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The d- and f-Block, Elements, Recap Notes, TRANSITION ELEMENTS, (d-BLOCK ELEMENTS), Elements in which the last electron enters, any one of the five d-orbitals of their, respective penultimate shell are known as, transition elements or d-block elements., Their general electronic configuration is, (n – 1)d1 – 10ns0 – 2., , Transition series : d-block consists of four, transition series,, 1st Transition series or 3d series 21Sc – 30Zn, 2nd Transition series or 4d series 39Y – 48Cd, 3rd Transition series or 5d series 57La, 72Hf, , – 80Hg, th, 4 Transition series or 6d series 89Ac, 104Rf, , – 112Cn, , General characteristics :, Melting and boiling points, Enthalpies of atomisation, Ionisation enthalpies, Oxidation states, Atomic radii, Complex formation, , Coloured compounds, Magnetic properties, , Catalytic behaviour, Interstitial compounds, Alloy formation, , High due to strong metallic bonding, High due to strong interatomic interactions, Generally increases from left to right in a series, Variable due to participation of ns and (n – 1)d electrons, Decrease from left to right but become constant when pairing of, electrons takes place, Form complexes due to high nuclear charge and small size and, availability of empty d-orbitals to accept lone pair of electrons, donated by ligands., Form coloured compounds due to d-d transitions, Transition metal ions and their compounds are paramagnetic, due to presence of unpaired electrons in the (n – 1)d-orbitals, and it is calculated by using the formula, m = n(n + 2) where, n, is the no. of unpaired electrons., Due to variable oxidation states and ability to form complexes, Due to empty spaces in their lattices, small atoms can be easily, accommodated, Due to similar atomic sizes, , INNER TRANSITION ELEMENTS, ( f-BLOCK ELEMENTS), X Lanthanoids : Last electron enters one, of the 4f-orbitals. Cerium (At. no. 58) to, lutetium (At. no. 71)., , X, , Actinoids : Last electron enters one of, the 5f-orbitals. Thorium (At. no. 90) to, lawrencium (At. no. 103)., , General electronic configuration :, (n – 2)f 1 – 14 (n – 1)d0 – 1 ns2

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General characteristics of lanthanoids :, Atomic and ionic Decrease on going from La to Lu., radii, Oxidation states, , Most common oxidation state of lanthanoids is +3. Some elements exhibit, +2 and +4 oxidation states due to extra stability of empty, half-filled or fullyfilled f-subshell, e.g., Ce4+ acts as an oxidising agent and gets reduced to Ce3+,, Eu2+, Yb2+ act as strong reducing agents and get oxidised to Eu3+ and Yb3+., , Action of air, , All the lanthanoids are silvery white soft metals and tarnish readily in, moist air. They burn in oxygen of air and form oxides (Ln2O3 type)., , Coloured ions, , They form coloured trivalent metal ions due to f-f transitions of unpaired, electrons. La3+ and Lu3+ are colourless ions due to empty (4f 0 ) or fully, (4f 14) orbitals., , Magnetic, properties, , La3+, Lu3+ are diamagnetic while trivalent ions of the rest of lanthanoids, are paramagnetic., , Reducing agents, , They readily lose electrons and are good reducing agents., , Electropositive, character, , Highly electropositive because of low ionisation energies., , Alloy formation, , They form alloys easily with other metals especially iron., , Tendency to form, complexes, , Lanthanoids do not have much tendency to form complexes due to low, charge density because of their large size. The tendency to form complexes, and their stability increases with increasing atomic number., , Lanthanoid contraction : In lanthanoid, series, with increasing atomic number, there, is progressive decrease in atomic/ionic radii, (M3+ ions) from La3+ to Lu3+., X Reason : Due to addition of new electrons, into f-subshell and imperfect shielding of, one electron by another in the f-orbitals,, there is greater effect of increased nuclear, , charge than screening, contraction in size occurs., X, , effect, , hence, , Consequences : Their separation is, difficult, they have small differences in, properties and 4d and 5d transition series, have almost same atomic radii (Zr and Hf, have similar properties due to same size).

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(c) Mn3+ is more stable than Mn2+ due to higher, oxidation state., (d) second ionisation enthalpy of Mn is higher, than third ionisation enthalpy., 13. I n t e r s t i t i a l c o m p o u n d s a r e n o n stoichiometric compounds formed by trapping, small atoms like C, H or N in crystal lattices, of transition metals. Which of the following, properties is not shown by these compounds?, (a) They have high melting points, higher than, those of pure metals., (b) They are very hard, some borides are, comparable to diamond in hardness., (c) They are chemically very reactive., (d) They retain metallic conductivity., 14. Reactivity of transition elements decreases, almost regularly from Sc to Cu because of, (a) lanthanoid contraction, (b) regular increase in ionisation enthalpy, (c) regular decrease in ionisation enthalpy, (d) increase in number of oxidation states., 15. For Zn2+, Ni2+, Cu2+ and Cr2+ which of the, following statements is correct?, (a) Only Zn2+ is colourless and Ni2+, Cu2+ and, Cr2+ are coloured., (b) All the ions are coloured., (c) All the ions are colourless., (d) Zn2+ and Cu2+ are colourless while Ni2+ and, Cr2+ are coloured., 16. Which of the following statement concerning, lanthanide elements is false?, (a) All lanthanides are highly dense metals., (b) More characteristic oxidation state of, lanthanide elements is +3., (c) Lanthanides are separated from one another, by ion exchange method., (d) Ionic radii of trivalent lanthanides steadily, increases with increase in the atomic, number., 17. Which of the following statements is not, correct about magnetic behaviour of substances?, (a) Diamagnetic substances are repelled by an, applied magnetic field., (b) Paramagnetic substances are attracted by, an applied magnetic field., (c) Magnetic moment of n unpaired electrons is, given by µ = n(n − 2) B.M., (d) Magnetic moment increases as the number, of unpaired electrons increases., , 18. The electronic configuration of Cu(II) is, 3d9 whereas that of Cu(I) is 3d10. Which of the, following is correct?, (a) Cu(II) is more stable., (b) Cu(II) is less stable., (c) Cu(I) and Cu(II) are equally stable., (d) Stability of Cu(I) and Cu(II) depends on, nature of copper salts., 19. Which of the following are basic oxides?, Mn2O7, V2O3, V2O5, CrO, Cr2O3, (a) Mn2O7 and V2O3, (b) V2O3 and CrO, (c) CrO and Cr2O3, (d) V2O5 and V2O3, 20. Which of the following catalysts is not, correctly matched with the reaction?, (a) Vanadium(V) oxide in contact process for, oxidation of SO2 to SO3., (b) Finely divided iron in Haber’s process in, conversion of N2 and H2 to NH3., (c) PtCl2 catalyses the oxidation of ethyne to, ethanal in the Wacker process., (d) Ni in presence of hydrogen for conversion of, vegetable oil to ghee., 21. Which is the non-lanthanide element?, (a) La, (b) Lu, (c) Pr, (d) Pm, 22. Magnetic moment of Ce3+ ion on the basis, of ‘spin-only’ formula will be _____ B.M., (a) 1.232, (b) 1.332, (c) 1.532, (d) 1.732, 23. In which of the following pairs of ions, the, lower oxidation state in aqueous solution is more, stable than the other?, (a) Tl+, Tl3+, (b) Cu+, Cu2+, 2+, 3+, (c) Cr , Cr, (d) V2+, VO2+ (V4+), 24. Which of the following compounds is expected, to be coloured?, (a) Ag2SO4, (b) CuF2, (c) MgF2, (d) CuCl, 25. Consider the following statements with, respect to lanthanides :, 1. The basic strength of hydroxides of, lanthanides increases from La(OH)3 to Lu(OH)3., 2. The lanthanide ions Lu3+, Yb2+ and Ce4+ are, diamagnetic., Which of the statement(s) given above is/are, correct?, (a) 1 only, (b) 2 only, (c) Both 1 and 2, (d) Neither 1 nor 2

