EG_Chemistry-12_(22-07-2022) a

activation energy barrier. Such reactants though would occur at the same temperature without the energetically (what is termed thermodynamically) presence of the catalyst. In this way the presence unstable but are kinetically stable at ordinary of a catalyst makes the reaction go faster, keeping temperature. other conditions remaining the same. • When potential of the system undergoing reaction Activated is plotted against the reaction coordinate (the Complex progress of the reaction), then the difference in the potential energy between the reactants and Potential Energy Ea Activated the activated complex is the activation energy, Ea. Ecat Complex Involving Catalyst • The reactants need to have this minimum energy to undergo the reaction through the transition Reactants DH state. Products Activation Energy and Catalysis Reaction Coordinate • The minimum amount of energy required to bring about a chemical reaction is known as the Collision Theory of Chemical Reactions activation energy of the reaction. The activated molecules on collision first forms an activated • Collision frequency: The number of collisions complex. As a result of breaking and formation per second per unit volume of the reaction mixture of new bonds, the activated complex dissociates to yield the product molecules. is collision frequency. • A catalyst can lower the activation energy • According to collision theory; rate =Zp—ZcAoBl e–Ea/RT, of a reaction by providing a new pathway where p—probability factor; lision (mechanism). frequency of reactants A and B, e–Ea/RT represents • Thus, larger number of effective collisions occuring in the presence of a catalyst than it the fraction of molecules with energy equal to or greater than Ea. ExERCisE 3.4 Multiple Choice Questions (MCQs) (1 Mark) (iii) Activation energy of both forward and 1. The role of a catalyst is to change .................... . mbaocrkewsatardblreetahcatinonpriosdEu1c+t.E2 and reactant is (i) gibbs energy of reaction. (ii) enthalpy of reaction. (iv) Activation energy of bstaacbklwe atrhdanreraecaticotnanist.E1 (iii) activation energy of reaction. and product is more (iv) equilibrium constant 2. Activation energy of a chemical reaction can be 4. Consider the Arrhenius equation given below and determined by .................... . mark the correct option. (i) determining the rate constant at standard temperature. k = Ae–Ea/RT (ii) determining the rate constants at two temperatures. (i) Rate constant increases exponentially with (iii) determining probability of collision. increasing activation energy and decreasing (iv) using catalyst. temperature. 3. Consider the figure and mark the correct option. (ii) Rate constant decreases exponentially with Activated complex increasing activation energy and decreasing temperature. E 1 (iii) Rate constant increases exponentially with decreasing activation energy and decreasing Products temperature. E (iv) Rate constant increases exponentially with 2 decreasing activation energy and increasing temperature. 5. Which of the following graphs represents exothermic reaction? Energy Energy Activated complex Reactants Reaction coordinate (i) Activation energy of forward trheaacntiroenacistaEn1t+. (a) E2 and product is less stable Reactants (ii) Activation energy of forward rtehaacntiroenaicstaEn1t+. Products E2 and product is more stable Reaction coordinate 100 Chemistry-12

Activated complex down the activation energy but the different in energy of reactant and product remains the same. (b) Energy 9. Assertion: All collision of reactant molecules lead Products to product formation. Reactants Reason: Only those collisions in which molecules Reaction coordinate have correct orientation and sufficient kinetic energy lead to compound formation. Activated complex 10. Assertion: Rate constants determined from (c) Energy Arrhenius equation are fairly accurate for simple as well as complex molecules. Reactants Products Reason: Reactant molecules undergo chemical Reaction coordinate change irrespective of their orientation during collision. (i) (a) only (ii) (b) only Very Short Answer Type Questions (1 Mark) (iii) (c) only (iv) (a) and (b) 11. For a reaction,the energy of activation is zero. 6. Which of the following statements is incorrect What is the value of rate constant at 300 K if k = about the collison theory of chemical reaction? 1.6 × 106 sec–1 at 280 K? (i) It considers reacting molecules or atoms to 12. fTehaesibreleacyteiotnalbloewtwinegenthHe2g(ga)seasndtoOst2(agn)disathrigohomly be hard spheres and ignores their structural temperature in the same vessel, does not lead to the formation of water. Explain? features. (ii) Number of effective collisions determines the [NCERT Exemplar] rate of reaction. (iii) Collision of atoms or molecules possessing Or sufficient threshold energy results into the eHx2o(gt)hearnmd iOc 2r(ge)acctoimonb.inTehs etno, form Hd2oOt(hl)ebyynaont product formation. why (iv) Molecules should collide with sufficient combine when kept together for any period of threshold energy and proper orientation for time? the collision to be effective. 13. Oxygen is available in plenty in air yet fuels do 7. Which of the following statement is not correct not burn by themselves at room temperature. for the catalyst? Explain. [NCERT Exemplar] (i) It catalyses the forward and backward 14. Why in thheeatreodxoaxlictiatrcaidtiosnoluotfioKnMbneOfo4rev/sstaorxtainligc reaction to the same extent. acid, we (ii) It alters DG of the reaction. the titration? [NCERT Exemplar] (iii) It is a substance that does not change the 15. Define temperature coefficient of a reaction. equilibrium constant of a reaction. 16. What is activated complex? 17. Define activation energy of a reaction. (iv) It provides an alternate mechanism by reducing activation energy between reactants [AI 2012, 11, 10, 09] and products 18. What is the relation between rate constant and activation energy? Assertion-Reason Type Questions (1 Mark) 19. Can activation energy of a reaction be zero? Note: In the following questions a statement of assertion followed by a statement of reason is given. 20. According to Arrhenius hypothesis ‘rate of a Choose the correct answer out of the following choices. reaction increases with increase in temperature.’ Give reason. (i) Both assertion and reason are correct and the reason is correct explanation of assertion. 21. A chemical reaction is exothermic in nature. Will it become faster or slower if temperature is (ii) Both assertion and reason are correct but reason increased? does not explain assertion. 22. Write the Arrhenius equation for the rate (iii) Assertion is correct but reason is incorrect. constants ikn1 taenrdmks2ooff a reaction at temperatures T1 and T2 activation energy. (iv) Assertion is incorrect but reason is correct. 8. Assertion: The enthalpy of reaction remains 23. If in the Arrhenius equation, T → ∞, what is constant in the presence of a catalyst. the value of rate constant? Reason: A catalyst participating in the reaction, forms different activated complex and lowers 24. Write the form of Arrhenius equation which relates the rate constant with temperature. ChemiCal KinetiCs 101

Short Answer Type Questions-I (2 Marks) 36. With the help of diagram explain the role of activated complex in a reaction. [Delhi 2013(C)] 25. The slope of line in the graph of log k v/s 1 T Short Answer Type Questions-II (3 Marks) for a reaction is 5841. Calculate the energy of 37. Rate constant, k of a reaction varies with temperature according to the equation activation. Ea 26. For raatreeacocntisotna,nEtsa = 80 kJ. What will be ratio of log k = Constant – 2.303 RT , the when temperature varies from 273 K to 323 K? Ea = activation energy. 27. The activation energy of a reaction is 75.2 kJ mol–1 When a graph is plotted for log k v/s  1  , a in the absence of a catalyst and 50.14 kJ mol–1  T  with a catalyst. How many times will the rate of reaction will be raised in the presence of catalyst straight line curve with slope equal to – 6670 is if the reaction proceeds at 25°C? (R = 8.314 JK–1 obtained. Calculate the energy of activation for mol–1) this reaction. (R = 8.314 JK–1 mol–1). 28. Rate of reaction increases four times when 38. All energetically effective collisions do not result the temperature changes from 300 K to in a chemical change. Explain with the help of an 320 K. Calculate the energy of activation of the example. [NCERT Exemplar] reaction assuming that it does not change with temperature (R = 8.314 JK–1 mol–1). 39. Describe how does the enthalpy of reaction remain unchanged when a catalyst is used in the reaction. [NCERT Exemplar] [Delhi 2010, Foreign 2008] 40. State the following: (i) Law of mass action 29. For a certain reaction large fraction of molecules has energy more than threshold energy, yet the (ii) Average rate of reaction rate of reaction is very slow. Why? (iii) Instantaneous rate of reaction [NCERT Exemplar] [Delhi 2013 C] (iv) Threshold energy 30. Why does the rate of reaction increase with rise (v) Activation energy in temperature? (vi) Temperature coefficient [NCERT Exemplar] 41. For a chemical reaction the energy of activation 31. What is Arrhenius equation? What is its is 85 kJ mol–1. If the frequency factor is 4.0 × 109 L mol–1 s–1, what is the rate constant at 400 K? importance? Long Answer Type Questions (5 Marks) 32. The value of rate constant of a reaction at 25°C and 35°C are respectively 3 × 10–6 atm–1 sec–1 and 42. The rate constant of a reaction is 0.01439 min–1 7.5 × 10–6 atm–1 sec–1. What is the order of reaction at 25°C and its activation energy is 70000 J mol–1. and what is the value of temperature coefficient? What is the value of rate constant at 40°C? 33. Which of the reaction in each pair listed below 43. The rate constants of a reaction at 700 K and will be faster? 760 K are 0.011 min–1 and 0.105 min–1 respectively. Calculate the values of Arrhenius parameter. 1 (i) (a) CO(g) + 2 O2(g) → CO2(g) at 100°C 44. A certain reaction is 50% completed in 20 minutes at 300 K and same reaction again get 50% (b) CO(g) + 1 O2(g) → CO2(g) at 1000°C completed in 5 minutes at 50K. Calculate the 2 activation energy if it is a first order reaction. (ii) (a) Na2SO4(s) + BaCl2(s) → 45. (i) Graphically represent the variation in BaSO4(s) + 2NaCl(s) at 100°C the concentration v/s time for a zero order reaction. (b) Na2SO4(aq) + BaCl2(aq) → BaSO4(s) + 2NaCl(aq) at 25°C (ii) A first order reaction takes 40 minutes for 30% decomposition. Calculat[eDte1l/h2 io2f0t0h8e] (iii) (a) N2 (10 atm) + 3H2 (10 atm) → 2NH3(g) reaction. (b) N2(10 atm) + 3H2 (5 atm) → 2NH3(g) (iii) Draw the distribution curve showing (iv) (a) 2K(s) + I2(s) → 2KI(s) temperature dependence on the rate of a (b) 2K(s) + I2(g) → 2KI(s) reaction. 34. State the role of activated complex in a reaction (iv) The activation energy of first order reaction and state its relation with activation energy. at 300 K is 60 kJ/mol. In the presence of a catalyst, the activation energy gets lowered 35. Write Arrhenius equation showing the effect of to 50 kJ/mol at 300 K. How many times the temperature on the reaction rate. How does it help reaction rate changes in the presence of a in calculating the activation energy of a reaction? catalyst at the same temperature? [NCERT] 102 Chemistry-12

Answers 3.4 1. (iii) 2. (ii) 3. (i) When log k is plotted against  1  we get a straight  T  4. (iv) 5. (i) 6. (iii) 7. (ii) 8. (i) 9. (iv) − Ea  2.303R  10. (iii) line curve with slope of 11. When activation energy of the reaction is zero, then rate will be independent of temperature, Ea \\ − 2.303R = 5841 \\ k (at 280 K) = 1.6 × 106 sec–1 = k (at 300K) Ea = – 5841 × 2.303 × 8.314 J mol–1 \\ k 300 K = 1.6 × 106 sec–1 = 1.118 × 105 J mol–1. 12. 2H2(g) + O2(g) → 2H2O(l) 26. log  k2  = Ea 1 − 1 The reaction does not take place at room  k1  2.303R   temperature because the activation energy of    T1 T2  reaction is very high. 13. Combustion of fuels require very high activation log k323 = 80 × 1000 1 − 1 energy, therefore they do not burn at room k273 2.303 × 8.314  273 323  temperature. = 2.3691 14. sRleoawc,ttihonerbeefotrweereenacKtiMonnmO4ixatnudreoixsailnicitiaaclildy is very heated k323 = Antilog (2.3691) = 233.94 k273 to speed up the process. 15. It is the ratio of the rate constant at two 27. Activation energy is lowered by the catalyst. So temperatures separated by 10°C (generally at 25°C and 35°C), i.e. DE = (75.2 – 50.14) = 25.06 kJ mol–1 m= k35°C . = 25.06 × 103 J mol–1 k25°C We know 16. It is believed that when reactant molecules kp = Antilog  ∆E  absorb energy, their bonds are weakened and new ka  2.303RT  arrangement of atoms depicting bond formation  25.06 × 103  takes place. This intermediate, thus formed is = Antilog     called the activated complex. 2.303 × 8.314 × 298 17. The minimum extra amount of energy absorbed = 24547.09 by the reactant molecules so that their energy becomes equal to the threshold value for the 28. log10 k2 = Ea 1 − 1 reaction to take place is called the activation k1 2.303R   energy.  T1 T2  Activation energy = Threshold energy– Average log10 (4) = Ea 1 − 1 energy of the reactant molecules. 2.303 × 8.314  300 320  18. k = Ae–Ea/RT 19. No Ea = 55327 J mol–1 = 55.327 kJ mol–1 29. Reactant molecules having energy greater than 20. The increase in the rate of reaction with increase the threshold energy may be converted into in temperature is mainly due to increase in products only when the collision among reactant the number of effective collisions between the molecules is in proper orientation. The rate may reactant molecules. be slow due to improper collisions among reactant 21. Rate of chemical reaction becomes faster. molecules. k2 Ea 22. log  k1  = 2.303R  1 − 1 30. When temperature of the reaction mixture is    T1  raised, the rate of reaction increases due to  T2  increase in collision frequency of the reaction, i.e. 23. k ≈ A (collision frequency) greater number of reactant molecules crossing the 24. k = Ae–Ea/RT energy barrier of the reaction. 25. log10 k = log10 A – Ea × 1 31. k = Ae–Ea/RT; it is used to calculate the relative rate 2.303R T and rate constants at different temperature. It It is an equation of type: explains the dependence of reaction rate on the y=C+Mx temperature. ChemiCal KinetiCs 103

32. Second order; Temp. coeff. = 7.5 # 10–6 = 2.5. 39. Enthalpy of a reaction may be given as 3 # 10–6 DH = (Ef – Eb) fWorhweraerdEfaannddbEacb kawreartdher activation energy of 33. (i) (b), (ii) (a), (iii) (a), (iv) (b) eactions. Activation 34. Activated complex is the unstable complex formed at the maxima of the energy barrier. It represents energy of forward and backward reactions both are equally affected by the catalyst, hence the energy barrier that reactant molecules must enthalpy of reaction remains unaffected. overcome in order form the products. 40. (i) Law of Mass Action states that the rate of a Activation energy of forward reaction chemical reaction is directly proportional to = Energy of activated complex the concentration of the reacting species. – energy of the reactant molecules (ii) Average rate of reaction is the rate of decrease 35. Arrhenius Equation: k = Ae−Ea /RT in the concentration of reactants/increase in Activation energy of the reaction can be calculated the concentration of products per unit time by using the following equation: interval. log  k2  = Ea  1 − 1 (iii) Instantaneous rate of reaction is the rate  k1  2.303R  T1  of reaction at a particular moment instant  T2  of time and measured as a very small 36. Activated complex is the intermediate compound concentration change with a very small time between reactants and products in which bonds are half broken and half formed. It is highly interval (Dt → 0). unstable and readily changes into products e.g. (iv) Threshold energy is the minimum energy that reactant molecules must posses in order H—H HH H to undergo effective collisions to form the + 2½ product. II I—I Activated complex I (v) Activation energy is the energy required by reactants molecules to form the intermediate (i.e. activated complex) during the course of (E ) (E ) reaction. af ab Energy (vi) Temperature coefficient is the ratio of specific rate constant for a reaction separated by 10°C (generally 25°C and 35°C). ∆H 41. We know, Progress of reaction log10k = log10 A − Ea 2.303RT 37. As we know, Given slope, m = – 6670 85000 = log(4 × 109) − 2.303 × 8.314 400 R = 8.314 JK–1mol–1 × Using tEhae=fo?rmula, = 9.602 – 11.0982 = – 1.4962 Ea k = 0.0319 L mol–1 s–1 2.303 RT log k = Constant – k = 3.19 × 10–2 L mol–1 s–1 On comparing with equation 42. log  k2  = Ea 1 − 1  k1  2.303R   y = mx + C, we get – Ea = – 6670  T1 T2  2.303 R or Ea = 127.711 kJ mol–1 k1 = 0.01439 min–1, k2 = ?, 38. Not only the energetically effective collisions but T1 = 298 K, T2 = 313 K also the collisions in proper orientation results Ea = 70000 J mol–1 into a chemical change. By putting these value in the above equation and taking log Example: N2O + CO → N2 + CO2 k2 70000 1 1 N + C O → No product log 0.01439 = 2.303 × 8.314  298 − 313 N O (Oxygen opposite to CO) = 1050000 2.303 × 8.314 × 298 × 313 N+ N C O → N + O C O k2 = 0.0557 min–1 O N 43. k1 = 0.011 min–1, (Oxygen near to CO) k2 = 0.105 min–1, 104 Chemistry-12

T1 = 700 K, T2 = 760 K (ii) k= 2.303 log10  1  By substituting the values in the equation 40 × 60  0.7  log  k2  = Ea  1 − 1 = 1.486× 10–4 s–1  k1  2.303R  T1     T2  t1/2 = 0.693 1.486 × 10−4 We get log 0.105 = Ea 1 − 1 = 4.66 × 103 seconds 0.011 2.303 × 8.314  700 760  (iii) For most of the reactions, the rate of reaction E = 166341 J mol–1 becomes nearly double or even more for 10°C rise in temperature. = 166.34 kJ mol–1 At a particular temperature, T all the As we know, reacting molecules do not have the same kinetic energy. The Maxwell’s distribution log k = log A − Ea curve for fraction of molecules and their 2.303 RT kinetic energies shows that above the peak of the graph, which shows the K.E. possessed by log 0.011 – log A = 166341 maximum fraction of molecules and is called 2.303 × 8.314 × 700 the most probable kinetic energy. – 1.958 = log A – 12.41 A = 2.83 × 1010 At higher temperature the curve shifts to the right indicating that the fraction of molecules 44. At 300 K: k1 = 0.693 min−1 having energy more than the threshold 20 energy have become double with every 10°C rise in temperature. At 350 K: k2 = 0.693 min−1 5 This rise in the rate of reaction with rise in We know temperature is due to increase in the number of effective collisions. log k2 = Ea  1 − 1 k1 2.303R  T1   T2  log  0.693 / 5  = Ea 1 − 1 Fraction of molecules t This area shows fraction  0.693 / 20  2.303 × 8.314  300 350  (t + 10) of additional molecules which react at (t + 10) log 4 = Ea  50  2.303 × 8.314 300 × 350  Energy of activation Ea = 0.602 × 2.303 × 8.314 × 300 × 350 This area shows 50 fraction of molecules reacting at t Kinetic energy = 24205.8 J mol–1 Show distribution curve showing temperature dependence on a rate of a reaction. = 24.21 kJ mol–1 50 45. (i) (iv) log K′ =log A – 2.303 RT log K = log A – 60 2.303 RT log K′ = 10 × 1000 K 2.303 × 8.314 × 300 Variation in the concentration vs time plot for a K′ = Antilog 1.7409 zero order reaction K Case based questions = 5.507 × 10 = 55 times 1. Case Study The saponification of poly(ethylene terephthalate) kinetics. An online conductivity measurement is (PET) is represented as an example of pseudo carried out to evaluate the second order reaction first order in the literature dealing with chemical rate constants with possibly lowest experimental error for saponification of PET waste powder ChemiCal KinetiCs 105

in order to obtain information about activation (c) It is the extra energy required by reactants energy, activation enthalpy, activation entropy to cross the barrier and form the products. and free energy of activation. Saponification of PET waste powder was carried out using various (d) It is the maximum energy possessed by amounts of PET waste powder, sodium hydroxide the transition state in a reaction. and particle size. The saponification was also OR studied at different temperatures. Conductometric kinetics of saponification of PET waste powder By altering what conditions a reaction of was undertaken on the basis of the hydroxide ion higher order can behaves as first order and terephthalate formed in the reaction product. reaction? With progress of reaction, highly conducting (a) By decreasing the concentration of all OH− ions in the reaction mixture were replaced by an identical number of very less conducting reactants to minimum terephthalate ions, resulting in continuous (b) By taking very high concentration of all decrease in conductivity of the reaction mixture. From the start of the reaction, the decrease in reactants. conductivity was continuously monitored online (c) By adding a suitable solvent in the using conductivity meter. In view of simplicity in experimental arrangement and measurement reaction. technique, the conductivity seems to be better (d) By taking concentration of one reactant in technique for this kinetic investigation. Early investigators found the reaction as first order by normal range while others in such a high different technique. The conductometric study concentration that there concentration do shows the second order kinetics since both the not change practically during the reaction. reactants were consumed in the reaction.The (iii) How the rate constant changes with increase reaction rate constant had also led to evaluating in temperature? thermodynamic parameter for this saponification (a) Nearly 10 times with increasing the reaction. The reported value on activation energy temperature two times. is lower and more precise than the value obtained (b) Nearly 2 times with increasing the by gravimetric technique. temperature two times. (c) Nearly 2 times with increasing the (Reference: Dilip B. Patil, Vijendra Batra, temperature with 10°C and Sushil B. Kapoor. Kinetic Studies on (d) Nearly 10 times with increasing the Saponification of Poly(ethylene terephthalate) temperature with 10°C Waste Powder Using Conductivity Measurements. (iv) The saponification reaction studied in the Journal of Polymers, Volume 2014, Article ID above passage was thought to follow first 321560, 7 pages) order kinetics but the conductometric study shows that the reaction follows– The following questions are multiple choice (a) Zero order kinetics questions. Choose the most appropriate (b) Second order kinetics answer: (c) Third order kinetics (d) Fourth order kinetics (i) In the saponification reaction of PET waste 2. Case Study powder, why the conductivity decreases with The galvanic replacement method was applied for time? the synthesis of gold nanotubes from sacrificial silver nanorods. The progress of the replacement (a) OH– ions is replaced with terephthalate reaction was monitored by obtaining UV-Vis ions. spectra and by measuring the silver and gold ion concentrations. The replacement reaction (b) Terephthalate ions is replaced with OH– was promoted by increased gold precursor ions. concentration as well as by increased reaction temperature. The dissolved silver concentration (c) OH– ions combine with Na+ ions to form increased linearly with the gold ion concentration. NaOH The replacement reaction was promoted with increased temperature, which enhances the (d) OH– ions is replaced with smaller H+ ions reaction rate constant. The reaction rate depends on the gold precursor concentration nonlinearly. (ii) Which of the following statement is correct This study determined kinetic information about regarding activation energy? gold replacement with sacrificial silver nanorod templates through the galvanic replacement (a) It is the energy possessed by the reactants reaction. as potential energy. (b) It is the energy possessed by the products of the reaction. 106 Chemistry-12

