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CBSE, New Pattern, Chemistry, Class 11, , (Term I)

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ARIHANT PRAKASHAN (School Division Series), , © Publisher, No part of this publication may be re-produced, stored in a retrieval system or by any, means, electronic, mechanical, photocopying, recording, scanning, web or otherwise, without the written permission of the publisher. Arihant has obtained all the information, in this book from the sources believed to be reliable and true. However, Arihant or its, editors or authors or illustrators don’t take any responsibility for the absolute accuracy of, any information published and the damage or loss suffered thereupon., , All disputes subject to Meerut (UP) jurisdiction only., Administrative & Production Offices, Regd. Office, ‘Ramchhaya’ 4577/15, Agarwal Road, Darya Ganj, New Delhi -110002, Tele: 011- 47630600, 43518550, , Head Office, Kalindi, TP Nagar, Meerut (UP) - 250002, Tel: 0121-7156203, 7156204, , Sales & Support Offices, Agra, Ahmedabad, Bengaluru, Bareilly, Chennai, Delhi, Guwahati,, Hyderabad, Jaipur, Jhansi, Kolkata, Lucknow, Nagpur & Pune., , ISBN : 978-93-25793-66-8, PO No : TXT-XX-XXXXXXX-X-XX, Published by Arihant Publications (India) Ltd., For further information about the books published by Arihant, log on to, www.arihantbooks.com or e-mail at [email protected], Follow us on, , CBSE, New Pattern

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Contents, Chapter, Some Basic Concepts of Chemistry, , -, , Chapter, Structure of Atom, , -, , Chapter, Classification of Elements and Periodicity, in Properties, , -, , Chapter, Chemical Bonding and Molecular Structure, , -, , Chapter, Redox Reactions, , -, , Chapter, Hydrogen, , -, , Chapter, Organic Chemistry: Some Basic Principles, and Techniques, , Practice Papers, , CBSE, New Pattern, , -, , -, , -

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Syllabus, (Term I), S.No., , Units, , ., , Some Basic Concepts of Chemistry, , ., , Structure of Atom, , ., , Classification of Elements and Periodicity in Properties, , ., , Chemical Bonding and Molecular Structure, , ., , Redox Reactions, , ., , Hydrogen, , ., , Organic Chemistry: Some Basic Principles and Techniques, , Periods, , Marks, , Total, , Chapter-, , Some Basic Concepts of Chemistry, General Introduction: Importance and scope of Chemistry. Atomic and, molecular masses, mole concept and molar mass, percentage composition,, empirical and molecular formula, chemical reactions, stoichiometry and, calculations based on stoichiometry., , Chapter-, , Structure of Atom, Bohr s model and its limitations, concept of shells and subshells, dual nature, of matter and light, de Broglie s relationship, Heisenberg uncertainty, principle, concept of orbitals, quantum numbers, shapes of s, p and d, orbitals, rules for filling electrons in orbitals - Aufbau principle, Pauli s, exclusion principle and Hund s rule, electronic configuration of atoms,, stability of half-filled and completely filled orbitals, , CBSE, New Pattern

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Chapter-, , Classification of Elements and Periodicity in Properties, Modern periodic law and the present form of periodic table, periodic trends, in properties of elements -atomic radii, ionic radii, inert gas radii, Ionization, enthalpy, electron gain enthalpy, electronegativity, valency. Nomenclature of, elements with atomic number greater than, ., , Chapter-, , Chemical Bonding and Molecular Structure, Valence electrons, ionic bond, covalent bond, bond parameters, Lewis, structure, polar character of covalent bond, covalent character of ionic bond,, valence bond theory, resonance, geometry of covalent molecules, VSEPR, theory, concept of hybridization, involving s, p and d orbitals and shapes of, some simple molecules, molecular orbital theory of homonuclear diatomic, molecules qualitative idea only , Hydrogen bond., , Chapter-, , Redox Reactions, Concept of oxidation and reduction, redox reactions, oxidation number,, balancing redox reactions, in terms of loss and gain of electrons and change, in oxidation number., , Chapter-, , Hydrogen, Position of hydrogen in periodic table, occurrence, isotopes, hydrides-ionic, covalent and interstitial; physical and chemical properties of water, heavy, water, hydrogen as a fuel., , Chapter-, , Organic Chemistry: Some basic Principles and Techniques, General introduction, classification and IUPAC nomenclature of organic, compounds. Electronic displacements in a covalent bond: inductive effect,, electromeric effect, resonance and hyper conjugation. Homolytic and, heterolytic fission of a covalent bond: free radicals, carbocations, carbanions,, electrophiles and nucleophiles, types of organic reactions., , CBSE, New Pattern

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1, , Some Basic Concepts of Chemistry, , 01, Some Basic Concepts, of Chemistry, Quick Revision, 1. All those things that have mass and a physical, appearance can be considered as matter., Matter, Physical classification, , Solid, , Liquid, , Gas, , Chemical classification, , Pure substances, , Mixtures, Heterogeneous, Homogeneous, , Elements, , Compounds, , 2. Some Physical Quantities, (i) Mass and weight Mass (m) is the, amount of matter present in a substance., Weight (w) is the force exerted by gravity, on an object., w =m × g, where, m = mass, g = gravity., (ii) Volume The space occupied by matter, (usually by liquid or a gas) is called its, volume., Volume = (length)3 = m 3, 1m 3 = (100 cm)3 = 10 6 cm 3, , Metals, , Non-metals, , Inorganic compounds, , Metalloids, , Organic compounds, , (i) Mixture is formed by mixing two or more, substances together in any proportion. It, can be homogeneous (uniform composition, throughout) or heterogeneous (different, composition throughout)., (ii) Pure substances have fixed composition and, non-variable properties. An element is a, substance that contains only one type of atoms,, whereas a compound is formed when atoms of, different elements combine in a fixed ratio., , = 10 3 dm 3 = 10 3 L, (iii) Density It is defined as the amount or, mass per unit volume and has units, kg m −3 or g cm −3 ., Density = Mass/Volume, (iv) Temperature It is defined as the degree, of hotness or coldness. Its unit are °C, °F, and K but its SI unit is K., 5, ° C = ( ° F) − 32, 9, 9, or, ° F = ( ° C) + 32, 5, ° C = K − 273.15

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CBSE New Pattern ~ Chemistry XI (Term-I), , 2, 3. Significant Figures, These are the total number of digits in a number, including the last digit whose value is uncertain., Rules for counting the number of significant, figures are as follows :, (i ) All digits are significant except zero at the, beginning of the number., (ii ) The zeroes to the left of the first non-zero, digit in a number are not significant., (iii ) The zeroes to the right of the decimal point, are significant., (iv ) The above rules propose that, the numbers, are expressed in scientific notation. In this, term, every number is written as N × 10n ., where,, N = a number with a single non-zero, digit to the left of the decimal point, n = an integer, Significant figures in operational calculation, are as follows :, ● The result of an addition or subtraction, should be reported to the same number of, decimal places as that of the term with least, number of decimal places., ● The result of a multiplication or division, should be reported to the same number of, significant figures as is possessed by the least, precise term used in the calculation., ● If a calculation involves a number of steps, the, result should contain the same number of, significant figures as that of the least precise, number involved other than the exact number., , 4. Accuracy and Precision, (i) Accuracy is the agreement of a particular, value to the true value of the result., Accuracy = Mean value − True value, (ii) Precision refers to the closeness of various, measurements for the same quantity., Precision = Individual value, − Arithmetic mean value, , 6. Different Masses, Mole Concept, and Formula of Compounds, (i) Atomic mass, Mass of an atom, =, (1/12) × Mass of a carbon atom (12 C), (ii) Average atomic mass, Σ pi A i, =, 100, p1 A 1 + p 2 A 2 + …, =, 100, where, pi is the percentage abundance of, isotope having atomic mass A i ., (iii) Gram atomic mass = Atomic mass expressed, in gram = gram atom, (iv) Molecular mass is defined as the sum of, atomic masses of all the elements present in, a molecule., (v) Gram molecular mass or gram molecule, = Molecular mass expressed in gram, (vi) Formula mass for an ionic compound, = number of cations × its atomic mass, + number of anions × its atomic mass, (vii) Relation between molecular mass and, vapour density of a gas, Molecular mass = 2 × vapour density, (viii) Mole concept and molar masses, ● Number of moles of atoms, Given mass of element, =, Atomic mass of element, ● Number of moles of molecules, Given mass of molecule, =, Molecular mass, ● Number of moles of gases, Volume of the gas (STP), =, 22.4 L, ●, , ●, , 5. Dimensional Analysis, In calculations, sometimes it becomes, necessary to convert units from one system to, another. This is achieved by factor label, method or unit factor method (CF) or, dimensional analysis., Information sought = Information given × CF, , ●, , 1 mole = 6.022 × 10 23 particles, = Gram atomic/molecular mass, Number of molecules, = number of moles × N A ,, where, N A = Avogadro’s number, = 6.022 × 10 23, Number of atoms = number of molecules, × atomicity (or number of atoms in, molecule of an element of a compound)

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3, , Some Basic Concepts of Chemistry, , 7. Percentage Composition, Mass % of an element, Mass of element in 1 mole, of compound × 100, =, Mass of 1 mole of compound, n=, , Molecular mass, Empirical formula mass, , Molecular formula = n × empirical formula, , 8. Empirical and Molecular Formula, An empirical formula represents the simplest, whole number ratio of various atoms present in a, compound, whereas the molecular formula, shows the exact number of different types of, atoms present in the molecule of a compound., , Short Trick to Find Empirical and, Molecular Formula, Step 1 Divide percentage composition by atomic, mass to obtain atomic ratio., Step 2 Divide atomic ratio by minimum value of, atomic ratio to obtain simplest ratio., Step 3 Multiply simplest ratio by integer to, obtain simplest whole number ratio., Step 4 Write symbols of various elements, present with their respective whole, number ratio as a subscript to the lower, hand corner of symbol to obtain empirical, formula, Step 5 Multiply empirical formula by n to obtain, molecular formula, , , Molecular mass, , n =, Empirical formula mass , , , 9. Stoichiometry and Stoichiometric, Calculation, , ●, ●, , The calculations involving the amount of, reactants and products in a given chemical, reaction is called the stoichiometry., The number before the formula unit or, molecules used to balance the equation are, called stoichiometric coefficients., In the balanced reaction,, CH 4 ( g ) + 2O2 ( g ) → CO2 ( g ) + 2H 2O ( g ) ,, , ●, , the coefficients 2 for O 2 and H 2 O are called, stoichiometric coefficients., , Similarly, for CH 4 and CO 2 , stoichiometric, coefficients are 1 only., ● According to the above chemical reaction,, (i) One mole of CH 4 ( g ) reacts with two moles, of O2 ( g ) to give one mole of CO 2 ( g ) and, two moles of H 2O ( g )., (ii) One molecule of CH 4 ( g ) reacts with 2, molecules of O2 ( g ) to give one molecule of, CO2 ( g ) and 2 molecules of H 2O ( g )., (iii) 22.7 L of CH 4 ( g ) reacts with 45.4 L of O2 ( g ), to give 22.7 L of CO2 ( g ) and 45.4 L of, H 2O ( g )., (iv) 16 g of CH 4 ( g ) reacts with 2 × 32 g of O2 ( g ), to give 44g of CO2 and 2 × 18 g of H 2O ( g ) ., Limiting reagent The reactant which, consumed completely in a reaction is called, limiting reactant / reagent. It decides the, amount of other reactants reacted or the, amount of products formed., ●, , 10.Concentration Terms of Solutions, The concentration of the solution is usually, expressed in the following ways :, (i) Mass per cent, Mass of solute, × 100, Mass per cent =, Mass of solution, (ii) Volume percentage Volume per cent of A, Volume of A, =, × 100, Volume of A + Volume of B, (iii) Part per million, (ppm), Mass of compound A, ppm of A =, × 10 6, Total mass of solution, (iv) Mole fraction In case of a solution of two, components A and B, mole fraction of A ,, Number of moles of A, χA =, Total number of moles of solution, nA, =, nA + nB, (where, n A and n B are the moles of A and B, respectively)., Also, remember that χ A + χ B = 1

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CBSE New Pattern ~ Chemistry XI (Term-I), , 4, (v) Molarity ( M ) It is defined as the number of, moles of the solute present in 1L of the, solution. Thus, molarity, Number of moles of solute, (M ) =, Volume of solution (in L), or M =, , Mass of solute × 1000, Molar mass of solute ×V (in mL), , The units of molarity is mol L−1 or, mol dm −3 ., M =, , % by weight × specific gravity × 10, Molar mass, , In case of dilution, M 1V 1 = M 2V 2, , where, M 1 and V 1 are the molarity and, volume before dilution and M 2 and V 2 are, the molarity and volume after dilution, respectively., (vi) Molality (m) It is defined as the number of, moles of solute present in 1 kg of solvent. It, is denoted by m., Thus, molality, Number of moles of solute, (m ) =, Mass of solvent (in kg), Mass of solute × 1000, or, m=, Molar mass of solute × Mass of, solvent (in g), The SI unit for molarity is mol kg − 1 ., , Objective Questions, Multiple Choose Questions, 1. If measured temperature on Fahrenheit, scale is 200°F then the reading on, Celsius scale will be …… ., (a) 40°C, (c) 93.3°C, , (b) 94°C, (d) 30°C, , only four significant figures?, (b) 285 cm, , (c) 0.0025 L, , (d) 0.200 g, , 10.3406 g is ……, (b) 3, , (c) 1, , 6. The least count of an instrument is 0.01, cm. Taking all precautions, the most, possible error in the measurement can, be …… ., , 3. The number of significant figures in, (a) 2, , (a) A→ significant figures, B → accuracy, (b) A→ accuracy, B → precision, (c) A→ precision, B → accuracy, (d) A→ significant figures, B → precision, , 2. Which one of the following data has, (a) 6 .023 × 1023, , the true value of the result. A and B, respectively are, , (d) 6, , 4. The result of which of the following, has/have least significant figure(s)?, , 0.02856 × 298.15 × 0.112, 0.5785, (b) 5 × 5.364, (c) 0.0125 + 0.7864 + 0.0215, (d) All have same number of significant figures., (a), , 5. A refers to the closeness of various, measurements for the same quantity. B, is the agreement of a particular value to, , (a) 0.005 cm, (c) 0.0001 cm, , (b) 0.01 cm, (d) 0.1 cm, , 7. Given that, the true value for a result is, 2.00 g. Three students A, B and C take, two measurements and report the, result, data to illustrate precision and, accuracy as given below., Student, Student A, , Measurement (in g), 1, , 2, , Average (in g), , 1.95, , 1.93, , 1.940, , Student B, , 1.94, , 2.05, , 1.995, , Student C, , 2.01, , 1.99, , 2.000

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5, , Some Basic Concepts of Chemistry, , Which of the following students got the, values which are both precise and, accurate?, (a) Student A, (c) Student C, , (b) Student B, (d) None of these, , 8. Which of the following statements, about the molecular mass is correct?, (a) Molecular formula shows the exact number, of different types of atoms present in a, molecule., (b) Molecular formula can be obtained from, empirical formula if molar mass is known., (c) Percentage composition of a compound, can be calculated from its molecular, formula., (d) All the above statements are correct., , 9. The weight of one molecule of a, compound C 60 H122 is …… ., −20, , (a) 1.3 × 10 g, (c) 3.72 × 1012 g, , 13. If 1 mL of water contains 20 drops then, number of molecules in one drop of, water is …… molecules, (a) 6.023 × 1023, (c) 1.344 × 10, , (b) 1.376 × 1026, , 18, , (d) 4.346 × 1020, , 14. In which case is the number of, molecules of water maximum ?, (a) 0.00224 L of water vapours at 1 atm and, 273 K, (b) 0.18 g of water, (c) 18 mL of water, (d) 10−3 mole of water, , 15. Match the following Column I with, Column II and choose the correct, codes from the option given below., Column I, , −21, , (b) 5.01 × 10 g, (d) 1.4 × 10−21 g, , 10. In the standardisation of Na 2 S 2 O 3, using K 2 Cr2 O7 by iodometry, the, equivalent weight of K 2Cr 2O 7 is, , molecular weight, 2, molecular weight, (b), 6, molecular weight, (c), 3, (d) same as molecular weight, (a), , 11. One mole of any substance contains, , 6.022 × 10 23 atoms/molecules. What, will be number of molecules of H 2 SO4, present in 100 mL of 0.02 M H 2 SO4, solution?, (NCERT Exemplar), , Column II, , A. 46 g of Na, , 1. 0.01 mol, , B. 6.022 × 1023, molecules of H 2O, , 2. 2 mol, , C. 0.224 L of O2 at STP 3. 1 mol, D. 84 g of N 2, , 4. 6.022 × 1023, atoms/molecules, , E. 1 mole of any gas, , 5. 3 mol, , Codes, A B C D, (a) 2 3 1 5, (b) 1 2 3 4, (c) 4 2 1 3, (d) 5 4 3 1, , E, 4, 5, 4, 2, , 16. The mass per cent of different, elements present in sodium sulphate,, (i.e. sodium, sulphur and oxygen), respectively are, (NCERT Exemplar), , (a) 12.044 × 1020 molecules, (b) 6.022 × 1023 molecules, , (a) 32.37 ; 45.06 and 22.57, , (c) 1 × 1023 molecules, (d) 12.044 × 1023 molecules, , (c) 45.06 ; 32.37 and 40.06, , (b) 22.57 ; 32.37 and 45.06, (d) 32.37 ; 22.57 and 45.06, , 12. The number of molecules in 18 mg of, water in terms of Avogadro number,, N A is …… ., (a) 10−3NA, , (b) 10−2NA, , (c) 10−1NA, , (d) 10 NA, , 17. What is the mass per cent of carbon in, carbon dioxide?, (a) 0.034%, (c) 3.4%, , (b) 27.27%, (d) 28.7%

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CBSE New Pattern ~ Chemistry XI (Term-I), , 6, , 18. The empirical formula and molecular, , 25. 1.0 g of magnesium is burnt with 0.56 g, , mass of a compound are CH2 O and, 180 g respectively. What will the, molecular formula of the compound?, , O 2 in a closed vessel. Which reactant is, left in excess and how much? (Atomic, weight, Mg = 24, O = 16), , (a) C 9 H 18 O 9, , (b) CH2 O, , (c) C 6 H 12 O 6, , (d) C2 H 4 O2, , (a) Mg, 0.16 g, (c) Mg, 0.44 g, , 26. For a reaction,, , 19. Find empirical formula of the, , compound if M = 68% (atomic mass, = 34) and remaining 32 % oxygen., , (a) MO, (c) MO2, , (b) M2O, (d) M2O 3, , 20. An organic compound on analysis was, found to contain 10.06% carbon, 0.84%, hydrogen and 89.10% chlorine. What, will be the empirical formula of the, substance?, (a) CH2Cl2, (c) CCl4, , (b) CHCl3, (d) CH3Cl, , 21. Out of two oxides of iron, the first, contained 22% and the second, contained 30% of oxygen by weight., The ratio of weights of iron in the two, oxides that combine with the same, weight of oxygen is …… ., (a) 3 :2, (c) 1 :2, , (b) 2 :1, (d) 1 :1, , (b) 3 : 1.5, (d) 2 : 3, , Ca in the reaction?, Ca + Al 3+ → Ca 2+ + Al, (c) 3, , permanganate in acidic medium, the, number of electrons involved in, producing one molecule of CO 2 is …… ., (b) 5, , (c) 1, , 0.20 mole of Na 3 PO 4 , the maximum, number of Ba 3 (PO 4 ) 2 that can be, formed is …… ., (a) 0.70, , (b) 0.50, , (c) 0.20, , (d) 0.10, , 28. What volume of oxygen gas (O 2 ), measured at 0°C and 1 atm, is needed, to burn completely 1 L of propane gas, (C 3 H 8 ) measured under the same, conditions?, (b) 6 L, , (c) 5 L, , (d) 10 L, , will be formed from a reaction between, 6.5 g of PbO and 3.2 g of HCl?, (a) 0.044, (c) 0.011, , (b) 0.333, (d) 0.029, , treatment with excess HCl produces, 0.01186 mole of CO 2 . The molar mass, of M 2 CO 3 in g mol −1 is …… ., (a) 1186, , (d) 4, , 24. In the reaction of oxalate with, , (a) 2, , 27. If 0.5 mole of BaCl 2 is mixed with, , 30. 1 g of a carbonate (M 2 CO 3 ) on, , 23. What is the stoichiometric coefficient of, , (b) 1, , (a) 56 g of N2 + 10 g of H2, (b) 35 g of N2 + 8 g of H2, (c) 14 g of N2 + 4 g of H2, (d) 28 g of N2 + 6 g of H2, , 29. How many moles of lead (II) chloride, , dihydrogen orthophosphate and sodium, hydrogen orthophosphate required for, the synthesis of Na 5 P3 O10 is, , (a) 2, , N 2 ( g ) + 3H 2 ( g ) → 2NH 3 ( g ), identify, dihydrogen ( H 2 ) as a limiting reagent in, the following reaction mixtures., , (a) 7 L, , 22. Stoichiometric ratio of sodium, , (a) 1.5 : 3, (c) 1 : 1, , (b) O2 , 0.16 g, (d) O2 , 0.28 g, , (d) 10, , (b) 84.3, , (c) 118.6, , (d) 11.86, , 31. How many grams of concentrated nitric, acid solution should be used to prepare, 250 mL of 2.0 M HNO 3 ? The, concentrated acid is 70% HNO 3 ., (a) 45.0 g conc. HNO 3, (c) 70.0 g conc. HNO 3, , (b) 90.0 g conc. HNO 3, (d) 54.0 g conc. HNO 3

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7, , Some Basic Concepts of Chemistry, , 32. The mass of potassium dichromate, , crystals required to oxidise 750 cm 3 of, 0.6 M Mohr’s salt solution is (Given,, molar mass : potassium dichromate, = 294, Mohr’s salt = 392), (a) 0.49 g, (c) 22.05 g, , (b) 0.45 g, (d) 2.2 g, , 33. The normality of 10% (w/V ) of acetic, acid is ……, (a) 1 N, (c) 1.7 N, , (b) 1.3 N, (d) 1.9 N, , 34. If 500 mL of a 5 M solution is diluted to, 1500 mL, what will be the molarity of, the solution obtained? (NCERT Exemplar), (a) 1.5 M, (c) 0.017 M, , (b) 1.66 M, (d) 1.59 M, , 35. The number of molecules in 100 mL of, 0.02 N H 2SO 4 is ……, (a) 6.023 × 1022, (c) 6.023 × 1020, , (b) 6.023 × 1021, (d) 6.023 × 1018, , 36. The molarity of one litre solution of, 22.2 g of CaCl 2 will be ……, (a) 0.4 M, (c) 0.8 M, , (b) 0.2 M, (d) 0.6 M, , 37. Dissolving 120 g of urea (mol. wt. 60) in, 1000 g of water gave a solution of, density 1.15 g/mL. The molarity of the, solution is ……, (a) 1.78 M, (c) 2.05 M, , (b) 2.00 M, (d) 2.22 M, , 40. The density of 2 M aqueous solution of, NaOH is 1.28 g/cm 3 . The molality of, the solution is ……, [Given that, molecular mass of, NaOH = 40 g mol –1 ], (a) 1.20 m, , (b) 1.56 m, , 41. A sample of nitric acid is 69% by mass, and it has a concentration of 15.44, moles per litre. Its density is ……, (a) 1.86 g/cc, (c) 2.60 g/cc, , (b) 1.41 g/cc, (d) 1.02 g/cc, , 42. Mole fraction of the solute in a 1.00, molar aqueous solution is ……, (a) 0.0177, (c) 1.7700, , (b) 0.0344, (d) 0.1770, , 43. 8 g of NaOH is dissolved in 18 g of H 2 O., Mole fraction of NaOH in solution and, molality (in mol kg −1 ) of the solution, respectively are ……, (a) 0.2, 11.11, (c) 0.2, 22.20, , (b) 0.167, 22.20, (d) 0.167, 11.11, , 44. Match the items of Column I with, Column II and choose the correct, codes from the options given below., Column I, , Column II, , A. Mole fraction 1. M 2 × V 2, B. Molarity, , 2. The solution of higher, concentration., , C. Molality, , 3. It is defined as the, number of moles of, solute present in 1 kg of, solvent., , D. M 1 × V1, , 4. It is the ratio of number of, moles of a partial, component to the total, number of moles of the, solution., , 38. What is the concentration of sugar, , (C12 H 22O11 ) in mol L−1 if its 20 g are, dissolved in enough water to make a, final volume up to 2 L in mol L −1 ?, , (a) 0.0592, (c) 0.0375, , (b) 0.0292, (d) 0.0711, , 39. What will be the molality of the, solution containing 18.25 g of HCl gas, in 500 g of water?, (NCERT Exemplar), (a) 0.1 m, (c) 0.5 m, , (b) 1 M, (d) 1m, , (c) 1.67 m (d) 1.32 m, , E. Stock solution 5. It is defined as the number, of moles of the solute in 1, L of the solution., , Codes, A B C D, (a) 1 2 3 4, (c) 1 2 5 4, , E, 5, 3, , A B C D E, (b) 4 5 3 1 2, (d) 3 2 1 4 5

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CBSE New Pattern ~ Chemistry XI (Term-I), , 8, , 45. Match the following physical quantities, with their units and choose the correct, codes from the options given below., Column I, (Physical quantity), , Column II, (Unit), , A. Molarity, , 1. mol, , B. Mole fraction, , 2. Unitless, , C. Mole, , 3. mol L−1, , D. Molality, , 4. mol kg−1, , Codes, A B C D, (a) 3 2 1 4, (c) 3 2 1 3, , A B C D, (b) 2 1 4 3, (d) 1 2 4 3, , Assertion-Reasoning MCQs, Directions In the following questions, (Q.No. 46-60) a statement of Assertion, followed by a statement of Reason is, given. Choose the correct answer out of, the following choices., (a) Both Assertion and Reason are correct, statements and Reason is the correct, explanation of the Assertion., (b) Both Assertion and Reason are correct, statements, but Reason is not the correct, explanation of the Assertion., (c) Assertion is correct, but Reason is incorrect, statement., (d) Assertion is incorrect but Reason is correct, statement., , 46. Assertion Significant figures for 0.200, is 3 where as 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., (NCERT Exemplar), , 47. Assertion A number 138.42 can be, , written as 1.3842 × 10 2 in scientific, notation., Reason In scientific notation, a number, is generally expressed in the form of, N × 10 n , where N is a number between, 1.00 ... and 9.999 ... and n is an exponent., , 48. Assertion 1.231 has three significant, figures., Reason All numbers right to the, decimal point are significant., , 49. Assertion Equivalent weight of Cu in, CuO is 31.8 and in Cu 2O is 63.6., Reason Equivalent weight of an element, Atomic weight of the element, =, Valency of the element, , 50. 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, (NCERT Exemplar), compound., , 51. Assertion Molar volume of gases, change considerably with temperature, and pressure., Reason Molar volume of a substance is, the volume occupied by 1 mole of that, substance., , 52. Assertion The balancing of chemical, equations is based on law of conservation, of mass., Reason Total mass of reactants is equal, to total mass of products., , 53. 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, abundant isotope of carbon and has been, (NCERT Exemplar), chosen as standard., , 54. Assertion Atomicity of oxygen is 2., Reason 1 mole of an element contains, 6.023 × 10 23 atoms., , 55. Assertion Number of g-molecules of, SO 2Cl 2 in 13.5 g of sulphuryl chloride, is 0.2.

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9, , Some Basic Concepts of Chemistry, , Reason Gram-molecules is equal to, those molecules which are expressed, in gram., , 56. Assertion One mole of SO 2 contains, , equal the number of molecules present, in one mole of O 2 ., Reason Molecular weight of SO 2 is half, to that of O 2 ., , 57. Assertion Molecular weight of a, compound is 44 if its vapour density, is 22., Reason Vapour density × 2 = Molecular, weight., , 58. Assertion Combustion of 16 g of, methane gives 18 g of water., Reason In the combustion of methane,, water is one of the products., (NCERT Exemplar), , 59. Assertion A reactant that is entirety, consumed when a reaction goes to, completion is known as limiting reaction., Reason The amount of limiting reactant, limits the amount of product formed., , 60. Assertion Molarity of a solution, represents its concentration., Reason Molarity is the number of, moles of solute per litre of solution., , Case Based MCQs, 61. Read the passage given below and, answer the following questions :, The identity of a substance is defined not, only by the types of atoms or ions it, contains, but by the quantity of each type, of atom or ion. The experimental, approach required the introduction of a, new unit for amount of substances, the, mole, which remains indispensable in, modern chemical science. The mole is an, amount unit similar to familiar units like, pair, dozen, gross, etc. It provides a, , specific measure of the number of atoms or, molecules in a bulk sample of matter., A mole is defined as the amount of, substance containing the same number of, discrete entities (atoms, molecules, ions, etc.), as the number of atoms in a sample of pure, 12, C weighing exactly 12g. One Latin, connotation for the word "mole" is “large, mass” or “bulk,” which is consistent with its, use as the name for this unit., The mole provides a link between an easily, measured macroscopic property, bulk mass,, and an extremely important fundamental, property, number of atoms, molecules and, so forth., The number of entities composing a mole, has been experimentally determined to be, 6.02214179 × 10236.02214179 × 1023, a, fundamental constant named Avogadro's, number (N A ) or the Avogadro constant in, honor of Italian scientist Amedeo Avogadro., This constant is properly reported with an, explicit unit of “per mole,” a conveniently, rounded version being 6.022×1023/mol, 6.022×1023/mol., Consistent with its definition as an amount, unit, 1 mole of any element contains the, same number of atoms as 1 mole of any, other element. The masses of 1 mole of, different elements, however, are different,, since the masses of the individual atoms are, drastically different., The molar mass of an element (or, compound) is the mass in grams of 1 mole, of that substance, a property expressed in, units of grams per mole (g/mol)., The following questions (i-iv) are multiple, choice questions. Choose the most, appropriate answer :, (i) A sample of copper sulphate, pentahydrate contains 8.64 g of oxygen., How many grams of Cu is present in the, sample ?, (a) 0.952g, (c) 3.782g, , (b) 3.816g, (d) 8.64g

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CBSE New Pattern ~ Chemistry XI (Term-I), , 10, (ii) A gas mixture contains 50% helium, and 50% methane by volume. What is, the per cent by weight of methane in, the mixture?, (a) 19.97%, (c) 50%, , (b) 20.05%, (d) 80.03%, , (iii) The mass of oxygen gas which, occupies 5.6 litres at STP could be, (a) gram atomic mass of oxygen, (b) one fourth of the gram atomic mass of, oxygen, (c) double the gram atomic mass of oxygen, (d) half of the gram atomic mass of oxygen, , (iv) What is the mass of one molecule of, yellow phosphorus? (Atomic mass of, phosphorus = 30), (a) 1.993 × 10−22 mg, (b) 1.993 × 10−19 mg, (c) 4.983 × 10−20 mg, (d) 4.983 × 10−23 mg, , Or, The number of moles of oxygen in 1L, of air containing 21% oxygen by, volume, in standard conditions is, (a) 0.186 mol, (c) 2.10 mol, , (b) 0.21 mol, (d) 0.0093 mol, , 62. Read the passage given below and, answer the following questions :, A chemical reaction is a change in which, one or more substance(s) react(s) to form, new substance(s) with entirely different, properties., A chemical equation is a brief, representation of a chemical reaction in, terms of symbols and formulae of, substances involved in it, e.g. the reaction, of silver nitrate with sodium chloride to, give silver chloride and sodium nitrate can, be represented as, AgNO 3 + NaCl → AgCl + NaNO 3, 1442, 443, 1442443, Reactants, , Products, , The substances which react among, themselves to bring about the chemical, changes are known as reactants whereas, the, substances which are produced as a result of, the chemical change, are known as, products., Reactants and products of a chemical, equation are separated by arrow pointing, towards the products., Quantitative information conveyed by a, chemical equation is as follows :, ● The relative number of reactant and, product species (atoms or molecules), taking part in the reaction., ● The relative number of moles of the, reactants and products., ● The relative masses of the reactants and, products., ● The relative volumes of gaseous reactants, and products., The reactant which is present in the lesser, amount and gets consumed after sometime,, i.e. which limits the amount of product, formed is called the limiting reagent., The reactant other than the limiting, reagent, which is in somewhat excess is called, the excess reagent. The remaining amount of, excess reagent is calculated by subtracting the, available amount of limiting reagent from the, amount of the excess reagent., Remember that in stoichiometric, calculations, it is very important to choose, the limiting reagent., The following questions (i-iv) are multiple, choice questions. Choose the most, appropriate answer :, (i) What will be the volume of the mixture, after the reaction?, , NH3 (g ) + HCl(g ) → NH4 Cl(s ), 1L, , (a) 1.5 L, (c) 1 L, , 1.5 L, , (b) 0.5 L, (d) 0 L

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11, , Some Basic Concepts of Chemistry, , (ii) For the formation of 3.65 g hydrogen, chloride gas, what volume of hydrogen, gas and chlorine gas are required at, NTP conditions?, (a) 1.12 L, 1.12 L, , (a) Assertion and Reason both are correct, statements and Reason is correct, explanation for Assertion., , (b) 1.12 L, 2.24 L, (c) 3.65 L, 1.83 L, (d) 1 L, 1 L, , (iii) A gas mixture contains 50% He and, 50% CH 4 by volume. What is the, percentage by weight of methane in, the mixture ?, (a) 19.97%, , (b) 20.05%, , (c) 50, , (d) 80.03%, , (iv) AgNO 3 will react with 5.85 g NaCl to, produce 14.35 g of AgCl and 8.5 g of, NaNO 3 . Than mass of AgNO 3 will, be …… ., (a) 1.7 g, (c) 170 g, , (b) 17.0 g, (d) 0.17 g, , Or, The percentage of nitrogen in urea is, (a) 46%, (c) 18%, , In these questions (i-iv) a statement of, Assertion followed by a statement of, Reason is given. Choose the correct answer, out of the following choices :, , (b) 85%, (d) 28%, , 63. Read the passage given below and, answer the following questions :, Stoichiometry is a section of chemistry that, involves a calculation based on chemical, equations. Chemical equations are, governed by laws of chemical, combination., The mass of reactants is equal to the mass, of products. The compound obtained from, different methods contains the same, elements in the fixed ratio by mass. A, mole is a counting unit, equal to, 6022, ., × 10 23 particles., One mole is also equal to molar mass, expressed in grams., One mole of every gas at STP has a, volume equal to 22.4 L., The reacting species which are consumed, in the reaction completely is called limiting, reagent which decides the amount of, products formed., , (b) Assertion and Reason both are correct, statements but Reason is not correct, explanation for Assertion., (c) Assertion is correct statement but Reason is, incorrect statement., (d) Assertion is incorrect statement but Reason, is correct statement., , (i) Assertion 22.4 L of N 2 at NTP and, 5.6 L of O 2 at NTP contain equal, number of molecules., Reason Under similar conditions of, temperature and pressure, all gases, contain equal number of molecules., (ii) Assertion A reactant that is entirely, consumed when a reaction goes to, completion is known as limiting, reagent., Reason The amount of limiting reactant, limits the amount of product formed., (iii) Assertion Both 44 g CO 2 and 16 g CH 4, have same number of carbon atoms., Reason Both contain 1 g atom of, carbon which contains 6.023 × 10 23, carbon atoms., (iv) Assertion As mole is the basic chemical, unit, the concentration of the dissolved, solute is usually specified in terms of, number of moles of solute., Reason The total number of molecules, of reactants involved in a balanced, chemical equation is known as, molecularity of the reaction., , Or, Assertion One mole of NaCl contains, 6023, ., × 10 23 molecules of sodium, chloride., Reason 58.5 g of NaCl also contains, 6.023 × 10 23 molecules of NaCl.

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CBSE New Pattern ~ Chemistry XI (Term-I), , 12, , 64. Read the passage given below and answer, the following questions :, Most of the reactions occuring in the, laboratories are carried out in solutions. In, solutions, generally two components are, present. The one which is user in amount is, called the solute and the other one which is, in higher amount is called the solvent. The, amount of solute present in a given quantity, of solvent or solution is expressed in term of, concentration., The concentration of the solution is, expressed in many ways, i.e. in mass percent,, volume percent, parts per million, mole, fraction, molarity, molality and normality., In these questions (i-iv) a statement of, Assertion followed by a statement of Reason, is given. Choose the correct answer out of the, following choices :, (a) Assertion and Reason both are correct, statements and Reason is correct explanation, for Assertion., (b) Assertion and Reason both are correct, statements but Reason is not correct, explanation for Assertion., (c) Assertion is correct statement but Reason is, incorrect statement., (d) Assertion is incorrect statement but Reason is, correct statement., , (i) Assertion The mass percentage of an, element is used to determine, percentage composition of each, element in a compound., Reason Mass percentage depends up, the molar mass of the compound., (ii) Assertion The sum of mole fraction, of all the components of a solution is, unity., Reason Mole fraction is a, temperature dependent mode of, concentration., (iii) Assertion Molarity of a solution, represents its concentration., Reason Molarity is the number of, moles of solute per litre of solution., (iv) Assertion The molality of solution, does not change with change in, temperature., Reason Molality depend on the mass, of the solvent., , Or, Assertion Molarity of solution, independent of temperature., Reason Volume of a solution is, temperature dependent, i.e. it, changes with change in temperature., , ANSWERS, Multiple Choice Questions, 1. (c), 11. (a), 21. (a), , 2. (a), 12. (a), 22. (b), , 3. (d), 13. (c), 23. (c), , 4. (a), 14. (c), 24. (c), , 5. (c), 15. (a), 25. (a), , 6. (b), 16. (d), 26. (a), , 7. (c), 17. (b), 27. (d), , 8. (d), 18. (c), 28. (c), , 9. (d), 19. (a), 29. (d), , 10. (b), 20. (b), , 31. (a), 41. (b), , 32. (c), 42. (a), , 33. (c), 43. (d), , 34. (b), 44. (b), , 35. (c), 45. (a), , 36. (b), , 37. (c), , 38. (b), , 39. (d), , 30. (b), 40. (c), , 49. (a), 59. (a), , 50. (a), 60. (a), , 51. (b), , 52. (a), , 53. (b), , 54. (b), , 55. (d), , Assertion-Reasoning MCQs, 46. (c), 56. (c), , 47. (a), 57. (a), , 48. (d), 58. (d), , Case Based MCQs, 61. (i)-(b), (ii)-(d), (iii)-(d), (iv)-(b) or-(d), , 62. (i)-(b), (ii)-(a), (iii)-(d), (iv)-(b) or-(a), , 63. (i)-(a), (ii)-(a), (iii)-(a), (iv)-(b) or-(b), , 64. (i)-(b), (ii)-(c), (iii)-(a), (iv)-(a) or-(d)

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EXPLANATIONS, 9, t ° C + 32, 5, 9, ( 200 − 32) = t ° C, 5, 9, ⇒, t ° C = 168, 5, 168 × 5, t°C =, = 93.3° C, 9, 2. 6. 023 × 10 23 has four significant figures. The, total number of digits in a number including the, last digit whose value is uncertain is called the, number of significant figures., , 1., , F=, , 3. All non-zero digits are significant. Hence, there, are 6 significant figures in 10.3406 g., 0.02856 × 29815, . × 0112, ., has least, 4. The result of, 0.5785, significant figures., (i) In multiplication and division, the least, precise term 0.112 has 3 significant figures., Hence, the answer should not have more, than 3 significant figures., (ii) In multiplication, 5 is the exact number and, the other number has 4 significant figures., Hence, the answer should have 4 significant, figures., (iii) In addition, the answer cannot have more, digits to the decimal point than either of the, original members. Hence, the answer should, have 4 significant figures., , 5. Precision refers to the closeness of various, measurements for the same quantity. Accuracy, is the agreement of a particular value to the true, value of the result., Thus, A and B are precision and accuracy, respectively., 6. As we know that, least count of an instrument is, equal to the most possible error of the, instrument, hence most possible error of the, instrument will be 0.01 cm., 7. Given that, the true value for a result is 2.00 g, and student ‘ A ’ takes two measurements and, report the result as 1.95 g and 1.93 g. The values, are precise as they are close to each other but, are not accurate., Another student ( B ) repeats the experiment and, obtains 1.94 g and 2.05 g as the results for two, measurements. These observations are neither, precise nor accurate., , The third student (C ) repeats these, measurements and reports 2.01 g and 1.99 g as, the result, these values are both precise and, accurate., , 8. All the given statements are correct about, molecular mass., , 9. The molecular weight of, C60H122 = (60 × 12) + (122 × 1), = 720 + 122 = 842 g/mol, Weight of 6.023 ×10 23 molecules of, C60H122 = 842 g, ∴ Weight of 1 molecule of C60H122, 842, g = 1 .39 × 10 –21 g, =, 6.023 × 10 23, ≈ 1.4 × 10, , −21, , g, , Thus, the weight of one molecule of C60 H122 is, 1.4 × 10 −21 g ., , 10. Cr2O72 − + 14H+ + 6e − → 2Cr 3 + + 7H2O, 6I– → 3I2 + 6e −, 2Na 2S2O3 + I2 → Na 2S4O6 + 2NaI, In this reaction, equivalent weight of K2Cr2O7, molecular weight, =, 3× 2, molecular weight, =, 6, , 11. One mole of any substance contains 6.022 × 10 23, atoms/ molecules., Hence, number of millimoles of, H2 SO4 = Molarity × Volume (in mL), = 0.02 × 100 = 2 × 10 −3 mol, Number of molecules = number of moles × N A, = 2 × 10 −3 × 6.022 × 10 23, = 12.044 × 1020 molecules, , 12. In 18 g of water, number of H2O molecules is, NA ., So, in 18 mg of water, number of H2O molecules, 18 mg, would be = N A ×, 18 × 10 3 mg, = 10 −3 × N A, , 13. Q 22400 mL water contains water molecules, = 6.023 × 10 23 molecules

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CBSE New Pattern ~ Chemistry XI (Term-I), , 14, ∴ In 1 mL, the number of water molecules, 6.023 × 10 23, molecules, =, 22400, Since, 1 mL contains 20 drops., Therefore, number of water molecules in 1 drop, 6.023 × 10 23, =, 22400 × 20, = 1.344 × 1018 molecules, , 14. Number of molecules = mole × Avogadro’s, number ( N A ), The number of molecules of water in each of the, given options is calculated as,, (a) (V (H 2 O) g )STP = 0.00224 L, 0.00224, V, =, = 0.0001, nH 2O =, 22.4 L, 22.4, ∴ Number of molecules of water, = 0.0001 × N A, (b) 0.18 g of water, w H 2O, 0.18, =, = 0.01, nH 2O =, 18, M H 2O, Number of molecules = 0.01 × N A, (c) 18 mL of water, number of moles (nH 2 O ), mass of substance in g (w H 2 O ), =, molar mass in g mol −1 (w H 2 O ), , w H 2 O = 18 g/mol, [Q Density of water ( d H 2 O ) = 1 g L −1 ], 18, ∴ nH 2O =, =1, 18, Number of molecules of water = 1 × N A, (d) nH 2 O = 10 −3, ∴Number of molecules of water, = 10 −3 × N A, Hence, maximum number of molecules are, present in 18 mL of water., , 15. A → (2); B → (3); C → (1); D → (5); E → (4)., 16. Mass per cent of an element, =, , Mass of that element in the compound × 100, Molar mass of the compound, , Molar mass of Na 2SO4, = ( 2 × 2299, . ) + 32.06 + ( 4 × 16.00 ), = 142.04 g, , . × 100, 4598, = 32.37, 142.04, 32.06 × 100, Mass per cent of sulphur =, = 22.57, 142.04, 64 × 100, Mass per cent of oxygen =, = 45.06, 142.04, Mass per cent of sodium =, , 17. Molecular mass of CO2 = 1 × 12 + 2 × 16 = 44 g, 1 g molecule of CO 2 contains 1 g atoms of, carbon, Q 44 g of CO 2 contain C = 12 g atoms of carbon, 12, % of C in CO 2 =, ∴, × 100 = 27.27%, 44, Hence, the mass per cent of carbon in CO 2 is, 27.27%., , 18. Empirical formula mass = CH2O, , = 12 + 2 × 1 + 16 = 30, Molecular mass = 180, Molecular mass, n=, Empirical formula mass, , = 180 / 30 = 6, Molecular formula = n × empirical formula, = 6 × CH2 O = C6 H12 O6, , ∴, , 19. Given, % of element ( M ) = 68%, and of oxygen ( O) = 32%, Atomic mass of M = 34, Atomic mass of O = 16, Thus, empirical formula of M xO y is, element → % of mass → moles (n ), M → 68 → 68/34 = 2, O → 32 → 32/16 = 2, Hence, empirical formula of given compound is, M 2O2 or M O., Thus, option (a) is correct., , 20. Element, , %, , Atomic, Simpler, Molar ratio, mass, molar ratio, , C, , 10.06%, , 12, , 10.06, = 0.84, 12, , 0.84, =1, 0.84, , H, , 0.84%, , 1, , 0.84, = 0.84, 1, , 0.84, =1, 0.84, , Cl, , 89.10%, , 35.5, , 89.10, = 2.5, 35.5, , 2.5, =3, 0.84, , Thus, the empirical formula of the substance is, CHCl 3.

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15, , Some Basic Concepts of Chemistry, , 21. For first oxide,, Element, , %, Atomic, amount weight, , Number of, Simple, moles, molar ratio, , O, , 22, , 16, , 22, = 1.375, 16, , 1.375, =1, 1.375, , Fe, , 78, , 56, , 78, = 1.392, 56, , 1.392, ≈1, 1.375, , ∴ The formula of first oxide is FeO., Similarly, for second oxide, , The two equations, 3Ca → 3Ca 2 + + 6e −, 2Al 3+ + 6e − → 2Al, Net reaction, 3Ca + 2Al 3+ → 3Ca 2+ + 2Al, Therefore, the stoichiometric coefficient of Ca in, the given reaction is 3., , 24. Reaction of oxalate with permanganate in acidic, medium., , 5C2 O24 − + 2MnO4 →10CO2 + 2Mn 2 + + 8H2 O, , No. of mol 5, , %, Atomic, Element, amount weight, , Number, of moles, , Simple molar, ratio, 1.875, ≈ 1.5, 1.25, , O, , 30, , 16, , 30, = 1.875, 16, , Fe, , 70, , 56, , 70, = 1.25, 56, , 1.25, =1, 1.25, , ∴ The formula of second oxide is Fe2O3., Suppose in both the oxides, iron reacts with x g, oxygen., ∴ Equivalent weight of Fe in FeO, weight of Fe (II), =, ×8, weight of oxygen, 56 weight of Fe (II), …(i), =, ×8, 2, x, Similarly, equivalent weight of Fe in Fe2O3, weight of Fe (III), =, ×8, weight of oxygen, 56 weight of Fe (III), …(ii), =, ×8, 3, x, From Eqs. (i) and (ii),, weight of Fe (II) 3, =, weight of Fe (III) 2, , 22. 2Na 2HPO4 + NaH2PO4 + 2(NH2 )2 CO, → Na 5P3O10 + 4NH3 + 2CO2, Hence, the stoichiometric ratio of, sodium dihydrogen orthophosphate and, sodium hydrogen orthophosphate is 2 : 1, or 3 : 1.5., , 23. Ca → Ca 2++ 2e −, Al, , 3+, , (Oxidation), + 3e → Al (Reduction), −, , Eq. (i) is multiplied by 3 and Eq. (ii) is, multiplied by 2., , …(i), …(ii), , 10, , −, 5C2O2−, 4 ions transfer 10e to produce 10, molecules of CO2 ., So, number of electrons involved in producing, 10 molecules of CO2 is 10. Thus, number of, electrons involved in producing 1 molecule of, CO2 is 1., , 25. The balanced chemical equation is, Mg +, , 1, O2 → MgO, 2, , 24 g, , 16 g, , 40 g, , From the above equation, it is clear that 24 g, Mg reacts with 16 g O2 ., 16, Thus, 1.0 g Mg reacts with,, O2 = 0.67 g O2, 24, But only 0.56 g O2 is available which is less than, 0.67 g. Thus, O2 is the limiting reagent., Further, 16 g O2 reacts with 24 g Mg., ∴ 0.56 g O2 will react with Mg, 24, =, × 0.56 = 0.84g, 16, ∴ Amount of Mg left unreacted, = 1 . 0 − 0 . 84, = 0.16 g Mg, Hence, Mg is present in excess and 0.16 g Mg is, left behind unreacted., , 26. When 56 g of N2 + 10 g of H2 is taken as a, combination then dihydrogen ( H2 ) act as a, limiting reagent in the reaction., …(I), N2 ( g ) + 3H2 ( g ) → 2NH3 ( g ), 2 × 14 g, 28 g, , 3 × 2g, 6g, , 2 (14 + 3 ), 34g, , 28 g N2 requires 6g H2 gas., 6g, 56 g of N2 requires, × 56 g = 12g of H2, 28 g

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17, , Some Basic Concepts of Chemistry, , 32. Mohr’s salt is [ FeSO4 ⋅ (NH4 )2 SO4 ⋅ 6H2O], 36. Molarity ( M ), , Only oxidisable part is Fe2+, 2+, , 3+, , [Fe → Fe, , −, , +e ]× 6, , Cr2O72– + 14H+ + 6 e − → 2Cr 3+ + 7H2O, 6Fe2+ + Cr2O72− + 14H+ → 6Fe3+ + 2Cr 3+ + 7H2O, Millimoles of Fe, , 2+, , Moles of Fe2+, , = 750 × 0.6 = 450, 450, =, = 0.450 mol, 1000, , 6 moles of Fe2+ ≡ 1 mol Cr2O72−, 0.450, = 0.075 mol Cr2O72−, 6, = 0.075 × 294, = 22.05 g, w × 1000, 33. Normality ( N ) =, M × V ( mL ), 10 × 1000, =, = 1.66, 60 × 100, ∴ 0.450 mole Fe2+ ≡, , ∴, , 34., , N = 1.66 N ⇒ N ≈ 1.7 N, M1V1 = M 2V 2, 5 × 500 = M 2 × 1500, 5 × 500, = M2, 1500, M 2 = 166, . M, , 35. As, normality = molarity × valence factor for, H2SO4,, valence factor for H2SO4 = 2, ∴ Normality, N = M × 2, N 0.02, or, M =, =, = 0.01 M, 2, 2, Moles of H2SO4, 0.01 × 100, =, = 1 × 10 −3 mol, 1000, [Q Molarity, Number of moles of solute, × 1000] ∴, =, Volume in solution (in mL), Number of molecules in 100 mL of 0.02 N, H2SO4 = 1 × 10–3 × N A, = 1 × 10 −3 × 6.023 × 10 23, = 6.023 × 10 20 molecules, , =, , 22.2 g, Mass of solute (in g), =, Molecular weight of solute 111 g mol −1 × 1 L, × volume of solution (in L), , [Q Molecular weight of CaCl 2 = 111 g mol −1 ], = 0.2 mol L−1 = 0.2 M., Moles of solute, 37. Molarity ( M ) =, Volume of solution (in L), Moles of urea = 120 /60 = 2, Weight of solution, = Weight of solvent +Weight of solute, = 1000 + 120 = 1120 g, 1120 g, 1, ×, = 0.973 L, ⇒Volume =, 115, . g / mL 1000 mL, ∴ Molarity = 2 / 0.973 = 2.05 M, , 38. Molar mass of the sugar, ( C12H22O11 ), m = (12 × 12.01) + ( 22 × 1.0079 ) + (11 × 16.00 ), = 342.2938 g mol − 1, ≈ 342 g mol −1, Given, w = 20 g , V = 2 L, w, Molarity =, m × V ( in L), 20, =, = 0.0292 mol L− 1, 342 × 2, = 0.0292 M, , 39. Molality is defined as the number of moles of, solute present in 1 kg of solvent. It is denoted, by m ., Moles of solute, ...(i), Thus, molality (m ) =, Mass of solvent (in kg), Given that, Mass of solvent ( H2O) = 500 g = 0.5 kg, Weight of HCl = 18.25 g, Molecular weight of HCl = 1 × 1 + 1 × 35.5 = 36.5 g, 18.25, Moles of HCl =, = 0.5, 36.5, 0.5, [from eq. (i)], m =, =1 m, 0.5, , 40. Given, molarity = 2M, [i.e. 2 moles NaOH is dissolved in 1 L solution], Density ( d ) = 1.28 g/cm 3, Molecular weight of NaOH = 40 g mol −1

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CBSE New Pattern ~ Chemistry XI (Term-I), , 18, , We know that,, Density =, , mass of solution, volume of solution, , ∴ Mass of solution = 1.28 g cm −3 × 1000 mL, = 1280 g, Moreover,, Number of moles of solute, Molarity =, Volume of solution (in L), mass of solute, ∴, 2=, 40 × 1, Mass of solute = 80 g, Now, mass of solution = mass of solvent + mass, of solute, 1280 g = mass of solvent +80, ∴Mass of solvent = 1280 − 80 = 1200 g, = 1.2 kg, number of moles of solute, Now, molality =, mass of solvent (in kg), 2 20 5, =, =, = = 167, . m, 1.2 12 3, Alternative method, Molality (m ) =, , M, × 1000, 1000d − Mwt ., , where, M = molarity, d = density of solution, Mwt. = molar mass of solute., On substituting the given values, 2, Molality =, × 1000, 1000 × 1.28 − 40, 2, 2000, m, =, × 1000 =, = 1612, ., 1280 − 40, 1240, Mass, 41. Volume of 100 g sample =, = 100/d, Density (d ), Mass of solute, Molarity =, Molar mass of solute, 1000, ×, Volume of solution (in mL), 69, 1, Molarity =, ×, × 1000, 63 100 /d, 69, 1, 15.44 =, ×, × 1000, 63 100 /d, 63 × 15.44, d =, = 1.409 g/cc ≈ 1.41 g / cc, 69 × 10, , 42. 1.00 molar aqueous solution, = 1.0 mole in 1000 g water, , ⇒, , nsolute = 1 ; w solvent = 1000 g, 1000, nsolvent =, = 55.56, 18, nsolute, χ solute =, nsolute + nsolvent, 1, χ solute =, = 0.0177, 1 + 55.56, , 43. Mole fraction of solute, number of moles of solute + number of moles, of solvent, =, number of moles of solute, w solute, nsolute, Mw solute, χ solute =, =, w solute, w solvent, nsolute + nsolvent, +, Mw solute Mw solvent, Given, w solute = w NaOH = 8 g, Mw solute = Mw NaOH = 40 g mol −1, , ∴, , w solvent = w H 2O = 18 g;, Mw solvent = 18 g mol −1, 8 / 40, χ solute = χ NaOH =, 8 / 40 + 18 /18, 0.2, 0.2, =, =, = 0167, ., 0.2 + 1 1.2, Moles of solute, Mass of solvent (in kg), w solute, Mw solute, =, × 100, w solvent ( in g), , Now, molality (m ) =, , 8 / 40, 0.2, × 1000 =, × 1000, 18, 18, = 1111, . mol kg −1, =, , 44. A → (4); B → (5); C → (3); D → (1); E → (2)., 45. A → (3); B → (2); C → (1); D → (4)., 46. Assertion is correct but Reason is incorrect., 0.200 contains 3 while 200 contains only one, significant figure because zero at the end or right, of a number are significant provided they are on, the right side of the decimal point., , 47. Both Assertion and Reason are correct and, Reason is the correct explanation of Assertion., , 48. Thus, Assertion is incorrect but Reason is correct., 1.231 has four significant figures. All numbers, from left to right are counted, starting with the

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19, , Some Basic Concepts of Chemistry, , first digit that is not zero for calculating the, number of significant figures., At. wt. 636, ., 49. Eq. wt. of Cu in CuO =, =, = 31.8, Valency, 2, and CuO is 63.6., Both Assertion and Reason are correct and, Reason is the correct explanation of Assertion., 50. Both Assertion and Reason are correct and, Reason is the correct explanation of Assertion., The molecular formula of ethene is C2H4 and its, empirical formula is CH2 ., Thus, molecular formula = empirical formula × 2, , 51. Volume occupied by a gas changes with change, in temperature and pressure. For example,, change in temperature increases the K.E. of the, molecules and gases expand and hence, occupy, more volume., Both Assertion and Reason are correct but, Reason is not the correct explanation of, Assertion., , 52. According to law of conservation of mass, in a, chemical reaction total mass of the products is, equal to the total mass of the reactants., Both Assertion and Reason are correct and, Reason is the correct explanation of Assertion., , 53. Both Assertion and Reason are correct but, Reason is not the correct explanation of, Assertion., Atomic masses of the elements obtained by, scientists by comparing with the mass of carbon, comes out to be close to whole number value., , 54. Number of atoms present in a molecule of a, gaseous element is called atomicity., e.g. O 2 has two atoms and hence, its atomicity, is 2. Both Assertion and Reason are correct but, Reason is not the correct explanation of, Assertion., , 55. No. of gram molecules, wt. of sub., (GMM = Gram molecular mass), GMM, ∴ GMM of SO 2Cl 2 = 135 g, Thus, Assertion is incorrect but Reason is correct., =, , 56. One mole of any substance corresponds to, , 6.023 × 10 23 entities irrespective of its weight., Molecular weight of, SO 2 = 32 + 2 × 16 = 64 g mol − 1, , Molecular weight of O 2 = 2 × 16 = 32 g mol − 1, Therefore, molecular weight of SO 2 is double to, that of O 2., Thus, Assertion is correct but Reason is incorrect., , 57. This is vapour density method for determination, of molecular weight., Both Assertion and Reason are correct and, Reason is the correct explanation of Assertion., , 58. Assertion is correct but Reason is incorrect., Combustion of 16 g of methane gives 36 g of, water., CH4 + 2O2 → CO2 + 2H2O, 1mol, = 16 g, , 2 mol, = 36 g, , 59. The amount of product formed will be decided by, the amount of limiting reagent., Both Assertion and Reason are correct and, Reason is the correct explanation of Assertion., , 60. Concentration means how much amount of, substance (solute) is present in a given volume of, a solution., Now, as amount can be measured in terms of, moles, so molarity means concentration of the, solution., Both Assertion and Reason are correct and, Reason is the correct explanation for Assertion., , 61. (i) CuSO4 ⋅ 5H2O has 1 mole of Cu and 9 moles, of oxygen atoms., , 63.5 g of Cu = 9 ×16 g of oxygen, 8.64 g of oxygen =, , 63.5, × 8.64 = 381, . g, 9 × 16, , (ii) Equal volumes contain equal number of moles., Hence, molar ratio of He : CH4 = 1 : 1, ∴ Ratio by weight = 4 : 16 = 1 : 4, 4, ∴ CH4 present by weight = × 100 = 80%, 5, (iii) 22.4 L of O2 gas at STP = 1 gram molecule, (1 mole), 1, 5.6 L of O2 gas at STP =, × 56, . of gram, 22.4, molecule, 1, 1, 1, = gram molecule = × 2 gram atoms =, 4, 4, 2, gram atom, (iv) Yellow phosphorus is an allotropic form of, phosphorus with molecular formula P4 .

