Match List I with List II: (A) K2[Ni(CN)4] (B) [Ni(CO)4] (C) [Co(NH3)6]Cl3 (D) Na3[CoF6] with (I) sp3 (II) sp3d2 (III)…

🧠 Hybridisation = Ligand Field Strength + d-Count

For each complex, identify the metal's d-count, the ligand-field strength, and apply VBT:

🗺️ Walk Each Complex

(A) $\mathrm{K_2[Ni(CN)_4]}$: Ni(II) $\mathrm{d^8}$ + strong-field CN⁻ → square-planar → $\mathrm{dsp^2}$ → (III).

(B) $[\mathrm{Ni(CO)_4}]$: Ni(0) $\mathrm{d^{10}}$ (CO neutral). All 10 d-electrons paired in inner d → use 4s + 4p → tetrahedral → $\mathrm{sp^3}$ → (I).

(C) $[\mathrm{Co(NH_3)_6}]\mathrm{Cl_3}$: Co(III) $\mathrm{d^6}$ + N-donor → low-spin octahedral → uses inner 3d → $\mathrm{d^2sp^3}$ → (IV).

(D) $\mathrm{Na_3[CoF_6]}$: Co(III) $\mathrm{d^6}$ + weak-field F⁻ → high-spin octahedral → uses outer 4d → $\mathrm{sp^3d^2}$ → (II).

A→III, B→I, C→IV, D→II → option (1).

⚡ The Inner vs Outer d Heuristic

$\mathrm{d^2sp^3}$ (inner-orbital) → strong-field, low-spin, paired-up d-electrons. $\mathrm{sp^3d^2}$ (outer-orbital) → weak-field, high-spin, unpaired d-electrons available.

For Co(III) $\mathrm{d^6}$:

• N-donor (NH₃, en, CN⁻) → $\mathrm{d^2sp^3}$, diamagnetic.
• F⁻, H₂O, Cl⁻ → $\mathrm{sp^3d^2}$, paramagnetic.

⚠️ Ni(0) ≠ Ni(II)

$[\mathrm{Ni(CO)_4}]$: Ni is 0 (CO neutral), $\mathrm{d^{10}}$ — completely paired. So no low-spin/high-spin question; geometry is purely tetrahedral $\mathrm{sp^3}$. Don't apply VBT pairing rules to a $\mathrm{d^{10}}$ ion.

$\boxed{\text{Answer: (1) A-III, B-I, C-IV, D-II}}$