Match the complex given in List-I with type of hybridisation given in…

The correct answer is Option (1) → (A)-(IV), (B)-(II), (C)-(III), (D)-(I)

List-I (Complex)	List-II (Hybridisation)
(A) $[Ni(CO)_4]$	(IV) $sp^3$
(B) $[Ni(CN)_4]^{2-}$	(II) $dsp^2$
(C) $[Co(NH_3)_6]^{3+}$	(III) $d^2sp^3$
(D) $[CoF_6]^{3-}$	(I) $sp^3d^2$

(A) $[\text{Ni}(\text{CO})_4] \rightarrow$ (IV) $sp^3$

Nickel is in the zero oxidation state with an electronic configuration of $[\text{Ar}]3d^84s^2$. Carbon monoxide $(\text{CO})$ is a very strong field ligand. Under its influence, the two electrons from the $4s$ orbital are forced into the $3d$ orbitals, fully pairing up the configuration to $3d^{10}$. This leaves the $4s$ and three $4p$ orbitals completely empty, which hybridize to form $sp^3$ hybridization (Tetrahedral geometry).

(B) $[\text{Ni}(\text{CN})_4]^{2-} \rightarrow$ (II) $dsp^2$

Nickel is in the +2 oxidation state with an electronic configuration of $[\text{Ar}]3d^84s^0$. Cyanide $(\text{CN}^-)$ is a strong field ligand that forces the two unpaired electrons in the $3d$ orbitals to pair up against Hund's Rule. This vacates one inner $3d$ orbital, leading to $dsp^2$ hybridization (Square planar geometry).

(C) $[Co(NH_3)_6]^{3+}$ (III): Cobalt is in the +3 oxidation state ($3d^6$). Ammonia ($NH_3$) acts as a strong field ligand here, pairing all six $d$-electrons. This leaves two inner $3d$ orbitals vacant, resulting in $d^2sp^3$ hybridization (Inner orbital octahedral geometry).

(D) $[CoF_6]^{3-}$ (I): Cobalt is in the +3 oxidation state ($3d^6$). Fluoride ($F^-$) is a weak field ligand and cannot cause electron pairing. Consequently, the metal uses outer $4d$ orbitals, leading to $sp^3d^2$ hybridization (Outer orbital octahedral geometry).