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CBSE New Pattern ~ Chemistry XI (Term-I), , 20, , ∴ Molecular mass of yellow phosphorus, ( P4 ) = 4 × 30, =120 g mol −1, 1 mole of P4 = 6.022 × 10 23 molecules = 120 g, ∴Mass of one molecule of P4, 120, =, 6.022 × 10 23, , Or, In 1 L of air, volume of O2 = 210 cc, 22400 cm3 = 1 mol, 210, 210 cm3 =, = 0.0093 mol, 22400, , 62. (i) 1 L NH3 will react with 1 L HCl to form, NH4Cl which has negligible volume., Hence, final mixture will contain only 0.5 L, HCl., (ii) H2 ( g ) + Cl 2 ( g ) → 2HCl( g ), 1 mol, , 2 mol, , 2 × 36.5 g, , 22.4 L, , For the formation of 3.65 g HCl, H2 or Cl 2, 22 .4, required =, × 365, . = 112, . L, 2 × 36.5, (iii) Equal volumes of gases contain equal, number of moles., Hence, molar ratio of He : CH4 = 1 : 1, Ratio by weight = 4 : 16 = 1 : 4, 4, ∴ CH4 present by weight = × 100 = 80%, 5, (iv) AgNO3 + NaCl → AgCl + NaNO3, x, , 5.85, , 14.35, , 63. (i) Both Assertion and Reason are correct and, Reason is the correct explanation of Assertion., Molar volume (at NTP) = 22.4 L, Now 22.4 L of N 2 = volume occupied by, 1 mole of N 2 = 28 g = 6.023 × 1023 molecules., , ∴22.4 L = 6.023 × 1023 molecules, , = 1993, ., × 10 −19 mg, , 22.4 L, , 28, × 100 = weight percent of N = 46%, 60, , Similarly, 1 mole of O 2 = 2 × 16 = 32 g, = 6.023 × 1023 molecules = 22.4 L, , = 1993, ., × 10 −22 g, , 1 mol, , ∴, , 8.5, , x + 585, . = 14.35 + 8.5, x = 17.0 g, Or, Urea is NH2CONH2, 1 mol = 60 g, There are two nitrogen atoms in urea., Weight of nitrogen = 28 g, , (ii) Both Assertion and Reason are correct and, Reason is the correct explanation of Assertion., (iii) Both Assertion and Reason are correct and, Reason is the correct explanation of Assertion., 44 g of CO 2 = 1 mole ≡ 1 g atom of C, 16 g of CH4 = 1 mole ≡ 1 g atom of C, 1 g atom of C = 12 g of C, 12 g of C contains 6.023 × 10 23 carbon atoms., (iv) Both Assertion and Reason are correct but, Reason is not the correct explanation of, Assertion., The number of moles of a solute present in a, litre of solution is known as molarity ( M )., The total number of molecules of reactants, present in a balanced chemical equation is, known as molecularity. For example,, PCl 5 → PCl 3 + Cl 3 (Unimolecular), (Bimolecular), 2HI → H2 + I2, Therefore, molarity and molecularity are used, in different sense., Or, Both Assertion and Reason are correct but, Reason is not the correct explanation of, Assertion., Molar mass or mass of 1 mole of NaCl, = 23 + 35.5 = 58.5 g, According to mole concept,, 1 mole of a compound, = 6.023 × 10 23 molecules, ∴58.5 g of NaCl also contains 6.023 × 10 23, molecules of NaCl., , 64. (i) Both Assertion and Reason are correct but, Reason is not the correct explanation of, Assertion.

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21, , Some Basic Concepts of Chemistry, , (ii) Assertion is correct but Reason is incorrect., nA, Mole fraction of A is solution ( χ A ) =, nA + nB, nB, Mole fraction of B is solution ( χ B ) =, nA + nB, n + nB, So, χ A + χ B = A, =1, nA + nB, The moles of any substances is related to its, mass and mass is independent of, temperature., Thus, Assertion is correct but Reason is, incorrect., (iii) Both Assertion and Reason are correct and, Reason is the correct explanation of Assertion., Concentration mean how much amount of, substance solute is present in a given volume, of a solution. Now, as amount can be, measured in terms of moles so molarity, means concentration of the solution., , (iv) Molality of solution does not change with, temperature as it depend on the mass, and mass remain unaffected with, temperature., Moles of solute, Molality, m =, × 1000, Weight of solvent (in g), Thus, both Assertion and Reason are correct, statements and Reason is the correct, explanation of Assertion., Or, Assertion is incorrect but Reason is the, correct statement., Molarity ( M ) is calculated by following, equation., Weight × 1000, Molarity =, Molecular weight × Volume (mL), Thus, molarity of solution depends upon, temperature as it depend on volume.

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CBSE New Pattern ~ Chemistry XI (Term-I), , 22, , 02, Structure of Atom, Quick Revision, Nature of electromagnetic radiation and, experimental results regarding atomic spectra, play an important role in development of Bohr’s, model., 1. According to Maxwell’s electromagnetic wave, theory, energy is emitted continuously from, any source in the form of radiations travelling, in wave form and associated with electric, and magnetic fields oscillating perpendicular to, each other and to the direction of radiation., Characteristics of a wave are, (i) Wavelength (λ) Distance between any two, consecutive crests or troughs and are, expressed in Å, nm, pm or m., (ii) Frequency ( ν ) Number of waves passing, through a point in one second. Units are, Hertz or s – 1 ., (iii) Velocity (c ) Distance travelled by the, wave in one second. Units are ms −1 ., c = ν × λ ⇒ λ = c /ν, (iv) Wave number ( ν ) Reciprocal of, wavelength., 1, ν=, λ, It’s commonly used unit is cm −1 (not SI unit)., , 2. Particle Nature of Electromagnetic, Radiation : Planck’s Quantum Theory, ●, , 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., , ●, , ●, , According to Planck’s quantum theory, the, radiant energy which is emitted or absorbed in, the atom of small discrete packets of energy, known as quantum and in case of light, the, quantum of energy is called photons., c, E = hν or E = h, λ, where, h = Planck’s constant = 6.63 × 10 − 34 Js, E = Energy of photon or quantum, If n is the number of quanta of a particular, frequency and E T be the total energy, then, E T = nhν, , 3. Photoelectric Effect, ●, , ●, , ●, , ●, , The phenomenon of ejection of electrons from, a metal surface when a light of certain, frequency strikes on its surface is called, photoelectric effect., For each metal, there is a characteristic, minimum frequency, known as threshold, frequency ( ν 0 ) below, which photoelectric, effect is not observed., The number of electrons ejected is proportional, to the intensity or brightness of light., Following the conservation of energy principle,, the kinetic energy of ejected electron is given, by the equation,, 1, hν = hν 0 + me ν 2, 2, Dual Nature of Electromagnetic Radiation, Light possesses both particle and wave like, properties.

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23, , Structure of Atom, , Whenever radiation interacts with matter, it, displays particles, like properties in contrast to the, wave like properties., , 4. Atomic Spectra, ●, , ●, , ●, , ●, , ●, , ●, , The pictorial representation of arrangement of, various types of EMR in their increasing order of, wavelength (or decreasing order of frequency) is, known as spectrum., The spectrum of radiation emitted by a substance, that has absorbed energy in increasing order of, wavelengths or decreasing frequencies is called as, an emission spectrum., An absorption spectrum is like the photographic, negative of an emission spectrum., The line spectra of hydrogen lies in three, regions of electromagnetic spectrum, i.e. infrared,, visible and UV-region., The set of lines in the visible region is known as, Balmer series., The general formula for wave number of lines of, different series of hydrogen spectrum is, 1, 1, ν = 109677  2 − 2  cm − 1, n1 n 2 , where, n1 = 1, 2 … n 2 = n1 + 1, n1 + 2 …, , ●, , ●, , The value 109677 cm − 1 is called the Rydberg, constant., The first five series of lines that correspond to, n1 = 1, 2, 3, 4 , 5 are known Lyman, Balmer,, Paschen, Bracket and Pfund series respectively., , 5. Bohr’s Model, The main postulates of this model are as follows :, ● Energy of an electron is quantised., The orbits or stationary states of electron are, numbered, n = 1, 2, 3 or K, L, M.... ., These integral numbers are known as principal, quantum number., ● The energy of an electron is certain in the orbit,, i.e. it does not change with time., ● The difference in energy ∆E is given by,, ∆E = hν, ⇒, , ν=, , ∆E E 2 − E1, =, h, h, , ●, , where, E 1 and E 2 are the energies of the, lower and higher allowed energy states, respectively., The angular momentum (mvr) of an electron,, in a given stationary states is given as, nh, mvr =, 2π, (n =1 , 2, 3, ...), , Significance of Bohr’s Model, This model is useful to explain the following, facts :, (i) Calculation of Bohr radius According, to Bohr’s model, radius of n th orbit is, given by the expression,, rn =, , 52.9 n 2, pm, Z, , 0.0529 n 2, nm, Z, where, Z is the nuclear charge., =, , (ii) Calculation of energy of the electron, The energy of the electron in the nth, orbit has been found to be, 1, E n = −R H  2 , (n = 1, 2, 3, ... ), n , 2 π 2me 4, where, R H = Rydberg constant =, h2, −18, = 2.18 × 10, J (for n = 1), For n th orbit,, 218, . × 10 −18 2, En = −, Z J per atom, n2, Z = atomic number of an atom, 13. 595 eV, , Z 2 per atom, n2, – 13 . 6 2, ≈, Z eV atom, n2, 1311.8 2, Z kJ mol −1, =−, n2, – 1312 2, Z kJ mol −1, ≈, n2, (iii) Calculation of frequency and wave, number for a transition The energy, gap between the two orbits is given by, equation,, En = −

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CBSE New Pattern ~ Chemistry XI (Term-I), , 24, , ●, , ∆E = E f − E i, ν =R H, , 1, 1, ν, = 109677  2 − 2  cm −1, , c, n f ,  ni, , where, ni and n f stands for initial orbit and, final orbit., For absorption spectrum, n f > ni ., For emission spectrum, ni > n f ., Limitations of Bohr’s Model, It fails to account for the finer details of the, hydrogen atom spectrum observed by using, spectroscopic technique. It could not explain, the ability of atoms to form molecules by, chemical bonds., , 6. Towards Quantum Mechanical, Model of an Atom, ●, , ●, , ●, , ●, , ●, , In view of shortcoming of the Bohr’s model,, two important developments, i.e. dual, behaviour of matter and Heisenberg, uncertainty principle came into existence., de-Broglie explains the dual nature of electron, (i.e. wave nature as well as particle nature), h, h, Hypothesis to be given, λ =, =, mv p, where, λ = wavelength, v = velocity of particle,, m = mass of particle, p = momentum, Heisenberg’s uncertainty principle This, principle is a result of dual nature of matter, and radiation., According to this principle, it is impossible to, determine at any given instant, both the, momentum and the position of subatomic, particles, like electron, simultaneously., h, h, or ∆x ⋅ ∆v x ≥, ∆x ⋅ ∆p ≥, 4π, 4 πm, Heisenberg uncertainty principle rules out the, existence of definite paths or trajectories of, electrons and other similar particles., The effect of Heisenberg uncertainty principle, is significant only for motion of microscopic, objects and is negligible for that of, macroscopic objects. This is because in dealing, with milligram sized or heavier objects, the, associated uncertainties are hardly of real, consequence., , ●, , In case of microscopic object like an electron,, ∆v ⋅ ∆x obtained is much larger and such, uncertainties are of real consequence., The precise statement of the position and, momentum of electrons have to be replaced, by the statements of probability that the, electron has at a given position and, momentum., , 7. Quantum Mechanical Model of Atom, ●, , ●, , The branch of science that takes into account, this dual behaviour of matter is (i.e. wave, nature as well as particle nature) called, quantum mechanics., Schrodinger derived an equation for an, electron which describes the wave motion of, an electron along any three axes., For a system (such as an atom or molecule, whose energy does not change with time), the, Schrodinger equation is written as H$ψ = Eψ., , where, H$ is a mathematical operator called, Hamiltonian., ● The important features of quantum mechanical, model of atom are :, (i) The energy of electron in atoms is quantised., (ii) Both the exact position and exact velocity, of an electron in an atom cannot be, determined simultaneously., (iii) An atomic orbital is the wave function ψ, for an electron in an atom., (iv) The probability of finding an electron at, a point within an atom is proportional, to the square of the orbital wave function, | ψ |2 at that point, which is known, as probability density and is always, positive., From the value of | ψ |2 at different points within an, atom, it is possible to predict the region around the, nucleus, where electron will most probably be, found., , 8. Atomic Orbitals and Quantum Numbers, The set of four numbers which give a complete, information about energy, shape, orientation and, also spin of electron in an atom, are called, quantum numbers.

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25, , Structure of Atom, , There are four types of quantum numbers:, (i) Principal quantum number (n) It determines, the size and to a large extent the energy of the, orbital. It is always a positive integer with value, of n = 1 , 2, 3 …. ., ● The principal quantum number also identifies, the shell., ● All the orbitals of a given value of ‘n’ constitute, a single shell of atom and are represented by, the following letters., n = 1, 2, 3, 4 ……, Shell = K L M N ……, ● The ‘sub-shells’ are the orientations and shapes, for your orbitals going in order by s, p, d, f., ● Number of sub-shells in a shell = shell number., ● The first sub shells has 1 orbital. Each succesive, sub shell adds 2 more orbital (1, 3, 5, 7, etc.), ● Each orbital can hold only 2 electrons of, opposite spin., (ii) Azimuthal quantum number (l) It determines, the angular momentum of the electron and is, also known as orbital angular momentum or, subsidiary quantum number., For a given value of n, the possible values of l, are l = 0, 1, 2, …, (n − 1)., e.g. If n = 1, value of l is only 0., , (iv) Spin quantum number (s ) It tells about the, spinning motion of the electron, i.e. clockwise, or anti-clockwise., For a given value of m,, –1, 1, s = + and ., 2, 2, It helps to explain magnetic properties of the, substances., h, Spin angular momentum = s (s + 1), 2π, ●, ●, ●, , ●, ●, , 9. Node and Nodal Plane, A node is a region of space where probability of, finding an electron is zero., (i) (n − l − 1) = radial nodes, (ii) l = angular or planar nodes, (iii) (n − 1) = total nodes, , 10. Shape of Atomic Orbitals, ●, ●, , Value of l, , 0, , 1, , 2, , 3, , 4, , 5, , Subshell, notation, , s, , p, , d, , f, , g, , h, , Number of, orbitals, , 1, , 3, , 5, , 7, , 9, , 11, , Angular momentum of the electron in an, h, orbital = l ( l + 1), = l ( l + 1) h, 2π, It also tells about the shape of orbitals of a, subshell., Energy of subshells : s < p < d < f ., , Number of subshells in n th shell = n, Number of orbitals in a subshell = 2l + 1 ,, Maximum number of electrons in a subshell, = 2 ( 2l + 1 ), Number of orbitals in n th shell = n 2, Maximum number of electrons in n th shell, = 2n 2 ., , ●, , s-orbitals Spherical with node., p-orbitals Dumb-bell in shape., d-orbitals Four d -orbitals have clover leaf, shape with four lobes., Three orbitals have lobes between the axis, and are called d xy , d yz , d xz ., Fourth has lobes along the axis and is called, d x 2 –y2 ., Fifth has doughnut shape, where electron cloud, is present at the centre and is called d 2 ., z, , 11. Energies of Orbitals, The lower the value of (n + l ) for an orbital, the, lower is its energy. If two orbitals have the same, value of (n + l ), the orbital with lower value of n, will have the lower energy., , (iii) Magnetic quantum number (ml ) It shows the, behaviour of electron in a magnetic field. For a, given value of l (subshell), 2 l + 1 possible, 12. Rules for Filling of Electrons in Orbitals, values for ml are given by, Filling of electrons into the orbitals of different, ml = − l, − ( l − 1), − ( l − 2 ) … 0,, atoms takes place according to the following, … ( l − 2 ), ( l − 1), l, discussed rules.

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CBSE New Pattern ~ Chemistry XI (Term-I), , 26, (i) Aufbau principle In the ground state of the, atoms, the orbitals are filled in order of their, increasing energies., 1s , 2s , 2 p , 3s , 3 p , 4 s , 3d , 4 p , 5s , 4 d , 5 p , 6s ,, 4 f , 5d , 6 p , 7s ., (ii) Pauli’s exclusion principle No two, electrons in an atom can have the same set, of four quantum numbers., It can also be stated as, “only two electrons, may exist in the same orbital and these, electrons must have opposite spin”. The, maximum number of electrons in the shell, with principal quantum number n is equal, to 2n 2 ., (iii) Hund’s rule of maximum multiplicity, states that ‘‘pairing of electrons in the, orbitals belonging to the same subshell, ( p , d or f ) does not take place until each, orbital belonging to that subshell has got, one electron each, i.e. it is singly occupied.’’, , 13. Stability of Half-filled and Completely, Filled Orbital, ●, , ●, , ●, , The ground state electric configuration of atom, corresponds to lowest energy state and gives, higher stability., The electronic configuration of most of the atoms, follows the basic rules. However, certain elements, such as Cr or Cu do not follow the rules., The extra stability of half-filled and fully-filled, electronic configuration can be explained in, terms of symmetry and exchange energy with, the help of Hunds rule., , 14. Electronic Configuration of Elements, ●, , ●, , The arrangement of electrons in various shells,, sub-shells and orbitals in an atom is termed as, electronic configuration., It is written in term of nl x ., where, n indicates the order of shell, l indicates, the subshell and x indicates the number of, electrons present in the sub-shell., , Objective Questions, Multiple Choose Questions, 1. The value of Planck's constant is, , 6 .63 × 10 −34 Js. The speed of light is, 3 × 10 17 nm s –1 ., Which value is closest to the, wavelength (in nm) of a quantum of, light with frequency of 6 × 10 15 s −1 ?, (a) 10, (c) 50, , (b) 25, (d) 75, , 2. Calculate the energy in joule, , corresponding to light of wavelength 45, nm (Planck’s constant, h = 6.63 × 10 −34, Js; speed of light, c = 3 × 10 8 nm s −1 )., , (a) 6.67 × 1015, (c) 4.42 × 1015, , (b) 6.67 × 1011, (d) 4.42 × 10−18, , 3. The ratio of the shortest wavelength of, two spectral series of hydrogen, spectrum is found to be about 9., , The spectral series are, (a), (b), (c), (d), , Lyman and Paschen, Brackett and Pfund, Paschen and Pfund, Balmer and Brackett, , 4. Which of the following series of, , transitions in the spectrum of hydrogen, atom fall in visible region?, (a) Balmer series, (b) Paschen series, (c) Brackett series, (d) Lyman series, , 5. Which transition in the hydrogen, , atomic spectrum will have the same, wavelength as the Balmer transition, (i.e. n = 4 to n = 2) of He + spectrum?, (a) n = 4 to n = 3, (c) n = 4 to n = 2, , (b) n = 3 to n = 2, (d) n = 2 to n = 1

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27, , Structure of Atom, , 6. What is the wave number of 4th line in, Balmer series of hydrogen spectrum?, ( R = 109, , ,677 cm −1 ), (b) 24,360 cm−1, (d) 24,372 cm−1, , (a) 24,630 cm−1, (c) 24,730 cm−1, , 7. The wavelength (in Å) of an emission, 2+, , line obtained for Li during an, electronic transition from n 2 = 2 to, n 1 = 1 is (R = Rydberg constant), 3R, (a), 4, 4, (c), 3R, , 27R, (b), 4, 4, (d), 27R, , (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, , 9. The expression for Bohr’s frequency, rule is …… ., , ∆E E2 − E1, =, h, h, (b) v = ∆E ⋅h = E2 − E1 ⋅h, v, (c) = ∆E + h = (E2 − E1) + h, h, ∆E E1 − E2, (d) v =, =, h, 2, , (a) v =, , orbits for which its angular momentum, is integral multiple of …… ., (b), , h, 2π, , (c), , h, 2π, , (d) h⋅2π, , 11. For which of the following species,, Bohr's theory is not applicable?, (a) Be 3+, (c) He2+, , (b) Li2+, (d) H, , (n = 1) is approximately 0.530Å. The, radius for the first excited state (n = 2) is, (in Å) …… ., (b) 1.06, , (b) 192 nm, (d) 9.1× 10−6 nm, , 14. What is the energy (in eV) required to, excite the electron from n = 1 to n = 2, state in hydrogen atom ?, (n = principal quantum number), (b) 3.4, (d) 10.2, , 15. 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 kJmol− 1, (b) − 1312 kJmol− 1, (c) − 164 kJmol− 1, (d) − 82 kJmol− 1, , (c) 4.77, , (a) it fails to account for the finer details of the, hydrogen atom spectrum observed by using, sophisticated spectroscopic techniques, (b) it is unable to explain the spectrum of, atoms other than hydrogen, (c) it is unable to explain the ability of atoms to, form molecules by chemical bonds, (d) All of the above, , 17. Relation between wavelength ( λ ) and, momentum (P ) of a material particle, is …… ., (a) λ = hP, (c) λ = h + P, , (b) λ = h / P, (d) λ = h − P, , 18. If travelling at same speeds, which of, , 12. Bohr radius for the hydrogen atom, , (a) 0.13, , (a) 91 nm, (c) 406 nm, , 16. Limitations of Bohr’s model is/are, , 10 An electron can move only in those, h, 4π, , emitted, when in a hydrogen atom, electron falls from infinity to stationary, state 1, would be …… ., (Rydberg constant = 1.097 × 10 7 m −1 ), , (a) 13.6, (c) 17.0, , 8. Bohr’s model can explain, , (a), , 13. The wavelength of the radiation, , (d) 2.12, , the following matter waves have the, shortest wavelength? (NCERT Exemplar), , (a), (b), (c), (d), , Electron, Alpha particle ( He2+ ), Neutron, Proton

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CBSE New Pattern ~ Chemistry XI (Term-I), , 28, 19. The momentum of a particle having, a de-Broglie wavelength of 10, is …… ., (Given, h = 6.625 ×10 −34 m ), −7, , −17, , m, , 25. Which of the following does not, , represent the mathematical expression, for the Heisenberg uncertainty principle?, , (a) 3.3125 × 10 kg m s, (b) 26.5 × 10 −7 kg m s –1, (c) 6.625 × 10 –17 kg m s –1, (d) 13.25 × 10 −17 kg m s −1, , –1, , 20. If E e , E α and E p represent the kinetic, , energies of an electron, α-particle and a, proton respectively, each moving with, same de-Broglie wavelength then …… ., (a) Ee = Eα = Ep, (c) Eα > Ep > Ee, , (b) Ee > Eα > Ep, (d) Ee > Ep > Eα, , 21. As per de-Broglie’s formula a, , macroscopic particle of mass 100 g and, moving a velocity of 100 cm s − 1 will, have a wavelength of …… ., (a) 6.6 × 10−29 cm, (b) 6.6 × 10−30 cm, (c) 6.6 × 10− 31 cm, (d) 6.6 × 10−32 cm, , 22. A body of mass x kg is moving with a, , velocity of 100 m s −1 . Its de-Broglie, wavelength is 6.62 × 10 −35 m. Hence, x is, (h = 6.62 × 10 −34 Js) …… ., (a) 0.25 kg, (c) 0.2 kg, , (b) 0.15 kg, (d) 0.1 kg, , 23. The wavelengths of electron waves in, two orbits is 3 : 5. The ratio of kinetic, energy of electrons will be ……… ., (a) 25 :9, , (b) 5 : 3, , (c) 9 :25, , (d) The position and velocity of the electrons, in the orbit cannot be determined, simultaneously, , (d) 3 :5, , 24. Which of the following statements is, , not correct regarding Bohr’s model of, hydrogen atom?, , (a) Energy of the electrons in the orbit is, quantised, (b) The electron in the orbit nearest to the, nucleus has the lowest energy, (c) Electrons revolve in different orbits around, the nucleus, , (a) ∆x ⋅ ∆p ≥ h / (4π), , (b) ∆x ⋅ ∆v ≥ h / (4πm), , (c) ∆E ⋅ ∆t ≥ h / (4π), , (d) ∆E ⋅ ∆x ≥ h / (4π), , 26. In atom, an electron is moving with a, , speed of 600 m/s with an accuracy of, 0.005%. Certainty with which the, position of the electron can be located is, …… . (h = 6.6 × 10 −34 kg m 2 s −1 , mass, of electron, e m = 6.6 × 10 −31 kg ), (a) 1.52 × 10−4 m, (c) 1.92 × 10−3 m, , (b) 5.10 × 10−3 m, (d) 3.84 × 10−3 m, , 27. Uncertainty in the position of an, , electron (mass = 9 .1 × 10 −31 kg ) moving, with a velocity 300 ms −1 , accurate upon, 0 .001% will be …… ., (h = 6 . 63 × 10 −34 Js ), (a) 19.2 × 10−2 m, (c) 1.93 × 10−2 m, , (b) 5.76 × 10−2 m, (d) 3.84 × 10−2 m, , 28. The uncertainties in the velocities of, , two particles A and B are 0.05 and, 0.02 ms −1 respectively. The mass of B is, five times to that of mass A. What is,  ∆x , the ratio of uncertainties  A  in,  ∆x B , their positions?, (a) 2, (c) 4, , (b) 0.25, (d) 1, , $ is, 29. In the Schrodinger equation, H, (a) a mathematical operator called Hamiltonian, operator, (b) introduced by Schrodinger from the, expression for the total energy of the system, (c) Both (a) and (b), (d) None of the above

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29, , Structure of Atom, , 30 The graph between | ψ | 2 and r (radial, distance) is shown below. This represents, , 36. Orbital angular momentum depends on, ......... ., , (a) l, (c) n and m, |ψ|2, , (b) n and l, (d) m and s, , 37. Two electrons occupying the same, orbital are distinguished by …… ., , r, , (a) 1s-orbital, (c) 3s-orbital, , (b) 2p-orbital, (d) 2s-orbital, , 31. Principal, azimuthal and magnetic, , quantum numbers are respectively, related to, (a) size, orientation and shape, (b) size, shape and orientation, (c) shape, size and orientation, (d) None of the above, , 32. The angular momentum of electrons in, d orbital is equal to, (a) 6 h, (c) 2 3 h, , (b) 2 h, (d) 0 h, , 33. The value of azimuthal quantum, , number (l ) is 2 then the value of, principal quantum number (n) is, (a) 2, (c) 4, , (b) 3, (d) 5, , 34. If azimuthal quantum number, l = 0, shape of orbital would be …… ., (a) spherical, (b) dumball, (c) complex, (d) double dumball, , 35. For n = 2, the correct set of azimuthal, and magnetic quantum numbers are, (a) l = 2 ; m = −2 , −1, 0, +1, +2, (b) l = 1 ; m = −2 , −1, 0, +1, +2, (c) l = 0 ; m = −1 , 0, +1, (d) l = 1 ; m = −1 , 0, +1, , (a) magnetic quantum number, (b) azimuthal quantum number, (c) spin quantum number, (d) principal quantum number, , 38. Match the quantum numbers given in, , Column I with the information, provided by these given in Column II., Column II, (Information, provided), , Column I, (Quantum, number), , A. Principal quantum 1. Orientation of the, number, orbital, B. Azimuthal, quantum number, , 2. Energy and size of, orbital, , C. Magnetic, quantum number, , 3. Spin of electron, , D. Spin quantum, number, , 4. Shape of the orbital, , Codes, A B C D, (a) 2 4 1 3, (c) 2 4 3 1, , (NCERT Exemplar), , A B C D, (b) 3 2 4 1, (d) 1 2 3 4, , 39. The number of orbitals associated with, quantum numbers n = 5, m s = +, (a) 25, , (b) 50, , (c) 15, , 1, is ….. ., 2, , (d) 11, , 40. The number of radial nodes for, 3p-orbital is....... ., (a) 3, , (b) 4, , (c) 2, , (d) 1, , 41. Total number of orbitals associated with, third shell will be ........... ., (a) 2, (c) 9, , (b) 4, (d) 3

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CBSE New Pattern ~ Chemistry XI (Term-I), , 30, 42. Match the Column I with Column II, , and choose the correct options from the, codes given below :, Column I, (Boundary surface, diagram), , Column II, (d-orbital), , 1. dz 2, , z, , 2. d xy, , B., x, , y, , (a) Pauli’s exclusion principle, (b) Aufbau principle, (c) Hund’s rule, (d) uncertainty principle, , x, z, , 4. d x 2, , D., , − y2, , x, , y, , 45. The representation of the ground state, , configuration 1s 2 , 2s 2 2 p x1 2 p 1y 2 p z1 and, not 1s 2 2s 2 2 p x2 2 p 1y 2 p z0 which is, determined by ……, , 3. d xz, y, , 47. Which one is a wrong statement?, (a) The electronic configuration of N-atom is, 1s2, , z, , 5. d yz, , E., x, y, , Codes, A B C D, (a) 4 5 2 1, (c) 5 3 1 4, , E, 3, 2, , A B C D E, (b) 3 4 5 2 1, (d) 5 4 1 2 3, , 43 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) 4 f < 5 p < 5 d < 6 s, (d) 5 p < 5 d < 4 f < 6 s, , (b) Co and Zn, (d) Cu and Cr, , 46. Nitrogen has the electronic, , z, , C., , (a) Cu and Zn, (c) Mn and Cr, , (a) Heisenberg’s uncertainty principle, (b) Bohr’s quantisation theory of angular, momenta, (c) Pauli exclusion principle, (d) Hund’s rule, , x, , xy, , correct arrangement of filling up of the, atomic orbitals in, , electronic configuration of He by box −, diagram as ↑↑ is wrong because it, violates, , z, , A., , 44. Aufbau principle does not give the, , 2s2, , 2p1x 2p1y 2p1z, , (b) An orbital is designated by three quantum, numbers while an electron in an atom is, designated by four quantum numbers, (c) Total orbital angular momentum of electron, in ‘s’ orbital is equal to zero, (d) The value of m for dz 2 is zero, , 48. The correct distribution of four, electrons in p-orbital would be, (a), , (b), , (c), , (d)

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31, , Structure of Atom, , 49. Match the following rules given in, Column I with their correct, statements given in Column II., Column I, (Rules), A. Hund’s rule, , Column II, (Statements), , 51 Out of the following electronic arrangements, for outer electronic configuration., , II., , B. Aufbau, principle, , 2. Half-filled and completely, filled orbitals have extra, stability., , III., , C. Pauli, exclusion, principle, , 3. Pairing of electrons in the, orbitals belonging to the, same subshell does not, take place until each, orbital is singly occupied., , IV., , 5. In the ground state of, atoms, orbitals are filled in, the order of their, increasing energies., , Codes, A B, (a) 3 5, (b) 1 2, (c) 2 1, (d) 1 2, , C, 1, 3, 3, 4, , D, 4, 4, 4, 3, , 50 Which one of the following is the, wrong statement?, (a) de-Broglie’s wavelength is given by, h, λ=, , where, m = mass of the particle, mv, v = group velocity of the particle, (b) The uncertainty principle is, ∆E × ∆t ≥ h /4π, (c) Half-filled and fully-filled orbitals have, greater stability due to greater exchange, energy, greater symmetry and more, balanced arrangement, (d) The energy of 2s-orbital is less than the, energy of 2p-orbital in case of hydrogen, like atoms, , 3d, , 4s, , 3d, , I., , 1. No two electrons in an, atom can have the same set, of four quantum numbers., , D. Heisenberg’s 4. It is impossible to, uncertainty, determine the exact, principle, position and exact, momentum of a subatomic, particle simultaneously., , 4s, , 4s, , 3d, , 4s, , 3d, , The most stable arrangement is, (a) I, , (b) II, , (c) III, , (d) IV, , 52. The correct set of four quantum numbers, , for the valence electron of rubidium atom, (Z = 37) is, (a) 5, 1, 1, +, , 1, 2, , (c) 5, 0, 0, +, , 1, 2, 1, (d) 5, 1, 0, +, 2, , (b) 6, 0, 0, +, , 1, 2, , 53. The number of d-electrons in Fe 2+ ( Z = 26) is, not equal to the number of electrons in which, one of the following?, (a) s-electrons in Mg (Z = 12), (b) p-electrons in Cl (Z = 17), (c) d-electrons in Fe (Z = 26), (d) p-electrons in Ne (Z = 10), , 54. Which of the following ion shows maximum, unpaired electrons ?, (a) Fe 3+, , (b) CO 3+, , (c) Ni2+, , (d) Cu2+, , 55. Match species given in Column I with the, , electronic configuration given in Column II., Column I, , Column II, , A. Cr, , 1., , [Ar]3d 8 4 s 0, , B. Fe 2 +, , 2., , [Ar]3d 10 4 s 1, , C. Ni 2 +, , 3., , [Ar]3d 6 4 s 0, , D. Cu, , 4., , [Ar]3d 5 4 s 1, , 5., , [Ar]3d 6 4 s 2, (NCERT Exemplar)

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CBSE New Pattern ~ Chemistry XI (Term-I), , 32, Codes, A B C D, (a) 4 3 1 2, (c) 2 1 3 4, , A B C D, (b) 1 2 4 3, (d) 1 2 4 3, , 56. Match the following species given in, , Column I with their corresponding, ground state electronic configuration, given in Column II., Column I, (Atom / Ion), , Column II, (Electronic, configuration), , A. Cu, , 1. 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10, , B. Cu 2 +, , 2. 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4 s 2, , C. Zn 2 +, , 3. 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4 s 1, , D. Cr 3 +, , 4. 1s 2 2s 2 2p 6 3s 2 3p 6 3d 9, 5. 1s 2 2 s 2 2p 6 3s 2 3p 6 3d 3, , Codes, A B C D, (a) 2 1 3 4, (c) 1 2 4 3, , A B C D, (b) 3 4 1 5, (d) 5 2 3 4, , Assertion-Reasoning MCQs, Directions In the following questions, (Q.No. 57-71) a statement of Assertion, followed by a statement of Reason is, given. Choose the correct answer out of, the following choices., (a) Both Assertion and Reason are correct, statements and Reason is the correct, explanation of the Assertion., (b) Both Assertion and Reason are correct, statements, but Reason is not the correct, explanation of the Assertion., (c) Assertion is correct, but Reason is incorrect, statement., (d) Assertion is incorrect but Reason is correct, statement., , 57. Assertion de-Broglie equation has, , significance for any microscopic, particles., Reason de-Broglie wavelength is, inversely proportional to the mass of, the object., , 58. Assertion The radius of the first orbit of, hydrogen atom is 0.529 Å ., , Reason Radius for each circular orbit is, (rn ) = 0.529 Å (n 2 / Z ), where n = 1, 2, 3, and Z = atomic number., , 59. Assertion Energies of the orbitals in, , hydrogen or hydrogen like species, depend only on the quantum number ‘n’., Reason Energies of the orbitals in, multi-electron atoms depend on, quantum numbers ‘n’ and ‘ l ’., , 60. 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., , 61. Assertion Energy of the orbital, , increases with increase of principal, quantum number., Reason Energy is required in shifting, away the negatively charged electron, from the positively charged nucleus., , 62. Assertion Spin quantum number can, , have the value +1/ 2 or −1/ 2., Reason Here, ( +) or (−) sign signifies the, wave function., , 63. Assertion Magnetic quantum number, can have the value m l = − l , ... 0 ..., l ., Reason Magnetic quantum number, specifies the shape or orbitals., , 64. Assertion For a given principal, , quantum number, s, p, d, f …, subshells,, all have different energies., Reason Mutual repulsion exists among, the electrons in a multi-electron atoms., , 65. Assertion Half-filled and fully-filled, , degenerate set of orbitals acquire extra, stability., Reason The reason for the above fact is, the symmetry of such orbitals.

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Structure of Atom, , 66. Assertion s-orbital electron will be, , more tightly bound to the nucleus than, p-orbital electron., Reason Z eff experienced by the, electron decreases with increase of, azimuthal quantum number (l )., , 67. Assertion Shapes of the orbitals are, , represented by boundary surface, diagrams of constant probability, density., Reason Boundary surface diagram, helps in interpreting and visualising an, atomic orbitals., , 68. Assertion 5s-orbital has greater energy, than 4s., Reason Energy of the orbital depends, on the azimuthal quantum number., , 69. Assertion Atomic orbital in an atom is, designated by n, l , m l and m s ., Reason These play no role in, designating electron present in an, orbital., , 70. Assertion In case of isoelectronic ions, the ionic size increases with the, increase in atomic number., Reason The greater the attraction of, nucleus, greater is the ionic radius., , 71. Assertion Total number of electrons is, , a subshell is disignated by ( 2l + 1),, where, l = azimuthal quantum number., Reason l can have value 1, 2...n + 1,, where, n is principal quantum number., , Case Based MCQs, 72. Read the passage given below and, answer the following questions :, , A total of four quantum numbers are used, to describe completely the movement and, trajectories of each electron within an, , 33, atom. Each electron in an atom has a unique, set of quantum numbers; according to the, Pauli Exclusion principle, no two electrons, can share the same combination of four, quantum numbers. Quantum numbers are, important because they can be used to, determine the electron configuration of an, atom and the probable location of the atom’s, electrons., In atoms, there are a total of four quantum, numbers; the principal quantum number (n ),, the orbital angular momentum quantum, number (l ), the magnetic quantum number, (m l ), and the electron spin quantum number, (m s ). The principal quantum number, (n ),, describes the energy of an electron and the, most probable distance of the electron from, the nucleus., In other, words, it refers to the size of the, orbital and the energy level an electron is, placed., The number of subshells, or l , describes the, shape of the orbital. It can also be used to, determine the number of angular nodes. The, magnetic quantum number, (m l ) describes, the energy levels in a subshell, and (m s ) refers, to the spin on the electron, which can either, be up or down. The value of the principal, quantum number n is the level of the, principal electronic shell (principal level). All, orbitals that have the same n value are in the, same principal level., The size of the atom and its atomic radius, increases. Because the atomic radius, increases, the electrons are farther from the, nucleus. Thus, it is easier for the atom to, expel an electron because the nucleus does, not have as strong a pull on it, and the, ionisation energy decreases. We can, designate a principal quantum number, n,, and a certain subshell by combining the, value of n and the name of the subshell, (which can be found using (l)., The s-orbital, because the value of m l can, only be 0, can only exist in one plane. The

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CBSE New Pattern ~ Chemistry XI (Term-I), , 34, p-orbital, however, has three possible, because the value of m l and so it has three, possible orientations of the orbitals, shown, by p x , p y and p z . The pattern continues,, with the d -orbital containing 5 possible, orbital orientations, and f has 7., In these questions (i-iv) a statement of, Assertion followed by a statement of, Reason is given. Choose the correct, answer out of the following choices :, (a) Assertion and Reason both are correct, statements and Reason is correct, explanation for Assertion., (b) Assertion and Reason both are correct, statements but Reason is not correct, explanation for Assertion., (c) Assertion is correct statement but Reason, is incorrect statement., (d) Assertion is incorrect statement but, Reason is correct statement., , (i) Assertion Only principal quantum, number determines the energy of, an electron in an orbital in sodium, atoms., Reason For one electron system, the, expression of energy is same as that, obtained in Bohr’s theory., (ii) Assertion s-orbital electron will be, more tightly bound to the nucleus, than p-orbital electron., Reason Z eff experienced by the, electron decreases with increase of, azimuthal quantum number (l )., (iii) Assertion Orbit and orbital are not, synonymous., Reason Orbital represents the circular, path along which the electron moves, but orbit represents the three, dimensional space around the, nucleus where probability of, finding the electron is maximum., (iv) Assertion Energy of electron is, largely determined by its principal, quantum number., , Reason Principal quantum number is a, measure of the most probable distance, of finding the electron around the nucleus., , Or, Assertion The clockwise or anti-clockwise, motion/movement of e − can be determined, with the help of spin quantum number., Reason Spin quantum number depends on, the value of principal quantum number., , 73. Read the passage given below and answer, the following questions :, In other words, kinetic energy of the, ejected electron is proportional to the, frequency of the electromagnetic, radiation., Since, the energy of striking photon is hν, and the minimum energy required to, eject the electron is hν 0 (also called work, function,W 0 ) then the difference in, energy (hν − hν 0 ) is transferred as the, kinetic energy of the photoelectron., Following the law of conservation of, energy principle, the kinetic energy of, the ejected electron is given by the, equation,, 1, hν = hν 0 + m e v 2, 2, where, m e is the mass of the electron, moving with velocity (v)., The following questions (i-iv) are multiple, choice questions. Choose the most, appropriate answer :, , (i) The work function of a metal is 4.2 eV., If radiation of 2000Å fall on the metal, then the kinetic energy of the fastest, photoelectron is, (a) 1.6 × 10−19 J, (b) 16 × 10−10 J, (c) 3.2 × 10−19 J, (d) 6.4 × 10−10 J

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35, , Structure of Atom, , (ii) The number of electron ejected in the, photoelectric experiment is, proportional to the, (a), (b), (c), (d), , intensity of light, brightness of light, Both (a) and (b), None of the above, , (iii) Kinetic energy of the ejected electron, is, (a) equal to the frequency of the, electromagnetic radiation, (b) proportional to the frequency of the, electromagnetic radiation, (c) more than the frequency of the, electromagnetic radiation, (d) inversely proportional to the frequency, of the electromagnetic radiation, , (iv) Magnitude of kinetic energy in an, orbit is equal to, (a) half of the potential energy, (b) twice of the potential energy, (c) one fourth of the potential energy, (d) None of the above, , Or, , Which of the following statements is, correct?, (a) The energy of a quantum of radiation is, proportional to its frequency (ν) is, expressed by equation, E = hν, (b) With the help of quantum theory, Plank, explained the distribution of intensity in, the radiation from black body as a, function of frequency or wavelength at, different temperatures, (c) In photoelectric effect, there is no time, lag between the striking of light beam, and the ejection of electrons from metal, surface, (d) All of the above, , 74. Read the passage given below and, , answer the following questions :, The characteristics of an orbital are, expressed in terms of three numbers,, called principal (n ), azimuthal (l ) and, magnetic quantum numbers (m l )., These numbers are obtained from the, solutions of the Schrodinger wave, , equation. Further, to represent the spin, (rotation) of the electron about its own axis,, a fourth quantum number, called spin, quantum number has been introduced., n defines the shell, determines the size of the, orbital and also to a large extent the energy, of orbitals. There are n subshells in the nth, shell. ‘l ’ identifies the subshell and, determines the shape of the orbital. m l, designates the orientation of orbital. For a, given value of l , m l has ( 2l + 1) values, the, same as the number of orbitals per subshell., m s refers to orientation of the spin of the, electron., In these questions (i)-(iv) a statement of, Assertion followed by a statement of Reason, is given. Choose the correct answer out of, the following choices :, (a) Assertion and Reason both are correct, statements and Reason is correct, explanation for Assertion., (b) Assertion and Reason both are correct, statements but Reason is not correct, explanation for Assertion., (c) Assertion is correct statement but Reason is, incorrect statement., (d) Assertion is incorrect statement but Reason, is correct statement., , (i) Assertion Energies of the orbitals in, hydrogen or hydrogen like species, depend only on the quantum number, ‘n’., Reason Energies of the orbitals in, multielectron atoms depend on, quantum numbers ‘n’ and ‘ l ’., (ii) Assertion Energy of the orbital, increases with increase of principal, quantum number., Reason Energy is required in shifting, away the negatively charged electron, from the positively charged nucleus., (iii) Assertion For a given principal, quantum number, s, p, d, f …, subshells,, all have different energies., Reason Mutual repulsion exists, among the electrons in a multielectron, atoms.

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CBSE New Pattern ~ Chemistry XI (Term-I), , 36, (iv) Assertion 2p-orbitals do not have any, spherical node., Reason The number of nodes in, p-orbitals is given by (n − 2) where, n is, the principal quantum number., Or, , Assertion Half-filled and fully filled, degenerate set of orbitals acquire, extra stability., Reason The reason for the above fact, is the symmetry of such orbitals., , 75. Read the passage given below and, answer the following questions :, , The Heisenberg’s uncertainty principle is a, result of dual nature of matter and, radiation. It was put forward by W., Heisenberg a German Physicist in 1927., According to this principle, it is impossible, to know simultaneously both the conjugate, properties accurately. Both the position, and the momentum of the particle at any, instant can’t be determined with absolute, exactness or certainty. The uncertainty in, measurement of position, ∆x and the, uncertainty of determination of, momentum, ∆P are related by, Heisenberg’s relationship., This principle rules out the existence of, definite paths or trajectories of electrons, and other similar particles. However, the, effect of this principle is significant only in, case of microscopic objects and is, negligible in case of macroscopic object., Since, the uncertainty principle is such a, basic result in quantum mechanics, typical, experiments in quantum mechanics, routinely observe aspects of it. Certain, experiments, however, may deliberately, test a particular form of the uncertainty, principle as part of their main research, program. These include, for example, tests, of number-phase uncertainty relations in, superconducting or quantum optics, systems., , Applications dependent on the uncertainty, principle for their operation include, extremely low-noise technology such as that, required in gravitational wave, interferometers., The following questions (i-iv) are multiple, choice questions. Choose the most, appropriate answer :, (i) If uncertainty in the position of electron, is zero, the uncertainty in its momentum, would be, h, 4π, (c) zero, , (a) <, , h, 4π, (d) infinite, , (b) ≥, , (ii) The uncertainty in position of an, electron (m = 91, . × 10 −28 g) moving with, a velocity 3 × 10 4 cm /s accurate upto, 0.001% will be, (a), (b), (c), (d), , 3.84 cm, 1.92 cm, 7.68 cm, 5.76 cm, , (iii) If the light of wavelength, λ is used to, observe an electron then uncertainty in, position of the electron would be, (a), (b), (c), (d), , less than λ, more than λ, equal to λ, equal to or greater than λ, , (iv) If uncertainties in the measurement of, position and momentum of an electron, are equal, the uncertainty in the, measurement of velocity is, (a) 8.0 × 1012 ms−1, (c) 8.5 × 1010 ms−1, , (b) 4.2 × 1010 ms−1, (d) 6.2 × 1010 ms−1, , Or, The uncertainty in momentum of an, electron is 1 × 10 −5 kg m / s. The, uncertainty in its position will be, (h = 6.62 × 10 −34 kg m 2 / s), (a), (b), (c), (d), , 2.36 × 10−28 m, 5.25 × 10−28 m, 2.27 × 10−30 m, 5.27 × 10−30 m

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37, , Structure of Atom, , 76. Read the passage given below and, , answer the following questions :, Orbitals are regions or spaces where the, probability of finding electrons is, maximum. Qualitatively, these orbitals can, be distinguished by their size, shape, and, orientation., An orbital of small size means there is, more chance of finding the electron near, the nucleus. Shape and orientation means, the direction in which the probability of, finding the electron in maximum. Atomic, orbitals can be distinguished by quantum, numbers., Each orbitals is designated by three, quantum numbers n, l and m l (magnetic, quantum number) which define energy,, shape, and orientation but these are not, sufficient to explain spectra of, multi-electrons atoms. Spin quantum, number (m s ) determines the spin of, electrons. Spin angular momentum of the, electron has two orientations relative to the, chosen axis which are distinguished by, spin quantum numbers m s which can take, 1, 1, values + and − ., 2, 2, The following questions (i-iv) are multiple, choice questions. Choose the most, appropriate answer :, (i) How many orbitals are associated with, n = 3?, (a) 3, (c) 9, , (b) 6, (d) 12, , (ii) What will be the representation of, n = 2, l = 1?, (a) 2s, , (b) 2p, , (c) 2sp, , (d) 3s, , (iii) How many electrons are possible in an, orbital of 2s?, (a) 1, 1, (c), 2, , (b) 2, (d) 4, , (iv) What is the shape of ‘p’ orbital?, y, , y, , x, , (a), , (b), , x, z, , y, , y, , z, , (c), , x, , (d), , x, z, , Or, What will be the value of m s of the, following?, 1s, , (a) 3, (c), , 1, 2, , 1, 2, , 2s, , 2p, , (b) 2, (d), , 7, 2, , 1, 2

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38, , CBSE New Pattern ~ Chemistry XI (Term-I), , ANSWERS, Multiple Choice Questions, 1. (c), 11. (c), 21. (c), , 2. (d), 12. (d), 22. (d), , 3. (a), 13. (a), 23. (a), , 4. (a), 14. (d), 24. (d), , 5. (d), 15. (d), 25. (d), , 6. (d), 16. (d), 26. (c), , 7. (d), 17. (b), 27. (c), , 8. (c), 18. (b), 28. (a), , 9. (a), 19. (c), 29. (c), , 31. (b), 41. (c), , 32. (a), 42. (d), , 33. (b), 43. (b), , 34. (a), 44. (d), , 35. (d), 45. (c), , 36. (a), 46. (c), , 37. (c), 47. (a), , 38. (a), 48. (b), , 39. (a), 49. (a), , 51. (a), , 52. (c), , 53. (b), , 54. (a), , 55. (a), , 56. (b), , 60. (c), 70. (d), , 61. (b), 71. (d), , 62. (c), , 63. (c), , 64. (a), , 65. (a), , 10. (b), 20. (d), 30. (d), 40. (d), 50. (d), , Assertion-Reasoning MCQs, 57. (a), 67. (a), , 58. (a), 68. (b), , 59. (b), 69. (d), , 66. (b), , Case Based MCQs, 72. (i)-(d), (ii)-(a), (iii)-(c), (iv)-(a, c), 74. (i)-(b), (ii)-(b), (iii)-(a), (iv)-(a, a), , 73. (i)-(c), (ii)-(c), (iii)-(b), (iv)-(a, d), 75. (i)-(d), (ii)-(b), (iii)-(a), (iv)-(a, d), , 76. (i)-(c), (ii)-(b), (iii)-(b), (iv)-(d, c), , EXPLANATIONS, 1. Given, Planck's constant, h = 6.63 × 10 −34 J s, 17, , Speed of light, (c ) = 3 × 10 nm s, , −1, , Frequency of quantam ( ν ) = 6 × 1015 s−1, Wavelength ( λ) = ?, c, c, ,λ =, λ, ν, 3 × 1017, = 0.5 × 102 nm = 50 nm, =, 6 × 1015, , We know that, ν =, , 2. Given, λ = 45 nm = 45 × 10−9 m [Q 1 nm = 10−9 m], The wavelength of light is related to its energy, by the equation,, E = hν, [Q where, ν = c / λ ], hc, E =, λ, 6.63 × 10−34 Js × 3 × 108 ms−1, Hence, E =, 45 × 10−9 m, = 4.42 × 10 −18 J, Hence, the energy corresponds to the light of, wavelength 45 nm is 4.42 × 10 −18 J., , 3. According to Rydberg’s equation,, 1 RH  1, 1, 1 1, 1, =, ∝ − ,  −  or, λ, λ n12 n22 , hc n12 n22 , , For shortest wavelength, i.e. highest energy, spectral line, n2 will be ( ∞ )., For the given spectral series, ratio of the shortest, wavelength of two spectral series can be, calculated as follows :, 1, 1, 1, −, −0, λ L 32 ∞ 2 9, 1, (a), =, =, =, 1, 1, λP, 1, −, 0, 9, −, 12 ∞ 2, 1, 1, − 2, 2, λ Bk, 1 16 16, ∞, 5, (b), =, =, ×, =, 1, 1, λ Pf, 25, 1, 25, −, 4 2 ∞2, 1, 1, − 2, 2, 1 9 9, λP, 5, ∞, (c), =, × =, =, 1, 1, λ Pf, − 2 25 1 25, 2, 3, ∞, 1, 1, − 2, 2, λB, ∞ = 1 ×4=1, (d), = 4, 1, 1, λ Bk, 16 1 4, −, 22 ∞ 2, Note Lyman = L (n1 = 1), Balmer = B (n1 = 2), Paschen = P (n1 = 3) , Brackett = Bk(n1 = 4 ), Pfund = Pf (n1 = 5)

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39, , Structure of Atom, , 4. Balmer series of transitions in the spectrum of, , hydrogen atom fall in visible region. Lyman, series fall in ultraviolet while Paschen, Brackett, and Pfund fall in infrared region., , 5. Q, , 1, 1, 1, =Z2 ⋅R  2 − 2, λ, n1 n2 , , ∆E E 2 − E 1, =, h, h, , where, E 1 and E 2 are the energies of the lower, and higher allowed energy states respectively., This expression is commonly known as Bohr's, frequency rule., , ∴ For He+ ion, Z = 2, , 10. An electron can move only in those orbits for, , 1, 1 , 1, = 22 ⋅ R 2 − 2, λ, 4 ,  2, 1, 3 3, =4×R×, = R, λ, 16 4, The same value for H-atom is possible when, electron jumps from n = 2 to n = 1,, , 11. Bohr’s atomic model is applicable only for, , 3, 1, 1 1 , ⇒, R, =1× R −, 1 4 , λ, 4, 1, 1, 6. ν = RH  2 − 2 , n, n, 1, 2, For Balmer series, n1 = 2 and for 4th line in, Balmer series, n2 = 6, i.e., , Q, , which its angular momentum is integral multiple, of h / 2π., This means angular momentum is quantised., hydrogen atom and H-like species, i.e. to the, one electron system., Since, He2+ does not contain one electron while, all other species have one electron each, thus, Bohr’s model is not applicable to He2+ ., , 12. r ∝ n 2 /Z, where,n = number of orbit, Z = atomic number, Q, r1 ∝ n12, r2 ∝ n22, , RH = 1, 09,677 cm −1, , 1, 1, −,  22 6 2 , 1 , 1, = 109677, −,  4 36 , , ν = 109677, , ν = 24, 372 cm −1, , 7., , v =, , 1, 1, 1, = R HZ 2  2 − 2 , λ, n1 n2 , Where, RH = R, Z = 3, (Q for Li 2+ , atomic number = 3), n2 = 2,n1 = 1,  1, 1, 1 , = R( 3) 2  2 −, , λ,  (1), ( 2) 2 , R ×9 × 3, =, 4, 4, λ=, 27R, , 8. As the Bohr’s model is applicable only to, , hydrogen atom and hydrogen like species, i.e. to, one electron system, thus Bohr’s model can easily, explain the spectrum of any atom or ion, containing one electron only., , 9. The frequency of radiation is absorbed or emitted, , when transition occurs between two stationary, states that differ in energy by ∆ E, is given by, , (Z = 1 for H-atom), , r1 n12, =, r2 n22, 2, , So,, , 0 . 530 1, = 2, r2, 2, ∴, r2 = 0 .530 × 4 = 2 .120 Å, , 1, 1 1, 13. = v H = RH  2 − 2 , λ, n1 n2 , = 1.097 × 107, , 1 , 1, −, 12 ∞ 2 , , 1, × 107 = 911, . × 10 −8 m, 1.097, = 91.1 nm• 91 nm, , λ=, , 1 1, − , n12 n12 , , 14. ∆E = 136, . Z2 , , 1 1, . × (1) 2  2 − 2 , = 136, 1, 2 , 3,  1, = 136, . 1 −  = 136, . × = 10.2 eV,  4, 4, , 15. The energy of second bohr orbit of H-atom ( E 2 ), is − 328 kJ mol −1 because, E2 =, , −1312, kJ mol −1, 22

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40, , CBSE New Pattern ~ Chemistry XI (Term-I), , 1312, kJ mol −1, n2, 1312, E n = − 2 kJ mol −1 = − 82 kJ mol −1, 4, 16. Limitations of Bohr’s model are :, (i) It fails to account for the finer details of the, hydrogen atom spectrum observed by using, sophisticated spectroscopic techniques., (ii) It is unable to explain the spectrum of atoms, other than hydrogen., (iii) It is unable to explain the ability of atoms to, form molecules by chemical bonds., En = −, , 17. Relation between wavelength ( λ ) and, momentum ( p ) of a material particle is, h, λ=, P, This is popularly called as de-Broglie's equation., 1, 18. From de-Broglie equation, λ ∝, m, Greater the mass of matter waves, lesser is, wavelength and vice-versa. In these matter waves,, alpha particle (He2+ ) has higher mass, therefore,, shortest wavelength., , 19. According to de-Broglie relation,, h, h, λ=, =, mv p, λ = wavelength, h = Planck’s constant, p = momentum, h = 6.625 × 10 −34 J s, , where,, , Here,, , λ = 10 −17 m, h 6.625 × 10 –34, p= =, λ, 10 −17, , ∴, , = 6.625 × 10 −34 × 1017, = 6.625 × 10 −17 kg m s –1, , 20. de-Broglie wavelength, h, h, ...(i), ⇒v =, mv, mλ, 1, ...(ii), KE = mv 2, 2, On putting the value of v in Eq. (ii), we get, λ=, , ⇒, , 1  h , KE = m , , 2 m ⋅ λ , 1  h2 , KE = , , 2 m ⋅ λ2 , , 2, , Hence, KE ∝ 1 /m (if λ is same) and order of, mass is as, me < m p < m α, Thus, the order of KE is, Ee > E p > E α, , 21. According to de-Broglie wavelength, λ =, Given, m = 100 g , v = 100 cm s − 1, , h, mv, , h = 6.6 × 10− 34 J -s, , and, , = 6.6 × 10 − 27 erg - s, On substituting values, we get, 6.6 × 10− 27, λ=, 100 × 100, = 6.6 × 10 − 31 cm, , 22. According to de-Broglie equation,, h, mv, λ = 6.62 × 10 −35 m, , λ=, Given,, , v = 100 m s−1, h = 6.62 × 10 −34 Js, m = x kg, 6.62 × 10 −34, ∴ 6.62 × 10 −35 =, x × 100, and, , x = 0.1 kg, , 23. de-Broglie equation is,, h, mv, 1, KE = mv 2, 2, 2KE, v =, m, h, λ=, 2m KE, λ=, , or, , 2, , 25, (KE)1 λ22  5, or 25 : 9, = 2 =  =, , , 9, (KE)2 λ 1, 3, Hence, the ratio of kinetic energy of electrons, will be 25 : 9., , 24. The position and velocity of the electrons in the, orbit cannot be determined simultaneously., This statement is not correct for Bohr’s model of, hydrogen atom. It is infact given by Heisenberg, in his uncertainty principle.