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26. Cu+ ion is not stable in aqueous solution, because, (a) second ionisation enthalpy of copper is less, than the first ionisation enthalpy, (b) large value of second ionisation enthalpy, of copper is compensated by much more, negative hydration energy of Cu2+(aq), (c) hydration energy of Cu +(aq) is much more, negative than that of Cu2+(aq), (d) many copper (I) compounds are unstable, in aqueous solution and undergo, disproportionation reaction., 27. Select the correct option, among Sc(III),, Ti(IV), Pd(II) and Cu(II) ions,, (a) all are paramagnetic, (b) all are diamagnetic, (c) Sc(III), Ti(IV) are paramagnetic and Pd(II),, Cu(II) are diamagnetic, (d) Sc(III), Ti(IV) are diamagnetic and Pd(II),, Cu(II) are paramagnetic., 28. Identify the species in which the metal atom, is in +6 oxidation state., –, 3–, (a) MnO4, (b) Cr(CN)6, 2–, , (c) NiF6, , (d) CrO2Cl2, , 29. Which of the following statements is correct, about stability of the complexes of lanthanoids?, (a) Stability of complexes increases as the size, of lanthanoid decreases., (b) Stability of complexes decreases as the size, of lanthanoid decreases., (c) Lanthanoids do not form complexes., (d) All the complexes of lanthanoids have same, stability., 30. Fe3+ ion is more stable than Fe2+ ion because, (a) more the charge on the atom, more is its, stability, (b) configuration of Fe2+ is 3d6 while Fe3+ is 3d5, (c) Fe2+ has a larger size than Fe3+, (d) Fe3+ ions are coloured hence more stable., 31. Transition metals make the most efficient, catalysts because of their ability to, (a) adopt multiple oxidation states and to form, complexes, (b) form coloured ions, (c) show paramagnetism due to unpaired, electrons, (d) form a large number of oxides., , 32. General electronic configuration of transition, metals is, (b) nd10ns2, (a) (n – 1)d1-10ns0-2, 10 2, (c) (n – 1)d ns, (d) (n – 1)d1-5ns2, 33. Consider the following statements, I. La(OH)3 is least basic among hydroxides of, lanthanides., II. Zr4+ and Hf 4+ possess almost the same ionic, radii., III. Ce4+ can act as an oxidizing agent., Which of the above is/are true?, (a) I and III, (b) II and III, (c) II only, (d) I and II, 34. Arrange the following in increasing value of, magnetic moments., (ii) [Fe(CN)6]3–, (i) [Fe(CN)6]4–, 3+, (iii) [Cr(NH3)6], (iv) [Ni(H2O)4]2+, (a) (i) < (ii) < (iii) < (iv), (b) (i) < (ii) < (iv) < (iii), (c) (ii) < (iii) < (i) < (iv), (d) (iii) < (i) < (ii) < (iv), 35. Fe3+ compounds are more stable than Fe2+, compounds because, (a) Fe3+ has smaller size than Fe2+, (b) Fe3+ has 3d5 configuration (half-filled), (c) Fe3+ has higher oxidation state, (d) Fe3+ is paramagnetic in nature., 36. Following order is observed in oxidising, power of certain ions:, +, , 2–, , –, , VO2 < Cr2O7 < MnO4, , The reason for this increasing order of oxidising, power is, (a) increasing stability of the lower species to, which they are reduced, (b) increasing stability of the higher species to, which they are oxidised, (c) increasing stability of the higher species to, which they are reduced, (d) increasing stability of the lower species to, which they are oxidised., 37. Which of the following transition metal ions, has highest magnetic moment?, (b) Ni2+, (a) Cu2+, (c) Co2+, (d) Fe2+

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39. Which one of the following is a ‘d-block, element’?, (a) Gd, (b) Hs, (c) Es, (d) Cs, 40. Which of the following lanthanide is, commonly used?, (a) Lanthanum, (b) Nobelium, (c) Thorium, (d) Cerium, 41. Transition elements form binary compounds, with halogens. Which of the following elements, will form MF3 type compounds?, (a) Cr, (b) Cu, (c) Ni, (d) All of these, 42. Although zirconium belongs to 4d and, hafnium to 5d-transition series even they show, similar physical and chemical properties because, both, (a) belong to d-block, (b) have same number of electrons, (c) have similar atomic radius, (d) belongs to the same group of the periodic, table., 43. Most of the transition metals exhibit, (i) paramagnetic behaviour, (ii) diamagnetic behaviour, (iii) variable oxidation states, (iv) formation of coloured ions, (a) (ii), (iii) and (iv), (b) (i), (iii) and (iv), (c) (i), (ii) and (iii), (d) (i), (ii) and (iv), 44. Which of the following has no unpaired, electrons but is coloured?, (a) K2Cr2O7, (b) K2MnO4, (c) CuSO4⋅5H2O, (d) MnCl2, 45. The second and third row elements of, transition metals resemble each other much, more than they resemble the first row because, of, (a) lanthanoid contraction which results in, almost same radii of second and third row, metals, (b) diagonal relationship between second and, third row elements, (c) similar ionisation enthalpy of second and, third row elements, (d) similar oxidation states of second and third, row metals., , 46. Which of the following compounds is not, coloured?, (b) Na2[CdCl4], (a) Na2[CuCl4], (c) K4[Fe(CN)]6], (d) K3[Fe(CN)6], 47. The salts of Cu in +1 oxidation state are, unstable because, (a) Cu+ has 3d10 configuration, (b) Cu+ disproportionates easily to Cu(0) and, Cu2+, (c) Cu + disproportionates easily to Cu2+ and, Cu3+, (d) Cu+ is easily reduced to Cu2+., 48. Colour of transition metal ions are due to, absorption of some wavelength. This results in, (a) d-s transition, (b) s-s transition, (c) s-d transition, (d) d-d transition., 49. The observed values and calculated values, of E° of various 3d-series elements are shown in, the figure., Standard electrode potential / V, , 38. Which of the following are amphoteric, oxides?, (i) Mn2O7, (ii) CrO3, (iii) Cr2O3, (iv) CrO, (vi) V2O4, (v) V2O5, (a) (i) and (iv), (b) (ii) and (iii), (c) (iii) and (v), (d) (ii) and (vi), , 0.5, 0, –0.5, –1, –1.5, –2, , Ti, , V, , Cr Mn, Observed, values, , Fe Co, , Ni Cu Zn, Calculated, values, , Which of the following facts cannot be explained, on the basis of the given graph?, (a) Inability of Cu to liberate H2 from acids, (b) Extra stability of d5(Mn 2+) and d 10(Zn 2+), configuration, (c) Extra stability of Ni2+ due to d10 configuration, (d) All of these., 50. For the given reactions :, –, X + Y + H2O, [Fe(H2O)6]2+ + NO3 + H+, Z + H2O, [Fe(H2O)6]2+ + X, The incorrect statement about X, Y and Z is, (a) Magnetic moment of Y is 5.9 B.M., (b) Oxidation state of Fe in Z is +1., (c) Complex Z is reddish-brown in colour., (d) X is an acidic oxide of nitrogen., 51. W, X, Y and Z are four consecutive members, of 3d-series., Trend in their melting point are shown in the, given figure.

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On the basis of these values, Krish concluded, , M.P. /103 K, , 3, W, , Y, , 2, , the following statements:, Z, , X, 1, , Atomic number, , Correct statement about W, X, Y and Z is, (a) magnetic moment of X in its +2 oxidation, state is 2.83 B.M., (b) W 3+ ion is green in colour., (c) Y3+ catalyses reaction between iodide and, persulphate ions., (d) stable oxidation states of Z are +1, +2 and, +6., , I., , Cr2+ is a reducing agent, , II. Mn3+ is an oxidizing agent, III. both Cr2+ and Mn3+ exhibit d4 electronic, configuration, IV. when Cr 2+ is used as a reducing agent,, the chromium ion attains d 5 electronic, configuration., The incorrect conclusion made by him is, (a) I, , (b) II, , (c) III, , (d) IV, , 52. Few electrode potential values are given, below:, , 53. Find the incorrect analogy for lanthanoids., , Cr3+/Cr2+ = – 0.41 V, , (b) Paramagnetic lanthanide ion : Yb2+, , Cr2+/Cr = – 0.90 V, , (c) Ions that can exist in aqueous solution :, , 3+, , Mn /Mn, , 2+, , = +1.57 V, , Mn2+/Mn = –1.18 V, , (a) Good oxidising agent : Ce4+, , Eu2+, Yb2+, (d) Colourless ions : Ce3+, Yb3+, , Case Based MCQs, Case I : Read the passage given below and, answer the following questions., The lanthanide series is a unique class of, 15 elements with relatively similar chemical, properties. They have atomic number ranging, from 57 to 71, which corresponds to the filling, of the 4f orbitals with 14 electrons. This, configuration leads to phenomenon known, as lanthanide contraction. The lanthanides, are sometimes referred to as the ‘rare earth, elements’, leading to misconception that they, are rare. In fact many of the rare earth elements, are more common than gold, silver and in some, cases, lead. The lanthanides are commonly, found in nature as a mixture in a number of, monazite (LnPO4) and bastnaesite (LnCO3F) in, the +3 oxidation state., The chemical and physical properties of, lanthanides provide the unique features that set, them apart from other elements. Lanthanides, are most stable in the +3 oxidation state. Yb, and Sm though stable in the +3 state, also, have accessible +2 oxidation states. The ease, of accessibility of both oxidation states is quite, , important in chemical synthesis and these, elements act as Lewis acid in the +3 oxidation, state and single electron reductant in the +2, oxidation state., In the following questions (Q. No. 54-58), a, statement of assertion followed by a statement, of reason is given. Choose the correct answer out, of the following choices on the basis of the above, passage., (a) Assertion and reason both are correct, statements and reason is correct explanation, for assertion., (b) Assertion and reason both are correct, statements but reason is not correct, explanation for assertion., (c) Assertion is correct statement but reason is, wrong statement., (d) Assertion is wrong statement but reason is, correct statement., 54. Assertion : The elements scandium and, yttrium are called “rare earths”., Reason : Scandium and yttrium are rare on, earth’s crust.

