Comprehensive CBSE Question Bank in Chemistry XI (Term II)

CBSE II Question Bank in Chemistry CLASS 11 Features Short Answer Type Questions Long Answer Type Questions Strictly Based on the Latest CBSE Term-wise Syllabus Case Study Based MCQs For Quick Revision Very Short Answer Type Questions



Comprehensive CBSE Question Bank in Chemistry Term–II (For Class XI) (According to the Latest CBSE Examination Pattern) By Dr. N.K. VErMa s.K. KhaNNa Formerly, Associate Professor Formerly, Associate Professor Chemistry Department Chemistry Department D.A.V. College D.A.V. College Chandigarh Chandigarh   laxmi Publications (P) Ltd (An iso 9001:2015 company) bengaluru • chennai • guwahati • hyderabad • jalandhar Kochi • kolkata • lucknow • mumbai • ranchi new delhi

Comprehensive CBSE Question Bank in chemistry–XI (Term-II) Copyright © by Laxmi Publications Pvt., Ltd. All rights reserved including those of translation into other languages. In accordance with the Copyright (Amendment) Act, 2012, no part of this publication may be reproduced, stored in a retrieval system, translated into any other language or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise. Any such act or scanning, uploading, and or electronic sharing of any part of this book without the permission of the publisher constitutes unlawful piracy and theft of the copyright holder’s intellectual property. If you would like to use material from the book (other than for review purposes), prior written permission must be obtained from the publishers. Printed and bound in India Typeset at: Goswami Associates, Delhi New Edition ISBN : 978-93-93738-14-1 Limits of Liability/Disclaimer of Warranty: The publisher and the author make no representation or warranties with respect to the accuracy or completeness of the contents of this work and specifically disclaim all warranties. The advice, strategies, and activities contained herein may not be suitable for every situation. In performing activities adult supervision must be sought. Likewise, common sense and care are essential to the conduct of any and all activities, whether described in this book or otherwise. Neither the publisher nor the author shall be liable or assumes any responsibility for any injuries or damages arising here from. The fact that an organization or Website if referred to in this work as a citation and/or a potential source of further information does not mean that the author or the publisher endorses the information the organization or Website may provide or recommendations it may make. Further, readers must be aware that the Internet Websites listed in this work may have changed or disappeared between when this work was written and when it is read. All trademarks, logos or any other mark such as Vibgyor, USP, Amanda, Golden Bells, Firewall Media, Mercury, Trinity, Laxmi appearing in this work are trademarks and intellectual property owned by or licensed to Laxmi Publications, its subsidiaries or affiliates. Notwithstanding this disclaimer, all other names and marks mentioned in this work are the trade names, trademarks or service marks of their respective owners. Bengaluru 080-26 75 69 30 Chennai 044-24 34 47 26 Branches Guwahati 0361-254 36 69 Hyderabad 040-27 55 53 83 Jalandhar 0181-222 12 72 Kochi 0484-405 13 03 Kolkata 033-40 04 77 79 Lucknow 0522-430 36 13 Published in India by Ranchi 0651-224 24 64 Laxmi Publications (P) Ltd. C— Printed at : (An ISO 9001:2015 Company) 113, GOLDEN HOUSE, GURUDWARA ROAD, DARYAGANJ, NEW DELHI - 110002, INDIA Telephone : 91-11-4353 2500, 4353 2501 www.laxmipublications.com [email protected]

Contents 1. States of Matter: Gases and Liquids .............................................. 1–18 2. Chemical Thermodynamics .......................................................... 19–39 3. Equilibrium ................................................................................... 40–68 4. The s-Block Elements ................................................................... 69–83 5. The p-Block Elements ................................................................... 84–95 6. Hydrocarbons .............................................................................. 96–114



SYLLABUS For Session 2021–22 (Code No. 043) Class XI (Term–II) S.No. Unit No. of Periods Marks States of Matter: Gases and Liquids 9 1. Chemical Thermodynamics 14 15 2. Equilibrium 12 3. s-Block Elements 5 11 4. Some p-Block Elements 9 9 5. Hydrocarbons 10 35 6. TOTAL 59 States of Matter: Gases and Liquids: Three states of matter, intermolecular interactions, types of bonding, melting and boiling points, role of gas laws in elucidating the concept of the molecule, Boyle's law, Charles law, Gay Lussac's law, Avogadro's law, ideal behaviour, empirical derivation of gas equation, Avogadro's number, ideal gas equation and deviation from ideal behaviour. Chemical Thermodynamics: Concepts of System and types of systems, surroundings, work, heat, energy, extensive and intensive properties, state functions. First law of thermodynamics—internal energy and enthalpy, measurement of U and H, Hess’s law of constant heat summation, enthalpy of bond dissociation, combustion, formation, atomization, sublimation, phase transition, ionization, solution and dilution. Second law of Thermodynamics (brief introduction). Introduction of entropy as a state function, Gibb's energy change for spontaneous and non-spontaneous processes. Third law of thermodynamics (brief introduction). Equilibrium: Equilibrium in physical and chemical processes, dynamic nature of equilibrium, law of mass action, equilibrium constant, factors affecting equilibrium— Le Chatelier's principle, ionic equilibrium—ionization of acids and bases, strong and weak electrolytes, degree of ionization, ionization of poly basic acids, acid strength, concept of pH, buffer solution, solubility product, common ion effect (with illustrative examples). s-Block Elements: Group 1 and Group 2 Elements—General introduction, electronic configuration, occurrence, anomalous properties of the first element of each group, diagonal relationship, trends in the variation of properties (such as ionization enthalpy, atomic and ionic radii), trends in chemical reactivity with oxygen, water, hydrogen and halogens, uses.

Some p-Block Elements: General Introduction to p-Block Elements Group 13 Elements: General introduction, electronic configuration, occurrence, variation of properties, oxidation states, trends in chemical reactivity, anomalous properties of first element of the group, Boron—physical and chemical properties. Group 14 Elements: General introduction, electronic configuration, occurrence, variation of properties, oxidation states, trends in chemical reactivity, anomalous behaviour of first elements. Carbon-catenation, allotropic forms, physical and chemical properties. Hydrocarbons: Classification of Hydrocarbons; Aliphatic Hydrocarbons: Alkanes—Nomenclature, isomerism, conformation (ethane only), physical properties, chemical reactions. Alkenes—Nomenclature, structure of double bond (ethene), geometrical isomerism, physical properties, methods of preparation, chemical reactions: addition of hydrogen, halogen, water, hydrogen halides (Markovnikov's addition and peroxide effect), ozonolysis, oxidation, mechanism of electrophilic addition. Alkynes—Nomenclature, structure of triple bond (ethyne), physical properties, methods of preparation, chemical reactions: acidic character of alkynes, addition reaction of— hydrogen, halogens, hydrogen halides and water. Aromatic Hydrocarbons: Introduction, IUPAC nomenclature, benzene: resonance, aromaticity, chemical properties: mechanism of electrophilic substitution. Nitration, sulphonation, halogenation, Friedel Craft's alkylation and acylation, directive influence of functional group in monosubstituted benzene. Carcinogenicity and toxicity.

Unit 2: Chemical Thermodynamics  FOR QUICK REVISION   Chemical Thermodynamics. Branch of science which deals with the energy changes associated with chemical reactions.  System. Part of the universe selected for investigations.  Surroundings. Part of the universe other than the system.  State of System. The conditions of existence of a system when its macroscopic properties have definite values.  State Function. The thermodynamic quantity which depends only on initial and final state of the system.  Heat (q). It is a random form of energy. It is one of the mode of transference of energy between system and surroundings across the boundary.  Work (w). It is organised form of energy. It is another mode of transference of energy between system and surroundings across the boundary.  Internal Energy (U). The energy associated with the system at particular conditions of temperature and pressure.  Internal Energy Change (U). It is the measure of heat change occurring during the process at constant temperature and constant volume.  Enthalpy (H). It is sum total of internal energy and pV-energy of the system at particular conditions of temperature and pressure. It is also called heat content of the system (H = U + pV).  Enthalpy Change (H). It is the measure of heat change taking place during the process at constant temperature and constant pressure.  Law of Conservation of Energy. It is also called first law of thermodynamics and states that energy of universe always remain constant during chemical or physical changes. Mathematically, U = q + w.  Zeroth Law. If two bodies A and B are in thermal equilibrium with body C, then A and B must also be in thermal equilibrium with each other.  Relation between H and U (H = U + pV) or (H = U + nRT)  Enthalpy of Reaction. The enthalpy change accompanying the chemical reaction in which number of moles of reactants consumed and those of products formed are the same as the stoichiometric coefficients.  Standard Enthalpy of Formation (fH). It is the enthalpy change accompanying the formation of 1 mole of the substance from its constituent elements in their standard state. fH can be > 0 or < 0. 19

20 Chemistry—XI  Standard Enthalpy of Combustion (cH). It is the enthalpy change occurring during the combustion of one mole of the substance in excess of oxygen. cH is always less than zero.  Enthalpy of Solution (sol.H). It is the enthalpy change taking place when 1 mole of the solute is dissolved in large excess of solvent so that on further dilution no enthalpy change occurs.  Standard Enthalpy of Fusion (fus.H). It is the enthalpy change taking place during the fusion of one mole of solid at its melting point.  Standard Enthalpy of Vaporisation (vap.H). It is the enthalpy change taking place during the vaporisation of one mole of liquid at its boiling point.  Enthalpy of Neutralisation (nH). It is the enthalpy change taking place during formation of 1 mole of water by combination of 1 mole of H+ ions and 1 mol of O–H ions in dilute solutions.  Enthalpy of Hydration (hyd.H) . It is the enthalpy change occurring during the hydration of one mole of an anhydrous salt by combining with specific number of moles of water.  Hess’s Law. The enthalpy change in a chemical or physical process is same whether the process is carried out in one step or in several steps.  Bond Enthalpy (bondH). The average amount of energy required to break one mole of the bonds of a particular type in gaseous molecules.  Enthalpy of Atomisation (aH). It is the enthalpy change accompanying the dissociation of one mole of a substance into gaseous atoms. In case of elements, it may be defined as the enthalpy change taking place in the formation of one mole of gaseous atoms of the element.  Spontaneous Process: A process which has an urge or a natural tendency to occur in a particular direction either of its own or after proper initiation under a given set of conditions.  Entropy (S): A thermodynamic quantity, the change in the value of which during the process is given by the ratio of heat absorbed in a reversible manner to the temperature at which it is absorbed.  Gibb’s energy (G): The amount of energy available from the system under particular set of conditions that can be converted into useful work.  Second Law of Thermodynamics: The entropy of universe always tends to increase during any spontaneous process.  Third Law of Thermodynamics: At absolute zero, the entropy of a perfectly crystalline substance is zero. SOME IMPORTANT FORMULAE  U = q + w or U = q – pV  H = U + pV or H = U + nRT  H = fH (Products) –  fH (Reactants)  H  Sum of bond enthalpies  Sum of bond enthalpies  of reactants   of products 

