for dissociation of NH 3, , 3 3 0 .5, 3, , 30., , (b) Let the amount of the K 2Cr2 O 7 in the mixture be x g,, then amount of KMnO4 will be (0.5 – x) g, x, 49, , p3H2, , p 2 NH 3, , 0.5 x, 31.6, , =, , 100 0.15, 1000, , where 49 is Eq. wt. of K 2Cr2 O 7 and 31.6 is Eq. wt. of, KMnO4 ., On solving, we get x = 0.073 g, % age of K 2Cr2 O 7 =, , 0.0732 100, 0.5, , 14.64%
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Mock Test-5, Time : 1 hr, 1., , 2., , 3., , 4., , 5., , Max. Marks -120, , An atom X belongs to 4th period of the periodic table and, has highest number of unpaired electrons in comparison, to the other elements of the period. The atomic number of, X is, (a) 23, (b) 25, (c) 24, (d) 33, In O2 , O 02 and O 22 molecular species, the total number, of antibonding electrons respectively are :, (a) 7, 6, 8, (b) 1, 0, 2, (c) 6, 6, 6, (d) 8, 6, 8, Aluminothermy used for on the spot welding of large iron, structure is based on the fact that(a) As compared to iron, aluminium has greater affinity, for oxygen., (b) As compared to aluminium, iron has greater affinity, for oxygen., (c) Reaction between aluminium and oxygen is endothermic., (d) Reaction between iron and oxygen is endothermic., Which of the following shows the tendency to form, peroxide?, (a) Lithium, (b) Magnesium, (c) Beryllium, (d) Radium, A 1.0 M solution with respect to each of the metal halides, , 8., , (b) 2B2H6 + 6NH3, (c) B2H6 + 2N(CH3)3, , E°, , 6., , 7., , C3+ /C, , = – 0. 74 V, E°, , A3+ / A, , D2+ /D, , = 1.50 V, E°, , B2+ /B, , = 0.3 V,, , 9., , 10., , 11., , 12., , = – 2.37 V., , The correct sequence in which the various metals are, deposited at the cathode is, (a) A, B, C, D, (b) A, B, C, (c) D, C, B, A, (d) C, B, A, Arrange hypophosphorous acid (H3PO2), phosphorous, acid (H3 PO 3 ) and Phosphoric acid (H3 PO 4 ) in the, decreasing order of acidic strength, (a) H3PO3 > H3PO4 > H3PO2, (b) H3PO4 > H3PO3 > H3PO2, (c) H3PO4 > H3PO2 > H3PO3, (d) H3PO4, H3PO3 H3PO2, Which of the following reactions corresponds to the, definition of enthalpy of formation ?, (a) C(diamond) + O2(g) CO2(g), (b) C(graphite) + O2(l) CO2(g), , B3N3H6 (borazine), 2(CH3)3 NBH3, H O, , AX 3 , BX 2 , CX 3 and DX 2 is electrolysed using, platinum electrodes. If E°, , (c) C(graphite) + O2(g) CO2(g), (d) C(g) + O2(g) CO2(g), Among the reactions given below for B2H6, the one which, does not take place is, (a) B2H6 + HCl, B2H5Cl + H2, , 13., , 3, (d) B2H6 + 6C2H4, 3C2H5OH + 2B(OH)3, The pure crystalline substance on being heated gradually, first forms a turbid liquid at constant temperature and still at, higher temperature turbidity completely disappears. The, behaviour is a characteristic of substance forming., (a) Allotropic crystals, (b) Liquid crystals, (c) Isomeric crystals, (d) Isomorphous crystals., Silver bromide when dissolve in hypo solution gives, complex ..... in which oxidation state of silver is ...., (a) Na3[Ag(S2O3)2], (I), (b) Na3[Ag(S2O3)3], (III), (c) Na3[Ag(S2O3)2], (II) (d) Na3[Ag(S2O3)4], (I), Which of the following can be termed as a mixed complex?, (a) K4 [Fe(CN)6], (b) [Cu(NH3)4] SO4, (c) [Co(NH3)4NO2Cl] Cl, (d) K2FeO4, The relationship between the values of osmotic pressures, of solutions obtained by dissolving 6.00 g L–1 of CH3COOH, ( 1) and 7.45 g L–1 of KCl ( 2) is, , (a), , 1, , 2, , (b), , (c), , 1, , 2, , (d), , 1, , 2, , 1, 1, , 2, 2, , 1, , 2, , In the reaction of KMnO4 with an oxalate in acidic medium,, MnO4 is reduced to Mn 2, , and C 2 O 24 is oxidised to, , CO2. Hence, 50 ml of 0.02 M KMnO4 is equivalent to, (a) 100 ml of 0.05 M H 2C 2O 4, (b) 50 ml of 0.05 M H 2C 2O 4, (c) 25 ml of 0.2 M H 2C 2O 4, (d) 50 ml of 0.10 M H 2C 2O 4
