Match List I with List II: (A) [Cr(H2O)6]3+ (B) [Fe(H2O)6]3+ (C) [Ni(H2O)6]2+ (D) [V(H2O)6]3+ with (I) t2g2 eg0 (II)…

🧠 Each Complex Is $[\mathrm{M(H_2O)_6}]$ → Weak Field → HS

| Complex | Metal | d-count | HS oct config | |---|---|---|---| | (A) $[\mathrm{Cr(H_2O)_6}]^{3+}$ | Cr(III) | $d^3$ | $t_{2g}^3 e_g^0$ → (II) | | (B) $[\mathrm{Fe(H_2O)_6}]^{3+}$ | Fe(III) | $d^5$ | $t_{2g}^3 e_g^2$ → (III) | | (C) $[\mathrm{Ni(H_2O)_6}]^{2+}$ | Ni(II) | $d^8$ | $t_{2g}^6 e_g^2$ → (IV) | | (D) $[\mathrm{V(H_2O)_6}]^{3+}$ | V(III) | $d^2$ | $t_{2g}^2 e_g^0$ → (I) |

🗺️ Why d³ and d⁸ Are Spin-Independent

For $d^3$: all three electrons enter separate $t_{2g}$ orbitals → 3 unpaired, no choice between HS/LS.

For $d^8$: $t_{2g}$ is fully paired (6 electrons), $e_g$ has 2 unpaired — same in any field strength.

So Cr³⁺ ($d^3$) and Ni²⁺ ($d^8$) octahedral always have the same configuration regardless of ligand.

⚡ The HS d⁵ Spread

$d^5$ HS = $t_{2g}^3 e_g^2$ — all five electrons unpaired across both sets. This is the most paramagnetic 3d configuration (μ ≈ 5.9 BM).

⚠️ V³⁺ vs V²⁺

V atomic config: $[\mathrm{Ar}]\,3d^3 4s^2$. V³⁺ removes both 4s + one 3d → $d^2$. V²⁺ removes only 4s → $d^3$. Always check OS carefully when working out d-count for V.

$\boxed{\text{Answer: (4) A-II, B-III, C-IV, D-I}}$