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41, , Structure of Atom, , 25. From Heisenberg uncertainty principle, h, …(i), 4π, h, ∆x ⋅m∆v ≥, [Q ∆p = m∆v ], 4π, h, …(ii), ∆x ⋅ ∆v ≥, 4πm, This principle is also applicable for pairs like, energy time ( ∆E ⋅ ∆t ) and angular moment-angle, ( ∆ω ⋅ ∆θ ) along with position-momentum, ( ∆x ⋅ ∆P ) ., h, Thus,, …(iii), ∆E ⋅ ∆t ≥, 4π, ∆x ⋅ ∆P ≥, , 29. In Schrodinger equation, H$ is a mathematical, , operator called Hamiltonian. It was introduced, by Schrödinger from the expression for the total, energy of the system., , 30. The graphs between | ψ |2 and r are radial, density plots having (n − l −1) number of radial, nodes. For 1s, 2s, 3s and 2p-orbitals these are, respectively., , |Ψ|2, , For 1s-orbital, number of radial node = 1– 0 –1=0, , r, , 26. By Heisenberg’s uncertainty principle,, ∆x ⋅m∆v =, , h, 4π, , 600 × 0.005, ∆v = 0.005% of 600 m/s =, = 0.03, 100, 6.6 × 10−34, ∆x × 9.1 × 10−31 × 0.03 =, 4 × 3.14, 6.6 × 10−34, 4 × 3.14 × 0.03 × 9.1 × 10−31, = 1.92 × 10−3 m, h, 27. ∆x ⋅ ∆v ≥, 4πm, Hence, ∆x =, , 6.63 × 10 −34, 4 × 3.14 × 9.1 × 10 −31 × 300 × 0.001 × 10 −2, = 0.01933 = 1.93 × 10 −2 m, , ∆x =, , |Ψ|2, , For 2s-orbital, number of radial node = 2– 0 –1=1, , r, , |Ψ|2, , For 3s-orbital, number of radial node = 3– 0 –1=2, , r, , |Ψ|2, , For 2p-orbital, number of radial node = 2–1–1=0, , 28. According to Heisenberg uncertainty principle,, h, 4π, where, ∆x = uncertainty in position, m = mass of particle, ∆v = uncertainty in velocity, According to question,, h, ...(i), ∆x A × m × 0.05 =, 4π, h, ...(ii), ∆x B × 5m × 0.02 =, 4π, On dividing Eq. (i) by Eq. (ii), we get, ∆x A × m × 0.05, =1, ∆x B × 5m × 0.02, ∆ x A 5m × 0.02, ∆x A, ⇒, =, =2, = 2 or, ∆xB, m × 0.05, ∆x B, ∆x × m × ∆v =, , r, , Thus, the given graph between | ψ |2 and r, represents 2s-orbital., , 31. Principal, azimuthal and magnetic quantum, , numbers are respectively related to size, shape, and orientation of the shell, subshell and, orbitals., , 32. Angular momentum of electrons in, d -orbital is, , = l ( l + 1), , h, ;, 2π, , for d -orbital, l = 2, = 2( 2 + 1) h, Angular momentum = 6 h, , h , , , Q h =, , 2π 

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42, , CBSE New Pattern ~ Chemistry XI (Term-I), , 33. The value of azimuthal quantum number is, given by l = n –1, n = l + 1, , Here, l = 2, thus,n = 2 + 1 = 3, Hence, the value of principal quantum number, (n ) is 3., , 34. For s-subshell, l = 0. Its shape is spherical., 35. For n = 2, l can be 0 and 1., For l = 0, m = 0 and l = 1, m = −1, 0, +1., h, 36. Orbital angular momentum, mvr =, l( l + 1)., 2π, Hence, it depends only on ‘l ’. l can have values, ranging from 0 to (n − 1)., , 37. Two electrons occupying the same orbital have, , equal spin but the directions of their spin are, opposite. Hence, spin quantum number,, s (represented +1/ 2 and − 1/ 2) distinguishes, them., 38. The correct match is :, A → (2); B → (4); C → (1); D → (3), (A) Principal quantum number is the most, important quantum number as it determines, the size and to large extent the energy of the, orbital., (B) Azimuthal quantum number determines the, angular momentum of the electron and, defines the three-dimensional shape of the, orbital., (C) Magnetic quantum number gives, information about the spatial orientation of, orbitals with respect to a standard set of, coordinate axes., (D) Spin quantum number arises from the, spectral evidence that an electron in its, motion around the nucleus in an orbit also, rotates or spin about its own axis., , 39 According to quantum mechanical atom model,, , for each value of n (principal quantum number),, there are ‘n’ different values of l (azimuthal, quantum number), i.e. l = 0, 1, 2, …, (n − 1)., And, for each value of l, there are 2l + 1 different, values of m l (magnetic quantum number), i.e., m l = 0, ±1, ±2 … ±l., ∴Total number of possible combinations of n, l, and m l , for a given value of n is n 2 , and each, such combination is associated with an orbital., Each orbital can occupy a maximum of two, , electrons, having a different value of spin, 1, 1, quantum number (m s ), which are + or − ., 2, 2, ∴ Number of orbitals associated with n = 5 is, n 2 = 25. Each of those orbitals can be associated, 1, 1, with m s = + as well as m s = − ., 2, 2, ∴ Answer = 25, , 40. For a hydrogen atom wave function, there are, , n − l − 1 radial nodes and (n − 1) total nodes., Number of radial nodes for 3p orbital = n − l − 1, = 3 −1 −1 =1, , 41. Total number of orbitals associated with n th, shell = n 2, ∴Total number of orbitals associated with third, shell = ( 3) 2 = 9, , 42. The correct match is :, , A → (5); B → (4); C → (1); D → (2); E → (3), , 43. According to Aufbau principle, the order in, , which the energies of the orbitals increases is as, follows:, 1s , 2s , 2 p, 3s , 3 p, 4s , 3d , 4 p, 5s , 4d ,, 5 p, 6s , 4 f , 5d , 6 p, 7s , 5 f , 6d , 7 p, 8s, Hence, the correct order of increasing energy of, orbital is 5 p < 6s < 4 f < 5d ., , 44. Aufbau principle does not give the correct, , arrangement of filling up of atomic orbitals in, copper and chromium because half-filled and, completely filled electronic configuration of Cr, and Cu have lower energy and therefore, more, stable., Cr(Z = 24 ) = 1s 2 , 2s 2 2 p 6 , 3 s 2 3 p 6 3 d 5 , 4s 1, Cu(Z = 29 ) = 1 s 2 , 1 s 2 2 p 6 , 3 s 2 3 p 6 3 d 10 , 4 s 1, , 45. According to Pauli exclusion principle, in any, orbital, maximum two electrons can exist,, having opposite spin., , 46. According to Hund’s rule, electron pairing in p,, d and f-orbitals cannot occur until each orbital of a, given subshell is singly occupied.’, Hence, 1s 2 , 2s 2 2 p x2 2 p1y 2 p z0 electronic, configuration is not possible for nitrogen., , 47. According to Hund’s rule “the pairing of, , electrons in the orbitals of a particular subshell, does not take place until, all the orbitals of a, sub-shell are singly occupied.

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43, , Structure of Atom, , Moreover, the single orbitals must have the, electrons with parallel spin. i.e., 1s2, , 2s2, , 1s2, , 2s2, , h, 4π, [Q ∆E = F ⋅ ∆x , E = energy], Thus, statement (b) is correct., (c) The half and fully-filled orbitals have greater, stability due to greater exchange energy,, greater symmetry and more balanced, arrangement. Thus, statement (c) is correct., (d) For a single electronic species like H, energy, depends on value of n and does not depend, on l. Hence, energy of 2s -orbital and, 2p-orbital is equal in case of hydrogen like, species. Therefore, statement (d) is incorrect., or, , 2p 1x 2p1y 2p1z, , or, , 2p 1x 2p1y 2p1z, , 48. According to Hund’s rule, in p-orbital the, distribution of electrons are as follows :, , 3p-orbital, , 51. Most stable arrangement is, , 49. The correct match is, A → (3); B → (5); C → (1); D → (4), (A) Hund’s rule states that pairing of electrons in, the orbitals belonging to the same subshell, ( p, d or f ) does not take place until each, orbital belonging to that subshell has got one, electron each, i.e. it is singly occupied., (B) Aufbau principle states that in the ground, state of the atoms, the orbitals are filled in, order of their increasing energies., (C) According to Pauli exclusion principle, no, two electrons in an atom can have the same, set of four quantum numbers., (D) Heisenberg’s uncertainty principle states that, it is impossible to determine the exact, position and exact momentum of a, subatomic particle simultaneously., , 50. (a) According to de-Broglie’s equation,, h, mv, where, h = Planck’s constant., Thus, statement (a) is correct., (b) According to Heisenberg’s uncertainty, principle, the uncertainties of position ( ∆x ), and momentum ( ∆p = m∆v ) are related as,, h, h, or, ∆x . m∆v ≥, ∆x ⋅ ∆p ≥, 4π, 4π, h, ∆x ⋅ m . ∆a ⋅ ∆t ≥, 4π, wavelength ( λ ) =, , ,  ∆v, = ∆a , a = acceleration ,  ∆t, , or ∆x ⋅ F ⋅ ∆t ≥, , h, 4π, , ∆E ⋅ ∆t ≥, , [Q F = m ⋅ ∆a ], , 4s, , 3d, , Because,, (i) According to Aufbau principle, electrons, enter in the subshell of an atom in the, increasing order of energy., (ii) According to Hund’s rule of maximum, multiplicity, pairing of electrons in the, orbitals of a subshell does not take place, until all orbitals of a subshell are singly, occupied., =36 [Kr]5s 1. Its valence electronic, configuration is 5s 1 ., So, quantum numbers for 5s 1 electron are n = 5, (for, s-orbital), l =0, (As, m = – l to +l ), m =0, s = +1 / 2, 53. Electronic configuration of Fe2+ is [ Ar ] 3d 6 4s 0 ., ∴Number of d -electrons = 6, Mg − 1s 2 2s 2 2 p 6 3s 2 (s -electrons are 6), , 52., , 37 Rb, , It matches with the d -electrons in Fe2+, Cl −1s 2 2s 2 2 p 6 3s 2 3 p 5 (p-electrons are 11), It does not match with the d -electrons in Fe2+, (d -electrons are 6), Fe − [ Ar ] 3d 6 4s 2, It matches with the d -electrons in Fe2+, (p-electrons are 6), Ne −1s 2 2s 2 2 p 6, It matches with the d -electrons in Fe2+ ., Hence, Cl has 11 electrons in p-subshell which, does not match with number of d -electrons in, Fe2+ .

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44, 54., , CBSE New Pattern ~ Chemistry XI (Term-I), , (a) Fe3+ = [Ar] 3d 5, (b) Co3+ = [Ar], , 3d 6, , (c) Ni2+ = [Ar] 3d 8, (d) Cu2+ = [Ar] 3d 9, Hence, the maximum unpaired electrons is, present in Fe3+ion, which is 5., , 55. The correct match is :, A → (4); B → (3); C → (1); D → (2), (A) Cr ( Z = 24) = 1s 2 2 s 2 2 p 6 3s 2 3 p 6 3 d 5 4s 1, = [Ar]3 d 5 4s 1, (B) Fe2+ ( Z = 26) = 1s 2 2 s 2 2 p 6 3s 2 3 p 6 3 d 6 4s 0, = [Ar]3d 6 4s 0, (C) Ni 2+ ( Z = 28) = 1s 2 2 s 2 2 p 6 3s 2 3 p 6 3 d 8 4s 0, = [Ar]3 d 8 4s 0, (D) Cu ( Z = 29) = 1s 2 2 s 2 2 p 6 3s 2 3 p 6 3 d 10 4s 1, = [Ar]3 d 10 4s 1, 56. The correct match is :, A → (3); B → (4); C → (1); D → (5), (A) Cu ( Z = 29) : 1s 2 2s 2 2 p 6 3s 2 3 p 6 3 d 10 4s 1, (B) Cu 2+ ( Z = 29) : 1s 2 2s 2 2 p 6 3s 2 3 p 6 3 d 9, (C) Zn 2+ ( Z = 30) : 1s 2 2s 2 2 p 6 3s 2 3 p 6 3 d 10, (D) Cr 3+ ( Z = 24) : 1s 2 2s 2 2 p 6 3s 2 3 p 6 3 d 3, , 57. According to de-Broglie equation,, , λ ( wavelength ) = h /mv ., Because of inverse relation of wavelength and, mass, the wavelength associated with ordinary, objects (microscopic object) are so short, (because of their large mass), that their wave, properties cannot be detected., Both Assertion and Reason are correct and, Reason is the correct explanation of Assertion., , n 2h 2, n2, =, × 0.529 Å, 2, 4 πe mZ Z, With increase of n, rn also increases, indicating a, greater separation between the orbit and the, nucleus., Both Assertion and Reason are correct and, Reason is the correct explanation of Assertion., , 58. Radius, r =, , 59. Energies of the orbitals in hydrogen or hydrogen, , like species depend only on the quantum number, ‘n’. Energies of the orbitals in multielectron atoms, depend on quantum numbers ‘n’ and ‘ l ’ , i.e., more than 1 quantum number., Both Assertion and Reason are correct, but, Reason is not the correct explanation of the, Assertion., , 60. Assertion is correct and Reason is incorrect., According to Heisenberg’s uncertainty principle,, the exact position and exact momentum of an, electron cannot be determined simultaneously., Thus, the path of electron in an atom is not, clearly defined., , 61. Both Assertion and Reason are correct but, , Reason is not the correct explanation of, Assertion., With the help of principal quantum number (n), the average distance of the electron can be, determined., Distance of the electron from the nucleus, energy, and size of orbital increase as the value of, principle quantum number ‘n’ increases., Shell = K L M N ………, n = 1 2 3 4 ………, , 62. Assertion is correct, but Reason is incorrect, , statement., Spin angular momentum of the electron, a vector, quantity, can have two orientations (represented, by + and − sign) relative to a chosen axis., These two orientations are distinguished by the, spin quantum numberm s equals to +1/ 2 or −1/ 2., These are called the two spin states of the, electron and are normally represented by the two, arrows ↑ (spin up) and (spin down) ↓, respectively., , 63. Magnetic quantum number m l can have values, , m l = −1, − ( l − 1), ... 0, . ..( l − 1), l (total 2l + 1, values). It gives the information about the, number of orbitals and its orientation., Thus, Assertion is correct, but Reason is incorrect., , 64. For a given principal quantum number, s, p, d, f,, , … subshells, all have different energies because, mutual repulsion exists among the electrons in a, multi-electron atoms., Thus, both Assertion and Reason are correct and, Reason is the correct explanation of Assertion.

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45, , Structure of Atom, , 65. Half-filled and fully-filled degenerate set of, , orbitals acquire extra stability because of the, symmetry., Thus, both Assertion and Reason are correct and, Reason is the correct explanation of Assertion., , 66. s-orbital electron will be more tightly bound to, , the nucleus than p-orbital electron because, s -orbital electrons are less shield as compared to, p-orbital electrons., Z eff experienced by the electron decreases with, increase of azimuthal quantum number ( l )., Thus, both Assertion and Reason are correct but, Reason is not the correct explanation of Assertion., , 67. Both Assertion and Reason are correct statements, and Reason is the correct explanation of the, Assertion., For a given orbital only that boundary surface, diagram of constant probability density | ψ |2 is, taken to be a good representation of the shape of, the orbital which encloses a region or volume in, which the probability of finding the electron is, very high, i.e. 90 %., , 68. According to (n + l ) rule, energy level also, depends on azimuthal quantum number., Only in unielectronic species which are very few, (H and related atoms), ‘l ’ is immaterial for, deciding energy level., Both Assertion and Reason are correct, statements, but Reason is not the correct, explanation of the Assertion., 69. Atomic orbital is designated by n, l and m l while, state of an electron in an atom is specified by four, quantum numbersn, l , m l and m s ., Thus, Assertion is incorrect but Reason is, correct statement., , 70. In case isoelectronic ions, i.e. ions having the, , same number of electrons and different nuclear, charge, the size decreases with increase in, atomic number., Ion, , At. no., , No. of, electrons, , Na +, , Ionic radii, , 11, , 10, , 0.95 Å, , 2+, , 12, , 10, , 0.65 Å, , Al 3 +, , 13, , 10, , 0.50 Å, , Mg, , Thus, Assertion is incorrect but Reason is, correct statement., , 71. ( 2l + 1) gives the value for number of orbitals in, that subshell not the electrons., ‘l ’ can have only values of 0, 1, 2, 3 for ‘s ’, ‘p ’, ‘, d ’ and ‘f ’ not only other value., Thus, Assertion is incorrect but Reason is correct, statement., , 72. (i) Assertion is incorrect but Reason is correct., For multielectron atom, such as sodium,, energy of electron is determined by both n, as well as l., (ii) Assertion and Reason both are correct, statement and Reason is correct explanation, of Assertion., (iii) Assertion is correct but Reason is incorrect., Its correct is :, Orbit represents the circular path along which, the electron moves but orbital represents the, three dimensional space around the nucleus, within which the probability of finding the, electron is maximum., (iv) Assertion and Reason both are correct and, Reason is the correct explanation of, Assertion., Or, Assertion is correct but Reason is incorrect., The value of spin quantum number can be, 1, 1, either + or − and it is independent of, 2, 2, value of principal quantum number., , 73. (i) Given, E 0 = 4.2 eV = 4.2 × 1.60 × 10 −19 J, = 6.72 × 10 −19 J, We know that, (c) = 3 × 10 8 m/s [1Å = 10 −10 m], hc, Q, E = hν =, λ, ∴, , E =, , 6.63 × 10 −34 Js × 3 × 10 8 ms −1, 2000 × 10 −10 m, , = 9.94 × 10 −19 J, ∴ Kinetic energy of electron emitted, = ( 9.94 − 6.72 ) × 10 −19 J, = 3.22 × 10 −19 J, (ii) Number of electron ejected depends upon, intensity and brightness of light., (iii) Kinetic energy of ejected electron is, proportional to frequency of, electromagnetic radiation.

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46, , CBSE New Pattern ~ Chemistry XI (Term-I), , Ze 2, …(i), 8πε 0r, Ze 2, …(ii), Potential energy in an orbit = −, 4πε 0 r, , (iv) Kinetic energy in an orbit =, , On comparing Eqs. (i) and (ii), we get, 1, KE = PE, 2, Hence, the magnitude of kinetic energy in, an orbit is equal to half of the potential, energy., Or, Option (d) is the correct answer as all these, three options are incorrect., , 74. (i) Energies of the orbitals in hydrogen or, , hydrogen like species depend only on the, quantum number ‘n’. Energies of the, orbitals in multielectron atoms depend on, quantum numbers ‘n’ and ‘ l ’ , i.e. more, than 1 quantum number., Thus, both Assertion and Reason are, correct but Reason is not the correct, explanation of Assertion., , (ii) Thus, both Assertion and Reason are, correct and Reason is the correct, explanation of Assertion., (iii) For a given principal quantum number, s, p,, d, f, … subshells, all have different energies, because mutual repulsion exists among the, electrons in a multi-electron atoms., Thus, both Assertion and Reason are, correct and Reason is the correct, explanation of Assertion., (iv) Both Assertion and Reason statements, are correct and Reason is the correct, explanation of Assertion., Or, Half-filled and fully-filled degenerate set of, orbitals acquire extra stability because of, the symmetry., Thus, both Assertion and Reason are, correct and Reason is the correct, explanation of Assertion., h, 75. (i) ∆x ⋅ ∆P ≥, 2π, when, ∆x = 0 ∆P becomes infinite., (ii) Uncertainty in velocity, 0.001, = 3 × 10 4 ×, = 0.3 cm/s, 100, , ∆x ⋅ ∆v =, , 6.6 × 10 −27, h, =, 4πm 4 × 3.14 × 9.1 × 10 −28, , 6.6 × 10 −27, 4 × 3.14 × 9.1 × 10 −28 × 0.3, = 1.92 cm, (iii) The uncertainty in the position of, electron would be less than λ., (iv) Given, ∆x = ∆P or ∆x = m ⋅ ∆v, From Heisenberg’s uncertainty principle,, h, h, ∆ x ⋅m ⋅ ∆ v =, ⇒ m ⋅ ∆v ⋅m∆v =, 4π, 4π, h, ( ∆v ) 2 =, 4 πm 2, 1 h, ∆v =, 2m π, ∆x =, , =, , 1, 2 × 9.1 × 10 −31, , 6.63 × 10 −34, 3.14, , = 7.98 × 1012 ms −1 ≈ 8 × 1012 ms −1, Or, h, ∆x ⋅ ∆P =, 4π, ∆x =, , 6.63 × 10 −34, 5.27 × 10 −35, =, 4 × 3.14 × 10 −5, 1 × 10 −5, , = 5.27 × 10 −30 m, , 76. (i) Number of orbitals = n 2 = ( 3) 2 = 9 orbitals., 9 orbitals are 3s , 3p x , 3p y , 3p z , 3d x 2 − y 2 ,, 3d z 2 , 3d xy , 3d yz and 3d x ., , (ii) n = 2 , l = 1 ⇒ 2p, (iii) Orbital can have a maximum of two, electrons which must be of opposite spin., (iv) The following shape is of p x -orbital., y, , x, z, , Or, 1s, , 2s, , 2p, , Number of unpaired electron = 1, , 1, 2, because spin of paired electrons is zero., , ∴Total spin of unpaired electrons =

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03, Classification of Elements, and Periodicity in Properties, Quick Revision, 1. Modern Periodic Law and the Present, Form of the Periodic Table, ●, , ●, , ●, , ●, , ●, , ●, , ●, , In 1913, the English physicist, Henry Moseley, modified the Mendeleev’s periodic table which, is known as modern periodic law., This law can be stated as, “the physical and, chemical properties of the elements are periodic, functions of their atomic numbers.”, The modern periodic law is essentially the, consequence of the periodic variation in, electronic configurations which determine the, physical and chemical properties of elements, and their compounds., Long form of periodic table consist of, horizontal rows called periods and the vertical, columns called groups or families. It contains, 7 periods and 18 groups., Elements having similar electronic configuration, in their atoms are arranged in groups. The, period number corresponds to the highest, principal quantum number (n ) of the elements., The first period contain 2 elements. The, subsequent periods consists of 8, 8, 18, 18 and, 32 elements respectively. The seventh period, is incomplete and like the sixth period would, have theoretical maximum of 32 elements., In this form of periodic table, 14 elements of, both sixth and seventh periods (lanthanoids, and actinoids respectively) are placed in, separate panels at the bottom., , 2. Nomenclature of Elements with, Atomic Numbers > 100, The names are derived by using roots for the, three digits in the atomic number of the, elements followed by adding ‘–ium’ at the end., The roots for the numbers are as, Digit, , Name, , Abbreviation, , 0, , nil, , n, , 1, , un, , u, , 2, , bi, , b, , 3, , tri, , t, , 4, , quad, , q, , 5, , pent, , p, , 6, , hex, , h, , 7, , hept, , s, , 8, , oct, , o, , 9, , enn, , e, , 3. Important Properties of s, p, d and f-block, Elements, (i) s-block elements General electronic, configuration of elements of this block is, ns 1 2 (where, n = 1, 2, K)., ● Group I (alkali metals) and group II, (alkaline earth metals) elements belongs to, this block., ● These are reactive metals with low, ionisation enthalpies., -

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CBSE New Pattern ~ Chemistry XI (Term-I), , 48, Their metallic and reactive character, increases as on going down in a group., ● The compounds of s-block elements with, exceptions of those of lithium and beryllium, are predominantly ionic., (ii) p-block elements The general configuration, for these elements can be written as ns 2 np 1- 6, (where, n = 1 , 2, K )., ● Group 13th to 18th excluding He, belongs to, this block. Their last electron enters in p-block., ● s and p-block elements are known as, representative elements or main group, elements., ● Group 15 members are called pnicogens,, group 16 members are called chalcogens, and group 17 members are called halogens., (iii) d-block elements, ● Elements of 3rd to 12th in periodic table, belongs to d -block., ● General electronic configuration of elements, of this block is (n - 1) d 1 10 ns ( 0 2) where,, (n = 4-7)., ●, , -, , -, , ●, , They are all metals. They mostly formed, coloured ions, exhibit variable valence,, paramagnetism and also used as catalyst., , ●, , Zn, Cd and Hg have the electronic, configuration (n - 1) d 10ns 2, do not show most, of the properties of transition elements., , They are also called as transition elements, as they form a bridge between the chemically, active metals of s-block elements and the less, active elements of group 13 and 14., (iv) f -block or inner transition elements, ● Last electron enters in f -orbital., ●, , ●, , General configuration is, ns 2 (n - 1) d 0, , ●, , -, , 10, , (n - 2 ) f, , 1 - 14, , where, n = 6-7, Two series 4 f (lanthanoids) and, 5 f (actinoids) belong to this block., , 4. Valence Electrons, The electrons present in the outermost shell are, called valence electrons., (i) For s-block elements, group number is equal, to the number of valence electrons., , (ii) For p-block elements, group number is equal, to 10 + number of valence electrons., (iii) For d-block elements, group number is equal, to number of e – in (n – 1 )d subshell + number of, electrons in valence shell., , 5. Periodic Trends in Properties of Elements, (i) Atomic radius The atomic radius is defined as, “the distance from the centre of the nucleus to, the outermost shell of electrons.’’, Depending upon the nature of combining, atoms, atomic radius can be of following types:, (a) Covalent radius, 1, rcovalent =, 2, [Internuclear distance between two covalently, bonded atoms], (b) Metallic radius The one-half of the, internuclear distance separating the metal, cores in the metallic crystal., (c) van der Waals’ radius One-half of the, distance between the nuclei of two identical, non-bonded isolated atoms., Variation of atomic radii in the periodic table are :, ● Variation along a period The atomic radii, generally decreases from left to right along a, period. This is because, within a period, the, electron is added to the same valence shell, due, to which the effective nuclear charge increases, and hence, the outer electron is held more tightly, by the nucleus which results in decreased radii., ● Variation in a group The atomic radii, generally increases on moving down the group., This is because the electrons are added in the, successive shells, i.e. principal quantum number, (n ) increases and thus, the electrons of, inner-shells shield the electrons of principal, energy level from the pull of the nucleus, due to, which the distance between the valence electron, and the nucleus increases and hence, the atomic, radii increases., (ii) Ionic radius The ionic radius is defined as the, effective distance from the centre of the nucleus, of an ion upto which it has an influence in the, ionic bond., Internuclear distance = radii of cation, + radii of anion

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Classification of Elements and Periodicity in Properties, , (a) Variation of ionic radii It shows same, trend in the periodic table as shown by the, atomic radii, i.e. decreases across a period, and increases while moving down the group., (b) Isoelectronic species and their radii These, are neutral or ionic species which have the, same number of electrons but different nuclear, charges., The ionic radii of isoelectronic ions, increase with the decrease in the magnitude, of the nuclear charge., e.g. Al 3+ < Mg 2+ < Na + < Ne., (iii)Ionisation enthalpy (IE) The minimum, amount of energy required to remove an, electron from an isolated gaseous atom (X ) in, its ground state is called the ionisation, enthalpy., X ( g ) ¾® X + ( g ) + e - ; D i H = + ve, or X ( g ) + D i H ¾® X + ( g ) + e ●, , ●, , ●, , Variation of IE in periodic table Generally,, on moving left to right in period, IE increases, and on moving down the group, it decreases., Half-filled orbitals and fully-filled orbitals are, more stable thus, have high IE., Various factors with which IE varies are, (a) Atomic size : varies inversely, (b) Screening effect : varies inversely, (c) Nuclear charge : varies directly, Exceptions IE of elements of 2nd group is, higher than the corresponding elements of, thirteen group because of fully-filled, configuration (ns 2 )., Similarly, IE of elements of group 15 is higher, than corresponding elements of group 16, because of half-filled configuration (np 3 )., , (iv) Electron gain enthalpy ( De g H ) When an, electron is added to a gaseous atom in its, ground state to convert it into a negative, ion, the enthalpy change accompanying, the process is called the electron gain, enthalpy (De g H )., X ( g ) + e - ¾® X - ( g ), Variation in periodic table Generally, on, moving left to right in period, electron gain, enthalpy increases and on moving down the, group, it decreases., , 49, , Various factors affecting electron gain enthalpy, are, ● Atomic size : varies inversely, ● Nuclear charge : varies directly, ● Configuration Half-filled orbitals and, fully-filled orbitals are stable form, therefore, electron gain enthalpy will be low or even, sometimes energy is required rather than, getting released., ● Exception Cl > F > S > O F and O-atom, have small size and high charge density, but, have lower electron gain enthalpy., ● Chlorine has highest electron affinity but, oxidising power of fluorine is larger than, chlorine., (v) Electronegativity It is the ability of an atom of, a compound to attract the shared pair of, electrons towards itself., 6. Variation of Electronegativity in the, Periodic Table, The variation of electronegativity in the periodic, table is shown as follows :, (i) Variation along a period On moving from left, to right across a period, as the effective nuclear, charge increases and size decreases, the value of, electronegativity increases due to increase in the, attraction between the shared pair of electrons, and the nucleus., (ii) Variation along a group On moving down a, group, as the atomic size increases, the force of, attraction between the shared pair of electron, and the nucleus decreases and, hence the, electronegativity decreases., Various factors which affect the magnitude of, electronegativity are as follows :, (i) Atomic radius As the atomic radius of the, elements increases, the electronegativity value, decreases, i.e., 1, Electronegativity µ, Atomic radius, (ii) Effective nuclear charge The electronegativity, value increases as the effective nuclear charge, on the atomic nucleus increase. i.e., Electronegativity µ Effective nuclear charge (Z eff ), (iii) Oxidation state The electronegativity, increases as the oxidation state (i.e. the number of, positive charge) of the atom increases.

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CBSE New Pattern ~ Chemistry XI (Term-I), , 50, (iv)s-character If the s-character in the, hybridisation state of the central atom, increases, electronegativity also increases., Different scales of calculating electronegativity:, (a) Pauling scale, (b) Mulliken-Jaffe scale, , 7. Valency or Oxidation States, The valency is the most characteristic property of, the elements and can be understood in terms of, their electronic configuration. It is the combining, power of an element., , Transition elements and actinoids also exhibit, variable valency., Variation of Valency in the Periodic Table, ● Along a period from left to right, valency, increases gradually from 1 to 4 with respect to, hydrogen and then, decreases to 1 with respect, to hydrogen., ● Valency of noble gases is taken as zero.In a, group, all the elements have same valency, because the number of valence electrons are, same in them., , Objective Questions, Multiple Choice Questions, 1. The long form of periodic table based, on, , 2. Elements having similar outer shell, electronic configuration in their atoms, are arranged in, (b) vertical columns, (d) All of these, , 3. The period number in the long form of, the periodic table is equal to, , (NCERT Exemplar), , (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, , 4. Successive filling of 3s and 3p-orbitals, give rise to the third period. The, number of elements present in this, period are, (a) 2, , (b) 4, , elements of 7th period in the periodic, table are termed as respectively, (a) lanthanoids, actinoids, (b) actinoids, lanthanoids, (c) chalcogens, halogens, (d) actinoids, halogens, , (a) atomic volume, (b) atomic mass, (c) electronic configuration, (d) effective nuclear charge, , (a) groups, (c) families, , 5. 14 elements of 6th period and 14, , (c) 6, , (d) 8, , 6. The elements in which electrons are, progressively filled in 4 f -orbital are, called, (NCERT Exemplar), (a) actinoids, (b) transition elements, (c) lanthanoids, (d) halogens, , 7. Predict the position of an element, having the electronic configuration, 1s 2 2s 2 2 p 6 3s 2 3 p 6 3d 5 4s 1 ., (a) Period 4, group 6, (c) Period 3, group 1, , (b) Period 6, group 4, (d) Period 4, group 5, , 8.The group of elements in which the, differentiating electron enters in the, anti-penultimate shell of atoms are called, (a) f-block elements, (c) s-block elements, , (b) p-block elements, (d) d-block elements, , 9. Cu ( Z = 29 ) is element of, (a) s-block, (c) d-block, , (b) p-block, (d) f-block

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51, , Classification of Elements and Periodicity in Properties, , 10. An atom has electronic configuration, 2, , 2, , 6, , 2, , 6, , 3, , 2, , 1s , 2s , 2 p , 3s , 3 p , 3d , 4s ,, you will place it in, (a) fifth group, (c) second group, , (b) fifteenth group, (d) third group, , 11. In the periodic table, metals usually, used as catalysts belong to, (a) f-block (b) d-block (c) p-block (d) s-block, , 12. The elements having characteristics of, both metals and non-metals can be, termed as, (a), (b), (c), (d), , semi-metals, metalloids, Either [(a) or (b)], amphoteric elements, , 13. The electronic configuration of four, elements are, I. [Xe] 6s 1, III. [Ar] 4s 2, 4 p 5, , II. [Xe] 4 f 14, 5d 1, 6s 2, IV. [Ar] 3d 7, 4s 2 ., , Which one of the following statements, about these elements is not correct ?, (a), (b), (c), (d), , I is a strong reducing agent, II is a d-block element, III has high electron affinity, IV shows variable oxidation state, , 14. The electronic configuration of, gadolinium (Atomic number 64) is … ., (a) [Xe] 4 f 3 5d 5 6 s 2, (c) [Xe] 4 f 7 5d 1 6 s 2, , (b) [Xe] 4 f 7 5d 2 6 s 1, (d) [Xe] 4 f 8 5d 6 6 s 2, , 15. Which of the following is not an, actinoid ?, (a) Curium (Z = 96), (c) Uranium (Z = 92), , (NCERT Exemplar), , (b) Californium (Z = 98 ), (d) Terbium (Z = 65), , 16. Match the Column I with Column II, and select the correct answer using, given codes., Column I, (Number of periods), , Column II, (Number of elements), , A. First period, , 1., , 14, , B. Third period, , 2., , 2, , C. Lanthanoids, , 3., , 8, , D. Actinoids, , 4., , 4, , Codes, A B C D, (a) 2 4 1 3, (c) 4 2 1 3, , A B C D, (b) 2 3 1 1, (d) 4 2 3 3, , 17. The name of the element with atomic, number 105 is …… ., (a) kurchatovium, (c) nobelium, , (b) dubnium, (d) holmium, , 18. An element with atomic number 112, has been made recently. It should be, …… ., (a) an actinide, (b) a transition metal, (c) a noble gas, (d) a lanthanide, , 19. Match the Column I with Column II, and choose the correct option using the, codes given below :, Column I, (Elements), , Column II, (IUPAC name), , A., , 109, , 1., , Ununbium, , B., , 112, , 2., , Unnilennium, , C., , 115, , 3., , Ununpentium, , D., , 118, , 4., , Ununoctium, , Codes, A B C D, (a) 1 2 3 4, (c) 1 2 4 3, , A B C D, (b) 2 1 3 4, (d) 2 1 4 3, , 20. If the bond distance in sodium, molecule (Na) is 3.72 Å, then the radius, of sodium is …… ., (a) 3.72Å, , (b) 1.86 Å, , (c) 7.44 Å, , (d) 1.24 Å, , 21. Which of the following orders of ionic, radius is correctly represented ?, (a) H- > H > H+, (c) F - > O 2 - > Na+, , (b) Na+ > F - > O 2 (d) Al3 + > Mg 2 + > N3 -, , 22. If the intermolecular distance between, two adjacent copper atoms in solid, copper is 256 pm, then the metallic, radius of copper is …… ., (a) 128 pm, (c) 74 pm, , (b) 12.87 Å, (d) 74 Å

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CBSE New Pattern ~ Chemistry XI (Term-I), , 52, 23. Match the correct atomic radius with, the element and choose the correct, option using the codes given below., Column I, (Element), , Column II, (Atomic radius (pm)), , A., , Be, , 1., , 74, , B., , C, , 2., , 88, , C., , O, , 3., , 111, , D., , B, , 4., , 77, , E., , N, , 5., , 66, , Codes, A B, (a) 3 5, (b) 1 4, (c) 1 2, (d) 5 2, , (NCERT Exemplar), , C, 4, 2, 3, 4, , D, 2, 5, 4, 3, , E, 1, 3, 5, 1, , of the following elements is …… ., (A) C (B ) O (C ) F (D ) Cl (E ) Br, (A) < (B) < (C) < (D) < (E), (C) < (B) < (A) < (D) < (E), (D) < (C) < (B) < (A) < (E), (B) < (C) < (D) < (A) < (E), , energy would be, (a) He, (c) N, , (b) Be, (d) F, , 28. The first ionisation enthalpies of, Na, Mg, Al and Si are in the order, (a) Na < Mg > Al < Si, (b) Na > Mg > Al > Si, (c) Na < Mg < Al < Si, (d) Na > Mg > Al < Si, , X + ( g ) ¾¾® X, , 2+, , ( g ) + e - represents, , (a) first ionisation enthalpy, (b) second ionisation enthalpy, (c) electronegative character, (d) electron gain enthalpy, , 30. Amongst the following, select the, , 25. Consider the isoelectronic species,, , Na + , Mg 2+ , F – and O 2– . The correct, order of increasing length of their radii, is …… ., (NCERT Exemplar), (a), (b), (c), (d), , 27. The element having highest ionisation, , 29. The energy required in the equation,, , 24. The increasing order of the atomic radii, (a), (b), (c), (d), , (c) X = alkali metal, Y = Noble gas,, Z = alkaline earth metal, (d) X = alkaline earth metal, Y = Noble gas,, Z = alkali metal, , F – < O 2 – < Mg 2 + < Na+, Mg 2 + < Na+ < F – < O 2 –, O 2 - < F – < Na+ < Mg 2 +, O 2 – < F – < Mg 2 + < Na+, , element having highest ionisation, enthalpy., (a) Sodium, (c) Beryllium, , (b) Potassium, (d) Magnesium, , 31. The correct order of electron affinities, of N, O, S and Cl is, , 26. Few elements with first ionisation, enthalpies are given in the table., Identify these elements., Elements, , IE 1 (kJ/mol), , X, , 520, , Y, , 2080, , Z, , 899, , (a) X = Noble gas, Y = alkali metal,, Z = alkaline earth metal, (b) X = Noble gas, Y = alkaline earth metal,, Z = alkali metal, , (a) N < O < S < Cl, (b) O < N < Cl < S, (c) O » Cl < N » S, (d) O < S <Cl < N, , 32. Which of the following pair contain will, have the most negative and least, negative electron gain enthalpy, respectively, P, S, Cl and F ?, (a) P and Cl, (c) Cl and F, , (b) S and Cl, (d) Cl and P, , 33. Electronic configuration of some, elements is given in Column I and their, electron gain enthalpies are given in, Column II. Match the electronic, configuration with electron gain enthalpy.

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53, , Classification of Elements and Periodicity in Properties, , Column II, (Electron gain, enthalpy/ kJ mol –1), , Column I, (Electronic, configuration), A., , 1s 2 2s 2 2p 6, , - 53, , B., , 1s 2 2s 2 2p 6 3s 1, , - 328, , C., , 1s 2 2s 2 2p 5, , - 141, , D., , 1s 2 2s 2 2p 4, , + 48, (NCERT Exemplar), , Codes, A B C D, (a) 1 2 3 4, (c) 3 2 1 4, , A B C D, (b) 4 1 2 3, (d) 1 2 4 3, , 34. The increasing order of the density of, , Electronegativity of hydrogen = 2 .1, (a) 0.498, (c) 2.134, , 39. The increasing order of electronegativity, of C, N, P and Si element will be …… ., (a) C, N, Si, P, (b) N, Si, C, P, (c) Si, P, C, N, (d) P, Si, N, C, , 40. Match Column I with Column II and, select the correct answer using given, codes., , alkali metals is …… ., (a) Li < K < Na < Rb < Cs, (c) Cs < Rb <Na < K < Li, (e) Li < Na < Rb < K < Cs, , (b) Li < Na < K < Rb <Cs, (d) Cs < Rb < K < Na < Li, , 35. The ability of an atom in a chemical, compound to attract shared electron is, termed as …… ., (a) electron affinity, (c) atomic attraction, , (b) ionisation enthalpy, (d) electronegativity, , 36. The correct order of decreasing, electronegativity values among the, elements, I-beryllium, II-oxygen, III-nitrogen and, IV-magnesium, is, (a), (b), (c), (d), , II > III > I > IV, III > IV > II > I, I > II > III > IV, I > II > IV > III, , (b) 0.598, (d) 2.598, , Column I, (Atoms), , Column II, (Properties), , A., , He, , 1. High electronegativity, , B., , F, , 2. Most electropositive, , C., , Rb, , 3. Strongest reducing agent, , D., , Li, , 4., , Codes, A B, (a) 4 2, (b) 1 4, (c) 4 1, (d) 4 1, , C, 3, 2, 3, 2, , Highest ionisation energy, , D, 1, 3, 2, 3, , 41. Match the element (in Column I) with, its unique properties (in Column II)., Column I, A., , 37. The correct order of electronegativity, , F, , Column II, 1. Maximum ionisation energy, , B., , Cl, , 2. Maximum electronegativity, , C., , Fe, , 3. Maximum electron affinity, , of N, O and F is …… ., , D., , He, , 4. Recently named by IUPAC, , (a) N > O > F, (c) O > N > F, , E., , Ds, , 5. Variable valence, , (b) O > F > N, (d) F > O > N, , 38. What will be the electronegativity of, carbon at Pauling scale?, Given that E H ¾ H = 104.2 kcal mol -1 ,, E C ¾ C = 83.1 kcal mol -1, E C ¾ H = 988, . kcal mol -1, , Codes, A, (a) 5, (b) 3, (c) 2, (d) 3, , B, 4, 4, 3, 1, , C D, 1 2, 2 1, 5 1, 4 2, , E, 3, 5, 4, 5

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CBSE New Pattern ~ Chemistry XI (Term-I), , 54, Assertion-Reasoning MCQs, Directions In the following questions, (Q.No. 42-55) a statement of Assertion, followed by a statement of Reason is, given. Choose the correct answer out of, the following choices., (a) Both Assertion and Reason are correct, statements and Reason is the correct, explanation of the Assertion., (b) Both Assertion and Reason are correct, statements, but Reason is not the correct, explanation of the Assertion., (c) Assertion is correct, but Reason is incorrect, statement., (d) Assertion is incorrect but Reason is correct, statement., , 42. Assertion Noble gases are highly, reactive., Reason Noble gases have stable outer, electronic configuration., , 43. Assertion Element in the same vertical, column have similar properties., Reason Elements have periodic, dependence upon the atomic number., , 44. Assertion The elements having, , 1s 2 , 2s 2 , 2 p 6, 3s 2 and 1s 2 , 2s 2 electronic, configuration belong to same period., Reason Both have same outermost, electronic configuration., , 45. Assertion Atomic number of the, element copernicium is 112., Reason IUPAC name of this element, is ununbium in which un-and bi-are, used for 1 and 2 respectively in Latin, words., , 46. Assertion The atomic radii of the, elements of the oxygen family are, smaller than the atomic radii of the, corresponding elements of the nitrogen, family., , Reason The members of the oxygen, family are more electronegative and, thus, have lower values of nuclear, charge than those of the nitrogen, family., , 47. Assertion Isoelectronic species have, same radii., Reason They contain different number, of electrons., , 48. Assertion The atomic and ionic radii, generally decrease towards right in a, period., Reason The ionisation enthalpy, decreases on moving towards left in a, period., , 49. Assertion Boron has a smaller first, ionisation enthalpy than beryllium., Reason The penetration of 2s electron, to the nucleus is more than the, 2p electron. Hence, 2p electron is more, shielded by the inner core of electrons, than the 2s electrons . (NCERT Exemplar), , 50. Assertion Ionisation enthalpy is the, energy released to remove an electron, from an isolated gaseous atom in its, ground state., Reason Element which has a tendency, to lose the electron to attain the stable, configuration., , 51. Assertion Generally, ionisation, enthalpy increases from left to right in a, period., Reason When successive electrons are, added to the orbitals in the same, principal quantum level, the shielding, effect of inner core of electrons does, not increase very much to compensate, for the increased attraction of the, electron to the nucleus.

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55, , Classification of Elements and Periodicity in Properties, , 52. Assertion Alkali metals have least value, of ionisation energy within a period., Reason They precedes alkaline earth, metals in periodic table., , 53. Assertion Electron gain enthalpy, becomes less negative as we go down a, group., Reason Size of the atom increases on, going down the group and the added, electron would be farther from the, nucleus., , 54. Assertion Cesium and fluorine both, reacts violently., Reason Cesium is most electropositive, and fluorine is most electronegative., , 55. Assertion Fluorine has a less negative, electron affinity than chlorine., Reason There is relatively greater, effectiveness of 2p electrons in the small, fluorine atom to repel the additional, electron entering the atom than to, 3p electrons in the larger Cl atom., , Case Based MCQs, 56. Read the passage given below and, answer the following questions:, When an electron is added to a gaseous, atom in its ground state to convert it into a, negative ion, the enthalpy change, accompanying the process is called the, electron gain enthalpy ( De g H ). It is a, direct measure of the ease with which an, atom attracts an electron to form anion., X ( g ) + e - ¾® X - ( g ); DH = De g H, The most stable state of an atom is the, ground state. If an isolated gaseous atom is, in excited state, comparatively lesser, energy will be released on adding an, electron. So, electron gain enthalpies of, gaseous atoms must be determined in their, ground states. Therefore, the terms ground, , state and isolated gaseous atom has been, also included in the definition of electron, gain enthalpy. Like ionisation enthalpy,, electron gain enthalpy is measure either in, electron volts per atom or kJ per mole., The following questions (i-iv) are multiple, choice questions. Choose the most, appropriate answer :, (i) Noble gases have positive electron, gain enthalpy due to, (a) stable configuration, (b) large size, (c) high reactivity, (d) unstable configuration, , (ii) The electron gain enthalpy of O or F is, less than that of S or Cl. It is due to, (a) small size, (b) less repulsion, (c) large size, (d) high electronegativity, , (iii) The electron gain enthalpy (in kJ/mol), of fluorine, chlorine, bromine and, iodine, respectively, are, (a), (b), (c), (d), , -333, - 325, - 349 and -296, -296, - 325, - 333 and -349, -333, - 349, - 325 and -296, -349, - 333, - 325 and -296, , (iv) Why beryllium has higher ionisation, enthalpy than boron ?, (a) More penetration of s-electron, (b) More penetration of p-electron, (c) Large size, (d) Small size, , Or, , Factors affecting electron gain, enthalpy is, (a) atomic size, (b) number of electrons, (c) number of neutron, (d) None of the above, , 57. Read the passage given below and, answer the following questions :, Comprehension given below is followed by, some multiple choice questions. Each, question has one correct option. Choose the, correct option.

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CBSE New Pattern ~ Chemistry XI (Term-I), , 56, In the modern periodic table, elements are, arranged in order of increasing atomic, numbers which is related to the electronic, configuration., Depending upon the type of orbitals, receiving the last electron, the elements in, the periodic table have been divided into, four blocks, viz., s, p, d and f., The modern periodic table consists of 7, periods and 18 groups. Each period begins, with the filling of a new energy shell. In, accordance with the Aufbau principle, the, seven periods (1 to 7) have 2, 8, 8, 18, 18,, 32 and 32 elements respectively., The seventh period is still incomplete. To, avoid the periodic table being too long, the, two series of f -block elements, called, lanthanoids and actinoids are placed at the, bottom of the main body of the periodic, table., The following questions (i-iv) are multiple, choice questions. Choose the most, appropriate answer:, (i) The element with atomic number 57, belongs to, (a) s - block, (c) d - block, , (b) p - block, (d) f - block, , (ii) The last element of the p-block in 6th, period is represented by the outermost, electronic configuration., (a) 7 s 2 7 p6, (c) 4f 14 5d 10 6s 2 6p6, , (b) 5f 14 6d 10 7 s 2 7 p0, (d) 4f14 5d10 6s 2 6p4, , (iii) Which of the following elements,, whose atomic numbers are given, below, cannot be accommodated in, the present set up of the long form of, the periodic table?, (a) 107, (c) 126, , (b) 118, (d) 102, , (iv) The electronic configuration of the, element which is just above the, element with atomic number 43 in the, same group is ......... ., (a) 1s 2 2s 2 2p6 3s 2 3p6 3d 5 4s 2, (b) 1s 2 2s 2 2p6 3s 2 3p6 3d 5 4s 3 4p6, (c) 1s 2 2s 2 2p6 3s 2 3p6 3d 6 4s 2, (d) 1s 2 2s 2 2p6 3s 2 3p6 3d 7 4s 2, , Or The elements with atomic numbers 35,, 53 and 85 are all ......... ., (a) noble gases, (c) heavy metals, , (b) halogens, (d) light metals, , 58. Read the passage given below and, answer the following questions :, Moseley modified Mendeleev periodic law., He stated “physical and chemical properties, of elements are the periodic function of their, atomic numbers.” It is known as modern, periodic law and considered as the basis of, modern periodic table., When the elements were arranged in, increasing order of atomic numbers, it was, observed that the properties of elements, were repeated after certain regular intervals, 01 2, 8, 8, 18, 18 and 32. These numbers are, called magic numbers and cause of, periodicity in properties due to repetition of, similar electronic configuration. Long form, of periodic table is called Bohr’s periodic, table. There are 18 groups and seven periods, in this periodic table. The horizontal rows, are called periods., First period ( 1 H - 2 He) contains 2 elements., It is the shortest period. Second period, ( 3 Li - 10 Ne) and third period ( 11 Na - 18 Ar), both contains 8 elements each. Fourth, period ( 19 K - 36 Kr ) and fifth period, ( 37 Rb - 54 Xe) contain 18 elements each. These, are long periods., Sixth period (55 Cs - 86 Rn) consists of, 32 elements and is the longest period., Seventh period starting with 87 Fr is, incomplete and consists of 19 elements., Elements of group 1 are called alkali metals., Elements of group 2 are called alkaline earth, metals. Elements of group 16 are called, chalcogens [ore forming elements).Elements, of group 17 are called halogens [sea salt, forming]. Elements of group 18 are called, noble gases., Anomalous behaviour of the first element of, a group. The first element of a group differs, considerably from its congeners (i.e. the rest, of the elements of its group).

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Classification of Elements and Periodicity in Properties, , This is due to (i) small size (ii) high electronegativity and (iii) non-availability of d-orbitals, for bonding. Anomalous behaviour is observed, among the second row elements (i.e. Li to F )., In these questions (i-iv) a statement of Assertion, followed by a statement of Reason is given., Choose the correct answer out of the following, choices :, (a) Assertion and Reason both are correct, statements and Reason is correct explanation, for Assertion., (b) Assertion and Reason both are correct, statements but Reason is not correct, explanation for Assertion., (c) Assertion is correct statement but Reason is, incorrect statement., (d) Assertion is incorrect statement but Reason is, correct statement., , (i) Assertion Zinc is a d-block element., Reason Zinc does not form, coordination compounds., (ii) Assertion The first ionisation enthalpy, of Be is greater that that of C., Reason 2p-orbital is lower in energy, than 2s-orbital., (iii) Assertion Outermost electronic, configuration of most electronegative, element is ns 2 np 7 ., Reason Most electronegative elements, are halogen., (iv) Assertion Mn has less favourable, electron affinity than its neighbours in, either side., Reason The magnitude of electron, affinity depends on the electronic, configuration of the atom., Or, Assertion Generally, ionisation enthalpy, increases from left to right in a period., Reason When successive electrons, are added to the orbitals in the same, principal quantum level, the, shielding effect of inner core of, electrons does not increase very much to, compensate for the increased attraction, of electrons to the nucleus., , 57, , 59. Read the passage given below and answer, the following questions:, In 1913, Henry Moseley, the English, physicist performed an experiment by, bombarding high speed electrons on 38, different elements starting from aluminium, and ending with gold in vacuum and, generated X-rays. He observed that the, square root of the frequency ( n) of the X-rays, emitted by a metal is proportional to the, atomic number and not to the atomic, weight of the element of the electron,, i.e., x = a (Z - b ), [where, a and b = constants]., Thus, when n is plotted with atomic number, (Z ), a straight line is obtained, but this is not, true when n is plotted with atomic mass. He, postulated that the atomic number is a more, fundamental property of an element than its, atomic mass., Thus, he modified the Mendeleev’s periodic, law and gave modern periodic law, which, states that., “The physical and chemical properties of the, elements are the periodic function of their, atomic numbers,” i.e. when elements are, arranged in increasing order of atomic, numbers, the elements having similar, properties are repeated after certain regular, intervals., In these questions (i -iv), a statement of, Assertion followed by a statement of Reason, is given. Choose the correct answer out of, the following choices., (a) Assertion and Reason both are correct, statements and Reason is correct explanation, for Assertion., (b) Assertion and Reason both are correct, statements but Reason is not correct, explanation for Assertion., (c) Assertion is correct statement but Reason is, incorrect statement., (d) Assertion is incorrect statement but Reason, is correct statement., , (i) Assertion Mendeleev’s arranged, elements in horizontal rows and vertical, columns.

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CBSE New Pattern ~ Chemistry XI (Term-I), , 58, Reason Mendeleev’s ignored the order, of atomic weight thinking that the, atomic measurements might be, incorrect., (ii) Assertion Mendeleev’s left the gap, under aluminium and silicon and called, these Eka-aluminium and Eka-silicon,, respectively., Reason Dobereiner arranged, elements on the basis of increasing, atomic number., (iii) Assertion The horizontal rows in the, periodic table are called periods or, Mendeleev’s series., Reason Elements having similar outer, electronic configurations in their atoms, are arranged in groups/families., (iv) Assertion Sixth period is the longest, period in the periodic table., Reason Sixth period involves the filling, of all the orbitals of sixth energy level., Or Assertion The elements having, 1s 2 , 2s 2 , 2p 6, 3s 2 and 1s 2 , 2s 2 electronic, configuration belong to same period., Reason Both have same outermost, electronic configuration., , 60. Read the passage given below and, answer the following questions :, Elements having a place within the group 13, (i.e. group IIIA) to group 17 (i.e. group, VIIA) of the periodic table alongside the, group 18, i.e. the zero group elements, together frame the p-block of the periodic, table., In the elements of p-block, the last electron, enters the furthest p-orbital. They have 3 to 8, electrons in the peripheral shell. As we, realise that the quantity of p-orbitals is three, and, therefore, the most extreme number of, electrons that can be obliged in an, arrangement of p-orbitals is six., Consequently, there are six groups of, p-block elements in the periodic table., , In the p-block, all the three sorts of, elements are available, i.e. the metals,, non-metals and metalloids. The crisscross, line in the p-block isolates the elements that, are metals from those that are non-metals., Metals are found on the left of the line and, non-metals are those on the right. Along, the line, we discover the metalloids., Because of the nearness of a wide range of, elements, the p-block demonstrates a great, deal of variety in properties., The general valence shell electronic design, of p-block elements is ns 2 np 1 - 6 (with the, exception of He). The internal core of the, electronic arrangement may although, contrast., In these questions (i -iv), a statement of, Assertion followed by a statement of, Reason is given. Choose the correct answer, out of the following choices., (a) Assertion and Reason both are correct, statements and Reason is correct, explanation for Assertion., (b) Assertion and Reason both are correct, statements but Reason is not correct, explanation for Assertion., (c) Assertion is correct statement but Reason, is incorrect statement., (d) Assertion is incorrect statement but Reason, is correct statement., , (i) Assertion The ionisation of s-electrons, requires more energy then that for the, ionisation of p-electrons of the same, shell., Reason s-electrons are closer to, nucleus than p-electrons and hence are, more strongly attracted by nucleus., (ii) Assertion The first ionisation energy, of Al is lower than that of Mg., Reason Ionic radius of Al is smaller, than that of Mg., (iii) Assertion F is more electronegative, than Cl., Reason F has higher electron affinity, than Cl.

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59, , Classification of Elements and Periodicity in Properties, , (iv) Assertion Boron differs from Al and, other members of group 13 in a, number of properties., Reason B does not show anomalous, behaviour., , Or Assertion Carbon forms large number, of stable compounds., Reason Carbon is less electropositive as, compared to other members of group 14., , ANSWERS, Multiple Choice Questions, 1. (c), 11. (b), 21. (a), , 2. (d), 12. (c), 22. (a), , 3. (c), 13. (b), 23. (a), , 4. (d), 14. (c), 24. (b), , 5. (a), 15. (d), 25. (b), , 6. (c), 16. (b), 26. (c), , 7. (a), 17. (b), 27. (a), , 8. (a), 18. (b), 28. (a), , 9. (c), 19. (b), 29. (b), , 10. (a), 20. (b), , 31. (a), 41. (c), , 32. (d), , 33. (b), , 34. (a), , 35. (d), , 36. (a), , 37. (d), , 38. (d), , 39. (c), , 30. (c), 40. (d), , 45. (b), 55. (a), , 46. (c), , 47. (d), , 48. (c), , 49. (a), , 50. (d), , 51. (a), , Assertion-Reasoning MCQs, 42. (d), 52. (b), , 43. (a), 53. (a), , 44. (d), 54. (a), , Case Based MCQs, 56. (i)-(a), (ii)-(a), (iii)-(c), (iv)-(a) or-(a), 58. (i)-(c), (ii)-(c), (iii)-(d), (iv)-(a) or-(b), 60. (i)-(a), (ii)-(b), (iii)-(c), (iv)-(c) or-(c), , 57. (i)-(c), (ii)-(c), (iii)-(b), (iv)-(a) or-(b), 59. (i)-(b), (ii)-(c), (iii)-(b), (iv)-(c) or-(d), , EXPLANATIONS, 1. The long form of periodic table is based on, electronic configuration of elements, i.e., Bohr-Bury concept., , 2. Similar outer configuration in their atoms are, arranged in vertical columns called groups or, families., , 3. Since each period starts with the filling of, electrons in a new principal quantum number,, therefore, the period number in the long form of, the periodic table refers to the maximum, principal quantum number of any element in, the period., Period number = maximum n of any element, (where, n = principal quantum number)., , 4. Successive filling of 3s and 3p-orbitals give rise, to the third period of 8 elements from sodium to, argon., , 5. 14 elements of 6th period are called lanthanoids, and those of 7th period are termed as actinoids., , 6. The elements in which electrons are progressively, filled in 4 f -orbital are called lanthanoids., Lanthanoids consist of elements from Z = 58, (cerium) to 71 (lutetium)., , 7. n = 4 hence, element lies in 4th period., Group = ns + (n - 1) d = 1 + 5 = 6, , 8. The group of elements in which the differentiating, electron enters in the anti-penultimate shell, inner, to the penultimate shell, i.e. (n - 2) shell is called, f - block elements or inner-transition elements., , 9. Electronic configuration of Cu, = 1s 2 , 2s 2, 2 p 6 , 3s 2, 3 p 6 , 3d 10 , 4s 1 ., Hence, in this element, the last electron enters in, one of the inner d -orbital. Thus, Cu is the element, of d -block.