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55. Assertion : Separation of lanthanide, elements is difficult., Reason : They have similar chemical properties., 56. Assertion : There is continuous increase in, size among lanthanides., Reason : Lanthanides show lanthanide contraction., 57. Assertion : Yb2+ is more stable than Yb3+., Reason : Electronic configuration of Yb 2+ is, [Xe]4f 7., 58. Assertion : All lanthanides have similar, chemical properties., Reason : Because the lanthanoids differ only in, the number of 4f - electrons., Case II : Read the passage given below and, answer the following questions from 59 to 63., The transition elements have incompletely filled, d-subshells in their ground state or in any of, their oxidation states. The transition elements, occupy position in between s- and p-blocks in, groups 3-12 of the Periodic table. Starting from, fourth period, transition elements consists of, four complete series : Sc to Zn, Y to Cd and, La, Hf to Hg and Ac, Rf to Cn. In general, the, electronic configuration of outer orbitals of these, elements is (n – 1)d 1–10 ns 0–2. The electronic, configurations of outer orbitals of Zn, Cd, Hg, and Cn are represented by the general formula, (n – 1)d 10 ns 2 . All the transition elements, have typical metallic properties such as high, tensile strength, ductility, malleability. Except, mercury, which is liquid at room temperature,, other transition elements have typical metallic, structures. The transition metals and their, compounds also exhibit catalytic property and, paramagnetic behaviour. Transition metal, also forms alloys. An alloy is a blend of metals, prepared by mixing the components. Alloys may, be homogeneous solid solutions in which the, atoms of one metal are distributed randomly, among the atoms of the other., 59. Which of the following characteristics of, transition metals is associated with higher, catalytic activity?, (a) High enthalpy of atomisation, (b) Variable oxidation states, (c) Paramagnetic behaviour, (d) Colour of hydrated ions, 60. Transition elements form alloys easily, because they have, , (a), (b), (c), (d), , same atomic number, same electronic configuration, nearly same atomic size, same oxidation states., , 61. The electronic configuration of tantalum, (Ta) is, (b) [Xe]4f14 5d2 6s2, (a) [Xe]4f 0 5d 1 6s2, 14, 3, 2, (c) [Xe]4f 5d 6s, (d) [Xe]4f 14 5d 4 6s2, 62. Which one of the following outer orbital, configurations may exhibit the largest number, of oxidation states?, (b) 3d 54s2, (a) 3d 54s1, 2 2, (c) 3d 4s, (d) 3d 34s2, 63. The correct statement(s) among the following, is/are, (i) all d- and f-block elements are metals, (ii) all d- and f-block elements form coloured, ions, (iii) all d- and f-block elements are paramagnetic., (a) (i) only, (b) (i) and (ii) only, (c) (ii) and (iii) only, (d) (i), (ii) and (iii), Case III : Read the passage given below and, answer the following questions from 64 to 68., Transition metal oxides are compounds formed, by the reaction of metals with oxygen at high, temperature. The highest oxidation number in, the oxides coincides with the group number. In, vanadium, there is a gradual change from the, basic V2O3 to less basic V2O4 and to amphoteric, V2O5⋅ V2O4 dissolves in acids to give VO2+ salts., Transition metal oxides are commonly utilized, for their catalytic activity and semiconductive, properties. Transition metal oxides are also, frequently used as pigments in paints and, plastic. Most notably titanium dioxide. One, of the earliest application of transition metal, oxides to chemical industry involved the use of, vanadium oxide for catalytic oxidation of sulfur, dioxide to sulphuric acid. Since then, many, other applications have emerged, which include, benzene oxidation to maleic anhydride on, vandium oxides; cyclohexane oxidation to adipic, acid on cobalt oxides. An important property of, the catalyst material used in these processes is, the ability of transition metals to change their, oxidation state under a given chemical potential, of reductants and oxidants., 64. Which oxide of vanadium is most likely to, be basic and ionic ?, (a) VO, (b) V2O3, (c) VO2, (d) V2O5

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65. Vanadyl ion is, (a) VO2+, (c) V2O, , +, , (b), (d), , +, VO2, VO43–, , Which of the following statements is false?, With fluorine vanadium can form VF5., With chlorine vanadium can form VCl5., Vanadium exhibits highest oxidation state, in oxohalides VOCl3, VOBr 3 and fluoride, VF5., (d) With iodine vanadium cannot form VI 5, due to oxidising power of V5+ and reducing, nature of I–., , 66., (a), (b), (c), , 67. The oxidation state of vanadium in V2O5 is, (a) +5/2, (b) +7, (c) +5, (d) +6, 68. Identify the oxidising agent in the following, reaction., V2O5 + 5Ca, 2V + 5CaO, (b) Ca, (a) V2O5, (c) V, (d) None of these, Case IV : Read the passage given below and, answer the following questions from 69 to 73., The unique behaviour of Cu, having a positive, E° accounts for its inability to liberate H2 from, acids. Only oxidising acids (nitric and hot, concentrated sulphuric acid) react with Cu, the, acids being reduced. The stability of the halffilled (d5) subshell in Mn2+ and the completely, filled (d 10) configuration in Zn 2+ are related, to their E°(M3+/M2+) values. The low value for Sc, reflects the stability of Sc3+ which has a noble, gas configuration. The comparatively high value, for Mn shows that Mn 2+ (d 5 ) is particularly, stable, whereas a comparatively low value for, Fe shows the extra stability of Fe3+(d 5). The, comparatively low value for V is related to the, stability of V2+ (half-filled t2g level)., 69. Standard reduction electrode potential of, Zn2+/Zn is – 0.76 V. This means, (a) ZnO cannot be reduced to Zn by H2 under, standard conditions, (b) Zn cannot liberate H 2 with concentrated, acids, (c) Zn is generally the anode in an electrochemical, cell, (d) Z n i s g e n e r a l l y t h e c a t h o d e i n a n, electrochemical cell., , 70. E° values for the couples Cr 3+ /Cr 2+ and, Mn3+/Mn2+ are –0.41 and +1.51 volts respectively., These values suggest that, (a) Cr2+ acts as a reducing agent whereas Mn3+, acts as an oxidizing agent, (b) Cr2+ is more stable than Cr3+ state, (c) Mn3+ is more stable than Mn2+, (d) Cr 2+ acts as an oxidizing agent whereas, Mn3+ acts as a reducing agent., 71. The reduction potential values of M, N and, O are +2.46, –1.13 and –3.13 V respectively., Which of the following order is correct regarding, their reducing property?, (a) O > N > M, (b) O > M > N, (c) M > N > O, (d) M > O > N, 72. Which of the following statements are true?, (I) Mn2+ compounds are more stable than Fe2+, towards oxidation to +3 state., (II) Titanium and copper both in the first series, of transition metals exhibits +1 oxidation, state most frequently., (III) Cu+ ion is stable in aqueous solutions., (IV) The E° value for the Mn3+/Mn2+ couple is, much more positive than that for Cr3+/Cr2+, or Fe3+/Fe2+., (a) (II) and (III), (b) (I) and (IV), (c) (I) and (III), (d) (II) and (IV), 73. The stability of Cu2+(aq) rather than Cu+(aq), is due to, (a) more negative Dhyd H° of Cu2+(aq), (b) less negative Dhyd H° of Cu2+(aq), (c) more positive Dhyd H° of Cu2+(aq), (d) less positive Dhyd H° of Cu2+(aq)., Case V : Read the passage given below and, answer the following questions from 74 to 78., The f-block elements are those in which the, differentiating electron enters the (n –2)f, orbital. There are two series of f-block elements, corresponding to filling of 4f and 5f-orbitals., The series of 4f-orbitals is called lanthanides., Lanthanides show different oxidation states, depending upon stability of f 0 , f 7 and f 14, configurations, though the most common, oxidation states is +3. There is a regular decrease, in size of lanthanides ions with increase in, atomic number which is known as lanthanide, contraction.

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74. The atomic numbers of three lanthanide, elements X, Y and Z are 65, 68 and 70 respectively,, their Ln3+ electronic configuration is, (b) 4f 11, 4f 8, 4f 13, (a) 4f 8, 4f 11, 4f13, 0, 2, 11, (c) 4f , 4f , 4f, (d) 4f 3, 4f 7, 4f 9, 75. Lanthanide contraction is observed in, (a) Gd, (b) At, (c) Xe, (d) Te, 76. Which of the following is not the configuration, of lanthanoid?, (a) [Xe]4f106s2, (b) [Xe]4f15d16s2, (c) [Xe]4d145d106s2, (d) [Xe]4f75d16s2, 77. Name a member of the lanthanoid series, which is well known to exhibit +4 oxidation, state., , (a), (b), (c), (d), , Cerium (Z = 58), Europium (Z = 63), Lanthanum (Z = 57), Gadolinium (Z = 64), , 78. Identify the incorrect statement among the, following., (a) Lanthanoid contraction is the accumulation, of successive shrinkages., (b) The different radii of Zr and Hf is due to, consequence of the lanthanoid contraction., (c) Shielding power of 4f electrons is quite, weak., (d) There is a decrease in the radii of the atoms, or ions as one proceeds from La to Lu., , Assertion & Reasoning Based MCQs, For question numbers 79-95, a statement of assertion followed by a statement of reason is given. Choose, the correct answer out of the following choices., (a) Assertion and reason both are correct statements and reason is correct explanation for assertion., (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion., (c) Assertion is correct statement but reason is wrong statement., (d) Assertion is wrong statement but reason is correct statement., 79. Assertion : Transition metals form, substitutional alloys., Reason : Alloys are made to develop some useful, properties which are absent in the constituent, elements., , Reason : It is radioactive and has been prepared, by artificial means., , 80. Assertion : Reduction potential of Mn (+3 to +2), is more positive than Fe (+3 to +2)., Reason : Ionisation potential of Mn is more, than that of Fe., , of H2SO4 by Contact process., , 81. Assertion : Mn2+ is more stable than Mn3+., Reason : Mn2+ has half-filled configuration., 82. Assertion : [Ti(H2O)6]3+ is a coloured ion., Reason : Ti shows +2, +3, +4 oxidation states., 83. Assertion : Chromium is hard but mercury, is soft., Reason : Chromium is a 3d transition element., , 87. Assertion : Transition metals are good, catalysts., Reason : V2O5 or Pt is used in the preparation, 88. Assertion : Europium (II) is more stable, than cerium (II)., Reason : Cerium salts are used as a catalyst in, petroleum cracking., 89. Assertion : When Zn is placed in a magnetic, field, it is feebly magnetised in a direction, opposite to that of the magnetising field., Reason : Zn has completely filled atomic, orbitals., 90. Assertion : The correct order of oxidising, +, , 2+, , 84. Assertion : Cu is paramagnetic., Reason : Cu+ is less stable than Cu2+., , power is : VO2 < VO < VO ., , 85. Assertion : Co (IV) is known but Ni (IV) is, not., Reason : Ni (IV) has d6 electronic configuration., , is +7., , +, , 86. Assertion : Promethium is a man-made, element., , Reason : The oxidation state of Mn in MnO4, 91. Assertion : Transition metals form a large, number of interstitial compounds., Reason : They have high melting point and, boiling point.

