CBSE II Question Bank in Chemistry CLASS 12 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 XII) (According to the Latest CBSE Examination Pattern) By S.K. Khanna Dr. N.K. VERMA Formerly, Associate Professor Chemistry Department Formerly, Associate Professor D.A.V. College Chemistry Department Chandigarh D.A.V. College Chandigarh laxmi Publications (P) Ltd (An iso 9001:2015 company) bengaluru • chennai • guwahati • hyderabad • jalandhar Kochi • kolkata • lucknow • mumbai • ranchi new delhi
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Contents 1. Electrochemistry ............................................................................. 1–19 2. Chemical Kinetics ......................................................................... 20–42 3. Surface Chemistry ........................................................................ 43–58 4. d- and f-Block Elements ............................................................... 59–74 5. Co-ordination Compounds ............................................................ 75–97 6. Aldehydes, Ketones and Carboxylic Acids ................................. 98–135 7. Amines ....................................................................................... 136–161
SYLLABUS For Session 2021–22 (Code No. 043) Class XII (Term–II) S. No. Unit No. of Periods Marks 13 1. Electrochemistry 7 2. Chemical Kinetics 5 9 3. Surface Chemistry 5 13 4. d-and f-Block Elements 7 35 5. Coordination Compounds 8 6. Aldehydes, Ketones and Carboxylic Acids 10 7. Amines 7 TOTAL 49 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.
Unit 2: Chemical Kinetics FOR QUICK REVISION Chemical Kinetics: The branch of chemistry which deals with the study of reaction rates and their mechanism. Rate of Reaction: It is the rate of change of concentration of any of the reactant or product with time at any particular moment of time. Qualitative Rate: The rate of a reaction when expressed on the basis of visual parameters like change in colour, effervescence, change in pressure, etc. Quantitative Rate: The true or actual rate of a reaction expressed on the basis of experimental or kinetic data. Average Rate: The rate of reaction measured over a long time interval is called average rate. It is given as x/t. Instantaneous Rate: It is the rate of a reaction at a given instant of time i.e., x t (average rate) becomes dx when t approaches zero. dt Stoichiometric Coefficients: The number of moles of species participating in a balanced chemical equation. Initial Rate: The rate at the beginning of a reaction when the conc. have not changed appreciably. Freezing of a Reaction: When a sample is placed in the freezing mixture of ice and sodium chloride to retard or almost stop a reaction. Law of Mass Action: The rate of a reaction is directly proportional to the product of conc. of reactants with each term raised to power number of times it is present in a balanced equation at a given temperature. Rate Expression: The mathematical expression giving the rate of a reaction in terms of conc. of reactants at a given temperature. Rate Constant (k): It is the rate of the reaction when the concentration of each of reacting species is unity. It is also called velocity constant or specific reaction rate of the reaction. Rate Law: The mathematical expression based on experimental fact, which describes the reaction rate in terms of concentrations of reacting species. It cannot be written from the balanced equation. Molecularity: The number of reacting particles which collides simultaneously to bring about the chemical change. It is a theoretical concept. Integrated Rate Equation: The differential rate equations which are integrated to give a relationship between rate constant and concentrations at different times. 20
Chemical Kinetics 21 Order of Reaction: The sum of the exponents of the concentration terms in the experimental rate law of reaction. It can be zero, 1, 2, 3 or and any fractional value. Zero Order Reaction: Reactions in which the rate of reaction does not change with the concentration of reactions, i.e., r = k [A]° = k First Order Reaction: A reaction in which the rate of reaction is directly proportional to the conc. of reacting substance, i.e., r = k[A] or kt = 2.303 log [A]0/[A] Pseudo First Order Reaction: Such reactions which appear to be of higher order but follows lower order kinetics. Elementary Process: One step reaction is known as elementary process. Complex Reactions: The reactions in which the balanced equation shows a molecularity of three or more. Rate Determining Step: The slowest step in the reaction mechanism is rate determining step. Half-life Period of Reaction (t1/2): The time taken for the concentration of reactants to be reduced to half of their initial concentration. Catalyst: A substance which enhances the rate of a reaction. Life Time of Reaction: The time in which 98% of reaction is complete. Mechanism of Reaction: The sequence of elementary steps leading to overall stoichiometry of a reaction. Temperature Coefficient: It is ratio of rate constant at 308 K and 298 K. IMPORTANT FORMULAE For the reaction aA + bB cC + dD Average Rate = 1 [A] 1 [B] = 1 [C] 1 [D] a t b t c t d t Instantaneous Rate = 1 d[A] 1 d[B] = 1 d[C] 1 d[D] a dt b dt c dt d dt Rate Law: For a general reaction, aA + bB products, Rate = k[A]m[B]n m and n are determined experimentally Order w.r.t. A = m; Order w.r.t. B = n Overall Order = m + n Units of Rate = mol L–1 s–1 or Atm s–1 Units of k: For reaction of nth order k = (mol L–1)1 – n s–1 or (atm)1– n s–1 Rate law for a zero order reaction = dx k [A]0 dt Units of k = mol L–1 s–1 The integrated rate law equation for a zero order reaction, R P is k = [R0 ] [R] t
22 Chemistry—XII Half-life Period (t1/2) of a zero order reaction is t1/2 = [R0 ] 2k Rate law for a Ist order reaction = dx = k [A]1 dt Units of k = s–1 or min–1 The integrated rate law equation for a Ist order reaction, A B is k = 2.303 log [A0 ] t [A] The plot of log [A] vs time gives a straight line whose slope = k 2.303 Half-life Period of a 1st order reaction, t1/2 = 0.693 k For a Ist order reaction t75% = 1.386 = 2 t1/2; t60% = 0.9165 k k t90% = 2.303 ; 6.909 k t99.9% = k = 10 t1/2 1 nth order reaction, t1/2 [A]0n 1 Amount of substance left after n Half-lives for 1st Order Reaction [A]n = [A]0 (1/2)n Time of nth Fraction of Ist Order t1/n = 2.303 log 1 1 / n k 1 VERY SHORT ANSWER QUESTIONS 1. What is the significance of minus sign in the instantaneous rate expression, 1 d[NO2 ] ? 2 dt Ans. The negative sign is used in the expression to get a positive rate of reaction because change in conc., A is negative. 2. Differentiate between specific reaction rate and rate of the reaction? Ans. Specific reaction rate is the rate of reaction at unit concentration of each of reactant. It is always constant for a particular reaction at a given temperature.
Chemical Kinetics 23 3. For a chemical reaction X Y, the rate of reaction increases by the factor 2.25 when concentration of X is increased by 1.5. Suggest rate law and order. Ans. r = k [X]2; order is 2. 4. For a chemical reaction 2A + B C, the rate of formation of C is 0.25 mol L–1 hr–1. What is the rate of disappearance of A and B? Ans. Rate of disappearance of A is twice of rate of appearance of C whereas rdiss B = rfor C. i.e., rdiss A = – 0.5 mol L–1 hr–1 rdiss B = – 0.25 mol L–1 hr–1. 5. What is the order of the reaction whose rate law expression is r = k[A]3/2[B]–1? Ans. 0.5 6. For a chemical reaction A H, the rate is found to increase by the factor 1.59 when conc. of A is doubled. What is the order? Ans. 2/3 7. The rate constant of a reaction is found to have dimensions of mol L–1 sec–1. What is its order? Ans. Zero 8. A reaction is found to be zero order. Will its molecularity be zero? Ans. No, molecularity cannot be zero. 9. The reaction A + 3B 2C obeys the rate equation : rate = k [A]1/2 [B]3/2. What is the order of this reaction? Ans. 2 10. For a reaction, X2(g) + 2Y2(g) 2XY2(g), write the rate equation in terms of the rate of disappearance of Y2. Ans. Rate = – 1 d[Y2 ] 2 dt 11. Write a chemical reaction in which the units of rate and rate constant are same. Ans. The units of rate and rate constant are same for zero order reaction. N2(g) + 3H2(g) Pt 2NH3(g) 1170 K 12. For a reaction NO2 + CO CO2 + NO, the rate law expression is rate = k [NO2]2. How many molecules are involved in slowest step? Ans. Two 13. The rate of formation of a dimer in second order dimerisation reaction is 9.1 × 10–6 mol L–1 s–1 at 0.01 mol. L–1 monomer concentration. What is the value of rate constant? Ans. 9.1 × 10–2 mol–1 L s–1 14. How is the rapid change in concentrations of reactants/products monitored for fast reactions? Ans. Flow methods or Relaxation method or spectrophoto-metric techniques.
24 Chemistry—XII 15. A substance with initial concentration ‘a’ follows zero order kinetics. In how much time will the reaction go to completion. Ans. For a zero order reaction, dx = k or dx = kdt dt The expression can also be written as, x = kt Hence, t = x k a For completion x = a t = k . 16. Write expression for half-life in case of a reaction between hydrogen and chlorine to form hydrochloric acid gas. Ans. Given reaction is a reaction of zero order. Hence, t1/2 = [A]0/2k. 17. For the assumed reaction X2 + 3 Y2 2 XY3, write the rate equation in terms of rate of disappearance of Y2. Ans. Rate = – d [X2 ] 1 d [Y2 ] 1 d [XY2 ] dt 3 dt 2 dt Rate of disappearance of Y2 = – d [Y2 ] 3 d [X2 ] 3 d [XY2 ] dt dt 2 dt 18. How does the value of rate constant vary with reactant concentration? Ans. For reaction of nth order, Rate = dx = k [conc.]n dt or k= dx 1 conc. 1 dt (conc.)n Time (conc.)n = 1 1 1 Time (conc.)n1 i.e. k (conc.)n1 . 19. For a reaction A B, the rate of reaction becomes twenty-seven times when the concentration of A is increased three times. What is the order of the reaction? Ans. r = k [A]n when conc. is 3 times, then 27r = k [3A]n Hence, 27r k[3A ]n or n = 3 r k[A]n 20. A reaction is 50% complete in 2 hours and 75% complete in 4 hours. What is the order of reaction? Ans. t1/2 of a first order reaction is 2t1/2. Hence, reaction is of first order. 21. For the reaction, Cl2 (g) + 2 NO (g) 2 NOCl (g), the rate law is expressed as, rate = k [Cl2] [NO]2 what is the overall order of the reaction. Ans. Order of reaction is 3.