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CBSE New Pattern ~ Chemistry XI (Term-I), , 60, 10. An atom has electronic configuration,, 2, , 2, , 6, , 2, , 6, , 3, , 2, , 1s , 2s , 2 p , 3s , 3 p , 3d , 4s ., It is a member of d-block elements because the, last electron is filled in d-subshell as 3d 3 . Group, number of a d-block element, = ns -electron + (n - 1) d-electrons = 2 + 3 = 5., Hence, it is a member of fifth group ., , 11. Metals which are usually used as catalysts, belong to d-block of the periodic table, e.g. Ni,, Pt, etc., as they have large surface area., , 12. The elements such as silicon, germanium,, arsenic, antimony and tellurium have, characteristics of both metals and non-metals, and are termed as semi-metals or metalloids., , 5d 1, 6s 2 is not a d-block element., It is lutetium and is present in lanthanide series., So, lutetium is a f-block element., [Xe] 6s 1 is a strong reducing agent as by losing, one electron it acquires stable electronic, configuration., [Ar]4s 2 4 p 5 has high electron affinity as by gaining, one electron, it acquires stable electronic, configuration., [Ar]3d 7 4s 2 is a d -block element and it shows, variable oxidation states., , 13. [Xe] 4 f, , 14, , Note The generalised electronic configuration, of f - block elements is (n - 2) f 1 - 14 (n - 1) d 0 - 1ns 2 ., , 14. The electronic configuration of La ( Z = 57 ) is, [Xe] 5 d 1 6s 2 . Therefore, further addition of, electrons occurs in the lower energy 4 f -orbital, till it is exactly half-filled at Eu ( Z = 63)., Thus, the electronic configuration of Eu is, [Xe] 4 f 7 6s 2 ., Thereafter, addition of next electron does not, occur in the more stable exactly half-filled 4 f 7, shell but occurs in the little higher energy, 5d-orbital., Thus, the electronic configuration of, Gd ( Z = 64 ) is [Xe] 4 f 7 5 d 1 6s 2 ., 15. Elements with atomic number, Z = 90 to 103, are called actinoids., Thus, terbium ( Z = 65) is not an actinoid., Terbium belong to lanthanoids., , 16. The correct match is, A ® 2; B ® 3; C ® 1, D; ® 1., (A) First period contains 2 elements., , (B) Third period contains 8 elements., (C) and (D). 14 elements of both sixth period, [from Z = 58 to Z = 71] and seventh period, [from Z = 90 to Z = 103] are known as, lanthanoids and actinoids respectively., , 17. The element with atomic number 105 is, dubnium (Db). In IUPAC nomenclature,, it is known as unnilpentium., Db(105) = [Rn ] 86 5 f 14 6d 3 7s 2, , 18. The electronic configuration of given element, with atomic number 112 is [Rn] 86, 5 f 14 , 7s 2,6d 10 . As its outermost electron enters, in d -subshell, thus it belongs to d -block or a, transition metal., , 19. Correct match is, A ® (2); B ® (1); C ® (3); D ® (4), 109 – Unnilennium, 112 – Ununbium, 115 – Ununpentium, 118 – Ununoctium, , 20. Radius of Na-atom, =, , Bond distance of Na 372, . Å, =, = 186, . Å, 2, 2, , 21. Option (a) is correct ., (a) H- > H > H+, It is known that radius of a cation is always, smaller than that of a neutral atom because, it has fewer electrons while its nuclear, charge remains the same., Whereas, the radius of anion is always, greater than neutral atom due to decrease in, effective nuclear charge., Hence, the given order is correct., (b) The given species, in option (b) are, isoelectronic as they contain same number, of electrons., For isoelectronic species,, 1, ionic radius µ, atomic number, Ion, –– Na + F– O2 –, Atomic number –– 11 9 8, Hence, the correct order of ionic radius is, O2 - > F - > Na +, (c) Similarly, the correct option is, O2 - > F - > Na + .

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61, , Classification of Elements and Periodicity in Properties, , (d) Ion, –– Al 3+ Mg 2+ N 3Atomic number –– 13, 12, 7, Hence, the correct order is, N 3 - > Mg 2+ > Al 3+ ., , 28. Follow the following steps to solve out such, problems, , Steps, , 22. Internuclear distance between Cu-atoms in, solid copper = 256 pm., 1, Metallic radius = ´ length between two atoms, 2, 1, = ´ 256 = 128 pm, 2, , 23. All the given elements are of same period and, along a period, atomic radii decreases because, effective nuclear charge increases., Thus, the order of atomic radii is, O < N < C < B < Be or, Be = 111 pm,, O = 66 pm, C = 77 pm, B = 88 pm, N = 74 pm., , Method, , Apply, , Step I, , Write the electronic, configuration to find, position in the, periodic table, , 1, ,, 11 Na = [Ne] 3s, 2, 12 Mg = [Ne] 3s, 2, 3p 1 ,, 13 Al = [Ne] 3s, 2, Si, =, [Ne], 3, s, 3, p2, 14, , Step II, , Arrange them in the 11 12 13 14, Na Mg Al Si, order as they are, placed in the, periodic table, , Step III Follow the general, trend and also keep, in mind the, exception, , The IP increases, along a period from, left to right but IP of, Mg is higher than, that of Al due to, completely filled 3s, orbital in Mg., , Step IV On the above basis, find the order, , The order of IP is, Na < Mg > Al < Si., Thus, option (a) is, the correct., , 24. Atomic radius generally decreases as we, compare elements in a period from left to right,, \, C>O>F, but elements present in next period are larger in, size,, \, Br > Cl > C > O > F ;, CNO F, , energy required to remove the second most, loosely bounded electron. Hence, the amount of, energy required in the given equation represents, second ionisation enthalpy., X + ( g ) ¾® X 2 + ( g ) + e -, , Size decreases, , Cl, Br, Size increases, , So, the correct increasing order of the atomic, radii, C < B < A < D < E ., , 25. In case of isoelectronic species ionic radii, 1, atomic number, The ionic radii increases as the positive charge, decreases or the negative charge increases., Ion, Mg 2+ < Na + < F - < O2 µ, , Atomic number (12), , (11), , 29. Second ionisation enthalpy is defined as the, , (9), , (8), , 26. I.E. of noble gases are maximum, whereas that, of alkali metal is least. I.E. of alkaline earth metal, is higher than alkali metal but lower than, subsequent elements, thus X is an alkali, metal, Y is a noble gas and Z is an alkaline, earth metal., , 27. Helium possesses a stable configuration 1s 2., In it, the K-shell is completely filled. This is why,, the ionisation energy of helium is much greater, than another elements., , 30. Ionisation enthalpy increases on moving from, left to right in a period and decreases on moving, down in a group. Thus, order of ionisation, enthalpy is Be > Mg > Na > K ., , 31. Generally, electron affinity increases on moving, from left to right in a period and decreases on, moving down in a group. Electron affinities of, second period elements (such as N, O) are less, negative as compared to corresponding third, period element. This is because of the small, atomic size of second period elements. Hence, the, correct order of electron affinities is, N < O < S < Cl., , 32. Electron gain enthalpy generally becomes more, negative across a period, as we move from left, to right. Within a group, electron gain enthalpy, becomes less negative down the group., However, adding an electron to the 2p orbital, leads to greater repulsion than adding an, electron to the larger 3p-orbital.

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CBSE New Pattern ~ Chemistry XI (Term-I), , 62, Chlorine (Cl) has most negative electron gain, enthalpy, while phosphorus (P) has least, negative electron gain enthalpy., , 38. c C - c H = 0.208 D, where,, , D = E C— H - E C— C ´ E H — H, , 33. A ® (4);, , B ® (1); C ® (2); D. ® (3), A. This electronic configuration corresponds to, the noble gas i.e., neon. Since, noble gases, have + De g H values, therefore, electronic, configuration (A) corresponds to the d, De g H =+48 kJ mol –1 ., B. This electronic configuration corresponds to, the alkali metal, i.e. potassium., Alkali metals have small negative De g H, values, hence, electronic configuration (B), corresponds to De g H = – 53 kJ mol –1 ., C. This electronic configuration corresponds to, the halogen, i.e. fluorine. Since, halogens, have high negative De g H values, therefore,, electronic configuration (C) corresponds to, De g H = - 328 kJ mol –1 ., , D. This electronic configuration corresponds to, the chalcogen, i.e. oxygen., Since, chalcogens have De g H values less, negative than those of halogens, therefore,, electronic configuration (D) corresponds to, De g H = - 141 kJ mol -1 ., , 34. On moving downward in a group, density, increases but the density of K is somewhat lesser, than that of Na. Due to abnormal increase in, size of K atom., Thus, the order of density is, Li < K < Na < Rb < Cs., Densities are, Li - 0.53 g/cc, Na - 0.97 g/cc,, K - 0.86 g/cc, Rb - 1.53 g/cc and, Cs - 1.90 g/cc., , 35. A qualitative measure of the ability of an atom, in a chemical compound to attract shared, electrons to itself is called electronegativity., , 36. Electronegativity increases along a period, and decreases in a group. Thus, the order is, II > III > I > IV., Electronegativity of O = 3.5,, N = 3.0, Be = 1.5, Mg = 1.2, , 37. Electronegativity increases on moving from left, to right in a period., So, the correct order of electronegativity of N, O, and F is F > O > N., , D = 98.8 - 83.1 ´ 104.2, \, , D = 5.75, c C - 2.1 = 0. 208 5.75 = 0.497, c C = 2.598, , 39. In general, the electronegativity increases on, moving from left to right in a period. Hence, the, increasing order of electronegativity is as follows, Si < P < C < N, , 40. The correct match is, A ® 4, B ® 1, C ® 2, D ® 3, Helium (He), 1s 2, , ¾® Highest ionisation energy, due to noble gas in nature, and small size., , Fluorine (F), 1s 2 , 2s 2 2p 5, , ¾® High electronegativity in, nature due to small size and, –1 oxidation state., , Rubidium (Rb) ¾® Most electropositive element, due to large atomic size., [Kr ] 5s 1, Lithium (Li), 1s 2 2s 1, , ¾® Strongest reducing agent, due to small size and, positive oxidation state (+1)., , 41. Correct match is A ® (2); B® (3) ; C ® (5);, D ® (1); E ® (4), F – Maximum electronegativity, l Cl – Maximum electron affinity, l Fe – Variable valence, l He – Maximum ionisation energy, l Ds – Recently named by IUPAC, l, , 42. Noble gases are very less reactive due to stable, , outer electronic configuration like ns 2np 6 or ns 2., Thus, R is correct but A is incorrect., , 43. Both (A) and (R) are correct statements and (R), is the correct explanation of (A)., Elements in same group have same number of, electrons in the outer orbitals and similar, electronic configuration. Therefore, they have, similar properties.., , 44. 2 p 6 , 3 s 2 ¾® Belongs to 3rd period, 1s 2, 2s 2 ¾® Belongs to 2nd period, Both have ns 2 , i.e. same electronic configuration., Hence, both belongs to different period., Thus, Reason is correct but Assertion is incorrect.

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63, , Classification of Elements and Periodicity in Properties, , 45. IUPAC name of element copernicium having, atomic number 112 is ununbium, for 1, suffix, ‘un’ and for 2 suffix ‘bi’ is used which are Latin, words., Thus, Both (A) and (R) are the correct statements, but (R) is not the correct explanation of (A)., , 46. The atomic radii of the elements of oxygen, family are smaller than atomic radii of the, corresponding elements of the nitrogen family, because of increase in effective nuclear charge, the results in the increased attraction of, electrons to the nucleus., Thus, (A) is correct statement but (R) is a, incorrect statement., , 47. Isoelectronic species have different radii because, of their different nuclear charges., Thus, (A) is incorrect statement but (R) is a correct, statement., , 48. The atomic and ionic radii decrease in a period, from left to right due to increase in effective, nuclear charge. The ionisation enthalpy, increases on moving left to right in period., Thus, Assertion is correct but Reason is incorrect., , 49. Boron has a smaller first ionisation enthalpy than, beryllium because the penetration of 2s electron, to the nucleus is more than the 2p electron., Hence, 2p electron is more shielded by the inner, core of electrons than the 2s electron., Thus, both (A) and (R) statements are correct and, (R) is the correct explanation of statement (A)., , 50. Ionisation enthalpy is the energy required to, remove an electron from an isolated gaseous, atom in its ground state. Every element does not, have tendency to lose electrons. Thus, (A), statement is incorrect but (R) statement is correct., , 51. Assertion and reason both are correct statements, and reason is correct explanation of assertion., lonisation enthalpy increases along a period, because effective nuclear charge increases and, atomic size decreases., Hence, both (A) and (R) are correct statement, and (R) is the correct explanation of (A)., , 52. Both (A) and (R) are correct statements but (R), is not the correct explanation of (A)., Alkali metals belong to first group and have, larger size in a period and hence have low I.E., , 53. Both (A) and (R) are correct statements and (R), is the correct explanation for (A)., Electron gain enthalpy becomes less negative as, the size of an atom increases down the group., , This is because within a group screening effect, increases on going down in a group and the, added electron would be farther away from the, nucleus., , 54. Both (A) and (R) are correct statements and (R) is, the correct explanation of (A)., Cesium and fluorine both reacts violently, because cesium is most electropositive and, fluorine is most electronegative., , 55. Fluorine has a less negative electron affinity, than chlorine. There is relatively greater, effectiveness of 2p electrons in the small fluorine, atom to repel the additional electron entering, the atom than to 3p electrons in the larger Cl atom., Hence, both (A) and (R) are correct statement, and (R) is the correct explanation of (A)., , 56. (i) Due to stable configuration noble gas do, not accept an electron and hence they have, positive electron gain enthalpy., (ii) There is more repulsion for the incoming, electron when the size of atom is small., When an electron is added to O or F, it, goes to a smaller (n = 2) level and suffers, more repulsion than the electron in S or Cl, in larger level (n = 3)., (iii) Electron gain enthalpy ( D e g H ) is the, enthalpy change for converting 1 mol of, isolated atoms to anions by adding, electrons. All halogens have negative D e g H, (exothermic) values. Generally, D e g H, becomes less negative when comparing, elements of the same group from top to, bottom., But among fluorine and chlorine there is, an anomaly because inter-electron, repulsion is stronger in fluorine due to its, extra small size., \ De g H is less exothermic than expected, for F-atom., Thus, the correct values of electron gain, enthalpies, Br, >, I, F <, Cl >, kJ mol - 1, , ( -333), , ( -349), , ( -325), , ( -296), , (iv) It is easier to remove electron from, 2p-orbital as compared to 2s -orbital due to, more penetration of s-electrons., Or, Atomic size and nuclear charge.

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CBSE New Pattern ~ Chemistry XI (Term-I), , 64, 57. (i) The element with atomic number 57, belongs to d -block element as the last, electron enters the 5d-orbital against the, aufbau principle. This anomalous, behaviour can be explained on the basis of, greater stability of the xenon (inert gas), core., After barium ( Z = 56 ) , the addition of the, next electron (i.e. 57th) should occur in, 4 f -orbital in accordance with aufbau, principle. This will however, tend to, destabilize the xenon core ( Z = 54 ), [Kr], ( 4d 10 4 f 0 5s 2 5 p 6 5d 0 ) since the 4 f -orbitals, lie inside the core., Therefore, the 57th electron prefers to, enter 5d -orbital which lies outside the, xenon core and whose energy is only, slightly higher than that of 4 f -orbital. In, doing so, the stability conferred on the, atom due to xenon core more than, compensates the slight instability caused by, the addition of one electron to the higher, energy 5d- orbital instead of the lower, energy 4 f - orbital., Thus, the outer electronic configuration of, La ( Z = 57 ) is 5 d 1 6 s 2 rather than the, expected 4 f 1 6 s 2 ., (ii) Each period starts with the filling of, electrons in a new principal energy shell., Therefore, 6th period starts with the filling, of 6s-orbital and ends when 6 p-orbitals are, completely filled., In between 4 f and 5d -orbitals are filled in, accordance with aufbau principle. Thus,, the outmost electronic configuration of the, last element of the p -block in the 6th, period is, 6s 2 4 f 14 5 d 10 6 p 6 or 4 f 14 5 d 10 6s 2 6 p 6 ., (iii) The long form of the periodic table contain, element with atomic number 1 to 118., (iv) The fifth period begins with Rb ( Z = 37 ), and ends at Xe., Thus, the element with Z = 43 lies in the, 5th period. Since, the 4th period has 18, elements, therefore, the atomic number of, the element which lies just above the, element with atomic number 43 is, 43 - 18 = 25 ., , Now, the electronic configuration of, element with Z = 25 is, 1s 2 2 s 2 2 p 6 3s 2 3 p 6 3 d 5 4s 2 (i.e. Mn)., Or, Each period ends with a noble gas. The, atomic number of noble gases (i.e. group 18, elements) are 2, 10, 18, 36, 54 and 86., Therefore, elements with atomic numbers, 35 ( 36 - 1) , 53 ( 54 - 1) , and 85 ( 86 - 1) , lie in a, group before noble gases, i.e. halogens (group, 17) elements., Thus, the elements with atomic number 35, 53, and 85 are all belongs to halogens., , 58. (i) Assertion is correct but Reason is incorrect., Zinc forms lesser coordination compounds as, compared to other elements of ‘d’-block., (ii) Assertion is correct but Reason is incorrect., Be has fully-filled configuration 1 s 2, 2s 2 and, that’s why removal of electron would require, higher energy as the configuration of B is, 1 s 2, 2s 2, 2 p 1 ., (iii) Assertion is incorrect while Reason is, correct. Outermost electronic configuration, of most electronegative elements is ns 2, np 5 ., (iv) Both Assertion and Reason are correct and, Reason is the correct explanation of the, Assertion., The magnitude of electron affinity depends, on electronic configuration., Or, Assertion and Reason both are correct but, Reason is not the correct explanation of, Assertion., The correct explanation would be :, Ionisation enthalpy increases along a period, due to increase in effective nuclear charge, which cause decreases in atomic size., , 59. (i) Mendeleev’s arranged elements in horizontal, rows and vertical columns in order of their, increasing atomic weights in such a way that,, the elements with similar properties, occupied the same vertical column or group., Thus, both A and R are correct but R is not, the correct explanation of A., (ii) Both gallium and germanium were unknown, at the time Mendeleev published his, periodic table.

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65, , Classification of Elements and Periodicity in Properties, , He left the gap under aluminium and a gap, under silicon and called these elements, Eka-aluminium and Eka-silicon. Dobereiner, arranged elements on the basis of, increasing atomic weights., Thus, A is correct but R is incorrect., (iii) Elements are arranged in horizontal rows are, called periods or Mendeleev’s series and, having similar outer electronic configuration, in their atoms are arranged in vertical, columns known as groups or families., Thus, both (A) and (R) correct and (R) is, not the correct explanation of (A)., (iv) (A) is correct but (R) is incorrect statement., The correct (R) statement is as follows :, Sixth period does not involve the filling, of all the orbitals of the sixth energy level, (6s , 4 f , 5d , 6 p-orbitals are filled)., Or, 2 p 6 , 3 s 2 ¾® Belongs to 3rd period, 1s 2, 2s 2 ¾® Belongs to 2nd period, Both have ns 2 , i.e. same electronic, configuration., , Hence, both belongs to different period., Thus, R is correct but A is incorrect., , 60., , (i) Assertion and Reason both are correct and, Reason is the correct explanation of, Assertion., (ii) Assertion and Reason both are correct and, Reason is not correct explanation of, Assertion. The correct explanation is : In, Al, the first ionisation energy is required, for removal of electron from 3s 1 and in Mg, electron is being removed from fully-filled, 2s 2 -orbital., (iii) Assertion is correct but Reason is incorrect., The correct reason is as follows :, F has greater tendency than Cl to attract, the shared pair of electrons of a covalent, bond., (iv) Assertion is correct but Reason is incorrect., B shows anomalous behaviour., Or, Assertion is correct but Reason is incorrect., Carbon is more electronegative as compared, to other members of group 14.

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04, Chemical Bonding and, Molecular Structure, Quick Revision, 1. Chemical Bond, It is defined as the force of attraction, which holds various constituents, (atoms, ions, etc.) together in different, chemical species. The combining tendency, of an element or ion is called its ‘valency’., Factors due to which atoms interact with, each other are, ●, , ●, , Forces of attraction overpowers the forces, of repulsion due to decrease in energy., Tendency to achieve nearest noble gas, configuration., , 2. Kossel-Lewis Approach to Chemical, Bonding (Octet Rule), According to this approach, the atoms of, different elements take part in chemical, combination in order to complete their octet, (to have eight electrons in the outermost, valence shell) or duplet (to have two, valence electrons) in some cases such as, H, Li, Be etc., or to attain the nearest, noble gas configuration. This is known as, octet rule., The group valency 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., , Limitations of octet rule are as follows, ●, , ●, , In accordance to this rule, the shape of the molecule, cannot be predicted., The relative stability of molecule cannot be known by, this rule., , Exceptions to Octet Rule, The octet rule is violated in a significant number of, cases. These are :, ●, ●, ●, , ●, ●, , Electron deficient compounds : BeCl 2, BF3, AlCl 3 etc., Hypervalent compounds : PCl 5, SF6 , IF7, H 2SO 4 etc., Compounds of noble gases : XeF2, XeF6 , XeF4 , KrF2, etc., Odd electron molecules : NO, NO 2, O -2 , O 3 etc., Others like : H2+ , He2+ , O2 , NO, NO2 , ClO2 are some, of the examples of stable molecules having odd, electron bonds (bonds formed by sharing of usually, one or three electrons)., , 3. Types of Chemical Bonds, The three types of bonds or linkages which hold the, atoms together in a molecule are as follows, ●, , ●, , Covalent bond It is formed by mutual sharing of, electrons between two atoms so as to attain nearest, noble gas configuration., Electrovalent bond It is formed as a result of the, electrostatic attraction between the positive and, negative ions.

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67, , Chemical Bonding and Molecular Structure, , é total number of, ù, ê, - non -bonding (lone pair) ú, ê, ú, êë electrons, úû, 1 é total number of bonding ù, - ê, ú, 2 ë(shared) electrons, û, 6. Ionic or Electrovalent Bond, It is formed by complete transfer of electrons, from one atom to another. e.g. Formation of, sodium chloride, ··, , N a´ + · Cl ·· ¾® Na +, ··, , (2, 8, 1), , (2, 8, 7), , (2, 8), , ´, ·, , ··, , Cl ·· ··, , (2, 8, 8), , Factors governing the formation of ionic bond are, ●, ●, , ●, , Low ionisation enthalpy of metal atom., High electron gain enthalpy of non-metal, atom., High lattice enthalpy of the ionic compound., , ●, , charge on ions (greater is the charge, greater is, the lattice enthalpy)., size of the ion (smaller is the size, greater is the, lattice enthalpy)., , 7. Bond Parameters, (i) Bond length The equilibrium distance between, the nuclei of two bonded atoms in a molecule is, called the bond length., The various factors that affect bond length are, as follows, (a) Size of the atoms The bond length, increases with increase in the size of the, atoms. Thus, the order of bond length of, H ¾ X bond is, HF < HCl < HBr < HI., Similarly, C ¾ C < Si ¾ Si < Ge ¾ Ge, (b) Bond multiplicity The bond length, decreases as the multiplicity of the bond, increases., Thus, the carbon-carbon bond lengths, follow the order, , C ºº C < C == C < C ¾ C, (ii) Bond angle The angle between the orbitals, containing bonded electron pairs around, central atom in a molecule/complex ion is, called the bond angle., e.g. C ¾ H bond angle in CH 4 , H ¾ N ¾ H, bond angle in NH 3 are shown below :, H, , 109° 28′, C, H, H, H, , N, H, H, 107°, H, , :, , 5. Formal Charge on Atom or Ion, It is defined as the difference between the, number of valence electrons of that atom in an, isolated or free state and the number of electrons, assigned to that atom in the Lewis structures., Formal charge on an atom in a molecule/ion, é total number of valence ù, =ê, ú, ë electron in the free atomû, , ●, , , , 4. Lewis Representation of Simple, Molecules (Lewis Structures), It is representation of valence electrons, done by, dots around the symbol of the element. e.g., Nitrite ion (NO -2 ), é · · · · · ·· ù, é· ·· · · ·· · · ·ù, or, ==, ¾, O, N, O, O, N, O, ·ú, ê ··, ê · ·· · · · ··ú, · ·, êë, úû, êë, úû, ·, ·, · ·ù, é ·· ·, or ê · ·O· ¾ N == O, · ·ú, êë, úû, , Lattice enthalpy, It is defined as the amount of energy required to, complete seperate one mole of a solid ionic, compound into a gaseous constituent ions. It, depends on, , , , Coordinate bond It is a type of covalent bond, in which shared pair of electrons come from, one atom only., , , , ●, , (iii)Bond enthalpy It is the amount of energy, required to break one mole of bonds of a, particular type between two atoms in gaseous, state. Its unit is kJ mol -1.

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CBSE New Pattern ~ Chemistry XI (Term-I), , 68, 1, Size of atoms, 1, µ, Number of lone pair(s) of electrons, , Bond enthalpy µ, , µ Multiplicity of bond, Greater the multiplicity of the bond, greater is, the bond enthalpy., e.g., H ¾ H < O == O < N ºº N, (iv) Bond order The number of bonds formed, between the two atoms in a molecule is called, the bond order., e.g. The bond order of a few molecules are, given below:, Molecule, , ¬, , H ¾ H O == O N ºº N C == O, , 3, 1, Bond order µ Bond enthalpy µ, Bond length, Bond order, , 1, , 2, , 3, , (v) Resonance structures According to the, concept of resonance, whenever a, single Lewis structure cannot describe a, molecule., Accurately, a number of structures with, similar energy, positions of nuclei, bonding, and non-bonding pairs of electrons are, taken as the canonical structures of the, hybrid which describes the molecule, accurately., (vi)Dipole moment It is a measurement of, polarity in any molecule/compound/ion. It is, defined as the product of charge (q ) and, distance between the charges (r )., Dipole moment (m ) = q ´ r . Its unit is Debye, (D). The dipole moment of symmetrical, molecule is always zero., , 8. Partial Ionic Character of Covalent Bond, ●, , ●, , ●, , Electronegative difference between combining, atoms = 1.7, then bond is 50% ionic and 50%, covalent., Electronegativity difference > 1.7, ionic, character in bond is more than 50%., Electronegativity difference < 1.7, ionic, character in bond is less than 50%., , 9. Partial Covalent Character in Ionic, Bonds : Fajans’ Rule, According to Fajans’ rule, the covalent character, will be favoured by, (i) small size of cation., (ii) large size of anion., (iii)high charge of cation and anion., (iv)the cation having 18-electrons shell., , 10.Valence Shell Electron Pair Repulsion, (VSEPR) Theory, ●, , ●, , ●, , According to this theory, all valence shell, electron pairs surrounding the central atom, arrange themselves in such a manner that they, are as far away from each other as possible., There are two types of electron pairs around the, central atom; bonding electron pairs, i.e. bond, pair (bp ) and non-bonding electron pairs, i.e., lone pair (lp )., The strength of repulsion between the electron, pairs varies as, , lp-lp > lp -bp > bp -bp, ●, , ●, , If central atom is linked to similar atoms and there, are no lone pairs, the shape is symmetrical, otherwise irregular., If A represents central atom,, B bond pair, L lone pair, then AB 2 ® linear

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69, , Chemical Bonding and Molecular Structure, , Shape of Some Molecules or Ions with Central Ions having One or More, Lone Pairs of e - on the Basis of VSEPR Theory, Molecule type, , AB 2 E, , No. of bonding No. of, pair, lone pair, , 2, , 1, , Arrangement of, electron pairs and, geometry of molecule, , A, B, , B, , Shape, , Examples, , Bent, , SO2 , O3 , SnCl 2 , PbCl 2 , NO+2, , Trigonal, pyramidal, , PH3 , BiCl 3 ,NH3 , PCl 3 , AsCl 3 ,, XeO3 , ClO3- , SO23 ,, P(CH3 )3, , Trigonal planar, , AB 3 E, , 3, , 1, , A, B, , B, , B, , Tetrahedral, , -, , AB 2 E 2, , 2, , H2 S, NH2 H2 O, OF2 , Cl 2 O, , A, , 2, , B, , B, , Bent, , Tetrahedral, , ·, , ·, , Cl 2 O, ClO–2 , ClOF,, SCl 2, , B, B, , AB 4 E, , 4, , A, , 1, , B, , B, , See-saw, , SF4 , SCl 4 , TeCl 4 ,, XeO2 F2, , Trigonal bipyramidal, , B, AB 3 E 2, , 3, , 2, , B, , A, , T-shape, , B, , ClF3 , IF3 , BrF3 ,, XeOF2, , Trigonal bipyramidal, , B, B, , B, , AB 5 E, , 5, , 1, , A, B, , B, , Square, pyramidal, , IF5 , BrF5 ,, 2SbF5, , Square, planar, , XeF4 , ICl -4, , Octahedral, , B, , B, , AB 4 E 2, , 4, , 2, , A, B, , B, , Octahedral, , 11. Valence Bond Theory (VBT), It explains bond formation in terms of overlapping of orbitals, e.g. the formation of H 2 molecule, from two hydrogen atoms involves the overlapping of 1s-orbital of two H-atoms which are singly, occupied. Because of orbital overlapping, the electron density between the nuclei increases which, helps in bringing them closer. The overlapping of orbitals may result in two types of bonds, sigma, (s ) and pi (p) bond.

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CBSE New Pattern ~ Chemistry XI (Term-I), , 70, 12. Concept of Hybridisation, (involving s , p , d and f -orbitals), ●, , ●, , ●, , Intermixing of orbitals of almost similar energy, belonging to the same atom to form same number, of orbitals of exactly equal energy is known as, hybridisation., The new orbital thus formed are known as hybrid, orbitals. The number of hybrid orbitals of the, molecule can be determined by, 1, (V + Y - C + A ), H =, 2, where,V = valence shell electrons of the central, metal atom, Y = number of monovalent atom, C = total positive charge, A = total negative charge on the molecule, 2, sp, , 3, sp, , 4, 2, , sp, , 3, , 5, , 6, , 3, , 3, , sp d, , sp d, , » p 2p y , p* 2 p x » p* 2 p y < s* 2 p z, However, the above sequence of energy, levels is not true in case of remaining, molecules, like Li 2, Be 2, B2, C 2, N 2., For these molecules, the increasing order of, energies of various molecular orbitals is, s1s < s* 1s < s 2s < s* 2s < p 2 p x » p 2 p y, < s 2 p z < p* 2 p x » p* 2 p y < s* 2 p z, , Stability and Magnetic Character of, Molecules/Molecular Ions, ●, , 7, 2, , sp 3d 3, , ●, , Geometry and Hybridisation of Molecules, Number of, Hybridisat, Bond Molecular, electron, ion, angle geometry, pairs, sp, , 2, , 180°, , Linear, , Examples, , ●, , If bond order is positive, molecule or ion is, stable otherwise not. Greater the bond, order, greater is the stability and shorter is, the bond length., Further, if all electrons are paired, the, species is diamagnetic and if some, unpaired electrons are present, the species, is paramagnetic., Bond order, , 1, (Number of electrons in BMOs), 2, – (Number of electrons in AMOs), 1, BO = ( N b – N a ), 2, =, , BeCl 2 , HgCl 2 ,, ZnCl 2, , 2, , 3, , 120°, , Trigonal, planar, , BF3 , AlF3, , sp 3, , 4, , 109.5 °, , Tetrahedral, , CH 4 , NH 4+ , CCl 4, , sp 3d, , 5, , 90° ,, 120°, , Trigonal, bipyramidal, , PCl 5, , sp 3d 2, , 6, , 90°, , Octahedral, , SF6, , sp, , In the case of O2 and F2,, s 1s < s* 1s < s 2s < s* 2s < s 2 p z < p 2 p x, , 14. Hydrogen Bonding, ●, , ●, , 13. Molecular Orbital Theory (MOT), (i) Bonding molecular orbitals The molecular, orbitals obtained by the addition of atomic, orbitals are called bonding molecular orbitals, (BMOs) and are represented by s and p., (ii) Anti-bonding molecular orbitals The molecular, orbitals obtained by the subtraction of atomic, orbitals are called anti-bonding molecular orbitals, (AMOs) and are represented by s* and p* ., The electronic configuration of molecular orbitals, (MO) are written in the following manner., , The attractive force which binds hydrogen, atom of one molecule with the, electronegative atom (F, O or N) of another, molecule, is called hydrogen bonding., It is represented by dotted lines, e.g., +, , -, , +, , -, , +, , K Hd ¾ F d ... K Hd ¾ F d ... K Hd ¾ F d, , -, , Types of H-Bonds, There are two types of H-bonds :, (i) Intermolecular hydrogen bond Such, bond exists between two different molecules, of the same or different compounds., (ii) Intramolecular hydrogen bond Such, bond exists when hydrogen atom is, internally bonded though H-bond with, highly electronegative (F, O, N) atoms, present within the same molecule.

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Objective Questions, Multiple Choice Questions, , 8. Ionic compounds are formed most easily, with, , 1. According to Lewis and Kossel, , (a) low electron affinity, high ionisation energy, (b) high electron affinity, low ionisation energy, (c) low electron affinity, low ionisation energy, (d) high electron affinity, high ionisation energy, , approach, which of the following, molecule has complete octet of the, central atom?, (a) LiCl, , (b) BeH2, , (c) BCl3, , (d) CO 2, , 9. Which of the following has electrovalent, , 2. Lewis dot structure of CO, NO -2 and, , linkage?, , CO 23 are I, II and III respectively, O, C, , O, , [ O—N, , I, , –, , O], , 2–, , O—C—O, , II, , III, , Which of the above structure(s) is/are, wrong?, (a) Only I, (c) Only III, , (b) Only II, (d) All of the above, , 3. Which of the following has octet, around central atom?, (a) PF5, , (b) SF6, , (c) CCl 4, , (d) BF3, , 4. Number of electrons surrounding Kr in, KrF2 is, (a) 10, , (b) 6, , (c) 4, , (d) 8, , 5. Which of the following species contains, equal number of s and p-bonds?, , (a) HCO -3, (c) (CN 2 ), , (b) XeO 4, (d) CH2 (CN) 2, , 6. In PO 34 ion the formal charge on the, oxygen atom of P ¾ O bond is, (a) + 1, , (b) - 1, , (c) - 0.75, , (d) + 0.75, , 7. Which of the following is not the, characteristic of a covalent compound?, (a) No definite geometry, (b) Insoluble in polar solvent, (c) Small difference in electronegativity, between the combining atoms, (d) Low melting point, , (a) CaCl 2, (c) SiCl 2, , (b) AlCl 3, (d) PCl 3, , 10. The correct order of the lattice energies, of the following ionic compounds is, ……, (a) NaCl > MgBr2 > CaO > Al2O 3, (b) NaCl > CaO > MgBr2 > Al2O 3, (c) Al2O 3 > MgBr2 > CaO > NaCl, (d) MgBr2 > Al2O 3 > CaO > NaCl, , 11. The higher lattice energy corresponds to, (a) MgO, (c) SrO, , (b) CaO, (d) BaO, , 12. The correct order of increasing bond, length of C¾H, C¾O, C¾ C and, C == C is, (a) C, (b) C, (c) C, (d) C, , ¾ H < C ¾ O < C ¾ C < C == C, ¾ H < C == C < C ¾ O < C ¾ C, ¾ C < C == C < C ¾ O < C ¾ H, ¾ O < C ¾ H < C ¾ C < C == C, , 13. Which of the following has lowest bond, angle?, (a) H2 S (2lp), (c) SO 2 (1lp), , (b) NH3 (1lp), (d) H2O (2lp), , 14. The bond angle in NF3 (102.3° ) is smaller, than NH 3 (107.2° ). This is because of, , (a) large size of F compared to H, (b) large size of N compared to F, (c) opposite polarity of N in the two molecules, (d) small size of H compared to N

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CBSE New Pattern ~ Chemistry XI (Term-I), , 72, 15. In case of H 2O molecule, the enthalpy, , needed to break the two O ¾ H bonds, is not the same, i.e., H 2O( g ) ¾® H( g ) + OH( g );, D a H 1s = 502 kJ mol -1, OH( g ) ¾® H( g ) + O( g );, D a H 2s = 427 kJ mol -1, What should be the mean bond, enthalpy of O—H bonds in case of H 2O, molecule?, (a) 75 kJ mol -1, (c) 464.5 kJ mol -1, , (b) –75 kJ mol -1, (d) 929 kJ mol -1, , 16. Which of the following molecules have, same bond order?, H2 , Cl 2 , CO, Br2 , N2, I, , II, , III, , IV, , V, , Choose the correct option., (a), (b), (c), (d), , I, II and IV have same bond order, III and V have same bond order, Both (a) and (b), None of the above, , 17. Dipole moment is usually expressed in, Debye units (D). From the formula, the, unit of dipole moment is Coulomb, metre (Cm). The relation between both, units of dipole moment is, (a), (b), (c), (d), , 1 D = 1 Cm, 1D = 3.33564 ´10-30 Cm, 1D = 2.22564 ´ 10-30 Cm, 1D = 1.11564 ´ 10-30 Cm, , (a) Bond order, 1, 1, µ, µ, µ stability, bond length bond enthalpy, (b) Stability, 1, 1, 1, µ, µ, µ, bond order bond length bond enthalpy, (c) Stability µ bond order µbond length µbond, enthalpy, (d) Stability µbond order µbond enthalpy, 1, µ, bond length, , 21. The most ionic one is ......... ., (a) P2O 5, , (b) MnO 2, , (c) Mn2O 7, , (d) P2O 3, , 22. The correct sequence of increasing, covalent character is represented by, (a) LiCl < NaCl < BeCl2, (c) NaCl < LiCl < BeCl2, , (b) BeCl2 < NaCl < LiCl, (d) BeCl2 < LiCl < NaCl, , 23. Which of the following has covalent, bond?, (a) Na2 S, , (b) AlCl3, , (c) NaH, , (d) MgCl2, , 24. Which of the following molecules can, be represented in terms of resonance?, (b) CO 23 (d) All of these, , (a) O 3, (c) CO 2, , 25. The bond order of the N¾O bonds in, NO -3 ion is, (a) 0.33, , 18. In case of polyatomic molecules, dipole, moment, (a) only depends upon the individual dipole, moments of bonds, (b) is vector sum of the dipole moments of, various bonds, (c) is substraction of dipole moments of, various bonds, (d) All of the above are correct, , 19. Which of the following molecules has, the maximum dipole moment ?, (a) CO 2, (c) NH3, , 20. Which of the following is correct?, , (b) CH4, (d) NF3, , (b) 1.00, , (c) 1.33, , (d) 1.50, , 26. The types of hybrid orbitals of nitrogen, in NO +2 , NO –3 and NH +4 respectively are, expected to be, (a) sp, sp 3 and sp 2, (c) sp 2, sp and sp 3, , (b) sp, sp 2 and sp 3, (d) sp 2, sp 3 and sp, , 27. The two sp-hybrid points in the, opposite direction along the z-axis with, projecting lobes are of size. (taking, positive sign as overlapping end), (a), (b), (c), (d), , more negative and less positive, same positive and negative, more positive and less negative, Both (a) and (b)

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73, , Chemical Bonding and Molecular Structure, , Column I, (Representation), , 28. Which of the following pair of ions are, isoelectronic and isostructural?, (a) CO 23 - , SO 23 (c) SO 23 - ,NO -3, , (b) ClO -3 , CO 32 (d) ClO -3 ,SO 23 -, , 29. Match the compounds given in Column, , + +, z, , C., , Column II, , XeF 6, , 1., , Distorted octahedral, , B., , XeO 3, , 2., , Square planar, , C., , XeOF 4, , 3., , Pyramidal, , D., , XeF 4, , 4., , Square pyramidal, , Codes, A B C D, (a) 1 2 4 3, (c) 4 1 2 3, , A B C D, (b) 4 3 1 2, (d) 1 3 4 2, , 30. The valence bond theory explains the, shape, the formation and directional, properties of bonds in polyatomic, molecules like CH 4 , NH 3 and H 2 O etc., in terms of, (a), (b), (c), (d), , overlapping of atomic orbitals, hybridisation of atomic orbitals, Both (a) and (b), None of the above, , 31. Match the following columns and, choose the correct option from the, codes given below., Column I, (Representation), , Column II, (Type), , Codes, A B C, (a) 1 2 3, (c) 2 3 1, , 32., , +, , A, , 0, , D, , B, , Distance of separation, Bond energy, , 435.8, Bond C, length 74 pm, , Internuclear, distance, , The above potential energy curve is, given for the formation of H 2 molecule, as a function of internuclear distance of, H-atoms. At what point in the curve H 2, is found in the most stable state?, (a) A, , (b) B, , (c) C, , (d) D, , 33. The pair of electron in the given, , carbanion, CH 3C ºº C - , is present in, which orbitals?, (a) sp 3, , z, , (b) sp 2, , (c) sp, , (d) 2p, , 1. Positive overlap, , Column I with the type of hybridisation, in Column II., , –, , Column I, , px, + –, z, , B., – +, py py, , A B C, (b) 2 1 3, (d) 3 1 2, , 34. Match the shape of molecules in, , +, , A., , py py, , Energy (kJ/mol), , Column I, , 3. Negative overlap, , – –, , I with the shape given in Column II, and mark the correct option., A., , Column II, (Type), , 2. Zero overlap, , Column II, , A., , Tetrahedral, , 1. sp 2, , B., , Trigonal, , 2. sp, , C., , Linear, , 3. sp 3, , Codes, A B C, (a) 2 3 1, (c) 1 2 3, , A B C, (b) 3 1 2, (d) 3 2 1

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CBSE New Pattern ~ Chemistry XI (Term-I), , 74, , 35. Which of the following statements are, , 41. The correct statement with regards to, , correct about CO 23 ?, , H +2 and H -2 is, , (a) The hybridisation of central atom is sp 3, (b) Its resonance structure has one C¾O single, bond and two C == O double bonds, (c) The average formal charge on each oxygen, atom is 0.67 units, (d) All C¾O bond lengths are not same, , (a) both H+2 and H-2 do not exist, (b) H-2 is more stable than H+2, (c) H+2 is more stable than H-2, (d) both H+2 and H-2 are equally stable, , 36. Match the species in Column I with the, type of hybrid orbitals in Column II., Column I, SF4, , 1., , sp 3d 2, , B., , IF5, , 2., , d 2 sp 3, , C., , NO+2, , 3., , sp 3d, , D., , NH +4, , 4., , sp 3, , 5., , sp, , A B C D, (b) 1 2 3 4, (d) 4 2 3 1, , 37. The electron probability distribution, around a group of nuclei in a molecule, is given by, (a), (b), (c), (d), , atomic orbital, molecular orbital, only antibonding molecular orbital, only bonding molecular orbital, , 38. The paramagnetic behaviour of B2 is, due to the presence of ……, (a) 2 unpaired electrons in p b MO, (b) 2 unpaired electrons in p* MO, (c) 2 unpaired electrons in s* MO, (d) 2 unpaired electrons in s b MO, , 39. Bond order of which among the, following molecules is zero?, (a) F2, (c) Be 2, , (b) O 2, (d) Li2, , 40. Which of the following diatomic, , molecular species has only p-bonds, according to molecular orbital theory?, (a) N2, (c) Be 2, , choose the correct option from the, codes given below., Column I, , Column II, , A., , Codes, A B C D, (a) 3 1 5 4, (c) 2 3 5 1, , 42. Match the following Columns and, , (b) C 2, (d) O 2, , A., , Column II, , He 2, , 1., , Paramagnetic, , B., , O2, , 2., , Diamagnetic, , C., , Li 2, , 3., , Does not exist, , D., , C2, , Codes, A B C D, (a) 1 2 3 1, (c) 3 1 2 2, , A B C D, (b) 1 1 2 3, (d) 3 3 1 2, , 43. Which of the following molecule does, not show hydrogen bonding?, (a) HF, (c) NH3, , (b) H2O, (d) H2 S, , 44. The higher boiling point of water is due, to which reason?, (a) Coordinate bonding, (b) Covalent bonding, (c) Electrostatic force of attraction, (d) Hydrogen bonding, , 45. The boiling point of a substance, increases with increase in, (a), (b), (c), (d), , intermolecular hydrogen bonding, intramolecular hydrogen bonding, molecular mass, Both (a) and (c), , Assertion-Reasoning MCQs, Directions In the following questions, (Q.No. 46-60) a statement of Assertion, followed by a statement of Reason is, given. Choose the correct answer out of, the following choices.

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75, , Chemical Bonding and Molecular Structure, , (a) Both Assertion and Reason are correct, statements and Reason is the correct, explanation of the Assertion., (b) Both Assertion and Reason are correct, statements, but Reason is not the correct, explanation of the Assertion., (c) Assertion is correct, but Reason is incorrect, statement., (d) Assertion is incorrect but Reason is correct, statement., , 46. Assertion SF6 and PF5 are examples of, expanded octed molecules., , Reason P in PF5 and S in SF6 are in sp 3 d, and sp 3 d 2 -hybridisation respectively., , 47. Assertion Octet theory cannot account, for the shape of the molecule., Reason Octet theory can predict, relative stability and energy of a, molecule., , 48. Assertion Sodium chloride formed by, the action of chorine gas on sodium, metal is a stable compound., Reason This is because sodium and, chloride ions acquire octet in sodium, chloride formation., (NCERT Exemplar), , 49. Assertion Ionic bond is non-directional., , 52. Assertion Bond energy and bond, dissociation energy has identical value, for diatomic molecules., Reason Greater the bond dissociation, energy, less reactive is the bond., , 53. Assertion BF 3 molecule has zero dipole, moment., Reason F is electronegative and B¾F, bonds are polar in nature, , 54. Assertion According to Fajan’s rule,, covalent charcacter is favoured by small, cation and large anion., Reason The magnitude of covalent, character in the ionic bond depends, upon the extent of polarisation., , 55. Assertion Shape of NH 3 molecule is, tetrahedral., Reason In NH 3 , nitrogen is, sp 3 -hybridised., , 56. Assertion Through the central atom of, both NH 3 and H 2 O molecules are, sp 3 -hybridised, yet H ¾ N ¾ H bond, angle is greater than that of H ¾ O ¾ H., , Reason Each ion is surrounded by a, uniformly distributed electric field, , Reason This is because nitrogen atom, has one lone pair and oxygen atom has, two lone pairs., , 50. Assertion Order of lattice energy for, , 57. Assertion The hybridisation possed by, , same halides are as LiF > NaF > KF., Reason Size of alkali metals increases, from Li to K., , 51. Assertion In NH 3 , N is sp 3 -hybridised,, , boron atom in BCl 3 molecule is sp 3 ., Reason sp 2 -hybrid structure have, trigonal planar geometry., , 58. Assertion Fluorine molecule has bond, , but angle is found to be 107°, , ordcer one., , Reason The decrease in bond angle is, due to repulsion between the lone pair, on nitrogen and bond pair between, N and H., , Reason The number of electrons in the, anti-bonding molecular orbitals is two, less than that in bonding molecular, orbitals.

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CBSE New Pattern ~ Chemistry XI (Term-I), , 76, 59. Assertion Among the two O ¾ H, bonds in H 2 O molecule, the energy, required to break the first O ¾ H bond, and the other O ¾ H bond is the not, same., Reason This is because the electronic, environment around oxygen is the, same even after breakage of one O ¾ H, bond., , 60. Assertion Water is one of the best, solvents., Reason H-bonding is present in water, molecules., , Case Based MCQs, 61. Read the given passage and answer the, 1 to 5 that follows :, Chemical bonding, involve interactions, that account for the association of atoms, into molecules, ions, crystals, and other, stable species that make up the familiar, substances of the everyday world. When, atoms approach one another, their nuclei, and electrons interact and tend to, distribute themselves in space in such a, way that the total energy is lower than it, would be in any alternative arrangement., If the total energy of a group of atoms is, lower than the sum of the energies of the, component atoms, then bond together, and the energy lowering is the bonding, energy., The ideas that helped to establish the, nature of chemical bonding came to, fruition during the early 20th century,, after the electron had been discovered and, quantum mechanics had provided a, language for the description of the, behaviour of electrons in atoms., However, even though chemists need, quantum mechanics to attain a detailed, quantitative understanding of bond, formation, much of their pragmatic, understanding of bonds is expressed in, , simple intuitive models. These models treat, bonds as primarily of two kinds—namely,, ionic and covalent., The type of bond that is most likely to, occur between two atoms can be predicted, on the basis of the location of the elements, in the periodic table, and to some extent the, properties of the substances so formed can, be related to the type of bonding., A key concept in a discussion of chemical, bonding is that of the molecule. Molecules, are the smallest unit of compounds that can, exist. One feature of molecules that can be, predicted with reasonable success is their, shape., Molecular shapes are of considerable, importance for understanding the reactions, that compounds can undergo, and so the, link between chemical bonding and, chemical reactivity is discussed briefly in, this article., The following questions are multiple choice, questions. Choose the most appropriate, answer:, (i) According to molecular orbital theory,, which of the following will not be, available molecule?, (a) He22 +, , (b) He+2, , (c) H-2, , (d) H22 -, , (ii) Which of the following compounds of, chlorine contains both ionic as well as, covalent bonds?, (a) NaCl, (c) PCl 3, , (b) NaClO4, (d) POCl 3, , (a) - 0.75, 1.25, (c) - 0.75, 0.6, , (b) - 0.75, 1.0, (d) - 3, 1.25, , (iii) In PO 34 - , the formal charge on each, oxygen atom and P ¾ O bond order, respectively are, , (iv) On the basis of valence bond theory,, the formation of H 2 molecules from two, H-atoms involves., (a) The overlap of vacant is orbitals of two, H-atom, (b) The lowering of potential energy of the, system as the two H-atom come near to, each other

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77, , Chemical Bonding and Molecular Structure, , (c) The maximum energy of the system at, the equilibrium internuclear distance, (d) Stabilisation of the molecule; when the, nuclei are brought still closer to each, other from the equilibrium inter-nuclear, distance, Or, , The dipole moment is highest for, (a) trans-2-butene, (c) acetophenone, , (b) 1, 3-dimethyl benzene, (d) ethanol, , 62. Read the passage given below and, answer the following questions :, In order to explain the shapes of, molecules adequately, Sidgwick and, Powell in 1940 proposed a theory based, on the repulsive interaction of the electron, pairs in the valence shell of the atoms., Nyholm and Gillespie (1950) further, developed and redefined the concept., The main postulates of this theory are as, follows :, The number of valence shell electron pairs, (bonded or non-bonded) present around, the central atom decides the shape of the, molecules. The shared electron pairs are, called bond pairs and unshared or, non-bonding electrons are called lone, pairs., Electron pairs of valence shell repel one, another because their electron clouds are, negatively charged., These electron pairs arrange themselves in, such a way so that there is minimum, repulsion and maximum distance in, between them. The valence shell is, considered as a sphere in which the, electron pairs are localised on the, spherical surface at maximum distance, from one another., A lone pair occupies more space than a, bonding pair, since it lies closer to the, central atom. This means that the, repulsion between the different electron, pairs follow the order :, , Lone pair-lone pair > lone pair-bond pair >, bond pair-bond pair, (lp - lp ) > (lp - bp ) > (bp - bp ), The following questions (i-iv) are multiple, choice questions. Choose the most, appropriate answer :, (i) Which of the following molecule has net, dipole moment zero?, (a) HF, (c) BF3, , (b) H2 O, (d) CHCl 3, , (ii) Which one of the following species, contains three bond pairs and one lone, pair around the central atom?, (a) H2 O, (c) NH-2, , (b) BF3, (d) PCl 3, , (iii) Why do the deviations occur from, idealised shape of H2 O and NH3, molecules?, (a) Same hybridisation, (b) Different hybridisation, (c) Repulsive effect, (d) None of the above, , (iv) The species, having bond angles of 120°, is, (a) PH3, (c) NCl 3, , (b) ClF3, (d) BCl 3, , Or, Which of the following structures is most, stable?, F, , F, , Cl, , F, , Cl, F, , F, (a), , (b), , F, Cl, , F, , F, F, , Cl, F, , F, F, (c), , (d)

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CBSE New Pattern ~ Chemistry XI (Term-I), , 78, , 63. Read the passage given below and, answer the following questions :, Lewis postulated that atoms achieve the, stable octet(or duplet in case of H and He), when they are joined through chemical, bonds or in other words atoms join, together either by transfer of valence, electrons from one atom to another, (gaining or losing) or by sharing of, valence electrons in order to complete, their octet (i.e. eight electrons in the, valence shell). This is known as octet rule., Octet rule has not been followed in all, molecules., Elements of the third and higher periods, of the periodic table, because of the, availability of d-orbitals can expand their, covalency and can accommodate more, than eight valence electrons around the, central atom. This is referred as, expanded octet., Here, also the octet rule is not applicable., e.g. PF 5 (10 electrons around P-atom), SF, (12 electrons around S-atom), H2 SO4, 6, (12 electron around S atom). Compound, having expanded octet are also termed as, hypervalent compounds. Formal charge is, a factor which is based on pure covalent, view of bonding in which electron pairs, are shared equally by neighbouring atom., It may be defined as the difference, between the number of valence electrons, of that atom in an isolated or free state, and the number of electrons assigned to, that atom in the Lewis structure. It is, represented as :, 1, FC =V - L - S, 2, where,, FC = formal charge on an atom in a Lewis, structure,, V = total number of valence electrons in, the free atom,, L = total number of non-bonding (lone, pair) electrons and, S = total number of bonding (shared), electrons., , In these questions (i-iv) a statement of, Assertion followed by a statement of Reason, is given. Choose the correct answer out of, the following choices :, (a) Assertion and Reason both are correct, statements and Reason is correct, explanation for Assertion., (b) Assertion and Reason both are correct, statements but Reason is not correct, explanation for Assertion., (c) Assertion is correct statement but Reason is, incorrect statement., (d) Assertion is incorrect statement but Reason, is correct statement., , (i) Assertion In case of polyatomic ions,, the net charge is possessed by the ion as, a whole and not by a particular atom., Reason The formal charge of an atom, in a polyatomic molecule or ion may be, defined as the difference between the, number of valence electrons of that, atom in an isolated or free state and the, number of electrons assigned to that, atom in the Lewis structure., (ii) Assertion The correct Lewis structure of, O3 may be drawn as :, O, 1, , O, 2, , O, , −, , 3, , Reason The formal charges on atom 1,, 2 and 3 are +1, 0 and –1, respectively., (iii) Assertion Formal charges do not, indicate real charge separation within, the molecule., Reason Formal charges help in the, selection of the lowest energy structure, from a number of possible Lewis, structures for a given species., (iv) Assertion H2 SO4 does not follow the, octet rule., Reason In sulphur atom 3d-orbitals also, available for bonding and hence, it have, 12 electrons around itself in H2 SO4 ., Or Assertion Sulphur forms many, , compounds in which the octet rule is, obeyed.

