Question

Geometrical shapes of the complexes formed by the reaction of $N{i}^{2+}withC{l}^{-},C{N}^{-}and{H}_{2}O$ , respectively, are

• A. Octahedral, tetrahedral and square planar
• B. Tetrahedral, square planar and octahedral
• C. Square planar, tetrahedral and octahedral
• D. Octahedral, square planar and octahedral

Answer 1

Answer: (B)

$N{i}^{2+}+4C{N}^{-}\to {\left(\left[Ni{\left(CN\right)}_4\right]\right)}^{2-}$
Here $N{i}^{2+}$ has ${\text{d}}^{\text{8}}$ -configuration with $C{N}^{-}$ as strong ligand.

${\text{d}}^{\text{8}}$ -configuration in strong ligand field gives ${\text{dsp}}^{\text{2}}$ hybridisation, hence square planar geometry.
$N{i}^{2+}+4C{l}^{-}\to {\left[NiC{l}_4\right]}^{2-}$
Here $N{i}^{2+}$ has ${\text{d}}^{\text{8}}$ -configuration with $C{l}^{-}$ as weak ligand.

$d^8$ -configuration in weak ligand field gives $s{p}^3$ hybridisation, hence tetrahedral geometry.
$N{i}^{2+}$ with $H_{2}O$ forms ${\left(\left[Ni{\left(H_{2}O\right)}_6\right]\right)}^{2+}$ complex and $H_{2}O$ is a weak ligand.

Therefore, ${\left(\left[Ni{\left(H_{2}O\right)}_6\right]\right)}^{2+}$ has octahedral geometry.
In octahedral complexes with ${\text{d}}^{\text{8}}$ configuration, there is always ${\text{sp}}^{\text{3}}{\text{d}}^{\text{2}}$ hybridisation (outer orbital complex) irrespective of strong ligand field or weak ligand field.