Chemical Kinetics 25 22. Define the term ‘order of reaction’ for chemical reactions. Ans. Sum of powers to which the concentration terms are raised in a rate law expression. 23. Express the rate of the following reaction in terms of disappearance of hydrogen in the reaction 3H2(g) + N2(g) 2NH3(g) Ans. Rate = – 1 d [H2 ] 3 dt 24. The reaction A + 2B C obeys the rate equation, Rate k [A]1/2 [B]3/2. What is the order of the reaction. Ans. The order of reaction is 1 3 = 2 22 25. The decomposition reaction of ammonia gas on platinum surface has a rate constant = 2.5 × 10–4 mol L–1 s–1. What is the order of the reaction. Ans. The decomposition of NH3 on platinum surface is a zero order of reaction. 26. What is the molecularity of the reaction, Cl 1 Cl2 (g)? 2 Ans. The molecularity is one. 27. Define order of a reaction. Ans. Powers to which concentration terms are raised in a rate law expression. 28. For what type of reactions the rate constant shall have the same units as the rate of reaction? Ans. Zero order reaction. 29. Why the rate of a chemical reaction do not remain uniform throughout the reaction? Is it possible to have reactions with a uniform rate? Ans. It is because rate of reaction depends on concentration which changes with passage of time. A reaction of zero order proceeds at a uniform rate. 30. Give units of specific rate constant for a zero order reaction. Ans. mol L–1 s–1. 31. What is meant by half-life period for a reaction? Ans. It is the time taken for completion of 50% of the reaction. It is denoted by t1/2. 32. The reaction A + B C has zero order. Write rate equation. Ans. Rate = k [A]0[B]0 = k. (rate constant) SHORT ANSWER QUESTIONS (TYPE–I) 1. Why can’t rate of a reaction be obtained by dividing the total amount of reactant consumed with time taken? Ans. It is because the rate calculated in this manner would be constant throughout the entire span of the reaction which is contrary to the actual observation. 2. What do you understand by the rate of a chemical reaction? Name the factor on which it depends. Ans. Rate depends upon R, conc. of reactants, temperature, nature of reactants, presence of a catalyst, etc.
26 Chemistry—XII 3. Express the relationship between rate of production of iodine and the rate of disappearance of hydrogen iodide in the following reaction. 2HI H2 + I2 Ans. d[I2 ] 1 d[HI] dt 2 dt 4. State a condition under which a bimolecular reaction is kinetically first order reaction. Ans. A bimolecular reaction is kinetically first order reaction if the conc. of one of the reactant is taken in excess. 5. Write the rate equation for the reaction 2A + B C if the order of the reaction is zero. Ans. rate = k [A]0[B]0 6. How can you determine the rate law of the following reaction? 2NO(g) + O2(g) 2NO2(g) Ans. The rate law can be determined by initial rate method. 7. For which type of reactions, order and molecularity have the same value? Ans. For an elementary reaction order and molecularity is same. 8. In a reaction if the concentration of reactant A is tripled, the rate of reaction becomes twenty seven times. What is the order of the reaction? Ans. r1 = k1[A]x When conc. is tripled, 27r1 = k1[3A]x Hence, 27r1 k1[3A ]x r1 k1[ A ]x 27 = (3)x or (3)x = (3)3 Hence, x = 3 i.e., order of reaction = 3 9. Write the general expression for the rate constant, k, for a first order reaction and show how to derive from it the expression k = 0.693/t1/2. Ans. k= 2.303 log [A]0 ; At t1/2, [A] = [A]0 t [A] 2 k = 2.303 log 2 = 2.303 0.3010 0.693 . t1/2 t1/2 t1/2 10. What is t1/2? Give expression relating time required for nth fraction of reactant of disappear to the rate constant of first order reaction. Ans. Time required to complete nth fraction of a first order reaction is given as: t1/n = 2.303 log [A]0 2.303 log n. k n 1 k [A]0 [A]0 /n 11. What is half life period of the reaction show that for the first order reaction t1/2 is independent of initial concentration? 0.693 Ans. t1/2 = k for first order reaction. As the concentration term does not appear in the expression, it is independent of concentration.
Chemical Kinetics 27 12. For a reaction A + B Products, the rate law is— Rate = k [A][B]3/2 can the reaction be an elementary reaction? Explain. Ans. Reactions with fractional orders cannot be elementary. 13. What is t1/2? How it is related to t3/4 and t7/8? Ans. t3/4 = 2 × t1/2; t7/8 = 3 × t1/2 14. For the reaction 2NO2 + F2 2NO2F, the rate law is given by r = k [NO2] [F2] (i) Is this reaction an elementary reaction? (ii) Had the reaction been the single step reaction what would have been its rate law? Ans. (i) It is not an elementary reaction because, the order of reaction on the basis of overall balanced equation is three whereas the rate expression shows it is to be second order. (ii) r = [NO2]2 [F2] If the reaction is elementary i.e., single step then r = k[NO2]2[F2]. 15. The rate expression for the chemical reaction 2NO2F 2NO2 + F2 is: r = k [NO2F] (i) write theoretical steps for the overall reaction. (ii) what should be the units of rate constant? Ans. The theoretical steps can be written as: NO2F Slow NO2 + F; NO2F + F Fast NO2 + F2 The validity of steps can be confirmed experimentally. Since the reaction is of Ist order, therefore units of k = s–1. 16. Why does the rate of any reaction, generally, decreases during the course of the reaction? Ans. Conc. of reactants falls with passage of time and hence the rate. 17. How can the plot of log [A] and t be used for the calculation of rate constant for the first order? Ans. The plot of log [A] v and t is a straight line with a slope = 2.303 . k 18. The gas phase decomposition of acetaldehyde CH3CHO(g) CH4(g) + CO(g) at 680 K is observed to follow the rate expression: rate = – d [CH3CHO]/dt = k [CH3CHO]3/2 If the rate of the decomposition is followed by monitoring the partial pressure of the acetaldehyde, we can express the rate as: – dPCH3CHO/dt = k[PCH3CHO]3/2 If the pressure is measured in atmosphere and time in minutes, then (i) what are the units of the rate of the reaction? (ii) what are the units of the rate constant, k? Ans. Units of rate = atm s–1; units of k = (atm)–1/2 s–1
28 Chemistry—XII 19. What is meant by the ‘rate constant, k’ of a reaction? If the concentration be expressed in mol L–1 units and time in seconds, what would be the units for k (i) for a zero order reaction and (ii) for a first order reaction. Ans. (i) For a zero order reaction r = k [A]° or r = k units of k = mol L–1 s–1 (ii) For a first order reaction r = k [A] or k = r mol L1 s1 = s–1. [A] mol L1 20. A reaction is of second order with respect to a reactant. How will the rate of reaction. (i) doubled, (ii) reduced to half? Ans. For a second order reaction r = k [A]2 (i) When conc. of A is doubled i.e., 2A then, r = k [2A]2 = 4. k[A]2 the rate increases by 4 times. A (ii) When conc. of A is reduced to 1/2 i.e., 2 then, r=k A 2 1 k [A]2 2 4 The rate decreases by 4 times. 21. A first order reaction has a rate constant of 0.0051 min–1. If we begin with 0.10 M concentration of the reactant, what concentration of reactant will remain in solution after 3 hours? Ans. For a first order reaction kt log [R]0 2.303 [R] or [R]0 = antilog kt antilog 0.0051 180 [R] 2.303 2.303 = antilog 0.3986 = 2.503 [R] (conc. after 3 hours) = 0.1 = 0.04 M. 2.503 22. The rate constant for a reaction of zero order in A is 0.0030 mol L–1 s–1. How long will it take for the initial concentration of A to fall from 0.10 M to 0.075 M. Ans. For a zero order reaction t= 1 ([A]0 – [A]) = (0.1 0.075) 0.025 = 8.33 sec. k 0.003 0.003 23. A reaction is of first order in reactant A and of second order in reactant B. How is the rate of this reaction affected when (i) the concentration of B alone is increased to three times (ii) the concentration of a as well as B are doubled? Ans. Rate = k [A] [B]2
Chemical Kinetics 29 when [B] = tripled then r = k [A] [3B]2 = 9k [A] [B]2 = 9 times more Similarly, r = k [2A] [2B]2 = 8k [A] [B]2 = 8 times more 24. What do you understand by the rate law and rate constant of a reaction? Identify the order of a reaction if the units of its rate constant are: (i) L–1 mol s–1 (ii) L mol–1 s–1 Ans. Units of rate constant for a reaction of nth order is k = (mol L–1)1 – n × s–1 Hence, (i) zero order (ii) second order 25. The thermal decomposition of HCO2H is a first order reaction with a rate constant of 2.4 × 10–3 s–1 at a certain temperature. Calculate how long will it take for three-fourths of initial quantity of HCO2H to decompose. (log 0.25 = –0.6021) Ans. For a first order reaction, t1/2 = 0.693 0.693 = 288.75 = 289 sec. k 2.4 103 t0.75 = 2 t1/2 = 289 × 2 = 578 sec. 26. A reaction is of second order with respect to a reactant. How is its rate affected if the concentration of the reactant is (i) doubled (ii) reduced to half? Ans. Let the concentration of the reactant be [A] = a Rate of reaction, R = k [A]2 = ka2 (i) If the concentration of the reactant is doubled, i.e., [A] = 2a, then the rate of the reaction would be k = 2a2 = 4ka2 = 4R Therefore, the rate of the reaction would increase by 4 times. (ii) If the concentration of the reactant is reduced to half, i.e., a/2, then the rate of the reaction would be = k(a/2)2 rate = k(a/2)2 = ¼ ka2 = ¼rate 27. How a catalyst only changes the speed of a reaction whereas a photosensitizer initiates the reaction? Ans. A catalyst does not chemically participate in a reaction. It simply provides a path of low activation energy. On the contrary a photosensitizer acts as the energy carrier for a reaction and hence helps in initiating a reaction. 28. The reaction 2NO(g) + Cl2(g) 2NOCl was studied at – 10°C, and the following data were obtained: Run Initial Initial Rate of Formation of 1 Concentrations NOCl (mol L–1 min–1) 2 (mol L–1) 3 0.18 [NO] [Cl2] 0.36 0.10 0.10 1.44 0.10 0.20 0.20 0.20 (i) what is the order of reaction with respect to NO and with respect to Cl2? (ii) what is the numerical value of the rate constant at – 10°C?
