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CBSE Term II, , 2022, , Chemistry, Class XII

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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 :, , 978-93-25796-90-4, , 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 Term II, , 2022

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Syllabus, CBSE Term II Class 12, S.No., , Unit, , ., , Electrochemistry, , ., , Chemical Kinetics, , ., , Surface Chemistry, , ., , d-and f-Block Elements, , ., , Coordination Compounds, , ., , Aldehydes, Ketones and Carboxylic Acids, , Periods, , Marks, , Amines, Total, , ., , Electrochemistry : Redox reactions, EMF of a cell, standard electrode potential, Nernst equation, and its application to chemical cells, Relation between Gibbs energy change and EMF of a cell,, conductance in electrolytic solutions, specific and molar conductivity, variations of conductivity, with concentration, Kohlrausch s Law, electrolysis., , ., , Chemical Kinetics : Rate of a reaction Average and instantaneous , factors affecting rate of, reaction: concentration, temperature, catalyst; order and molecularity of a reaction, rate law and, specific rate constant, integrated rate equations and half-life only for zero and first order reactions ., , ., , Surface Chemistry : Adsorption - physisorption and chemisorption, factors affecting adsorption of, gases on solids, colloidal state: distinction between true solutions, colloids and suspension;, lyophilic, lyophobic, multi-molecular and macromolecular colloids; properties of colloids; Tyndall, effect, Brownian movement, electrophoresis, coagulation., , ., , d-and f-Block Elements : General introduction, electronic configuration, occurrence and, characteristics of transition metals, general trends in properties of the first row transition metals, metallic character, ionization enthalpy, oxidation states, ionic radii, colour, catalytic property,, magnetic properties, interstitial compounds, alloy formation. Lanthanoids - Electronic, configuration, oxidation states and lanthanoid contraction and its consequences., , ., , Coordination Compounds : Coordination compounds - Introduction, ligands, coordination, number, colour, magnetic properties and shapes, IUPAC nomenclature of mononuclear, coordination compounds. Bonding, Werner s theory, VBT, and CFT., , ., , Aldehydes, Ketones and Carboxylic Acids : Aldehydes and Ketones: Nomenclature, nature of, carbonyl group, methods of preparation, physical and chemical properties, mechanism of, nucleophilic addition, reactivity of alpha hydrogen in aldehydes, uses. Carboxylic Acids:, Nomenclature, acidic nature, methods of preparation, physical and chemical properties; uses., , ., , Amines : Nomenclature, classification, structure, methods of preparation, physical and chemical, properties, uses, identification of primary, secondary and tertiary amines., , CBSE Term II, , 2022

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CHAPTER 01, , Electrochemistry, In this Chapter..., l, , Cells, , l, , Electrode Potential, , l, , Electrochemical Series, , l, , Nernst Equation, , l, , Conductance of Electrolytic Solutions, , Electrochemistry is the study of production of electricity from, energy released during spontaneous chemical reactions and the, use of electrical energy to bring about non-spontaneous chemical, transformations., , Electrode Potential, l, , Cells, Cells are the devices that involves the interconversion of, electrical energy and chemical energy. These are of two types, (i) Electrochemical cell In this cell, chemical reactions are, carried out that result in production of electricity. It is also, called as voltaic or galvanic cell. e.g. Daniell cell., , l, , l, , (ii) Electrolytic cell In this cell, electricity is used to produce, non-spontaneous chemical changes. As there occur, breaking, down of a molecule with the help of electricity, the process is, called electrolysis., , E °cell = E °oxidation + E °reduction, = E °cathode - E °anode, = E °right - E °left, , Electrolytic cell and Electrochemical cell, Electrolytic cell, , Electrochemical cell, (Galvanic or voltaic cell), , Non-spontaneous, redox reaction is made to, occur., , Gibbs energy of the spontaneous, redox reaction is converted into, electrical work. e.g. Daniell cell, , Occur in same electrolytic, solution., , Occur in different electrolytic, solutions., , No salt-bridge is required., , Salt-bridge is essential to, maintain electroneutrality., , Cathode is negative and, anode is positive electrode., , Cathode is positive and anode is, negative electrode., , Potential difference between the metal and the, metal ion, in which electrode is dipped is called electrode, potential. At 1 atm pressure and 1M concentration of, electrolyte, electrode potential is also known as standard, electrode potential (E°)., According to the IUPAC convention, standard, reduction potential of cell is now called standard, electrode potential., The difference between electrode potentials (reduction, potential) of the two half-cells of a cell is known as cell, potential., , l, , When switch is in on position, the flow of electrons is from, negative electrode to the positive electrode and the, direction of current flow is opposite to that of electron, flow., Electromotive force (emf) of a cell is the potential, difference, when no current is drawn through the cell., It is measured in volts (V)., It depends upon the following factors, (i) nature of metals and its ions., (ii) concentrations of electrolysis used., (iii) temperature.

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02, , CBSE Term II Chemistry XII, , Standard Hydrogen Electrode (SHE), The construction of the SHE is shown in the figure below, H2 (g) at, 1 bar, , 1.00 M H+, Finely divided platinum, coated on platinum foil, , Standard Hydrogen Electrode (SHE), , The standard hydrogen electrode consists of a platinum, electrode coated with platinum black. The electrode is, dipped in an acidic solution and pure hydrogen gas is, bubbled through it., At 298K, the emf is measured by combining the SHE of the, anode and another half-cell of the cathode, resulting in the, standard reduction potential of the other half-cell (cathode)., ° = E° - E°, EMF of the cell = Ecell, R, L, , As E L° for SHE is zero, E ° = E R° - 0 = E R°, , The arrangement of various standard half cells in the order of, their decreasing values of standard reduction potential is, called electrochemical series., Standard electrode potentials at 298K, (Electrochemical series), Reaction (oxidised form + ne - ¾® reduced form), E- / V, F2 ( g ) + 2e -, , ¾® 2F-, , 2.87, , ¾® Co2 +, , 1.81, , H2O 2 + 2H+ + 2e -, , ¾® 2H2O, , 1.78, , MnO -4 + 8H+ + 5 e -, , ¾® Mn 2 + + 4H2O, , 1.51, , ¾® Au ( s ), , 1.40, , ¾® 2Cl -, , 1.36, , Co, , 3+, , Au, , +e, , 3+, , -, , + 3e, , -, , Cl 2 (g ) + 2e Cr2O 27 -, , +, , + 14H + 6 e, , -, , O 2 ( g ) + 4H+ + 4 e +, , ¾® 2Cr, , MnO 2 ( s ) + 4H + 2e ¾® Mn, NO -3 +, 2+, , +, , 4H + 3 e, , -, , + 2e -, , 2Hg, +, , -, , + 7H2O, , ¾® 2H2O, -, , Br2 + 2e –, , 3+, , 2+, , 1.33, 1.23, , + 2H2O, , 1.23, , ¾® 2Br -, , 1.09, , ¾® NO ( g ) + 2H2O, , 0.97, , ¾® Hg22 +, , 0.92, , ¾® Ag(s ), , 0.80, , Fe3 + + e -, , ¾® Fe2 +, , 0.77, , O 2 ( g ) + 2H+ + 2e -, , ¾® H2O 2, , 0.68, , Ag + e, , I 2 + 2e, , -, , ¾® 2l, , Cu + + e Cu, , 2+, , + 2e, , -, , AgCl ( s ) + e AgBr (s ) + e, 2H+ + 2e -, , -, , -, , 0.54, , ¾® Cu(s ), , 0.52, , ¾® Cu ( s ), , 0.34, , ¾® Ag( s ) + Cl ¾® Ag ( s ) + Br, , 0.22, -, , 0.10, , ¾® H2 ( g ), , 0.00, , -, , ¾® Pb ( s ), , -0.13, , Sn 2 + + 2e -, , ¾® Sn ( s ), , -0.14, , Pb, , 2+, , -, , ¾® Ni ( s ), , -0.25, , Fe2 + + 2e -, , ¾® Fe( s ), , -0.44, , Cr 3 + + 3e -, , ¾® Cr (s ), , -0.74, , Zn 2 + + 2e -, , ¾® Zn ( s ), , Ni, , 2+, , + 2e, + 2e, , 2H2O + 2e, , -, , Al 3 + + 3 e 2+, , -, , -0.76, -, , ¾® H2 (g) + 2OH (aq), , -0.83, , ¾® Al ( s ), , -1.66, , ¾® Mg( s ), , -2.36, , Na + + e -, , ¾® Na ( s ), , -2.71, , Ca 2 + + 2e –, , ¾® Ca (s ), , -2.87, , ¾® K ( s ), , -2.93, , ¾® Li ( s ), , -3.05, , Mg, , +, , + 2e, , K +e, , –, , Li + + e -, , Increasing strength of reducing agent, , The EMF of the cell can be represented as, E cell = E right - E left, For cell reaction;, Cu( s ) + 2Ag + ( aq) ¾® Cu 2+ ( aq) + 2Ag( s ), E cell = E Ag + /Ag - E Cu 2 + /Cu, , Electrochemical Series, , Increasing strength of oxidising agent, , While representing the galvanic cell, the anode is written on, the left and the cathode on the right. A vertical line separates, the metal from the metal ion (electrolyte solution) and a, double vertical line indicates a salt bridge connecting the two, half-cells (two electrolytes)., The state of the metal and the solution is written with the, concentration of the electrolyte in parenthesis., M1 ( s )| M1n + ( aq)|| M 2n + ( aq)|M 2 ( s ), Let us consider an example to understand the concept., Step I Write the chemical reaction, Cell reaction, Cu( s ) + 2Ag + ( aq) ¾® Cu 2+ ( aq) + 2Ag( s ), Step II Split the reaction into two half-cell reactions., Half-cell reactions, At cathode (reduction), 2Ag + ( aq) + 2e - ¾® 2Ag( s ), At anode (oxidation), Cu( s ) ¾® Cu 2+ ( aq) + 2e Step III Represent the complete reaction according to the, conversion discussed above., Cu( s )|Cu 2+ ( aq)|| Ag + ( aq)| Ag( s )., , Some metals like platinum or gold are used as inert, electrodes. They do not participate in the reaction but, provide their surface for oxidation or reduction and for the, conduction of electrons., , ¾, ¾¾¾¾¾¾¾¾¾¾¾¾¾¾¾¾¾¾¾¾¾¾¾¾¾¾¾¾¾¾¾¾¾¾¾®, , Cell Representation

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03, , CBSE Term II Chemistry XII, , Note, , ˜, , ˜, , Negative E° means that the redox couple is stronger reducing, agent than H + / H 2, e.g. Zn ( -0.76)., Positive E° means that the redox couple is a weaker reducing, agent than H + / H 2, e.g. Ag ( + 0.80)., , From top to bottom in the table, the standard electrode, potential decreases and with this, decreases the oxidising, power of the species on the left and increases the reducing, power of the species on the right hand side of the reaction., Electrochemical series also helps to predict the feasibility of a, redox reaction, thermal stability of metal oxides and, spontaneity of an electrochemical cell by choosing the, half-cells with appropriate E° values., If the standard electrode potential of an electrode is greater, than zero, then its reduced form is more stable as compared, to hydrogen gas and vice-versa., The standard electrode potential for flourine is highest which, indicates that fluorine gas (F2 ) has maximum tendency to get, reduced to fluoride ion (F - ). In case of lithium, lithium ion is, weakest oxidising agent while lithium metal is the most, powerful reducing agent in an aqueous solution., , Nernst Equation, l, , For a general electrochemical reaction of the type, -, , aA + bB ¾ne, ¾® cC + dD, 2.303 RT, [ C ] c [ D] d, log, nF, [ A] a [ B] b, , where,, , l, , 2, , 2, , 1, [H + ], , = 0 - 0.0591 ( - log[ H + ]) = - 0.0591 pH, [Q pH = - log[H + ]], , Conductance of Electrolytic Solutions, (i) The electrical resistance ( R) of any object is directly, proportional to its length l and inversely proportional to, its area of cross-section A., l, l, Thus,, R µ or R = r ., A, A, where, r is resistivity (or specific resistance) of the object., The SI unit of resistance is ohm (W) and it can be, measured with the help of a Wheatstone bridge., (ii) Resistivity (specific resistance) is defined as the, resistance offered by a substance of 1 m length with area, of cross-section to 1 m 2 ., Its SI unit is ohm metre (W-m)., l, The quantity is called the cell constant., A, It will be given by,, G* =, , l, =R´ k, A, , The unit of cell constant (G* ) is m - 1 ., , At 298 K,, °, E cell = E cell, -, , EH + / H = E °H + /H - 0.0591 log, , It is denoted by the symbol G* ., , Nerst equation is written as, °, E cell = E cell, , Relation between E cell and pH, , c, , d, , 0.0591, [ C ] [ D], log, n, [ A] a [ B] b, , E = electrode potential, E° = standard electrode potential, R = gas constant, T = temperature, F = Faraday constant, n = number of electrons involved in reaction, Nernst equation is used to find equilibrium constant and, Gibb’s free energy change by using the formula’s, 0.0591, [QAt equilibrium , Ecell = 0], log KC, n, where, K = equilibrium constant, , °, (i) Ecell, =, , °, (ii) DGr° = - nFEcell, , where, DG°= standard Gibbs free energy change, (iii) DrG ° = - 2.303 RT log KC, If two half reactions having potentials E1° , E2° are combined, to give a third half reaction having a potential E3° ,, then, DG3° = DG1° + DG2° or E3° =, , n1 E1° + n2 E2°, n3, , (iii) Conductance (G) is the reciprocal of resistance and, defined as the ease with which the electric current flows, 1 A, through the conductor, G = = ., R rl, Here, R = Resistance, SI unit of G = Siemen (S) or ohm -1, (iv) Conductivity or specific conductance (k) of an, electrolytic solution may be defined as the conductance, of a solution of 1 m length with its area of cross-section, equal to 1 m 2 ., k=, , 1, ´ cell constant, R, , Its unit is Sm -1 or ohm -1 m -1 ., , (v) Molar conductivity (L m ) is defined as the conducting, power of all the ions produced by dissolving one mole of, an electrolyte placed between two electrodes with area, of cross-section ‘A’ and at a distance of 1 cm., k, It is given by, L m = ,, C, Here, k is expressed in Sm -1 and C in mol m -3 thus,, unit of L m is Sm 2 mol -1 or ohm -1 m 2 mol -1 .

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04, , CBSE Term II Chemistry XII, , Lm =, , or, , k ´ 1000, M, , (ii) Dissociation constant can be determined as, Ca2, (for weak electrolyte), Ka =, (1 - a ), , where, M = molarity of solution, If k is S cm -1 , then unit is W -1 cm 2 mol -1 or S cm 2 mol -1 ., , The molar conductivity can also be written as,, Lm =, , kA, =k, l, , Since, l = 1 and A = V (volume containing 1 mole of, electrolyte), , \, , Lm = kV, , Variation of Conductivity and Molar, Conductivity with Concentration, l, , l, , l, , Conductivity always decreases with decrease in, concentration both for weak and strong electrolyte. This is, due to the fact that the number of ions that carry current in a, unit volume of solution always decreases with decrease in, concentration., Molar conductivity increases with decrease in concentration, because the total volume V of solution containing 1 mole of, electrolyte increases. When the concentration approaches, zero, the molar conductivity reaches a limiting value, known, as limiting molar conductivity and is denoted by L°m ., , l, , l, , Products of Electrolysis, Products of electrolysis depend upon the nature of material, being electrolysed, the type of electrodes being used,, different oxidising and reducing species present in the, electrolytic cell and their standard electrode potentials,, over voltage (kinetic barrier)., Some important examples of electrolysis are as follows :, l, , L°m, , Electrolysis of molten sodium chloride, At cathode Na + ( aq) + e - ¾® Na( s ), At anode, , l, , Electrolysis of aqueous sodium chloride, 1, At cathode H 2O ( l ) + e - ¾® H 2 ( g ) + OH - ( aq), 2, 1, At anode, Cl ( aq) ¾® Cl 2 ( g ) + e 2, 1, 1, Na + ( aq) + OH - ( aq) + H 2 ( g ) + Cl 2 ( g ), 2, 2, , -A C, , The value of constant ‘A’ depends upon the type of, electrolyte, i.e. the charges on the cation and anion produced, on the dissociation of electrolyte in the solution. Therefore,, NaCl, CaCl 2 , MgSO 4 are known as 1 : 1, 2 : 1 and 2 : 2, electrolytes, respectively., In case of weak electrolytes, like acetic acid, a lower value of, degree of dissociation is observed at higher concentration and, hence, there is a rapid increase in the value of L°m with dilution., The limiting molar conductivity can be determined by using, Kohlrausch law of independent migration of ions., Kohlrausch’s law states that limiting molar conductivity of an, electrolyte is the sum of individual contributions of, conductivity of cation and anion of the electrolyte., L°m = x l°m + y l°m, (x and y = number of cations and anions), Some important applications of Kohlrausch law are as follows, (i) Degree of dissociation of weak electrolyte can be, L, determined as, a = m, L°m, [L m = molar conductivity at a given concentration,, L°m = limiting molar conductivity]., , 2Cl - ( aq) ¾® Cl 2 ( g ) + 2e -, , Net reaction NaCl( aq) + H 2 O( l ) ¾®, , For strong electrolytes, L m increases slowly with dilution., This can be represented by Debye-Huckel Onsager equation., Lm =, , l, , [C = concentration, a = degree of dissociation], , Note At cathode, H 2 is produced due to higher value of E ° and at, anode due to oxygen over potential, liberation of chlorine gas is, preferred., l, , Electrolysis of sulphuric acid, At anode, °, 2H 2 O ( l ) ¾® O 2( g ) + 4H+ ( aq ) + 4 e - ; Ecell, = 1.23 V …(i), , 2SO24 - ( aq ) ¾® S2 O28 - ( aq ) + 2e - ;, , °, Ecell, = 1.96 V …(ii), , For dilute sulphuric acid, reaction (i) is preferred but at, higher concentration of H 2 SO 4 , reaction (ii) is, preferred., 1, ° = 0.00 V, At cathode H + + e - ¾® H 2 ( g ); E cell, 2, l, , Electrolysis of brine using, Hg as a cathode,, At cathode, , Na + + e - ¾® Na, Na + Hg ¾® Na ¾ Hg, , 2Na ¾ Hg + 2H 2O ¾® 2NaOH + Hg + H 2 ( g ), 1, At anode Cl - ( aq) ¾® Cl 2 ( g ) + e - , E°cell = 1 . 36 V, 2, 2H 2O( l ) ¾® O 2 ( g ) + 4H + ( aq) + 4 e - ; E°cell = 1 . 23 V`

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Solved, Examples, Example 1. (i) Formulate the electrochemical cell, representing the reaction, 2Cr( s ) + 3Fe 2+ ( aq) ¾® 2Cr 3+ ( aq) + 3Fe ( s ), ° ., (ii) Calculate E cell, [Given, E°Cr 3+ /Cr = -0.74 V, E°Fe2+ / Fe = -0.44 V]., Sol., , (i) Representation of the given cell,, Cr( s ) | Cr 3+ ( aq ) || Fe2 + ( aq ) | Fe( s ), (ii) Ecell, ° = Eright, ° - Eleft, °, As, ECr, ° 3 + / Cr = -0.74 V ; E°Fe2 + / Fe = -0.44 V, = - 0.44 - (- 0.74), = - 0.44 + 0.74 = 0.30 V, , Example 2. Calculate the potential for half-cell, 3+, containing 0.10 M Cr 2O 27 ( aq), 0.20 M Cr ( aq) and, 1.0 ´ 10 -4 M H + (aq)., The half-cell reaction is, +, 3+, Cr 2O 2–, 7 ( aq) +14H ( aq) + 6e ¾® 2Cr ( aq) + 7H 2O( l ), , and the standard electrode potential is given as E° =1.33 V., , Sol. Zn ( s )+Cu 2 + ( aq ) ¾¾® Zn 2 + ( aq ) + Cu( s ), E° = + 1.1 V , DG° = ?, 1F = 96500 C mol -1 , n = 2, DG° = - nFE° = - 2 ´ 96500 ´1.1, = -212300 J mol -1, , Example 4. Calculate emf of the following cell at 25 °C., Fe|Fe 2+ (0.001M)|| H + (0.01M)|H 2 ( g ) (1bar)|Pt( s ), E(Fe, ° 2+ /Fe) = - 0.44 V; E°(H+ / H ) = 0.00V, 2, , Sol. For the given cell representation, the cell reaction will be, Fe ( s ) + 2H+ ( aq ) ¾® Fe 2 + ( aq ) + H 2 ( g ), The standard emf of the cell will be,, Ecell, ° = E°H + /H - E°Fe 2 + /Fe , E°cell = 0 - ( - 0.44) = 0.44 V, 2, , The Nernst equation for the cell reaction at 25°C,, 0.0591, [Fe 2 + ], Ecell = Ecell, ° log + 2, n, [H ], 0.0591, [0.001], = 0.44 log, 2, [0.01] 2, = 0.44 - 0.02955 (log10), = 0.44 - 0.02955 (1 ), = 0.41045 V = 0.41 V, , Sol. Given, [ Cr2O72- ] = 0.10 M, [Cr 3+ ] = 0.20 M and, ° = 1.33 V, n = 6, [ H+ ] = 1 ´ 10-4 M, Ecell, ° Ecell = Ecell, , 0.0591 V, [Cr 3+ ] 2, log, n, [Cr2O 27 – ] [H+ ]14, , = 1.33 V -, , 0.0591 V, (0.2) 2, log, 6, (0.1) (1.0 ´ 10-4 )14, , 0.0591 V, 4 ´10-2, log, 6, 1.0 ´ 10-57, 0.0591 V, = 1.33 V log 4 ´ 1055, 6, 0.0591 V, = 1.33 V ´ 55.6021, 6, = 1.33 V -, , = 1.33 V - 0.5477 V, = 0.7823 V, , Example 3. The standard electrode potential (E°) for, Daniell cell is +1.1 V. Calculate the D G° for the reaction., Zn( s ) + Cu 2 + ( aq) ¾¾® Zn 2+ ( aq) + Cu( s ), (1 F = 96500 C mol -1 ), , (Delhi 2015), , [Q log10 = 1], , Example 5. Calculate the cell emf and D r G ° for the cell, , reaction at 25° C., Zn( s ) | Zn 2+ (0.1M)|| Cd 2+ (0.01M) |Cd(s ), [Given, E°Zn 2+ / Zn = - 0.763 V, E°Cd 2+ /Cd = - 0.403 V, 1 F = 96500 C mol -1 , R = 8.314 JK -1 mol -1 ], , ° = E°, Find E cell, cathode - E° anode then, D r G° by using, ° ., formula, D r G ° = - nFE cell, , Sol. E°cell = E°right - E °left = - 0.403 - (- 0.763) = 0.36 V, 0.059, [Zn 2+ ( aq )], log, n, [ Cd 2+ ( aq )], 0.059, 0.1, = 0.36 log, [ n = 2], 2, 0.01, , Ecell = E°cell Ecell, , Ecell = 0.36 - 0.0295 log 10, = 0.36 - 0.0295 ´ 1= 0.3305 V, D r G° = - nFEcell, ° = - 2 ´ 96500 ´ 0.36, = - 69480 J mol -1 = -69.48 kJ mol -1

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06, , CBSE Term II Chemistry XII, , Example 6. Conductivity of two electrolyte solutions A and Example 8. The electrical resistance of a column of, , B each having a concentration of 0.1 M are 8.5 ´ 10 -2 S cm -1, and 4.1 ´ 10 -4 S cm -1 respectively. Which of the two offers, less resistance to the flow of current ?, 1, R, Hence, B will offer greater resistance., , Sol. k µ, , 0.05 M NaOH solution of diameter 1 cm and length 50 cm, is 5 .55 ´ 10 3 W. Calculate its resistivity, conductivity and, molar conductivity., (i) Calculate area from radius by using, A = pr 2, (ii) Calculate resistivity from the formula, r =, , Example 7. Conductivity of 0.00241 M acetic acid, , RA, l, , solution is 7.896 ´ 10 -5 S cm -1 ., , (iii) Calculate conductivity ( k ) from resistivity as k =1/ r, , Calculate its molar conductivity in this solution. If L°m for, acetic acid be 390.5 S cm 2 mol -1 , what would be its, dissociation constant?, (NCERT), , (iv) Calculate molar conductivity, L m by using the formula, , 2, , (ii) Then, find degree of dissociation (a) and, dissociation constant (K a ) by using formula,, LCm, Ca 2, and Ka =, , respectively., L°m, 1-a, , Sol. Given, k = 7.896 ´ 10 -5 S cm -1 ,, L°m (CH COOH) = 390.5 S cm 2 mol -1, , = 0.01148 W -1cm -1, , 3, , = 1.148 ´ 10–2 S cm –1, Molar conductivity,, k ´ 1000, Lm =, M, 1.148 ´ 10-2 ´ 1000, =, = 229.6 S cm 2 mol -1, 0.05, , and Molarity (M) = 0.00241 M, k ´ 1000, Molar conductivity, LCm =, Molarity, =, , 7.896 ´ 10 -5 S cm -1 ´ 1000, 0.00241 M (mol cm -3 ), , LCm = 32.76 S cm 2 mol -1, Degree of dissociation, a =, , LCm, L°m, , =, , Example 9. Calculate L°m for acetic acid,, , 32.76, = 8.4 ´ 10 -2, 390.5, , Given that, L°m, , CH3 COOH w, , CH3 COO - + H+, , C, C - Ca, , After time t,, , 0, , 0, , Ca, , Ca, , Dissociation constant,, [CH3 COO -] [H+ ] Ca 2, Ka =, =, [CH3 COOH], 1-a, Ka =, , -2 2, , 0.00241 (8.4 ´ 10 ), = 1.86 ´ 10 -5, (1 - 0.084), , (HCl), , = 426 S cm 2 mol -1, , L°m (NaCl) = 126 S cm 2 mol -1, , For dissociation constant of acetic acid, CH3 COOH, Initial conc., , k ´ 1000, M, , æ 1ö, Sol. Area, A = pr 2 = 3.14 ´ ç ÷ cm 2 = 0.785 cm 2, è 2ø, RA, Resistivity, r =, l, 5.55 ´ 103W ´ 0.785 cm 2, =, 50 cm, = 87.135 W cm, 1, 1, Conductivity, k = =, W -1 cm -1, r 87.135, , (i) First, find molar conductivity using the formula,, k ´ 1000, ., LCm =, C, , a=, , Lm =, , L°m, Sol. L°m, , (CH 3COONa), , (CH3 COOH), , = 91 S cm 2 mol -1, , = L°H+ + L°CH, , –, 3 COO, , = L°H+ + L°Cl – + L°CH, , –, 3 COO, , + L°Na+, , - L°Cl – - L°Na+ = L°m (HCl) + L°m (CH3 COONa) - L°m, 2, , = (426 + 91 - 126) S cm mol, = 391 S cm 2 mol -1, , -1, , (NaCl )

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Chapter, Practice, PART 1, Objective Questions, l, , If half-cell potentials are :, Zn 2+ ( aq) + 2e - ¾® Zn( s ); E° = + 0.76 V, , Multiple Choice Questions, , 1. The electrode potential of a half-cell depends upon, (a) nature of metal, (b) concentration of metal ions in solution, (c) temperature of the solution, (d) All of the above, , (b) 0.42 V, (d) 1. 34 V, , tubes containing A, B, C solutions. XB react with A, and C. XA does not react with any of these. XC, reacts with A., Arrange the anion in the decreasing order of their, oxidation., , Ag | AgCl| KCl || AgNO 3 | Ag + , the overall cell, reaction is, (a) Ag+ + KCl ¾® AgCl(s ) + K+, 1, (b) Ag + AgCl ¾® 2Ag + Cl 2, 2, (c) AgCl(s ) ¾® Ag+ + Cl -, , (a), (b), (c), (d), , (d) Ag+ + Cl – ¾® AgCl(s ), °, 3. For the cell, Cu |Cu 2+||Ag +|Ag, Ecell, = + 0.46 V, , If concentration of Cu, will be, , (a) 1.1 V, (c) 0.84 V, , 6. Compounds XA, XB, XC are added into separate test, , 2. For the electrochemical cell,, , 2+, , Ag 2O ( s ) + H 2O( l ) + 2e - ¾® 2Ag ( s ) + 2OH - ;, E° = 0.34V., The cell potential will be, , ions is doubled, then, , °, E cell, , A- > B- > C B- > C - > AC - > A- > BB- > A- > C -, , 7. Using the data given below find out the strongest, reducing agent., E sCr2O27- /Cr 3+ = 1.33V; E sCl 2 /Cl - = 1.36V, E s MnO-4 / Mn 2+ = 1.51V; E sCr 3+ /Cr = - 0.74V, , (a) halved, (b) doubled, (c) four times, (d) remains the same, , (a) Cl (c) Cr 3+, , (b) Cr, (d) Mn 2 +, , 8. In the given reaction,, , 4. In the electrolysis of aqueous sodium chloride, solution, which of the half cell reaction will occur at, anode?, (a) Na + ( aq ) + e - ¾® Na ( s ); Escell = - 2.71 V, +, , -, , (b) 2H2O ( l ) ¾® O 2 (g) + 4H ( aq ) + 4e ; Ecell = 1.23 V, 1, ° = 0.00 V, (c) H+ ( aq ) + e- ¾® H2 ( g ); Ecell, 2, 1, (d) Cl - ( aq ) ¾® Cl 2 ( g ) + e - ; Escell = 1.36 V, 2, s, , 5. A button cell used in watches, functions as following :, Zn( s ) + Ag 2O ( s ) + H 2O( l ) 3, 2Ag( s ) + Zn 2+ ( aq) + 2OH - ( aq), , 2Cu + ( aq) s, , Cu 2+ ( aq) + Cu ( s ), , °, °, ECu, = 0.6 V and ECu, = 0.41 V, +, 2+, /Cu, /Cu +, , The equilibrium constant for this reaction will be, (a) 2.76 ´ 102, (b) 2.76 ´ 104, (c) 2.76 ´ 106, (d) 2.76 ´ 108, , 9. Electrode potential for Mg electrode varies, according to the equation, 0.059, 1, °, E Mg 2+ / Mg = E Mg, log, ., 2+, / Mg, 2, [Mg 2+ ]

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8, , CBSE Term II Chemistry XII, , (d), log[Mg 2+]®, , log[Mg 2+]®, , 10. Which of the following statement is correct?, (a) Ecell and D r G of cell reaction both are extensive properties, (b) Ecell and D r G of cell reaction both are intensive properties, (c) Ecell is an intensive property while D r G of cell reaction is, an extensive property, (d) Ecell is an extensive property while D r G of cell reaction is, an intensive property, , (a) 4.5V, , (b) 3.0V, , (c) 2.5V, , (d) 5. 0V, , Here, the limiting molar conductivity is, (a) 148.6, (c) 87.46, , solution is 0.002765 mho cm -1 , if the resistance of, cell with solution is 400 W then the cell constant (G ), will be, (a) 0.533 cm -1, (c) 2.212 cm -1, , (NCERT Exemplar), , (a) Conductivity of solution depends upon size of ions., (b) Conductivity depends upon viscosity of solution., (c) Conductivity does not depend upon solvation of ions, present in solution., (d) Conductivity of solution increases with temperature., , 14. The graph of molar conductivity ( L m ) versus root of, , concentration C 1/ 2 is plotted. Which type of electrolyte, are used in A and B ?, Lm /(S cm2 mol–1), , 400, , (A), , (b) 1.106 cm -1, (d) None of these, , 17. L°m (NH 4OH) is equal to ......... ., , (NCERT Exemplar), , (a) L°m (NH4 OH) + L°m (NH4 Cl) - L°m (HCl ), (b) L°m (NH4 Cl) + L°m (NaOH) - L°m (NaCl ), (c) L°m (NH4 Cl) + L°m (NaCl ) - L° m (NaOH), , (b) DG° > 0; K eq > 1, (d) DG° < 0; K eq < 1, , electrolytes is not correct?, , (b) 150, (d) 147, , 16. At, 25°C the specific conductivity of N/50 KCl, , then which of the following gives the correct, relationships for the values of DG° and K eq ?, , 13. Which of the following statements about solutions of, , 0.005 0.010 0.015 0.020, ÖC, , °, 12. If the Ecell, for a given reaction has a negative value,, , (a) DG°> 0; K eq < 1, (c) DG° < 0; K eq > 1, , 149.8, 149.4, 149.0, 148.6, 148.2, 147.8, 147.4, 147.0, O, , 11. The Gibbs energy for the decomposition of Al 2O 3 at, 500°C is as follows, 2, 4, Al 2O 3 ¾® Al + O 2 ; D r G = + 960 kJ mol -1, 3, 3, The potential difference needed for the electrolytic, reduction of aluminium oxide (Al 2O 3 ) at 500°C is, atleast, , B, Strong electrolyte, Weak electrolyte, Weak electrolyte, Strong electrolyte, , 15. Consider the following graph,, , EMg 2+/Mg, , (c), , log[Mg 2+]®, , EMg 2+/Mg, , log[Mg 2+]®, , A, Weak electrolyte, Strong electrolyte, Weak electrolyte, Strong electrolyte, , Lm /(S cm2 mol–1), , (b), , (a), (b), (c), (d), , EMg 2+/Mg, , (a), , EMg 2+/Mg, , The graph of E Mg 2+ / Mg vs log [Mg 2+ ] is, , (d) L°m (NaOH) + L°m (NaCl) - L°m (NH4 Cl ), l, , Assertion-Reasoning MCQs, Direction (Q. Nos. 18-22) Each of these questions, contains two statements Assertion (A) and Reason (R)., Each of these questions also has four alternative choices,, any one of which is the correct answer. You have to select, one of the codes (a), (b), (c) and (d) given below., (a) Both A and R are true and R is the correct, explanation of A., (b) Both A and R are true, but R is not the correct, explanation of A., (c) A is true, but R is false., (d) A is false, but R is true., , 18. Assertion Current stops flowing when Ecell = 0., , 200, , Reason Equilibrium of the cell reaction is attained., , (B), , (NCERT Exemplar), , 19. Assertion Copper sulphate can be stored in zinc, 0, , 0, , 0.2, , 0.4, , C1/2/(mol/L)1/2, , vessel., Reason Zinc is less reactive than copper., (NCERT Exemplar)

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9, , CBSE Term II Chemistry XII, , 20. Assertion EAg + / Ag increases with increase in, +, , concentration of Ag ions., Reason E Ag + / Ag has a positive value., , (NCERT Exemplar), , 21. Assertion Electrolysis of NaCl solution gives chlorine, at anode instead of O 2 ., Reason Formation of oxygen at anode requires over, voltage., (NCERT Exemplar), , 22. Assertion pH of brine solution increases during, electrolysis., Reason H 2 is liberated at cathode and Cl 2 is at anode, during electrolysis of NaCl solution., l, , Case Based MCQs, , 23. Read the following and answer the questions from (i), to (iv) given below, Molar conductivity of a solution is the conductance of, solution containing one mole of electrolyte, kept, between two electrodes having unit length between, them and large cross-sectional area, so as to contain the, electrolyte. In other words, molar conductivity is the, conductance of the electrolytic solution kept between, the electrodes of a conductivity cell at unit distance, but having area of cross-section large enough to, accommodate sufficient volume of solution that, contains one mole of the electrolyte. It is denoted by, Lm ., The molar conductivity is related to conductivity as, 1000, 1000, Lm = k ´V =, ´ k= k´, C, Molarity, Unity of L m (molar conductivity) shall be, ohm -1 cm -1 mol -1 or Scm 2 mol -1 ., Thus, knowing molar concentration (C ) and, conductivity ( k), L m can be calculated. L°m is called, molar conductivity at infinite dilution. The molar, conductivity of strong electrolytes is found to vary, with concentration according to the equation,, LCm = L°m - A C, This equation is called Debye-Huckel Onsager, equation., Here, A is constant depending upon the type of, electrolyte taken and nature of solvent and, temperature., (i) If conductivity of 0.00241 M acetic acid is, 7.896 ´ 10 -5 S cm -1 , the molar conductivity of, the solution shall be (in S cm -1 mol -1 ), (a) 3.276, (c) 32.76, , (b) 0.3276, (d) 327.6, , (ii) Molar conductivity of a solution is, 1.26 ´ 10 2 W -1cm 2mol -1 . Its molarity is 0.01., Its specific conductivity will be, (a) 1. 26 ´ 10-5, , (b) 1.26 ´ 10-3, , -4, , (d) 12.6 ´ 10-3, , (c) 1. 26 ´ 10, , (iii) The increase in molar conductivity of HCl with, dilution is due to, (a) increase in self-ionisation of water, (b) hydrolysis of HCl, (c) decrease in self-ionisation of water, (d) decrease in interionic forces, , (iv) Which of the following is wrong about molar, conductivity ?, (a) The solution contains Avogadro number of, molecules of the electrolyte, (b) It is the product of specific conductivity and, volume of solution in cc containing 1 mole of, electrolyte, (c) Its units are ohm -1 cm 2 mol -1, (d) Its value for 1 M NaCl is same as that of 1 M, glucose solution, , Or, , Highest molar conductivity will be exhibited by, (a) 0.005 M NaCl, (c) 0.5 M NaCl, , (b) 0.1 M NaCl, (d) 0.01 M NaCl, , 24. Read the following and answer the questions from, (i) to (iv) given below, The potential difference between two half-cell is, called the cell-potential. The calculation of cell, potential for emf requires only the addition of the, emf values for each half reaction, while the same, cell potential calculation using standard potentials, requires the usage of the following convention:, °, °, °, = E cathode, - E anode, E cell, Each half-cell reaction has a specific standard, potential reported as the potential of the reduction, reaction vs the normal hydrogen electrode (NHE)., In an electrochemical cell, there is a half-cell, corresponding working electrode (WE), where the, reactions under study take place, and a reference, half-cell. Experimentally, the cell potential is, measured as the difference between the potentials, of the WE half-cell and the reference, electrode/reference half-cell. The archetypal, reference electrode is the NHE, also known as the, standard hydrogen electrode (SHE) and is defined,, by convention, as 0 V for any temperature. The, arrangement of various standard half-cells in the, order of their decreasing values of standard, reduction potentials is called electrochemical, series.

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10, , CBSE Term II Chemistry XII, , (i) On which of the following, emf of a cell does, not depend?, (a) Temperature of electrolyte, (b) Nature of electrolyte, (c) Concentration of electrolyte in two half-cells, (d) Size of the plates, , (ii) Electrical conductance increases with ........, temperature., (a) increasing, (c) constant, , (b) decreasing, (d) nullifies, , (iii) Which of the following is correct about a, salt-bridge?, (a) It maintains the resistivity of both half-cells, (b) It maintains the electrical neutrality between, solutions of both the half-cells, (c) It maintains the electricity flow between both, half-cells, (d) None of the above, , (iv) The reduction of 1 mole of Zn 2 + to given 1 mole, of Zn would need how many coulombs?, (a) 193000 C, (c) 56900 C, , (b) 96500 C, (d) 487500 C, , Or, The equilibrium can be achieved in chemical, reaction at what value of emf?, (a) + 1, (c) 0.5, , (b) - 1, (d) Zero, , PART 2, Subjective Questions, l, , Short Answer Type Questions, , 1. Define electrochemical cell. What happens if, external potential applied becomes greater than, °, of electrochemical cell?, E cell, , 2. For a cell, Ag ( s )| AgNO 3 (0.01M)|| AgNO 3 (1.0M)| Ag( s), (i) Calculate the emf of the cell at 25°C., (ii) Write the net cell reaction., (iii) Will the cell generate emf, when two, concentrations become equal?, , 3. Write the cell reaction and Nernst equation for the, cell reaction in the Daniell cell. How will the E cell, be affected when concentration of Zn 2+ ions is, increased?, , 4. Calculate the potential of hydrogen electrode, which is in contact with a solution whose pH is 10., , 5. Write the Nernst equation and emf of the following cell, at 298 K., (i) Mg ( s ) | Mg 2+ (0.001 M) | | Cu 2+ (0.0001 M)|Cu ( s ), (ii) Fe ( s ) | Fe 2+ (0.001 M)| | H+ (1 M)| H2 ( g ), (1 bar) | Pt ( s ), , Given that,, E°Mg 2+ / Mg = - 2.36 V,, E°Cu2+ / Cu = 0.34 V, E°Fe2+ / Fe = -0.44 V, , 6. Use the data to answer the following and also justify, giving reason, Cr, , Mn, , Fe, , Co, , °, EM, 2+, /M, , - 0.91, , - 1.18, , - 0.44, , - 0.28, , °, EM, 3+, / M 2+, , - 0.41, , + 1.57, , + 0.77, , + 1.97, , (i) Which is a stronger reducing agent in aqueous, medium, Cr 2+ or Fe 2+ and why?, (ii) Which is the most stable ion in +2 oxidation and why?, , 7. Calculate the emf of the cell in which the following, reaction takes place., Ni ( s ) + 2Ag + (0.002 M) ¾® Ni 2+ (0.160 M) + 2Ag ( s ), Given that, E°cell = 1.05 V, , 8. (i) The cell in which the following reaction occurs,, 2Fe 3+ ( aq) + 2I - ( aq) ¾® 2Fe 2+ ( aq) + I 2( s ), has E°cell = 0.236 V at 298 K. Calculate the standard, Gibbs energy of the cell reaction., (Given, 1 F = 96500 C mol -1 ), (ii) How many electrons flow through a metallic wire, if, a current of 0.5 A is passed for 2 hours ?, (Given, 1 F = 96500 C mol -1 ), [All India 2017], , 9. Calculate D r G° and log K C for the following reaction, Cd 2+ ( aq) + Zn( s ) ¾® Zn 2+ ( aq) + Cd ( s ), °, Given, E Cd, = - 0.403 V, 2+, / Cd, °, E Zn, = - 0.763 V, 2+, / Zn, , 10. The resistance of a conductivity cell, when filled with, 0.05 M solution of an electrolyte x, is 100 W at 40°C., The same conductivity cell when filled with 0.01 M, solution of electrolyte y, has a resistance of 50 W., The conductivity of 0.05 M solution of electrolyte x is, 1.0 ´ 10 -4 S cm -1 ., Calculate, (i) cell constant., (ii) conductivity of 0.01 M y solution., (iii) molar conductivity of 0.01 M y solution.

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11, , CBSE Term II Chemistry XII, , 11. What will be the limiting molar conductivity of, , 20. Following reactions can occur at cathode during the, , CH 3COOH if the limiting molar conductivity of, CH 3COONa is 91 S cm 2 mol -1 ? Limiting molar, conductivity for individual ions are given in the, following table., S. No., , electrolysis of aqueous silver nitrate solution using Pt, electrodes:, (Delhi 2015), +, Ag ( aq) + e ¾® Ag( s ); E° = 0.80 V, 1, H + ( aq) + e - ¾® H 2 ( g ); E° = 0.00 V, 2, On the basis of their standard electrode potential, values, which reaction is feasible at cathode and why?, , Ions Limiting molar conductivity/S cm 2 mol -1, , 1, , H+, , 349.6, , 2, , Na +, , 50.1, , 3, , K+, , 73.5, , 4, , OH-, , 199.1, , l, , 12. Suggest a way to determine the L°m for water.(NCERT), 13. Calculate the degree of dissociation (a ) of acetic, acid, if its molar conductivity ( L m ) is 39.05 S cm 2, mol -1 ., Given : L° (H + ) = 349.6 S cm 2 mol –1, and L° (CH 3COO - ) = 40.9 S cm 2 mol -1, , 14. The conductivity of 0.20M solution of KCl at 298K, is 0.0248 S cm -1 . Calculate its molar conductivity., , 15. The conductivity of 0.001 M acetic acid is, , 4 ´ 10 -5 S /cm. Calculate the dissociation constant of, acetic acid, if molar conductivity at infinite dilution, for acetic acid is 390 S cm2/mol., , 16. The molar conductivity of 0.025 mol L -1 methanoic, acid is 46.1 S cm 2 mol -1 . Calculate its degree of, dissociation and dissociation constant., Given, L° (H + ) = 349.6 S cm 2 mol -1, and L° (HCOO - ) = 54.6 S cm 2 mol -1, , 17. Conductivity of 2.5 ´ 10 -4 M methanoic acid is, , 5.25 ´ 10 -5 S cm -1 . Calculate its molar conductivity, and degree of dissociation., Given: L°(H + ) = 349.5 S cm 2 mol - 1 and, L°(HCOO - ) = 50.5 S cm 2 mol - 1 ., (All India 2015), , 18. When acidulated water (dil. H 2SO 4 solution) is, electrolysed, will the pH of the solution be affected?, Justify your answer., (NCERT Exemplar), , 19. Predict the products of electrolysis in each of the, following, (i) An aqueous solution AgNO 3 with silver, electrodes., (ii) An aqueous solution AgNO 3 with platinum, electrodes., , Long Answer Type Questions, , 21., , (i) Calculate E°cell for the following reaction at 298 K:, 2Al( s ) + 3Cu2+ (0.01 M) ¾®, 2Al3 + (0.01 M) +3Cu( s ), Given, Ecell = 1.98 V, (ii) Using the E° values of A and B, predict which is better, for coating the surface of iron, [E°(Fe2+/ Fe) = -0.44V] to prevent corrosion and, why?, Given : E °( A2+ / A) = -2.37 V,, E °( B 2+ / B) = -0.14 V, (Delhi 2016), , 22. (i) The emf of the following cell at 298 K is 0.1745 V., Fe( s )|Fe 2+ (0.1 M)|| H + (x M)|| H 2 ( g )(1 bar)| Pt( s), °, Given, E Fe, = - 0.44 V, 2+, / Fe, Calculate the H + ions concentration of the, solution at the electrode where hydrogen is being, produced., (ii) Why on dilution the L m of CH 3COOH increases, drastically, while that of CH 3COONa increases, gradually?, (NCERT Exemplar), , 23. (i) State Kohlrausch law ., (ii) Calculate the emf of the following cell at 298 K:, Al ( s )|Al 3+ ( 015, . M)||Cu 2+ 0.025 M| Cu(s), (Given : E(°Al 3+ / Al ) = - 1.66 V, E(°Cu 3+ / Cu ) = 0.34 V,, log 0.15 = - 0 . 8239, log 0.025 = -1.6020), , 24. (i) On the basis of E° values identify which amongst, the following is the strongest oxidising agent?, Cl 2 (g) + 2e - ¾® 2Cl - ; E° = +1.36 V, MnO 4- + 8H+ + 5 e - ¾® Mn 2+ +4H2 O;, E° =1.51V, Cr2 O 72 - + 14H+ + 6e - ¾® 2Cr 3+ + 7H2 O;, E° = +1.33 V, (ii) The following figure 2, represents variation of, (Lm ) vs C for an electrolyte. Here, L m is the, molar conductivity and C is the concentration of, the electrolyte.

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12, , CBSE Term II Chemistry XII, , ° +, ° 2+, [Given: ECu, = + 0.34; E Ag, = + 0.80 V,, /Cu, / Ag, , 150.0, 148.8, , ° 3+, E Au, = + 1.40 V], / Au, , Lm/(S cm mol–1), , 149.4, , (ii), , 149.0, 148.6, , Zinc, plate, , 148.2, 147.8, , Silver, plate, Salt-bridge, , 147.4, , Ag+(aq), , 147.0, 0, , .005 .010 .015 .020 .025 .030 .035, C1/2/(mol/L)1/2, , Zn2+(aq), , Consider the figure given above and answer the, following questions:, (a) What is the direction of flow of electrons?, (b) Which one is anode and which one is cathode?, (c) What will happen if the salt bridge is removed?, (d) How will concentration of Zn 2+ and Ag + ions be, affected when the cell functions?, (e) How will concentration of these ions be affected, when the cell becomes dead?, , (a) Define molar conductivity., (b) Identify the nature of electrolyte on the basis of, the above plot. Justify your answer., (c) Determine the value of L m for the electrolyte., (d) Show how to calculate the value of A for the, electrolyte using the above graph., , 25. (i) Calculate emf of the following cell :, Zn( s ) / Zn 2 + (0.1 M) || (0.01 M) Ag + / Ag ( s ), °, Given E Zn, = - 0.76 V,, 2+, / Zn, °, E Ag, +, / Ag, , Case Based Questions, 28. Read the following and answer the questions from (i), , l, , = + 0.80 V, , [Given, log 10 = 1], (ii) X and Y are two electrolytes. On dilution molar, conductivity of ‘X’ increases 2.5 times, while that, of Y increases 25 times. Which of the two is a, (All India 2020), weak electrolyte and why?, , 26. (i) The electrical resistance of a column of 0.05 M, KOH solution of length 50 cm and area of, cross-section 0.625 cm 2 is 5 ´ 10 3 ohm., Calculate its resistivity, conductivity and molar, conductivity., (ii) Predict the products of electrolysis of an aqueous, solution of CuCl 2 with platinum electrodes., °, (Given : ECu, = + 0.34 V, E °(1/ 2 Cl, 2+, / Cu, °, EH, +, /H, , 2 ( g ), Pt, , 2 /Cl, , -, , ), , = + 1.36 V, , = 0.00 V, E °(1/ 2 O2 / H 2 O) = + 1.23 V), (All India 2020), , 27. (i) When a bright silver object is placed in the, solution of gold chloride, it acquires a golden, tinge but nothing happens when it is placed in a, solution of copper chloride. Explain this, behaviour of silver., , to (iv) given below, Molar conductivity, L m is defined as the conducting, power of all the ions produced by dissolving one gram, mole of an electrolyte in solution. It is related to, conductivity (or specific conductance), k as follows :, k, Molar conductivity, L m =, C, In the above equation, if k is expressed in S m -1 and, the concentration C in mol m -3 then the units of L m, are in Sm 2 mol -1, Now, 1 mol m -3 = molarity (mol/L)´100 L / m 3, Hence, L m (S m2 mol -1 ), = k (S m -1 ) /molarity (mol L -1 ) ´1000 Lm -3, If k is expressed in Scm -1 and C in mol cm -3 , then, the units for L m are S cm 2 mol -1 . It can be, calculated by using the equation, k (S cm -1 ) ´100 (cm 3 /L), L( S cm 2 mol -1 ) =, molarity (mol/L), Both type of units are used in literature and are, related to each other by the equations

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CBSE Term II Chemistry XII, , 1 S m 2 mol -1 = 10 4 S cm 2 mol -1, 1 S cm 2 mol -1 = 10 -4 S cm 2 mol -1, Equivalent conductance, L (lambda). It is defined as, the conducting power of all the ions produced by, dissolving one gram equivalent of an electrolyte in, solution., (i) The molar conductivity of a 1.5 M solution of an, electrolyte is found to be 138.9 Scm 2 mol -1 ., Calculate the conductivity of the solution., (ii) How is the unit of molar conductivity arrived ?, (iii) Calculate the degree of dissociation (a) of, acetic acid if its molar conductivity( L m ) is, 40.05 S cm2 mol -1 ., Given: l°(H + ) = 349.6 S cm2 mol -1 and, l°( A - ) = 50.9 S cm2 mol -1, (iv) Why does the conductivity of a solution decrease, with dilution?, Or, What is the effect of decreasing concentration on, the molar conductivity of a weak electrolyte?, 29. Read the following and answer the questions from (i), to (iv) given below, Standard potentials are physical electrochemical, quantities that represent essential properties of, systems that involves metals and gases on the one, hand and ions in solution on the other., , 13, The standard potentials are always measured under, standard conditions, i.e. when the pressure of a gas is, 1 atm, concentration of ions is 1 mol dm -3 and the, temperature is 298 K. However, practically standard, conditions are hardly achieved or used. The potential, of the electrode whose reaction can be written (in the, form or reduction) as, M n+ ( aq) + ne - ¾® M( s ), which is measured against standard hydrogen, electrode, is written as, [M], RT, ln, E( M n+ / M ) = E(-M n+ / M ) nF [ M n+ ], This is known as the Nernst equation., (i) Write Nernst equation for the following cell, reaction:, Zn | Zn 2+ (aq)||Cu 2+ (aq)|Cu(s ), (ii) What is the effect of increase in concentration of, zinc ions on the electrode potential of zinc, electrode for which E Zn 2+ / Zn equals - 0.76 V ?, (iii) How can the reduction potential of an electrode, be increased?, (iv) What does the standard electrode potential of a, metal being negative( E -Zn 2+ /Zn = - 0.7632), indicate ?, Or, How does temperature affect Nernst equation?

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EXPLANATIONS, Objective Questions, 1. (d) Electrode potential of any half-cell depends all of the, given options, i.e. on nature of metal, concentration of metal, ions in the solution and temperature of solution., 2. (c) For the electrochemical cell,, Ag | AgCl | KCl || AgNO3 | Ag+, The cell reaction is, AgCl (s)+ e - ¾® Ag + Cl – ( aq ), , 8. (c) Overall cell reaction,, 2Cu + (aq ) ¾® Cu (s ) + Cu 2+ ( aq ), ° ) = E° +, ° 2+ +, \ Cell potential ( Ecell, - ECu, Cu / Cu, / Cu, = 0.60 - 0.41 = 0.19 V, As we know that,, - nFE° = - RT ln K eq, nE°, 2 ´ 0.19 V, Þ, log K eq =, =, = 6.44, ( 2.303RT / F ), 0.059 V, , Ag ¾® Ag+( aq ) + e –, Overall cell reaction,, AgCl(s ) ¾® Ag+ + Cl –, ° = Ecathode, °, °, 3. (d) Ecell, - Eanode, i.e. it does not depend upon the concentration of ions., Hence, if the concentration of Cu 2 + ions is doubled, then, ° will remains the same., Ecell, 4. (d) In case of electrolysis of aqueous NaCl, oxidation, reaction occurs at anode as follows, 1, °, Cl - ( aq ) ¾® Cl 2 ( g )+ e - ; Ecell, = 1.36 V, 2, °, 2H2O (l) ¾® O 2 (g ) + 4H+ (aq ) +4 e - ; Ecell, = 1.23 V, °, But due to lower Ecell, value, water should get oxidised in, preference of Cl (aq)., , K eq = 106.44 = 2.76 ´ 106, æ 0.059 ö, °, 9. (b) EMg 2 + / Mg = ç, ÷ log [ Mg2 + ] + EMg, 2+, / Mg, è 2 ø, This equation represents equation of straight line. It can be, correlated as, æ 0.059 ö, °, EMg 2+ /Mg = ç, ÷ log [Mg2+ ] + EMg, 2+, /Mg, è 2 ø, ­, ­, ­, ­, Y, M, X, +C, So, intercept (C) = E° 2+, Mg, , Thus, the correct option is (d) not (b)., 5. (a) Anode is always the site of oxidation and cathode is the, site of reduction., Thus, anode half-cell is, Zn( s ) ¾® Zn 2+ (aq ) + 2e - ; E° = -0.76 V, and cathode half-cell is, Ag2O(s ) + H2O( l ) + 2e - ¾® 2Ag ( s ) + 2OH- ( aq ) ;, E° = 0.34 V, ° = E°, °, Q, Ecell, cathode - Eanode, \, E° = 0.34 - ( - 0.76) = 1.1 V, cell, , 1, electrode potential, XA ¾® No reaction, XB ¾® A, C, XC ¾® A, Order of electrode potential is XB < XC < XA., Order of oxidation of anion B - > C - > A - ., , 6. (b) Oxidising tendency µ, , 7. (b) Here, out of given four options standard reduction, potential of chromium has highest negative value hence,, most powerful reducing agent is chromium., , EMg 2+/Mg, , However, the actual reaction taking place in the, concentrated solution of NaCl is (d) and not (b), i.e. Cl 2 is, produced and not O 2 ., This unexpected result is explained on the basis of the, concept of ‘overvoltage’, i.e. water needs greater voltage for, oxidation to O 2 (as it is kinetically slow process) than that, needed for oxidation of Cl - ions to Cl 2 ., , /Mg, , 0.059, Slope =, 2, Thus, equation can be graphically represented as,, , Slope =, , 0.059, 2, , ° 2+/Mg, C = E Mg, log[Mg 2+]®, , 10. (c) Ecell is an intensive property as it does not depend upon, mass of species (number of particles) but D r G of the cell, reaction is an extensive property because this depends upon, mass of species (number of particles)., 11. (c) The half-cell reactions are, 4Al 3+ ¾® 4Al + 12e - [Reduction], 12e - + 6O2 - ¾® 3O2, or, , [Oxidation], , 4 3+, 4, Al ¾® Al + 4e - [Reduction], 3, 3, [Oxidation], 4e - + 2O2- ¾® O2, , As we know that,, DG° = - nFE°, Given,, , DG = + 960 kJ mol -1 = 960 ´ 1000 J mol -1, n = 4F = 4 ´ 96500 C, , \, , 960 ´ 1000 = - 4 ´ 96500 ´ E°, , 960000, = - 2.48 V, 4 ´ 96500, Potential difference » 2.5 V, 12. (a) The standard free energy ( DG° ) is related to standard, electrode potential ( E° ) and equilibrium constant as, DG° = - nFE° and DG° = - RT loge K eq, E° = -

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15, , CBSE Term II Chemistry XII, , 13., , 14., , 15., , 16., , ° is - ve, then DG° is + ve, i.e. DG° > 0., Thus, if Ecell, Hence, K eq would be negative and less than 1; K eq < 1., (c) Solution consists of electrolytes is known as electrolytic, solution and conductivity of electrolytic solution depends, upon the following factors, (i) Size of ions As ion size increases, ion mobility, decreases and conductivity decreases., (ii) Viscosity of solution Greater the viscosity of the, solvent, lesser will be the conductivity of the solution., (iii) Solvation of ions Greater the solvation of ions of an, electrolyte, lesser will be the electrical conductivity of, the solution., (iv) Temperature of medium Conductivity of solution, increases with increase in temperature., (a) In the given graph, ‘A’ represents weak electrolyte like, CH3COOH as Lm is increasing steeply on dilution at low, concentration region and ‘B’ represents strong electrolyte, like NaCl as Lm is increasing slowly with dilution., (b) When concentration approaches zero, the molar, conductivity is known as limiting molar conductivity., So, here limiting molar conductivity is near about 150., (b) Given, k = 0.002765, R = 400W, [Here, R = resistance, k = specific conductivity], Cell constant ( G) = R ´ k, , 17. (b), , = ( 400W ) ´ ( 0.002765 mho cm -1 ) = 1.106 cm -1, = L°m (NH +) + L°m ( Cl - ), , L°m (NH 4 Cl), , 4, , L°m (NaOH), , L°m (Na+ ), , =, , + L°m (OH- ), , L°m (NaCl) = L°m (Na+ ) + L°m (Cl - ), L°m (NH4 Cl) + L°m (NaOH) - L°m (NaCl) = L°m (NH4 OH), Hence, option (b) is correct choice., 18. (a) Both A and R are true and R is the correct explanation of A., Current stop flowing when Ecell = 0, As at Ecell = 0, reaction reaches the equilibrium., 19. (c) A is true, but R is false., Copper sulphate can’t be stored in zinc vessel as zinc is more, reactive than copper due to negative value of standard, reduction potential of Zn., 20. (b) Both A and R are true but R is not the correct explanation, of A., 0.0591, 1, E = Elog, 1, [Ag+ ], E = E° + 0.059 log [Ag+ ], Thus, EAg + /Ag increases with increase in concentration of Ag+ ., 21. (a) Both A and R are true and R is the correct explanation of A., Explanation Electrolysis of NaCl is represented by, following chemical reactions, 1, At cathode H+ ( aq ) + e- ¾¾® H 2 ( g ), 2, , 1, ° = 1 .36V, At anode Cl -( aq ) ¾, ¾® Cl 2 + e - ; Ecell, 2, ° = 1.23 V, 2H2O ( aq ) ¾, ¾® O 2 ( g ) + 4H+ ( aq ) + 4e - ; Ecell, °, for this reaction has lower value but formation of, Ecell, oxygen at anode requires over potential., 22. (b) Both A and R are true but R is not the correct explanation, of A. pH of brine solution increases due to formation of, NaOH at cathode during electrolysis., At cathode, 2Na ¾ Hg + 2H2O ¾® 2NaOH + Hg + H2( g ), 23. (i) (c) Here, L = k ´ 1000, m, M, 7.896 ´ 10-5 ´ 1000, =, 0.00241, = 32.76 S cm -1 mol -1, , (ii) (b) Lm =, , =, , 1000 ´ k, Molarity, , or, , k=, , Lm ´ Molarity, 1000, , 1. 26 ´ 10 -2 ´ 0.01, = 1.26 ´ 10 -3, 1000, , (iii) (d) The increase in molar conductivity of any strong, electrolyte like HCl with dilution is due to decrease in, the interionic forces., (iv) (d) Statement (d) is wrong about molar conductivity., Glucose is a non-electrolyte and hence question of, having molar conductivity does not arise., Or, (a) Lm = k ´ V, where V is the volume of solution, containing 1 mole of the solute. Specific conductivity, ( k ) decreases with dilution but volume increases much, more on dilution. Hence, Lm increases with dilution., As volume containing 1 mole of NaCl will be highest, for 0.005 M NaCl solution, the product of k ´ V will be, highest and hence Lm will also be highest., 24. (i) (d) Emf of a cell depends on the temperature of, electrode, concentration of electrolyte in two half-cells,, nature of the electrode and the distance of separation, between two electrodes. It does not depend on its, shape and size., (ii) (a) The electrical conductance increases with increasing, temperature. As the temperature increases, the, mobility of ions present in the solution increases and as, a result, conductance increases., (iii) (b) A salt-bridge is used to connect the oxidation and, reduction half-cells of a galvanic cell and it maintains, the electrical neutrality between both half-cells. It, completes the cycle of the charge flowing through the, cell., (iv) (a) The reduction of Zn 2 + involves 2 electrons., Zn 2 + + 2e - ¾® Zn( s ), So, 2 Faradays or 2 ´ 96500 = 193000 C charge is, required., Or, (d) When the cell reaction attains equilibrium, then the, value of emf of a cell becomes zero.

