CBSE Class 12 Chemistry: Marking Scheme Solution

(a) With the same $d^4$ configuration, $\text{Mn}^{3+}$ is an oxidising agent whereas $\text{Cr}^{2+}$ is a reducing agent. [1 Mark]

• Step 1 (Governing Principle): [0.5 Mark] The stability of transition metal ions depends on the electronic configuration, specifically the stability of the $t_{2g}$ and $e_g$ orbitals in an octahedral field.
• Step 2 (Step-by-Step Reasoning): [0.25 Mark]
  • $\text{Mn}^{3+}$ ($d^4$) reduces to $\text{Mn}^{2+}$ ($d^5$), which has a stable half-filled $t_{2g}^3 e_g^2$ configuration. Thus, it acts as an oxidising agent.
  • $\text{Cr}^{2+}$ ($d^4$) oxidises to $\text{Cr}^{3+}$ ($d^3$), which has a stable $t_{2g}^3$ configuration (half-filled $t_{2g}$ level). Thus, it acts as a reducing agent.
• Step 3 (Conclusion): [0.25 Mark] $\text{Mn}^{3+}$ gains an electron to achieve stability, while $\text{Cr}^{2+}$ loses an electron to achieve stability.

(b) Actinoid contraction is greater from element to element than that among lanthanoids. [1 Mark]

• Step 1 (Governing Principle): [0.5 Mark] The extent of contraction depends on the shielding effect provided by the inner orbitals ($f$-orbitals).
• Step 2 (Step-by-Step Reasoning): [0.25 Mark] The $5f$ orbitals in actinoids have poorer shielding effect compared to the $4f$ orbitals in lanthanoids.
• Step 3 (Conclusion): [0.25 Mark] Due to the very poor shielding of the $5f$ electrons, the effective nuclear charge increases more significantly in actinoids, leading to a greater contraction of the atomic/ionic radii.

(c) Transition metals form a large number of interstitial compounds with H, B, C, and N. [1 Mark]

• Step 1 (Governing Principle): [0.5 Mark] Interstitial compounds are formed when small atoms (H, B, C, N) are trapped inside the crystal lattice of transition metals.
• Step 2 (Step-by-Step Reasoning): [0.25 Mark] Transition metals have a specific crystal structure (hcp, ccp, etc.) with vacant spaces known as interstitial sites.
• Step 3 (Conclusion): [0.25 Mark] Because the atomic radii of H, B, C, and N are small, they easily fit into these voids without distorting the metallic lattice, resulting in the formation of non-stoichiometric interstitial compounds.

💡 EL10 Concept Insight:

Think of atoms like a crowded room: (a) $\text{Mn}^{3+}$ and $\text{Cr}^{2+}$ are just trying to reach a "comfortable" half-full state to be stable. (b) Actinoids have "lazy" $5f$ electrons that don't block the nucleus well, so the nucleus pulls the outer shell in much tighter than lanthanoids do. (c) Transition metals are like a sponge with tiny holes; small atoms like Hydrogen or Carbon are just the right size to hide in