Chemical Thermodynamics 21  S = q(rev) ; vapS  vapH ;  fusS  fusH T Tb Tf  For process to be spontaneous, SSys + SSurr > 0 at constant T and p.  G = H – TS or G = U + pV – TS  G = H – TS  GSys = – TSTotal  rG = fG (Products) – fG (Reactants)  GT, P = wuseful  GT, P < 0 refers to spontaneous process  GT, P = 0 refers to equilibrium conditions.  G = – 2.303 RT log Keq  log K2  vapH  T2  T1  K1 2.303 R  T1T2     VERY SHORT ANSWER QUESTIONS  1. What type of system is constituted by coffee placed in thermos flask? Ans. Isolated system 2. Under what conditions does heat absorbed or evolved by the system represent internal energy change? Ans. At constant temperature and constant volume 3. Although heat is a path function but heat absorbed by the system under certain specific conditions is independent of path. What are those conditions. Ans. At constant volume, heat absorbed (qv) is a measure of internal energy change (U). At constant pressure, heat absorbed (qp) is a measure of enthalpy change (H). Since U and H are state functions, thus, heat absorbed by the system under these conditions are state functions. 4. Heat capacity (C) is an extensive property but specific heat (c) is an intensive property. What is the relationship between them for 1 mole of water? Ans. For 1 mole of water, the molar heat capacity = 18 × specific heat or Cm = 18 × c c for water = 4.18 Jg–1 K–1  Cm = 18 × 4.18 = 75.3 JK–1 5. H for mixing of two gases A and B is zero. Explain whether the diffusion of these gases into each other in a closed container is a spontaneous process or not? Ans. It is spontaneous process. Although H = 0 but disorder (S) is increasing. Thus, G = 0 – TS = –ve. Hence, the process is spontaneous.

22 Chemistry—XI 6. Heat has a randomising influence on the system and temperature is the measure of average chaotic motion of particles in a system. Write the mathematical relation which relates these three parameters. Ans. S = q(rev) T 7. Which thermodynamic parameters are taken to be zero at reference state of elements? Ans. fH and fG. 8. If fH of Freon, CHClF2 is – 480.0 kJ mol–1, what is the thermochemical equation for this? Ans. C(s) + 1 H2(g) + 1 Cl2(g) + F2(g)  CHClF2; fH = – 480.0 kJ 2 2 9. If fH of H(g) is 218 kJ mol–1, what is the value of enthalpy of atomisation of dihydrogen? Ans. Atomisation of hydrogen is represented as H2(g)  2H(g)  aH = 2 × fH H(g) = 2 × 218 = 436 kJ mol–1 10. When H > 0 and S < 0, the reaction is never spontaneous. Why? Ans. G = H – TS If H > 0 and S < 0, then G is always positive. Hence reaction is always non- spontaneous. 11. What type of enthalpy change is represented by the following thermochemical equation? 1 Cl2(g) + e– + aq  Cl–(aq)? H = – x kJ. 2 Ans. Standard enthalpy of formation of Cl–(aq) ion. 12. At 298 K the value of Kp for reaction, N2O4(g) 2NO2(g) is 0.98. Predict whether the reaction is spontaneous or not? Ans. The reaction is spontaneous. We know G = – RT ln Kp. Since Kp is + ve. Therefore, Gis – ve. 1 13. A sample of 1.0 mol of monoatomic gas is taken through p a cyclic process of expansion and compression as shown in figure. What will be value of H for the cycle as a 3 2 whole? Ans. Hcyclic= 0 v 14. Should the following reaction be favourable at high temperature or at low temperature? SO2(g) + 1 O2(g)  SO3(g). 2 Ans. Low temperature, because it is an exothermic process. 15. For a reaction, both H and S are positive, under what conditions will the reaction occur spontaneously? Ans. When TS > H.

Chemical Thermodynamics 23 16. Give the relation between Suniverse and G. Ans. – G = TSuniverse 17. Will melting of ice be spontaneous below 273 K at one atmospheric pressure? Ans. non-spontaneous. 18. Out of, 2 mol of O2 at 100°C, 1 atm and 2 mol of O2 at 100°C, 10 atm pressure which one has higher entropy? Ans. At a given temperature, the entropy change is given by S = 2 × 2.303 Cv log p1 = 2 × 2.303 Cv log 1 = – 4.606 Cv. p2 10 Thus, entropy of 2 mole of O2 at 100° and 1 atm pressure is higher. 19. Give the relationship between Gibb’s energy change and equilibrium constant. Ans. G = – 2.303 RT log K. 20. Identify state functions and path functions out of the following: entropy, temperature, enthalpy, work, free energy. Ans. Enthalpy, Entropy, Temperature and free energy are state functions. Work is path dependent. 21. Which quantity out of rG and rG is zero at equilibrium? Ans. rG will be zero rG is zero for K = 1 (G = –RT ln K) For other values of K, G will be non-zero. 22. Increase in enthalpy of surroundings is equal to decrease in enthalpy of system. Will the temperature of system and surroundings be the same when they are in thermal equilibrium? Ans. Yes 23. Enthalpy diagram for certain reaction is given in Products figure. Is it possible to decide the spontaneity of a HP reaction from a given diagram. DrH Ans. No. Enthalpy change is not a sole criterion of spontaneity. We have to consider the value of entropy change in order to HR Reactants know the correct value of G. Reaction-co-ordi 24. The value of fH for NH3 is – 91.8 kJ mol–1. Calculate enthalpy change for the following reaction 2NH3(g)  N2(g) + 3H2(g) Ans. – 183.6 kJ mol–1. 25. In each of the following processes predict the mode of energy change. (i) Radio (ii) Sky-jump (iii) Electrical toaster (iv) Flourescent lamp. Ans. (i) Electrical energy to sound (ii) Potential energy to kinetic energy. (iii) Electrical energy to heat energy (iv) Electrical energy to radiant energy.

24 Chemistry—XI 26. Why do you feel cold on touching a block of ice? Ans. Ice has a lower temperature. The loss of heat from your fingers makes you feel cold. 27. What is the mode of transference of energy when petrol is subjected to combustion in internal combustion engine of a scooter? Ans. Partly as work and partly as heat. 28. HI is put in a glass bulb which is sealed and is then heated to decompose HI into H2 and I2. What type of system does the reaction mixture represent? Ans. Closed system.  SHORT ANSWER QUESTIONS (TYPE–I)  1. Identify from the following: Open, closed or nearly isolated systems (i) the earth (ii) human beings (iii) can of tomato soup (iv) ice cube tray filled with water (v) satellite in orbit (vi) helium filled balloon. Ans. Open systems (i), (ii), (iv), (v) Closed system, (iii), (vi) 2. State whether each of the following will increase or decrease the total energy content of the system: (i) work done by the system (ii) heat transferred to the surroundings (iii) work done on the system. Ans. (i) Decreases (ii) Decreases (iii) Increases 3. 1.0 mol of a monoatomic ideal gas is expanded State (1) from state (1) to state (2) as shown in diagram. 2.0 Calculate the work done for the expansion of gas from state (1) to state (2) at 298 K. p(bar) Ans. Figure reveals that the process of expansion is carried out in infinite number of steps. Hence, it State (2) represents isothermal reversible expansion. 1.0  wrev = – 2.303 n RT log p1 22.7 p2 V(L) = – 2.303 × 1 × 8.314 × 298 log 2 = – 1717.46 J 1 4. An ideal gas is allowed to expand against a constant pressure of 2 bar from 10 L to 50 L in one step. Calculate the amount of work done by the gas. If the same expansion were carried out reversibly, will the work done be higher or lower than the earlier case? (Given that 1 L bar = 100 J) Ans. w = – pext (V2 – V1) = – 2 × 40 = – 80 L-bar = – 8 kJ – ve sign represents that work is done by the system. wrev > w

Chemical Thermodynamics 25 5. What will be the work done on an ideal gas enclosed in a cylinder, when it is compressed by a constant external pressure, pext in a single step as shown in the figure given below. Explain graphically. Pext V1 V2 Ans. The work done can be calculated with the help of p-V plot. A p-V plot for work of compression carried out in infinite number of steps is as follows. Area under the curve represents the work done. p2 p p1 V2 V1 V 6. Which of the following reactions are exothermic or endothermic: (i) 4C(s) + 5H2(g)  C4H10(g); H = – 123 kJ (ii) HgO(s) + 180.4 kJ  Hg(l) + O2(g) (iii) 0.5N2(g) + 1.5H2(g)  NH3(g) + 45 kJ (iv) SnO2(s) + 2CO(g)  Sn (s) + 2CO2 – 360 kJ. Ans. Reactions (i) and (iii) are exothermic because in (i) H = –ve and in (iii) the heat evolved is represented in products with +ve sign. Similarly, the reaction (ii) and (iv) are endothermic, because heat is absorbed by the reactants. 7. The enthalpy change for a reaction CH4(g) + Cl2(g)  CH3Cl(g) + HCl(g) is – 115 kJ mol–1. How much heat is evolved/absorbed during the formation of 2.525 kg of CH3Cl by this reaction? How many moles of Cl2 is consumed? Ans. CH4(g) + Cl2(g)  CH3Cl(g) + HCl(g); H = – 115 kJ mol–1 (1 mol) (50.5 g) When 50.5 g of CH3Cl(g) is formed heat produced = – 115 kJ

