Question

The hybridization involved in complex $\left[Ni\right(CN{)}_4{]}^{2-}$ is :
[Atomic number of $Ni$=28]

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

Answer 1

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

$\left[Ni\right(CN{)}_4{]}^{2-}$ is an square planar complex with $ds{p}^2$ hybridization.
The outer electronic configuration of $N{i}^{2+}\text{ion is}3d^{8}4s^0$
One 3d, one 4s and two 4p orbitals undergo $ds{p}^2$ hybridization to give four equivalent $ds{p}^2$ hybrid orbitals which are occupied by 4 electron pairs from 4 cyanide ligands.

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