CBSE Class 12 Chemistry 2026: question 32(a)

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Step-by-step solution

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1. (i)(I) The energies of the $(n-1)d$ and $ns$ orbitals are very close, so electrons from both can take part in bonding. This gives transition metals variable oxidation states.

   1 mark

2. (i)(II) $\mathrm{M}{\mathrm{n}}^{2+}$ is more paramagnetic.

   ½ mark

3. $\mathrm{M}{\mathrm{n}}^{2+}$ is $3d^5$ with 5 unpaired electrons, whereas $\mathrm{T}{\mathrm{i}}^{2+}$ is $3d^2$ with only 2 unpaired electrons.

   ½ mark

4. (i)(III) $\mathrm{M}{\mathrm{n}}^{3+}$ is the strongest oxidising agent.

   ½ mark

5. $\mathrm{M}{\mathrm{n}}^{3+}$ ($3d^4$) readily gains an electron to become $\mathrm{M}{\mathrm{n}}^{2+}$, which has the extra-stable half-filled $3d^5$ configuration.

   ½ mark

6. (ii)(I) $2Mn{\mathrm{O}}_2+4KOH+{\mathrm{O}}_2\to 2{\mathrm{K}}_2\mathrm{Mn}{\mathrm{O}}_4+2{\mathrm{H}}_2\mathrm{O}$

   1 mark

7. (ii)(II) $5{\mathrm{C}}_2{\mathrm{O}}_4^{2-}+2Mn{\mathrm{O}}_4^{-}+16{\mathrm{H}}^{+}\to 2M{\mathrm{n}}^{2+}+10C{\mathrm{O}}_2+8{\mathrm{H}}_2\mathrm{O}$

   1 mark

Answer: (i)(I) similar energies of ns and (n−1)d; (II) Mn²⁺ (5 unpaired e⁻); (III) Mn³⁺ (d⁴ → stable d⁵); (ii)(I) 2K₂MnO₄ + 2H₂O; (II) 2Mn²⁺ + 10CO₂ + 8H₂O

Revise the chapter

• Chapter: The d- and f-Block Elements (notes, formulas, where marks are lost and every past question)