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Class 12 Chemistry Part 1

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Previous year questions: The d- and f-Block Elements

The d- and f-Block Elements30 min readLesson 6 of 6

Q1 (CBSE Board 2023, 5 marks)

Give a reason for each of the following. (i) Manganese shows the highest oxidation state of +7 among the 3d series. (ii) Transition metals and their compounds are good catalysts. (iii) Cr2+ is reducing while Mn3+, with the same d4 configuration, is oxidising. (iv) Zinc has the lowest enthalpy of atomisation in the 3d series. (v) Cu+ is unstable in aqueous solution.

Answer: (i) Manganese has the configuration 3d5 4s2, so it can lose all seven outer electrons and reach +7, the largest number available in the row. (ii) Transition metals easily change oxidation state, adsorb reactants on their surfaces and form intermediate compounds, all of which lower the activation energy of a reaction. (iii) Cr2+ loses an electron to reach Cr3+, which has the stable half filled t2g3 configuration; Mn3+ gains an electron to reach Mn2+, which has the stable d5 configuration. Each ion therefore moves towards a more stable configuration by the change shown. (iv) Zinc has a completely filled 3d10 subshell, so its d electrons take no part in metallic bonding and less energy is needed to atomise the metal. (v) In water 2Cu+ disproportionates to Cu and Cu2+, and the large hydration energy of Cu2+ makes the +2 state more stable.

Q2 (CBSE Board 2024, 2 marks)

(a) What is the lanthanoid contraction? (b) Why is the actinoid contraction greater from element to element than the lanthanoid contraction?

Answer: (a) The lanthanoid contraction is the steady decrease in atomic and ionic radius across the lanthanoid series. As the atomic number rises, the 4f subshell is filled, and these 4f electrons shield the nucleus poorly, so the effective nuclear charge acting on the outer electrons grows and draws them closer. (b) The 5f electrons of the actinoids shield the nuclear charge even less effectively than the 4f electrons, so the effective nuclear charge grows faster across the actinoid series and the radii contract more sharply from element to element.

Q3 (CBSE Board 2024, 4 marks)

A case study states that transition metals have partly filled d orbitals and therefore show several oxidation states, coloured ions, catalytic action and paramagnetic behaviour. (a) On what basis can we say that copper is a transition element but zinc is not? (b) Why do transition elements show a variety of oxidation states? (c) Explain why the M2+/M standard potentials do not follow a smooth pattern from vanadium to zinc, and point out how the range of oxidation states available to a transition metal differs from that of a non-transition element.

Answer: (a) Copper has the atom configuration 3d10 4s1, but its Cu2+ ion is 3d9 with a partly filled d subshell, so copper qualifies. Zinc is 3d10 4s2 and Zn2+ is 3d10, so its d subshell is never partly filled and it is not a transition element. (b) The ns and (n-1)d electrons have comparable energies, so electrons from both can be used for bonding and the metal can show several oxidation states. (c) The M2+/M potential depends on sublimation energy, ionisation enthalpy and hydration enthalpy, and these three quantities vary irregularly across the series, so the potential does not follow a smooth curve. Transition metals differ from non-transition elements because their oxidation states commonly differ by one, while non-transition elements show states that differ by two and are fewer in number.

Q4 (CBSE Board 2025, 3 marks)

Give an explanation for each observation. (a) With the same d4 configuration, Mn3+ is an oxidising agent while Cr2+ is a reducing agent. (b) The actinoid contraction is greater than the lanthanoid contraction. (c) Transition metals form a large number of interstitial compounds with H, B, C and N.

Answer: (a) Mn3+ gains an electron to become Mn2+, which has the extra stable d5 configuration, so the reduction is favourable and Mn3+ acts as an oxidiser. Cr2+ loses an electron to become Cr3+, which has the stable d3 (t2g3) configuration, so the oxidation is favourable and Cr2+ acts as a reducer. (b) The 5f electrons shield the nuclear charge less effectively than the 4f electrons, so the effective nuclear charge increases faster across the actinoid series and the contraction is larger. (c) The small atoms of hydrogen, boron, carbon and nitrogen fit into the voids of the metal lattice, so they form interstitial compounds that are non-stoichiometric, hard and high melting.

Q5 (CBSE Board 2026, 5 marks)

(a)(i) Why do transition metals show variable oxidation states? (ii) Out of Mn2+ and Ti2+, which is more paramagnetic and why? (iii) Which of Cr3+, V3+ and Mn3+ is the strongest oxidising agent, and why? (iv) Complete and balance the equations: (I) MnO2 plus KOH plus O2, and (II) C2O4(2-) plus MnO4- plus H+.