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92. Assertion : Members of 4d and 5d series, of transition elements have nearly same atomic, radii., , Reason : The effective nuclear charge felt by, (n – 1)d electrons is higher as compared to that, by ns electrons., , Reason : Atomic and ionic radii for transition, elements are smaller than their corresponding, s-block elements., , 94. Assertion : The maximum oxidation state of, chromium in its compounds is +6., Reason : Chromium has only six electrons in ns, and (n – 1)d orbitals., , 93. Assertion : In transition elements, ns, orbital is filled up first and (n – 1)d afterwards,, during ionization ns electrons are lost prior to, (n – 1)d electrons., , 95. Assertion : Transition metals are poor, reducing agents., Reason : Transition metals form numerous, alloys with other metals., , SUBJECTIVE TYPE QUESTIONS, , Very Short Answer Type Questions (VSA), 1. Give reasons for the following :, Eu2+ is a strong reducing agent., , 7., , 2. Write the formula of an oxoanion of, manganese (Mn) in which it shows the oxidation, state equal to its group number., 3. Name a member of the lanthanoid series, which is well known to exhibit +4 oxidation state., 4. Assign reason for the following :, Copper (I) ion is not known in aqueous solution., , How would you account for the following :, , Among lanthanoids, La(III) compounds are, predominant. However, occasionally in solutions, or in solid compounds, +2 and +4 ions are also, obtained., 8., , Account for the following :, , Zr and Hf have almost similar atomic radii., , 5. Name a member of the lanthanoid series, which is well known to exhibit +2 oxidation state., , 9. Write the formula of an oxoanion of chromium, (Cr) in which it shows the oxidation state equal, to its group number., , 6. Account for the following :, Zn is not considered as a transition element., , 10. Zn2+ salts are white while Cu2+ salts are, coloured. Why?, , Short Answer Type Questions (SA-I), 11. Why is europium (II) more stable than, cerium (II)?, , (ii) Which is the most stable ion in +2 oxidation, state and why?, , 12. The magnetic moment of a transition metal, ion is found to be 3.87 BM. How many number of, unpaired electrons are present in it ?, , 14. How would you account for the following :, (i) The E°M 2+/M for copper is positive (+0.34 V)., Copper is the only metal in the first series of, transition elements showing this behaviour., (ii) The metallic radii of the third (5d) series of, transition metals are nearly the same as those, of the corresponding members of the second (4d), series., , 13. Use the data to answer the following and, also justify giving reasons :, Cr, , Mn, , Fe, , Co, , E°M2+/M, , –0.91, , –1.18, , –0.44, , –0.28, , E°M3+/M2+, , –0.41, , +1.57, , +0.77, , +1.97, , (i) Which is a stronger reducing agent in, aqueous medium, Cr2+ or Fe2+ and why?, , 15. Write the electronic configuration of Ce3+, ion, and calculate the magnetic moment on, the basis of ‘spin-only’ formula. [Atomic no. of, Ce = 58]

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16. (i) Which metal in the first transition series, (3d-series) exhibits +1 oxidation state most, frequently and why?, (ii) Which of following cations are coloured in, aqueous solutions and why?, Sc3+, V3+, Ti4+, Mn2+, (At. Nos. Sc = 21, V = 23, Ti = 22, Mn = 25), 17. How would you account for the following :, (i) Cr2+ is reducing in nature while with the, same d-orbital configuration (d4), Mn3+ is an, oxidising agent., (ii) In a transition series of metals, the, metal which exhibits the greatest number of, , oxidation states occurs in the middle of the, series., 18. What is lanthanoid contraction and what is, it due to? Write two consequences of lanthanoid, contraction., 19. What is meant by ‘disproportionation’? Give, an example of a disproportionation reaction in, aqueous solution., 20. (i) Write two characteristic of the transition, elements., (ii) Which of the 3d-block elements may not be, regarded as the transition elements and why?, , Short Answer Type Questions (SA-II), 21. (i) Explain the cause of paramagnetism in, lanthanoid ions., (ii) Nb and Ta exhibit similar properties. Give, reason., (iii) Among the ionic species, Sc3+, Ce4+ and Eu2+,, which one is a good oxidising agent., 22. Following are the transition metal ions of 3d, series :, Ti4+, V2+, Mn3+, Cr3+, (Atomic numbers : Ti = 22, V = 23, Mn = 25, Cr = 24), Answer the following :, (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?, 23. Compare qualitatively the first and second, ionisation potentials of copper and zinc. Explain, the observation., 24. (i) Ce (IV) is a good analytical reagent., Why?, (ii) Account for the following : Copper(I), compounds are white whereas copper(II), compounds are coloured., , (ii) The E° value for the Mn3+/Mn2+ couple is, much more positive than that for Fe3+/Fe2+, couple., (iii) The highest oxidation state of a metal is, exhibited in its oxide or fluoride., 27. (a) Assign reasons for the following :, (i) Cu(I) ion is not known to exist in aqueous, solutions., (ii) Transition metals are much harder than the, alkali metals., (b) Name the lanthanoids which show, abnormally low value of third ionisation enthalpy., 28. Account for the following :, (i) The transition metals and their compounds, act as good catalysts., (ii) The lowest oxide of transition metal is basic,, the highest is amphoteric/acidic., (iii) The magnetic moment (B.M.) of Fe2+ ion, is 24 ., , 25. How do the oxides of transition elements in, lower oxidation states differ from those in higher, oxidation states and why?, , 29. (a) Account the following :, (i) Transition metals form large number of, complex compounds., (ii) E°° value for the Mn3+/Mn2+ couple is highly, positive (+1.57 V) as compared to Cr3+/Cr2+., (iii) Which of following cations are coloured in, aqueous solutions and why?, Sc3+, V3+, Ti4+, Mn2+, (At. Nos. Sc = 21, V = 23, Ti = 22, Mn = 25), , 26. How would you account for the following :, (i) The atomic radii of the metals of the third, (5d) series of transition elements are virtually, the same as those of the corresponding members, of the second (4d) series., , 30. (i) Out of the ions Ag+, Co2+ and Ti4+ which, will be coloured in aqueous solution?, (ii) If each one of the above ionic species is placed, in a magnetic field, how will they respond and, why?

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31. (a) Explain the following :, The enthalpies of atomization of transition, metals are quite high., (b) Explain the following observations., (i) With the same d-orbital configuration, (d4), Cr2+ is a reducing agent while Mn3+ is an, oxidising agent., (ii) There is hardly any increase in atomic size, with increasing atomic numbers in a series of, transition metals., 32. (i) Transition metals have very high melting, and boiling points. Why?, (ii) In d-block element, ionic radii of ions of, the same charge decreases progressively with, increasing atomic number in a series. Why?, 33. (a), , , E°(M2+/M), , Cr, –0.91, , Mn, –1.18, , Fe, Co, Ni, Cu, –0.44 –0.28 –0.25 +0.34, , From the given data of E° values, answer the, following questions :, , (i) Why is E°(Cu2+/Cu) value exceptionally positive?, (ii) Why is E°(Mn2+/Mn) value highly negative as, compared to other elements?, (b) Give reason and select one atom/ion which, will exhibit asked property :, (i) Sc3+ or Cr3+ (exhibit diamagnetic behaviour), (ii) Cr or Cu (high melting and boiling point), 34. Give reasons for the following :, (i) Mn3+ is a good oxidising agent., (ii) EM2+/M values are not regular for first row, transition metals (3d-series)., (iii) d-block elements exhibit more oxidation, states than f-block elements., 35. How would you account for the following :, (i) The oxidising power of oxoanions are in the, order, VO+2 < Cr2O72– < MnO4–, (ii) The third ionization enthalpy of manganese, (Z = 25) is exceptionally high., (iii) Cr2+ is a stronger reducing agent than Fe2+., , Long Answer Type Questions (LA), , (iii) Co (II) is easily oxidised in the presence of, strong ligands., , 38. (a) The elements of 3d transition series are, given as, Sc Ti V Cr Mn Fe Co Ni Cu Zn, Answer the following :, (i) Write the element which shows maximum, number of oxidation states. Given reason., (ii) Which element has the highest melting, point?, (iii) Which element shows only +3 oxidation, state?, (iv) Which element is a strong oxidising agent in, +3 oxidation state and why?, (v) Why Mn2O3 is basic whereas Mn2O7 is, acidic?, , 37. Compare the general characteristics of, the first series of the transition metals with, those of the second and third series metals in, the respective vertical columns. Give special, emphasis on the following points :, (i) electronic configuration, (ii) oxidation states, (iii) ionisation enthalpies and, (iv) atomic sizes., , 39. (a) Account for the following :, (i) Transition metals show variable oxidation, states., (ii) Zn, Cd and Hg are soft metals., (b) Give reason :, Iron has higher enthalpy of atomization than, that of copper., (c) What are interstitial compounds? Write their, properties., , 36. (a) Give reason :, (i) Sc (21) is a transition element but Ca (20) is, not., (ii) The Fe2+ is much more easily oxidised to, Fe3+ than Mn2+ to Mn3+., (b) How would you account for the following :, (i) Metal-metal bonding is more extensive in, the 4d and 5d series of transition elements than, the 3d series., (ii) Mn (III) undergoes disproportionation reaction, easily.