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16, , CBSE Term II Chemistry XII, , Subjective Questions, , (ii) Fe ( s )| Fe2 + (0.001 M) || H+ (1 M)| H2( g ) (1 bar) | Pt ( s ), , 1. The device through which chemical energy changes to, electrical energy is called electrochemical cell. The chemical, reaction of the cell is reversed and current flows in opposite, direction when opposing emf is greater than that of the cell., 2. Electrochemical reaction is, Ag + Ag+ ¾® Ag+ + Ag, (1 M), °, Ecell = Ecell, , The overall cell reaction will be, Fe(s ) + 2H+ ( aq ) ¾® Fe2 + ( aq ) + H2( g ), The Nernst equation is, 0.059, [ Fe2 + ], °, Ecell = Ecell, log + 2, n, [H ], 0.059, ( 0.001), = [ 0 - ( - 0.44)] log, 2, (1)2, 0.059, = 0.44 log( 10- 3 ), 2, = 0.44 - 0.0295 (- 3) = 0.44 + 0.0885, = 0.528 = 0.53 V (Approximately), , ( 0. 01), , 0.059, [ 0.01 ], log, n, [1 ], , °, ° +, ° +, = - 0.059 log10-2 [Q Ecell, = EAg, - EAg, = 0], / Ag, / Ag, , (i) Ecell = 2 ´ 0.059 = 01182, V, ., (ii) Net cell reaction is AgNO 3( 1M ) ¾® AgNO 3( 0.01 )M, (iii) The cell would not generate emf., If the concentrations are equal, then, 0.0591, [ Ag+ ], °, Ecell = Ecell, log, n, [ Ag+ ], 0.0591, = 0log 1 = 0, n, 3. The cell reaction of Daniell cell is, Zn ( s ) + Cu 2 + ( aq ) ¾® Zn 2 + ( aq ) + Cu ( s ), The Nernst equation is, 0.059, [ Zn 2 + ], °, Ecell = Ecell, log, 2, [ Cu 2 + ], According to above equation, Ecell decreases when, concentration of Zn 2 + ions is increased as Ecell is directly, dependent upon concentration of Cu 2 + and inversely, dependent upon the concentration of Zn 2 + ions., 1, 4. For hydrogen electrode, H+ + e - ¾® H2, 2, Applying Nernst equation,, 0.0591, 1, o, E(H + /1/ 2 H ) = E(H, log +, +, /1/ 2 H 2 ), 2, n, [H ], 0.0591, 1, =0log, [pH = 10 ;[H+ ] = 10-10 M], 1, ( 10-10 ), 0.0591, =0´ ( 10 log10) = - 0591, ., V, 1, E(H + /1/ 2 H ) = - 0.591V, 2, , 5., , (i) For the given cell, the half-cell reactions will be given, as below, At anode, Mg ¾® Mg 2 + + 2 e 2+, At cathode Cu + 2e - ¾® Cu, Therefore, the overall cell reaction will be, Mg + Cu 2+ ¾® Mg2+ + Cu, The Nernst equation is, 0.0591 [Mg2+ ], Ecell = E°cell log, n, [Cu 2+ ], 0.0591, (0.001), \, Ecell = [0.34 - (- 2.36)] log, 2, (0.0001), = 2.71 - 0.0295 log(10) = 2.7 - 0.0295, = 2.68 V, , 6., , °, °, (i) The ECr, value is - 0.41 V and EFe, is 0.77 V., 3+, 3+, / Cr, / Fe, 2+, This means Cr can be easily oxidised to Cr 3+ than, Fe2 + as Fe2 + does not oxidised easily to Fe3+ . Cr 2 + is a, stronger reducing agent., , (ii) Among the given ions, the ion with more (–) ve value of, E° (red.) will loose the electron more easily, thus is, more stable in (+) 2 oxidation state. Hence, Mn in, (+) 2 oxidation state is the most stable species., 7. From the given cell reaction and Nernst equation,, 0.0591, [Ni 2 + ], Ecell = Ecell, ° –, log, n, [Ag+ ] 2, 0.0591, (0.160), = 1.05 log, 2, (0.002) 2, 0.0591, = 1.05 log (4 ´ 104 ), 2, 0.0591, = 1.05 (4.6021), 2, = 1.05 - 0.135 = 0.915 V, = 0.915 V, °, 8. (i) Ecell, = 0.236 V, °, The standard Gibbs energy, DG = - nFEcell, = 2 ´ 96500 ´ 0.236, = - 45.548 kJ, (ii) Here, current ( i) = 0.5 A, t = 2 hours, Quantity of current passed ( Q) = Current ´ Time, = ( 0.5 A) ´ (2 ´ 60 ´ 60 sec), = 3600 A sec = 3600 C, Now, Q = ne where, n = number of electrons,, e - = charge on electron, Q, 3600 C, n= -=, = 2250 ´ 1019, e, 1.6 ´ 10- 19 C, \Number of electron flow = 2.250 ´ 1022, 9. Given,, , ° 2+, ECd, = - 0.403 V, / Cd, ° 2+, EZn, = - 0.763V, / Zn, , \, , ° = - 0.403 - (-0.763), Ecell, ° = 0.36 V, Ecell, , Q, , °, D r G°`= - nFEcell

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17, , CBSE Term II Chemistry XII, , where, n = number of moles of electrons used, n = 2 (Here), F = 96500 C mol -1, , 15. Concentration of acetic acid = 0.001 M, k ´ 1000, Molar conductivity, Lm =, M, 4 ´ 10- 5 ´ 1000, =, 0.001, = 40 S cm 2 mol -1, , \, , D r G° = - 2 ´ 96500 ´ 0.36 = 69480 J / mol, ° = 0.0591 log K, Also, Ecell, C, n, °, n ´ Ecell, 2 ´ 0.36, \ log K C =, =, = 12.18, 0.0591, 0.0591, log K C = 12.18, , L°m ( CH3COOH) = 390 S cm 2 mol -1, L, 40, So, a = m =, = 0.102, L°m 390, , 10. (i) Cell constant, G* = resistance ( R ) ´ conductivity ( k), = 100 ´ 1.0 ´ 10-4 = 10-2 cm -1, (ii) Conductivity of solution y,, Cell constant 10-2, k=, =, = 2 ´ 10-4 S cm -1, Resistance, 50, (iii) Molar conductivity of solution y,, k ´ 1000 2 ´ 10-4 ´ 1000, = 20 S cm 2 mol –1, Lm =, =, Molarity, 0.01, 11. Given, , L°m (H+ ), , 2, , ¾® 349.6 S cm mol, , Now, dissociation constant ( K a ) for CH3COOH =, =, , 16., , L°m (HCOOH) = L°m (HCOO- ) + L°m (H+ ), , = (54.6 + 349.6) S cm 2 mol -1, = 404.2 S cm 2 mol -1, , a=, , L°m(Na+ ) ¾® 50.1 S cm 2 mol -1, , HCOOH( aq ) s, , L(CH3 COONa) ¾® 91 S cm 2 mol -1, , Initial conc., , C, , Equilibrium conc., , C (1 - a ), , HCOO- ( aq ) + H + ( aq ), 0, , 0, , Ca, , Ca, , Dissociation constant,, , Complete reaction is as follows, CH3COONa + HCl ¾® NaCl + CH3COOH, , Ka =, , L°m (CH3COOH) =L°m (CH3COONa) - L°m (Na+ ) + L°m (H+ ), = 91 - 50.1 + 349.6 Þ 390.5 S cm 2 mol -1, 12. The molar conductance of water at infinite dilution can be, obtained from the knowledge of molar conductances at, infinite dilution of sodium hydroxide, hydrochloric acid and, sodium chloride (all strong electrolytes). This is in, accordance with Kohlrausch’s law., L°m ( H 2 O ) = L°m ( NaOH ) + L°m ( HCl ) - L°m (NaCl), 13. According to Kohlrausch’s law,, L°m (CH3 COOH) = L°(H+ ) + L° ( CH3 COO – ), = 349.6 S cm 2 mol –1 + 40.9 S cm 2 mol -1, = 390.5 S cm 2 mol -1, Lm 39.05 S cm 2 mol –1, Þ a = 0.1, =, 2, –1, L°m 390.5 S cm mol, , 14. Conductivity ( k ) = 0.0248 S cm -1 = 0.0248 ohm -1 cm -1, Molar concentration (C) = 0.20 mol L-1, (0.2 mol), =, = 2.0 ´10-4 mol cm -3, (1000 cm 3 ), Molar conductivity ( Lm ) =, , Lm, (46.1) S cm 2 mol -1, =, = 0.1140, L°m, (404.2) S cm 2 mol -1, Water, , Lm (K + ) ¾® 73.5 S cm 2 mol -1, , a=, , 0.001(0.102)2, (1 - 0.120), , = 1.15 ´ 10- 5, , -1, , Lm(OH- ) ¾® 199.1 S cm 2 mol -1, , Ca 2, 1-a, , -1, , -1, , k (0.0248 ohm cm ), =, C (2.0 ´ 10-4 mol cm -3 ), , = 124 ohm -1 mol -1 cm 2, or 124 S mol -1 cm 2, Molar conductivity = 124 S mol -1 cm 2, , 17. Given,, , [HCOO- ] [H+ ] C a ´ C a, C a2, =, =, [HCOOH], C (1 - a ) (1 - a ), , =, , (0.025 mol L-1 ) ´ (0.114)2, (1 - 0.114), , =, , (3.249 ´ 10-4 mol L-1 ), = 3.67 ´10-4 mol L-1, (0.886), , K = 5.25 ´ 10- 5 S cm -1, , Molarity ( M ) = 2.5 ´ 10- 4 M, L°(H+ ) = 349.5 S cm 2 mol -1, L°( HCOO- ) = 50.5 S cm 2 mol -1, 1000 ´ K, Molar conductivity, Lcm =, S cm 2 mol -1, Molarity, =, , 1000 ´ 5.25 ´ 10- 5, S cm 2 mol -1, 2.5 ´ 10- 4, , = 210 S cm 2 mol -1, L°m ( HCOOH) = l °( HCOO - ) + l°(H + ), = (50.5 + 349.5) S cm 2 mol -1, = 400 S cm 2 mol -1, L°m, L°m, 210, =, = 0.525, 400, , a=

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18, , CBSE Term II Chemistry XII, , 18. pH of the solution will not be affected as [ H+ ] remains, constant., At anode 2H2O ¾® O2 + 4H+ + 4e At cathode 4H+ + 4e - ¾® 2H2, 19., , (i) An aqueous solution of AgNO3 with silver electrodes., In aqueous solution, ionisation of AgNO3 and H2O, takes place., ( aq ), , Ag+ ( aq ) + NO-3 ( aq ), , AgNO3( s ) s, , H+ ( aq ) + OH- ( aq ), , H2O (l ) s, , At cathode Ag+ ions has less discharge potential than, H+ ions so silver will be deposited at cathode., Ag+ ( aq ) + e - ¾® Ag( s ), At anode An equivalent amount of silver will be, oxidised to Ag+ ions by releasing electrons., Ag( s ) ¾® Ag+ ( aq ) + e Ag anode is attacked by NO-3 ions, so it will also, produce Ag+ in the solution., (ii) An aqueous solution of AgNO3 with platinum electrodes., In aqueous solution ionisation of AgNO3 and H2 O, occurs, ( aq ), , AgNO3( s ) s, , +, , Ag ( aq ) +, , NO3-( aq ), -, , = 1.98 +, , 0.0591, 0.0591, log100 = 1.98 +, ´2, 6, 6, , [Q log100 = log 102 = 2 log 10 = 2 as log 10 = 1], = 1.9997 V » 2.0 V, (ii) Coating the surface of iron with a more active metal, prevents the iron from losing electrons and hence from, corrosion. As reduction potential of A is lower than that, of iron, therefore A would be used for coating the, surface of iron to prevent it from corrosion. This is, because lower the reduction potential of an element,, more reactive is the element. Therefore, A would be, more reactive than iron as well as B and hence, A, protects the iron metal from corrosion., 22., (i) Given, Ecell = 01745, V, ., EFe2+ / Fe = - 0.44, [H+ ] = ?, The reaction for the given cell is written as,, Fe( s ) + 2H+( aq ) ¾® H2 ( g ) + Fe2+( aq ), Nernst equation for the above reaction is represented as,, 2.303RT, [Fe2 + ], °, …(i), Ecell = Ecell, log + 2, nF, [H ], °, °, °, Also, Ecell, = EH, - EFe, +, 2+, /H, / Fe, 2, , As platinum electrodes are non-attackable electrodes,, they will not be reacted upon by NO-3 ions., , = 0 - ( -0.44) = + 0.44 V, °, Put Ecell, value in Eq. (i),, 0.0591, [ 01, . ], \ 01745, ., = 0.44 log 2, 2, [ x], , At cathode Ag will be deposited at cathode., , or, , H2 O( l ) s, , +, , H ( aq ) + OH ( aq ), , Ag+ ( aq ) + e - ¾® Ag( s ), of NO -3, , -, , OH- ( aq ) ¾® OH + e 4 OH ¾® 2 H2 O( l ) + O2 ( g ), So, oxygen gas is produced at anode. The solution, remains acidic due to the presence of HNO3., H+ ( aq ) + NO-3 ( aq ) ¾® HNO3 ( aq ), +, 20. Ag ( aq ) + e - ¾® Ag( s ); E° = 0.80 V., Reaction is feasible at cathode because it has higher, reduction potential., 21. (i) 2Al(s ) + 3Cu 2+ (0.01M) ¾® 2Al 3+ (0.01M) + 3Cu(s ), , E cell = 1.98 V, Applying Nernst equation for the given cell reaction,, , ° =E, Ecell, cell +, , 0.0591, [Al 3+ ]2, log, n, [Cu 2+ ] 3, 3+ 2, , 0.0591, [Al ], log, n, [Cu 2+ ] 3, , Here, n = 6 (n = number of electrons transferred), = 1.98 +, , 0.0591, (0.01)2, log, 6, (0.01)3, , log x = - 5, log [H+ ] = - 5, , -, , At anode Out, and OH ions, only OH ions will, be oxidised (due to less discharge potential), preferentially and NO-3 ions will remain in the solution., , ° Ecell = Ecell, , [Q[ H+ ] = x (given)], , [H+ ] = 10-5, , \, , (ii) In the case of CH3COOH, which is a weak electrolyte,, the number of ions increases on dilution due to an, increase in degree of dissociation., CH3COOH + H2 O s CH3COO- + H3O+, In case of strong electrolyte, the number of ions remains, the same but the inter ionic attraction decreases., 23. (i) Kohlrausch law It states that limiting molar, conductivity of an electrolyte is the sum of the, individual contributions of the cation and the anion of, the electrolyte., Kohlrausch examined L°m values for a number of strong, electrolytes and found certain regularities. Also,, Kohlrausch law of independent migration of ions is, used for determination of limiting molar conductivity, for weak electrolyte., (ii) E° = E°, - E°, cell, , cathode, , anode, , = 0.34 - ( -1.66) = 2.00 V, According to Nernst equation,, 3+ 2, ° = E°cell - 0.059 log [Al ], Ecell, n, [Cu 2+ ] 3, Here, n = 6

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19, , CBSE Term II Chemistry XII, , 0.059, [ 015, . ]2, log, 6, [ 0.025 ] 3, 0.059, =2( 2 log 015, . - 3 log 0.025 ), 6, 059, ., =2( -1.6478 + 4.8062), 6, = 2 - 0.0311 = 1.9689 V, 24. (i) MnO 4 is the strongest oxidising agent. Oxidising power, of metals decreases with a increase in their E° values., (ii) (a) Molar conducivity of a solution is the conductance of, that volume of solution containing one mole of, electrolyte, kept between two electrodes having unit, length joining them and large cross sectional areas so, as to contain the electrolyte., (b) The nature the electrolyte on the basis of plot is, strong electrolyte. For strong electrolyte, L°m, increases slowly with dilution., (c) Using Debye-Huckel Onsager equation,, , = 1.56 - 0.0295 log, , Ecell = 2 -, , = 1.56 - 0.0295 log 103, = 1.56 - 0.0295 ´ 3 log 10, = 1.56 - 0.0295 ´ 3, = 1.56 - 0.0885 = 1.4815 V, (ii) Y is a weak electrolyte. On dilution, complete, dissociation of weak electrolyte occurs and thus a step, increase in molar conductivity. However, in case of, strong electrolyte it has already dissociated completely, therefore on dilution, the rise in conductivity is not, significant., , 150.0, 149.8, Lm /(S cm mol–1), , 149.4, 149.0, 148.6, 148.2, 147.8, 147.4, 147.0, 0, , .005 .010 .015 .020 .025 .030 .035, C1/2/(mol/L)1/2, , æy -y ö, (149 - 147.8), A = - slope = çç 2 1 ÷÷ = x, x, (0.010, - 0.022), è 2 1ø, = 100 S cm 2 mol -1 / (mol / L-1 )1/ 2, 25., , (i) Using Nernst equation,, Ecell, °, Ecell, , 0.059, =, log K, n, °, °, = EAg, - EZn, +, 2+, / Ag, / Zn, °, Ecell, , = 0.80 - ( - 0.76), = 0.80 + 0.76 = 1.56 V, At anode, Zn ¾® Zn 2 + + 2e At cathode, 2Ag+ + 2e - ¾® 2Ag, On combining the net equation is, Zn( s ) + 2Ag+ ( aq ) ¾® Zn 2 + ( aq ) + 2Ag( s ), Ecell = 1.56 -, , 0.059, [Zn 2 + ], log, 2, [Ag+ ] 2, , For CH3COOH (weak electrolyte), , Lm (S cm2 mol–1), , L m = L°m - A C, (d) L° value from graph is 150 S cm 2 mol -1., , 10-1, (10-2 ) 2, , For KCl (strong electrolyte), , 1/2, , C (mol/L ), , 26., , (i) Given,, Molarity = 0.05 M, Length of column = 50 cm, Area of cross-section = 0.625 cm 2, Resistance = 5 ´ 103 W, rl, A, We know, R =, or r = R, A, l, 5 ´ 103 ´ 0.625, \ Resistivity =, = 62.5 W cm, 50, 1 æ 1 ö, Conductivity ( k ) = = ç, ÷ S cm = 0.016 S cm -1, r è 625, . ø, k ´ 1000, Molar conductivity, ( Lm ) =, C, 0.016 ´ 1000, =, 0.05, = 320 S cm 2 mol -1, °, (ii) Given : ECu, = 0.34 V,, 2+, / Cu, , E°(1/ 2Cl, , 2 / Cl, , °, EH, +, /H, , -, , ), , 2 ( g ), Pt, , = + 1.36 V, = 0.0 V, E1/° 2O2 / H2 O = + 1.23 V, , The reaction with a higher value of E° takes place at the, cathode. Therefore, deposition of copper will take place, at cathode., At cathode, following reduction reaction take place as:, Cu 2 + + 2e - ¾® Cu( s ); E° = 0.34 V, 1, H+ ( aq ) + e - ¾® H2 ; E° = 0.0 V, 2, At anode, the reaction with a lower value of E° is, preferred. But due to over potential of oxygen, Cl - gets, oxidised at anode to produce Cl 2 gas. The reactions, taking place at anode is

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20, , CBSE Term II Chemistry XII, , = 349.6 S cm 2 mol -1 + 50.9 S cm 2 mol -1, , 1, Cl 2 + e - ; E° = 1.36 V, 2, 2H2O ¾® O 2 + 4H+ + 4e - ; E° = +1.23 V, Cl - ¾®, , 27., , = 400.5 S cm 2 mol -1, L, a= m, L°m, , (i) E° value of silver (E°Ag + / Ag = + 0.80 V) is lower, than that of gold, hence silver displaces gold, ( E°Au 3 + / Au = + 1.40 V), which gets deposited on the, silver object., , E° value of copper ( E°Cu 2 + / Cu = +0.34) is lower than, that of silver, hence silver cannot displace copper from, its solution., (ii) (a) Electrons flow from Zn to Ag plate., (b) Zn acts as anode and Ag acts as cathode, (c) Cell will stop functioning, (d) Concentration of Zn 2 + ions will increase and that of, Ag + ions will decrease., (e) No change, 28. (i) Molar conductivity, Lm = 138.9 S cm 2 mol -1, Molarity = 1.5 M; conductivity k = ?, k ´ 1000, Molar conductivity, Lm =, Molarity, L ´ Molarity, or, k= m, 1000, 138.9 S cm2 mol -1 ´ 1.5 mol L-1, =, 1000 cm3 L-1, = 0.208 S cm-1, 100 ´ k 1000 1 l, (ii) Lm =, =, ´ ´, M, M, R a, 1000cm3, 1, cm, =, ´, ´, ohm cm2, mol -1, = ohm-1 cm2mol -1, = S cm2mol -1, 2, , (iii) Given, Lm = 40.0.05 S cm mol, l°( H+ ) = 349.6 S cm 2 mol -1, l°( A - ) = 50.9 S cm 2 mol -1, L°m (H)( A ) = l °( H+ ) + l °( A - ), , -1, , 40.05 S cm 2 mol -1, 400.5 S cm 2 mol -1, Þ, a = 0.1, (iv) Conductivity of a electrolyte solution decreases with, dilution because the number of ions per unit, volume furnished by an electrolyte decreases with, dilution., Or, On decreasing concentration, molar conductivity of a, weak electrolyte increases., 29. (i) Zn | Zn 2 + (aq)||Cu 2+(aq )|Cu(s ), =, , Ecell = Ecell -, , 2.303RT, [Zn 2+], log, 2F, [Cu 2+], , (ii) According to Nernst equation,, EZn2+ / Zn = EZn, + 0.0591 log [Zn 2+ ], 2+, / Zn, Hence, electrode potential will increase with increase, in concentration of Zn 2 + ions., (iii) For the electrode reaction,, M n+ + ne - ¾® M, RT, 1, EM n+ / M = EM, ln n+, n+, /M, nF, [M ], RT, ln [Mn+], = E - n+ +, M /M, nF, \Electrode potential can be increased either by, increasing metal ion concentration, [ M n+ ] or, temperature T., (iv) E 2+, = -0.7632 V indicates that zinc electrode acts, Zn, , / Zn, , as anode when coupled to NHE. Since, oxidation takes, place at anode, therefore oxidation potential is positive, but reduction potential is negative., Or, Temperature does not affect the Nernst equation.

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Chapter Test, , 6., , Assertion Conductivity decreases but molar, conductivity increases with increase in dilution., Reason Molar conductivity is the conductance of, electrolytic solution which kept between the electrodes, of conductivity cell at unit distance., , 7., , Assertion KCl solution is generally used to determine, the cell constant., Reason The conductivity of KCl is known accurately at, various concentration and different temperatures., , Multiple Choice Questions, , 1., , The positive value of the standard electrode potential, of Cu 2+ /Cu indicates that, (a) this redox couple is a stronger reducing agent than the, H + /H 2 couple, (b) this redox couple is a stronger oxidising agent than H + /H 2, couple, (c) Cu can displace H 2 from acid, (d) Both (b) and (c), , 2., , The electrode potentials for, Cu 2+ ( aq ) + e- ¾® Cu + ( aq ) and, Cu + ( aq ) + e- ¾® Cu ( s) are + 0.15V and + 0.50V, respectively. The value of E°Cu 2+ /Cu will be, (a) 0.150 V, (c) 0.325 V, , 3., , (b) 0.50 V, (d) 0.650 V, , 8., , If the molar conductivity of Ca2 + and Cl- ions are, 119 and 71 S cm 2 mol-1 respectively, then the molar, conductivity of CaCl2 at infinite dilution is, (a) 215 S cm 2mol -1, (b) 340 S cm 2mol -1, (c) 126 S cm 2mol -1, (d) 261 S cm 2mol -1, , s, , [Ans. 22.70], , 9., 10., , [Ans. 0.473 V], , 11., , Long Answer Type Questions, , Reason The oxidation of Cu to Cu 2+ ion is done by, hydrogen ion., , Using the standard electrode potentials, predict if the, reaction between the following is feasible., (i) Fe 3+( aq ) and I -( aq ), (iii) Fe 3+( aq ) and Br - ( aq ), , 1. (b), , 2. (c), , 3. (d), , 4. (d), , Assertion-Reasoning MCQs, 5. (c), , 6. (b), , 7. (a), , (iv) Ag( s) and Fe 3+( aq ), , (v) Br2 ( aq ) and Fe (aq), Given standard electrode potentials,, E°1 /2 I 2 , I - = 0.541 V, E°Cu 2+, Cu = 0.34 V,, E°1 /2 Br, , 2 , Br, , -, , = 1.09 V,, , E°Ag+, Ag = 0.80 V, , E°Fe 3+, Fe 2+ = 0.77 V, , 13., , (i) Calculate the degree of dissociation of 0.0024 M, acetic acid if conductivity of this solution is, 8.0 ´ 10 -5 S cm -1 ., Given : L°H+ = 349.6 S cm 2 mol-1 ;, , [Ans. 0.085], L°CH3COO = 40.9 S cm 2mol-1, (ii) Solutions of two electrolytes ‘A’ and ‘B’ are diluted., The limiting molar conductivity of ‘B’ increases to a, smaller extent, while that of ‘A’ increases to a much, larger extent comparatively. Which of the two is a, strong electrolyte? Justify your answer., , Answers, Multiple Choice Questions, , (ii) Ag+( aq ) and Cu ( s), , 2+, , Direction (Q. Nos. 5-7) Each of these questions, contains two statements Assertion (A) and Reason (R)., Each of these questions also has four alternative, choices, any one of which is the correct answer. You, have to select one of the codes (a), (b), (c) and (d), given below., , Assertion Cu is oxidised to Cu 2+ in the reaction of, copper with nitric acid., , The resistance of 0.2 M solution of an electrolyte is 50 W., The specific conductance of the solution is 1.3 S m - 1 . If, resistance of 0.4 M solution of same electrolyte is 260 W,, then what will be its molar conductivity?, [Ans. 6.25 ´ 10 -4 Sm2 mol -1], , Assertion-Reasoning MCQs, , 5., , °, For a cell involving one electron, Ecell, = 0.59 V at 298 K,, what will be the equilibrium constant for the cell, reaction?, [Ans. 1.0 ´1010], Given the equilibrium constant ( KC ) of the reaction:, , Cu( s) + 2 Ag+ ( aq ) ¾® Cu 2 + ( aq ) + 2Ag( s), °, is 10 ´ 10 15 . Calculate the Ecell, of this reaction at 298 K., , 12., , (a) Both A and R are true and R is the correct, explanation of A., (b) Both A and R are true, but R is not the correct, explanation of A., (c) A is true, but R is false., (d) A is false, but R is true., , Calculate the equilibrium constant for the reaction., Fe( s) + Cd2 + ( aq ), Fe 2 + ( aq ) + Cd( s), [Given, E°Cd2+/Cd = - 0.40V, EFe° 2+/Fe = - 0.44 V], , The cell constant of a conductivity cell, (a) changes with change of electrolyte, (b) changes with change of concentration of electrolyte, (c) changes with temperature of electrolyte, (d) remains constant for a cell, , 4., , Short Answer Type Questions, , For Detailed Solutions, Scan the code

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22, , CHAPTER 2, , Chemical, Kinetics, In this Chapter..., l, , Chemical Reactions, , l, , Rate Law and Rate Constant, , l, , Order of Reaction, , l, , Molecularity of Reaction, , l, , Half-Life and Pseudo First Order Reaction, , Chemical kinetics is the branch of chemistry which deals with, the study of rates of the chemical reactions and their, mechanism. It also describes the conditions in which rates can, be altered., , Chemical Reactions, When one or more substances undergo a change which results, in the formation of a new product, it is called chemical reaction., On the basis of their speeds, chemical reactions are of three, types which are as follows, (i) Very fast reactions Some reactions (such as ionic reactions), occur very fast, e.g. precipitation of AgCl from AgNO3 and, NaCl., (ii) Very slow reactions Some reactions are very slow, e.g., rusting of iron in the presence of air and moisture., (iii) Moderately slow reactions Some reactions occur at, moderate speed, e.g. inversion of cane sugar and hydrolysis, of starch., On the basis of number of steps, chemical reactions are of two, types as follows, (i) Elementary reactions A balanced chemical equation does, not give a true picture of how a reaction takes place. It is very, rare that the reaction gets completed in one step., The reactions taking place in one step zare called the, elementary reactions., , (ii) Complex reactions When a sequence of elementary, reactions (called mechanism) gives us the products,, the reactions are called complex reactions. Each step, in a complex reaction is called elementary step of a, reaction. The slowest step is called rate determining, step., , Rate of a Chemical Reactions, Rate of a chemical reaction is the change in molar, concentration of the species taking part in the chemical, reaction per unit time., For the reaction, A ® B, Rate of disappearance of A, Decrease in concentration of A -D[ A], =, =, Time taken, Dt, Rate of appearance of B, Increase in concentration of B D [ B], =, =, Time taken, Dt, and Rate may be expressed as, -D [ A] D [ B], =, Dt, Dt, As D t ® 0, Instantaneous rate = Average rate, The equations given above represent the average rate of, reaction (rav )., Units of rate = mol L-1 s -1 or atm s -1 (in gaseous state), Rate =

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23, , CBSE Term II Chemistry XII, , Instantaneous rate of reaction is the rate of change of, concentration of any one of the species involved in the, reaction at a particular instant of time., For a general reaction, aA + bB ¾® cC + dD, 1 d[ A], 1 d[ B] 1 d[C ] 1 d[ D], Instantaneous rate = ==, =, a dt, b dt, c dt, d dt, At Dt ® 0, Instaneous rate = Average rate, , Rate Law and Rate Constant, Rate law is the expression in which reaction rate is given in, terms of molar concentration of reactants with each terms, raised to some power, which may or may not be same as, the stoichiometric coefficient of the reacting species in a, balanced chemical equation. It is also called as rate equation, or rate expression., For a general reaction, aA + bB ¾® cC + dD, Rate = k[ A] x [ B] y , where x and y may or may not be equal to, the stoichiometric coefficients (a and b) of the reactants., d[ R], = k[ A] x [ B] y, dt, The above equation is known as differential rate equation,, where k is a proportionality constant called rate constant., Rate law for any reaction cannot be predicted by merely, looking at the balanced chemical equation, i.e. theoretically, but must be determined experimentally., Characteristics of Rate Constant, Rate constant is a measure of the rate of reaction. Greater, is the value of the rate constant, faster is the reaction., Each reaction has a definite value of the rate constant at a, particular temperature and its value for the same reaction, changes with change in temperature., Its value does not depend upon the concentrations of the, reactant., The units of rate constant depend upon the order of reaction., , Some enzyme catalysed reactions, reactions which occur on, metal surfaces, decomposition of gaseous ammonia on a hot Pt, surface, thermal decomposition of HI on gold surface are the, examples of zero order reaction., Hydrogenation of ethene, all natural and artificial, radioactive decay of unstable nuclei, decomposition of, N 2 O5 and N 2 O are the examples of first order reaction., Units of Rate Constant, The units of rate constant of different orders are different., This can be easily illustrated below, For a general reaction,, a A + bB ¾® cC + dD, Rate = k [ A] x [ B] y, where, x + y = n = order of the reaction, Rate, concentration, 1, k=, =, ´, x, y, time, [ A] [ B], (concentration) n, = (concentration)1- n time -1, , Considering SI units of concentration, mol L-1 and time s, the, units of k = ( mol L-1 ) 1- n s -1 . Thus, the units of k for different, order of reaction are given below, Units of rate constants, Reaction, , l, , l, , l, , l, , Order of Reaction, , (where, [A] = [B]), , Reaction in, solutions, , Gaseous, reactions, , Zero order, , mol L-1s -1, , atm s -1, , First order, , s -1, , s -1, , Second order, , mol -1L s -1, , atm -1 s -1, , nth order, , (mol L-1 )1- ns -1, , (atm)1- n s -1, , Molecularity of a Reaction, It is the number of the reacting species taking part in an, elementary reaction, which must colloide simultaneously in, order to bring about a chemical reaction.e.g., Unimolecular reaction, NH4 NO 2 ¾® N 2 + 2H 2O, 2HI ¾® H 2 + I 2, , It is the sum of powers of the concentration of the reactants, in the rate law expression., Order of a reaction = x + y; if rate = k[ A] x [ B] y, For zero order, first order and second order reaction, value of, n is 0, 1, 2., , Bimolecular reaction,, , Characteristics of Order of Reaction, Order of a reaction is an experimental quantity. It can be zero, and even a fraction., Order is applicable to elementary as well as complex reactions., For a complex reaction, order is given by the slowest step., , Characteristics of Molecularity of a Reaction, Molecularity cannot be zero or a non-integer., It is applicable only for elementary reactions., In case of complex reaction, molecularity of the slowest, step is same as the overall order of the reaction., , Trimolecular reaction, 2NO + O 2 ¾® 2NO 2, Note The probability that more than three molecules can collide and, react simultaneously is very small. Thus, reactions with the, molecularity three are very rare and slow to proceed., , l, , l, , l

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24, , CBSE Term II Chemistry XII, , Difference between Molecularity and Order of Reaction, Molecularity, , Order of Reaction, , The number of reacting species which collide simultaneously in The sum of powers of the concentrations of the reactants in the rate, order to carry out a chemical reaction is called molecularity of a law expression is called the order of that chemical reaction., reaction., Molecularity is always a whole number value (except zero),, i.e. 1, 2, 3 ....... ., , Order of reaction can be a whole number or even a fractional value., , Molecularity is a theoretical concept., , Order of reaction is determined experimentally., , Molecularity is applicable to elementary reactions. For complex Order of reaction is applicable to elementary as well as complex, reactions, it has no meaning., reactions., For simple reactions, molecularity can be obtained from the, stoichiometry of the equations., , For simple reactions, order of reaction may not be equal to the number, of molecules of the reactants as seen from the balanced equation., , Molecularity of a reaction cannot be zero., , Order of a reaction can be zero., , Differential and Integrated Rate Equations, The differential and integrated rate equations for zero and first order reactions are tabulated below, Reaction, , (i), , A ¾® Product, a = initial, concentration of, reactants, , Order, , ( a), , 0, , Differential, equation, , dx, =k, dt, , Integral equation, , k=, , x, t, , Unit of, k, , t 1/ 2, , Plot, , mol L-1s -1 a, 2k, , (a–x), , t, , (ii), , A ¾® P, , 1, , dx, 2.303, a, = k( a - x ) k =, log, dt, t, a-x, , s-1, , 0.693, k, , (a–x), , t, , The time in which the concentration of a reactant is reduced, to one-half of its initial concentration is called the half-life of a, reaction. It is represented as t 1/ 2 ., For the first order reaction,, [ R] 0, 1 [ R] 0, 2.303, k = ln, =, log, t, [ R], t, [ R], R0, At t 1/ 2 ,, [ R] =, 2, [ R] 0, 2.303, Therefore,, k=, log, t 1/ 2, [ R] 0 / 2, 2.303, 2.303, or, k=, log 2 =, ´ 0.301, t 1/ 2, t 1/ 2, 0.693, or, t 1/ 2 =, k, Thus, for first order reaction, t 1/ 2 is independent of [ R] 0, and hence, constant., The half-life of a reaction with nth order is given as, 1, t 1/ 2 µ, [ R] n0 - 1, , (a–x), , t1/2, , dx, dt, , log, (a–x), , Half-life of a Reaction, , t1/2, , dx, dt, , x, , (a–x), , Therefore, for zero order reaction t 1/ 2 µ [ R] 0 . For first order, reaction, t 1/ 2 is independent of [ R 0 ] ., 1, For second order reaction t 1/ 2 µ, and so on., [ R0 ], , l, , l, , Pseudo First Order Reaction, The reaction which is not actually of first order but behave, so due to altered conditions is called pseudo first order, reaction. In such a chemical reaction, between two, substances, one reactant is present in large or excess, amounts.e.g., Acid hydrolysis of ethyl acetate, , l, , H+, , CH 3COOC 2H5 + H 2O ¾® CH 3COOH + C 2H5 OH, Rate = k¢ [CH 3COOC 2H5 ] [H 2O], but [H 2O] can be taken, as constant., l, , \ Rate = k[CH 3COOC 2H5 ], where, k = k¢ [H 2O], Inversion of cane sugar, H+, , C12H 22O 11 + H 2O ¾® C 6H12O 6 + C 6H12O 6, Rate = k [C12H 22O 11 ]

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Chapter, Practice, PART 1, Objective Questions, l, , 4. A graph of volume of hydrogen released vs time for, the reaction between zinc and dil. HCl is given in, figure. On the basis of this mark the correct option., V5, V4, , Multiple Choice Questions, , 1. The unit of rate and rate constant is same for which, order of reaction?, , V3, , (a) Zero, (c) Second, , (b) First, (d) Third, , V2, V1, , 2. Which of the following expressions is correct for, the rate of reaction given below?, 5 Br - ( aq) + BrO 3 ( aq) + 6 H + ( aq) ¾® 3 Br 2 ( aq), + 3 H 2O( l ), +, D[Br ], D[H ], (a), =5, Dt, Dt, D[Br- ] 6 D[H + ], (b), =, Dt, 5 Dt, , (c), , D[Br- ] 5 D[H + ], =, Dt, 6 Dt, , (d), , D[Br- ], D[H + ], =6, Dt, Dt, , (NCERT Exemplar), , 20 30, , 40, , 50, , V - V2, (a) Average rate upto 40 s is 3, 40, V3 - V2, (b) Average rate upto 40 s is, 40 - 30, V, (c) Average rate upto 40 s is 3, 40, V3 - V1, (d) Average rate upto 40 s is, 40 - 20, , (NCERT Exemplar), , 5. Contact process is used in the formation of sulphur, , 3. Consider the reaction A ¾® B. The concentration, of both the reactants and the products varies, exponentially with time. Which of the following, figures correctly describes the change in, concentration of reactants and products with time?, (NCERT Exemplar), [B], , [B], , (a), , (b), [A ], , [A ], , Time, , Time, [B], [A ], , (c), , (d), , trioxide,, 2SO 2 ( g ) + O 2 ( g ), 2SO 3 ( g ), The rate of reaction can be expressed as, -D [O 2 ], = 2.5 ´ 10 -4 mol L-1 s -1 ., Dt, Then rate of disappearance of [SO 2 ] will be, , º, , (a) 50.0 ´ 10 -5 mol L-1 s -1 (b) 3.75 ´ 10 -4 mol L-1 s -1, (c) 2.5 ´ 10 -4 mol L-1 s -1 (d) 4.12 ´ 10 -4 mol L-1 s -1, , 6. The rate law for a reaction between the substances, , P and Q is given by rate = k[ P ] a [Q ] b . On doubling, the concentration of P and reducing the, concentration of Q to one-half, the ratio of the new, rate to the earlier rate of the reaction will be, (a), , [B], , [A ], Time, , O, , Time, , 1, (a + b), , 2, (c) ( a - b ), , (b) 2( a - b ), (d) ( a + b )

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26, , CBSE Term II Chemistry XII, , 7. In acidic medium, the rate of reaction between, -, , -, , [BrO 3 ] and Br ions is given as, , (a), (b), (c), (d), , –, , d[BrO 3 ], = k[BrO -3 ][Br - ][H + ]2, dt, It means that,, -, , (a) the change in pH of the solution will not affect the rate, of reaction, (b) rate constant of the reaction is 2s -1, (c) the reaction rate increases by 4 times on doubling the, concentration of H+ ions, (d) All of the above, , 8. Rate law for the reaction, A + 2B ¾® C is found to, be, Rate = k[ A][ B], Concentration of reactant B is doubled, keeping the, concentration of A constant, the value of rate, constant will be.......... ., (NCERT Exemplar), (a) the same, , (b) doubled, , (c) quadrupled, , (d) halved, , (d) r = k¢ [O2 ]-1, , 10. The unit of velocity constant for a zero order, reaction is, (b) L mol –1, (d) mol -1, , 11. For the reaction, I - + OCl - ¾® IO - + Cl - in, aqueous medium, the rate of a reaction is given by, d[IO - ], [I - ][OCl - ], the overall order of the, =k, dt, [OH - ], reaction is, (b) 1, (d) 2, , 12. The rate constant is numerically equal for three, reactions of first, second and third order, respectively. Which of the following is true?, (a) If [ A] > 1 ; r3 > r2 > r1, (b) If [ A] = 1 ; r1 = r2 = r3, (c) If [ A] < 1 ; r1 > r2 > r3, (d) All of the above, , 2, , is, , bimolecular and first order, unimolecular and second order, bimolecular and second order, unimolecular and first order, , about order of a reaction?, , (NCERT Exemplar), , (a) The order of a reaction can be a fractional number., (b) Order of a reaction is experimentally determined, quantity., (c) The order of a reaction is always equal to the sum of, the stoichiometric coefficients of reactants in the, balanced chemical equation for a reaction ., (d) The order of a reaction is the sum of the powers of, molar concentration of the reactants in the rate law, expression., , 15. The half-life period for a zero order reaction is equal to, 0.693, k, , (b), , 2k, [ R]0, , (c), , 2.303, k, , (d), , [ R]0, 2k, , (where, [ R ] 0 is initial concentration of reactant and, k is rate constant.), , 16. The value of rate constant of a pseudo first order, reaction ......... ., , (b) r = k¢ [O3 ][O2 ], (c) r = k¢ [O3 ]2, , (a) -1, (c) 0, , 2, , (All india 20201), , …(i), …(ii), , (a) r = k¢ [O3 ]2 [O2 ]–1, , (a) L s -1, (c) mol L -1 s -1, , º 4N O + O, , 14. Which of the following statements is not correct, , (a), , 9. Consider the following reactions,, O 3 r O 2 + O (fast), O + O 3 ¾® 2O 2 (slow), The rate law expression should be, , 13. The reaction, 2N 2O5, , (NCERT Exemplar), , (a) depends on the concentration of reactants present in, small amount, (b) depends on the concentration of reactants present in, excess, (c) is independent of the concentration of reactants, (d) depends only on temperature, , 17. In a first order reaction, the time taken to complete, half of the reaction, (a) depends upon its initial concentration, (b) is inversely proportional to its initial concentration, (c) does not depend upon its initial concentration, (d) depends upon square root of its initial concentration, , 18. If a reaction obeys the following equation, k=, , 2.303, a, log 10, t, ( a - x), , The order of reaction will be, (a) zero, (c) second, , (b) first, (d) third, , 19. The half-life period of a first order reaction is, 4 min, the time after which 99.9% reaction gets, completed is, (a) 16 min, (c) 32 min, , (b) 8 min, (d) 40 min

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27, , CBSE Term II Chemistry XII, , 23. Assertion D [R ] is multiplied with - 1 to make the, , 20. Which of the following graph is correct for a zero, , (b), , 24. Assertion Zero order reactions are relatively, uncommon but these occur under special conditions., Reason Thermal decomposition of HI on gold, surface is a zero order reaction., , (d), , Concentration, of reactant, , Time, , Reaction rate, , Time, , (c), , rate of the reaction a positive quantity., Reason D[R ] is a negative quantity in the, - D [ R], expression, rate of disappearance of R =, ., Dt, , Concentration, of reactant, , (a), , Reaction rate, , order reaction?, , Time, , 25. Assertion Order and molecularity are same., Reason Order is determined experimentally and, molecularity is the sum of the stoichiometric, coefficient of rate determining elementary step., , Time, , (NCERT Exemplar), , 21. Consider a first order gas phase decomposition, reaction given below, A( g ) ¾® B( g ) + C( g ), The initial pressure of the system before, decomposition of A was p i . After lapse of time ‘t’, total pressure of the system increased by x units, and became ‘p t ’. The rate constant k for the, reaction is given as ......... ., (NCERT Exemplar), (a) k =, , 2.303, pi, log, t, pi - x, , (b) k =, , 2.303, pi, log, t, 2pi - pt, , (c) k =, , 2.303, pi, log, t, 2pi + p t, , (d) k =, , 2.303, pi, log, t, pi + x, , 22. Some reactions alongwith the units of their rate, constants are given below, S.No, , Reaction, , Unit of rate constant, s -1, , 1., , SO2Cl 2 ¾® SO2 + Cl 2, , 2., , NO + O3 ¾® NO2 + O2 L mol -1s -1, , 3., , 2NO + Cl 2 ¾® 2NOCl L2 mol -2s -1, , Which of the following represents a third order reaction?, (a) SO2 Cl 2 ¾® SO2 + Cl 2 (b) NO + O3 ¾® NO2 + O2, (c) 2NO + Cl 2 ¾® 2NOCl (d) None of these, l, , Assertion-Reasoning MCQs, Direction (Q. Nos. 23-27) Each of these questions, contains two statements Assertion (A) and Reason (R)., Each of these questions also has four alternative choices,, any one of which is the correct answer. You have to select, one of the codes (a), (b), (c) and (d) given below., (a) Both A and R are true and R is the correct explanation, of A., (b) Both A and R are true, but R is not the correct, explanation of A., (c) A is true, but R is false., (d) A is false, but R is true., , 26. Assertion Hydrolysis of ethyl acetate in the presence, of acid is a reaction of first order whereas in the, presence of alkali, it is a reaction of second order., Reason Acid acts as catalyst only whereas alkali act, as one of the reaction., 27. Assertion The t1/ 2 of zero order reaction depends, on concentration of reactants., Reason t1/ 2 and rate constant of zero order reaction, [ R], are related to each other as, t1/ 2 =, ×, 2k, l, , Case Based MCQs, , 28. Read the following and answer the questions from, (i) to (iv) given below, The instantaneous rate can be measured by, determination of slope of the tangent at point ‘t’ in, concentration vs time plot. This makes it difficult to, determine the rate law and hence the order of the, reaction., In order to avoid this difficulty, we can integrate, the differential rate equation to give a, relation between directly measured experimental, data, i.e. concentrations at different times and rate, constant., The integrated rate equations are different for the, reactions of different reaction orders. We shall, determine these equations only for zero and first, order chemical reactions., Zero order reaction means that the rate of the, reaction is proportional to zero power of the, concentration of reactants., e.g. Consider the reaction,, d[ R], R¾, ¾® P, Rate =, = k[ R] 0, dt

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28, , CBSE Term II Chemistry XII, , First order reaction means that, the rate of the, reaction is proportional to the first power of the, concentration of reactions R. e.g., d[ R ], For the reaction, R ¾, ¾® P, Rate =, = k[ R ], dt, 1 [ R] 0, and the integrating rate equation is k = ln, ,, t, [ R], where [R] is final concentration and [ R ] 0 is initial, concentration of reactant, respectively., (i) For a hypothetical reaction, R ¾® products;, rate = - k[ R]. The negative sign used in the rate, expression indicates, (a) decrease in the concentration of reactants with time, (b) decrease in the rate with time, (c) reaction is reversible, (d) None of the above, , (ii) For a reaction, P + Q ¾® R + S, The curve which depicts the variation of the, concentration of products is, P, Q, Conc., , Time, , (ii) Which of the following reactions is of zero order?, , (c) R, , ((d) S, , (iii) For the reaction, Hg( l ) + Cl 2 ( g ) ¾® HgCl 2 ( s ), The rate of reaction is given as, D [HgCl 2 ], Dt, D[Cl 2 ], (c) Dt, , (b) -, , (a), , D [Hg], Dt, , What will be the expression for instantaneous rate, of the reaction?, -d[ X ], 2dt, , (c) -, , (b) +, , 1 d[ X 2 Y ], 2 dt, , hn, (b) H2 + Cl 2 ¾¾, ® 2HCl, 1, (c) N2O5 ¾® 2NO2 + O2, 2, (d) 2SO2 + O2 ¾® 2SO3, , (iii) The decomposition of ammonia gas on a hot platinum, surface is, , (d) All of these, , (iv) For the reaction, 2X + Y ¾® X 2 Y, , (a), , (a) CH3Cl + H2O ¾® CH3OH + HCl, , 1 d[ Y ], 2 dt, , (a) zero order reaction, (c) second order reaction, , (a), , (d) None of these, , (b) first order reaction, (d) third order reaction, , (iv) Which of the following graph is correct for zero order, reaction?, , Rate, , ((b) Q, , (iv) given below, The zero order kinetics is an artifact of the conditions, under which the reaction is carried out. Therefore,, reactions that follow zero-order kinetics are considered, as pseudo-zero order reaction., Zero-order systems provide an interesting opportunity, for students to think about the underlying mechanism, behind the physical phenomena being modelled., Thirty-six general chemistry students, five physical, chemistry students, and three chemical engineering, students were asked to think aloud as they responded to, an interview prompt about the half-life of a, catalyst-driven zero-order reaction., Our findings revealed that students often described, zero-order in mathematical terms (i.e. the zero-order rate, law, integrated rate law, and graphical representation),, but lacked a clear understanding of the particulate nature, of zero-order systems. Actually, zero-order reactions are, the curiosities of chemical kinetics., (i) For zero order reaction, the rate constant can be, calculated from, (a) reactant concentration vs rate of reaction graph, (b) product concentration vs rate of reaction group, (c) concentration vs time graph, (d) reaction rate vs concentration graph, , R, S, , (a) P, , 29. Read the following and answer the questions from (i) to, , (b) t1/2, , Or Contact process is used in the formation of sulphur, Time, 3, , The rate of reaction can be expressed as, -D [O 2 ], = 1.5 ´ 10 -4 mol L-1 s -1 ., Dt, Then rate of disappearance of [SO 2 ] will be, -5, , -1, , -1, , (a) 30.0 ´ 10 mol L s, (c) 2.5 ´ 10-4 mol L-1 s -1, , [A ] 0, , º 2SO (g ), -4, , -1, , (c) t1/2, , (d), , Rate, , trioxide,, 2SO 2 ( g ) + O 2 ( g ), , [A ] 0, -1, , (b) 3.75 ´ 10 mol L s, (d) 50.0 ´ 10-4 mol L-1 s -1, , 1/[A]0, , Or The units of rate constant in zero order reaction is, (a) mol L-1 s -1 (b) mol -1 L s -1 (c) s -1, , (d) mol 2L-2s -1

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29, , CBSE Term II Chemistry XII, , 30. Read the following and answer the questions from, (i) to (iv) given below, The rate of a reaction, which may also be called its, velocity or speed, can do defined with relation to, the concentration of any of the reacting substances,, or to that of any product of the reaction., If the species chosen is a reactant which has a, dc, concentration c at time t the rate is , while the, dt, rate with reference to a product have a, dx, concentration x at time t is ., dt, Any concentration units may be used for expression, the rate, thus if mole per litre are expressing the, rate; thus, if moles per litre are employed for, concentration and seconds for the time, the units, for the rate are moles litre - 1 sec - 1 ., For gas reactions pressure units are sometimes used, to place of concentrations, so that units for the rate, would be (mm Hg) sec - 1 and atm sec - 1 ., The order of a reaction concerns the dependence of, the rate upon the concentrations of reaching, substances thus, if the rate is found experimentally, to be proportional to the a th power of the, concentration of one of the reactants A, to the b th, power of the concentration of a second reactant B,, and so forth via.,, rate = kC AaC Bb, , ...(ii), , Such a reaction is said to be of the a th order with, respect to the substance A, the b th order with, respect to B., (i) Which of the following statement is correct?, (a) Rate of reaction is measure of change is concentration, of reactant only with respect to time, (b) Rate of reaction is measure of change in, concentration of reactant and product with respect to, time, (c) Concentration of reactant increases with increase in, time, (d) Both (b) and (c), , (ii) For a reaction,, P + 2Q ¾® Products, The order of reaction depends on the, (a) concentration of products, (b) rate constant, (c) concentration of reactants, (d) speed of the reaction, , (b) (mol s) + n, (d) (mol L)1- n s -1, , (a) (mol L) 1-n, (c) (mol L -1s)1+ n, , (iv) Rate of reaction is also called ........, (a) rate law of reaction, , (b) speed of reaction, , (c) momentum of reaction (d) concentration gradient, , Or For gaseous reaction, the unit of rate constant for, which order reaction is atm -1s -1 ?, (a) Zero order, (c) Second order, , (b) First order, (d) nth, , PART 2, Subjective Questions, l, , Short Answer Type Questions, , 1. (i) Consider the following, ln [ R ] vs time (min) plot., Y, , ln [R], Time (min), , ...(i), , the overall order of the reaction is simply, n=a+b+K, , (iii) The units of rate constant ‘k’ for n th order reaction is, , X, , (a) What is the order of the reaction?, (b) What are the units of rate constant, ( k ) for the, reaction?, , (ii) For the given reaction, N2 (g) + 3H2 (g) ¾® 2NH3 (g), If D [NH3 ] / Dt = 4 ´ 10 -8 mol L-1s -1 ,, , then what is the value of - D [H2 ] / Dt ?, , 2. For a chemical reaction R ¾® P, the variation in the, [ R] 0, concentration of log, vs time ‘ t ’ plot is given, [ R], here., log, , [R]0, [R], , t (s), , For this reaction, (i) What is the order of the reaction?, (ii) What is the unit of rate constant?, (iii) What is the slope of the curve?

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30, , CBSE Term II Chemistry XII, , 3. For a reaction, the rate law expression is represented, as follows Rate =, , 1, k[ A] [ B] 2, , (i) Interpret whether the reaction is elementary or, complex. Give reason to support your answer., (ii) Write the units of rate constant for this reaction,, if concentration of A and B is expressed in moles/L., , 4. The following results have been obtained during the, kinetic studies of the reaction:, P + 2Q ¾® R + 2S, Exp., , 9. The following data were obtained during the first, order thermal decomposition of SO 2 Cl 2 at a, constant volume., SO2 Cl2 (g) ¾® SO2 (g) + Cl2 (g), Experiment, , Time (s) -1, , Total pressure (atm), , I, , 0, , 0.5, , II, , 100, , 0.6, , Calculate the rate constant of the reaction when, total pressure is 0.65 atm., (Delhi 2017) (NCERT), , Initial P, (mol/L), , Initial Q, (mol/L), , Initial rate of, formation of R, (M min -1), , 1, , 0.10, , 0.10, , 3.0 ´ 10- 4, , 2, , 0.30, , 0.30, , 9.0 ´ 10- 4, , How much time will it take to reduce the, initial concentration of the reactant to its 1/16th, (NCERT), value?, , 3, , 0.10, , 0.30, , 3.0 ´ 10- 4, , 11. For the first order thermal decomposition reaction,, , 0.40, , -4, , 4, , 0.20, , 6.0 ´ 10, , 10. The rate constant for a first order reaction is 60 s - 1 ., , Determine the rate law expression for the reaction., 5. The chemical reaction, 2O 3 ¾® 3O 2 proceeds as, follows, (Fast), O3 q O2 + O, (Slow), O + O 3 ¾® 2O 2, What should be the rate law expression?, , 6. In a reaction between A and B, the initial rate of, reaction ( r0 ) was measured for different initial, concentrations of A and B as given below, A/ mol L-1, , 0.20, , -1, , 0.30, , B/ mol L, , -1, , r0 / mol L, , s, , 0.20, , -1, , 5.07 ´ 10, , 5.07 ´ 10, , 1.43 ´ 10, , 7. The reaction between A and B is of first order with, respect to A and of zero order with respect to B. Fill, in the blanks in the following table., [B]/M Initial rate/M min - 1, , [A]/M, , I., , 0.1, , 0.1, , 2.0 ´ 10-2, , II., , —, , 0.2, , 4.0 ´ 10-2, , III., , 0.4, , 0.4, , —, , IV., , —, , 0.2, , 2.0 ´ 10-2, , 0, , 0.30, , 300, , 0.50, , Calculate the rate constant., (Given log 2 = 0.3010, log 3 = 0.4771,, log 4 = 0.6021), , (All India 2016), , 13. The half-life period of a radioactive element is, -4, , What is the order of the reaction with respect to, A and B?, (NCERT), , Experiment, , Total pressure/atm, , 1.15 ´ 10 - 3 s -1 . How long will 5 g of this reactant, take to reduce to 3 g?, , 0.05, -5, , Time/s, , 12. A first order reaction has a rate constant, , 0.40, , 0.10, -5, , the following data obtained, C 2 H 5 Cl(g ) ¾® C 2 H 4 (g ) + HCl(g ), , (NCERT), , 8. Consider a certain reaction, A ¾® products, with, , k = 2.0 ´ 10 - 2 s - 1 . Calculate the concentration of A, remaining after 100 s, if the initial concentration of, A is 1.0 mol L - 1 ., , 140 days. After 650 days, what amount of 1 g of the, element will be reduced ?, , 14. The half-life for radioactive decay of 14 C is 5730 yr., An archaeological artefact contained wood had, only 80% of the 14 C found in a living tree., Estimate the age of the sample., (NCERT), , 15. (i) The decomposition of dimethyl ether leads to the, formation of CH4 ,H2 and CO and the reaction, rate is given by rate = k[CH3 OCH3 ]3/2 ., The rate of reaction is followed by increase in, pressure in a closed vessel, so the rate can also be, expressed in terms of the partial pressure of, dimethyl ether, i.e. rate = k [ pCH 3 OCH 3 ]3/2 ., If the pressure is measured in bar and time in, minutes, then what are the units of rate and rate, constant?, (ii) What is the order of a reaction that is 50%, complete after 2h and 75% complete after 4h?

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31, , CBSE Term II Chemistry XII, , 16. For a first order reaction, show that the time required, for 99% completion is twice the time required for the, completion of 90% of reaction., (NCERT; All India 2017C, Delhi 2013), , 17. A first order reaction is 25% complete in, 40 minutes. Calculate the value of rate, constant. In what time will the reaction be 80%, completed?, (Delhi 2020), , 18. The rate of a first order reaction is 0.04 mol L-1s -1, at 10 min and 0.03 mol L-1s -1 at 20 min after, initiation. Find the half-life of reaction., , 23. Consider the following reaction,, ¾® 2NOCl( g ), 2NO( g ) + Cl2 ( g ) ¾, The rate of reaction becomes doubled when the, concentration of Cl2 is doubled. However, when the, concentration of both the reactants are doubled, the, rate becomes eight times. What is the order with, respect to NO and with respect to reactant, chlorine? What is the overall order of reaction?, , 24. For the decomposition of azoisopropane to hexane, and nitrogen at 543 K, the following data is obtained:, , 19. The conversion of molecules x to y follows second, order kinetics. If concentration of x is increased to, three times, how will it affect the rate of formation, of y ?, (NCERT), l, , Long Answer (LA) Type Questions, 20. For the reaction,, The following data were collected at 270 K., Initial [NO] Initial [Cl 2 ], Exp. No. mol L-1, mol L-1, , Initial rate of, disappearance, of Cl 2 (mol/min), , 1, , 0.15, , 0.15, , 0.60, , 2, , 0.15, , 0.30, , 1.20, , 3, , 0.30, , 0.15, , 2.40, , 4, , 0.25, , 0.25, , x, , (i) Write expression for rate law., (ii) Calculate rate constant and write its unit., (iii) Find initial rate of disappearance of Cl 2 in, experiment number (4)., 21. (i) In a first order reaction, the reactant concentration, decreases from 0.8 M to 0.4 M in 15 min. What is, the time taken for the concentration to change from, 0.1 M to 0.025 M?, (ii) A first order reaction takes 100 min for completion, of 60% of the reaction. Find the time when 90% of, the reaction will be completed., 22. (i) For a reaction A + B ¾® P, the rate is given by, Rate = k[ A] [ B] 2, (a) How is the rate of reaction affected, if the, concentration of B is quadrupled?, (b) What is the overall order of reaction, if A is, present in large excess ?, (ii) A first order reaction takes 23.1 min for 50%, completion. Calculate the time required for 75%, completion of the reaction., , p(mm of Hg), , 0, , 35.0, , 360, , 54.0, , 720, , 63.0, , Calculate the rate constant., , (NCERT), , 25. (i) Following rate data were obtained at 300 K for the, reaction, , 2NO(g) + Cl2 (g) ¾® 2NOCl(g), , t (s), , 2A + B ¾® C + D, , Initial rate of, Exp., [A] mol L -1 [B] mol L -1 formation of [C], No., mol L -1 min -1, 1, , 0.3, , 0.2, , 7.2 ´ 10-2, , 2, , 0.3, , 0.4, , 2.88 ´ 10-1, , 3, , 0.4, , 0.1, , 2.4 ´ 10-2, , 4, , 0.1, , 0.1, , 6.0 ´ 10-3, , If the concentration of A = 0.5 mol L -1 and, B = 0.2 mol L -1 , then calculate the rate of formation, of D., (ii) Decomposition of NH3 on platinum surface is zero, order reaction. What are the rates of production of, N 2 and H2 , if k = 2.5 ´ 104 Ms -1 ., , 26. (i) The decomposition of phosphine, PH3 proceeds, according to the following equation, , 4PH3 ( g ) ¾® P4 ( g ) + 6H2 ( g ), It is found, that the reaction follows the following, rate equation, Rate = k(PH3 ), Half-life of PH3 is 37.9 s at 120°C., (a) How much time is required for 3/4th of PH 3 to, decompose?, (b) What fraction of the original sample of PH 3, remains behind after 1 minute?, (ii) Consider a certain reaction,, A ® products with k = 2.0 ´ 10 -2 s -1 ., Calculate the concentration of A remaining after, 100 s, if the initial concentration of A is 1.0 mol L -1 .

