Match List-I with List-II: (A) Ni(CO)4 (B) [Ni(CN)4]2- (C) [Co(CN)6]3- (D) [CoF6]3- with (I) sp3 (II) sp3d2 (III) d2sp3…

🧠 Hybridisation From d-Count + Field Strength

🗺️ Walk Each

(A) $\mathrm{Ni(CO)_4}$: Ni(0), $\mathrm{d^{10}}$. CO is neutral; tetrahedral 4-coordinate. Use 4s + 4p → $\mathrm{sp^3}$ → (I).

(B) $[\mathrm{Ni(CN)_4}]^{2-}$: Ni(II), $\mathrm{d^8}$ + strong CN⁻ → square-planar → $\mathrm{dsp^2}$ → (IV).

(C) $[\mathrm{Co(CN)_6}]^{3-}$: Co(III), $\mathrm{d^6}$ + strong CN⁻ → low-spin octahedral → $\mathrm{d^2sp^3}$ → (III).

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

A→I, B→IV, C→III, D→II → option (2).

⚡ The Four Common Hybridisations

| Geometry | Hybridisation | Use case | |---|---|---| | Tetrahedral | $\mathrm{sp^3}$ | CN = 4, e.g. Ni(CO)₄, ZnCl₄²⁻ | | Square-planar | $\mathrm{dsp^2}$ | d⁸ + strong field | | Octahedral, low-spin | $\mathrm{d^2sp^3}$ | Co(III) with N-donor, CN⁻ | | Octahedral, high-spin | $\mathrm{sp^3d^2}$ | weak-field ligands |

⚠️ Co(III) Switches Hybridisation Based on Ligand

Same Co(III), different ligands → different hybridisation:

• $[\mathrm{Co(CN)_6}]^{3-}$: $\mathrm{d^2sp^3}$ (inner-orbital).
• $[\mathrm{CoF_6}]^{3-}$: $\mathrm{sp^3d^2}$ (outer-orbital).

Same metal, same OS, opposite hybridisation — driven entirely by ligand strength.

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