Match List-I with List-II (tetrahedral crystal field d-electron configurations): (A) TiCl4 (B) [FeO4]2- (C) [FeCl4]-…

🧠 Use Td Splitting (e Lower, t₂ Upper)

In tetrahedral geometry, fill $e$ first (lower), then $t_2$ (upper).

| Complex | Metal | d-count | Td filling | |---|---|---|---| | (A) $\mathrm{TiCl_4}$ | Ti(IV) | $d^0$ | $e^0 t_2^0$ → (III) | | (B) $[\mathrm{FeO_4}]^{2-}$ | Fe(VI) | $d^2$ | $e^2 t_2^0$ → (I) | | (C) $[\mathrm{FeCl_4}]^-$ | Fe(III) | $d^5$ | $e^2 t_2^3$ → (IV) | | (D) $[\mathrm{CoCl_4}]^{2-}$ | Co(II) | $d^7$ | $e^4 t_2^3$ → (II) |

🗺️ Why Each Configuration

• d⁰: nothing to place; both sets empty.
• d²: both go in lower $e$ set (Hund's rule, both unpaired across two $e$ orbitals).
• d⁵ Td: 2 in $e$ (unpaired), 3 in $t_2$ (unpaired) → 5 unpaired (HS, since Δ_t is small).
• d⁷ Td: $e$ filled (4 electrons, 2 pairs), $t_2$ has 3 unpaired → 3 unpaired total.

⚡ Td Is Always HS for First-Row Metals

$\Delta_t \approx \frac{4}{9}\Delta_o$ — too small to overcome pairing energy in 3d metals. So tetrahedral d⁵ stays as $e^2 t_2^3$ (5 unpaired) and d⁷ stays as $e^4 t_2^3$ (3 unpaired). Don't even bother checking ligand strength for Td.

⚠️ Don't Mix Up Td Order with Oct

In octahedral, $t_{2g}$ is lower and $e_g$ is upper — opposite of tetrahedral. Any time you see "$t_2$" without the $g$ subscript, it's tetrahedral, and the order is reversed.

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