26 Chemistry—XI When 2.525 × 103 g of CH3Cl is formed heat produced is = 115  2.525  103 = 5750 kJ 50.5 The moles of Cl2 consumed during the process = 1  2.525  103 mol = 50 mol 50.5 8. Will the process, Mg(g)  Mg2+(g) + 2e– be exothermic or endothermic? Ans. The process is endothermic because energy is required to dislodge electrons from the gaseous magnesium atom. 9. Do you expect bond dissociation enthalpies of all C—H bonds in ethane molecule to be same? Justify. Ans. The bond dissociation enthalpies of various C—H bonds are not equal because when the bonds are cleaved one after the other, then each time the dissociation products are different and thus associated with different energy content. 10. 500 g of diamond and 500 g of graphite are separately burnt in oxygen. The amount of heat produced will be same or different. Why? Ans. Different because, the reference state chosen for carbon is graphite and graphite to diamond transition is endothermic. 11. Which of the following processes are accompanied by an increase of entropy (i) Dissolution of iodine in solvent (ii) HCl added to AgNO3 solution and precipitate of AgCl is obtained. (iii) A partition is removed to allow two gases to mix. Ans. (i) Dissolution of iodine leads to a more random state. Hence, there is increase in entropy. (ii) When the two gases are allowed to mix both acquire more disordered state. Hence there is increase in entropy. (iii) During precipitation, the particles move from less ordered to a more ordered state. Hence, the precipitation involve decrease of entropy. 12. Predict the sign of entropy change for each of the following changes of state (i) Hg(l)  Hg(g) (ii) AgNO3(s)  Ag+ (aq) + NO3– (aq) (iv) C(graphite)  C(Diamond) (iii) I2(g)  I2(s) Ans. (i) Hg(l)  Hg(g) This is evaporation process, in which randomness increases. Therefore, S = +ve (ii) AgNO3(s)  Ag+(aq) + NO3(aq) This represents dissolution of salt which is a more disordered state. Hence, S = +ve (iii) I2(g)  I2(g) Since solid state is more ordered state than the gaseous state, hence, S = –ve (iv) Graphite  Diamond Diamond is a more ordered state as compared to graphite. Therefore, S = –ve

Chemical Thermodynamics 27 13. Calculate the entropy change involved in the following process: (i) Fusion of ice at 0°C. Latent heat of ice is 334.72 Jg–1. (ii) Vaporisation of water at 373 K. The latent heat of vaporisation of water is 2.52 kJ g–1. Ans. (i) Latent heat of ice = 334.72 Jg–1  fusH of ice = 334.72 (Jg–1) × (18 g mol–1) = 6024.96 J mol–1 fusS = fusH = 6024.96 = 22.06 JK–1 Tm 273 (ii) Latent heat of vaporisation of water = 2.52 Jg–1 vapH of water = 2.52 × 103(Jg–1) × 18(g mol–1) = 45.36 × 103 J mol–1 vapS =  vapH = 45.36  103 = 121.60 JK–1 Tb 373 14. Lifting of water to the top of the hill is quite possible. Why can’t this be considered as spontaneous process? Ans. This can be done with the help of a pump which is an external agency. As such water has no tendency to flow up the hill. Hence, it is a non-spontaneous process. 15. Calculate the entropy change in the surroundings when 1.0 mol of H2O(l) is formed under standard conditions at 298 K. fH = – 286 kJ mol–1. Ans. H2(g) + 1 O2(g)  H2O(l); rH = – 286 kJ mol–1 2 The reaction is exothermic and heat transferred to surrounding is equal to 286 kJ. Heat gained by surroundings = 286 kJ  SSurr = q  286  103(J mol1 ) = 959.73 JK–1 mol–1 T 298(K) 16. Calculate the Gibb’s energy change for the formation of propane, C3H8(g) at 298 K. Given that f H for propane = – 103.85 kJ mol–1 S for the formation reaction of propane is – 269.74 JK–1. Ans. The formation of C3H8(g) is represented by 3C(Graphite) + 4H2(g)  C3H8(g) S = – 269.74 JK–1 fG = fH – TS = – 103.85 × 103(J) – 298(K) × (– 269.74 JK–1) = – 23467.48 J or – 23.467 KJ. 17. For the reaction, N2(g) + 3H2(g)  2NH3(g) H = – 95.4 kJ and S = – 198.3 JK–1 Calculate the temperature at which Gibb’s energy change (G) is equal to zero. Predict the nature of the reaction at this temperature and above it. Ans. G = H – TS When G = 0 H – TS = 0 or H = TS

28 Chemistry—XI T= H   95.4  1000 J = 481 K. S  198.3 JK1 At this temperature the reaction would be in equilibrium. With increase in temperature the opposing factor TS would become more and hence, G would become positive and the reaction would become non-spontaneous. The reaction would be spontaneous at the temperatures below 481 K. 18. For an isolated system, U = 0 what will be S? Ans. For an isolated system with U = 0, the spontaneous change will occur if S > 0. For example, in isolated system involving inter mixing of gases U = 0 but S > 0 because of increase of disorder.  SHORT ANSWER QUESTIONS (TYPE–II)  1. What are the differences between Reversible and Irreversible Processes? Ans. Reversible Process Irreversible Process 1. It is process which follows reversible path. 1. It is process which follows irreversible path. 2. It is an ideal process and takes infinite time. 2. It is a spontaneous process and takes finite time. 3. It occurs through infinite number of steps under equilibrium conditions at each step. 3. Equilibrium exists only in the begining and at the completion stage. 4. In this process, the opposing force and driving force differ only by infinitesimally 4. In this process, there is large difference small magnitude. between driving and opposing force. 5. Work obtained is maximum. 5. Work obtained is not maximum. 6. It is only an imaginary process and cannot 6. It is a natural process and occurs in a be realised in actual practice. particular direction under given set of conditions. 2. (a) 2.5 mol of ideal gas at 2 atm and 27°C expands isothermally to 2.5 times of its original volume against the external pressure of 1 atm. Calculate work done. (b) If the same gas expands isothermally in a reversible manner, then what will the value of work done? Ans. (a) w = – pex(V2 – V1) Now initial volume (V1) = nRT = 2.5  0.082  300 = 30.75 L p 2 Now Final volume (V2) = 30.75 × 2.5 = 76.87 L w = – 1(76.87 – 30.75) = – 46.37 L atm = – 46.37 × 101.325 J = – 4698.44 J.

Chemical Thermodynamics 29 (b) For isothermal reversible expansion of ideal gas wrev = – 2.303 nRT log V2 V1 FGH KIJ= – 2.303 × 2.5 × 8.314 × 300 log 76.87 = – 5714.15 J. 30.75 3. Define the terms: (a) Exothermic reactions (b) Endothermic reactions Ans. (a) Exothermic reactions. The chemical reactions which proceed with the evolution of heat energy are called exothermic reactions. The heat energy produced during the reaction is indicated by writing + q or more precisely by giving the actual numerical value along with products. In general, exothermic reactions may be represented as: A + B  C + D + q (heat energy) (b) Endothermic reactions. The chemical reactions which proceed with the absorption of heat energy are called endothermic reactions. The heat energy absorbed during the reaction can be indicated by writing + q (or the actual numerical value) with the reactants. It can be indicated by writing – q (or the actual numerical value) with the products. In general, an endothermic reaction can be represented as: A + B + q (heat)  C + D or A + B  C + D – q (heat) where q is the heat absorbed. 4. Derive the relationship between H and U. Ans. In order to derive the relationship between H and U, let us consider a reaction involving gases. Let the process be isothermal and carried out at constant pressure (p). If VA be the total volume of the gaseous reactants and VB be the total volume of the gaseous products, also nA be the number of moles of gaseous reactants and nB be the number of moles of gaseous products. Then pVA = nART and pVB = nBRT pVB – pVA = (nB – nA)RT or p(VB – VA) = (nB – nA)RT or pV = (n)RT The change in enthalpy during the process at constant pressure is given by the expression H = U + pV Substituting the value of PV, we have H = U + (n)RT 5. Standard enthalpy of vaporization of benzene at the boiling point is 30.8 kJ mol–1. For how long would a 100 W electric heater have to operate in order to vaporise a 100 g sample at that temperature. Ans. Molar mass of benzene (C6H6) = 78 g mol–1 100 ( g) Moles of benzene in 100 g sample = 78(g mol1 ) = 1.282 mol

30 Chemistry—XI Heat required for evaporating 1 mole of benzene = 30.8 kJ Heat required for evaporating 1.282 mole of benzene = 30.8 × 1.282 = 39.485 kJ Now, Power = Energy or 1 W = 1 Js–1 Thus, time Time = Energy  329.485  103 (J) power 100 (Js1 ) = 394.85s or 394.85 min = 6.58 min. 60 6. The standard enthalpy of combustion of methane (CH4) is given to be – 890.3 kJ mol–1. Calculate (i) mass of methane required to produce 445.15 kJ of heat. (ii) Amount of heat produced by burning 8.0 kg of methane. Ans. CH4(g) + 2O2(g)  CO2(g) + 2H2O(l); cH = – 890.3 kJ mol–1  16 g (i) 890.3 kJ of heat is produced from CH4 = 16 g 445.15 kJ of heat is produced from CH4 = 16  445.15 = 8.0 g 890.3 (ii) 16 g of CH4 produces heat = 890.3 kJ 8 × 103 g of CH4 produces heat = 890.3  8  103 kJ = 4.45 × 105 kJ 16 7. Calculate the enthalpy of combustion of glucose from the following data: (i) C(graphite) + O2(g)  CO2(g); H = – 395.0 kJ (ii) H2(g) + 1 O2(g)  H2O(l); H = – 269.4 kJ 2 (iii) 6C(graphite) + 6H2(g) + 3O2(g)  C6H12O6(s); H = – 1169.8 kJ. Ans. The required equation is: C6H12O6(s) + 6O2(g)  6CO2(g) + 6H2O(l) Multiply equations (i) and (ii) by 6 and add them up: (iv) 6C(graphite) + 6H2(g) + 9O2(g)  6CO2(g) + 6H2O(l); H = – 3986.4 kJ Subtracting equation (iii) from equation (iv), we get C6H12O6(s) + 6O2(g)  6CO2(g) + 6H2O(l); H = – 2816.6 kJ  Enthalpy of combustion of glucose = – 2816.6 kJ. 8. Predict in which of the following, entropy increases or decreases. (i) A liquid crystallises into solid (ii) Temperature of crystalline solid is raised from 0 K to 115 K (iii) 2NaHCO3(s)  Na2CO3(s) + CO2(g) + H2O(g) (iv) H2(g)  2H(g) Ans. (i) During crystalisation, the molecules attain more ordered state. Hence, the process involves decrease of entropy. (ii) At 0 K, the entropy is lowest. As the temperature is raised, there is increase in disorder. Therefore, entropy increases.