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32, , CBSE Term II Chemistry XII, , 27. During nuclear explosion, one of the products is, 90, , 90, , Sr with half-life of 28.1 yr. If 1 m g of Sr was, absorbed in the bones of a newly born baby instead, of calcium, how much of it will remain after 10 yr, and 60 yr, if it is not lost metabolically?, (NCERT), l, , Case Based Questions, , 28. Read the following and answer the questions from, (i) to (iv) given below, The rate of a chemical reaction at a given, temperature may depend on the concentration of, one or more reactants and sometimes even on the, concentration of products or some foreign, substances. The representation of rate of a reaction, in terms of the concentration of the reactants is, called rate law., The rate law for a given reaction has to be, established by experimental study of the rate of, reaction over a wide range of concentration of the, reactants and products. The rate law thus, established is also called differential rate equation, or rate expression., (i) The reaction 2A + B + C ¾® 2E is found to be first, order in A , second order in B and zero order in C,, then write the rate expression., (ii) Express the rate of the following reaction, H 2 + I2, 2HI, (iii) Why does the rate of a reaction not remain constant, throughout the reaction process?, (iv) The reaction A + B ¾® C has zero order. Write, rate equation., Or, Is there any reaction for which reaction rate does, not decrease with time?, , -, , 29. The sum of the exponents of the concentration of, the reactants in the rate law expression is called the, order of that chemical reaction., For a general reaction,, aA + bB ¾® cC + dD ; Rate = k [ A] x [ B] y, , Here, x and y, calculated experimentally, indicate, how sensitive the rate is to the change in, concentrations of A and B. Sum of these exponents,, i.e. x + y gives the overall order of a reaction, whereas, x and y represent the order with respect, to the reactants A and B, respectively., Order of a reaction can be 0, 1, 2, 3 and even a, fraction., Molecularity is an another property of a reaction,, which helps in understanding its mechanism. The, number of reacting species (atoms, ions or, molecules) taking part in an elementary reaction,, which must collide simultaneously in order to bring, about a chemical reaction is called molecularity of, the reaction., e.g. If a reaction involves the decomposition of only, a single species, the molecularity is one and it is, called unimolecular reaction., Molecularity has no meaning for complex reactions, as these reactions are supposed to take place in a, sequence of a number of steps. It can be, determined only for the elementary reactions., KClO 3 + 6FeSO 4 + 3H 2SO 4 ¾®, KCl + 3Fe 2 (SO 4 ) 3 + 3H 2O, This reaction seems to be a tenth order reaction but, actually it is a second order reaction. This indicates, that the reaction takes place in several steps., (i) Find the order of the reaction whose rate constant, is 2.5 ´ 10 -2 min -1 ., (ii) For which order of reaction, the units of rate, constant and rate of the reaction are same?, (iii) For the given rate expression = k[ A] 3/ 2 [ B] -1 , the, overall order of a reaction ?, (iv) What will be the order of the reaction given, below?, 2N2O5, , º 2N O + O, 2, , 4, , 2, , Or, , A reaction is second order with respect to a, reaction. How is the rate of reaction affected if the, concentration of the reactant is doubled ?

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EXPLANATIONS, Objective Questions, 1. (a) Rate =, , 7. (c) Since, the rate law given is, - d[ BrO3 ] = k[ BrO-3 ] [ Br - ] [ H+ ] 2, dt, Thus, the reaction rate increases by 4 times on doubling the, concentration of H+ ions. Moreover, since it contains [ H+ ], term, this means that, rate changes with variation in the, value of pH., 8. (b) Rate law for the given reaction, A + 2B ¾® C, , - d[ A ] M, =, = mol L-1 s -1, dt, t, , k = (Conc.)1 - n s -1, If n = 0,, k = (mol L–1 )1 - 0 , s -1 = mol L-1 s -1, Rate = k for zero order reaction., 2. (c) Given, chemical reaction is, 5 Br - ( aq ) + BrO 3- ( aq ) + 6H+ ( aq ) ¾® 3Br2 ( aq ) + 3H2O( l ), Rate law expression for the above equation can be written as, 1 D[Br - ], D[BrO 3- ] -1 D[H+ ] +1 D[Br2 ], =, ==, 5 Dt, 3 Dt, Dt, 6, Dt, D[Br - ], D[BrO 3- ] -5 D[ H+ ], Þ ==, Dt, Dt, 6, Dt, D[Br - ] 5 D[ H+ ], Þ, =, Dt, 6 Dt, -, , Concentration of reactants and products varies exponentially, w.r.t time., (i) Concentration of reactant (here, A) decreases, exponentially w.r.t time., (ii) Concentration of product (here, B) increases, exponentially w.r.t time. Correct graph representing the, above reaction is (b)., 4. (c) Zn + Dil. HCl ¾® ZnCl 2 + H2 ­, Change in concentration of H2, Average rate of reaction =, Change in time, V3 - 0 V3, =, =, 40 - 0 40, , b, , [ Q] b, é Qù, \ New rate = k[ 2P ] a ê ú = k × 2a[ P ] a × b, 2, ë 2û, (a - b), =2, × k[ P ] a[ Q] b = 2( a - b ) × rate, New rate, Thus,, = 2( a - b ), Rate, , R 2 = k [ A ] [ 2B ] = 2k [ A ] [ B ] = 2 R1, Therefore, as concentration of B is doubled keeping the, concentration of A constant, rate of reaction doubles., 9. (a) Given,, …(i), O3 r O2 + O ( fast ), …(ii), O + O3 ¾® 2O2 (slow), From Eq. (i), K eq =, , 3. (b) In the reaction, A ¾® B, , 5. (a) For the reaction,, 2SO2 ( g ) + O2 ( g ) r 2SO3( g ), the rate of reaction can be written as, - D[SO2 ], æ D[O2 ]ö, = 2 ç÷, Dt, Dt ø, è, - D[ O2 ], Q, = 25, . ´ 10-4 mol L -1 s -1, Dt, - D[SO2 ], \, = 2 ´ 2.5 ´ 10-4 mol L-1 s -1, Dt, = 5.0 ´ 10-4 mol L-1 s -1, = 50.0 ´ 10-5 mol L-1 s -1, a, 6. (b)Q Rate = k[ P ] [ Q] b, , can be written as, Rate = k[ A ][ B ], Rate of reaction with respect to B is of first order., R1 = k [ A ] [ B ], When concentration of reactant ‘B’ is doubled then rate ( R 2 )., , or, , [O ] =, , [O2 ][O], [O3 ], K eq [ O 3 ], [O 2 ], , From Eq. (ii),, Rate = k[O3 ][O], k[O3 ][O3 ], or, r=, K eq, [O2 ], or, , r = k¢ [ O3 ] 2 [ O2 ] -1 [ k' = k ´ K eq ], , 10. (c) Unit of rate constant = (mol L–1 )1- n s -1, For zero order reaction, n = 0, \ Unit of rate constant = ( mol L-1 )1- 0 = mol L-1 s -1, 11. (b) The rate of reaction is given by,, d[ IO- ] k[ I- ] [ OCl - ], =, ×, dt, [ OH- ], Therefore, overall order of reaction = 1 + 1 - 1 = 1, 12. (d) Consider a reaction A ¾® Products, , (Given), , Let r1, r2 and r3 are the rate for first, second and third order, respectively., …(i), \, r1 = k[ A ]1 for 1st order, r2 = k[ A ] 2 for 2nd order, 3, , r3 = k[ A ] for 3rd order, Q k is same in all the three cases., Therefore, If [ A ] = 1, r1 = r 2 = r 3, If [ A] < 1, r1 > r 2 > r 3, If [ A ] > 1, r 3 > r 2 > r1, 13. (a) The given reaction,, i.e., , 2N2O5 r 2N2O4 + O2; Rate = k[N 2O5 ] 2, , …(ii), …(iii)

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34, , CBSE Term II Chemistry XII, , Slope (m) = - k (rate constant), , The reaction occurs in two steps, Step I, , Intercept (c) = [ R ] 0 (initial concentration), , Slow, , N 2O5 ¾¾® NO 2 + NO 3, , Thus, the graph between conc. of reactant and time is given, as,, , Fast, , Step II N 2O5 + NO 2 ¾¾® 3NO 2 + O 2, As the reaction, hence bimolecular reaction and the rate, determining step is the slowest step which is of first order., 14. (c) Out of the given four statements, option (c) is not correct., Order of reaction is equal to the sum of powers of, concentration of the reactants in rate law expression., For any chemical reaction, x A + yB ¾® Product, Rate = k [ A ] x[ B ] y, , Moreover, on rearranging Eq. (i), , Order = x + y, Order of reaction can be a fraction also. Order of reaction is, not always equal to sum of the stoichiometric coefficients of, reactants in the balanced chemical equation. For a reaction,, it may or may not be equal to sum of stoichiometric, coefficient, of reactants., [ R ]0, 15. (d) For a zero order reaction, t1/ 2 =, ., 2k, , [ R ] - [ R ]0, = -k, t, [ R ] - [ R ]0, = - kt 0, t, Rate µ t 0, Therefore, the graph between reaction rate and time is, given as,, , 18. (b) For a first order reaction, the integrated rate equation is, written as,, 2.303, a, k=, log, t, a-x, 19. (d) Given,, , t 1/ 2 = 4 min, , a = 100; a - x = 01, ., 0.693, [For first order reaction], Now, t1/ 2 =, k, 0.693, ...(i), \, k=, 4, As we know that,, 2.303, a, k=, log, t, a-x, 2.303 ´ 4, 100, [From Eq. (i)], t=, log, 0.693, 01, ., t = 39.88 min » 40 min, 20. (d) For a zero order reaction,, [ R ] = ( - k) t + [ R ] 0, On comparing with equation of straight line;, y = mx + c, y = [ R ] concentration, x = t (time), , …(i), , Concentration, of reactant, , Slope = – k, , Time, , Reaction rate, , 16. (a) Pseudo first order reaction is a chemical reaction in which, rate of reaction depends upon concentration of one reactant, (present in small amount) while concentration of another, reactant has no effect on rate of reaction (present in excess)., Hence, (a) is the correct choice., 17. (c) Half-life period of a first order reaction is, 0. 693, t1/ 2 =, k, Equation of half-life period of a first order reaction does not, involve any concentration term. Thus, it is independent of, initial concentration of reactants., , [R]0, , Time, , A( g ) ¾® B( g ) + C( g ), , 21. (b), Initially, At time, t, , pi, pi - x, , 0, x, , 0, x, , p t = pi - x + x + x, = pi + x, For first order reaction x = p t - pi, 2.303, pi, k=, log, t, pi - x, 2.303, pi, =, log, t, pi - ( p t - p i), 2.303, pi, =, log, t, 2p i - p t, 22. (c) The reaction 2NO + Cl 2 ¾® 2NOCl represents a third, order reaction as its unit of rate constant is L 2 mol - 2 s -1., 23. (a) Both A and R are true and R is the correct explanation of A., D[ R ], Dt, Since, D[ R ] is a negative quantity, it is multiplied with -1 to, make the reaction rate a positive quantity., Rate of disappearance of R = -, , 24. (b) Both A and R are true but R is not the correct explanation, of A., Zero order reactions are not so common and they occur, under special conditions, like some enzyme catalysed, reactions and reactions which occur on metal surfaces are a, few examples of zero order reactions. Moreover, thermal, decomposition of HI on gold surface is also zero order, reaction.

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35, , CBSE Term II Chemistry XII, , 25. (d) A is false, but R is true., Order and molecularity may or may not be same as the order, of reaction is sum of power of reactant which can be, determined experimentally. But molecularity is sum of, stoichiometric coefficient of rate determining elementary, step., 26. (a) Both A and R are true and R is the correct explanation of A., Hydrolysis of ester in the presence of acid is pseudo first, order reaction., 27. (a) Both A and R are true and R is the correct explanation of A., t1/ 2 of zero order reaction is, t1/ 2 = [ R ] / 2k., Therefore, t1/ 2 of zero order reaction depends on, concentration of reactants., 28. (i) (a) For a hypothetical reaction, R ® products;, rate = - k[ R ]. The negative sign used in the rate, expression indicates decrease in the concentration of, reactants with time., (ii) (a) The curve P tells, that the gradual increase in, concentration with the passage of time and, concentration of product increases in the course of, reaction., (iii) (d) Rate of reaction, D[Hg], D[ Cl 2 ] D[HgCl 2 ], ==, Dt, Dt, Dt, (iv) (a) Complete expression for instantaneous rate of, reaction is as follows, d[ X ] - d [ Y ], d[ X 2 Y ], =, =+, 2dt, dt, dt, =-, , Or, - D[ SO 2 ], æ D[ O 2 ] ö, = 2ç÷, Dt, Dt ø, è, - D[SO 2 ], \, = 2 ´ 1.5 ´ 10-4 mol L -1 s -1, Dt, = 3.0 ´ 10-4 mol L -1 s -1, (a), , = 30.0 ´ 10-5 mol L -1 s -1, 29. (i) (c) In case of zero order reaction, the slope of the, concentration vs time graph is equal to negative of the, reaction constant., (ii) (b) Among the given reactions, the reaction between, hydrogen and chlorine under influence of UV-light is an, example of zero-order reaction., hn, H2 + Cl 2 ¾¾, ® 2HCl, , (iii) (a) The decomposition of NH3 gas on a hot platinum, surface is zero order reaction because the rate of this, reaction is independent on the concentration of, ammonia., , Thus, a plot of t1/ 2 vs [ A ] 0 will give a straight line graph, starting from the origin, i.e., , t1/2, , [A] 0, , Or, (a) For a zero order reaction, units of k = molL-1s -1., 30., , (i) (b) Rate of reaction is measure of change in, concentration of reactant and product with respect to, time. As concentration of product increases with, increase in time, concentration of reactant deceases., For reaction R ® P, - D[ R ], D[ P ], Rate =, =+, Dt, Dt, (ii) (c) Order of reaction depends on the concentration of, reacting substances. Thus, if the rate of reaction found, experimentally proportional to the a th power of, concentration of one of the reactant A, to the b th power, of the concentration of a second reactant B. Then,, Rate = kC AaC Bb, Overall order of the reaction is simply, n=a+b, (iii) (d) The unit of k depends on order of reaction and it is, (mol L -1)1-n s -1, where n = order of reaction., (iv) (b) Rate of reaction can be different for every reaction., Rate of reaction is also known as speed of reaction or, velocity of reaction., Or, (c) For second order reaction, the unit of k is atm -1s -1 in, gaseous medium., , Subjective Questions, 1. (i) (a) The given plot shows the first order reaction because, according to integrated rate law of first order reaction,, ln[ R ] = ln[ R ] 0 - kt, where, slope is -k which is represented in graph., (b) Unit of k = (mol L- 1 )1 - n s - 1, = (mol L- 1 )1 - 1 s - 1 = s -1, (ii) N 2 ( g ) + 3H2 ( g ) ¾® 2NH3( g ), D[N 2 ], 1 D[H2 ], =Dt, 3 Dt, 1 D[NH3 ], =+, 2, Dt, , Rate of reaction = -, , K, 2NH3 ( g ) ¾1130, ¾¾, ® N2 ( g ) + 3H2 ( g ), Pt catalyst, , Rate = k[ NH3 ], , \, , 0, , Hence, this reaction is a zero-order reaction., (iv) (b) For zero order reaction, t1/ 2 =, , [ A] 0, 2k, , -, , 1 D[H2 ], 1 D[NH3 ], =+, 3 Dt, 2, Dt, - D [H2 ] 3, = ´ 4 ´ 10- 8, Dt, 2, = 6 ´ 10-8 mol L-1s -1

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36, , CBSE Term II Chemistry XII, , On substituting the above value of [ O ] in Eq. (i),, [ O ] [ O3 ], r = k × K eq × 3, [O2 ], , 2. (i), log, , [R]0, [R], , =, t (s), , where, k¢ = k × K eq, 6. Rate law states that rate = k [ A ] x[ B ] y, , [ R ]0, kt, It is a first order reaction, log, =, [ R], 2.303, , (Rate)1 = k [0.20]x [ 0.30]y = 5.07 ´ 10 -5, , -1, , (ii) Unit of rate constant k is s ., k, (iii) Slope will be =, 2.303, 3. (i) For an elementary reaction, order of reaction must be, equal to the molecularity but molecularity should be, integral., For the given rate law, order of reaction comes out to, 3é, 1ù, be ê1 + ú . As molecularity cannot be fractional,, 2ë, 2û, therefore, order is not equal to molecularity. Hence, the, given rate law do not belong to an elementary reaction., (ii) Units of rate constant are mol -1/ 2 L1/ 2 s -1., 4. For reaction, P + 2Q ¾® R + 2S, Let the rate law expression be, Rate = k[ P ] X [ Q] Y, ...(i), , For experiment 2,, Rate 2 = 9.0 ´ 10- 4 = k( 0.30) x( 0.30) y, , ...(ii), , For experiment 3,, Rate 3 = 3.0 ´ 10- 4 = k( 010, . ) x( 0.30) y, , ...(iii), , Equating (i) and (iii),, y, , y, , O + O 3 ¾® 2O 2, , [ O] =, , K eq [ O 3 ], [ O2 ], , On dividing Eq. (i) by Eq. (ii), we get, (Rate)1, k [ 0.20] x[ 0.30] y 5.07 ´10 -5, =, =, =1, (Rate) 2 k [ 0.20] x[ 010, . ] y 5.07 ´10 -5, [ 3] y = [ 3] 0, y=0, Now, on dividing Eq. (iii) by Eq. (ii), we get, (Rate) 3 k ( 0.40) x( 0.05 ) 0 1.43 ´ 10 - 4, =, =, (Rate) 2, k ( 0.20) x( 010, . ) 0 5.07 ´ 10 - 5, 0, , æ1 ö, 2.8205 = 2x ç ÷ = 2x, è 2ø, log 2 . 8205 = x log 2, 0.4503, = 1.5, 0.3010, Thus, order of the reaction with respect to A and B are, 1.5 and 0, respectively., 7. The rate law is,, Rate = k[ A ]1[ B ] 0 = k[ A ], Since, the reaction is of zero order w.r.t. B, hence any change, in the concentration of B has no effect on the rate of, reaction., For experiment I,, 2.0 ´ 10- 2 M min -1 = k ´ (0.1 M)1, k=, , 4.0 ´ 10-2 M min -1, = 0.2 M, 0.2 min -1, For experiment III,, Rate = 0.2 min -1 ´ (0.4 M)1, = 0.08 M min -1, For experiment IV,, [ A] =, , (fast), (slow), …(i), , Fast step is in equilibrium, hence the concentration of [O] is, calculated from this step., [ O 2 ][ O ], K eq =, [ O3], \, , …(iii), , 2.0 ´ 10- 2 M min -1, = 0.2 min -1, 0.1 M, For experiment II,, 4.0 ´ 10-2 M min -1 = 0.2 min -1 ´[ A ], , As slowest step is rate determining step, rate of the reaction is, r = k[ O ][ O 3 ], , …(ii), , (Rate) 3 = k [0.40]x[0.05]y = 1.43 ´ 10 - 4, , y, , (Rate) 2 = k [0.20] [ 010, . ] = 5.07 ´ 10, , So,, , So, x = 1, Rate law = k[ P ], \, 5. The given reaction is, 2O 3 ¾® 3O 2, The reaction proceeds as follows, O3 ® O2 + O, , …(i), , -5, , x, , x=, , For experiment 1,, Rate 1 = 3.0 ´ 10- 4 = k( 010, . ) x( 010, . )y, , Rate (1) æ 1 ö, æ1 ö, =ç ÷ Þ1 = ç ÷, Rate (3) è 3 ø, è 3ø, Rate (2), So, y= 0, and, = (3) x Þ 3 = ( 3) x, Rate (3), , k ¢[ O 3 ] 2, = k ¢[ O 3 ] 2 [ O 2 ] - 1, [ O2 ], , 2.0 ´ 10-2 M min -1 = 0.2 min -1 ´ [A], 2.0 ´ 10-2 M min -1, or, [A] =, = 0.1 M, 0.2 min -1, 8. k = 2.0 ´ 10- 2 s- 1; [ A ] 0 = 1.0 mol L- 1; t = 100 s, 2.303, [ A] 0, k=, log, t, [ A]

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37, , CBSE Term II Chemistry XII, , Therefore,, , 2.303, ´ 4 ´ 0 . 3010, 60 s -1, Time ( t ) = 4.62 ´ 10-2 s, =, , -1, , 2.303, 1.0 mol L, log, 100 s, [ A], 2.303, -2 -1, 2.0 ´ 10 s =, (log 1 - log [ A ]), 100 s, 2.303, 2.0 ´ 10 - 2 s - 1 =, ( 0 - log [ A ]), 100 s, 2.0 ´ 10- 2 s - 1 ´ 100 s, - log [ A ] =, 2.303, æ 2.0 ´ 10- 2 s -1 ´ 100 s ö, ÷, or, [ A ] = antilog çç ÷, 2.303, è, ø, 2.0 ´ 10- 2 s - 1 =, , or, , 11. C 2H5 Cl ( g ) ¾® C 2H4 ( g ) + HCl ( g ), Initial pressure at t = 0 is 0.30 atm, After 300s, ptotal = 0.30 - p + p + p atm, = 0.30 + p atm, ptotal (given) = 0.50 = 0.30 + p, Þ, p = 0.50 - 0.30 = 0.20 atm, Pressure of C 2H5 Cl ( g ) after 300 sec, ( pC2 H5 Cl ) = 0.30 - 0.20 atm = 010, . atm, , [ A ] = 0.135 mol L- 1, , k=, , SO 2Cl 2 ( g ) ¾® SO 2 ( g ) + Cl 2 ( g ), , 9., Initially, After time t,, , pi, , 0, , 0, , pi – p, , p, , p, , 2 .303, 2 .303, log 3 =, ´ 0.4771 = 3.66 ´ 10-3s -1, 300, 300, 2.303, [ R ]0, 12. For a first order reaction, t =, log, k, [ R], 2.303, (5 g), =, ´ log, (3 g), (1.15 ´ 10- 3 s- 1 ), 2.303, =, (log 5 - log 3), (1.15 ´ 10- 3 s- 1 ), 2.303, =, ( 0.6990 - 0.4771), (1.15 ´ 10- 3 s- 1 ), =, , Total pressure after time t, i.e., p t = pi – p + p + p = pi + p, So,, a = pi, a - x = pi - ( pt - pi ), = pi – pt + pi = 2pi - p t, Substitutions of the value of a and ( a - x ) in first order, integrated rate equation gives., 2.303, pi, k=, log, t, ( 2pi – p t ), , = 2.0 ´ 103 ´ 0.2219 s = 443.8 s, , Calculation of rate constant, k, Given, pi = 0.5 atm; pt = 0.6 atm, k=, , ( pC2 H5 Cl ) 0, 2 .303, 2 .303, 0.30 atm, log, =, log, ( pC2 H5 Cl ) 300, t, 300, 010, . atm, , time = 444 s, 0.693 0.693, k=, =, t1/ 2, 140, , or, 13., , 2.303, pi, log, t, (2pi – pt ), , t=, , \, , 2.303, 0.5 atm 2.303, =, log, =, log 1.25, (100 s), 0.4 atm (100 s), 2.303, =, ´ 0.0969 = 2.23 ´ 10–3s –1, (100 s), , Þ log, , Calculation of reaction rate when total pressure is 0.65 atm., p SO2 Cl 2 = 0.50 - (0.65 - 0.50), = (1 - 0.65) = 0.35 atm, , Rate = 7.8 ´ 10– 4 atm s –1, 10. For first order reaction,, 2.303, a, t=, log, k, ( a - x), a, -1, Given, ( a - x ) =, ; k = 60 s, 16, On putting the values in Eq. (i), , Þ, , 2.303, a ´ 16 2.303, log, =, log 16, a, 60 s- 1, 60 s- 1, 2.303, 2.303, =, ´ log 24 =, ´ 4 log 2, -1, 60s, 60s -1, , t=, , Þ, , …(i), , [ A] 0, t ´ k 650 ´ 0.693, =, =, [ A], 2.303 2.303 ´ 140, , [ A] 0, = 1.397, [ A], [ A] 0, = antilog 1.397, [ A], [ A] 0, = 24.94 `, [ A], 1, [ A] =, 24.94, [ A ] = 0.04 g `, µ 5730 yr, , log, , k = 2.23 ´ 10-3s -1, Rate = k ´ p SO2 Cl 2, , = (2.23 ´ 10-3s -1 ) ´ (0.35 atm), , 2.303, [ A] 0, log, k, [ A], , Þ, 14. t1/ 2, , If a is the initial 14 C activity in a living tree, then, 80, Activity in the dead wood, a t =, ´a, 100, The nuclear l radioactive decays follow the first order, kinetics. So,, 2.303, a, t=, log, k, at, 0.3693, and, k=, t1/ 2

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38, , CBSE Term II Chemistry XII, , K = 0.0072 min -1, 2.303, 100, t=, log, 0.0072, 20, = 223.6 min, , 2.303, a, log, 80, 0.693 / t1/ 2, a, 100, 2.303 ´ 5730, 100, =, log, 0.693, 80, So,, t = 1845.4 yr, 15. (i) Given that, rate = k[CH3OCH3 ] 3/ 2, In this problem, pressure is in bar and time is in, minutes. So, the unit of rate = bar min -1, Also, unit of k = unit of rate/unit of [CH3OCH3 ] 3/ 2, , Therefore, t =, , 18. For a first order reaction, A ¾® Products,, For concentration of the reactant at two different times, the, rate constant is, 2.303, [ A ]1, k=, log, t 2 - t1, [ A] 2, , \ Unit of k = bar min -1 / bar 3/ 2 = bar -1/ 2 min -1., (ii) It is a first order reaction because for 75% completion of, reaction two half-lives are required (as, t1/ 2 = 2 h), which, suggests that t1/ 2 is independent of initial concentration., -8, , 16. Case I, , -1, , But as, We have,, , rate = k [ A ],, (Rate)1 [ A ]1, =, (Rate) 2 [ A ] 2, 2.303, (Rate)1, log, t 2 - t1, (Rate) 2, 2.303, 0.04, =, log, ( 20 - 10) min, 0.03, k=, , Hence,, , -1, , = 6 ´ 10 mol L s, If a = 100 , ( a - x ) = ( 100 - 99) = 1, , For 99% completion of the reaction,, 2.303, 100, t 99% =, log, k, 1, 2.303, 2.303 ´ 2, =, log 102 =, k, k, 4.606, t 99% =, k, , = 2.88 ´ 10-2 min -1, 0.693, 0.693, t1/ 2 =, =, k, 2.88 ´ 10- 2 min - 1, = 24.06 min or 1443.6 s, …(i), , Reaction rate, ( r ) = k [ x ] 2 (for II order), , Case II If a = 100, ( a - x ) = ( 100 - 90) = 10, For 90% completion of the reaction,, 2. 303, 100, t 90% =, log, k, 10, 2. 303, 2. 303, =, log10 =, k, k, On dividing Eq. (i) by Eq. (ii), we get, , ...(i), , If the concentration of x is increased three times, then, Reaction rate, ( r ¢) = k [ 3x ] 2 = k ´ [ 9x 2 ], …(ii), , ...(ii), , t 99% 4. 606, k, =, ´, =2, t 90%, k, 2. 303, It means that time required for 99% completion of the, reaction is twice the time required to complete 90% of the, reaction., 17. For first order reaction,, 2.303, a, k=, log, t, a-x, The given first order reaction completes 25% in 40 min., Let,, a = 100, a - x = 100 - 25, a - x = 75, 2.303, 100, \, k=, log, 40, 75, 2.303, =, (log 4 - log 3), 40, k = 0.0072 min -1, For 80% completion of reaction,, a = 100, a - x = 100 - 80 = 20, , 19. For the reaction, x ® y, , On dividing Eq. (ii) by Eq. (i), we get, r ¢ k ´ [ 9x 2 ], =, =9, r, k ´ [ x2 ], It means that the rate of formation of y will increase by nine, times., 20. For the reaction,, , 2NO( g ) + Cl 2 ( g ) ¾® 2NOCl( g ), r (rate) = k[NO ]x × [Cl 2 ] y, (i) Comparing experiment number (1) and (2), r1 = k[015, . ]x × [ 015, . ] y = 0.60, x, , y, , r2 = k[ 015, . ] [ 0.30] = 1 . 20, , …(i), …(ii), , On dividing Eq. (ii) by Eq. (i),, r2 k[ 015, . ] x × [ 0.30] y 1 .20, =, =, 0.60, r1, k[ 015, . ] x[ 015, . ]y, Þ, , 2y = 21, , [On comparing the powers], \, y=1, Comparing experiment number (1) and (3), …(i), r1 = k[ 015, . ] x × [ 015, . ] y = 0.60, r3 = k[ 0.30] x × [ 015, . ] y = 2.40, On dividing Eq. (iii) by Eq. (i),, r3 k[ 0.30] x × [ 015, . ] x 2.40, =, =, x, r1 k[ 015, . ] × [ 015, . ] y 0.60, 2x = 22, , …(iii)

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39, , CBSE Term II Chemistry XII, , [On comparing the powers], \, x=2, Hence, rate law, r = k × [NO] 2 × [Cl 2 ]1, , =, , 0.693, 2.303, =, ´ 0.6020, 231, . min, t, 2.303 ´ 0.6020, t=, ´ 231, . min Þ 46.2 min, 0.693, , (ii) From rate law,, Rate, 0.60, k=, =, [NO] 2 × [Cl 2 ] ( 015, . ) 2 ( 015, . ), k = 177.77 mol -2 L 2 min -1, (iii) Initial rate of disappearance of Cl 2 in experiment number, four., r = k ×[0.25] 2 [0.25], = 177.77( 0. 25 ) 2 ( 0. 25 ), r = 2.78 mol/min, 21. (i ) Let, a = 0.8 M, a - x = 0.4 M, t = 15 min, For first order reaction,, 2.303, a, k=, log, t, a-x, 2.303, 0.8, =, log, = 0.462 min -1, 15, 0.4, \ Time taken for 0.1 M to 0.025 M can be calculated as, 2.303, a, t=, log, k, a-x, 2.303, 01, ., =, log, = 30 min, 0.0462, 0.025, (ii ) If a first order reaction is 60% complete in 100 min., Let, [ x ] 0 = 100 M, [ x ] 60% = 100 - 60 = 40 M, and t = 100 min, 2.303, x, 2.303, 100, k=, log 0 =, log, t, x, 100, 40, = 0.02303 log 25, ., = 0.02303 ´ 0.397 = 0.009 min -1, Time taken to complete 90% reaction = ?, [ x ] 0 = 100 M, [ x ] 90% = 100 - 90 = 10 M, k = 0.009 min -1, 2.303, x, 2.303, 100, t=, log 0 =, log, k, x, 0.009, 10, = 255.8 log10 = 255.8 min, 22. (i) Rate = k[ A ] [ B ] 2, (a) When the concentration of B is quadrupled, then, Rate = k[ A ] [ 4( B )] 2, = k[ A ] 16 × [ B ] 2 = 16 k [ A ] [ B ] 2, The reaction rate increases sixteen times., (b) If A is present in large excess, its concentration remains, almost unchanged., Hence, order with respect to A can be taken as zero., Under these conditions, the overall order of the, reaction is two w.r.t B., 0.693, (ii) For a first order reaction, t1/ 2 =, = 231, . min, k, For 75% completion of the reaction, we can write, k=, , 2.303, [ A] 0, log, t, [ A] t, , 2.303, [ A] 0, 2.303, log, =, log 4, t, [ A] 0 / 4, t, , 23. 2NO( g ) + Cl 2( g ) ¾, ¾® 2NOCl ( g ), Let the rate of reaction be, r1 = k[ NO] a [ Cl 2 ] b, , ...(i), , a, , r2 = 2r1 = k[ NO] [ 2Cl 2 ], a, , b, , r3 = 8r1 = k[ 2NO] [ 2Cl 2 ], , ...(ii), b, , ...(iii), , On dividing Eq. (ii) by Eq. (i), we get, r2 2r1 k[ NO] a [ 2Cl 2 ] b, =, =, r1, r1, k[ NO] a [ Cl 2 ] b, ( 2)1 = ( 2) b Þ b = 1, , \, , Similarly, on dividing Eq. (iii) by Eq. (i), we get, r3 8r1 k[ 2NO] a [ 2Cl 2 ] b, =, =, r1, r1, k[ NO] a [ Cl 2 ] b, 8 = ( 2) a ( 2) b, 8 = ( 2) a ( 2)1 ; 4 = ( 2) a, ( 2) 2 = ( 2) a ; a = 2, r = k[ NO] 2 [ Cl 2 ]1, , \, , As, order with respect to NO = 2, Order with respect to Cl 2 = 1, Overall order of reaction = 2 + 1 = 3., 24. Let the pressure of a azoisopropane decreases by x atm., As 1 mole of azoisopropane decompose to give 1 mole of N2, and 1 mole of hexane (C6H14 ), the pressure of N2 and hexane, increases by x atm., (CH3 )2 CHN == NCH(CH3 )2 (g ) ¾® N2 (g ) + C6H14(g ), Azoisopropane, , At t = 0, , 35.0 atm, , 0 atm, , 0 atm, , At time, t, , ( 35.0 - x ) atm, , x atm, , x atm, , Þ, \, , pt = pazoisopropane+ pN 2 + pC6 H14, pt = 35.0 - x + x + x = 35 + x, x = pt - 35, p azoisopropane = 35.0 - ( pt - 35 ) = 70 - pt, , Rate constant after 360 s = k1, At t = 360 s, pi = 35 atm, pt = 54 atm, \ pazoisopropane = ( 70 - 54) atm = 16 atm, \Rate constant,, 2.303, 35 2.303, =, log, =, log 2.187, 360, 360, 16, 2.303 ´ 0.33984, =, = 2.17 ´ 10-3s-1, 360, · Rate constant after 720s = k2, pi = 35 atm, pt = 63.0 atm, \, pazoisopropane = (70 - 63) atm = 7 atm, ·

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40, , CBSE Term II Chemistry XII, , \ Rate constant,, , 26. (i) (a) Decomposition of PH3 is a first order reaction, [PH3 ] 0, 0.693 ´ t, 0.3010 ´ t, log, =, =, [PH3 ] t 2.303 ´ t1/ 2, t1/ 2, , 2.303, 35, k2 =, log, 720, 7, 2.303, 2.303 ´ 0.6990, =, log5 =, 720, 720, -3 -1, = 2.24 ´ 10 s, (2.17 + 2.24), Average rate constant, k =, ´ 10-3, 2, = 2.21 ´ 10-3 s-1, , For t = t 3/ 4 , log, , 37.9 s, log 4 = 75.8 s, 0.3010, (b) If [ PH3 ] t is the concentration of PH3 left behind after, 1 min, then, [PH3 ] 0, kt, 0.693 ´ 60 s, log, =, == - 0.476, [PH3 ] t 2.303, 2.303 ´ 37.9 s, t=, , 25. (i) Let, rate of formation of D = k[ A ] n[ B ] n, Then,, Exp. 4 6.0 ´ 10-3 = k[ A ] n[ B ] m = k( 01, . ) n( 01, . )m, Exp. 3 2.4 ´ 10, , -2, , n, , m, , n, , = k[ A ] [ B ] = k( 0.4) ( 01, . ), , m, , …(a), , (ii) k = 2.0 ´ 10- 2 s - 1; [ A ] 0 = 1.0 mol L- 1; t = 100 s, k=, , 4 = 4n Þ n = 1, , Exp. 2 2.88 ´ 10, , = k( 0.3) ( 0.4), , …(c), , m, , …(d), , 2.0 ´ 10- 2 s - 1 =, - log [ A ] =, , Þ, m=2, \Rate of formation of D = k[ A ]1[ B ] 2 = k[ A ] [ B ] 2, -1 2, , = k( 05, . mol L ) (0.2 mol L ), , …(e), , Using data of exp. 4, 6.0 ´ 10-3 mol L -1 min -1 = k( 01, . ) ( 01, . )2, Dividing Eq. (e) by Eq. (f), Rate of formation of D, k( 05, . ) ( 0.2) 2, =, -3, -1, -1, 6.0 ´ 10 mol L min, k( 01, . ) ( 01, . )2, 5 ´ ( 2) 2 Þ 20, , Þ, , Rate of formation of D = 6 ´ 10- 3 mol L -1 min -1 ´ 20, -1, , = 0.12 mol L min, (ii) For reaction,, 2NH3 ¾® N 2 + 3H2, , -1, , Reaction rate = k[NH3 ] 0 = k, Rate of production of N 2, d[N 2 ], =, = k = 25, . ´ 10-4 Ms -1, dt, Rate of production of H2, d[H2 ], = 3k, dt, = 3 ´ 25, . ´ 10-4 Ms -1, , =, , = 75, . ´ 10-4 Ms -1, , 2.303, 1.0 mol L- 1, log, 100 s, [ A], 2.303, 2.0 ´ 10- 2 s - 1 =, (log 1 - log [ A ]), 100 s, 2.0 ´ 10- 2s - 1 =, , Dividing Eq. (d) by Eq. (c), 2.88 ´ 10-1 k( 0.3) n ( 0.4) m, =, = 2m, 7.2 ´ 10-2, k( 0.3) n( 0. 2) m, 4 = 2m, , -1, , 2.303, [ A] 0, log, t, [ A], , Therefore,, , Similarly, from, Exp. 1 7. 2 ´ 10-2 = k( 0.3) n( 0. 2) m, n, , [PH3 ] t, 1, = 0.334 =, [PH3 ] 0, 3, , …(b), , Dividing Eq. (b) by Eq. (a), 2.4 ´ 10-2 k( 0.4) n ( 01, . ) m ( 0.4) n, =, =, = 4n, -3, n, 6.0 ´ 10, k( 01, . ) ( 01, . ) m ( 01, . )n, , -1, , [PH3 ] 0, 0.3010 ´ t, =, [PH3 ] 0 / 4, 37.9 s, , …(f), , 2.303, ( 0 - log [ A ]), 100 s, 2.0 ´ 10- 2 s - 1 ´ 100 s, 2.303, , æ 2.0 ´ 10- 2 s -1 ´ 100 s ö, ÷, [ A ] = antilog çç ÷, 2.303, è, ø, or, [ A ] = 0.135 mol L- 1, 0.693 0.693 -1, 27. t1/ 2 =28.1 yr, k =, =, yr, t1/ 2, 28.1, or, , For first order reaction,, 2.303, a, t=, log, k, ( a - x), 0.693 -1, yr, 28.1, 2.303, a, 10 yr =, ´ 28.1 yr log, 0.693, ( a - x), , Case I, , log, , a = 1 m g, t = 10 yr, k =, , a, 10 yr ´ 0.693, =, = 0.107, ( a - x ) 2.303 ´ 28.1, a, = antilog (0.107) = 1.279, ( a - x), ( a - x) =, , a, ( 1 mg ), =, = 0.7819 mg, 1.279 1.279, , Amount left after 10 yr = 0.7819 mg, Case II a = 1 mg, t = 60 yr, 0.693 -1, yr, k=, 28.1

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41, , CBSE Term II Chemistry XII, , 60 yr =, , 2.303, a, ´ log, -1, (, a, x), 0.693 /28.1 yr, , a, ( 60 yr ) ´ 0.693 /28.1 yr -1, =, = 0.642, ( a - x), 2.303, a, = antilog 0.642 = 4.385, ( a - x), a, ( 1 mg ), ( a - x) =, =, = 0.2280 mg, 4.385 4.385, Amount left after 60 yr = 0.2280 mg, log, , 29., , (i) The value of k = 2.5 ´ 10 -2 min -1 carries the unit of, time alone and therefore, it is a first order reaction., (ii) The units of rate constant and rate of a reaction are, same for zero order reaction, i.e. molL-1s -1., (iii) Rate = k[ A ] x [ B ] y, Order = x + y, æ 3ö, \Order is ç ÷ + ( -1 ) = 1 / 2 , i.e. half-order., è 2ø, (iv) 2N 2 O5 r 2N 2 O 4 + O 2 is a first order reaction., , 28. (i) Rate = k[ A ]1 [ B ] 2 [ C ] 0, d[ H2 ], d[ I2 ], 1 d[ HI], ==+, dt, dt, 2 dt, change in concentration, (iii) Rate µ, time, The rate of a reaction decreases as time progresses as it, is inversely proportional to time., (iv) For the zero order reaction, rate equation = k[ A ] 0 [ B ] 0, (ii) Rate = -, , Or, For a zero order reaction, the reaction rate does not, decrease with time because it is independent of, concentration of the reactants., , Rate = k[ N2O5 ] 2, This reaction occurs in two steps as, Step I, Step II, , N2O5 ¾Slow, ¾, ¾® NO2 + NO, N2O5 + NO3 ¾Fast, ¾® 3NO2 + O2, , The slow step is unimolecular which is rate determining, steps. Hence, it is first order reaction., Or, A reaction is second order with respect to a reaction., Rate = k[ A ] 2, If the concentration of the reactant is doubled, the rate of, reaction becomes 4 times.

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Chapter Test, , 7. Assertion In the reaction, N2 + 3H2 ¾® 2NH3 , the rate of, reaction is different in terms of N2 , H2 and NH3 ., Reason Rate of reaction is equal to the rate of, disappearance or formation divided by the stoichiometric, coefficient., , Multiple Choice Questions, , 1. For the given rate expression= k[ A ] 1 [B] 1 , units of rate, (a) mol -1 Ls -1, (c) mol -1 L-1, , (b) mol -1 Ls, (d) None of these, , Reason Order of the given reaction is 2., , 2. For the given rate expression = k[ A ] 3 / 2 [B] -1 , the, overall order of a reaction is, (a) zero, , 8. Assertion The reaction, H2 + Br2 ¾® 2HBr, has molecularity of two., , constant is, , (b) half, , (c) one, , varies with time ‘ t’ as, 1, t, , (b) log C decreases with, , 1, increases linearly with t, C, (d) log C decrease linearly with t, , 1, t, , (c), , 5, for life of a first order reaction?, 6, k, 6, log, 2.303, 5, 2.303, (c), log 6, k, , 2.303, 4, log, k, 6, 2.303, (d), log 5, k, , 5. The rate constant ‘k’ for pseudo first order reaction, is, 2.303, log C0 - C, t, 2.303, (c), log C0 + C, t, , [Ans. 8.0 ´ 10 -9mol L-1s -1], , (ii) Calculate the rate of reaction after [ A] is reduced to, 0.06 mol L- 1 ., [Ans. 3.89 ´ 10 -9mol L-1s -1], following results were obtained :, 0, 30, t /s, , 2.303, C, log, t, C0, 2.303, C, (d), log 0, t, C, (b), , Assertion-Reasoning MCQs, Direction (Q. Nos. 6-8) Each of these questions, contain two statements Assertion (A) and Reason (R)., Each of these questions also has four alternative, choices, any one of which is the correct answer. You, have to select one of the codes (a), (b), (c) and (d), given below., (a) Both A and R are true and R is the correct explanation, of A., (b) Both A and R are true, but R is not the correct, explanation of A., (c) A is true, but R is false., (d) A is false, but R is true., , 6. Assertion Rate of reaction can be expressed as rate, of change in partial pressure of the gaseous, reactants or products., Reason Partial pressure of a gas is equal to its, concentration., , Multiple Choice Questions, 2. (b), , 3. (d), , 4. (c), , Assertion-Reasoning MCQs, 6. (c), , 7. (d), , 8. (c), , 90, , [Ester]/mol L, 0.55, 0.31, 0.17, 0.085, (i) Calculate the average rate of reaction between the time, interval 30 to 60 s., [Ans. 4.67 ´ 10 -3mol L-1s -1], (ii) Calculate the pseudo first order rate constant for the, hydrolysis of an ester., [Ans. 1.98 ´ 10 -2s -1], , 11. For a first order reaction, calculate the ratio between, time taken to complete three fourth of the reaction and, time taken to complete half of the reaction., [Ans. 2], , 12. A reaction is of second order with respect to a reactant., How is the rate of reaction affected, if the concentration, of the reactant is reduced to half? What is the unit of rate, constant for such a reaction?, , Long Answer Type Questions, , 13. (i) A first order reaction is 25% complete in 40 minutes., Calculate the value of rate constant. In what time will, the reaction be80% completed?, , [Ans. 0.0072 min-1, 223.6 min]], , (ii) Define order of reaction. Write the condition under, which a bimolecular reaction follows first order kinetics., , 14. (i) In a reaction, if the concentration of the reactant R is, quadrupled, the rate of reaction becomes sixty four, times. What is the order of reaction?, [Ans. 3], (ii) A first order reaction has a rate constant of 0.0051 min -1 ., If we begin with 0.10 M concentration of reactant, what, concentration of reactant will remain in solution after 3, hours?, [Ans. 0.04 mol L-1 ], , Answers, 1. (a), , 60, , -1, , (b), , (a), , (i) Calculate the initial rate of reaction when, [ A] = 0.1 mol L-1 , [B] = 0.2 mol L-1, , 10. In a pseudo first order hydrolysis of an ester in water, the, , 4. Which of the following represents the expression, , (a), , 9. For the reaction, 2 A + B ¾® A2B ;, , The rate = k [ A ] [B] 2 with k = 2.0 ´ 10 -6 mol-2 L2 s -1 ., , (d) two, , 3. In a first order reaction, reactant concentration ‘ C’, (a) C decreases with, , Short Answer Type Questions, , 5. (d), , For Detailed Solutions, Scan the code

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CHAPTER 03, , Surface, Chemistry, In this Chapter..., l, , Adsorption, , l, , Adsorption Isotherms, , l, , Colloidal State, , Surface chemistry deals with phenomena which occurs at the, surfaces or interfaces. It is represented by separating the bulk, phases by a hyphen or a slash., , If, and, DH = -ve, DS = -ve, \, DG = -ve - [ T( -ve)] = -ve + T, i.e. DG is negative only when DH has sufficiently high, negative value, and temperature is not very high., , Adsorption, l, , l, , l, , l, , It is the phenomenon of attracting and retaining the, molecules of a substance on the surface of a solid resulting, into a higher concentration on the surface than in the bulk., e.g. water vapour on silica gel., Adsorbent is a material on the surface of which, adsorption, takes place. Absorbate is the species which is adsorbed on, the surface of adsorbent., Adsorption arises due to the fact that the surface particles of, the adsorbent are not in the same environment as the, particles inside the bulk., The unbalanced or residual attractive forces of absorbent, molecules are responsible for attracting the adsorbate, particles on is surface., , l, , DG value is negative means, Gibb’s free energy decreases, and when DG becomes zero, a state of adsorption, equilibrium is established., Distinction between Absorption and Adsorption, Absorption, l, , l, , Thermodynamics of Adsorption, During adsorption, following thermodynamical changes occur, Heat is evolved, i.e. adsorption is an exothermic process and, that’s why it’s, DH value is negative., When adsorbate is adsorbed, freedom of movement of its, molecules gets restricted. It results in decrease in entropy of, adsorbate, i.e. DSis negative for adsorption., From Gibbs-Helmholtz equation,, l, , l, , l, , It is a bulk phenomenon,, e.g. water vapours are, absorbed by anhydrous, calcium chloride., The concentration of solute, is uniform throughout the, bulk of the solid, e.g. when, cotton is dipped in blue, ink, it gets blue throughout., , The concentration of solute, is uniform throughout the, bulk of the solid., , Note, , l, , l, , DG = DH - TDS, , l, , Adsorption, l, , l, , l, , It is a surface phenomenon,, e.g. water vapours are, adsorbed by silica gel on its, surface., The concentration of, substance (adsorbate) is only, at the surface and it does not, penetrate to the bulk,, e.g. when a chalk stick, is dipped in ink, it is blue, on the surface and white in, bulk., The concentration of the, adsorbate increases only at, the surface of the adsorbent., , When adsorption and absorption occur side by side it is known, as sorption., Desorption is the process of removing an adsorbed substance,, i.e. adsorbate from the surface of adsorbent.

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44, , CBSE Term II Chemistry XII, , Types of Adsorption, There are two types of adsorption of gases on solids,, i.e. physisorption and chemisorption., , l, , Difference between Physisorption and Chemisorption, Physisorption, l, , It arises because of van der, Waals’ forces., , Chemisorption, l, , l, , It is non-specific in nature., , l, , l, , It is reversible in nature., , l, , l, , l, , l, , l, , l, , l, , l, , It depends on the nature of, gas. More easily liquefiable, gases are adsorbed readily., Enthalpy of adsorption is low, (20-40 kJ mol -1), Low temperature is, favourable for this, adsorption. It decreases with, increases of temperature., No appreciable activation, energy is needed., No compound formation, takes place., It results into formation of, multimolecular layers on, adsorbent surface under high, pressure., , l, , l, , l, , l, , l, , l, , It is caused by chemical bond, formation, covalent or ionic., It is highly specific in nature., , l, , It is irreversible in nature, It also depends on the nature, of gas. Gases which can react, with the adsorbent show, chemisorption., Enthalpy of adsorption is, high. (80-240 kJ mol -1), This type of adsorption first, increases with increase of, temperature and then, decreases., High activation energy is, sometimes needed., Surface compounds are, formed., It result into the formation of, unimolecular layer., , Similarity between chemisorption and physisorption, Both are directly depends on the surface area,, i.e. increase with an increase in surface of the adsorbent., , l, , divided metals (Pt, Ni) and porous substance (charcoal, silica, gel) are best solid adsorbents., Temperature Adsorption is an exothermic process (taking, place with evolution of heat), therefore in accordance with, Le-Chatelier’s principle, the magnitude of adsorption, should increase with fall in temperature. This actually, happens except in chemisorption where extent of adsorption, first increases and then decreases with increases in, temperature., Pressure of the Adsorbate Gas Since, adsorption of a gas, leads to decrease of pressure, the magnitude of adsorption, increases with increase in pressure. At low temperature, the, extent of adsorption increases very rapidly with pressure., Over a small range of pressure, the extent of adsorption is, found to be directly proportional to the pressure., However, at higher pressures (closer to the saturation, point of the vapour pressure of the gas), the adsorption, tends to achieve a limiting value. At this point, formation, of multilayers starts and a stage of desorption may be, observed., Activation of Adsorbent Activation of an adsorbent refers to, increasing the adsorbing capacity of the adsorbent through, different methods. For instance making the surface of solid, rough or breaking it into small pieces increases surface area, hence the extent of adsorption., , Adsorption Isotherms, The relationship between extent of adsorption (x/m) and, pressure of the gas at constant temperature is known as, adsorption isotherm., It is generally represented in the form of a curve as shown below, 195 K, , Factors Affecting the Adsorption, of Gases on Solids, The extent of adsorption is affected by the following factors, Nature of Adsorbent A gas is adsorbed in different, amounts on different adsorbents. e.g. Hydrogen is, strongly adsorbed on nickel surface, while it is weakly, adsorbed on alumina surface under identical conditions., Nature of the Gas In general, the more liquefiable a gas, is, the more readily will it be adsorbed. Easily liquefiable, gases such as NH 3, HCl , Cl 2, SO 2, CO 2 , etc., are, readily adsorbed than the permanent gases, such as, O 2, N 2, H 2, etc. Higher the critical temperature of the, gas, greater is the extent of adsorption., l, , x, m, , 244 K, 273 K, , p, O, Adsorption isotherm, , l, , Thus, the amount of adsorption of gas on same mass of, charcoal follows the order, SO 2 ( Tc = 157 ° C) > CH4 ( Tc = 83 °C ) > H 2 (Tc = -120 ° C)., l, , Surface Area Increase in the surface area of the, adsorbent, increases the total amount of the gas adsorbed, because number of adsorbing sites increases. Thus, finely, , Freundlich Adsorption Isotherm, In 1909, Freundlich gave the relationship between x / m and p, at particular temperature as :, x, ...(i), = kp1/ n ( n > 1 ), m, where, m = mass of adsorbent, x = mass of the gas adsorbed, p = pressure, k = constant, n = constant/integer

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45, , CBSE Term II Chemistry XII, , Taking logarithm of whole Eq. (i), we get, x, 1, log = log k + log p, m, n, It now becomes similar to straight line equation as, y = mx + c, x, A graph of log is plotted against log p as shown below, m, Y, , log, , Q, a, , x, m, , }, O, , (ii) Colloidal solution Here, the size of dispersed, particles range between 1nm and 1000 nm, ( i. e. 10 -9 m -10 -6 m ). The dispersed particles of, colloids could be macromolecular or association of, many molecules. e.g. Protein globules in blood, plasma, fat droplets in milk., (iii) Suspension It has size of dispersed particles greater, than 1000 nm (i. e. >10 -6 m), which are aggregates of, millions of molecules. e.g. Soil particles in water, soot, particles in smoke., , b, Slope =, , Classification of Colloids, , 1, n, , l, , log k=intercept, log p, , X, , Dispersed Dispersion Types of, medium, phase, colloid, , The slope of line and its y-axis intercept correspond to 1 / n and, log k respectively., 1, The factor can have values between 0 and 1., n, 1, x, When = 0, = k, adsorption is independent of pressure., n, m, 1, x, When = 1, = kp, adsorption varies directly with pressure., n, m, Adsorption from Solution Phase, Freundlich adsorption isotherm gives approximate idea about, the adsorption from solution. Here, in place of pressure, the, concentration of adsorbate in solution are used., x, = kC 1/ n ( n > 1 ), m, Here, C is the equilibrium concentration, i.e. when adsorption, is complete., , Applications of Adsorption, The adsorption phenomena finds a number of applications such, as in production of high vacuum, in controlling humidity, in, heterogeneous catalysis, chromatographic analysis, froth floatation, process, in removal of colouring matter from solutions, in curing, diseases, separation of noble gases, in making gas masks etc., , Colloidal State, l, , l, , A colloid is a heterogeneous system in which one substance is, dispersed (dispersed phase) as very fine particles in another, substance (dispersion medium)., Colloids are sometimes also referred to as solutions. But essential, difference between a solution and a colloid is that of particle size, (dispersed phase). Thus, on the basis of the size of the dispersed, phase, solutions are divided into following three classes, (i) True solution It has dispersed particles with size less than, 1 nm (i.e. < 10 -9 m). Here, the dispersed particles constitute, single molecules or ions. e.g. Sugar in water or salt in water., , Depending upon the phases of dispersed phase and, dispersion medium, colloids are of following types, , l, , Solid, , Solid, , Solid, , Liquid, , Examples, , Solid sol Some coloured glasses,, gem stones, Sol, , Paints, cell fluids, , Solid, , Gas, , Liquid, , Solid, , Aerosol Smoke, dust, , Liquid, , Liquid, , Liquid, , Gas, , Aerosol Fog, mist, cloud,, insecticide sprays, , Gas, , Solid, , Solid sol Pumice stone, foam, rubber, , Gas, , Liquid, , Gel, , Cheese, butter, jellies, , Emulsion Milk, hair cream, , Foam, , Froth, whipped cream,, soap lather, , Out of the various types of colloids given above on the, table, the most common are sols, gels and emulsions., Based on the nature of interaction between dispersed, phase and dispersion medium, colloidal sols are divided, into following two categories, Lyophilic colloid, , Lyophobic colloid, , A colloidal solution in which, A colloidal solution in which, dispersed phase and dispersion dispersed phase and dispersion, medium attract each other., medium repel each other., It is prepared by simple mixing It is prepared by special, of the dispersed phase and, methods., dispersion medium., It is reversible sol., , It is irreversible sol., , No stabilising agent is required. Stabilising agent is required., Viscosity increases., , Viscosity remains same., , Surface tension decreases., , Surface tension remains same., , Quite stable and cannot be, easily coagulated., e.g. Starch sol, albumin sol., , Not stable and can be easily, coagulated by heating, shaking,, etc. e.g. Gold sol., , Note Lyophilic means solvent attracting and lyophobic means, solvent repelling. If water is the dispersion medium, the terms, used are hydrophilic and hydrophobic.

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46, l, , CBSE Term II Chemistry XII, , Based on the type of particles of the dispersed phase colloids, are classified as, Multimolecular, colloids, , Macromolecular, colloids, , Associated colloids, (micelles), , Aggregates of atoms, or molecules with, diameter less than, 1 nm is called, multimolecular, colloids., , These particles are, large molecules of, colloidal range., e.g. Starch,, cellulose, nylon,, proteins,, polythene, etc., , These colloids are, produced by the, aggregates of a large, number of ions because of, the attraction towards, oppositely charged ions in, concentrated solution., e.g. Soap solution,, detergent., , e.g. Sulphur sol,, gold sol., , They have generally They have, lyophilic character. lyophobic, character., , These molecules, contain both lyophilic and, lyophobic group., , These particles are, held together by, weak van der, Waals’ forces., , Greater van der Waals’, forces are observed at, higher concentration of, ions/molecules., , Due to long chain,, van der Waals’, forces are, comparatively, stronger., , Double, , As 2 O 3 + 3H 2 S ¾¾¾¾® As 2 S 3 (sol) + 3H 2 O, decomposition, , (ii) Electrical disintegration method (Bredig’s arc, method) It involves dispersion as well as, condensation. By this method, colloidal solutions of, metals like gold, silver, etc., are obtained., (iii) Peptisation is defined as the process of converting, freshly prepared precipitate into colloidal sol by, shaking it with dispersion medium in the presence of, small amount of electrolyte. The electrolyte added is, called peptising agent., , Purification of Colloidal Solution, The process used for reducing the amount of impurities to, the required extent is known as purification of colloidal, solution. It is carried out by the following methods, l, , l, , Micelle Formation, l, , l, , l, , l, , The formation of micelles takes place only above a particular, temperature known as Kraft temperature(Tk ) and above a, particular concentration known as Critical Micelle, Concentration (CMC)., The hydrophilicCOO - group (also called polar-ionic ‘head’), interacts with water molecule while the hydrophobic long, non-polar hydrocarbon chain R (also called non-polar ‘tail’) does, not interact., At CMC, the anions are pulled into the bulk of the solution and, aggregate to form a spherical shape in which carboxylate groups, forms a negatively charged spherical surface with the hydrocarbon, chains inside the sphere. Thus, the aggregate formed is known as, ‘ionic micelle’., The cleansing action of soap is due to the fact that soap, molecules form micelle around the oil droplet in such a way that, its hydrophobic part is in the oil and the hydrophilic part project, outside the grease due to which it interacts water with the, molecules present which is around the oil droplet is pulled from, the surface of cloth into water., , l, , A few important methods for the preparation of colloids are as, follows, (i) Chemical methods includes double decomposition,, oxidation reduction or hydrolysis reactions., Oxidation, , 2H 2S + SO 2 ¾¾¾® 2S + 2H 2O, (Sol), , Reduction, , 2 AuCl 3 + 3 HCHO + 3H 2O ¾¾® 2 Au + 3 HCOOH + 6 HCl, (Sol), , Hydrolysis, , FeCl 3 + 3 H 2O ¾¾¾® Fe(OH) 3 + 3 HCl, Red sol, , Electro-dialysis is a dialysis process carried out under, the influence of an electric field. It speeds up the, migration of ions to the opposite electrodes., Ultra filtration is a process of separation of, colloidal particles from the solvent and soluble, solutes present in the colloidal solution through an, ultrafilter paper., , Note The pores of the filter paper can be reduced in size by, impregnating with colloidion solution to stop the flow of, colloidal particles., , Properties of Colloidal Solution, The important properties of colloidal solutions are given, below, The value of colligative properties (osmotic pressure,, lowering in vapour pressure, depression in freezing, point and elevation in boiling point) are of small order, as compared to values shown by true solutions at same, concentrations., When light passes through a colloidal sol, its path, become visible due to scattering of light by particles. It, is called Tyndall effect and the illuminated path of the, beam is called Tyndall cone., Movement of colloidal particles at random in zig-zag, motion is known as Brownian movement. This, movement is due to the bombardment of dispersion, medium particles over dispersed phase particles. It helps, in providing stability to colloidal sol by not allowing, them to settle down., Colloidal particles always carry an electric charge. The, nature of this charge (either +ve or -ve) is the same on, all the particles in a given colloidal solution., l, , l, , Preparation of Colloids, , Dialysis is a process of removing small molecules or, ions from a colloidal solution by diffusion through a, semipermeable membrane. The apparatus used for this, purpose is known as dialyser., , l, , l

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47, , CBSE Term II Chemistry XII, , The presence of charge is due to preferential adsorption of, ions present in the dispersion medium., The combination of the two layers (i.e. fixed layer and, diffused layer) of opposite charges around the sol particles, is called Helmholtz electrical double layer., The potential difference between fixed layer and, diffused layer is called the zeta potential or electrokinetic, potential., The presence of equal and similar charges on colloidal, particles is largely responsible in giving stability to the, system, as the repulsive forces between charged particles, having same charge prevent them from coalescing or, aggregating when they come closer to one another., Following are the significance of charge., (i) Electrophoresis (Cataphoresis) The existence of charge, on colloidal particle is confirmed by electrophoresis, experiment. The migration of colloidal particles toward, the oppositely charged electrodes under the influence of, applied electric field., (ii) When electrophoresis is prevented by some suitable, means, it is seen that the dispersion medium begins to, move in an electric field. This phenomenon is called, electroosmosis., (iii) The process of settlings of colloidal particles is called, coagulation or precipitation of the sol. Coagulation of a, lyophobic sols can be carried by electrophoresis, by, mixing two oppositely charged sols, by boiling,, persistent dialysis and by addition of electrolytes., l, , Note, , l, , l, , l, , l, , l, , l, , l, , l, , The coagulating power of electrolyte is governed by Hardy-Schulze, rule. It state that ‘‘greater the valency of the flocculating ion added, the greater is its power to cause precipitation.’’, The coagulating power of an electrolyte is directly proportional to, the valency of the effective ion., , The minimum concentration of the electrolyte in millimoles, that must be added to one litre of the colloidal sol. So, as to, bring about complete coagulation is called the coagulation, value of flocculation value of the electrolyte., The smaller the coagulation or flocculating value of the, electrolyte, greater is its coagulating power., Stability of lyophillic sols depends upon the two factors,, i.e. charge and solvation of the colloidal particles. When, these two factors are removed, lyophillic sols can be, coagulated. This can be done by (i) adding a suitable solvent, to lyophilic solution, (ii) adding an electrolyte., , Protection of Colloids, Lyophillic sols are more stable than lyophobic sols because, lyophillic sols are extensively solvated. They have a unique, property of protecting lyophobic colloids and thus are called, as protective colloids., The protecting power of lyophillic sols are expressed in, terms of gold number. Gold number of a protective colloid, is the minimum weight in milligrams which must be, added to 10 mL of a standard red gold sol, so that no, coagulation of gold sol takes place when 1 mL of 10% NaCl is, added to it.

