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

The correct option is C $ds{p}^2$
In $\left[Ni\right(CN{)}_4{]}^{2-}$, the oxidation state of $Ni$ is $+2$ .
$x-4=2$
$x=+2$
Now,
$N{i}^{2+}\to \left[Ar\right]3d^{8}4s^0$

Due to the presence of a strong field ligand ($C{N}^{-})$ all unpaired electrons are paired up.

Hybridisation of $\left[Ni\right(CN{)}_4{]}^{2-}$ is $ds{p}^2$