Unit 4: The s-Block Elements  FOR QUICK REVISION   Elements in which the last electron enters the s-orbital are called s-block elements. The elements of group-1 (alkali metals) and group-2 (alkaline earth metals) constitute s-block elements. The elements of these groups contain one or two electrons in their outermost s-orbital. The electronic configuration of group-1 (alkali metals) may be represented as [Noble gas] ns1 while that of group-2 (alkaline earth metals) may be represented as [Noble gas] ns2 where n is the number of outermost shell.  Alkali metals. The group-1 elements are lithium, sodium, potassium, rubidium, caesium and francium. Lithium is the lightest metal. Lithium shows diagonal relationship with Mg. Sodium is the most abundant alkali metal in the earth crust.  Alkali metals have lowest ionization enthalpies and largest atomic radii in their respective periods. The hydration enthalpies of alkali metal ions are in the order Li+ > Na+ > K+ > Rb+ > Cs+ Alkali metals behave as strong reducing agents due to their low ionization enthalpies. All alkali metals impart characteristic colour to the flame. Li—caramine red, Na—golden yellow, K—pale violet, Rb—reddish violet and Cs—sky-blue. Caesium, due to its very low ionization enthalpy, is used in photoelectric cells. Alkali metals, due to their large atomic radii, have very weak metallic bond. That is why they are soft and can be cut with the help of a knife.  The reactivity of alkali metals increases from lithium to caesium. Alkali metal hydrides are ionic in nature and are strong reducing agents. Lithium forms normal oxide, Li2O even with excess of oxygen. Na forms peroxide while K, Rb and Cs form superoxides on reaction with excess oxygen. The basic character of alkali metal hydroxides increases on descending the group. Alkali metals dissolve in liquid ammonia giving deep blue solutions which contain ammoniated cations and ammoniated electrons.  Lithium exhibits anomalous behaviour due to its small size, high ionization enthalpy and greater polarizing power.  Alkaline earth metals. The group-2 elements are beryllium, magnesium, calcium, strontium, barium and radium. Calcium is the most abundant element of group-2. Ionization enthalpies of group-2 elements are larger than those of group-1 elements. Hydration enthalpies of group-2 elements are in the order: Be2+ > Mg2+ > Ca2+ > Sr2+ > Ba2+ 69

70 Chemistry—XI The metallic bond in alkaline earth metals is stronger than in alkali metals. As a consequence, they are harder, have higher melting points and higher densities as compared to those of alkali metals. Be and Mg do not impart colour to the flame because their valence electrons cannot be excited by the energy of the flame.  The solubility and basic character of hydroxides increases in the order: Be(OH)2 < Mg(OH)2 < Ca(OH)2 < Sr(OH)2 < Ba(OH)2. Be(OH)2 is amphoteric in nature. An aqueous solution of calcium hydroxide is known as lime water while that of barium hydroxide is known as baryta.  Thermal stability of carbonates is in the order: MgCO3 < CaCO3 < SrCO3 < BaCO3. Solubility of sulphates is in the order: MgSO4 > CaSO4 > SrSO4 > BaSO4.  Beryllium exhibits diagonal relation with aluminium.  VERY SHORT ANSWER QUESTIONS  1. Explain why alkali metals do not occur in free state in nature? Ans. Alkali metals are highly reactive metals. Hence, they do not occur in free state in nature. 2. Name any two minerals of (i) Sodium (ii) Potassium Ans. (i) Rock salt (NaCl) and Chile saltpetre (NaNO3). (ii) Sylvine (KCl) and Carnallite (KCl. MgCl2 . 6H2O). 3. Why does sodium form Na+ ions and not Na2+? Ans. After losing one electron, sodium atom changes into Na+ ion and attains noble gas configuration i.e., complete octet. Further loss of electrons requires a very high ionization enthalpy. Hence, sodium does not form Na2+. 4. Explain why the colour imparted to the flame by sodium salts is different from the colour imparted by potassium salts. Ans. Due to loosely bound valence electron, lesser energy is required to excite the valence electron of potassium as compared to that of sodium. 5. Which out of sodium and lithium has higher melting point and why? Ans. Lithium has higher melting point due to its small size as a result of which the metallic bond is stronger in lithium. 6. Why do alkali metals behave as reducing agents? Ans. Because they have great tendency to lose electrons due to their low ionization enthalpy. 7. Explain why alkali metals have low densities. Ans. Due to their large atomic size and weak metallic bond. 8. Explain why the deep blue solution obtained by dissolving sodium in liquid ammonia is good conductor of electricity.

The s-Block Elements 71 Ans. Deep blue solution obtained by dissolving sodium in liquid ammonia contains ammoniated cation and ammoniated electron which make the solution good conductor of electricity. 9. Explain why alkali metals have least value of first ionization enthalpies in their respective periods. Ans. Due to large atomic size alkali metals have the least value of first ionization enthalpy in their respective periods. 10. What is the common oxidation state exhibited by alkali metals? Ans. + 1. With loss of valence electron alkali metals acquire a stable noble gas configuration. 11. Out of Li and Na which is softer and why? Ans. Na is softer than Li due to weaker metallic bond. 12. List the three factors on which reduction potential of an alkali metal depends. Ans. Reduction potential means, Mn+(aq) + ne–  M(s). The three factors governing it are: (i) Ionization enthalpy (ii) Sublimation enthalpy and (iii) Hydration enthalpy. 13. Why are potassium and caesium, rather than lithium, used in photoelectric cells? Ans. Potassium and caesium lose electron easily due to their low ionization enthalpy. 14. What is the colour imparted by caesium metal to the flame of a Bunsen burner? Ans. Sky blue. 15. What is the order of reactivity of alkali metals? Ans. The reactivity increases down the group. The order of increasing reactivity is Li < Na < K < Rb < Cs < Fr 16. Give equations for the reactions of Na and K with excess oxygen on heating. Ans. 2Na + O2  Na2O2; peroxide K + O2  KO2 superoxide 17. How does NaH react with water? Ans. NaH + H2O  NaOH + H2. 18. How does basic character of hydroxides of alkali metals vary on moving down the group from top to bottom? Ans. Basic character increases from top to bottom because the strength of metal-oxygen bond decreases due to decrease in ionization enthalpy. 19. Find the oxidation state of oxygen in Na2O2. Ans. In Na2O2 oxidation state of Na is + 1. Let the oxidation state of O be x 2(+1) + 2(x) = 0 x = –1

72 Chemistry—XI 20. Explain why a freshly cut piece of sodium soon loses its shine when exposed to air. Ans. On exposure, sodium metal reacts with oxygen, CO2 and moisture of air and hence its surface gets tarnished. 21. Name the alkali metals which form superoxide when heated in excess of air. Ans. Potassium, rubidium and caesium 22. Give chemical equations for the reactions taking place when the following compounds are heated? (i) Li2CO3 (ii) Na2CO3 (iii) LiNO3 (iv) KNO3. Ans.   Li2CO3  Li2O + CO2; Na2CO3  no reaction. 4LiNO3  2Li2O + 4NO2 + O2; 2KNO3  2KNO2 + O2. 23. Explain why alkaline earth metals are harder than alkali metals? Ans. Alkaline earth metals have stronger metallic bond than alkali metals due to smaller size and greater charge on ions. 24. Out of Na and Mg which has higher second ionization enthalpy and why? Ans. Na, has higher second ionization enthalpy because Na+ has stable noble gas configuration. 25. What is the common oxidation state exhibited by group-2 metals? Ans. + 2 26. Which element among group-2 metals has the highest melting point? Ans. Beryllium 27. Beryllium and magnesium do not give colour to flame whereas other alkaline earth metals do so. Why? Ans. Because the energy required to promote an electron from lower energy level to higher energy level in Be and Mg is not available in the flame. In Be and Mg electrons are held more tightly. 28. Which colours are imparted to the flame when the following elements are introduced in the flame one by one? (i) Strontium (ii) Barium (iii) Calcium. Ans. (i) Crimson red (ii) grassy green (iii) brick red. 29. Which is more electropositive; Na or Mg? Ans. Na is more electropositive because of its larger atomic size and low ionization enthalpy. 30. What is hydrolith? How is it prepared? Ans. Hydrolith is CaH2. It is obtained by heating calcium metal in the presence of hydrogen. Ca(s) + H2(g)  CaH2(s) 31. Give equations to show that amphoteric character of beryllium oxide. Ans. As base: BeO + 2HCl  BeCl2 + H2O As acid: BeO + 2NaOH + H2O  Na2 [Be(OH)4] 32. How does the basic character of hydroxides of group-2 metals vary on moving down the group? Ans. Basic character of hydroxides of group-2 metals increases on going down the group from Be to Ba.

The s-Block Elements 73 33. What happens when water is added to quick lime (CaO)? Ans. Calcium hydroxide (Slaked lime) is produced. The reaction is highly exothermic. CaO(s) + H2O(l)  Ca(OH)2(s) Slaked lime 34. What is baryta? Give its one use. Ans. Baryta is an aqueous solution of barium hydroxide. It is used for detection of CO2 (just like lime water).  SHORT ANSWER QUESTIONS (TYPE–I)  1. Explain why sodium is less reactive than potassium. Ans. The reactivity of a metal depends on its ionization enthalpy. A metal having lower value of ionization enthalpy is more reactive. Sodium is less reactive than potassium because its ionization enthalpy is more than that of potassium. 2. Why caesium can be used in photoelectric cell while lithium cannot be? Ans. Caesium has the lowest while lithium has the highest ionization enthalpy among all the alkali metals. Hence, on exposure to light caesium can lose electron very easily while lithium cannot. 3. Why do alkali metals impart colour to the flame? Ans. Alkali metals have low ionization enthalpies. Their valence electrons absorb energy from the flame and get excited to higher energy levels. When they return to the ground state, the energy is emitted back in the form of radiations, frequencies of which fall in the visible region imparting a characteristic colours to flame. 4. Why alkali metals are normally kept in kerosene oil? Ans. Alkali metals, when exposed to atmosphere, react with oxygen, moisture and carbon dioxide present in the air forming oxides, hydroxides and carbonates. In order to prevent the reaction, the alkali metals are stored under kerosene. 5. In aqueous solution, Li+ ion has the lowest mobility. Why? Ans. Li+ ions are maximum hydrated in aqueous solution which results in decrease in its mobility. 6. Lithium has highest ionization enthalpy in group I elements, yet it is strongest reducing agent. Why? Ans. The strength of an element as reducing agent depends not only on ionization enthalpy but also on enthalpy of hydration and enthalpy of sublimation of the element. Li+ ion has very high hydration enthalpy due to its small size which more than compensates for the higher value of ionization enthalpy of lithium. As a result lithium behaves as the strongest reducing agent. 7. Explain, why lithium is kept wrapped in paraffin wax and not stored in kerosene oil? Ans. It is because lithium is a light metal and therefore it floats at the surface of kerosene oil. In order to prevent its exposure to air it is kept wrapped in paraffin wax.