30 Chemistry—XII Ans. (i) 2nd order in NO and 1st order in Cl2; (ii) 180 mol–2 L2 s–1 29. How is rate of reaction related to concentration of the reactants? Ans. Rate of reaction increases with increase in concentration except in zero order reaction. As the concentration increases total number of collisions will increase. Therefore probability of effective collisions will increase. 30. What is the effect of concentration on the decomposition of NH3 regarding order of reaction? Ans. The decomposition of NH3 on finely divided platinum surface is first order when the pressure and concentration of NH3 in the system is low. At higher concentration when the surface of the catalyst is fully covered with layer of NH3, the reaction become zero order. 31. For a reaction, the rate law is: Rate = k = [A] [B]1/2.. Can this reaction be an elementary reaction? Ans. For an elementary reaction, order of reaction should be equal to molecularity and further molecularity should be integral. For the given reaction, order of reaction = 1 + 1/2 = 3/2. Since molecularity cannot be fractional, therefore, for the given reaction, order is not equal to molecularity. Hence, given reaction cannot be an elementary reaction. 32. List two differences between rate of reaction and reaction rate constant. Ans. Rate of Reaction Rate Constant 1. It varies with the concentration 1. At a given temperature, it is constant of reactant. for a particular reactions. 2. The unit of rate of reaction is always 2. The unit of rate constant varies with mol L–1 s–1. the order of reaction. 33. List the points of difference between order of reaction and molecularity. Ans. Order of Reaction Molecularity 1. It is the sum of the power of the molar 1. It is the total number of molecules of concentrations of the reactants in the the reactant which take part in a rate law. single step chemical reaction. 2. Order of reaction may be zero and 2. Molecularity always a whole number fractional. and cannot be zero. 3. It is an experimental value. 3. It is a theoretical value. SHORT ANSWER QUESTIONS (TYPE–II) 1. A chemical reaction R P, has half life (t1/2 ) depending on single power of concentration of reactant. If the concentration of reactant is increased by 5 times then how many times the reaction rate increases? Ans. For the reaction, R P t0.5 [R]1.
Chemical Kinetics 31 In general, the half life of reaction of nth order is t0.5 1 (Here, n = order of reaction) [ R] n1 0 For the given reaction, n = 0 i.e., the order of reaction is zero. For zero order reaction, rate = k[R]0 = k Hence, the rate is independent of concentration of reactants. 2. The concentration of a reactant following first order kinetics, decreases from 10 g to 2.5 g in 60 minutes. What is the rate constant of this reaction? Ans. For a first order of reaction, the concentration of substance left after n half lives is given as, [R] (Here, n = number of half lives) [R]0 = 2n Given, [R]0 = 10 g, [R] = 2.5 g 2n = 10 = 4 = (2)2 or n = 2, i.e., number of half lives = 2 2.5 Hence, 60 t1/2 = 2 = 30 min For a first order reaction, k = 0.693 0.693 = 0.0231 min–1. t1/2 30 min 3. The rate constant of a zero order reaction is 1 × 10–3 mol L–1 s–1. Starting with 50 moles, calculate the time in minutes in which the concentration decreases to 10 moles. Ans. For a zero order of reaction x = kt [x = change in concentration in time t] t= x 50 10 (mol L1 ) = 40 × 103 s. k 1 103 mol L1 s1 t = 40 103 s 1 mm = 667 min. 60 s 4. How a catalyst increases the rate of reaction? Ans. The energy of activation of the In absence reaction decreases because of the of catalyst presence of catalyst, i.e., an alternative path is provided which Potential energy Ea (I) Ea (II) In presence has lower energy barrier. Due to Reactants of catalyst decrease in activation energy more reactant molecule will cross over the Products energy barrier and converted into products. Ultimately products will increase and rate of reaction also increase. Reaction coordinate
32 Chemistry—XII 5. Derive an expression for the rate constant and half-life period of a zero order reaction. Ans. For zero order reaction d[R] dt = k[R]0 – d[R] = kdt – [R] = kt + I ...(I) At initial condition – [R]0 = k × 0 + I – [R]0 = I The value of “I” put in equation (I) – [R] = kt + [R]0 or K= [R]0 [R] . t Half-life period for zero order t = t1/2 [R] [R]0 2 [R]0 [R]0 2 t1/2 = k t1/2 = [R]0 . 2k In this case the half-life depends upon the initial concentration of reactant. 6. Calculate the half-life period of a first order reaction from their rate constants given below: (a) 200 s–1 (b) 2 min–1 (c) 5 year–1. Ans. For the first order reaction, t1/2 = 0.693 k (a) k = 200 s–1 t1/2 = 0.693 = 3.465 × 10–3 s 200 (s–1 ) (b) k = 2 min–1 0.693 t1/2 = 2 (min1 ) = 0.3465 min (c) k = 4 year–1 0.693 t1/2 = 4 (year1 ) = 0.1732 year. 7. The half life period of a reaction of first order is 100 sec. Calculate its rate constant. Ans. For a first order reaction, t1/2 = 0.693 k
Chemical Kinetics 33 0.693 k= 0.693 = 6.93 × 10–3 (s–1). Now, k= 102 t1/2 t1/2 = 100 (s) LONG ANSWER QUESTIONS 1. (a) The rate constant of a reaction A B is 0.5; when initial conc. of A is 1 mol L–1. Calculate the degree of conversion of substance A within 1 hour if the reaction is of zeroth and first order. How does the degree of conversion depend on the order of reaction ? (b) From the kinetic data [A] mol/L [B] mol/L Rate mol L–1 s–1 1. 1.0 2.0 2.25 × 10–3 2. 2.0 2.0 4.5 × 10–3 3. 2.0 1.63 3.0 × 10–3 Predict the order to reaction w.r.t reactant B. Ans. (a) (i) For a zero order reaction: k = 0.5 mol L–1h–1 t0.5 = R [A]0 1 = 1 hour 2k 2 0.5 Hence, the conversion in 1 hour = 50% (ii) For a first order reaction k = 0.5 h–1 k = 2.303 log [A]0 t [A] or log [A]0 = kt 0.5 1 = 0.217 = [A]0 = antilog 0.217 = 1.648 [A] 2.303 2.303 [A] Hence, conc. after 1 hour = [A]0 = 0.606 1.648 Amt. of substance reacted or converted = 1.000 – 0.606 = 0.394 mol L–1 Conversion in 1 hour = 0.394 × 100 = 39.4%. (b) The rate law expression, r = k[A]x [B]y Applying initial rate method for calculating order divide exp. run 2 with exp. run 3 4.5 103 = (2.0)x (2.0)y 3.0 103 (2.0)x (1.63)y or 1.5 = 2.0 y = (1.226)y 1.63
34 Chemistry—XII Since, 1.5 = 1.225 or s . (1.225)2 = 1.5 (1.225)2 = (1.226)y Hence, y=2 Order w.r.t. B = 2. 2. Consider the data for the reaction between A and B [A] [B] Initial Rate (mol L–1) (mol L–1) (mol L–1 s–1) At 300 K At 320 K 2.5 × 10–4 3.0 × 10–5 5 × 10–4 2 × 10–3 5.0 × 10–4 6.0 × 10–5 4 × 10–3 — 1.0 × 10–3 6.0 × 10–5 1.6 × 10–2 — Calculate (i) order w.r.t. A and B (ii) rate constant at 300 K (iii) the pre-exponential factor if Ea = 55.3 kJ Ans. (i) From the given data, the rate law comes out to be R = k1 [A]2[B] Order w.r.t. A = 2 Order w.r.t. B = 1 (ii) Now k1 = Rate [A]2[B] = 5.0 104 mol L1 s1 (2.5 104 )2 (mol L1 )2 (3 105 ) mol L1 = 2.67 × 108 L2 mol–2 s–1 (iii) Calculation of A k = A.eEa /RT log k = log A – Ea 2.303 R log A = log (2.67 × 108) – 5.53 104 J 2.303 8.314 J mol1 k1 300 k = 18.0565 or A = 1.145 × 1018. 3. For the reaction, A + B products, the following initial rates were obtained at various given initial concentrations S.No. [A] mol L–1 [B] mol L–1 Rate (mol L–1 s–1) 1 0.1 0.1 0.05 2 0.2 0.1 0.10 3 0.1 0.2 0.05 Write the rate law and find the rate constant of the above reaction.
Chemical Kinetics 35 Ans. Suppose rate = k[A]x[B]y ...(i) for case I, 0.05 = k[0.1]x[0.1]y ...(ii) for case II, 0.10 = k[0.2]x[0.1]y ...(iii) for case III, 0.05 = k[0.1]x[0.2]y From equation (ii) and (i) we get 0.10 = (2)x or 2 = 2x or x = 1 0.05 Similarly from equation (iii) and (i) we get, 0.05 = (2)y or 1 = (2)y or 0.05 log 1 = y log 2 0 = y × 0.3010 or y = 0 The rate law expression is, r = k[A][B]0 = k[A] The value of k = r 0.05 mol L1 s1 = 0.5 s–1. [A] 0.1 mol L1 4. For a first order reaction, show that time required for completion of 99.9% of reaction is 3 times the time required for completion of 90% of the reaction. OR Show that, t99.9% = 3t90% for a first order reaction. Ans. For a first order reaction, we know that t = 2.303 log a k ax t99.9% = 2.303 log a k a .999a = 2.303 log 103 3 2.303 ...(i) kk and t90% = 3 2.303 log a k a 0.90a = 2.303 log 10 2.303 ...(ii) kk Now divide (i) by (ii) t99.9% = 3 2.303 k 3 t90% k 2.303 1 t99.9% = 3t90%.
36 Chemistry—XII 5. The following rate data was obtained for the first order thermal decomposition of SO2Cl2(g) at a constant volume. SO2Cl2(g) SO2(g) + Cl2(g) Exp. Time (s) Total pressure (atm) 1. 0 0.5 2. 100 0.6 Calculate the reaction rate when total pressure is 0.65 atmosphere. Ans. Let us say that the pressure of SO2Cl2 decreases by x atm, then the increase of pressure of SO2 and Cl2 = x atm each. [ 1 mole of SO2Cl2 decomposes to give 1 mole of SO2 and 1 mole of Cl2] SO2Cl2(g) SO2(g) + Cl2(g) Pressure at t = 0 0.5 atm 0 0 Pressure at time t, (0.5 – x atm) x atm x atm Since total pressure (PT) = p p p = (0.5 – x) + x + x SO2Cl2 SO2 Cl 2 PT = 0.5 + x or x = PT – 0.5 Hence, pSO2Cl2 = 0.5 – (PT – 0.5) = 0.5 – PT + 0.5 = 1.0 – PT Since, at t = 100 sec., PT = 0.6 atm pSO2Cl2 = 1.0 – 0.6 = 0.4 atm. (a) Evaluation of k k = 2.303 log Initial pressure t Pressure at time t = 2.303 log 0.5 atm 2.303 log 1.25 100 s 0.4 atm 100 = 2.303 × 0.0969 = 2.23 × 10–3 s–1. 100 s (b) Rate at pT = 0.65 atm pSO2Cl2 at total pressure of 0.65 atm = 1.0 – 0.65 = 0.35 atm Rate = k[N2O5] = 2.23 × 10–3 s–1 × 0.35 atm = 7.8 × 10–4 atm sec–1. 6. Nitric oxide, NO, reacts with oxygen to produce nitrogen dioxide: 2NO(g) + O2(g) 2NO2(g) The rate law for this reaction is rate = k[NO]2[O2]. Propose a mechanism for the above reaction.