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40. Give reasons for the following :, (i) Silver bromide is used in photography., (ii) Most transition metal compounds are, coloured., (iii) Zinc and not copper is used for the recovery, of metallic silver from complex, , [Ag(CN)2]–. Explain., (iv) The colour of mercurous chloride, Hg2Cl2,, changes from white to black when treated with, ammonia., (v) The species [CuCl4]2– exists while [CuI4]2–, does not., , OBJECTIVE TYPE QUESTIONS, , 11. (c) : Zn 2+ → 3d 10 has no unpaired electrons to be, excited. Hence, it is colourless., , 1. (b) : The overall decrease in atomic and ionic radii from, La3+ to Lu3+ is called lanthanoid contraction. Hence, the, correct order is, Yb3+ < Pm3+ < Ce3+ < La3+, 2. (a) : Cu 2+ has 3d 9 configuration i.e., one unpaired, electron, hence, it is paramagnetic while Zn2+ has 3d 10, configuration i.e., all orbitals are filled, hence it is diamagnetic, in nature., 3. (c) : Fe3+ – 3d 5, Cr3+ – 3d 3, Ni2+ – 3d 8, Cu2+ – 3d 9, , No. of unpaired electrons = 5, No. of unpaired electrons = 3, No. of unpaired electrons = 2, No. of unpaired electrons = 1, , 4. (a) : Acidic strength of oxides of transition metals, increases with increase in oxidation number., +2, , +8 /3, , +3, , +4, , +7, , Hence acidic strength is of the order of, MnO < Mn3O4 < Mn2O3 < MnO2 < Mn2O7, 5., , Acidic, , Amphoteric, , (d) :, , 24 Cr → 1s, , 2, , 1 5, 2s 2 2p 6 3s 2 3p 6 4, s, 3, d, , half -filled, , 6. (c) : E° values for M 2+/M with negative signs are, Cr = – 0.91 V, Mn = – 1.18 V, Fe = – 0.44 V, Co = – 0.28 V, Thus, the order is Mn > Cr > Fe > Co., +6, , +7, , +4, , → 2MnO4− + MnO2 + 2H2O shows, 7. (b) : 3MnO24− + 4H+ , disproportionation since the oxidation state of Mn changes, from +6 to +7 (MnO4– ) and +4 (MnO2)., 8., , (d) : Hg is the last element of third transition series., , 9. (a) : µ = n (n + 2), Electronic configuration of ion (25) = 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 3 d 5 4 s 2, No. of unpaired electrons (n) = 5, µ = 5(5 + 2) = 35 = 5.9 B.M., 10. (d), , 13. (c) : They are not chemically reactive. They are chemically, inert., 14. (b) : Reactivity of transition elements decreases almost, regularly from Sc to Cu because of regular increase in, ionisation enthalpy., 15. (a) : Zn2+(3d10) has zero unpaired electron (colourless)., Ni2+(3d 8) has 2 unpaired electrons (coloured)., Cu2+(3d 9) has 1 unpaired electron (coloured)., Cr2+(3d 4) has 4 unpaired electrons (coloured)., 16. (d) : Ionic radii of trivalent lanthanides decrease with, increase in atomic number., 17. (c) : µ = n (n + 2) BM, . ., , MnO, Mn3 O4 ,Mn2 O3 ,MnO2 ,Mn2 O7, , Basic, , 12. (b) : The third ionisation energy of Mn required to, change Mn2+(d 5) to Mn3+(d 4) is much larger due to stable, half-filled d 5 electronic configuration., , 18. (a) : Though Cu(I) has 3d10 stable configuration while, Cu(II) has 3d 9 configuration, yet Cu(II) is more stable due, to greater effective nuclear charge of Cu(II) (i.e., to hold 17, electrons instead of 18 in Cu(I))., 19. (b) : In case of transition metal oxides, the oxides with, metals in lower oxidation state are basic in nature., O.S. of Mn in Mn2O7 = +7; V in V2O3 = +3; V in V2O5 = +5;, Cr in CrO = +2; Cr in Cr2O3 = +3, Thus in V2O3, CrO and Cr2O3 transition metal ion is in lower, oxidation state but Cr2O3 is amphoteric in nature. Hence V2O3, and CrO are basic in nature., 20. (c) : In the Wacker process, the oxidation of ethyne to, ethanal is catalysed by PdCl2., 21. (a) : Lanthanum is a d-block element which resembles, lanthanides., 22. (d) : The electronic configuration of Ce3+ is 4f 1, Hence, µ = n (n + 2) = 11, ( + 2) = 1.732 B.M., 23. (a) : Tl+ is more stable than Tl3+ due to inert pair effect., Cu2+ is more stable than Cu+. Cr3+ is more stable than Cr2+., V4+ in aqueous solution is more stable than V2+.

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24. (b) : Ag2SO4 → Ag+ (4d10) – colourless, CuF2 → Cu2+ (3d 9) – coloured, MgF2 → Mg2+ (no d-electrons) – colourless, CuCl → Cu+ (3d10) – colourless, 25. (b) : Basic strength decreases from La(OH)3 to Lu(OH)3., Hence, (1) is incorrect., 0, Ce : [Xe] 4f 1 5d1 6s2; Ce4+ : [Xe] 4f, Yb : [Xe] 4f 14 6s2; Yb2+ : [Xe] 4f 14, Lu : [Xe] 4f 14 5d1 6s2; Lu3+ : [Xe] 4f 14, The given ions contain no unpaired electrons and therefore,, are diamagnetic., 26. (b), 27. (d) : Sc3+ (3d 0), Ti 4+ (3d 0) are diamagnetic while Pd2+, (4d 8) and Cu2+ (3d 9) are paramagnetic., 28. (d) : In CrO2Cl2, O.S. of Cr = +6, MnO4–, O.S. of Mn = +7, 3–, Cr(CN)6 , O.S. of Cr = +3, 2–, NiF6 , O.S. of Ni = +4, 29. (a), 30. (b) : Fe2+ − 1s 2 2s 2 2p 6 3s 2 3p 6 3d 6, Fe3+ − 1s 2 2s 2 2p 6 3s 2 3p 6 3d 5, In Fe3+, orbital is half-filled hence provides extra stability., 31. (a) : The transition metals and their compounds are, known for their catalytic activity because of their ability to, adopt multiple oxidation states and to form complexes., 32. (a) : (n – 1)d1-10 ns0 –2, 33. (b) : La(OH)3 is most basic. Hence, (I) is wrong. (II) is, correct due to lanthanoid contraction. (III) is correct because, Ce4+ tends to change to stable Ce3+., 34. (b) : [Fe(CN)6]4– ; No. of unpaired electrons = 0, [Fe(CN)6]3– ; No. of unpaired electrons = 1, [Ni(H2O)4]2+ ; No. of unpaired electrons = 2, [Cr(NH3)6]3+ ; No. of unpaired electrons = 3, 35. (b) : 3d 5 configuration is more stable due to singly, occupied half-filled orbitals., 36. (a) : The increasing order of oxidising power is due to, increasing stability of the lower species to which they are, reduced., 37. (d) : More the number of unpaired d-electrons, more is, the magnetic moment., Cu2+ – 3d 9 ; No. of unpaired electrons = 1, Ni2+ – 3d 8 ; No. of unpaired electrons = 2, Co2+ – 3d 7 ; No. of unpaired electrons = 3, Fe2+ – 3d 6 ; No. of unpaired electrons = 4, , 38. (c) : Cr2O3 and V2O5 are amphoteric oxides., 39. (b) : Hs (Z = 108) belongs to 6d series with electronic, configuration – [Rn]5f 146d 67s 2., 40. (d) : Ce is most commonly used lanthanoid, nobelium, (No) and Th (thorium) are actinoids., 41. (a) : Cr forms CrF3 whereas Cu and Ni do not form CuF3, and NiF3., 42. (c) : Due to lanthanoid contraction, Zr and Hf have, nearly equal size., 43. (b) : Due to presence of unpaired electrons in (n – 1)d, orbitals, the most of the transition metal ions and their, compounds are paramagnetic. They form coloured ions and, show variable oxidation states due to presence of vacant, d-orbitals., 44. (a) : The electronic configuration of Cr is, 5, 1, 24Cr → [Ar] 3d 4s (six unpaired electrons), In K2Cr2O7, the oxidation number of Cr = +6. So it has no, unpaired electron., 45. (a) : Due to lanthanoid contraction, the atomic radii of, second and third row transition elements is almost same., Hence, they resemble each other much more as compared, to first row elements., 46. (b) : Na2[CuCl4] ; Cu = +2 or, Na2[CdCl4] ; Cd = +2 or, K4[Fe(CN)6] ; Fe = +2 or, K3[Fe(CN)6] ; Fe = +3 or, Since Cd 2+ has completely filled, colourless., , Cu2+ → 3d 9, Cd2+ → 4d 10, Fe2+ → 3d 6, Fe3+ → 3d 5, d-subshell hence it is, , 47. (b) : Cu+ ions undergo disproportionation,, 2Cu+ → Cu2+ + Cu, 48. (d) : The colour of transition metal ions is due to d-d, transitions., 49. (c) : Negative value of E° for Ni2+/Ni is related to the, highest negative DH°hyd ., 50. (d) : 3[Fe(H2O)6]2+ + NO3– + 4H+, , NO + 3[Fe(H2O)6]3+ + 2H2O, , , [Fe(H2O)6]2+ + NO, , , (X )   , , (Y ), , [Fe(H2O)5NO]2+ + H2O, Brown (Z), , In Y, Fe3+ is present : [Ar]3d 5, µ = n (n + 2) = 5.9 B.M., NO is a neutral oxide., 51. (c) : From the melting point trend it is clear that, W → Cr, X → Mn, Y → Fe, Z → Co