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48, , Chapter, Practice, 6 The correct option representing a Freundlich, , PART 1, Objective Questions, Multiple Choice Questions, , 1. Which of the following is an example of absorption?, Water on silica gel, Water on calcium chloride, Hydrogen on finely divided nickel, Oxygen on metal surface, , (NCERT Exemplar), , 7. Which of the following curves is in according with, , 2. At the equilibrium position in the process of, adsorption ......... ., , Freundlich adsorption isotherm?, , (b) D H = TDS, (d) D H < TDS, , (a), , 3. Activation of adsorbent means its adsorbing power is, increased by, , (b), , log x/m, , (NCERT Exemplar), , (a) D H > 0, (c) D H > TDS, , log x/m, log p, , (a) making the surface rough, (b) dipping the surface in acid to make it smooth, (c) dissolving in water, (d) All of the above, , (c), , 4. On the basis of data given below predict which of the, following gases shows least adsorption on a definite, amount of charcoal?, (NCERT Exemplar), Gas, , CO 2, , SO 2, , CH4, , H2, , Critical temp./K, , 304, , 630, , 190, , 33, , (a) CO 2, , (b) SO 2, , (c) CH4, , (d) H2, , 5. Freundlich adsorption isotherm is represented at, temperature T1 , T2 and T3 ., T2, , x/m, , T3, , (d), , log p, , log p, , 8. Butter is a colloidal solution of, (a) solid-solid, (b) liquid-solid, (c) solid-liquid, (d) gas-solid, , 9. An example of solid sol is, (a), (b), (c), (d), , T1, , log p, , log x/m, , (a), (b), (c), (d), , x, = k p0. 3, m, x, (b) = k p2.5, m, x, (c) = k p-0.5, m, x, (d) = k p-1, m, (a), , log x/m, , l, , adsorption isotherm is, , gem stones, hair cream, butter, paint, , 10. Among the colloids cheese (C ), milk ( M ) and smoke, (S), the correct combination of the dispersed phase, and dispersion medium, respectively is, , p, , Arrange the temperature in the increasing order., (NCERT Exemplar), , (a) T1 < T2 < T3, , (b) T3 < T2 < T1, , (c) T2 < T3 < T1, , (d) T3 < T1 < T2, , (a), (b), (c), (d), , C : liquid in solid; M : liquid in liquid; S : solid in gas, C : solid in liquid; M : liquid in liquid; S : gas in solid, C : liquid in solid; M : liquid in solid; S : solid in gas, C : solid in liquid; M : solid in liquid; S : solid in gas

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49, , CBSE Term II Chemistry XII, , 11. Method of formation of solution is given in Column I., , 15. What is the reason of Brownian movement ?, , Match it with the type of solution given in Column II., Column I, , (a) Variation of temperature in liquid state, (b) Attraction and repulsion of colloidal particles due to, presence of charge on it, (c) Impact of dispersion medium particles on colloidal particles, (d) Size of particles, , Column II, , A. Sulphur vapours passed 1. Normal electrolyte, through cold water., solution, B. Soap mixed with water, above critical micelle, concentration., , 2. Molecular colloids, , C. White of egg whipped, with water., , 3. Associated colloid, , D. Soap mixed with water, below critical micelle, concentration., , 4. Macromolecular, colloids, , 16. Freshly prepared precipitate sometimes gets, converted to colloidal solution by ......... ., (NCERT Exemplar), , (a) coagulation, (c) diffusion, , 17. Which of the following processes is involved in, Bredig’s arc method?, (a), (b), (c), (d), , Choose the correct code that are given below., Codes, A, (a) 2, (c) 3, , (NCERT Exemplar), , B, 3, 2, , C, 4, 4, , D, 1, 1, , A, (b) 1, (c) 2, , B, 2, 3, , C, 3, 1, , D, 4, 4, , (a), (b), (c), (d), , protected., By addition of oppositely charged sol (NCERT Exemplar), By addition of an electrolyte, By addition of lyophilic sol, By boiling, , coagulation power of an ion depend?, (a) Both magnitude and sign of the charge on the ion, (b) Size of the ion alone, (c) The magnitude of the charge on the ion alone, (d) The sign of charge on the ion alone, , phenomenon given in Column II., Column II, , A. Dispersion medium moves in an, electric field., , 1. Osmosis, , B. Solvent molecules pass through, semipermeable membrane, towards solvent side., , 2. Electrophoresis, , C. Movement of charged colloidal 3. Electroosmosis, particles under the influence of, applied electric potential towards, oppositely charged electrodes., D. Solvent molecules pass through, semipermeable membranes, towards solution side., , Codes, A, (a) 3, (b) 4, (c) 1, (c) 2, , C, 2, 2, 3, 4, , D, 1, 1, 4, 3, , 14. Which property of colloidal solution is independent, of charge on the colloidal particle?, (a) Coagulation, (c) Electroosmosis, , maximum coagulating value for Ag/Ag + sol?, , (NCERT Exemplar), , (a) Na 2S, (c) Na 2SO 4, , (b) Electrophoresis, (d) Tyndall effect, , (b) Na 3PO 4, (d) NaCl, , 21. Gold number of A, B, C and D sol is respectively, 0.001, 0.15, 20 and 25, most effective protective, colloidal is, (a) A, (c) C, l, , (NCERT Exemplar), , B, 4, 3, 2, 1, , 20. Which of the following electrolytes will have, , 4. Reverse-osmosis, , Choose the correct code that are given below., , negatively and positively charged sols, respectively, negatively charged sols, positively charged sols, positively and negatively charged sols, respectively, , 19. On which of the following properties does the, , 13. Match the statement given in Column I with the, Column I, , Only dispersion, Only condensation, Dispersion as well as condensation, Diffusion, , 18 Haemoglobin and gold sol are examples of, , 12. Select the method by which lyophobic sol can be, (a), (b), (c), (d), , (b) electrolysis, (d) peptisation, , (b) B, (d) D, , Assertion-Reasoning MCQs, Direction (Q. Nos. 22-26) Each of these questions, contains two statements Assertion (A) and Reason (R)., Each of these questions also has four alternative choices,, any one of which is the correct answer. You have to, select one of the codes (a), (b), (c) and (d) given below., (a) Both A and R are true and R is the correct, explanation of A., (b) Both A and R are true, but R is not the correct, explanation of A., (c) A is true, but R is false., (d) A is false, but R is true.

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50, , CBSE Term II Chemistry XII, , 22. Assertion Chemical adsorption of molecules on, surface requires activation energy., Reason Because the bonds of the adsorbed, molecules are broken., , 23. Assertion The vegetable tanning materials have, smaller surface areas., Reason The capillaries present in leather are, reduced to smaller diameters., 24. Assertion Fe 3+ can be used for coagulation of As 2S 3, sol., Reason Fe 3+ reacts with As 2S 3 to give Fe 2S 3 ., , 25. Assertion Aqueous gold colloidal soultion is red in, colour., Reason The colour arises due to scattering of light by, colloidal gold particles., , 26. Assertion Micelles are formed by surfactant, molecules above the critical micellar concentration, (CMC)., Reason The conductivity of a solution having, surfactant molecules decreases sharply at the CMC., l, , Case Based MCQs, , 27. Read the following and answer the questions from, (i) to (iv) given below, The state of the substance in which dispersed phase, (solute) and dispersion medium (solvent) floats with, each other is called colloidal state., Depending on the size of particles of the dispersed, phase, colloids are classified as multimolecular., macromolecular and associated colloids. Associated, colloids are very important in our day-to-day life., There are some substances which at low, concentrations behave as normal strong, electrolytes, but at higher concentrations exhibit, colloidal behaviour due to the formation of, aggregates. The aggregated particles thus formed, are called micelles. These are also known as, associated colloids., The formation of micelles takes place only above a, particular temperature and above a particular, concentration. On dilution, these colloids revert, back to individual ions. Surface active agents such, as soaps and synthetic detergents belong to this, class. For soaps, the CMC is 10 -4 to 10 -3 mol L-1 ., These colloids have both lyophobic and lyophilic, parts. Micelles may contain as many as 100, molecules more., Taking an example of soap solution in which soap is, sodium salt of higher fatty acid and may be, , represented RCOO - Na + (e.g. sodium stearate, CH 3 (CH 2 )16 COO - Na + or sodium palmitate, CH 3 (CH 2 )14 COO - Na + ). When soap is dissolved in, water, it dissociates into RCOO - and Na + ions. The, RCOO - ions consist of two parts hydrocarbon chain, R (non-polar tail) which is hydrophobic, (water-repelling), and a polar group COO (polar, ionic) which is hydrophilic (water-loving)., (i) A large number of atoms or molecules of a substance, aggregate together to form species having size in the, colloidal range, these species are called, (a) multimolecular colloids, (b) macromolecular colloids, (c) solid sol, (d) emulsion, , (ii) Gold sols and sulphur sols are the examples of, (a), (b), (c), (d), , multimolecular colloids, macromolecular colloids, associated colloids, All of the above, , (iii) Which of the following colloids resemble to the true, solutions?, (a), (b), (c), (d), , Micelles, Macromolecular colloids, Lyophobic colloids, All of the above, , Or, The formation of micelles takes place above the, particular range of temperature termed as, (a) critical temperature, (c) CMC, , (b) Kraft temperature, (d) Micelles temperature, , (iv) Soaps and synthetic detergents belong to the class of, (a), (b), (c), (d), , macromolecular colloids, associated colloids, multimolecular colloids, Both (b) and (c), , 28. Read the following and answer the questions from, (i) to (iv) given below, A colloid is one of the three primary types of mixtures,, with the other two being a solution and suspensions. A, colloid is a mixture that has particles ranging between, 1 to 1000 nanometers in diameter, yet are still able to, remain evenly distributed throughout the solution., These are also known as colloidal dispersions because, the substances remain dispersed and do not settle to, the bottom of the container. In colloids, one substance, is evenly dispersed in another. The substance being, dispersed is referred to as being in the dispersed, phase, while the substance in the dispersed phase, must be larger than the size of a molecule but smaller, than what can be seen with the naked eye . This can be, more precisely quantified as one or more of the

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51, , CBSE Term II Chemistry XII, , substance’s dimensions must be between 1 and, 1000 nanometers., If the dimensions are smaller than this substance is, considered a solution and if they are larger than the, substance is a suspension. A common method of, classifying colloids is based on the phase of the, dispersed substance and what phase it is dispersed, in. The types of colloids includes sol, emulsion,, foam and aerosol., (i) The disperse phase, dispersion medium and nature of, colloidal solution (lyophilic or lyophobic) of ‘Gold sol’, respectively are, (a) solid, solid, lyophobic, (c) solid, liquid, lyophobic, , (b) liquid, liquid, lyophobic, (d) solid, liquid, lyophilic, , (ii) Which of the following make the lyophilic solution, unstable?, (a) Dialysis, (b) Addition of electrolyte, (c) Addition of alcohol, (d) Addition of alcohol and electrolyte both, , (iii) The colloidal solution of mercury in water can be, easily obtained by, (a) mechanical precipitation (b) Bredig’s Arc method, (c) repeated washing, (d) ultrasonic dispersion, (b) liquid in liquid, (d) solid in gas, , Or, Which property of colloidal solutions is, independent of charge on the colloidal particles ?, (a) Coagulation, (c) Electroosmosis, , chemisorption with respect to the following., (i) Specificity, (ii) Temperature dependence, (iii) Reversibility and, (iv) Enthalpy change, , 6. What are the three factors which influence the, adsorption of the gas on a solid?, , 7. What do you understand by activation of adsorbent?, How is it achieved? Give one example., , 8. What is an adsorption isotherm? Describe Freundlich, adsorption isotherm., , 9. Comment on the statement that colloid is not a, substance but a state of substance., , (NCERT), , 10. (i) What type of colloid is formed when a gas is, dispersed in a solid ?, (All India, 2017), (ii) Based on the type of dispersed phase, what type of, colloid is micelles ?, , 11. Define the following terms by giving an example of, each., (i) Hydrosol, , (ii) Aerosol, , 12. Write the dispersed phase and dispersion medium of, , (iv) Foam is a type of colloidal of, (a) liquid in gas, (c) gas in liquid, , 5. Write the differences between physisorption and, , (b) Electrophoresis, (d) Tyndall effect, , the following colloidal systems., (i) Smoke, (ii) Milk, , 13. Write one difference in each of the following:, (i) Lyophobic sol and lyophilic sol., (ii) Solution and colloid., (Delhi, 2017), (iii) Multimolecular colloid and associated colloid, (All India, 2020; Delhi, 2017), , 14. How are the following colloidal solutions prepared?, , PART 2, Subjective Questions, l, , Short Answer Type Questions, 1. Distinguish between the meaning of the terms, adsorption and absorption. Give one example of, each., (NCERT), , 2. Why do physisorption and chemisorption behave, differently with rise in temperature?, (NCERT Exemplar), , 3. Why physisorption forms multimolecular layers, whereas chemisorption forms unimolecular layers?, , (i) Sulphur in water, , (ii) Gold in water, , 15. What is meant by coagulation of a colloidal, solution? Describe briefly any three methods by which, coagulation of lyophobic sols can be carried out., , 16. Explain the cleaning action of soap. Why do soaps not, work in hard water?, , 17. Explain what is observed,, (i) when a beam of light is passed through a colloidal, solution?, (All India), (ii) when an electrolyte, NaCl, is added to hydrated ferric, oxide solution?, (iii) electric current is passed through a colloidal, solution?, , 4. Why are substances like platinum and palladium 18. What modification can you suggest in the, often used for carrying out electrolysis of aqueous, solutions ?, , Hardy-Schulze rule?, , (NCERT)

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52, , CBSE Term II Chemistry XII, , (c) Artificial rain is caused by spraying salt over, clouds., (ii) Define the following terms with an example in each, case, (a) Macromolecular colloids, (b) Peptisation, , 19. Give reasons for the following observations:, (i) Leather gets hardened after tanning., (ii) Lyophilic sol is more stable than lyophobic sol., (Delhi 2015), , 20. How are colloids classified on the basis of, (i) physical state of components?, (ii) nature of dispersed phase?, (iii) interaction between dispersed phase and dispersion, medium?, (NCERT), , 25. (i) What happens, when, (a) a freshly prepared precipitate of Fe(OH) 3 is, shaken with a small amount of FeCl 3 solution?, (b) persistent dialysis of a colloidal solution is, carried out?, (CBSE 2018), (ii) How can a colloidal solution and true solution of the, same colour be distinguished from each other?, (iii) Distinguish between micelles and colloidal, particles., , Long Answer Type Questions, 21. Answer the following, , l, , (i) Out of NH3 and N 2 , which gas will be adsorbed, more readily on the surface of charcoal and why?, (ii) Give reason why a finely divided substance is more, effective as an adsorbent?, (iii) What is the role of activated charcoal in gas masks, used in coal mines?, (iv) Why does the white precipitate of silver, halide becomes coloured in the presence of the dye, eosin?, (v) Write one example of best semipermeable, membrane., , 22. (i) Write one similarity between physisorption and, chemisorption., (Delhi 2017), (ii) In physisorption or chemisorption, which has a, higher enthalpy of adsorption?, (iii) Give reason for the following, (a) Alum can be used to purify water., (b) Bleeding can be stop by rubbing moit alum., (c) A delta is formed at the meeting point of sea, water and river water., (All India 2015), , 23. (i) Out of MgCl 2 and AlCl 3 , which is more effective, in causing coagulation of negatively charged, solution and why?, (ii) Out of sulphur sol and proteins, which one forms, multimolecular colloidal., (Delhi 2016), (iii) Define the following terms, (a) Desorption, (b) Critical micelle concentration, (c) Dialysis, , 24. (i) Give reason for the following, (a) It is essential to wash the precipitate with water, before estimating it quantitatively., (b) Same substance can acts as colloids and, crystalloids, , 26. (i) What type of forces is responsible for the occurrence, of physisorption?, (ii) What is aquasol and alcosol?, (iii) What are the dispersed phase and dispersion, medium in milk?, (iv) Give one example of sol and gel., (v) Why is a colloidal sol stable?, (Delhi 2017), l, , Case Based Questions, , 27. Read the following and answer the questions from (i), to (iv) given below, A colloid is a heterogeneous system in which one, substance is dispersed (dispersed phase) as very fine, particle in another substance (dispersion medium), colloids are sometimes also referred to as solutions., But the essential difference between a solution and a, colloid is that of particle size., In a true solution, the constituent particles are ions or, small molecules while in a colloid, the dispersed, phase may consist of particle of single macromolecule, or an aggregate of many ions, atoms or molecules., Thus, on the basis of the size of dispersed phase,, solutions are divided as: true solution, colloidal and, suspension., (i) What is the diameter of colloidal particles?, (ii) What type of colloid is formed when a liquid is, dispersed in a solid? Give an example., (iii) What type of colloid is formed when a gas is, dispersed in a liquid? Give an example., (iv) What is collodion?, Or What is the reason for stability of colloidal sols?

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53, , CBSE Term II Chemistry XII, , 28. Read the following and answer the questions from (i), to (iv) given below, The boundary which separates the two bulk phases, (may be compounds or solutions) is called surface or, interface., The interface or surface is represented by separating, the bulk phases by a hyphen or a slash, e.g., solid-solid, solid-liquid interface etc., The interface is usually few molecules thick but it, area depends on the size of the particles of bulk, phases. As gases are completely miscible, no, interfaces exists between them., Many important phenomenon like corrosion, electrode processes, dissolution and, crystallisation occur at interfaces. Surface chemistry, deals which such phenomenon occuring at an, interface., Solid surface has the tendency to attract and retain to, attract and attain the molecules of the phase with, which it comes in contact. Those molecules, remain only at the surface and do not go deeper into, the bulk., The process in which molecular species are, accumulated at the surface rather than in the bulk of, a solid or liquid phase is termed as adsorption., Whereas, the process in which molecular species are, go deeper into the bulk of a solid or liquid phase is, termed as absorption., (i) Out of NH 3 and CO 2 , which gas will be adsorbed, more readily on the surface of activated charcoal, and why?, , (ii) Adsorption of a gas on surface of solid is generally, accompanied by a decrease in entropy, still it is a, spontaneous process. Explain., (iii) Describe the application of adsorption in, controlling humidity., (iv) Why is adsorption always exothermic?, Or, Name two type of adsorption phenomenon., , 29. Read the following and answer the questions from (i), to (iv) given below, Colloidal particles always carry an electric charge, which may be either positive or negative. For, example, when AgNO 3 solution is added to KI, solution, a negatively charged colloidal sol is, obtained. The presence of equal and similar charges, on colloidal particles provide stability to the colloidal, sol and if, somehow, charge is removed, coagulation, of sol occurs. Lyophobic sols are readily coagulated, as compare to lyophilic sols., (All India, 2020), (i) What is the reason for the charge on sol particles ?, (ii) Why the presence of equal and similar charges on, colloidal particles provide stability ?, (iii) Why a negatively charged sol is obtained on adding, AgNO 3 solution to KI solution ?, (iv) Name one method by which coagulation of lyophobic, sol can be carried out., Or, Out of KI or K 2SO 4 , which electrolyte is better in, the coagulation of positive sol ?

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54, , CBSE Term II Chemistry XII, , EXPLANATIONS, Objective Questions, , So, the value of 1/ n lies between 0 and 1 in all cases., , log x/m, , 1. (b) Absorption means penetration of adsorbate, molecules into the bulk of the adsorbent., From the given option (b) is an example of absorption, because, when water is spread over calcium chloride,, water gets penetrated into bulk of the calcium chloride., 2. (b) As we know that, at equilibrium DG = 0, DH - TDS = 0, D H = TD S, Hence, at equilibrium enthalpy change is equal to, product of temperature and entropy change., 3. (a) Activation of an adsorption refers to increasing the, absorbing capacity of the adsorbent through different, methods., For instance, breaking it into small pieces increases the, surface area, hence, increases the extent of adsorption., 4. (d) Lesser the value of critical temperature of gases, lesser will be the extent of adsorption. Here, H2 has, lowest value of critical temperature, i.e. 33., Hence, hydrogen gas shows least adsorption on a, definite amount of charcoal., x, 5. (a) shows the extent of adsorption., m, Extent of adsorption increases by decrease of, temperature., Therefore, the order of temperature is T1 < T2 < T3., 6. (a) According to Freundlich adsorption isotherm., x, = k × p1/ n, m, x, where,, = amount of the gas adsorbed per unit mass of, m, adsorbent, p = pressure, k and n are the constants., The value of n lies in between 0 to 1., x, Thus, = k × p0. 3 is the correct option., m, 7. (c) According to the Freundlich adsorption isotherm, 1, , x, = kp n, m, x æ1 ö, Taking log on both side log, = ç ÷ log p + log k, m è nø, ­, ­, ­, ­, Y, = m, X + C, On comparing it with equation of straight line and drawing, x, the graph log versus log p , we get a straight line with, m, intercept log k and slope of the straight line gives the value, 1, of ., n, The value of n is always greater than 1., , Slope = 1/n, log k (intercept), log p, , 8. (b) Butter is a colloidal solution of liquid-solid,, , i.e. the dispersed phase is liquid and dispersion medium is solid., This type of colloid is called gel., 9. (a) Solid sol consists of solid as both dispersed phase and, dispersion medium. In gemstones, metal crystals (salt and, oxides of metals) are dispersed in solid (stone) medium. Hair, cream is an emulsion (liquid in liquid). Butter is a colloidal, solution of liquid in solid. Paint is also sol (solid in liquid)., 10. (a), Dispersed, phase, , Dispersion, medium, , Type of, colloid, , Liquid, , Solid, , Gel, , Liquid, , Liquid, , Emulsion Milk (M), hair, cream, , Solid, , Gas, , Aerosol, , Examples, Cheese (C),, butter, jellies, , Smoke (S), dust, , Thus, option (a) is correct, i.e. C : liquid in solid, M : liquid in, liquid and S : solid in gas., 11. (a) A® (2); B ® (3); C ® (4); D ® (1), A. When sulphur vapours passed through cold water it leads to, formation of molecular colloids., B. When soap is mixed with water above critical micelle, concentration it lead to formation of associated colloids., C. White of egg whipped with water is an example of, macromolecular colloids in which proteneous molecule,, having high molecular mass, acts as a colloidal particle., D. Soap mixed with water below critical micelle concentration, is known as normal electrolyte solution., 12. (c) Lyophobic sol can be protected by addition of lyophilic sol. As, lyophobic sols are readily precipitated on addition of small amount, of electrolytes or shaking, or heating hence they are made stable by, adding lyophillic sol which stabilises the lyophobic sols., 13. (a) A ® (3);, B ® (4);, C ® (2);, D ® (1), A. Dispersion medium moves in an electric field is known as, electroosmosis., B. Solvent molecules pass through semipermeable membrane, towards solvent side is known as reverse-osmosis., C. Movement of charged colloidal particles under the, influence of applied electric potential towards oppositely, charge electrodes is known as electrophoresis., D. Solvent molecules pass through semipermeable membranes, towards solution side is known as osmosis.

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55, , CBSE Term II Chemistry XII, , So, both A and R are true and R is the correct explanation, of A., , 14. (d) Tyndall effect is the scatttering of light by sol particles,, which cannot be affected by charge on them., , Other given options such as coagulation, electrophoresis and, electroosmosis depend on charge particles., Thus, option (d) is correct., 15. (c) The molecules of dispersion medium due to their kinetic, motion strike against the colloidal particles from all sides, with different forces causing them to move., 16. (d) Freshly prepared precipitate sometimes gets converted, to colloidal solution by peptisation., Peptisation is a process in which precipitate gets converted, into colloidal solution by addition of a suitable peptising, agent., 17. (c) Bredig’s arc method involves the process of both, dispersion as well as condensation. Colloidal sols of metals, such as gold, silver, platinum etc., can be prepared by this, method., 18. (d) Haemoglobin and gold sol both are colloids and always, carry an electric charge. Haemoglobin is a positively, charged sol, because in haemoglobin, Fe2+ ion is the central, metal ion of the octahedral complex., All metal sols like, Au-sol, Ag-sol etc; are negatively charged, sols., 19. (a) Coagulation power of an ion depends both on magnitude, and sign of the charge (positive or negative) on the ion. This, fact can be explained by Hardy-Schulze rule. According to, this rule “greater the valency of the coagulation, ion/flocculating ion (appositively charged ion) added, the, greater is its power to coagulation., , 24. (c) When oppositely charged sols are mixed in almost equal, proportions, their charges are neutralised. Both sols may be, partially or completely precipitated., When negatively charged As 2S 3 sol is added to positively, charged Fe(OH) 3 sol in suitable amounts, precipitation of, both the sols take place simultaneously. Thus, A is true but R, is false., 25. (a) Both A and R are true and R is the correct explanation of A., The colour of colloidal solution depends on the wavelength, of light scattered by the dispersed particles. The wavelength, of light further depends on the size and nature of the, particles., Finest gold sol in red in colour. As size of the particles, increases, it becomes purple, then blue and finally golden, yellow., 26. (b) At a certain concentration, surfactant molecules start to, aggregate and from micelle. This concentration is called, critical micellar concentration (CMC)., Moreover, the conductivity of a solution having surfactant, molecules decreases sharply at the CMC., Thus, both A and R are true but R is not the correct, explanation of A., 27., , (i) (b) A large number of atoms or molecules of a substance, aggregates together to form species having size in the, colloidal range. These species are called, macromolecular colloids., (ii) (a) Gold sols and sulphur sols are the examples of, multimolecular colloids., (iii) (b) Macromolecular colloids are quite stable and, resemble to the true solutions in many respects., Or, (b) The formation of micelles takes place above the, particular range of temperature termed as Kraft, temperature., (iv) (b) Soaps and synthetic detergent belongs to the class of, association colloids., , 28., , (i) (c) In gold sol, gold particles (solid) are dispersed in, water (liquid). Gold sol cannot be prepared directly by, shaking with water as it is a lyophobic sol., (ii) (d) Addition of alcohol is similar to replacement of the, dispersion medium and addition of electrolyte may, cause coagulation. Hence, both these make lyophilic, solution unstable., (iii) (d) Colloidal solution of mercury in water is easily, obtained by ultrasonic dispersion., (iv) (c) When gas is dispersed in liquid medium then foam, is formed., Or, (d) Tyndall effect is the scattering of light by sol, particles, which cannot be affected by charge on them., , 20. (b) According to Hardy-Schulze law, greater the charge on, anion greater will be its coagulating power., , Electrolytes Anionic part Charge on anion, Na 2S, , S2 -, , 2, , Na 3PO 4, , PO 34, , 3, , Na 2SO 4, , SO 24 -, , 2, , NaCl, , Cl -, , 1, , 3Here, PO 34 have highest charge. Hence, PO 4 have highest, coagulating power., 1, 21. (a) As Protective power µ, Gold number, Therefore, sol A is most effective protective colloidal., , 22. (a) Both A and R are true and R is the correct explanation of A., Enthalpy of chemisorption is high as it involves chemical, bond formation., 23. (a) The vegetable tanning materials have smaller surface, areas because the capillaries are reduced to smaller, diameters.

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56, , CBSE Term II Chemistry XII, , Adsorption, , Absorption, l, , l, , l, , It is a bulk phenomenon,, e.g. water vapours are, absorbed by anhydrous, calcium chloride., The concentration of solute, is uniform throughout the, bulk of the solid, e.g. when, cotton is dipped in blue, ink, it gets blue, throughout., , The concentration of solute, is uniform throughout the, bulk of the solid., , l, , l, , l, , It is a surface phenomenon,, e.g. water vapours are, adsorbed by silica gel on its, surface., The concentration of, substance (adsorbate) is, only at the surface and it, does not penetrate to the, bulk, e.g. when a chalk, stick is dipped in ink, it is, blue on the surface and, white in bulk., The concentration of the, adsorbate increases only, at the surface of the, adsorbent., , 2. Behaviour of physisorption and chemisorption on increase in, temperature can be explained on the basis of nature of forces, present to bind their particles., Physisorption involves weak van der Waals’ forces which, weakens with increase in temperature. Chemisorption, involves formation of chemical bond which requires activation, energy., 3. Chemisorption occurs when reactant molecules form, covalent or ionic bonds with the surface of adsorbent., Once the layer of reactant molecules has been formed, on the surface of adsorbent, no other active site is available, for bonding of reactant-adsorbent to form further layers., Physisorption is simply driven by van der Waals' forces of, attraction. So, number of layers may be formed one over the, other on the surface of the adsorbent., , (i) Effect of pressure on adsorption of gases on solids At, constant temperature, the adsorption of gases increases, with increase of pressure. (This is further explained by, Freundlich’s adsorption isotherm.), (ii) Effect of temperature Adsorption is an exothermic, process. Therefore, in accordance with Le-Chatelier’s, principle, the magnitude of adsorption increases with, decrease in temperature. Actually, physisorption, increases with decrease in temperature but, chemisorption first increases and then decreases with, increase in temperature., , Adsorption, , 6., , 1. Distinction between Absorption and Adsorption, , Adsorption, , Subjective Questions, , T, Physisorption, , (ii) Temperature, dependence, , (iii) Reversibility, (iv) Enthalpy, change, , l, , l, , l, , l, , It is non-specific, in nature., Low temperature, is favourable for, adsorption., It decreases with, increase of, temperature., It is reversible in, nature., Enthalpy of, adsorption is low, (20-40 kJ mol -1 )., , l, , l, , l, , l, , It is highly, specific in nature., High temperature, is favourable for, adsorption. It, increases with the, increase in, temperature and, then decreases., It is irreversible, in nature., Enthalpy of, adsorption is high, (80-240 kJ mol -1)., , T, Chemisorption, , (iii) Nature of adsorbent A gas is adsorbed in different, amounts on different adsorbents., 7. Activation of an adsorbent refers to increasing the adsorbing, capacity of the adsorbent through different methods. For, instance making the surface of solid rough or breaking it into, small pieces increases surface area hence the extent of, adsorption. That is why, metals in finely divided state act as, good adsorbents. Certain specific treatment also activate an, adsorbent. e.g. Heating wood charcoal between 600 - 1200°C, activate it., 8. Adsorption isotherm The variation in the amount of gas, adsorbed by the adsorbent with pressure at constant, temperature can be expressed by means of a curve called, adsorption isotherm., 195 K, , 4. Due to the following features, (i) Chemical inertness Both form chemically inert, electrodes., (ii) Adsorption capacity Platinum and palladium both have, good adsorptive capacity for hydrogen., 5., Physisorption, Chemisorption, Properties, (i) Specificity, , p-constant, , x, m, , 244 K, 273 K, , p, Adsorption isotherm, , These curves indicated that at a fixed pressure, there is a, decrease in physical adsorption with increase in temperature., Freundlich’s adsorption isotherm It is an empirical, relationship between the quantity of gas adsorbed by unit mass, of solid adsorbent and pressure at a particular. temperature., , x, = kp1/ n ( n > 1 ), m, , …(i), , x, x, = kp or, µp, m, m, where, x is the mass of gas adsorbed on mass m of the, adsorbent at pressure p, k and n are constants which depend, on the nature of the adsorbent and the gas at a particular, temperature., when n = 1,

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57, , CBSE Term II Chemistry XII, , Taking log in Eq. (i), gives, x, 1, log = log k + log p, m, n, The validity of Freundlich isotherm can be verified by, x, plotting log on Y-axis and log p on X-axis., m, If it comes to be a straight line, the Freundlich isotherm is, valid., , (iii) Multimolecular colloid These are aggregates of atoms, or molecules with diameter less than 1 nm., Associated colloid These colloids are produced by the, aggregates of a large number of ions because of the, attraction towards oppositely charged ions in, concentrated solution., 14., , (i) Sulphur sol is prepared by the oxidation of H 2 S with SO 2 ., Oxidation, , SO 2 + 2 H 2 S ¾¾¾® 3S + 2H 2 O, (Sol), , log, , x, m, Slope =, , (ii) Gold sol is prepared by Bredig’s arc process or by the, reduction of AuCl 3 with HCHO., , 1, n, , Reduction, , e.g. 2AuCl 3 + 3HCHO + 3H2O ¾¾¾® 2Au, , (Sol), , log p, Freundlich isotherm, , 9. Colloid is not a substance but a state of substance is a, true statement because some substances which are, crystalloids under certain conditions can be colloids under, the other, e.g. NaCl is a crystalloid in aqueous medium, but, when mixed with benzene, it behaves as a colloid. It is the size, of the particles of that matter, i.e. the state in which the, substance exists. If size lies in range of 1-1000 nm, it is in the, colloidal state. So, above statement is true., 10., , (i) Foam, Example : Shaving cream., (ii) Associated colloids from micelles There are some, substances which at low concentration behave as normal, strong electrolyte but at higher concentration exhibit, colloidal behaviour due to the formation of aggregates., The aggregated particles thus formed are called, micelles., , 11., , (i) Hydrosol When solid is dispresed in water, it is called, hydrosol, e.g. starch sol., (ii) Aerosol It is a type of colloidal system in which, dispersion medium is gas and dispersed phase is either, solid or liquid, e.g. fog, smoke, mist, etc. Colloidal, suspensions in the air are also called aerosols., , 12., , (i) Dispersed phase of smoke = Solid, Dispersion medium of smoke = Gas, (ii) Dispersed phase of milk = Liquid, Dispersion medium of milk = Water (liquid), , 13., , (i) Lyophilic sol The colloidal solution in which the, particles of the dispersed phase have a great affinity for, the dispersion medium is called lyophilic sol. e.g. gum,, geletin, starch etc., Lyophobic sol The colloidal solution in which there is, no affinity between particles of the dispersed phase, and the dispersion medium is called lyophobic sol., Lyophobic sols need stabilising agents for their, preservation., (ii) Solution It contains small solute particles dispersed, throughout the solvent. The particle size is less than 1 nm., Colloid It contain particles of intermediate size. It is a, heterogeneous solution. The particles of colloid have, diameters between 1 to 1000 nm., , +3HCOOH + 6HCl, 15. The process of settling of colloidal particles is called, coagulation of the sol. It is also known as precipitation., Following are the three methods by which coagulation of, lyophobic sols can be carried out:, (i) Electrophoresis In this process, the colloidal particles, move towards oppositely charged electrodes and get, discharged resulting in coagulation., (ii) Mixing of two oppositely charged sols (or mutual, coagulation) When equal proportions of oppositely, chaged sols are mixed, they neutralise each other, resulting in coagulation., (iii) Prolonged dialysis By this method, electrolytes present, in sol are removed completely and colloid becomes, unstable resulting in coagulation., 16. The cleansing action of soap is due to the fact that soap, molecules form micelle around the oil droplet in such a way, that hydrophobic part of the stearate ions [ C17H35 COO - ] is in, the oil droplet and hydrophilic part interact with water, the, oil droplet surrounded by stearate ions is now pulled in water, and removed from the dirty surface., Thus, soap helps in emulsification and washing away of oils, and fats. The negatively charged sheath around the globules, prevents them from coming together and forming aggregates., , (a), Stearate on cloth, , (c), (b), Stearate ions, Grease droplet, arranging around the, surrounded by, grease droplets, stearate ions, (micelle formed), , Hard water contains calcium and magnesium ions. These ions, form insoluble calcium and magnesium when sodium or, potassium soaps are disoolved in hard water. These insoluble, soaps separate as scum in water and are useless as cleansing, agent., 17., , (i) When a beam of light is passed through a sol, scattering, of light takes place and the path of light becomes visible, (Tyndall effect).

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58, , CBSE Term II Chemistry XII, , (ii) Charge on ferric ion is neutralised by chloride ions. Thus,, coagulation takes place., , 21., , (iii) When electric current is passed through a sol, the, colloidal particles move towards the oppositely charged, electrode and the phenomenon is called electrophoresis., 18. According to Hardy-Schulze rule, greater the valency of, flocculating ion, greater is its power to cause precipitation., This rule takes into consideration only the charge not the size, of an ion. Since, both the size and the charge of an ion, determine polarising power hence, this rule can be modified, as greater the polarising power of flocculating ion, greater is, its power to cause precipitation., 19., , 20., , (iii) It adsorbs poisonous gases present in coal mines., (iv) Eosin dye gets adsorbed on the surface of white, precipitate of silver halide, making it coloured., (v) Cellophane., 22., , (i) The process of hardening of leather is known as tanning., Animal skin is colloidal in nature and has positively, charged particles. When they are soaked in tannin, their, mutual coagulation takes place and, thus leather becomes, hard after tanning., Note Tannin, which is obtained from a mixture of, polyhydroxy benzoic acids and contains negatively, charged colloidal particles., (ii) Lyophilic sols are more stable than lyophobic sols. This is, due to the fact that lyophilic colloids are extensively, solvated, i.e. colloidal particles are covered by a sheath of, the liquid in which they are dispersed., , (b) Blood is a colloidal sol. When we rub the injured, part with moist alum then coagulation of blood, takes place. Hence, clot is formed, which stops, the bleeding., (c) River water is colloidal solution of clay and sea, water contains a number of electrolytes. When, river water meets the sea water, the electrolytes, present in sea water coagulates the colloidal, solution of clay resulting in its deposition with, the formation of delta., , Examples, , Solid, , Solid, , Solid, , Some coloured, glasses, gem, stones, , Solid, , Liquid, , Sol, , Paints, cell fluids, , Solid, , Gas, , Aerosol, , Smoke, dust, , Liquid, , Solid, , Gel, , Cheese, buter,, jellies, , Liquid, , Liquid, , Emulsion Milk, hair cream, , Liquid, , Gas, , Aerosol, , Gas, , Solid, , Solid sol Pumice stone,, foam rubber., , Gas, , Liquid, , Foam, , Fog, mis, cloud,, insecticide sprays, , Froth, whipped, cream, soap lather., , Note Ninth possibility, i.e. a gas mixed with another gas is a, homogeneous system and not a colloid., , (ii) On the basis of nature of dispersed phase, collids are, classified as:, (a) Multimolecular collids, e.g. gold sol, sulphur etc., (b) Macromolecular colloids, e.g. cellulose, starch, etc., (c) Associated colloids, e.g. soaps, synthtic detergents., (iii) On the basis of interaction between dispersed phase and, dispersion medium, colloids are of two types:, (a) Lyophilic colloids, e.g. gum, gelatin etc., (b) Lyophobic colloids, e.g. metals, their sulphides etc., , (i) Both physisorption and chemisorption directly, depends on the surface area i.e. increases with an, increase in surface area of the adsorbent., (ii) Chemisorption has higher enthalpy of adsorption, i.e., 80-240 kJ mol -1 as it involves chemical bond formation, (in case of physisorption, it is 20- 40 kJ mol -1)., (iii) (a) We add alum to purify water as, it coagulates the, colloidal impurities present in water so that these, impurities settle down and get removed by, decantation or filtration., , (i) Depending on the basis of the physical states of the, dispersed phase and dispersion medium, eight types of, colloidal systems are formed. These are as follow, , Dispersed Dispersion Type of, phase, medium, colloid, , (i) NH3, adsorbed more readily than N 2 on the surface of, charcoal because, it is more easily liquefiable as, compared to N 2 and has a greater molecular size too., (ii) A finely divided substance is more effective as an, adsorbent because of the availability of its greater, surface area per unit mass of the adsorbent leading to, greater adsorption., , 23., , (i) According to Hardy-Schulze rule, ‘‘greater the, valency of the flocculating ion added, the greater is its, power to cause precipitation.’’, As Al 3+ ion due to greater valency has more, flocculating power than Mg2 + ion. Therefore, AlCl 3 is, more effective than MgCl 2 in causing coagulation of, negatively charged sol., (ii) Sulphur sol forms multimolecular colloids. It consists, of particles containing a thousand or more S 8 sulphur, molecules which associate together to form a, multimolecular colloid., (iii) (a) Desorption is the process of removing an, adsorbed substance from the surface to which it is, adsorbed., (b) Critical micelle concentration (or CMC) is the, minimum concentration required for the, formation of micelle., (c) It is a process of removing a dissolved substance, from a colloidal solution by means of diffusion, through a suitable membrane. For dialysis, process, colloidal solution is taken in a parchment, paper or cellophane membrane and this bag is, suspended in water. Impurities (molecules or ions, as small as solute of true solution) diffuse in water, through the membrane and colloids become pure., The apparatus as shown in figure, used for this, purpose is called dialyser.

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59, , CBSE Term II Chemistry XII, , 24., , 25., , (i) (a) Some of the reactant ions may be adsorbed or may, adhere to the surface of the particles of the precipitate, formed during an ionic reaction. In order to remove, these reactant ions, the precipitate should be washed, with water. If this is not done, an error may be, produced during quantitative analysis., (b) The nature of the substance whether colloid or, crystalloid depend upon size of the solute particles., When size of solute particles lies between 1 to, 1000 nm, it behaves as a colloid and when the size is, less than 1 nm it behaves as crystalloid., (c) As we know rainfall occurs when oppositely charged, clouds meet, since, clouds are colloidal in nature and, carry charge. Spray of silver iodide, an electrolyte,, leads to artificial rain. Sometimes electrified sand is, also used for this purpose., (ii) (a) Macromolecular colloids When substances which, possess very high molecular masses are dispersed in, suitable dispersion medium, the colloidal solution,, thus, formed are called macromolecular colloids., These colloids are quite stable, e.g. cellulose, starch, etc., (b) The process of conversion of a fress precipitate, into colloidal sol by shaking it with dispersion, medium in the presence of electrolyte is called, peptisation., , (iv) Sol - paints, Gel - butter, (v) Following method can be used to stabilised the, colloidal sol., l, , l, , 27., , (iv) 4% solution of nitrocellulose in a mixture of alcohol, and ether is called collodion., Or, The presence of equal and similar charges on colloidal, particles is largely responsible in providing stability to, the colloidal solution because having same charge, present them from coalescing or aggregating or, coagulating when they come to one another., 28., , (i) Easily liquefiable gases, i.e. with higher critical, temperatures are readily adsorbed as van der Waals’, forces are stronger near the critical temperature., Therefore, out of NH3 and CO 2 gas, NH3 gas will be, adsorbed more readily on the surface due to its high, critical temperature than CO 2 gas., (ii) Adsorption is an exothermic proces, i.e. energy factor, favours the process. As, DG = DH - TDS, in adsorption,, through DS is negative but DH is also negative and, DH > TDS in magnitude so that DG is negative. Hence,, the process is spontaneous., (iii) Application of adsorption in controlling humidity, Silica and aluminium gels are used as adsorbents for, removing moisture and controlling humidity., (iv) During adsorption, there is always a decrease in, residual or unbalanced forces of the surface. This, results in decrease in sufrace energy which appears, as heat, therefore adsorption in an exothermic process., Or Physisorption and chemisorption., , 29., , (i) Origin of charge on colloidal particles is due to, preferential adsorption of ions from the solution., (ii) The presence of equal and similar charges on colloidal, particles is largely responsible for providing stability to, the collodial solution, as the repulsive force between, charged particles having same charge prevent them, from coalescing or aggregating when they come close, to one another., (iii) When AgNO 3 is added to KI solution, the precipitated, AgI adsorbs iodide ions from the dispersion medium, and negatively charged colloidal sol results., (iv) Electrophoresis, Or Out of KI and K 2SO 4 , latter one is a better electrolyte, for coagulation of positive sol according to, Hardy-Schulze rule., , The resultant particles formed after adsorption are, of colloidal size. Thus, forming sol with remaining, dispersion medium. The complete reaction can be, seen as, Fe (OH) 3 + x Fe 3 + ¾® Fe(OH) 3 xFe 3+, Colloidal sol, , 26., , (i) van der Waals’ forces., (ii) If the dispersion medium is water, the sol is called, aquasol and if it is alcohol then called as alcosol., (iii) Dispersed phase - liquid, Dispersion medium - liquid, , (i) Diameter of colloidal particles vary between, 1-1000 nm, i.e. 10-9 - 10-6 m., (ii) The colloid which is formed when a liquid is dispersed, in a solid is gel, e.g. butter., (iii) Foam is the colloid that the colloid that obtained when, gas is dispersed in liquid, e.g. whipped cream., , (i) (a) When a freshly prepared precipitate of Fe(OH) 3 is, shaken with a small amount of FeCl 3 solution, sol, formation takes place. It is due to the preferential, adsorption of Fe 3 + of FeCl 3 over Fe(OH) 3 ., , (b) Persistent and prolonged dialysis of colloidal sol, remove even traces of electrolytes completely and, colloids become unstable and ultimately coagulate., (ii) When a beam of light is passed through true and, colloidal solutions each kept in a different glass vessel,, colloidal solution exhibits Tyndall effect whereas true, solution does not., (iii) Micelles Some substances (like soap) at higher, concentrations (10-4 to 10-3 mol/L in case of soap) show, colloidal behaviour due to the formation of aggregates., These aggregates particles are called micelles or, associated colloids., Colloidal solution Those are the solution in which the, diameter of dispersed phase particles may range from 1, to 100 nm. These are intermediate of true solution and, suspensions. The colloidal particles do not settle down, under the force of gravity even on long standing., , Solvation-colloidal particles are covered by a, sheath of liquid in which they are extensively, solvated, thereby providing stability., Electrostatic stabilisation-presence of equal and, similar charges on the colloidal particles prevents, coagulation of the colloidal sol.

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Chapter Test, , 6. Assertion Most of the medicines are more effective, in the colloidal form., Reason Colloidal form exhibit large surface area and, medicines are easily assimilated in this form., , Multiple Choice Questions, , 1. Adsorption of oxalic acid can be takes place on activated, charcoal. It this case, the activated charcoal is act as, (a) adsorbed, (c) adsorbent, , (b) adsorbate, (d) occlusion, , 2. Which of the following process is responsible for the, formation of delta at a place where rivers meet the sea?, (a) Colloid formation, (c) Peptisation, , (b) Coagulation, (d) None of these, , Short Answer Type Questions, , (a) It is not specific in nature, (b) It arises because of van der Waals’ forces, (c) It is reversible in nature, (d) Enthalpy of adsorption is in range 80-240 kJ mol -1, , 9. (i) Explain the following terms with suitable examples:, , 4. The protecting power of lyophilic colloidal sol is expressed, in terms of, , passed through a colloidal solution?, (ii) What is Helmholtz electrical double layer?, , 11. (i) Explain the electrophoresis., , 5. Match the items of Column I and Column II., Column I, , Column II, , A. Butter, , 1. Dispersion of liquid in liquid, , B. Pumice stone, , 2. Dispersion of solid in liquid, , C. Milk, , 3. Dispersion of gas in solid, , D. Paint, , 4. Dispersion of liquid in solid, , Codes, , A, (b) 2, (d) 3, , B, 3, 2, , C, 1, 1, , (ii) What happens when dialysis is prolonged?, (iii) Why is Fe(OH) 3 colloid positively charged, when, prepared by addingFe(OH) 3 to hot water?, , 12. (i) Write an important characteristic of lyophilic sols., (ii) Based on the type of particles of dispersed phase,, give one example each of associated colloid and, multimolecular colloid., , 13. (i) How can a colloidal solution and true solution, , Choose the correct code that are given below, D, 2, 3, , (a) Aerosol, (b) Hydrosol, (ii) Why the colour of water is blue?, , 10. (i) What is especially observed when a beam of light is, , (a) coagulating value, (b) gold number, (c) critical micelle concentration, (d) oxidation number, , C, 1, 2, , 8. Assertion Substances whose solutions cannot pass, Reason The size of colloidal particles are smaller, than the size of suspension particles., , physisorption., , B, 3, 4, , Na+ ., Reason Greater the valency of the flocculating ion, added, greater is its power to cause precipitation, (Hardy-Schulze rule)., throughout filter paper are called as colloids., , 3. Choose the incorrect statement with respect to, , A, (a) 4, (c) 1, , 7. Assertion Coagulation power of Al3+ is more than, , D, 4, 4, , of the same colour be distinguished from each, other?, (ii) Why is water alone not very effective for removing, dirt from clothes?, , Long Answer Type Questions, , Assertion-Reasoning MCQs, Direction (Q. Nos. 6-8) Each of these questions contains, two statements Assertion (A) and Reason (R). Each of, these questions also has four alternative choices, any one, of which is the correct answer. You have to select one of, the codes (a), (b), (c) and (d) given below., (a) Both A and R are true and R is the correct explanation of A., (b) Both A and R are true, but R is not the correct explanation, of A., (c) A is true, but R is false., (d) A is false, but R is true., , 14. Write any three methods for preparation of colloid, and explain the various methods used in the, purification of colloid., , 15. (i) What are the factors which influence the, adsorption of a gas on a solid?, (ii) What is meant by critical micelle concentration?, (iii) Which will be adsorbed more readily on the surface, of charcoalNH3 orO 2 and why?, (iv) What is common in aqua and solid sols?, (v) How is adsorption of a gas related to its critical, temperature?, , Answers, Multiple Choice Questions, 1. (c), , 2. (b), , 3. (d), , 4. (b), , Assertion-Reasoning MCQs, 6. (a), , 7. (a), , 8. (d), , 5. (a), , For Detailed Solutions, Scan the code

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CHAPTER 04, , The d - and f- Block, Elements, In this Chapter..., l, , d -Block Elements, , l, , f-Block Elements, , The d-block of the periodic table includes the elements of, the groups 3 to12, in which the d-orbitals are, progressively filled in each of the four long periods., The elements constituting the f- block are those in which, the 4 f- and 5 f- orbitals are progressively filled in the latter, two long periods., , d-Block Elements, (Transition Elements), l, , l, , l, , l, , l, , The elements in which the differentiating electron, enters in ( n -1 ) d-sub level are called d-block, elements (group 3-12)., The d-block elements in which the atoms or ions, have incomplete d-orbitals are called transition, elements., There are mainly four series of the transition metals,, 3d-series (Sc to Zn), 4d- series (Y to Cd), 5 d-series, (La and Hf to Hg) and 6d-series which has Ac and, elements from Rf to Cn., Zn, Cd and Hg have completely filled ( n - 1 ) d-orbitals,, so they do not show much resemblance with other, transition elements., The d-block occupies the large middle section of the, periodic table flanked between s and p -blocks in the, periodic table., , Electronic Configuration, In general, the electronic configuration of outer orbitals of these, elements is ( n - 1 )d 1 - 10 ns 1 - 2 . Half and completely filled sets of, orbitals are relatively more stable., e.g. Cr ( 3 d 5 4 s 1 ) and Cu ( 3 d 10 4 s 1 )., Exception, Last members of different series such as Zn, Cd and Hg have, completely filled d-orbitals ( n -1 ) d 10 ns 2 in the ground state as, well as in their common oxidation states. These do not show, any characteristic properties of transition elements except, complex formation. Hence, these elements are not considered, as transition elements., Some of the exceptions are observed in electronic, configuration due to very little energy difference between, ( n -1 )d and ns-orbitals such as, Cr : 3 d 5 , 4 s 1 instead of 3 d 4 , 4 s 2, Cu : 3 d 10 , 4 s 1 instead of 3 d 9 , 4 s 2, l, , l, , General Properties of Transition Elements, The general properties of transition elements are as follows, Physical Properties, l, , Nearly, all the transition elements (except Zn, Cd, Hg and Mn), display typical metallic properties such as high tensile strength,, ductility, malleability, high thermal and electrical conductivity, and metallic lustre.

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62, l, , The transition metals (except Zn, Cd and Hg) are very hard, and have low volatility., , CBSE Term II Chemistry XII, , l, , Melting and Boiling Points, , Their melting and boiling points are high due to involvement of, greater number of electrons from ( n -1 ) d in addition to ns, electrons in the interatomic metallic bonding., In any row, the melting point of these metals rise to a maximum, at d 5 except for anomalous values of Mn and Tc and fall regularly, as the atomic number increases., Enthalpies of Atomisation, , They have high enthalpies of atomisation. The metals of the, second and third series have greater enthalpies of atomisation, than the corresponding elements of the first series., , l, , l, , l, , l, , Atomic and Ionic Radii, l, , l, , l, , l, , In general, ions of the same charge in a given series show, progressive decrease in radius with increasing atomic number., The shielding effect of a d-electron is not that effective,, hence, the net electrostatic attraction between the nuclear, charge and the outermost electron increases and the ionic, radius decreases., The filling of 4 f -orbitals before 5 d-orbitals result in a regular, decrease in atomic radii called lanthanoid contraction which, essentially compensates for the expected increase in atomic size, with increasing atomic number., Density The decrease in metallic radius coupled with increase, in atomic mass results in a general increase in the density of, these elements., , Ionisation Enthalpies, l, , l, , l, , There is an increase in ionisation enthalpy along each series of, the transition elements from left to right due to an increase in, nuclear charge which accompanies the filling of inner d-orbitals., The first ionisation enthalpy, in general increases, but the, magnitude of the increase in the second and third ionisation, enthalpies for the successive elements is much higher along a, series. Zn, Cd and Hg have highest ionisation energy., The trend of steady increase in second and third ionisation, enthalpy breaks for the formation of Mn 2+ and Fe 3+, respectively., , Oxidation States, l, , l, , l, , l, , The transition elements show variety of oxidation states due to, very small difference between ( n -1 ) d and ns-orbitals., The elements which give the greatest number of oxidation, states occur in or near the middle of the series., e.g. Mn show +2 to +7 oxidation state., Low oxidation states are found when a complex compound has, ligands capable of p-acceptor character in addition to, s -bonding. e.g. Ni(CO) 4 and Fe(CO) 5 ., The highest oxidation numbers are achieved in TiX 4 , VF5 and, CrF6 . Beyond Mn, no metal has a trihalide except FeX 3 and, CoF3 ., , The ability of fluorine to stabilise the highest oxidation, state is due to either higher lattice enthalpy or higher, bond enthalpy for higher covalent compounds., Cu 2+ is more stable than Cu + due to lower reduction, potential and highly negative hydration lattice energy of, Cu 2+ compounds., The highest oxidation number in the oxides coincide, with the group number and is attained in Sc 2 O 3 to, Mn 2 O 7 . Beyond group 7, no higher oxides of Fe above, Fe 2 O 3 are known., Besides the oxides, oxocations stabilise V v as VO+2 ,, V iv as VO2+ and Ti iv as TiO 2+ ., The ability of oxygen to stabilise these high, oxidation states exceeds that of fluorine. Thus, the, highest Mn fluoride is MnF4 , whereas the highest oxide, is Mn 2O 7 ., , M 2+ /M Standard Electrode Potential, l, , l, , l, , l, , There is no regular trend for M 2+ /M standard electrode, potential due to the irregularities in ionisation, enthalpies and enthalpy of atomisation., The general trend towards less negative E° values, across the series is related to the general increase in the, sum of the first and second ionisation enthalpies., The unique behaviour of Cu having a positive E° is due, to its inability to liberate H 2 from acids., The value of E° for Mn, Ni and Zn are more negative, than expected from the trend., , M 3+ / M 2+ Standard Electrode Potential, l, , l, , l, , There is no regular trend for M 3+ / M 2+ standard, electrode potential., ° 3+ 2+, The highest EM, for Zn is due to the removal of, /M, an electron from the stable d 10 configuration of Zn 2+ ., Similarly, low value for Sc reflects the stability of Sc 3+ ,, which has a noble gas configuration., The comparatively high value for Mn shows that M 2+, ( d 5 ) is stable, whereas low value of Fe shows the extra, stability of Fe 3+ ( d 5 ). The low value for V is related to, the stability of V 2+ ( half-filled t 2 g level)., , Chemical Reactivity and E ° Values, l, , l, , l, , The metals of the first series with the exception of, copper are relatively more reactive and are oxidised, by 1M H + ., The E° values for M 2+ / M decreases to form divalent, cations across the series. An examination of the E° values, for the redox couple M 3+ / M 2+ shows that Mn 3+ and, Co 3+ ions are the strongest oxidising agents in aqueous, solutions., The ions Ti 2+ , V 2+ and Cr 2+ are strong reducing, agents and will liberate hydrogen from dilute acid.

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63, , CBSE Term II Chemistry XII, , Magnetic Properties, l, , l, , Alloy Formation, , Diamagnetic substances are repelled by the applied, field, while the paramagnetic substances are attracted., Substances which are attracted very strongly by the, applied field are said to be ferromagnetic. Many of the, transition metal ions are paramagnetic., The magnetic moment is determined by, m = n( n + 2), where, n = number of unpaired electrons and m is the, magnetic moment in units of Bohr Magneton (BM)., , l, , l, , l, , l, , Brass, Bronze, Bell metal, Gun metal, Monel metal, , The magnetic moment increases with the increasing, number of unpaired electrons., , Formation of Coloured Ions, l, , l, , If unpaired electron is present, complex is coloured due, to d-d transition and also, paramagnetic in nature., If unpaired electron is absent, complex or compound is, colourless due to the absence of d-d transition and, diamagnetic in nature., , l, , Transition metals due to their small size, high charge, density,presence of vacant d-orbital for bond formation, can form complex., Transition elements and their compounds show catalytic, properties, due to variable oxidation states and ability to, form complex., , 70-80% Cu + 30-20% Zn, 80-90% Cu + 20-10% Sn, 72-80% Cu + 28-20% Zn, 85% Cu + 5% Sn + 10% Zn, 70% Cu + 30% Ni, , (Inner Transition Elements), l, , l, , Formation of Complex Compounds, l, , –, –, –, –, –, , f-block Elements, , 0, , Note Few compounds in which transition metal has d, configuration, may also exhibit colour due to charge transfer., e.g. CrO 3 has, d 0 configuration of Cr and shows orange colour due to charge, transfer., , Alloys are homogeneous solid solutions in which the atoms of, one metal are distributed randomly among the atoms of the, other., Because of similar radii and other characteristics of transition, metals, alloys are readily formed by these metals. Alloys so, formed are hard and often have high melting points., Some alloys of Cu,, , l, , The f -block consists of elements in which 4 f -and 5 f -orbitals, are progressively filled. They are placed in a separate panel at, the bottom of the periodic table. They are known as inner, transition elements., The two series of the inner transition metals are 4 f (Cl to Lu), and 5 f (Th to Lr) are known as lanthanoids and actinoids, respectively., Lanthanum and actinium does not belong to f -block, elements as they do not have any electron in 4 f and, 5 f - subshell respectively. However, their properties resemble, with elements of 4 f and 5 f -series., , Lanthanoids, , Catalytic Properties, , Some important properties of lanthanoids are discussed below., , Transition metals have ability to adopt multiple oxidation, states and to form complexes., For example, Vanadium (V) oxide (in contact process),, finely divided iron (in Haber’s process) and nickel (in, catalytic hydrogenation)., They can change their oxidation states and act as effective, catalysts. e.g., 2Fe 3+ + 2I - ¾® 2Fe 2+ + I 2, 2Fe 2+ + S 2O 28- ¾® 2Fe 3+ + 2SO 42-, , Electronic Configuration, The lanthanoids include Ce to Lu and have electronic, configuration with 6 s 2 common but with variable occupancy of 4, f level., , Formation of Interstitial Compounds, Transition metals also have a tendency to form, interstitial compounds with H, C, B or N-atoms., They are usually non-stoichiometric and are neither, typically ionic nor covalent. e.g. TiC, Mn 4 N, Fe 3 N,, VH 0.56 and TiH1 .7 etc., The interstitial compounds are very hard, retain metallic, conductivity have high melting points and are, chemically inert., l, , l, , l, , Atomic and Ionic Radii, The atomic and ionic radii decrease from lanthanum to lutetium., This is due to the unique property of lanthanoids which is known, as lanthanoid contraction., Lanthanoid Contraction, The filling of 4f-orbital before 5d-orbital results in a regular, decrease in atomic radii called lanthanoid contraction. The, factor responsible for the lanthanoid contraction is similar to that, of ordinary transition series, i.e. the imperfect shielding of one, electron by another in the same set of orbitals., However, the shielding of one 4f-electron by another is less than, that of one d-electron by another and as the nuclear charge, increases along the series, there is fairly regular decrease in the, size of the entire 4 f n orbitals.