Chemical Kinetics 37 Ans. The rate law indicate that order of reaction is 2 w.r.t. NO and 1 w.r.t. O2. The possible mechanism for given reaction may be, K1 NO3 ...(fast step) NO + O2 ...(slow step) K2 NO3 + NO NO2 + NO2 Overall reaction, (by addition of two steps) 2NO + O2 2NO2 As slowest step of mechanism of reaction determine the rate of reaction, rate = k2 [NO3] [NO] But [NO3] = k1 [NO][O2] ( NO3 is an intermediate specie, and its formation is in equilibrium state) rate = k1 k2 [NO][O2][NO] = k[NO]2[O2] ...(where k is rate constant and k = k1k2) The above expression of rate law derived from proposed mechanism is same as in given data. 7. A gaseous hypothetical chemical reaction 2A 4B + C is carried out in a closed vessel. The concentration of B is found to increase by 5 × 10–3 mol L–1 in 10 seconds. Calculate (i) Rate of appearance of B (ii) Rate of reaction and (iii) Rate of disappearance of A. Ans. (i) Rate of appearance of B = Increase in concentration of B Time taken = 5 103(mol L1 ) = 5 × 10–4 mol L–1 s–1 10 (s) (ii) Rate of reaction = 1 × Rate of appearance of B 4 = 1 × 5 × 10–4 (mol L–1 s–1) = 1.25 × 10–4 mol L–1 s–1 4 (iii) Rate of disappearance of A = 1 × Rate of appearance of B 2 = 1 × 5 × 10–4 (mol L–1 s–1) = 2.5 × 10–4 mol L–1 s–1. 2 8. If the rate of formation of oxygen gas is 48 g min–1 in the reaction below: 2N2O5(g) 4NO2(g) + O2(g) (i) What is the rate of disappearance of N2O5? (ii) What is the rate of appearance of NO2? Ans. The rate of a reaction is always expressed in terms of mol L–1, therefore, rate of formation of oxygen gas, O2 = 48 (g min1 ) = 1.5 mol min–1 32 (g)
38 Chemistry—XII (i) rate of disappearance of N2O5, – 1 [N2O5 ] = [O2 ] 2 t t = 2 × 1.5 (mol min–1) = 3.0 mol min–1 (ii) rate of appearance of NO2, 1 [NO2 ] = [O2 ] 4 t t = 4 × 1.5 (mol min–1) = 6.0 mol min–1. 9. (a) Draw a schematic graph showing how the rate of a first order reaction changes with change in concentration of the reactant. (b) Rate of a reaction is given by the equation: Rate = k[A]2 [B]2 What are the units for rate and rate constant for this reaction. Ans. (a) Rate of a first order reaction is directly proportional to molar concentration of the reactant i.e., rate conc. So the graph is a straight line as shown in the following figure: Rate Conc. (b) Unit of rate = mol L–1 s–1 rate = Change in conc. time Unit of rate constant, k = rate mol L1 s1 [A]2[B] (mol L1 )2(mol L1 ) = mol–2 L2 s–1 In terms of molarity, (M), these units can also be written as: Unit of rate of reaction = M s–1 (here M = mol L–1) Unit of rate constant = M–2 s–1 10. (i) Explain the difference between instantaneous rate of a reaction and average rate of a reaction. (ii) With the help of an example explain what is meant by Pseudo first order reaction. Ans. (i) Differance between average rate of reaction and instantaneous rate of reactions are: Average Rate: The average rate of a reaction is defined as the rate of change of concentration per unit time. It is calculated by dividing the total change in concentration of any one of the reactant or product by the total time taken to do so. For a reaction R P (reactant) (product)
Chemical Kinetics 39 Average rate = Change in concentration in given time Time taken i.e., = – R or =+ P or = x t t t The Average rate of reaction changes with passage of time hence to know the true rate, it is expressed as instant rate. Instantaneous Rate: The instantaneous rate of a reaction is defined as the decrease in concentration of any one of the reactants or increase in concentration of any one of the product at a particular instant of time for a given temperature. Mathematically, the instantaneous rate may be written as, rinst = d[R] or d[P] dx * dt + dt or = dt Here, dt is the infinitesimally small time interval and dx is the change in concentration of any of the species in time interval dt. (ii) Pseudo first order reaction: The order of a reaction is sometimes altered by conditions. For example, during the hydrolysis of 0.01 mol of ethyl acetate with 10 mol of water, amounts of the various constituents at the beginning (t = 0) and completion (t) of the reaction are given as under. H CH3COOC2H5 + H2O CH3COOH + C2H5OH t=0 0.01 mol 10 mol 0 mol 0 mol t 0 mol 9.9 mol 0.01 mol 0.01 mol In the given reaction, which is bimolecular, the concentration of water does not get altered much during the course of the reaction. So, in the rate equation, Rate = k [CH3COOC2H5] [H2O] the term [H2O] can be taken as constant. The equation, thus, becomes Rate = k [CH3COOC2H5] (where k = k[H2O]) Accordingly, the reaction which appears to be of second order behaves as first order reaction. Such reactions are called pseudo first order reactions. Inversion of canesugar is another pseudo first order reaction. C12H22O11 + H2O H C6H12O6 + C6H12O6 Canesugar Glucose Fructose Rate = k [C12H22O11]
40 Chemistry—XII CASE BASED QUESTIONS CASE STUDY–I Read the passage given below and answer the following questions: A reaction of the type, aA + bB products, was studied by the initial rate method. The following kinetic data was obtained. exp. [A]0 mol L–1 [B]0 mol L–1 Initial Rate mol L–1 s–1 1. 0.10 0.10 2. 0.20 0.30 3.0 × 10–4 3. 0.30 0.10 7.2 × 10–3 4. 0.20 0.60 8.1 × 10–3 ? 1. The order of reaction with respect to A and B are respectively (a) 1, 2 (b) 2, 0 (c) 3, 1 (d) 3, 2. Ans. (c) Let the rate of reaction be expressed as, r = k[A]x[B]y Accordingly, exp1; 3.0 × 10–4 = k[0.1]x[0.1]y exp2; 7.2 × 10–3 = k[0.2]x[0.3]y exp3; 8.1 × 10–3 = k[0.3]x[0.1]y Now, exp3 = 8.1 103 k[0.3]x [0.1]y or exp1 3.0 104 k[0.1]x [0.1]y 2.7 × 10–1 = (3)x or (3)3 = (3)x. Hence x = 3 Similarly, y=1 Order w.r.t A = 3, B = 1 2. The rate constant of the reaction at 300 K when concentrations are in mol L–1 and time in seconds will be (a) 3.33 × 10–2 (b) 3.0 × 10–4 (c) 30 (d) 3.0. Ans. (d) Rate = k[A]3[B]1 or 3.0 × 10–4 = k[0.1]3[0.1] or k= 3.0 104 3.0 104 3. The initial rate in the experiment 4 is (a) 3.6 × 10–4 mol L–1 s–1 (b) 7.2 × 10–3 mol L–1 s–1 (c) 1.44 × 10–2 mol L–1 s–1 (d) 1.8 × 10–4 mol L–1 s–1. Ans. (c) rinitial = 3.0 × (0.2)3(0.6)1 or 3.0 × 0.2 × 0.2 × 0.2 × 0.6 = 1.44 × 10–2 mol L–1s–1 4. The pre-exponential factor will be of the order of (a) 108 (b) 1012 (c) 1015 (d) 1018. Ans. (d) Pre-exponential factor is 1018
Chemical Kinetics 41 CASE STYDY–II Read the passage given below and answer the following questions: In a parallel reaction, the reactant A (radioactive metal) decays by two parallel pathways to form product B and C k1 B ; conc. at time t = x A k2 C ; conc. at time t = y The ratio of concentration of [B] : [C] are 2 : 98 to and the t1/2 of the reaction is 22 years. 1. The decay constant for B is (c) 126 × 10–5 y–1 (d) 189 × 10–5 y–1. (a) 31.5 × 10–5 y–1 (b) 63 × 10–5 y–1 Ans. (b) For the given reaction, t1/2 = 0.693 k or k = 0.693 0.693 = 0.0315 y–1 k 22 Since, k = k1 + k2 or k2 = 0.0315 – k1 Now we know that for the parallel reactions, k1 = x 2 k2 y 98 k1 2 or 98k1 = 0.063 – 2k1 = 0.0315 k1 98 or k1 = 63 × 10–5 y–1 2. The decay constant for c is correctly given as (a) 0.03 y–1 (b) 0.06 y–1 (c) 0.09 y–1 (d) 1.5 y–1. Ans. (a) We know that, k = k1 + k2 k2 = k – k1 = 0.0315 – 63 × 10–5 = 0.03087 = 0.031 y–1 3. Starting with 2.0 kg of A the amount of C that shall be obtained of the 44 years will be (b) 4.9 × 10–3 kg (c) 1.47 kg (d) 1.5 kg. (a) 0.5 kg Ans. (c) t1/2 = 22 yrs (given) Cn = C0 2 (n = number of self lines in 44 yrs = 2) 2n 22 or Cn = 2 = 0.5 kg (Amt. of reactant left) 4 Amt. of produced formed after 44 yrs = 1.5 kg 98 Hence amt. of C formed = 1.5 × 100 = 1.47 kg.