(Reference: : Sunil Kwon, Hyunbae Dong, and (ii) Assertion: The reaction rate depends on the Sang-Yup Lee . Study of the Reaction Rate of gold precursor concentration linearly. Gold Nanotube Synthesis from Sacrificial Silver Nanorods through the Galvanic Replacement Reason: The dissolved silver ion Method . Journal of Nanomaterials Volume 2010, concentration increased linearly with gold Article ID 819279, 7 pages) In these questions, a statement of assertion ion concentration. followed by a statement of reason is given. Choose the correct answer out of the (iii) Assertion: Arrhenius equation, k = Ae–Ea/RT following choices. shows the temperature dependence of rate (a) Assertion and reason both are correct constant. statements and reason is correct explanation for assertion. Reason: The graph between log k and 1 −Ea T (b) Assertion and reason both are correct 2.303 R statements but reason is not correct has slope of value explanation for assertion. (iv) Assertion: According to collision theory, rate (c) Assertion is correct statement but reason is is equal to PZABe–Ea/RT wrong statement. Reason: The factor e–Ea/RT represents the fraction of molecules with energy less than Ea. (d) Assertion is wrong statement but reason is OR correct statement. Assertion : Rate of an irrevesible reaction (i) Assertion: Rate of reaction increases with depends on concentration of products. increase in temperature. Reason: Rate of a reaction depends on Reason: Kinetic energy of molecules and presence of catalyst. effective collisions between molecules increases on increasing temperature. Answers 1. (i) (a) (ii) (c) OR (d) (iii) (c) (iv) (b) 2. (i) (a) (ii) (d) (iii) (b) (iv) (c) OR (d) Analogy based questions (ii) t½ for zero order : [R]° :: t½ for I order : 0.693 1. Complete the following analogy: 2k k time–1 : x : : mol lit –1 time–1 : y (i) x : unit of zero order, y : unit of first order (iii) Order : Found by stoichiometry :: Molecularity (ii) x : unit of zero order, y : unit of pseudo first order : Found experimentally (iii) x : unit of first order, y : unit of zero order (iv) Average rate : dx :: Instantaneous rate : ∆x (iv) x : unit of second order, y : unit of first order dt ∆t 2. Which of the following analogy is correct? Answers (i) Order of reaction : Cannot be fractional :: Molecularity : Can be fractional 1. (iii) 2. (ii) Multiple type questions Column II 1. Match the items of Column I and Column II. (1) rate constant Column I (2) rate law A. Mathematical expression for rate of reaction (3) order of slowest step B. Rate of reaction for zero order reaction is equal to (4) rate of a reaction C. Units of rate constant for zero order reaction is same as that of D. Order of a complex reaction is determined by ChemiCal KinetiCs 107

Code: (iv) A (2) B (4) C (1) D (3) (i) A (4) B (3) C (2) D (1) Answers (ii) A (2) B (1) C (4) D (3) B (4) C (1) D (2) 1. (ii) (iii) A (3) Quick revision notes • Chemical kinetics constitutes an important role in physical chemistry. It concerns with the Order of Value Unit of rate constant measurement of rate of reactions proceeding Reaction of n under given conditions of temperature, pressure (Conc)1 time–1, i.e., mole lit–1 time–1 and concentration of reactants (or products 0 0 (Conc)0 time–1, i.e., time–1 formed). 1 1 (Conc)–1 time–1, i.e., lit. mol–1 time–1 2 2 (Conc)–2 time–1, i.e., lit2 mol–2 time–1 • A chemical reaction involves breaking of bonds of 3 3 reactant molecules and formation of new bonds forming product molecules. Since the number The zero order rate constant has the same unit and nature of bonds are different in different as the rate of a reaction. substances, the rate of chemical reactions differ a lot from one another. • Expression for rate constant— (i) Zero order: • For a general reaction, aA + bB → cC + dD, the k= ∆x = C0 − C ∆t t rate of the reaction may be expressed in terms of wcohnecreenCtr0aitsiothneaifnteitriatilmcoen‘cte’ fnrtormatitohne, C is the concentration of reactants A or B. Thus, the rate start. A may be expressed iansdi−cdadt[teAs] tohrat−cddo[ntBc]e.ntration plot of C versus t will be a straight line with The negative sign a negative slope = –k. • of (ii) First order reaction: the reactants decreases with time. 1 C0 2.303 C0 • The unit of rate of reaction is concentration time–1 k= t ln C = t log10 C or mole lit–1 time–1. • The order of a reaction with respect to a particular A plot of ‘log C’ versus ‘t’ gives a straight line reactant is the power (or exponent) to which the concentration of the reactant is raised in the rate with a negative slope = −k expression (this is called partial order). The sum 2.303 of the order with respect to the various reactants is called the overall order or total order of the The first order rate constant can also be reaction. Thus for the reaction, aA + bB → cC + dD, if rate a [A]x [B]y, the total order of the expressed in the exponential form C = C0 e–kt. reaction is (x + y). (iii) Second order reaction: • The order of a reaction is an experimental quantity. k= 1 1 − 1  It may be zero or an integer or a fraction. t  C0   C  i.e. 1 = (k) t + 1 C C0 • Molecularity of any step of a reaction is the number of atoms, molecules or ions that participate in that Therefore, a plot of 1 versus time, t gives a step. Molecularity is the number of such species, C which forms the activated complex. Very often straight line with a positive slope = k. molecularity is assigned to the rate determining step. Thus, molecularity is a theoretical concept (iv) In general for a nth order reaction: and it depends on the mechanism that is adduced for a reaction. Molecularity always has integral k= 1 1 − 1 value.   t(n − 1)  Cn −1 C0n−1  This a general expression applicable to all • The rate constant is the rate of reaction, when the values of n, except 1. concentration of each of the reacting species are unity, since rate = k (concentration)n • Half life of a reaction: This is the time during which the concentration gets reduced to one half Rate Concentration time−1 of the initial value. (Concentration)n (Concentration)n k= = This has a special significance for the first order = (Concentration)1–n time–1 reactions. The rate of these reactions and the concentration of the reactants vary exponentially 108 Chemistry-12

with time. Hence, theoretically it will take infinite where the proportionality constant A is called the time for the reaction to go to completion (though apcrtei-veaxtpioonne. nTtaiaklinfagcltoogra. rEitahims c, alled the energy of after a finite period, the concentration becomes nearly zero). However, the half life of the reaction ln k = ln (A) –  Ea  , i.e. is finite.  RT  Consider a reaction at two different temperatures (i) For 1st order reaction, T1 and T2, t1/2 = 2.303 log10 2 = 0.693 i.e., log k2 = Ea  T2 − T1  k k k1 2.303R  T1T2    ot1f/2tihseinrdeeapcteinndgesnpteocfieths.e initial concentration 1 tewexhmperper seesrikao1nt uairsnedsiTmk12ilaaarnr deto the rate constants at (ii) For 2nd order reaction, t1/2 = C0 k Tth2 erveasnp’et ctively. This Hoff Reaction Isochore. (iii) For a nth order reaction, • Significance of energy of activation: In order t1/2 = 1 2n−1 − 1 that molecules may react, it is necessary that they k(n − 1) C0n−1 must overcome an energy barrier and for this C0 purpose the molecules must possess a minimum (iv) For a zero order reaction, t1/2 = energy. 2k • The factors which influence reaction rate are: • Catalyst: (i) Does not alter the Gibbs energy, DG of a (i) Concentration (ii) Catalyst reaction. (iii) Temperature (iv) Intensity of radiation (ii) Catalyst only alters the rate of the reaction. (v) Surface area (iii) Does not change the equilibrium constant of reaction • Temperature coefficient of reaction rate and activation energy: It is experimentally • Collision frequency refers to the number of observed that for several reactions, an increase of collisions per second per unit volume of the temperature by about 10 degrees Kelvin increases reaction mixture. therateconstanttonearlydouble,usually kT+10 = 2. kT • Whether a reaction will proceed in a particular direction and to what extent is decided by the • There is no rigidity about this ratio of 2, it may be tendency to decrease the free energy and to less or more than 2, depending on the nature of increase the disorder. There should be a decrease the reaction and the temperature range at which in free energy (DG < 0) for the reaction to be the difference of 10 K is considered. spontaneous. A spontaneous reaction may be slow or it may be fast. • Arrhenius equation: It was suggested that the rate constant k increases exponentially with • It is of great importance in selecting optimum temperature. conditions for an industrial process so that it proceeds at a rate to give maximum yield. k ∝ e–Ea/RT or k = A e–Ea/RT Important Formulae [R]0 − [R] 1. Integrated rate law for zero order reaction: k = t 2. Integrated rate law for Ist order reaction: k = 2.303 log [R] 0 t [R] [R] 0 3. t1/2 for zero order reaction: = 2k 4. t1/2 for Ist order reaction: = 0.693 k 5. Arrhenius equation: k = Ae−Ea/RT Ea ln k = ln A – RT TRSSSSSSSSSSsfolorptehoefplliontebet2w.3ee0En3aRlog WWWWWWWWWVXW log k = n = logA – Ea 1 2.303 RT T k and log k2 Ea 1 − 1 k1 2.303 R d T1 T2 6. For first order reaction: t75% = 1.386 t60% = 0.9165 t99% = 4.606 t90% = 2.303 t99.9% = 6.909 k k k k k ChemiCal KinetiCs 109

Common errors errors CorreCtIons (i) Students often write wrong units for rate (i) The general formula (mol L–1)1–n s–1 can be given constant. to write units for rate constant of any order (n is the order of the reaction) (ii) Student often confused between order and (ii) The order can only be found by experimentation, molecularity. whereas molecularity can be identified by the stoichiometric coefficients. (iii) Students often commit mistakes in taking right (iii) Writing of proper integrated rate equations can concentration terms. avoid this error. In the first order reactions, [R] in the denominator is the conc. of the reactant present at that given time. (iv) Students often take temperature in Celsius scale (iv) In Arrhenius equation, temperature must always instead of Kelvin scale. be taken in Kelvin Scale. 110 Chemistry-12

revIsIon Chart Average rate = MeCHanisM of reaCtion: The rate Constant is equal to rate of exact path followed by reac- reaction when molar conc. of the tants to undergo the reaction reactants is unity. and form products. Zero order: The rate does not Rate of reaction is the Instantaneous rate = dx change in conc. of reactants depend upon concentration of dt or products per unit time. reactant Sun MOLECULARITY is the no. of light molecules taking part in an H2(g) + Cl2(g) 2HCl(g) elementary reaction. order is sum of powers first order: The sum of power to to which conc. terms are which concentration of reactants raised in the rate law. are raised in rate law is equal to unity, e.g., 1 O2 N2O5 Æ 2NO2 + 2 DH = (Ea)f – (Ea)b Ea = Threshold energy – DH nature of reaCtants: Rate depends upon the reactivity ARRHENIUS EQUATION factorS affecting rate of reactants. k = Ae–Ea/RT of reaction PHysiCal states: Gases react CATALYST: It increases the rate faster than liquids and solids. of reaction by lowering down the activation energy. It does not ConCentration: Rate of reac- affect DH, DG, DS and K. tion increases with increase in concentration of reactants. teMPerature: Rate of reac- tion generally increases with increase in temperature. Rate = ZAB e–Ea/RT effeCtive Collisions: The collisions between reactant molecules having energy ≥ Ea and Proper orientation of reactant collides in proper orientation leads to the molecules leads to bond forma- formation of product molecules. tion. colliSion tHeory NE is the fraction of molecules Nt possessing activation energy. It is equal to e−Ea /RT ChemiCal KinetiCs 111

Chapter trend—Based on past Years’ CBSe exams ➣ It has been observed from this chapter that the weightage of topics ‘Integrated Rate Equations and Pseudo First Order Reactions’ (especially, numericals based on I order reaction using formulae, k = 2.303 log C0 and t1/2 = 0.693 ) t C k and ‘Factors Influencing Rate of a Reaction’ are maximum. Hence, these are most important topics. ➣ From this chapter , generally 3 marks questions were asked from the topic ‘Integrated Rate Equations and Pseudo First Order Reactions’. ➣ Most of the 1 and 2 marks questions asked from this chapter belong to the topic ‘Factors Influencing Rate of a Reaction’. Questions For practice Very Short Answer Types Questions (1 Mark) 1. For a reaction R Æ P, half life (t1/2) is independent of the initial concentration of reactants. What is the order of reaction? [Delhi 2017] 2. Consider the following reaction: 2NO(g) + 2H2(g) → N2(g) + 2H2O(g) The rate law for the reaction is first order with respect to H2 and second order with respect to NO. Write the rate law for this reaction. Short Answer Types Questions-I (2 Marks) 3. Find the overall order for the following reactions: (i) A + 3B → 2C (ii) 2H2(g) + 2NO(g) → 2H2O(g) + N2(g) Rate = k[A]1/2 [B]3/2 Rate = k[H2] [NO]2 What are the dimensions of rate constant in each case? Short Answer Types Questions-II (3 Marks) 4. A first order reaction is taken 10 minutes for 25% decomposition. Calculate t1/2 for the reaction. (Given log 2 = 0.3010, log 3 = 0.4771, log 4 = 0.6021) [Foreign 2015] 5. The data below are for the reaction of NO and Cl2 to form NOCl at 295 K. Expt. No. [Cl2] (mol L–1) [NO] (mol L–1) Initial Rate (mol L–1 s–1) 1. 0.05 0.05 1.0 × 10–3 2. 0.15 0.05 3.0 × 10–3 3. 0.05 0.15 9.0 × 10–3 (i) What is the order with respect to NO and Cl2 in the reaction? (ii) Calculate the rate constant. (iii) Determine the reaction rate when concentrations of Cl2 and NO are 0.2 M and 0.4 M respectively. 6. A first order reaction has a specific rate of 10–3 sec–1. How much time will it take from 10 g of the reactant to be reduced to 7.5 g? [log 2 = 0.3010, log 4 = 0.6020 and log 6 = 0.7782] 7. A first order reaction is 40% complete within 8 min. How long will it take before it is 90% completed? What is the value of rate constant? 8. A solution of cane sugar was hydrolysed to the extent of 25% in 1 hour. Calculate the time taken for the sugar to be 50% hydrolysed assuming that the reaction is of first order. Answers 3. (i) Order = 2, (ii) Order = 3 4. t1/2 = 24.08 min 5. (i) Order w.r.t. NO = 2; order w.r.t. Cl2 = 1; (ii) k = 8.0 L2 mol–2 s–1; (iii) Rate = 0.256 mol L–1 s–1 6. 287.7 s 7. k = 0.06386 min–1; t = 36.06 min 8. 2.408 hours 112 Chemistry-12

assIgnment Time: 45 Minutes M.M.: 25 Multiple Choice Questions (MCQs) (1 Mark) 1. The possible unit of rate constant for first order reaction is (i) mol L–1 s–1 (ii) s–1 (iii) mol s–1 (iv) mol–1 L s 2. In the presence of a catalyst, the heat evolved or absorbed during the reaction .................... . (i) increases. (ii) decreases. (iii) remains unchanged. (iv) may increase or decrease. Assertion Reason Type Questions (1 Mark) In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. (i) Assertion and reason both are correct statements and reason is correct explanation for Assertion. (ii) Assertion and reason both are correct statements but reason is not correct explanation for Assertion. (iii) Assertion is correct statement but reason is wrong statement. (iv) Assertion is wrong statement but reason is correct statement. 3. Assertion: Order of the reaction can be zero or fractional. Reason: We cannot determine order from balanced chemical equation. 4. Assertion: Rate constants determined from Arrhenius equation are fairly accurate for simple as well as complex molecules. Reason: Reactant molecules undergo chemical change irrespective of their orientation during collision. Very Short Answer Types Questions (1 Mark) ∆[O3 ] ∆[O2 ] during this time period. 5. For the reaction, 2O3  3O2 , − ∆t = 4.0 × 10–4 atm s–1. Determine the value of ∆t 6. The rate constant of a reaction is 3 × 102 hr–1. What is the order of the reaction? 7. In some cases it is found that large number of colliding molecules have energy more than threshold energy but yet the reaction is slow, why? Short Answer Types Questions-I (2 Marks) 8. A substance decomposes following the first order reaction. If the half-life period of the reaction is 5 minutes, what is the rate constant of this reaction? 9. A first order reaction is 20% complete in 10 minutes. Calculate the time taken for the reaction to go 80%. Short Answer Types Questions-II (3 Marks) 10. For a first order reaction, the rate constant is 0.1 s–1. How much time will it take to reduce the concentration from initial value of 0.6 mol L–1 to 0.06 mol L–1? 11. What will be the initial rate of a reaction if its rate constant is 10–3 min–1 and the concentration of the reactant is 0.2 mol dm–3? How much reactant will get converted into products in 200 minutes? 12. The rate constant for first order reaction increase from 2 × 10–2 × 8 to 10–2 when the temperature changes from 300 K to 320 K. Calculate the energy of activation (Ea) (log 2 = 0.3010, log 3 = 0.4771, log 4 = 0.6021) [AI 2015] Long Answer Types Questions (5 Marks) 13. Consider the graph plotted between energy v/s progress of reaction as shown alongside for a reaction: Are2a+ctBio2n. 2AB (i) Calculate DH of the Energy (kJ/mol) 140 130 (ii) Calculate the energy of activation for forward and backward reaction. 120 (iii) The dotted line curve is the path in the presence of a catalyst. What is the 110 energy of activation for the forward and backward reactions in the presence of 100 the catalyst? What is the lowering in activation energy in the presence of the catalyst? Progress of reaction (iv) Will the catalyst change the extent of the reaction? 1. (ii) 2. (iii) Answers 3. (ii) 4. (iii) 5. 6.0 × 10–4 atm s–1 6. First order 7. 1.38 × 10–1 min–1 8. 72.18 min 10. 23.0 seconds 11. Initial rate = 2 × 10–4 mol dm–3 min–1; 0.036 mol dm–3 12. EEba== 55.3364 kJ mol–1 13. (i) DH = –10 kJ/mol; (ii) Ef = 30 kJ/mol; Eb = 40 kJ/mol; (iii) Ef = 10 kJ/mol; 20 kJ/mol; 20kJ/mol; (iv) No ChemiCal KinetiCs 113

4 The d- and f-Block Elements Topics covered 4.1 Transition Elements: General Characteristics and Uses 4.2 Preparation and Properties of K2Cr2O7 and KMnO4 4.3 Lanthanoids and Actinoids: Characteristics and Uses C hapter map d AND f-BLOCK ELEMENTS d-Block Elements f-Block Elements Transition Elements Lanthanoids Actinoids lies in between General Characteristics and Uses s and p-block elements Lanthanoid Actinoid contraction contraction General Characteristics Atomic radii Differences between Ionisation enthalpy Lanthanoids and Actinoids Metallic character Melting and boiling points KMnO4 Preparation, Properties Oxidation states K2Cr2O7 and Uses Magnetic properties Complex formation Catalytic property Alloy formation Interstitial compounds Important Transition Element Compounds 114