Answer: (a)(i) The ns and (n-1)d electrons have comparable energies and both can be lost or shared, so the metal can show several oxidation states. (ii) Mn2+ is d5 with five unpaired electrons and has the larger spin only moment, while Ti2+ is d2 with two unpaired electrons, so Mn2+ is more paramagnetic. (iii) Mn3+ is the strongest oxidiser. It gains an electron to reach the stable d5 configuration of Mn2+, while Cr3+ and V3+ are already in stable d3 and d2 configurations and resist further reduction. (iv)(I) 2MnO2 plus 4KOH plus O2 gives 2K2MnO4 plus 2H2O. (II) 5C2O4(2-) plus 2MnO4- plus 16H+ gives 2Mn2+ plus 10CO2 plus 8H2O.

Q6 (CBSE Board 2026, 1 mark)

Assertion A: Transition metals show their highest oxidation state with oxygen. Reason R: Oxygen has the ability to form multiple bonds with metals. Choose the correct option: (A) both true and R explains A, (B) both true but R does not explain A, (C) A true and R false, (D) A false and R true.

Answer: Option A. The assertion is correct because the highest oxidation states of the transition metals appear in oxides and oxoanions, such as +7 for manganese in permanganate and +6 for chromium in dichromate. The reason is also correct, because oxygen can form strong multiple bonds with the metal, and this extra bonding stabilises the small, highly charged metal centre, making the high oxidation state accessible. The ability to form multiple bonds is therefore the explanation behind the assertion.

Q7 (CBSE SQP 2022-23, 1 mark)

KMnO4 is coloured due to: (a) d-d transitions, (b) charge transfer from ligand to metal, (c) unpaired electrons in the d orbital of manganese, (d) charge transfer from metal to ligand.

Answer: Option (b). Manganese in permanganate is in the +7 state, so the ion has a d0 configuration. With no d electrons there can be no d-d transition, and the unpaired electron explanation fails because there are no unpaired d electrons at all. The deep purple colour comes from charge transfer from the oxygen ligands to the empty d orbitals of manganese, and that absorption lies in the visible region. Dichromate is coloured for the same reason.

Q8 (CBSE SQP 2024-25, 1 mark)

Match the properties in column I with the metals in column II. (i) An actinoid with configuration [Rn] 5f7 6d1 7s2, (ii) a lanthanoid with 4f14 configuration in the +3 state, (iii) a lanthanoid that shows the +4 oxidation state. Metals: (A) Ce, (B) Lu, (C) Cm.

Answer: The correct pairing is (i) with Cm, (ii) with Lu and (iii) with Ce. Curium has seven 5f electrons and is an actinoid, so it matches (i). Lutetium after losing three electrons has a filled 4f14 core, which is the stable arrangement that makes it match (ii). Cerium can lose its single 4f electron and reach the stable empty 4f0 configuration, which is why Ce4+ is important and Ce matches (iii).

Q9 (CBSE SQP 2024-25, 3 marks)

Account for the following. (a) The lowest oxide of a transition metal is basic while the highest oxide is acidic. (b) Chromium is a hard metal while mercury is a liquid. (c) The ionisation energy of the 3d series does not vary much with increasing atomic number.

Answer: (a) In the lowest oxide the metal is in a low oxidation state, so the oxide behaves like a metal oxide and reacts with acids to give salts, which makes it basic. In the highest oxide the metal is in a high oxidation state, so the oxide behaves like an acidic oxide and reacts with bases, as CrO3 and Mn2O7 do. (b) Chromium has many partly filled d electrons that take part in strong metallic bonding, giving a hard solid with a high melting point. Mercury has a filled d subshell, and its electrons are held in a way that leaves very weak metallic bonding, so it is a liquid at room temperature. (c) Across the 3d series the nuclear charge increases, but the extra electrons enter the inner 3d subshell and shield the outer 4s electrons effectively, so the net increase in effective nuclear charge is small and the ionisation energy changes only slowly.

Q10 (CBSE SQP 2025-26, 3 marks)

(a) The enthalpies of atomisation of zinc and copper are listed as 339 and 130 kJ mol^-1. Are they matched correctly? Justify your answer. (b) Out of sulphuric acid and hydrochloric acid, which acid would you prefer for acidifying a solution of potassium permanganate, and why?

Answer: (a) No, the two values are interchanged. The textbook value for zinc, the lowest in the 3d series, is about 126 kJ mol^-1, whereas copper is close to 339 kJ mol^-1. Zinc carries a completely filled 3d10 subshell whose electrons stay out of metallic bonding, so atomising the metal costs less energy. Copper contributes its 4s electron and part of its d band to bonding, which raises the energy needed. The pair in the question has the numbers swapped. (b) Dilute sulphuric acid is preferred. Hydrochloric acid would be oxidised by permanganate to chlorine, which would consume some of the permanganate and change the titration result. Sulphuric acid is not oxidised under these conditions, so it only supplies the hydrogen ions needed for the reduction.