42 Chemistry—XII 4. If the concentration of B and C of the time t is x and y moles respectively than and rB and rC is the rate of formation of B and C then which of the following is correct? d[A] d[A] (a) – dt = k1[A] + k2[B] (b) – dt = k1[B] + k2[C] 0.693 (d) rB x . rC y (c) t0.5 = (k1 k2) Ans. (d) For the given reaction, k1 B ; conc. at time t = x A k2 C ; conc. at time t = y (i) For conversion of A B – d[A] d[B] = k1[A] dt dt (ii) For conversion of A C – d[A] d[C] = k2[A] dt dt Hence Nitrate of disappearance of d[A] A = – dt = k1[A] + k2[A] = (k1 + k2) [A] d[A] d[A] or – dt = (k1 + k2) [A] or – [A] = (k1 + k2) dt On integrating the above equation we get (k1 + k2) t = 2.303 log [A0 ] [ A] The rate of formation of B, rB = d[B] = k1[A] = x dt dt Also d[C] x rC = dt = k2 [A] = dt Hence, rB = k1 x . rC k2 y
Unit 4: d- and f-Block Elements FOR QUICK REVISION Transition Elements: Elements having incomplete d-orbitals in their atoms or in their simple ions. Four Transition Series: The series involving filling of 3d-, 4d-, 5d- and 6d-subshell respectively. All transition elements are metals. They possess characteristic properties of metals such as high tensile strength, malleability, ductility and thermal and electrical conductivity. Transition elements have high melting and boiling points due to stronger interatomic bonding. The atomic and ionic radii of transition elements are smaller than those of s-block elements but larger than those of p-block elements. Transition elements exhibit higher enthalpies of atomisation due to stronger interatomic bonding. Among the elements of the first transition series, zinc has the least enthalpy of atomisation. In the 3d-transition series, Cu is the only element which has positive electrode potential for the M2+/M couple. Variable Oxidation States: Transition elements exhibit variable oxidation states due to participation of (n – 1)d-electrons in bonding. Among the elements of the 3d-transition series scandium does not show variable oxidation states while manganese shows maximum number of oxidation states. The most common oxidation state of the elements of the first transition series is +2. Coloured Complexes: Transition elements form coloured complexes. Colour of the complexes is due to absorption of light energy due to d-d transitions in partially filled d-orbitals. Paramagnetic Nature: Most of the transition element compounds are paramagnetic due to the presence of unpaired electrons in them. The magnetic moment is due to orbital motion and spin of the unpaired electrons. The ‘spin only’ magnetic moment is given by the relation: = n (n 2) BM. Zinc, cadmium and mercury do not show characteristic properties of transition elements because they do not contain partially filled d-orbitals in their atoms or common oxidation states. 59
60 Chemistry—XII Inner Transition Elements: Elements in which the last electrons enters the f-subshell of anti-penultimate energy level. Lanthanoids: A series of elements which involve filling of 4f-subshell. These are fourteen elements following lanthanum from Cerium to Lutetium. Lanthanoid Contraction: The steady decrease in size of lanthanoid ions (M3+) with increase in atomic number. VERY SHORT ANSWER QUESTIONS 1. Which three elements belonging to d-block are usually not considered as transition elements? Ans. Zinc, cadmium and mercury. 2. Which transition metal has highest density? Ans. Iridium has the highest density among all the transition metals. 3. Which transition metal has highest melting point and which has lowest melting point? Ans. Tungsten has the highest melting point and mercury has the lowest melting point. 4. What is the shape of chromate ion? Ans. Tetrahedral. 5. Out of ferrous and ferric salts which possesses higher magnetic moment? Ans. Ferric salts have higher magnetic moment because Fe3+ ions (3d5) have more number of unpaired electrons than Fe2+ ions (3d6). 6. Copper(I) is diamagnetic whereas copper(II) is paramagnetic. Give reason. Ans. Cu(I) (3d10) has all its electrons paired and hence is diamagnetic whereas Cu(II) (3d9) has an unpaired electron in 3d-subshell and hence is paramagnetic. 7. Why Ti4+ complexes are diamagnetic? Ans. It has no unpaired electron Ti4+ : [Ar] 3d0. 8. Which of the following divalent ions is likely to have maximum magnetic moment? Cr2+, Mn2+, Fe2+, Co2+, Ni2+ Ans. Mn2+ : [Ar] 3d5 Mn2+ has maximum number of unpaired electrons. 9. What happens when chromates are kept in acidic solution and dichromates in the alkaline solution? Ans. Chromate ions in acidic medium change into dichromate ions while dichromate ions in basic medium change into chromate ions. There is always an equilibrium between chromate and dichromate ions. + H Cr2O72 H2O 2 CrO24 + Yellow OH Orange 10. Why is copper (At. No. 29) considered a transition metal? Ans. Cu(II) has partially filled d-orbitals.
d- and f-Block Elements 61 11. Why does vanadium pentoxide act as a catalyst? Ans. Due to variable oxidation states of vanadium. 12. Which of the two, V(IV) or V(V), is paramagnetic and why? Ans. V(IV) has one electron in 3d-subshell and hence would be paramagnetic. 13. Why do Zr and Hf exhibit similar properties? Ans. Because they have almost equal atomic radii and ionisation enthalpies. 14. Which element of the first transition series shows the highest number of oxidation states? Ans. Manganese. 15. Name a transition metal which does not exhibit variable oxidation states in its compounds. Ans. Scandium. 16. Why Cd2+ salts are white? Ans. Cd2+ (4d10) has fully filled d-subshell. Hence, no d-d transitions are possible in it. 17. What is the oxidation state of Cr in chromate ion and in dichromate ion? Ans. Oxidation state of Cr is +6 in CrO42– as well as Cr2O72–. 18. What is the most characteristic oxidation state of lanthanoids? Ans. +3 is the most common oxidation state of lanthanoids. 19. Why lanthanoids are called f-block elements? Ans. In lanthanoid elements, the last electron enters the f-subshell of their atoms. 20. In the transition series, starting from lanthanum (57La) the next element hafnium (72Hf) has an atomic number of 72. Why do we observe this jump in atomic number? Ans. In between lanthanum (Z = 57) and hafnium (Z = 72), there are 14 lanthanoids. 21. The size of the trivalent cations in the lanthanoids series decreases steadily as the atomic number increases. What is this known as? Ans. Lanthanoid contraction. 22. Name the lanthanoid element which forms tetrapositive ions in aqueous solution. Ans. Cerium (Ce). 23. Write the electronic configuration of the element with atomic number 102. Ans. [Rn] 5f 14 7s2. 24. Why does copper not liberate hydrogen from acids? Ans. Copper shows positive value for electrode potential and hence it cannot reduce H+ ions to hydrogen. 25. Out of copper(I) chloride and copper (II) chloride which is more stable and why? Ans. Copper (II) chloride is more stable. It is due to greater value of lattice enthalpy in solid state (and enthalpy of hydration in solution form) for Cu2+ ion than for Cu+ ion. 26. Although Zr and Hf belong to different transition series but it is difficult to separate them. Explain why? Ans. Zr and Hf have nearly same atomic radii (or ionic radii) due to lanthanoid contraction. 27. Explain why cerium exhibits +4 oxidation state. Ans. Ce(IV) has stable 4f 0 configuration.
62 Chemistry—XII 28. What is the general electronic configuration of transition elements? Ans. (n – 1) d1–10 ns1–2. 29. Name the first and the last element in the second transition series. Ans. The first element of second transition series is Yttrium and the last element is Cadmium. 30. Name the d-block elements which do not have partially filled d-orbitals in their atoms or in their simple ions. Ans. Zinc, cadmium and mercury. 31. Name the three factors which determine the stability of particular oxidation state. Ans. (i) Enthalpy of atomization, (aH) (ii) Ionization enthalpy, (iH) (iii) Hydration enthalpy, (hydH). 32. What is the most common oxidation state of first transition series? Ans. +2 is the most common oxidation state. 33. Name a transition element which does not exhibit variable oxidation state. Ans. Scandium (Z = 21) does not exhibit variable oxidation states. It exhibits oxidation state of only +3. 34. Name a member of the lanthanoid series which is well known to exhibit +4 oxidation state. Ans. Cerium. 35. How is it that several transition metals act as catalysts? Ans. Transition metals can act as catalysts due to their ability to exist in variable oxidation states. They have also good adsorbing power. 36. Of the ions Co2+, Sc3+ and Cr3+ which ones will give coloured aqueous solutions and how will each of them respond to a magnetic field? (Atomic numbers: Co = 27, Sc = 21, Cr = 24) Ans. Co2+ and Cr3+ will give coloured solutions and will be paramagnetic. 37. Why is the third ionization energy of manganese (At. no. = 25) unexpectedly high? Ans. Mn2+ (3d5) has stable configuration and hence removal of electron from it is difficult. SHORT ANSWER QUESTIONS (TYPE–I) 1. Compare the magnetic moments of the following trivalent ions. Cr3+, V3+, Fe3+ Ans. Cr3+ : 3d3, V3+ : 3d2, Fe3+ : 3d5 Fe3+ > Cr3+ > V3+. Greater the number of unpaired electrons, more the magnetic moment. 2. Decide giving reason which one of the following pairs exhibits the property indicated. (i) Sc3+ or Cr3+ exhibits paramagnetism (ii) V or Mn exhibits more number of oxidation states. (Atomic numbers: Sc = 21, Cr = 24, V = 23, Mn = 25)
d- and f-Block Elements 63 Ans. (i) Cr3+ has unpaired electrons and hence exhibits paramagnetism. (ii) Mn exhibits more number of oxidation states. 3. The sums of first and second ionization enthalpies and those of third and fourth ionization enthalpies of nickel and platinum are: iH1 + iH2 (MJ mol–1) iH3 + iH4 (MJ mol–1) Ni 2.49 8.80 Pt 2.66 6.70 Based on this information, write: (i) The most common oxidation states of Ni and Pt and why? (ii) Name of the metal (Ni or Pt) which can more easily form compounds in its + 4 oxidation state and why? Ans. (i) + 2 (ii) Pt, because sum of first four ionisation enthalpies is less for it. 4. (a) Of the ions Ag+, Co2+ and Ti4+, which one will be coloured in aqueous solutions. (Atomic numbers: Ag = 47, Co = 27, Ti = 22). (b) If each one of the above ionic species is in turn placed in a magnetic field, how will it respond and why? Ans. (a) Co2+ will be coloured. (b) Co2+ will be attracted while Ag+ and Ti4+ will be repelled. 5. Explain why: (i) The scandium salts are white (At. no. of Sc = 21) (ii) Es for Mn3+/Mn2+ couple is more positive than that for Fe3+/Fe2+? Ans. (i) Sc3+ : [Ar] 3d0. (ii) Mn3+ (3d4), Mn2+ (3d5) Fe3+ (3d5), Fe2+ (3d6) Mn2+ to Mn3+ change is difficult whereas Fe2+ to Fe3+ change is easier because d5 configuration is stable. 6. Account for the following: Of the d4 species Cr(II) is strongly reducing but Mn(III) is strongly oxidizing. (Atomic numbers: Cr = 24, Mn = 25, Co = 27) Ans. Cr(II) has great tendency to go over to Cr(III) (d5) by losing one electron and hence acts as reducing agent Mn(III) has great tendency to go over to Mn(II) (d5) by gaining one electron and hence acts as oxidising agent. Cr(III) and Mn(II) have stable d5 electronic configuration. 7. Give the electonic configuration of the (a) d-orbitals of Ti in [Ti(H2O)6]3+ ion in an octahedral crystal field. (b) Why is this complex coloured? Explain on the basis of distribution of electrons in the d-orbitals. (c) How does the colour change on heating [Ti(H2O)6]3+ ion? Ans. (a) In [Ti (H2O)6]3+ ion, Ti is in +3 oxidation state. There is only 1 electron in the d-orbital and its configuration is t2g1 eg0.
