Question

Hybridisation and geometry of $[Ni(CN{)}_4{]}^{2-}$ are

• A. $s{p}^{2}d$ and tetrahedral
• B. $s{d}^3$ and square planar
• C. $s{p}^3$ and tetrahedral
• D. $ds{p}^2$ and square planar

Answer 1

Answer: (D)

The oxidation state of $Ni$ in $[Ni(CN{)}_4{]}^{2-}$ is $+2.$ The electronic configuration of $N{i}^{2+}is[Ar]3d^{8}4s^0.$
$\stackrel{N{i}^{2+}}{\text{Ground state}}$
$AsC{N}^{-}$ is a strong field ligand, so it will pair up electrons
$[Ni(CN{)}_4{]}^{2-}$
The hybridisation of Ni in $[Ni(CN{)}_4{]}^{2-}$ is $ds{p}^2$ and thus, it will a square planar geometry