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Mn2+ ⇒ [Ar]3d 5, µ = 5 (5 + 2) = 5.9 B.M., Cr3+ → Violet, Fe3+ catalyses the given reactions :, I2 + 2SO42–, 2I– + S2O82–, Co shows +2 and +3 oxidation states., 52. (d) : (a) Cr2+ is a reducing agent, it gets oxidised to, Cr3+ (3d 3 or t 32g, stable half-filled configuration)., (b) Mn3+ is an oxidizing agent, it gets reduced to Mn2+, (3d 5, most stable, half-filled configuration)., Mn (25) : 3d 54s2, (c) Cr (24) : 3d 44s2, Cr2+ : 3d 4, Mn3+ : 3d 4, 2+, 3+, 4, Both Cr and Mn exhibit d electronic configuration., (d) When Cr2+ is used as a reducing agent, the chromium, ion attains d 3 electronic configuration., 53. (b) : Ce3+ and Yb3+ are colourless despite having one, unpaired electron., The lanthanide ions, other than 4f 0 type (La3+ and Ce4+) and, the 4f 14 type (Yb2+ and Lu3+) are all paramagnetic., 54. (c) : The elements scandium and yttrium are called “rare, earths” because they were originally discovered together, with lanthanides in rare minerals and isolated as oxides or, “earths”. Collectively, these metals are also called rare earth, elements., 55. (a), , 64. (a) : Oxide of V in lowest oxidation state, i.e., VO is basic, and ionic in character., 65. (a) : Vanadyl ion is VO2+ where V is in +4 oxidation, state., 66. (b) , 67. (c), 68. (a) , 69. (a), 70. (a) : Lesser and negative reduction potential indicates, that Cr2+ is a reducing agent. Higher and positive reduction, potential indicates that Mn3+ is a stronger oxidizing agent., 71. (a) : The electrode which has more reduction potential, is a good oxidizing agent and has least reducing power., 72. (b) : (I) It is because Mn 2+ has 3d 5 electronic, configuration which has extra stability., (II) Not titanium but copper, because with +1 oxidation, state an extra stable configuration, 3d10 results., (III) It is not stable as it undergoes disproportionation;, 2Cu+(aq) → Cu2+(aq) + Cu(s). The E° value for this is favourable., (IV) Much larger third ionisation energy of Mn (where the, required change is d 5 to d 4) is mainly responsible for this., +, 73. (a) : The stability of Cu2+, (aq) rather than Cu (aq) is due to, +, than Cu+, which, the much more negative DHydH° of Cu2(aq), more than compensates for the second ionisation enthalpy, of Cu., , 56. (d) : In lanthanide series, with increasing atomic number,, there is a progressive decrease in the atomic as well as on, radii of trivalent ions form La3+ to Lu3+., 57. (c) : Yb 2+ is more stable than Yb3+ because it will, acquire stable configuration of completely filled 4f subshell, after losing 2 electrons. Electronic configuration of Yb2+ is, [Xe]4f14., , 74. (a) : Terbium (Tb3+) : 4f 8, Erbium (Er3+) : 4f 11, Ytterbium, (Yb) : 4f 13., 75. (a) , 76. (c), , 58. (a), , 79. (b) : Transition metals form substitutional alloys since, they have nearly the same size, they can substitute one, another in the crystal lattice., , 59. (b) : The transition metals and their compounds are, known for their catalytic activity. This activity is ascribed, to their ability to adopt multiple oxidation states to form, complexes., 60. (c) : Because of similar radii and other characteristics of, transition metals, alloys are readily formed by these metals., 61. (c), 62. (b) : Greater the number of valence electrons, more will, be the number of oxidation states exhibited by the element., 63. (a) : All the d-block elements are metals, they exhibit, most properties of metals like lustre, malleability, ductility,, high density, high melting and boiling point, hardness,, conduction of heat and electricity, etc. All the f-block elements, are also metals but they are not good conductors of heat and, electricity., , 77. (a), 78. (b) : The almost identical radii of Zr (160 pm) and, Hf (159 pm), a consequence of lanthanoid contraction., , 80. (c) : Mn2+ = [Ar]3d 5, Mn3+ = [Ar]3d 4, Fe2+ = [Ar]3d 6, Fe3+ = [Ar]3d 5, Thus, Mn2+ has more stable configuration than Mn3+ while, Fe3+ has more stable configuration than Fe2+., Hence, reduction potential for Mn3+/Mn2+ couple is more, positive than Fe3+/Fe2+., As we move across the period, ionisation potential increases,, thus, ionisation potential of Fe is more than that of Mn., 81. (a) : A half-filled or fully-filled orbital is more stable than, incompletely filled orbital., Hence Mn2+ (3d 5) is more stable than Mn3+ (3d 4).

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82. (b) : Ti3+ has [Ar]3d1 configuration. Thus, d–d transition, is possible and thereby it shows colour., 83. (b) : Chromium has maximum number of unpaired, d electrons while Hg do not have any unpaired, d electrons. Thus, Cr is hard but Hg is soft metal., 84. (d) : Even though Cu+ has completely filled 3d orbitals,, [Ar]3d10 the nuclear charge in Cu is not enough to hold a, core of 18 electrons in Cu+ and thus, Cu+ is unstable in, comparison to Cu2+. Cu+ is diamagnetic in nature., 85. (d) : Both Co and Ni have (IV), has 3d 6 electronic configuration., Metals Outer electronic, configuration, Co 3d 74s2 , Ni 3d 84s2 , , oxidation state. Ni (IV), Oxi. states, +2, +3, +4, +2, +3, +4, , 86. (a), 87. (b) : Due to large surface area and variable valencies,, transition metals form intermediate activated complexes, easily, hence they are used as good catalysts., 88. (b) : The electronic configurations of europium (II) and, cerium (II) are Eu2+ : [Xe] 4f 7 and Ce2+ : [Xe] 4f 1 5d1, In Eu2+, f-subshell is half-filled thus, it is more stable., 89. (a) : Zinc has all electrons paired [Ar]3d104s2. So, it is, diamagnetic in nature., 90. (d) : The oxidation states of the given compounds are, the following,, VO+2 : x + 2(–2) = +1, ⇒ x = +5, VO : x –2 = 0, ⇒ x = +2, VO2+ : x + 1(–2) = +2, ⇒ x=+4, The correct order of oxidising power is :, +, VO < VO2+ < VO 2, –, In MnO4 : x + 4 (–2) = –1, ⇒ x – 8 = –1 ⇒ x = – 1 + 8 = +7, 91. (b) : Some non-metallic atoms (e.g., H, B, C, N, etc.) are, able to fit in the interstitial sites of transition metals lattices, to form interstitial compounds., 92. (b) : It is due to lanthanide contraction., 93. (a) , 94. (a), , 95. (b) : In actual practice transition metals react with acid, very slowly and act as poor reducing agents. This is due to, the protection of metal as a result of formation of thin oxide, protective film. Further, their poor tendency as reducing agent, is due to high ionisation energy, high heat of vapourization, and low heat of hydration., , SUBJECTIVE TYPE QUESTIONS, 1. Eu2+ has a strong tendency to lose electrons to attain, the more stable +3 oxidation state of lanthanoids hence, it, is a strong reducing agent., 2. Formula of oxoanion of manganese is MnO–4., Oxidation state of Mn in this oxoanion = + 7, Group number of Mn is 7., 3., , Lanthanoids showing +4 oxidation state are, 58Ce, 59Pr, 65Tb., , 4. In aqueous solutions, Cu+ undergoes disproportionation, to form a more stable Cu2+ ion., 2Cu+(aq) → Cu2+(aq) + Cu(s), Cu2+ in aqueous solutions is more stable than Cu+ ion, because hydration enthalpy of Cu2+ is higher than that of, Cu+. It compensates the second ionisation enthalpy of Cu, involved in the formation of Cu2+ ions., 5. Europium (Eu) is well known to exhibit +2 oxidation, state due to its half-filled f orbital in +2 oxidation state., 6. In the electronic configuration of Zn the d-orbitals are, completely filled in the ground state as well as in its common, oxidation state. So, it is not regarded as transition metal., 7. Lanthanum and all the lanthanoids predomi-nantly, show +3 oxidation state. However, some of the lanthanoids, also show +2 and +4 oxidation states in solution or in solid, compounds. This irregularity arises mainly due to attainment, of stable empty (4f 0), half-filled (4f 7) and fully filled (4f14), sub shell., e.g., Ce4+ : 4f 0 , Eu2+ : 4f 7, Tb4+ : 4f 7 , Yb2+ : 4f 14, 8. Due to lanthanoid contraction the elements of 4d and, 5d-series have similar atomic radii e.g., Zr = 145 pm and, Hf = 144 pm., 9. Oxoanion of chromium in which it shows +6 oxidation, state equal to its group number is Cr2O72– (dichromate ion)., 10. Zn2+ ion has completely filled d-subshell and no d-d, transition is possible. So zinc salts are white. Configuration, of Cu2+ is [Ar] 3d 9. It has partly filled d-subshell and hence, it is coloured due to d-d transition., 11. Europium (II) has electronic configuration [Xe]4f 75d 0, while cerium (II) has electronic configuration [Xe] 4f 1 5d1. In, Eu2+, 4f subshell is half filled and 5d-subshell is empty. Since,, half filled and completely filled electronic configurations are, more stable, so Eu2+ ions is more stable than Ce2+ in which, neither 4f subshell nor 5d subshell is half filled or completely, filled.

