Match List I with List II (CFSE in units of 0): (A) [Cu(NH3)6]2+ (B) [Ti(H2O)6]3+ (C) [Fe(CN)6]3- (D) [NiF6]4- with (I)…

🧠 Compute CFSE for Each

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

| Complex | Metal | d-count | Spin/Field | Config | CFSE / $\Delta_o$ | |---|---|---|---|---|---| | (A) $[\mathrm{Cu(NH_3)_6}]^{2+}$ | Cu(II) | $d^9$ | (no choice) | $t_{2g}^6 e_g^3$ | $6(-0.4) + 3(0.6) = -0.6$ → (I) | | (B) $[\mathrm{Ti(H_2O)_6}]^{3+}$ | Ti(III) | $d^1$ | (no choice) | $t_{2g}^1 e_g^0$ | $1(-0.4) = -0.4$ → (IV) | | (C) $[\mathrm{Fe(CN)_6}]^{3-}$ | Fe(III) | $d^5$ | LS (CN⁻ strong) | $t_{2g}^5 e_g^0$ | $5(-0.4) = -2.0$ → (II) | | (D) $[\mathrm{NiF_6}]^{4-}$ | Ni(II) | $d^8$ | (no choice) | $t_{2g}^6 e_g^2$ | $6(-0.4) + 2(0.6) = -1.2$ → (III) |

So A-I, B-IV, C-II, D-III.

🗺️ The $d^5$ Spin-State Decision

$[\mathrm{Fe(CN)_6}]^{3-}$ is the textbook example of d⁵ low-spin: CN⁻ is so strong-field that all 5 electrons pile into $t_{2g}$ (one pair must form). CFSE = −2.0Δ — the largest possible for any d-count.

By contrast, the corresponding HS d⁵ (e.g. $[\mathrm{FeCl_6}]^{3-}$) has CFSE = 0 (3 in t₂g = −1.2Δ, 2 in eg = +1.2Δ, cancel exactly).

⚡ CFSE Quick-Reference (without P correction)

| d | HS | LS | |---|---|---| | 1 | −0.4 | — | | 2 | −0.8 | — | | 3 | −1.2 | — | | 4 | −0.6 | −1.6 | | 5 | 0 | −2.0 | | 6 | −0.4 | −2.4 | | 7 | −0.8 | −1.8 | | 8 | −1.2 | — | | 9 | −0.6 | — | | 10 | 0 | — |

⚠️ Don't Forget Pairing-Energy Corrections (When Asked)

JEE usually asks for CFSE without the pairing-energy correction, but be prepared to add $nP$ if a question explicitly includes pairing terms.

$\boxed{\text{Answer: (2) A-(I), B-(IV), C-(II), D-(III)}}$