Question

$N{i}^{2+}\stackrel{KCN}{\to }\text{complex 1}$
$N{i}^{2+}\underset{KCl}{\overset{\text{Excess}}{\to }}\text{complex 2}$

Choose the correct option considering the complexes formed.

• A. $ds{p}^2$ in both
• B. $s{p}^2$ in both
• C. $ds{p}^2$ in cyano and $s{p}^3$ in chlorine complex
• D. $s{p}^3$ in cyano complex and $ds{p}^2$ in chlorine complex

Answer 1

The correct option is C $ds{p}^2$ in cyano and $s{p}^3$ in chlorine complex

$N{i}^{2+}\stackrel{KCN}{\to }\left[Ni\right(CN{)}_4{]}^{2-}$
$N{i}^{2+}\underset{KCl}{\overset{\text{Excess}}{\to }}[NiC{l}_4{]}^{2-}$

Strong field ligand causes pairing in $3d$ orbital. As $C{N}^{-}$ is strong filled ligand, so $\left[Ni\right(CN{)}_4{]}^{2-}$ shows $ds{p}^2$ hybridisation while in $[NiC{l}_4{]}^{2-}$, $C{l}^{-}$ being weak field ligand shows $s{p}^3$ hybridisation.