The molecular shape of is tetrahedral while that of is square planar. Justify the shape of the two…

Oxidation state of Ni is zero in $[\mathrm{Ni(CO)_4}]$. CO is a stronger field ligand, and it can pair the unpaired electrons. The 3d orbital is now totally filled, while the 4s and 4p orbitals remain open. Therefore, exhibits $sp^3$ hybridization and tetrahedral geometry. The oxidation state of Ni in $[\mathrm{Ni(CN)_4}]^{2-}$ is +2. While in presence of $\mathrm{CN^-}$ being a stronger field ligand all the electrons are paired up. The empty 3d, 4s and two 4p orbitals undergo $dsp^2$ hybridization to make bonds with $\mathrm{CN^-}$ ligands in square planar geometry.

Ni has 0 oxidation state in complex $[\mathrm{Ni(CO)_4}]$ and +2 oxidation state in $[\mathrm{Ni(CN)_4}]^{2-}$