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64, , CBSE Term II Chemistry XII, , The decrease in atomic radii is not regular but the decrease, in ionic radii of their M 3 + ions are quite regular as, represented by the following graph., , Consequences of Lanthanoid Contraction, , (i) Basic character of oxides and hydroxides Due to the, lanthanoid contraction, the covalent nature of, Ln¾OH bond increases as the basic character of oxides, and hydroxides decreases from La(OH) 3 to Lu(OH) 3 ., (ii) Separation of lanthanoids Due to lanthanoid contraction,, there is a small difference in the size of lanthanoids. There, is the difference in some properties of lanthanoids, likesolubility, degree of hydration and complex formation., These differences enable the separation of lanthanoids by, ion exchange method., , Sm2+, , 110, , Eu2+, La3+, Ce3+, Pr3+, Nd3+, Pm3+, Sm3+, Eu3+, Tm2+, Gd3+, 3+, Yb2+, 4+, Tb, Ce, 3+, Dy, Pr4+, Ho3+, Er3+, Tm3+, 4+, Tb, Yb3+, Lu3+, , Ionic radii/pm, , 100, , 90, , 57, , 59 61 63 65 67 69, Atomic number, Trends in ionic radii of lanthanoids, , 71, , Oxidation States, In the lanthanoids, La (II) and Ln (III) compounds are, predominant species. However, occasionally +2 and +4 ions, in solution or in solid compounds are also obtained. This, irregularity arises mainly from the extra stability of empty,, half-filled or fully-filled f -subshell., e.g. Ce 4 + acts as an oxidising agent by converting into Ce 3+ ,, whereas Eu 2+ and Yb 2+ acts as reducing agent by, converting into Eu 3+ and Yb 3+ respectively., l, , l, , Pr, Nd, Tb and Dy also exhibit +4 state but only in oxides,, MO 2 . The behaviour of samarium is very much like, europium exhibiting both +2 and +3 oxidation states.

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65, , Chapter, Practice, 6. Although zirconium belongs to 4d-transition series, , PART 1, Objective Questions, l, , and hafniun to 5 d-transition series even then they, show similar physical and chemical properties, because ............. ., (NCERT Exemplar), , Multiple Choice Questions, , 1. The electronic configuration of Cu(II) is 3d 9 where, , (a) both belong to d-block, (b) both have same number of electrons, (c) both have similar atomic radius, (d) both belong to the same group of the periodic table, , as that of Cu(I) is 3d 10 . Which of the following is, correct?, (NCERT Exemplar), , 7. Which of the following is a paramagnetic ion?, , (a) Cu(II) is more stable, (b) Cu(II) is less stable, (c) Cu(I) and Cu(II) are equally stable, (d) Stability of Cu(I) and Cu(II) depends on nature of, copper salts, , 8. Identify the configuration of transition element,, , 2. Transition element which does not show variable, oxidation state is, (a) Ti, (c) Fe, , +4 (MnF4 ) but highest oxidation state in oxides is, +7(Mn 2O 7 ) because, (NCERT Exemplar), , (a) fluorine is more electronegative than oxygen, (b) fluorine does not possess d-orbitals, (c) fluorine stabilises lower oxidation state, (d) in covalent compounds, fluorine can form single bond, only while oxygen forms double bond, , 4. Metallic radii of some transition elements are given, below. Which of these elements will have highest, density?, (NCERT Exemplar), Element, , Fe, , Co, , Ni, , Cu, , Metallic radii/pm, , 126, , 125, , 125, , 128, , (b) Ni, , (b) Ni 2+, (d) Ag +, , which shows highest magnetic moment., (a) 3 d 7, (c) 3 d 8, , (b) 3 d5, (d) 3 d 2, , (NCERT Exemplar), , 9. Arrange the following in increasing order of number, , (b) V, (d) Zn, , 3. Highest oxidation state of manganese in fluoride is, , (a) Fe, , (a) Zn 2+, (c) Cu +, , (c) Co, , (d) Cu, , 5. Which one of the following does not correctly, represent the correct order of the property indicated, against it?, (a) Ti < V < Cr < Mn, increasing number of oxidation states, (b) Ti 3+ < V3+ < Cr 3+ < Mn 3+ , increasing magnetic moment, (c) Ti < V < Cr > Mn, increasing melting points, (d) Ti < V < Mn < Cr, increasing 2nd ionisation enthalpy, , of unpaired electrons., I. Cu 2+ II. Cr 3+ III. Ni 2+ IV. Fe 3+, (a) I > III > II > IV, (c) IV > II > III > I, , (b) III > I > IV > II, (d) II > IV > III > I, , 10. The ‘spin-only’ magnetic moment [in units of Bohr, magneton, ( m b )] of Ni 2+ in aqueous solution would, be (atomic number of Ni = 28), (a) 2.84, (c) 0, , (b) 4.90, (d) 1.73, , 11. Generally, transition elements form coloured salts, due to the presence of unpaired electrons. Which, of the following compounds will be coloured in, solid state?, (NCERT Exemplar), (a) Ag 2 SO4, (c) ZnF2, , (b) CuF2, (d) Cu 2 Cl 2, , 12. Out of TiF62- , CoF63- , Cu 2Cl 2 and NiCl 24 (at. no. of, Ti = 22, Co = 27, Cu = 29 and Ni = 28 ), the colourless, species are, (a) TiF62- and CoF63–, (b) Cu 2 Cl 2 and NiCl42(c) TiF62- and Cu 2 Cl 2, (d) CoF63– and NiCl42-

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66, , CBSE Term II Chemistry XII, , 13. Which of the following is not the characteristic, property of interstitial compounds?, , (a) They have high melting points in comparison to pure, metals, (b) They are very hard, (c) They retain metallic conductivity, (d) They are chemically very reactive, , 14. Outermost electronic configuration of lanthanide is, (a) 4 f 1 - 14 , 5 d 0 , 6 s 2, , (b) 4 f 1 - 14 , 5 d 0 - 1 , 6 s 2, , (c) 4 f 0 -14 , 5 d 0 - 2 , 6 s 2, , (d) 4 f 0 - 14 , 5 d1 , 6 s 2, , 15. Lanthanoid for which + II and + III oxidation states, are common, (a) La, , (b) Nd, , (c) Ce, , (d) Eu, , 16. Which of the following oxidation state is common for, all lanthanoids?, (a) +2, , (b) +3, , (NCERT Exemplar), , (c) +4, , (d) +5, , 17. Although +3 is the characteristic oxidation state for, , lanthanoids but cerium also shows +4 oxidation state, because, I. it has variable ionisation enthalpy, II. it has a tendency to attain noble gas configuration, III. it has a tendency to attain f 0 configuration, IV. it resembles Pb 4+, (a) Only II, , (b) II and III, , (c) I and IV (d) II and IV, , 18. As the atomic number of lanthanoid increases, the, atomic radius decreases, but exception is, (a) Ga and Eu, (c) Nd and Ho, l, , (b) Eu and Yb, (d) Dy and Ho, , Assertion-Reasoning MCQs, Direction (Q. Nos. 19-23) Each of these questions, contains two statements Assertion (A) and Reason (R)., Each of these questions also has four alternative choices,, any one of which is the correct answer. You have to select, one of the codes (a), (b), (c) and (d) given below., (a) Both A and R are true and R is the correct, explanation of A., (b) Both A and R are true, but R is not the correct, explanation of A., (c) A is true, but R is false., (d) A is false, but R is true., , 19. Assertion Cr and Cu are extra stable elements in the, 3d-series., Reason Half and completely filled set of orbitals are, relatively more stable., , 20. Assertion Cu, Reason Cu, , 2+, , 22. Assertion Separation of Zr and Hf is difficult., Reason Because Zr and Hf lie in the same group of, the periodic table., (NCERT Exemplar), , (NCERT Exemplar), , 2+, , iodide is not known., oxidises I - to iodine., (NCERT Exemplar), , 21. Assertion The highest oxidation state of osmium is +8 ., Reason Osmium is a 5d-series element.(NCERT Exemplar), , 23. Assertion La(OH) 3 is more basic than Lu(OH) 3 ., Reason Size of Lu 3 + ions increases along the, lanthanide series shows less covalent character., , l, , Case Based MCQs, , 24. Read the following and answer the questions from, (i) to (iv) given below, “The elements in which the last differentiating, electrons enters into d-orbital of the penultimate, shell, i.e. ( n - 1)d where, n is the last shell are, called d-block elements”., A transition element may also be defined as the, element which has partially filled d-orbital in their, ground state or most stable oxidation state., The properties of these elements are intermediate, between the properties of s- and p-block elements., The atomic radii of the transition metals lie in, between those of s- and p-block elements. Except, mercury, all the transition elements have typical, metallic structure. They have high melting point., Their first ionisation enthalpies lies between, s-block and p-block elements., They are higher than those of s-block and are lesser, than those of p-block elements with exception of, few elements, most of the d-block elements shows, more than one oxidation state., Thermodynamic stability of the compounds of, transition elements can be evaluated in terms of, the magnitude of ionisation enthalpies of the, metals. Most of these transition elements and their, compounds have good catalytic properties. Their, ions forms a large number of complex compounds., (i) The atomic radii decreases less rapidly in case of, 3d-series because, (a) 4s-electrons shield 3d-electrons more effectively, than the other outershell electrons, (b) 3d-electrons shield 4s-electrons more effectively, than the other outershell electrons, (c) 3d-electrons shield 4s-electrons less effectively than, the other outershell electrons, (d) 4s-electrons shield 3d-electrons less effectively than, the other outershell electrons, , (ii) The third ionisation enthalpy is minimum for, (a) Mn, , (b) Ni, , (c) Co, , (d) Fe, , (iii) Four successive members of the first series of the, transition metals are listed below. For which one, °, of them, the standard potential (E M, ) value, 2+, /M, has a positive sign?

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67, , CBSE Term II Chemistry XII, , (a) Co (Z = 27), (c) Cu ( Z = 29), , Or E °, , (b) Ni ( Z = 28), (d) Fe ( Z = 26), , Mn 3+ / Mn 2 +, , ° 3+ 2 +, > E Cr, / Cr, , This trend is found because, , (iv) Cuprous ion is colourless, while cupric ion is, coloured because, , (a) Mn 2 + can be easily oxidised to Mn 3+ due to low, ionisation enthalpy, (b) third IE of Mn is much larger due to its stable, configuration, (c) Mn 3+ is more stable than Mn 2 + due to its higher, oxidation state, (d) second IE of Mn is higher than its third IE, , (a) both have unpaired electrons in the d-subshell, (b) cuprous ion has a complete d-subshell and cupric, ion has an incomplete d-subshell, (c) both have half-filled p- and d-subshell, (d) cuprous ion has incomplete d-subshell and cupric, ion has a complete d-subshell, , Or, ° 2+ / M for which of the following, The value of EM, transition elements are more negative than, expected?, , PART 2, Subjective Questions, , (a) Cr, Mn and Zn, (c) V, Mn and Cu, , l, , (b) Mn, Ni and Zn, (d) Fe, Co and Ni, , 25. Read the following and answer the questions from (i), , to (iv) given below, The elements which have incompletely filled d-orbitals, in its ground state or in any of its oxidation state are, called d-block elements. (transition elements)., As zinc (Zn), cadmium (Cd), mercury (Hg) of group 12, have full d 10 -configuration it their ground state as well, as in their common oxidation states, thus they are not, regarded as transition metals, but they are studied, along with the chemistry of the transition metals., Various precious metals like silver, gold and platinum, and industrially important metals like iron, copper and, titanium form part of the transition metals., (i) Ground state configuration of element having atomic, number = 24 is 3d 5 4s 1 . This is because,, (a) the energy of 3d and 4 s-orbitals is very low, (b) 3d and 4s-orbitals are of comparable energy, (c) energy of 3d-orbital is lower than the energy of, 4s-orbital, (d) Energy gap between third and fourth shell is very small, , (ii) The filling of 4 f before 5 d-orbital results in a, regular decrease in atomic radii called, lanthanoid contraction, shielding effect, actinoids contraction, None of the above, , (a) Sc ( Z = 21 ), (b) La 3+ ( Z = 57), (c) Ti 3+( Z = 22), (d) La 3+ ( Z = 71 ), , (NCERT Exemplar), , 5. Why E° values for Mn, Ni and Zn are more negative, than expected?, 6. Explain the following, (i) The second and third rows of transition elements, resemble each other much more than they resemble, the first row., (NCERT Exemplar), (ii) Copper can not replace hydrogen from acids., (NCERT Exemplar), , electrode potentials of elements from Ti to Zn in the, first reactivity series are depicted in figure, , (b) scandium, (d) zinc, , (iv) Which of the following ions will exhibit colour in, aqueous solutions?, 3+, , abnormal electronic configuration and explain why?, 2. Explain, why density of transition elements increases, from left to right in a period?, 3. Why do transition elements show variable oxidation, states?, (i) Name the element showing maximum number of, oxidation states among the first series of transition, metals from Sc ( Z = 21) to Zn ( Z = 30 )., (ii) Name the element which shows only +3 oxidation, state., (Delhi 2020), 4. Explain briefly how + 2 state becomes more and more, stable in the first half of the first row transition, elements with increasing atomic number?, , 7. Observed and calculated values for the standard, , (iii) The transition element which does not exhibit, variable oxidation states is, (a) titanium, (c) copper, , 1. Name the two transition elements which have, , Standard electrode potential/V, , (a), (b), (c), (d), , Short Answer Type Questions, , 0.5, 0, –0.5, –1, –1.5, –2, Ti, , V, , Cr Mn Fe Co Ni Cu Zn, , Observed values, , Calculated values

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68, , CBSE Term II Chemistry XII, , (iii) Zn, Cd and hg are quite soft and have low melting, points., (Delhi 2020), , Explain the following observations, (i) The general trend towards less negative E° values, across the series., (ii) The unique behaviour of copper., (iii) More negative E° values of Mn and Zn., , 17. (i) Consider the hydrated ions of Ti 2+ , V 2+ , Ti 3+ and, Sc 3+ . Write the correct order of their spin-only, magnetic moment, (ii) What are misch metal?, , 8. Explain the following observations., (i) Copper atom has completely filled d-orbitals ( 3 d 10 ), in ground state, yet it is regarded as a transition, element., (ii) Cr 2+ is a stronger reducing agent than Fe 2+ in, aqueous solution ., , 18. Give reasons for the following, (i) Transition metals form alloys., (ii) Eu 2+ is a strong reducing agent., , 19. What are inner-transition elements? Describe which, of the following are the atomic numbers of the, inner-transition elements : 29, 59, 74 ?, , 9. Calculate the number of unpaired electrons in the, , following gaseous ions Mn 3+ , Cr 3+ , V 3+ and Ti 3+., Which one of these is the most stable in aqueous, solution?, (NCERT Exemplar), , 20. Given reason, (i) Although Zr belongs to 4d and Hf belongs to 5 d, transition series but it is quite difficult to separate, them., (NCERT Exemplar), (ii) Although +3 oxidation states is the characteristic, oxidation state of lanthanoids but cerium shows +4, oxidation state also. Why?, (NCERT Exemplar), , 10. Use Hund’s rule to derive the electronic, , configuration of Ce 3+ ion and calculate its magnetic, moment on the basis of ‘spin only’ formula., (NCERT Exemplar), , 11. Calculate the ‘spin only’ magnetic moment of, , M 2+ ( aq) ion. ( Z = 27), (NCERT Exemplar), 12. (i) Calculate the magnetic moment of a divalent ion in, aqueous solution [Fe(H 2O) 6 ] 2+, if atomic number, of Fe is 26., (ii) Although fluorine is more electronegative than, oxygen, but the ability of oxygen to stabilise higher, oxidation states exceeds that of fluorine. Why?, (iii) Although Cr 3+ and Co 2+ ions have same number of, unpaired electrons but the magnetic moment of, Cr 3+ is 3.87 BM and that of Co 2+ is 4.87 BM. Give, reason., , 13. Give reasons, (i) E° value for Mn 3 + / Mn 2 + couple is much more, positive than that for Fe 3 + / Fe 2 + ., (ii) Iron has higher enthalpy of atomisation than that of, copper., (iii) Sc 3 + is colourless in aqueous solution whereas, Ti 3 + is colured., (CBSE 2018), , 14. Predict which of the following will be coloured in, aqueous solution?, Ti 3+ , V 3+ , Cu + , Sc 3+ , Mn 2+ , Fe 3+ and Co 2+. Give, reasons for each., (NCERT Exemplar), 15. (i) Why Zn 2+ salts are white while Ni 2+ salts are blue?, (ii) Out of [Sc(H 2O) 6 ] 3+ and [Ti(H 2O) 6 ] 3+ ions, which, is coloured and why?, , 16. Give reasons for the following, (i) Iron is a transition metal while sodium is not., (ii) Cd 2+ salts are white., , Long Answer (LA) Type Questions, 21. Account for the following., , l, , (i) Mn 2+ is more stable than Fe 2+ towards oxidation to, +3 state., (ii) The enthalpy of atomisation is lowest for Zn in, 3d-series of the transition elements., (iii) Identify the metal in MO 3 F and justify your answer., (iv) The E° value for the Mn 3+ / Mn 2+ couple is much, more positive than that for Cr 3+ / Cr 2+ couple., (v) Transition metals form a large number of complexes., 22. (i) Answer the following questions, (a) Which element of the first transition series has, highest second ionisation enthalpy?, (b) Which element of the first transition series has, highest third ionisation enthalpy?, (c) Which element of the first transition series has, lowest enthalpy of atomisation?, (ii) Identify the metal and justify your answer., Carbonyl M(CO)5, (iii) Although F is more electronegative than O, the, highest fluoride of Mn is MnF4 , whereas the, highest oxide is Mn 2O 7 . Give reason., , 23. Explain giving reasons., (i) The enthalpies of atomisation of the transition, metals are high., (ii) Transition metals and many of their compounds, show paramagnetic behaviour., (iii) The transition metals generally form coloured, compounds.

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69, , CBSE Term II Chemistry XII, , (iv) Transition metals and their many compounds act as, good catalyst., (v) Ni 2 + is more stable than Pt 2 + whereas Pt 4 + is, more stable than Ni4 + . (NCERT Exemplar) (Delhi 2020), , 24. (i) Name a compound of, (a) transition metal which is used in the, manufacture of sulphuric acid., (b) transition metal which is used in Haber’s, process for the manufacture of ammonia., (c) transition metal that has light sensitive, properties and act as valuable source in, photographic industry., (ii) Reactivity of transition elements decreases almost, regularly from Sc to Cu. Explain. (NCERT Exemplar), 25. Mention the type of compounds formed when small, , atoms like H, C and N get trapped inside the crystal, lattice of transition metals. Also give physical and, chemical characteristics of these compounds., (NCERT Exemplar), , 26. (i), , (a) Which transition element in 3d series has positive, E °M 2 + / M value and why?, (b) Name a member of lanthanoid series which is well, known to exhibit +4 oxidation state and why?, , (ii) Given reason for the following observations., (a) The highest oxidation state is exhibited in oxonions, of transition metals., (b) Transition elements show high melting points., (c) First ionisation enthalpy of Cr is lower than that of, Zn?, , 27. On the basis of lanthanoid contraction, explain the, , following, (i) Nature of bonding in La 2O 3 and Lu 2O 3 ., (ii) Trends in the stability of oxo salts of lanthanoids, from La to Lu., (iii) Stability of the complexes of lanthanoids., (iv) Radii of 4d and 5 d-block elements., (v) Trends in acidic character of lanthanoid oxides., l, , Case Based Questions, , 28. Read the following and answer the questions from (i), to (iv) given below, General electronic configuration of transition metals, is ( n - 1)d 1 - 10 n 1 s - 2 , where n the outermost shell, which may have one or two electrons, whereas ( n - 1), stands for the inner d-orbitals which may have one to, ten electrons., Transition elements display typically metallic, property such as high tensile strength, ductility,, malleability and lustre. With the exceptions of Zn,, , Cd, Hg and Mn, they have one or more typical, metallic structures at normal temperature., Transition metals shows a great variety of oxidation, states in its compounds, except few elements. The, elements, which give the greatest number of, oxidation states occur in or near the middle of the, series. Stability of higher oxidation states increases in, a group form top to bottom due to increase in the, distance from the outermost shell., (i) Why are Mn 2+ compounds more stable than Fe 2+, towards oxidation to their +3 state?, (ii) Why is the highest oxidation state of a metal, exhibited in its oxide or fluoride only?, (iii) Explain, why Cu + ion is not stable in aqueous, solution?, (iv) Which metal in the first series of transition metals, exhibit + 1 oxidation state most frequently and why?, Or, 3+, Explain why Mn is a good oxidising agent?, , 29. Read the following and answer the questions from (i), to (iv) given below, The d-block occupies the large middle section, flanked between s- and p-blocks in the periodic table., The name ‘transition’ given to the elements of, d-block is only because of their position between sand p-block elements., The d-orbitals of the penultimate energy level in, their atoms receive electrons giving rise to the three, rows of the transition metals, i.e. 3d, 4d, and 5d. The, fourth row of 6d is still incomplete. First transition, series or 3d-series: Scandium (Sc) to Zinc (Zn)., Second transition series or d-series: to cadmium (Cd)., Third transition series or 5d-series: Lanthanum (La)., to Mercury (Hg). Excluding Ce to Lu, Fourth transition series or 6d-series: Actinium (Ac), to Copernicium (Cu) excluding Th to Lr., (i) Name the element which shows outer electronic, configuration 3 d 2 4 s 2 ., (ii) Silver atom has completely filled d-orbitals ( 4 d 10 ) in, its ground state. How can you say that it is a, transition element?, (iii) In the series Sc (Z = 21) to Zn (Z = 30 ), the enthalpy, of atomisation of zinc is lowest, i.e. 126 kJ mol -1 ., Why?, (iv) Which transition metal of the 3d-series exhibits the, largest number of oxidation states and why?, Or, o, The E M, value, of, copper is positive (+ 0.34 V)., 2+, /M, What is the possible reason for this?

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70, , CBSE Term II Chemistry XII, , EXPLANATIONS, 10. (a) Ni 2+ = [Ar] 3d 8, , Objective Questions, 1. (a) Cu(II) is more stable than Cu(I). As it is known that,, Cu(I) has 3 d10 stable configuration while Cu(II) has 3 d 9, configuration. But Cu(II) is more stable due to greater, effective nuclear charge, i.e. it holds 17 electrons instead of, 18 in Cu(I)., 2. (d) Due to completely filled d-orbital Zn shows only + 2, oxidation state., 3. (d) Highest oxidation state of manganese in fluoride is, +4 (MnF4 ) but highest oxidation state in oxides is +7(Mn 2O 7 )., The reason is that in covalent compounds fluorine can form, single bond while oxygen forms double bond., 4. (d) On moving left to right along period, metallic radius, decreases while mass increases. Decrease in metallic radius, coupled with increase in atomic mass results in increase in, density of metal., Hence, among the given four choices Cu belongs to right, side of periodic table in transition metal, and it has highest, density (89 g/cm 3)., 5. (c) The correct order of melting point is Ti < V < Cr > Mn., The melting point of transition elements first increases to a, maximum extent and then fall as atomic number, increases. Mn, however, has abnormally low melting, point, due to stable configuration and weak intermetallic, bonding., 6. (c) Due to lanthanoid contraction, Zr and Hf possess, nearly same atomic and ionic radii, i.e. Zr = 160 pm and, Hf = 159 pm, Zr 4+ = 79 pm and Hf 4+ = 78 pm. Therefore,, these two elements show similar properties (physical and, chemical properties)., , Hence, magnetic moment of Ni 2 + in aqueous solution, m = n( n + 2) = 8 = 2.84 BM, 11. (b) Transition elements form coloured salt due to the, presence of unpaired electrons. In CuF2 , Cu(II) contain one, unpaired electron hence, CuF2 is coloured in solid state., 12. (c) Oxidation state of Ti in TiF62- = Ti +4, = [Ne]3 s 2 3p6 (no unpaired electrons), Oxidation state of Cu in Cu 2Cl 2 = Cu +, = [ Ar ] 3d10 (no unpaired electron), \ Both are colourless species as they do not have unpaired, electrons., 13. (d) Interstitial compounds are formed when small atoms are, trapped inside the crystal lattice of metals. Some of their, important characteristics are as follows, (i) They are very hard and rigid., (ii) They have melting point which are even higher than, those of the pure metals., (iii) They show conductivity like that of the pure metal., (iv) They acquire chemical inertness., 14. (b) Lanthanoids exhibit 4 f 1 - 14 , 5 d 0 - 1, 6s 2 electronic, configuration., 15. (d) Lanthanoids are f -block elements starting from Ce(58) to, Lu(71)., , Elements Atomic number Oxidation state, , 7. (b), Zn2+ =, , Number of unpaired electrons = 2, , [Ar], , 3d10, , (0 unpaired electron), , Ni2+ = [Ar] 3d8, , (2 unpaired electrons), , Cu+ = [Ar] 3d10, , (0 unpaired electron), , Ag+ = [Kr] 4d10, , (0 unpaired electron), , Thus, due to presence of two unpaired electrons Ni2 + is, paramagnetic ion., 8. (b) Greater the number of unpaired electron, higher will be, the value of magnetic moment., Since, 3 d 5 has 5 unpaired electrons hence highest magnetic, moment., m = 5( 5 + 2) = 35 = 5.95 BM, 9. (c) I. Cu 2+ - 3d 9 4 s0 (Number of unpaired electrons 1), II. Cr 3+ - 3d 3 (Number of unpaired electrons 3), III. Ni 2 + - 3d 8 (Number of unpaired electrons 2), IV. Fe3+ - 3d5 (Number of unpaired electrons 5), So, the correct order of number of unpaired electrons is, IV > II > III > I., , La, , 57, , +3, , Ce, , 58, , + 3, + 4, , Nd, , 60, , + 2, + 3, + 4, , Eu, , 63, , + 2, + 3, , Thus, for Eu common oxidation states are + 2 and + 3. Nd, also exhibits + 2 and + 3 oxidation state but + 3 is its, common oxidation state., 16. (b) Lanthanoids show common oxidation state of +3. Some of, which also show +2 and +4 stable oxidation state alongwith, +3 oxidation state.These are shown by those elements which, on losing 2 or 4 electrons acquire a stable configuration of, 2+, 7, 2+, 14, 4+, f 0 , f 7 or f 14 . e.g. Eu is [Xe] 4 f , Yb is [ Xe] 4 f , Ce is, 0, 4+, 7, [Xe]4 f and Tb is [Xe]4 f ., 17. (b) Electronic configuration of 58Ce = 54 [Xe] 4 f 2 5 d 0 6 s 2 ., Therefore, electronic configuration of Ce4+ = 54 [Xe] 4 f 0 ., Thus, it has a tendency to attain noble gas configuration and, attain f 0 configuration., 18. (b) In Eu and Yb as atomic number increases atomic radius, increases, which is an exception.

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71, , CBSE Term II Chemistry XII, , Therefore, it must have positive sign for standard, potential [EM° 2+ /M ]., , 19. (a) Both A and R are true and R is the correct explanation of A., Half and completely filled set of orbitals are relatively more, stable. Thus, electronic configuration of Cr is 3 d5 4s1 instead, of 3 d 4 4s 2, the energy gap between the two sets (3d and 4s ) of, orbitals is small enough to prevent electron entering the, 3d-orbitals. Similarly, in the case of Cu, the configuration is, 3 d10 4s1 and not 3 d 9 4s 2 ., 20. (a) Both A and R are true and R is the correct explanation of A., Copper (II) iodide (CuI2 ) is not known because Cu 2+ oxidises, I- to iodine., , (iv) (b) Cu (Z = 29) : [Ar]18 4s1 3 d10, Cu + (cupric ion) = [Ar]18 3 d 10, Cu 2+ (cuprous ion) = [Ar]18 3d 9, Cu + has completely filled d-subshell, while Cu 2+ has, incomplete d-subshell. Thus, Cu + (cuprous) is, colourless and Cu 2+ (cupric) is blue., Or, (b) The value of E° for Mn, Ni and Zn are more, negative than expected from the trend. The stability of, the half-filled d-subshell in Mn 2 + and the completely, filled d10 -configuration in Zn 2 + are related to their E°, values, whereas E° for Ni is related to the highest, negative D hyd H °., , 21. (a) Both A and R are true but R is not the correct explanation, of A., The highest oxidation state of osmium is +8 due to its ability, to expand their octet by using all its 8 electrons (2 from 6s, and 6 from 5 d)., 22. (b) Both A and R are true but R is not the correct, explanation of A., Separation of Zr and Hf is difficult; it is not because of they, lie in the same group of periodic table. It is due to, lanthanoid contraction which causes almost similar radii of, both of them., 23. (c) A is true, but R is false., Due to lanthanoid contraction, the size of Lu 3 + is decreases,, then covalent character of hydroxides increases and hence, the basic strength decreases. Thus, La(OH)3 is more ionic, and thus, more basic., 24., , (i) (b) As we move along the period in a 3d-series, nuclear, charge increases from Sc to Zn but electrons are added, to the orbital of inner subshell, i.e. 3d-orbitals. These, 3d-electrons shield the 4s-electrons from the increasing, nuclear charge more effectively than the outershell, electrons can shield one another., Therefore, the atomic radii decreases less rapidly in, case of 3d-series., (ii) (d) Fe has minimum value of third ionisation enthalpy., Ground state electronic configuration of Fe is, [Ar]3d 6 4s 2 ., Fe2 + has [Ar ]3d 6 configuration, whereas Fe3+ has, [Ar ]3d5 . The latter is a stable configuration and easier, to ionise Fe2 + to Fe3+ than expected. Hence possess least, value., Ni has highest value of third ionisation enthalpy due to, its greater nuclear charge and smaller size., Due to the same reason, Co > Fe > Mn should be the, order for the remaining elements, but the anomalous, order is due to greater stability of Mn 2 + having [Ar ]3d5, configuration than Mn 3+ with [Ar ]3d 4 ., Whereas for Co2 + , the electronic configuration is, [Ar ]3d 7 , Thus, ionisation enthalpy needed to, remove third electron is less as compared to that, of Mn 2+ ., (iii) (c) In electrochemical series, metals with positive, standard potential are placed below hydrogen. Out of, given transition metals, only Cu is placed below, hydrogen in electrochemical series., , 25., , (i) (b) One electron in case of atomic number 24, moves, from 4s to 3d-orbital to achieve more stability. This is, because, energy of 3d and 4s-orbitals are of comparable, energy., (ii) (a) The filling of 4 f -orbital before 5 d-orbital results in a, regular decrease in atomic radii called lanthanoid, contraction which essentially compensates for the, expected increase in atomic size with increasing atomic, number., (iii) (b) Among the given elements, Sc and Zn do not show, variable oxidation state but Zn is not a transition, element., (iv) (c) Electronic configuration of Ti 3+ is 3d 1 , thus, Ti 3+, contains an unpaired electron, so it will exhibit colour, in aqueous solution due to d -d transition., , Or, (b) Third ionisation enthalpy of Mn is much larger, because configuration of Mn 2+ (3d 5 ) is stable and a, large amount of energy is required to convert it into, less stable Mn 3+ (3d 4 )., , Subjective Questions, 1. Some of the exceptions are observed in electronic, configuration due to very little energy difference between, ( n - 1 ) d and ns orbitals such as, Cr : 3d 5 4s1 instead of 3d 4 4s 2, Cu : 3d10 4s1 instead of 3d 9 4s 2, This can be explained on the basis of electronic repulsion, and energy exchange which proves that half-filled and, completely filled d-orbital are more stable configuration., In the transition elements, the d-orbitals project to the, periphery of an atom more than the other orbitals (i.e. s and p)., Thus, they are more influenced by the surroundings as well as, affecting the atoms or molecules surrounding them., 2. While going from left to right in a period, the radii of, transition metals remains similar but their atomic masses, increases gradually. As a result their density increases, correspondingly.

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72, , CBSE Term II Chemistry XII, , 3. Transition elements show variable oxidation state due to, their ability of utilising single or multiple 4s and 3d, electrons., (i) Mn (manganese) shows the maximum number of, oxidation states., (ii) Scandium shows only +3 oxidation state., 4. Elements (+ 2 state), , 21 Sc, , 2+, , , 22 Ti, , 2+, , , 23 V, , 2+, , , 24 Cr, , Ti 3+ ( Z = 22) = [Ar] 3d1, Cr 3 + is the most stable among these in aqueous solution, because it has half-filled t 2 g level ( = t 23 g )., , 2+, , 2+, , and 25 Mn have more stable +2 oxidation state and the, outer electronic configuration are 3d1, 3d 2 , 3d 3, 3d 4 and 3d5, respectively. In all the elements listed, the removal of two 4s, electrons (in Cr 2 + , 1 e - from 4s and 1 e - from 3d), the, 3d-orbitals get gradually occupied. Since, the number of, empty d-orbitals decreases or the number of unpaired, electrons in 3d-orbitals increases with increase in atomic, number of cations, so the stability of the cations (M 2+ ), increases from Sc 2 + to Mn 2 +., 2+, , V3+ ( Z = 23) = [ Ar] 3d 2, , 10. Ce ( Z = 58) = [ Xe] 4 f 1 5 d16s 2, Ce3 + = 54[Xe] 4 f 1(only one unpaired electron), i.e. n = 1, According to ‘spin-only’ formula,, Magnetic moment of Ce3+ (m ) = n ( n + 2) BM, = 1 (1 + 2) BM = 3 BM = 1.73 BM, 11. Magnetic moment ( m ) = n( n + 2) BM, of M (27) =[ Ar ] 3d 7 4s 2, , 2+, , 5. Negative values of Mn and Zn are related to stabilities, of half-filled and completely filled configuration, respectively. But for Ni 2+, E° value is related to the highest, negative enthalpy of hydration., Hence, E s values for Mn, Ni and Zn are more negative than, expected., 6., , 7., , 8., , Fe2 + =[ Ar ] 3d 7 4s 0, m = n( n + 2), = 3( 3 + 2) = 15, = 3.8 BM, 12., , (i) Due to lanthanoid contraction, the atomic radii of the, second and third rows of transition elements is almost, same., So, they resemble each other much more as, compared to first row elements and show similar, characteristics., (ii) Copper cannot replace hydrogen from acids because, Cu has positive E° value, i.e. less reactive than, hydrogen which has electrode potential 0.00V., (i) The general trend towards less negative E°V values, across the series is related to the general increase in the, sum of the first and second ionisation enthalpies., (ii) The high energy to transform Cu(s) to Cu 2 + (aq) is not, balanced by its hydration enthalpy., Most of Cu (I) compound are unstable in aqueous, solution and undergo disproportionation., 2Cu+ ¾® Cu 2+ + Cu, (iii) The stability of the half-filled d-subshell in Mn 2 + and, the completely filled d10 configuration in Zn 2 + are, related to their more negative E° V values., (i) Because it has incompletely filled d-orbitals in one of, its oxidation states, i.e. Cu 2 + ([ Ar ] 3d 9 )., (ii) Cr 2 + ([Ar] 3d 4 ) changes to Cr 3+ ([ Ar ] 3d 3 ) while Fe2 +, ([ Ar ] 3d 6 ) changes to Fe3+ ([Ar] 3d 5 )., In aqueous medium, the configuration [Ar] 3d 3 is more, stable than the configuration [Ar] 3d5 . Hence, Cr 2 + is a, stronger reducing agent., , 9. Mn 3 + ( Z = 25 ) = [ Ar] 3d 4, Cr 3 + (Z = 24) = [Ar] 3d 3, , (i) Magentic moment ( m ) = n( n + 2) BM, Atomic number of Fe = 25 =[Ar] 3d 6 4s 2, Fe2 + = [Ar] 3d 6 4s 0, , So, m = n( n + 2) = 4( 4 + 2) = 24 = 4.89 BM, (ii) Oxygen can form multiple bonds with metals, while, fluorine can not form multiple bond with metals. Hence,, oxygen has more ability to stabilize higher oxidation state, rather than fluorine., (iii) Magnetic moment of any metal ion can be decided on, the basis of spin as well as orbital contribution of, electron. Due to symmetrical electronic configuration,, there is no orbital contribution in Cr 3+ ion. However,, appreciable orbital contribution takes place in Co2+ ion., 13., , (i) Mn 2 + compounds are more stable due to half-filled, d-orbitals. Fe 2 + compounds are comparatively less stable, as they have six electrons in their orbitals. So, they, tend to lose one electron from Fe 2 + and get stable 3d 5, configuration in Fe 3 + ., Therefore, comparatively high positive value of E° for, Mn 3 + / Mn 2 + indicates the stability of Mn 2 + ( d5 ) whereas, comparatively low value for Fe 3 + / Fe 2 + indicates the extra, stability of Fe 3 + ( d5 )., (ii) Energy required to convert metallic crystal into individual, atom is enthalpy of atomisation. In transition row, elements, it first increases and reaches to maximum upto, middle element and then decreases. This is because of, strong inter atomic interaction due to unpaired electron., Greater the number of unpaired electron, stronger will, be bonding and thus enthalpy of atomisation will also be, more. Since iron has more unpaired electron than copper, therefore its enthalpy of atomisation is more.

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73, , CBSE Term II Chemistry XII, , (iii) The metal ions with partially or incomplete, filled d-orbitals will be coloured in aqueous, solution. While the metal ions having either empty, or completely filled d-orbitals are colourless., The colour will be due to d-d transition of electrons., Thus, the outer electronic configuration of metal ions, are, Sc 3 + : 3d 0 , Ti 3 + : 3d1, Hence, among the given ions, Ti 3 + will exhibit colour, in aqueous solution while Sc 3 + will be colourless., , 3+, , = [Ar] 3d, , Green, , Cu = [Ar] 3d, , 10, , Colourless, , V, , +, , Sc, , 3+, , = [Ar], , Colourless, , Mn 2 + = [Ar] 3d 5, Fe, , 3+, , Co, , 2+, , = [Ar] d, , 5, , = [Ar] 3d, , Pink, Pink, , As Sc 3+ and Cu + have 3d 0 and 3d 0 configuration in their, valence shell so their aqueous solutions are colourles. All, other, i.e. Ti 3+ , V 3+ , Mn 2 + , Fe3+ and Co2+ are coloured in, aqueous medium., 15., , 18., , Brown, 7, , (i) Electronic configuration of Zn 2 + = 3d10 4s 0, , ¼ ¼ ¼ ¼ ¼, , (i) The spin only magnetic moment ( m ) of each ion can be, calculated as, , m = n ( n + 2) BM, , [Q n = No. of unpaired electron(s)] Þ m µ n, i.e. higher, the number of unpaired electron, higher will be the, value of m., , 8, , Paramagnetic, , V2 +, , 23, , 3 ( 3d 3 ), , 15, , Paramagnetic, , 3+, , 1, , 22, , 1 ( 3d ), , 3, , Paramagnetic, , Sc3+, , 21, , 0 ( 3d 0 ), , 0, , Diamagnetic, , (i) Atoms of transition metal can easily take place in the, crystal lattice of another metal in the molten state and, are miscible with each other and thus forms alloys., (ii) Eu 2 + having electronic configuration [Xe]4 f 7 is a, strong reducing agent because in the aqueous solution,, it events back to the most stable +3 oxidation., , 74 = [ Xe] 6s 2 4 f 1 3d 3 (d-block element or inner transition, element)., , ¼ ¼ ¼, , 17., , 2 (3d 2 ), , 59 = [Ar] 6s1 4 f 3 (d-block element or inner transition element)., , ¼ ¼ ¼ ¼ ¼, , (i) This is because iron has unpaired electrons in the, d-subshell while sodium does not., (ii) This is because Cd 2 + ion has completely filled, d-orbitals ( 4d10 ). Hence, in the absence of d-d, transition, they show only white colour., (iii) Zn, Cd, Hg metals have completely filled d-orbitals, ( d10 ). It means that d-electrons are not readily available, for metallic bond formation. Since, the metallic bonds, are weak. Therefore, these metals are quite soft and, also have low melting points., , 22, , 29 = [Ar] 4s1 3d10 (d-block element or transition element)., , Electronic configuration of Ni 2 + = 3d 8 4s 0, , 16., , Ti 2 +, , 19. Inner-transition elements inner transition elements (or, f -block elements) include lanthanoids (Z = 58 to 71) and, actinoids ( Z = 90 to 103). These elements have incomplete, f -orbital in penultimate shell of their atoms., The electronic configuration of given atomic number are as, , ¼ ¼ ¼ ¼ ¼, , Compounds that contains unpaired electrons are, coloured., (ii) Out of [Ti(H2O) 6 ] 3 + and [Sc(H2O) 6 ] 3+ , latter is, colourless because it cannot undergo d-d transition as, do not have unpaired electrons. [Ti(H2O) 6 ] 3 + has, unpaired electron for d-d transition which is, responsible of colourness of a complex., , n, M, (for metal ion) (BM), , Thus, the correct order of spin only magnetic moments, of given hydrated ions will be, Sc3+ < Ti 3+ < Ti 2 + < V2 + V2 +, (ii) Misch metal is an alloy which consists of a lanthanoid, metal (~ 95%) iron (~5%) and traces of S, C, Ca and Al., It is used in Mg based alloy to produce bullets, shell, and lighter-flint., , Purple, , 2, , Z, , Ti, , 14. Only those ions will be coloured which have incomplete, d-orbitals. Ions which has complete or vacant d-orbitals are, colourless., Ti 3 + = [Ar] 3d 1, , Nature, , Metal ion, , 20., , (i) Separation of Zr and Hf are quite difficult because of, lanthanoid contraction. Due to lanthanoid contraction,, they have almost same size (Zr = 160 pm and, Hf = 159 pm) and thus, similar chemical properties., That’s why it is very difficult to separate them by, chemical methods., (ii) It is due to the fact that after losing one more electron, Ce acquires stable 4 f 0 electronic configuration. So, Ce, shows +4 oxidation state along with +3 oxidation state., , 21., , (i) Electronic configuration of Mn 2 + = [ Ar ]18 3d5, Electronic configuration of Fe 2 + = [ Ar ]18 3d 6, Mn 2 + having half-filled d-orbitals will be more stable, than Fe 2 + , as it has partially filled d-orbitals., (ii) Zinc has completely filled d-orbitals, which limits its, tendency to form metallic bonds. Thus, it requires least, enthalpy to get atomised., (iii) MO3F is MnO 3 F., In MO3F,, Let the oxidation state of M in x., x + 3 ´ ( - 2) + ( - 1 ) = 0, or, x = + 7, i.e. M is in oxidation state of +7. Hence, the, given compound is MnO 3 F.

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74, , CBSE Term II Chemistry XII, , (v) Ni 2 + compounds have the sum of first two ionisation, enthalpies to be lower than that of Pt 2 + compounds and, are thermodynamically stable. However, the sum of, first four IEs of Pt 4 + is lower than that of first four IEs, of Ni 4 + and is relatively stable., , (iv) E°Cr 3 + / Cr 2 + is negative (–0.4 V). It shows the stability, of Cr 3+ ions i.e. Cr 3+ in solution cannot be reduced to, Cr 2 + ions. Mn 3+ has high positive value of E° due to, extra stability of half-filled electronic configuration., Thus, Cr 3+ is the most stable, Mn 3+ is least stable., (v) Transition elements form a large number of complexes, due to the comparatively smaller sizes of the metal ions,, their high conic charges and the availability of, d-orbitals for bond formation., 22., , 23., , (i) (a) Cu, because the electronic configuration of Cu is, 3d10 4s1. So, second electron needs to be removed, from completely filled d-orbital which is very, difficult., (b) Zinc, because of electronic configuration of, Zn = 3 d10 4s 2 and Zn 2 + = 3 d10 which is fully filled, and hence is very stable. Removal of third electron, requires very high energy., (c) Zinc, because of it has completely filled 3d subshell, and no unpaired electron is available for metallic, bonding., (ii) Carbonyl M(CO)5 is Fe(CO)5 ., According to EAN rule, the effective number of a metal, in a metal carbonyl is equal to the atomic number of, nearest inert gas EAN is calculated as, EAN = number of electrons in metal +2 ´ (CO), = atomic number of nearest inert gas, In M(CO)5 = x + 2 ´ ( 5 ) = 36 (Kr is the nearest inert, gas), x = 26 (atomic number of metal), So, the metal is Fe(iron)., (iii) It is because oxygen can form multiple bonds, whereas, fluorine can only form single bonds with metals., (i) Energy required to convert metallic crystal into, individual atom is known as enthalpy of atomisation. The, transition metals have high enthalpy of atomisation., It first increases, reaches to the maximum in the, middle of each series and then decreases. It can be, explained on the basis of strong interatomic interaction, due to unpaired electrons. Greater the number of, unpaired electrons, stronger is the resultant bonding., (ii) Paramagnetic behaviour arises due to presence of, unpaired electrons in d-subshell., (iii) Most of the compounds of transition metals are, coloured due to d-d transition. Transition metal ions, having d 0 configuration are colourless. When ligand has, free electrons, the transfer of electron from atom to, cation takes place. In this case, energy is absorbed, during this process which is responsible for colour., (iv) Catalytic activity of transition metal is due to their, ability to adopt multiple oxidation states and to form, complexes. Transition metals because of their variable, valencies sometimes form unstable intermediate, compounds and provide a new path with lower, activation energy for the reaction., , 24., , (i) (a) Vanadium, (b) Iron, (c) Silver, (ii) Reactivity of transition elements depends mostly upon, their ionisation enthalpies. As we move from left to, right in the periodic table (Sc to Cu), ionisation, enthalpies increase almost regularly., Hence, their reactivity decreases almost regularly from, Sc to Cu., , 25. When small atoms like H, C and N get trapped inside the, crystal lattice of transition metals., (i) Such compounds are called interstitial compounds., (ii) Their characteristic properties are, (a) They have high melting points, higher than those of, pure metals., (b) They are very hard., (c) They retain metallic conductivity., (d) They are chemically inert., 26., , (i) (a) Copper; Due to high D aH ° and low D hyd H °, (b) Cerium; Due to stable 4 f 0 configuration/Tb; Due, to stable 4 f 7 configuration, (ii) (a) Due to ability of oxygen to form multiple bond to, metal.), (b) Transition elements show high melting point, due, to involvement of greater number of electrons in, the interatomic bonding from ( n - 1 ) d-orbitals in, addition to ns electrons in forming metallic bond., Thus, large number of electrons participate, forming large number of metallic bond., (c) Ionisation enthalpy of Cr is less than that of Zn, because Cr has stable d 5 configuration. In case of, zinc, electron comes out from completely filled, 4s-orbital. So, removal of electron from zinc requires, more energy as compared to the chromium., , 27., , (i) As the size decreases covalent character increases., Therefore, La 2O 3 is more ionic and Lu 2O 3 is more, covalent., (ii) As the size decreases from La to Lu, stability of oxo, salts also decreases., (iii) Stability of complexes increases as the size of, lanthanoids decreases., (iv) Radii of 4d and 5 d-block elements will be almost same., (v) Acidic character of oxides increases from La to Lu., , 28., , (i) Mn 2 + compounds are more stable due to half-filled, d-orbitals. Fe 2 + compounds are comparatively less, stable as they have six electrons in their orbitals. So,, they tend to lose one electron from Fe 2 + and get stable, 3d5 configuration., (ii) Oxygen and fluorine both have small size and high, value of electronegativity. So, they can oxidise the metal, to their highest oxidation states.

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75, , CBSE Term II Chemistry XII, , (iii) In aqueous solution, copper (I) undergoes, disproportionation reaction., 2Cu + ( aq ) ¾® Cu 2 + + Cu (s ), The highest stability of Cu 2 + ion in aqueous solution is, due to negative enthalpy of hydration., Hence, Cu + is not known (or unstable) in aqueous, solution., (iv) Electronic configuration of Cu is [Ar] 3d 10 4s1. When, copper atom loses 1 electron, it exhibits + 1 oxidation, state and forms Cu + ion with stable 3d10 configuration., So, copper metal in the first transition series tends to, have + 1 oxidation state to achieve a stable, configuration., , Or, 3+, , 2+, , Mn /Mn has large positive E° value. Hence,, Mn 3+ can be easily reduced to Mn 2 + because Mn 2 +, has half-filled electronic configuration, so it is stable, and Mn 3 + is least stable. Therefore, it is a good, oxidising agent., , 29., , (i) Titanium, (ii) Silver in its + 1 oxidation state, exhibits 4d 10 5 s 0, configuration. But in some compounds, it also shows, +2 oxidation state, so the configuration becomes 4d 9 5 s 0 ., Here, d-orbital is not completely filled. Therefore, silver, is a transition element., (iii) Zinc has stable ground state due to its completely filled, d-orbitals. It therefore, has least tendency to form, metallic bonds, in the series, thus requires least enthalpy, of atomisation to get atomised., (iv) Manganese shows maximum number of oxidation states, ( EC : 3d 5 , 4s 2 ), i.e. + 2, + 3, + 4, + 5, + 6, + 7 in its, compounds due to maximum number of unpaired, electrons., Or, o, value, of, copper, is positive as value of hydration, EM, 2+, /M, enthalpy is less than the sum of values of ionisation, enthalpy and enthalpy of ionisation.

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Chapter Test, Multiple Choice Questions, , 7. Assertion Cr 2+ is reducing, while Mn 3+ is oxidising even, , 1. Which of the following elements does not have, ( n - 1 ) d 10 ns2 electronic configuration?, (a) Zn, (c) Hg, , (b) Cd, (d) Cu, , 2. Which element can have oxidation state from 4 to 6?, , though both have d 4 -configuration., , Reason Configuration of Cr changes from d 3 to d 4 ., , Short Answer Type Questions, , 8. Why there is striking similarities (horizontal and vertical), among successive members of the transition series?, , (a) Fe, (b) Mg, (c) Co, (d) Cr, , 9. The E (oM 2+ / M ) value for copper is positive (+0.34 V). What, is the possible reason for this?, , 3. Which one of the following ions is the most stable in, aqueous solution?[Atomic number of Ti = 22, V = 23,, Cr = 24 and Mn = 25], (a) Cr 3+, (c) Ti 3+, , (b) V 3+, (d) Mn 3+, , 4. The highest value of the calculated spin only, magnetic moment (in BM) among all the transition, metal complexes is, (a) 5.92, (c) 6.93, , (b) 3.87, (d) 4.90, , Assertion-Reasoning MCQs, Direction (Q. Nos. 5-7) Each of these questions contains, two statements Assertion (A) and Reason (R). Each of, these questions also has four alternative choices, any, one of which is the correct answer. You have to select, one of the codes (a), (b), (c) and (d) given below., (a) Both A and R are true and R is the correct explanation of A., (b) Both A and R are true, but R is not the correct, explanation of A., (c) A is true, but R is false., (d) A is false, but R is true., , 5. Assertion Radii of the members of the third, transition series is very similar to those of the, corresponding members of the second series., Reason It is due to lanthanoid contraction., , 6. Assertion First ionisation enthalpy of Cr is lower than, that of Zn., , 10. (i) The second and third rows of transition elements, resemble each other much more than they resemble the, first row., (ii) Identify the metal MO 3F and justify your answer., , 11. How would you account for the following?, (i) Of the d 4 species,Cr 2+ is strongly reducing agent while, manganese (III) is strongly oxidising agent., (ii) Cobalt (II) is stable in aqueous solution but in the, presence of complexing reagents, it is easily oxidised., (iii) The d 1 configuration is very unstable in ions., , Long Answer Type Questions, , 12. (i) Transition metals can act as catalysts because these, can change their oxidation state. How does Fe (III), catalyse the reaction between iodide and persulphate, ions?, (ii) Mention any three processes where transition metals, act as catalysts., , 13. (i) Explain the following observation., (a) Which element of the first transition series has, highest second ionisation enthalpy?, (b) Write down the electronic configuration of, gadolinium (Gd). Its atomic number is 64., (c) What are the different oxidation states exhibited, by lanthanoids?, (ii) What is meant by disproportionation? Give two, examples of disproportionation reaction in aqueous, solution., , Reason I.E. of Cr is lower due to it’s stable half-filled, configuration., , Answers, Multiple Choice Questions, 1. (d), , 2. (d), , 3. (a), , 4. (a), , Assertion-Reasoning MCQs, 5. (a), , 6. (a), , 7. (c), , For Detailed Solutions, Scan the code

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CHAPTER 05, , Coordination, Compounds, In this Chapter..., l, , Double Salt and Complex Compounds, , l, , Werner’s Theory of Coordination Compounds, , l, , Terms Used in Coordination Compounds, , l, , Nomenclature of Coordination Compounds, , l, , Bonding in Coordination Compounds, , Transition metals form a large number of complex compounds in, which the metal atoms are bound to a number of anions or neutral, molecules by sharing of electrons. Such compounds are known as, coordination compounds., Coordination compounds are those in which a central metal atom, or ion is attached with a fixed number of groups or molecules, (ligands) through coordinate bonds., , Double Salts and Complex Compounds, Double salts These are the addition compounds which are stable, inthe solid state but when dissolved in water they dissociate, completely into ions,, e.g. Mohr’s salt : [FeSO 4 (NH 4 ) 2 SO 4 × 6H 2 O]., Coordination compounds They do not dissociate completely into, simple ions and do not completely lose their identity in solution., e.g. K 4 [Fe(CN) 6 ]., , Werner’s Theory of Coordination, Compounds, (i) Alfred Werner, a Swiss chemist was the first to formulate his idea, about the structures of coordination compounds. He proposed, the concept of primary valence and secondary valence., (ii) Postulates of Werner’s Theory, l, , Metals possess two types of valencies, called primary or, ionisable valency (oxidation number) and secondary or, non-ionisable valency (coordination number)., , l, , l, , l, , l, , Every metal atom must satisfy both of its valencies., Primary valencies are satisfied by negative ions, whereas secondary valencies are satisfied by neutral, molecules or negative ions., Primary valencies are non-directional. Secondary, valencies are directed in space towards fixed, position giving a definite shape to the complex., The ions or groups bound by the secondary linkages, to the metal have characteristic spatial, arrangements corresponding to coordination, numbers. In modern formulations such spatial, arrangements is called coordination polyhedra., , Terms Used in Coordination, Compounds, Definitions of some important terms used in, coordination compounds are as follows, (i) Coordination entity It constitutes a central metal, atom or ion bonded to a fixed number of ions or, molecules., (ii) Central metal atom or ion In a coordination entity,, the atom or ion around which a fixed number of ions, or groups are bound in a definite geometrical, arrangement is called the central atom or ion. In, complex compounds, the central metal atom or, ion can act as Lewis acid and ligands act as Lewis, base.

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78, , CBSE Term II Chemistry XII, , (iii) Ligands Ions or molecules which can donate a lone pair of, electron to the central atom or ion are called ligands., When a ligand is bound to a metal ion through a single, donor atoms (e.g. Cl - , H 2O), the ligand is said to be, unidentate., When a ligand can bind through two donor atoms as in, oxalate, the ligand is said to be didentate and when, several donor atoms are present in a single ligand as in, N(CH 2CH 2NH 2 ) 3 , the ligand is said to be polydentate., When a di or polydentate ligand uses its two or more, donor atoms simultaneously to bind a single metal ion, it is, said to be a chelating ligand., The number of such ligating groups is called the, denticity of the ligand., Chelating ligand is a neutral molecule, free radical or ion, with two or more lone pairs of electrons on different atoms., Ligand which has two different donor atoms and either of, the two ligates in the complex is called ambidentate, ligands. Examples are NO -2 and SCN - ion., (iv) Coordination number For a metal ion in a complex it can, be defined as the number of ligand donor atoms to which, the metal is directly bonded., (v) Counter ions The part of the coordination which is present, outside the square brackets is known as ionisation sphere., These are also known as counter ions., (vi) Coordination sphere The central atom/ion and the, ligands attached to it are enclosed in the square brackets, are collectively known as coordination sphere., (vii) Coordination polyhedron The spatial arrangement of the, ligand atoms which are directly attached to the central, atom/ion defines a coordination polyhedron about the, central atom., (viii) Oxidation number The charge present on central metal, atom or ion, if all the ligands are removed alongwith the, electron pairs that are shared with it is called oxidation, number of central atom., (ix) Charge on the complex ion The charge carried by a, complex ion is the algebraic sum of the charges carried by, the central metal ion and the total ligands attached to the, central metal ion., , l, , l, , l, , l, , l, , l, , l, , l, , l, , l, , l, , l, , l, , l, , Homoleptic Complexes and Heteroleptic Complexes, Complexes in which metal is bound to only one kind of donor, groups are known as homoleptic complexes. e.g. [Co(NH 3 ) 6] 3+ ., Complexes in which a metal is bound to more than one, kind of donor groups are known as heteroleptic complexes., e.g. [Co(NH 3 ) 4 Cl 2 ] + ., , Nomenclature of Coordination, Compounds, The rules for the nomenclature of coordination compounds are, as follows, 1. Formulae of Mononuclear Coordination Compounds, , The following rules are applied, while writing the formulae, , l, , The central atom is listed first., The ligands are then listed in alphabetical order. The, placement of a ligand in the list does not depend on its, charge., Polydentate ligands are also listed alphabetically., In case of abbreviated ligand, the first letter of the, abbreviation is used to determine the position of the, ligand in the alphabetical order., The formula for the entire coordination entity, whether, charged or not is enclosed in square brackets. When, ligands are polyatomic, their formulae are enclosed in, parenthesis. Ligand abbreviations are also enclosed in, parenthesis., There should be no space between the ligands and the, metal within a coordination sphere., When the formula of a charged coordination entity is to, be written without that of the counter ion, the charge is, indicated outside the square brackets as a right, superscript with the number before the sign. e.g., [Co(CN) 6 ] 3- , [Cr(H 2O) 6 ] 3+ etc., The charge of the cation(s) is balanced by the charge of, the anion(s)., , 2. Naming of Mononuclear Coordination Compounds, , The following rules are used while naming the, coordination compounds, The cation is named first in both positively and, negatively charged coordination entities., The ligands are named in an alphabetical order before, the name of the central atom or ion. (This procedure is, reversed from writing formula)., Name of the anionic ligands end in -‘o’, those of neutral, and cationic ligands are the same except aqua for H 2 O,, ammine for NH3 , carbonyl for CO and nitrosyl for NO., These are placed within enclosing marks ( )., If ambidentate ligand is present, then it is named by, placing the symbol of donor atom or by using different, names. e.g. ONO (nitrito-O), ¾NO 2 (nitrito-N)., Prefixes mono, di, tri, etc. are used to indicate the number, of the individual ligands in the coordination entity. When, the names of the ligands include a numerical prefix, then, the terms, bis, tris, tetrakis are used, the ligand to which, they refer is placed in parenthesis., Oxidation state of the metal in cation, anion or neutral, coordination entity is indicated by Roman numeral in, parenthesis., If the complex ion is a cation, the metal is named same, as the element, e.g. Co in a complex cation is called, cobalt and Pt is called platinum., If the complex ion is an anion, the name of the metal, ends with the suffix—‘ate’, e.g. Co in a complex anion,, [Co(SCN)4 ] 2- is called cobaltate. For some metals, the, Latin names are used in the complex anions, e.g. ferrate, for F., The neutral complex molecule is named similar to that, of the complex cation., l, , l, , l, , l, , l, , l, , l, , l

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80, , CBSE Term II Chemistry XII, , Inner and Outer Orbital Complexes, A complex compound may be inner orbital complex or, outer orbital complex depending upon whether d-orbitals, of the inner shell or outer shell are involved in, hybridisation. Inner orbital complex is generally low spin, complex, while outer orbital complex is generally high, spin complex., , The splitting of d-orbitals in an octahedral crystal field can be, seen in following diagram., L, , Y, , L, , dx2–y2 dz2, , eg, 3/5Do, , X, Bary centre, , Metal, d-orbitals, , 2/5Do, dxy dxz dyz, , dx2–y2 dz2 dxy dxz dyz Average energy, Free metal ion of the d-orbitals in, spherical crystal field, , Do, t2g, , Splitting of d-orbitals, in octahedral, crystal field, , d-orbital splitting in an octahedral crystal field, , Here, CFSE can be calculated as,, CFSE = [ - 0.4 x + 0.6 y ] D0, where, D0 = CFSE in octahedral complex, x = number of electrons in t 2 g -orbitals, y = number of unpaired electrons in eg - orbitals., , l, , l, , l, , l, , l, , l, , l, , l, , Crystal Field Theory (CFT), , l, , l, , l, , L, , L, Z, , l, , It is an electrostatic model which considers the, metal-ligand bond to be ionic arising purely from, electrostatic interactions between the metal ion and the, ligand., Ligands are treated as point charges in case of anions or, point dipoles in case of neutral molecules., According to CFT, under the influence of ligand field,, degeneracy of the d-orbital is disturbed and it splits into, two or more energy levels., Octahedral complexes yield three orbitals of lower, energy, t 2 g set and two orbitals of higher energy, e g set., This splitting of degenerate levels due to the presence of, ligands in a definite geometry is termed as crystal field, splitting and the energy separation is denoted by D o ., , M, , Energy, , l, , l, , L, L, , Magnetic Properties of, Coordination Compounds, If no unpaired electron is present, the complex is, diamagnetic. If unpaired electron(s) is/are present,, complex is paramagnetic., Its magnetic moment m can be calculate of as, m = n( n + 2)BM where, n is the number of unpaired, electrons present., Limitations of Valence Bond Theory, Although, valence bond theory is advantageous to explain, formation, structures and magnetic properties of, coordination compounds but it suffers with few limitations,, listed as follows, It is based on a number of assumptions., It does not distinguish between weak and strong ligands., It does not give quantitative interpretations of magnetic, data., It does not give a quantitative interpretation of the, thermodynamic or kinetic stabilities of coordination, compounds., It fails to explain relative energies of different shapes., It does not explain the colour and spectral properties of, the 4-coordination compounds., It fails to predict the exact tetrahedral and square planar, structures of 4-coordinate complexes., , M, , M, , l, , l, , l, , The extent of crystal field splitting depends on the strength of, ligand and charge on metal ion., A strong ligand causes greater splitting, while a weak ligand, causes smaller splitting., Ligands can be arranged in a series in the order of increasing, field strength as given below, I - < Br - < SCN - < Cl - < S 2- < F - < OH - < C 2 O 2–, 4, < H 2 O < NCS - < NH 3 < en < NO 2- < CN - < CO, This series is called spectrochemical series., The single d-electron of metal ion in octahedral coordination, entities occupies one of the lower energy t 2 g -orbitals. In, d 2 and d 3 coordination entities, the d-electrons occupy the, single t 2 g - orbitals in accordance with the Hund’s rule., For metal ion complex, d 4 configuration, d-orbital occupancy, depends on the relative magnitude of the crystal field, splitting, D o and the pairing energy, P (P represents the, energy required for electron pairing in a single orbital)., If D o < P, the fourth electron enters one of the eg - orbitals giving, the configuration t 23g e1g . Ligands for which D o < P are known as, weak field ligands and form high spin complexes., If D o > P, it becomes more energetically favourable for the, fourth electron to occupy t 2 g orbital with configuration t 42 g e g0 ., Ligands which produce this effect are known as strong field, ligands and form low spin complexes.