74 Chemistry—XI 8. Explain why lithium chloride has more covalent character than potassium chloride. Ans. Lithium ion, due to its small size, has high polarising power. Thus, it can polarise the electron cloud of chloride ion to much greater extent as compared to potassium ion. Hence, LiCl has more covalent character. 9. (i) Which alkali metal ion has the maximum polarizing power and why? (ii) Lithium is the only alkali metal to form a nitride directly. Why? Ans. (i) Li+ ion has the maximum polarizing power among all the alkali metal ions. This is due to small size of Li+ ion as a result of which it has maximum charge/radius ratio. (ii) Lithium is smallest in size and best reducing agent, therefore, it forms nitride with N2 6Li + N2  2Li3N. 10. What happens when: (i) NaH reacts with water. (ii) Water is dropped over sodium peroxide. Write balanced chemical equation for each. Ans. (i) NaH reacts with water to form sodium hydroxide and hydrogen gas. NaH + H2O  NaOH + H2 (ii) Na2O2 on reaction with water is hydrolysed to sodium hydroxide and hydrogen peroxide Na2O2 + 2H2O  2NaOH + H2O2. 11. Explain why (i) Lithium on being heated in air mainly forms the monoxide and not peroxide. (ii) An aqueous solution of sodium carbonate gives alkaline tests. Ans. (i) Li+ ion is smaller in size. It is stabilized more by smaller anion, oxide ion (O2–) as compared to peroxide ion (O22–). (ii) An aqueous solution of sodium carbonate gives alkaline tests because Na2CO3 undergoes hydrolysis forming sodium hydroxide. Na2CO3 + H2O  NaHCO3 + NaOH 12. The E0 for Cl–/Cl2 is + 1.36, for I–/I2 is + 0.53, for Ag+/Ag is + 0.79, Na+/Na is – 2.71 and for Li+/Li is – 3.04. Arrange the following ionic species in decreasing order of reducing strength: I–, Ag, Cl–, Li, Na. Ans. The species having greater negative reduction potential (E0) is stronger reducing agent. Therefore, the decreasing order of reducing strength is Li > Na > I– > Ag > Cl–. 13. Why is KO2 paramagnetic? Ans. The superoxide ion contains one unpaired electron in one of the  antibonding orbitals. Due to the presence of this unpaired electron KO2 is paramagnetic. The electronic configuration of superoxide ion is: O2–: (1s)2 (1s*)2 (2s)2 (2px)2 (2py)2 (2px*)2 (2py*)1.

Unit 5: The p-Block Elements  FOR QUICK REVISION  p-Block of the periodic table is unique block as it has all types of elements i.e., metals, non-metals and metalloids. There are six groups of p-block elements in the periodic table numbering from 13 to 18. The general valence shell electronic configuration of the elements is ns2np1–6 (except for He). There is lot of variation in the physical and chemical properties of the elements belonging to the p-block. This is because there is a difference in the inner core of their electronic configuration. As a consequence of this, a lot of variation in properties among these elements is observed.  The elements belonging to a particular group show the group oxidation state; as well as other oxidation states differing from the total number of valence electrons by unit of two. The group oxidation state is the most stable for the lighter elements of the group; lower oxidation states become progressively more stable for the heavier elements. This is because of the inert pair effect.  The elements of the p-block have a strong ability to form -bonds. This is attributable to the combined effect of (i) size and (ii) availability of d orbitals. While the lighter elements form p –p bonds, the heavier ones form d – p or d – d bonds.  The absence of d-orbital in second period elements limits their maximum covalence to 4 while heavier ones can exceed this limit. GROUP-13  Boron It is a typical non-metal and the other members are metals. Boron (2s22p1) has 3 valence electrons for covalent bond formation. The presence of empty vacant orbital in boron makes them good electron acceptors and thus boron compounds behave as Lewis acids. GROUP-14 The members of the carbon family mainly exhibit +4 and +2 oxidation states; compounds in +4 oxidation states are generally covalent in nature. The lower oxidation state of +2 becomes more favorable as we move down the group. For example, lead in +2 state is highly stable whereas in +4 oxidation state it is a strong oxidizing agent. Carbon also exhibits negative oxidation states.  Carbon It is typical non-metal forming covalent bonds employing all its four valence electrons (2s22p2). (i) It shows the property of catenation i.e., the ability to form chains or rings, not only with C–C single bonds but also with multiple bonds (C=C or CC). This tendency considerably decreases from C to Si to Ge to Sn to Pb. (ii) Carbon exhibits allotropy. Three important allotropes of carbon are diamond, graphite and fullerenes. 84

The p-Block Elements 85  VERY SHORT ANSWER QUESTIONS  1. Name two amorphous allotropes of carbon. Ans. Charcoal, lamp black. 2. What is common oxidation state of group 13 elements? Ans. + 1 and + 3. 3. Why BF3 acts as a Lewis acid? Ans. It is an electron deficient compound. Boron atom in BF3 has only six electrons around it and hence can accept a pair of electrons and behaves as a Lewis acid. 4. Arrange boron halides in the order of decreasing Lewis acid character. Ans. BI3 > BBr3 > BCl3 > BF3. 5. What is buckminster fullerene? Ans. Buckminster fullerene is C60 allotrope of carbon. 6. Which out of PbCl2 and PbCl4 can act as a good reducing agent? 2 4 Ans. Pb can exist in + 2 and + 4 oxidation state. As PbCl2 can change to PbCl4 , therefore it can act as a better reducing agent out of the two. 7. Which element among group 13 elements has the highest value of ionization enthalpy? Ans. Boron. 8. Which element of group 13 forms acidic oxide? Ans. Boron forms acidic oxide, B2O3. 9. Among the trihalides of boron given below, which is the strongest Lewis acid? BF3, BCl3, BBr3. Ans. BBr3. 10. What are the oxidation states exhibited by group 14 elements? Ans. + 4 and + 2. 11. Out of group 14 element which element shows catenation to maximum extent? Ans. Carbon. 12. Name the two crystalline allotropic forms of carbon. Ans. Diamond and Graphite. 13. Among group 14 elements which element has the highest tendency to form p-p bonds? Ans. Carbon. 14. Name the purest form of silica. Ans. Quartz. 15. Why Ge, Sn and Pb show divalency? Ans. Due to inert pair effect Ge, Sn and Pb exhibit divalency or + 2 state. 16. Select the member(s) of group 14 that (i) forms the most acidic dioxide (ii) is commonly found in +2 oxidation state (iii) used as semiconductor. Ans. (i) carbon (ii) lead (iii) silicon and germanium.

86 Chemistry—XI 17. What is the hybrid state of carbon in graphite and in CO2? Ans. Carbon in graphite is sp2-hybridised and in CO2 it is sp-hybridised. 18. Which allotrope of carbon is used as moderator in atomic reactors? Ans. Graphite. 19. What is the formula of buckminster fullerene? Ans. C60 20. What is the state of hybridisation of C in (i) CO (ii) HCN (iii) CO32– (iv) CaC2. Ans. (i) sp (ii) sp (iii) sp2 (iv) sp  SHORT ANSWER QUESTIONS (TYPE–I)  1. Why oxidation state of + 1 becomes more predominant in moving from B to Tl in boron family? Ans. The boron family exhibits two oxidation states of + 3 and + 1 in their compounds. On moving down the group from boron to thallium the stability of lower oxidation increases i.e., + 1 oxidation state becomes more prominent due to ‘inert pair effect’ i.e., the electrons in s-orbitals of the valence shell do not participate in bond formation because they are more strongly held to the nucleus due to poor shielding effect of intervening electrons. 2. Why does boron not form B3+ ions? Ans. Boron has three electrons in the valence shell. Due to its small size and high sum of first three ionisation enthalpies i.e., iHI + iHII + iHIII, it cannot lose all the three electrons to form B3+ ions. 3. Why boron forms electron deficient compounds? Ans. Boron has only three electrons in the outermost shell which it can share with other atoms. Hence in the compounds, there are only 6 electrons present around boron atom i.e., octet is not complete. 4. How do you explain that BBr3 is a stronger Lewis acid than BF3. Ans. This is due to back donation of electrons into empty 2p orbital of boron atom from filled p-orbital of Br atom is much less than that by F atom due to larger size of Br atom than F atom. 5. Is boric acid a protic acid? Explain. Ans. A protic acid is the one that gives hydrogen ions in solutions. Boric acid is a weak monobasic acid. It is not a protonic acid but acts as a Lewis acid by accepting electrons from a hydroxyl ion. B(OH)3 + H2O  [B(OH)4]– + H3O+ 6. Why trihalides of group 13 elements fume in the moist air? Ans. They are hydrolysed by water forming hydrogen halides (MX3 + 3H2O  M(OH)3 + 3HX). 7. Write reactions to justify amphoteric nature of Gallium. Ans. An amphoteric substance is the one which can react both with acids as well as bases. The following reactions justify the amphoteric nature of aluminium.

The p-Block Elements 87 (i) Reaction with acid: 2Ga(s) + 6HCl(aq)  2Ga3+ + 6Cl–(aq) + 3H2(g) (ii) Reaction with alkali: 2Ga(s) + 2NaOH(aq) + 6H2O (l)  2Na+ [Ga(OH)4]–(aq) + 3H2(g) sod. tetrahydroxyaluminate (III) 8. What is catenation? Explain. Ans. Catenation. The property of forming the chains of identical atoms is called catenation. The property of catenation depends upon the strength of atom-atom bond. The greater the strength of the atom-atom bond, greater is the extent of catenation. The elements of group 14 exhibit catenation. It is most extensive in the case of carbon because C—C bond is very strong. On going down the group, the atom-atom bond strength decreases and hence the extent of catenation also decreases. It varies in the order C > > Si > Ge ~_ Sn > > Pb lead does not show catenation. 9. Compare the properties of CO2 and SiO2 in tabular form. Ans. CO2 SiO2 1. It exists as gas. 1. It is a solid. 2. It is a linear, monomeric non-polar 2. It is a three dimensional network solid. molecule. 3. Insoluble in water. 3. Appreciably soluble in water. 4. On reduction with coke in electrical 4. On reduction with coke gives carbon furnace gives silicon carbide monoxide SiO2 + 3C SiC + 2CO CO2 + C 2CO 10. Why CCl4 is resistant to hydrolysis but SiCl4 is readily hydrolysed? Ans. CCl4 does not hydrolyse because carbon cannot exceed its co-ordination number beyond four due to absence of vacant d-orbitals in its valence shell. On the other hand SiCl4 can easily hydrolyse because Si has vacant d-orbitals in its valence shell and can therefore extend its co-ordination number beyond four. 11. Why carbon forms covalent compounds whereas lead forms ionic compounds? Ans. Carbon cannot lose electrons to form C4+ because the ionization energy required is very high. It cannot gain electrons to form C4– because it is energetically not favourable. Hence C forms covalent compounds. Down the group, the ionization enthalpy decreases. Pb, being the last element has so low ionization enthalpy that it can lose electrons to form ionic compounds. 12. CO2 is a gas while SiO2 is a solid. Explain. Ans. Carbon atom being small in size forms double bonds with O-atoms (p-p bonding) and hence exists as monomeric linear molecules, silicon atom being large in size forms single bonds and is linked to four O-atoms forming a three-dimensional network structure.