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79, , Chemical Bonding and Molecular Structure, , Reason Due to the absence of d-orbitals, in sulphur, it follows the octet rule., , 64. Read the passage given below and, answer the following questions :, Dipole moment is the measure of degree of, polarity and is defined as the product of the, magnitude of the charge and the distance, between the centres of positive and, negative charge. It is usually represented, by a Greek letter m (mu)., It is a result of polarisation., Mathematically, it is expressed as :, Dipole moment (m) = charge (Q ) ´, distance of separation (r ), It is usually expressed in Debye units (D)., 1 D = 3.33564 ´ 10 -30 C m, where, C is coulomb and m is metre., or 1 D = 1 ´ 10 -18 esu (or, stat C m), Here, esu = electrostatic unit, stat C cm = static coulomb centimeter, It is a vector quantity and is represented, by the crossed arrow ( -¾®, |, ) pointing, towards the more electronegative atom., -|-®, , ··, , H ¾ F ·· ® More electronegative atom, ··, , This arrow symbolises the direction of the, shift of electron density in the molecule., The following questions (i-iv) are multiple, choice questions. Choose the most, appropriate answer :, (i) The relation between resultant dipole, moment of NH 3 and NF3 is, (a) m NH 3 = m NF 3, , (b) m NH 3 > m NF 3, , (c) m NH 3 < NF3, , (d) None of these, , (ii) In a diatomic molecule, the bond, distance is 1 ´ 10 -8 cm and its dipole, moment is 1.2 D. What is the, fractional electronic charge on each, atom?, (a) 0.25 (b) 2.5, , (c) 1, , (d) 0.5, , (iii) In water molecule, the two O ¾ H, bonds are oriented at an angle of, 104.5°. In BF3 , the three B ¾ F bonds, are oriented at an angle of 120°. In, , BeF2 , the two Be¾F bonds are, oriented at an angle of 180°. Which, will have highest dipole moment?, (a) BeF2, (b) BF3, (c) H2 O, (d) All have equal dipole moment, , (iv) CO 2 molecule is, (a) polar, (c) linear, , (b) non-polar, (d) Both (b) and (c), , Or, , Dipole moment is given by, (a) m = q ´ d, qd, (c) m =, 2, , (b) m = 2qd, , (d) None of these, , 65. Read the passage given below and, answer the following questions :, The formation of molecular orbitals, may be described by the linear, combination of atomic orbitals that can, take place by addition and by, subtraction of wave function. The, molecular orbital formed by the, addition of atomic orbital is called, bonding melecular orbital., The molecular orbitals obtained by the, subtraction of atomic orbitals are called, antibonding molecular orbitals (AMOs), and are represented by s* and p*. In, the formation of AMOs, the electron, waves cancel each other due to, destructive interference., In these orbitals, most of the electron, density is located away from the space, in between the nuclei., Infact, they have a nodal plane (on, which the electron density is zero) in, between the nuclei and hence, the, repulsion between the nuclei is high., In contrast to the bonding molecular, orbitals, the electrons placed in these, orbital destabilise the molecule because, of more mutual repulsion of the, electrons in this orbital as compared to

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CBSE New Pattern ~ Chemistry XI (Term-I), , 80, the attraction between the electrons, and the nuclei which cause a net, increase in energy., In these questions (i)-(iv) a statement of, Assertion followed by a statement of, Reason is given. Choose the correct, answer out of the following choices :, (a) Assertion and Reason both are correct, statements and Reason is correct, explanation for Assertion., (b) Assertion and Reason both are correct, statements but Reason is not correct, explanation for Assertion., (c) Assertion is correct statement but Reason, is incorrect statement., (d) Assertion is incorrect statement but, Reason is correct statement., , (i) Assertion The total energy of two, molecular orbitals, however, remains, the same as that of two original, atomic orbitals., Reason The energy of the antibonding, orbital is raised above the energy of, the parent atomic orbitals that have, combined and the energy of the, , bonding orbital has been lowered than the, parent orbitals., (ii) Assertion The combining atomic orbitals, must have the same or nearly the same, energy., Reason 1s -orbital can combine with, another 2s-orbital but not with 1s-orbital, because the energy of 2s-orbital is, appreciably higher than that of 1s-orbital., This is not true if the atoms are very, different., (iii) Assertion Bond energy and bond, dissociation energy have identical value, for diatomic molecules., Reason Greater the bond dissociation, energy, less reactive is the bond., (iv) Assertion Bond energy has order like, C ¾ C < C == C < C ºº C, Reason Bond energy increases with, increase in bond order., Or Assertion The bond order of helium is, , zero., Reason The number of electrons in, bonding molecular orbital and, antibonding molecular orbital is equal., , ANSWERS, Multiple Choice Questions, 1. (d), 11. (a), 21. (d), , 2. (a), 12. (b), 22. (c), , 3. (c), 13. (a), 23. (b), , 4. (a), 14. (c), 24. (d), , 5. (b), 15. (c), 25. (c), , 6. (c), 16. (c), 26. (b), , 7. (a), 17. (b), 27. (c), , 8. (b), 18. (b), 28. (d), , 9. (a), 19. (c), 29. (d), , 10. (c), 20. (d), , 31. (c), 41. (c), , 32. (c), 42. (c), , 33. (c), 43. (d), , 34. (b), 44. (a), , 35. (c), 45. (d), , 36. (a), , 37. (b), , 38. (a), , 39. (c), , 30. (c), 40. (b), , 49. (a), 59. (d), , 50. (b), 60. (b), , 51. (a), , 52. (b), , 53. (b), , 54. (a), , 55. (d), , Assertion-Reasoning MCQs, 46. (b), 56. (a), , 47. (c), 57. (b), , 48. (a), 58. (a), , Case Based MCQs, 60. (i)-(d), (ii)-(b), (iii)-(a), (iv)-(b, c), 63. (i)-(a), (ii)-(a), (iii)-(b), (iv)-(a, c), 65. (i)-(b), (ii)-(c), (iii)-(b), (iv)-(a, a), , 61. (i)-(c), (ii)-(d), (iii)-(c), (iv)-(d, a), 64. (i)-(b), (ii)-(a), (iii)-(c), (iv)-(d, a)

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EXPLANATIONS, 1. According to Lewis and Kossel approach, CO2, molecule has complete octet of central atom, LiCl, BeH2 and BCl 3 are the molecules with, incomplete octet of central atom. Lewis, structure of CO 2 can be represented as :, O, , C, , O, , Column I, (a), , O, , O, , 2. Lewis dot structure (I) of CO, is wrong., , O, , or, , 3. Carbon tetrachloride ( CCl 4 ) has octet around, , Cl, Cl, , C, , Cl, , 4. Krypton (Kr) forms two covalent bonds with two, fluorine atoms (one with each fluorine)., Kr, , F, , (d), , Xe, O, , O, CC, , s-bond-3, p-bond-4, , N, , s-bond-6, p-bond-4, , H, N, , CC C, , N, , H, , 6. In PO34 ion, formal charge on each O-atom of, P¾ O bond, total charge, 3, =, = - = - 0.75, Noumber of O - atom, 4, , 7. Covalent bonds are directional. Therefore,, covalent compounds have definite shape, (geometry), e. g. acetylene is linear, methane is, tetrahedral in shape., , Cl, , F, , N, , , , , , central atom. The number of bonded, electrons in other given compounds are as, follows :, PF5 (10e - ), SF6 (12e - ), BF3 (6e - ), Structural representation of CCl 4 is as follows :, , s-bond-4, p-bond-4, , , , C O, (Correct), , (c), , O H, , O, , Complete octet, , C, , C, , , (b), , It may be correctly represented as :, , Column II, s-bond-4, p-bond-1, , O, , or, , F  Kr  F, , Number of electrons in valence shell of Kr = 8, Number of electrons its shares with two F-atoms, =2, \Total number of electrons found around Kr in, KrF = 8 + 2 = 10, 5. XeO4 contain equal number of s and p-bonds., number of s and p-bonds present in species are, as follows:, , 8. Formation of ionic bond is favoured by, (i) low ionisation potential of metal., (ii) greater value of electronaffinity of non-metal., (iii) higher value of lattice energy of resulting, ionic compound., , 9. In CaCl 2 , Ca¾Cl bonds are electrovalent or, ionic bonds because of the higher, electronegativity difference between Ca and, Cl-atom., , 10. Lattice energy depends upon, (i) the product of ionic charges., (ii) the interionic distance., (iii) nature of the crystal structure.

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CBSE New Pattern ~ Chemistry XI (Term-I), , 82, Hence, more is the ionic charge, more will be the, lattice energy., So, the correct order of lattice energy is, Al 2 O3 > MgBr2 > CaO > NaCl., 1, 11. Lattice energy µ charge of ion µ, size of ion, On moving downward, the size of alkaline earth, metals increases due to increase in number of, shells. Thus, Mg has the smallest size., Hence, lattice energy is higher for MgO., , 12. The C ¾ H is the shortest due to small size of, both atoms and large electronegativity, difference. The double bonds are shorter than, single bonds. Hence, C == C bond is shorter than, C ¾ O and C ¾ C. Due to electronegativity, difference between C and O, the C ¾ O bond, length is shorter than C ¾ C bond. Hence, the, correct order is, C — H < C == C < C — O < C — C, 13. H2S has the lowest bond angle because two lone, pairs are present in it and S has low electro, negativity., H2S < H2O < NH3 < SO2, 92 °, , 104 °, , 107 °, , 119 °30 ¢, , 14. In NF3 , N is less electronegative as compared, to F but in NH3 , it (N) is more electronegative, than H. In case of same central atom, as the, electronegativity of other atoms increases,, bond angle decreases. Thus, bond angle in NF3, is smaller than that in NH3 because of the, opposite polarity of N in these molecules., , N, F, , H, H, , 15. In case of H2 O molecule, the enthalpy needed to, break the two O ¾ H bonds is not the same, i.e., H2 O(g ) ¾® H(g ) + OH(g );D a H 1s = 502 kJ mol -1, OH(g ) ¾® H(g ) + O(g ); D a H, , s, 2, , and O) the bond order is 3. For N 2 , bond order, is 3. H2 , Cl 2 and Br2 have bond order 1., Hence, both (a) and (b) options are correct., , 17. 1D = 3.33564 ´ 10 -30 cm, 18. In case of polyatomic molecule dipole moment, is the vector sum of the dipole moments of, various bonds., , 19. CO2 and CH4 have zero dipole moment as these, are symmetrical in nature. Between NH3 and, NF3 , NH3 has greater dipole moment though in, NH3 and NF3 both, N possesses one lone pair of, electrons. This is because in case of NH3 , the net, N ¾ H bond dipole is in the same direction as, the direction of dipole of lone pair but in case of, NF3 , the direction of net bond dipole moment of, three N ¾ F bonds is opposite to that of the, dipole moment of the lone pair., H, µres µ4, O ==C== O, µ1 C µ3, µnet = 0, H, µ2 H, H, µ=0, Q, \, , m res = m 1 + m 2 + m 3 = - m 4, m net = m 1 + m 2 + m 3 + m 4, = - m4 + m4 = 0, µ4, N µ1, Hµ, H, 2, H, µ3, , Resultant of, 3 N—H bond lie in, the same direction, as m 4., , N, F H, , F, , 16. In CO (three shared electron pairs between C, , = 427 kJ mol -1, , The mean or average bond enthalpy of, O¾H bonds in case of H2O molecule is, obtained by dividing total bond dissociation, enthalpy by the number of bonds broken as, explained below in case of water molecule., 502 + 427, Average bond enthalpy =, 2, = 464.5 kJ mol -1, , µ3, F, , µ4, µ, N 1, F, µ2, F, , Resultant of 3 N—F bond lie, opposite to m 4 m net = 0.2 D., , 20. A general correlation useful for understanding, the stabilities of molecules is that, with increase, in bond order, bond enthalpy increases and, bond length decreases. Stability increases with, increase in bond order and bond enthalpy., Therefore, stability µ bond order µ bond, enthalpy and stability increases with decrease in, bond length., 1, Therefore,, stability µ, bond length, , 21 According to Fajan’s rule, compounds having, large charge on cation are more covalent or, least electrovalent.

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83, , Chemical Bonding and Molecular Structure, , NO+2 = 2 bp + 0 lp = linear = sp-hybridised, NO-3 = 3 bp + 0 lp Þ sp 2 -hybridised, NH+4 = 4 bp + 0 lp Þ sp 3 -hybridised, , So, compounds having low charge on cation, will be more electrovalent or ionic., +7, , +4, , +5, , +3, , MnO2 Mn 2 O7 P2 O5 P2 O3, So, P2 O3 will be most ionic., , 27. The two sp-hybrid points in the opposite, direction along the z-axis with projecting more, positive lobes a less negative lobes, which, provides more effective overlapping resulting in, the formation of stronger bond., , 22 On the basis of Fajan’s rule, lower the size of, cation, higher will be its polarising power and, higher will be covalent character., 1, \ Polarising power µ, size of cation, Covalent character µ polarising power, , 28. ClO-3 , == SO23Cl, , (The order of size of cation is Na + > Li + > Be2+ ), So, the correct order is NaCl < LiCl < BeCl 2 ., , 23 Al 3+ cation is smaller than Na + (because of, , O, , I, , O, , (b), , C, , –, , O, , O –, , –, O, , I, , (c) O, , C, , O, C, , I, , s, O, , C, , 25 The resonance structures of NO ions are as :, O, , ⊕, , O, , III, , II, , O–, , →→, , F, , O O, , O, , –, O, , The N ¾ O bond order, number of bonds between two atoms, =, total number of resonating structure, 4, = = 1.33, 3, , 26. The type of hybrid orbitals of nitrogen can be, decided by using VSEPR theory counting bp, and as lp in, , F, , F, Xe, F, , Xe, , F, , F, Distorted, octahedral XeF6, , ↑, , N, , F, Xe, , O, , ↑, , N, , N, , F, , 3, , O–, , →→, N, –, O→, O O, , O, , O, , F, , III, , s, O, , C, , O, , –, O, , O, , O, II, , O, , O, , –, , A ®1, B ® 3, C ® 4, D ® 2, The structure of the xenon compounds are, represented below:, , –, , –, , C, , ⊕, , O–, , O–, , 29. The correct match is, , O, II, , O, , O, , Hence, ClO-3 and SO23 are isoelectronic and are, pyramidal in shape., , O, O, , O, , CO 23 - = 6 + 2 + 24 = 32, SO23 = 16 + 2 + 24 = 42, ClO-3 = 17 + 24 + 1 = 42, NO-3 = 7 + 1 + 24 = 32, , terms of resonance. These are shown below :, O, , O, , Number of electrons, , 24. All the given molecules can be represented in, , O, , S, –, , O–, , C, , O, , greater nuclear charge). According to Fajans’, rule, small cation polarises anion upto greater, extent., Hence, Al 3+ polarises Cl - ions up to greater, extent, therefore AlCl 3 has covalent bond, between Al and Cl-atoms., , (a), , O, , O, , O, O, Pyramidal XeO3, , F, , F, Xe, , F, , Square pyramidal, XeOF4, , F, F, Square planar, XeF4, , 30. The valence bond theory explains the shape, the, formation and directional properties of bonds in, polyatomic molecules like CH4 , NH3 and H2 O, etc., in terms of overlap and hybridisation of, atomic orbitals.

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CBSE New Pattern ~ Chemistry XI (Term-I), , 84, 31. The correct match is (A) ® (2), (B) ® (3), (C) ®(1)., +, , Zero overlap (Out of, , z phase due to, , +, , A., , –, , different orientation, or direction of, approach), , s, , px, , 36. The correct match is, , +, , –, Negative overlap, , z (Orbitals forming, , B., , bond have different, sign and orientation), , – +, py py, , –, , Positive overlap, z (Orbitals forming, bond have same sign, and orientation in, space), , –, , py py, , (d), , stable state due to minimum energy, as the, energy gets released when bond is formed, between two hydrogen atoms., , Number of s - electrons, 2, 2 + 2 (negative ion), =, 2, = 2 = sp, Hence, in the carbanion, CH3C ºº C! , pair of, electron as negative charge is present in, sp-hybridised orbital., , 33 Hybridisation =, , 34. A ® (3);, , B ® (1); C ® (2), A. Tetrahedral shape – sp 3 -hybridisation, B. Trigonal shape – sp 2 -hybridisation, C. Linear shape – sp-hybridisation, , 2, 35. The hybridisation of central atom in CO23 is sp ., , Hence, (a) is wrong., O, , O––, , –, O–, , C, , C, , C, , O–, –, , O–, , (B) IF 5 number of bp (5) + number of lp (1), = sp 3d 2 -hybridisation, , (D) NH+4 = number of bp (4) + number of lp (0), = sp 3 -hybridisation., , 37. The electron probability distribution on around, a group of nuclei in a molecule is given by, molecular orbital., , 38. B2 = 5 + 5 = 10e -, , 32. At C point in the curve, H2 is found in the most, , –O, , (A) ® (3); (B) ® (1); (C) ® (5); (D) ® (4)., (A) SF 4 = number of bp (4) + number of lp (1), = sp 3d -hybridisation, , (C) NO+2 = number of bp (2) + number of lp (0), = sp-hybridisation, , (d), , + +, C., , Formal charge on each O-atom, Total charge, =, Number of O - atoms, -2, =, = - 0.67 units., 3, All C¾ O bond lengths are equal as mentioned, above., , O, , O, , O, , –, , Due to resonance all C¾O bond lengths are, equal., , = s1s 2 s* 1s 2, s 2s 2 s* 2s 2, p 2 p x1 » p 2 p 1y, Due to the presence of 2 unpaired bonding, electrons in p b MO, B2 shows paramagnetic, behaviour., , 39. (a) F2 molecule, = s1s 2, s* 1s 2, s 2s 2 s* 2s 2, s 2 p z2 ,, p 2 p x2 » p 2 p 2y , p* 2 p x2 » p* 2 p 2y, Bond order =, , 1, 1, ( N b - N a ) = (10 - 8 ) = 1, 2, 2, , (b) O2 molecule, = s1s 2, s* 1s 2, s 2s 2 s* 2s 2, s 2 p z2 ,, p 2 p x2 » p 2 p 2y , p* 2 p x1 » p* 2 p 1y, 1, Bond order = (10 - 6 ) = 2, 2, (c) Be2 molecule = s1s 2, s* 1s 2, s 2s 2, s* 2s 2, 1, Bond order = ( 4 - 4 ) = 0, 2, (d) Li 2 molecule = s1s 2, s* 1s 2, s 2s 2, 1, Bond order = ( 4 - 2) = 1, 2, Therefore, Be2 molecule has zero bond order.

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85, , Chemical Bonding and Molecular Structure, , 40. According to MO theory, C2 is the diatomic, , molecular species that has only p-bonds., The molecular orbital configuration of C2 is, C2( Z = 12 ) Þ s1s 2 , s *1s 2 , s 2s 2 , s * 2s 2 ,, p 2 p x2 = p 2 p 2y ., Double bond in C2 consists of both p-bonds, because of the presence of last (valence) four, electrons in two p-molecular orbitals., The configuration of N 2 , Be2 and O 2 are as, follows:, N2 (Z = 14) - s1s 2 , s *1s 2 , s 2s 2 , s * 2s 2 ,, p 2 p x2 = p 2 p 2y , s 2 p z2 (1 s and 2p-bonds), Be2 ( Z = 8 ) - s1s 2 , s *1s 2 , s 2s 2 , s * 2s 2 (s bonds, only), O2 ( Z = 16 ) - s1s 2 , s *1s 2 , s2s 2 , s * 2 s 2 , s * 2 p z2 ,, p 2 p x2 = p 2 p 2y , p * 2 p x1 = p * 2 p z1 (s, p and, 1p * -bond), , 41., , H+2 : ( s1s 1 ), 1, 1, \ BO = (1 - 0) =, 2, 2, H-2 : ( s1s ) 2( s* 1s 1 ), 1, 1, \ BO = (2 - 1) =, 2, 2, Even though, the bond order of H+2 and H-2 are, equal but H+2 is more stable than H-2 as in the, latter, an electron is present in the higher energy, antibonding ( s* 1s ) orbital., , 42. C2 , Li 2 are diamagnetic., O2 is paramagnetic, He2 does not exist., Hence, correct match is, (A) ® (3), (B) ® (1), (C) ® (2), (D) ® (2)., , 43. H2 S does not show hydrogen bonding, that’s why, it exists as a gas, other given molecules such as, HF, H 2 O and NH3 show hydrogen bonding., , 44. The higher boiling point of water is due to, hydrogen bonding. O-atom of each H2O, molecule is covalently linked with two H-atoms, of its own molecule and with another H-atom of, adjacent H2O molecule by H-boiling., , 45. The boiling point of a substance increases with, increase in intermolecular hydrogen bonding, and molecular mass. Association of molecules, takes place due to intermolecular hydrogen, bonding which increases the boiling point of a, substance., , 46. In such type of compounds, participation of, d -orbitals in bonding increase the number of, , vlence electrons more than eight. Thus, it is, termed as expanded octet., Both Assertion and Reason are correct but, Reason is not the correct explanation of the, Assertion., , 47. Octet theory is totally silent about the energy of, a molecule and about its relative stability, \Assertion is correct but Reason is incorrect., , 48. Both Assertion and Reason are correct and, Reason is correct explanation of Assertion., , 49. Each ion, because of uniformly distributed, electric field, is non directional., Both Assertion and Reason are correct and, Reason is correct explanation of Assertion., , 50. Both Assertion and Reason are correct, but, Reason is not the correct explanation of, Assertion., , 51. Lone pair lone pair repulsion > lone pair bond pair repulsion > bond pair - bond pair, repulsion. In NH3 there are three bond pairs, and one lone pair. The three N¾ H bond pairs, are pushed closer because of the lone pair bond pair repulsion and N¾ H¾ H bond, angle gets reduced from 109° 28 (the, tetrahedral angle) to 107°., Both Assertion and Reason are correct and, Reason is correct explanation of Assertion., , 52. Bond dissociation energy is the energy required, to break one mole of bond of a particular type., So as to separate them into gaseous atoms., Bond energy of the same type of bond present, in the molecule. Bond energy of C¾ H bond in, methane is 99.2. kcal/mol., Thus, for polyatomic molecules, average bond, energy is taken as the dissociation bond energy., Both Assertion and Reason are correct but, Reason is not the correct explanation of the, Assertion., , 53. BF3 is sp 2-hybridised. 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, , Resultant dipole moment = 0, , F, Both Assertion and Reason are correct but, Reason is not the correct explanation of the, Assertion., , 54. Assertion and Reason both are correct and, Reason is the correct explanation of Assertion.

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CBSE New Pattern ~ Chemistry XI (Term-I), , 86, 55. NH3 molecule is pyramidal in shape and, 3, , sp -hybridised because out of four electron pairs,, three are bonding pairs and one is lone pair., Thus, Assertion is incorrect but Reason is, correct., , 56. Assertion and Reason both are correct and, Reason is the correct explanation of Assertion., O, H 104.5° H, sp 3-hybridised, , N, , H, H, sp 3-hybridised, H, , 107°, , 1, 2, 1, = ( 3 + 3 - 0 ) = 3 Þ sp 2, 2, Both Assertion and Reason are correct but, Reason is not the correct explanation of the, Assertion., , 57. Hybridisation = (VE + MA - c + a ), , 58. Electronic configuration of F 2 molecule is, s1s 2s *1s 2s * 2s 2s * 2s 2 s 2 p z2 p 2 p x2 p 2 p 2y p * 2 p x2 p * 2 p 2y, No. of electron in bonding MOs, No. of electrons in anti - bonding MOs, 2, 10 - 8, =, =1, 2, Both Assertion and Reason are correct but, Reason is the correct explanation of Assertion., =-, , 59. Correct Assertion The bond enthalpies of the, two O ¾ H bonds in H ¾O ¾ H are not equal., Correct Reason This is because electronic, environment around O is not same after, breakage of one O ¾ H bond., Thus, Assertion is incorrect but Reason is, correct., , 60. Water is an excellent solvent because it has high, value of dielectric constant., Due to high value of dielectric constant, the, electrostatic forces of attraction between the, ions decrease and these ions get separated, and ultimately get solvated by the solvent, molecules., Both Assertion and Reason are correct but, Reason is not the correct explanation of the, Assertion., , 61. (i) If bond order of any species is zero, that, can not exist., , 2-2, 2, * 2, So, bond order of H2=0, 2 = s 1s , s 1s =, 2, , (ii) NaClO4 contains two parts: It have ionic, bonding as Na + and ClO-4 ions. However,, in ClO-4 (perchlorate ion) there is, sp 3 -hybridisation and Cl ¾O bonds are, covalent., (iii) The formal charge on each O-atom in, 3, PO34 - = - = - 0.75., 4, P ¾O bond order in PO34 Total number of bonds between atoms, =, Total number of resonating structures, 5, = = 1.25, 4, (iv) On the basis of VB theory, the formations, of H2 molecule from two H-atoms involves, lowering of potential energy of the, system as the two H-atoms come near to, each other. This energy will be minimum, at the equilibrium internuclear distance., Or, The dipole moment of acetophenones is, 3 Debye which is highest amongst the, given compounds, as it has strong, electron withdrawing group., , 62. (i) In BF 3, the dipole moment is zero, , although the B—F bonds are oriented at, an angle of 120° to one another, the three, bond moments give a net sum of zero as, the resultant of any two is equal and, opposite to the third. Structure of BF 3, molecule is as follows :, F, F, , B, F, , (ii) PCl 3 contains three bond pairs and one, lone pair around the central atom., PCl3 ⇒, , Cl, , P, , Cl, , Cl, , (3bp + 1lp), , (iii) Repulsive effects result in deviations from, idealised shapes and alterations in bond, angles in the molecules. Presence of lone, pair of electrons causes repulsion between, lone pair-bond pair., · ·, , In H ¾·O· ¾H, two lone pairs of oxygen, repel two bond pairs of ¾ OH.

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87, , Chemical Bonding and Molecular Structure, , ··, , N, ƒ ‚, In H ½ H one lone pair of nitrogen, H, repel three bond pairs of N¾H., (iv) The species having bond angles of 120° is, BCl 3 . It is sp 2 -hybridised and central atom, does not have any lone pair of electrons., Cl, B, Cl, , Cl, (BCl3), , Or, Structure (a) is the most stable geometry, because both the lone pairs are present at, equatorial position., Due to which repulsion is minimum in, molecule as compared to the repulsion in, other molecules where lone pair is at axial, position., , 63. (i) Both Assertion and Reason are correct and, , (iii) Real formal charge helps in the selection, of lowest energy structure from a number, of possible structures for a given species., Formal charge do not indicate real charge, separation within the molecule., Hence, both Assertion and Reason, are correct but Reason is not the correct, explanations of Assertion., (iv) Both Assertion and Reason are correct, and Reason is the correct explanation of, Assertion., Due to presence of vacant d-orbital on, S-atom it can expand its octet as in the, case of H2 SO4 ., Or, Sulphur has vacant d-orbital so it can, expand its octet. In general, it forms, many compounds which obey octet rule., Hence, Assertion is correct but Reason is, incorrect., , 64. (i) The lone pair of nitrogen opposed the, dipole moment of NF3 while it is added to, the dipole moment of NH3 ., So, m NH 3 > m NF3, , Reason is the correct explanation of, Assertion., éFormal ù é Total valence ù, êcharge on ú = êelectrons in ú, ê, ú ê, ú, êëatom, úû êë free atom, úû, é Total non - bonding ù, -ê, ú, ëelectrons (lone pair) û, ù, 1 é Total, – ê, 2 ë shared electrons úû, (ii) Both Assertion and Reason are correct and, Reason is the correct explanation of, Assertion., , N, H, H, , 2, , O, , (ii) Partial charge =, =, , Dipole moment, ⇒ 0.80 × 10–30 cm, , Dipole moment, Bond distance, 1.2 ´ 10 -18 esu cm, 1 ´ 10 -8 cm, , = 1.2 ´ 10 -10 esu, 1.2 ´ 10 -10, Fractional charge =, = 0.25, 4.8 ´ 10 -10, (4.8 ´ 10 -10 is a theoretical value of m), , 3, , O, , 1, Formal charge on O1 = 6 – 2 – ( 6 ) = + 1, 2, 1, Formal charge on O2 = 6 – 4 – ( 4 ) = 0, 2, 1, Formal charge on O3 = 6 – 6 – ´ 2 = – 1, 2, Hence, correct representation of O3 is, , (iii), , BeF 2 : m =0, , O, , O, , O, , –1, , Be, F, , F, , BF 3 : m =0, , F, , B, F, , +1, , 0, , F, F, , Dipole moment, ⇒ 4.90 × 10–30 cm, , 1, , O, , N, F, , H, , H 2O : m =1.84 D, , F, O, , H, , H

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CBSE New Pattern ~ Chemistry XI (Term-I), , 88, C, O is a linear molecule., (iv) O, It’s dipole moment is zero and hence, non-polar., Or, Dipole moment = charge ( q ) ´ distance of, separation ( d ), , 65. (i) According to assumptions of MOT, the, atomic orbitals which undergo LCAO, must have the comparable energy because, each specific orbital have its own specific, energy depending upon various quantum, number and their sum., Thus, both Assertion and Reason are, correct, Reason is not the correct, explanation of Assertion., (ii) The linear combination of atomic orbitals, to form molecular orbitals takes place, if, the combining atomic orbitals must have, the same or nearly the same energy., 1s-orbital can not combine with another, 2s-orbital but with 1s-orbitals because the, energy of 2s-orbitals is appreciably higher, than that of 1s-orbital. This is not true, if, the atoms are very different., Thus, Assertion is correct but Reason is, incorrect., , (iii) Bond dissociation energy is the energy, required to break one mole of bonds of a, particular type. So as to separate them, into gaseous atoms. Bond energy is the, average value of dissocation energies of, the same type of bond present in the, molecule., Bond energy of C ¾ H bond in methane is, 99.2 kcal/mol., Thus, for polyatomic molecules, average, bond energy is taken as the dissociation, bond energy., Hence, both Assertion and Reason are, correct and Reason is not he correct, explanation of Assertion., (iv) With increase in bond order, bond length, decreases and hence bond energy, increases., So, both Assertion and Reason are correct, and Reason is the correct explanation of, Assertion., Or, Both Assertion and Reason are correct, but Reason is the correct explanation of, Assertion.

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89, , Redox Reactions, , 05, Redox Reactions, Quick Revision, The reactions that involve oxidation and reduction, as its two half reactions are called redox reactions., , 1. Classical Idea of Redox Reactions, According to classical concepts, oxidation and, reduction are defined as the process that involve:, Oxidation, (i) Addition of oxygen, (ii) Addition of electronegative element, (iii) Removal of hydrogen, (iv) Removal of electropositive element, Reduction, (i) Removal of oxygen, (ii) Removal of electronegative element, (iii) Addition of hydrogen, (iv) Addition of electropositive element, , 2. Oxidising and Reducing Agents, In a redox reaction, the substance which, oxidises the other species or itself undergo, reduction is called the oxidising agent. The, substance that reduces the other species and, itself undergo oxidation is called the reducing, agent., , 3. Redox Reactions in Term of Electron, Transfer, Loss of electrons or an increase in oxidation, state is called oxidation. Gain of electrons or a, decrease in oxidation state is called reduction., Because of simultaneous loss and gain of, electrons in oxidation-reduction processes, the, redox reactions, (or the oxidation-reduction, reactions) are also called electron transfer, reactions., , 4. Oxidation Number, It is defined as ‘‘the charge that an atom of the, element possesses in its ion or appear to have, when present in the combined state with, other atoms.’’, , 5. Rules for Calculating Oxidation Number, These rules are given below:, Rule 1 The oxidation number of an atom in its free, or elementary state or in any of its allotropes, is zero. e.g. The oxidation state of H in H 2 , S, in S 8 , P in P4., Rule 2 In case of ions having only one kind of, atoms, the oxidation number of each atom, is equal to charge present on the ion. e.g. In, case of Na + , Mg 2+ , Fe 3+ , Cl − and O2− , the, oxidation state is respectively + 1 , + 2 , + 3 ,, − 1, − 2 ., Rule 3 The oxidation state of alkali metals in all, their compounds is always + 1. Similarly,, in case of alkaline earth metals, it is always, +2. For aluminium, oxidation state is, always + 3., Rule 4 The oxidation state of oxygen in most of its, compounds is − 2, with an exception of, peroxides and superoxides in which the, oxidation state of oxygen is respectively − 1, and − 1/ 2 ., Rule 5 The oxidation state of hydrogen is generally, + 1 with an exception of metallic hydrides, like NaH, CaH 2 etc. In these hydrides,, oxidation state of hydrogen is −1 .

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CBSE New Pattern ~ Chemistry XI (Term-I), , 90, Rule 6 The oxidation state of fluorine in all of its, compounds is always − 1. Other halogens (i.e., chlorine, bromine and iodine) also, exhibit − 1 oxidation state but it is not, always true., In case of oxoacids and oxoanions, halogens, (except fluorine) exhibit positive oxidation, state., Rule 7 The algebraic sum of the oxidation numbers of, all the atoms present in a compound must be, equal to zero., Rule 8 In case of polyatomic ions, the algebraic, sum of oxidation number of all the atoms, present in the ion must be equal to the, charge on the ion. e.g. In case of carbonate, ion (CO2−, 3 ), it is equal to −2., , 6. Paradox of Fractional Oxidation Number, −2, , +2, , 0, , O ==, , O, +6, , == ==, , ●, , +2, , −2, , In C 3O2 , O == C == C == C == O, two carbon atoms are present in +2 oxidation, state each whereas third one is present in zero, oxidation state., The average of oxidation states of 3 C-atoms in, 2+ 2+ 0 4, C 3O2 =, =, 3, 3, Average oxidation state of Br in Br3O8 is 16/3, while the oxidation states of three Br atoms are, + 6, + 4 and + 6 as shown below:, +4 +6, , ==, , ●, , O, , O== Br — Br —Br=== O, , =, , O=, , ●, , O, , =, , O, , The average oxidation state of four S-atoms in, S 4O2–, 6 is 2.5 while the actual oxidation state of, the four S-atoms are +5, 0, 0 and +5 as shown, below:, , O, O, , +5, , , 0, 0, +5, – O  S  S  S  S  O–, , , O, O, , 7. Balancing of Redox Reactions, (i) Ion electron method The method, involves the following steps:, (a) Write redox reaction in ionic form., (b) Split redox reaction into oxidation half, and reduction half reactions. Balance, atoms of each half reactions by using, simple multiples., For balancing H and O, add H + ion and, H 2O to the appropriate sides, similarly, add OH − and H 2O to the appropriate, sides., Balance the charge on both sides and, multiply one or both half reactions by, suitable number to equalise number of, electrons in both equations. Add the two, balance half reactions and cancel, common terms., (ii) Oxidation number method The method, involves the following steps:, (a) Assign oxidation number to the atoms in, the equation and write separate equations, for atoms undergoing oxidation and, reduction., (b) Find the change in oxidation number in, each equation and make the change, equal in both the equations by, multiplying with suitable integers., After adding both the equations, complete the balancing (by balancing H, and O).

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Objective Questions, Multiple Choose Questions, 1. Which of the following reactions, represent(s) redox process?, (a) Electrochemical process for extraction of, highly reactive metals and non-metals, , Which of the following processes takes, place?, (a), (b), (c), (d), , (b) Manufacturing of caustic soda, (c) Corrosion of metals, , 7. In the following reaction,, , (d) All of the above, , 2. Which of the following processes take, place in oxidation?, (a) Addition of oxygen, (c) Removal of oxygen, , (b) Addition of hydrogen, (d) Removal of chlorine, , 3. Which of the following processes take, place in reduction?, , 2FeCl 3 (aq ) + H 2 ( g ) → 2FeCl 2 (aq ), +2HCl(aq ), , reduction process, oxidation process, addition process, All of the above, , 5. In the reaction given below, identify the, species undergoing oxidation and, reduction, respectively, H 2 S + Cl 2 → 2HCl + S, (a), (b), (c), (d), , the process taking place w.r.t. Mg is, known as, (a) oxidation, (c) redox reaction, , (b) reduction, (d) None of these, , 8. In the given reaction,, , (a), (b), (c), (d), , 4. In the given reaction,, ferric chloride undergoes, , 2Mg (s ) + O 2 ( g ) → 2MgO(s ), , CH 2 == CH 2 + H 2 → H 3C — CH 3, there occurs, , (a) Removal of oxygen, (b) Addition of hydrogen, (c) Removal of hydrogen, (d) Both (a) and (b), , (a), (b), (c), (d), , Oxidation due to removal of potassium, Oxidation due to removal of iron, Reduction due to removal of potassium, Oxidation due to removal of, electronegative element, , H2 S is oxidised and Cl2 is reduced, H2 S is reduced and Cl2 is oxidised, Both H2 S and Cl2 are oxidised, Both H2 S and Cl2 are reduced, , 6. In the given reaction,, , 2K 4 [Fe(CN) 6 ](aq ) + H 2O 2 (aq ) →, 2K 3 [Fe(CN) 6 ] (aq ) + 2KOH (aq ), , oxidation of ethylene, reduction of ethylene, Both (a) and (b), None of the above, , 9. In oxidation process,, (a) oxidation number decreases, (b) number of electrons decreases, (c) oxygen content decreases, (d) number of ions decreases, , 10. Given the reaction for the discharge of, a cobalt-cadmium battery, 2Co(OH) 3 + Cd +2H 2O → 2Co(OH) 2, + Cd(OH) 2, Which species is oxidised during the, discharge of the battery?, (a) Co3+, , (b) Co2+, , (c) Cd, , (d) Cd2+, , 11. Both oxidation and reduction takes, place in, (a) NaBr + HCl → NaCl + HBr, (b) HBr + AgNO 3 → AgBr + HNO 3, (c) H2 + Br2 → 2HBr, (d) CaO + H2 SO 4 → CaSO 4 + H2O

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CBSE New Pattern ~ Chemistry XI (Term-I), , 92, 12. In the reaction,, 2KClO 3 → 2KCl + 3O 2 ,, the elements which have been oxidised, and reduced respectively are, (a), (b), (c), (d), , chlorine and oxygen, oxygen and chlorine, potassium and oxygen, oxygen and potassium, , (b) H2O2, (d) K2Cr2O 7, , 14. Water molecule is formed by the reaction,, 2H 2 + O 2 → 2H 2 O, What does happen in this reaction?, Electrons are transferred from H to O-atom, Electrons are transferred from O to H-atom, Electrons are accepted by H from O-atom, Electrons are donated by O to H-atom, , 15. Which of the following is not an example, of redox reaction?, (a), (b), (c), (d), , (NCERT Exemplar), , CuO + H2 → Cu + H2O, Fe2O 3 + 3CO → 2Fe + 3CO2, 2K + F2 → 2KF, BaCl2 + H2 SO 4 → BaSO 4 + 2HCl, , 16. Match the Column I (Reaction with, underlined species) with Column II, (Type of change shown by underlined, species) and choose the correct option, from the codes given below., Column I, (Reactions), , (b) Mg, , (c) Al, , (d) Ba, , 18. In the reaction,, , oxidising and reducing agent is, , (a), (b), (c), (d), , A B C D, (b) 3 4 1 2, (d) 3 2 1 4, , 17. Strongest reducing agent is, (a) K, , 13. The compound that can work both as an, (a) KMnO 4, (c) Fe2 (SO 4 ) 3, , Codes, A B C D, (a) 2 3 4 1, (c) 3 4 2 1, , Column II, (Type of change), , A., , 1. Removal of, 2Mg + O2 →, hydrogen., 2MgO, , B., , 2. Removal of, Mg + Cl 2 →, electropositive, MgCl 2, element., , C., , 3. Addition of, 2H2S+O2 →, oxygen., 2S+2H2O, , D. 2KI +H2O + O3 → 4. Addition of, electronegative, 2KOH + I2 + O2, element., , 4Na + O 2 → 2Na 2O,, sodium acts as a/an, (a) oxidising agent, (c) complexing agent, , (b) reducing agent, (d) None of these, , 19. Which reaction indicates the action of, HNO3 as oxidising agent?, (a) NaOH + HNO 3 → NaNO 3 + H2O, (b) Ca(OH)2 + 2HNO 3 → Ca(NO 3)2 + 2H2O, (c) C 6H6 + HNO 3 → C 6 H 5 NO2 + H2O, (d) NaCl + HNO 3 → HCl + NaNO 3, , 20. In the following reaction reducing, agent is, 14H + + Cr2 O 72– +3Ni → 2Cr 3+, + 7H 2 O + 3Ni 2+, (a) H+, , (b) Cr2O2−, 7, , (c) H2O, , (d) Ni, , 21. Which is the best description, behaviour of bromine in the given, equation?, H2O + Br2 → HBr + HOBr, (a) Proton acceptor, (b) Both oxidised and reduced, (c) Oxidised, (d) Reduced, , 22. In which of the following compounds,, nitrogen exhibits highest oxidation, state?, (a) N2H4, (c) N3H, , (b) NH3, (d) NH2OH, , 23. Oxidation states of X, Y, Z are +2, +5, and –2 respectively. The formula of, the compound formed by these will be, (a) X2YZ6, , (b) XY2 Z6, , (c) XY5, , (d) X 3YZ4

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93, , Redox Reactions, , 24. In which of the following, Fe exhibits, minimum oxidation state?, (a) K 4Fe(CN) 6, (b) Fe 3 O 4, (c) Fe(CO) 5, (d) FeSO 4 (NH4 )2 SO 4 .6H2O, , (a) Na2 S4O 6 > H2 S2O 7 > Na2 S2O 3 > S8, (b) H2 SO 4 > SO2 > H2 S > H2 S2O 8, , (a) 3d 14 s 2, (c) 3d 5 4 s 1, , (d) H2 SO 5 > H2 SO 3 > SCl2 > H2 S, , 26. In which of the following compounds,, an element exhibits two different, oxidation states?, (b) NH4NO 3, (d) N3H, , 27. State the oxidation number of carbonyl, carbon in methanal and methanoic acid, respectively., (b) 0 and + 2, (d) + 1 and + 3, , 28. +3 oxidation state is most common in …, (b) Fe (26), (d) Cu (29), , 29. In the reaction, 4Na + O 2 → 2Na 2 O,, sodium acts as a/an, (a) oxidising agent, (c) complexing agent, , in Cl 2 , NaOCl and ClO −3 are, respectively., , (b) 0, + 2, + 5, (d) 0, + 1, + 5, , 31. What is the average oxidation number, of carbon in carbon suboxide?, (a) +, , 4, 3, , (b) +, , 10, 4, , (c) +2, , (d) +, , 2, 3, , (b) 3d 34 s 2, (d) 3d 5 4 s 2, , 34. Magnesium reacts with an element ( X ), to form an ionic compound. If the, ground state electronic configuration of, ( X ) is 1s 2 2s 2 2 p 3 , the simplest formula, for this compound is, (a) Mg2 X, (c) Mg2 X 3, , (b) MgX2, (d) Mg 3X2, , 35. The valency of Cr in [Cr(H 2 O)4 Cl 2 ] +, ion is, (a) 3, (c) 6, , (b) 1, (d) 5, , 36. Match the Column I with Column II, and select the correct option for, oxidation number of N-atom from the, codes given below., Column I, (Compounds), , (b) reducing agent, (d) None of these, , 30. The value of oxidation numbers of Cl,, (a) + 2, 0, + 5, (c) + 2, + 3, + 5, , by an element depends on its outer, electronic configuration. With which, of the following outer electronic, configurations the element will, exhibit largest oxidation number?, , (NCERT Exemplar), , (c) SO22 − > SO24− > SO23− > HSO 4−, , (a) Ni (28), (c) Zn (30), , (b) +6, + 6, + 4, (d) −4, + 6, + 6, , 33. The largest oxidation number exhibited, , arranged in the decreasing order of, oxidation number of sulphur?, , (a) 0 and 0, (c) + 1 and + 2, , H 2 SO4, HSO −3 and SO 2 Cl 2, respectively are, (a) +6, + 4, + 6, (c) +6, − 6, + 4, , 25. Which of the following have been, , (a) NH2OH, (c) N2H4, , 32. The oxidation states of sulphur in, , Column II, (Oxidation number), , A., , NH2OH, , 1., , –1, , B., , Mg3N2, , 2., , −1, , C., , N2O, , 3., , +5, , D., , N2O5, , 4., , −3, , Codes, A B C D, (a) 1 3 4 2, (b) 2 4 3 1, (c) 2 4 1 3, (d) 4 2 1 3

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CBSE New Pattern ~ Chemistry XI (Term-I), , 94, 37. In oxygen difluoride (OF2 ) and, , dioxygen difluoride (O 2 F2 ), the oxygen, is assigned an oxidation number of, (a), (b), (c), (d), , +1 and +2 respectively, +2 and +2 respectively, +1 and +1 respectively, +2 and +1 respectively, , 38. Which of the following arrangements, represent increasing oxidation number, of the central atom ? (NCERT Exemplar), (a) CrO , ClO , CrO24− , MnO −4, (b) ClO −3 , CrO24− , MnO 4− , CrO2−, (c) CrO2− , ClO 3− , MnO 4− , CrO24−, (d) CrO24− , MnO 4− , CrO2− , ClO 3−, −, 2, , −, 3, , 39. Which of the following reactions is, represented in basic medium?, (a) MnO −4 (aq) + SO2 (g) → Mn2 + (aq) + HSO 4− (aq), (b) H2O2 (aq) + Fe2 + (aq) → Fe 3+ (aq) + H2O (l), , (c) MnO –4 (aq) + I− (aq) → MnO2 (s ) + I2 (s ), , 3+, 2−, (d) Cr2O2–, 7 + SO2 (g) → Cr (aq) + SO 4 (aq), , 40. In the following redox reaction,, xUO 2+ + Cr 2O 27− + yH + →, 3+, aUO 2+, + bH 2 O, 2 + z Cr, the values of coefficients x , y and z, respectively, are, (a) 3, 8, 2, (c) 3, 2, 4, , (b) 3, 8, 7, (d) 3, 1, 8, , acid and sulphuric acid is heated., During the reaction which element, undergoes maximum change in, oxidation number?, (b) H, , (c) Cl, , 42. Assertion KMnO 4 is a stronger, , oxidising agent than K 2 Cr2 O7 ., Reason This is due to the increasing, stability of the lower species to which, they are reduced., , 43. Assertion In the reaction between, potassium permanganate and potassium, iodide, permanganate ions act as, oxidising agent., Reason Oxidation state of manganese, changes from +2 to +7 during the, reaction., , 44. Assertion Average oxidation number of, , I in KI 3 is − 1/3., Reason KI 3 is made up of KI and I 2 . Each, species have different oxidation number., , 45. Assertion PbCl 4 is a powerful oxidising, agent., Reason PbCl 4 is more stable than PbCl 2 ., , 41. A mixture of potassium chlorate, oxalic, , (a) S, , (a) Both Assertion and Reason are correct, statements and Reason is the correct, explanation of the Assertion., (b) Both Assertion and Reason are correct, statements, but Reason is not the correct, explanation of the Assertion., (c) Assertion is correct, but Reason is incorrect, statement., (d) Assertion is incorrect but Reason is correct, statement., , (d) C, , Assertion-Reasoning MCQs, Directions In the following questions, (Q.No. 42-55) a statement of Assertion, followed by a statement of Reason is, given. Choose the correct answer out of, the following choices., , 46. Assertion In some cases oxygen shows, positive oxidation number though it is, an electronegative element., Reason Fluorine is more electronegative, than oxygen., , 47. Assertion H 2SO 4 cannot act as a, , reducing agent., Reason Sulphur cannot increase its, oxidation number beyond + 6., , 48. Assertion Among halogens fluorine is, the best oxidant.

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95, , Redox Reactions, , Reason Fluorine is the most, electronegative atom. (NCERT Exemplar), , 49. Assertion The two Fe atoms in Fe 3 O 4, have different oxidation numbers., Reason Fe 2 + ions decolourise KMnO 4, solution., , 50. Assertion In the species, Br3 O 8 each, of two extreme bromine exhibits, oxidation state of + 6 and the middle, bromine of + 4., Reason The average of three oxidation, numbers of bromine of the Br3 O 8 is16 / 3., , 51. Assertion Amongst the halogens,, fluorine cannot oxidise the element to, the highest oxidation state., Reason Due to small size of fluoride, ion, it is difficult to oxidise flouride ion, to fluorine. Hence, reverse reaction, takes place more easily., , 52. Assertion The oxidation number of O, in O 3 is zero and the oxidation number, of S in SO 3 is + 4., Reason O 3 can act as an oxidising, agent as well as a reducing agent but, SO 2 can act only as an oxidising agent., , 53. Assertion The formal oxidation, number of sulphur in Na 2S4 O 6 is 2.5., Reason Two S-atoms are not directly, linked with O-atoms., , 54. Assertion MnO −4 is reduced to Mn 2 + in, acidic medium., Reason In acidic medium, following, reaction takes place., KMnO 4 + H 2O → MnO 2, , 55. Assertion Cl 2 gas bleaches the articles, permanently., Reason Cl 2 is a strong reducing agent., , Case Based MCQs, 56. Read the passage given below and, answer the following questions :, The oxidation state of an individual atom, is 0. The total oxidation state of all atoms, in a neutral species is 0 and in an ion is, equal to the ion charge. Group 1 metals, have an oxidation state of + 1 and group 2, an oxidation state of + 2., The oxidation state of fluorine is − 1 in, compounds. Hydrogen generally has an, oxidation state of + 1 in compounds., Oxygen generally has an oxidation state of, − 2 in compounds., In binary metal compounds, group 17, elements have an oxidation state of − 1,, group 16 elements of − 2, and group 15, elements of − 3. The sum of the oxidation, states is equal to zero for neutral, compounds and equal to the charge for, polyatomic ion species. An atom is, oxidised if its oxidation number increases, and an atom is reduced if its oxidation, number decreases., The atom that is oxidised is the reducing, agent and the atom that is reduced is the, oxidising agent., (Note the oxidising and reducing agents, can be the same element or compound)., Redox reactions are comprised of two, parts, a reduced half and an oxidised half,, that always occur together., The reduced half gains electrons and the, oxidation number decreases, while the, oxidised half loses electrons and the, oxidation number increases., The ion or molecule that accepts electrons, is called the oxidising agent, by accepting, electrons it causes the oxidation of another, species. conversely, the species that, donates electrons is called the reducing, agent; when the reaction occurs, it reduces, the other species.

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CBSE New Pattern ~ Chemistry XI (Term-I), , 96, The following questions (i-iv) are multiple, choice questions. Choose the most, appropriate answer :, (i) One mole of acidified K2 Cr2 O7 on, reaction with excess KI will liberate n, mole of I 2 , then the value of n is, (a) 6, (c) 3, , (b) 1, (d) 7, , (ii) When electrons are transferred from, Zn to Cu2+ in copper sulphate, solution, the energy (heat) is, (a) absorbed, (b) evolved, (c) consumed, (d) Both (a) and (b), , (iii) Negative E s indicates that redox, couple is, , (a) weaker reducing agent than H+ / H2 couple, (b) stronger reducing agent than H+ / H2 couple, (c) sronger oxidising agent than H+ / H2 couple, (d) weaker oxidising agent than H+ / H2 couple, , (iv) Which of the following statements, is/are incorrect?, (a) The reactants, which undergo oxidation, and reduction are called reductant and, oxidant respectively, (b) In redox reaction, the oxidation number, of oxidant increases, while that of, reductant decreases, (c) HNO2 acts as an oxidising as well as, reducing agent, (d) Oxidation is the process, in which, electrons are lost, , Or, In alkaline medium, ClO 2 oxidises, H2O 2 to O 2 and itself gets reduced, to Cl − . How many moles of H2O 2, are oxidised by 1 mole of ClO 2 ?, (a) 1, (c) 2.5, , (b) 1.5, (d) 5, , 57. Read the passage given below and, answer the following questions :, The concept of electron transfer is found, unable to explain the redox changes or, electron shift in case of covalent, compounds., , To explain these changes a new concept,, called oxidation number is introduced., Oxidation number is defined as the charge, that an atom of the element has in its ion or, appear to have when present in the, combined state with other atoms., In other words, it is also defined as the, charge that an atom appear to have in a, compound when all other atoms are, removed as ions from the compound., , The following steps are involved while, calculating the oxidation number of an, atom in a given compound/ion., Step I Write down the formula of given, compound/ion leaving some space, between the atoms., Step II Write the oxidation state of each, element above its atoms. Write down, x above the atom, oxidation state of, which we have to find out., Step III Multiply the oxidation numbers of, each element with the number of, atoms of that element present in the, compound., Enclose the product in a bracket., Step IV Equate the algebraic sum of the, oxidation numbers of all the atoms, present in a compound to zero or to, the charge in case of ionic species, charge on the ion., Step V Solve the equation obtained for the, value of x., , The following questions (i-iv) are multiple, choice questions. Choose the most, appropriate answer:, (i) Highest oxidation state of Mn is present, in, (a) KMnO4, (c) Mn2O3, , (b) K2MnO4, (d) MnO2, , (ii) Identify the element which never has, positive oxidation number in any of its, compound?, (a) Oxygen, (b) Chlorine, (c) Fluorine, (d) Bromine

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97, , Redox Reactions, , (iii) When a manganous salt is fused with a, mixture of KNO 3 and solid NaOH, the, oxidation number of Mn changes,, from + 2 to, (a) + 4, (c) + 6, , (b) + 3, (d) + 7, , (iv) The brown ring complex compound is, formulated as [Fe(H 2O) 5 NO]SO 4 ., What will be the oxidation state of, iron in the given complex?, (a) + 2, (c) + 4, , (b) + 3, (d) + 1, , Or, In which of the following reactions,, there is no change in valency?, , (a) SO2 + 2H2S → 2H2O + 3S, (b) 2Na + O2 → 2Na2O2, (c) Na2O2 + H2SO4 → Na2SO4 + H2O2, (d) 4KClO3 → 3KClO4 + KCl, , 58. Read the passage given below and, answer the following questions :, In a redox reaction, the substance which, oxidises the other or which itself, undergoes reduction is called the oxidising, agent and the substance that reduces the, other and itself undergoes oxidation is, called the reducing agent. e.g., 2HgCl 2 + SnCl 2 → Hg 2 Cl 2 + SnCl 4, Thus, in the above example, HgCl 2 is an, oxidising agent (as it reduces) and SnCl 2 is, a reducing agent (as it oxidises)., In these questions (i-iv) a statement of, Assertion followed by a statement of, Reason is given. Choose the correct, answer out of the following choices :, (a) Assertion and Reason both are correct, statements and Reason is correct, explanation for Assertion., (b) Assertion and Reason both are correct, statements but Reason is not correct, explanation for Assertion., (c) Assertion is correct statement but Reason, is incorrect statement., (d) Assertion is incorrect statement but, Reason is correct statement., , (i) Assertion This reaction is redox, reaction,, K 2Cr2O7 (aq ) + 3SO2 ( g ) + H 2 SO4 (aq ) →, Cr2 (SO4 )3 (aq ) + K 2 SO4 (aq ) + H 2O(l ), , Reason Both oxidation and reduction, take place in the given reaction., (ii) Assertion The decomposition of, hydrogen peroxide to form water and, oxygen is an example of, disproportionation reaction., Reason The oxygen of peroxide is in, −1 oxidation state and it is converted to, zero oxidation state in O2 and −2, oxidation state in H2 O., (iii) Assertion In the species, Br3O 8 each, of two extreme bromine exhibits, oxidation state of + 6 and the middle, bromine of + 4., Reason The average of three oxidation, numbers of bromine of the Br3O 8 is, 16/ 3., (iv) Assertion In the reaction between, potassium permanganate and, potassium iodide, permanganate ions, act as oxidising agent., Reason Oxidation state of manganese, changes from +2 to +7 during the, reaction., , Or, Assertion A negative value of E ° means, that the redox couple is a weaker, reducing agent than the H + /H2 couple., Reason A negative E ° means that the, redox couple is stronger reducing agent, than the H + /H2 ., , 59. Read the passage given below and, answer the following questions :, The real or imaginary charge which an, atom appears to have in its combined state, is called oxidation state or oxidation, number of that atom. Fraction oxidation, states are often used to represent the, average oxidation states of several atom in

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CBSE New Pattern ~ Chemistry XI (Term-I), , 98, a structure. These oxidation states are very, helpful in finding the oxidation and, reduction process in redox reactions., Redox reactions are of two main, i.e, intermolecular redox reactions and, intramolecular redox reactions., The elements that show an increase in, oxidation number (hydrogen and chlorine, in the above reaction) are oxidised, while, the elements that are reduced (oxygen and, chlorine in the above reaction) show a, decrease in their oxidation numbers from, their initial values., In these questions (i-iv) a statement of, Assertion followed by a statement of, Reason is given. Choose the correct, answer out of the following choices :, (i) Assertion Oxidation state of nitrogen, 1, in N 3 H is − ., 3, Reason Nitrogen is less, electronegative than hydrogen., (ii) Assertion The decomposition of, hydrogen peroxide to form water and, oxygen is an example of, disproportionation reaction., , Reason The oxygen of peroxide is in, −1 oxidation state and it is converted to, zero oxidation state in O2 and −2, oxidation state in H2 O., (iii) Assertion The electrons are, transferred from zinc to copper, through the wire which connects the, two rods., Reason Electricity flows through the, salt-bridge by migration of ions from, one beaker to other., (iv) Assertion Redox couple is the, combination of oxidised and reduced,, form of a substance involved in an, oxidation or reduction half-cell., Reason In the representation, E s 3+ 2+ and E s 2+, , Fe 3+ / Fe 2+ and, Fe, , / Fe, , Cu, , / Cu, , Cu 2+ / Cu are redox couples., , Or, Assertion Oxidation number of, hydrogen − 1 in CaH 2 ., Reasons CaH 2 is a metal hydrides and, for hydrides, hydrogen is assigned the, oxidation number of −1., , ANSWERS, Multiple Choice Questions, 1. (d), 11. (c), 21. (b), , 2. (a), 12. (b), 22. (c), , 3. (d), 13. (b), 23. (b), , 4. (a), 14. (a), 24. (c), , 5. (a), 15. (d), 25. (d), , 6. (a), 16. (b), 26. (b), , 7. (a), 17. (a), 27. (b), , 8. (b), 18. (b), 28. (b), , 9. (b), 19. (c), 29. (b), , 10. (c), 20. (d), , 31. (a), 41. (c), , 32. (a), , 33. (d), , 34. (d), , 35. (a), , 36. (c), , 37. (d), , 38. (a), , 39. (c), , 30. (d), 40. (a), , 45. (c), 55. (c), , 46. (a), , 47. (a), , 48. (b), , 49. (b), , 50. (b), , 51. (b), , Assertion-Reasoning MCQs, 42. (a), 52. (c), , 43. (c), 53. (c), , 44. (a), 54. (d), , Case Based MCQs, 56. (i)-(c), (ii)-(b), (iii)-(b), (iv)-(b) or-(c), 58. (i)-(a), (ii)-(a), (iii)-(b), (iv)-(c) or-(d), , 57. (i)-(a), (ii)-(c), (iii)-(c), (iv)-(b) or-(c), 59. (i)-(c), (ii)-(a), (iii)-(b), (iv)-(b) or-(b)

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EXPLANATIONS, 1. Electrochemical processes for the extraction of, highly reactive metals and non-metals,, manufacturing of chemical compounds like, caustic soda, operation of dry and wet batteries, and corrosion of metals fall within the range of, redox processes., , 9. In oxidation process, oxidation number, increases and number of electrons decreases., +3, , 0, , 10. 2 Co(OH)3 + Cd + 2H2O →, , + 2, , Increase in ON, , 2. Addition of oxygen/electronegative element and, removal of hydrogen/electropositive element, takes place in oxidation., , 3. Reduction is a process which involves addition, of hydrogen or electropositive elements to a, substance or removal of oxygen or, electronegative element from a substance., , Here, oxidation number of Cd increases from 0, to +2, hence Cd is oxidised., , 11. Both oxidation and reduction are taking place in, the following reaction,, Oxidised (reducing agent), , 4. In the given reaction, removal of, , 0, , Reduced (oxidising agent), , H2 -reducing agent; Br2 -oxidising agent., In other options, neither oxidation nor, reduction takes place because oxidation, number of elements involved in reaction, remain same., , Removal of chlorine, (electronegative element), , 5. H2 S + Cl 2 → 2HCl + S, , +1 + 5 − 2, , +1 −1, , Loss of e −, (oxidation), , Thus, H2 S is oxidised and Cl 2 is reduced., , Gain of e − (reduction), , 6. In the given reaction,, , 13. Oxygen in H2O2 has oxidation number −1 which, can increase or decrease., Hence, it can act as both oxidising agent or, reducing agent., , Removal of potassium (electropositive element), , So, here oxidation takes place due to removal of, one potassium atom., , 7. Oxidation is a process, which involves addition, of oxygen/electronegative element to a, substance or removal of hydrogen/electropositive, element from a substance., e.g. 2Mg + O 2 → 2MgO, Addition of oxygen = Oxidation, , 8. CH2==CH2 + H2 → H3C — CH3, (Addition of hydrogen), , Reduction of ethylene occurs due to the, addition of hydrogen., , 0, , 12. 2KClO3 → 2 KCl + 3 O2, , Removal of hydrogen (oxidation), , 2K4 [Fe(CN)6 ]( aq ) + H2O2( aq ) →, 2K3[Fe(CN)6 ] ( aq ) + 2KOH( aq ), , + 1 −1, , 0, , H2 + Br2 → 2H Br, , electronegative element, i.e. chlorine from ferric, chloride takes place. Hence, it is an example of, reduction process., 2FeCl 3 ( aq ) + H2 ( g ) → 2FeCl 2 ( aq ) + 2HCl( aq ), , Addition of hydrogen, (reduction), , + 2, , 2Co(OH)2 + Cd(OH)2, , 14., , Hydrogen is oxidised by loss of 1 electron, 0, , 0, , +1 –2, , 2 H2 + O2 → 2 H2 O, , Oxygen is reduced by gain of 2 electrons, , In this reaction, hydrogen (H) has transferred, electrons to oxygen (O)., , 15. Following are the examples of redox reaction :, (a) CuO + H2 → Cu + H2O, (b) Fe2O3 + 3CO → 2Fe + 3CO2, (c) 2K + F 2 → 2KF, Option (d) is not an example of redox reaction.