Topic 1. Transition Elements: General Characteristics and Uses • Transition elements: Elements in which last in a series which causes increase in density. electron enters in d-orbitals of penultimate shell are called d-block elements. These elements take • Ionisation enthalpy: It generally increases from position in between s- and p-block elements thus, left to right in a series. Exceptions are Mn and Zn they are called transition elements. having higher value than expected due to stable • Electronic configuration: General electronic configuration of these elements is (n-1)d1–10 ns1–2. half and fully filled orbitals respectively. Also, Cr Half-filled and completely filled orbitals have extra stability thus, Cr and Cu have exceptional and Cu have lower value than expected in verge configuration of 3d54s1 and 3d104s1 respectively. Zn, Cd and Hg have completely filled orbitals to attain half and fully filled orbital. (n-1) d10ns2 thus, not regarded as transition elements. Those metal, atoms or ions with partly • Oxidation states: It varies due to involvement filled d-orbitals show variety of oxidation states, of (n-1)d and ns electrons. Mn shows maximum formation of coloured ions (due to d-d transition) number of oxidation states from +2 to +7. catalytic properties and formation of complexes. • Electrode potentials (M2+/M): It generally becomes less negative from left to right across a series. Cu having unique positive value of E° while rest of elements in the series have negative E° value. Thus, Cu is unable to liberate H2 from dilute acids at room temperature. General Properties • Magnetic properties: Due to unpaired electrons they have magnetic moment associated with them • Melting point: They have high melting point which can be calculated using ‘spin only formula’ due to large number of electrons [(n-1)d and ns for 3d series elements. electrons] form strong interatomic metallic bonds. Magnetic moment, µ = n(n + 2) • Enthalpy of atomisation: High value due to strong forces of attraction. Where, n = number of unpaired electrons. • Formation of interstitial compounds: Small • Atomic size: With in a series it varies quite small. The corresponding members of 4d and 5d atoms like H, C or N trapped inside the crystal series have almost same size due to lanthanoid lattices of metals to form interstitial compounds. contraction. Lanthanoid contraction is the These compounds are very hard, chemically inert decrease in size of 5d series due to filling of 4f and have high melting point. orbitals before 5d orbitals which have ineffective shielding effect. Ionic or metallic radius decreases • Alloys formation: Dut to similar size and other characteristics they readily form alloys. ExErcisE 4.1 Multiple Choice Questions (MCQs) (1 Mark) (i) Fe (ii) Ni 1. Electronic configuration of a transition element X (iii) Co (iv) Cu in +3 oxidation state is [Ar]3d5. What is its atomic 4. Generally transition elements form coloured salts number? due to the presence of unpaired electrons. Which of the following compounds will be coloured in (i) 25 (ii) 26 solid state? (iii) 27 (iv) 24 (i) Ag2SO4 (ii) CuF2 2. The electronic configuration of Cu (II) is 3d9 (iii) ZnF2 (iv) Cu2Cl2 whereas that of Cu (I) is 3d10. Which of the 5. Interstitial compounds are formed when small following is correct? atoms are trapped inside the crystal lattice (i) Cu (II) is more stable of metals. Which of the following is not the (ii) Cu (II) is less stable characteristic property of interstitial compounds? (iii) Cu (I) and Cu (II) are equally stable (i) They have high melting points in comparison to pure metals. (iv) Stability of Cu (I) and Cu (II) depends on nature of copper salts (ii) They are very hard. 3. Metallic radii of some transition elements are (iii) They retain metallic conductivity. given below. Which of these elements will have (iv) They are chemically very reactive. highest density? Assertion-Reason Type Questions (1 Mark) Element Fe Co Ni Cu Note: In the following questions a statement Metallic radii/pm 126 125 125 128 of assertion followed by a statement of reason the d- and f-BloCK elements 115

is given. Choose the correct answer out of the ions of the elements of first transition series? Calculate the theoretical value of its magnetic following choices. moment. (i) Both assertion and reason are true, and reason is 23. Which among Fe2+ and Fe3+ ions are more the correct explanation of the assertion. paramagnetic and why? (ii) Both assertion and reason are true but reason is 24. The sum of first and second and those of third and not the correct explanation of assertion. fourth ionisation energies for nickel and platinum metals are given below: (iii) Assertion is true but reason is false. (iv) Assertion is false but reason is true. 6. Assertion: Cu2+ iodide is not known. Metal IE1 + IE2 IE3 + IE4 Ni (kJ mol–1) (kJ mol–1) Reason: Cu2+ oxidises I– to iodine. 2.49 × 103 8.80 × 103 7. Assertion: Cu cannot liberate hydrogen from Pt 2.66 × 103 6.70 × 103 acids. Reason: Because it has positive electrode Taking these values into account, write: potential. (i) The most common oxidation state for nickel 8. Assertion: The highest oxidation state of osmium and platinum metals. is +8. (ii) Write the name of metal which form Reason: Osmium is a 5d-block element. compounds in +4 oxidation state easily and 9. Assertion: Mercury is not considered as why is it so? transition element. 25. Transition metal fluorides are ionic whereas Reason: Mercury is liquid. bromides and chlorides are usually covalent in Very Short Answer Type Questions (1 Mark) nature. Explain. 10. Copper does not react readily with dilute acids to 26. Transition metals and their compounds act as liberate hydrogen gas. Explain. good catalysts. Explain. 11. Explain why chromium is a typical hard metal [Delhi 2012 C; AI 2009, 2014; Foreign 2010] while mercury is a liquid? 27. First ionisation energies of 5d series elements are 12. Why the melting point and boiling point of Zn, Cd higher than those of 3d and 4d series elements. and Hg are low? Why? 13. Copper, silver and gold have completely filled 28. In the transition metals series, with an increase in d-orbitals, yet they are considered as transition atomic number the atomic radius does not change metals. Explain why? appreciably. Explain. 14. Explain the fact that transition elements have high enthalpies of hydration. 29. What are transition elements? Write any two characteristics of transition elements? 15. Why do transition elements exhibit higher enthalpies of atomisation? [NCERT] [Delhi 2015] 16. Name a transition element which does not exhibit 30. Assign the reason for each of the following: variable oxidation state. [NCERT] (i) Transition metals generally form coloured compounds. 17. In the series Sc (Z = 21) to Zn (Z = 30), the enthalpy of atomisation of zinc is the lowest, i.e., (ii) Manganese exhibits the highest oxidation state, i.e. +7 amongst the 3d series transition 126 kJ mol–1. Why? [NCERT] elements. [Delhi 2011] 18. Why is the highest oxidation state of a transition 31. Give reasons for the following: metal exhibited in its oxide or fluoride only? (i) Cu(I) ion is not stable in aqueous solution. [NCERT] OR 19. Which is a stronger reducing agent Cr2+ or Fe2+ and why? [AI 2010] [NCERT] Explain why Cu+ ion is not stable in aqueous 20. Which metal in the first series of transition metals solution? [DSB 2009; AI 2011] [NCERT] exhibits +1 oxidation state most frequently and (ii) Unlike Cr3+, Mn2+, Fe3+ and the subsequent why? [NCERT] other M2+ ions of the 3d series metals, the 4d Short Answer Type Questions-I (2 Marks) and the 5d series metals generally do not have tendency to form stable cationic species. 21. On what ground can you say that scandium (z = 21) is a transition metal but zinc (Z = 30) is not a [AI 2011] transition metal. [NCERT] 32. How would you account for the following: 22. W h y d o e s M n ( I I ) i o n s s h o w m a x i m u m (i) Most of the transition elements are known paramagnetic character amongst the bivalent to form interstitial compounds. 116 Chemistry-12

(ii) The metallic radii of the 5d series metals are 48. How is the variability in oxidation states of virtually the same as those of corresponding transition metals different from that of the non- group members of 4d series elements. transition metals? Illustrate with examples. [NCERT] 33. Explain the following observation by giving an appropriate reason: There occurs much 49. (i) Of the ions Ag+, Co2+ and Ti4+ which one will more frequent metal-metal bonding in heavier be coloured in an aqueous solution? transition metals compounds. (ii) If each one of the above ionic species is placed 34. Complete the following reactions: in a magnetic field, how will it respond and why? (i) Cu2+(aq) + I– (aq) → (ii) AgCl(s) + NH3 (aq) → [AI 2012] 50. Write down the electronic configuration of: (i) Cr3+ 35. Why is Cr2+ reducing and Mn3+ oxidising even (ii) Cu+ (iii) Co2+ (iv) Mn2+ ions. [NCERT] though both have d4 configuration? Short Answer Type Questions-II (3 Marks) [NCERT] [AI 2008, DSB 2009] 51. Transition metals form complex compounds. Give 36. How would you account for the increasing three reasons. [AI 2010] oxidising power in the series: 52. Why in any transition metal series the melting VO2+ < Cr2O72− < MnO4− ? [NCERT] point first increases, then decreases and also shows a dip in the middle? 37. Why is E° value for the Mn3+/Mn2+ couple much more positive than that for Cr3+/Cr2+ or Fe3+/Fe2+ 53. The elements of 3d series are given as Sc, Ti, V, couple? Explain. Cr, Mn, Fe, Co, Ni, Cu and Zn. [Delhi 2017(C)][CBSE 2022][NCERT] Answer the following: 38. What is meant by ‘disproportionation’ of an (i) Name the element which exhibits maximum oxidation state? Give an example. [NCERT] number of oxidation states. Give reason. 39. Silver atom has completely filled d-orbitals (4d10) (ii) Which element has the highest melting point? in its ground state. How can you say that it is a transition element? [Delhi 2017(C)][NCERT] (iii) Which element shows only one oxidation state of + 3? [AI 2016] 40. Which of the 3d series transition metal exhibits 54. (i) Name the 3d transition series element which the largest number of oxidation states and why? exhibits maximum number of oxidation [NCERT] states and why does it shows so? 41. WThheaEt i0(Ms2p+/oMs) sviablluyethfoerrceoapspoenrfoisr positive (0.34 V). (ii) Which 3d series transition metal has positive this? [NCERT] E°(M2+/M) value and why? 42. Why are Mn2+ compounds more stable than (iii) Amongst Cr3+ and Mn3+ ions, which is a Fe2+ compounds towards oxidation to their stronger oxidising agent and why? +3 state? [AI 2012] [NCERT] 55. Account for the following: OR (i) Zn is not considered as the transition element. [CBSE 2022] Explain the following observation; giving an appropriate reason for each: Mn2+ ion is much (ii) Transition metals forms a large number of more resistant towards oxidation than Fe2+ ion. complexes. [Delhi 2012] (iii) The E° value for Mn3+/Mn2+ couple is highly 43. Explain briefly how +2 oxidation state becomes positive as compared to that of Cr3+/Cr2+ more and more stable in the first half of the first row transition elements with increasing atomic couple. [AI 2012] number. [NCERT] 56. For the first row transition metals, the E° values are: 44. Compare the stability of +2 oxidation state for the elements of the first transition series. [NCERT] E°M2+ / M V Cr Mn Fe Co Ni Cu 45. To what extent do the electronic configurations −1.18 −0.91 −1.18 −0.44 −0.28 −0.25 +0.34 decide the stability of oxidation states in the first Explain the irregularity in these values. series of the transition elements? Illustrate your [NCERT] answer with examples. [NCERT] 57. Calculate the magnetic moment of a divalent ion in aqueous solution if its atomic number is 25. 46. What may be the stable oxidation states of the [NCERT] transition elements with the following electronic configurations in the ground state of their atoms: 58. How would you account for the irregular variation 3d3, 3d5, 3d8 and 3d4? [NCERT] of ionisation enthalpies (first and second) in the first series of the transition elements? [NCERT] 47. In what way the electronic configuration of the transition elements different from that of the 59. Calculate the ‘spin only’ magnetic moment of non-transition elements? [NCERT] M2+(aq) ion (Z = 27). [NCERT] the d- and f-BloCK elements 117

60. What are interstitial compounds? Why are such Example Magnetic moment (BM) compounds well known for transition metals? [NCERT] K4[Mn(CN)6] 2.2 [Fe(H2O)6]2+ 5.3 61. For M2+/M and M3+/M2+ system, the E° values for K2 [MnCl4] 5.9 some of the metals are as follows: [NCERT] Cr2+/Cr – 0.9 V Cr3+/Cr2+ – 0.4 V Mn2+/Mn – 1.2 V Mn3+/Mn2+ + 1.5 V Long Answer Type Questions (5 Marks) Fe2+/Fe – 0.4 V Fe3+/Fe2+ + 0.8 V 67. How many unpaired electrons are there in each of the following: Using this data to comment upon: (i) The stability of Fe3+ in acid solution as (i) Cr2+ (z = 24) (ii) Fe2+ (z = 26) compared to Cr3+ or Mn3+ and (ii) The ease with which iron can be oxidised (iii) Mn2+ (z = 25) (iv) Ti2+ (z = 22) as compared to a similar process for either (v) Zn2+ (z = 30) chromium or manganese metal. [NCERT] 68. Which of the following ions are expected to be 62. Predict which of the following will be coloured in coloured and why? Explain. aqueous solution? Cu+, Fe2+, Mn2+, Cr3+, Sc3+, Ti4+. Ti3+, V3+, Cu+, Sc3+, Mn2+, Fe3+ and Co2+ 69. What are the characteristics of transition Give reasons for each. [NCERT] elements and why are they called transition 63. How would you account for the following: elements? Which d-block elements may not be regarded as the transition elements? [NCERT] (i) Amongst the d4 species, Cr2+ is strongly reducing while Mn3+ is strongly oxidising. 70. Explain giving reasons: [AI 2011] (i) Transition metals and many of their (ii) Cobalt (II) is quite stable in aqueous solution compounds show paramagnetic behaviour. but in the presence of a complexing agent it (ii) The enthalpies of atomisation of transition can be easily oxidised. metals are generally high. (iii) The d1 configuration is not unstable in (iii) The transition metals generally form coloured transistion metal ions. compounds. 64. Give examples and suggest reasons for the (iv) Transition metals and many of their following features of the transition metal compounds act as good catalysts. chemistry: [AI 2011] [NCERT] (i) The lowest oxide of a transition metal is basic, 71. Calculate the number of unpaired electrons in the the highest is amphoteric/acidic. following gaseous ions: (ii) A transition metal exhibits highest oxidation Mn3+, Cr3+, V3+ and Ti3+. state in oxides and fluorides. Which one of these is most stable in aqueous (iii) The highest oxidation state is exhibited in oxoanions of a metal. [NCERT] solution? [NCERT] 65. Comment on the statement that elements of the 72. Write down the number of 3d electrons in each of first transition series possess many properties the following ions: different from those of heavier transition Ti2+, V2+, Cr3+, Mn2+, Fe2+, Fe3+, Co2+, Ni2+ and Cu2+. elements. [NCERT] Indicate how would you expect the five 3d orbitals to be occupied for these hydrated ions 66. What can be inferred from the magnetic moment values of the following complex species? (octahedral)? [NCERT] Answers 4.1 1. (ii) 2. (i) 3. (iv) a very hard metal. On the contrary in mercury the 4. (ii) 5. (iv) 6. (i) 3d subshell is fully filled (5d10 6s2), i.e. electrons are not available for bonding and thus bonding is 7. (i) 8. (ii) 9. (ii) very weak in case of mercury. 10. Copper has positive reduction potential for Cu2+ 12. In Zn, Cd and Hg, the (n – 1)d orbitals are fully to Cu. Thus, it cannot reduce H+ ions of acids to filled. These d-electrons does not take part in the metallic bonding. Thus, the extent of metallic hydrogen gas. bonding is weaker in Zn, Cd and Hg. Hence, they 11. In chromium, the interatomic bonding is very strong due to the presence of five unpaired have low melting and boiling points. electrons in 3d subshell (3d5 4s1). Therefore, it is 118 Chemistry-12

13. In the common oxidation states exhibited by Cu, Magnetic moment (Fe3+) 5×7 Au and Ag metals, they have partially filled (n – 1)d = n(n + 2) = orbitals: Cu2+ (3d9), Ag2+ (4d9), Au3+ (5d8). On this = 5.91 B.M. basis, these metals are considered as transition metals. Fe2+ (3d6) 14. Due to small size of ions and high ionic charge, Magnetic moment (Fe2+) there is more tendency of hydration in transition elements. This gives rise to high enthalpy of = n(n + 2) = 4 × 6 hydration. = 4.9 B.M. 15. Transition elements have unpaired electrons 24. (i) For nickel, +2 is the common oxidation state involved in strong interatomic bonding. This as the sum of only first and second ionisation results their higher enthalpies of atomisation. Fenoerrpgileastin(IuEm1 ,++I4E2i)s for Ni has lower value. the common oxidation 16. oSxci2d1 atio→n 3d1, 4s2; and it exhibits only a fixed state as the sum of all the first, second third state of +3. IaEn3d+foIuEr4t)hfoiornPistahtaiosnloewntehravlapliuese.(IE1 + IE2 + 17. In case of zinc, 3d-electrons are not involved in (ii) Platinum can easily form compounds in +4 metal bonding as all the electrons are paired. oxidation state because the sum of ionisation Because of poor metal-metal bonding, the eansecrogmiepsa(rIeEd1 t+oINEi2 +mIeEta3 l+. IE4) is lower for Pt enthalpy of atomisation of zinc is the lowest. 18. Because of small size and high electronegativity of 25. Since electronegativity of halogens decreases in oxygen and fluorine, they oxidise metals to their the order: highest oxidation state. 19. Cr2+ ion is a stronger reducing agent than Fe2+ F > Cl > Br ion. Therefore, the ionic character of transition metal Reason: d4 → d3 occurs in case of Cr2+ to Cr3+ halides decreases in the order as follows: but d6 → d5 occurs in case of Fe2+ to Fe3+. In a neutral medium (like water) d3 configuration is M – F > M – Cl > M – Br more stable than d5. Hence, fluorides are ionic while chlorides and 20. Copper with electronic configuration [Ar] 3d10 4s1 bromides are covalent in nature. exhibits +1 oxidation state. Copper easily loses 4s1 electron and achieves a stable (3d10) configuration 26. This can be accounted on the following two by forming the cation, Cu+. reasons: 21. Scandium is regarded as a transition element on (i) They have variable valencies and exhibit the basis of incompletely filled 3d-orbitals in its multiple oxidation states and transition ground state (3d1). metals sometimes form unstable intermediate compounds which provides a new pathway On the other hand, zinc atom has completely filled with lower activation energy for the reaction. penultimate d-orbital (3d10) configuration in its ground state as well as in its +2 oxidation state (ii) In some cases transition elements provide a (Zn2+), hence it is not considered as a transition suitable surface area for the reaction to occur. element. 27. Because of weaker shielding (or screening) effect 22. The extent of paramagnetic nature depends upon of 4f electrons, the effective nuclear charge on the number of unpaired electrons present. Mn(II) the valence electrons in 5d series elements is ions possesses the maximum number of unpaired higher. Hence, the first ionisation energies for electrons i.e. 5, amongst the divalent ions of the 5d elements are higher than those of 3d and 4d first transition series and thus it possesses the series elements. maximum paramagnetic character. Theoretical value of magnetic moment: 28. As we move from left to right along the transition metals series, the nuclear charge increases but (For Mn(II)) = n(n + 2) due to ineffective shielding of d electrons net electrostatic attraction between the nucleus and 5(5 + 2) = 35 valence electrons increases. Thus atomic radius does not change appreciably across the series. = 5.91 BM 29. Those atoms or ions having d-orbitals incompletely 23. Fe3+ ion is more paramagnetic than Fe2+ ion, since or partially filled are called transition elements it consists of five unpaired electrons while Fe2+ ion or d-block elements. The general valence shell possess only 4 unpaired electrons. electronic configuration of transition elements is: (n – 1)d1–10 ns1–2 Fe3+ (3d5) Two characteristics shown by transition elements are: the d- and f-BloCK elements 119

