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

${\left[\mathrm{N}\mathrm{i}{\left(\mathrm{C}\mathrm{N}\right)}_4\right]}^{2-}$

Oxidation state of Ni is +2

$\mathrm{x}-4=2$

$\mathrm{x}=+2$

Now, ${\mathrm{Ni}}^{2+}=\left[\mathrm{Ar}\right] 3{\mathrm{d}}^8 4\mathrm{s}°$

therefore, ${\mathrm{CN}}^{-}$ is a strong field ligand. Hence, all unpaired electrons are paired up.

therefore, hybridisation of ${\left[\mathrm{Ni}(\mathrm{CN}{)}_4\right]}^{2-}$ is ${\mathrm{dsp}}^2$