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CBSE New Pattern ~ Chemistry XI (Term-I), , 100, , Here, oxidation number of bromine increases as, well as decreases, i.e. bromine is oxidised as, well as reduced., , 16. The correct match is, , A → 3, B → 4, C → 1, D → 2, Oxidation (addition of oxygen), , 22. Let, the oxidation state of nitrogen in the given, compounds be x., , (A) 2Mg + O2 → 2MgO, , x +1, , (a) N 2H4 , 2x + ( + 1) 4 = 0, 2x = − 4, x=−2, , Oxidation (addition of electronegative element), , (B) Mg + Cl 2 → MgCl 2, , x +1, , (C) 2H2S + O2 → 2S + 2H2O, , (b) NH3 , x + ( + 1) 3 = 0, x=−3, , Reduction (removal of hydrogen), x +1, , (D) 2KI + H2O + O3 → 2KOH + I2 + O2, , (c) N 3H, 3x + ( + 1) = 0, 3x = − 1, x = − 1/ 3, , Oxidation (removal of electropositive element), , 17. Since, K contains only one electron in its, outermost shell, it has higher tendency to donate, it, i.e. has higher tendency to get oxidised., Therefore, it is the strongest reducing agent, among the given elements., , 18. 4 Na + O2 → 2Na 2O, Loss of e − (oxidation), , In the above reaction, Na converts into (Na + ), ion, i.e. Na donates its electron to oxygen atom., So, it behaves as reducing agent., +5, , +3, , 19. C6H6 + HNO3 → C6H5 — NO2 + H2O, Reduction (gain of e − ), , In this reaction, HNO3 behaves as an oxidising, agent while in rest of the reactions such as,, (i) NaOH + HNO3 → NaNO3 + H2O, (ii) Ca(OH)2 + 2HNO3 → Ca(NO3 )2 + 2H2O, (iii) NaCl + HNO3 → HCl + NaNO3, HNO3 neither behaves as oxidising agent nor as, reducing agent., , 20., , Increase in ON (oxidation +2), +6, , +, , 3+, , 2–, , 14H + Cr2O7 + 3Ni, , +2, , x +1 − 2 + 1, , (d) NH2OH,, x + ( + 1) 2 + ( − 2) + ( + 1) = 0, x + 2 − 2 +1= 0, x + 1 = 0 ⇒ x = −1, The oxidation state of nitrogen is highest in N3H., , 23. We know that, the algebraic sum of the oxidation, states is always zero in neutral compound., Oxidation states of X = + 2, Y =+5, Z = −2, So, the algebraic sum of total X, Y and Z should, be equal to zero which is found in XY 2Z 6 ., XY 2Z 6 = + 2 + ( 5 × 2) + ( −2 × 6 ), = + 2 + 10 − 12 = 0, , 24. (a) K4 Fe(CN) 6 ⇒1 × 4 + x + ( −1)6 ⇒ 0 ; ∴x = 2, , (b) Fe 3 O 4 ⇒ FeO. Fe 2 O 3 ⇒ + 2, + 3, (c) Fe(CO) 5 ⇒ x + 0( 5) = 0 ; ∴x = 0, (d) FeSO 4 (NH4 ) 2 SO 4 .6H 2 O ⇒ x + (–2) + 0 + 0 = 0, ∴, x =+2, Thus, in Fe(CO)5 , Fe shows minimum, oxidation number., , 2Cr + 7H2O + 3Ni, , 2.5, , +6, , +2, , 0, , 25. (a) Na 2 S4 O6; H 2 S2 O7 ; Na 2 S2 O3; S8, Decrease in ON (reduction), , As in this reaction Ni is oxidised to Ni 2+ and, reduced to Cr2 O 72 – , thus it acts as reducing agent., −, , Loss of e (oxidation), 0, , +, , –, , – +, , 21. H2O + Br2 → H Br + H O Br, Gain of e − (reduction), , + 6, , + 4, , −2, , +6, , (b) H2 SO 4; SO 2; H2S; H2 S2 O8, + 2, , + 6, , + 4, , +6, , (c) SO22; SO24; SO23− ; H SO−4, + 6, , + 4, , + 2, , −2, , (d) H2 SO5; H2 SO3; SCl 2; H2 S, , Thus, in option (d), compounds are arranged in, the decreasing order of oxidation number of, sulphur.

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101, , Redox Reactions, , 26. NH4 NO3 is actually NH+4 and NO−3 . It is an, ionic compound., , ⇒ x + ( 4 + 1) = +1 or x + 4 = +1, or, x = −3, Let, oxidation number of N in NO−3 is x, ⇒ x + ( 3x − 2) = −1 or x − 6 = −1 or x = +1, , 27. Let the oxidation number of carbonyl carbon in, methanal (HCHO) and methanoic acid (HCOOH), is x and y respectively., In HCHO, 2 ( + 1) + x + ( − 2) = 0, 2 + x − 2= 0 ⇒x = 0, In HCOOH, 2 ( + 1) + y + 2 ( − 2) = 0, 2+ y − 4= 0 ⇒y = + 2, , º, º, , 3d, , 28. (a) Ni(III) = 3d 7 4s 0 =, , 4s, , ¼ ¼¼, 3d, , (b) Fe(III) = 3d 5 4s 0 =, , 4s, , ¼ ¼ ¼¼¼, , + 3 oxidation state of Fe provides the extra, stability due to half-filled d -orbitals., (c) Zn(III) = 3d 4s =, , 4s, , º, º, º, º, , 3d, 0, , ¼, , (d) Cu(III) = 3d 4s =, (0), , 0, , º, º, º, , 3d, 8, , 4s, , ¼¼, , +1, , 29. 4 Na+ O 2 → 2 Na 2 O, Loss of e − (oxidation), , In this reaction,, Na converts into (Na + ) ion, i.e. Na donates its, electron to oxygen atom. So, it behaves as an, reducing agent., , 30. Oxidation number of Cl in molecular state, (i.e. in Cl 2 ) is zero., Let, oxidation number of Cl in NaOCl is x., ∴, 1 + ( −2) + x = 0, x = +1, Let, oxidation number of Cl in ClO –3., ∴, , 0, , +2, , Carbon suboxide, , The oxidation number of nitrogen in the two, species is different as shown below:, In NH+4 ,, , 9, , +2, , 31. O == C == C == C == O, , x + 3( −2) = − 1, x = +5, , In C3O2 , two C-atoms linked with oxygen atoms, are present in +2 oxidation state and central, carbon has zero oxidation state., 4, So, the average oxidation state of carbon is + ., 3, , 32. (i) Let oxidation number of S in H 2 SO 4 is x., , ∴ +1 × 2 + x + ( −2) × 4 = 0, x = 8 − 2 or x = + 6, Therefore, oxidation number of S in H 2 SO 4, is +6., (ii) Let oxidation number of S in HSO −3 is x., , ( +1) + x + ( −2) 3 = − 1, x = − 1 + 5 or x = + 4, Therefore, oxidation number of S in HSO−3, is +4., (iii) Let the oxidation number of S in SO2Cl 2, be x., x + 2( −2) + 2( −1) = 0, x = +6, Therefore, in SO 2 Cl 2 the oxidation number of, sulphur is +6, , 33. Highest oxidation number of any transition, , element = (n – 1) d electrons + ns electrons., Therefore, larger the number of electrons in the, 3d -orbitals, higher is the maximum oxidation, number., (a) 3 d 1 4s 2 = 3, (b) 3 d 3 4s 2 = 3 + 2 = 5, (c) 3 d 5 4s 1 = 5 + 1 = 6 and, (d) 3 d 5 4s 2 = 5 + 2 = 7, Thus, option (d) is correct., , 34. Given, electronic configuration of X =1s 2 2s 2 2 p 3, ∴ The valency of X will be 3., The valency of Mg is +2., ∴ Magnesium reacts with element X to form an, ionic compound with formula Mg 3 X 2 ., , 35. The valency of Cr in [Cr(H 2 O) 4 Cl 2 ]+ ion is x., , x + 4 × ( 0 ) − 2 ( −1) = + 1, x =1 + 2, x =3, So, the valency of Cr in [Cr(H 2 O) 4 Cl 2 ]+ ion is 3.

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CBSE New Pattern ~ Chemistry XI (Term-I), , 102, , Thus, Cl is the element which undergoes, maximum change in the oxidation state from +5, to −1., , 36. The correct match is, A → 2, B → 4, C → 1, D → 3., The oxidation number of N-atom in given, compounds are shown below :, –1+1–2+1, , +2, , –3, , +1 –2, , +5 –2, , NH2OH, Mg3N2, , N2O, , N2O5, , –1, , –1, , +5, , 42. Both Assertion and Reason are correct, explanation and Reason is the correct explanation, for Assertion., , 43. The reaction of potassium permanganate and, –3, , potassium iodide is as follows :, , 37. Electronegativity of fluorine is more than that of, oxygen atom, so F gains electron with negative, charge., In oxygen difluoride (OF2 ) and dioxygen, difluoride (O2F2 ), oxygen transfers electron to, fluorine atom. Thus,, Oxidation number of oxygen in OF 2 = + 2., Oxidation number of oxygen in O2F 2 = + 1., , 38. Writing the oxidation number (O.N.) of Cr, Cl, and Mn on each species in the four set of ions,, then,, +3, , +5, , +5, , +6, , +3, , +5, , +6, , +7, , –, –, (b) Cl O–3 ,Cr O2–, 4 , Mn O4 ,Cr O2, +6, , +7, , +3, , +5, , Only in the arrangement (a), the ON of central, atom increases from left to right, therefore,, option (a) is correct., +4, , −1, , 0, , 39. Mn O–4 ( aq ) + I−( aq ) → MnO2 ( s ) + I 2 ( s ), This reaction is represented in basic medium, +7, , because in basic medium MnO−4 is reduced to, +4, , +7, , +4, , MnO2 (i.e. Mn to Mn ), while in acidic medium,, MnO−4 is reduced from Mn7+ to Mn 2+ ., , 40. The balanced chemical reaction is given as, 3UO2+ + Cr2O72− + 8H+ →, , 3UO22+ + 2Cr 3+ + 4H2O, Hence, the value of x , y and z are, respectively 3, 8 and 2., , 41. When a mixture of potassium chlorate, oxalic, acid and sulphuric acid is heated, the following, reaction occurs :, +1 +5 − 2, , + 1+ 3− 2, , +1 + 6 −2, , ∆, , KClO3 + H2C2O4 + H2 SO4 →, + 1+ 6− 2, , ∴Average oxidation number of I, −1 × 1 + 0 × 2 −1, =, =, 1+ 2, 3, Thus, either two values are reported separately or, one value is reported., , Assertion is correct but Reason is incorrect, statement., , 3–, –, –, (d) Cr O2–, 4 , Mn O4 ,Cr O2 ,Cl O3, , +7, , 44. KI 3 dissociates into KI and I 2 ., , (Oxidising agent), , (c) Cr O–2 ,Cl O–3 , Mn O–4 ,Cr O2–, 4, +6, , Oxidation state of Mn decreases from +7 to +2., Thus, Assertion is correct but Reason is incorrect., , stable than Pb 4 + (due to the inert pair effect), 2e − + Pb 4 + → Pb 2 + (Reduction), , +3, , +7, , +2, , 2 Mn SO4 + 6 K 2 SO4 + 8 H2O + 5 I 2, , 45. PbCl 2 is more stable than PbCl 4 or Pb 2 + is more, , –, (a) Cr O–2 ,Cl O–3 ,Cr O2–, 4 , Mn O4, +7, , +7, , 10KI + 2K MnO4 + 8 H2 SO4 →, , + 1− 1, , + 4− 2, , + 1− 2, , K 2 SO4 + KCl + CO2 + H2O, , 46. Oxygen is the most electronegative element after, fluorine. Therefore, in the compounds between, oxygen and fluorine, oxygen is found to show, positive oxidation state., Both Assertion and Reason are correct statements, and Reason is the correct explanation of the, Assertion., 47. Maximum oxidation state of S is + 6, it cannot, exceed beyond, it. Therefore it, cannot be further, oxidised., Both Assertion and Reason are correct statements, and Reason is the correct explanation of the, Assertion., , 48. Among halogen F 2 is the best oxidant because it, , has the highest E ° value., Both Assertion and Reason are correct, but, Reason is not the correct explanation of Assertion., + 2, , + 2, , 49. Fe3O 4 ≡≡ (FeO ⋅ Fe2 O 3 ), The oxidation states of Fe in FeO and Fe2O 3 are, + 2 and + 3., Fe2 + + MnO −4 → Fe3 + + Mn 2 +, (Pink), , (Colourless)

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103, , Redox Reactions, , Both Assertion and Reason are correct, statements, but Reason is not the correct, explanation of the Assertion., , 50. The structure of Br3O8 (tribromooctaoxide) is, O–2, , –2, , O, +6, , –2, , Br, , O, –2, , O, , +4, , Br, , O–2, +6, , Br, , O–2, , O–2, O–2, , Thus, oxidation state of two corner Br atoms is, +6 and of middle one is +4. The difference in, oxidation states is due to difference in bonding, situations., + 6 + 4 + 6 16, Average oxidation state =, =, 3, 3, Thus, both Assertion and Reason are correct, but Reason is not the correct explanation of, Assertion., , 51. Due to high electronegativity and high heat of, association, fluorine oxidises the elements to, their highest oxidation state., Both Assertion and Reason are correct, statements, but Reason is not the correct, explanation of the Assertion., 52. SO 2 can act both as an oxidising and a reducing, agent. O 3 can act only as an oxidising agent., The oxidation number of O in O 3 is zero. It can, only decrease from zero to − 1 or − 2 but cannot, increase to + 2. Therefore, it can act as an, oxidising agent only. In SO 2, the oxidation, number of S is + 4. It can have a minimum, oxidation number of − 2 and maximum of + 6., Its oxidation number either decreases or, increases and hence, it can act both as an, oxidising and a reducing agent., Assertion is correct, but Reason is incorrect, statement., O, O, , , , + −, 0, 0, 5, +5 − +, 53. NaO  S +, S  S  S  ONa, , , O, O, Formal oxidation number of sulphur, 2×5+ 2×0, =, = 2.5, 4, Assertion is correct, but Reason is incorrect, statement., , 54. MnO −4 is reduced to MnO2 in a mild basic, , medium or neutral medium whereas in an acidic, medium,, MnO−4 , is reduced to Mn 2 + and in a strong basic, medium, it is reduced to MnO24 − ., So, Assertion is incorrect but Reason is correct., 55. Cl 2 is an oxidising agent. It bleaches the articles, permanently by oxidation in presence of moisture., Assertion is correct, but Reason is incorrect, statement., , 56. (i) K 2Cr2O7 + Excess KI → Cr 3+ + H2O + nI 02, 6+, , +3e −, , 3+, , Cr → Cr, Cr 6+ accepts 3e − , so mole of I 2 = 3., (ii) If zinc rod is dipped in copper sulphate, solution, then due to transfer of electron, from zinc to copper ion, heat is evolved., (iii) A negative E s means that the redox couple, is a stronger reducing agent than H+ /H2, couple., − E s = strong reducing agent, + E s = weak reducing agent, , (iv) Oxidant = oxidising agent = reduces itself by, increasing oxidation number., Reductant = reducing agent = oxidising itself,, by decreasing oxidation number., Or, In alkaline medium, the balanced equation is,, 2ClO2 + 5H2O2 + 2OH– → 2Cl – + 5O2, + 6H2O, 2 moles of ClO2 oxidise 5 moles of H2O2, 5, ∴ 1 mole of ClO2 will oxidise = × 1 = 2.5., 2, , 57. (i) Let the oxidation state of Mn be x., (a) KMnO 4 ,, + 1 + x + ( − 2) × 4 = 0, x −7 = 0, x =+7, (b) K 2 MnO 4 ,, ( + 1) × 2 + x + ( − 2) × 4 = 0, x −6 = 0, x =+6, (c) Mn 2 O 3 ,, x × 2 + ( − 2) × 3 = 0, 2x = + 6, x =+3

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CBSE New Pattern ~ Chemistry XI (Term-I), , 104, (d) MnO 2 , x + ( − 2) × 2 = 0, x =4, Thus, oxidation state of Mn is highest in, KMnO 4., (ii) Fluorine is the most electronegative, element., Hence, does not possess positive oxidation, number in any of its compound., ( +2 ), , (iii) Mn 2 + + 2NO −3 + 2OH− →, (+ 6), , MnO 24 − + 2H2O + 2NO, , The oxidation number of Mn changes from, + 2 to + 6., (iv) Let oxidation state of Fe in complex is x., ∴ x + 5 ( 0 ) + ( − 1) = − 2, or, x =+3, Or, −2, , 0, , (a) SO 2 + 2H2 S → 2H2O + 3 S, 0, , +1, , 0, , Thus, both Assertion and Reason are correct, and Reason is the correct explanation of, Assertion., (iii) The structure of Br3O8 (tribromooctaoxide) is, O, , O–2, O–2, Thus, oxidation state of two corner Br atoms, is +6 and of middle one is +4. The, difference in oxidation states is due to, difference in bonding situations., + 6 + 4 + 6 16, Average oxidation state =, =, 3, 3, Thus, both Assertion and Reason are correct, but Reason is not the correct explanation of, Assertion., (iv) The reaction of potassium permanganate, and potassium iodide is as follows :, O, , +7, , +2, , 2 Mn SO4 + 6 K 2 SO4 + 8 H2O + 5 I 2, , −1, , Oxidation state of Mn decreases from +7 to +2., Thus, Assertion is correct but Reason is, incorrect., Or, As we know H+ / H2 couple has zero, standard reduction potential so, ions having, positive E ° value are weaker reducing agent,, while ions having negative E ° value are, stronger reducing agent. Thus, Assertion is, incorrect but Reason is correct., , −1, , +7, , (d) 4K ClO 3 → 3KClO 4 + KCl, , 58. (i) The redox change in the given reaction is as, follows :, Reduction, , K2Cr2O7 + 3SO2 + H2SO4, (+6), , (+4), , Cr2(SO4)3, (+3) (+6), , Oxidation, , + K2SO4 + H2O, , 59. (i) Let, oxidation state of N in N 3H be x., 3x + 1 = 0, , (+6), , 1, 3, Nitrogen is more electronegative than, hydrogen., ∴Assertion is correct but Reason is, incorrect., x =−, , ∴ Both Assertion and Reason are correct, and Reason is the correct explanation of, Assertion., (ii), –1, 2H2O2, , Oxidation, , –2, , O–2, , Br, , –2, , (c) 2 Na 2 O 2 + H2SO 4 → Na 2 SO 4 + H2O 2, + 5, , Br, , O–2, +6, , 10KI + 2K MnO4 + 8 H2 SO4 →, , +1, , (no change in valency), , +4, , Br, , O, , −1, , −1, , +6, , –2, , (b) 2 N a + O 2 → 2 Na 2 O 2, +1, , O–2, , –2, , 0, , 2H2 O + O2, Reduction, , Thus, the above reaction is an example of, disproportionation reaction., , (ii), –1, 2H2O2, , Oxidation, , –2, , 0, , 2H2 O + O2, Reduction

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Redox Reactions, , Thus, the above reaction is an example of, disproportionation reaction., Both Assertion and Reason are correct and, Reason is the correct explanation of, Assertion., (iii) The electrons are transferred from Zn to, Cu 2+ through the metallic wire which, connects the two rods., While electricity flows through the, salt-bridge by migration of ions from one, beaker to other., Both Assertion and Reason are correct, statements but Reason is not the correct, explanation of Assertion., (iv) Redox couple is the combination of oxidised, and reduced form of a substance., , 105, In the representation E °Fe 3 + / Fe 2 + and, E °Cu 2 + / Cu , Fe3 + /Fe2 + and Cu 2+ /Cu are, redox couples., Both Assertion and Reason are correct, statements but Reason is not the correct, explanation of Assertion., Or, Oxidation number of elements in their, compounds or ions is obtained using some, rules, e.g. hydrogen is assigned oxidation, number of + 1 in general and − 1 for metal, hydrides., Both Assertion and Reason are correct, statements but Reason is not the correct, explanation of Assertion.

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CBSE New Pattern ~ Chemistry XI (Term-I), , 106, , 06, Hydrogen, Quick Revision, 1. Occurrence of Hydrogen, Dihydrogen (H 2 ) is the most abundant element, in the universe (70% of the total mass of the, universe) and is the principal element in the solar, atmosphere. The giant planets Jupiter and Saturn, consist of mainly hydrogen., , 2. Position of Hydrogen in the Periodic Table, ●, , ●, , ●, , Hydrogen is the first element in the periodic, table with electronic configuration1s 1 ., In atomic form, it consists of one proton and, one electron and in elemental form it exists as, a diatomic (H 2 ) molecule and is called, dihydrogen., Its position is not certain because it has, resemblance to alkali metals which lose one, electron to form unipositive ions as well as, with halogens which gain one electron to form, uninegative ion., , 3. Isotopes of Hydrogen, (i) Hydrogen has three isotopes: protium (11 H),, deuterium or heavy hydrogen (12 H or D) and, tritium (13 H or T)., (ii) Tritium is radioactive and emits low energy, β − -particles., (iii) These three isotopes have different masses, hence, their rates of reaction and equilibrium, constants are different. This is known as, isotopic effect. Because of the extreme, temperature of sun fusion of hydrogen atoms, occurs, which liberates large amount of, energy., 4 11 H → 42 He + 210 e + Energy, Positron, , 4. Compounds of Hydrogen, (i) Hydrides Hydrogen combines with almost, all the elements except noble gases under, certain reaction conditions to give binary, compounds. These binary compounds are, called hydrides. Hydrides can be classified, into following categories, (a) Ionic, saline or salt like hydrides, They are formed by elements of group 1, and group 2., They are crystalline, non-volatile and, non-conducting in solid state., e.g. Li +H − , Na +H − , Ca 2 +H −2 , etc., (b) Covalent or molecular hydrides, They are formed by p-block and s-block, elements, e.g. HF. They are covalent,, volatile and non-conducting. These are, further classified as (i) electron deficient, (ii) electron precise and (iii) electron rich, hydrides., (c) Metallic or non-stoichiometric, hydrides d -block elements of group 3,, 4, 5, 10, 11, 12 and f -block elements on, heating with H 2 under pressure form, these hydrides., They conduct heat and electricity and, are mostly non-stoichiometric. e.g., VH 0.56 , NiH 0.6-0.7 etc., Hydride gap In group 6, only Cr forms, the hydride CrH. The metals of group 7,, 8 and 9 do not form hydrides.

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107, , Hydrogen, , (ii) Water It is a compound of hydrogen and, oxygen, in which these are present in, 1 : 8 by weight. Its formula is H 2O. It is, very essential for existence of all forms, of life., Structure of water in gaseous state, , 7. Chemical Properties of Water, (i) Amphoteric nature, H 2O( l ) + NH 3 (aq ), , +, 4, , −, , sNH (aq ) + OH (aq ), H O( l ) + H S (aq ) sH O (aq ) + HS (aq ), Acid 1, , Base 2, , Acid 2, , Base 1, , +, , 2, , 2, , 3, , Base 2, , Acid 2, , Acid 1, , −, , Base 1, , (ii) Auto-protolysis or self-ionisation, 2δ–, , O 95.7 pm, δ+, , H, , 104.5°, , H 2O(l ) + H 2O(l ), , H, , Acid1, (acid), , δ+, , H, , H, , (b), (a), (a) The bent structure of water with dipole, (b) The orbital overlap picture in water, , 5. Structure of Ice, Here each oxygen atom is surrounded, tetrahedrally by four other oxygen atom. Due, to hydrogen bonding, ice has open structure, with wide holes., , Base 2, , +, , −, , sH O (aq ) + OH (aq ), 3, , Acid 2, (conjugate, acid), , (base), , Base 1, (conjugate, base), , (iii) Redox reactions involving water, 2Na(s ) + 2H 2O(l ) → 2NaOH(aq ) + H 2 ( g ), 6CO2 ( g ) + 12H 2O(l ) → C 6H12O6 (aq ), + 6H 2O(l ) + 6O2 ( g ), (iv) Hydrolysis reactions, SiCl 4 (l ) + 2H 2O(l ) → SiO2 (s ) + 4HCl(aq ), Silicon, tetrachloride, , AlN, , Aluminium, nitride, , Silicon, dioxide, , + 3H 2O → Al(OH)3 + NH 3, Aluminium, hydroxide, , Ammonia, , 8. Hard and Soft Water, , =O, =H, Three dimensional highly ordered hydrogen, bonded structure of ice., , 6. Physical Properties of Water, ●, , ●, , ●, , ●, , It is transparent, colourless, tasteless and, odourless substance., It is freezing point, boiling point, heat of, vapourisation and heat of fusion is higher than, hydrides of other elements of group 16., It has high specific heat thermal conductivity,, surface tension and dielectric constant., It is regarded as universal solvent., , Water that forms lathers with soap is called soft, water, e.g. rain water, distilled water etc., and, that does not do so is called hard water, e.g. sea, water, well water, etc., Types of hardness The hardness of water is of, two types :, (i) Temporary hardness It is because of the, presence of bicarbonates (hydrogen, carbonates) of calcium and magnesium, so it, is also called carbonate hardness., The methods used to remove temporary, hardness are as follows:, (a) By boiling, Heating, , Mg(HCO3 )2 → Mg(OH)2 ↓ + 2CO2 ↑, Magnesium, bicarbonate, , Magnesium, hydroxide, , Heating, , Ca(HCO3 )2 → CaCO3 ↓ + H 2O, Calcium, bicarbonate, , Calcium, carbonate, , + CO2 ↑, (b) Clark’s method, Ca(HCO3 )2 + Ca(OH)2 → 2CaCO3 ↓, Calcium, bicarbonate, , Calcium, hydroxide, , Calcium, carbonate, , + 2H 2O

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CBSE New Pattern ~ Chemistry XI (Term-I), , 108, Mg(HCO3 )2 + 2Ca(OH)2 → 2CaCO3 ↓, Magnesium, bicarbonate, , Calcium, hydroxide, , The water obtained from this process is called, demineralised water., 2R  H(s ) + M 2 + (aq ) → MR 2 (s ) + 2H + (aq ), (where, M = Ca or Mg), , Calcium, carbonate, , + Mg(OH)2 ↓ + 2H 2O, Magnesium, hydroxide, , (ii) Permanent hardness It is because of the, presence of chlorides and sulphates of, calcium and magnesium in the water., Following methods can be employed to, remove this type of hardness:, (a) By using washing soda, (sodium carbonate), M Cl 2 + Na 2CO3 → M CO3 ↓ + 2NaCl, , RNH 2 (s ) + H 2O( l ) → RNH +3 OH − ( l ), , RNH +3 OH − (s ) + X − (aq ) →, , RNH +3 X − (s ) + OH − (aq ), , 9. Heavy Water, It is the oxide of heavy hydrogen, i.e. deuterium, and thus, has the formula D2O. Its molecular, mass is 20. It was discovered by Harold C., Urey, an American chemist in 1932., (i) Physical properties Some important, properties are as follows :, ● It is colourless, odourless, tasteless liquid., It has maximum density −1.1073 g mL −1, at 11.6°C (water at 4°C)., ● Solubility of salts in heavy water is less, than in ordinary water because it is more, viscous than ordinary water., (ii) Chemical properties Heavy water gives all, the reactions that are shown by ordinary, water but the rate of these reactions is slower., This is called isotopic effect ., e.g. 2Na + 2D2O → 2NaOD + D2, , Sodium, chloride, , Sodium, carbonate, , (where, M = Mg, Ca), , M SO4 +Na 2CO3 → M CO3 ↓ +Na 2 SO4, Sodium, sulphate, , Sodium, carbonate, , (b) Calgon's method (sequestration), Na 6P6O18, , Sodium, hexametaphosphate, , M, , 2+, , 2−, → 2Na + + Na 4P6O18, , 2−, + Na 4P6O18, , (where, M = Mg, Ca), → [Na 2 M P6O18 ]2 −, + 2Na +, , Sodium, , The trade name for sodium, hexametaphosphate is calgon (which, means calcium gone), thus, the process is, called Calgon process., (c) Ion-exchange method When sodium, aluminium silicate (NaAlSiO4 ) is added, in hard water, then exchange reactions, take place., 2NaZ (s ) + M 2 + (aq ) →, Sodium, MZ 2 (s ) + 2Na + (aq ), aluminium, silicate, , (where, M = Mg, Ca), where, Z = Al 2 Si 2O8 ⋅ xH 2O, (d) Synthetic resin method In this, method, all types of cations, (Na 2 + , Ca 2 + , Mg 2 + etc.) and anions, (Cl − , SO24 − , HCO3− etc.) can be removed., , CaC 2, Calcium, carbide, , Sodium, deuteroxide, , + 2D2O → Ca(OD)2 + C 2D2, Calcium, deuteroxide, , NaOH + D2O, , Sodium, hydroxide, , CHCl 3, Chloroform, , + D2O, , a NaOD, , Sodium, deuteroxide, , 3 +, Deutero, chloroform, , a CDCl, , Deutero, acetylene, , + HOD, , Semiheavy, water, , HOD, , 10.Hydrogen as a Fuel : Hydrogen Economy, ●, , ●, , Dihydrogen is an abundant element in the, nature and on combustion produces a large, amount of heat energy which is much more, than that is obtained by combustion of fuels like, methane, LPG, petrol., Hydrogen economy involves safe transportation, and storage of energy in the form of H 2 .

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109, , Hydrogen, , Objective Questions, Multiple Choose Questions, 1. Which of the following statements is, incorrect?, (a) Hydrogen is the most abundant element in, the universe, (b) The giant planets Jupiter and Saturn, consist mostly hydrogen, (c) The isotopes of hydrogen have different, physical properties, (d) Hydrogen is used to reduce lighter metal, oxides (more active than iron) to metals, , 2. In which of the following respect,, electronic configuration of hydrogen, has resemblance to alkali metals and, halogens respectively?, (a) It lose one electron to form unipositive ion, and gain one electron to form uninegative, ion, (b) It gain one electron to form uninegative ion, and lose one electron to form unipositive, ion, (c) It has the ability to gain one electron only, (d) None of the above, , 3. Which of the following properties of, hydrogen does not resemble with that, of halogens ?, (a) It forms a diatomic molecule, (b) It combines with elements to form hydrides, (c) It forms large number of covalent, compounds, (d) It has same reactivity as halogens, , 4. Most common isotope of hydrogen, (non-radioactive) is, (a) protium, (c) tritium, , (b) deuterium, (d) All of these, , 5. Which of the following does not react, with hydrogen even at high, temperature to form corresponding, hydrides ?, (a) Alkali, (c) Transition metals, , (b) Noble gases, (d) All of these, , 6. Deuterium atom contains how many, proton and neutron(s) respectively?, (a) 0, 1, , (b) 1, 2, , (c) 1, 0, , (d) 1, 1, , 7. Dihydrogen, under certain reaction, conditions, combines with almost all, elements except noble gases to form, binary compounds. The binary, compounds are called ……, (a) oxides, (c) carbides, , (b) halides, (d) hydrides, , 8. Saline hydrides are known to react with, water violently producing fire. Can, CO 2 , a well known fire extinguisher, be, used in this case ?, (a) No, because CO2 gets oxidised by metal, hydride, (b) Yes, because CO2 gets oxidised by metal, hydride, (c) No, because CO2 gets reduced by metal, hydride, (d) Yes, because CO2 gets reduced by metal, hydride, , 9. The saline hydrides, remove traces of, water from organic compounds because, (a) in saline hydrides, the H − ion is a strong, Bronsted base, (b) in saline hydrides, the H − ion is a weak, Bronsted base, (c) in saline hydrides, the M + ion is a strong, Bronsted acid, (d) in saline hydrides, the M + ion is a weak, Bronsted acid, , 10. Which of the following is covalent and, polymeric in structure?, (a) LiH, (c) MgH2, , (b) BeH2, (d) Both (b) and (c), , 11. The d-block elements forms ……, (a), (b), (c), (d), , ionic hydrides, non-stoichiometric hydrides, molecular hydrides, covalent hydrides

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CBSE New Pattern ~ Chemistry XI (Term-I), , 110, , 12. Recent studies show that the metallic, hydrides have different lattice from that, of the parent metal except …… ., (a) Cu, , (b) Pt, , (c) K, , (d) Ac, , 13. Which of the following statements is, correct regarding metallic or, non-stoichiometric hydrides?, (a) These are formed by all d and f-block, elements, (b) These hydrides conduct heat and electricity, (c) Like saline hydrides they are almost always, stoichiometric, (d) None of the above, , 14. Metal hydrides are ionic, covalent or, molecular in nature. Among LiH, NaH,, KH, RbH, CsH, the correct order of, increasing ionic character is, , (NCERT Exemplar), , (a) In gas phase,H O H bond angle is 109°, 28′, (b) O H bond length is 95.7 pm, (c) In liquid phase, there occurs intramolecular, H-bonding, (d) All of the above, , 18. The density of water is less in its solid, state because, , (a) in solid state, water molecules are arranged, in highly ordered open cage like structure, (b) more extensive hydrogen bonding is present, in solid state, (c) the water molecules are closest in solid, state of water, (d) water is a rigid crystalline, closely packed, structure in its solid state, , 19. In which of the following reactions H 2O, acts only as a Bronsted acid?, (a) H2O (l) + NH3 (aq), (b) H2O (l) + H2 S (aq), , −, , +, 4, −, , 5 OH (aq) + NH (aq), 5H O (aq) + HS (aq), (c) H O (l) + H O (l) 5 H O (aq) + OH (aq), 2, , 15. Which of the following is incorrect, statement ?, (a) s-block elements, except Be and Mg form, ionic hydrides, (b) BeH2 ,MgH2 , CuH2 , ZnH2 , CaH2 and HgH2 are, intermediate hydrides, (c) p-block elements form covalent hydrides, (d) d- and f-block elements form ionic hydrides, , 16. Match the Column I with Column II, and choose the correct option from the, codes given below., Column II, , A. Electron-deficient, molecular hydride, , 1., , CH4, , B. Electron-precise, molecular hydride, , 2., , B2 H6, , C. Electron-rich, molecular hydride, , 3., , NH3, , Codes, A B C, (a) 1 3 2, (c) 3 2 1, , statement about the structure of water?, , (NCERT Exemplar), , (a) LiH > NaH > CsH > KH > RbH, (b) LiH < NaH < KH < RbH < CsH, (c) RbH > CsH > NaH > KH > LiH, (d) NaH > CsH > RbH > LiH > KH, , Column I, , 17. Which of the following is the correct, , A B C, (b) 2 1 3, (d) 1 2 3, , 2, , +, , 3, , 3, , +, , −, , (d) None of the above, , 20. Choose the incorrect statement ., (a) The H  O H angle in water molecule is, 104.5°, (b) The maximum number of hydrogen bonds, formed by a water molecule in ice is 2, (c) Each oxygen in ice crystal is surrounded, tetrahedrally by four other O-atoms, (d) The density of liquid water is higher than, that of ice due to hydrogen bonding, , 21. What is reason of temporary hardness, of water ?, (a) Na2 SO 4, (c) NaCl, , (b) CaCl2, (d) Ca(HCO 3)2, , 22. The temporary hardness of a water, sample is due to compound X . Boiling, this sample converts X to compound, Y . X and Y , respectively, are, (a), (b), (c), (d), , Mg(HCO 3)2 and Mg(OH)2, Ca(HCO 3)2 and Ca(OH)2, Mg(HCO 3)2 and MgCO 3, Ca(HCO 3)2 and CaO

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111, , Hydrogen, , 23. The method used to remove temporary, hardness of water is, (a) Clark’s method, (b) ion-exchange method, (c) synthetic resins method, (d) Calgon’s method, , 29. Match the reactions given in Column I, with the suitable method given in, Column II. Select the correct option, from the codes given below., Column I, , Column II, , Heating, , 24. 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, , 25. The formula of Calgon, used for water, softening is …… ., (a) Na2 [Na4 (PO 3 ) 6], (b) Na4 [Na2 (PO 3 ) 6], (c) Na2 [Na4 (PO 4 ) 5], (d) Na4 [Na4 (PO 4 ) 6], , A. Mg(HCO3 )2 →, Mg(OH)2 ↓ + 2CO2 ↑, B., , 2. Calgon’s, Ca(HCO3 )2 + Ca(OH)2, method, → 2CaCO3 + 2H2O, , 2–, C. M 2 + + Na 4 P 6O18, →, , [Na 2 MP 6O 18 ], , (a), (b), (c), (d), , H+ ions, H+ and Ca2+ ions, OH− and Mg2+ ions, Ca2+ and Mg2+ ions, , 27. In comparison to the zeolite process for, the removal of permanent hardness,, the synthetic resins method is, (a) more efficient as it can exchange only, cations, (b) less efficient as it exchange only anions, (c) less efficient as the resins cannot be, regenerated, (d) more efficient as it can exchange both, cations as well as anions, , 28. Which of the following soften hard, water?, (a), (b), (c), (d), , Passing through cation exchange resin, Passing through lime water, Passing through sand, Passing through alumina, , 2−, , + 2Na, , 3. Boiling, +, , D. 2NaZ ( s ) + M 2 + ( aq ) → 4. Ion-exchange, method, MZ 2 ( s ) + 2 Na + ( aq ), , Codes, A B C D, (a) 3 1 2 4, (c) 4 1 2 3, , A B C D, (b) 4 3 2 1, (d) 3 2 4 1, , 30. Match the items given in Column I with, the relevant item given in Column II., , 26. When zeolite is treated with hard water,, the sodium ions are exchanged with, , 1. Clark’s, method, , Column I, , Column II, , A. Hydrogen peroxide, is used as a, , 1. Zeolite, , B. Used in Calgon, method, , 2. Perhydrol, , C. Permanent hardness 3. Sodium, of hard water is, hexametaphosphate, removed by, 4. Propellant, , Codes, A, (a) 2, 4, (b) 1, (c) 2, 1, (d) 3, , B, 3, 3, 4, 1, 2, , C, 1, 3, 2, 3, 4, , 31. The degree of hardness of water is, usually expressed in terms of, (a) ppm by weight of MgSO 4, (b) g/L of CaCO 3 and MgCO 3 present, (c) ppm by weight of CaCO 3 irrespective of, whether it is actually present or not, (d) ppm of CaCO 3 actually present in water

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CBSE New Pattern ~ Chemistry XI (Term-I), , 112, , 32. The hardness of water is estimated by, ……, (a) EDTA method, (b) titrimetric method, (c) conductivity method, (d) distillation method, , (b) D2O, (d) None of these, , 34. Which of the following is/are the use(s), of heavy water?, (a) It is extensively used as a moderator in, nuclear reactor, (b) It is used in exchange reactions for the, study of reaction mechanism, (c) It is used for the preparation of other, deuterium compounds, (d) All of the above, , 35. Which compound is formed when, calcium carbide reacts with heavy water?, (a) CH4, , (b) C2H2, , (c) C2HD, , (d) C2D2, , 36. Which one of the following statements, is correct about D 2O and H 2O ?, D2O has lower dielectric constant than H2O, NaCl is more soluble in D2O than in H2O, Both (a) and (b) are correct, None of the above, , 37. Select the correct statement., (a) Melting point of H2O is greater than that of, D2O, (b) Hardness of water depends on the soap, consuming power, (c) Marine species can survive in distilled water, (d) Permanent hardness is due to soluble, sulphates chlorides and nitrates of Ca, and Mg, , 38. The fuel used for running the, automobiles first time in the history of, India during October 2005 is …… ., (a) D2O, (c) D2, , (b) 11.6 °C and 4 °C, (d) 12.5 °C and 4 °C, , 40. Advantage of hydrogen economy is the, , is, , (a), (b), (c), (d), , H 2O and D 2O is respectively …… and, ……, (a) 4 °C and 11.6 °C, (c) 4 °C and 12.5 °C, , 33. The moderator used in nuclear reactor, (a) TEL, (c) R  O  R, , 39. Temperature of maximum density of, , (b) H2O2, (d) H2, , (a) transmission of energy in the form of, electric power, (b) transmission of energy in the form of, chemical energy, (c) transmission of energy in the form of, dihydrogen and not as electric power, (d) transmission of mechanical energy, , Assertion-Reasoning MCQs, Directions In the following questions, (Q.No. 41-52) a statement of Assertion, followed by a statement of Reason is, given. Choose the correct answer out of, the following choices., (a) Both Assertion and Reason are correct, statements and Reason is the correct, explanation of the Assertion., (b) Both Assertion and Reason are correct, statements, but Reason is not the correct, explanation of the Assertion., (c) Assertion is correct, but Reason is incorrect, statement., (d) Assertion is incorrect but Reason is correct, statement., , 41. Assertion Hydrogen shows resemblance, with alkali metals as well as halogens., Reason Hydrogen exists in atomic form, only at high temperature., , 42. Assertion Deuterium is also known as, heavy hydrogen., Reason Deuterium is a good conductor, of heat and electricity., , 43. Assertion Saline hydrides are, non-volatile, non-conducting and, crystalline solids., Reason Saline hydrides are compounds, of hydrogen with most of the p-block, elements.

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113, , Hydrogen, , 44. Assertion Ammonia and water are, electron rich hydrides., Reason They have electrons more than, required for bonding., , 45. Assertion Lithium hydride is used in, the synthesis of other useful hydrides., Reason Lithium hydride is unreactive at, moderate temperature with O 2 or Cl 2 ., , 46. Assertion Water has high boiling point., Reason Water shows hydrogen bonding., , 47. Assertion Water is an amphoteric, substance., Reason Water has a tendency to accept, and donate a proton easily., , 48. Assertion Temporary hardness can be, removed by boiling., Reason On boiling, the soluble, bicarbonates change to carbonates, which being insoluble, get precipitated., , 49. Assertion An ice cube floats on water., Reason Density of ice is less than that, of water., , 50. Assertion In winter season, ice formed, on the surface of a lake provides, thermal insulation., Reason It ensures the death of the, aquatic life and this fact is of great, ecological significance., , 51. Assertion Heavy water is used as a, moderator in nuclear reactors., Reason Heavy water is made up of an, isotope of helium., , 52. Assertion Permanent hardness of water, is removed by treatment with washing, soda., Reason Washing soda results with, soluble magnesium and calcium, sulphate to form insoluble carbonates., , Case Based MCQs, 53. Read the passage given below and, answer the following questions :, Hydrogen can exist in three isotopic forms,, viz., protium, deuterium and tritium, which, differ from each other in the number of, neutrons., Out of these three isotopes, tritium is, formed in the upper atmosphere by, reactions induced by cosmic rays. It decays, to emit low energy β-particles., 3, 3, 0, 1 H → 2 He + −1 e, β -particle, , Tritium, , Tritium is used for making thermonuclear, devices and for carrying out researches in, fusion reactions as a source of energy. It is, also used as a radioactive tracer as it is, relatively cheap and easy to work with., The following questions (i-iv) are multiple, choice questions. Choose the most, appropriate answer :, (i) The relative atomic mass of isotopes of, hydrogen is, (a) 1 :2 : 3, (c) 2 :4 :5, , (b) 1 :1 :2, (d) 1 :2 :4, , (ii) The n / p ratio for 1 H 2 is, (a) 1 :2, (c) 2 :1, , (b) 1 :1, (d) 2 : 3, , (iii) Which is the most reactive isotope of, hydrogen?, (a) Tritium, (b) Deuterium, (c) Protium, (d) All are equally reactive, , (iv) What type of reactions are generated, by tritium?, (a) Chemical reaction, (b) Radioactive reaction, (c) Addition reaction, (d) All of these, , Or The isotope of hydrogen are differs in, (a) chemical properties, (b) physical properties, (c) Both (a) and (b), (d) None of these

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CBSE New Pattern ~ Chemistry XI (Term-I), , 114, , 54. Read the passage given below and, answer the following questions:, Water is the main constituent of earth’s, hydrosphere and fluids of all known living, organisms. It is vital for all known forms of, life, even though it provides no chlorines, or organic nutrients., Water cover approximately 70.9% of, earth’s surface, mostly in seas and, oceans. Water plays an important role in, the world economy. Approximately 70%, of the fresh water used by humans goes to, agriculture., Water is the excellent solvent for a wide, variety of substance both mineral and, organic; as such it is widely used in, industrial processes and in cooking and, washing., Water, ice and snow are also central to, many sports and other forms of, entertainment pure water has a low, electrical conductivity, which increases, with the dissolution of a small amount of, ionic material such as common salt., The following questions (i-iv) are multiple, choice questions. Choose the most, appropriate answer :, (i) Which one of the following statements, about water is incorrect?, (a) Water can act both as an acid and as a, base, (b) Water can be easily reduced to, dihydrogen by highly electronegative, elements., (c) Ice formed by heavy water sinks in, normal water, (d) Presence of water can be detected by, adding a drop to anhydrous CuSO 4, , (ii) In nuclear reactors, ordinary water is, not used as a moderator because, (a) it cannot slow down the fast moving, neutrons, (b) it cannot remove the heat from the, reactor core, (c) it has corrosive action on the metallic, parts of the nuclear reactor, (d) None of the above, , (iii) Consider the following statements, about intermolecular and, intramolecular hydrogen bonding., I. Both types of H-bonds are temperature, dependent., II. Water exhibits amphoteric nature., III. The boiling points of compounds, having intramolecular H-bond are, lower than those having intermolecular, H-bond., Which of the statements given above, are correct?, (a) I and III, (c) I and II, , (b) Both II and III, (d) All of these, , (iv) Consider the following statements, regarding water., I. There is extensive hydrogen bonding, between water molecules ., II. Water has high melting point in, comparison to H2 S and H2 Se., III. High heat of vaporisation and heat, capacity of water are responsible for, moderation of climate and body, temperature of living beings., IV. Covalent compounds like alcohol and, carbohydrates dissolve in water., Select the correct statements among, above., (a) Both I and II, (c) I, II and III, , (b) Both II and IV, (d) All of these, , Or, Some of the properties of water are, described below. Which of them is/are, not correct?, I. Water is known to be a universal solvent., II. Hydrogen bonding is present to a large, extent in liquid water., III. There is no hydrogen bonding in the, frozen state of water., IV. Frozen water is heavier then liquid water., Choose the correct option., (a) I and II, (c) III and IV, , (b) II and III, (d) II and III

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Hydrogen, , 55. Read the passage given below and, answer the following questions :, Hydrogen combines with almost all the, elements except noble gases under certain, reaction condition to give binary, compounds. These binary compounds are, called hydrides. These are generally, represented by the formula EH x (where E, is the other element and x is the number of, hydrogen atoms e.g. CaH2 ) or E m Hn (here, also E is the element and m and n show the, number of E and H atoms respectively), e.g., B2 H 6 ., Depending upon the nature of element with, which dihydrogen combine to form hydride,, hydrides can be classified into following, types, i.e., (i) Ionic (ii) Covalent (iii) Metallic, Ionic hydrides are stoichiometric, compounds of hydrogen and the elements, which are more electropositive than, hydrogen. These hydrides are formed by, the transfer of electrons from metal to, hydrogen, thus they have hydrogen in the, form of hydride ion (H− )., Metallic hydrides are the hydrides of, elements of d and f -blocks., In other words, these hydrides do not, follow law of constant composition., When dihydrogen combines with p-block, elements or non-metals, it results in the, formation of covalent hydrides. Further, these hydrides have molecular lattices., In these questions (i-iv) a statement of, assertion followed by a statement of reason, is given. Choose the correct answer out of, the following choices., (a) Assertion and Reason both are correct, statements and Reason is correct, explanation for Assertion., (b) Assertion and Reason both are correct, statements but Reason is not correct, explanation for Assertion., (c) Assertion is correct statement but Reason is, incorrect statement., (d) Assertion is incorrect statement but Reason, is correct statement., , 115, (i) Assertion LiH is an ionic hydride., Reason Lithium is more electropositive, than hydrogen and there is transfer of, electron from metal to hydrogen., , Or, Assertion Ionic hydrides are good, conductor of electricity in molten state., Reason Free ions are present in ionic, hydrides in molten state., (ii) Assertion Covalent hydrides are, volatile compound with lower melting, and boiling point., Reason Covalent hydrides are held, together by weak var der Waals' force., (iii) Assertion Metallic hydrides are always, stoichiometric., Reason Metallic hydrides do not, follow law of constant composition., (iv) Assertion Elements of groups 7, 8, 9, forms by hydrides., Reason This region of periodic table is, called hydride gap., 56. Read the passage given below and, answer the following questions:, Water is classified as either soft or hard., Soft water contains relatively few minerals, and hard water is rich in minerals. Water, hardness is usually expressed as the, number of part per million (ppm) of, calcium carbonate. Regions with soft water, include the Pacific Northwest from Oregon, up through British Columbia. The hard, water region (100 + pm) include the, Canadian Prairies, the U.S Midwest, and, the Southwester states of New Mexico and, Arizona., In a sense, the hardness of water is the, other side of the coin to alkalify. In general, terms, rainy climates such as the Pacific, Northwest have acid water. Rain leaches, out much of the minerals ions in the soil,, replacing them with hydrogen ions. The, result is that the water is rich in hydrogen, and thus acidic (soft). The reverse is the, case in the dry regions, where moisture

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CBSE New Pattern ~ Chemistry XI (Term-I), , 116, evaporates, leaving the minerals intact., The result is water rich in minerals and, thus alkaline (hard)., , In the questions (i-iv), a statement of, assertion followed by a statement of, reason is given. Choose the correct, answer out of the following choices on, the basis of the above passage., (a) Assertion and Reason both are correct, statements and Reason is correct, explanation for Assertion., (b) Assertion and Reason both are correct, statements but Reason is not correct, explanation for Assertion., (c) Assertion is correct statement but Reason, is wrong statement., (d) Assertion is incorrect statement but, Reason is correct statement., , (i) Assertion Pure demineralised water is, obtained by passing water successively, through a cation exchange and an, anion exchange resins., Reason In cation exchange process, H +, exchanges for Na + , Ca 2+ , Mg 2+ and, other cations present in water. In anion, exchange process, OH − exchanges, , anions like Cl − , HCO −3 , SO 24− present in, water., (ii) Assertion Hard water does not lather, with soap., Reason Hard water contains calcium, and magnesium salts in the form of, hydrogen carbonate, chloride and, sulphate., (iii) Assertion Soft water is free from the, soluble salts of calcium and magnesium., Reason It does not lather with soap easily., (iv) Assertion Calgon is used for removing, Ca 2 + and Mg 2 + ions from hard water., Reason Calgon forms precipitate with, Ca 2 + and Mg 2 + ions., Or, Assertion Permanent hardness of water, is removed by treatment with washing, soda., Reason Washing soda reacts with, soluble calcium and magnesium, chlorides and sulphates in hard water to, form insoluble carbonates., , ANSWERS, Multiple Choice Questions, 1. (d), 11. (b), 21. (d), , 2. (a), 12. (d), 22. (a), , 3. (d), 13. (b), 23. (a), , 4. (a), 14. (b), 24. (c), , 5. (b), 15. (d), 25. (c), , 6. (d), 16. (b), 26. (d), , 7. (d), 17. (b), 27. (d), , 8. (c), 18. (a), 28. (a), , 9. (a), 19. (b), 29. (a), , 10. (d), 20. (b), , 31. (c), , 32. (a), , 33. (b), , 34. (d), , 35. (d), , 36. (a), , 37. (b), , 38. (d), , 39. (a), , 30. (a), 40. (c), , 44. (a), , 45. (a), , 46. (a), , 47. (a), , 48. (a), , 49. (a), , 50. (c), , Assertion-Reasoning MCQs, 41. (b), 51. (c), , 42. (c), 52. (a), , 43. (c), , Case Based MCQs, 53. (i)-(a), (ii)-(b), (iii)-(a), (iv)-(b or b), , 54. (i)-(b), (ii)-(d), (iii)-(d), (iv)-(c or a), , 55. (i)-(a or a), (ii)-(a), (iii)-(a), (iv)-(d), , 56. (i)-(a), (ii)-(a), (iii)-(c), (iv)-(a or a)

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117, , Hydrogen, , EXPLANATIONS, 9. Saline hydrides such as NaH, CaH2, etc., react, , 1. Statement (d) is incorrect., It’s correct form is as follows :, Hydrogen is used in metallurgical processes,, it is used to reduce heavy metal oxides to metals., Rest other statements are correct., , 2. Hydrogen has resemblance to the alkali metals,, which lose one electron to form unipositive ions,, as well as with halogens, which gain one, electron to form uninegative ion., , 3. The property of hydrogen which does not, resemble with that halogen is given in, statement (d)., Because it is very low reactive as compared to, halogens., Rest other statements are correct., , 4. Protium ( 11 H) , deuterium or heavy hydrogen, , ( 12 H or D) and tritium ( 13 H or T ) are the isotopes, of hydrogen., Protium or ordinary hydrogen has one proton, and no neutron in the nucleus and one electron, revolves around the nucleus. It is the most, abundant and common form of hydrogen, (approximately 99.98%)., , 5. Noble gases do not react with dihydrogen even at, higher temperature to yield the corresponding, hydrides., , 6. Deuterium, , nucleus ( 12 D ), , has 1-proton, 1-neutron, , and 1-electron., , 7. Dihydrogen, under certain reaction conditions,, combine with almost all elements except noble, gases to form binary compounds, called hydrides., If ‘E ’ is the symbol of an element then hydride, can be expressed as EHx (e. g. MgH2 ) or E m Hn, (e. g . B2H6 )., 8. Saline hydrides (such as NaH, CaH2 etc.) react, with water violently to form the corresponding, metal hydroxides with the evolution of, dihydrogen., NaH ( s ) + H2O(l ) → NaOH ( aq ) + H2( g ), CaH2( s ) + 2H2O ( l ) → Ca(OH)2( aq ) + 2H2( g ), These reactions are so much exothermic that the, evolved H2 catches fire. This type of fire cannot, be extinguished by CO2 because it gets reduced, by the hot metal hydride to form sodium formate., − +, , NaH + CO2 → HCOONa, , with traces of water present in organic, compounds and form their corresponding metal, hydroxides with the evolution of hydrogen gas., NaH(s ) + H2O(aq ) → NaOH(aq ) + H2 (g ), This is because in saline hydrides ( M + H− ),, the H− ion is a strong Bronsted base and, thus it, reacts with water easily., 10. LiH, BeH 2 and MgH 2 are all covalent in nature., LiH molecules do not associate with each other,, while the molecules of BeH 2 and MgH 2 exhibit, aggregation among themselves and form, polymeric chains of molecules, thus BeH 2 and, MgH 2 are polymeric in structure., , 11. d -block elements forms non-stoichiometric, hydrides. These hydrides are formed by, adsorbing hydrogen directly at appropriate, temperatures by metal. The composition of, these hydrides may not correspond to simple, whole number ratio. Their composition varies, with temperature and pressure., , 12. Except Ni, Pd, Ce and Ac, other metallic, hydrides have different lattice from that of the, parent metal., , 13. Statement (b) is correct, while the other, statements are incorrect., Corrected form are as follows :, (a) Metallic hydrides are formed by many d and, f -block elements except group 7, 8 and 9, elements., However, even from group 6, chromium, forms CrH., (c) Unlike saline hydrides, they are almost, always non-stoichiometric, being deficient in, hydrogen., , 14. Ionic character increases as the size of the metal, atom increases or the electronegativity of the, metal atom decreases., The correct order of increasing ionic character is,, LiH < NaH < KH < RbH < CsH, , 15. Statement (d) is incorrect., It’s correct form is as follows :, d and f-block elements form metallic hydrides., Rest other statements are correct., , 16. A → (2); B → (1); C → (3).

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CBSE New Pattern ~ Chemistry XI (Term-I), , 118, 17. Statement (b) is correct, while the other, statements are incorrect., Corrected form are as follows :, (a) In gas phase, H  O  H is a bent molecule, with a bond angle of 104.5., (c) In liquid phase, intermolecular H-bonding, takes place., , 18. When water is converted to ice, an open cage, like three dimensional structure is formed which, has void filled with air. That’s why, density of, ice is less than that of water., , 19. H2O has the ability to act as an acid as well as, base, i.e. it behaves as an amphoteric substance., In the context of Bronsted theory, it acts as an, acid with NH3 and a base with H2S as shown, below :, (a) H2O(l ) + NH3(aq ), OH– (aq ) + NH+4 (aq ), (b) H2O(l ) + H2S(aq ), , l, l, , +, , –, , H3O (aq ) + HS (aq ), , 20. Statement (b) is incorrect., , It’s correct form is as follows :, In liquid state, water molecules form two, hydrogen bonds with their neighbouring water, molecules. In solid state (ice), it contains four, H-bonds due to spatial arrangement of, molecules to form an open cage like structure., Rest other statements are correct., , 21. Temporary hardness of water is due to, Ca ( HCO3 ) 2 ., , In this method, calculated amount of lime is, added to hard water. It precipitates out calcium, carbonate and magnesium hydroxide which can, be filtered off., Ca(HCO3 )2 + Ca(OH)2 → 2CaCO3 ↓ + 2H2O, Mg(HCO3 )2 + 2Ca(OH)2 → 2CaCO3 ↓, + Mg(OH)2 ↓ + 2H2O, Besides this, temporary hardness can also be, removed by boiling. All the other given methods, are used to remove permanent hardness of water., , 24. Permanent hardness of water is due to the, presence of chlorides and sulphates of calcium, and magnesium. Hardness can be removed by, adding Na 2CO3 or by passing the water through, an ion-exchange column., , 25. Calgon is used for water softening. Its formula is, Na 2[Na 4 (PO3 )6 ]. It removes dissolved minerals, of water which reduces the cleaning efficiency of, water., , 26. When zeolite is treated with hard water, the, sodium ions are exchanged with Ca 2+ and, Mg2+ ions., Na 2Z + M 2+ → 2Na + + MZ, , (M = Ca or Mg), , Zeolite, , 27. Zeolites exchange their Na + ions with Ca 2+ or, Mg2+ in hard water. But, they can’t exchange, anions., 2NaZ (s ) + Mg2+ (or Ca 2+ ) (aq ), From hard water, , 22. The temporary hardness of a water sample is, due to compound X [i.e. Mg( HCO3 ) 2 ]. Boiling of, this sample converts X [i.e. Mg( HCO3 ) 2 ] to, compoundY [i.e. Mg(OH)2 ]., Generally,temporary hardness is due to, presence of magnesium and calcium hydrogen, carbonates. It can be removed by boiling., During boiling, the soluble Mg(HCO3 ) 2 is, converted into insoluble Mg(OH) 2 and, Ca( HCO3 ) 2 changed to insoluble CaCO3 ., These precipitates can be removed by, filteration., Heating, , Mg( HCO3 ) 2 → Mg(OH)2 ↓ + 2CO2 ↑, Heating, , Ca(HCO3 ) 2 → CaCO3 ↓ + H2 O + CO2 ↑, , 23. Temporary hardness in water is due to presence, of magnesium and calcium hydrogen, carbonates., Temporary hardness in water can be removed, by Clark’s method., , → MgZ 2(s )(or CaZ 2)(s ) + 2Na + ( aq ), In synthetic resin method, all types cations, (Na + , Ca 2+ , Mg2+ etc.) and anions, (Cl – , SO2–, 4 etc.) can be removed., Thus, this method is more efficient than zeolite, process., 28. Cation exchange resins are used to remove, Na + , Ca 2 + , Mg 2 + electrons these contain large, organic molecules with reactive acidic group, which are water insoluble., , 29. A → (3); B → (1); C → (2); D → (4)., 30. A → (2, 4); B → (3); C → (1, 3)., 31. Degree of hardness of water is measured in, terms of ppm by weight of CaCO3 irrespective of, whether it is actually present or not., , 32. Ethylene diamine tetraacetate (EDTA) when, treated with water forms stable complex with, metal ions and hence, it is used to measure, hardness of water.