(i) Transition elements can exhibit variable Mn2+ has a stable half-filled d-subshell oxidation states. configuration. Hence Mn3+ undergoes reduction and it acts as an oxidising agent. (ii) They can act as catalysts in many reactions. 36. The given sequence of oxidising behaviour can be accounted on the basis of increasing stability of 30. (i) Transition metals generally form coloured the species with lower oxidation state of metal to compounds and this can be accounted on the which they are reduced. basis of d-d transitions. Due to partially filled d-orbitals, there is a transition of electrons among the d-orbitals by absorption of visible light. (ii) Maximum number of unpaired electrons are exhibited by Mn i.e. 5 unpaired electrons in 3d orbitals also it has 2 electrons in 4s orbitals. Mn can lose all its valence electrons to show +7 oxidation state. 31. (i) Cu2+(aq) ion is much more stable than 37. eEx0(Mtrna3+/Mstn2a+b) iislitmyodreuepotsoithivaelfbfeilclaeudsde -Mornb2i+t(adl5s) have Cu+(aq) ion. Though second ionisation than enthalpy is large but the hydration enthalpy of Cu2+ is highly negative and it compensate Mn3+ while in case of Cr, Cr3+(d3) is more stable idnuceatsoehoaflFf efi,llFeed3+t(2dg 5o)ribsitmalosrethsatanbtlheadtuoef Cr2+ and for large second ionisation enthalphy. Thus to stable Cu2+ (aq) ion is more easier to form. Cu(I) half filled d-orbitals than that of Fe2+. compounds are a bit unstable in aqueous 38. When an element as a molecule or an ion solution and undergoes disproportionation simultaneously undergoes oxidation as well as follows: as reduction, this phenomenon is called 2Cu+ → Cu2+ + Cu disproportionation. It takes place when the (ii) Due to lanthanoid contraction, the size of 4d particular oxidation state of an element becomes and 5d elements decreases which increases less stable relative to the other oxidation states, their ionisation energy. Thus, they cannot one lower and one being higher. For example, easily lose electrons to form cationic species. 3MVnIO42– + 4H+ 2MVInIO4– + IV 32. (i) Because small non-metallic atoms like H,C, + 2H2O N can easily get trapped inside the crystal MnO2 oxidation lattice of the transition metals. reduction (ii) This is because of lanthanoid contraction. 39. Silver exhibits +2 oxidation state. In this oxidation state, 4d subshell contains nine electrons, i.e. 33. This is because of the fact that the filling of partially filled 4d-orbitals. Hence, it can be electrons in 4f orbital starts in 3rd transition series regarded as a transition element [Ag2+ ion, 4d9]. elements, as a result of which the elements have large number of unpaired electrons alongwith 40. The electronic configuration of Mn atom is 3d5 4s2, i.e. all the d-orbitals are singly occupied by ns-orbitals electrons which take part in metallic electrons. Thus, Mn exhibits a large number of bonding. oxidation states ranging from +2 to +7 by losing all its valence electrons. 34. (i) 2Cu2+(aq) + 4I–(aq) → Cu2I2 + I2(s) (ii) AgCl(s) + 2NH3(aq) → Ag (NH3)2 Cl 41. eEn0(Mth2+a/Mlp) ifeosr any metal depends on the sum of (soluble) of atomisation, ionisation enthalpies 35. Cr2+ → Cr3+ + e– 3d4 4s0 3d3 4s0 and hydration enthalpies. Copper has high enthalpy of atomisation and low hydration energy. Hence, E°(Cu2+/Cu) is positive. 42. The electronic configuration of Mn2+ ion is [Ar] 3d5, i.e. all the five d-orbitals are singly occupied. Thus, this configuration is highly stable and it resists further oxidation, i.e. loss of further electrons requires very high energy for Cr3+ is highly stable due to the presence of half- this process. On the other hand, the electronic ofixllidedatito2gn orbitals. Thus Cr2+ easily undergoes configuration of Fe2+ ion is [Ar] 3d6. Hence, it can and it acts as a reducing agent. easily lose an electron and achieve the stable half Mn3+ + e− → Mn2+ filled electronic configuration, i.e. oxidation of Fe2+ 3d4 4s0 3d5 4s0 into Fe3+ ions can be easily achieved. 120 Chemistry-12

43. In M2+ ions, 3d-orbitals gets occupied gradually reduction potential is due to increase in the sum as the atomic number increases. of first and second ionisation enthalpies. Since, the number of empty d-orbitals decreases, 45. In a transition metal series, the oxidation state the stability of the cation (M2+) eventually which leads to exactly half filled or completely increases from Sc2+ to Mn2+. Mn2+ ion is the filled d-orbital configuration are more stable and most stable as all its five 3d-orbitals are singly thus decides the stability of oxidation states. For occupied. example, the electronic configuration of Fe (Z = 26) is [Ar] 3d6 4s2. It shows various oxidation 44. The stability of +2 oxidation state decreases states but Fe (III) is the most stable because in from left to right except for Mn and Zn. Stability +3 oxidation state, Fe has the stable half filled decreases towards right due to decreasing configuration: [Ar] 3d5. Similarly, Mn2+([Ar]3d54s0) negative value of standard reduction potential. is more stable than Mn3+. The decrease in the negative value of standard 46. Electronic configuration (in ground state) Element Most stable oxidation state 3d3 Vanadium (3d3 4s2) +5 out of +2, +3, +4, +5 3d5 Chromium (3d5 4s1) +3, +6 out of +2, +3, +4, +5, +6 Manganese (3d5 4s2) +2, +7 out of +2, +3, +4, +5, +6, +7 3d8 Nickel (3d8 4s2) +2 out of +2, +3, +4 3d4 (There is no d4 configuration in the ground state) 47. Electronic configuration of transition metals is 51. Transition metals have marked ability to form (n-1)d1–10ns2. Here, d-orbitals of penultimate shell complex compounds. This ability can be accounted is progressively filled while for non-transition on the basis of the following three reasons: elements the electrons are filled in valence shell. (i) Small size 48. In transition elements, the successive oxidation states differ by unity. For example, Mn shows all (ii) High Nuclear charge the oxidation states ranging from +2 to +7. On the other hand, non-transition metals exhibit variable (iii) Availability of vacant degenerate d-orbitals. oxidation states which differ by only two units. For example, Pb(II), Pb(IV); Sn (II), Sn (IV); Sb (III), 52. Melting point first increases as the number of Sb (V), etc. In p-block elements lower oxidation state unpaired electrons increases which thereby is more stable in heavy elements and in d-block, increases the strength of metallic bonding. heavy elements show higher oxidation state. After reaching the maximum possible value, the melting point starts decreasing as pairing of 49. (i) The ionic species which possesses unpaired electrons starts in d-orbitals and thus metallic electron in (n-1) d-subshell will exhibit bonding gets weakened. The dip in the middle colour. Of the ions Ag+ (4d10), Co2+ (3d7) and is due to half filled electronic configuration of Ti4+ (3d0), Co2+ will be coloured as it contains d-subshell, which has higher stability. three unpaired electrons while other ions Ag+, Ti4+ with all the electrons paired up will be 53. (i) Mn exhibits greatest no. of oxidation states colourless. as it has maximum number of unpaired electrons (five unpaired electrons in d (ii) When placed in a magnetic field, Co2+ orbital). ions will be attracted by it, because of its paramagnetic nature due to the presence of (ii) Cr has the highest melting point. As the unpaired electrons. Ag+ and Ti4+ ions will be number of unpaired electrons increases upto repelled by the magnetic field as they are d5 configuration, this results in the increase diamagnetic in nature due to the absence of of strength of metallic bond. To break the unpaired electrons. metallic bond, significant amount energy is required. After Cr, the electrons start pairing 50. Element At. Ion Electronic up and hence melting point decreases. (i) Chromium No. configuration (iii) Scandium shows only +3 oxidation state, 24 Cr3+ [Ar] 3d3 because it possesses only one electron in 3d orbital. So, it can lose two electrons from 4s (ii) Copper 29 Cu+ [Ar] 3d10 orbital and one electron from 3d orbital to exhibit +3 oxidation state. (iii) Cobalt 27 Co2+ [Ar] 3d7 54. (i) Mn, because of the presence of maximum (iv) Manganese 25 Mn2+ [Ar] 3d5 number of five unpaired electrons i.e. 5 electrons. the d- and f-BloCK elements 121

(ii) Cu, because enthalpy of atomisation and usually non-stoichiometric with formulae like enthalpy of ionisation is not compensated by each other. VH0.56, TiH1.7, etc. Transition metals have such a size which creates (iii) Mn3+ ion, because Mn2+ ion is more stable due to highly stable half filled (3d5) configuration. the interstitial sites of size in which small atoms, get fitted comfortably thus these compounds have 55. (i) Zn/Zn2+ has completely filled d-orbitals. known for transition metals. (ii) This is due to smaller size/higher ionic charge Due to the formation of interstitial compounds, and availability of the vacant d-orbitals. transition metal crystals become more hard and (iii) Because Mn2+ is more stable (3d5) than Mn3+ rigid. Melting points of such non-stoichiometric (3d4). Cr3+ is more stable than Cr2+ due to compounds is usually higher than their stable t2g3(d3) configuration. corresponding parent metals. 56. Standard reduction potential (E°) values are the 61. (i) As E° value (Cr3+/Cr2+ couple) is negative sum of the following thermodynamic data: (–0.4 V), so Cr3+ ions in solution cannot be (i) Sublimation enthalpy reduced to Cr2+ easily, i.e. Cr3+ ions are highly Mvstaanlub3+elei(o.+n1Ss.i5cnaVcne),btEeh(°eMaann3s+t/iMhlyna2+rt)eoidfsuEch(°Feiedg3+/htFoee2+rf)othrpmeorseMitfoinvr2ee+ (ii) Ionisation enthalpy (iii) Hydration enthalpy The irregularity in the values of standard ions as compared to the reduction of Fe3+ reduction potential is due to irregular variation of o(IfEs1u+blIiEm2a) taionnd also due to relatively lower value ions to Fe2+ ions. Thus, the order of relative enthalpies for Mn and V. stability in acidic solution of these ions is: 57. Mn25 → 3d5 4s2 Mn3+ < Fe3+ < Cr3+ Mn2+ → 3d5 4s0 (Least stable) (Most stable) Number of unpaired electrons (n) = 5. (ii) Oxidation potential for the given pairs will be negative of reduction potential and thus, Spin only magnetic moment, the value becomes +0.9 V, +1.2 V and 0.4 V. µ = n(n + 2) = 5(5 + 2) BM = 5.92 BM Thus, the order in which they are getting 58. Irregular variation of ionisation enthalpies (first and second) is mainly due to difference in oxidised to +2 oxidation state will be: stability of different 3d-configurations. d0, d5 and d10 configurations have extra stability and hence Mn > Cr > Fe in such cases the values for ionisation enthalpies are usually high. 62. The ions having one or more unpaired electrons For example, Cr has low first ionisation enthalpy will be coloured in aqueous solution. The unpaired because electron needs to be removed from 4s-orbital but the second ionisation enthapy is electrons can undergo d-d transitions after very high as Cr+ has a stable 3d5 configuration. Zn absorbing radiations from white light. The ions has a very high first ionisation enthalpy because electron needs to be removed from the stable with d0 or d10 configuration are colourless. electronic configuration: 3d10 4s2. However the second ionisation enthalpy is not very high, since a Ion Configuration Coloured/ stable 3d104s0 configuration can be obtained when Ti3+ 3d1 Colourless Zn converts to Zn2+ state. Coloured 59. Electronic configuration of metal-atom (M) with atomic number 27 is: [Ar] 3d7 4s2 V3+ 3d2 Coloured \\ Electronic configuration of M2+ ion Cu+ 3d10 Colourless = [Ar] 3d7, i.e. Three orbitals have unpaired electrons, i.e. n = 3 Sc3+ 3d0 Colourless Mn2+ 3d5 Coloured Fe3+ 3d5 Coloured Co2+ 3d7 Coloured 63. (i) Cr2+ ion is reducing in nature, since its configuration changes from d4 to d3 (a stable configuration with half filled ito2ng-.orObnitatlhse) Applying µ = n(n + 2) BM formula for ‘spin only’ when Cr2+ is oxidised to Cr3+ magnetic moment. We have other hand, Mn3+ is oxidising in nature as the m = 3 × 5 = 15 = 3.87 BM configuration changes from d4 to d5 (a stable configuration with half filled d-orbitals), 60. Small atoms like B, C, N, etc., can occupy the when Mn3+ is reduced to Mn2+ ion. positions in the voids in the lattice spaces among the atoms of transition metals. Such combinations (ii) Strong ligands forces cobalt (II) ions to lose are termed as interstitial compounds. They are one more electron from 3d-subshell and thereby induces d2sp3-hybridisation. 122 Chemistry-12

(iii) The ions with d1 or 3d104s1 configuration electrons present in the complex, i.e. at the tends to lose the only electron present in d- time of the formation of complex, there is no or 4s-subshell in order to attain the stable rearrangement of electrons in 3d-orbitals. electronic configuration. Thus +1 oxidation Hence, it is an outer orbital octahedral state is exhibited by some of the transition complex involving sp3d2-hybridisation with elements. electronic configuration as shown below: 64. (i) In case of metal forming a number of oxides, the acidic strength increases with the increase in oxidation state of the metal, i.e. acidic strength increases with the increase in percentage of oxygen in the oxide. For example, MnO is basic while Mn2O7 is an (iii) sKt2a[tMe noCf lM4]n: In this complex, the oxidation acidic oxide. is +2. The value of magnetic moment explains that there are five unpaired (ii) A metal can exhibit higher oxidation state in electrons present in the complex, i.e. at the the form of its oxides and fluorides as oxygen and fluorine are highly electronegative time of the formation of complex, there is no rearrangement of electrons in the 3d-orbitals. elements and they are small in size, i.e. they Hence, it is an outer orbital tetrahedral act as strong oxidising agents. For example, chromium shows an oxidation state of +6 in complex involving sp3-hydribisation with electronic configuration as shown below: oCfr+F86 and osmium shows an oxidation state in OsO4. (iii) This is because of the high electro-negativity of oxygen atom and its ability to form double bond with metals. For example, chromium can show an oxidation state of +6 in sCtra2tOe72o−f 67. (i) Cr2+ (z = 24) and manganese shows an oxidation Electronic configuration of Cr2+ = 3d4 +7 in MnO4− . 4, 65. The following points justify that the given i.e. 4 unpaired electrons statement holds true: (ii) Fe2+ (z = 26) (i) Ionisation enthalpies of heavier transition Electronic configuration of Fe2+ = 3d6 elements is comparatively higher than the elements belonging to 3d series. 6, Consequently, heavier transition elements are less reactive as compared to 3d-series i.e. 4 unpaired electrons elements. (iii) Mn2+ (z = 25) (ii) Melting point of heavier transition elements Electronic configuration of Mn2+ = 3d5 is comparatively quite higher than the elements belonging to 3d-series. 5, (iii) Higher oxidation states in heavier transition i.e. 5 unpaired electrons elements are stable whereas lower oxidation (iv) Ti4+ (z = 22) states are more stable in 3d-series elements. Electronic configuration of Ti4+ = 3d0 66. (i) sKt4a[tMe nof(CMNn)6i]s: In this complex, the oxidation = 1s2, 2s2, 2p6, 3s2 3p6, 3d0, 4s0 +2 and the value of magnetic So, no unpaired electron moment explains that there is only one (v) Zn2+ (z = 30) unpaired electron. In the formation of this Electronic configuration of Zn2+ = 3d10 complex, the electrons are therefore arranged = 1s2, 2s2, 2p6, 3s2, 3p6, 3d10, 4s0 in 3d-orbitals in the following way as shown: So, no unpaired electron 68. An ion of a transition element which possesses 3d 4s 4p unpaired electrons in d-orbitals, i.e. in which d-d transition is possible, shows a characteristic d2sp3-hybridisation colour. The d0 or d10 configuration does not exhibit any d-d transition as there are no unpaired (ii) t[Fhies(Hco2mOp)6l]e2x+: The oxidation state of iron in electrons and hence they are colourless. is +2. The value of magnetic Fe2+, Mn2+ and Cr3+ are coloured while Cu+, Sc3+ and Ti4+ are colourless. moment explains that there are four unpaired the d- and f-BloCK elements 123

Ion Configuration of transition elements owing to completely filled Cu+ 3d10 (colourless) d-orbital in their ground state and in its stable oxidation state. Sc3+ 3d0 (colourless) 70. (i) Transition metals and most of their Ti4+ 3d0 (colourless) compounds show paramagnetic character Fe2+ 3d6 (coloured) because one or more of their d-orbitals are Mn2+ 3d5 (coloured) singly occupied. Cr3+ 3d3 (coloured) (ii) In transition element atoms, there are large 69. The general characteristics of transition elements number of unpaired electrons and thus are as follows: they have strong interatomic interactions (i) They all are metals. and thereby stronger bond between their (ii) They all are hard and brittle in nature (Hg atoms. Due to this very fact they have high is an exception). enthalpies of atomisation. (iii) They are good conductors of heat and (iii) The colour of transition element compounds electricity. arises due to the presence of unpaired (iv) They possess high densities. electrons and vacant (n–1)d subshells. With (v) They possess high melting and boiling points, the absorption of energy from light, there is with high heat of atomisation. excitation of electrons and these electrons can (vi) They are appreciably electropositive, dissolve shift from lower energy d-orbitals to higher in mineral acids, however a few of them are noble in nature. energy d-orbitals, i.e. a d-d transition can occur. The rest of the energy of white light (vii) Ionisation energy values (first) are higher is transmitted and the compound appears than the corresponding s-block elements but it is less than that for the p-block elements. to be coloured. The observed colour of the (viii) With only a few exceptions, they exhibit a compound is always the complimentary wide range of oxidation states. colour of the colour which is absorbed. (iv) Because of the availability of the vacant (ix) Most of them can form coloured ions. d-orbitals, they can easily form intermediate (x) Their compounds are paramagnetic in nature. (xi) They have a good tendency to form complexes. activated products. The size of transition (xii) They can act as good catalysts and form metal atoms and ions are also favourable for interstitial compounds. the formation of transitory complex with the reactants and hence transition metals and d-block elements represent the change or transition their compounds can act as good catalysts. in properties from the most electropositive s-block elements to the least electropositive p-block 71. Mn3+ : [Ar] 3d4 –Four unpaired electrons elements. Therefore, they are known as transition elements. Cr3+ : [Ar] 3d3 –Three unpaired electrons V3+ : [Ar] 3d2 –Two unpaired electrons Ti3+ : [Ar] 3d1 –One unpaired electron The elements of group 12, i.e., Zinc metal group Cr3+ is the most stable, due to stable half filled t2g3 (Zn, Cd and Hg) are not included in the category orbital configuration. 72. S. Ion Electronic No. of 3d-orbital electronic No. configuration 3d electrons arrangement (i) Ti2+ [Ar] 3d2 2 (ii) V2+ [Ar] 3d3 3 (iii) Cr3+ [Ar] 3d3 3 124 Chemistry-12

(iv) Mn2+ [Ar] 3d5 5 eg (v) Fe2+ [Ar] 3d6 6 t2g (vi) Fe3+ [Ar] 3d5 5 eg [Ar] 3d7 7 t2g (vii) Co2+ [Ar] 3d8 8 eg t2g eg t2g (viii) Ni2+ (ix) Cu2+ [Ar] 3d9 9 Topic 2. Preparation and Properties of K2Cr2O7 and KMnO4 • Preparation of KMnO4: KMnO4 is prepared OO by fusion of MnO2 with an alkali (KOH) and an oxidising agent like KNO3. This produces green Mn Mn K2MnO4 which disproportionates in a neutral or acidic medium to give permanganate ions. which O O– O– O O– is purple in colour. O 2MnO2 + 4KOH + O2 → K2MnO4 + 2H2O Manganate Permanganate 3MnO42– + 4H+ → 2MnO4– + MnO2 + 2H2O Commercially, MnO42– is electrolysed to oxidise to • Chemical properties: give MnO4–: ã It decomposes on heating at 513 K. MnO42– Electrolytic reduction MnO4– + e– • Physical properties: 2KMnO4 → K2MnO4 + MnO2 + O2 ã tAIo2c, ifSdo2ir–fmiteodCSOKa2Mn, dFnOeS2O4+ s3t2–ooltuFoteSi3o+On, 42N–o.xOi2d– itsoesNoOx3–a,lIa–tetos ã It is a dark purple coloured crystalline solid. KneMutnrOal4, can act as an oxidising agent in ã acidic as well in basic medium. ã It is not very soluble in water. 2MnO4– + 5COO– + 16H+ → 2Mn2+ + 10CO2 ã It is stored in dark coloured bottles because it | + 8H2O gets decomposed in the presence of sunlight. COO– ã MnO42– is paramagnetic due to the presence of MnO4– + 5Fe2+ + 8H+ → Mn2+ + 5Fe3+ + 4H2O unpaired electrons but MnO4– is diamagnetic. 2MnO4– + 10I– + 16H+ → 2Mn2+ + 5I2 + 8H2O The p-bonding takes place by overlap of 2MnO4– + 5NO2– + 6H+ → p-orbitals of oxygen and d-orbitals of Mn. 2Mn2+ + 5NO3– + 3H2O 5H2S + 2MnO4– + 6H+ → 2Mn2+ + 8H2O + 5S ã Permanganate and manganate both are 5SO32– + 2MnO4– + 6H+ → tetrahedral. 2Mn2+ + 3H2O + 5SO42– the d- and f-BloCK elements 125