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81, , CBSE Term II Chemistry XII, , l, , l, , Calculations show that d 4 to d 7 coordination entities are, more stable for strong field as compared to weak field, cases., In tetrahedral coordination entity formation, the d-orbital, splitting is inverted and is smaller as compared to the, octahedral field splitting. It can be easily illustrated from, the given figure., dxy , dxz , dyz, , Energy, , 2D, —, 5 t, , t2, Dt, , 3D, —, 5 t, e, Average energy, 2, 2, 2, d-orbitals of the d-orbitals dx –y dz, free ion, in spherical Splitting of, crystal field d-orbital in, tetrahedral, crystal field, , l, , Limitations of Crystal Field Theory, CFT has the following limitations :, As the ligands are point charges, the anionic ligands should, exert the greatest splitting effect. Actually,, the anionic ligands are found at the low end of the, spectrochemical series., CFT treats the metal-ligand bond as purely ionic and it does, not take into account the covalent character of bonding, between the ligand and the central atom., l, , l, , Note Valence bond theory consider the M—L bond as covalent while, crystal field theory consider the M—L bond to be ionic., , Colour of Coordination Compounds, l, , d-orbital splitting in a tetrahedral crystal field, , Here, CFSE can be calculated as,, CFSE = [ - 0.61 + 0.4y ] Dt, Dt = CFSE in tetrahedral in e-orbitals, x = number of electrons in e-orbitals, y = number of electrons in t 2 -orbitals, l, , For the same metal, the same ligands and metal ligand, distances, it can be shown as D t = 4 / 9 D o ., , The orbital splitting energies are not sufficiently large to, force pairing and therefore low spin are rarely observed in, tetrahedral complexes., , l, , l, , In complex compounds, d-orbitals split into two sets t 2 g and, e g . These have different energies. The difference in energies, lies in visible region and electron jump from ground state t 2 g, level to higher state e g level., This is known as d - d transition and is responsible for colour, of coordination compounds. d - d transition takes place in d 1, to d 9 ions, so the ions having d 1 to d 9 configuration are, coloured., On the other hand, the ions with d 0 and d 10 configuration do, not show d-d transition.

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Chapter, Practice, 7. The correct structure of ethylenediamminetetraacetic, , PART 1, Objective Questions, l, , Multiple Choice Questions (MCQs), , acid (EDTA) is, HOOCCH2, (a), HOOCCH2, HOOC, (b), HOOC, , 1. Primary valency of binary compound CrCl 3 , CoCl 2, and PdCl 2 are, (a) 2, 2 and 3 respectively, (c) 3, 3 and 2 respectively, , 2. When 1 mole of CrCl 3 × 6H 2O is treated with excess of, AgNO 3 , 3 moles of AgCl are obtained.` The formula, of the complex is, (a), (b), (c), (d), , [CrCl 3 (H2O)3 ]× 3H2O, [CrCl 2 (H2O)4 ]Cl × 2H2O, [CrCl(H2O)5 ]Cl 2 × H2O, [Cr(H2O)6 ]Cl 3, , excess of AgNO 3 , 0.2 mole of AgCl are obtained. The, conductivity of solution will correspond to, (b) 1 : 2 electrolyte, (d) 3 : 1 electrolyte, , 4. The oxidation number of the central atom in a, complex is defined as the charge it would carry, (a) if all the ligands are removed along with the electron, pairs that are donated by the central atom, (b) if all the ligands are removed along with the electron, pairs that are shared with the central atom, (c) if all the ligands are removed from the central metal atom, (d) if all the electrons pairs are shared with the central atom, , 5. Coordination number of Co in [Co(en) 2 Cl 2 ]Cl is, (a) 3, , (b) 4, , 3+, , (c), HOOCCH2, , CH2COOH, CH2COOH, , COOH, COOH, , ‚, ƒ, N — CH2 — CH2 — N, ‚, ƒ, , CH2COOH, CH2COOH, , H2C ¾ COOH, HOOC ¾ H2C, H, ½, ‚, ƒ, (d), N— CH— CH— N, ƒ, ‚, H, ½, CH COOH, CH2COOH 2, , 8. IUPAC name of [Pt(NH 3 ) 2 Cl(NO 2 )] is, , 3. When 0.1 mole of CoCl 3 (NH 3 ) 5 is treated with, (a) 1 : 3 electrolyte, (c) 1 : 1 electrolyte, , ‚, ƒ, N — CH2 — CH2 — N, ‚, ƒ, , HOOCCH2, (b) 2, 3 and 2 respectively, (d) 3, 2 and 2 respectively, , ƒ, ‚, N — CH == CH — N, ƒ, ‚, , (c) 5, , (d) 6, , +, , 6. [Co(NH 3 ) 6 ] and [Co(NH 3 ) 4 Cl 2 ] are the, examples of, (a) homoleptic complexes, (b) heteroleptic complexes, (c) [Co(NH3 )6 ]3+ homoleptic complex, [Co(NH3 )4 Cl 2 ]+, heteroleptic complex, (d) [Co(NH3 )6 ]3+ heteroleptic complex, [Co(NH3 )4 Cl 2 ]+, homoleptic complex, , (a), (b), (c), (d), , platinum diamminechloronitrite, (NCERT Exemplar), chloronitrito-N-ammineplatinum (II), diamminechloridonitrito-N-platinum (II), diamminechloronitrito-N-platinate (II), , 9. The correct IUPAC name of [Pt(NH 3 ) 2 Cl 2 ] is, (a), (b), (c), (d), , diamminedichloridoplatinum (II), diamminedichloridoplatinum (IV), diamminedichloridoplatinum (0), dichloridodiammineplatinum (IV), , (NCERT Exemplar), , 10. Write the formula for coordination compound, diamminesilver (I) dicyanoargentate (I)., (a) [Ag(NH3 )2 ][Ag(CN)2 ], (c) [Ag(NH3 )2 ][Ag(CN)3 ], , (b) [Ag(NH3 )3 ][Ag(CN)2 ], (d) [Ag(NH2 )2 ][Ag(CN)2 ], , 11. Match the complex ions given in Column I with the, hybridisation and number of unpaired electrons, given in Column II and assign the correct code., Column I, (Complex ion), , Column II, (Hybridisation, number, of unpaired electrons), , A. [Cr(H2O) 6 ] 3+, , 1., , dsp2 , 1, , B. [Co(CN) 4 ] 2 -, , 2., , sp3d 2 , 5, , C. [Ni(NH3 ) 6 ] 2+, , 3., , d 2 sp3, 3, , D. [MnF6 ] 4 -, , 4., , sp3 d 2 , 2

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83, , CBSE Term II Chemistry XII, , Codes, A B, (a) 3 1, (c) 3 2, , C, 4, 4, , D, 2, 1, , A, (b) 4, (d) 4, , B, 3, 1, , C, 2, 2, , Reason Unpaired electrons are present in their, (NCERT Exemplar), d-orbitals., 320. Assertion [Fe(CN) 6 ] ion shows magnetic moment, corresponding to two unpaired electrons., Reason Because it has d 2 sp 3 type hybridisation., , D, 1, 3, , 12. Which of the following complex ions is diamagnetic, in nature?, , (a) [CoF 6 ] 3-, , (NCERT Exemplar), , (b) [NiCl 4 ] 2 - (c) [Ni(CN) 4 ] 2 - (d) [CuCl 2 ] 2 -, , -, , 21. Assertion [CuCl 2 ] has linear shape with sp, , 13. Which one of the following is an outer orbital, , hybridisation., Reason [CuCl 2 ] - has zero unpaired electrons., 22. Assertion d 4 and d 7 configuration is more stable in, high spin state., Reason High spin are stable when D o < P., , complex and exhibits paramagnetic behaviour?, (a) [Ni(NH3 ) 6 ] 2 +, , (b) [Zn(NH3 ) 6 ] 2 +, , 3+, , (d) [Co(NH3 ) 6 ] 3+, , (c) [Cr(NH3 ) 6 ], , 14. For d 4 ions, the fourth electron enters one of the, eg -orbitals giving the configuration t 23g e1g , when, (a) D o > P, , (b) D o < P, , (c) D o = P, , (d) D o ³ P, , 15. What is the correct electronic configuration of the, central atom in K 4 [Fe(CN ) 6 ] based on crystal field, theory?, (a), , t62g, , e g0, , (b), , e 3t 23, , (c), , e 4 t 22, , (d), , t 24 g e g2, , 16. Colourless complex ion in the following is, (a) [Cu(NH3 ) 4 ] 2+, 3+, , (c) [Fe(H 2 O) 6 ], l, , (b) [Zn(NH3 ) 4 ] 2+, (d) [Fe(CN) 6 ]3 -, , Assertion-Reasoning MCQs, Direction (Q. Nos. 17-22) Each of these questions, contains two statements Assertion (A) and Reason (R)., Each of these questions also has four alternative, choices, any one of which is the correct answer. You, have to select one of the codes (a), (b), (c) and (d) given, below., (a) Both A and R are true and R is the correct, explanation of A., , (b) Both A and R are true, but R is not the correct, explanation of A., (c) A is true, but R is false., (d) A is false, but R is true., , 17. Assertion In the coordination compound, [Co ( H 2 NCH 2CH 2 NH 2 ) 3 ]2 , ethane-1,2-diammine is, a neutral molecule., Reason Oxidation number of Co in the complex ion, is +3., , 18. Assertion When EDTA solution is added to Mg 2+, ion solution then four coordinate site of Mg 2+ are, occupied by EDTA and remaining two sites are, occupied by water molecules., Reason EDTA is a hexadentate ligand., , 19. Assertion [Cr(H 2O) 6 ]Cl 2 and [Fe(H 2O) 6 ]Cl 2 are, reducing in nature., , l, , Case Based MCQs, 23. Read the following and answer the questions from (i), to (iv) given below, The crystal field theory (CFT) is an electrostatic, model which considers the metal-ligand bond as, ionic arising purely from electrostatic interactions, between the metal ion and the ligand., In an octahedral coordination entity with six ligands, surrounding the metal atom or ion, there will be, repulsion between the electrons in metal d-orbitals, and the electrons (or negative charges) of the ligands., Such a repulsion is more when the metal d-orbital is, directed towards the ligand than when it is away, from the ligand., Thus, the dx 2 -y 2 and dz 2 orbitals, which point, towards the axes along the direction of the ligand will, experience more repulsion and will be raised in, energy and the dxy , dz x and dyz orbitals which are, directed between the axes will be lowered in energy, relative to the average energy in the spherical crystal, field., Thus, the degeneracy of the d-orbitals has been, removed due to ligand-metal electron repulsions in, the octahedral complex to yield three orbitals of, lower energy, t 2g set and two orbitals of higher, energy, eg set. This splitting of the degenerate levels, due to the presence of ligands in a definite geometry, is termed as crystal field splitting., The difference of energy between the two sets of, degenerate orbitals as a result of crystal field splitting, is known as Crystal Field Stabilisation Energy, (CFSE), it is denoted by D o (the subscript o is for, octahedral). Thus, the energy of the two e g orbitals, will increase by (3/5) D o and that of the three t 2g will, decrease by (2/5)D o .

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84, , CBSE Term II Chemistry XII, , (i) Which one of these statements about [Co(CN) 6 ] 3is correct?, , Secondary valency is equal to the coordination, number of central atom or ion. It is directional and, non-ionisable. It is commonly satisfied by neutral, molecules or negative ions and is always fixed for a, metal. Each metal ion has a fixed number of, secondary valency., (i) When 1 mole of CrCl 3 × 6H2 O is treated with excess, of AgNO3 , 3 moles of AgCl are obtained. The, formula of the complex is, , (a) [Co(CN)6 ]3- has no unpaired electrons and will be in a, low-spin configuration, (b) [Co(CN)6 ]3- has four unpaired electrons and will be in, a low-spin configuration, (c) [Co(CN)6 ]3- has four unpaired electrons and will be, in a high-spin configuration, (d) [Co(CN)6 ]3- has no unpaired electrons and will be in, a high-spin configuration, , (a) [CrCl 3 (H 2 O) 3 ] × 3H 2 O, (b) [CrCl 2 (H 2 O) 4 ]Cl × 2H 2 O, (c) [CrCl(H 2 O) 5 ]Cl 2 × H 2 O, (d) [Cr(H 2 O) 6 ]Cl 3, , (ii) Among the ligands NH 3 , en, CN - and CO, the, correct order of their increasing field strength, is, (a) CO < NH3 < en < CN (b) NH3 < en < CN - < CO, (c) CN - < NH3 < CO < en, (d) en < CN - < NH3 < CO, , (ii) The primary and secondary valencies of chromium, in the complex ion, dichlorodioxalato chromium, (III) respectively are, , (iii) Identify the correct trend given below : (Atomic, number, Ti= 22, Cr = 24 and Mo = 42), , (a) 3 and 4, (c) 6 and 3, , (a) D o of [Cr(H2O)6 ]2+ < [Mo(H2O)6 ]2+, and D o of [Ti(H2O)6 ]3+ < [Ti(H2O)6 ]2+, , (iii) Which one of the following complexes will consume, more equivalents of aqueous solution of AgNO3 ?, , (b) D o of [Cr(H2O)6 ]2 + > [Mo(H2O)6 ]2+, and D o of [Ti(H2O)6 ]2+ > [Mo(H2O)6 ]2+, (c) D o of [Cr(H2O)6 ]2+ > [Mo(H2O)6 ]2+, and D o of [Ti(H2O)6 ]2+ < [Mo(H2O)6 ]2+, (d) D o of [Cr(H2O)6 ]2+ < [Mo(H2O)6 ]2+, , (a) Na 2[CrCl6 ], (c) [Cr(H2O)6 ] Cl 3, , (a) 3, , (a) 18000 cm -1, (c) 8000 cm -1, , (b) 16000 cm -1, (d) 20000 cm -1, , to (iv) given below, Alfred Werner (1866-1919), a Swiss chemist was the, first to formulate his ideas about the structures of, coordination compounds in 1898. He proposed the, concept of a primary valence and a secondary, valence for a metal ion., Binary compounds such as CrCl 3 , CoCl 2 and PdCl 2, have primary valence of 3, 2 and 2 respectively. In a, series of compounds of cobalt (III) chloride with, ammonia, it was found that some of the chloride ions, could be precipitated as AgCl on adding excess, AgNO 3 solution in cold but some remained in the, solution., Primary valency is equal to the oxidation number of, central ion. It is non-directional and ionisable. It is, satisfied by negative ions, e.g. in CrCl 3 , the primary, valency is three., , (c) 5, , (d) 4, , Or Consider the formula [Co(NH3 ) 6 ] (3Cl – ). In this, formula the species within the square bracket and the, ions outside the square bracket respectively, (a), (b), (c), (d), , (iv) The CFSE for octahedral [CoCl 6 ] 4- is 18,000 cm -1 ., The CFSE for tetrahedral [CoCl 4 ] 2- will be, , 24. Read the following and answer the questions from (i), , (b) 2, , 3+, , Or Crystal field stabilisation energy for high spin d 4, octahedral complex is, (b) -1.8D o, (d) -1.2 D o, , (b) [Cr(H2O)5 Cl] Cl 2, (d) Na 2[CrCl 5 (H2O)], , (iv) How many ions are obtained after dissociation of, the complex [ Co(NH3 ) 6 Cl 3 ?, , and D o of [Ti(H2O)6 ]3+ > [Ti(H2O)6 ]2+, , (a) -0.6D o, (c) -1.6 D o + P, , (b) 4 and 3, (d) 3 and 6, , Coordination entity and counter ions, Counter ions and coordination entity, Counter ions and counter entity, Coordinated ions and counter entity, , PART 2, Subjective Questions, l, , Short Answer Type Questions, , 1. Write the IUPAC name of the following coordination, compounds., (i) [Cr(NH 3 )3 Cl 3 ], (iii) [Co(NH 3 ) 5 (CO 3 )]Cl, , (Delhi 2017), , (ii) K 3 [Fe(CN)6 ], , 2. Using IUPAC norms, write the formulae for the, following, (i) Tetrahydroxozincate(II), (ii) Potassium tetrachloridopalladate(II), (iii) Diamminedichloridoplatinum(II), (iv) Potassium tetracyanonickelate(II), (v) Pentaamminenitrito-O-cobalt(III), (vi) Hexaamminecobalt(III) sulphate, , (NCERT)

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85, , CBSE Term II Chemistry XII, , 3. (i) Write the electronic configuration of iron ion in the, , 14. What is meant by crystal field splitting energy?, , following complex ion and predict its magnetic, behaviour : [Fe(H2 O) 6 ] 2+ ., (ii) Write the IUPAC name of the coordination complex:, [CoCl 2 (en) 2 ]NO3 ., , How does the magnitude of splitting decide the, actual configuration of d-orbitals in an octahedral, field for a coordination entity?, , 15. Using crystal field theory, draw energy level, diagram, write electronic configuration of central, atom or ion and determine the magnetic moment, (i) [FeF6 ] 3- (ii) [Fe(H 2O) 6 ] 2+, (iii) [Fe(CN) 6 ]4 -, , 4. Specify the oxidation number of metal in the following, coordination entities., (i) [Co(H 2O)(CN)(en) 2 ] 2 +, (iii) [PtCl 4 ] 2 -, , (ii) [CoBr2 (en) 2 ] +, (NCERT), , 16. Account for the following :, , 5. (i) Write the systematic names of the following :, (a) [Co(NH 3 ) 6 ] Cl 3, (b) [Pt(NH 3 ) 2 Cl (NH 2CH 3 )] Cl, (ii) Write the formula of the following compounds:, (a) Sodium hexanitrito-N-cobaltate (III), (b) Tetraaquadichlorochromium (III) chloride, (iii) Calculate the oxidation state of Cu in, (All India 2017), [Cu(NH 3 )4 ] SO 4 ., 6. Give the oxidation state, d-orbital occupation and, coordination number of the central metal ion in the, following complexes:, (i) K 3 [Co(C 2O 4 ) 3 ], (ii) (NH4 ) 2 [CoF4 ], (iii) cis-[Cr(en) 2 Cl 2 ]Cl (iv) [Mn(H 2O) 6 ]SO 4, , (i) Which of the two K4 [Fe(CN)6 ] or K3 [Fe(CN)6 ], is more stable and why ?, (ii) What are the various factors affecting crystal field, splitting?, , 17. Answer the following, (i) Why Co 2+ is easily oxidised to Co 3+ in presence, of a strong field ligand., (ii) Arrange the following complex ions in increasing, order of crystal field splitting energy ( D o )., [CrCl 6 ] 3- , [ Cr(CN)6 ] 3- , [Cr (NH3 ) 6 ] 3+, , 18. Account for the following, Co(III) forms paramagnetic octahedral complex, with weak field ligands, whereas it forms, diamagnetic octahedral complex with strong field, ligands., , 7. (i) Write the IUPAC nomenclature of [Cr(NH 3 )4 Cl 2 ] +, and [Co( en ) 3 ] 3+ complexes., (ii) Magnetic moment of [MnCl4 ] 2- is 5.92 BM. Give, reason., , 8. Discuss the nature of bonding in the coordination, 3-, , entity, [Co(C 2 O 4 ) 3 ] on the basis of valence bond, theory. Also, comment on the hybridisation and, magnetic behaviour of the given entity., 9. [Cr(NH3 )6 ]3 + is paramagnetic while[Ni(CN)4 ]2 - is, diamagnetic. Explain, why?, , 10. Explain, why [Co(NH3 )6 ]3 + is an inner orbital, 2+, , complex, whereas[Ni(NH3 )6 ] is an outer orbital, complex., (NCERT), 11. Why are different colours observed in octahedral and, tetrahedral complexes for the same metal and same, ligands ?, 12. Why do compounds having similar geometry have, different magnetic moment?, (NCERT Exemplar), 13. (i) Arrange the following complexes in the increasing, order of conducting nature of their solution., [Co(NH3 )3 Cl3 ],[Co(NH3 )4 Cl2 ]Cl,, [Co(NH3 )6 ]Cl3 ,[Co(NH3 )5 Cl]Cl2, (ii) Explain, why [Fe(H2 O)6 ]3+ has magnetic moment, value of 5.92 BM, whereas [Fe(CN)6 ]3- has a value, of only 1.73 BM?, , 19. Both [Co(NH 3 ) 6 ] 3+ and [CoF6 ] 3- are octahedral, complexes. Predict the magnetic nature of the two, complexes using CFT., 20. A solution of [Ni(H2 O)6 ]2 + is green but a solution, of [Ni(CN)4 ]2 - is colourless. Explain. (Delhi 2017), l, , Long Answer (LA) Type Questions, , 21. Discuss the nature of bonding in the following, coordination entities on the basis of valence bond, theory., (NCERT), (i) [Fe(CN) 6 ]4 (ii) [FeF6 ] 3(iii) [Co(C 2O 4 ) 3 ] 3(iv) [CoF6 ] 3-, , 22. Write the important postulates and the limitations, of valence bond theory., 23. Using valence bond theory, explain the following in, relation to the complexes given below, [Mn(CN) 6 ]3- , [Co(NH 3 ) 6 ]3+ , [Cr(H 2O) 6 ]3+, (i), (ii), (iii), (iv), , Type of hybridisation., Inner or outer orbital complex., Magnetic behaviour., Spin only magnetic moment value., , (CBSE 2019)

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86, , CBSE Term II Chemistry XII, , 24. (i) On the basis of crystal field theory explain why, Co (III) forms paramagnetic octahedral complex with, weak field ligands whereas it forms diamagnetic, octahedral complex with strong field ligands., (ii) Give the electronic configuration of the following, complexes on the basis of crystal field splitting theory, [CoF6 ] 3- , [Fe(CN) 6 ] 4- and [Cu(NH3 ) 6 ] 2+ ., , 25. The hexaaquamanganese (II) ion contains five, unpaired electrons, while the hexacyano ion contains, only one paired electron. Explain using crystal field, theory., (NCERT), , 26. Using crystal field theory, draw energy level diagram,, write electronic configuration of the central metal, atom or ion and determine the magnetic moment value, in the following., (NCERT Exemplar), (i) [CoF6 ], (ii) [FeF6 ], , 3-, , 3-, , , [Co(H2 O) 6 ], , [Fe(H2 O) 6 ], , 2+, , 2+, , and [Co(CN) 6 ], , 3-, , and [Fe(CN) 6 ]4 -, , 27. (i) [Cr(NH3 ) 6 ] 3+ is paramagnetic while [Ni(CN) 4 ] 2- is, diamagnetic. Explain, why?, (ii) [Fe(CN) 6 ] 4- and [Fe(H2 O) 6 ] 2+ are of different, colours in dilute solutions. Why?, (NCERT), l, , Case Based MCQs, 28. Read the following and answer the questions from (i), to (iv) given below, Valence bond theory describes the bonding in, complexes in terms of coordinate covalent bonds, resulting from overlap of filled ligand orbitals with, vacant metal hybrid orbitals. This theory explains, magnetic behaviour and geometrical shape of, coordination compounds., Magnetic moment of a complex, compound can be, determined experimentally and theoretically by using, spin only formula., Magnetic moment n( n + 2) BM, (where, n = no. of unpaired electrons), , (i) Write the state of hybridisation, shape and the, magnetic behaviour of the [Cr(H 2O) 2 (C2O 4 ) 2 ] - ., (ii) Why s- orbital does not show preference to any, direction?, (iii) Why is [CoF6 ]3- is paramagnetic but, [Co(NH 3 ) 6 ]3+ is diamagnetic in nature?, (iv) Describe the type of hybridisation, shape and, magnetic properites of [Co(NH3 ) 4 Cl 2 ] Cl, [Given : Atomic number of Co = 27], Or, Explain the geometry and magnetic character of, complex [Cr(NH3 ) 6 ]3+ ., , 29. Read the following and answer the questions from, (i) to (iv) given below, The properties of coordination compounds are, different from their constituents. In coordination, compounds, the central metal atom or ion is linked, to a number of ions or neutral molecules, called, ligands, by coordinate bonds. For example,, dimethyl glyoxime (dmg) is a bidendate ligand, chelating large amounts of metals., The ligands which contains three or more donor, atoms, are called polydentate ligands. When a di-or, polydentate ligand uses its two or more donor sites, to bind a single metal ion, it is called a chelating, ligand., The number of such liganting groups is called the, denticity of the ligand., (i) What is the structure of [ Pt(en)Cl 2 ]?, (ii) Give an example of a coordination complexes, where central metal atom has coordination, number 8., (iii) Name any one bidentate ligand., (iv) Categorised ethylenediamminetetra-acetate on, the basis of denticity., Or, Categorise (ethylene diammine) on the basis of, denticity.

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87, , EXPLANATIONS, 1. (d) Primary valency corresponds to the oxidation state of, metal. Thus, in the given binary compounds CrCl 3, CoCl 2, and PdCl 2 , the primary valencies are 3, 2 and 2 respectively., 2. (d) 1 mole of AgNO 3 precipitates one free chloride ion (Cl - )., Here, 3 moles of AgCl are precipitated by excess of AgNO 3., Hence, there must be three free Cl - ions., So, the formula of the complex can be [Cr(H2O) 6 ]Cl 3., 3. (b) One mole of AgNO 3 precipitates one mole of chloride, ion. In the given reaction, when 0.1 mole CoCl 3(NH3 )5 is, treated with excess of AgNO 3, 0.2 mole of AgCl are obtained, thus, there must be two free chloride ions in the solution of, electrolyte., So, molecular formula of complex will be [Co(NH3 )5 Cl] Cl 2, and electrolytic solution must contain [Co(NH3 )5 Cl] 2+ and, two Cl - as constituent ions. Thus, it is 1 : 2 electrolyte., [Co(NH3 )5 Cl]Cl 2 ¾® [Co(NH3 )5 Cl] 2 Å (aq ) + 2Cl - ( aq ), Hence, option (b) is the correct., 4. (b) The O.N. of central atom in complex is the charge it, would carry if all ligands are removed along with electron, pairs that are shared with central atom., 5. (d) Coordination number of Co in [Co(en)2Cl2]Cl is 6,, because it is comprise of 2 en ligands (bidentate ligands) and, 2 Cl (unidentate ligand)., 6. (c) Homoleptic complexes are those complexes which have, one type of ligands. e.g. [Co(NH3 )6 ]3+ , whereas heteroleptic, complexes are those which have more than one kind of, ligand, e.g. [Co(NH3 )4 Cl 2 ]+ ., 7. (c) The correct structure of ethylenediaminetetracetic acid is, as follows:, HOOCCH2, CH2COOH, N — CH2 — CH2 — N, HOOCCH2, CH2COOH, , The ligands present in the compound are, (i) NH3 - neutral ligand represented as amine., (ii) Cl - anion ligand (ending with-o-) represented as, chlorido and di prefixed is added to represent two ligands., The oxidation number of platinum in the compound is 2., , (ii) Weak field ligand forms outer orbital complex with, hybridisation sp3d 2 ., According to VBT, hybridisation and number of unpaired, electrons of coordination compounds can be calculated as :, A. [Cr(H2O) 6 ] 3+, MOEC (Molecular orbital electronic configuration) of Cr 3+, in [Cr(H2O) 6 ] 3+is, 3d, ×× ××, , 4s, , 4p, , ××, , ×× ×× ××, , H2O H2O H2O, 2, , H2O H2O H2O, , 3, , Hybridisation = d sp, n (number of unpaired electrons) = 3, 12. (c) Ni has dsp2 -hybridisation where CN- is a strong field, ligand., dsp 2 -hybridisation, , CN –, , CN – CN – CN –, , Since, all the electrons are paired, thus it is diamagnetic., 13. (a) Outer orbital complex utilises d-orbitals for bonding and, exhibit paramagnetic behaviour, only if there are unpaired, electrons., In [Ni(NH3 ) 6 ] 2 + , Ni 2+ = [Ar] 3d 8 4s 0, [Ni(NH3 ) 6 ] 2 + =, 3d, , 4s, , 4p, , 4d, , ®, , 8. (c) Correct IUPAC name can be written as, The ligands present in the given coordination compound are:, (i) (NH3 ) represented as ammine, (ii) Cl s represented as chlorido, (iii) NO s2 represented as nitrito-N, According to IUPAC rule, ligands are named in an, alphabetical order before central atom. Prefex di-will be, used to indicate the number of NH 3 ligands present., Oxidation state of metal is indicated by Roman numeral in, parenthesis., So, IUPAC name will be, diamminechloridonitrito-N-platinum (II)., Hence, option (c) is correct., 9. (a) The complex compound is [Pt(NH3 ) 2 Cl 2 ] ., , Hence, correct IUPAC name of [Pt (NH3 ) 2 Cl 2 ] is, Diammine dichloridoplatinum (II)., So, (a) option is correct., 10. (a) The formula of compound diammine silver (I), dicyanoargentate (I) is [Ag(NH3 )2 ][Ag(CN)2 ]., x + 0=1 x - 2= -1, x =1, x =1, And oxidation state of silver in both is + 1., 11. (a) A ® (3); B ® (1); C ® (4); D ® (2), Formation of inner orbital complex and outer orbital complex, determines hybridisation of molecule which inturn depends, upon field strength of ligand and number of vacant d-orbitals., (i) Strong field ligand forms inner orbital complex with, hybridisation d 2 sp3., , Two unpaired, electrons, , sp 3d 2 -hybridisation, , So, this is an outer orbital complex having paramagnetic, character., 14. (b) The fourth electron enters one of the e g orbitals giving the, configuration t 23 g e1g when D o < P., 15. (a) In K4[Fe(CN)6 ], Fe is present as Fe2+ = [Ar]3 d 6 4s 0

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88, , CBSE Term II Chemistry XII, , CN- is a strong field ligand and as it approaches the metal, ion, the electrons must pair up., The splitting of the d-orbitals in two sets orbitals in an, octahedral complex, K4[Fe(CN)6 ] may be represented as, , 22. (d) A is false, but R is true., d 4 and d 7 are stable in low spin state. When D 0 > P, pairing, is not favoured and electron excite to e g level forming high, spin complex., 23. (i) (a) Statement (a) is correct. [Co(CN) 6 ] 3-, , dx2–y2 dz2, , eg, , Co 3+=1 s 2 , 2s 2 2p6 , 3s 2 3p6 3d 6, , dxy dxz dyz, Average energy of the, d-orbitals in spherical, crystal field, , d-orbitals, free ion, , -, , t2g, Splitting of d-orbitals, in octahedral, crystal field, , CN is a strong field ligand and as it approaches the metal, ion, the electron must pair up. The splitting of the d-orbitals, into two sets of orbitals in as octahedral complex. Splitting, of [Co(CN) 6 ] 3- can be shown as, dx2– y2,dz2, , Hence, the electronic configuration of the central atom in, K4[Fe(CN)6 ] is t 26 g e g0 ., 2+, , 16. (b) [Zn(NH3 ) 4 ] is a colourless complex as, Zn, unpaired electron., Zn = 3d 10 , 4 s 2, , 2+, , ion has no, , Zn 2+ = 3 d 10, , i.e., , eg -orbitals, , 3d, ×× ××, , Do, t2g -orbitals, dxy dyz dxz, , 6, , 3d, , », », », », », (no unpaired electron), , 17. (b) Both A and R are true but R is not the correct explanation, of A., Ethane-1, 2-diammine is a neutral molecule. As it, carries no charge., Oxidation number of Co in the complex ion is +3., 18. (d) A is false, but R is true., , Here, for d ions, three electrons first enter orbitals with, parallel spin. The remaining may pair up in t 2 g -orbital, giving rise to low spin complex (strong ligand)., \ [Co(CN) 6 ] 3- has no unpaired electrons and will be in low, spin configuration., (ii) (b) Based on spectrochemical series, ligands arranged in, increasing order of crystal field strength are as, , NH 3 < en < CN - < CO, , (iii) (b) For [Cr(H2O) 6 ] 2 +, , Outer configuration of Cr 2 + = [Ar], 3 d 4, , When EDTA solution is added to Mg 2 + ion solution, then, six coordinate site of Mg 2 + are occupied by EDTA., EDTA has a hexadentate ligand., 19. (b) Both A and R are true but R is not the correct explanation, of A. Correct R is; [Cr(H2O 6 )]Cl 2 and [Fe(H2O) 6 ]Cl 2 are, reducing in nature due to formation of more stable complex, ion after gaining of electron., 20. (d) A is false, but R is true., According to VBT, MOEC (Molecular Orbital Electronic, Configuration) of Fe3+ in [Fe(CN) 6 ] 3- is, 3d, [Ar], , 2, , 4s, , 4p, , ×× ××, , ××, , ×× ×× ××, , CN CN, , CN, , CN CN CN, , 3, , Hybridisation = d sp, n =1, Hence, correct A is; [Fe(CN) 6 ] 3- ion shows magnetic, moment corresponding to one unpaired electron., i.e., , m = n( n + 2), = 1( 1 + 2), , = 3 = 1.73 BM, 21. (d) A is false, but R is true., [ CuCl 2 ] - has d10 configuration so zero unpaired electron but, exist in tetrahedral geometry with sp3-hybridisation., , eg, 0.6 Do, Energy, Average energy of, d-orbitals in symmetrical, crystal field, , –0.4 Do, t2g, Splitting in an, octahedral crystal field, , As H2O is a weak field ligand, so pairing of electrons does, not occur., CFSE for [Cr(H2O) 6 ] 2 + = 3( -0.4D o ) + 1(0.6D o ) = -0.6D o, D o depends on Zeff and for 3d-series, Zeff is less than, 4d-series., Hence, D o of [Cr(H2O) 6 ] 2 + > [Mo(H2O) 6 ] 2 +, D a charge on metal ion, The crystal field splitting energy, o, size of metal ion, 3+, \, D o of [Ti(H2O) 6 ] > D o of [Ti(H2O) 6 ] 2 +, , Or, (a) CFSE = 3 ( - 0.4) + 1 (0.6) = - 0.6D o, (iv) (c) CFSE for octahedral and tetrahedral complexes are, 4, closely related to each other by formula D t = D o., 9, where, D o = CFSE for octahedral complex,, D t = CFSE for tetrahedral complex, According to question, D o = 18000 cm -1

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89, , CBSE Term II Chemistry XII, , \, , Dt =, , 4, 4, D o = ´ 18000 cm -1, 9, 9, = 4 ´ 2000 cm -1, = 8000 cm -1, , Hence, correct choice is option (c)., 24. (i) (d) 1 mole of AgNO 3 precipitates one free chloride ion (Cl - )., Here, 3 moles of AgCl are precipitated by excess of AgNO 3 ., Hence, there must be three free Cl - ions., So, the formula of the complex can be [Cr(H 2 O) 6 ]Cl 3 and, correct choice is (d)., (ii) (d) The primary and secondary valencies of chromium in the, complex ion, dichlorodioxalatochromium (III), [ CrCl 2( Ox ) 2 ] 3 - are 3 and 6 respectively., (iii) (c), Complex, , Dissociation in aqueous, solution of AgNO 3, , (a) Na 3[CrCl6 ], , 3Na + + [CrCl6 ]3-, , (b) [Cr(H2O)5 Cl] Cl 2, , [Cr(H2O)5 Cl]2+ + 2Cl -, , (c) [Cr(H2O)6 ]Cl 3, , [Cr(H2O)6 ]3+ + 3Cl -, , (d) Na 2[CrCl 5( H2O)], , 2Na + + [CrCl 5 (H2O)]2-, , [Cr(H2O)6 ]Cl 3 contains maximum number of ionisable anion, (3Cl - )., Hence, it will consume more equivalents of aqueous solution, of AgNO3 and form three mole of AgCl (white ppt.)., (iv) (d) Primary valency of coordination compounds are, ionisable. Thus, dissociation of the complex, [Co(NH3 )6 ] Cl 3, takes place as follows, [Co(NH3 )6 ] Cl 3 ¾® [Co(NH3 ) 6 ] 3+ + 3Cl Therefore, total number of ions obtained after dissociation, are 4., Or, (a) In the complex, [Co(NH3 )6 ]3+ (3Cl - ) the species within, the square bracket is coordination entity and the ions, outside the square bracket is counter ions., Subjective Questions, (i), (ii), (iii), 2. (i), 1., , Triamminetrichloridochromium(III), Potassium hexacyanoferrate(III), Pentamine carbonatocobalt (III) chloride, (ii) K 2 [PdCl 4 ], [Zn(OH) 4 ] 2 -, , (iii) [Pt(NH3 ) 2 Cl 2 ], (iv) K 2 [Ni(CN) 4 ], (v) [Co(NH3 ) 5 (ONO)] 2 +, (vi) [Co(NH3 ) 6 ] 2 (SO 4 ) 3, 2+, 3. (i) In [Fe(H2O) 6 ] , the oxidation state of Fe is + 2 ., Electronic configuration of Fe(26) = 3 d 6 4s 2 and, Fe2 + = 3 d 6 4s 0 or Fe2 + = t 24 g e g 2, The number of unpaired electrons are four., Hence, it is paramagnetic., (ii) The IUPAC name of the coordination complex, [CoCl 2 (en) 2 ]NO 3 is dichlorido-bis (ethane - 1,, 2-diamine) cobalt (III) nitrate., , 4. Coordination entities, Oxidation state, (i) [ Co( H2O )( CN )( en ) 2 ] 2 +, +3, (ii) [ CoBr2 ( en ) 2 ] +, , +3, , (iii) [ PtCl 4 ] 2 +2, 5. (i) (a) Hexaammine cobalt (III) chloride., (b) Diammine chloro (methylamine) platinum (II), chloride., (ii) (a) Na[Co(NO 2 ) 6 ], (b) [CrCl 2 (H2O) 4 ] Cl, (iii) Oxidation state of Cu in [Cu(NH3 ) 4 ] SO 4 is calculated as,, Let oxidation state of Cu = x, x + ( 0 ´ 4) ´ - 2 = 0, x=+2, 6. (i) Oxidation state Co, In K 3[Co(C 2O 4 ) 3 ] , is calculated as,, Let oxidation state of Co = x, ( + 1 ) 3 + x + ( - 2) 3 = 0, Þ, x=+3, Thus, Co is present as Co3+., 7, 2, 27 Co = [Ar] 3d , 3s, \, , Co3+ = [Ar] 3d6, , So, d-orbital occupation is d 6 or t 62 g e g0., (As C 2O 42 - is strong field ligand), C.N. of Co = 3 ´ denticity of C 2O 4 = 3 ´ 2 = 6, (ii) (NH4 ) 2 [CoF4 ], ( + 1) 2 + x + ( - 1) 4 = 0 Þ x = + 2, Thus, Co is present as Co2+., Co2 + = [ Ar] 3d 7, So, the configuration is d 7 or t52 g e g2., As F- is a weak field ligand., C.N. of Co = 4, (iii) cis- [Cr(en) 2 Cl 2 ]Cl, x + (0) 2 + ( -1)2 + ( -1) = 0Þ x = +3, Thus, Cr is present as Cr 3+., Cr 3+ = [Ar] 3d 3, 3, , So, the configuration is d 3 or t 2 g e g0., C.N. of Cr = 2 ´ 2 + 2 = 6 (as en is didentate.), (iv) [Mn(H2O) 6 ]SO 4, x + ( 0) 6 + ( - 2) = 0Þ x = + 2, Thus, Mn is present as Mn 2 +., Mn 2 + = [Ar] 3d5 or t 23 g e g2., C.N. of Mn = 6, 7. (i) (a) [ Cr ( NH3 ) 4 Cl 2 ] +, IUPAC name Tetraammine dichlorido chromium (III) ion., (b) [ Co( en ) 3 ] 3+, , IUPAC name tris-(ethan-1,2-diamine) cobalt (III) ion.

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90, , CBSE Term II Chemistry XII, , (ii) The magnetic moment 5.92 BM shows that there are, five unpaired electrons present in the d-orbitals of, Mn 2 + ion. As a result, the hybridisation involved is sp3, rather than dsp2 . Thus, tetrahedral structure of, [MnCl 4 ] 2 - shows 5.92 BM magnetic moment value., , Co-atom, Z=27, Orbitals of, Co 3+ ion, , 8. [Co(C 2O4 )3 ] 3In this complex, the oxidation state of Co is +3., Outer configuration of Co = 3 d 7 4s 2, Co3+ = 3 d 6 4s 0, 3d6, , 3d, , 4s, , 4p, , 3d, , 4s, , 4p, , [Co(NH3)6] 3+, , ´´ ´´ ´´ ´´ ´´ ´´, 3d, , Co3+=, , In [Ni(NH3 ) 6 ] 2 + ,, , Oxalate ion being a strong field ligand pairs up the, 3d-electrons, thus two out of the five 3d-orbitals are, available for oxalate ions., [Co(C 2O4 )3 ]3- =, , 3d, , 4d, , ´´ ´´ ´´ ´´ ´´ ´´, sp 3d 2, , 3d, ×× ×× ××, , d 2sp3, , Thus, the hybridisation of given entity is d 2 sp3. Since, all the, electrons are paired, the complex is diamagnetic., 9. Hexaammine chromium (III) ion, [Cr(NH3 ) 6 ] 3 + ion, , Since, ( n - 1 ) d-orbitals are not available so, the nd-orbitals, are used in bond formation and form outer orbital complex., æ 4ö, 11. D t = ç ÷ D o, è 9ø, So, higher wavelength of light is absorbed in octahedral, complexes than tetrahedral complexes for same metal and, ligands. Thus, different colours are observed., , Oxidation state of Cr = + 3, 24 Cr =, , [Ar] 3 d5 4s1, , Cr 3 + = [Ar] 3 d 3, 3+, , Cr provides six empty orbitals to accommodate six, electron pairs from six NH3 molecules. It involves, d 2 sp3-hybridisation and hence, octahedral., It is paramagnetic due to the presence of three unpaired, electrons., 3d, , Cr, , 4s, , 12. Compounds having similar geometry may have different, magnetic moment it is due to presence of weak or strong, field ligands in complexes. e.g. [CoF6 ] 3 - and [Co(NH3 ) 6 ] 3 + ,, the former is paramagnetic, and the latter is diamagnetic, because F- is a weak field ligand and NH3 is a strong field, ligand while both have similar geometry., F, , =, , F, Co, , [Cr(NH3 )6 ] 3+=, , F, 3d, , 4p, , 4s, , ´´ ´´, , ´´, , ´´ ´´ ´´, , NH3 NH3, , NH3, , NH3 NH3 NH3, , d 2sp 3-hybridisation, , Tetracyanonickelate (II) ion, [Ni(CN) 4 ] 2 Ni = [Ar] 3 d 8 4s 2 , Ni 2 + = [Ar] 3 d 8, 3d, , Ni, , 3–, , F, , 4p, , 3+, , Three unpaired, electron, , 4s, , 4p, , 2+, , =, , (CN - being strong field ligand causes pairing of 3d electrons, thus, one of the 3d- orbital becomes empty for CN - ion.), Unpaired electron is not present, so it is diamagnetic., 3d, , 4s, , ´´, All electrons, are paired, , 3, , Complex has octahedral ( d sp ) geometry and is diamagnetic., Since, ( n-1 ) d orbitals are available so forms inner orbital, complex., , 3s0, , ×× ××, , d 2sp 3, 2, , ´´, , 4p, , ´´ ´´, , dsp 2-hybridisation, , 10. In [Co(NH3 ) 6 ] 3 + , cobalt is in + 3 oxidation state having, electronic configuration [Ar]18 3d 6 ., , 13., , F, , NH3, , NH3, and, , F, , 3+, , NH3, , Co, NH3, NH3, , NH3, , (i) The increasing order of conducting nature of given, complexes in their solution is, [Co(NH3 ) 3Cl 3 ] < [Co(NH3 ) 4 Cl 2 ] Cl <, [Co(NH3 )5 Cl] Cl 2 < [Co(NH3 ) 6 ]Cl 3, This is because ions produced from these complexes, are 0, 2, 3 and 4 respectively., (ii) In both the complex, Fe is in +3 oxidation state with, configuration 3d5 ,i.e. it has five unpaired electrons. In, the presence of weak H2O ligand, the unpaired, electrons do not pair up but in the presence of strong, ligand, i.e. CN - , they get paired up., 3d 5, [Fe (CN)6]3– =, , ×× ×× ×× ×× ×× ××, , CN– CN– CN– CN– CN– CN–, 2, , 3, , d sp -hybridisation, (Square planar structure), , Magnetic moment ( m ) = n( n + 2), = 3 = 1.73 BM

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91, , CBSE Term II Chemistry XII, , 16., , 3d 5, [Fe(H2O)6]3+ =, , ×× ×× ×× ×× ×× ××, , (i) In K 4 [Fe(CN) 6 ], oxidation state of Fe is + 2., In K 3 [Fe(CN) 6 ], oxidation state of Fe is + 3., Therefore, outer electronic configuration of, Fe2 + = 3 d 6 = t 26 g and of, , 3 2, , sp d -hybridisation, , Magnetic moment ( m ) = n( n + 2), , Fe3+ = 3 d5 = t52 g, , = 5( 5 + 2), = 35 = 5.92 BM, , C.F.S.E of K 4 [Fe(CN) 6 ] = - 0.4 ´ 6D o = -2.4D o, , 14. In complexes, the splitting of degenerate orbitals into, two sets, t 2 g and eg occurs due to the presence of ligands, in a definite geometry. This is known as crystal field, splitting and in case of octahedral complexes, the energy, separation is known as crystal field splitting energy, (C.F.S.E). It is denoted by D o., The magnitude of splitting decides the actual, configuration of d-orbitals in an octahedral field for a, coordination entity., If D o < P ( P is the energy required for pairing of electrons), then fourth electron enters in one of the e g orbitals, ( t 23 g e1g ). It means first five orbitals are individually, occupied and then pairing will take place. It is the case of, weak field ligand and a high spin complex., If D o > P, t 2 g set will have d 1 to d 6 -electrons (because, pairing occurs in t 2 g ), whereas e g level orbitals remain, unoccupied upto d 6 . It is the case of strong field ligand and, low spin complex., 15. We know that magnetic moment,, , More is the C.F.S.E, more is the stability of the complex., Therefore, K 4 [Fe(CN) 6 ] is more stable than K 3 [Fe(CN) 6 ]., (ii) The crystal field splitting, D o depends upon the field, produced by the ligand and charge on the metal ion., Some ligands are able to produce strong fields in that, case, the splitting will be large, whereas others produce, weak field and consequently results in small splitting of, d-orbitals., 17., , (i) Co2 + is a 3 d 7 species. In the presence of a strong field, ligand, the distribution of these electrons is t 26 g e1g . For, greater stability thus Co2 + can lose one unpaired, electron from higher ( e g ) energy level., (ii) Crystal field splitting energy (C.F.S.E) is higher, when the complex contains strong field ligand, The order of field strength of given ligand is, Cl - < NH3 < CN -, , Therefore, crystal field splitting energy increases in the order, [CrCl 6 ] 3 - < [Cr(NH3 ) 6 ] 3 + < [Cr(CN) 6 ] 3 18. With weak field ligands, D o < P, (pairing energy) so, pairing, does not occurs the electronic configuration of Co (III) will be, t 24 g e g2 , i.e. it has 4 unpaired electrons and is paramagnetic., , m = n( n + 2), where, n is the unpaired electrons., eg, , dz2 dx2–y2, , C.F.S.E of K 3 [Fe(CN) 6 ] = - 0.4 ´ 5D o = -2.0D o, , (i) [FeF6]3–, Fe3+= 3d 5, , d5, , dxy, , dyz, , eg, , t2g, , yzx, , dx 2–y 2 dz 2, Do, , Number of unpaired electrons = 5, Degenerate, d-orbital, , m = 5( 5 + 2) = 35 = 5.92 BM, , dz2 dx2–y2, , eg, , With strong field ligands D o > P (pairing energy), so pairing, occurs thus, the electronic configuration will be t 26 g e g0 . It has, , (ii) [Fe(H2O)6]2+, Fe2+= 3d 6, , dxy, , dyz, , dzx, , t2g, , no unpaired electrons and is diamagnetic., , Number of unpaired electrons = 4, , 3d, , dz 2 dx2– y2, , Co, Þ, (Do > P), eg, , (iii) [Fe(CN)6]4–, , No unpaired electrons, so diamagnetic., , dx, , dy, , 6, , 4s0, , 3+, , Magnetic moment = 4( 4 + 2) = 24 = 4.9 BM, , Fe2+= 3d 6, , t2g, dxy dyz d zx, , dz, , t2g, , No unpaired electron, , 19. In [ Co( NH3 ) 6 ] 3+ , Co has oxidation state + 3 and its outer, outer electronic configuration is 3 d 6 . Since, NH3 is a strong, field ligand and D o > P, the pairing of electrons occur and, electronic configuration is t 26 g e 0g .