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EBD_7207, CHEMISTRY, , MT-28, , 14., , In which reaction, there is change in oxidation number of N, , (b) purple colour of CCl4, , (a) 2NO2, , (c) brown colour in the organic layer of CCl4, , N2O4, , 15., , 16., , 2HNO3, , (d) 2NO2 + H2O, , HNO3 + HNO2, , Potassium permanganate acts as an oxidant in neutral,, alkaline as well as acidic media. The final products obtained, from it in the three conditions are, respectively, (a) MnO42–, Mn3+ and Mn2+, (b) MnO2, MnO2 and Mn2+, (c) MnO2, MnO2+ and Mn3+, (d) MnO, MnO2 and Mn2+, , (a) 10.376, , (b) 5.19, , (c) 7.289, , (d) 2.144, , An organic compound A (C4H10O) has two enantiomeric, forms and on dehydration it gives B(major product) and C, (minor product). B and C are treated with HBr/ Peroxide, and the compounds so produced were subjected to alkaline, hydrolysis then(a) B will give an isomer of A, , (a) 4 NA, , (b) 0.4 NA, , (c) Neither of them will give isomer of A, , (c) 0.1 NA, , (d) 0.2 NA, , (d) Both B and C will give isomer of A, , C–C H, , (b), , (c), , (d), , 20., , 22., , An element (atomic mass =100g/mol) having bcc structure, has unit cell edge 400pm. The density (in g/cm3) of the, element is, , The number of electrons present in 6.4 g of calcium, carbide is – (NA = Avagadro’s number), , (a), , 19., , 21., , (c) N2O5 + H2O, , 17., , 18., , (d) deep blue colour in CCl4, , NH4+ + OH–, , (b) NH4OH, , C, , C, , CH 3 CH 2 MgBr gives, CMgBr, , CH3, , (b) C will give an isomer of A, , 23., , CH2, , A reaction is found to be second order w.r.t. one of the, reactants & has rate constant of 0.5 mol–1 dm3 min–1. If, initial concentration is 0.2 mol dm–3 then t1/2 of reaction is, (a) 5 min, , (b) 10 min, , (c) 15 min, , (d) 20 min, , CH 2, , CMgBr C2H6, , 24., MgBr CH3, , C2H5, , HC, , CH2, , C, , CH 2, , (b) CH3Br, , (c) (CH3)2CHBr, , (d) (CH3)3CBr, , Two elements A & B form compounds having molecular, formulae AB2 and AB4. When dissolved in 20.0 g of benzene, 1.00g of AB2 lowers f.p. by 2.3°C whereas 1.00g of AB4, lowers f.p. by 1.3°C. The molal depression constant for, benzene in 1000g is 5.1. The atomic masses of A and B are, (a) 52, 48, , (b) 42, 25, , (c) 25, 42, , (d) None, , O, , The above shown polymer is obtained when a carbonyl, compound is allowed to stand. It is a white solid. The, polymer is, , CMgBr, , (a) C6H5CH2Br, , CH 2, , O, , CH, , The rate of SN1 reaction is fastest in the hydrolysis of, which of the following halides ?, , O, , 25., , 26., , (a) trioxane, , (b) formose, , (c) paraformaldehyde, , (d) metaldehyde., , Concentration of NH4Cl and NH4OH in a buffer solution is, in the ratio of 1 : 1, Kb for NH4OH is 10–10. The pH of the, buffer is, (a) 4, , (b) 5, , (c) 9, , (d) 11, , Aniline when diazotized in cold and when treated with, dimethyl aniline gives a coloured product. Its structure, would be, (a) CH3NH, , N=N, , To detect iodine in presence of bromine, the sodium extract, , (b) CH3, , is treated with NaNO 2 + glacial acetic acid + CCl4 . Iodine, is detected by the appearance of, , (c) (CH3)2N, , N=N, , (a) yellow colour of CCl4 layer, , (d) (CH3)2N, , NH, , N=N, , NHCH3, NH2
