Hybridisation of Nickel in Tetracarbonylnickel and Tetracyanonickelate Complexes

To determine the hybridization of nickel in both complex ions, we analyze their oxidation states, electronic configurations, and the nature of the ligands involved:

1. For $[\text{Ni}(\text{CO})_4]$:

   • Oxidation state of $\text{Ni}$: Carbon monoxide ($\text{CO}$) is a neutral ligand, so the oxidation state of nickel is $0$.
   • Electronic configuration of $\text{Ni}(0)$:
     $\text{Ni} = [\text{Ar}]\, 3\text{d}^8 4\text{s}^2$
   • Effect of ligand: $\text{CO}$ is a strong field ligand. It causes pairing of electrons, pushing the two $4\text{s}$ electrons into the $3\text{d}$ subshell:
     $\text{Ni in } [\text{Ni}(\text{CO})_4] = [\text{Ar}]\, 3\text{d}^{10} 4\text{s}^0 4\text{p}^0$
   • Since all five $3\text{d}$ orbitals are completely filled, inner $3\text{d}$ orbitals cannot participate in hybridization.
   • The complex utilizes one $4\text{s}$ orbital and three $4\text{p}$ orbitals to accept electron pairs from four $\text{CO}$ ligands.
   • Therefore, the hybridization of $\text{Ni}$ in $[\text{Ni}(\text{CO})_4]$ is $\mathbf{sp^3}$ (tetrahedral geometry).

2. For $[\text{Ni}(\text{CN})_4]^{2-}$:

   • Oxidation state of $\text{Ni}$: Let the oxidation state be $x$.
     $x + 4(-1) = -2 \implies x = +2$
   • Electronic configuration of $\text{Ni}^{2+}$:
     $\text{Ni}^{2+} = [\text{Ar}]\, 3\text{d}^8 4\text{s}^0$
   • Effect of ligand: Cyanide ($\text{CN}^-$) is a strong field ligand. It forces the two unpaired electrons in the $3\text{d}$ subshell to pair up:
     $\text{Ni}^{2+} \text{ in } [\text{Ni}(\text{CN})_4]^{2-} = [\text{Ar}]\, 3\text{d}^8 \quad (\text{with 1 empty } 3\text{d} \text{ orbital})$
   • The available empty orbitals for coordinate bonding are one inner $3\text{d}$ orbital, one $4\text{s}$ orbital, and two $4\text{p}$ orbitals.
   • Therefore, the hybridization of $\text{Ni}$ in $[\text{Ni}(\text{CN})_4]^{2-}$ is $\mathbf{dsp^2}$ (square planar geometry).

Thus, the hybridisations of nickel in $[\text{Ni}(\text{CO})_4]$ and $[\text{Ni}(\text{CN})_4]^{2-}$ are $\text{sp}^3$ and $\text{dsp}^2$, respectively.

Correct Option: (B)