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119, , Hydrogen, , 33. D2O is used as moderator in nuclear reactor., Uses of other given options are as follows :, (a) TEL (tetraethyl lead) is used as antiknock agent, in petrol engine., (c) R  O  R (ether) is used as a solvent., , 34. All the given statements are the uses of heavy, water., , 35. D2O is used for the preparation of deuterium, containing compounds. The reaction is similar to, H2 O ., CaC2 + 2D2O → C2D2 + Ca(OD)2, , 36. Statement (a) is correct, while statement (b) is, incorrect. It’s correct form is as follows :, The molecular mass of H2 O is slightly less than that, of D2 O, hence rate of diffusion of NaCl (i.e., solubility) is slightly higher in the case of H2 O., , 37. (b) Among the given option, option (b) is the, correct statement. Correct statements of other, options are as follows, (a) Melting point of D 2O is higher than H2O., (b) Marine species cannot servive in distilled water., (d) Permanent hardness is due to soluble sulphates, of calcium and magnesium., , 38. Hydrogen ( H2 ), 39. H2O and D2O have maximum density at 4°C and, 11.6°C respectively., , 40. Advantage of hydrogen economy is the, transmission of energy in the form of dihydrogen, and not as electric power., , 41. Both Assertion and Reason are correct but Reason, is not a correct explanation of Assertion., Hydrogen can gain an electron to form the H − ion to, get stable noble gas configuration of helium. It can, also lose its electron to give H + ion, i.e. univalent ion, like alkali metals., Hydrogen, therefore has resemblance to the, halogens as well as to the alkali metals which gain, or lose an electron respectively to form univalent, negative and positive ions with noble gas, configurations., Hydrogen exists in atomic form at high, temperatures. At normal temperature is exists as a, diatomic molecule., , 42. Assertion is correct but Reason is incorrect. It’s, correct form is as follows :, Deuterium is a bad conductor of heat and, electricity., , 43. Assertion is correct but Reason is incorrect. It’s, correct form is as follows., Saline or ionic hydrides are compounds of, hydrogen with most of the s-block metals., However, with p-block metals hydrogen forms, molecular or covalent hydrides., 44. Both Assertion and Reason are correct and, Reason is the correct explanation of Assertion., The excess electrons being present as lone, pairs of electrons (ammonia—one lone pair and, water - two lone pairs)., , 45. Lithium hydride is rather unreactive at, moderate temperature with O2 or Cl 2. It is, therefore, used in the synthesis of other useful, hydrides., e.g. 8LiH + Al 2Cl6 → 2LiAlH4 + 6LiCl, 2LiH + B2H6 → 2LiBH4, Thus, both Assertion and Reason are correct, and Reason is the correct explanation of, Assertion., , 46. Both Assertion and Reason are correct and, Reason is the correct explanation of Assertion., The high boiling point of water is due to, hydrogen bonding which holds the water, molecules together rather than leaving them, free., Molecules having hydrogen, bonded to an, electronegative atom, can, interact with another, electronegative atom having a lone pair, leading, to the formation of a hydrogen bond., , 47. Both Assertion and Reason are correct and, Reason is the correct explanation of the, Assertion., According to Bronsted, a substance having an, ability to accept or donate a proton is known, as amphoteric substance., As an acid, H2O + NH3 → OH− + NH4+, As a base, H2O + H2S → H3O + HS−, , 48. Both Assertion and Reason are correct and, Reason is the correct explanation of Assertion., Temporary hardness is due to presence of, bicarbonates of calcium and magnesium which, can be removed by boiling., , 49. Density of ice is less than that of water because, it has open cage like structures which consists, of air filled voids., Therefore, an ice cube floats on water., Thus, both Assertion and Reason are correct, and Reason is the correct explanation of, Assertion.

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CBSE New Pattern ~ Chemistry XI (Term-I), , 120, 50. In winter season, ice formed on the surface of a, lake provides thermal insulation which ensures, the survival of the aquatic life. This fact is of, great ecological significance., Thus, Assertion is correct but Reason is incorrect., 51. Assertion is a correct but Reason is incorrect., It’s correct form is as follows :, Heavy water is made up of an isotope of, hydrogen, i.e. deuterium., 52. Both Assertion and Reason are correct and, Reason is the correct explanation of Assertion., Na 2CO3 + MgSO4 → Na 2 SO4 + MgCO3, or CaSO4, (from hard, water), , or CaCO3, (insoluble), , 53. (i) 1 : 2 : 3, (ii) Number of protons in 1 H2 = 1, Number of neutrons = 1, ∴ n/p ratio = 1 : 1, (iii) Tritium, (iv) Radioactive reactions, Or, The isotopes of hydrogen differ in their, physical properties due to large difference in, atomic masses. This type of difference in, properties due to difference in atomic, masses is called isotope effect., , 54. (i) Statement (b) is incorrect., It’s correct form is as follows :, Water can be easily reduced to dihydrogen, by highly electropositive metals as shown, below :, 2H2O(l ) + 2Na(s ) → 2NaOH(aq ) + H2 (g ), Rest other statements are correct., (ii) Ordinary water stops the nuclear fission by, absorbing the fast moving neutrons., (iii) All the given statements are correct., (iv) I, II and III statements are correct., Statement IV is incorrect covalent, compounds like alcohol and carbonates, doesnot dissolves in water., Or, Water is a universal solvent and exists to a, larger extent in liquid state. It also contains, , H-bonding in frozen state. The density of ice is, lower than that of water due to the presence of, vacant spaces., , 55. (i) Both Assertion and Reason are correct and, Reason is the correct explanation of Assertion., Or, Both Assertion and Reason are correct and, Reason is the correct explanation of Assertion., (ii) Both Assertion and Reason are correct and, Reason is the correct explanation of Assertion., (iii) Assertion is incorrect but Reason is correct., Metallic hydrides are almost always, non-stoichiometric. They do not follow law of, constant composition., (iv) Assertion is incorrect but Reason is correct. It’s, correct form is as follows :, The elements of group 7, 8, 9 do not forms, hydrides but Reason is correct with hydrogen., , 56. (i) Both Assertion and Reason are correct and, Reason is the correct explanation of Assertion., (ii) Presence of calcium and magnesium salts in, the form of hydrogen carbonate, chloride and, sulphate in water makes water ‘hard’. These, salts react with soap molecules to form a, precipitate called scum., That’s why, hard water does not lather with, soap., Thus, both Assertion and Reason are correct, and Reason is the correct explanation of, Assertion., (iii) Soft water gives lather with soap easily as it is, free from soluble salts of calcium and, magnesium., Thus, Assertion is correct but Reason is, incorrect., (iv) Calgon is used for making Ca 2 + and Mg 2 +, ions present in hard water ineffective. It forms, soluble complex with Ca 2 + and Mg 2 + ions., Thus, both Assertion and Reason are correct, and Reason is the correct explanation of, Assertion., Or, MCl 2 + Na 2CO 3 → MCO 3 ↓ + 2NaCl, MSO 4 + Na 2CO 3 → MCO 3 ↓ + Na 2SO 4, (where, M = Mg, Ca)

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121, , Organic Chemistry : Some Basic Principles and Techniques, , 07, Organic Chemistry :, Some Basic Principles and, Techniques, Quick Revision, 1. Classification On the basis of functional groups, organic compounds are classified into following, families or homologous series:, Organic Compounds, Single C—C Double/triple, Halogen, Oxygen containing, Nitrogen containing, bond containing derivatives compounds, bond, compounds, containing, compounds, Alcohol (R—OH), Amines, compound, Aldehyde, Ether (R OR), Haloalkanes Haloarenes, 1° amine (RNH2), (RCHO), C6H5X, R—X, Carbonyl compounds, Alkyne, 2° amine (R2NH), (Alkanes) Alkene, C C —C C — (where, X = F, Cl, Br, I), Ketone, 3° amine (R3N), Carboxylic acids, (R COR), Aromatic amine (ArNH2), (RCOOH), Ester, Nitro compounds (R NO2), (RCOOR), Amide (RCONH2), Cyanide (RCN), Carboxylic acid, Anhydride (RCO)2O Isocyanides (RNC), derivatives, Acid halide (RCOCl), , 2. Nomenclature of Organic Compounds, Steps involved in IUPAC nomenclature are as, follows:, Step 1 Locate the longest chain with as many, possible secondary functional groups, along with multiple bonds., Step 2 Select the root word corresponding to the, length of longest possible chain., Step 3 Number the longest possible chain, according to the rules to give an, identification number to each C-atom of, longest possible chain., , Step 4 Add suitable prefixes and suffixes along, with numerals to indicate the number and, position of each side chain substituent or, functional group present in compound., Hence, according to these rules the IUPAC, name of a compound can be written as :, Prefix, Tells us about the, branches or, substituents except, in the case of, multiple bonds., , + Root word, Tells us about, the length of, longest possible, chain., , + Suffix, Tells us about the, principal functional, groups and multiple, bonds mostly except, in case of halogens.

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122, , CBSE New Pattern ~ Chemistry XI (Term-I), , 3. Fission of a Covalent Bond, (i) Heterolytic cleavage or heterolysis, Heterolysis, X Y    → X + + Y −, (when X is less electronegative than Y ), Heterolysis, X Y    → X − + Y +, (when Y is less electronegative than X ), Heterolytic cleavage results in formation of, two types of species. These are :, (a) Carbocation or carbonium ion If the, carbon has sextet of electrons and a, positive charge, then the species is called, carbocation (earlier called carbonium, ion), e.g. CH +3 (methyl carbonium ion)., It is trigonal planar with C-atom, carrying-positive charge is sp 3 -hybridised., The order of stability of carbocations is, +, , +, , +, , CH 3 < CH 3 CH 2 < (CH 3 )2 CH < (CH 3 )3 C +, or methyl carbocation < 1° < 2 ° < 3 °., (b) Carbanion If the carbon of the, obtained species is negatively charged, and have octet of electrons in its valence, shell, then the species is called, carbanion. It is pyramidal in shape, with-atom carrying negative charge is, sp 3 -hybridised., The order, of stability of carbanions is :, –, –, (C 6H 5 )3 C > (C 6H 5 )2 CH – > C 6H 5 CH 2, , 4. Nucleophiles and Electrophiles, (i) Nucleophiles or nucleophilic reagents An, electron rich species having a lone pair of, electrons or is negatively charged species, which attacks on electron dificient areas is, called a nucleophile (Nu•• ). e.g., Negatively charged species like, –, –, H − , Cl − , Br− , I− , carbanions, O H, − OR , S R ,, CN− . Nucleophiles can also be seen in the, form of neutral molecules with lone pair. e.g., •• ••, , Light, , Alkyl free, radical, , The neutral chemical species, thus formed, are called the free radicals. It is a planar, species in which C atom bearing odd electron, is sp 2 -hydridised. Like carbocations, the order, of stability of alkyl free radicals is, •, , CH3, , •, , •, , •, , < CH2CH3 < CH(CH3 ) 2 < C(CH3 ) 3, , Methyl, free radical, (1°), , Ethyl, free radical, (1°), , Isopropyl, free radical, (2°), , tert -butyl, free radical, (3°), , ••, , ••, , ••, , ••, , ••, , (ii) Electrophiles or electrophilic reagents, An electrophile is defined as electron, deficient species which attacks on electron, rich areas. e.g. : Positively charged species, like H + , Cl + , Br+ , I+ , NO+2 , SO+3 and, carbocations., The electrophiles can also be seen in the, form of neutral molecules,, e.g. carbenes, ••, (•• CR 2 ), nitrenes (N R ), BF3 , etc., 5. Inductive Effect (I-effect), In a polar covalent bond, the electrons are shifted to, the more electronegative atom and hence, a polarity, is developed in the bond. e.g. In a chain of carbon, atoms having C  Cl bond, the C attached directly to, the Cl-atom because of its less electronegativity, acquires some positive charge (δ + )., δδδ +, , δδ +, , δ+, , δ−, , C H 3  C H 2 → CH 2 → Cl, 3, , –, , > allyl carbanions > CH 3 > (1° > 2 ° > 3 °, carbanions), (ii) Homolytic cleavage or homolysis This, cleavage involves movement of only one, electron, which is represented by half, headed (fish hook) curved arrow., Heat or, e.g. R——X ¾¾®, R• + X •, , ••, , H 2 O, N H 3 , R N H 2 , R OH, R S H, etc., , 2, , 1, , This induction of polarity due to presence of, polar bold in an organic molecule is called, inductive effect., , Types of Inductive Effect, (i) Negative inductive effect Atoms or groups, having greater electron affinity than hydrogen, are said to have − I -effect (electron attracting)., Some of the − I -effect producing groups in, decreasing order of inductive effect are, +, , +, , R 3N > NH 3 >  NO2 >  CN >,  SO3H >  CHO >  CO > COOH >,  COCl >  COOR >  CONH 2 > F >,  Cl >  Br >  I >  OH >  OR >,  NH 2 > C 6H 5 > H, (−I power in decreasing order with respect, to H.)

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123, , Organic Chemistry : Some Basic Principles and Techniques, , (ii) Positive inductive effect Atoms or groups, having lower electron affinity than hydrogen, are said to have + I effect (electron, withdrawing)., The order of +I-effect of some groups is as, follows : 3 ° > 2 ° > 1° or CH 3 ., 6. Resonance Structure, These structures are hypothetical and individually, don’t represent any real molecule. Benzene is a, resonance hybrid of following two canonical forms, 6, , 6, 5, , 1, 2, , 4, 3, I, , ←→, , 5, , 1, 2, , /, , 4, 3, II, , Resonance, hybrid, , (ii) Negative electromeric effect (− E -effect), If the electrons of the double ( π ) bond are, transferred to an atom of the double bond, other than the one to which the reagent, gets finally attached, then that effect is, known as − E -effect., C, , –, O + CN ¾®, , C—O, , –, , CN, , 9. Hyperconjugation, The stabilising interaction that involves, delocalisation of σ-electrons of C—H bond of, an alkyl group linked directly to an atom, of unsaturated system or to an atom having, unshared p -orbital, is called hyperconjugation., , 7. Resonance Effect (R-effect), The polarity developed with in a molecule due to, +, interaction of two π-bond or of a π-bond with the, CH 3 CH 2 has empty p -orbital, with which one, lone pair of electrons present on a fix atom is, of the CH bonds of the methyl group align, called resonance effect., in its plane and thus, the electrons of the, (i) Positive resonance effect (+ R-effect), CH bonds can be delocalised into the, When the electrons move away from the, empty p -orbital as shown below., atom or substituent group attached to the, Hyperconjugation, conjugated system, the effect is called positive, C, —H, 3, H, sp, 1s, resonance effect.In this effect, electron density, bond, increases at certain places. e.g. Groups like, H, +, halogen, —OH, —OR,—OCOR,—NH 2 ,, C, C, NHR, —NR 2 show + R-effect., H, H, (ii) Negative resonance effect (−R-effect), Empty 2p-orbital, H, of carbon, When the electrons are transferred towards, Orbital diagram showing, the atom or substituent group attached to the, hyperconjugation in ethyl cation, conjugated system, the effect is called, negative resonance effect., 10. Types of Organic Reactions and Mechanism, e.g.  C == O,  CN and  NO2 show, The organic reactions can be categorised into, —R-effect., following five categories:, 8. Electromeric Effect (E -effect), (i) Substitution reactions, The polarity developed due to complete transfer, Substituent, of a shared pair of π-electrons to any of the atoms, hν or, R — CH3 + X2, R—CH2 X + H X, linked directly to the multiple bond in the, 520-670 K, presence of an attacking reagent is called the, Substitution product, electromeric effect., (i) Positive electromeric effect (+ E -effect), These are further classified as follows :, If the electrons of the π-bond are transferred, (a) Free-radical substitution The, to that atom of the double bond to which the, reaction brought by free radicals are called, reagent gets finally attached, then that effect is, free radical substitutions. e.g. Halogenation, known as + E -effect. e.g., of alkenes., CH3  CH == CH2 + H+ →, , +, , CH3  CH CH3, , hν, , CH 4 + Cl 2 →, Methane, , HCl, , CH 3Cl, Chloromethane

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124, , CBSE New Pattern ~ Chemistry XI (Term-I), , (b) Electrophilic substitution reactions, + R CH2 X, , CH2R, , Anhydrous, AlCl3, Friedel-Crafts´, alkylation, , + HX, , (c) Nucleophilic substitution reactions These, are usually written as S N (S-substitution,, N-nucleophile) and are of two types, i.e., unimolecular (shown as S N1) and bimolecular, (shown as S N 2)., These reactions are generally seen in alkyl, halides and alcohols., e.g. RX + KOH(aq ) → ROH + KX, (ii) Elimination reactions An elimination reaction, is one that involves the loss of two atoms or, groups of atoms from the same or adjacent atoms, of a substance leading to the formation of a, multiple (double or triple) bond., Elimination reaction is given by those, compounds which have a nucleophile as a, leaving group, i.e., r, r, r, R r, – –, –, —, —R,—S —R, X, OH, OR, N2, N3, H3O, —N—, R, R, , Elimination reactions are of two types :, (a) α-elimination reactions, (b) β-elimination reactions, e.g. (CH3 ) 3 COH, , (i) H3O +, , → (CH3 ) 2 C == CH2 + H2O, , (ii) Heat, 2-methylpropan-2-ol, , (iii)Addition reactions Reactions which involve, combination between two reacting molecules, to give a single molecule of the product are, called addition reactions., Such reactions are of typical compounds, containing multiple bonds., R  CH == CH 2 + HX → R  CH  CH 3, , X, (iv)Addition-elimination reactions These, reactions involve addition of two molecules, with elimination of smaller molecules like, H 2O, e.g. esterification., C 2H 5OH + CH 3COOH, Ethanol, , Ethanoic acid, , H+, , a, , CH 3COOC 2H 5 + H 2O, Ethyl acetate, , (v) Rearrangement reactions These reaction, involve the migration of an atom or a group, from one atom to another within the same, molecule., e.g. CH 3 CH 2 CH 2 CH 2 Br, 1- bromobutane, , Br, , Anhyd. AlCl 3, CH 3 CH 2 CH  CH 3, →, 575 K, , 2- bromobutane, , 2-methyl propene, , Objective Questions, Multiple Choose Questions, 1. Which of the following is a correct, representation of condensed formula for, HOCH 2 CH 2 CH 2 CH(CH 3 )CH, (CH 3 )CH 3 ?, (a) HO(CH2 )2 CHCH3CH(CH3)2, (b) HO(CH2 ) 3CH(CH3)CH(CH3)2, (c) HOCH2CHCH3CH(CH3)2, (d) None of the above, , 2. Consider the following methane, molecule on paper., H, C, H, H, , X, H, Y, , Here, X and Y are respectively, (a) dashed wedge, solid wedge, (b) solid wedge, dashed wedge, (c) both are bonds in the plane of paper, (d) both are dashed wedge

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125, , Organic Chemistry : Some Basic Principles and Techniques, , 3. In cyclic compounds, the bond-line, formula for chlorocyclohexane is, represented by which of the following, representations?, , 9. The IUPAC name of the compound, , Cl, , Cl, , (a), , (b), , O, H, , Cl, , Cl, (c), , O, , C, , is ......... ., , (d), , 4. Which of the following is a homocyclic, alicyclic compound ?, (a), , (b), , (d), , (c), , O, , O, , S, , 5. Which of the following compounds, is/are heterocyclic aromatic compound?, (a) Furan, (c) Pyridine, , (b) Thiophene, (d) All of these, , 6. Which type of compound is shown by, the following structure, (a), (b), (c), (d), , (a) 1-chloro-2-nitro-4-methylbenzene, (b) 1-chloro-4-methyl-2-nitrobenzene, (c) 2-chloro-1-nitro-5-methylbenzene, (d) m-nitro-p-chlorotoluene, , ?, (Naphthalene), , Alicyclic compound, Benzenoid aromatic compound, Non-benzenoid aromatic compound, Acyclic compound, , 7. In the following compounds, number of, 1° (primary) carbons is, CH 3, , CH 3  CH  C  CH 2  CH 3, , , CH 3 CH 3, (a) 3, , (b) 4, , (c) 5, , 8. The IUPAC name for, Cl, , 10. The IUPAC name of following, compound is, CH 3  CH  CH  CH  CH 2  CH 3, , , , CH 3 CH 3 CHO, (a), (b), (c), (d), , 2-ethyl-3,4-dimethylpentanal, 2,3-dimethyl-4-aldohexane, 3-aldol-4,5-dimethyl hexane, 1,3,4,5-tetraethyl butanal, , 11. The IUPAC name for, , O, O, ||, ||, CH 3  C  CH 2  CH 2  C  OH is, (a) 1-hydroxypentane-1,4-dione, (b) 1,4-dioxopentanol, (c) 1-carboxybutan-3-one, (d) 4-oxopentanoic acid, , 12. The IUPAC name of the following, compound., HOCH 2 (CH 2 ) 3 CH 2 COCH 3 is, (a) 7-hydroxyheptan-2-ol, (b) 7-hydroxyheptan-2-one, (c) 2-oxoheptan-7-ol, (d) Heptan-2-oxo-7-ol, , 13. Neo-pentyl group is named as based on, NO2, , CH3, , (d) 6, , (a) 3-keto-2-methylhex-4-enal, (b) 5-formylhex-2-en-3-one, (c) 5-methyl-4-oxohex-2-en-5-al, (d) 3-keto-2-methylhex-5-enal, , IUPAC system on, (a), (b), (c), (d), , iso-pentyl, neo-pentyl, 2,2-dimethyl propyl, 1,1-dimethyl

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127, , Organic Chemistry : Some Basic Principles and Techniques, , 23. The most stable carbocation, among the, following is, , ⊕, , 29. The most stable carbocation, among the, , (a) (CH3) 3C CH CH3, , following is, , ⊕, , (b) CH3 CH2 CH CH2 CH3, , ⊕, , (a) (CH3) 3C CH CH3, , ⊕, , (c) CH3  CH CH2 CH2 CH3, , ⊕, , (b) CH3 CH2 CH CH2 CH3, , ⊕, , (d) CH3  CH2  CH2, , ⊕, , (c) CH3 CH CH2 CH2 CH3, , 24. Which of the following free radicals is, the most stable ?, (a) Primary, (c) Secondary, , (b) sp2, (d) None of these, , (a) sp 3, (c) sp, , (b) Methyl, (d) Tertiary, , ⊕, , (d) CH3 CH2 CH2, , 30. Among the following carbocations, , 25. When CH3 Cl undergoes homolytic, bond-fission, (a) carbon undergoes a geometric change from, tetrahedral to planar, (b) hybridisation changes from sp 3 to sp2, (c) Both (a) and (b), (d) None of the above, , 26. Compare stability of free radicals., •, , Ph 2C + CH 2 Me (I), PhCH 2CH 2CH + Ph (II) ,, Ph 2CHCH 2+ (III), Ph 2C(Me)• CH ⊕, 2 (IV), the order of stability is, (a) IV > II > I > III, (c) II > I > IV > III, , (b) I > II > III > IV, (d) I > IV > III > II, , 31. What is the correct order of decreasing, stability of the following cations?, ⊕, , ⊕, , CH 3 — CH— CH 3 CH 3 — CH— OCH 3, , I. CH 3  CH  CH 3, , I, , II, , ⊕, , II., , CH 3 — CH— CH 2 — OCH 3, , CH2, , III, , •, , III. CH3  C(CH 3)2, , (a) II > I > III, (c) III > I > II, , •, , IV. CH 2  CH 3, (a) II > I > III > IV, (c) I > II > III > IV, , (b) II > III > I > IV, (d) IV > III > I > II, , 27. Which is the most stable carbocation ?, Å, , (b), , (a) (CH3)3C, Å, , (c), , (d) (CH3)2CH, , 28. Look at the figure below and find the, hybridisation of carbon atom., H, +, , C, , H, , H, Shape of methyl cation, , 32. Which one of the following is an, intermediate in the reaction of benzene, with CH 3Cl in the presence of, anhydrous AlCl 3 ?, , (a) Cl+, , (b) CH3, +, , Å, Å, , —CH2, , (b) II > III > I, (d) I > II > III, , (c) CH+3, , (d), , 33. Match the intermediates given in, Column I with their probable structure, in Column II., Column I, , Column II, , A. Free radical, , 1. Trigonal planar, , B., , 2. Pyramidal, , Carbocation, , C. Carbanion, , 3. Linear

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128, , CBSE New Pattern ~ Chemistry XI (Term-I), , Codes, A B C, (a) 1 1 2, (c) 1 3 2, , A B C, (b) 1 2 3, (d) 2 1 3, , 34. Select the incorrect comparison as, nucleophile., s, , s, , (a) (CH3) 3CO < OH, (c) ROs < RSs, , (b) H2O < NH3, (d) R3P < R3N, , 35. Which has the highest nucleophilicity ?, (a) F −, (c) CH−3, , (b) OH−, (d) NH2−, , react with the Lewis base Cl − in a Lewis, acid-base reaction ?, (b) AlCl3, (d) CH−3, , 37. An example of electrophile is, (b) NH3, , (a NO2, , +, , (c) NO2, , (d) H2O, , 38. Electrophiles are electron seeking, species. Which of the following groups, contain only electrophiles ?, (b) AlCl3, SO 3, NO2+, (c) NO2+ , CH+3 , CH3 —C + == O, (d) C2 H−5 , C2H5 , C2H+5, , with the terms in Column II., Column I, , Column II, 2, , A. Carbocation, , 1. sp hybridised carbon, with empty p-orbital, , B. Nucleophile, , 2. Ethyne, , C. sp-hybridisation, , 3. Species that can receive, a pair of electrons, , D. Electrophile, , 4. Species that can supply, a pair of electrons., , D, 3, 2, 2, 1, , O, s, C—O, (b), , (a), s, , s, , (d) CH3 CH2, , (c) CH2 == CHCH2, , 41. Observe the effect carefully and predict, ‚, ‚+, ƒ, ƒ, C == C, + H + →, C—C, ƒ, ‚, ƒ, ‚, , (Attacking reagent), H, Choose the correct option., (a), (b), (c), (d), , Negative electromeric effect, Resonance effect, Positive electromeric effect, Inductive effect, , 42. Ethynic hydrogen is released most, easily in, C, , (b) O2N, , 39. Match the terms mentioned in Column I, , C, 2, 4, 4, 4, , Os, , (a), , +, , (a) BF3,CH3, NO2, , Codes, A B, (a) 1 4, (b) 1 3, (c) 1 3, (d) 2 3, , energy as a result of resonance ?, , the nature of it., , 36. Which of the following compounds can, (a) CH4, (c) OH−, , 40. Which has maximum stabilisation, , CH, C, , C, , (c), , CH, , CH, , (d) O2N, , C, , CH, , 43. Observe the orbital diagram showing, hyperconjugation in ethyl cation., Here A, B and C refer to, C, , H, , A, , H, C, , +, , H, , C, , H, , H, B, , (a) A→ hyperconjugation, B → empty 2p -orbital, of carbon, C → C 3— H1s bond, sp

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129, , Organic Chemistry : Some Basic Principles and Techniques, , (b) A→ inductive effect, B → i 2p-orbital of, carbon, C → C sp 2 — H1s bond, , (c) A→ electromeric effect, B → empty, 2p-orbital of carbon, C → C sp 3 — H1s bond, , (d) A→ resonance effect, B → i 2p-orbital of, carbon, C → C 3 — H1s bond, sp, , 44. Choose the correct order of stability of, carbocation using the concept of, hyperconjugation., CH3, CH3, C3H—C, , r, , , CH3—C, , CH3, , H, , I, , II, , (a) I < II < III < IV, (c) III < IV < II < I, , r, , r, , III, , IV, , 45. The order of relative stability of the, contributing structures are, ••, • O •s, • •, , , ⊕, , CH2 == CH—C — H ←→ C H2 —CH == C— H, I, II, • •+, • O•, , (a) II > I > III, (c) III > II > I, , , CH2 — CH == C — H, III, (b) I > II > III, (d) I = II = III, , −, •, •, , 46. Consider the following compounds,, NH2, , NO2, , (b) H2C—, , (c) H2C—, , (d) H2C==, , (a), (b), (c), (d), , inductive effect of the Me group, resonance effect of the Me group, hyperconjugative effect of the Me group, resonance as well as inductive effect of the, group, , 49. Hyperconjugation involves overlapping, of which of the following orbitals., (a) σ - σ, (c) p -p, , (b) σ -p, (d) π -π, , 50. Formic acid is a stronger acid than, acetic acid. This can be explained using, (a) +M effect, (b) −I effect, (c) +I effect, (d) −M effect, , 51. The arrangement of (CH 3 ) 3 C  ,, , (CH 3 ) 2 CH  , CH 3CH 2 — when, attached to benzene or unsaturated, group in increasing order of inductive, effect is, (a) (CH3) 3C — < (CH3)2 CH — < CH3CH2 —, , and, Aniline, (I), , (a) H2C==, , than that of MeCH 2CH == CH 2 due to, , r, , (b) IV < III < II < I, (d) None of these, , ←→, , structure ?, , 48. The stability of Me 2C == CH 2 is more, , , CH3 CH2 , CH3, , • O•, • •, , 47. Which is the incorrect resonating, , Nitrobenzene, (II), , Which of the following option is/are, true regarding I and II ?, (a) I shows + R -effect, whereas II shows, –R-effect, (b) I shows –R -effect, whereas II shows, + R-effect, (c) Both I and II show +R -effect, (d) Both I and II show −R -effect, , (b) CH3CH2 — < (CH3)2 CH — < (CH3) 3C —, (c) (CH3)2 CH — < (CH3) 3C — < CH3CH2 —, (d) (CH3) 3C — < CH3CH2 — < (CH3)2 CH —, , 52. Which of the following does not show, electromeric effect ?, (a) Alkene, (b) Ether, (c) Aldehyde, (d) Ketone

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130, , CBSE New Pattern ~ Chemistry XI (Term-I), , 53. Match the ions given in Column I with, their nature given in Column II., Column I, A. CH 3 NO2, ⊕, B. F3  C, , 2. Destabilised due, to inductive effect, ⊕, , 3. Stabilised by, hyperconjugation, , A B C, (b) 2 3 1, (d) 3 2 1, , 54. Ionic species are stabilised by the, dispersal of charge. Which of the, following carboxylate ion is the most, stable?, , O, ||, (c) F—CH2 —C—O −, , O, ||, (b) Cl—CH2 —C—O −, O, F, , (d), , CH—C—O–, , F, , 55. In which of the following compounds, the carbon marked with asterisk is, expected to have greatest positive, charge ?, (a) *CH3  CH2  CI, (b)*CH3 CH2 Mg +Cl−, (c) *CH3  CH2 Br, (d) *CH3  CH2  CH3, , 56. Dichlorocarbene is generated by the, action of potassium-1- butoxide on, chloroform. This is an example of, (a), (b), (c), (d), , (c), , ¾ COCH3, , (b), , CHO, (d), , CH3, , NO2, , 58. Electrophilic addition reactions proceed, , Codes, A B C, (a) 1 2 3, (c) 3 1 2, , O, ||, (a) CH3 — C —O −, , ¾ CHO, CHO, , Column II, 1. Stable due to, resonance, , C. CH 3  CH  CH 3, , (a), , α-elimination, β-elimination, addition reaction, rearrangement reaction, , 57. Which one is most reactive towards, nucleophilic addition reaction ?, , in two steps. The first step involves the, addition of an electrophile. Name the, type of intermediate formed in the first, step of the following addition reaction., H 3C  HC == CH 2 + H + → ?, (a) 2° carbanion, (c) 2° carbocation, , (b) 1° carbocation, (d) 1° carbanion, , 59. Which one is a nucleophilic substitution, reaction among the following ?, (a) CH3CHO + HCN → CH3CH(OH)CN, H+, , (b) CH3 CH == CH2 +H2O → CH3  CH CH3, , OH, (c) RCHO + R′ MgX → R —CH— R′, , OH, CH3, , (d) CH3 CH2 CH  CH2Br + NH3 →, CH3, , CH3  CH2 CH  CH2NH2, , 60. Dehydration of alcohol is an example, of which type of reaction?, (a) Substitution, (c) Addition, , (b) Elimination, (d) Rearrangement, , Assertion-Reasoning MCQs, Directions In the following questions, (Q.No. 61-75) a statement of Assertion, followed by a statement of Reason is, given. Choose the correct answer out of, the following choices.

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131, , Organic Chemistry : Some Basic Principles and Techniques, , (a) Both Assertion and Reason are correct, statements and Reason is the correct, explanation of the Assertion., (b) Both Assertion and Reason are correct, statements, but Reason is not the correct, explanation of the Assertion., (c) Assertion is correct, but Reason is incorrect, statement., (d) Assertion is incorrect but Reason is correct, statement., , 61. Assertion Acyclic compounds consist, of straight or branched chain compounds., Reason Alicyclic compounds contain, carbon atoms joined in the form of a, ring., , 62. Assertion When a covalent bond is, formed between atoms of different, electronegativity, the electron density is, more towards the more electronegative, atom of the bond., Reason Shift of electron density results, in a polar covalent bond., , 63. Assertion All the carbon atoms in, , H 2C == C == CH 2 are sp 2 - hybridised., , Reason In this molecule, all the carbon, atoms are attached to each other by, double bonds., CH3, , 64. Assertion, , is, , 3-methylcyclohexene., Reason In numbering, double bonded, carbon atoms get preference to the alkyl, group in cycloalkenes., , 65. Assertion In bond-line structural, representations, only atoms specifically, written are oxygen, chlorine, nitrogen etc., Reason In bond-line structural, representations, line junctions denote, carbon atoms bonded to appropriate, number of hydrogens required to satisfy, the valency of the carbon atoms., , 66. Assertion The names of straight chain, compounds are based on their chain, structure and carry a prefix indicating, the number of carbon atoms present in, the chain., Reason From CH 4 to C 4 H10 , the, prefixes are derived from common or, trivial names., , 67. Assertion 2, 3-dimethylhept-5-ene is, CH 3, , CH 3  CH ==CH  CH , , CH 3, , CH 2  CH  CH 3, , Reason The alkyl group do not get, preference to the double bond., , 68. Assertion Heterolytic fission involves, the breaking of a covalent bond in such, a way that both the electrons of the, shared pair are carried away by one of, the atoms., Reason Heterolytic fission occurs, readily in polar covalent bonds., , 69. Assertion A covalent bond may be, cleaved either by heterolytic cleavage, or by homolytic cleavage., Reason Heterolytic cleavage of CH 3 Br, +, −, will give C H 3 and Br., , 70. Assertion Carbanion has an octet of, electorns., Reason In carbanion, negatively, charged carbon atom is sp 3 -hybridised., , 71. Assertion Free radicals are short lived, and highly reactive., Reason Free radicals are highly unstable., , 72. Assertion Inductive effect and, resonance effect cause permanent, polarisation of bond.

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132, , CBSE New Pattern ~ Chemistry XI (Term-I), , Reason These involve electron, displacements due to the influence of an, atom or a substituted group present in, the molecule., , 73. Assertion Conjugated system show, abnormal behaviour, it contains, alternate single and double bonds in an, open chain or cyclic system., Reason In conjugated system,, π-electrons are delocalised and the, system develops polarity., , 74. 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., , 75. Assertion Polarisation of one σ bond, caused by the polarisation of adjacent, σ bond is referred to as the inductive effect., Reason The substituents can be, classified as electron withdrawing or, electron donating groups relative to, hydrogen., , When the group or atom release electron, density then electron density of conjugated, system increases while the group or atom, attract/withdraw electron density then, electron density of conjugated system, decreases., The following questions (i-iv) are multiple, choice questions. Choose the most, appropriate answer :, (i) In which molecule dipole moment is, the maximum?, (a) CH3CH2 CH2  Cl, (b) CH3CH2 CH2  NO2, (c) CH3CH2 CH2  OH, (d) CH3  CH2 NH  C  CH3, , O, , (ii) In which benzene ring electron density, is maximum?, O, OCH3, (a), , O—C—CH3, (b), , O, C—O—CH3, (c), , CH==O, (d), , Case Based MCQs, 76. Read the passage given below and, answer the following questions :, The electron displacements due to the, influence of an atom or a substituent group, present in the molecule cause permanent, polarisation of the bond (called electronic, δδ+, , δ+, , (iii) Which of the following system show, abnormal behaviour in their properties, (like-stability, polarity … etc.) ?, NH2, , NH2, , (a), , (b), , δ–, , effect), e.g. CH3 CH2 CH2  F ., In above example, polar C—F bond, induce polarity in the adjacent bonds., Such polarisation of adjacent σ- bond is, referred to as the inductive effect. This, effect decreases rapidly as the number of, intervening bonds increases., The resonance effect is defined as the, polarity produced in the molecule by the, interaction of two π-bonds or in conjugated, system., , OCH3, , (c), , (d), , (iv) The permanent displacement of, electron through a chain involving, only σ-bonds is called, (a) inductive effect, (b) hyperconjugation effect, (c) electromeric effect, (d) mesomeric effect

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133, , Organic Chemistry : Some Basic Principles and Techniques, , Or, Which of the following is correct with, respect to −I -effect of the substituents?, (R = alkyl), (a) NH2 >OR > F, (b) NR2 < OR < F, (c) NH2 > OR < F, (d) NH2 > OR > F, , the question accordingly., An intermediate is a molecular entity, that, is formed from the reactants and reacts, further to give the directly observed, products of a chemical reaction. Most, chemical reactions are stepwise, that is, they take more than one elementary step, to complete. An intermediate is the, reaction product of each of these steps,, except for the last one, which forms the, final very isolated. Also, owing to the short, lifetime, they do not remain in the product, mixture., In certain cases, they are separated and, stored. For example matrix isolation and, low temperature., Matrix isolation is a technique that is used, experimentally in physics and chemistry, that includes a material that has been, trapped with in an unreactive material., Host matrix generally comprises guest, particles that are generally embedded., Guest particles can be molecules, atoms, and ions. The guest is isolated within the, host matrix., The following questions (i-iv) are multiple, choice questions. Choose the most, appropriate answer:, (i) Relative stabilities of the following, carbocations will be in the order, ⊕, , ⊕, , CH 3 , CH 3CH 2 , CH 2 OCH 3, (A ), , (a) C > B > A, (b) C < B < A, (c) B > C > A, (d) C > A > B, , (B ), , •, , •, , ⊕, , ⊕, , ⊕, , (b) CH3 CH2 and Cl −, , (a) CH3 CH2 and Cl, , •, , (d) CH3 CH2 and Cl −, , (c) CH3 CH2 and Cl, , (iii) The shape of carbocation is, , 77. Read the following passage and answer, , ⊕, , (ii) CH3 CH2 Cl undergoes homolytic fission,, produces, , (C ), , (a) square planar, (b) trigonal planar, (c) octahedral, (d) trigonal pyramidal, , (iv) Compare stability of free radicals, •, , I. CH3  CH  CH3, II., , CH2, •, , III. CH2  CH(CH3 ) 2, •, , IV. CH 2 CH 3, (a) II > I > III > IV, (c) I > II > III > IV, , (b) II > I > IV > III, (d) IV > III > I > II, , Or, The radical is aromatic because it has, , · 2, CH, (a) 6p-orbitals and 6 unpaired electrons, (b) 7p-orbitals and 6 unpaired electrons, (c) 7p-orbitals and 7 unpaired electrons, (d) 6p-orbitals and 7 unpaired electrons, , 78. Read the passage given below and, answer the following questions :, When a single Lewis dot structure is, unable to explain all the properties of a, compound, two or more structures, called, the canonical forms or resonating, , structures, are drawn to explain all the, properties of that compound, then the, actual structure of the compound is in, between these canonical form and is, called the resonance hybrid of these, canonical forms. This phenomenon is, called resonance., The resonating structure are hypothetical,, i.e. have no real existence. The energy of, actual structure or resonance hybrid is, lower than, that of any of the canonical

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134, , CBSE New Pattern ~ Chemistry XI (Term-I), , forms. The difference between hybrid and, the energy of most stable (lowest energy), canonical form is called the resonance, stabilisation energy or resonance energy., The more the number of important, canonical forms more is the resonance, energy. In case of resonating structures of, equivalent energy, resonance is of great, importance., In these questions (i-iv) a statement of, Assertion followed by a statement of, Reason is given. Choose the correct, answer out of the following choices :, (i) Which rules of resonating structures is, true ?, I. The overall charge of system, remains same., II. The arrangement of atoms must be, identical or almost same in, resonance structure., (a) Rule I, (b) Rule II, (c) Both the rules, (d) Neither rule I nor rule II, , (ii) There are how many resonance, structure of phenoxide ion ?, (a) 4, (c) 6, , (b) 5, (d) 3, , (iii) Which is more stable among the, following pair of resonance, contributing structures., r, , r, , R¾ C — O, , R—C, , (I), , O, , (II), , (a) Structure I, (b) Structure II, (c) Both are equally stable, (d) Can’t say, , (iv) The relation between resonance and, bond order is, no. of bonds, no. of resonating structure, no. of π - bond, (b) B.O. =, no. of σ - bond, no of σ - bond, (c) B.O. =, no. of resonating structure, (d) B.O.= no. of (σ + π) bond, (a) B.O. =, , Or, Which of the following show −R-effect ?, (b) NHCOR, (d) COOH, , (a) OR2, (c) NH2, , 79. Read the passage given below and, answer the following questions:, In a polar covalent bond, the electrons are, shifted to the more electronegative atom, and hence, a polarity is developed in the, bond. e.g. In a chain of carbon atoms, having C—Cl bond, the C attached, directly to the Cl-atom because of its less, electronegativity acquires some positive, charge (δ + ). The carbon ( C2 ) that is, +, , bonded directly to the Cδ also loose, some of its electron density due to more, +, polar Cδ and hence, acquires somewhat, less positive charge (δ ++ ). Similarly, the, next carbon ( C3 ) also acquires some, positive charge but less than (δ ++ ) as, shown below :, δδδ +, , δδ +, , δ+, , 3, , 2, , 1, , δ−, , CH3  CH2 → CH2 → Cl, , This induction of polarity, due to the, presence of polar bond in an organic, molecule, is called inductive effect., In these questions (i-iv) a statement of, Assertion followed by a statement of, Reason is given., Choose the correct answer out of the, following choices :, (a) Assertion and Reason both are correct, statements and Reason is correct, explanation for Assertion., (b) Assertion and Reason both are correct, statements but Reason is not correct, explanation for Assertion., (c) Assertion is correct statement but Reason, is incorrect statement., (d) Assertion is incorrect statement but, Reason is correct statement., , (i) Assertion Polarisation of one σ-bond, caused by the polarisation of adjacent, σ-bond is referred to as the inductive, effect.

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135, , Organic Chemistry : Some Basic Principles and Techniques, , Reason The substituents can be, classified as electron withdrawing, or electron donating groups relative to, hydrogen., (ii) Assertion Formic acid is a stronger acid, than benzoic acid., Reason +R-effect phenyl group is, greater than −I -effect., (iii) Assertion Allyl and benzyl carbonium, ions are stable than propyl carbonium, ions., Reason Electron releasing groups, stabilise carbonium ions., (iv) Assertion When a covalent bond is, formed between atoms of different, electronegativity, the electron density, is more towards the more, electronegative atom of the bond., Reason Shift of electron density, results in a polar covalent bond., , Or, Assertion Inductive effect and, resonance effect cause permanent, polarisation of bond., Reason These involve electron, displacements due to the influence of, an atom or a substituted group present, in the molecule., , 80. Read the following passage and answer, the questions accordingly., The name carbon is derived from the latin, word ‘carbo’ which means coal. This is, because it is the main constituent of coal., The amount of carbon present in the earth’s, crust and atmosphere is very small. In fact,, all the living things, plants and animals, are, made up of carbon compounds., The compounds having organic origin are, known as organic compounds. In morden, days organic compounds are not restricted, to living systems., , Various organic compounds are known, which came from non-living systems., Carbon is known to form more than five, million compounds. The existence of such a, large number of compounds is due to some, unique properties of carbon., In these questions (i -iv), a statement of, Assertion followed by a statement of Reason, is given. Choose the correct answer out of, the following choices., (i) Assertion Saturated hydrocarbons are, chemically less reactive., Reason All isomeric paraffins have same, parent name., (ii) Assertion Olefins are unsaturated, hydrocarbons containing two hydrogen, atoms less than the, corresponding paraffin., Reason There is one triple bond, between two carbon atoms in olefin, molecules., (iii) Assertion The IUPAC name for, CH 3CH == CH C ≡≡ CH is, pent-3-en-1-yne and not pent-2-en-4-yne., Reason While deciding the locants for, double and triple bonds, lowest sum rule, is always followed., (iv) Assertion, CH 3 CH CH 2 CH  CHO, , , CH 3, Cl, is 4-chloro-2-methylpentanal., Reason Numbering is to be given, according to the alphabetical order of, substituents., , Or, , Assertion neo-hydrocarbons contain a, quaternary carbon atom., Reason Whenever a carbon atom is, bounded to four carbon atoms it is, quaternary.

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136, , CBSE New Pattern ~ Chemistry XI (Term-I), , ANSWERS, Multiple Choice Questions, 1. (b), 11. (d), 21. (a), , 2. (a), 12. (b), 22. (d), , 3. (c), 13. (c), 23. (c), , 4. (b), 14. (d), 24. (d), , 5. (d), 15. (a), 25. (c), , 6. (b), 16. (b), 26. (b), , 7. (c), 17. (c), 27. (c), , 8. (b), 18. (a), 28. (b), , 9. (a), 19. (a), 29. (c), , 10. (a), 20. (c), , 31. (a), 41. (c), , 32. (c), 42. (d), , 33. (a), 43. (a), , 34. (d), 44. (b), , 35. (c), 45. (b), , 36. (b), 46. (a), , 37. (c), 47. (b), , 38. (b), 48. (c), , 39. (a), 49. (b), , 51. (a), , 52. (b), , 53. (a), , 54. (d), , 55. (a), , 56. (a), , 57. (d), , 58. (c), , 59. (d), , 50. (c), 60. (b), , 64. (a), 74. (d), , 65. (b), 75. (b), , 66. (b), , 67. (d), , 68. (b), , 69. (b), , 70. (b), , 30. (b), 40. (b), , Assertion-Reasoning MCQs, 61. (b), 71. (b), , 62. (b), 72. (a), , 63. (d), 73. (a), , Case Based MCQs, 76. (i)-(b), (ii)-(a), (iii)-(b), (iv)-(a) or-(b), , 77. (i)-(a), (ii)-(a), (iii)-(b), (iv)-(b) or-(c), , 78. (i)-(c), (ii)-(b), (iii)-(a), (iv)-(a) or-(d), , 79. (i)-(b), (ii)-(a), (iii)-(b), (iv)-(b) or-(b), , 80. (i)-(b), (ii)-(c), (iii)-(a), (iv)-(c) or-(b), , EXPLANATIONS, not have aromatic character. Homocyclic means, it contains only C-atoms in ring, e.g., , 1. The correct condensed formula for, HOCH2CH2CH2CH(CH3 )CH(CH3 )CH3, is HO(CH2 ) 3 CH(CH3 )CH(CH3 ) 2 ., , 2. X is dashed wedge (bond away from observer), and Y is solid wedge (bond towards observer)., In these formulas, the solid wedge is used to, indicate a bond projecting out of the plane of, paper towards the observer., The dashed wedge is used to depict the bond, projecting out of the plane of paper and away, from the observer., 3. In cyclic compounds, the, bond-line formula for, chloro-cyclohexane is represented, by following formula :, , Cl, , 4. Alicyclic (aliphatic cyclic) compounds contain, carbon atoms joined in the form of a ring but do, , Cyclopentane, , Cyclohexane, , Cyclohexene, , Cyclobutane, , 5. All the given compounds, i.e. furan, thiophene,, pyridine are known as heterocyclic aromatic, compounds., , 6. Since, the given compounds contains benzene, nucleus, so naphthalene is a benzenoid aromatic, compound.

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137, , Organic Chemistry : Some Basic Principles and Techniques, , A compound containing both an alcohol and a, keto group is named as hydroxy alkanone since, the keto group is preferred over hydroxy group., , 7. 1° carbons are attached directly to one carbon, CH3, , 3°, , 1°, 2°, , CH3—CH—C—CH2—CH3, , 13., , CH3, , 4°, 1°, , CH3 CH3, , 1°, , CH3 —C—CH2 —, 2, , 3, , 1°, , 1°, , So, there are 5 primary carbons., , 8. For tri or higher substituted benzene derivatives,, the compounds are named by identifying, substituent, positions on the ring by following, the lowest locant rule., Substituent of the base compound is assigned, number 1 and then the direction of numbering, is chosen such that the next substituent gets the, lowest number., The substituents appear in the name in, alphabetical order., , 2, 2-dimethyl propyl, , Longest chain is of three carbons; two alkyl, groups at C 2 ., , 14. The IUPAC name of compound is, 2-(4-chlorophenyl) propanoic acid., 2, , 3, , 1, 2, 3, 4, , Cl, , 1, , NO2, , 15 ‘iso’ means one Me group is present at side chain., Hence, the structure of iso-butyl group is, , 2, 3, , CH3, , CH3 C CH2 , , H, , CH3, , 9., , O, H, , O, , C, 1, , 2, , 4, , 3, , 5, 6, ,  CHO group gets higher priority (act as, principal group) over > C == O and C == C, group in numbering of principal carbon chain., IUPAC name 3-keto-2-methylhex-4-enal., , 10., , 5, , 4, , 3, , 1, , CH3—CH—COOH, , Cl, , 4, , 1, , CH3, , ‘yl’ suffix is used to represent one  H less than, the parent hydrocarbon., , 16. Option (b) is the correct structure of 2-methyl, 2-cyclohexen-1-ol., OH, 1, , CH3, , 2, , 6, , 2, , CH3CH—CH—CH CH2CH3, CH3 CH3 CHO, 1, , IUPAC name is 2-ethyl-3, 4-dimethylpentanal., , 11. When more than one functional group lie in the, main chain, nomenclature is done according to, that functional group which has higher priority., Carboxylic acid ( COOH) has more priority, than ketone ( > C == O)., , 12. The correct IUPAC name of the given, compound HOCH2 (CH2 )3 CH2COCH3 is, 7-hydroxyheptan-2-one., , 5, , 3, 4, , 17. trans-2-chloro-3-iodo-2-pentene, Cl, 1, , H3C, , 4, , 2, , 3, , 5, , CH2CH3, , C == C, I, , 18. The structure, CH3, , CH3 CH2 CH2 C  CH  CH2 CH3,  , CH3 C2H5

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138, , CBSE New Pattern ~ Chemistry XI (Term-I), , While writing IUPAC name, alkyl groups are, written in alphabetical priority, thus lower, locant 3 is assigned to ethyl., Note Prefix di, tri, tetra are not included in, alphabetical order., , greater is the delocalisation of odd electron and, hence, more stable is the free radical., , 25. CH3 •• Cl →, , 19. The IUPAC name of given compound is, , H, , 3, 5-dimethyl cyclohexene., H, , 1, 2, , 6, 5, , 4, , 3, , 2, , CH2 is most stable due to resonance., , , CH3, , IUPAC name is 1-bromo-3-methylbutane., 21. Option (a) is not in accordance with IUPAC, system. In IUPAC system of nomenclature,, preference is given to multiple bond than, halogen substituent, so the correct name is, 2, , H, , 26. Among the given free radicals,, 1, , 20. C H3  CH  CH 2  CH 2  Br, , 3, , 1, , In the remaining free radicals, more the alkyl, groups bonded to the electron deficient carbon,, the more stable is the radical., Thus, the correct order of stability of given free, radicals is, CH2 > CH3, , CH3 > CH2, , CH3 > CH2, (I), , CH3, , (IV), , CH3, , (III), , 3- bromoprop-1-ene, , 22. Heterolytic fission of Cl 2 gives one carbocation, and one carbanion due to transfer of electron, from shared pair of bonded electrons., +, , 23. CH3 C H CH2 CH2 CH3 is the most stable, carbocation among the given carbocations. It is, because the number of α-H atom attached to, +, , carbocation is maximum in CH3 CHCH2CH2CH3., Thus, it has maximum number of, hyperconjugating structure hence, it is most, stable., , 27. Benzyl carbocation is most stable due to, resonance, (a) 3° alkyl, (b) 3° vinylic least stable, (c) 1° benzyl most stable, (d) 2° alkyl, Order is : (b) < (d) < (a) < (c), Å, Å, , —CH2, , 24. Stability order of free radical is as, CH3, H3C, , •, •, ( C• H > CH — CH, •, > CH3, CH3 — C >, 3, 2, ', Methyl, , H3C, free radical, CH3, Secondary, free radical, , C, , CH3, , (II), , Br  CH2  CH == CH2, , Tertiary, free radical, , Bond angle 120°, with three co-planar (C—H), bonds), , sp 3 -hybridised carbon, tetrahedral bond angle 109°28′, , CH3, , 4, , •, , + Cl, , Methyl free radical, sp 2 -hybridised (with singly, occupied p -orbital), , Cl, , 3, , H3C, , C, , •, , CH3, , Primary, free radical, , Greater the number of alkyl groups attached to, the carbon atom carrying the odd electrons,, , Å, , CH2, , CH2, , CH2, Å, , 28. The shape of CH+3 may be considered as being, derived from the overlap of three equivalent, C-( sp 2 )-hybridised orbitals with 1s-orbital of, each of the three hydrogen atoms., Each bond may be represented as C( sp 2 ) —H (1s), sigma bond.