ã Permanganate ions at [H+] = 1 oxidises 2Na2Cr2O4 + 2H+ →Na2Cr2O7 + 2Na+ + H2O Na2Cr2O7 + 2KCl → K2Cr2O7 + 2NaCl water slowly but the reaction becomes faster Properties: in the presence of Mn2+ ions and at high temperature. (i) K2Cr2O7 is an orange crystalline solid. (ii) It is soluble in water. ã In neutral or faintly basic Im– teodIiOum3– a, nKdMMnnO2+4 otoxiMdinseOs2.S2O32– ions to SO42–, (iii) Its colour depends on pH of the solution. In 2MnO4– + H2O + I– → 2MnO2 + 2OH– + IO3– the presence of [H+], it forms iCt rfo2Orm72–,s orange 8MnO4– + 3S2O32– + H2O → and in the presence of (OH–) yellow 8MnO2 + 6SO42– + 2OH– CrO42–. 17p9m O 2MnO4– + 3Mn2+ + 2H2O → 5MnO2 + 4H+ O Cr 126° O O Cr p1m63 Uses: O O– O– Cr (i) It is used as an oxidising agent in the presence O O– O– O o(f dHilC. Hl w2SilOl 4gebtuotxniodtisiendtthoeCplr2easnendcHe 2oOf )H. Cl • Chemical Properties: It oxidises Fe2+ to Fe3+, (ii) It is used in analytical chemistry. I– to I2, S2–to S, Sn2+ to Sn4+ in acidic medium. In acidic medium, its oxidising action is represented (iii) It is used as disinfectant. (iv) It is used in bleaching wool, silk and other as: textile fibres. Cr2O72– +14H+ + 6e– → 2Cr3+ + 7H2O (E0 = 1.33 V) (v) It is also used in decolourisation of oils. Uses: • PwchrirtehopmNaaarta2eCt:iOo3ninotfhKe2pCrre2sOen7:ceCohfrOom2 tiotefoorrme issofduisuemd (i) It is used as an oxidising agent. (ii) It is used for tanning of leather. 4FeCr2O4 + 8Na2CO3 + 7O2 → (iii) It is used as a disinfectant. Sodium 8Na2CrO4 + 2Fe2O3 + 8CO2 (iv) It is used in volumetric analysis as a primary then on cthrreoamtmateenotnwaictihdiKficCaltiiotncghiavnesgeNsat2Co rg2Oiv7e, standard. (v) It is used in chromyl chloride test for Cl– ions K2Cr2O7 as follows: in qualitative analysis. ExErcisE 4.2 Multiple Choice Questions (MCQs) (1 Mark) 4. Which of the following is amphoteric oxide? 1. oOcobnntacaeinndetddriatwtieohdnicHohf2SissOm4h,aigallhglayrmeeeoxnuponloitlsyiovcfeoKminMponnuaOntud4 rteios. Mn2O7, CrO3, Cr2O3, CrO, V2O5, V2O4 (i) V2O5, Cr2O3 (ii) Mn2O7, CrO3 (iii) CrO, V2O5 (iv) V2O5, V2O4 Identify the compound from the following: 5. KmMedniuOm4 .aWctsheans an oxidising agent in alkaline KI, iodide ion is (i) Mn2O7 (ii) MnO2 alkaline tKoM...n..O...4..i.s...t.r..e..a.t.ed with (iii) MnSO4 (iv) Mn2O3 oxidised 2. When oKnM, tnhOe4 solution is added to oxalic acid (i) I2 (ii) IO– soluti decolourisation is slow in the (iii) IO3– (iv) IO4– beginning but becomes instantaneous after some 6. Highest oxidation state of manganese in fluoride time because is +4 ((MMnnF2O4)7)bubet chaiughsees..t..o..x.i..d..a..t.i.o..n...s.tate in oxides is +7 (i) CO2 is formed as the product. (ii) Reaction is exothermic. (i) fluorine is more electronegative than oxygen. (iii) MnO4– catalyses the reaction. (ii) fluorine does not possess d-orbitals. (iv) Mn2+ acts as autocatalyst (iii) fluorine stabilises lower oxidation state. 3. KmbeMendneiueOmd4e.daTcthtoesrneauasmctabwneirtoohxfoimdnoiesleminsogolfeaKogMfesnnutOlp4ihntihdaaetciiwodniilscl (iv) in covalent compounds fluorine can form single bond only while oxygen forms double bond. in acidic solution is Assertion-Reason Type Questions (1 Mark) (i) 2 (ii) 3 Note: In the following questions a statement 5 5 of assertion followed by a statement of reason is given. Choose the correct answer out of the (iii) 4 (iv) 1 following choices. 5 5 126 Chemistry-12

(i) Both assertion and reason are true, and the oxidation state equal to its group number. reason is the correct explanation of the assertion. [NCERT] (ii) Both assertion and reason are true but reason 19. What is meant by disproportionation? Give is not the correct explanation of assertion. two examples of disproportionation reaction in aqueous solution. [NCERT] (iii) Assertion is true but reason is false. Short Answer Type Questions-II (3 Marks) (iv) Assertion is false but and reason is true. 20. How does acidified KMnO4 solution reacts with: 7. Aagsesnetrtiniopnr:esKeMncneOo4f is not used as an oxidising (i) Iodide ion (ii) H2S HCl. (iii) H2O2 Reason: KMnO4 is not very soluble in water. 21. Complete the following reactions: 8. Assertion: In acidic medium, K2Cr2O7 is orange coloured. (i) Cr2O72– + 6Fe2+ + 14H+ → (ii) 2CrO42– + 2H+ → Reason: In basic medium, K2Cr2O7 forms yellow (iii) 2MnO4– + 5C2O42– + 16H+ → [Delhi 2013] coloured C r O42− . 22. Explain the following: 9. Assertion: Permanganate and manganate both are tetrahedral. (i) AsocdidiuifmiedsuKlp2Chrit2Oe 7issoaldudteiodnttouirtn. s green when Reason: Permanganate is square planar and (ii) A ferrous salt decolourises acidified KMnO4 manganate is tetrahedral. solution. 10. Assertion: MnO42− is paramagnetic. 23. Why only dilute sulphuric acid and not dilute HCl or dilute HNO3 is used to acidify the permaganate Reason: MnO4− is paramagnetic. solution during volumetric analysis? 11. Assertion: KMnO4 is stored in dark coloured 24. When chromite ore, (A) is fused with NaOH in bottles. the presence of air, a yellow coloured compound Reason: KMnO4 decomposes on heating at 513K. (B) is obtained which on acidification using dilute Very Short Answer Type Questions (1 Mark) sulphuric acid forms a compound (C). Compound 12. Name the following: (C) when treated with KCl forms an orange (i) Transition metal compound in which metal is in +7 oxidation state. coloured crystalline compound (D). (ii) Ore which is used in the preparation of (i) Write the formulae of the compounds (A), (B), potassium dichromate. (C) and (D). 13. Complete the following reaction: (ii) Give one use of compound (D). [Delhi 2016] [NCERT Exemplar] Cr2O72– (aq) + H2S (g) + H+ (aq) → [AI 2012] 14. Complete the following reaction: 25. Describe the oxidising nature of potassium dichromate and write the ionic equations for its MnO4– + 8H+ + 5e– → [Delhi 2014] Short Answer Type Questions-I (2 Marks) reaction with: 15. The yellow colour of an aqueous solution of (i) iodide ion (ii) iron (II) solution and NExap2ClariOn.4 changes to orange on passing CO2 gas. (iii) H2S [AI 2009] [NCERT] Long Answer Type Questions (5 Marks) 16. Complete the following chemical equations: 26. Describe the process for preparation of potassium permanganate from pyrolusite ore. Write a (i) MnO4– (aq) + S2O32–(aq) + H2O(l) → (ii) Cr2O72–(aq) + Fe2+(aq) + H+(aq) → balanced chemical equation for at least one [AI, Delhi 2012] reaction to show that potassium permanganate 17. Complete the following equation: is oxidising in nature. [AI 2013(C)] (i) Cr2O72– + 2OH– → 27. D e s c r i b e t h e p r e p a r a t i o n o f p o t a s s i u m (ii) MnO4– + 4H+ + 3e– → permanganate. How does the acidified 18. Name the oxometal anions of the first series of the transition metals in which the metal exhibits permanganate solution react with (i) iron (II) ieoqnusa(tiiio)nSsOfo2ratnhde (iii) oxalic acid. Write the ionic reactions. [NCERT] Answers 4.2 1. (i) 2. (iv) 3. (i) is not the correct explanation of assertion. In 4. (i) 5. (iii) 6. (iv) presence of HCl, HCl will get oxidised to Cl2 and H2O. 7. (ii) Both assertion and reason are true but reason the d- and f-BloCK elements 127

8. (ii) Both assertion and reason are true but reason 22. (i) Nacaid2SifiOed3 is a reducing agent. It reduces is not the correct explanation of assertion. which is sulphate Kgr2Ceern2Oi7nscoolulotuiornatso chromic 9. (iii) Assertion is true but reason is false. follows: 10. (iii) Assertion is true but reason is false. K2Cr2O7 + 4H2SO4 + 3Na2SO3 → + 4H2O K2SO4 + 3Na2SO4 + Cr2 (SO4)3 11. (ii) Both assertion and reason are true but reason Green is not the correct explanation of assertion. (ii) Ferrous salt acts as a good reducing agent. 12. (i) KMnO4 (or any other suitable example) It reduces acidified KcMolonuOr4leisnstosMoluntSioOn4, and (ii) Chromite ore Kde2ScoOlo4 uwrhisiachtiofonrmofs i.e. place: 13. Cr2O72–(aq) + 3H2S(g) + 8H+(aq) → KMnO4 solution takes 2Cr3+(aq) + 3S(s) + 7H2O(l) 2KMnO4 + 10FeSO4 + 8H2SO4 → 14. MnO4– + 8H+ + 5e– → Mn2+ + 4H2O K2SO4 + 2MnSO4 + 5Fe2(SO4)3 + 8H2O 15. sCoOlu2toionn.dissolution in water produces an acidic 23. Volumetric titrations involving potassium CO2 + H2O → H2CO3  2H+ + CO32– permanganate solution are usually carried out mNae2dCiurOm4: changes to Na2Cr2O7 (orange) in acidic iitsnhebthepceraepusrseeenscoeexnyocefgedoniflupdtrieloudHtueCcHel do2SrbOyd4iltuohnteelycHoamNnbOdi3nn.aoTttihoiinns 2CrO42– + 2H+ → Cr2O72– + H2O 16. (i) 8MnO4–(aq) + 3S2O32–(aq) + H2O(l) → KorexMdyugnceOinn4go+nagdietinlstHo. wM2SnoOrse4ooivsaesru,stHoed2oSxoOind4lidysoeteotshnoeoxtirdepidrsouedcutinhcgee 8MnO2(s) + 6SO42– + 2OH– (ii) C r O 2 –( a q ) + 6Fe2+(aq) + 14H+(aq) → 2 7 2Cr3+(aq) + 6Fe3+(aq) + 7H2O(l) aagnedncto, mwhpielteeHs wNiOth3 dKoMesnpOro4.duce O2 gas on its own 2KMnO4 + 3H2SO4 → K2SO4 + 2MnSO4 17. (i) Cr2O72– + 2OH– → 2CrO42– + H2O (ii) MnO4– + 4H+ + 3e– → MnO2 + 2H2O + 3H2O + 5[O] 18. Cr2O72– (dichromate) and CrO42– (chromate) ions, Reducing agent + [O] → Oxidised product Group number of Cr = Oxidation state of Cr = 6 In case if HCl is being used, the oxygen so MnO4– (Permanganate), Group number of Mn = Oxidation state of Mn = 7 pHthrCeodlouaxnciedddissfeordmomperCKodlM2ugncatO.s4iissaplasortplyroudtuilciseeddatloonoxgiwdiitshe VO43– (Vanadate), 24. (i) 4FeCr2O4 [A] + 16NaOH + 7O2 → Group number of V = Oxidation state of 8Na2CrO4 [B] + 2Fe2O3 + 8H2O V=5 2Na2Cr2O4 [B] + H2SO4 → Na2Cr2O7 [C] + Na2SO4 + H2O 19. Those reactions are termed dispropor-tionation Na2Cr2O7 [C] + 2KCl → reactions in which the same substance undergoes K2Cr2O7 [D] + 2NaCl oxidation as well as reduction, i.e. oxidation number of an element increases as well as (ii) Cstormonpgouoxniddi[sDin],gi.aeg. eKn2tCirn2Oin7 dcaunstbrieesu.sed as a decreases simultaneously. Example: VI + 4H+ VII + IV + 2H2O 25. CaKhthcer2iiCn2sdOcrirce27e2O–amK7cie2toaCidncoirtnus2s,OmcCa7asr(nada(ci+lttus6pat)okeawiessHeraur2efSnpudOluot4cxoh)e.xieddiIdinetsioilsaneiCcncgirtgdrai(oacg+nge3mesn).neta.TtdnihIiannde, 3MnO42− → 2MnO4− MnO2 I II + 0 2 Cu + → Cu2+ Cu 20. (i) 2MnO4− + 10I− + 16H+ → chemical equation for the reduction process is: 2Mn2+ + 5I2 + 8H2O Cr2O72– + 14H+ + 6e– → 2Cr3+ + 7H2O (E° = 1.3 V) (ii) 2MnO4− + 5H2S + 6H+ → (i) Iodine is liberated from the solution 2Mn2+ + 5S + 8H2O containing iodide ions: (iii) 2MnO4− + 5H2O2 + 6H+ → Cr2O72– + 14H+ + 6I– → 2Cr3+ + 3I2 + 7H2O 2Mn2+ + 8H2O + 5O2 (ii) Ferrous ions is oxidised to form ferric ions: 21. (i) Cr2O72– + 6Fe2+ + 14H+ → Cr2O72– + 14H+ + 6Fe2+ → 2Cr3+ + 6Fe3+ + 7H2O 2Cr3+ + 6Fe3+ + 7H2O (ii) 2CrO42– + 2H+ → Cr2O72– + H2O (iii) H2S gas is oxidised to give sulphur: (iii) 2MnO4– + 5C2O42– + 16H+ → Cr2O72– + 8H+ + 3H2S → 2Cr3+ + 3S + 7H2O 2Mn2+ +8H2O + 10CO2 128 Chemistry-12

26. Formation of KMnO4: [2FeSO4 + H2SO4 + [O] → Iwtictahncabuesotbictapionteadshb(yKfOusHin) gorppyortoalusssiiutemocraer(bMonnOat2e) Fe2(SO4)3 + H2O] × 5 in the presence of air which gives a green mass due to the formation of potassium manganate. 2KMnO4 + 8H2SO4 + 10FeSO4 → K2SO4 + 2MnSO4 + 5Fe2(SO4)3 + 8H2O 2MnO2 + 4KOH + O2 Heat → + H2O 2K2 MnO4 27. Formation of KMnO4: Refer to the solution 26. Potassium manganate (Green mass) (i) Iron (II) get oxidised to form iron (III) ions. 2MnO2 + 2K2CO3 + O2 → + 2CO2 MnO4– + 8H+ + 5Fe2+ → 2K2 MnO4 Mn2+ + 5Fe3+ + 4H2O Potassium manganate (ii) SO2 is oxidised to form SO42– ions. 2MnO4– + 5SO2 + 2H2O → The green mass oafcKid2Mic nsOol4uutniodnertgooegsivdeispporotapsosrituiomn 2Mn2+ + 5SO42– + 4H+ in a neutral or (iii) Oxalic acid is oxidised to form CO2 gas. permanganate. 3K2MnO4 + 2H2SO4 → 2KMnO4 + MnO2 + 2H2O + 2K2SO4 Oxidising action of acidified KMnO4 on iron 2MnO4– + 6H+ + 5 → (II) ions: 2KMnO4 + 3H2SO4 → 2Mn2++ 10CO2 + 8H2O K2SO4 + 2MnSO4 + 3H2O + 5[O] Topic 3. Lanthanoids and Actinoids: Characteristics and Uses Lanthanoids [ce(58) to Lu(71)] to form nitride. They liberate hydrogen with dilute acids and form oxides and hydroxides • Electronic configuration: They have electron which are basic in nature. configuration with 6s2 common but variable 4f ã Lanthanoids are used for the production of occupancy. For tripositive ions it is of the form alloy steels for plates and pipes. Mischmetal 4f 1–14 (varies regularly with increasing atomic consist of lanthanoid metals (95%) and iron number). (5%) and traces of S, C, Ca and Al. It is used in Mg based alloy to produce bullets, shells • Atomic and ionic sizes: Atomic and ionic radii and lighter flint. decreases with atomic number due to ineffective ã Mixed oxides of lanthanoids are used as shielding of 4f electrons (even less effective catalysts for petroleum cracking. Some oxides than d-electrons. Decrease in atomic radii is not are used as phosphors in T.V. screens and in regular as it is in M3+ ions. fluorescing surfaces of computers. • Oxidation states: Common oxidation state is Actinoids [Th (90) to Lr (103)] +3 however, +2 and +4 ions in solution or in • Electronic configuration: They have electronic solid compounds are also found. +2 and +4 ions configuration with 7s2 common and variable are formed due to extra stability of empty, Half occupancy of 5f and 6d. The 5f orbitals are not as filled and fully filled orbitals. For example, Ce4+, buried as 4f orbitals and hence 5f electrons can Tb4+, Yb2+ and Eu2+. Eu2+, Yb2+ are good reducing participate in bonding. agents, Ce4+ and Tb4+ are good oxidising agents. • Ionic sizes: Size of atoms or M3+ ions gradually decreases due to actinoid contraction. This • General characteristics: contraction is greater due to poor shielding of 5f electrons. ã All lanthanoids are silvery white soft metals • Oxidation states: Greater range of oxidation and tarnish rapidly in air. state is due to comparable energies of 5f, 6d and 7s orbitals. Th, Pa, U, Np show +4, +5, +6, ã The melting point of lanthanoids ranges +7 oxidation states respectively, +3 is the most between 1000 to 1200 K but Samarium melts common oxidation state. at 1623 K. • General Characteristics: ã All actinoids are radioactive. The later ones ã They have metallic lustre and are good have very short half lives, Lr has t1/2 = 3 min. conductors of heat and electricity. ã Most of the trivalent ions are paramagnetic and they form coloured ions. ã These metals csuomlpbhiunretowfiothrmHs2utlpohfiodrems, hydrides, with with halogens to form halides, with nitrogen the d- and f-BloCK elements 129

ã Chemistry of actinoids is difficult to study as ã They plriboeteracttieveHo2xoidnerweaitchtioHnNwOi3t.h HCl but these are radioactive with short half-lives. form ã There are irregularities in metallic radii of ã They do not react with alkalies. actinoids more than lanthanoids. ã They are also paramagnetic and form ã They are highly reactive metals. coloured ions due to f-f transitions like ã Actinoids react with boiling water to form lanthanoids. oxides and hydroxides. ã Ionisation enthalpies of early actinoids are ã They react with non-metals at moderate temperature. not accurately known as these are radioactive but it is lower than lanthanoids due to outer 5f electrons are less firmly held. ExErcisE 4.3 Multiple Choice Questions (MCQs) (1 Mark) (iii) Assertion is true but reason is false. 1. Zr and Hf have almost equal atomic and ionic (iv) Assertion is false but reason is true. radii because 5. Assertion : Separation of Zr and Hf is difficult. (i) Both belong to same transition series Reason : Because Zr and Hf lie in the same group of the periodic table. (ii) Of lanthanoid contraction 6. Assertion : Actinoids form relatively less stable (iii) Of diagonal relationship complexes as compared to lanthanoids. (iv) Of actinoid contraction Reason : Actinoids can utilise their 5f orbitals along with 6d orbitals in bonding but lanthanoids 2. Which of the following oxidation state is common for all lanthanoids? do not use their 4f orbital for bonding. (i) +2 (ii) +3 Very Short Answer Type Questions (1 Mark) (iii) +4 (iv) +5 7. Name a member of lanthanoid series which is well known to exhibit +4 oxidation state. [NCERT] 3. Although Zirconium belongs to 4d transition series and Hafnium to 5d transition series even 8. How would you account for the fact that lanthanides primarily form +3 ions while actinoids then they show similar physical and chemical properties because .................... . usually exhibit higher oxidation states in their (i) both belong to d-block. compounds, +4 or +6 and even the typical ones. (ii) both have same number of electrons. [Delhi 2012] (iii) both have similar atomic radius 9. Name a member of lanthanide series which is well known to show +2 oxidation state. (iv) both belong to the same group of the periodic 10. Assign reason for the fact that: Ce3+ ions can be table. easily oxidised to Ce4+ ions. 4. Although +3 is the characteristic oxidation 11. Why are lanthanides considered as f-block state for lanthanoids but cerium also shows +4 elements? oxidation state because .................... . 12. Among lanthanides, Ln (III) compounds are more predominant. However, occasionally in solutions (i) it has variable ionisation enthalpy (ii) it has a tendency to attain noble gas or in solid compounds, +2 and +4 ions are also configuration seen. Explain. [AI 2012] (iii) it has a tendency to attain f 7 configuration 13. Actinoid contraction is greater from element to (iv) it resembles Pb4+ element than lanthanoid contraction. Why? Assertion-Reason Type Questions (1 Mark) [DSB 2009] [NCERT] Note: In the following questions a statement of 14. What are the different oxidation states exhibited assertion followed by a statement of reason by lanthanoids? [NCERT] is given. Choose the correct answer out of the Short Answer Type Questions-I (2 Marks) following choices. 15. The outer electronic configuration of two members of lanthanide series are: (i) Both assertion and reason are true, and reason is the correct explanation of the 4f 15d16s2 and 4f 7 5d0 6s2 assertion. What are their atomic numbers? Predict the (ii) Both assertion and reason are true but reason oxidation states shown by these elements in their is not the correct explanation of assertion. compounds. [AI 2008] 130 Chemistry-12

