The correct statement with respect to these complexes

$\mathrm{Ni}(\mathrm{CO}{)}_4$ and ${\left[\mathrm{Ni}(\mathrm{CN}{)}_4\right]}^{2-}$ is

Hybridization of nickel in $\mathrm{Ni}(\mathrm{CO}{)}_4$ is ${\mathrm{sp}}^3$ therefore geometry of  $\mathrm{Ni}(\mathrm{CO}{)}_4$ is tetrahedral whereas hybridization of nickel in

${\left[\mathrm{Ni}(\mathrm{CN}{)}_4\right]}^{2-}$ is ${\mathrm{dsp}}^2$ therefore geometry of ${\left[\mathrm{Ni}(\mathrm{CN}{)}_4\right]}^{2-}$ is square planar.