Match List-I with List-II (CFSE values in units of 0): (A) [Ti(H2O)6]2+ (B) [V(H2O)6]2+ (C) [Mn(H2O)6]3+ (D)…

🧠 Compute CFSE for Each (HS, since H₂O Is Weak Field)

\mathrm{CFSE} = (\text{# in } t_{2g})(-0.4\Delta_o) + (\text{# in } e_g)(+0.6\Delta_o).

| Complex | Metal | d-count | HS config | CFSE / $\Delta_o$ | |---|---|---|---|---| | (A) $[\mathrm{Ti(H_2O)_6}]^{2+}$ | Ti(II) | $d^2$ | $t_{2g}^2 e_g^0$ | $2(-0.4) = -0.8$ → (IV) | | (B) $[\mathrm{V(H_2O)_6}]^{2+}$ | V(II) | $d^3$ | $t_{2g}^3 e_g^0$ | $3(-0.4) = -1.2$ → (I) | | (C) $[\mathrm{Mn(H_2O)_6}]^{3+}$ | Mn(III) | $d^4$ HS | $t_{2g}^3 e_g^1$ | $3(-0.4) + 1(0.6) = -0.6$ → (II) | | (D) $[\mathrm{Fe(H_2O)_6}]^{3+}$ | Fe(III) | $d^5$ HS | $t_{2g}^3 e_g^2$ | $3(-0.4) + 2(0.6) = 0$ → (III) |

🗺️ The HS d⁵ Special Case

HS d⁵ in oct gives CFSE = 0 because the stabilisation from 3 electrons in $t_{2g}$ (−1.2Δ) is exactly cancelled by destabilisation from 2 electrons in $e_g$ (+1.2Δ). This is why HS d⁵ ions (Fe³⁺, Mn²⁺) are colourless or very pale — minimal crystal-field stabilisation, weak d–d transitions.

⚡ CFSE Maxima for HS

Among HS oct, max stabilisation is at $d^3$ and $d^8$ (CFSE = $-1.2\Delta_o$). $d^5$ has zero CFSE. $d^{10}$ also zero.

⚠️ Always Identify Spin State First

CFSE depends on filling. With weak-field $\mathrm{H_2O}$, all four complexes here are HS — but if any had been with CN⁻ or CO, you'd recompute as LS.

$\boxed{\text{Answer: (3) A-IV, B-I, C-II, D-III}}$