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12. Magnetic moment, meff = 3.87 BM corresponds to the, number of unpaired electrons, n = 3 by applying the formula., meff =, , n(n + 2) BM, , For n = 1, m = 1.73 B.M, for n = 2, m = 2.83 BM, For n = 3, m = 3.87 B.M and so on., 13. Cr2+ is a stronger reducing agent than Fe2+., E° Cr3+/Cr2+ is negative (–0.41 V) whereas E° Fe3+/Fe2+ is, positive (+ 0.77 V). Thus Cr2+ is easily oxidized to Cr3+ but, Fe2+ cannot be easily oxidized to Fe3+. Hence, Cr2+ is stronger, reducing agent than Fe2+., (ii) More positive is the value of E°, reaction will be more, feasible., As E°Co3+/Co2+ is maximum, thus Co2+ ion is most stable., 14. (i) Electrode potential (E°) value is the sum of three, factors :, (a) Enthalpy of atomisation ; DaH for Cu(s) → Cu(g), (b) Ionisation enthalpy ; DiH for Cu(g) → Cu2+, (g), Cu2+, (g), , Cu2+, (aq), , (c) Hydration enthalpy ; DhydH for, →, In case of copper the sum of enthalpy of atomisation and, ionisation enthalpy is greater than enthalpy of hydration., This is why E°M2+/M for Cu is positive., (ii) This is due to lanthanoid contraction., 15. Ce (Z = 58) = [Xe] 4f1 5d1 6s2, \ Ce3+ = [Xe] 4f1 5d 0 6s0, Therefore, it has only one unpaired electron. i.e., n = 1, \, , µ = n (n + 2) = 11, ( + 2) = 3 = 1.73B.M., , 16. (i) Copper exhibits +1 oxidation state in its compounds., Electronic configuration of Cu in the ground state is 3d10 4s1., So, Cu can easily lose 4s1 electron to attain a stable 3d10, configuration. Thus, it shows +1 oxidation state., (ii) Only those ions will be coloured which have partially filled, d-orbitals facilitating d-d transition. Ions with d0 and d10 will, be colourless., From electronic configuration of the ions, V3+(3d 2) and, Mn2+(3d 5), are all coloured. Ti4+(3d 0) and Sc3+(3d 0) are, colourless., 17. (i) E° values for the Cr3+/Cr2+ and Mn3+/Mn2+ couples, are, –, Cr3+, Cr2+, (aq) + e, (aq); E° = –0.41 V, –, Mn3+, (aq) + e, , Mn2+, (aq) ; E° = +1.551 V, , These E° values indicate that Cr2+ is strongly reducing while, Mn3+ is strongly oxidising agent., (ii) Middle of the transition series contains greater number of, unpaired electrons in (n –1)d and ns orbitals., , 18. Lanthanoid contraction : The steady decrease in the, atomic and ionic radii of lanthanoid elements with increase, in atomic number is called lanthanoid contraction., It is caused due to imperfect shielding of nuclear charge by, 4f-electrons., Consequences of lanthanoid contraction :, (i) The basic strength of oxides and hydroxides of, lanthanoids decrease with increasing atomic number., (ii) Atomic and ionic sizes of 4d transition series elements, and 5d series elements are similar. e.g., atomic radii of, zirconium(Zr) is same as that of hafnium (Hf)., 19. Disproportionation reaction involves the oxidation, and reduction of the same substance. The examples of, disproportionation reaction are :, (i) Aqueous NH3 when treated with Hg2Cl2 (solid)forms, mercury aminochloride disproportionatively., Hg + Hg(NH2)Cl + NH4Cl, Hg2Cl2 + 2NH3, +, Cu + Cu2+, (ii) 2Cu, 20. (i) Elements which have incompletely filled d-orbitals in, their ground state or in any one of their oxidation states are, called transition elements., Characteristics of transition elements :, (a) They show variable oxidation states., (b) They exhibit catalytic properties., (ii) Zn, Cd, Hg are considered as d-block elements but not, as transition elements because they do not have partly filled, d-orbitals in their atomic state or their common oxidation, states (i.e., Zn2+, Cd2+, Hg2+)., 21. (i) Due to presence of unpaired electrons in f-orbital, lanthanoid ions are paramagnetic in nature., (ii) Due to lanthanoid contraction, their sizes are same., Hence, their properties are similar., (iii) Ce4+. The stable oxidation state of lanthanoids is +3., Ce4+ tends to accept an electron to change to +3 state., Hence, it acts as a good oxidising agent., 22. (i) Ti4+ has highest oxidation state among the given ions., Ti4+ has stable inert gas configuration and hence, most stable, in aqueous solution., On the other hand, V2+, Mn3+, Cr3+ have unstable electronic, configuration and hence, are less stable., (ii) Due to presence of highest oxidation state of Ti, it acts, as the strongest oxidising agent among the given ions., (iii) Due to absence of unpaired electron in Ti4+, it is a, colourless ion., E.C. of Ti4+ : [Ar]3d 04s0, 23. The electronic configuration of Zn and Cu are :, Zn: 1s2 2s2 2p6 3s2 3p6 3d10 4s2, Cu: 1s2 2s2 2p6 3s2 3p6 3d10 4s1

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From the above configuration it is clear that first ionisation, energy of Zn is greater than that of Cu (because of 4s2 and, 4s1 configuration of Zn and Cu respectively). More energy is, needed to remove an electron from 4s2 than from 4s1., The second I.E. of Cu is higher than that of Zn because for, Cu+ the configuration is 1s2 2s2 2p6 3s2 3p6 3d10 and for, Zn+ the configuration is 1s2 2s2 2p6 3s2 3p6 3d10 4s1, it is, easier to remove 4s1 electron of Zn+ than a 3d-electron from, 3d10 (stable configuration)., 24. (i) The E° value for Ce4+/Ce3+ is 1.74 V which suggests, that it can oxidise water. However the reaction rate is very, slow and hence Ce (IV) is a good analytical reagent., (ii) Cu(I) compounds have completely filled d-orbitals and, there are no vacant d-orbitals for promotion of electrons, whereas in Cu(II) compounds have one unpaired electron, which is responsible for colour formation., 25. In the lower oxidation state, the transition metal, oxides are basic and they are acidic if the metal is in higher, oxidation state. The oxides are amphoteric when the metal, is in intermediate oxidation state., For example,, +3, , Mn2O3, Basic, , +4, , MnO2, , Amphoteric, , +7, , Mn2O7, Acidic, , In case of lower oxide of a transition metal, the metal atom, has a low oxidation state. This means some of the valence, electrons of the metal atom are not involved in bonding,, hence, these can be used for donation. Thus, these are act, as bases., 26. (i) This is due to lanthanoid contraction., (ii) Much larger third ionisation energy of Mn(where change, is d 5 to d 4) is mainly responsible for this. This also explains, that +3 state of Mn is of little importance., From the relation, DG° = –nFE°, More positive is the value of E°, reaction will be feasible., –, + e–, Mn3+, Mn2+, ;, Fe3+ + e Fe2+, 3d 4, 3d 5, 3d 5, 3d 6, more stable, , more stable, , (half filled), , (half filled), 3+, , 2+, , Hence, E°value for Mn /Mn couple is much more positive, than that for Fe3+/Fe2+., (iii) Manganese can form pp-dp bond with oxygen by utilising, 2p-orbital of oxygen and 3d-orbital of manganese due to, which it can show highest oxidation state of +7. While with, fluorine it cannot form such pp - dp bond thus, it can show, a maximum oxidation state of +4., 27. (a) (i) In aqueous solutions, Cu + undergoes, disproportionation to form a more stable Cu2+ ion., 2Cu+(aq) → Cu2+(aq) + Cu(s), , Cu 2+ in aqueous solutions is more stable than Cu + ion, because hydration enthalpy of Cu2+ is higher than that of, Cu+. It compensates the second ionisation enthalpy of Cu, involved in the formation of Cu2+ ions., (ii) This is attributed to the involvement of greater number, of electrons from (n –1)d in addition to the ns electrons in, the interatomic metallic bonding., (b) Third ionization enthalpy of lanthanoid is low if it leads, to stable empty, half filled or completely filled configuration,, as indicated by the abnormally low third ionization enthalpies, of La, Gd, and Lu., 28. (i) The transition metals and their compounds, are known, for their catalytic activity. This activity is ascribed to their, ability to adopt multiple oxidation states, ability to adsorb, the reactant(s) and ability to form complexes. Vanadium (V), oxide (in Contact Process), finely divided iron (in Haber’s, Process), and nickel (in catalytic hydrogenation) are some of, the examples., (ii) Lowest oxidation compounds of transition metals are basic, due to their ability to get oxidised to higher oxidation states., Whereas, the higher oxidation state of metal and compounds, gets reduced to lower ones and hence acts as acidic in nature., e.g., MnO is basic whereas Mn2O7 is acidic., (iii) :, Fe2+ ⇒ 3d 6, \ n = 4, Magnetic moment =, , n(n + 2) B.M. = 24 B.M., , 29. (a) (i) Transition metals form a large number of complex, compounds due to following reasons :, – Comparatively smaller size of metal ions., – High ionic charges., – Availability of d-orbitals for bond formation., (ii) Because Mn2+ is stable due to half filled configuration., Thus Mn3+ has high tendency to form Mn2+ while Cr3+ is, more stable than Cr2+., (iii) Only those ions will be coloured which have partially, filled d-orbitals facilitating d-d transition. Ions with d 0 and, d10 will be colourless., From electronic configuration of the ions, V3+(3d 2) and, Mn2+(3d 5), are all coloured. Ti4+(3d 0) and Sc3+(3d 0) are, colourless., 30. (i) The ionic species which possesses unpaired electron, or electrons in (n – 1)d-subshell will show colour. Out of, the ions Ag+(4d10), Co2+(3d7) and Ti4+(3d 0), Co2+ will be, coloured as it contains three unpaired electrons, Ag+ and, Ti4+ will be colourless., (ii) When placed in magnetic field, Co2+ will be attracted, because it is paramagnetic due to unpaired electrons. Ag+, and Ti4+ ions will be repelled by the magnetic field as they, are diamagnetic.

