Geometrical Shapes of Nickel II Complexes with Different Ligands

To determine the geometrical shapes of the complexes formed by $\text{Ni}^{2+}$ with $\text{Cl}^-$, $\text{CN}^-$, and $\text{H}_2\text{O}$, we analyze each complex individually based on Valence Bond Theory (VBT) and Crystal Field Theory (CFT):

1. Reaction with $\text{Cl}^-$:

   • The complex formed is tetrachloridonickelate(II), $[\text{NiCl}_4]^{2-}$.
   • Oxidation state of $\text{Ni}$ is $+2$, giving the electronic configuration $[\text{Ar}] 3\text{d}^8 4\text{s}^0$.
   • $\text{Cl}^-$ is a weak field ligand and does not cause pairing of the unpaired $3\text{d}$ electrons.
   • The complex utilizes one $4\text{s}$ and three $4\text{p}$ orbitals, leading to $\text{sp}^3$ hybridization.
   • Geometry: Tetrahedral.

2. Reaction with $\text{CN}^-$:

   • The complex formed is tetracyanidonickelate(II), $[\text{Ni(CN)}_4]^{2-}$.
   • Oxidation state of $\text{Ni}$ is $+2$ ($3\text{d}^8$ system).
   • $\text{CN}^-$ is a strong field ligand and causes forced pairing of the two unpaired $3\text{d}$ electrons.
   • This leaves one inner $3\text{d}$ orbital vacant. The hybridization involved is $\text{dsp}^2$ using one $3\text{d}$, one $4\text{s}$, and two $4\text{p}$ orbitals.
   • Geometry: Square planar.

3. Reaction with $\text{H}_2\text{O}$:

   • In aqueous solution, $\text{Ni}^{2+}$ forms the hexaaquanickel(II) ion, $[\text{Ni(H}_2\text{O)}_6]^{2+}$.
   • The coordination number is $6$.
   • Hybridization involved is $\text{sp}^3\text{d}^2$ using $4\text{s}$, $4\text{p}$, and outer $4\text{d}$ orbitals.
   • Geometry: Octahedral.

Therefore, the geometrical shapes of the complexes are tetrahedral, square planar, and octahedral, respectively.

Correct Answer: (B)