Question

Hybridization and geometry of $\left[Ni\right(CN{)}_4{]}^{2-}$ 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

The correct option is D $ds{p}^2$ and square planar

$\left[Ni\right(CN{)}_4{]}^{2-}$ is a square planar geometry formed by $ds{p}^2$ hybridisation. $\left[Ni\right(CN{)}_4{]}^{2-}$ is diamagnetic, so $N{i}^{2+}ionhas$3d^8$ outer configuration with two unpaired electrons.

For the formation of the square planar structure by $ds{p}^2$ hybridisation, two unpaired d-electrons are paired up due to energy made available by the approach of ligands, making one of the 3d orbitals empty.

Hence, option D is correct.