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MOCK TEST 5, 27., , (c) three phosphate groups, , (b) CH3CH2NH2, (d) (CH3 CH2)3N, , (d) three carboxylic acid residues, , C N, + C6H5MgBr, , The standard reduction potential of, Li+/Li, Ba2+/Ba, Na+/Na and Mg2+/Mg are –3.05, –2.73, –, 2.71 and –2.37 volts respectively. Which one of the following, is strongest oxidising agent?, , 29., , (b) one carboxylic acid residue and two phosphate, groups, , An organic amino compound reacts with aqueous nitrous, acid at low temperature to produce an oily nitrosoamine., The compound is:, (a) CH3 NH2, (c) CH3CH2NH.CH2CH3, , 28., , MT-29, , (a) Na+, , (b) Li+, , (c) Ba2+, , (d) Mg2+, , 30., , Ether, , O, , H3O, , C, , (b), , Phospholipids are esters of glycerol with, (a) two carboxylic acid residues and one phosphate, group, , N–OH, , N–H, , (c), , C, , Ba, , N–MgBr, , C, , (a), , A, , C, , (d)
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EBD_7207, CHEMISTRY, , MT-30, , ANSWER KEY, 1. (c), , 2. (a), , 3. (a), , 4. (d), , 5. (b), , 6. (d), , 7. (c), , 8. (d), , 9. (b), , 10. (a), , 11. (c), , 12. (a), , 13. (b), , 14. (d), , 15. (b), , 16. (b), , 17. (b), , 18. (a), , 19. (c), , 20. (b), , 21. (b), , 22. (b), , 23. (b), , 24. (a), , 25. (a), , 26. (c), , 27. (c), , 28. (d), , 29. (a), , 30. (a), , HINTS&SOLUTIONS, 1., , 2., , 3., 4., 5., , 6., , 7., 8., , (c) The configuration of atom of 4th period with maximum, unpaired electrons is 1s2 2s2 2p6 3s2 3p6 3d5 4s1 ., Hence its atomic number is 24., (a) The total numberof electrons in the molecular species, given, respectively are 17, 16 and 18. Write down the, electronic configuration of the molecular species and, observe the number of electrons in antibonding, orbitals which are respectively are 7, 6 and 8., (a) Aluminium has greater affinity for oxygen and the, reaction is highly exothermic., (d) Among alkaline earth metals, barium and radium have, the tendency to form peroxides., (b) The more the reduction potential, the more is the, deposition of metals at cathode. Cation having E° value, less than – 0.83 V (reduction potential of H2O) will not, deposit from aqueous solution., (d) There is very little difference in acid strength in the, series H3 PO 4 , H3 PO 3 and H3 PO 2 because the, hydrogen in these acids are not all bonded to, phosphorus., In the above three acids although the number of –OH, groups (ionisable hydrogen increases, yet the acidity, does not increase very much. This is due to the fact, that the number of unprotonated oxygen, responsible, for the enhancement of acidity due to inductive effect,, remains the same with the result dissociation constant, also remains nearly same., (c) C(graphite) is the stable state of aggregation of, carbon., (d) Reaction between diborane and alkene are carried