64 Chemistry—XII (b) Due to d-d transition, configuration becomes t2g0 eg1. (c) On heating, [Ti (H2O)6]3+ ion becomes colourless as there is no ligand (H2O) left on heating. In the absence of ligand, crystal field splitting does not occur. 8. Write down the electronic configurations of the following ions: (i) Mn2+ (ii) Fe3+ (iii) Cr3+ (iv) Ni2+ (v) Zn2+. Ans. (i) Mn2+ : [Ar] 3d5 or 1s2, 2s2, 2p6, 3s2, 3p6, 3d5 (ii) Fe3+ : [Ar] 3d5 (iii) Cr3+ : [Ar] 3d3 (iv) Ni2+ : [Ar] 3d8 (v) Zn2+ : [Ar] 3d10. 9. On what ground can you say that scandium (Z = 21) is a transition element but zinc (Z = 30) is not. Ans. Sc: [Ar] 3d14s2, Scandium has incompletely filled 3d-orbitals whereas zinc has completely filled 3d orbitals in its ground state as well as common oxidation state. 10. Explain why transition metals have higher density than alkaline earth metals. Ans. Transition metals have smaller atomic size and stronger interparticle attractive forces (metallic bond) than alkaline-earth and hence have higher density than alkaline earth metals. 11. Which of the following ions would form white complexes? Cu2+, Zn2+, Ti3+, Ti4+, Cd2+. Ans. Zn2+ (3d10), Ti4+ (3d0) and Cd2+ (4d10) would form white complexes because d-d transition cannot take place in them due to either fully filled or vacant d-subshell. 12. Explain why transition elements are less electropositive than alkaline earth metals. Ans. Transition metals, due to their smaller atomic size and greater effective nuclear charge, have higher ionization energy than alkaline earth metals and hence are less electropositive. 13. Why do transition elements exhibit higher enthalpies of atomisation? Ans. Transition elements have very strong interatomic bonds due to their small size and also due to the presence of large number of unpaired electrons in their atoms. As a result they have very high enthalpies of atomisation. 14. Calculate the magnetic moment of a divalent ion in aqueous solution of its atomic number is 25. Ans. The divalent ion of the element with atomic number would have the electronic configuration [Ar] 3d5. Thus, it would have five unpaired electrons. The magnetic moment, is = n (n 2) 5 (5 2) = 5.92 BM 15. Explain why the E value for the Mn3+/Mn2+ couple is much more positive than that for Cr3+/Cr2+ or Fe3+/Fe2+. Ans. The high reduction potential of Mn3+/Mn2+ couple is due to the reasion that Mn3+ (3d4) by accepting an electron acquires very stable 3d5 electron configuration. Mn3+ + e– Mn2+ [Ar] 3d4 [Ar] 3d5
d- and f-Block Elements 65 16. Explain why (i) Sm(II) in aqueous solutions is a good reducing agent? (ii) Ce(IV) in aqueous solution is a good oxidizing agent? Ans. (i) Lanthanoids have tendency to attain +3 oxidation state. When Sm(II) changes to Sm(III), it undergoes oxidation and behaves as a reducing agent. (ii) Ce(IV) has tendency to attain Ce(III) state. Therefore, it acts as a good oxidizing agent. 17. La(OH)3 is more basic than Lu(OH)3. Explain why? Ans. Atomic size of Lu is smaller than La due to lanthanoid contraction. As a result M—OH bond in Lu(OH)3 has more covalent character and hence dissociates to smaller extent. 18. What are the different oxidation states exhibited by lanthanoids? Ans. +2, +3 and + 4. +3 in the most common oxidation state. 19. Explain why the +3 oxidation states of lanthanum, gadolinium and lutetium are especially stable? Ans. Gd (III): 4f 7 Lu (III): 4f 14 Gadolinium and lutetium have exceptionally stable +3 oxidation states because Gd3+ and Lu3+ have exactly half-filled and fully filled f-subshell respectively. 20. What is the name, symbol and the electronic configuration of (i) last lanthanoid (ii) last actinoid? Ans. (i) Last lanthanoid: Lutetium (Lu): [Xe] 4f 14 5d1 6s2. (ii) Last actionoid: Lawrencium (Lr): [Rn] 5f 14 6d1 7s2. 31. Silver atom has completely filled d-orbitals (4d10) in its ground state. How can you say that it is a transition element? Ans. For an element to be classified as a transition element, it should have partially filled d-orbitals in its atom or common oxidation states. Silver can exhibit + 2 oxidation state wherein it has incompletely filled d-orbitals, hence we can say it is a transition element. 22. Explain why? (i) Sc3+ salts are diamagnetic. (ii) Ti3+ salts are coloured whereas Ti4+ salts are white. Ans. (i) Sc3+: 1s2, 2s2, 2p6, 3s2, 3p6 In Sc3+, all the electrons are paired. Hence, Sc3+ salts are diamagnetic. (ii) Ti3+ has one electron in 3d-subshell whereas Ti4+ has no electron in 3d-subshell. Therefore, in Ti3+ (3d1) d-d transitions are possible whereas in Ti4+ (3d0) d-d transitions are not possible. Hence ,Ti3+ salts are coloured whereas Ti4+ salts are white. 23. What may be the stable oxidation state of the transition element with the following d-electron configurations in the ground state of their atoms: 3d3, 3d5, 3d8, 3d4? Ans. For 3d34s2, the most stable oxidation state is + 5 For, 3d5 4s2, the stable oxidation states are + 2 and + 7 For 3d8 4s2, the stable oxidation state is + 2 For 3d4 4s2, the stable oxidation states are + 3 and + 6.
66 Chemistry—XII 24. Which metal in the first series of transition metals exhibits +1 oxidation state most frequently and why? Ans. In the first transition series copper exhibits +1 oxidation state most frequently. This is due to the reason that Cu(I) has stable electronic configuration. Cu(I): [Ar] 3d10. 25. Calculate the number of unpaired electrons in following gaseous ions: Mn2+, Cr3+, V3+, and Ti4+. Ans. Mn3+: [Ar] 3d5; five unpaired electrons Cr3+: [Ar] 3d3; three unpaired electrons V3+: [Ar] 3d2; two unpaired electrons Ti4+: [Ar] 3d0; There is no unpaired electron 26. The Es(M2+/M) value for copper is positive (+ 0.34 V). What is possibly the reason for this? Ans. E 2 / M ) value for copper is positive because the sum of the first and second ionisation (M enthalpies for copper is very large due to exceptionally high second ionization enthalpy. This is not compensated by the hydration enthalpy (hyd H) of Cu2+ ion. 27. Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only? Ans. Due to very high electronegativity and small size oxygen and fluorine can oxidize the metal to its highest oxidation state. 28. Which is stronger reducing agent Cr2+ or Fe2+ and why? Ans. Cr2+ is stronger reducing agent that Fe2+. This can be explained in terms of E(M3+/M2+) values. E(Cr3+/Cr2+) has negative value (– 0.41 V), indicating that Cr3+ would change into Cr2+ with difficulty whereas reverse would take place readily. 29. Calculate the ‘spin only’ magnetic moment of M2+ (aq) ion (Z = 27). Ans. The electronic configuration of the M2+ ion (Z = 27) would be M2+(aq): [Ar] 3d7 It would contain three unpaired electrons. The ‘spin only’ magnetic moment is given by the relation: = n (n 2) BM = 3 (3 2) BM = 3.87 BM 30. Ti3+ salts are coloured while Sc3+ salts are colourless. Explain. Ans. Ti3+ has one electron in the d-orbital (3d1) which can absorb a part of visible light and undergoes d-d transition. But Sc3+ (3d0) has no electron in the d-orbital and hence cannot undergo d-d transitions. As a result Sc3+ salts are colourless. 31. Why does a transition metal form alloys with other transition metals easily? Ans. The transition elements have almost similar atomic radii. Therefore, these elements can mutually substitute their positions in the metallic lattice and thus form alloys. For example, brass is an alloy of copper and zinc. 32. Why are Ni2+ compounds thermodynamically more stable than Pt2+ compounds whilst Pt4+ compounds are relatively more stable than Ni4+ compounds? Ans. The sum of first two ionization enthalpies (iH1 + iH2) is less for Ni than Pt whereas sum of first four ionization enthalpies is less for Pt than Ni. Hence, Ni2+ compounds are more stable than Pt2+ while Pt4+ compounds are more stable than Ni4+.
d- and f-Block Elements 67 33. Explain why: Ce4+ is a good oxidizing agent whereas Sm2+ is a good reducing agent. Ans. Among lanthanoids +3 state is most stable. Ce4+ tends to go over to +3 state by undergoing reduction and hence it acts as oxidizing agent. On the otherhand, Sm2+ tends to go over to +3 state by undergoing oxidation and hence it acts as reducing agent. 34. Explain why: La(OH)3 is stronger base than Lu(OH)3. Ans. Lu3+ is smaller in size than La3+ due to lanthanoid contraction. Due to smaller size of Lu3+, Lu–O bond is stronger than La–O bond in the respective hydroxides. Due to weaker La–O bond, La(OH)3 behaves as stronger base. SHORT ANSWER QUESTIONS (TYPE–II) 1. Give reasons (i) K2PtCl6 is a well known compound whereas the corresponding Ni compound does not exit (ii) [Ti(H2O)6]3+ is coloured while [Sc(H2O)6]3+ is colourless. Ans. (i) Pt(IV) is stable whereas Ni(IV) due to high third and fourth ionisation enthalpies is not stable. Hence, K2NiCl6 does not exist. (ii) Ti(III) has one electron in 3d-subshell whereas Sc(III) has no electron in 3d-subshell Ti(III): [Ar] 3d1 Sc(III): [Ar] 3d0. 2. What is meant by disproportionation? Give two examples of disproportionation reactions in aqueous solutions. Ans. Sometimes a particular oxidation state becomes less stable relative to other oxidation states, one lower and the other higher. In such a situation a part of the species undergoes oxidation while a part undergoes reduction. Such a species is said to undergo disproportionation. For example, Cr(V) and Mn(VI) undergo disproportionation as shown below: 3CrO43– + 8H+ 2 CrO42– + Cr3+ + 4H2O (+ V) (+ VI) (+ III) 3MnO42– + 4H+ 2MnO4– + MnO2 + 2H2O (+ VI) (+ VII) (+ IV) Thus, Cr(V) undergoes disproportionation to Cr(VI) and Cr(III) while Mn(VI) undergoes disproportionation to Mn(VII) and Mn(IV). 3. For the first row transition metals the E values are E V Cr Mn Fe Co Ni Cu (M2+/M) (Volts) – 1.18 – 0.91 – 1.18 – 0.44 – 0.28 – 0.25 + 0.34 Explain the irregularity in the above values.
