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

Choose the correct answer from the options given below:

(A)-(IV), (B)-(II), (C)-(III), (D)-(I)

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

(A) $[Ni(CO)_4]$ (IV): Nickel is in the zero oxidation state ($3d^8 4s^2$). Carbon monoxide ($CO$) is a very strong field ligand, causing the $4s$ electrons to pair into the $3d$ orbitals. This results in a completely filled $3d^{10}$ configuration, leaving the $4s$ and three $4p$ orbitals available for $sp^3$ hybridization (Tetrahedral geometry).

(B) $[Ni(CN)_4]^{2-}$ (II): Nickel is in the +2 oxidation state ($3d^8$). Cyanide ($CN^-$) is a strong field ligand that causes the two unpaired electrons in the $3d$ subshell to pair up. This vacates one $3d$ 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).