Question

The hybridization involved in complex $\left[Ni\right(CN{)}_4{]}^{2-}$ is:

[Atomic number of Ni $=28]$

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

Answer 1

The correct option is C $ds{p}^2$

$\left[Ni\right(CN{)}_4{]}^{2-}$ is an square planar complex with $ds{p}^2$ hybridization.

The outer electronic configuration of $N{i}^{2+}$ ion is $3d^{8}4s^0$.

One 3d, one 4s and two 4p orbitals undergo $ds{p}^2$ hybridization four equivalent $ds{p}^2$ hybrid orbitals which are occupied by 4 electron pairs of 4 cyanide ligands.

This forms $\left[Ni\right(CN{)}_4{]}^{2-}$.

Hence, option C is correct.