Match List - I with List - II.

Choose the correct answer from the options given below :

(A) $$[\text{CoF}_6]^{3-}$$

• Oxidation State: $$\text{Co}^{3+} \; (3d^6)$$
• Ligand Nature: $$\text{F}^-$$ is a weak field ligand (WFL), so no electron pairing occurs.
• Hybridization: The six $$3d$$ electrons remain unpaired in their original configuration. To accommodate 6 lone pairs from ligands, the outer orbitals (4s, 4p, and 4d) are used, resulting in an outer orbital complex with $$\text{sp}^3\text{d}^2$$ hybridization.
• Match: (A) $$\rightarrow$$ (III)

(B) $$[\text{NiCl}_4]^{2-}$$

• Oxidation State: $$\text{Ni}^{2+} \; (3d^8)$$
• Ligand Nature: $$\text{Cl}^-$$ is a weak field ligand (WFL), so the two unpaired electrons in the 3d subshell do not pair up.
• Hybridization: The inner 3d orbitals are completely full or occupied. The four ligand lone pairs occupy the 4s and 4p orbitals, leading to a tetrahedral geometry with $$\text{sp}^3$$ hybridization.
• Match: (B) $$\rightarrow$$ (II)

(C) $$[\text{Co(NH}_3)_6]^{3+}$$

• Oxidation State: $$\text{Co}^{3+} \; (3d^6)$$
• Ligand Nature: $$\text{NH}_3$$ behaves as a strong field ligand (SFL) with $$\text{Co}^{3+}$$, forcing the six 3d electrons to pair up completely.
• Hybridization: This pairing clears out two vacant inner 3d orbitals. The six coordination slots are filled using these inner orbitals along with 4s and 4p, forming an inner orbital complex with $$\text{d}^2\text{sp}^3$$ hybridization.
• Match: (C) $$\rightarrow$$ (I)

(D) $$[\text{Ni(CN)}_4]^{2-}$$

• Oxidation State: $$\text{Ni}^{2+} \; (3d^8)$$
• Ligand Nature: $$\text{CN}^-$$ is a very strong field ligand (SFL). It forces the two unpaired electrons in the $3d$ orbital to pair up.
• Hybridization: This pairing vacates a single inner 3d orbital. The four ligand pairs fill this 3d orbital, the 4s orbital, and two 4p orbitals, giving a square planar geometry with $$\text{dsp}^2$$ hybridization.
• Match: (D) $$\rightarrow$$ (IV)