CBSE Class 12 Chemistry 2025: question 31(a)

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Step-by-step solution

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1. (i)(I) Mn²⁺ has a stable half-filled $3d^5$ configuration. Mn²⁺ is therefore hard to reduce to Mn, so $E_{M{n}^{2+}/Mn}^{\circ }$ is highly negative. Mn³⁺ ($3d^4$) readily gains an electron to become stable $d^5$ Mn²⁺, so $E_{M{n}^{3+}/M{n}^{2+}}^{\circ }$ is highly positive.

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2. (i)(II) In actinoids, the 5f, 6d and 7s orbitals have comparable energies. Electrons from all of them can take part in bonding, so actinoids show a wide range of oxidation states.

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3. (i)(III) Transition metals have many unpaired electrons. Electrons from the (n−1)d orbitals, along with the ns electrons, take part in metallic bonding. This gives strong interatomic bonding and hence high melting points.

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4. (ii)(I) $5S{\mathrm{O}}_3^{2-}+2Mn{\mathrm{O}}_4^{-}+6{\mathrm{H}}^{+}\to 2M{\mathrm{n}}^{2+}+5S{\mathrm{O}}_4^{2-}+3{\mathrm{H}}_2\mathrm{O}$

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5. (ii)(II) $2Mn{\mathrm{O}}_4^{-}+{\mathrm{H}}_2\mathrm{O}+{\mathrm{I}}^{-}\to 2Mn{\mathrm{O}}_2+2O{\mathrm{H}}^{-}+\mathrm{I}{\mathrm{O}}_3^{-}$

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Answer: (i) (I) Stable d⁵ Mn²⁺; (II) comparable energies of 5f, 6d, 7s; (III) strong metallic bonding involving (n−1)d electrons. (ii) (I) 2Mn²⁺ + 5SO₄²⁻ + 3H₂O; (II) 2MnO₂ + 2OH⁻ + IO₃⁻

Revise the chapter

• Chapter: The d- and f-Block Elements (notes, formulas, where marks are lost and every past question)
• Topic: Oxidation States and Electrode Potentials of Transition Metals