88 Chemistry—XI 13. Why does elemental silicon not form graphite like structure as carbon does? Explain. Ans. This is due to reluctance of silicon to form p-p multiple bonds because of large size of silicon atom. Hence silicon exists only in the diamond structure. 14. Account for the following: (i) PbX2 is more stable than PbX4 (X = Cl, Br). (ii) PbCl4 is less stable than SnCl4 but PbCl2 is more stable than SnCl2. Ans. (i) Due to inert pair effect, Pb shows an oxidation state of + 2. Hence PbX2 is more stable than PbX4. (ii) Stability of + 4 oxidation state decreases down the group while that of + 2 oxidation state increases due to inert pair effect. 15. Explain why silicon shows a higher covalency than carbon. Ans. It is because carbon does not have vacant d-orbitals in its valence shell whereas silicon has vacant d-orbitals in its valence shell. As a result of this silicon can extend its co-ordination number beyond four. 16. [SiF6]2– is known whereas [SiCl6]2– not. Give possible reasons. Ans. The main reasons are: (i) six large chloride ions cannot be accomodated around silicon atom due to limitation of its size. (ii) interaction between lone pair of chlorine atom and silicon atom is not very strong. 17. What is allotropy? Name the alltropic forms of carbon. Ans. Allotropy is a phenomenon of existence of different forms of element, having different physical properties. The various form are called allotropes or allotropic forms. Carbon exists in crystalline allotropes as well as amorphous allotropes. Crystalline allotropes: diamond, graphite and fullerenes Amorphous allotropes: coal, charcoal and lampblock. 18. Why does boron trifluoride behave as a Lewis acid? Ans. The B atom in BF3 has only 6 electrons in the valence shell and thus needs two more electrons to complete its octet. Therefore, it easily accepts a pair of electrons .. from nucleophiles such as F –, NH3 (C2H5)2O, RCH2OH etc. and thus behaves as a Lewis acid. 19. Si and hydrogen compounds, silanes, are few in number while alkanes are large in number. Ans. Carbon has the maximum tendency for catenation due to stronger CC (355 kJ mol–1) bonds. As a result, it forms a large number of alkanes. Silicon, on the other hand, bonds has much lesser tendency for catenation due to weak SiSi (200 kJ mol–1) and hence it forms only a few silanes.

The p-Block Elements 89  SHORT ANSWER QUESTIONS (TYPE–II)  1. Aluminium chloride exists as a dimer, whereas boron does not, why? Ans. Aluminium in AlCl3 has got six electrons in its valance shell. To complete its octet it requires two more electrons. To do so it accepts a lone pair of electrons from Cl atom of other neighbouring AlCl3. Cl Cl Cl Al Al Cl Cl Cl In turn one of the chlorine atom of this molecule donates an electron pair to electron deficient Al atom of neighbouring AlCl3 molecule. Hence a dimer is formed. On the contrary boron due to its small size cannot arrange four large size chlorine atoms around it and hence exists as monomer. 2. Suggest reason why B-F bond length in BF3 (130 pm) and BF4– (143 pm) differ? Ans. B—F bond length in BF3 is 130 pm because boron in BF3 is sp2 hybridised and has a planar triangular structure. As a result of deficiency of electrons on the boron atom the excessive electron density on the fluorine atoms (due to its small size and strong interelectronic repulsions) is pushed on to the boron atom i.e., back donation or back bonding occurs. This introduces a partially double bond character shortening the B—F bond. However, in BF4– the boron atom is sp3 hybridised with B—F bond length of 143 pm. 3. Describe the shape of BF3. Assign hybridisation of boron and explain why B—X distance in BF3 is shorter than theoretically expected value. Ans. In BF3 the hybridisation of boron is sp2 and hence the molecule has a planar triangular shape. In BF3, boron has six electrons in its valance shell. Fluorine atoms F due to their small size and strong inter electronic repulsion push BF the electrons from their filled orbitals into the vacant orbitals on boron atom i.e., undergo p-p backbonding. As a result of this F the B—X distance becomes shorter than theoretically expected Planar triangular value. 4. Give any four uses of boron. Ans. (i) Boron fibres are used in making bullet-proof vests and light composite materials for aircraft. (ii) The boron-10 (10B) isotope has high ability to absorb neutrons and, therefore, metal borides are used in nuclear industry as protective shields and control rods. (iii) As aqueous solution of orthoboric acid is generally used as a mild antiseptic. (iv) Boron is used in steel industry for increasing hardness of steel. 5. Rationalise the given statements and give chemical reaction: (i) lead (II) chloride reacts with Cl2 to give PbCl4 (ii) lead (IV) chloride is highly unstable towards heat

Unit 6: Hydrocarbons  FOR QUICK REVISION   Hydrocarbons are the compounds of carbon and hydrogen only. Hydrocarbons are the major sources of energy. LPG (liquefied petroleum gas) and CNG (compressed natural gas), the main sources of energy for domestic fuels and the automobile industry, are obtained from petroleum. Hydrocarbons are classified as open chain saturated (alkanes) and unsaturated (alkenes and alkynes), cyclic (alicyclic) and aromatic, according to their structure.  Alkanes are saturated hydrocarbons having general formula CnH2n+2. They contain only C—C and C—H sigma bonds. Alkanes show conformational isomerism due to free rotation along the C—C sigma bonds. Out of staggered and the eclipsed conformations of ethane, staggered conformation is more stable as hydrogen atoms are farthest apart. Alkanes can be prepared from alkyl halides by Wurtz reaction or by reduction or through Grignard reagents. R – X + 2Na Ether R – R + 2NaX  R – X + Mg  RMgX H2O R – H R – X + 2H  R – H + HX Alkanes can also be prepared from unsaturated hydrocarbons by hydrogenation.  Alkenes are unsaturated hydrocarbons having general formula CnH2n. They contain a carbon-carbon double bond. Alkenes exhibit geometric isomerism due to restricted rotation about carbon-carbon double bond. The simplest alkene that can exhibit geometric isomerism is but-2-ene. The cis isomer is more polar and has higher boiling point than the trans isomer. On the other hand, the melting point of trans isomer is higher. Alkenes can be obtained by dehydrohalogenation of alkyl halides, by dehydration of alcohols or by partial hydrogenation of alkynes using Lindlar’s catalyst.  Alkynes are unsaturated hydrocarbons having general formula CnH2n–2. They contain a carbon-carbon triple bond. Alkynes are prepared by dehydrohalogenation of dihalides or dehalogenation of tetrahalides. Calcium carbide on reaction with water liberates ethyne. The important reactions of alkanes are free radical substitution, combustion, oxidation and aromatization. Alkenes and alkynes undergo addition reactions, which are mainly electrophilic additions. Markownikov’s Rule. During electrophilic addition across unsymmetrical double bond, the negative part of the adding molecule goes to that carbon which has less 96

Hydrocarbons 97 number of hydrogen atoms. Kharasch Effect. Anti-Markownikov addition of HBr to alkenes in the presence of organic peroxides. Terminal alkynes are acidic in character.  Arenes are aromatic hydrocarbons. They contain at least one benzene ring. Aromatic hydrocarbons undergo mainly electrophilic substitution reactions inspite of high degree of unsaturation present in them. Aromaticity in aromatic compounds is due to the presence cyclic, delocalized system of (4n + 2) pi electrons.  Electron releasing groups increase the reactivity of benzene ring towards electrophilic substitution reactions. Some examples are —R, —OH, —NH2, —OR, —NHCOCH3, etc. Electron withdrawing groups decrease the reactivity of benzene ring towards electrophilic substitution reactions. For example, OO —NO2, —CHO, —C—, —C—OR, etc. Benzene and many polynuclear hydrocarbons containing fused benzene rings are known to possess carcinogenic (cancer causing) properties.  VERY SHORT ANSWER QUESTIONS  1. What are hydrocarbons? Ans. The compounds which are made up of only carbon and hydrogen elements are called hydrocarbons. 2. Name the simplest alkane which can exhibit isomerism. Ans. Butane (C4H10). 3. Arrange the three isomers of C5H12 in the increasing order of their boiling points. Ans. neo-Pentane < iso-Pentane < n-Pentane. 4. Which simplest alkene can exhibit geometric isomerism? Ans. But-2-ene. 5. What is the C—C—C bond angle in propyne? Ans. 180° 6. Arrange the three isomeric xylenes in the increasing order of their polarity. Ans. p-Xylene < m-Xylene < o-Xylene 7. Draw structure and IUPAC name of the starting compound used for the manufacture of teflon. F F   Ans. F— C C —F, Tetrafluoroethene 8. How would you convert 2-butyne to trans-2-butene? Ans. 2-Butyne on reaction with sodium and liquid ammonia yields trans-2-butene.