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139, , Organic Chemistry : Some Basic Principles and Techniques, , The remaining carbon orbital is perpendicular to, the molecular plane and contains no electrons., H, +, , C, , H, , H, , = 3σ-bonds, ⇒ sp2-hybridisation, , Shape of methyl cation, +, , 29. CH3C HCH2CH2CH3 is the most stable, carbocation among the given carbocations., It is because the number of α-H atom attached, to carbocation is maximum in, ⊕, , CH3CHCH2CH2CH3. Thus, it has maximum, number of hyperconjugating structure., Hence, it is most stable., , 33. A → (1);, , B → (1);, , C → (2), , Column I Column II, A. Free radical Trigonal, planar, , Free radicals are formed, by homolytic fission e.g., °, CH 3 hybridisation sp 2, , B. Carbocation Trigonal, planar, , Formed by heterolytic, fission when carbon is, attached to a more, electronegative atom e.g., +, , CH 3 hybridisation sp 2, C. Carbanion, , Pyramidal, , 30. The order of stability of carbocation is, 3° benzylic > 2° benzylic >3° >2° > 1°., Thus, the correct order of stability will be, , then, R3 P > R3 N, but given comparison R3 P < R3 N is incorrect., , +, , (I), , (II), , H, Me, , , +, > Ph 2CH  C+ > Ph  C  CH2, , , Ph, H, (III), , (IV), , 31. Stability of the given cations can be understood, , 1, nucleophilicity, ∴ CH−3 has the highest nucleophilicity., , 35. Electronegativity ∝, , 36. (a) CH4 + Cl – → No reaction, (b), , +, , CH, I, , +, , CH3 ; CH3 — CH — O — CH3 ;, II, , Weak + I-effect of, the two methyl groups, stabilises carbocation (I), , Strong + R-effect of, —OCH3 group stabilises, carbocation (II), +, , CH3 — CH — CH2, III, , OCH3, , –I-effect of —OCH3, group destabilises the, carbocation (III), , Hence, the stability of carbocation decreases, II > I > III., +, , 32. CH3 group acts as an intermediate in the, reaction of benzene with CH3Cl in the presence, of anhydrous AlCl 3 . This reaction is called as, Friedel-Craft alkylation., , AlCl 3, Electron -deficient, Al (Lewis acid), , by the following structures, CH3, , Formed by heterolytic, fission when carbon is, attached to more, electropositive atom e.g., CH −3 hybridisation sp 3, , 34. Nucleophilic strength increases down the group, , Ph, H, , , Ph  C  CH2CH3 > PhCH2CH2  C  Ph, +, , Explanation, , +, , Cl –, Lewis base, , → AlCl –4, , (c) OH– + Cl – → No reaction, (d) CH–3 + Cl – → No reaction, , 37. Electrophiles are electron deficient species., Hence, among the given species, nitronium ion, +, , (NO2 ) due to lack of electrons act as, electrophile while all other given species due to, the presence of lone pair of electrons, act as, nucleophile., 38. AlCl 3 , SO3 (Lewis acids), NO+2 , CH+3 ,, +, , CH3  C ==O are electron deficient species., Hence, these acts as electrophiles., , 39. A → (1);, , B → (4);, , C → (2);, , D → (3)

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140, , CBSE New Pattern ~ Chemistry XI (Term-I), , Column I, , Column II, , Explanation, , A. Carbocation sp 2 -hybridised H 3C + is carbocation., carbon with, Loss of e − makes its, empty p-orbital p-orbitals empty, (sp 2 -hybridised, carbon), B. Nucleophile Species that can Nucleus loving, i.e., supply a pair of having negative, charge or excess of, electron, electrons., HC ≡≡ CH, (sp-hybridisation), , D. Electrophile Species that, receive a pair, of electron, , Electron loving, i.e., positive charge or, lack of electrons, , 40., , O, , (I ), , ( II ), , ( III ), , 45. The order of relative stability of the, contributing structures are I > II > III., , :O:–, , :O:, +, , CH2==CH — C — H, , C  O–, , ( IV ), , :, , C. sp-hybridisa Ethyne, -tion, , 44. 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, the correct order of stability of carbocation using the concept of hybridisation is, CH3, |, +, +, +, CH3 — C+ > (CH3 )2 CH > CH3 CH2 > CH3, |, CH3, , CH2 — CH == C —H, , I, , II, , :O:+, , C, , +, , C + H+, , C, , (Attacking, reagent), , C, H, , 42. Ethynic H is attached to (— C ≡≡ C —) system., ,  NO 2 is electron-withdrawing and (C— H) bond, strength is weakened to a greater extent. Since,, benzene nucleus is also resonance stabilised,, hence (d) is correct option., +, , 43. In CH3 C H2 (ethyl cation), the positively charged, carbon atom has an empty p -orbital . One of the, C—H bonds of the methyl group can align in the, plane of this empty p-orbital and the electrons, constituting the C—H bond in plane with this, p-orbital can then be delocalised into the empty, p-orbital as depicted below :, H, C, , +, , H, C, , (B), H empty 2p-orbital, of carbon, , r, , r, , r, , NH2, , NH2, , NH2, , NH2, , NH2, , –, , :, , H, H, , 46 Aniline (I) shows + R -effect, whereas, nitrobenzene (II) shows −R -effect. In +R, -effect, the transfer of electrons is away from, an atom or substituent group attached to the, conjugated system., This electron displacement make certain, positions in the molecule of high electron, densities., This effect in aniline is shown as :, , :, , bond, (C), , The reason for this order is, I. The resonating structure which is most, stable has more number of covalent, bonds, each carbon and oxygen atom has, complete octet and no separation of, opposite charges., II. Negative charge on more electronegative, atom and positive charge on more, electropositive atom., III. Structure does not contribute as oxygen, has positive charge and carbon has, negative charge, hence least stable., , :, , Hyperconjugation, (A), , Csp3 — H1s, , III, , :, , 41. In the given reaction, π-electrons of the multiple, bond are transferred to that atom on which the, attacking reagent get attached. So, it is an example, of positive electromeric effect, (+E -effect). e.g., , –, , : CH2 — CH == C —H, , :, , Resonance in benzene nucleus as well as in, carboxylate ion. Thus, maximum stable., Thus, maximum resonance energy., , –, , –

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141, , Organic Chemistry : Some Basic Principles and Techniques, , −R -effect is observed when the transfer of, electrons is toward the atom or substituent, group is attached to the conjugated system, e.g., in nitrobenzene this electron displacement can, be depicted as., , :, , :, , N, , –:O, , :, O – O, N, , N, , saturated carbon chain follows the order, (CH3 )3 C— > (CH3 )2 CH — > CH3CH2—, , However, when an alkyl group is attached to an, unsaturated system such as double bond or a, benzene ring, the order of inductive effect is, actually reversed., This effect is called hyperconjugation effect or, Baker-Nathan effect. Now, the reactivity order is, (CH3 )3 C — < (CH3 )2 CH — < CH3 — CH2 —, , O, , :, , :, , O, , :, , –:O, , :, , O, , :, , :, , O, , 51. The inductive effect of the alkyl group on a, , N, , r, , r, , 52. Electromeric effect implies complete transfer of, , r, , π-electrons in presence of a reagent. Since,, simple ethers do not contain any multiple bond,, hence they do not show electromeric effect., , 47., H2C, , H2C, (a), , 53. A → (1);, (d), , A. CH 3 NO2, , H2C—, , H2C, , (c), , 48., , Me, , Column I, , B., , F3  C +, , H, C==C, , H, Me, has 6- hyperconjugate forms while, H, MeCH2, C==C, H, H, has 2- hyperconjugate forms, , Therefore, Me2C ==CH2 is more stable than, MeCH2CH ==CH2 ., , 49. Hyperconjugation involves overlapping of the, , σ-p-orbitals. It is a general stabilising interaction., It involves delocalisation of σ-electrons of C—H, bond of an alkyl group directly attached to an, atom of unsaturated system or to an atom with, an unshared system or to an atom with an, unshared p -orbital., The σ-electrons of C—H bond of the alkyl, group enter into partial conjugation with the, attached unsaturated system or with the, unshared p-orbital. It is a permanent effect., , B → (2);, , Column II, , Explanation, , Stable due to, resonance, Destabilised due – I effect of F, to inductive effect creates electron, deficiency at, carbon C + ., +, , ⊕, , D. CH 3 —CH Stabilised due to CH is attached to, —CH 3 hyperconjugation two carbon. It can, also be stabilised by, hyperconjugation., , 54. In all the given carbocations, the negative, charge is dispersed which stabilises these, carbocations., Here, the negative charge is dispersed by two, factors, i.e. +R-effect of the carboxylate ion, (conjugation) and I-effect of the halogens., These effects are shown below in the, carbocations, , (a) CH3, , O, ||, C—O, , CH3, , 50. Due to + I effect, electron density increases,, which results in decrease in acidic character., Acetic acid, CH3  COOH has —CH3 group, that shows + I effect, whereas no such group is, present in formic acid. Therefore, formic acid is, a stronger acid than acetic acid., , C → (3), , (b) Cl, , CH2, , Cl, , CH2, , O, |, C==O / CH3, O, ||, C—O, O, |, C ==O / Cl, , O, C, O, , O, CH2, , C, O

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142, , CBSE New Pattern ~ Chemistry XI (Term-I), , (c) F, , F, , CH2, , O, ||, O—O, , CH2, , O, |, C==O / F, , F, (d), , CH, , O, CHO, O, CH2, , CH, , O, CH, , C, O, , As it is clearly evident from the above, structures, that +R-effect is common in all the, four structures, therefore, overall dispersal of, negative charge depends upon the number of, halogen atoms and electronegativity. Since, F, has the highest electronegativity and two, F-atoms are present in option (d), thus, dispersal, of negative charge is maximum in option (d)., Note In above structure (a), methyl group (CH 3 ), increases the density on C-atom., , 55. Electronegativity of Cl, Br, C and Mg follows, the order Cl > Br > C > Mg, *CH3 →,  CH2 →,  Cl, *CH3 , ← CH2 , ← Mg +Cl −, , ( − I-effect), , ( − I-effect), *CH3 →,  CH2 →,  Br, (+ I-effect), *CH3  CH2  CH3, −l effect of Cl > Br., Hence, CH3 CH2 Cl has the greatest, positive charge., K (t -butoxide), , Chloroform, , >, , >, (+R), , NO2, , CH3, , (–I, –M), , (+I), , 58. When electrophile attacks CH3  CH ==CH2, , O, F, |, C == O /, F, , 56. CHCl 3 →, , C  CH3, , CHO, , O, , F, , F, , >, , C, , O, ||, C—O, , F, , CHO, , •, • CCl 2, , Dichlorocarbene, , Since, the dominating group, i.e. hydrogen and, chlorine, here are lost from the same carbon, atom, it is an example of α-elimination., , 57. Reactivity of carbonyl compounds towards, nucleophilic addition reactions depends on the, presence of substituted group. Electron, withdrawing (– I , – M ) groups increases, reactivity towards nucleophilic addition, reactions., , delocalisation of electrons can take place, in two, possible ways, ⊕, , CH3  CH ==CH2 + H+ → CH3  CH  CH3, (2° carbocation), ⊕, , CH3  CH2  CH2, , (1° carbocation), , As 2° carbocation is more stable than 1°, carbocation thus first addition is more feasible., Note Stability of carbocations is the basis of, Markownikoff's rule., CH3, , 59. CH3  CH2  CH  CH2Br + NH3 →, CH3, , CH3 — CH2 — CH — CH2NH2, It is an example of nucleophilic substitution, reaction., , 60. Dehydration of alcohol involves the loss of two, atoms or groups from the adjacent carbon, atoms, hence it is an example of β-elimination, reaction., , 61. Acyclic or open chain compounds are, also called aliphatic compounds and, consist of straight or branched chain of carbon, atoms., CH3, , CH3CH3 , CH3  CH  CH3,, Ethane, , iso - butane, , O, , CH3  C  H ,, , O, , CH3  C  OH, , Acetaldehyde, , Acetic acid

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143, , Organic Chemistry : Some Basic Principles and Techniques, , Alicyclic or closed chain or ring compounds, contain carbon atoms joined in the form of a, ring (homocyclic). Some examples of this type, of compounds are, , Cyclopropane, , Cyclohexane, , Cyclohexene, , Thus, both Assertion and Reason are correct, and Reason is not the correct explanation of, Assertion., , 62. When a covalent bond is formed between atoms, of different electronegativity, the electron, density is more towards the more, electronegative atom of the bond. Such a shift of, electron density results in a polar covalent bond., Bond polarity leads to various electronic effects, in organic compounds., Thus, both Assertion and Reason are correct, and Reason is not the correct explanation of, Assertion., 63. Assertion is not correct but Reason is correct., Hybridisation can be determined by counting, σ-bond, 3σ 2σ 3σ, H2C== C == CH2, 3σ - sp 2 hybridisation, 2σ - sp hybridisation, Correct assertion In H2C == C == CH2 ,, the central carbon is sp-hybridised whereas the, terminal carbons are sp 2 -hybridised., , 64. In naming cycloalkenes, number the ring to, specify the position of the double bonded, carbons and choose the direction of numbering, so that the substituents get the lowest number., The position of the double bond is not indicated, because it is known to be between C -1 and C - 2., CH3, So,, is 3-methycyclohexene., , 65. In bond-line structural representations only, atoms specifically written are oxygen, chlorine,, nitrogen etc. Also the terminal position denote, methyl ( CH3 ) group while line-junctions, denote carbon atoms bonded to appropriate, number of hydrogens required to satisfy the, valency of the carbon atoms., , Thus, bond Assertion and Reason correct and, Reason is not the correct explanation of Assertion., , 66. The names of straight chain compounds are, based on their chain structure and end with suffix, ‘-ane’, carry a prefix indicating the number of, carbon atoms present in the chain (except, from CH4 to C4H10 where the prefixes are, derived from trivial names)., Thus, both Assertion and Reason are correct and, Reason is not the correct explanation of Assertion., 67. In numbering to carbon chain, the parent carbon, chain is numbered in a manner so as to give, lowest number to that carbon atom linked by, double or triple bond even it violates the rules of, saturated hydrocarbons so,, CH3, 1, , CH, 2, , CH, 3, , CH, , 4, , CH2, 5, , CH3, is 4, 6-dimethylhept-2-ene., , CH, , 6, , CH3, 7, , CH3, , 68. Heterolytic fission occurs when the two atoms, differ considerable in their electronegativities, and shared pair of electrons is carried by more, electronegative atom., 69. A covalent bond may be cleaved either by, heterolytic cleavage or by homolytic cleavage., In heterolytic cleavage, the bond breaks in such, a fashion that the shared pair of electrons, remains with one of the fragments., s, r, H3C — Br → H3 C + Br, In homolytic cleavage, one of the electrons of the, shared pair in a covalent bond goes with each of the, bonded atoms., R—X, , Heat or light, , R•, , Alkyl free, radical, , •, , + X, , Thus, both Assertion and Reason are correct and, Reason is not the correct explanation of Assertion., 70. Both Assertion and Reason are correct but Reason, is not the correct explanation of Assertion., 71. Since free radicals contain odd electrons, so they, are short lived and they readily try to pair up the, odd electrons to form neutral molecules, that is, why they are highly reactive., , 72. The electron displacements in an organic, molecule due to the influence of an atom or a, substituted group present in the molecule cause, permanent polarisation of the bond. Inductive

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144, , CBSE New Pattern ~ Chemistry XI (Term-I), , effect and resonance effect are examples of this, type of electronic displacement., Thus, both Assertion and Reason are correct, and Reason is the correct explanation of, Assertion., , 73. Presence of alternate single and double bonds, in an open chain or cyclic system is termed as, conjugated system. These systems often show, abnormal behaviour as in them the π- electrons, are delocalised and the develops polarity., Examples are 1, 3- butadiene, aniline and, nitrobenzene etc., Thus, both Assertion and Reason are correct, and Reason is the correct explanation of, Assertion., , 74. Assertion is incorrect but Reason is correct., Resonance hybrids are always more stable than, any of the canonical structures would be, if they, existed. The delocalisation of the electrons, lowers the orbital energies, imparting stability., The gain in stability of the resonance hybrid, over the most stable of the canonical structure is, called resonance energy., A canonical structure that is lower in energy, makes a relating greater contribution to, resonance hybrid., Thus, the correct assertion will be energy of, resonance hybrid is equal to the sum of energies, of all canonical forms in proportion of their, contribution towards the resonance hybrid., , 75. When a covalent bond is formed between atoms, of different electronegativity, the bonding, electrons are shifted towards the more, electronegative atom. This is known as inductive, effect and this unequal sharing of electrons, results in polar covalent bond., , (iv) The permanent displacement of electron, through a chain involving only σ-bonds is, called inductive effect. It involves the, electron displacement along the chain of, saturated carbon atoms due to the presence, of a polar covalent bond at one end of the, chain., Or, – I -effect is related to the ability of, substituent for the electron attraction, capacity from the attached carbon atom, i.e., it is based on electronegativity of an, atom. This effect increases with increase in, the electronegativity of an atom., From above, we can conclude that option (b), is correct., — NR2 < — OR < — F (−I -effect), , 77. (i) The dispersal of the charge stabilises the, carbocation. More the number of alkyl, groups, the greater is the dispersal of, positive charge and therefore, more is the, stability of carbocation., Thus, the correct order is, ⊕, , ⊕, , C, B, A, (ii) Homolytic cleavage results in the formation, of free radical., Homolytic, fission, , (iii) Aniline is a conjugated system and it is more, polar than expected because NH2 group, release electron density towards benzene, nucleus rather than attract from benzene., , •, , CH3CH2 — Cl → CH3 C H2 + • Cl, (iii) The shape of carbocation is trigonal planar, with positively charged carbon being, sp 2 -hybridised., It is shown below :, H, , 76. (i) Maximum polarity present in C  NO2, , bond so dipole moment of nitropropane is, the maximum amongst the given molecules., (ii) Electron donor tendency :, O, ,  OCH3 > O  C  CH3, , ⊕, , CH2OCH3 > CH3 CH2 > CH3, , +, , C, , H, , H, , (iv) Among the given free radicals,, is most stable due to resonance. In the, remaining free radicals, more the alkyl, groups bonded to the electron deficient, carbon, more stable is the radical., , CH2

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145, , Organic Chemistry : Some Basic Principles and Techniques, , Thus, the correct order of stability of given, free radicals is, , (I), , > CH2, , CH3 > CH2 —CH(CH3)2, , (IV), , 78. (i) Both the rules are correct for resonating, structure., I. The overall charge of system must remain, same., II. The arrangement of atoms must be, identical or almost same in every, resonance structures., (ii), s, , O, , O, , s, , s, , O, , Or, In compounds exhibiting resonance, bond, order can be given by the formula., , 79. (i) When a covalent bond is formed between, , (III), , Or Because of the presence of 7 p -orbitals and, 7 unpaired electrons. It is aromatic in nature, as these unpaired electrons delocalise in, p-orbitals., The structure of benzyl free radical is, , O, , −R-effect is shown :, ‚, COOH, CHO, ƒ C ==O, CN,, NO2 etc., , CH2 > CH3 CHCH3, (II), , ●, , s, , O, , s, , (Phenoxide ion), , (iii) Structure I is more stable than structure II as, in structure II, the positive charge is present, on the highly electronegative oxygen., (iv) + R-effect is shown by :, Halogen, OH, OR, OCOR, NH2 ,, COR , OR2, , atoms of different electronegativity, the, bonding electrons are shifted towards the more, electronegative atom. This is known as, inductive effect and this unequal sharing of, electrons results in polar covalent bond., Thus, both (A) and (R) are correct but (R) is, not the correct explanation of (A)., (ii) Formic acid (HCOOH) is a stronger acid than, benzoic acid (PhCOOH) although (Ph-) group, is e − - attracting (−I -effect). This is due to the, + R effect of phenyl group which is greater than, −I -effect., Hence, it decreases the acidic nature of benzoic, acid., (iii) The stability of carbonium ions is influenced, by both resonance and inductive effects. Allyl, and benzyl show resonance effect., Thus, both (A) and (R) are correct and (R) is, the correct explanation of (A)., (iv) When a covalent bond is formed between, atoms of different electronegativity, the, electron density is more towards the more, electronegative atom of the bond., Such a shift of electron density results in a, polar covalent bond. Bond polarity leads to, various electronic effects in organic, compounds., Thus, both Assertion and Reason are correct, and Reason is not the correct explanation of, Assertion., Or, The electron displacements in an organic, molecule due to the influence of an atom or a, substituted group present in the molecule cause, permanent polarisation of the bond. Inductive, effect and resonance effect are examples of this, type of electronic displacement., Thus, both Assertion and Reason are correct, and Reason is the correct explanation of, Assertion.

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146, , CBSE New Pattern ~ Chemistry XI (Term-I), , 80. (i) Both Assertion and Reason are correct but, Reason is not the correct explanation of, Assertion. Less reactivity of saturated, hydrocarbons is due to the presence of, single bonds between carbon atoms., (ii) Assertion is correct but Reason is incorrect., There is one double bond between two, carbon atoms in their molecules and they, have the general formula C n H2n ., (iii) Both Assertion and Reason are correct and, Reason is the correct explanation of, Assertion. When double and triple bonds, are present in a single compound lowest sum, rule is followed. Thus in, 5, , 4, , 3, , 2, , 1, , C H3 C H == C H C ≡≡ C H, Sum of locants (1 + 3) = 4., In C H3 C H == C H C ≡≡ C H, 1, , 2, , 3, , 4, , 5, , Sum of locants ( 2 + 4 ) = 6, So it is named as pent-3-en-1-yne., , (iv) Assertion is correct but Reason is incorrect., IUPAC name of, 5, , 4, , 3, , 2, , 1, , C H3  C H  C H2  C H  C HO, , , Cl, CH3, is 4-chloro-2-methylpentanal, the compound, consists of  CHO as functional group. The, numbering is done from the functional, group. The numbering is done from the, functional group side, Cl and CH3 are, substituents., When writing the names of the compounds,, it is written in alphabetical order., Thus, the name is given as 4-chloro-2-methyl, pentanal., Or Assertion and Reason both are correct, statement but Reason is not the correct, explanation of Assertion. neo-hydrocarbons, contain a quaternay carbon atom, i.e., ( CH3 ) 3 C  .

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PRACTICE PAPER 1, Chemistry Class 11th (Term I), Instructions, 1. This paper has 35 questions., 2. All questions are compulsory, each question carry 1 mark., 3. Answer the questions as per given instructions., Time : 90 Minutes, , Max. Marks : 35, , Multiple Choice Questions, , 6. The IUPAC name of the following, , Direction (Q. Nos. 1-20) Each of the, , compound is, , question has four options out of which only, one is correct. Select the correct option as, your answer., , CH2 == CH ¾ CH ¾ CH2 ¾ CH3, ½, CH2 ¾ CH2 ¾ CH3, (a) 3-propylpent-1-ene (b) 3-ethylpent-1-ene, (c) 4-ethylhex-1-ene, (d) 3-ethylhex-1-ene, , 1. The effect that makes 2, 3-dimethyl-2butene more stable than 2-butene is, (a) resonance, (c) electrometric, , (b) hyperconjugation, (d) inductive effect, , 2. Which of the following has octet around, central atom ?, (a) PF5, , (b) SF6, , (c) CCl4, , (d) BF3, , 3. NaH is an example of, (a), (b), (c), (d), , metallic hydride, electron-rich hydride, saline hydride, molecular hydride, , decreasing order of metallic characters., Si, Be, Mg, Na and P. Select the correct, answer., (a), (b), (c), (d), , Na > Be > Mg > P > Si, Na > Mg > Be > P > Si, Na > Be > P > Mg > Si, Na > Mg > Be > Si > P, , 8. 4d , 5 p, 5 f and 6 p -orbitals are arranged, in the order of decreasing energy., The correct option is, , 4. A compound contains 69.5% oxygen,, 30.5% nitrogen and its molecular weight, is 92. The formula of compound is, (a) N2O, (c) N2O 4, , 7. Arrange the following elements in the, , (b) NO 2, (d) N2O 5, , 5. In which of the following reactions, the, underlined substance has been oxidised?, (a) Br2 + H2 S ¾® 2HBr + S, , (b) 2HgCl2 + SnCl2 ¾® Hg 2Cl2 + SnCl4, (c) Cl2 + 2KI ¾® 2KCl + I2, (d) 2Cu 2 + + 4I- ¾® Cu 2I2 + I2, , (a) 6 p > 5 f > 5 p > 4 d, (c) 5 f > 6 p > 4 d > 5 p, , (b) 5 p > 5 f > 4 d > 5 p, (d) 5 f > 6 p > 5 p > 4 d, , 9. The correct order in which the O—O, bond length increases is, (a) H2O 2 < O 2 < O 3, (c) O 2 < O 3 < H2O 2, , (b) O 3 < H2O 2 < O 2, (d) O 2 < H2O 2 < O 3, , 10. Which of the following groups form, electron rich hydrides ?, (a) Chalcogens, (b) Alkali metals, (c) Noble gases, (d) Alkaline earth metals

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150, , CBSE New Pattern ~ Chemistry XI (Term-I), , (c) (i) n = 4, l = 1, ml = 1, ms = +, , 11. The most stable carbocation is, r, , r, , (a), , (ii) n = 3, l = 2 , ml = 1, ms = +, , (b), r, , 1, 2, 1, (ii) n = 3, l = 2 , ml = + 2 , ms = +, 2, , (d), , 12. Which of the following is a incorrect, match for hybridisation and geometry?, Hybridisation, (a) dsp 2, (b) d 3 s and sp 3, (c) d 2 sp 3 and sp 3d 2, (d) d 3 s, , Geometry, — Planar, — Tetrahedral, — Octahedral, — Planar, , 13. 2.76 g of silver carbonate on being, strongly heated yield a residue of, weighing …… ., (a) 2.16 g, (c) 2.64 g, , 18. In general, the properties that decrease, and increase down a group in the, periodic table, respectively are, (a) electronegativity and atomic radius, (b) electronegativity and electron gain enthalpy, (c) electron gain enthalpy and electronegativity, (d) atomic radius and electronegativity, , 19. The radius of the second Bohr orbit in, terms of the Bohr radius, a 0 , in Li 2+ is, 2a0, 3, 4a0, (c), 9, , 4a0, 3, 2a0, (d), 9, , (b), , (a), , (b) 2.48 g, (d) 2.32 g, , 14. Which of the following has minimum, -I-effect?, , 20. What volume of water is to be added to, , (a) —NO 2, , (b) —COOH, , (c) —F, , (d) — N R3, , 100 cm 3 of 0.5 M NaOH solution to, make it 0.1 M solution?, , +, , 15. The oxidation number of Cr in K 2Cr 2O 7, is …… ., (a) + 2, , 1, 2, , (d) (i) n = 3, l = 2 , ml = + 2 , ms = -, , r, , (c), , 1, 2, , (b) + 4, , (c) + 6, , (d) + 7, , 16. For the second period elements, the, correct increasing order of first, ionisation enthalpy is …… ., (a) Li < B < Be < C < O < N < F < Ne, (b) Li < B < Be < C < N < O < F < Ne, (c) Li < Be < B < C < O < N < F < Ne, (d) Li < Be < B < C < N < O < F < Ne, , 17. Out of the following pairs of electrons,, identify the pairs of electrons present in, degenerate orbitals., 1, 2, 1, (ii) n = 3, l = 2 , ml = - 1, ms = 2, 1, (b) (i) n = 3, l = 1, ml = 1, ms = +, 2, 1, (ii) n = 3, l = 2 , ml = 1, ms = +, 2, (a) (i) n = 3, l = 1 , ml = - 1 , ms = -, , (a) 200 cm 3, (c) 500 cm, , 3, , (b) 400 cm 3, (d) 100 cm 3, , Assertion-Reasoning MCQs, Direction In the following questions, (Q.No. 21-27) a statement of Assertion, followed by a statement of Reason is given., Choose the correct answer ouf of the, following choices., (a) Both Assertion and Reason are correct, statements and Reason is the correct, explanation of the Assertion., (b) Both Assertion and Reason are correct, statements, but Reason is not the correct, explanation of the Assertion., (c) Assertion is correct, but Reason is incorrect, statement., (d) Assertion is incorrect but Reason is correct, statement., , 21. Assertion H—S—H bond angle in H 2S, is closer to 90° but H—O—H bond, angle in H 2O is 104.5°.

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151, , Practice Paper 1, , Reason Lone pair-lone pair repulsion is, stronger in H 2S than in H 2O., 22. Assertion Permanent hardness of water, is removed by treatment with washing, soda., Reason Washing soda reacts with, soluble magnesium and calcium, sulphate to form insoluble carbonates., 23. Assertion Equivalent weight of a base, molecular weight, =, acidity, Reason Acidity is the number of, replaceable hydrogen atom in one, molecule of the base., , 24. Assertion The free gaseous Cr atoms, has six unpaired electrons., Reason Half-filled s-orbital has greater, stability., , 25. Assertion Shielding effect increases as, we go down the group., Reason More is the number of, electrons in the penultimate shell, more, is shielding., , 26. Assertion The stability of carbonium, ion is influenced by both resonance and, inductive effect., Reason More the number of alkyl, groups, greater the dispersal of positive, charge and therefore, more the stability, of carbonium ion is observed., , 27. Assertion Fluorine does not show, , Schrodinger equation for the hydrogen atom., It is a mathematical function of the three, coordinates of the electron (r , q, j) and can be, factorised into three separate parts each of which, is a function of only one coordinate:, y(r , q, j) = R (r ) Q(q) F( j), where, R (r ) is the radial function which gives the, dependence of orbital upon the distance r of the, electron from the nucleus and Q(q) and F( j) are, the angular functions giving the angular, dependence of orbital on q and j, respectively., Further, the radial function depends upon the, quantum numbers n and l , whereas the angular, part depends upon the quantum numbers l and m l, and is independent of n.The total wave function y, may, therefore, be more explicitly written as :, y(r , q, j) = R n , l (r ) Q l , m (q) Fm ( j), Angular part, , Radial part, , The orbital wave function y has no physical, significance. It is the square of the absolute value, of the orbital wave function | y |2 which has a, physical significance—it measures the electron, probability density at a point in an atom. It would,, therefore, be interesting to know how y and | y |2, vary as a function of the three coordinates r , q,, and j for different orbitals. Such a representation, of the variations of y or | y |2 in space would,, however, need a four-dimensional graph-three, dimensions for the coordinates and the fourth for, y or | y |2. It is not possible to show such a, variation in a single diagram since we can draw, only two-dimensional diagrams on paper, We can, get around this difficulty by drawing separate, diagrams for (i) variation of radial function and (ii), angular function., , 28. In Schrodinger wave mechanical mode,, , disproportionation tendency., , y 2 (r, q, f) represents:, , Reason Oxidation state of fluorine is, always negative., , (a) total probability of finding electron around, uncleus, (b) probability density of electron, (c) amplitude of electron wave, (d) orbit, , Case Based MCQs, Direction (Q. No. 28-31) Read the, following passage and answer the following, questions., An atomic orbital is a one-electron wave function, y(r , q, j) obtained from the solution to the, , 29. Radial amplitude of electron wave can, be represented by:, (a) R (r), (c) 4 pr 2, , (b) R 2 (r), (d) 4 pr 2 R 2 (r)

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152, , CBSE New Pattern ~ Chemistry XI (Term-I), , 30. The variation of radial probability, , density R 2 (r ) as a function of distance r, of the electron from the nucleus for, 3p-orbital., (a) R2(r), , (b) R2(r), , (c) R2(r), , (d) R2(r), , 31. Which orbital has only positive values, of wave function at all distances from, the nucleus ?, (a) 1s, , (b) 2s, , (c) 2p, , (d) 3d, , Or Which of the following orbitals has two, spherical nodes?, , Bond order (BO) = 1/ 2( N b - N a ), If N b > N a Þ molecule is stable., If N b < N a or N b = N a Þ molecule is unstable., If bond order is 1, 2 or 3, it shows the presence of, single, double or triple bonds respectively. Bond, length is measured with the help of bond order., All the molecular orbitals in a molecule are, doubly occupied, i.e. contains two electrons, than, substance is diamagnetic (repelled by magnetic, field)., However, if one or more molecular orbitals are, singly occupied, the molecule is paramagnetic, (attracted by magnetic field). e.g. O2 molecule., The following questions (i-iv) are multiple choice, questions. Choose the most appropriate answer :, , 32. Assertion The bond order of helium is, zero., Reason The number of electrons in, bonding molecular orbital and, antibonding molecular orbital is equal., , 33. Assertion Fluorine molecule has bond, R, , +_, –, , (a) 1s, (c) 3s, , +, r, , (b) 2s, (d) 2p, , Direction (Q. No. 32-35) Read the, following passage and answer the following, questions., The distribution of electrons in various molecular, orbitals is called the electronic configuration of, the molecule. If the number of electrons, occupying bonding orbitals are represented by N b, and the number of electrons occupying the, antibonding orbitals by Na then, (i) the molecule is stable if N b > N a and, (ii) the molecule is unstable if N b < N a ., Bond order is one half of the difference between, the number of electrons present in the bonding, and the antibonding orbitals, i.e., , order one., Reason The number of electrons in the, antibonding molecular orbitals is less, than that in bonding molecular orbitals., , 34. Assertion O 2 is paramagnetic., Reason It has one unpaired electron., , 35. Assertion Bond order can assume any, value including zero., Reason Higher is the bond order,, shorter is the bond length and greater is, the bond energy., Or, Assertion Both p(2 p x ) and p * ( 2p x ),, MO’s have one nodal plane each., Reason All MO’s formed by side way, overlapping of 2p-orbitals have one, nodal plane.

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PRACTICE PAPER 1, OMRSHEET, Instructions, Use black or blue ball point pens and avoid gel pens and fountain, pens for filling the sheets, Darken the bubbles completely. Don't put a tick mark or a cross, mark half-filled or over-filled bubbles will not be read by the software., , ü, , û, , Incorrect, , Incorrect, , Incorrect, , Correct, , Do not write anything on the OMR Sheet, Multiple markings are invalid, 1, , a, , b, , c, , d, , 19, , a, , b, , c, , d, , 2, , a, , b, , c, , d, , 20, , a, , b, , c, , d, , 3, , a, , b, , c, , d, , 21, , a, , b, , c, , d, , 4, , a, , b, , c, , d, , 22, , a, , b, , c, , d, , 5, , a, , b, , c, , d, , 23, , a, , b, , c, , d, , 6, , a, , b, , c, , d, , 24, , a, , b, , c, , d, , 7, , a, , b, , c, , d, , 25, , a, , b, , c, , d, , 8, , a, , b, , c, , d, , 26, , a, , b, , c, , d, , 9, , a, , b, , c, , d, , 27, , a, , b, , c, , d, , 10, , a, , b, , c, , d, , 28, , a, , b, , c, , d, , 11, , a, , b, , c, , d, , 29, , a, , b, , c, , d, , 12, , a, , b, , c, , d, , 30, , a, , b, , c, , d, , 13, , a, , b, , c, , d, , 31, , a, , b, , c, , d, , 14, , a, , b, , c, , d, , Or, , a, , b, , c, , d, , 15, , a, , b, , c, , d, , 32, , a, , b, , c, , d, , 16, , a, , b, , c, , d, , 33, , a, , b, , c, , d, , 17, , a, , b, , c, , d, , 34, , a, , b, , c, , d, , 18, , a, , b, , c, , d, , 35, , a, , b, , c, , d, , Or, , a, , b, , c, , d, , Students should not write anything below this line, , SIGNATURE OF EXAMINER WITH DATE, , MARKS SCORED

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PRACTICE PAPER 2, Chemistry Class 11th (Term I), Instructions, 1. This paper has 35 questions., 2. All questions are compulsory, each question carry 1 mark., 3. Answer the questions as per given instructions., Time : 90 Minutes, , Multiple Choice Questions, , Max. Marks : 35, , 4. Which of the following is not in, , Direction (Q. Nos. 1-20) Each of the, , accordance to IUPAC system?, , question has four options out of which only, one is correct. Select the correct option as, your answer., , (a) Br ¾ CH2 ¾ CH == CH2, , 1. An example of a disproportionation, reaction is, (a) 2MnO -4 + 10 I- + 16H+ ¾®, 2Mn2 + + 5I2 + 8H2O, (b) 2NaBr + Cl2 ¾® 2NaCl + Br2, (c) 2KMnO 4 ¾® K 2MnO 4 + MnO 2 + O 2, (d) 2CuBr ¾® CuBr2 + Cu, , 2. Among the following compounds, the, most basic is, (a) aniline, (b) acetanilide, (c) p-nitroaniline, (d) benzylamine, , 3. The element Z = 114 has been, discovered recently. It will belong to, which of the following family/group, and what will be its electronic, configuration?, (a) Halogen family, [Rn] 5 f 14 6 d 10 7s 2 7p 5, (b) Carbon family, [Rn] 5 f 14 6 d 10 7s 2 7p 2, (c) Oxygen family, [Rn] 5 f 14 6 d 10 7s 2 7p 4, (d) Nitrogen family, [Rn] 5 f 14 6 d 10 7s 2 7p 6, , 1-bromoprop-2-ene, , CH3, ½, (b) CH3 ¾CH2 ¾ C ¾ CH2 ¾ CHCH3, ½, ½, Br, CH3, 4-bromo-2, 4-dimethyl hexane, , (c) CH — CH — CH — CH — CH, 3, 2, 3, , CH3, 2-methyl-3- phenylpentane, , (d) CH3 ¾ C ¾CH2 ¾ CH2CH2COOH, ½½, O, 5-oxohexanoic acid, , 5. The oxidation number of Mn and S in, KMnO 4 and Na 2S 2O 3 respectively are, (a) +7 and +2, (c) +7 and +5, , (b) +7 and - 2, (d) +5 and +7, , 6. Which of the following orders of ionic, radius is correctly represented?, (a) H - > H > H+, , (b) Na+ > F - > O2 -, , (c) F - > O2 - > Na+, , (d) Al3+ > Mg2 + > N3-

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155, , Practice Paper 2, , 7. An electron is moving in Bohr’s fourth, , orbit. Its de-Broglie wavelength is l., What is the circumference of the fourth, orbit?, 2, (a), l, , (b) 2l, , (c) 4l, , 4, (d), l, , 8. Inductive effect involves, , (b) Na+, (d) F -, , 8.95 ´ 10 and substraction of numbers, 2.5 ´10 -2 - 4.8 ´10 -3 , respectively in, terms of scientific notation will be, 3, , 7.545 ´ 104, 2.02 ´ 10-2, 75.43 ´ 103 , 2.02 ´ 10-3, 745.5 ´ 102 , 2.02 ´ 10-1, 75.45 ´ 100 , 2.02 ´ 100, , (c) 4d, , (d) 2s, , 12. Among CaH 2 , BeH 2 , BaH 2 , the order of, ionic character is …… ., , (a) BeH2 < BaH2 < CaH2, (b) CaH2 < BeH2 < BaH2, (c) BeH2 < CaH2 < BaH2, (d) BaH2 < BeH2 < CaH2, , (d) NO–3, , and (C). If the number of neutron(s) in, (A), (B) and (C) respectively, are (x ), ( y ), and (z ), the sum of (x ), ( y ) and (z ) is, (b) 3, , (c) 2, , (d) 1, , 18. Which one of the following pairs of, species have the same bond order?, (a) CO, NO, , (b) O 2 , NO +, , (c) CN- , CO, , (d) N2 , O -2, , (a) SF6, (c) CH 4, , (b) PCl5, (d) BF3, , 20. CH 4 , NH 3 , H 2O and HF are the, examples of, (a) molecular hydrides, (c) ionic hydrides, , (b) metallic hydrides, (d) Both (a) and (c), , Assertion-Reasoning MCQs, Direction In the following questions, , 13. Which one of the following has a, magnetic moment of 1.75 BM?, (c) Fe 3 +, , (d) Ti3 +, , 14. Consider the following ionic equations,, Fe 2+ ¾® Fe 3+ + e - ,, MnO -4 + 5e - ¾® Mn 2 +, The ratio of stoichiometric coefficient of, Fe 2+ and MnO -4 is, (a) 1 :5, (c) 2 : 3, , (c) NO+2, , trigonal bipyramidal geometry with, bond angles 120° and 90°?, , is …… ., , (b) Cr 3 +, , (b) NO–2, , 19. Which of the following molecules has, , 11. Impossible orbital among the following, , (a) V 3 +, , 16. In which of the following compounds of, , (a) 4, , 10. The addition of number 6.65 ´ 10 4 and, , (b) 2p, , (b) 4.06 ´ 1024, (d) 3.24 ´ 1024, , 17. Hydrogen has three isotopes (A), (B), , minimum ionic radii?, , (a) 3f, , (a) 12.04 ´ 1024, (c) 2.01 ´ 1024, , (a) NO2, , 9. Which of the following species has, , (a), (b), (c), (d), , calcium carbonate is, ( NA = 6.023 ´ 10 23 ), , nitrogen, the maximum bond angle of, O ¾ N ¾ O is present in, , (a) displacement of s-electrons, (b) delocalisation of p-electrons, (c) delocalisation of s-electrons, (d) displacement of p-electrons, , (a) O 2 (c) Mg 2 +, , 15. Total number of protons in 10 g of, , (b) 5 :1, (d) 6 :1, , (Q.No. 21-27) a statement of Assertion, followed by a statement of Reason is given., Choose the correct answer ouf of the, following choices., (a) Both Assertion and Reason are correct, statements and Reason is the correct, explanation of the Assertion., (b) Both Assertion and Reason are correct, statements, but Reason is not the correct, explanation of the Assertion., (c) Assertion is correct, but Reason is incorrect, statement., (d) Assertion is incorrect but Reason is correct, statement.

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156, , CBSE New Pattern ~ Chemistry XI (Term-I), , 21. Assertion Calgon is used as water, softening agent., Reason Calgon form soluble, complexes with cationic species., , 22. Assertion It is impossible to determine, the exact position and exact momentum, of an electron, simultaneously., Reason The path of an electron is, clearly defined., , 23. Assertion Helium has the highest value, of ionisation energy among all the, elements known., Reason Helium has the highest value of, electron affinity among all the elements, known., 24. Assertion A compound with, delocalised electron is more stable than, that compound would be if all its, electrons were localised., Reason The extra stability, a compound, gains as a result of having delocalised, electrons, is called delocalisation energy., 25. Assertion Benzyl carbanion is more, stable than propyl carbanion., Reason The carbon atom in carbanion, is trivalent., 26. Assertion This sulphate reacts, differently with iodine and bromine in, the reaction given below., 2S 2 O 32 - + I 2 ¾® S 4 O 26 - + 2I S 2 O 23 - + 2Br2 + 5H 2 O ¾® 2SO 42 - + 2Br + 10H +, Reason Bromine is a stronger oxidant, than iodine., 27. Assertion The central atom of both, NH 3 and H 2 O molecules are, sp 3 -hybridised, yet H ¾ N ¾ H. bond, angle is greater than that of H ¾ O ¾ H., Reason In NH 3 , N-atom has one lone, pair of electron whereas in H 2 O, oxygen, atom has two lone pairs of electrons., , Case Based MCQs, Direction (Q. No. 28-31) Read the, following passage and answer the following, questions., Chemical reactions involve the interaction of, atoms and molecules. A large number of a atoms/, molecules (approximately 6.023 ´ 10 23) are, present in few grams of any chemical compound, varying with their atomic/molecular masses. To, handle such large numbers conveniently, the, mole concept was introduced. This concept has, implications in diverse areas such as analytical, chemistry, biochemistry, electrochemistry and, radiochemistry., The following example illustrates a typical case,, involving chemical/electrochemical reaction,, which requires a clear understanding of the mole, concept. In SI system, mole (symbol, mol) was, introduced as the seventh base quantity for the, amount of a substance., One gram atom of any element contains the same, number of atoms and one gram molecule of any, substance contains the same number of molecules., Thus, one mole is the amount of a substance that, contains as many entities or particles as there are, atoms in exactly 12 g (or 0.012 kg) of the 12C, isotope., , 28. Which of the following has least mass?, (a) 1 mole of S, (b) 3 ´ 1023 atoms of C, (c) 2 g atom of nitrogen, (d) 7.0 g of Ag, , 29. The mass of 1 molecule of water (H 2 O ) is, (a) 3.0 ´ 10- 23, (c) 2.0 ´ 10- 23, , (b) 3.0 ´ 10- 22, (4) 2.0 ´ 10- 22, , 30. The number of molecules in 16 g of, sulphur dioxide (SO 2 ) are, (a) 0.28, (c) 0.26, , (b) 0.27, (d) 0.25, , 31. The weight of one mole of sodium, carbonate (Na 2 CO 3 ) is, (a) 110 g, (c) 108 g, , (b) 109 g, (d) 106 g

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157, , Practice Paper 2, , r, , Or Moles and g-equivalents in 196 g of, Ca(OH)2 are, (a) 2.65 and 5.3, (c) 2.57 and 5.14, , 32. Assertion, , (b) 2.6 and 5.2, (d) 2.54 and 5.08, , is resonance, , stabilised., , Direction (Q. No. 32-35) Read the following, , Reason It follows Huckel’s rule., , passage and answer the following questions., r, , In case of certain molecules, a single Lewis, structure cannot explain all the properties of the, molecule. The molecule is then supposed to have, many structures, each of which can explain most, of the properties of the molecule but cannot, explain all the properties of the molecules. The, actual structure is in between of all these, contributing structures and is called resonance, hybrid and the different individual structures are, called resonating structures or canonical forms., This phenomenon is called resonance., Different resonating structures of a substance have, ● same positions of atoms, ● same number of paired and unpaired electrons, ● almost equal energy, ● they differ only in the arrangement of electrons, in different resonating forms., Resonance structures are represented by double, headed arrow («). Examples of resonance, structures of some molecules and ions are given, below:, Carbon dioxide (CO 2 ), O, , I, , C, , O, , +, , O, , II, , C, , r, , O, , r, , r, , O, , C, , O, III, , 33. Assertion, , r, , are correct, , pair of resonating structure., Reason Resonating structures have, almost equal energy., , 34. Assertion 2° carbocation is stable than, 1° carbocation., Reason Stability of carbocation is due to, resonance., , 35. Assertion Stability of carbanion is, influenced by resonance., Reason More the number of alkyl, groups, greater is the dispersal of, positive charge., Or Assertion The correct decreasing of, stability of resonating structure, È, Å, CH 2 == CH ¾ C(I) « CH 2 ¾ CH == Cl(II), Å, , È, , « CH 2 ¾ CH == Cl(III) is I > II > III., Reason Resonance hybrid has highest, stability.

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PRACTICE PAPER 2, OMRSHEET, Instructions, Use black or blue ball point pens and avoid gel pens and fountain, pens for filling the sheets, Darken the bubbles completely. Don't put a tick mark or a cross, mark half-filled or over-filled bubbles will not be read by the software., , ü, , û, , Incorrect, , Incorrect, , Incorrect, , Correct, , Do not write anything on the OMR Sheet, Multiple markings are invalid, 1, , a, , b, , c, , d, , 19, , a, , b, , c, , d, , 2, , a, , b, , c, , d, , 20, , a, , b, , c, , d, , 3, , a, , b, , c, , d, , 21, , a, , b, , c, , d, , 4, , a, , b, , c, , d, , 22, , a, , b, , c, , d, , 5, , a, , b, , c, , d, , 23, , a, , b, , c, , d, , 6, , a, , b, , c, , d, , 24, , a, , b, , c, , d, , 7, , a, , b, , c, , d, , 25, , a, , b, , c, , d, , 8, , a, , b, , c, , d, , 26, , a, , b, , c, , d, , 9, , a, , b, , c, , d, , 27, , a, , b, , c, , d, , 10, , a, , b, , c, , d, , 28, , a, , b, , c, , d, , 11, , a, , b, , c, , d, , 29, , a, , b, , c, , d, , 12, , a, , b, , c, , d, , 30, , a, , b, , c, , d, , 13, , a, , b, , c, , d, , 31, , a, , b, , c, , d, , 14, , a, , b, , c, , d, , Or, , a, , b, , c, , d, , 15, , a, , b, , c, , d, , 32, , a, , b, , c, , d, , 16, , a, , b, , c, , d, , 33, , a, , b, , c, , d, , 17, , a, , b, , c, , d, , 34, , a, , b, , c, , d, , 18, , a, , b, , c, , d, , 35, , a, , b, , c, , d, , Or, , a, , b, , c, , d, , Students should not write anything below this line, , SIGNATURE OF EXAMINER WITH DATE, , MARKS SCORED

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PRACTICE PAPER 3, Chemistry Class 11th (Term I), Instructions, 1. This paper has 35 questions., 2. All questions are compulsory, each question carry 1 mark., 3. Answer the questions as per given instructions., Time : 90 Minutes, , Max. Marks : 35, , Multiple Choice Questions, Direction (Q. Nos. 1-20) Each of the question, has four options out of which only one is correct., Select the correct option as your answer., , 1. Which of the following compounds of, chlorine contains both ionic and covalent, bonds ?, (a) NaCl, (c) AlCl3, , (b) NaClO 4, (d) POCl3, , 2. An organic compound containing C, H, and O has 49.3% carbon, 6.84% hydrogen, and its vapour density is 73. Molecular, formula of the compound is, (a) C 3H 5 O 2, (c) C 6H10O 4, , (b) C 4H10O 2, (d) C 3H10O 2, , 3. Consider the following reaction,, H 2O(l ) + H 2O(l ) ¾® H 3O + (aq ) + OH – (aq ), Acid-1, (Acid), , Base-2, (Base), , Acid-2, (Conjugate, acid), , Base-1, (Conjugate, base), , ratio of uncertainties ( Dx A / Dx B ) in, their positions ?, (a) 2, (c) 4, , (b) 0.25, (d) 1, , 5. When the first electron gain enthalpy, ( D e g H ) of oxygen is -141 kJ/mol, its, second electron gain enthalpy is, (a) a positive value, (b) a more negative value than the first, (c) almost the same as that of the first, (d) negative, but less negative than the first, , 6. I 2 + 2S 2O 23- ¾® S 4O62- + 2I In the above reaction,, (a) iodine is reduced; sulphur is reduced, (b) iodine is reduced; sulphur is oxidised, (c) iodine is oxidised; sulphur is reduced, (d) iodine is oxidised; sulphur is oxidised, , 7. The IUPAC name of the compound is, OMe, , The above reaction is known as, (a), (b), (c), (d), , auto-protolysis of water, self-ionisation of water, hydration of water, Both (a) and (b), , 4. The uncertainties in the velocities of two, particles A and B are 0.05 and, 0.02 ms -1 respectively. The mass of B, is five times to that of mass A. What is the, , Cl, , CH3, (a) 2-chloro-4-methyl anisole, (b) 3-chloro-1-methyl anisole, (c) 1-methoxy-2-chloro toluene, (d) 1-methoxy-4-methyl chlorobenzene

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160, , CBSE New Pattern ~ Chemistry XI (Term-I), , 8. The presence of unpaired electron in, phosphorus atom is explained by which, principle?, (a) Aufbau principle, (b) Pauli’s exclusion principle, (c) Hund’s rule, (d) Heisenberg’s principle, , 9. Which of the following will have the, maximum dipole moment ?, (a) CH3F, (c) CH3Br, , (b) CH3Cl, (d) CH3I, , 10. In which case is the number of, molecules of water maximum ?, (a) 0.00224 L of water vapours at 1 atm and, 273 K, (b) 0.18 g of water, (c) 18 mL of water, (d) 10 -3 mol of water, , 11. The ascending order of stability of the, -, , -, , carbanion CH 3 (P ), C6 H 5 CH 2 (Q ),, -, , -, , (CH 3 ) 2CH ( R ) and H 2 C — CH ==CH 2 ( S ), is, (a) P <, (b) R <, (c) R <, (d) P <, , R <S < Q, P<S<Q, P<Q<S, R<Q<S, , 12. Which of the following statements is, incorrect ?, (a) Ionic hydrides are also known as saline or salt, like hydrides, (b) Covalent hydrides are also known as, molecular hydrides, (c) Metallic hydrides are also known as, stoichiometric hydrides, (d) None of the above, , 13. Consider the following statements, concerning the quantum numbers,, I. Angular quantum number determines, the three dimensional 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 statements are, (a) I and II, (c) III and IV, , (b) I and IV, (d) II, III and IV, , 14. The acidic, basic and amphoteric, oxides, respectively are, (a) Cl2O, CaO, P4O 10, (c) Na2O, SO 3 , Al2O 3, , (b) N2O 3 , Li2O, Al2O 3, (d) MgO, Cl2O, Al2O 3 ,, , 15. Highest oxidation state of Mn is present, in …… ., (a) KMnO 4, (c) Mn2O 3, , (b) K 2MnO 4, (d) MnO 2, , 16. Chloroacetic acid is a stronger acid than, acetic acid. This can be explained using, …… ., (a) -M effect, (c) + M effect, , (b) -I effect, (d) + I effect, , 17. The percentage of s -character in the, hybridised orbitals of B in BF3 is …… ., (a) 25, (c) 75, , (b) 50, (d) 33.3, , 18. 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, (c) 4.680 g, , (b) 4680 ´ 10-3 g, (d) 46.80 g, , 19. What is the maximum number of, electrons that can be associated with, the following set of quantum numbers?, n = 3, l = 1 and m = - 1, (a) 10, (c) 4, , (b) 6, (d) 2, , 20. Outer electronic configuration of, f-block elements is, (a), (b), (c), (d), , (n + 1) f 1 - 14 (n - 1) d 0 - 1ns 2, (n - 2) f 1 - 14 (n + 1) d 0 - 1ns 2, (n - 2) f 1 - 14(n - 1) d 0 - 1ns 2, None of the above

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161, , Practice Paper 3, , Assertion-Reasoning MCQs, Direction In the following questions (Q.No., 21-27) a statement of Assertion followed by a, statement of Reason is given. Choose the, correct answer ouf of the following choices., (a) Both Assertion and Reason are correct statements, and Reason is the correct explanation of the, Assertion., (b) Both Assertion and Reason are correct, statements, but Reason is not the correct, explanation of the Assertion., (c) Assertion is correct, but Reason is incorrect, statement., (d) Assertion is incorrect but Reason is correct, statement., , 21. Assertion CN - ion is an ambident, nucleophile., Reason Nucleophiles are electron rich, species., , 22. Assertion Removal of s-electron is, relatively difficult than removal of, p-electron of the same main shell., Reason s-electrons are closer to the nucleus, than p-electrons of the same shell and hence,, are more strongly attracted by nucleus., , 23. Assertion Electromeric effect is brought, into play only at the requirement of the, reagent., Reason It is a temporary effect in which, bond pair is shifted to one of the, constituent atoms., , 24. Assertion Number of radial and angular, nodes for 3p-orbital are 1, 1 respectively., Reason Number of radial and angular, nodes depends only on principal quantum, number., 25. Assertion All F ¾ S ¾ F angle in SF 4 is, greater than 90° but less than 180°., Reason The lone pair-bond pair repulsion is, weaker than bond pair-bond pair repulsion., , 26. Assertion Hydrogen has only three, isotopes namely protium, tritium and, deuterium., Reason Protium is radioactive in nature., , 27. Assertion Equivalent weight of NH 3 in, the reaction, N 2 ® NH 3 is 17/3 while, that of N 2 is 28/6., Reason Equivalent weight, Molecular weight, =, Number of e - s lost or gained per mole, , Case Based MCQs, Direction (Q. No. 28-31) Read the, following passage and answer the, following questions., Hydrogen accounts for approximately 75% of the, mass of the universe. Hydrogen serves as the, nuclear fuel of our Sun and other stars, and these, are mainly composed of hydrogen. On the earth,, though hydrogen is rarely found in the, uncombined state. Since the earth’s gravity is too, weak to hold such light molecules, nearly all the, H 2 originally present in the earth’s atmosphere, has been lost to space. In the earth’s crust and, oceans, hydrogen is found in water, petroleum,, proteins; carbohydrates and other compounds, and it is the ninth most abundant element on the, basis of mass., Hydrogen has three isotopes: hydrogen or, protium), deuterium or heavy hydrogen, tritium., The physical properties of the three isotopes are, different due to the difference in their masses,, i.e. isotope effect. The chemical properties of the, three isotopes are similar as they have the same, electronic configuration. Reaction between, hydrogen and oxygen is highly exothermic, and, gas mixtures that contain as little as 4% by, volume hydrogen in oxygen (or in air) are highly, flammable and potentially explosive., 12 H 2( g ) + O 2( g ) ¾® 2 H 2O ( g );, DH È = - 484 kJ mol - 1, As hydrogen is environmentally clean it is an, enormously attractive fuel. ‘Hydrogen economy’, is an emerging field in which it is thought that, our energy needs can be met by gaseous, liquid, and solid hydrogen., As hydrogen is not a naturally occurring, substance such as coal, oil or natural gas; energy, must be expanded to produce hydrogen before it, can be used.

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162, , CBSE New Pattern ~ Chemistry XI (Term-I), , 28. If an isotope of hydrogen has one, neutron in its atom, its atomic number, and atomic mass will respectively be, (a) 1, 2, (c) 1, 1, , (b) 1, 3, (d) 2, 1, , 29. Which of the following is radioactive in, nature?, (a) Hydrogen only, (b) Deuterium only, (c) Tritium only, (d) Deuterium and tritium, , 30. Hydrogen, H 2 is very less abundant in, the atmosphere due to, , (a) inflammable nature of H2, (b) weak earth’s gravity which is not able to hold, light H2 molecules, (c) diatomic nature of hydrogen, (d) very rapid reaction between hydrogen and, atmospheric oxygen, , 31. Liquid H 2 has been used as rocket fuel as, (a) its reaction with oxygen is highly exothermic, (b) it occupies small space, (c) it has high thrust, (d) All of the above, , Or, Which of the following is the lightest gas?, (a) Hydrogen, (c) Nitrogen, , (b) Oxygen, (d) Helium, , Direction (Q. No. 32-35) Read the following, passage and answer the following questions., Modern periodic table arranges the elements in the, increasing order of atomic number. It has 18, groups and 7 periods. Atomic numbers are, consecutive in a period and increases in the group, in a pattern. Elements are divided into four blocks,, s-block, p-block, d -block, and f -block based on, their electronic configuration. 78% of elements are, metals, about 20 elements are non-metals and few, elements like B, Si, Ge, As are metalloids., , Metallic character increases down the group but, decreases along the period from left to right. The, physical and chemical properties vary, periodically with their atomic numbers., Periodic trends are observed in atomic size, ionisation valences. Oxides of metals are basic,, some are amphoteric. Non-metals form acidic, oxides, some form neutral oxides. s-block, elements are soft, highly reactive, do not show, variable oxidation states. p-block elements are, metals, non-metals as well as metalloids, show, variable oxidation states, exist as solids, liquids,, and gases. d -block elements are metals form, coloured ions, show variable oxidation state, have, high melting and boiling points., Lanthanoids and actinoids are f -block elements,, form coloured ions. All actinoids are radioactive., , 32. Assertion The highest oxidation state of, Os is +5., Reason Osmium is a 5d -block element., , 33. Assertion Sixth and seventh periods in, the periodic table contains 14 elements., Reason In the periodic table, 14 elements, each of sixth [from Z = 58 to Z = 71] and, seventh periods [from Z = 90 to Z = 103], respectively are known as lanthanoids and, actinoids., 34. Assertion Helium is placed in group 18, along with p-block elements., Reason Helium is much less reactive like, other inert gases., , 35. Assertion Chemistry of actinoids is, more complicated than lanthanoids., Reason Actinoid element are radioactive., Or, , Assertion Properties of atom and its, corresponding ions are different., Reason Electronic configurations of both, atom and ion are different.

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PRACTICE PAPER 3, OMRSHEET, Instructions, Use black or blue ball point pens and avoid gel pens and fountain, pens for filling the sheets, Darken the bubbles completely. Don't put a tick mark or a cross, mark half-filled or over-filled bubbles will not be read by the software., , ü, , û, , Incorrect, , Incorrect, , Incorrect, , Correct, , Do not write anything on the OMR Sheet, Multiple markings are invalid, 1, , a, , b, , c, , d, , 19, , a, , b, , c, , d, , 2, , a, , b, , c, , d, , 20, , a, , b, , c, , d, , 3, , a, , b, , c, , d, , 21, , a, , b, , c, , d, , 4, , a, , b, , c, , d, , 22, , a, , b, , c, , d, , 5, , a, , b, , c, , d, , 23, , a, , b, , c, , d, , 6, , a, , b, , c, , d, , 24, , a, , b, , c, , d, , 7, , a, , b, , c, , d, , 25, , a, , b, , c, , d, , 8, , a, , b, , c, , d, , 26, , a, , b, , c, , d, , 9, , a, , b, , c, , d, , 27, , a, , b, , c, , d, , 10, , a, , b, , c, , d, , 28, , a, , b, , c, , d, , 11, , a, , b, , c, , d, , 29, , a, , b, , c, , d, , 12, , a, , b, , c, , d, , 30, , a, , b, , c, , d, , 13, , a, , b, , c, , d, , 31, , a, , b, , c, , d, , 14, , a, , b, , c, , d, , Or, , a, , b, , c, , d, , 15, , a, , b, , c, , d, , 32, , a, , b, , c, , d, , 16, , a, , b, , c, , d, , 33, , a, , b, , c, , d, , 17, , a, , b, , c, , d, , 34, , a, , b, , c, , d, , 18, , a, , b, , c, , d, , 35, , a, , b, , c, , d, , Or, , a, , b, , c, , d, , Students should not write anything below this line, , SIGNATURE OF EXAMINER WITH DATE, , MARKS SCORED

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164, , CBSE New Pattern ~ Chemistry XI (Term-I), , ANSWERS, Practice Paper 1, 1. (b), 11. (d), 21. (c), 31. (a, c), , 2. (c), 12. (d), 22. (a), , 3. (c), 13. (a), 23. (c), , 4. (c), 14. (c), 24. (c), , 5. (c), 15. (c), 25. (a), , 32. (a), , 33. (a), , 34. (a), , 35. (b, a), , 6. (d), 16. (a), 26. (b), , 7. (d), 17. (d), 27. (a), , 8. (d), 18. (a), 28. (b), , 9. (c), 19. (c), 29. (a), , 10. (a), 20. (b), 30. (c), , Practice Paper 2, 1. (d), 2. (d), 11. (a), 12. (c), 21. (a), 22. (c), 31. (d,a) 32. (b), , 3. (b), 13. (d), 23. (c), 33. (b), , 4. (a), 14. (b), 24. (b), 34. (c ), , 5. (a), 6. (a), 15. (d), 16. (c), 25. (b), 26. (a), 35. (d, b), , 7. (c), 17. (b), 27. (a), , 8. (c), 18. (c), 28. (b), , 9. (c), 19. (b), 29. (a), , 10. (a), 20. (a), 30. (d), , 2. (c), 12. (c), 22. (a), , 3. (d), 13. (b), 23. (a), , 4. (a), 14. (b), 24. (c), , 5. (a), 15. (a), 25. (c), , 7. (a), 17. (d), 27. (a), , 8. (c), 18. (a), 28. (a), , 9. (b), 19. (d), 29. (c), , 10. (c), 20. (c), , 31. (d,a) 32. (b), , 33. (d), , 34. (b), , 35. (b, a), , Practice Paper 3, 1. (b), 11. (b), 21. (b), , 6. (b), 16. (b), 26. (c), , 30. (b)