Unit 6: Aldehydes, Ketones and Carboxylic Acids FOR QUICK REVISION Aldehydes, ketones and carboxylic acids are important class of organic compounds which contain carbonyl group (C O). The compounds with such groups form highly polar molecules. As a result, the compounds formed by them boil at higher temperatures than the hydrocarbons and weakly polar compounds like ethers of comparable molecular masses. The lower members are more soluble in water because they form hydrogen bonds with water. The higher members, because of large size of hydrophobic chain of carbon atoms, are insoluble in water but soluble in common organic solvents. Physical properties of aldehydes and ketones: Boiling points. Due to dipole-dipole interactions in aldehydes and ketones, they have higher boiling points than hydrocarbons and ethers of comparable molecular masses. Due to the presence of intermolecular. H-bonding in alcohols, the boiling points of aldehydes and ketones are lower than those of alcohols of comparable molecular masses. Solubility. The solubility of aldehydes and ketones in water decreases with increase in the length of the alkyl chain. Preparation of aldehydes. Aldehydes are prepared by dehydrogenation or controlled oxidation of primary alcohols and controlled or selective reduction of acyl halides. Aromatic aldehydes may also be prepared by oxidation of (i) methylbenzene with chromyl chloride or CrO3 in the presence of acetic anhydride, (ii) formylation of arenes with carbon monoxide and hydrochloric acid in the presence of anhydrous aluminium chloride, and (iii) cuprous chloride or by hydrolysis of benzal chloride. Preparation of ketones. Ketones are prepared by oxidation of secondary alcohols and hydration of alkynes. Ketones are also prepared by reaction of acyl chloride with dialkylcadmium. A good method for the preparation of aromatic ketones is the Friedel- Crafts acylation of aromatic hydrocarbons with acyl chlorides or anhydrides. Both aldehydes and ketones can be prepared by ozonolysis of alkenes. Reactivity: Due to steric and electronic reasons, aldehydes are more reactive than ketones towards nucleophilic addition reactions. Chemical reactions of aldehydes and ketones: Nucleophilic addition reactions. Aldehydes and ketones undergo nucleophilic addition reactions onto the carbonyl group with a number of nucleophiles such as, HCN, NaHSO3, alcohols (or diols), ammonia derivatives, and Grignard reagents. The -hydrogens in aldehydes and ketones are acidic. Therefore, aldehydes and ketones having at least one -hydrogen, undergo Aldol condensation in the presence of a base to give -hydroxy-aldehydes (aldol) and -hydroxy-ketones (ketol), respectively. Aldehydes having no -hydrogen undergo Cannizzaro reaction in the presence of concentrated alkali. Aldehydes and ketones are reduced to alcohols with 98
Aldehydes, Ketones and Carboxylic Acids 99 NaBH4, LiAlH4, or by catalytic hydrogenation. The carbonyl group of aldehydes and ketones can be reduced to a methylene group by Clemmensen reduction or Wolff- Kishner reduction. Aldehydes are easily oxidized to carboxylic acids by mild oxidizing reagents such as Tollens’ reagent and Fehling’s reagent. These oxidation reactions are used to distinguish aldehydes from ketones. SUMMARY OF REACTIONS OF ALDEHYDES AND KETONES Aldehydes Chemical properties HC º N OH CH3C—CN NaHSO3 CH3CHSO3Na Methods of preparation OH CH3MgI, H2O CH3CH(OH)CH3 C2H5OH, HCl(g) CH3CH(OC2H5)2 NH2OH CH3CH = NOH HC º CH Conc. H2SO4, 313 K NH2–NH2 CH3CH = N—NH2 NH3 CH3CH(OH)NH2 HgSO4 CH3COCl H2, Pd—BaSO4 C2H5OH [O] CH3CHO NH2NHCONH2 CH3CH = NNHCONH2 K2Cr2O7—H2SO4 CH3COOH MnO, 575 K NH2NHC6H5 CH3CH = NNHC6H5 + D HCOOH I2/NaOH (CH3COO)2Ca D CHI3 + (HCOO)2Ca Al(OC2H5)3 CH3COOC2H5 CH3C º N SnCl2/HCl OH—/H2O—H2O2 CuO CH3COOH + Cu2O H2O/H+ Fehling solution CH3MgBr + [O], Ag(NH3)2OH Tollens reagent CH3COOH + Ag HC º N 10% NaOH CH3CH(OH)CH2CHO (i) NH2—NH2, KOH CH3—CH3 or (ii) ZN—Hg, HCl HI, red P4 CH3—CH3
100 Chemistry—XII Ketones Chemical properties OH HC º N (CH3)2—CH—CN (CH3)2C—SO3Na NaHSO3 CH3MgI, H2O OH Methods of preparation CH3C(OH)CH3 Conc. H2SO4, 313 K C2H5OH, HCl(g) CH3 HgSO4 (CH3)2C(OC2H5)2 CH3C º CH NH2OH CH3C = NOH CH3COCl CH3MgBr or NH2–NH2 CH3 (CH3)2Cd NH3 (CH3)2C = N—NH2 CH3CHCH3 [O] (CH3)2CNH2 OH K2Cr2O7—H2SO4 CH3COCH3 CH2COCH3 2(CH3COOH) MnO, 575 K NH2NHCONH2 D NH2NHC6H5 (CH3)2C = NNHCONH2 I2/NaOH (CH3)2C = NNHC6H5 (CH3COO)2Ca D NaOH CHI3 + CH3COONa (CH3)2CCH2COCH3 CH3C º N CH3MgBr H2O, H+ HCl H3C OH CH3 OH CH3CHCH3 Cu, 575 K C = CH—C—CH = C H3C O CH3 CH3 Conc. H2SO4 NH2OH H3C CH3 Conc. H2SO4 (i) NH2—NH2, KOH CH3—NH—COCH3 or (ii) Zn—Hg, HCl CH3—CH2—CH3 HI, red P4 CH3—CH2—CH3 Carboxylic acids. Carboxylic acids are prepared by the oxidation of primary alcohols, aldehydes and alkenes by hydrolysis of nitriles, and by treatment of Grignard reagents with carbon dioxide. Aromatic carboxylic acids are also prepared by side-chain oxidation of alkyl benzenes. Physical properties: Boiling points. Due to more extensive association of molecules of carboxylic acid through intermolecular H-bonding, they have higher boiling points than aldehydes, ketones and alcohols of comparable molecular masses. Solubility. The solubility of carboxylic acid in water decreases with increase in number of carbon atoms. Acidic nature. Carboxylic acids are considerably more acidic than alcohols and most of simple phenols.
Aldehydes, Ketones and Carboxylic Acids 101 Chemical reactions. Carboxylic acids are reduced to primary alcohols with LiAlH4, or better with diborane in ether solution and also undergo -halogenation with Cl2 and Br2 in the presence of red phosphorus (Hell-Volhard Zelinsky reaction). Methanal, ethanal, propanone, benzaldehyde, formic acid, acetic acid and benzoic acid are highly useful compounds in industry. VERY SHORT ANSWER QUESTIONS 1. Name the reagent used to convert acetone into 2-methyl-2-propanol? Ans. Grignard’s reagent 2. What is Tollen’s reagent? Ans. Ammoniacal AgNO3 3. Write the chemical formula of Fehling’s solution. Ans. CuSO4 / NaOH / CH(OH)COONa CH(OH)COOK 4. What type of ketones will undergo iodoform test? O Ans. Ketone with CH3— C — type of grouping 5. What type of aldehydes and ketones undergo aldol condensation? Ans. Aldehydes and ketones containing an –H atom undergo aldol condensation. 6. Name the aldehyde that does not give Fehling’s solution test. Ans. Benzaldehyde 7. Why there is large difference in the boiling points of butanal and butan-1-ol? Ans. Due to H–bonding in butan-1-ol. 8. Give chemical equations when ethanal is heated with HI/P4 red under high pressure. Ans. CH3CHO + 4HI P442(5reKd) CH3—CH3 + H2O + 2I2 9. An organic acid X is heated with soda lime to produce ethane as the product. The acid X is? Ans. X = CH3CH2COONa 10. Name test to differentiate between pentan-2-one and pentan-3-one. Ans. Iodoform test 11. How will you change acetic acid to ethanol? Ans. CH3COOH CuCr2O4 CH3CH2OH 12. What is alternation effect? Ans. It is irregular increase in m.p/b.p of acids while increasing in number of carbon atoms in a chain from even to odd to even. 13. What do you understand by trans-esterification reaction? Ans. It is process of exchanging the organic group R of an ester with organic group of an alcohol.