16. Chemistry of all the lanthanides is quite similar. OR Explain by giving suitable reasons. What is lanthanoid contraction? Write two of its 17. Which element is the last member of lanthanide series? Write its electronic configuration and consequences. [Delhi 2013(C)] comment on the oxidation state exhibited by it. 26. Which is the last element in the series of actinoids? Write the electronic configuration of this element. 18. Comment on the statement that actinoids can exhibit a large number of oxidation states while Comment on the possible oxidation state of this the corresponding lanthanoids does not. element. [NCERT] [AI 2008, 09, 10, 12] 27. (i) With reference to structural variability and chemical reactivity, what are the 19. Give one similarity and one difference between the chemistry of lanthanoids and actinoids. main differences between lanthanoids and 20. What are inner transition elements? Decide which actinoids? of the following atomic numbers are the atomic (ii) Name a member of lanthanoid series which numbers of the inner transition elements: 29, 59, can exhibit +4 oxidation state. 74, 95, 102, 104 [NCERT] 28. Name the members of the lanthanoid series which exhibit +4 oxidation states and those which 21. The chemistry of the actinoid elements is not so smooth as that of the lanthanoids. Justify this exhibit +2 oxidation states. Try to correlate this statement by giving some examples from the oxidation state of these elements. type of behaviour with electronic configurations [AI 2011] [NCERT] of these elements. [NCERT] OR 29. What are alloys? Name an important alloy which contains some of the lanthanoid metals. Mention The chemistry of actinoids is not as smooth as that of lanthanoids. Explain giving reason(s). its uses. [NCERT] [AI 2011] Long Answer Type Questions (5 Marks) 22. Use Hund’s rule to derive the electronic 30. Explain: configuration of Ce3+ ion and calculate its (i) How Sm2+, Eu2+ and Yb2+ are good reducing magnetic moment on the basis of ‘spin only’ agents? formula. [NCERT] 23. Write the electronic configurations of the elements (ii) Why lanthanum (z = 57) cannot exhibit +4 with the atomic numbers 61, 91, 101 and 109. oxidation state? [NCERT] (iii) Why +3 oxidation state of gadolinium (z = 24. Write down the electronic configuration of: (i) 64) and lutetium (z = 71) is stable? Pm3+, (ii) Ce4+, (iii) Lu2+, (iv) Th4+ [NCERT] (iv) Why Zr and Hf exhibits similar properties? Short Answer Type Questions-II (3 Marks) (v) Why the actinoid contractions is greater 25. What is lanthanoid contraction? What are the consequences of lanthanoid contraction? from element to element than lanthanoid [AI 2011] [NCERT] contraction? Answers 4.3 1. (ii) 2. (ii) 3. (iii) stable configuration (4f 0 5d0 6s0). Thus, Ce3+ can easily be oxidised to give Ce4+. 4. (ii) 5. (ii) 6. (iii) 11. Because the valence electrons in lanthanoids enter in the f-orbitals progressively. 7. Ce (Z = 58) exhibits +4 oxidation state because it 12. Lanthanoids exhibits +2, +4 oxidation state is highly stable in +4 oxidation state, as it has f 0 occasionally because of the extra stability of empty, half filled or fully filled f-orbitals. configuration. 13. This is due to poorer shielding effect of 5f electrons in actinoids than that of 4f electrons in Ce58 → 4f 1 5d1 6s2 lanthanoids. Ce4+ → 4f 0 5d0 6s0 14. The common stable oxidation state of lanthanoids is +3. However, some members can also show +2 8. This is because of very small energy gap between and +4 oxidation states in their compounds. 5f, 6d and 7s orbitals in actinoids electrons that 15. The complete electronic configuration for first element is: in these orbitals can lose to show high oxidation states. 9. Eu+2 (Eu) and Yb2+ (Yb). 10. Ce3+ has the electronic configuration: 4f 1 5d0 6s0. It can easily lose an electron and aquire a more the d- and f-BloCK elements 131

[Xe]54 4f1 5d1 6s2 to uneven distribution of oxidation states among Atomic number = 58, i.e. the element is Cerium. actinoids, their chemistry is not very smooth. The oxidation states in its compounds are +3 and +4. 22. Electronic configuration of Cerium = [Xe] 4f1 5d1 6s2 The complete electronic configuration for second Ce3+ ion = [Xe] 4f1, element is: i.e. one unpaired electron is present [Xe]54 4f 7 5d0 6s2 Magnetic moment, Atomic number = 63, i.e. the element is Europium. m = n(n + 2) = 3 = 1.73 BM The oxidation states in its compounds are +2 and +3. 23. Z = 61 (Pm) [Xe] 4f 5 5d0 6s2 16. The valence shell electronic configuration of all Z = 91 (Pa)[Rn] 5f 2 6d1 7s2 the lanthanoids is the same, as all the electrons are added into the inner penultimate 4f-subshell. Z = 101 (Md) [Rn] 5f 13 6d0 7s2 Hence, they show similar chemical properties. Moreover, change in the size of the lanthanoids Z = 109 (Mt) [Rn] 5f 14 6d7 7s2 is very small (non-appreciable) due to lanthanoid contraction. Therefore their chemical properties 24. Element At. Ion Electronic are almost similar. No. configura- tion 17. Lutetium (Z = 71), [Xe] 4f 14 5d1 6s2; Two electrons from valence shell i.e. (6s) can be (i) Promethium 61 Pm3+ [Xe] 4f 4 easily ejected out to give oxidation state of +2. Also, one electron from 5d can also be ejected (ii) Cerium 58 Ce4+ [Xe] 4f 0 out to form common oxidation state of +3. At +3 oxidation state, it forms a stable fully filled (iii) Lutetium 71 Lu2+ [Xe] electronic configuration. (iv) Thorium 4f 14 5d1 18. In actinoids, 5f orbitals which are extended into 90 Th4+ [Rn] 4f 14 the space beyond 6s and 6p orbitals participate in bonding whereas in lanthanoids, the 4f orbitals 25. In lanthanoid series, there is a regular decrease are buried deep inside the atom and thus are in the atomic and ionic radii of trivalent ions as not being able to participate in the bonding. The the atomic number increases, which is termed as participation of 5f-orbitals explains the fact that lanthanoid contraction. higher oxidation states are shown by actinoids. Consequences of lanthanoid contraction: 19. Similarity: Both lanthanoids and actinoids show (i) The change in atomic and ionic size of contraction in size/irregularity in their electronic lanthanoids is very small. Thus, all the configuration and both of them are stable in +3 members of lanthanoid series shows similar oxidation state. chemical properties. Thus, it is very difficult to separate these elements and obtain them Difference: Actinoids are radioactive, however in pure state. lanthanoids are not/Actinoids can show wide range of oxidation states, but lanthanoids does not. (ii) In vertical column of transition elements, there is a very small change or even no 20. Elements in which the last electron is accomodated change in the size of atoms from second in the f-subshell are called inner transition to third member. Thus, Zr—Hf, Mo—W, elements. These include elements from atomic Nb—Ta pairs possesses similar chemical numbers 58 to 71, i.e. lanthanoids and from 90 properties. to 103, i.e. actinoids. (iii) As the size of M3+ ions decreases from Ce3+ Elements with atomic numbers 59, 95 and 102 to Lu3+, so the covalent nbaatsuircesotrfeMng(OthHo)f3 are the inner transition elements. increases and hence the M(OH)3 decreases. 21. Lanthanoids exhibit limited number of oxidation 26. Last element of the actinoid series = Lawrencium states, +2, +3 and +4. Out of these +3 oxidation state is the most common. This occurs because of (Z = 103). the large energy gap between 4f, 5d and 6s energy subshells. Its electronic configuration: The general oxidation state of actinoids is also +3 = [Rn] 5f14 6d1 7s2 but they show a wide range of oxidation states too. For example, neptunium and plutonium can Possible oxidation state = +3 exhibit +3, +4, +5, +6 and +7 oxidation states. This is because of the small difference in energy 27. (i) Lanthanoids Actinoids between 5f, 6d and 7s-subshells in actinoids. Due (a) Atomic/ionic Atomic/ionic radii gradually radii shows decreases across much variation. the series though It decreases to not regular in case a greater extent of atoms. as compared to lanthanoids. 132 Chemistry-12

(b) +3 is the most Oxidation state An alloy of magnesium and about 3% mischmetal common oxidation ranges from +3 is used for making jet engine parts. state alongwith to +7. +2 and +4 in few 30. (i) The most stable oxidation state among cases. lanthanoids is +3, hence ions in +2 oxidation state tends to change to +3 oxidation state by (c) They are not They are highly loss of an electron and they act as a reducing agent. radioactive except reactive and (ii) La3+ has a stable configuration of an inert gas Pm. radioactive. [(Xe) 5d0 6s0], i.e. Xenon (54 → 2, 8, 18, 18, 8). To obtain +4 oxidation state, the stable (ii) Cerium (Ce+4) configuration needs to be disturbed, which is not possible under ordinary conditions and 28. +4 = 58 Ce, 59 Pr, 60 Nd, 65 Tb, 66 Dy hence, La+4 does not exist. +2 = 60 Nd, 62 Sm, 63 Eu, 69 Tm, 70 Yb These oxidation states are stable due to their (iii) It is because gadolinium in +3 oxidation state electronic configuration i.e., extra stability due has a stable half filled 4f subshell (4f 7) and to half filled orbitals in Eu2+ ([Xe]4f 7) and Tb4+ lutetium in +3 oxidation state has completely ([Xe]4f 7) while extra stability due to fully filled filled stable 4f subshell (4f 14). orbitals in Ce4+ ([Xe]4f 0) and Yb+2 ([Xe]4f 14). (iv) Due to lanthanoid contraction, the atomic 29. An alloy is a homogeneous mixture of two or more size of Hf does not increase as expected which metals or metals and non-metals. Mischmetal results the atomic radii of both Zr and Hf does is an alloy consisting of lanthanoid metals (94– not vary much. Hence, their properties are 95%), iron (5%) and traces of sulphur, carbon, also similar. silicon, calcium and aluminium atoms. The main lanthanoid metal present is cerium (about 40%), (v) In actinoids, the electrons are progressively lanthanum and neodymium (about 44%) are the filled in 5f orbitals which have poorer other two important constituents. This alloy shielding effect than that of 4f orbitals. Thus, can be used for making ignition devices such as actinoid contraction is greater from element bullets, shells, flints or lighters. to element than lanthanoid contraction. Case based questions The following questions are multiple choice 1. Case Study questions. Choose the most appropriate answer: Transition metals (Fe, Co, or Ni) modified K/ Mo2C catalysts were prepared and investigated (i) The main catalyst used for CO hydrogenation as catalysts for CO hydrogenation.The addition to form alcohol is of Fe, Co, or Ni to K/Mo2C catalyst led to a sharp increase in both the activity and selectivity of (a) Mo2C (b) Fe C2+OH, but the promotion effects were quite (c) Co (d) Ni different and followed the sequence: Ni > Co > Fe for the activity and Fe > Co > Ni for the alcohol (ii) Which of the following promotes selectively selectivity. For the products distributions, it also for the formation of C5+ hydrocarbon to displayed some differences; Co promoter showed maximum extent? much higher C5+ hydrocarbon selectivity than Fe or Ni promoter, but Fe or Co promoter gave lower (a) Fe (b) Ni methane selectivity than Ni promoter, and Fe promoter showed the highest C2+ – C4+ selectivity. (c) Co (d) K In terms of characterization results, the different interactions between transition metals (Fe, Co, (iii) What is the reason for different activity and or Ni) and Mo might account for the different selectivity of different transition metals viz. catalytic performances over K/Mo2C catalyst. (Source: Minglin Xiang and Juan zou. CO Fe, Co and Ni? Hydrogenation over Transition Metals (Fe, Co, or Ni) Modified K/Mo2C Catalysts. Journal (a) Different transition metals interact of Catalysts, Volume 2013, Article ID 195920, differently with K of K/Mo2C catalyst. 5 pages) (b) Different transition metals interact differently with Mo of K/Mo2C catalyst. (c) Different transition metals interact differently with C of K/Mo2C catalyst. (d) None of these. (iv) Why transition metals and their compounds are known for their catalytic activity? the d- and f-BloCK elements 133

(a) They are large in size Kinetics, Quantification, and Implications for Effective Drinking Water Treatment. Journal of (b) They are small in size Toxicology Volume 2019, Article ID 3231473, 13 pages) (c) They can form alloys In these questions, a statement of assertion followed by a statement of reason is given. (d) They can adopt multiple oxidation state Choose the correct answer out of the following choices. OR (a) Assertion and reason both are correct Which of the following statement is correct? statements and reason is correct explanation for assertion. (a) Hydrated Fe2+ ion has two unpaired electrons (b) Assertion and reason both are correct statements but reason is not correct (b) Hydrated Co2+ ion has three unpaired explanation for assertion. electrons (c) Assertion is correct statement but reason is (c) Hydrated Ni2+ ion has four unpaired wrong statement. electrons (d) Assertion is wrong statement but reason is (d) Hydrated Zn2+ ion has five unpaired correct statement. electrons (i) A s s e r t i o n : P e r m a n g a n a t e i o n i s 2. Case Study tetrahedral in shape. Permanganate pretreatment of drinking water is effective in transforming dissolved, noxious Reason: Permanganate ion is diamagnetic contaminants and in reducing halogenated in nature. by-products. Permanganate targets specific compounds such as taste and odour compounds, OR disinfection precursors, manganese, and natural Assertion: Pretreatment of drinking organic contaminants that are not removed readily water from permanganate is an effective by conventional treatment alone. Cyanobacterial way to disinfect water. blooms (cHABs) can increase disinfection by- product precursors as well as the cyanotoxin, Reason: Permanganate transforms microcystin (MC), a potent liver toxin. MC dissolved and noxius contaminants, toxicity is conferred by a unique, conserved amino reduces halogenated by-products. acid, Adda, that inhibits protein phosphatase 1 and 2A. The results presented provide a better (ii) Assertion: Permanganate titrations in understanding of how permanganate reduces presence of HCl are unsatisfactory. MC toxicity and identifies several practical implications. Chemical oxidation of MC with Reason: HCl is oxidised to chlorine. 2.0 mg/L permanganate with sufficient contact (iii) Assertion: Microcystin (MC) is a potent times can be an effective first barrier for the more hydrophilic MCs, MC-RR, and MC-LR that oxidize liver toxin. rapidly. Reason: MCs, MC-RR, MC-LR oxidise (Reference: : David C. Szlag , Brian Spies, rapidly with permanganate. Regina G. Szlag, and Judy A. Westrick. Permanganate Oxidation of Microcystin-LA: (iv) Assertion: Permanganate targets taste and odour compounds in water. Reason: Permanganate increases natural organic contaminants in water. 1. (i) (a) (ii) (c) (iii) (b) (iv) (d) OR (b) Answers 2. (i) (b) OR (a) (ii) (a) (iii) (b) (iv) (c) Analogy based questions (iv) EO (M2+ /M) for Cu : Negative : : EO (M2+/M) 1. Which of the following analogies is correct? for Ni : Positive (i) Ka MPrnimO4ar: yPSritmanadryarsdtandard :: K2Cr2O7 : Not 2. Complete the following analogy. (ii) Half filled sets of orbital : More stable : : Lanthanides : X :: Actinides : Y Completely filled sets of orbital : Less stable (i) X : One stable oxidation state, Y : Wide range of oxidation state. (iii) Cr2+ : Reducing :: Mn3+ : Oxidising 134 Chemistry-12

(ii) X : Half of them are radioactive, (iv) X : Binding energy of 4f orbitals is lower : : Y : All are radioactive Y : Binding energy of 5f orbitals is higher (iii) X : Form complexes easily 1. (iii) Answers Y : Do not form complexes 2. (i) Multiple type questions 1. Match the catalysts given in Column I with the processes given in Column II. Column I (Catalyst) Column II (Processes) A. Ni in the presence of hydrogen (1) Zieglar Natta cataylst B. Cu2Cl2 (2) Contact process C. V2O5 (3) Vegetable oil to ghee D. Finely divided iron (4) Sandmeyer reaction E. TiCl4 + Al(CH3)3 (5) Haber’s Process Code: (i) A (3) B (4) C (2) D (5) E (1) (ii) A (4) B (3) C (1) D (5) E (2) (iii) A (2) B (1) C (5) D (4) E (3) (iv) A (1) B (2) C (3) D (4) E (5) 2. Match the statements given in Column I with the oxidation states given in Column II. Column I Column II A. Oxidation state of Mn in MnO2 is (1) +2 B. Most stable oxidation state of Mn is (2) +3 C. Most stable oxidation state of Mn in oxides is (3) +4 D. Characteristic oxidation state of lanthanoids is (4) +5 (5) +7 Code: (i) A (3) B (5) C (1) D (2) (ii) A (2) B (4) C (1) D (3) D (2) (iii) A (3) B (1) C (5) D (2) (iv) A (4) B (5) C (1) 3. Match the property given in Column I with the element given in Column II. Column I (Property) Column II (Elements) A. Lanthanoid which shows +4 oxidation state (1) Pm B. Lanthanoid which can show +2 oxidation state (2) Ce C. Radioactive lanthanoid (3) Lu D. Lanthanoid which has 4f 7 electronic configuration in +3 oxidation state (4) Eu E. Lanthanoid which has 4f 14 electronic configuration in +3 oxidation state (5) Gd Code: (ii) A (2) B (4) C (1) D (5) E (3) (i) A (5) B (2) C (3) D (1) E (4) (iv) A (2) B (5) C (3) D (1) E (4) (iii) A (4) B (3) C (2) D (5) E (1) 1. (i) 2. (iii) Answers 3. (ii) the d- and f-BloCK elements 135

Quick revision notes Mn has exceptionally low melting point due • The elements belonging to groups 3 to 12 of the to broken crystal lattice. modern periodic table are called d-block elements. ã Among the transition metals, the lightest They are all metals. These are in middle of the is Sc (density 3.4 g cm–3) and heaviest is Os periodic table. (density 22.59 g cm–3). ã Hg is known to exist in liquid form under • They are less electropositive than s-block ordinary conditions. elements due to small size, higher ionisation • Generally, the atomic radii of d-block elements enthalpy but more electropositive than p-block in a series decreases with increase in atomic elements, therefore they are called transition number but the decrease in atomic size is very elements. Their general electronic configuration small after midway. is: (n – 1)d1–10 ns0–2. ã In the first series, the atomic radii decreases upto Cr but after that it remains almost the • All transition metals are d-block elements but all same in Fe, Co, Ni. Copper and Zinc are d-block elements are not transition metals. For bigger in size. example, the elements of group 12, namely Zn, Cd, ã The atomic radii of second and third transition Hg and Cn, though belongs to d-block elements series in the same vertical columns are but are not considered as transition metals as almost the same. This is due to lanthanoid these elements do not have partially filled (n – 1) contraction and increase in effective nuclear d-subshell either in ground state or as an ion. charge. ã The atomic radii of the pairs such as Zr-Hf, • There are four series of transition elements which Nb-Ta, Mo-W, Ag-Au, etc. are almost the involves the filling of (n – 1) d-subshell, e.g. 3d, same due to lanthanoid contraction. 4d, 5d, 6d subshell with each series consisting • The ionisation energy gradually increases of ten elements. e.g. 3d series starts from Sc(21) with increase in atomic number along a given to Zn(30), 4d series starts from Y(39) to Cd(48). transition series but variation is less due to less difference in their atomic radii. • Certain elements show exceptional electronic ã The ionisation energies of 3d and 4d series configuration. In the first transition series, Cr(3d5 elements are irregular due to less variation of 4s1) and Cu(3d10 4s1) have exceptional electronic size and stability of electronic configuration. configuration. ã The first ionisation energy of 5d series elements is higher than 3d series and 4d • The elements of first (3d) transition series are: series elements belonging to the same group Sc (21), Ti (22), V (23), Cr (24), Mn (25), Fe (26), due to lanthanoid contraction which leads to Co (27), Ni (28), Cu (29) and Zn (30). increase in effective nuclear charge. • Transition metals show variable oxidation state • All the transition elements are metals. (except first and last member of the series). The most common oxidation state of the first ã Iron is the most abundant transition metal transition series is +2, except scandium. ‘Sc’ in the earth’s crust. Technetium is a man- shows +3 oxidation state only. Zn does not show made element and all the members of fourth variable oxidation states. transition series are radioactive. ã The maximum oxidation state increases with atomic number upto manganese in ã Technetium is the first synthetic element. the first transition series. This is due to the involvement of 4s electrons and unpaired ã The elements after uranium (Atomic number electrons of 3d-subshell. 92) are called transuranic elements. ã Manganese (3d54s2) shows a maximum oxidation state of +7, Osmium shows +8 ã Transuranic elements have been obtained oxidation state in OsO4 in 5d-series. artificially by nuclear transmutations. ã Compounds of transition metals with lower oxidation states are ionic while higher • Transition elements have high density, hardness, oxidation states are covalent due to high high tensile strength, malleability, ductility, high polarising power. thermal and electrical conductivity, high energy of atomisation, high melting and boiling points, form alloys, complexes, coloured ions and act as catalysts. ã Larger the number of unpaired electrons, stronger is the metallic bond and hence higher is the energy of atomisation and the melting point. That is why melting point increases in each series upto the fourth member, i.e. among Cr, Mo and W. W has the highest melting point–3683 K). 136 Chemistry-12