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31. (a) As transition metals have a large number of unpaired, electrons in the d-orbitals of their atoms they have strong, interatomic attraction or metallic bonds. Hence, they have, high enthalpy of atomization., (b) (i) E° values for the Cr3+/Cr2+ and Mn3+/Mn2+ couples, are, –, Cr3+, Cr2+, (aq) + e, (aq); E° = –0.41 V, –, Mn3+, (aq) + e, , Mn2+, (aq) ; E° = +1.551 V, , These E° values indicate that Cr2+ is strongly reducing while, Mn3+ is strongly oxidising agent., (ii) As one proceeds along a transition series, the nuclear, charge increases which tends to decrease the size but the, addition of electrons in the d-subshell increases the screening, effect which counterbalances the effect of increased nuclear, charge. As a result, the atomic radii remain practically same, after chromium., 32. (i) The high melting and boiling points of transition, metals are attributed to the involvement of greater number, of electrons from (n – 1) d-orbital in addition to the ns, electrons in the interatomic metallic bonding (d-d overlap)., (ii) As the atomic number increases the new electron enters, the d-orbital and expected to increase in atomic size, but, due to poor shielding effect of d-orbitals the electrostatic, attraction between nucleus and outermost orbital increases, and hence, the ionic radii decreases., 33. (a) (i) Copper has high energy of atomisation and low, energy of hydration., (ii) Mn2+ ion has stable half-filled (3d5) electronic configuration., Its ionisation enthalpy value is lower in comparison to hydration, enthalpy. Hence E °Mn2+/Mn is more negative., (b) (i) Sc3+ has 3d 0 outer electronic configuration, therefore, it is diamagnetic in nature whereas Cr 3+ has 3d 3 outer, electronic configuration. So, it is paramagnetic due to, presence of unpaired electrons., (ii) In a particular series, the metallic strength increases, upto middle with increasing number of unpaired electrons,, i.e., upto d 5 configuration. After Cr, the number of unpaired, electrons goes on decreasing. Accordingly, the m.pt and b.pt., decrease after middle (Cr) because of increasing pairing of, electrons., 2+, , 5, , 34. (i) Mn is more stable due to half filled d configuration, and Mn3+ easily changes to Mn2+ hence, it is a good oxidising, agent., (ii) The E° M 2+ / M values are not regular which can be, explained from the irregular variation of ionisation enthalpes, i.e., IE1 + IE2 and also the sublimation enthalpies which are, relatively much less for manganese and vanadium., , (iii) All transition elements except the first and the last, member in each series show a large number of variable, oxidation states. This is because difference of energy in the, (n – 1)d and ns orbitals is very little., Hence, electrons from both the energy levels can be used, for bond formation., 35. (i) Change in Cr2O72– to Cr(III) is 3 and in MnO4– to, Mn (II) is 5., Change in oxidation state is large and the stability of reduced, products in V(III) < Cr(III) < Mn(II). This is why oxidising, power of VO2+ < Cr2O72– < MnO4–., (ii) Third ionization enthalpy of Mn is very high because the, third electron has to be removed from the stable half-filled, 3d-orbitals [Mn2+ (Z = 25) = 3d 5]., (iii) Cr 2+ is a stronger reducing agent than Fe 2+. E°Cr3+/, Cr2+ is negative (–0.41 V) whereas E°Fe3+/Fe2+ is positive, (+ 0.77 V). Thus Cr 2+ is easily oxidized to Cr 3+ but Fe2+, cannot be easily oxidized to Fe3+. Hence, Cr2+ is stronger, reducing agent than Fe2+., 36. (a) (i) Sc(21) is a transition element but Ca(20) is not, because of incompletely filled 3d orbitals., (ii) Electronic configuration of Mn2+ is 3d 5 which is half, filled and hence stable. Therefore, third ionization enthalpy, is very high, i.e., 3rd electron cannot be lost easily. In case of, Fe2+, electronic configuration is 3d 6. Hence, it can lose one, electron easily to give the stable configuration 3d 5., (b) (i) The metals of 4d and 5d-series have more frequent, metal bonding in their compounds than the 3d-metals, because 4d and 5d-orbitals are more exposed in space than, the 3d-orbitals. So the valence electrons are less tightly held, and form metal-metal bonding more frequently., (ii) Mn3+ is less stable and changes to Mn2+ which is more, stable due to half-filled d-orbital configuration. That is why,, Mn3+ undergoes disproportionation reaction., (iii) The tendency to form complexes is high for Co(III) as, compared to Co(II). Co2+ ions are very stable and are difficult, to oxidise. Co3+ ions are less stable and are reduced by water., In contrast many Co(II) complexes are readily oxidised to, Co(III) complexes and Co(III) complexes are very stable, e.g.,, [Co(NH3 )6 ]2+  , → [Co(NH3 )6 ]3+, Air, , This happens because the crystal field stabilisation energy of, Co(III) with a d 6(t26g) configuration is higher than for Co(II), , with a d 7 (t 62g eg1) arrangement., , 37. (i) E l e c t r o n i c c o n f i g u r a t i o n : E l e c t r o n i c, configuration of group 3 elements (Sc, Y, La) is [Noble gas], (n – 1)d1 ns2.

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Elements of group 6 (Cr, Mo and W) show exception in, electronic configuration. For Cr and Mo [Noble gas] (n – 1), d 5 ns1 and for W it is [Noble gas] 4f 14 5d 4 6s2., Group 11 elements (Cu, Ag and Au) also show exceptional, electronic configuration :, Cu : [Ar] 3d10 4s1, Ag : [Kr]4d10 5s1,, Au : [Xe]4f 14 5d 10 6s1., Group - 10 (Ni, Pd and Pt) also show anomalous electronic, configuration:, Ni : [Ar] 3d 8 4s2, Pd : [Kr] 4d10 5s0, Pt : [Xe] 4f 14 5d 9 6s1, (ii) Oxidation states : Elements within the same group, show similar oxidation states. Highest number of oxidation, states are shown by the elements lying in the middle of the, transition series. Minimum oxidation states are shown by the, elements lying near to left and far right side of the series., Stability of higher oxidation states increases from first to third, series., (iii) Ionisation enthalpies : Ionisation enthalpies generally, decrease down a group. This trend is followed from 3d to, 4d-elements but the ionisation enthalpies either remain same, or increase in going from 4d to 5d-series with the same group., This reverse trend is due to the poor shielding of the nuclear, charge by the inner 4f-electrons. This increases the Zeff and in, turn increases the ionisation enthalpy., (iv) Atomic size : Due to poor shielding of nuclear charge, by 4f-electrons, increase in Zeff decreases the size. So, the, atomic size increase from 3d to 4d but decrease or remain, almost the same from 4d to 5d., 38. (i) Mn shows maximum no. of oxidation states from, +2 to +7 because Mn has maximum number of unpaired, electrons in 3d sub-shell., (ii) Cr has maximum melting point, because it has 6 unpaired, electrons in the valence shell, hence it has strong interatomic, interaction., (iii) Sc shows only +3 oxidation state because after losing, 3 electrons, it has noble gas electronic configuration., (iv) Mn is strong oxidising agent in +3 oxidation state, because change of Mn3+ to Mn2+ give stable half filled (d 5), electronic configuration,, E°(Mn3+/Mn2+) = 1.5 V., , (v) Basic nature of oxides decreases and acidic nature, increases with increase in oxidation state of the metal., Oxidation state of Mn in Mn2O3 is +3 while in Mn2O7 is +7., 39. (a) (i) Transition elements can use their ns and, (n – 1)d orbital electrons for bond formation therefore, they, show variable oxidation states., For example, Sc has ns2(n – 1) d1 electronic configuration., It utilizes two electrons from its ns subshell then its oxidation, state = +2. When it utilizes both the electrons then its, oxidation state = +3., (ii) In Zn, Cd and Hg, all the electrons in d-subshell are, paired. Hence, the metallic bonds are weak. That is why they, are soft metals with low melting and boiling points., (b) Greater the number of unpaired electrons, stronger is, the metallic bond and therefore, higher is the enthalpy of, atomisation. Since, iron has greater number of unpaired, electrons than copper hence has higher enthalpy of, atomisation., (c) When small atoms of non metals like H, C, B, N etc can, occupy vacant interstitial spaces in condition metals it give, size to interstitial compound like hydrides, carbides. Few, properties are as follow, (i), , They have high melting points then pure metals., , (ii) They are conductive., (iii) They are chemically inert., 40. (i) Silver bromide is used in photography because of its, sensitivity to sunlight. In light, AgBr reduces to metallic silver., (ii) The colour of transition metal compound is due to, the presence of incompletely filled d-orbitals in transition, metal ions/atoms, because of this d-d transition can occur, in them. The colour is due to d-d transition for which the, energy is absorbed from visible region. The visible colour of, a compound is the complementary colour of the absorbed, light., (iii) Zinc is a cheaper and stronger reducing agent as, compared to copper., (iv) Mercurous chloride (white) changes to black on, treatment with ammonia because of the formation of finely, divided mercury (grey)., (v) Cu2+ is reduced to Cu+ by I– and thus CuI2 gets converted, to Cu2I2. This change cannot be brought about by Cl–., , 