out, in dry ether under an atmosphere of N2 because B2H6, and the products are very reactive. The products, further treated with alkaline H2O2 to convert into, alcohols., B2 H 6, , 9., 10., , 6C2 H 4, , B(C2 H4 )3, reactive, , alkaline, H 2O 2, , 3CH3CH 2OH H3BO3, (b) Liquid crystals on heating first become turbid and, then clear., (a) AgX + 2 Na2 S2O3, Na3 [Ag(S2O3)2]+ NaX, Sodium argento thiosuphate, (soluble complex), , 11., 12., , 13., 14., 15., , (c) By definition, a mixed complex contains more than, one type of ligands., (a) Osmotic pressure, = CRT ;, 6, 7.45, n CH 3COOH, , n KCl, 60, 74.5, Since KCl ionises. Therefore its effective conc., in, solution increases., (b), (d) In reaction (d) oxidation number changes from + 4 in, NO2 to + 3 in HNO2 and + sin HNO3, (b) In neutral and alkaline medium, MnO 4, , 2H 2 O 3e, , MnO 2, , 4OH, , In acidic medium:, MnO 4, , 8H, , 5e, , Mn 2, , 4H 2 O, , C, , 16., , (b) Ca, , , 6.4g of CaC2 contain -electron = 4NA, C, , C CH, , 17., , (b), , + CH 3– CH2MgBr, C CMg Br, , C2 H6 +, , 18., , (a) Because of the formation of the most stable, carbonium ion, C6 H5 C H 2, , 19., , (c) Let atomic masses of A and B be a and b amu, respectively, Molar mass of AB2 = (a + 2b) g mol–1, and Molar mass of AB4 = (a + 4b) g mol–1, For compound AB2, Tb = Kb WB 1000/ WA MB, 2.3 = 5.1 1 1000/ 20.0 (a + 2b)....I, For compound AB4, 1.3 = 5.1 1 1000/ 20.0 (a + 4b)....II, Solving (I) and (II),, a = 25.49 b = 42.64
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MOCK TEST 5, 20., , MT-31, , (b) 2NaI + 2 NaNO 2 + 4 CH 3COOH, , I 2 + 2NO +, , 4 CH 3COONa + 2 H 2 O, , 21., , n M, a3, , N 2 Cl, , |, , |, , C, , 27., CH 2 CH 3, , |, , H, , 28., , H, , OH, CH 3 CH 2, , |, , 29., , H 2SO 4, , C H CH 3, , A, , CH 3 CH, , CH CH 3 CH 2, , Major ( B ), , (b) In general t1/ 2 of reaction, , NMe 2, , N=N, , HO, CH3 H 3C, , |, , +, , H, , (4 10 8 cm)3 6.02 1023, , No, , C, , 0 5 C, , NMe 2, , = 200/38.528 = 5.19 g/cc, (b) Entiomers of C4H10O are, H3CH 2C, , NaNO 2 ,HCl, , NH 2, , 2 100, , OH, , 23., , (c), , The colour of CCl4 layer turns purple due to liberated, I2., (b) For bcc lattice, number of atoms per unit cell = 2, Now d, , 22., , 26., , (c) The secondary amines react with HNO2 to give the, oily nitroso derivative. Amongst the options, (c) is, the secondary amine., (d) The strongest oxidising agent is one which has, maximum tendency to gain electrons, i.e. whose E°Red, is maximum, (a) Phospholipids - Phosphate + glycerol + fatty acids +, a nitrogen containing base., General formula : CH 2 O . COR ', , CHCH 2 CH 3, , |, CHO COR ' ', |, OH, , Minor ( C ), , 1, (a 0 ), , CH 2 O, , n 1, , 1, K(a 0 )n, , 1, K(a 0 ), , 1, , On keeping, , 24., , (a), , 3 HCHO, aq. solution, , 25., , (a), , pOH, , 0.5 0.2, CH2, , X, , Ethanolamine, , 30., , (a) Alkyl or Aryl cyanide react with grignard reagent to, form ketones, , O, , C N, , CH2, O, Trioxane, (meta formaldehyde), , +C6H5MgBr, , log 1 10; pH 14 10, , Ether, , O, , salt, base, , 10, , O, , X = OHCH 2CH 2 NH 2 ,, , 10 min, , CH2, , pK b log, log 10, , 1, , O, , ||, , O, , For a second order reaction,, , t1/ 2, , |, , P, , 4, , C6 H5– C=NMgBr, C6H5, or, , C6H5, , C, |, , H 3O+, , O MgBrNH 2, , C6 H 5, , C