ã The highest oxidation state shown by any ã In the first transition series, Cr has the transition metal is +8 (by Ru and Os). maximum paramagnetism and amongst ions, Mn2+ has the maximum magnetic moment ã Metal ions in a high oxidation state tends to 5.92 BM. be a good oxidising agent l(oCwr2Oox72i–d, MatnioOn4– ) and • Most of the transition metal compounds (ionic as oxo–anions of metals in state well as covalent) are coloured, both in solid state as well as in aqueous solution, except Sc3+, Ti4+, are good reducing agents (V2+, Cr2+). V5+, Cu+, Zn2+ ions due to the absence of unpaired electrons. • Most of the transition metals and their compounds exhibits paramagnetism. Substances containing ã Colour is due to the presence of incomplete unpaired electrons are paramagnetic, i.e. these d-subshell, i.e., there is d-d transition are attracted by the magnet. by absorbing visible light and radiating complementary colour. ã Those containing no unpaired electrons are diamagnetic (Zn, Cd, Hg, Zn2+, Cd2+, Hg2+, ã The compounds with ions having (n – 1)d0 or Cu+, Sc3+, Ti4+, V5+, etc). These are repelled (n – 1)d10 configuration are colourless. Sc3+, by the magnet. Te4+, Cu+, Zn2+, etc. are colourless because they have either 3d0 or 3d10 configuration. ã Magnetic moment is given as m = n(n + 2) ã Some of the colours observed for the transition BM, where n = number of unpaired electrons. metal ions are: BM is Bohr’s magneton, unit of magnetic moment. Ti3+ V3+ Mn3+ Fe3+ Fe2+ Cu2+ Co2+ Ni2+ Purple Green Violet Yellow Green Blue Pink Green • Transition metals have a tendency to form sodium carbonate and the mixture is fused in a complex compounds due to high ionic charge, small reverberatory furnace in excess of air. The roasted size, vacant d-orbitals. They form interstitial mass is extracted with water and the solution is compounds and alloys due to similar atomic size. ltersesatseodluwbliethcrcyosntca.llHise2SsOo4u.Nt aan2SdOi4s.1f0ilHte2rOedb. eHinogt They itself and their compounds can act as good catalysts as they show variable oxidation states. solution of NKaC2Cl rw2Oh7eirse treated with calculated amount of K2Cr2O7 crystals are • Iron, cobalt and nickel are termed as ferrous obtained: metals. 4FeO.Cr2O3 + 8Na2CO3 + 7O2 → 8Na2 CrO4 + 2Fe2 ã Copper, silver and gold are called coinage O3 + 8CO2 metals. Soluble Insoluble • Transition metals vary widely in their chemical 2Na2CrO4 + H2SO4 → Na2Cr2O7 + Na2SO4 + H2O Na2Cr2O7 + 2KCl → K2Cr2O7 + 2NaCl behaviour. Many of them are sufficiently ã Its acidic solution contains dichromate ions electropositive to dissolve in mineral acids although a few are noble, e.g. Au, Ag and Pt. while alkaline solution contains chromate ã Of the first transition metals series, with ions. It is a strong oxidising agent. Dichromate the exception of copper, all the metals are ions are orange, while chromate ions are relatively reactive. yellow. • The transition metals react with a number of non-metals like oxygen, sulphur, halogens, etc. to form binary compounds which are ionic in lower oxidation state and covalent in higher oxidation state. ã The first series transition metal oxides are formed by the reaction of metals with oxygen Cr2 O72– + 2OH– → 2CrO24– + H2 O at high temperature because they are less Orange Yellow reactive. 2CrO24– + 2H+ → Cr2 O72– + H2 O ã These oxides dissolve in acids and bases to Yellow Orange form oxo-metallic salts. ã pArciimdiafrieydstKa2nCdra2rOd7 solution is used as a in volumetric analysis ã Potassium dichromate and potassium permanganate are a few examples. for the estimation of reducing agents such • KCr2C2Or23O): 7Tisheprfeinpaelryedpforwomdetrhede chromite ore (FeO. as Ferrous sulphate (Mohr’s salt), Oxalic ore is mixed with acid, etc. the d- and f-BloCK elements 137

• TKhMenfOin4eilsy prepared from pyrolusite ore K(MOnHOo2)r: called Lanthanides or Lanthanoids or powdered ore is fused with Lanthanons as these show similarities with Lanthanum, La (57). Only promethium (61) Kfo2rCmOK3 i2nMtnhOe4p(rpeosteanscseiuomf amiraonrgoaxnidaitsei)n.g agent to is radioactive and man-made. 2MnO2 + 4KOH + O2 → 2K2MnO4 + 2H2O • It is then oxidised chemically by uKsMinngOH42:SO4, CO2 ã 5f-series: 5f-orbitals are gradually filled or Cl2 or by electrolysis to form up. These are also 14 elements from Th (90) to Lr (103) in 5f series and these are called MnO42– Electrolysis MnO4– + e– Actinides or Actinoids or Actinons as • KmModneOra4tieslyasodleuebplepiunrpwlaetcerry. stalline compound, these show similarities with actinium, Ac (89). These are radioactive elements. Most ã It decomposes when heated to give out O2 gas: of them are man-made. 2KMnO4 heat K2MnO4 + MnO2 + O2 • There is gradual decrease in the atomic and ã It acts as a strong oxidising agent in neutral, ionic sizes of these elements along the series. This contraction is called lanthanide (or alkaline and acidic medium. lanthanoid) contraction and actinide (or actinoid) contraction, for lanthanoids and ã TishuesaecdidaisfiaedvKolMumneOt4ri(cinrdeailg.eHn2tS. O4) solution actinoids respectively. It is due to poor shielding O– O of 4f-electrons and 5f-electrons or increase in effective nuclear charge. Mn Mn • The principal oxidation state of lanthanoids is O O– O O– +3, although +4, +2 oxidation states are also OO exhibited. Those having +2(i.e., Sm2+, Eu2+ and Yb2+) oxidation state can change their oxidation ã It is not a primary standard due to less state to +3 and acts as reducing agent. Eu2+ is more stable due to half filled d-orbitals. Those solubility. having +4 oxidation state tends to change to +3 oxidation state and acts as an oxidising agent. ã It gains 5e– in acidic conditions to form Mn2+ ions, e.g. MnO4– + 8H+ + 5e– → Mn2+ + 4H2O. • The maximum oxidation state shown by actinides Alkaline KanMdniOt 4issoulusetidoninisocragallnedicBaaneaylyesri’ss. is +7. However, all the actinides show a general ã reagent oxidation state of +3. Mi.en. Osp4–3ahnydbMridniOsa42–tihoanv. e tetrahedral structure • Most of the trivalent lanthanide ions are coloured. This is due to f-f transitions. Actinide cations are • The f-block elements (inner transition elements) also generally coloured. The cations having 5f 0 or are placed outside the periodic table. Their 5f 7 or 5f 14 configuration are colourless. general configuration is: (n – 2)f1–14 (n – 1)d0–1 ns2. • aLma(oOnHg)l3ains tmhoasntidbaess.ic while Lu(OH)3 is least basic • There are two series of inner-transistion elements: ã 4f-series: 4f-orbitals are gradually filled up. These are 14 elements from Ce (58) to Lu (71) ImporTanT rEacTIons 1. 2Cu2+ + 4I– → Cu2I2 + I2 2. 2Cr2+(aq) + 2H+(aq) → 2Cr3+(aq) + H2(g) 3. 2I– + S2O82– → I2 + 2SO42– 4. 2Fe3+ + 2I– → 2Fe2+ + I2 5. 2Fe2+ + S2O82– → 2Fe3+ + 2SO42– 6. 3MnO42– + 4H+ → 2MnO4– + MnO2 + 2H2O 7. 4FeCr2 O4 + 8Na2CO3 + 7O2 → 8Na2 CrO4 + 2Fe2O3 + 8CO2 Iron Chromite Sodium chromate 8. 2Na2CrO4 + 2H+ → Na2 Cr2 O7 + 2Na+ + H2O 9. Na2Cr2O7 + 2KCl → K2Cr2O7 + 2NaCl Sodium dichromate 10. 2CrO42– + 2H+ → Cr2O72– + H2O 11. Cr2 O72– + 2OH– → 2CrO24– + H2O Orange Yellow 12. Cr2O72– + 14H+ + 6e– → 2Cr3+ + 7H2O 13. Cr2O72– + 6I– + 14H+ → 2Cr3+ + 3I2 + 7H2O 14. Cr2O72– + 3H2S + 8H+ → 2Cr3+ + 3S + 7H2O 15. Cr2O72– + 14H+ + 6Fe2+ → 2Cr3+ + 6Fe3+ + 7H2O 16. Cr2O72– + 14H+ + 3Sn2+ → 3Sn4+ + 2Cr3+ + 7H2O 17. 2MnO2 + 4KOH + O2 → 2K2MnO4 + 2H2O 138 Chemistry-12

18. MnO2 Fused with KOH, MnO24– with air or KNO3 Manganate ion 19. MnO24– Electrolytic reduction in alkaline solution MnO4– Manganate ion Permanganate ion 20. 2Mn2+ + 5S2O82– + 8H2O → 2MnO4– + 10SO42– + 6H+ 21. 2KMnO4 heat K2MnO4 + MnO2 + O2 22. 2MnO4– + 5COO– + 16H+ → 2Mn2+ + 10CO2 + 8H2O | COO– 23. MnO4– + 5Fe2+ + 8H+ → Mn2+ + 5Fe3+ + 4H2O 24. 2MnO4– + 10I– + 16H+ → 2Mn2+ + 5I2 + 8H2O 25. 2MnO4– + 5NO2– + 6H+ → 2Mn2+ + 5NO3– + 3H2O 26. MnO4– + 4H+ + 3e– → MnO2 + 2H2O 27. MnO4– + 8H+ + 5e– → Mn2+ + 4H2O 28. 5H2S + 2MnO4– + 6H+ → 2Mn2+ + 8H2O + 5S 29. 5SO32– + 2MnO4– + 6H+ → 2Mn2+ + 3H2O + 5SO42– 30. 2MnO4– + H2O + I– → 2MnO2 + 2OH– + IO3– 31. 8MnO4– + 3S2O32– + H2O → 8MnO2 + 6SO42– + 2OH– 32. 2MnO4– + 3Mn2+ + 2H2O → 5MnO2 + 4H+ 33. 2Ln + 3S → Ln2S3, where Ln stands for lanthanoid. 34. 2Ln + 3O2 → Ln2O3 35. Ln + 2C ' LnC2 2773 K 36. Ln + X → LnX3 38. Ln + N → LnN 37. Ln + H2O → Ln(OH)3 + H2 39. Ln + H+(acids) → H2 common Errors Errors corrEcTIons (i) Students generally write incorrect electronic Due to stability of half-filled and fully-filled orbitals electronic configuration of Cr is [Ar]3d54s1 configuration of Cu, Cr. and Cu is [Ar]3d104s1. (ii) While writing electronic configuration of ions Students should write the electronic configuration arrange in increasing order of “n”. Then remove for any transition element, students incorrectly the electrons from valence shell ‘ns’. remove the electrons from the penultimate shell. 3d subshell has five orbitals, so maximum (iii) In case of 3d6 some students incorrectly write the number of unpaired electrons is 5 and after that number of unpaired electrons as 6. the next electron will pair up in a singly occupied orbitals to give number of unpaired electrons is 4. (iv) Students have confusion on the reducing and An oxidising agent itself gets reduced to oxidise oxidising nature. other species while a reducing agent itself gets oxidised to reduce other species. (v) Students think that the lanthanoids and actinoids These two rows are placed at the end of periodic form the last two rows of the periodic table. table for the sake of convenience otherwise it is difficult to arrange the elements according to (vi) Both La and Ac has d1 configuration, therefore their similarity in properties. students wrongly consider them as d block Both La and Ac in their +3 oxidation state behave elements. similar to that of lanthanoids and actinoids. the d- and f-BloCK elements 139

rEvIsIon charT d-block elements Elements in which last electron enters in a d-orbital of the respective penultimate shell are known as d-block elements. General electronic configuration: General characteristics : (n–1)d1–10 ns0–2 • Show variable oxidation states though most common There are 4-transition metal series; oxidation state is +2. (i) 3d-series–21Sc–30Zn • High m.p. and b.p. due to strong metallic bonding. (ii) 4d-series–39Y–48Cd (iii) 5d-series–57La, 72Hf–80Hg • High enthalpies of atomisation. (iv) 6d-series–89Ac, 104Rf–112Cn • Ionisation enthalpies increases from left to right in a series. Important transition element compounds • Atomic radii decreases from left to right in a series. • Easily form complexes due to high nuclear charge, small size and availability of empty d-orbitals. • Form coloured compounds due to d-d-electronic transitions. • Many transition metal ions and their compounds are paramagnetic due to unpaired e–s in (n–1)d-orbitals. • Can easily form complexes, thus exhibits catalytic behaviour. • Due to empty space in lattice sites, form interstitial compounds and alloys. • Magnetic moment, µ = n]n + 2g BM Where, n = no. of unpaired e–s K2Cr2O7 KMnO4 Preparation: Preparation: • 4FeCr2O4 + 8Na2CO3 + 7O2 → 8Na2CrO4 + 2Fe2O3 + 8CO2 • 2MnO2 + 4KOH + O2 → 2K2MnO4 + 2H2O • 2Na2CrO4 + 2H+ → Na2Cr2O7 + 2Na+ + H2O • 3K2MnO4 + 2H2SO4 → 2KMnO4 + MnO2 + 2K2SO4 + 2H2O • Na2Cr2O7 + 2KCl → K2Cr2O7 + 2NaCl Properties: Purple crystalline solid and act as oxidising Properties: Orange red crystalline solid and act as agent in acidic, neutral and faintly alkaline medium oxidising agent in acidic medium as: Cr2O7–2 + 14H+ + 6e– → 2Cr3+ + 7H2O Uses: As a disinfectant, germicide and as a Bayer’s reagent (alk. KMnO4) Uses: In dyes, photography, leather and tannning industry 140 Chemistry-12

rEvIsIon charT f-block elements These are the elements in which the last electron enters (n–2)f-subshell. These are characterised by filling up of f-orbitals with general electronic configuration (n–2)f1–14 (n–1)d10 ns0–2 lanthanoids Actinoids It involves filling up of 4f-orbitals starting from 57La to 71Lu It involves filling up of 5f-orbitals starting from 89Ac to 103Lr General characteristics : General characteristics : • Highly dense metals with high melting point. • Highly dense metals with high melting point. • Form alloys with other metals. • Form alloys with other metals, especially with iron. • Soft, malleable, ductile with low tensile strength • Silvery white metals and are highly reactive. • Most common oxidation state is +3. But, also exhibit +2 and • Tarnished on exposure to alkalies, however they are less +4 oxidation states. reactive towards acids. • Some trivalent ions are coloured due to (f–f)e–-transitions. • Most common oxidation state is +3. But also exhibit +4, +5, +6 and +7 oxidation states. • La3+ and Lu3+ is diamagnetic, all other are paramagnetic due to unpaired electrons. • Some of the ions are coloured due to (f–f)e–-transistions. • Highly electropositive metals with similar chemical reactivity. • Most of the actinoid ions are coloured due to unpaired electrons in 5f-orbital. • Due to empty space in lattice sites, form interstitial compounds. • They are highly electropostive and form salts as well as complexes. • Also form alloys due to similar atomic size. • All the actinoids are radioactive. lanthanoid contraction: Actinoid contraction: • Among lanthanoids the decrease in atomic/ionic radii with • Among actinoids the decrease in atomic/ionic radii increase in atomic number is very small. with increasing atomic number is greater than in • This is due to poor sheilding effect of 4f-orbitals. lanthanoids. • This is due to the poor sheilding effect of the 5f-orbitals. Uses: Highly useful in making alloys, as catalysts, in dyes, Uses: bullets and shells. • Thorium is used in treatment of cancer. • U and Pu are used as a fuel in nuclear reactors. • Also U is used in glass, textiles and medicines. the d- and f-BloCK elements 141

chapTEr TrEnd—Based on past Years’ cBsE Exams ➣ It has been observed from this chapter that the weightage of the topics‘Transition Elements: General Characteristics and Uses’and‘Preparation and Properties of K2Cr2O7 and KMnO4’are maximum. Hence, these are the most important topics. ➣ From this chapter, generally 2 marks questions were asked from the topics‘Transition Elements: General Characteristics and Uses’ and ‘Preparation and Properties of K2Cr2O7 and KMnO4’. ➣ Most of the 3 marks questions asked from this chapter belong to the topics ‘Transition Elements: General Characteristics and Uses’ and ‘Preparation and Properties of K2Cr2O7 and KMnO4’. ➣ Most of the 1 mark questions asked from this chapter belong to the topic ‘Lanthanoids and Actinoids: Characteristics and Uses’. Questions For practice Very Short Answer Type Questions (1 Mark) 1. Why do actinoids show a wide range of oxidation states. 2. Although Zr belongs to 4d and Hf belongs to 5d transition series but it is quite difficult to separate them. Why? Short Answer Type Questions-I (2 Marks) 3. Account for the following: (i) The highest oxidation state is exhibited in oxo-cations of transition metals. (ii) HCl is not used to acidify KMnO4 solution. 4. Copper atom has completely filled d-orbitals in its ground state but it is a transition metal. Why? [Delhi 2015] 5. (i) Calculate the spin magnetic moment of M2+(aq) ion. Atomic number (z) = 27 [Delhi 2015] (ii) Chromium is typical hard metal while mercury is liquid, explain. 6. Although fluorine is more electronegative than oxygen, yet the ability of oxygen to stabilise the higher oxidation state exceeds than that of fluorine. Why? 7. Although Cr3+ and Co2+ ions have same number of unpaired electrons but the magnetic moment of Cr3+ is 3.87 BM and that of Co2+ is 4.87 BM. Why? Short Answer Type Questions-II (3 Marks) 8. Following are the transition metal ions of 3d series: Ti4+, V2+, Mn3+, Cr3+ [Delhi 2017] [Atomic number: Ti = 22, V = 23, Mn = 25, Cr = 24] Ni Cu Answer the following questions: (i) Which ion is most stable in aqueous solution and why? (ii) Which ion is strong oxidising agent and why? (iii) Which ion is colourless and why? 9. Cr Mn Fe Co E0 –0.91 V –1.18 V –0.44 V –0.28 V –0.25 V +0.34 V (M2+/M) From the given E0 values, answer the following questions: (i) Why is values E0 exceptionally +ve? (Cu2+/Cu) (ii) Why is E0 value is highly negative as compared to other elements? (Mn2+/Mn) (iii) Which one is a stronger reducing agent: Cr2+ or Fe2+? 10. A blackish brown solid ‘A’ when fused with alkali metal hydroxide in presence of air, produces a dark green coloured compound ‘B’, which on electrolytic reduction in alkaline medium to give dark purple coloured compound ‘C’. Identify A, B and C and write the reactions involved. 142 Chemistry-12

assIgnmEnT Time: 45 Minutes M.M.: 25 Multiple Choice Questions (MCQs) (1 Mark) 1. Electronic configuration of a transition element X in +3 oxidation state is [Ar]3d5. What is its atomic number? (i) 25 (ii) 26 (iii) 27 (iv) 24 2. On addition of small amount Iodf eKnMtifnyOt4hteoccoomnpceonutnrdatferodmHt2hSeOf4o, lalogwrienegn: oily compound is obtained which is highly explosive in nature. (i) Mn2O7 (ii) MnO2 (iii) MnSO4 (iv) Mn2O3 Assertion-Reason Type Questions (1 Mark) Note: In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. (i) Both assertion and reason are true, and reason is the correct explanation of the assertion. (ii) Both assertion and reason are true but reason is not the correct explanation of assertion. (iii) Assertion is true but reason is false. (iv) Assertion is false but reason is true. 3. Assertion: KMnO4 is not used as an oxidising agent in presence of HCl. Reason: KMnO4 is not very soluble in water. 4. Assertion : Separation of Zr and Hf is difficult. Reason : Because Zr and Hf lie in the same group of the periodic table. Very Short Answer Type Questions (1 Mark) 5. Why copper does not replace hydrogen from acids? 6. Out of Cu2Cl2 and CuCl2, which one is more stable and why? 7. Although +3 oxidation state is the characteristic oxidation state of lanthanoids but cerium shows +4 oxidation state also. Why? Short Answer Type Questions-I (2 Marks) 8. The second and third row of transition elements resemble each other to a greater extent than they resemble the first row. Explain why? 9. Why E° values for Mn, Ni and Zn are more negative than their expected values? 10. Why first ionisation enthalpy of Cr is lower than that of Zn? Short Answer Type Questions-II (3 Marks) 11. While filling up of electrons in the atomic orbitals, the 4s orbital is filled before the 3d-orbital but reverse happens during the ionisation of the metal atom. Explain Why? 12. Reactivity of transition elements decreases almost regularly from Sc to Cu. Explain. 13. When a brown compound of manganese (A) is treated with HCl it gives a gas (B). The gas taken in excess, reacts with NH3 to give an explosive compound (C). Identify the compounds A, B and C. 14. sAolsuotliuotniodnepoef nKdMinngOu4poonn reduction yields either a colourless solution or a brown precipitate or a green the pH of the solution. What different stages of the reduction do these reactions represent and how are they carried out? 1. (ii) Answers 3. (ii) 4. (ii) 2. (i) the d- and f-BloCK elements 143