98 Chemistry—XI 9. Give the structure of the alkene (C4H8) which adds on HBr in the presence and in the absence of peroxide to give the same product, C4H9Br. Ans. CH3—CH CH—CH3 10. Arrange the following: HCl, HBr, HI and HF in the order of decreasing reactivity towards alkenes. Ans. HI > HBr > HCl > HF 11. How will your distinguish between 1-butyne and 2-butyne? Ans. By reaction with ammoniacal AgNO3 solution 1-butyne would give white ppt. whereas 2-butyne does not react. 12. How many monochlorination products are possible for (i) neo-pentane (ii) n-Pentane? Ans. (i) Only one (ii) Three 13. Explain why alkanes with three or more than three carbon atoms contain zig-zag chains of carbon atoms. Ans. Because of sp3-hybrid state of carbon atoms C—C—C bond angle will be about 109° 28. 14. Draw the structure of 2, 2-dimethylbutane and indicate in it the primary, secondary, tertiary and quaternary carbon atoms. p CH3 p ts p Ans. CH3—C—CH2—CH3 Cp H3 Here primary, secondary, tertiary and quaternary carbon atoms are indicated by p, s, t and q respectively. 15. Is it possible to have pure staggered ethane or pure eclipsed ethane? Explain your answer. Ans. No, it is not possible to have pure staggered or pure eclipsed ethane because they are easily interconvertible due to almost free rotation about C—C bond. 16. 2-Iodopropane is subjected to Wurtz reaction. Give structure and IUPAC name of the alkane formed. CH3 CH3 Ans. CH—CH CH3 CH3 2, 3-Dimethylbutane 17. Methyl magnesium bromide is treated with ethanol, what is the formula of the alkane produced? Ans. CH4. 18. Out of n-butane and iso-butane which has higher boiling point? Ans. n-butane has higher boiling point. 19. How do you account for the formation of ethane during chlorination of methane? Ans. Ethane is formed due to side reaction in the termination step. During termination step two methyl free radicals may combine to form ethane.

Hydrocarbons 99 20. Why is it not easy to convert one geometric isomer into the other? Ans. Because rotation about carbon-carbon double bond is hindered. 21. Why is rotation about carbon-carbon double bond hindred? Ans. Rotation about carbon-carbon double bond is hindred because this would involve breaking of -bond which requires quite a large amount of energy. 22. Which alkene on reductive ozonolysis will produce propanone only? H3C CH3 Ans. CC , 2, 3-Dimethylbut-2-ene. H3C CH3 23. (CH3)2C = CH2 + HI  A. What is the IUPAC name of A? Ans. 2-Iodo-2-methylpropane. 24. Can HCl produce 1-chloropropane from propene in the presence of peroxide? Ans. No, peroxide effect is only for addition of HBr. 25. Which reaction is helpful in locating the position of double bond in alkenes? Ans. Ozonolysis. 26. What happens when water is added to calcium carbide? Ans. When water is added to calcium carbide, ethyne gas is produced. CaC2 + 2H2O  Ca(OH)2 + C2H2 27. What is the product of ozonolysis of ethyne? Ans. OHC—CHO, Ethanedial 28. Which alkyne on oxidation with acidic KMnO4 gives ethanoic acid and CO2? Ans. Propyne (CH3—CCH). 29. Out of ethyne and propyne which is stronger acid? Ans. Ethyne is stronger acid. Electron releasing inductive effect of —CH3 group increases the electron density at carbon and makes the loss of H+ difficult. 30. How would you distinguish between pent-1-yne and pent-2-yne? Ans. On treatment with ammoniacal solution of AgNO3, pent-1-yne would give white ppt whereas pent-2-yne does not react. 31. Suggest name of another Lewis acid instead of anhydrous aluminium chloride which can be used during ethylation of benzene. Ans. FeCl3, Ferric chloride. 32. What is the hybrid state of carbon in benzene? Ans. The hybrid state of carbon atoms in benzene is sp2. 33. Why is benzene exceptionally stable though it contains three double bonds? Ans. Because of resonance or because of delocalisation of pi-electrons.

100 Chemistry—XI 34. Explain why the following systems are not aromatic? CH2 (i) (ii) (iii) Ans. Because these molecules donot have cyclic delocalised system of (4n + 2) electrons. 35. What is the general formula of alkanes? Ans. The general formula of alkanes is CnH2n+2 36. What is the general formula of alkenes? Ans. The general formula of alkenes is CnH2n 37. What is the general formula of alkynes? Ans. The general formula of alkynes is CnH2n–2 38. What is the full form of (i) CNG (ii) LNG? Ans. CNG — Compressed Natural Gas LNG — Liquefied Natural Gas 39. What are main sources of alkanes? Ans. Petroleum and natural gas are the main sources of alkanes.  SHORT ANSWER QUESTIONS (TYPE–I)  1. Draw the structure of 2, 2,4-trimethyl-hexane and find out how many each of 1°, 2°, 3° and 4° carbon atoms are there in it? CH3 1 carbon atoms  5 2 carbon atoms  2  3 carbon atoms  1 Ans. CH3— CCH2  CHCH2 CH3 4 carbon atoms  1  CH3 CH3 2. Which aliphatic hydrocarbon series is represented by CnH2n? Write structural formulae of the members of this series with four carbon atoms. Ans. Alkenes. CH3CH2CH CH2 CH3— C CH2 CH3—CH CH—CH3 But-1-ene  But-2-ene (cis and trans) CH3 2-Methylpropene 3. An alkane has a molecular mass equal to 72. Give the possible structural isomers along with their IUPAC names. Ans. Alkanes have general formula CnH2n+2 12 × n + 1 × (2n + 2) = 72 12n + 2n + 2 = 72 n=5 Thus, the molecular formula of alkane is C5H12

Hydrocarbons 101 There are three isomeric alkanes having molecular formula C5H12. These are: (i) CH3CH2CH2CH2CH3, Pentane CH3 (ii) H3C—CH—CH2—CH3, 2-Methylbutane CH3 (iii) H3C—C—CH3, 2, 2-Dimethylpropane. CH3 4. Write IUPAC names of the following compounds? (i) CH3CH = C(CH3)2 (ii) CH2 = CH—C  C—CH3 (iii) (iv) CH2—CH2—CH = CH2 Ans. (i) 2-Methylbut-2-ene (ii) Pent-1-ene-3-yne (iii) Buta-1, 3-diene (iv) 4-Phenylbut-1-ene 5. Draw cis and trans isomers of the following compounds. Also write their IUPAC names: (i) CHCl = CHCl (ii) C2H5(CH3)C = C(CH3)C2H5. HH H Cl CC Ans. (i) CC Cl Cl Cl H cis-1, 2-Dichloroethene trans-1, 2-Dichloroethene CH3 CH3 CH3 C2H5 (ii) C C CC C2H5 C2H5 C2H5 CH3 cis-3, 4-Dimethylhex-3-ene trans-3, 4-Dimethylhex-3-ene 6. Write chemical equations for combustion reaction of the following hydrocarbons. (i) Butane (ii) Pentene (iii) Hexyne (iv) Toluene. Ans. (i) C4H10 + 13 O2  4CO2 + 5H2O 2 (ii) C5H10 + 15 O2  5CO2 + 5H2O 2 (iii) C6H10 + 17 O2  6CO2 + 5H2O 2 (iv) C7H8 + 9O2  7CO2 + 4H2O

102 Chemistry—XI 7. For the following compounds, write structural formulas and IUPAC names for all possible structural isomers having the number of double or triple bond as indicated: (i) C4H8 (one double bond) (ii) C5H8 (one triple bond). Ans. (i) CH2 CH—CH2—CH3 CH3—CH CH—CH3 CH3—C CH2 But-1-ene But-2-ene ½ CH3 (ii) CH3—CH2—CH2—C CH 3-Methylpropene Pent-1-yne CH3—CH2—C C—CH3 Pent-2-yne CH3—CH C CH  CH3 3-Methylbut-1-yne 8. What is the product obtained by reaction of 2-butyne with sodium in liquid ammonia? How would you obtain the geometric isomer of this product from 2-butyne? Ans. 2-Butyne on reduction with sodium and liquid ammonia yields trans-2-butene. CH3—C C—CH3 + H2 Na/Liq. NH3 The cis isomer of this product can be obtained by reduction with hydrogen in the presence of Lindlar’s catalyst. Lindlar's catalyst CH3—C C—CH3 + H2  9. Ethyl bromide is heated with magnesium in the presence of dry ether and the product so obtained is boiled with water to give a gas X. What is X? Also give sequence of reactions. Ans. X is ethane (C2H6). C2H5Br + Mg  C2H5 Mg Br C2H5MgBr + H2O  C2H6 + Mg(OH) Br. 10. Complete the following: (i) C2H2 + HCl (1 mol)  ........ (ii) C2H2 + HCN (1 mol) Ba(CN)2 ......  (iii) CH3—C  CH + NaNH2  ........ ? (iv) CH3CHBrCH2Br  CH3CH = CH2.

Hydrocarbons 103 Ans. (i) HC CH + HCl  H2C CHCl (ii) HC CH + HCN Ba(CN)2 H2C CH—CN – (iii) CH3—C CH + NaNH2  CH3—C CNa + NH3 (iv) CH3— CH— CH2 Zn CH3—CH CH2 + ZnBr2    Br Br 11. Draw cis and trans isomers of the following compounds: (i) 1, 2-Dichloroethene (ii) 1-Phenylprop-1-ene Cl CC Cl H Cl Ans. (i) cis H CC H Cl H trans H3C CC C6H5 H3C CC H (ii) cis H H trans C6H5 H 12. Write structural formulas for the following compounds: (i) 3, 4, 4, 5—Tetramethylheptane (ii) 2, 5—Dimethyhexane CH3 Ans. (i) CH3—CH2—CH—C–—CH—CH2—CH3 CH3 CH3 CH3 CH3 CH3 (ii) CH3—CH—CH2—CH2—CH—CH3 13. How would you obtain butane by (i) Wurtz reaction (ii) Kolbe’s electrolysis. Ans. (i) Ethyl bromide on reaction with sodium in the presence of dry ether yields butane. dry ether 2C2H5Br + 2Na  CH3CH2CH2CH3 + 2NaBr Butane (ii) An aqueous concentrated solution of sodium propanoate on electrolysis yields butane at anode. 2CH3CH2COONa(aq) + 2H2O(l) Electrolysis  CH3CH2CH2CH3 + 2CO2 + H2 + 2NaOH.

