Chemistry - Coordination Chemistry Question with Solution | TestHub

$\mathrm{N}{\mathrm{i}}^{2+}+4\mathrm{C}{\mathrm{N}}^{-}\to {\left[\mathrm{N}\mathrm{i}{\left(\mathrm{C}\mathrm{N}\right)}_4\right]}^{2-}$

Here$\mathrm{N}{\mathrm{i}}^{2+}$has${\text{d}}^{\text{8}}$-configuration with$\mathrm{C}{\mathrm{N}}^{-}$as strong ligand.



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



$d^8$-configuration in weak ligand field gives$s{p}^3$hybridisation, hence tetrahedral geometry.
$\mathrm{N}{\mathrm{i}}^{2+}$with${\mathrm{H}}_2\mathrm{O}$forms${\left[\mathrm{N}\mathrm{i}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{2+}\mathrm{ }$complex and${\mathrm{H}}_2\mathrm{O}$is a weak ligand.



Therefore,${\left[\mathrm{N}\mathrm{i}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{2+}$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.