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93, , CBSE Term II Chemistry XII, , orbitals of equal energy and definite geometry, such as, linear, tetrahedral, square planar and octahedral shapes., These hybridised orbitals are allowed to overlap with, ligand orbitals that can donate electron pairs for, bonding., (iii) Ligands must have one sigma orbital containing lone, pair of electrons., The limitations of valence bond theory are as follows, (i) It is based on a number of assumptions., (ii) It does not distinguish between weak and strong, ligands., (iii) It does not give quantitative interpretations of, magnetic data., (iv) It does not give a quantitative interpretation of the, thermodynamic or kinetic stabilities of coordination, compounds., (v) It fails to explain relative energies of different shapes., (vi) It does not explain the colour and spectral properties of, the compounds with C.N. = 4., (vii) It fails to predict the exact tetrahedral and square, planar structures of complexes with C.N. = 4., 23. [Mn(CN)6 ]3Outer configuration of, , Mn 3+ = 3d 4, 3d, , Outer configuration of, 3d3, , 4s0, , [Cr3+=3d3 4s0=, 3d, [Cr(H2O)6]3+=, , ´´ ´´, , 4s, ´´, , 4p, ´´ ´´ ´´, d2sp3, , (i) d 2 sp3-hybridisation, (ii) Inner orbital complex [as ( n -1 ) d-orbital take part.], (iii) Paramagnetic (as three unpaired electrons are, present.), (iv) m = n ( n + 2) = 3 ( 3 + 2) = 15 = 3.87 BM, 24., , (i) With weak field ligands; D o < P, (pairing energy) so, the, electronic configuration of Co (III) will be t 24 g e g2 , i.e. it, has 4 unpaired electrons and is paramagnetic., eg, , Degenerate, d-orbital, , =, 3d, [Mn(CN)6]3– =, , [Cr(H 2O)6 ]3+, , ´´ ´´, , 4s, , 4p, , ´´, , ´´ ´´ ´´, , CN– CN– CN– CN– CN– CN–, , t2g, , With strong field ligands, D o > P (pairing energy),, so pairing occurs thus, the electronic configuration, will be t 26 g e g0 . It has no unpaired electrons and is, diamagnetic., eg, , d2sp3-hybridisation, , (i) d 2 sp3-hybridisation, (ii) Inner orbital complex because ( n - 1 ) d-orbitals are, used., (iii) Paramagnetic, as two unpaired electrons are present., (iv) Spin only magnetic moment, ( m ) = 2 ( 2 + 2) = 8 = 2.82 BM, [Co(NH 3 )6 ]3+, Outer configuration of Co3+ = 3d 6 4s 0, 3d, , Degenerate, d-orbital, , (ii) CN - and NH 3 being strong field ligand pair up the t 2 g, electrons before filling e g set., But this is not the case with F- ligand as it is a weak, field ligand., [CoF6 ]3- ; Electronic configuration Co3+ =( d 6 ) t 24 g e g2, , 6, , eg, , =, 3d, [Co(NH3)6]3+ =, , t2g, , ´´ ´´, , 4s, ´´, , 4p, ´´ ´´ ´´, , d2sp3, , (NH 3 pair up the unpaired 3d-electrons.), (i) d 2 sp3-hybridisation, (ii) Inner orbital complex because of the involvement of, ( n - 1 ) d-orbital in bonding., (iii) Diamagnetic, as no unpaired electron is present., (iv) m = n ( n + 2) = 0 ( 0 + 2) = 0 (Zero), , Degenerate d-orbitals, , t2g, , [Fe(CN)6 ]4- ; Electronic configuration Fe2 + = ( d 6 ) t 26 g e g0, eg, , Degenerate d-orbitals, , t2g

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94, , CBSE Term II Chemistry XII, , [Cu(NH3 )6 ]2+ ; Electronic configuration Cu 2 + = ( d 9 ) t 36g e g3, , eg, , eg, Degenerate, d-orbitals, , Degenerate, d-orbitals, , t2g, , Number of unpaired electrons ( n ) = 3, m = 3 ( 3 + 2) = 15 = 3.87 BM, , 25. Because of the presence of six ligands the complex is, octahedral and thus, d-orbital split up into lower energy, t 2 g -orbitals and higher energy eg- orbitals., , [Co(CN)6 ] , i.e. Co 3 +, 3-, , Electronic configuration of Co3+ = 3d6( or t 26 g e g0 ), , CN - is a strong field ligand (thus, causes pairing) and H2O is, a weak field ligand. Mn (II) ion has 3d5 configuration. In the, presence of H2O molecules (acting as weak field ligand), the, distribution of these five electrons is t 23 g , e g2 , i.e. all the, , eg, , Degenerate, d-orbitals, , electrons remain unpaired to form a high spin complex., eg, , t2g, , There is no unpaired electron, so it is diamagnetic., (ii) [FeF6 ] 3-, , d x2 _ y2 d z2, Do, Degenerate, d-orbitals, , t2g, , Electronic configuration of Fe3+ = 3 d5 ( or t 23 g e g2 ), eg, , t 2g, dxy dyz dzx, Splitting in the presence of, octahedral weak field, , Degenerate, d-orbitals, , However, in the presence of CN - (acting as strong field, ligands), the distribution of these electrons is t52 g , e g0 , i.e. two, t 2 g orbitals contain paired electrons, while the third t 2 g, orbital contains one unpaired electron.The complex formed, is a low spin complex., , t2g, , Number of unpaired electrons, n = 5, m = 5 ( 5 + 2) = 35 = 5.92 BM, [Fe(H 2O)6 ], , 2+, , Electronic configuration of Fe2+ = 3 d 6 ( or t 24 g e g2 ), , eg, , eg, , Degenerate, d-orbitals, Do, Splitting in the, presence of strong, field ligand, , 26., , Degenerate, d-orbitals, , t 2g, , t2g, , Number of unpaired electrons, n = 4, m = 4 ( 4 + 2), , (i) [CoF6 ]3- , i.e. Co 3 +, Configuration of Co3+ = 3d 6 (or t 24g e g2 ), , Z = 24 = 4.98 BM, [Fe(CN)6 ]4-, , eg, , Electronic configuration of Fe2+ = 3 d 6 ( or t 26 g e g0 ), eg, , Degenerate, d-orbitals, , t2g, , Number of unpaired electrons ( n ) = 4, Magnetic moment ( m ) = n ( n + 2), , Degenerate, d-orbitals, , = 4 ( 4 + 2), , Since, CN- is a strong field ligand, all the electrons get, paired. Hence, there is no unpaired electron, so it is, diamagnetic in nature., , = 24 = 4. 9 BM, 2+, , [Co(H 2O)6 ] , i.e. Co, , 2+, , Configuration of Co2+ = 3d 7 ( or t52 g e g2 ), , t2g, , 27., , (i) Hexaammine chromium (III) ion, [Cr(NH 3) 6] 3+ ion, Oxidation state of Cr = +3

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Chapter Test, , 7. Assertion Removal of water from [ Ti(H2O) 6 ] Cl3 on, heating renders it colourless., , Multiple Choice Questions, , 1. According to postulates of Werner’s theory for, coordination compounds, which of the following is true?, (a), (b), (c), (d), , Primary valencies are ionisable., Secondary valencies are ionisable., Only primary valencies are non-ionisable, Both primary and secondary valencies are non-ionisable., , 2. A complex compound in which the oxidation number of a, metal is zero, is, (a), (b), (c), (d), , Short Answer Type Questions, , 8. For the complex [Fe(en) 2 Cl2 ]Cl, identify the, following :, (i) Name of the complex and oxidation number of iron, (ii) Hybrid orbitals and shape of the complex, (iii) Magnetic behaviour of the complex, , 9. Predict the number of unpaired electrons and, , [Ni(CO) 4 ], [Pt(NH 3 ) 4 ]Cl 2, K 3[Fe(CN) 6 ], K 4 [Fe(CN) 6 ], , magnetic nature in the square planar [Pt (CN) 4 ]2ion., , 10. On the basis of crystal field theory, explain why, , 3. Which among the following will be named, as dibromidobis (ethylenediammine), chromium (III) bromide ?, (a) [Cr(en) 3 ] Br3, (b) [Cr(en) 2Br2 ]Br, (c) [Cr(en)Br4 ](d) [Cr(en)Br2 ]Br, , Co(III) forms paramagnetic octahedral complex with, weak field ligands whereas it forms diamagnetic, octahedral complex with strong field ligands., , 11. What is the relationship between observed colour, of the complex and the wavelength of light, absorbed by the complex?, , 4. For d 4 ions, the fourth electron enters in one of the eg, orbitals giving the, , Reason In the absence of ligand, crystal field, splitting does not occur., , configuration t 23g eg1 ,, , (a) D o > P, (c) D o = P, , when, , (b) D o < P, (d) D o ³ P, , Assertion Reason, Direction (Q. Nos. 5-7) Each of these questions contains, two statements Assertion (A) and Reason (R). Each of, these questions also has four alternative choices, any one, of which is the correct answer. You have to select one of, the codes (a), (b), (c) and (d) given below., (a) Both A and R are true and R is the correct explanation of A., (b) Both A and R are true, but R is not the correct explanation, of A., (c) A is true, but R is false., (d) A is false, but R is true., , 5. Assertion The ligands of nitro and nitrito are called, , Long Answer Type Questions, , 12. PtCl4 and NH3 may form five complexes,, A (PtCl4 ×6NH3 ), B (PtCl4 ×5NH3 ) , C (PtCl4 × 4NH3 ),, D (PtCl4 × 3NH3 ) and E(PtCl4 × 2NH3 ). One mole of, each A , B, C , D and E reacts with excess of AgNO 3 to, yield 4, 3, 2 and 1 mole(s) of AgCl respectively, while E, gives, no AgCl. The conductance of their solutions are, in the order A > B > C > D > E. On the basis of, Werner’s theory, write their structure and give the, total number of ions given by one complex., , 13. Write down the IUPAC name for each of the, following complexes and indicate the oxidation, state, electronic configuration and coordination, number. Also give stereochemistry and magnetic, moment of the complex., , ambidentate ligands., , (i) K [Cr(H2O) 2 (C 2O 4) 2 ] × 3H2O, , Reason These ligands give linkage isomer., , (ii) [Co(NH3 )5 Cl]Cl2, , 3-, , ion shows magnetic moment, corresponding to two unpaired electrons., , 6. Assertion [Fe(CN) 6 ], , Reason [Fe(CN) 6 ] 3- has d 2 sp 3 type hybridisation., , (iii) [CrCl3 (py) 3 ], (iv) Cs[FeCl4 ], (v) K 4 [Mn(CN) 6 ], , Answers, Multiple Choice Questions, 1. (a), , 2. (a), , 3. (b), , 4. (b), , Assertion-Reasoning MCQs, 5. (a), , 6. (d), , 7. (a), , For Detailed Solutions, Scan the code

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97, , CHAPTER 06, , Aldehydes, Ketones, and Carboxylic Acids, In this Chapter..., , l, , H, , Aldehyde, , l, , l, , Preparation of Aldehydes and Ketones, , l, , Chemical Reaction of Aldehydes and Ketones, , l, , Carboxylic Acid, , The organic compounds containing carbon-oxygen, double bond ( C == O ), called carbonyl group, are, known as carbonyl compounds. Their general formula, is C n H 2 n O ., In aldehydes, the carbonyl group is bonded to a, carbon and a hydrogen, while in ketones, it is bonded, to two carbon atoms., O, O, R, , l, , Nomenclature and Structure of Carbonyl Group, , R, , R´, , Ketone, , The carbonyl compounds in which carbonyl group is, bonded to an oxygen atom of hydroxyl moiety ( ¾ OH), are known as carboxylic acids., In compounds, where carbonyl carbon is attached to a, nitrogen and to halogens, are called amides and acyl, halides respectively., , Nomenclature and Structure, of Carbonyl Group, , l, , l, , l, , l, , The common names of ketones are derived by naming two alkyl, or aryl groups bonded to the carbonyl group., The locations of substituents are indicated by Greek letters, aa ¢, bb¢ and so on and beginning with the carbon atoms next to, the carbonyl group, indicated as aa ¢., The IUPAC names of open chain aliphatic aldehydes and ketones, are derived from the names of the corresponding alkanes by, replacing the ending -‘e’ with -‘al’ and -‘one’, respectively., The substituents are prefixed in alphabetical order along with, numerals indicating their positions in the carbon chain. When an, aldehyde group is attached to a ring, the suffix carbaldehyde is, added after the full name of the cycloalkane. The numbering of, the ring carbon atoms start from the carbon atom attached to the, aldehyde group., , Structure, The C-atom of carbonyl group is sp 2 -hybridised with bond angle, 120° and possessing trigonal co-planar structure., The carbonyl carbon and the three atoms attached to it lie in the, same plane and the p-electron cloud lies above and below this plane., , Nomenclature, l, , Common names of aldehydes are derived from, corresponding carboxylic acids by replacing the, ending -ic of acid with aldehyde. The location of, substituent in the carbon chain is indicated by Greek, letters a, b, g, etc., , p-bond, sp2, sp2, , C, , s-bond, , O, , sp2, sp2, , C, , 120°, , O, , 120°, , l, , l, , C, 120°, , Orbital diagram for the formation of carbonyl group, , O

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101, , CBSE Term II Chemistry XII, , Chapter, Practice, 6. Alcohol vapours are passed over which of the, , PART 1, Objective Questions, , following catalysts to give aldehydes and ketones?, (a) S or Pd, (c) F or Cl, , 7. Consider the following reaction,, , 1. The bond angle and structure of carbonyl, , compounds respectively are, (a), (b), (c), (d), , (b) Ag or Cu, (d) Li or K, CHC6H5, , (i) O3, , 120° and trigonal planar, 109°28¢ and tetrahedral, 120° and tetrahedral, 109°28¢ and trigonal planar, , (ii) Zn / H2O, , O, +X, , 2. The orbital diagram for the formation of carbonyl, , group is given as follows, , The compound X is, CHO, , sp, , 2, , sp 2, , II, 2, sp 2 C, , I, , 120°, , sp, , C, , O, , 120° C, 120°, , O, , III, O, , CHO, , (a), , (b), COOH, , OH, (d), , (c), , (A), , (B), , (C), , Which type of bond is involved in I and II?, , 8. Rosenmund reduction gives, , I, (a) s, (c) s, , 9. In the reaction,, , II, p, s, , I, (b) p, (d) p, , II, p, s, , 3. The most suitable reagent for the conversion of, , R ¾ CH 2 ¾ OH ¾® R ¾ CHO is, , (ii) H 2 O, , Identify the product formed in the given reaction., (a) H3C ¾ CH2CH2CH2CH2CN, , major product is, , (c), , (i) NaOCl, , (b), , CH2OH, , (ii) H3O+, , CH2OH, PCC, , (Pyridinium, chlorochromate), , (d), , (b) H3C ¾ CH == CH ¾ CH2CH2CHO, (c) H3CCH2CH2CH2CH2CHO, (d) None of the above, , COCH3, K2Cr2O7, , H 3C ¾ CH == CH ¾ CH 2 ¾ CH 2 ¾ CN, , (i) AlH(i —Bu)2, , 4. The reaction that does not give benzoic acid as the, CH2OH, , (b) ether, (d) hydrocarbon, , ¾¾¾¾¾¾®, , (a) KMnO4, (b) K2Cr2O7, (c) H2 SO4, (d) PCC (Pyridinium Chlorochromate), , (a), , (a) aldehyde, (c) carboxylic acid, , 10. Select the structure of chromium complex formed, , when toluene reacts with chromyl chloride to give, benzaldehyde on hydrolysis., , KMnO4/H+, , CH(OCrCl2)2, (a), , CH2(OCrOHCl2), (b), , 5. Which of the following compounds will give, , butanone on oxidation with alkaline KMnO 4, solution?, (NCERT Exemplar), (a) Butan-1-ol, (c) Both (a) and (b), , (b) Butan-2-ol, (d) None of these, , CH(OCrOHCl2)2, (c), , (d), , CH(OCrOH2Cl)2

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102, , CBSE Term II Chemistry XII, , 11. The reagent which does not react with both, acetone, , and benzaldehyde ?, , 18. The carboxylic carbon is less electrophilic than, , carbonyl carbon because of, , (NCERT), , (a), (b), (c), (d), , (a) Sodium hydrogen sulphite, (b) Phenyl hydrazine, (c) Fehling's solution, (d) Grignard reagent, , 19. Which of the following types of hydrolysis of esters, , 12. The most polar compound among the given carbonyl, , give directly carboxylic acids and carboxylates, respectively?, , compounds is, (a), (b), (c), (d), , propanone, formaldehyde, propanal, hexan-3-one, , 13. Arrange the following compounds in the increasing, , order of their boiling points., CH 3CH 2CH 2CHO, CH 3CH 2CH 2CH 2OH,, (I), , (a) Acidic hydrolysis and, (b) Basic hydrolysis and, (c) Acidic hydrolysis and, (d) Basic hydrolysis and, , CH 3CH 2CH 2CH 2CH 3, , (III), , (IV), , (a) IV < I < II < III, (c) IV < III < I < II, , esterification of carboxylic acids., I. The protonated esters loses a proton to give the, ester., II. Protonation of the carbonyl oxygen activates the, carbonyl group towards nucleophilic addition of, the alcohol., III. Proton transfer in the tetrahedral intermediate, occurs., Arrange the following steps in their correct sequence., , (b) IV < III < II < I, (d) IV < II < III < I, , 14. Which of the following compounds is most reactive, , towards nucleophilic addition reactions?, (NCERT Exemplar), , O, ||, (a) CH 3 ¾ C ¾ H, (c), , C, , (a) I, II and III, (c) II, III and I, , O, ||, (b) CH 3 ¾ C ¾ CH 3, , O, , C, , H, , (b) I, III and II, (d) III, II and I, , 21. Name the product formed during the, , decarboxylation of malonic acid., , O, , (d), , basic hydrolysis, acidic hydrolysis, acidic hydrolysis, basic hydrolysis, , 20. The following steps are given for the mechanism of, , (II), , H 5C 2O — C 2 H 5 ,, , the possible resonating structure, Lewis structure, the possible hyperconjugative structure, Both (a) and (b), , (a) Acetic acid, (c) Propanone, , CH3, , (b) Ethanone, (d) Formic acid, , 22. The product formed during Hell-Volhard-Zelinsky, , reaction is, , 15. Which of the following compound does not reduce, , (a) R ¾ CH— COOH, ½, X, X, ½, (c) R ¾ C — COOH, ½, X, , Fehling solution?, (a) CH 3COOH, (c) HCHO, , (b) HCOOH, (d) CH 3CHO, , 16. Which of the following compounds do not undergo, , aldol condensation?, CH3, ½, (a) CH3 ¾ C ¾ CHO, ½, CH3, O, ½½, (c) CH3 ¾ C ¾ CH3, , l, , (b) CH3 ¾ CHO, , (d) CH3CH2 CHO, , 17. Which one of the following aldehydes does not give, , Cannizzaro reaction?, (a), (b), (c), (d), , Formaldehyde, Acetaldehyde, Trimethyl acetaldehyde, Benzaldehyde, , (NCERT), , (b) R — CH2 ¾ COX, , (d) R ¾ CH— CH2 ¾ COOH, ½, X, , Assertion-Reasoning MCQs, Direction (Q. Nos. 23-29) Each of these questions, contains two statements Assertion (A) and Reason (R)., Each of these questions also has four alternative, choices, any one of which is the correct answer. You, have to select one of the codes (a), (b), (c) and (d) given, below., (a) Both A and R are true and R is the correct, explanation of A., (b) Both A and R are true, but R is not the correct, explanation of A., (c) A is true, but R is false., (d) A is false, but R is true.

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103, , CBSE Term II Chemistry XII, , 23. Assertion IUPAC name of, , 24., , 25., , 26., , 27., , 28., , 29., , l, , HOOC —, ( CH 2 —, ) 2 COOH is butane - dioic acid., Reason In compounds containing more than one, carboxyl group, ‘-e’ of the alkane is retained, and, prefix ‘di’ is added to the term-‘oic’., Assertion The boiling points of aldehydes and, ketones are greater than hydrocarbons and ethers of, comparable masses., Reason It is due to weak molecular association in, aldehydes and ketones arising out of the, dipole-dipole interactions., Assertion Cross aldol condensation of ethanol and, propanal gives a mixture of four products., Reason Ethanal and propanal both contain, a-hydrogen atom., Assertion Oxidation of ketones is easier than, aldehydes., Reason C ¾ C bond of ketones is stronger than, C ¾ H bond of aldehydes., Assertion Carboxylic acids have higher boiling, liquids than aldehydes, ketones and even alcohols of, comparable molecular masses., Reason More extensive association of carboxylic, acid molecules through intermolecular hydrogen, bonding is responsible for the high boiling point of, carboxylic acid., Assertion Nitration of benzoic acid gives, m-nitrobenzoic acid., Reason Carboxyl group increases the electron, density at the meta-position., Assertion Benzoic acid does not undergo, Friedal-Crafts reaction., Reason The carboxyl group is activating and undergo, electrophilic substitution reaction., , Case Based MCQs, , 30. Read the following and answer the questions from (i), , to (iv) given below, Carboxylic acids are organic acids identified by a, carboxyl group (—COOH). They are widespread in, nature and are typically weak acids, meaning they, can only dissociate into H + cations and RCOO anions in aqueous solutions. The presence of, electronegative groups (such as OH - or Cl - ) next to, be carboxylic acid increases the acidity through, inductive effects. The carboxylic acids react with, alcohols or phenols in the presence of a mineral acid, such as conc. H 2SO 4 or HCl gas (as a catalyst) to form, an ester., , However, on heating with only mineral acids,, carboxylic acids gives corresponding anhydride. Acid, anhydrides are obtained by the elimination of water, molecules., O, ½½, R, H+ , D, 2R ¾ C ¾ OH ¾ ¾¾®, or P2O5 , D, , R, , O, , C, C, , O + H 2O, , O, , (i) Which of the following is the strongest acid?, (a) CH3COOH, (c) Cl 2CHCOOH, , (b) ClCH2COOH, (d) Cl 3C ¾ COOH, , (ii) Which of the following reagents is/are used for the, conversion of ethanoic acid to ethanoic anhydride?, (a) SOCl 2, D, (c) P2O5, D, , (b) PCl 3, D, (d) All of these, , (iii) Identify the ‘X’ and ‘Y’ from following reaction., CH3 COOH + CH3 OH, X+Y, , W, , (a) X = CH3COOCH3, Y = H2O, (b) X = CH3CH2OH, Y = CO2, (c) X = CH3CH2COOH, Y = H2, (d) X = CH3COCH3, Y = CO2, , (iv) What is the name of the above reaction?, (a) Saponification, (b) Hell-Volhard-Zelinsky reaction, (c) Kolbe’s reaction, (d) Esterification, , Or, What will be the by products of the following, reaction?, R ¾ COOH + SOCl 2 ¾® RCOCl + ? + ?, (a) CO 2 , HCl, , (b) SO 2 , H2O, , (c) SO 2 , HCl, , (d) CO 2 , H2, , 31. Read the following and answer the questions from (i), , to (iv) given below, Nucleophilic addition reactions are encountered in, compounds containing polar functional groups, (C==O, CººN, C==S). In the first step, a nucleophile, with its pair of electrons attacks the carbon atom of a, double or triple bond, forming a carbanion. It is, followed by a second step in which this carbanion, reacts with a positive species., Y, Step I, , C, , C, , + Y, , C, , Cs, , + Wr, , s, , Cs, , C, , C, , Y, , Y, Step II, , C, , W, , When the olefin contains a good leaving group (as, defined for nucleophilic substitution), substitution is

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105, , CBSE Term II Chemistry XII, , 3. Draw the structures of the following derivatives., (i) 2,4-dinitrophenylhydrazone of benzaldehyde, (ii) Cyclopropanone oxime, (iii) Acetaldehyde dimethylacetal, (iv) Semicarbazone of cyclobutanone, (v) Ethylene ketal of hexan-3-one, (vi) Methyl hemiacetal of formaldehyde, , 11. How do you convert the following?, (i) Ethanal to propanone, (ii) Toluene to benzoic acid, , 12. Identify the compounds A, B and C in the following, , reaction, Mg / Ether, , (i) CO, , 2, CH 3 ¾ Br ¾¾¾® [ A] ¾¾¾®, [ B], , (ii) Water, , 4. Name the electrophile produced in the reaction of, , CH3OH / H +, , ¾¾¾¾® [ C ], , benzene with benzoyl chloride in the presence of, anhydrous AlCl3 . Name the reaction also., , (NCERT Exemplar), , (NCERT Exemplar), , 13. Account for the following, (i) Aromatic carboxylic acids do not undergo, Friedel-Craft’s reaction., (ii) pK a value of 4-nitrobenzoic acid is lower than that of, benzoic acid., (CBSE 2018), , 5. Arrange the following compounds in the increasing, , order of their reactivity towards nucleophilic, addition reactions., (i) Ethanal, propanal, propanone, butanone, (ii) Benzaldehyde, p-tolualdehyde,, (NCERT), p-nitrobenzaldehyde, acetophenone, 6. Complete the following reaction sequence., , 14. Account for the following., (i) Cl ¾ CH 2COOH is a stronger acid than CH 3COOH ., (ii) Carboxylic acids do not give reactions of carbonyl, group., , O, (i) CH MgBr, , CH3 — Br, 3, Na metal, CH3 ¾ C ¾ CH3 ¾¾¾¾, ® A ¾¾¾®, B ¾¾¾®, C, Ether, , 15. Write the products of the following reactions:, , (ii) H2O, , (NCERT Exemplar), , O, , 7. An organic compound with molecular formula, , C 9H10O, forms 2,4-DNP derivative, reduces Tollen’s, reagent and undergoes Cannizzaro reaction., On vigorous oxidation, it gives 1,2-benzene, dicarboxylic acid. Identify the compound. (NCERT), 8. An organic compound contains 69.77% carbon,, 11.63% hydrogen and rest oxygen. The molecular, mass of the compound is 86. It does not reduce, Tollen’s reagent but forms an addition compound, with sodium hydrogen sulphite and gives positive, iodoform test., On vigorous oxidation, it gives ethanoic acid and, propanoic acid. Write the possible structure of the, (NCERT), compound., 9. A, B and C are three non-cyclic functional isomers, of a carbonyl compound with molecular formula, C 4 H 8O., Isomers A and C give positive Tollens’ test whereas, isomer B does not give Tollens’ test, but gives, positive iodoform test. Isomers A and B on, reduction with Zn-Hg/conc. HCl give the same, product D., (a) Write the structures of A, B, C and D., (b) Out of A, B and C isomers, which one is least, reactive towards addition of HCN?, (CBSE 2018), 10. Write the equations involved in the following, reactions, (i) Wolf-Kishner reduction, (ii) Etard reaction, (Delhi 2017), , (CBSE 2018), , (i), , +, , H, == O + NH2 ¾ NH ¾ C ¾ NH2 ¾®, Dry ether, , (ii) CH 3MgBr + CO 2 ¾ ¾ ¾ ¾®, H 3 O+, , Red phosphorus, , (iii) CH 3CH 2COOH + Br2 ¾¾¾¾¾®, l, , (Delhi 2017C), , Long Answer Type Questions, 16. (i) An organic compound (A) having molecular formula, C4 H 8O gives orange red precipitate with 2, 4-DNP, reagent. It does not reduce Tollen’s reagent but gives, yellow precipitate of iodoform on heating with, NaOH and I 2 . Compound (A) on reduction with, NaBH 4 gives compound (B) which undergoes, dehydration reaction on heating with conc. H 2SO 4, to form compound (C). Compound (C) on ozonolysis, gives two molecules of ethanol., Identify (A), (B) and (C) and write their structures., Write the reactions of compound( A) with (i), NaOH / I 2 and (ii) NaBH4 ., (ii) Give reasons, (a) Oxidation of propanal is easier than propanone., (b) a-hydrogen of aldehydes and ketones is acidic in, nature., 17. An aldehyde A (C11H 8 O) which does not undergo self, , aldol condensation but gives benzaldehyde and two, moles of B on ozonolysis. Compound B on oxidation, with silver ions gives oxalic acid. Identify the, compounds A and B.

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106, , CBSE Term II Chemistry XII, , 18. (i) Draw structures of the following derivatives, (a) Cyanohydrin of cyclobutanone, (b) Hemiacetal of ethanal, (ii) Write a major product(s) in the following, (i) DIBAL - H, , (a) CH 3 ¾ CH == CH ¾ CH 2 ¾ CN ¾¾¾¾¾®, (ii) H 3 O+, , formula C 3 H 6 O. E neither gives Tollen’s test nor, reduces, Fehling’s solution but forms 2,4-dinitrophenyl, hydrazone. Identify A,B,C,D and E., 23. (i) Write the products of the following reactions, (a), , CrO3, , (b) CH 3 ¾ CH 2 ¾ OH ¾¾®, (iii) How can you distinguish between propanal and, propanone?, , (b) 2 C 6H5 CHO + Conc. NaOH ¾®, 2 /P®, (c) CH 3COOH ¾Cl, ¾¾, , 19. A ketone A (CH 4 H 8O) which undergoes a haloform, , reaction gives compound B on reduction, heating, with sulphuric acid gives a compound C which forms, mono-ozonide D. D on hydrolysis with zinc dust, gives only E. Identify A, B, C, D and E write the, reactions involved., How does compound A gives iodoform reaction?, 20. (i) Write the final products in the following, CH3, (a), , C, , O, , CH3, , (b), (c), , Zn/Hg, Conc. HCl, , —COONa, CH2, , NaOH/CaO, D, , CH—CH2—CN, , I. DIBAL-H, II. H3O+, , (ii) Arrange the following in the increasing order of, their reactivity towards nucleophilic addition,, reaction, CH3COCH3,HCHO, CH3CHO,, , —COCH3, , (iii) Draw the structure of 2, 4-DNP derivative of, acetaldehyde., 21. (i) Write the products formed when benzaldehyde, reacts with the following reagents, (a) CH 3CHO in presence of dilute NaOH, (b) H2N—NH—, , +, , H, == O + H2N ¾ OH ¾®, , (ii) Give simple chemical tests to distinguish between, the following pairs of compounds., (a) Benzaldehyde and benzoic acid., (b) Propanal and propanone., l, , Case Based Questions, , 24. Read the following and answer the questions from (i), , to (iv) given below, The reduction of aldehydes and ketones can be, carried out under different conditions. Aldehydes, and ketones are reduced to primary and secondary, alcohols respectively by NaBH 4 or LiAlH 4 as well as, catalytic hydrogenation., The carbonyl group of aldehydes and ketones is, reduced to CH2 group on treatment with Zn-Hg, and concentrated HCl or with hydrazine followed by, NaOH or KOH in highly boiling solvent such as, ethylene glycol., The aldehydes and ketones can be differ by oxidation, reactions. Aldehydes are easily oxidised to carboxylic, acids on treatment with HNO 2 , KMnO 4 , K 2Cr 2O 7 ,, etc. Even mild oxidising agents mainly Tollen’s, reagent and Fehling’s solution also oxidise aldehydes, whereas ketones are generally oxidised under, vigorous conditions, i.e. strong oxidising agents and, at elevated temperatures, to give mixture of, carboxylic acids having lesser number of C-atom, than the parent ketone. Benedict’s solution reacts in, the same way as Fehling’s solution., O, , (c) Conc. NaOH, (ii) Distinguish between following, (a) CH 3 ¾ CH == CH ¾ CO ¾ CH 3 and, CH 3 ¾ CH 2 ¾ CO ¾ CH == CH 2, (b) Benzaldehyde and benzoic acid., , (i) What will be the products if, is oxidised by, acidified K 2 Cr2 O7 ?, (ii) Name the appropriate reagents for the following, transformation., O, , 22. An organic compound A on treatment with ethyl, , alcohol gives carboxylic acid B and compound C., Hydrolysis of C under acidic conditions gives B, and D. Oxidation of D with KMnO 4 also gives B., B on heating with Ca(OH) 2 gives E with molecular, , CH2 CH3, , ?, , CH3, HO, , HO, , Or

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107, , CBSE Term II Chemistry XII, , Or, , Give the simple chemical test to distinguish between, ethanal and propanal., (iii) Complete the following reaction, O, ½½, 4, CH3 ¾ C ¾ CH3 ¾LiAlH, ¾¾, ¾, ®, (iv) The compound ‘X’ on treating with conc. H2 SO4, gives C4 H8 and water. What is the compound ‘X’?, 25. Read the following and answer the questions from (i), , to (iv) given below, The preparation of carboxylic acids can be done by the, oxidation of a number of functional groups using a, wide variety of reagents., The preparation of carboxylic acids is primarily, characterised by the structure of the acid produced,, the functionality of the starting material and finally, by, the oxidising agent used., Primary alcohols and aldehydes are readily oxidised to, corresponding carboxylic acids with common oxidising, agents such as potassium permanganate ( KMnO 4 ) in, , neutral, acidic or alkaline medium or by potassium, dichormate ( K 2Cr 2O 7 ) and chromium trioxide, (CrO 3 ) in acidic medium., Carboxylic acids are also prepared from aldehydes, by the use of mild oxidising agents like Tollen’s, reagent., (i) What happens when methyl benzene is oxidised, with alkaline KMnO4 ?, (ii) Complete the following reaction, 3®, CH3 ( CH2 )4 CH2 OH ¾CrO, ¾¾, H2SO4, , (iii) Name the oxidising agent which is used in the, conversion of toluene to potassium benzoate., (iv) When acetaldehyde is treated with Tollen’s, reagent, then the compound ‘X’ is formed. Write, the name and formula of ‘X ’., , Or, Write the reagents required in the following, reaction., CH 3COOH ¾, ¾? ® CH 3CONH 2

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108, , CBSE Term II Chemistry XII, , EXPLANATIONS, Objective Questions, 1. (a) The bond angles are approximately 120° with trigonal, planar structure in the carbonyl compounds., O, , 8. (a) Acyl chloride is hydrogenated over catalyst, palladium, barium sulphate to give aldehyde. This reaction is called, Rosenmund reaction., COCl, , 1 20, °, , C, , R´/ H, , R, , Benzoyl chloride, , Trigonal planar, , 2. (a) The type of bond involved in I and II are as follows, I ® Sigma (s), II ®Pi ( p), It can be explained as, In I, the carbonyl carbon is sp2 -hybridised and forms, 3s -bonds., In II, the fourth valence electron of carbon remains in its, p-orbital and forms a p-bond with oxygen by overlap with, p-orbital of an oxygen atom., 3. (d) The most suitable reagent for the conversion of alcohol, ( RCH2 ¾ OH) to aldehyde (R—CHO) is PCC, PCC, , R ¾ CH2OH ¾¾® R ¾ CHO, Pyridinium chlorochromate (PCC) is a mild oxidising, agent which causes the conversion of alcohol to aldehyde, and not carboxylic acid., Whereas, K2Cr2O7 and KMnO4 are strong oxidising agents, which cause conversion of alcohol directly to acid and H2SO4, acts as a dehydrating agent., Thus, they cannot be used for the conversion., 4. (c) Primary aromatic alcohols on reaction with pyridinium, chlorochromate (PCC), which is a mild oxidising agent, form, aromatic aldehydes., Thus, this reaction will not give benzoic acid., CH2OH, CHO, PCC, (Pyridinium, chlorochromate), , Benzyl alcohol, , Benzaldehyde, , 5. (b) Butan-2-ol on oxidation with alkaline KMnO4 solution, produces butanone complete reaction is as follows, , CH3, , CH3, , [O], Alkaline KMnO4, , OH, , Butan-2-ol, , O, , Butan-2-one, , 6. (b) Alcohol vapours are passed over heavy metal catalysts (Ag, or Cu) to give aldehydes and ketones. Primary and, secondary alcohols give aldehydes and ketones respectively., 7. (a), , CHC6H5, , (i) O3, , CHC6H5, O O, CHO, , O, (ii) Zn / H2O, Cyclohexanone, , Pd-BaSO4, , Benzaldehyde, , 9. (b) Diisobutylaluminium hydride is used to selectively, reduce nitriles aldehydes complete reaction is as follows :, (i) AlH( i - Bu) 2, , CH3 ¾ CH === CH ¾ CH2 ¾ CH2 ¾ CN ¾¾¾¾®, (ii) H 2O, , CH3 ¾ CH === CH ¾ CH2 ¾ CH2 ¾ CHO, 10. (c) Chromyl chloride oxidises methyl group to a chromium, complex, which on hydrolysis gives corresponding, benzaldehyde., The complete reaction is shown below, CH3, CH(OCrOHCl2)2, + CrO2Cl2, , H3 O +, , CS2, , Chromium, complex, , CHO, , Benzaldehyde, , This reaction is called Etard reaction., 11. (c) Acetone and benzaldehyde both do not react with, Fehling’s solution. Fehling’s solution do not react with, ketone as acetone is a ketone while benzaldehyde is an, aromatic aldehyde having absence of a-hydrogen., 12. (b) Among the given compounds, formaldehyde is the most, polar due to lowest electron density on C of carbonyl group,, whereas in all other cases, the electron density on C is high, due to +I effect of alkyl groups., 13. (c) The boiling point of butan-1-ol (II) would be the highest,, due to extensive intermolecular hydrogen bonding. Butanal, (I) is more polar than ethoxyethane (III). So, the, intermolecular dipole-dipole attraction is stronger in butanal, and will have higher boiling point than ethoxy ethane., n-pentane (IV) have only weak van der Waals’ forces and, thus will have the lowest boiling point., Therefore, the increasing order of boiling point of the given, compounds are, IV < III < I < II, , O, , Ozonide, , +, X, Benzaldehyde, , Here, x is benzaldehyde., , CHO, , H2, , 14. (a) Reactivity of carbonyl compounds can be decided by two, factors., (i) Steric factor Lesser the steric factor, greater will be its, reactivity., (ii) Electronic factor Greater the number of alkyl group,, lesser will be its electrophilicity., Hence, CH3 ¾ CHO is most reactive towards, nucleophilic addition reaction.

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109, , CBSE Term II Chemistry XII, , 15. (a) Only compounds having —CHO group can reduce, Fehling solution. Thus, CH3COOH does not react with, Fehling solution., HCOOH have —CHO group present in it. Thus, it also, reduces Fehling solution., , 25. (a) Both A and R are true but R is the correct explanation of A., 26. (d) A is false, but R is true. Oxidation of ketones is rather, difficult than aldehydes because C ¾ C bond of ketones is, stronger than C ¾ H bond of aldehydes. Therefore, ketones, cannot be oxidised by weak oxidising agent., , 16. (a) The necessary condition for aldol condensation is the, presence of atleast one a-H atom in aldehydes and ketones., CH3, |, Thus, among the given options, CH3 — C — CHO does not, |, CH3, undergo aldol condensation as it does not contains any, a-hydrogen atom., , 27. (a) Both A and R are true and R is the correct explanation of, A. Carboxylic acids are higher boiling liquids than, aldehydes, ketones and even alcohols of comparable, molecular masses. This is due to more extensive association, of carboxylic acid molecules through intermolecular, hydrogen bonding., , 17. (b) Cannizzaro reaction is given by the aldehydes that do not, contain a-H-atom. The aldehydes like acetaldehyde,, CH3CHO possess a-H-atom and hence, does not undergo, Cannizaro reaction., , 29. (c) A is true, but R is false., Benzoic acid do not undergo Friedel-Crafts reaction because, the carboxyl group is deactivating and the catalyst, aluminium chloride (Lewis acid) gets bonded to the carboxyl, group., , 18. (a) The carboxylic carbon is less electrophilic than carbonyl, carbon because of the possible resonating structures as, shown below :, –, O, O, —C, —C, O—H, O —H, , 28. (c) A is true, but R is false. Carboxyl group only marginally, decreases the electron density at m-position relative to oand p-positions., , 30., , ], , +, , e, , 19. (a) Acidic hydrolysis of esters directly give carboxylic acids, while basic hydrolysis of esters gives carboxylates, which on, acidification give corresponding carboxylic acids., COOC2H5, H3O, , Acetic acid, , + C2H5OH, , 21. (a) The decarboxylation of malonic acid is shown below, b, , a, , D, , Malonic acid, , Therefore, acetic acid is formed by the decarboxylation of, malonic acid., 22. (a) Carboxylic acids having an a-hydrogen are halogenated, at the a-position on treatment with chlorine or bromine in, the presence of small amount of red phosphorus to give, a-halocarboxylic acids. This reaction is known as, Hell-Volhard-Zelinsky reaction., 23. (a) Both A and R is true and R is the correct explanation of A., IUPAC name of HOOC —, ( CH2—, )2 COOH is butane-dioic, acid. For compounds containing more than one carboxyl, group, the ending ‘-e’ of the alkane is retained. The number, of carboxyl groups are indicated by adding the multiplicative, prefix, di, tri etc., to the term ‘oic’., 24. (a) Both A and R are true and R is the correct explanation of, A. The boiling points of aldehydes and ketones are greater, than hydrocarbons and ethers of comparable molecular, masses. It is due to weak molecular association in aldehydes, and ketones arising out of the dipole-dipole interactions., , ( Y), , Methanol, , e, , Methyl acetate, , Water, , This is an esterification reaction., Or, (c) In the given reaction,, RCOOH + SOCl 2 ¾® RCOCl + SO 2 ­ + HCl ­, The byproducts formed are SO 2 and HCl., , 20. (c) The correct arrangement of steps given in the, esterification of carboxylic acid is II, III and I., HO — C — CH2 — COOH ¾® H3C— COOH + CO2 ­, Acetic acid, ½½, O, , (X), , In the given reactions X = CH3COOCH3 and Y = H2O, (iv) (d) The reaction given in previous question, i.e., CH3COOH + CH3OH, CH3COOCH3 + H2O, , COOH, , +, , (i) (d) Cl 3C ¾ COOH is the strongest acid due to presence, of three electron withdrawing chlorine group ( ¾ Cl ),, next to ¾COOH group., (ii) (c) Carboxylic acids on heating with mineral acids such, as H2SO4 or P2O5 give corresponding anhydride., (iii) (a) CH3COOH + CH3OH, CH3COOCH3 + H2O, , 31., , (i) (d) The addition of NH3 to carbonyl compound is an, example in which nucleophilic addition is followed by, elimination reaction occur respectively., (i) NH3, ‚, ‚, C ==O ¾¾¾®, C ==NH+H2O, ƒ, ƒ, (ii) D, i.e. ammonia is added to carbonyl group and water, molecules gets eliminated., (ii) (a) In ethanal ® propanal ® propanone ® butanone,, the electron density on the carbon atom of the carbonyl, group increases due to the increase of +I-effect of the, alkyl group., Thus, the attack by the nucleophile becomes slower, and slower. The reactivity increases in the reverse, order, Butanone <propanone < propanal < ethanal, (iii) (b) The position of the equilibrium lies largely to the, right hand side for most aldehydes and to the left for, most ketones due to steric reasons., (iv) (c) Ketones react with ethylene glycol in the presence, of dry HCl to form cyclic products known as ethylene, glycol ketals.

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112, , CBSE Term II Chemistry XII, , 12. Complete chemical conversion can be done as, CH3 ¾ Br, , Mg / Ether, , ¾¾®, , H 3O, , ¾¾®, , CH3MgBr, , Bromomethane, , Dry ether, (ii) CH3MgBr + CO2 ¾¾¾¾®, CH3COOH, +, , (i) CO 2, , (ii) Water, , (A), , Red, phosphorus, , Methyl magnesium, bromide, , CH OH/ H+, , 3, CH3COOH ¾¾¾¾®, CH3COOCH3, (C ), Methyl ethanoate, , ( B), , Ethanoic acid, , Hence, A = CH3MgBr, B = CH3COOH,, , O, ½½, C = CH3 ¾ C ¾ O ¾ CH3, (i) The ¾ COOH group attached to benzene ring is electron, withdrawing in nature and thus deactivates ring. Secondly,, the catalyst used in the reaction, i.e. AlCl 3 is a Lewis acid, which has the tendency to make bond with carboxyl group., That’s why aromatic carboxylic acid does not undergo, Friedel-Craft’s reaction., (ii) Lower pK a value means stronger acid. pK a value of, 4-nitro benzoic acid is lower than benzoic acid due to, following reasons:, (a) Due to – I and – R effect of ¾ NO2 group, electron, density in O ¾ H bond decreases, thus it becomes, weak and easily loses a proton and therefore it is, more acidic., (b) Due to – I and – R effect of ¾ NO2 group,, delocalisation of -ve charge occurs and, p-nitrobenzoate ion becomes more stable than, benzoate ion., 14. (i) Cl ¾ CH2COOH is a stronger acid than CH3COOH. This, is because ¾ Cl group exhibits -I-effect which makes the, carboxylate ion more stable., Higher the stability of carboxylate ion, easier is the, removal of proton from the carboxylic acid and stronger is, the acid., In CH3COOH, ¾ CH3 group has + I-effect which, destabilised it. Hence, CH3COOH is a weaker acid., (ii) Carboxylic acids contain carbonyl group but do not show, the reactions of carbonyl group such as nucleophilic, addition reaction which are shown by aldehydes, and ketones., It is because of the presence of lone pair of electrons on, the O-atom of ¾OH group, the electrophilic character of, carbonyl carbon by resonance decreases, hence the, partial positive charge on carbonyl carbon atom is, reduced and therefore, they do not show nucleophilic, addition reactions., , 13., , O, , O-, , ¾ C ¾ O ¾ H ¬® ¾ C == O ¾ H, O, (i), , O + NH2 — NH — C—NH2, , CH3 ¾ CH2 ¾ C ¾ CH3 + H2NHN, , H+, , NNHCONH2, , NO2, , (A), Butanone, , NO2, –H2O, , ¾ N ¾ NH, CH3 ¾ CH ¾ CH¾, , NO2, , CH3, 2, 4-DNP derivative, O, , O, , CH3CH2¾C¾ CH3, NaBH4, , NaOH/I2, , CH3CH2¾ C¾ O–Na+, , CHI3 ¯ +, Haloform, Iodoform, test, (yellow ppt.), , OH, CH3¾CH2¾CH¾ CH3, , H3C, , Conc. H2SO4, , H, , Butan-2-ol, , 2CH3CHO, , C¾C, (C), But-2-ene, , H, CH3, , O3, Zn—CH3COOH, , (ii) (a) The ‚, ƒ C ==O bond present in carbonyl compounds, have electron withdrawing nature. The C ¾ H bond, in aldehyde is weak and thus it can be easily, oxidised to corresponding carboxylic acid., The C ¾ C bond in ketone is stronger than C ¾ H, bond of aldehyde. Hence, oxidation is easier in, propanal than propanone., (b) The acidity of a-hydrogen atom of carbonyl carbon, is due to the strong withdrawing effect of the, carbonyl group and resonance stabilisation of the, conjugate base., s, , O, , O, , ¾C¾C¾, , +, , 15., , (iii) CH3CH2COOH + Br2 ¾¾¾® CH3 — CH—COOH, |, Br, 16. (i) A forms 2, 4-DNP derivative. Therefore, it is an, aldehyde or a ketone. Since, it does not reduce Tollen’s, reagent. Therefore, it should be a ketone. Given,, compound also gives yellow precipitate of iodoform on, heating with NaOH and I2. So, it must be methyl, ketone., NO2, O, , O, , ¾C¾C¾, s, , ¾ C == C ¾, , H:B, , 17. As the compound A (C11 H8O) gives benzaldehyde as one of, the products on ozonolysis, it must have a benzene nucleus, or phenyl group ( C 6H5 )., Since, the compound ‘A’ gives ozonolysis products, it is an, aldehyde with C ==C attaches with the ring.

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Chapter Test, , 7. Assertion Compounds containing — CHO group, are easily oxidised to corresponding carboxylic, acids., Reason Carboxylic acids can be reduced to, alcohols by treatment with LiAlH 4 ., , Multiple Choice Questions (MCQs), , 1. The hydrogenation of benzoyl chloride in the presence of, Pd/BaSO 4 produces, (a) benzyl alcohol, (c) benzoic acid, , Short Answer Type Questions, , 8. Identify the missing compounds in the following, reaction sequence :, , (b) benzaldehyde, (d) phenol, , COOH, , 2. Consider the reaction given below., Conc. HNO3, , ?, , CH2 == CH ¾ CHO ¾¾® CH2 == CH ¾ CH2 ¾ OH, Which of the following is the suitable reagent for the, conversion of above reaction?, (a) NaBH 4, (c) Zn-Hg / HCl, , CHO, , (i), O, , (ii), CH3, , 4. Cannizzaro’s reaction is not given by ........ ., , (iii) CH2 == CHCHO, , CHO, (a), , (b), , CHO, , CH3, , 10. Convert, (i) Acetaldehyde to formaldehyde, , (c) HCHO, , (d) CH3CHO, , Assertion-Reasoning MCQs, Direction (Q. Nos. 5-7) Each of these questions contains two, statements Assertion (A) and Reason (R). Each of these, questions also has four alternative choices, any one of, which is the correct answer. You have to select one of the, codes (a), (b), (c) and (d) given below., Both A and R are true and R is the correct explanation of A., Both A and R are true, but R is not the correct explanation of A., A is true, but R is false., A is false, but R is true., , 5. Assertion Carboxylic acids are more acidic than phenols., Reason The carboxylate ion is less stabilised than phenoxide, ion., , 6. Assertion Formaldehyde is a planar molecule., Reason Formaldehyde contains sp 2 hybridised carbon atom., , (ii) Propanal to butanone., , 11. (i) Give chemical tests to distinguish between the, following pairs of compounds., (a) Propanal and propanone, (b) Acetophenone and benzophenone., (ii) Arrange the following compounds in increasing, order of reactivity towards HCN., Acetaldehyde, acetone, di- tert- butylketone,, methyl-tert-butylketone., , Long Answer Type Question, , 12. A compound A (C2H4O) on oxidation gives B, (C2H4O2 ) . A undergoes iodoform reaction. On, treatment with HCN, A forms a product C which, on hydrolysis gives 2-hydroxy propanoic acid., (i) Identify A, B and C., (ii) Write down the equations for the, reactions involved., (iii) Name the product when A reacts with dil. NaOH., , Answers, Multiple Choice Questions, 2. (a), , 3. (a), , C, , CHO, , Zn-Hg with HCl, LiAlH 4, H 2 and Pt as a catalyst, glycol with KOH, , 1. (b), , B, , following., , presence of, , (a), (b), (c), (d), , A, , (i) NaBH4, (ii) H3O+, , 9. Write down the IUPAC and common name for the, , (b) Ni / H 2, (d) Red P + HI, , 3. The Clemmensen reduction of ketones is carried out in the, (a), (b), (c), (d), , Conc. H2SO4, , SOCl2, , 4. (d), , Assertion-Reasoning MCQs, 5. (c), 6. (a), 7. (b), , For Detailed Solutions, Scan the code

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CHAPTER 07, , Amines, In this Chapter..., l, , Structure of Amines, , l, , Classification of Amines, , l, , Nomenclature of Amines, , l, , Identification of Amines, , l, , Uses of Amines, , Amines are the derivatives of ammonia obtained by replacing, either one, two or three H-atoms by alkyl or aryl groups., For example, CH 3 ¾ NH 2 , C 6 H5 ¾ NH 2 ,, , Classification of Amines, , CH 3, CH 3 ¾ NH ¾ CH 3 , CH 3 ¾ N ƒ, ‚ CH, 3, , Structure of Amines, l, , l, , l, , Depending upon the number of hydrogen atoms replaced by, alkyl or aryl groups in ammonia molecule., Amines are classified as primary (1° ), secondary ( 2° ) and, tertiary ( 3° )., , Nitrogen orbitals in amines are sp -hybridised and geometry, is pyramidal., The ground state electronic configuration of N is 1 s 2 ,, 2 s 2 , 2p1x 2p1y 2p1z . Here, one 2s-orbital and three 2p-orbitals, are hybridised to form four sp 3 -hybridised orbitals, three, of them have one electron and the fourth orbital has one, lone pair of electrons., Each of the three sp 3 -hybridised orbitals of nitrogen overlap, with orbitals of hydrogen or carbon depending upon the, composition of amines. The fourth orbital of nitrogen in all, amines contain an unshared pair of electrons., Unshared, electron pair, , N, H, , 108°, , H, Pyramidal, , sp3-hybridised, , H, , Replacement, , Replacement, , of H by R, , of H by R¢, , NH 3 ¾¾¾¾® RNH 2 ¾¾¾¾®, , 3, , Ammonia, , (1 ° ), , R‚, N¾ H, R¢ ƒ, (2 ° ), , R‚, ¾¾¾¾®, N ¾ R¢¢, of H by R¢¢, R¢ ƒ, Replacement, , (3 °), , Degree and Functional Group of Various Amines, Amine, , n (no. of H-atoms, attached to N), , Type of, amine, , H, |, , 2, , 1° (primary), , R‚, NH, Rƒ, , 1, , 2° (secondary), , 0, , 3° (tertiary), , R ¾ N¾ H, , R‚, R¾N, ƒ, R, , Functional, group, , ¾ NH2, -- NH -‚, ¾N, ƒ

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118, , CBSE Term II Chemistry XII, , Nomenclature of Amines, l, , l, , l, , l, , In common system of nomenclature, an aliphatic amine is named by prefixing alkyl group to amine, i.e. alkyl amine., In secondary and tertiary amines, when two or more groups are same, the prefix di or tri is added before the name of alkyl group., In IUPAC system, amines are named as alkanamines., To name 2° and 3° amines, we use locant ‘N’ to designate substituent attached to a nitrogen atom while, naming aryl amines,, suffix ‘e’ of arene is replaced by ‘amine’., , Methods of Preparation of Amines, Following methods can be used for the preparation of amines, Methods, , Important points regarding reaction, , Reactions involved, , By reduction of, (i) Alkyl nitrile, , RCH 2NH 2, , lH 4, LiA, , RCN, , (ii) Nitroalkane, , l, , Eth Na, ano, l, , RCH 2NH 2, , Raney Ni / 250°C or, , l, , R / ArNO2 + 3H2 ¾¾¾¾¾¾® R / ArNH2, Fe/ HCl or H 2 / Pt, , O, ½½, LiAlH 4, R C NH2 ¾¾® RCH2NH2 + H2O, , (iii) Amides, , Hofmann ammonolysis, method, , l, , NH 3 / 393K, , ··, , ··, , l, , · · RX, , ··, , RX ¾¾¾® R N H2 ¾® R2 NH ¾® R3 N ¾® [ R4 N] X (1°), , RX, , RX, , ( 2 °), , l, , ( 3°), , l, l, , Order of reactivity of halides with amines, RI > RBr > RCl, , Gabriel phthalimide, synthesis, , l, , O, COOH, , C, NH, C, , (i) KOH, (ii) RX, (iii) H 2O/OH –, , l, , COOH, + RNH 2, , O, D, , RCONH2 + Br2 + 4NaOH ¾® RNH2 + 2NaBr, + Na 2CO3 + 2H2O, , Hofmann-bromamide, degradation, , Reaction is used for ascent of amine series,, i.e. for preparation of amines containing, one C-atom more than starting amine., , l, , Reduction with Fe scrap and HCl is, preferred because FeCl 2 formed get, hydrolysed to release HCl during reaction., Reduction of amides give amine with same, number of carbons., It yields a mixture of 1°, 2° and 3° amines, and also quaternary ammonium salt., 1° amine is obtained as major product,, when excess of NH 3 is taken., Only 1° amines can be synthesised by this, method. Exception : t-butyl amine cannot, be prepared by this method as in this case,, elimination takes place., Aromatic 1° amines cannot be prepared by, this method as aryl halides do not undergo, nucleophilic substitution reaction with, potassium phthalimide., Amine formed contains one carbon less, than that present in the amide., , Physical Properties, , Chemical Properties and Reactions, , Physical state The lower aliphatic amines are gases with fishy, odour. 1° amines with three or more carbon atoms are liquid, and the higher ones are solids., Colour Aniline and other arylamines are usually colourless, but get coloured on storage due to atmospheric oxidation., Solubility It decreases with increase in molar mass of amines., Amines are soluble in organic solvents like alcohol, ether and, benzene., Boiling point Primary and secondary amines are engaged in, intermolecular association due to hydrogen bonding, while, tertiary amines do not have intermolecular association (due to, absence of H-atom)., Therefore, the order of boiling points of isomeric amines is as, follows, , Basic Character of Amines, , 1 ° > 2°> 3 °, , The reaction of amines with mineral acids to form ammonium, salts show that, they are basic in nature. Their basicity is due, to lone pair of electrons on N atom. Basic character can be, better understood by their K b and pK b values. Larger the, value of K b or smaller the value of pK b , stronger is the base., , Structure-Basicity Relationship of Amines, 1. Alkanamines vs Ammonia, Alkyl amines are stronger bases than NH 3 ., Basic nature of aliphatic amines increases with increase in, the number of alkyl groups, i.e. +I-effect of alkyl group, increases their basicity., l, , l

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119, , CBSE Term II Chemistry XII, , l, , l, , l, , l, , 2. Alkyl Amines vs Ammonia, Aniline is more stable than anilinium ion. Hence,, the proton acceptability (basic nature) of aniline or, aryl amines would be less than that of ammonia., The less basicity of aniline as compared to NH 3 is, attributed to - I-effect of benzene (sp 2 -hybridised), ring and delocalisation of lone pair of N., In substituted aniline, ERGs ( ¾ OCH 3 , ¾ CH 3 ), increase basic strength, whereas EWG, ( ¾ NO 2 , ¾ SO 3 H) decrease basic strength., , The order of basicity of amines in gaseous phase is, 3° amine > 2° amine > 1° amine > NH 3 ., In aqueous phase, solvation effect and steric hindrance, besides, inductive effect of alkyl group decides the basic strength of alkyl, amines., The order of basicity of amines, when alkyl group is ethyl group is,, Amines :, (C 2 H5 ) 2 NH > (C 2 H5 ) 3 N > C 2 H5 NH 2 > NH 3, 3.00, 3.25, 3.29, 4.75, pK b :, The order of basicity of amines, when alkyl group is methyl group, is,, Amines :, (CH 3 ) 2 NH > CH 3 NH 2 > (CH 3 ) 3 N > NH 3, 3.27, , pK b :, , 3.38, , 4.22, , l, , l, , l, , 4.75, , Important Chemical Reactions of Amines, Method, , Reaction involved, , Important points related to reaction, , Aliphatic amines, Alkylation, , HX, , RX, , l, , R ¾ X + R ¢ NH2 ¾® R ¾ NHR ¢ ¾®, ( 2 °), , (1°), , - HX, , +, , RY, , R 2 NR ¢ ¾® R 3 N R ¢ X ( 3°), , Acylation, , ··, , Quaternary, ammonium salt, ··, , CH 3 COCl or, , l, , It yields a mixture of 1°, 2° and 3° amines, and also quaternary ammonium salt., Major product is obtained when amine is, taken in excess., , R / Ar ¾ N ¾ H ¾¾¾¾® R / Ar ¾ N ¾ C CH3, (CH 3 CO 2 )O, ½, ½ ½½, Base, H, H O, , l, , N -alkyl/arylethanamide, , The reaction is carried out in presence of a, base stronger than amine (e.g. pyridine), which removes HCl so formed and shifts, equilibrium to right side., , + CH3COOH, Benzoylation, , RNH2 + C6H5COCl ¾® RNHCOC6 H5 + HCl, , Carbylamine reaction, , RNH2 + CHCl 3 + 3KOH ¾® R ¾ NC+3KCl + 3H2O, , D, , l, , l, , l, , Reaction with nitrous, acid, , HNO 2, , RNH2 ¾¾® ROH + H2 ­ + HCl, , l, , (1°), , Substituted aryl amides are obtained by this, method., Secondary and tertiary amines do not show, this reaction., It is used as a test for 1° amines., Reaction is used to distinguish three classes, of amines, i.e. 1°, 2° and 3° amines., , HNO2, , R 2 NH ¾¾® R 2 N ¾ N == O + H2O, ( 2 °), , D, , HNO 2, , R 3N ¾® [ R 3NH]+ NO-2 ¾® ROH + R 2N ¾ N == O, ( 3°), , Reaction with aryl, sulphonyl chloride, (Hinsberg’s reagent,, C6H5SO2Cl), , RNH2 + C6H5SO2Cl ¾® C6H5SO2NH ¾ R, - HCl, , (1°), , (Soluble in alkali), , l, , Reaction is used to distinguish three classes, of amines, i.e. 1°, 2° and 3° amines., , R 2 NH + C6H5SO2Cl ¾® C6H5SO2NR 2, , - HCl (Insoluble in alkali), , ( 2 °), , R 3N + C6H5SO2Cl ¾® No reaction, ( 3°), , Aromatic amines, Electrophilic substitution, reactions, , l, , l, , NH 2 group is ortho and para directing and powerful activating group., Activating effect of —NH 2 group in aniline can be controlled by protecting the —NH 2 group by, acetylation with acetic anhydride and then carrying out the desired substitution followed by hydrolysis., O, ½, ½, (CH 3 CO) 2 O, C6H5NH2 ¾¾¾® C6H5NH C CH3, Aniline, , Pyridine, , Acetanilide

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120, , CBSE Term II Chemistry XII, , Method, , (i) Bromination, , Reaction involved, NH 2, Br, , Br, , Br 2/H 2O, , + 3Br 2, , Important points related to reaction, , To get monosubstituted aniline derivative, path II is, followed., , NH 2, + 3HBr, Br, , O, NH 2, , NHCCH 3, , O, C 6H 5NHCCH 3, , 3OH – or H +, , Br 2, CH 3COOH, , Br, , (ii) Nitration, , NH 2, , NH 2, , Br, NH 2, , NH 2, NO 2, , HNO 3, H 2SO 4, , +, , +, , 288 K, , NO 2, NO 2, (51%), , C 6H 5NHCOCH 3, , (47%), , NHCOCH 3, , HNO 3, H 2SO 4, , OH –, , 288 K, , or OH +, , NO 2, , (iii) Sulphonation, , +, , NH 2, , –, , NH 3HSO 4, H 2SO 4, , To get monosubstituted p-nitro derivatives, path II, is followed., , (2%), , NH 2, , NO 2, NH 2, , Å, , l, , NH 3, , 453-473 K, , l, , SO 3H, , p-aminobenzene sulphonic acid is the major product, because distance between —NH 2 and —SO 3H is, maximum., Sulphanilic acid exist in Zwitter ion form as shown a, side., , SO –3, Zwitter ion, , Identification of Amines, 1°, 2° and 3° amines can be identified by the following tests, (i) Carbylamine test Only 1° amines give foul smelling isocyanide whereas 2° and 3° amine do not give this test., (ii) Reaction with nitrous acid Aliphatic 1° amine gives alcohol, aromatic 1° amine gives diazonium salt while 2° amine gives, a yellow green oily layer of N-nitrosoamine., (iii) Hinsberg test Primary amine gives monoalkyl sulphonamide which is soluble in alkali 2° amine gives dialkyl, sulphonamide which is insoluble in KOH whereas 3° and aromatic amines does not give this test., , Uses of Amines, l, , l, , l, , l, , l, , Aliphatic amines are used as solvents and intermediates in the manufacture of drugs., They are used as reagents in organic synthesis., Aromatic amines are used for the formation of polymers, dyes and drugs., Quaternary salts of amines are formed from long chain tertiary amines which are used as detergents,, e.g. n-hexadecyl trimethylammonium chloride., N, N-dimethylaniline (DMA) is used in the preparation of dyes.

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Chapter, Practice, PART 1, Objective Questions, l, , 4. How many primary amines are possible for the, formula C 4 H11N ?, (a) 4, (c) 3, , CH 2 == CHCH 2 NHCH 3 is, , (NCERT Exemplar), , (a) allyl methylamine, (b) 2-amino-4-pentene, (c) 4-aminopent-1-ene, (d) N-methylprop-2-en-1-amine, , 2. The IUPAC name of the compound, , following conversion ?, C 6H5 CH 2Cl ¾® C 6H5 CH 2 ¾ N ¾ CH 3, ½, CH 3, (a) CH3 NH 2 , CH3 Cl, (b) CH3 Cl, RNH 2, (c) NH3 , CH3 Cl, (d) CH3 CH 2 NH 2, , 6. Which of the following reagents is/are used in the, , N(CH3)2, , given reaction?, NO2, , is, N-phenyl-N-propene, N, N-methyl benzene, N, N-dimethyl benzenamine, N, N-dimethyl benzene, , Column II, , NH2, , is ......... ., (a) sodium azide, NaN3, (b) sodium nitrite, NaNO2, (c) potassium cyanide, KCN, (d) potassium phthalimide C6H4 (CO2 )N- K+, , 8. Which of the following can be prepared using, 1. Primary amine, , Gabriel phthalimide synthesis?, (a) Primary aromatic amines, (c) Primary aliphatic amines, , 2. Secondary amine, , choice for reducing an aryl nitro compound to an, amine?, (a) H2 (excess)/Pt, (c) Fe and HCl, , C. (C2 H5 )2 NH, , 3. Tertiary amine, , (b) LiAlH4 in ether, (d) Sn and HCl, , 10. The most suitable reagent used for the conversion of, 2-phenyl propanamide into 2-phenyl propanamine is, , Codes, , A B C, (a) 1 2 3, (c) 2 1 3, , (b) Secondary amines, (d) Tertiary amines, , 9. Which of the following reagents would not be a good, , N(CH3)2, , B., , (b) Sn + HCl, (d) All of these, , 7. The source of nitrogen in Gabriel synthesis of amines, , classification in the Column II and choose the, correct option from the codes given below., , A., , NH2, , ?, , (a) H2 /Pd, ethanol, (c) Fe + HCl, , 3. Match the amines given in Column I with their, , Column I, , (b) 2, (d) 5, , 5. Which of the following reagent is required for the, , Multiple Choice Questions, 1. The correct IUPAC name for, , (a), (b), (c), (d), , (CBSE 2018), , A B C, (b) 1 3 2, (d) 2 3 1, , (a) LiAlH4 /ether, (b) Br2 in aqueous NaOH, (c) excess H2, (d) I2 , red P, , (NCERT Exemplar)

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122, , CBSE Term II Chemistry XII, , 11. The correct order of increasing boiling points for, , 18. Consider the following reaction :, , the bases, CH 3 NH 2 , (CH 3 ) 2 NH, (CH 3 ) 3 N is, , O, , O, , (a), (b), (c), (d), , S—Cl + 'X', , S—N—C2H5 + HCl, , O, , O H, , CH3NH2 < (CH3 )2 NH < (CH3 )3 N, CH3NH2 < (CH3 )3 N < (CH3 )2 NH, (CH3 )3 N < (CH3 )2 NH < CH3NH2, (CH3 )3 N < CH3NH2 < (CH3 )2 NH, , Here, substance ‘X’ is, (a) C2H5NH2, (c) (C2H5 )3 N, , 12. Amongst the following, the strongest base in, aqueous medium is ......... ., (a) CH3 NH 2, (c) (CH3 ) 2 NH, , (b) NCCH 2 NH 2, (d) C 6 H5 NHCH3, , 13. The, , most reactive amine, hydrochloric acid is ......... ., (a) CH3 — NH2, , (b), , towards, , CH3 ‚, CH3 ƒ, , l, , CH3 ‚, CH3 ƒ, , NH ¾ CH3, , NH, , (a) Both A and R are true and R is the correct explanation of A, (b) Both A and R are true, but R is not the correct, explanation of A, (c) A is true, but R is false, (d) A is false, but R is true, , (d), (NCERT Exemplar), , 14. Hofmann bromamide degradation reaction is, , 19. Assertion The angle of C—N—C in trimethylamine is, 108°., Reason There is unshared pair of electron in, trimethylamine which makes the angle less than 109.5°., , shown by ......... ., (a) ArNH2, , (b) ArCONH2 (c) ArNO2 (d) ArCH2NH2, , 15. Benzylamine may be alkylated as shown in the, following equation,, C 6 H 5CH 2 NH 2 + R — X ¾® C 6 H 5CH 2 NHR, Which of the following alkyl halides is best suited, for this reaction through S N 1 mechanism ?, (a) CH3Br, , (b) C6H5Br, , (c) C6H5CH2Br (d) C2H5Br, , 16. In the nitration of benzene using a mixture of conc., H 2SO 4 and conc. HNO 3 , the species which, initiates the reaction is ......... ., (a) NO2, , (b) NO+, , (c) NO+2, , (d) NO-2, , 17. Match the reactions given in Column I with the, statements given in Column II and mark the, correct code that are give below., Column I, A. Ammonolysis, , Column II, 1. Amine with lesser number of, carbon atoms, , B. Gabriel phthalimide 2. Detection test for primary, synthesis, amines., C. Hofmann, 3. Reaction of phthalimide with, bromamide reaction, KOH and R—X, D. Carbylamine, reaction, , 4. Reaction of alkyl halides with, NH3, , Codes, , A B C D, (a) 1 4 3 2, (c) 4 3 1 2, , A B C D, (b) 2 3 1 4, (d) 2 3 4 1, , Assertion-Reasoning MCQs, Direction (Q. Nos. 19-23) Each of these questions, contains two statements Assertion (A) and Reason (R)., Each of these questions also has four alternative choices, any, one of which is the correct answer. You have to select one of, the codes (a), (b), (c) and (d) given below., , dilute, , NH2, , (c), , (b) (C2H5 )2 NH, (d) (C2H5 )4 N+, , 20. Assertion In order to convert R ¾Cl to pure, , R ¾ NH 2 . Gabriel phthalimide synthesis can be used., Reason With proper choice of alkyl halides, phthalimide, synthesis can be used to prepare 1°, 2° or 3° amines., 21. Assertion MeNH 2 is the weaker base than MeOH., Reason N is less electronegative than O, lone pair of, electrons on N is more easily available for donation in, MeNH 2 ., , 22. Assertion In Hofmann bromamide reaction, the amine, formed has one carbon atom less than the parent 1° amide., Reason N-methyl acetamide undergoes Hofmann, bromamide reaction., 23. Assertion N-ethyl benzene sulphonamide is soluble in, alkali., Reason Hydrogen attached to nitrogen in sulphonamide, is strongly acidic., (NCERT Exemplar), , Case Based MCQs, 24. Read the following and answer the questions from (i) to, , l, , (iv) given below, Amines are the derivatives of ammonia, obtained by the, replacement of one, two or three hydrogen atoms by, alkyl/aryl group. These are classified as primary,, secondary and tertiary depending upon the number of, hydrogen atoms in ammonia molecule replaced by alkyl, or aryl groups.