104 Chemistry—XI 14. Arrange 2, 2-dimethylbutane, 3-methyl-pentane and n-hexane in the ascending order of their boiling point, also explain the basis for your order. Ans. 2, 2-Dimethylbutane < 3-methylpentane < n-hexane. All the three are isomers having molecular formula C6H14. n-Hexane has the longest chain and therefore has the highest boiling point. 2, 2-Dimethylbutane is the most spherical and has the smallest surface area. In this case the van der Waal’s forces will be weakest and so it has the lowest boiling point. 3-Methylpentane will have boiling point in between these two extremes. 15. Write IUPAC names of the products obtained by addition reactions of HBr to hex-1-ene (i) in the absence of peroxide and (ii) in the presence of peroxide. Ans. (i) CH2 ==CHCH2CH2CH2CH3 Absence of CH3 CH CH2CH2CH2CH3 peroxide  Hex-1-ene + H—Br  Br 2-Bromohexane Peroxide (ii) CH2==CHCH2CH2CH2CH3 + H—Br  CH2 CH2CH2CH2CH2CH3  Br 1-Bromohexane 16. Why do arenes prefer to undergo electrophilic substitution reactions rather than electrophilic addition reactions? Ans. Arenes are highly stabilised by resonance. Addition to the double bond of an arene results in a product in which resonance stabilised ring is destroyed. Hence, addition is difficult in arenes. On the other hand, during substitution resonance stabilised ring remains intact and therefore, arenes undergo substitution reactions. 17. Arrange the following compounds in the order of their decreasing reactivity towards electrophilic substitution: Br NO2 OCH3 CH3 OCH3 CH3 Br NO2 > > Ans. > 18. How would you convert ethyne into toluene? Red hot iron tube CH3 ¾C¾H3¾Cl® Ans. 3 HC CH ¾¾¾¾¾¾¾® AlCl3 Toluene

Hydrocarbons 105 19. Explain why a halogen atom, bonded to a benzene ring, deactivates it towards electrophilic substitution and directs the substitution at ortho and para positions. Ans. Halogen atom bonded to a benzene ring decreases the electron density at the ring by –I effect and thus deactivates it towards electrophilic substitution. However, it increases the electron density at ortho and para positions relative to the meta position by resonance effect (+R effect). 20. What are the necessary conditions for any compound to show aromaticity? Ans. The conditions for a compound to show aromaticity are: (i) The molecule must be cyclic (ii) It must have a conjugated system of (4n + 2) -electrons (iii) The molecule must be planar so that ample delocalization of -electrons can take place. 21. Predict the major product in the following reactions: H2SO4 (i) C6H6 + (CH3)2CHCH2OH  Lindlar’s Catalyst (ii) R—C  C—R + H2  . CH3 R R Ans. (i) H3C C CH3 (ii) C = C H H cis-Alkene tert-Butylbenzene The product is formed through formation of tert-butyl carbocation as intermediate which then brings about electrophilic substitution on benzene ring. 22. Complete the following: Br2 (i) NaNH2 (3 equi)   C6H5CH = CH2 [A] [B]. (ii) CH3I Br2 (i) NaNH2 (3 equi) C6H5—C  Ans. C6H5CH = CH2  C6H5— CH CH2  C—CH3  (ii) CH3I Br Br [A] [B] Two moles of NaNH2 convert dibromo derivative to alkyne whereas the third mole of NaNH2 converts the terminal alkyne formed into its sodium salt which then reacts with CH3Br to form (B). 23. Give a simple chemical test to distinguish between benzene and cyclohexene. Ans. Cyclohexene decolourises bromine solution in CCl4 whereas benzene does not. Cyclohexene contains isolated double bond and thus it undergoes addition reactions readily.

106 Chemistry—XI Br Br + Br2 CCl4 Benzene is stabilised due to delocalisation of -electrons and does not give addition reactions readily.  SHORT ANSWER QUESTIONS (TYPE–II)  1. Why do alkynes undergo nucleophilic addition reactions while simple alkenes do not? Ans. Nucleophilic addition proceeds via carbanion as intermediate. The intermediate carbanion formed from nucleophilic attack on alkyne is more stable than formed from alkene. This is due to greater electronegativity of the sp2-hybridized carbon than the sp3-hybridized carbon. Therefore, alkynes undergo nucleophilic addition reactions while simple alkenes do not. .. .. RC  CH + Nu  R— C=C —H more stable   Carbon atom carrying negative  Nu  charge is sp2 -hybridized  .. .. RCH = CH2 + Nu  R C— C H2 less stable   Carbon atom carrying negative  Nu  charge is sp3 -hybridized  2. Describe the orbital structure of benzene. Ans. According to orbital structure, each H carbon atom in benzene assumes sp2-hybrid state. Each carbon has Is three sp2-hybrid orbitals lying in one plane and oriented at an angle 2 of 120°. There is one unhybridised p-orbital having two lobes lying sp perpendicular to the plane of hybrid orbitals. Each carbon atom 120° 120° uses two hybrid orbitals for axial overlap with similar orbitals of two Hs 2 C s s Is H adjacent carbon atoms on either side to form C—C sigma bonds. The Is 2 s sp 2 2 2 remaining one sp-hybrid orbitals on sp each carbon atom overlaps axially 2 120° sp sp sp with 1s orbitals of hydrogen atoms to form six C—H sigma bonds. The sp axial overlapping of hybrid orbitals to form C—C and C—H bonds. As CC 2 2 sp sp s s 2 2 sp sp s 2 C C s sp 2 2 2 sp sp sp H Is 2 2 Is H s sp sp C 2 sp Is H

Hydrocarbons 107 it clear, the framework of carbon and H C H hydrogen atoms is coplanar with H—C—C H——C C C or C—C—C bond angle as 120°. H C C——H The unhybridised p-orbital on each carbon atom can overlap to a small but equal extent H C with the p-orbitals of the two adjacent carbon H atoms on either side to constitute  bonds. H The molecular orbital containing  electrons spreads uniformly over the entire carbon C 109 pm skeleton and embraces all the six carbons as 120° shown in adjacent figure. This spreading of -electrons in the form of HC C CH ring of -electrons above and below the plane 139 pm of carbon atoms is called delocalisation of C H H -electrons. This delocalisation of -electrons, results, in the decrease in energy and hence, accounts for the stability of benzene molecule. C—C bond length in benzene is 139 pm and C—H bond length is 109 pm. 3. Predict which of the following systems would be aromatic and why? (i) (ii) (iii) (iv) (v) (vi) Ans. Compounds (ii), (iii) and (v) are aromatic. The compounds (ii) and (v) are aromatic because each of them contains a delocalised system of 10  electrons and thus it should be aromatic according to Huckel rule. The compound (iii) is aromatic because it contains two rings, each containing 6 -electrons Compounds (i), (iv) and (vi) are not aromatic. Compound (i) has 8  electrons while compound (iv) contains 4  electrons. Hence, these are not aromatic because they do not have (4n + 2)  electrons. Compound (vi) has (4n + 2)  electrons but is not aromatic because the system lacks complete conjugation due to the presence of one sp3-hybridized carbon. 4. Arrange benzene, n-hexane and ethyne in decreasing order of acidic behaviour. Also give reason for this behaviour. Ans. The decreasing order of acidic behaviour is ethyne > benzene > n-hexane. The C—H bonds in ethyne, benzene and hexane are formed by sp-s, sp2-s and sp3-s overlap respectively. Since sp-hybridized carbon is more electronegative then sp2-hybridized carbon which in turn is more electronegative than sp3-hybridized carbon, therefore, the polarity of C—H bond is in the order: ethyne > benzene > hexane

108 Chemistry—XI Consequently, the acidity is in the order: ethyne > benzene > hexane. sp sp2 H sp3 H—C º C—H Benzene CH3—(CH2)4—CH3 Acetylene Hexane 5. Why is Wurtz reaction not preferred for the preparation of alkanes containing odd number of carbon atoms? Illustrate your answer by taking one example. Ans. When preparation of an alkane is attempted through Wurtz reaction, a number of side products are formed. It is very difficult to separate the mixture of these products. For example, when synthesis of propane is attempted through Wurtz reaction by reaction between ethyl bromide and methyl bromide, butane and ethane are formed along with propane. Na /ether CH3Br + C2H5 Br  CH3—CH3 + CH3—CH2—CH3 + CH3—CH2—CH2—CH3 6. Complete the following: Ba(CN)2 Ni(CN)2 (i) H—C  C—H + HCN  ? (ii) H—C  C—H  ? – High pressure KOCH3 (iii) H—C  C—H + CH3OH  ? Ba(CN)2 Ans. (i) H—C  C—H + HCN  H C= C H  H CN Acrylonitrile Ni(CN)2 (ii) 4H—C  C—H  High pressure 1, 3, 5, 7-Cyclooctatetraene (iii) H—C  C—H + CH3OH – H C=CH K OCH3   H OCH3 Methoxyethene Methyl vinyl ether 7. Describe the mechanism of addition of bromine to alkenes. Ans. Addition of bromine to alkenes takes place through the following steps: Step I. The first step involves the electrophilic attack of bromine molecule on the double bond resulting in the formation of cyclic bromonium ion.

Hydrocarbons 109 d+ d– CC – + Brd+ C C + Br—Br Br + Br CC Bromonium ion Brd– p-complex Step II. The second step involves the attack of Br– ion on bromonium ion to form the 1, 2-dibromo compound. + Br —C—C— Br – C C + Br Br 1, 2-Dibromo compound 8. Give the structures of the major products from 3-ethylpent-2-ene under each of the following reaction conditions. (i) HBr in the presence of peroxide (ii) Br2/H2O. C2H5 C2H5 Br  HBr/Peroxide  CH3—CH2— CH CHCH3 Ans. (i) CH3CH2— C = CHCH3  (Anti Markownikov addition) 3-Ethylpent-2-ene 2-Bromo-3-ethylpentane C2H5 C2H5  Br2 /H2O  (ii)  CH3—CH2— C CHCH3 CH3CH2  CH = CHCH3 (HOBr)  OH Br 3-Ethylpent-2-ene (Markownikov’s Addition) 9. In alkyl halide X of formula C6H13Cl on treatment with potassium tertiary butoxide gives two isomeric alkenes Y and Z (C6H12). Both alkenes on hydrogenation, give 2, 3-dimethylbutane. Predict the structures of X, Y, Z. Ans. CH3 CH3 (CH3)3COK CH3 CH3 CH3 CH3 CH3—C——C——CH3 – HCl CH3—C = C—CH3 + CH2 = C——C—CH3 (Elimination Cl H 2, 3-Dimethylbut-2-ene 2, 3-Dimethylbut-1-ene 2-Chloro-2, 3-dimethyl- occurs) (Y) (Z) H2/Pt butane CH3 CH3 (X) CH3—CH—CH—CH3 2, 3-Dimethylbutane

ISBN: 978-93-93738-14-1 789393 738141 T11-8896-149-COMP.CBSE QB CHEM T-II XI