5 Coordination Compounds Topics covered 5.1 Terminology, Nomenclature and Isomerism in Coordination Compounds 5.2 Bonding in Coordination Compounds 5.3 Stability, Importance of Coordination Compounds and Organometallic Compounds C hapter map 144

Topic 1. Terminology, Nomenclature and Isomerism in Coordination Compounds Coordination compounds ll CH2COO– CH2COO– Coordination compounds are those having such H2C — N coordination entity which can retain their identity in solid state as well as in dissolved state, e.g. CH2COO– CH2COO– K4[Fe(CN)6]. aq 4K+ + [Fe(CN)6]4– ll K4[Fe(CN)6] H2C — N • Ambidentate ligands: These are the monodentate ligands which can ligate through two different NO2–, CN–, sites, but only one at a time, e.g. SCN–, etc. • Chelating ligands: Bidentate or polydentate ligands causes cyclisation around the central metal atom / ion are known as chelating ligands Double Salts and such complexes are known as chelate These are the addition molecular compounds which complexes. are stable in solid state but dissociate into constituent SK(ioNO+nH,24sA–4i)lo2i3Snn+,OsS,4tPO.h6o4e2Ht–a2isOsohonlgsau.eltutiomdni,,sKsoe2Sc.giOa. t4e.AMdlo2i(nhStrOo’s4F)3es.22a+4l,tH,N2HOF+e4gSiavOne4ds. • More the number of chelate rings, more is the stability of complex. • The stabilisation of coordination compounds due to chelation caused by chelating ligands is known as chelate effect. Coordination number: It is defined as the number of coordinate bonds formed by central metal atom Terms Related to Coordination Compounds with the ligands, e.g. in [PtCl6]2–, Pt has coordination number 6. Complex ion or coordination entity: It is the species in which central metal atom or ion is • In case of monodentate ligands: Coordination surrounded by number of ions or neutral molecules (ligands). For example, [Ni(CO)4], [Fe(CN)6]4–, etc. number = Number of ligands Central atom or ion: The atom or ion to which a • In polydentate ligands: Coordination number = fixed number of ions or groups are bound is termed as Number of ligands × Denticity central atom or ion. It is also referred as Lewis acid. Coordination sphere: The central metal ion and the ligands attached to it are enclosed in a square Ligands: Ligands are electron donating species (ions bracket, which is known as coordination sphere. The or molecules) bound to the central metal atom in the ionisable group written outside the square bracket is coordination entity. e.g. NH3, H2O, CN–, Cl– etc. These may be charged or neutral. Ligands are of the known as counter ion. following types : Coordination polyhedron: The spatial arrangement of the ligands which are directly • Unidentate: It is a ligand which has one donor attached to the central metal atom or ion is called site, i.e. the ligand bound to a metal ion through the coordination polyhedron around the central atom a single donor site, e.g. H2O (O as donor site), or ion. NH3 (N as donor site), etc. • Bidentate: It is the ligand which have two donor Oxidation number of central atom: The charge on the central atom in the complex if all the ligands are removed along with the electron pairs that are sites, e.g. shared with the central atom, is called oxidation [rCepur(CesNen)4t]e3–d, COO– ll number of the central metal atom, e.g. oxidation number of copper is +1, and is CH2—NH2 COO– ll as Cu(I). CH2—NH2 (oxalate ion) (ethylene diammine) Types of Complexes (ox) (en) Homoleptic complexes: Complexes in which • Polydentate: It is the ligand which have several the metal atom or ion is linked to only one kind of donor sites, e.g. [EDTA]4– is a hexadentate donor atoms are called homoleptic complexes, e.g. [Co(NH3)6]3+. ligand. Coordination Compounds 145

Heteroleptic complexes: Complexes in which the ambidentate ligands, e.g. [Co(NH3)5(NO2)]Cl metal atom or ion is linked to more than one kind of (pentaamminenitrito-N-cobalt(III) chloride) and donor atoms are called heteroleptic complexes, e.g. [Co(NH3)4Cl2]+. [Co(NH3)5(ONO)]Cl (pentaamminenitrito-o-co- balt(III) chloride) . IUPAC Naming of Complex Compounds • Coordination isomerism: This type of isomerism Naming of complex compounds is based on set of rules given by IUPAC. arises from the interchange of ligands between • Name of the compound is written in two parts (i) cationic and anionic complexes of different metal name of the cation, and (ii) name of the anion. ions present in a complex, e.g. [ C r ( N H 3 ) 6 ) ] • The cation is named first in both positively and [Co(CN)6] and [Co(NH3)6] [Cr(CN)6] negatively charged coordination complexes. • Ionisation isomerism: This isomerism arises due • The dissimilar ligands are named in alphabetical to exchange of ionisable anion with the ligand, order before the name of central metal atom or ion. e.g. [Co(NH3 )5SO4 ]Br and [Co (NH3)5 Br] SO4 (Violet) • For more than one similar ligands, the prefix di, (Red) tri, tetra, etc are added before its name. If the • Solvate isomerism: This is also known as hydrate prefixes di, tri, etc already appear in the name of isomerism when water is taken as solvent. It has ligand then bis, tris, tetrakis are used as prefix and the name of ligand placed in parentheses. different number of water (solvent) molecules • Names of the anionic ligands end in ‘– o’, names in the coordination sphere and outside it, of positive ligands end with ‘ium’ and names of neutral ligands remains as such. But exception [eC.go.(H2[OC)o3(CHl32]O. )36H]C2lO3, [Co(H2O)4 Cl2]Cl⋅2H2O, are there as we use aqua for H2O, ammine for Stereoisomerism: Stereoisomers have the same NH3, carbonyl for CO and nitrosyl for NO. chemical formula and chemical bonds but they have • Oxidation state of the metal in cation, anion or different spatial arrangement. These are of two types: in a neutral coordination compound is indicated by Roman numeral in parenthesis. • Geometrical isomerism: Geometrical isomers are of two types, i.e. cis-trans and fac-mer isomers. • The name of the complex part is written as one This isomerism is common in heteroleptic word. complexes with coordination number 4 and 6. • If the complex part is an anion, the name of the central metal ends with suffix ‘ate’. Geometrical isomerism in complexes with coordination number 4 • If the complex ion is a cation, the metal is named (i) Tetrahedral complexes do not show same as the element. geometrical isomerism because relative • The neutral complex molecule is named similar position of ligand is the same. to that of the complex cation. (ii) Square planar complexes of formula • Some examples are: g[MeoXm2Let2r]ic(aXl and L are unidentate) show (i) t[Crira(mNmH3in)3e(Htr2iaOq)u3]aCclh3 romium(III) chloride isomerism. The two X ligands (ii) t[Crios((NetHha2CneH-12C,2H-d2NiaHm2i)n3e]2)(cSoOba4)l3t(III) sulphate may be arranged adjacent to each other in (iii) [dAiagm(NmHin3)e2]si[lAvegr(C(IN) d)2i]cyanidoargentate(I) (iv) K4[Fe(CN)6] a cis isomer, or opposite to each other in a trans isomer, e.g. potassium hexacyanidoferrate(II) Cl NH3 H3N Cl Isomerism in Coordination Compounds Pt Pt Coordination compounds exhibit the following types of isomerism: Cl NH3 NH3 Cl Structural Isomerism: In this isomerism, isomers have different bonding pattern. Different types of cis trans structural isomers are: (iii) Square planar complex of the type [MABXL] • Linkage isomerism: This type of isomerism is (where A, B, X, L, are unidentate ligands) shown by the coordination compounds having shows three isomers, two cis and one trans, Geome.ge.t[rPict(aNl His3o)(mBre)r(Cislm)(Pyin)].complexes with coordination number 6 146 Chemistry-12

OwchtiachhetdhrealtwcoomXplliegxaensdsofmfaoyrmbue loari[eMntXe2dLc4]i,s in of the octahedron, it is known as meridional or • (Ompetri)caisl omiseorm, eer.gis.m[C: o(TNhHes3e)3(NarOe2)t3h].e complexes trans to each coaothmuepnrl,iedexe.egns.t,a[tCoeof (lNifgHoar3nm)d4uCllaa2n]+d.[MAX2iAs 2]a, Octahedral which have chiral structures. It arises when where X is mirror images cannot be superimposed on bidentate ligand, forms cis and trans isomers, one another. These mirror images are called e.g. [CoCl2(en)2]. enantiomers. The two forms are called dextro (d) and laevo (l) forms. NH3 NO2 Octahedral complexes (cis form) and octahedral H3N NO2 H3N NH3 complexes with didentate ligands exhibits Pt Co optical isomerism e.g. [Co(en)3]3+ has two optical isomers: H3N NO2 H3N NO2 NO2 NO2 fac mer Ithnroecetadhoendorraal tcoommsploefxethseofsafomremulilgaan[MdsAo3Xcc3u],piyf adjacent positions at the corners of an octahedral face, it is known as facial (fac) isomer. When the positions are non-adjacent around the meridian ExERCISE 5.1 Multiple Choice Questions (MCQs) (1 Mark) 8. Which of the following species is not expected to 1. How many coordination sites are there in an be a ligand? ethylenediaminetetraacetate ion? (ii) NH4+ (i) NO (iv) CO (i) 2 (ii) 4 (iii) 6 (iv) 8 (iii) NH2CH2CH2NH2 2. Give the linkage isomer of [Pt(SCN)(NH3)3] SCN. Assertion-Reason Type Questions (1 Mark) (i) [Pt (SCN)2(NH3)2]NH3 Note: In the following questions a statement (ii) [Pt (SCN)(NH3)3]NCS of assertion followed by a statement of reason (iii) [Pt (NCS)2(NH3)2]NH3 is given. Choose the correct answer out of the (iv) [Pt (NSC)(NH3)3]SCN following choices. 3. How many ions will be produced from the complex [Co(NH3)6]Cl2 in solution? (i) Assertion and reason both are true, reason is the correct explanation of assertion. (i) 1 (ii) 8 (iii) 9 (iv) 3 (ii) Assertion and reason both are true but reason is not the correct explanation of assertion. 4. What is the oxidation number of cobalt in K[Co(CO)4]? (i) +1 (ii) –1 (iii) +3 (iv) –3 (iii) Assertion is true but reason is false. 5. Wcoof nhAdegnuNc0tO.i1v3i,mty0o.ol2fCsomoCloull3t(iNoofnHAw3)gi5lCilslctoarrrereaetsoepbdotnwadiinttheode.xcTehses (iv) Assertion is false, reason is true. (i) 1:3 electrolyte (ii) 1:2 electrolyte 9. Assertion: Linkage isomerism arises in coordination compounds containing ambidentate ligand. (iii) 1:1 electrolyte (iv) 3:1 electrolyte Reason: Ambidentate ligand has two different donor atoms. 6. WAofghtNheOne 3c1,om3momplloeClxroCifslA:3 ·g6CHl 2aOreisotbrteaaitneeddw. Tithheefxocremssuloaf 10. Assertion: Complexes odfoMnXot6 ashnodwMgXe5oLmteytpreic(aXl and L are unidentate) (i) [CrCl3(H2O)3] · 3H2O isomerism. (ii) [CrCl2(H2O)4]Cl · 2H2O (iii) [CrCl(H2O)5]Cl2 · H2O Reason: Geometrical isomerism is not shown by (iv) [Cr(H2O)6]Cl3 complexes of coordination number 6. 7. The correct IUPAC name of [Pt(NH3)2Cl2] is (i) Diamminedichloridoplatinum (II) Very Short Answer Type Questions (1 Mark) 11. Why tetrahedral complexes do not show geometrical isomerism? (ii) Diamminedichloridoplatinum (Iv) 12. Which among the following two coordination entities is chiral (optically active)? (iii) Diamminedichloridoplatinum (0) (i) cis-[CrCl2(ox)2]3– (ii) trans-[CrCl2(ox)2]3– (iv) Dichloridodiammineplatinum (Iv) Coordination Compounds 147

13. What is meant by coordination number? (ii) Chlorobis(ethane-1,2-diamine)nitrito-N- 14. Write two examples of ambidentate ligands. cobalt(III) ion. [Foreign 2010] (iii) Bis(acetylacetonato)oxovanadium(Iv). [Co[CAlI2(2e0n1)22]+] 15. Which among the possible isomers of 35. What is meant by the unidentate, bidentate and does not show optical isomerism? ambidentate ligands? Give two examples of each. [NCERT 2014] 16. Give the formulae of the following compounds: (i) Potassium trioxalatoaluminate(III) 36. Draw geometrical and optical isomers of the (ii) Copper (II) hexacyanidoferrate(II) following: 17. Write the IUPAC name of the linkage isomer of (i) [CoCl2(en)2]+ (ii) [Co(NH3)2Cl2(en)]+ 18. W[Nhi(yNNHH3)4+2(iHon2Od)o(eNsOn2o)]tBarc.t as ligand. Explain? 19. Give name of the coordination isomer of the (iii) [Co(NH3)Cl(en)2]2+ [NCERT, Delhi 2014(C)] Long Answer Type Questions-I (5 Marks) 37. Give the IUPAC names of the following complex: [Co(en)3] [Cr(CN)6]. coordination compounds: 20. Give the name of the coordination isomer of [Co(NH3)6][Cr(CN)6]. [AI 2010(C), 2013] (i) [Pt(NH3)2Cl(NO2)] (ii) K3[Cr(C2O4)3] 21. How many possible isomers will be there for the (iii) [CoCl2(en)2]Cl (iv) [Co(NH3)5CO3] Cl] complex: [Co(NH3)4Cl2]Cl? 22. Give the IUPAC name of Na[Au(CN)2]. (v) Hg[Co(SCN)4] 23. How many ions are produced from the complex 38. Give the formulae of the following coordination compounds: [Co(NH3)6]Cl2 in solution? [CBSE 2022] (i) Pentacarbonyltriphenylphosphinechromi- Short Answer Type Questions-I (2 Marks) um(0) 24. Arrange the following complexes in the decreasing order of their electrical conductivity: (ii) Ferric hexacyanidoferrate(II) [Co(NH3)4Cl2] Cl; K2[PtCl6]; (iii) Tetraamminedichloridoplatinum(Iv) [Ni(NH3)4Cl2]; K4[Fe(CN)6] tetrachloridoplatinate(II) 25. Draw the structures of possible geometrical isomers of [Fe(NH3)2(CN)4]–. (iv) Chlorido bis(ethylenediamine)nitrocobalt(III) 26. Define ligand and give an example. ion (v) Tetracarbonylnickel (0) 27. What is meant by chelate? Give one example. 39. Using IUPAC norms, write the formulae of the 28. s1Foe:lSu1Otmi4oonsloalmur itrxiaoetndiomwgiiixtvehedsawtqhiuteheto(euNsstHoa4fm)2FSmeO2o+4nbsiaoultuinCtiuo1Sn:Oin44 following: [NCERT] molar ratio does not give the test of Cu2+ ion. (i) Tetrahydroxozincate(II) (ii) Potassium tetrachloridopalladate(II) Explain why? [NCERT] (iii) Diamminedichloridoplatinum(II) 29. Write the name of the following complexes (iv) Potassium tetracyanidonickelate(II) according to IUPAC system of nomenclature: (v) Pentaamminenitrito-O-cobalt(III) (i) [Co(NH3)4(H2O)Cl]Cl2 (vi) Hexaamminecobalt(III) sulphate (ii) [CrCl2(en)2]Cl [Delhi 2010] (vii) Potassium trioxalatochromate(III) 30. Give an evidence t hioantis[aCtoio(nNiHso3m)5Cerls].SO4 and (viii) Hexaammineplatinum(Iv) [Co(NH3)5SO4]Cl are 31. How many geometrical isomers are possible in (ix) Tetrabromidocuprate(II) the following coordination entities? (x) Pentaamminenitrito-N-cobalt(III) (i) [Cr(C2O4)3]3– [Delhi 2014(C)] 40. Using IUPAC norms, write the systematic names of the following: (ii) [Cr(NH3)3Cl3] [Delhi 2013] (i) [Co(NH3)6]Cl3 32. Write all the gheoowmmetarnicyaolfistohmeseerws oilfl[sPhto(NwHop3)t(iBcar)l (ii) [Co(NH3)4ClNO2]Cl (Cl)(py)] and (iii) [Ni(NH3)6]Cl2 isomerism? [NCERT] (iv) [Pt(NH3)2Cl(NH2CH3)]Cl (v) [Mn(H2O)6]2+ 33. Write the IUPAC names of the following complexes: (vi) [Co(en)3]3+ (a) Na3[AlF6]; (b) Ni(CO)4 (vii) [Ti(H2O)6]3+ (viii) [NiCl4]2– Short Answer Type Questions-II (3 Marks) (ix) [Ni(CO)4] 34. Give the formula of the following coordination compounds. (i) Sodium ethane-1,2-diaminetetracetatochro- [NCERT] mate(II). 148 Chemistry-12

41. Specify the oxidation numbers of the metals in (iii) [Cr(NH3)3Cl3] the following coordination entities: (iv) [CoBr2(en)2]+ [[PCtoC(Hl42]O2–)(CN)(en)2]2+ (v) K3[Fe(CN)6] (i) [NCERT] (ii) Answers 5.1 1. (iii) 2. (iv) 3. (iv) 4. (ii) 25. The structure of two geometrical isomers are: 5. (ii) 6. (iv) 7. (i) 8. (ii) (i) NC NH3 – (ii) CN – 9. (i) 10. (ii) CN NC CN 11. In tetrahedral geometry, the relative position of the ligands attached to the central metal atom is Fe Fe same with respect to each other. That is why no NC CN NC NH3 geometrical isomers are possible. NH3 NH3 12. Two entities are represented as follows: trans–form cis–form 26. The neutral molecules, anions or cations which are directly bonded to the central metal atom or ion in a complex ion or compound are called ligands. Free ligands contains atleast one pair of electrons which can be donated to the metal atom or ion. Examples: NH3, H2O, CN–, Cl–, etc. Among them, the cis form is chiral. 27. When a bidentate or polydentate ligand is attached via two or more donor atoms to the 13. The number of atoms of the ligand that are directly bounded to the central metal atom or ion same central metal atom or ion, forming a ring through coordinate bond is known as coordination structure, then the ligand is called the chelating number. ligand and the complex is called a chelate. An 14. SCN– or NCS–; NO–2 or (O—N—O)– example of chelate is [Cu(en)2] SO4. 15. Trans-isomer as it has superimposable mirror 28. The stability of complex when a bidentate or a polydentate ligand attacks on metal ion in such image. a way that after coordinating with the central 16. (i) K3[Al(C2O4)3] (ii) Cu2[Fe(CN)6] metal atom it forms a ring, then such an effect is 17. Diammineaquanitrito-o-nickel(II)bromide called ‘chelate effect’, eg. 18. Nitrogen atom does not have lone pair of electrons in NH4+ ion which it can donate to the central metal atom. Thus, it cannot act as a ligand. 19. [Cr(en)3][Co(CN)6]— Bis(ethane-1,2-diamine)copper(II)ion Tris(ethylenediamine) chromium(III) 29. (i) Tetraammineaquachloridocobalt(III) chloride hexacyanidocobaltate(III) (ii) D i c h l o r i d o b i s ( e t h a n e - 1 , 2 - d i a m i n e ) 20. [Cr(NH3)6] [Co(CN)6]—Hexaamminechro- mium(III) hexacyanidocobaltate(III) chromium(III) chloride 21. Two geometrical isomers. Among them cis-isomer 30. Ionisation isomers when dissolved in water exhibits optical activity (i.e. has two optical produces different ions which can be tested. isomers). [Co(NH3)5SO4]Cl + AgNO3 → [Co(NH3)5 SO4] NO3 + AgCl (ppt.) 22. Sodium dicyanidoaurate(I) If white precipitate is formed on reaction with 23. [Co(NH3)6]Cl2Water→[Co(NH3)6]2++ 2Cl– \\ Three ions are produced in the solution. AhagsNCOl3– solution, then it indicates that the isomer ion outside the coordination sphere. 24. Conductivity of a complex depends upon the number of ions produced by it in the solution. [Co(NH3)5Cl] SO4 + BaCl2 → Hence the order of conductivity is: [Co(NH3)5Cl]Cl2 + BaSO4 (ppt.) K4 6Fe^CN6h@ > K2 6PtCl6@ > 6Co^NH3h4 Cl2@Cl If on rietaicntdioincawteitshtBhaaCt lS2,Ow42h– iitoenpipst.oiustosbidtaeinthede 5 ions 3 ions 2 ions then > 6Ni^NH3h4 Cl2@ coordination sphere. 0 ions Coordination Compounds 149