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123, , CBSE Term II Chemistry XII, , replacement of one, two or three hydrogen atoms by, alkyl/aryl groups. Amines are very reactive due to the, difference in electronegativity between nitrogen and, hydrogen atoms and due to the presence of unshared, pair of electrons over N-atom., The number of hydrogen atoms attached to the, N-atom decides the course of reactions of amine, that, is why amines differ in many reactions. In aromatic, amines like aniline, electron density at ortho and, para-positions with respect to —NH 2 group is high., Therefore, this group is ortho or para directing and a, powerful activating group, (i) Carbylamine test is done for, , Amines are prepared by the reduction of nitro, compounds, Hofmann bromamide degradation, reaction and ammonolysis of alkyl halide. Amines are, basic in nature. Aniline show electrophilic, substitution reactions at o and p-position., Halogenation, nitration and sulphonation are some, common reaction that are shown by aniline., (i) Acid anhydrides on reaction with primary amines, give ......... ., (a) amide, (c) secondary amine, , (b) imide, (d) imine, , (ii) Which of the following should be most volatile?, II. (CH 3 ) 3 N, I. CH 3CH 2CH 2 NH 2, CH 3CH 2 ‚, III., IV. CH 3CH 2CH 3, NH, ƒ, CH 3, Choose the correct option, (a) II, , (b) IV, , (c) I, , (a) detection of NO 2 group, (b) confirming the presence of secondary amine, (c) confirming the presence of primary amine, (d) for confirming the basic nature of ¾NH2 group, , (d) III, , (ii) Aniline will not show which of the following, reaction?, , (iii) Which of the following species are involved in, the carbylamine test?, I. R—NC, III. COCl 2, (a) I and II (b) I and IV, , (a) Isocyanide test, (b) Diazotisation, (c) Schotten Baumann reaction, (d) Friedel-Crafts reaction, , II. CHCl 3, IV. NaNO2 + HCl, (c) II and IV (d) III and IV, , (iv) Which of the following arenium ion is not involve, in the bromination of aniline ?, +, , NH2, , NH2, H, Br, , (a), , (iii) The solubility of water for C 6 H5 NH2 (I),, (C 2 H5 )2 NH(II) and C 2 H5 NH2 (III) increases in, the order., (a) II < III < I, (c) III < II < I, , H, Br, , (b), , (iv) Which of the following on reduction with lithium, aluminium hydride yields secondary amine?, , +, +, , NH2, , (a) Methyl cyanide, (c) Methyl isocyanide, , NH2, , (d), , (c), , (b) I < II < III, (d) II < I < III, , +, , Or, Which of the following amines cannot be prepared, by Gabriel phthalimide synthesis?, , H, Br, , H Br, , Or Under which of the following reaction conditions,, aniline gives p-nitro derivative as the major, product ?, I. Acetyl chloride/pyridine followed by reaction with, conc. H2SO4 + conc. HNO3., II. Acetic anhydride/pyridine followed by conc., H2SO4 + conc. HNO3., III. Dil. HCl followed by reaction with conc., H2SO4 + conc. HNO3., IV. Reaction with conc. HNO3 + conc. H2SO4., (a) I and II, (b) II and III, (c) III and IV, (d) II and IV, , (a) Ethylamine, (c) Propylamine, , (b) Isopropyl amine, (d) Ethyl methyl amine, , PART 2, Subjective Questions, Short Answer Type Questions, 1. Classify the following amines as primary, secondary, , l, , or tertiary., , NH2, , N(CH3)2, , 25. Read the following and answer the questions from (i), to (iv) given below, Amines constitute one of the most important class of, organic compounds. In nature, they occur among, vitamins, proteins, alkaloids and hormones. These are, the derivatives of ammonia, obtained by the, , (b) Nitroethane, (d) Acetamide, , (i), , (iii) (C 2H5 ) 2 CHNH2, (iv) (C 2H5 ) 2 NH, , (ii)

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124, , CBSE Term II Chemistry XII, , 2. How will you convert the following?, (i) Nitrobenzene into aniline, (ii) Ethanoic acid into methanamine, , (NCERT), , 3. (i) Which type of reagent is a good choice for, reducing an aryl nitro compound to an amine ?, (ii) In the reduction of nitro compounds, why, Fe scrap and HCl is preferred over other, metals ?, 4. Give the structures of A, B and C in the following, reaction., NaCN, OH CH 3CH 2I ¾¾® A ¾¾¾¾¾¾®, B, Partial hydrolysis, NaOH + Br, , 2, ¾¾¾¾®, C, , 5. (i) Name the reagent used in the following reactions., (a) R -- C ºº N ¾, ¾A ® RCH 2NH 2, (b) C 6 H5 NO 2 + 3H 2 ¾, ¾C® C 6 H5 NH 2 + 2H 2O, (ii) Complete the following reaction., CH3 CH2 NH 2 + CHCl 3 + alc. 3KOH ¾®, , 6. Give the structures of A, B and C in the following, reactions., LiAlH 4, HNO 2, KCN, (i) CH3 CH2 Br ¾¾® A ¾¾® B ¾¾® C, NH, , NaOBr, , 0 °C, NaNO 2 / HCl, , 3, (ii) CH 3COOH ¾¾®, A ¾¾® B ¾¾¾¾® C, , D, , (NCERT), , 7. C5 H13N reacts with HNO 2 to give an optically active, , alcohol. What is this compound? Give its IUPAC, name., 8. Write an isomer of C 3H 9N which gives foul smell of, isocyanide when treated with chloroform and, ethanolic NaOH., (All India, 2020), , (ii) In the increasing order of basic strength, C 6H5 NH 2 , C 6H5 N(CH 3 ) 2 , (C 2H5 ) 2 NH and, CH 3NH 2 ., (iii) In the increasing order of basic strength., (a)Aniline, p-nitroaniline and p-toluidine, (b) C 6H5 NH 2 , C 6H5 NHCH 3 , C 6H5 CH 2NH 2, , 12. Give reasons for the following., , (NCERT), , (i) Aniline does not undergo Friedel-Crafts reaction., (ii) (CH 3 ) 2 NH is more basic than (CH 3 ) 3 N in an, aqueous solution., (iii) Primary amines have higher boiling point than, tertiary amines., (All India, 2016), 13. Why is —NH 2 group of aniline acetylated before, carrying out nitration?, , 14. Account for the following., (i) Diazonium salts of aromatic amines are more, stable than those of aliphatic amines., (ii) Gabriel phthalimide synthesis is preferred for, synthesising primary amines., (NCERT), , 15. Give reasons for the following., (i) Acetylation of aniline reduces its activation effect., (ii) CH 3NH 2 is more basic than C 6H5 NH 2 ., (iii) Although, ¾NH 2 is o/p-directing group, yet, aniline on nitration gives a significant amount of, (All India, 2017), m-nitroaniline., , 16. Identify A and B in the following reaction., O, Cl, KCN, , A, , H2/Pd, , B, , 9. Arrange the following., (i) In the decreasing order of basic strength in gas, phase, C 2H5 NH 2 , (C 2H5 ) 2 NH , (C 2H5 ) 3 N and NH3 ., (ii) In the increasing order of boiling point C 2H5 OH ,, (CH 3 ) 2 NH , C 2H5 NH2 ., (iii) In the increasing order of solubility in water, C 6H5 NH 2 , (C 2H5 ) 2 NH , C 2H5 NH 2 .(All India, 2015), , 10. Complete the following acid-base reactions and, name the products., (i) CH 3 ¾ CH 2 ¾ CH 2 ¾ NH2 + HCl ¾®, (ii) (C 2H5 ) 3 N + HCl ¾®, , 11. Arrange the following., (i) In the decreasing order of pK b values, C 2H5 NH 2 , C 6H5 NHCH 3 , (C 2H5 ) 2 NH and, C 6H5 NH2 ., , 17. How will you convert, (i) ethanamine into methanamine?, (ii) methanamine into ethanamine?, , (NCERT), , 18. Give one chemical test to distinguish between the, following pairs of compounds., (i) Aniline and benzylamine, (ii) Aniline and N-methyl aniline, , (All India, 2015), , 19. (i) What is Hinsberg reagent?, (ii) In the following reaction,, O, N, H, , Conc. HNO +, , 3, ¾¾¾¾, ®, X, Conc. H SO, 2, , 4, , What is the structure of product X ?, (iii) What is the role of pyridine in the acylation, reaction of amines?

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125, , CBSE Term II Chemistry XII, , 20. (i) Give a chemical test to distinguish between, , 25. A hydrocarbon ‘A’ (C 4 H 8 ) on reaction with HCl, , N-methylethanamine and N,N-dimethyl ethanamine., (ii) Write the reaction for catalytic reduction of, nitrobenzene followed by reaction of product so, formed with bromine water., (iii) Out of butan-1-ol and butan-1-amine, which will be, more soluble in water and why? (CBSE Sample Paper), , Long Answer Type Questions, 21. (i) Draw the structures of different isomers, , gives a compound ‘B,’ (C 4 H 9Cl), which on, reaction with 1 mol of NH 3 gives compound ‘C,’, (C 4 H11N). On reacting with NaNO 2 and HCl, followed by treatment with water, compound ‘C’, yields an optically active alcohol, ‘D’., Ozonolysis of ‘A’ gives 2 moles of acetaldehyde., Identify compounds ‘A’ to ‘D’. Explain the, reactions involved., (NCERT Exemplar), , l, , corresponding to the molecular formula, C 3H 9N., Write the IUPAC names of the isomers which will, liberate nitrogen gas on treatment with nitrous acid., (ii) Complete the following reactions., , 26. An aromatic compound ‘A’ of molecular formula, C 7H 6O 2 , undergoes a series of reactions as shown, below. Write the structures of A, B, C, D and E in, the following reactions., D, , (a) Aniline + 3Br2 (aq ) ¾®, (b) C 6H5 NH2 + (CH3CO) 2 O ¾®, , NH3/Heat, , LiAlH4/Ether, , (C7H6O2) ¾¾¾® C6H5CONH2, A, Br2, (CH3CO)2O, ¾¾¾® B ¾¾¾¾¾®C, , 22. (i) Suggest a route by which the following conversion, can be accomplished., O, , + NaOH, , NH2, , ¾¾®, , NH ¾ CH3, , Br2/(aq), , E, , (ii) Arrange the following in the increasing order of, their basic strength., , 27. (i) Write the main products of the following, reactions., HNO, , 2, (a) CH 3 —C H 2 —NH 2 ¾¾¾®, , (a) C2 H5 NH2 , C 6H5 NH2 , NH3 , C 6H5 CH2NH2 and, (C 2H5 ) 2 NH, , (0° C), , O, , (b) C 2H5 NH2 , (C 2H5 ) 2 NH, (C 2H5 ) 3N, C 6H5 NH2, , (b), , (c) CH3NH2 , (CH3 ) 2NH, (CH3 ) 3N, C 6H5 NH2 ,, C 6H5 CH2NH2, , O, , 23. (i) Write the structure of main products when aniline, reacts with the following reagents :, (a) Br2 water, (b) HCl, (c) (CH 3CO) 2 O /pyridine, (ii) Arrange the following in the increasing order of, their boiling point :, C 2H5 NH2 , C 2H5 OH, (CH 3 ) 3 N, (iii) Give a simple chemical test to distinguish between, the following pair of compounds (CH 3 ) 2 NH and, (Delhi, 2015), (CH 3 ) 3 N., , 24. (i) Account for the following., (a) Methylamine in water reacts with ferric chloride to, precipitate hydrated ferric oxide., (b) Aniline does not undergo Friedel-Crafts reaction., , (ii) How will you convert, (a) ethanoic acid into methanamine?, (b) hexane nitrile into 1-aminopentane?, (c) methanol into ethanoic acid?, , (All India, 2019), , —S—Cl + H—N—C2H5, H, O, , N—H, (c), , CH3 —C—Cl, Base, , H, , (ii) Accomplish the following conversions., (a) Benzyl chloride to 2-phenylethanamine, (b) Chlorobenzene to p-chloroaniline, , Case Based Questions, 28. Read the following and answer the questions from, , l, , (i) to (iv) given below, Amines are the derivatives of ammonia, obtained, by the replacement of one, two or three hydrogen, atom by alkyl/aryl group. Pure amines are, colourless but develop colour on exposure to air, or on storage due to atmospheric oxidation. Lower, aliphatic amines can form hydrogen bonds with, water molecules and hence, are soluble in water., They have high boiling points.

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126, , CBSE Term II Chemistry XII, , Amines are very reactive due to the difference in the, electronegativity between nitrogen and hydrogen, atoms. Most of the amines are basic, their basic, character depends upon the ease for the formation of, cation by accepting a proton from acid., Identification of 1°, 2° and 3° amines are done by, the reaction with nitrous acid and aryl sulphonyl, chloride., (i) Consider the following statement in considering, whether amine is 1°, 2° or 3°. “The amine is, gaseous and smells like ammonia”., (ii) Why are the higher aliphatic amines not soluble in, water ?, (iii) Give reason why amines are reactive?, (iv) Give the chemical test to distinguish between the, following pair of compounds: ethylamine and, dimethylamine., Or, Arrange the following compounds in the, increasing order of dipole moment., CH 3 CH 2CH 3 , CH 3 CH 2NH 2 , CH 3 CH 2OH, , 29. Read the following and answer the questions from (i), to (iv) given below, Amines constitute one of the most important class of, organic compounds. In nature they occur among, vitamins, proteins, alkaloids and hormones. Synthetic, examples includes drugs, polymers and dyestuffs., Biologically active compounds such as adrenaline and, , ephedrine, both containing secondary amino group, are used to increase the blood pressure., Amines are the derivatives of ammonia obtained by, the replacement of one, two or three hydrogen, atoms by alkyl/aryl groups. Like ammonia, nitrogen, atom of amines is trivalent and carries an unshared, pair of electrons. Amines are classified as primary, secondary and tertiary amine, depending upon, the number of hydrogen atoms in ammonia, molecule which get replaced by alkyl (B) or aryl (Ar), groups., Amines can be prepared by the reduction of nitro, compounds, by the hofmann bromamide, degradation, by Gabriel phthalimide synthesis etc., (i) Give the name of any two naturally occurring, amines., (ii) Draw the structure of N-methylethanamine., (iii) Write the IUPAC name of the following, compound and classify it into primary, secondary, and tertiary amine., m -BrC 6H 4 NH 2, (iv) Write the main product of the following reaction., Br2 + NaOH, , CH 3 ¾ C ¾ NH 2 ¾¾¾¾®, ½½, O, Or, What is the best reagent to convert nitrile to, primary amine?

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128, , CBSE Term II Chemistry XII, , 17. (c) A ® (4), B ® (3), C ® (1), D ® (2), , 12. (c), Compound, , Factors responsible for, basic character are, , (a) CH3 ¾ NH2, , Inductive effect ( + I), , (b) NC ¾ CH2 ¾ NH2 Inductive effect ( - I), Inductive effect ( + I)and solvation, , (c) (CH3 ) 2 NH, H, , Statements, , A. Ammonolysis Reaction of alkyl halide with NH3., R — X + NH3 ¾® RNH2 + HCl., B. Gabriel, Reaction of phthalimide with KOH and, phthalimide R—X., O, O, synthesis, , - I-effect and resonance., , N, , (d), , Reaction, , CH3, , NH, , Since, + I effect and solvation increases basic character while, -I effect and resonance decreases basic character. Hence,, correct choice is (c)., 13. (b) Greater the strength of base, greater will have its, reactivity towards dilute HCl., Hence, (CH3 ) 2 NH will have highest basic strength due to, highest reactivity., H3C, , N—H, , HCl, , H3C, , H, , H3C, , Cl, , N, H3C, , H, , 14. (b) Hofmann-bromamide degradation is shown by, , O, ½½, Ar — C— NH2 by which amide is converted into amine, via intramolecular migration of phenyl group., O, , OH, , KOH, R—X, , O, , OH, O, , C. Hofmann, bromamide, reaction, , Amine with lesser number of carbon atoms., NaOX, RCONH2 ¾ ¾ ¾, ¾® RNH2, , D. Carbylamine, reaction, , Detection test for primary amines., , 18. (a) The reaction of benzene sulphonyl chloride with primary, amine yields N-ethyl benzene sulphonamide., , O, —S—Cl + C2H5NH2, O, , (X), , O, —S—N—C2H5 + HCl, , NH2, , O H, NH2, , Br2/NaOH, , + K2CO3 + KBr + 2H2O, Aniline, , Benzamide, , 15. (c) S N1 reaction proceeds through formation of carbocation., Hence, more stable be the carbocation, more is the, reactivity towards SN 1 mechanism., Alkyl halides, (a), (b), (c), (d), , CH3Br, C 6H5 Br, C 6H5 CH2Br, C 2H5 Br, , Intermediate, ¾¾¾®, ¾¾¾®, ¾¾¾®, ¾¾¾®, , CH3Å, C 6H5 Å, , C 6H5 — CH2 Å ( more stable ), C 2H5 Å, , Hence, the reaction will proceed through S N 1 mechanism, when, C6H5CH2Br is the substrate. because on ionisation it, gives a resonance stabilised carbocation (C6H5CH2)., 16. (c) Nitration of benzene using a mixture of conc. H2SO 4 and, conc. HNO 3 proceeds as,, O, HO—N, , H3O + NO2 + 2H2O +SO2, , + H2SO4, O—H, Å, , NO2, , NO2, , This reaction is known as electrophilic substitution reaction., , N-ethyl benzene sulphonamide, (Soluble in alkali), , 19. (a) Both A and R are true and R is the correct explanation, of A., The angle of C—N¾ C in trimethyl amine is less than, 109.5°, i.e. 108° due to the presence of unshared pair of, electrons., 20. (c) A is true but R is false., Only primary aliphatic amines can be prepared by Gabriel, phthalimide reaction., 21. (d) A is false but R is true, MeNH2 is a stronger base than MeOH because N is less, electronegative than O and lone pair of electrons on N is, more easily available for the donation in MeNH2., 22. (c) A is true but R is false., Only primary amines can be prepared from amides, (RCONH2 ) by treating with Br2 and KOH., Thus, N-methyl acetamide, i.e. CH3CONHCH3 does not, undergo Hofmann bromamide reaction., 23. (a) Both A and R are true and R is the correct explanation, of A., N-ethylbenzene sulphonamide is soluble in alkali, because hydrogen attached to nitrogen in sulphonamide is, strongly acidic and forms a salt during reaction between, these two.

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Chapter Test, Short Answer Type Questions, , Multiple Choice Questions, , 1. The IUPAC name of C 2H5 ¾ N ¾ (CH2 ) 3 ¾ CH3 is, , 8. (i) Why amines are basic substances while amides are, neutral?, (ii) Draw the structure fo N, N-bis (1-bromo ethyl), ethanamine., , ½, C 2H5, , (a) N-ethylbutan-2-amine, (b) N-diethylbutan-1-amine, (c) N, N-diethylbutan-1-amine (d) N, N-diethylbutan-2-amine, , 9. Explain the following observations., (i) Electrophilic substitution in case of aromatic, amines, take place more readily than benzene., , 2. N-methylbenzamide is formed when methanamine reacts, with benzoyl chloride. This reaction is known as, (a) benzoylation, (c) Schmidt reaction, , (b) Hinsberg test, (d) Curtius reaction, , 3. The best reagent for converting 2-phenylpropanamide, into 2-phenylpropanamine is ......... ., (a), (b), (c), (d), , excess H 2, Br2 in aqueous NaOH, iodine in the presence of red phosphorus, LiAlH 4 in ether, , 4. What would be the side products formed with primary, amine in the Hofmann bromamide degradation reaction?, (a) Na 2CO 3 + NaBr, (c) NaBr + H 2O + Na 2CO 3, , (b) NaBr + H 2O + NaOH, (d) Br2 + H 2O + Na 2CO 3, , Assertion-Reasoning MCQs, , (ii) Tertiary amines do not undergo acylation., , 10. (i) How will you convert the following? Aniline into, N-phenylethanamide (write the chemical, equations involved), (ii) Complete the following reaction., C 6H5NH2 + H2SO 4 (conc. ) ¾®, 11. A compound Z with molecular formula C 3H9N reacts, with C 6H5SO 2Cl to give a solid, insoluble in alkali., Identify Z., , Long Answer Type Questions, , 12. (i) Write the chemical reaction of aniline with, benzoyl-chloride and also write the name of the, product obtained., (ii) Complete the following reactions., (a)C 6H5 NH2 + CHCl3 + alc. 3KOH ¾®, , Direction (Q. Nos. 5-7) Each of these questions contains, two statements Assertion (A) and Reason (R). Each of, these questions also has four alternative choices, any, one of which is the correct answer. You have to select, one of the codes (a), (b), (c) and (d) given below., (a) Both A and R are true and R is the correct explanation of A., (b) Both A and R are true, but R is not the correct explanation, of A., (c) A is true, but R is false., (d) A is false, but R is true., , 5. Assertion Ammonloysis of alkyl halides involves the, , HNO, , Fe/HCl, , 2, (b)C 6H5 NO 2 ¾¾® A ¾ ¾¾, ®B, , 273 K, , (iii) Arrange the following compounds in the, increasing order of basic strength., C 6H5NH2 , C 6H5NHCH3 , C 6H5CH2 NH2, , 13. Predict the reagent or the product in the following, reaction sequence., CH3, , CH3, , CH3, , reaction between aryl halides and alcoholic ammonia., Reason Reaction can be used to form 1°, 2°, 3° amines, and finally quaternary ammonium salts., , 1, , (CH3CO2)2O, , HNO3, , pyridine, , H2SO4, , 6. Assertion p-fluoro anilinium ion is more acidic than, NO2, , anilinium ion., Reason Electron density in the N—H bond of, p-fluoroanilinium ion decreases and release of a proton, from p-fluoroanilinium ion is much easier than from, anilinium ion., , 7. Assertion N, N-diethylbenzene sulphonamide is insoluble, in aklali., , NH2, , NHCOCH3, , 5, , 3. (d), , 4. (c), , Assertion-Reasoning MCQs, 6. (a), , NaNO2/HCl, , NO2, NH2, , Multiple Choice Questions, , 5. (a), , 4, , NO2, , Answers, 2. (a), , 3, , CH3, , CH3, , Reason Sulphonyl gorup attached to nitrogen atom is, strong electron withdrawing group., , 1. (c), , 2, , 7. (b), , For Detailed Solutions, Scan the code

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136, , CBSE Term II Chemistry XII, , Chemistry, Class 12t h ( Term II), , Practice Paper 1, , *, , (Solved), Time : 2 Hours, Max. Marks : 35, , General Instructions, , 1. There are 9 questions inthe questionpaper. All questions are compulsory., 2. Questionno. 1is a Case BasedQuestions, whichhas four MCQs. Eachquestioncarries one mark., 3. Questionno. 2 to 6 are Short Answer Type Questions. Eachquestioncarries 3 marks., 4. Questionno. 7to 9 are Long Answer Type Questions. Eachquestioncarries 5 marks., 5. There is no overall choice. However, internal choices have beenprovidedinsome questions. Students have to attempt, onlyone of the alternatives insuchquestions., , * As exact Blue-print and Pattern for CBSE Term II exams is not released yet. So the pattern of this, paper is designed by the author on the basis of trend of past CBSE Papers. Students are advised, not to consider the pattern of this paper as official, it is just for practice purpose., , 1. Read the following and answer the questions from (i) to (v) given below, Reductive alkylation is the term applied to the process of introducing alkyl groups into ammonia or a, primary or secondary amine by means of an aldehyde or ketone in the presence of a reducing agent. The, present discussion is limited to those reductive alkylations in which the reducing agent is hydrogen and a, catalyst or “nascent” hydrogen, usually from a metalacid combination; most of these reductive alkylations, have been carried out with hydrogen and a catalyst., The principal variation excluded is that in which the reducing agent is formic acid or one of its derivatives;, this modification is known as the Leuckart reaction. The process of reductive alkylation of ammonia consists, in the addition of ammonia to a carbonyl compound and reduction of the addition compound or its, dehydration product., The reaction usually is carried out in ethanol solution when the reduction is to be effected catalytically., RCHO + NH3, , RCHOHNH2, , 2[H], , RCH2NH2, , 2[H], , RCH, , NH, , Since, the primary amine is formed in the presence of the aldehyde it may react in the same way as, ammonia, yielding an addition compound, a Schiff ’s base ( RCH == NCH 2 R ) and finally, a secondary amine., Similarly, the primary amine may react with the imine, forming an addition product which also is reduced to, a secondary amine., Finally, the secondary amine may react with either the aldehyde or the imine to give products which are, reduced to tertiary amines., RCH == NH + RCH 2 NH 2 a, , 2[H], , RCHNHCH 2 R ¾® ( RCH 2 ) 2 NH + NH 3, ½, NH 2

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137, , CBSE Term II Chemistry XII, , ( RCH 2 ) 2 NH + RCHO a, , ( RCH 2 ) 2 NCHR, ½, OH, 2[H], ¾® ( RCH 2 ) 3 N + H 3O, , ( RCH 2 ) 2 N + RCH == NH a, 2[H], , ( RCH 2 ) 2 NCHR ¾® ( RCH 2 ) 3 N + NH 3, ½, NH 2, Similar reactions may occur when the carbonyl, compound employed is a ketone., (i) Ethanal on reaction with ammonia forms an imine, (X) which on reaction with nascent hydrogen gives, (Y). Identify ‘X’ and ‘Y’ ., (a) X is CH3CH == NH and Y is CH3NH2, (b) X is CH3CHOHNH2 and Y is CH3CH2 NH2, (c) X is CH3CHOHNH2 and Y is CH3NH2, (d) X is CH3CH == NH and Y is CH3CH2 NH2, , (ii) Acetaldehyde is reacted with ammonia followed by, reduction in presence of hydrogen as a catalyst. The, primary amine so formed further reacts with, acetaldehyde. The Schiff ’s base formed during the, reaction is, (a) CH3CH == NHCH3, (c) CH3 == NHCH2 CH3, , (b) CH3CH == NHCH2 CH3, (d) CH3CH2 CH == NHCH3, , (iii) The reaction of ammonia and its derivatives with, aldehydes is called, (a) nucleophilic substitution reaction, (b) electrophilic substitution reaction, (c) nucleophilic addition reaction, (d) electrophilic addition reaction, 2[ H ], , (iv) ( CH3 CH2 CH2 ) 2 NH + CH3 CH2 CHO ¾® P ¾® Q, The compound Q is, (a) ( CH3CH2 CH2 ) 3 N, (b) ( CH3CH2 CH2 ) 2 N(CH2 CH3 ), (c) ( CH3CH2 ) 3 N, (d) ( CH3CH2 ) 2 NH, , (i) Aniline does not undergo Friedel-Crafts reaction., (ii) (CH3 ) 2 NH is more basic than (CH3 ) 3 ×N in an, aqueous solution., (iii) Primary amines have higher boiling point than, tertiary amines., , 4. (i) Predict the products of electrolysis in each of the, following, (a) An aqueous solution of H2 SO4 with platinum, electrodes., (b) An aqueous solution of AgNO3 with, platinum electrodes., (ii) Estimate the minimum potential difference, needed to reduce Al 2O 3 at 500°C. The Gibbs, energy change for the decomposition reaction, 2, 4, Al 2O3 ¾® Al + O 2 is 960 kJ., 3, 3, ( F = 96500 C mol -1 ), Or, (i) Calculate D rG° and log K C for the following, reaction at 298 K., 2Cr( s ) + 3Cd 2+ ( aq) ¾® 2Cr 3+ ( aq) + 3Cd( s ), ° = + 0.34 V, F = 96500 C mol -1 ], [Given, E cell, , (ii) Resistance of a conductivity cell filled with, 0.1 mol L-1 KCl solution is 100 W. If the, resistance of the same cell when filled with, 0.02 mol L-1 KCl solution is 520 W, calculate, the conductivity and molar conductivity of, 0.02 mol L-1 KCl solution., The conductivity of 0.1 mol L-1 KCl solution is, 1.29 ´ 10 -2 W -1 cm -1., , 5. What can be inferred from the magnetic moment, values of the following complex species?, , Or Reductive alkylation of ammonia by means of an, aldehyde in presence of hydrogen as reducing, agents results in formation of, (a) primary amines, (b) secondary amines, (c) tertiary amines, (d) mixture of all three amines, , 2. Write structures of compounds A, B and C in each of, the following reactions, (g ), dry ether, 5, (i) C6H5 Br ¾Mg/, ¾¾, ¾ ¾® A ¾(a), ¾CO, ¾2¾, ¾® B ¾PCl, ¾¾, ®C, (b) H3 O+, , D, 2/ HCl, (ii) CH3 CN ¾(a), ¾SnCl, ¾¾, ¾, ¾® A ¾Dil.NaOH, ¾ ¾ ¾® B ¾¾, ®C, (b) H3 O+, , 3. Give the reasons for the following., , Examples, , Magnetic moment (BM), , K 4 [Mn(CN) 6 ], , 2.2, , [Fe(H2O) 6 ] 2 +, , 5.3, , Or, Write down the number of 3d-electron in each of, the following ions; Ti 2+ , V 2+ , Cr 3+ , Mn 2+ , Fe 2+ ,, Fe 3+ , Co 2+ , Ni 2+ and Cu 2+ . Indicate how would, you expect the five 3d-orbitals to be occupied for, these hydrated ions (octahedral)?, , 6. (i) Assign reason for, (a) amines are less acidic than alcohols of, comparable molecular masses., (b) aliphatic amines are stronger bases than, aromatic amines.

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138, , CBSE Term II Chemistry XII, , (ii) Give the structures of A, B and C in the following, reactions, HNO2, , LiAlH, , KCN, , 4, (a) CH3 Br ¾¾® A ¾¾®, B ¾¾¾® C, , 273 K, , Br + KOH, , NH, , CHCl, , + NaOH, , 2, 3, 3, (b) CH3 COOH ¾®, A ¾¾¾®, B ¾¾¾¾®, C, , D, , Or, An aromatic compound ‘A’ of molecular formula, C 7 H 6 O2 undergoes a series of reactions as shown, below. Write the structures of A, B, C, D and E in the, following reactions., D, LiAlH4/Ether, NH /Heat, , 3, (C7H6O2)A ¾¾¾®, C6H5CONH2, , NaOH Br2, , (CH CO) O, , 3, 2, ¾® B ¾¾¾¾®, C, , Br2(aq), , 7. A reaction is second order w.r.t A and first order w.r.t. B., (i) Write the differential rate equation., (ii) How is the rate affected on increasing the concentration, of A three times?, (iii) How is the rate affected when the concentrations of, both A and B are doubled?, , Or, Consider the reaction, 2A + B ¾® C + D, Following results were obtained from experiments, designed to study the rate of reactions., Initial concentration (mol L -1 ) Initial rate of formation, [A ], , [B ], , [ D ] (M/min), -3, , 1., , 0.10, , 0.10, , 1 .5 ´ 10, , 2., , 0.20, , 0.20, , 3.0 ´ 10-3, , 0.40, , -3, , 3., , 0.20, , (i) Write the rate law for the reaction., , I. A + B ¾® C + E (slow), A + E ¾® D (fast), II. B ¾® C + E (slow), A + E ¾® F (fast), A + F ¾® D (fast), , 8. What is the difference between multimolecular, and macromolecular colloids? Give one, example of each., Or, Differentiate among a homogeneous solution,, a suspension and a colloidal solution, give a, suitable example of each., , 9. Account for the following, , E, , Exp., No., , (ii) Calculate the value of rate constant for the, reaction., (iii) Which of the following possible reaction, mechanisms is consistent with the rate law?, , 6.0 ´ 10, , (i) The enthalpies of atomisation of the transition, metals are high., (ii) The lowest oxide of a transition metal is basic,, the highest is amphoteric or acidic., (iii) Cobalt (II) is stable in aqueous solution but in, the presence of complexing agents, it is easily, oxidised., , Or, From the given data of E° values, answer the, following questions, E (°M 2 + / M ), , Cr, , Mn, , Fe, , Co, , Ni, , Cu, , -0.91 -1.18 -0.44 -0.28 -0.25 +0.34, , ° 2+, (i) Why is E Cu, value exceptionally, / Cu, positive?, (ii) Why is E (°Mn 2 + / Mn ) value highly negative as, compared to other elements?, (iii) Which is a stronger reducing agent Cr 2+ or, Fe 2+ ? Give reason.

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140, , CBSE Term II Chemistry XII, , In [Fe(H2O)6 ]2+ , Fe is in +2 oxidation state. Given, m = 5.3, BM ( » 5 ), shows that there are four unpaired electrons., Since, H2O is a weak ligand. Hence, the electrons in 3d do, not pair up when the ligands, i.e. H2O molecules, approach., To accommodate the electrons donated by H2O molecules,, the hybridisation is sp3d 2 . Hence, it will be an outer orbital, octahedral complex., , (ii) Reaction involved, 2, 4, Al 2O3 ¾® Al + O2, 3, 3, Here, n = 4, Given, DG = 960 k J, DG = - nFE°, -960 kJ = -4 ´ 96500 ´ E°, 960000 J, E° =, = 2.48 V » 25, . V, 4 ´ 96500 J, Or, °, (i), D r G° = - nFEcell , n = 6, , 3d, , 4s, , 3d, , 4s, , Fe =, 26, Fe2+, , =, 3d, , = - 6 ´ 96500 C/mol ´ 0.34 V, = - 196860 J/mol or - 196.860 kJ / mol, , [Fe(H2O)6], , 2+, , ° = 0.059 V/ n ´ log K C, \ Ecell, 0.34 V ´ 6, log K C =, = 34.5762, 0.059 V, , =, Four unpaired, electrons, , 4d, , ´´´´ ´´´´ ´´´´, , (ii) (a) Cell constant,, G*= conductivity ´ resistance, = 1.29 S m -1 ´ 100W = 129 m -1 = 1.29 cm -1, (b) Conductivity of 0.02 M KCl solution,, Cell constant 1.29 cm -1, k=, =, Resistance, 520 W, -3, -1, = 2.48 ´ 10 S cm, k ´ 1000, (c) Molar conductivity, Lm =, M, 2.48 ´ 10-3 ´ 1000, = 124 S cm 2 mol -1, =, 0.02, 5. Magnetic moment ( m ) = n ( n + 2) BM, , Six pairs of electrons in, sp 3d 2 -hybrid from six H2 O, , Or, All the ions form complex of the type [ M (H2O)6 ] n+ . As water, is a weak field ligand, therefore, in case of octahedal, complex, pairing takes place, when each t 2 g and eg orbital is, singly occupied [ t 2g ( dxy , dyz , dzx ) set is of lower energy and, eg ( dz 2 , dx2 - y 2 ) is of higher energy]., , Ions, , Number, Configuration, of, 3d electrons, , Number, of, unpaired, electrons, , Occupancy, of, 3d-orbitals, , Ti 2+, , 3 d2, , 2, , 2, , t 22g, , V 2+, , 3 d3, , 3, , 3, , t 23g, , For n = 4, m = 4( 4 + 2) = 24 = 4.90 BM » 5, , Cr 3+, , 3 d3, , 3, , 3, , t 23g, , For n = 5, m = 5 ( 5 + 2) = 35 = 5.92 BM » 6, , Mn 2+, , 3 d5, , 5, , 5, , t 23g e g2, , Fe 2+, , 3 d6, , 6, , 4, , t 42 g e g2, , Fe 3+, , 3 d5, , 5, , 5, , t 23g e g2, , Co 2+, , 3 d7, , 7, , 3, , t 52 g e g2, , Ni 2+, , 3 d8, , 8, , 2, , t 26g e g2, , Cu 2+, , 3 d9, , 9, , 1, , t 26g e g3, , For n = 1, m = 1 ( 1 + 2) = 3 = 1.73 BM » 2, For n = 2, m = 2( 2 + 2) = 8 = 2.83 BM » 3, For n = 3, m = 3( 3 + 2) = 15 = 3.87 BM » 4, , In K4[Mn(CN)6 ] , Mn is in +2 oxidation state, i.e. given, value of m (2.2 BM » 2) indicates the presence of one, unpaired electron. Since, CN– is a strong field ligand., Hence, when CN– ligands approach Mn 2 + ion, the, electrons in 3d pair up. Hybridisation involved is d 2 sp3, forming inner orbital octahedral complex., , 3d, 25, , 4s, , Mn =, 3d, , Mn2+, , 4s, 6., , =, 3d, , [Mn(CN)6 ] 4- =, , ´´ ´´ ´´ ´ ´ ´ ´ ´ ´, One unpaired Six pairs of electrons, electron, in d 2 sp 3 -hybrid, –, from six CN, , (i) (a) Loss of proton from amines give RNH- ion whereas, loss of proton from alcohol forms alkoxide ion. Since, O, is more electronegative than N therefore, RO- can, accommodate the negative charge more easily than, RNH- . Further, O—H bond is more polar than N—H bond, as O is more electronegative than N. Hence, amines are less, acidic than alcohol.

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141, , CBSE Term II Chemistry XII, , (b) In aromatic amines, the lone pair of electrons present on nitrogen takes part in resonance and hence, not available for, donation. However, in aliphatic amines, the lone pair is easily available for donation. That’s why, aliphatic amines are more, basic than aromatic amines., HNO 2, LiAlH, (ii) (a) CH3Br ¾KCN, ¾, ¾® CH3CN ¾ ¾¾4 ® CH3CH2NH2 ¾273, ¾¾, ® CH3CH2OH + N2 + H2O, K, (A), , (b), , (B), , Br2 + KOH, , NH, , CHCl 3 + NaOH, , ¾3 ® CH3CONH2 ¾¾¾® CH3NH2 ¾ ¾ ¾ ¾ ¾, ¾® CH3NC + 3NaCl + 3H2O, CH3COOH ¾ ¾, D, , (A), , D, , (B), , (C), , Or, , NH2, , COOH, A is, , B is, , benzoic acid,, , aniline,, , ¾ NHCOCH3, , C is, , N-phenylethanamide, (Acetanilide), , NH2, Br, D is, , C6H5CH2NH2 ,, , Br, , E is, , Benzylamine, , Br, 2, 4, 6-tribromoaniline, , NHCOCH3, (C), , Acetanilide, (CH3CO)2O, NH3/Heat, , C7H6O2 (A) ¾¾¾®, , Br2 + NaOH, C6H5CONH2 ¾¾¾¾®, Hofmann, Benzamide, bromamide, degradation, Reduction, , C6H5NH2, Aniline, , NH2, , Br, , Acetylation, , Br, , Br2 (aq), Bromination, , (B), , Br, , LiAlH4 /ether, , 2, 4, 6-tribromo, aniline, , C6H5CH2NH2, , (E ), , Benzylamine, , ( D), , 7. A reaction is second order w.r.t. A and first order w.r.t B., -d[ R ], (i) Differential rate equation = (Rate)1 =, = k[ A ] 2 [ B ], dt, (ii) When the concentration of A is increased three times,, i.e. 3A then, (Rate) 2 = k [3 A ] 2 [ B ] = 9k[ A ] 2 [ B ] = 9 ( Rate )1, This shows that rate will be increased 9 times to the, initial rate., (iii) When concentrations of both A and B are doubled then,, (Rate) 3 = k [2 A ] 2 [ 2B ] = k 8[ A ] 2 [ B ], This shows that rate will be increased 8 times to the, initial rate., Or, Rate law may be written as, Rate = k [ A ] p [ B ] q, , (Rate)3 = k (0.2)p (0.4)q = 6 ´ 10- 3, Dividing Eq. (ii) by Eq. (i), we get, , or, , 2q=2 , q = 1, , Comparing experiments (i) and (ii), (Rate)1 = k (0.1)p (0.1)q = 1.5 ´ 10-3, p, , q, , (Rate)2 = k (0. 2) (0. 2) = 3 . 0 ´ 10, , …(iii), -3, , …(iv), , Dividing Eq. (iv) by Eq. (iii), we get, 3 ´ 10-3 k (0.2)p (0.2)q, =, 1.5 ´ 10-3 k (0.1)p (0.1)q, 2 = (2)p (2)q, , = 8k [ A ] 2 [ B ] = 8 (Rate)1, , Comparing experiments (ii) and (iii), we get, (Rate)2 = k (0.2)p (0. 2)q = 3 ´ 10-3, , (Rate)3 k (0.2)p (0.4)q 6 ´ 10-3, =, =, (Rate)2 k (0.2)p (0.2)q 3 ´ 10-3, , …(i), …(ii), , where, q = 1, 1 = 2 p, 2 0 = 2 p, p=0, (i) Thus, the rate law is, rate = k [ A ] 0 [ B ] = k [ B ], (ii) Rate = k [ B ], Rate 3 ´ 10-3, = 15 ´ 10-3 min -1, k=, =, [ B], 0. 2, (iii) B ¾® C + E (slow) is the possible reaction which is, consistent with the rate law, rate = k [ B ].

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142, , CBSE Term II Chemistry XII, , 8. Differences between multimolecular and macromolecular colloids, , Multimolecular colloids, , Macromolecular colloids, , When a large number of small, molecules or atoms (diameter <, 1 nm) of a substance combine, together in a dispersion medium to, form aggregates, which have size in, the colloidal range, the colloidal, solutions, thus formed, are known, as multimolecular colloids., , When substances which possess, very high molecular masses are, dispersed in suitable dispersion, medium, the colloidal solutions,, thus formed, are called, macromolecular colloids., , These colloids are not very much, stable., , These colloids are quite stable., , e.g. gold sol, sulphur sol, etc., , e.g. Cellulose, starch, etc., Or, , Homogeneous, solution, , Colloidal solution, , Suspension, , Particle size < 1 nm., , Particle size between 1 Particle size > 1000, to 1000 nm., nm., , Constituent particles, are ions or small, molecules., , Constituent, particles are single, macromolecule or, aggregates of many, atoms, ions or, molecules., , Constituent particles, are molecules., , e.g. Paint, , e.g. Sand solution, , e.g. Sugar solution, , 9., , (i) Transition metals have high enthalpies of, atomisation due to strong metallic bonding and, additional covalent bonding. Metallic bonding is, due to their smaller size while covalent bonding is, due to d-d overlapping., (ii) In lower oxidation states, transition metals behave, like metals and metal oxides are basic in nature., Thus, in lower oxidation states, transition metal, oxides are basic. As the oxidation state increases,, its metallic character decreases due to decrease in, size, thus, it becomes less metallic or more, non-metallic., Oxides of a non-metal may be acidic or neutral., Thus, in higher oxidation states, transition metal, oxides are amphoteric or acidic., (iii) In the presence of complexing agents, cobalt gets, oxidised from + 2 to + 3 state because it provides, energy to remove an electron from Co2 + ., Moreover, Co (III) is more stable than Co (II)., Or, (i) Sum of enthalphy of atomisation and ionisation is, greater than enthalphy of hydration. Thus,, E°M 2 + / M is positive for copper., (ii) Ionisation enthalpy is lower than hydration, enthalphy due to stable 3d5 configuration. Thus,, E°Mn 2 + /Mn is more negative., (iii) Cr 2 + is a stronger reducing agent because it can, lose electron to form Cr 3+ which has stable 3d 3, configuration (half-filled t 23 g ). Further is more, negative than Fe2 + .

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143, , CBSE Term II Chemistry XII, , Chemistry, , Class 12t h ( Term II), , Practice Paper 2, , *, , (Unsolved), General Instructions, , Time : 2 Hours, Max. Marks : 35, , 1. There are 9 questions inthe questionpaper. All questions are compulsory., 2. Questionno. 1is a Case BasedQuestions, whichhas four MCQs. Eachquestioncarries one mark., 3. Questionno. 2 to 6 are Short Answer Type Questions. Eachquestioncarries 3 marks., 4. Questionno. 7to 9 are Long Answer Type Questions. Eachquestioncarries 5 marks., 5. There is no overall choice. However, internal choices have beenprovidedinsome questions. Students have to attempt, onlyone of the alternatives insuchquestions., , * As exact Blue-print and Pattern for CBSE Term II exams is not released yet. So the pattern of this, paper is designed by the author on the basis of trend of past CBSE Papers. Students are advised, not to consider the pattern of this paper as official, it is just for practice purpose., , 1. Read the following and answer the questions from (i) to (v) given below., Some colloids are stable by their nature, i.e. gels, alloys, and solid foams. Gelatin and jellies are two, common examples of a gel. The solid and liquid phases in a gel are interdispersed with both phases being, continuous., In most systems, the major factor influencing the stability is the charge on the colloidal particles. If a, particular ion is preferentially adsorbed on the surface of the particles, the particles in suspension will repel, each other, thereby preventing the formation of aggregates that are larger than colloidal dimensions., The ion can be either positive or negative depending on the particular colloidal system, i.e. air bubbles, accumulate negative ions, sulphur particles have a net negative charge in a sulphur sol, and the particles in a, metal hydroxide sol are positively charged. Accumulation of charge on a surface is not an unusual, phenomenon-dust is attracted to furniture surfaces by electrostatic forces., When salts are added to lyophobic colloidal systems the colloidal particles begin to form larger aggregates, and a sediment forms as they settle. This phenomenon is called flocculation, and the suspension can be, referred to as flocculated, or colloidally unstable. If the salt is removed, the suspension can usually be, restored to its original state; this process is called deflocculation or peptisation. The original and restored, colloidal systems are called deflocculated, peptised, or stable sols., Why does a small amount of salt have such a dramatic effect on the stability of a lyophobic colloidal system?, The answer lies in an understanding of the attractive and repulsive forces that exist between colloidal, particles. van der Waals’ forces are responsible for the attractions, while the repulsive forces are due to the, surface charge on the particles. In a stable colloid, the repulsive forces are of greater magnitude than the, attractive forces., The magnitude of the electrical repulsion is diminished by addition of ionised salt, which allows the, dispersed particles to aggregate and flocculate. River deltas provide an example of this behaviour. A delta is, formed at the mouth of a river because the colloidal clay particles are flocculated when the freshwater mixes, with the salt water of the ocean.

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144, , CBSE Term II Chemistry XII, , (i) Gelatin is a ………… colloidal system., , 4. (i) For a chemical reaction R ® P, the variation in the, , (a) solid in solid, (b) solid in gas, (c) liquid in solid, (d) liquid in gas, , concentration, ln [R] vs time(s) plot is given as, , (ii) Colloidal solutions are stable due to, , In [R], , (a) presence of charges on the colloidal particles, (b) formation of aggregates by colloidal particles, (c) preferential adsorption on the surface, (d) preferential absorption on the surface, , t(s), , (a) Predict the order of the reaction., (b) What is the slope of curve?, (c) Write the unit of rate constant for this reaction., , (iii) Settling down of colloidal particles to form a, suspension is called, (a) flocculation, (b) peptisation, (c) aggregation, (d) deflocculation, , (ii) Show that the time required for 99% completion is, double of the time required for the completion of, 90% reaction., , (iv) When van der Waals’ forces are greater than other, forces due to the surface charge on the particles,, (a) flocculation occurs., (b) the colloid is stable., (c) peptisation takes place., (d) deflocculation occurs., , Or, The particles in suspension will repel each other,, thereby preventing the formation of aggregates, that are larger than colloidal dimensions. This, statement explains:, (a) formation of delta, (b) river water is a colloidal of clay particles, (c) effect of salt on lyphobic colloid, (d) phenomenon of flocculation, , 2. (i) Write the structures of A and B in the following, reactions, H 2, Pd -BaSO 4, , H 2 N -OH, , (a) CH3 COCl ¾¾® A ¾¾® B, (i) CO2, , (b) CH3 ¾ COOH and HCOOH, , Or, Write the structures of A, B, C, D and E in the, following reactions., CH3COCl, Anhyd. AlCl3, , A, , Zn-Hg/Conc. HCl, , NaOI, , D+E, , N 2O5 (mol L-1 ), , Rate of disappearance, of N 2O5 (mol L-1 min -1 ), , 1., , 1.13 ´ 10-2, , 34 ´ 10-5, , 2., , 0.84 ´ 10-2, , 25 ´ 10-5, , 3., , 0.62 ´ 10-2, , 18 ´ 10-5, , Determine for this reaction,, (i) order of reaction, (ii) rate law, (iii) rate constant, (ii) A first order reaction takes 10 min for 25%, decomposition. Calculate t 1/ 2 for the reaction., (Given : log 2 = 0.3010, log 3 = 0.4771,, log 4 = 0.6021), (a) Propanone oxime, (b) Semicarbazone of CH3 CHO, , +, , (ii) Distinguish between, (a) C6 H5 ¾ COCH3 and C6 H5 ¾ CHO, , C6H6, , S. No., , 5. (i) Draw the structures of the following derivatives, , PCl 5, , (b) CH3 MgBr ¾® A ¾® B, (ii) H 3O, , Or, The data given below is for the reaction,, 2N 2O5 ( g ) ¾® 4NO2 ( g ) + O2 ( g ), , (ii) How will you convert ethanal into the following, compounds? Give the chemical equations involved:, (a) CH3 ¾ CH3, (b) CH3 ¾ CH ¾ CH 2 ¾ CHO, ½, OH, (c) CH3 CH 2 OH, , B, , (i) KMnO4-KOH, D, (ii) H O+, , C, , 3, , 3. Why does bromination of aniline, even under very, mild conditions, gives 2, 4, 6- tribromoaniline, instantaneously?, , Or, A ketone A(C4 H 8 O) which undergoes a haloform, reaction and gives compound B on reduction. B on, heating with sulphuric acid gives a compound C, which forms mono-ozonide D. D on hydrolysis with, zinc dust gives only E. Identify A, B, C, D and E., Write the reactions involved.

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145, , CBSE Term II Chemistry XII, , ° = 0. 236 V at 298 K. Calculate the standard, has E cell, Gibbs energy of the cell reaction. (Given, 1 F = 96500, C mol -1 ), (ii) How many electrons flow through a metallic wire if a, current of 0.5 A is passed for 2 h? (given, 1 F =, 96500 C mol -1 ), , 6. (i) Why aniline does not undergo Friedel-Crafts, reaction?, (ii) Why can primary aromatic amines be not, prepared by Gabriel phthalimide synthesis?, , Write the main products of the following, reactions, , Or, Calculate emf of the following cell at 298 K., , HNO2, , (i) CH3 CH 2NH 2 ¾¾¾¾®, 0°C, , (ii), , Mg( s )|Mg 2 + (0.1 M)||Cu 2 + (0.01)|Cu( s ), , O, , Given, E°cell = +2.71 V, 1F = 96500 C mol -1, , —S—Cl + H—N—C2H5 ¾®, O, , 8. The first few members of the series are quite reactive., , H, , They usually form oxides and hydroxides. Complete, the following reactions of lanthanoids., , (iii), O, , Heated with N, , (i) Ln ¾¾¾¾®, , CH ¾ C ¾ Cl, , 3, N—H ¾¾¾¾¾®, , With halogens, , (ii) Ln ¾¾¾¾®, , H, , Burns in O, , 2, (iii) Ln ¾¾¾¾®, , Or, An aromatic compound ‘A’ of molecular formula, C 7 H 7 ON undergoes a series of reactions as, shown below. Write the structures of A, B, C, D, and E in the following reactions, , 9. Give the formula of each of the following coordination, , D, CHCl3 + NaOH, , (C7H7ON)A, , Br2 + KOH, Anhyd. AlCl3, , C6H5NH2, , CH3 COCl, , NaNO2 + HCl, 273 K, , E, B, , Br2, , C, , H 2O, , 7. (i) The cell in which the following reaction occurs, 2Fe3+ ( aq) + 2I - ( aq) ¾® 2Fe2+ ( aq) + I 2 ( s ), , entities., (i) Co 3+ ion is bound to one Cl - , one NH molecule and, two bidentate ethylenediamine (en) molecules., (ii) Ni 2+ ion is bound to two water molecules and two, oxalate ions., Write the name and magnetic behaviour of each of the, above coordination entities., (Atomic number of Co= 27, Ni= 28), Or, Explain the following., (i) Low spin octahedral complexes of nickel are not, known., (ii) CO is a stronger ligand than NH 3 for many metals., , Answers, 1. (i) - (c); (ii) - (a); (iii) - (a); (iv) - (d, b)

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Chemistry, Class 12t h ( Term II), , Practice Paper 3, , *, , (Unsolved), Time : 2 Hours, Max. Marks : 35, , General Instructions, , 1. There are 9 questions inthe questionpaper. All questions are compulsory., 2. Questionno. 1is a Case BasedQuestions, whichhas four MCQs. Eachquestioncarries one mark., 3. Questionno. 2 to 6 are Short Answer Type Questions. Eachquestioncarries 3 marks., 4. Questionno. 7to 9 are Long Answer Type Questions. Eachquestioncarries 5 marks., 5. There is no overall choice. However, internal choices have beenprovidedinsome questions. Students have to attempt, onlyone of the alternatives insuchquestions., , * As exact Blue-print and Pattern for CBSE Term II exams is not released yet. So the pattern of this, paper is designed by the author on the basis of trend of past CBSE Papers. Students are advised, not to consider the pattern of this paper as official, it is just for practice purpose., , 1. Read the following and answer the questions from (i) to (iv) given below., Molar conductivity of a solution is the conductance of solution containing one mole of electrolyte, kept, between two electrodes having unit length between them and large cross- sectional area, so as to contain the, electrolyte. In other words, molar conductivity is the conductance of the electrolytic solution kept between, the electrodes of a conductivity cell at unit distance but having area of cross-section large enough to, accommodate sufficient volume of solution that contains one mole of the electrolyte. It is denoted by L m ., The molar conductivity is related to conductivity as, 1000, 1000, Lm = k ´V =, ´ k= k´, C, Molarity, Unity of L m (molar conductivity) shall be ohm -1cm -1 mol -1 or S cm 2 mol -1 ., Thus, knowing molar concentration (C ) and conductivity ( k), L m can be calculated. L°m is called molar, conductivity at infinite dilution. The molar conductivity of strong electrolytes is found to vary with, concentration according to the equation,, LCm = L°m - A C, This equation is called Debye-Huckel Onsager equation., Here, A is constant depending upon the type of electrolyte taken and nature of solvent and temperature., (i) If conductivity of 0.00241 M acetic acid is 7.896 ´ 10 -5 S cm -1 , the molar conductivity of the solution shall be (in, S cm -1 mol -1 ), (a) 3.276, , (b) 0.3276, , (c) 32.76, , (d) 327.6, , (ii) Molar conductivity of a solution is 1.26 ´ 10 2 W -1cm 2 mol -1 . Its molarity is 0.01. Its specific conductivity will, be, (a) 1. 26 ´ 10-5, , (b) 1.26 ´ 10-3, , (c) 1. 26 ´ 10-4, , (d) 12.6 ´ 10-3

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147, , CBSE Term II Chemistry XII, , (iii) The increase in molar conductivity of HCl with, dilution is due to, (a) increase in self-ionisation of water, (b) hydrolysis of HCl, (c) decrease in self-ionisation of water, (d) decrease in interionic forces, , Or, Highest molar conductivity will be exhibited by, (a) 0.005 M NaCl, (c) 0.5 M NaCl, , (b) 0.1 M NaCl, (d) 0.01 M NaCl, , (iv) Which of the following is wrong about molar, conductivity ?, (a) The solution contains Avogadro number of molecules of, the electrolyte., (b) It is the product of specific conductivity and volume of, solution in cc containing 1 mole of electrolyte., (c) Its units are ohm -1 cm 2 mol -1., (d) Its value for 1 M NaCl is same as that of 1 M glucose, solution., , 2. Write down the IUPAC and common name for the, , Or, (i) Convert, (a) Benzoic acid to benzaldehyde, (b) Propanone to propane, (ii) Arrange the following comopunds in the, decreasing order of their acidic strength., CH3 CH2 COOH, CH3 CH2 CH(F)COOH,, CH3 CH(F)CH2 COOH, ( CH3 ) 2 CHCOOH, , 6. Electrophilic substitution in case of aromatic, amines takes place more readily than benzene., Identify A and B in the following reactions., Br × NaOH, , (i) CH3 ¾ CH2 ¾ CHNH2 ¾ ¾2 ¾ ¾¾®, ½, CH3, 2, A ¾HNO, ¾¾, ®B, 3+, , Fe, 2®B, (ii) C6 H5 NO2 ¾Br, ¾2 /¾, ¾, ¾® A ¾HNO, ¾¾, +, , O, 4®B, (iii) CH3 CN ¾H¾/H, ¾2¾, ® A ¾LiAlH, ¾¾, ¾, , following., O, , CHO, (ii), , (i), CHO, , CH3, , (iii) CH 2 == CHCHO, , Or, Give names of the reagents to bring about the, following transformations, (i) Benzoyl chloride to benzaldehyde, (ii) Benzene to benzaldehyde, (iii) Benzene to acetophenone, , 3. What happens when, (i) toluence is treated with conc. HNO3 and conc., H2 SO4 at 293 K?, (ii) nitrobenzene is treated with conc. HNO3 and conc., H2 SO4 at 363 K?, , 4. Although Cu + does not exist in solution state, but, CuCl( s ) is formed in the presence of Cl - ions in, aqueous solution of Cu(s) and Cu 2+ . Explain., , 5. (i) Write the chemical equations when acetic acid, reacts with, (a) Br2 in the presence of phosphorus., (b) Cl 2 in the presence of phosphorus., (ii) Give two reactions in which aliphatic aldehydes, differ from aromatic aldehydes?, , Or, Answer the following questions, (i) The compound of A( C3 H9 N) reacts with benzene, sulphonyl chloride to give a solid, insoluble in, alkali. Identify ‘A’., (ii) Outline a synthesis of p-bromonitrobenzene from, benzene in two steps., , 7. Following data are obtained for the reaction, 1, N 2 O 5 ¾® 2NO 2 + O 2, 2, 0, , t/s, -1, , [N 2O5 ] /mol L, , 1 .6 ´ 10, , 300, -2, , 0.8 ´ 10, , 600, -2, , 0.4 ´ 10-2, , (i) Show that it follows first order reaction., (ii) Calculate the half-life., (Given: log 2 = 0 .3010 , log 4 = 0 .6021), , Or, For the first order thermal decomposition reaction,, the following data obtained:, C 2H5 Cl ( g ) ¾® C 2H 4 ( g ) + HCl( g ), Time/s, 0, 300, , Total pressure/atm, 0.30, 0.50, , Calculate the rate constant. (Given : log 2 = 0.3010,, log 3 = 0.4771, log 4 = 0.6021)

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148, , CBSE Term II Chemistry XII, , (ii) Predict the number of unpaired electrons, the, magnetic moments for each of the following, complex., (i) [Fe(CN) 6 ]4 -, , 8. Write one difference in each of the following, (i) Lyophobic sol and lyophilic sol., (ii) Solution and colloid., (iii) Homogeneous catalysis and heterogeneous, catalysis., , (ii) [ Cr(NH3 ) 6 ] 2 +, (iii) [ Sc(H2 O) 6 ] 3 +, , Or, What happen, when, (i) a freshly prepared precipitate of Fe(OH) 3 is, shaken with a small amount of FeCl 3 solution?, (ii) presistent dialysis of a colloidal solution is carried, out?, (iii) an emulsion is centrifuged?, , 9. (i) [Ti(H2 O) 6 ] 3 + absorbs light of wavelength 500 nm., Name one ligand which would form Ti(III), complex absorbing light of lower wavelength than, 500 nm and one ligand which would form a, complex absorbing light of wavelength higher than, 500 nm., , Or, PtCl 4 and NH 3 may form five complexes,, A(PtCl 4 × 6NH 3 ), B(PtCl 4 × 5NH 3 ), C(PtCl 4 × 4NH 3 ),, D( PtCl 4 × 3NH 3 ) and E( PtCl 4 × 2NH 3 ). One mole of, each A, B, C, D and E reacts with excess of AgNO 3, to yield 4, 3, 2 and 1 mole(s) of AgCl respectively,, while E gives, no AgCl. The conductance of their, solutions are in the order A > B > C > D > E. On, the basis of Werner’s theory, write their structure, and give the total number of ions given by one, complex., , Answers, 1. (i) - (c); (ii) - (b); (iii) - (b, a); (iv) - (d)