102 Chemistry—XII 14. Name the two functional isomers that could be obtained from the compound with formula C3H6O2. Ans. Propanoic acid and methylethanoate. 15. What is glacial acetic acid? Ans. Undiluted acetic acid is some times called glacial acetic acid. 16. Why formic acid is more stronger than acetic acid? Ans. It does not contain +I group. 17. Why acids evolved CO2 on reaction with carbonates? Ans. Acids decompose carbonates to give CO2 18. Name a reagent used to oxidise propionic acid to -hydroxy propionic acid. Ans. H2O2 19. Benzoic acid is stronger than acetic acid, why? Ans. Due to resonance stabilisation of benzoate ion. 20. What is quick vinegar process? Ans. A dilute aq. solution of C2H5OH is oxidised in presence of enzyme microderma aceti to acetic acid. 21. Convert benzoyl chloride to benzoic acid. O Ans. CH3— CCl H2O/H+ CH3COOH 22. How is acidity constant, Ka expressed? Ans. Ka = RCOO ][H ] [RCOOH] 23. Name the product obtained by reaction of sodium acetate with acetyl chloride. CH3CO O Ans. Acetic anhydride CH3CO CONH2 24. Write IUPAC name of CH3— CH—CH2CH2OH . Ans. 4-Hydroxy-2-methyl butanamide. 25. What happens when ethanol is treated with methyl magnesium iodide and hydrolysed? OC2H5 Ans. CH3CH2OH + CH3MgI ¾H¾2O¾® CH4 + Mg I 26. How can acetaldehyde be obtained from acetic acid? Ans. CH3—COOH (i(ii))HPC2/lP5d/BaSO4 CH3CHO
Aldehydes, Ketones and Carboxylic Acids 103 27. Draw the structural formula of 1-phenyl propan-1-one molecule. O Ans. —CH2—C—CH3 28. Name chemical test to distinguish between ethanal and propanal. Ans. Tollen’s reagent test 29. Convert propanone to 2-methyl propan-2-ol. O OH Ans. CH3—C—CH3 ¾(i)¾C¾H3M¾g®+I CH3—C—CH3 (ii) H2O/H CH3 30. Write the structure of p-methylbenzaldehyde. CHO Ans. CH3 31. Name the reaction and the reagent used for the conversion of acid chlorides to the corresponding aldehydes. Ans. Name: Rosenmund’s reaction Reagent: H2 in the presence of Pd (supported over BaSO4) and partially poisoned by addition of Sulphur or quinoline. 32. What type of aldehydes undergo Cannizzaro reaction? Ans. Aromatic and aliphatic aldehydes which do not contain a hydrogens. 33. What makes acetic acid a stronger acid than phenol? Ans. Greater resonance stabilization of acetate ion over phenoxide ion. 34. Give the composition of Fehling A and Fehling B? Ans. Fehling A = aq. CuSO4 Fehling B = alkaline sodium potassium tartarate (Rochelle Salt) 35. Why does methanal undergoes Cannizzaro’s reaction? Ans. Because it does not possesses a hydrogen atom. Only those aldehydes can undergo Cannizzaro reaction which do not possess hydrogen atoms. 36. Why does solubility decreases with increasing molecular mass in carboxylic acid? Ans. It is because of increase in alkyl chain length which is hydrophobic in nature. 37. Give isomers of the compound with formula C2H4O2. O Ans. CH3COOH, HCOOCH3, HO—CH2— C— H. 38. Name the kind of isomerism that can be shown by an acid with formula C4H8O2. Ans. Functional, chain.
104 Chemistry—XII 39. Which bond C—OH or CO—H of carboxylic acids is broken when: (i) Acid chloride is formed (ii) Acid reacts with alcohol (iii) Acid reacts with Na. Ans. (i) C—OH (ii) C—OH (iii) CO—H. 40. Which acid is formed when p-bromotoluene is oxidised? Ans. 4-bromobenzoic acid. 41. Which out of -nitro butyric acid and -nitro butyric acid is a more stronger acid? Ans. -nitrobutyric acid. 42. Do aliphatic aldehyde show position isomerism? Ans. No. Because —CHO group always present at the end of the carbon chain. 43. Why benzoic acid less soluble in water than acetic acid? Ans. Due to larger hydrocarbon part in benzoic acid. 44. Mention a use of formalin in industry. Ans. Formalin is used as disinfectant and preservative for biological samples. 45. Give an example of a compound in which hydrogen-bonding results in the formation of a dimer. OHO Ans. Acetic acid, H3C—C C—CH3 (Intermolecular hydrogen bonding) OHO 46. Why formic acid is stronger acid than acetic acid? Ans. In acetic acid, the electron releasing —CH3 group increases the negative charge on the carboxyl group, making loss of proton more difficult. 47. Why ketones cannot be obtained by the reaction of acid chloride with Grignard’s reagent? Ans. It is because the ketone obtained would further react with Grignard reagent to form tert-alcohol. 48. Name two carbonyl compounds which do not produce crystalline products with sodium bisulphite. Ans. Diethyl ketone and acetophenone 49. Name the product obtained when calcium acetate is heated strongly and distilled. Ans. Acetone 50. Arrange HCHO, CH3COCH3, CH3CHO in order of increasing reactivity towards HCN. Ans. CH3COCH3 < CH3CHO < HCHO 51. Name the reagent that can be used to separate a mixture of ethyl alcohol and acetaldehyde. Ans. Sodium bisulphite 52. Acids change the colour of litmus, why? Ans. Because acids furnish a good concentration of H+ ions in solution.
Aldehydes, Ketones and Carboxylic Acids 105 53. Carboxylic acids decompose sodium bicarbonate, why? Ans. It is because carboxylic acids are stronger acids (Ka = 10–5) than carbonic acid (Ka = 10–7) SHORT ANSWER QUESTIONS (TYPE–I) 1. What product will be formed on reaction of propanal with 2-methylpropanal in the presence of NaOH? Write the name of the reaction also. O CH3 OH CH3 NaOH Ans. CH3—CH2—C—H + CH3—CH—CHO CH3—CH2—CH—CH2—CH—CHO CH3 and CH3—CH—CH—CH2CH2—CHO OH The reaction is a crossed or mixed Aldol condensation reaction. 2. Benzaldehyde can be obtained from benzal chloride. Write reactions for obtaining benzal chloride and then benzaldehyde from it. CH3 CHCl2 CH(OH)2 CHO Ans. + CH3Cl anhy. AlCl3 Cl2 aq. KOH –H2O 3. Name the electrophile produced in the reaction of benzene with benzoyl chloride in the presence of anhydrous AlCl3. Name the reaction also. Ans. C6H5COCl anhy. AlCl3 O C6H5—CÅ (Acylium ion) (Electrophile) 4. Oxidation of ketones involves carbon-carbon bond cleavage. Name the products formed on oxidation of 2, 5-dimethylhexan-3-one. CH3 O CH3 CH3 Ans. CH3—CH—C—CH2—CH—CH3 Oxidation 2 CH3—CH—COOH (isobutyric acid) 5. How will you obtain acetaldehyde from methyl magnesium iodide in one step? O OC2H5 Ans. CH3MgI + HC—OC2H5 ¾® CH3CHO + Mg I 6. Alkenes and carbonyl compounds both contain a bond but alkenes show electrophilic addition reactions whereas carbonyl compounds show nucleophilic addition reactions. Explain.
106 Chemistry—XII Ans. In alkenes C C is a source of electrons and hence electrophilic addition occurs. In carbonyl compounds due to C O group, – E effect operates and hence nucleophilic addition reactions. 7. Can Gattermann-Koch reaction be considered similar to Friedel Craft’s acylation? Discuss. Ans. Both Friedal craft’s acylation and Gattermann’s Koch reaction occur in the same manner i.e., the attack of an electrophile. In case of Friedal craft’s acylation the O electrophile CH3—CÅ is generated from CH3COCl whereas in Gattermann’s-Koch OO reaction the electrophile H—CÅ is generated from H—C—Cl (CO + HCl) 8. Formaldehyde gives Cannizaro’s reaction whereas acetaldehyde does not. Ans. Aldehydes which do not contain -hydrogen give Cannizzaro’s reaction. 9. Identify A and B NH3 Ni/H2 R2CO A B R O NH3 R NH Ni/H2 R C Ans. C CH2—NH2 R R R (A) (B) 10. How does formic acid react with Tollen’s reagent? Ans. HCOOH + 2[Ag(NH3)2] OH CO2 + 2H2O + 2Ag + 4NH3 11. Arrange the following in decreasing order of their acidic strength. CH3CH2OH, CH3COOH, ClCH2COOH, FCH2COOH, C6H5CH2COOH. Ans. FCH2COOH > ClCH2COOH > C6H5CH2COOH > CH3COOH > CH3CH2OH 12. Arrange in the order of ease of hydrolysis CH3COOC2H5, CH3COCl, (CH3CO)2O, CH3CONH2. Ans. CH3CONH2 < CH3COOC2H5 < (CH3CO)2O < CH3COCl 13. An acid on reduction with HI gave propane. What will be the products if sodium salt of the same acid is subjected to (i) soda lime decarboxylation (ii) Kolbe’s electrolytical decarboxylation? Ans. The acid is CH3CH2COOH (i) CH3CH3 is produced (ii) C2H5—C2H5 is produced 14. Compound ‘A’ was prepared by oxidation of compound ‘B’ with alkaline KMnO4. Compound ‘A’ on reduction with lithium aluminium hydride gets converted back to compound ‘B’. When compound ‘A’ is heated with compound B in the presence of H2SO4 it produces fruity smell of compound C. To which family the compounds ‘A’, ‘B’ and ‘C’ belong to? Ans. A = Aldehyde; B = Alcohol; C = Ester
Aldehydes, Ketones and Carboxylic Acids 107 15. Compare the strength of following acids: (i) Formic acid (ii) Acetic acid (iii) Benzoic acid. Ans. HCOOH > C6H5COOH > CH3COOH 16. Carboxylic acids contain carbonyl group but do not show the nucleophilic addition reaction like aldehydes or ketones. Why? Ans. It is because of resonance, the CO group of carboxylic acids do not have a complete C O character. 17. Identify the compounds A, B and C in the following reactions: (i) CH3—Br Mg/ether (A) (i) CO2 (B) CH3OH/H (C) (ii) Water O (i) MeMgBr Na metal Me-Br (ii) CH3—C—CH3 (A) (B) (C) (ii) H2O Et2O Ans. (i) A = CH3MgBr; B = CH3CO2MgBr; C = CH3COOH CH3 –+ OH ONa (ii) A = CH3—C—CH3 ; B = CH3—C—CH3 ; C = CH3—C—OMe Me Me Me 18. Suggest a reason for the large difference in the boiling points of butanol and butanal, although they have same solubility in water. Ans. The boiling point of butanol is higher than that of butanal because butanol has strong intermolecular H-bonding while butanal has weak dipole-dipole interaction. However both of them form H-bonds with water and hence are soluble. 19. Give Fehling solution test for identification of aldehyde group (only equations). Name the aldehyde which does not give Fehling’s solution test. Ans. R—CHO + 2 Cu2+ + 5OH– RCOO– + Cu2O + 3H2O Benzaldehyde does not give Fehling solution test. (Aromatic aldehydes do not give this test.) 20. Why HCOOH does not give HVZ (Hell Volhard Zelinsky) reaction but CH3COOH does? Ans. CH3COOH contains a hydrogens and hence give HVZ reaction but HCOOH does not contain -hydrogen and hence does not give HVZ reaction. 21. How will you distinguish between methanol and ethanol? Ans. By Iodoform test: Ethanol having -methyl group will give yellow ppt. of iodoform whereas methanol does not have -methyl group will not give ppt. of iodoform. 22. Why PCC cannot oxidize methanol to methane and while KMnO4 can? Ans. This is because pcc is a mild oxidizing agent and can oxide methanol to methanal only. While KMnO4 being strong oxidizing agent oxidizes it to methanoic acid.
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