Question

State a reason for each of the following situations :
The molecular shape of $Ni(CO{)}_4$ is not the same as that of $\left[Ni\right(CN{)}_4{]}^{2-}$.

Answer 1

${\left[Ni{\left(CN\right)}_4\right]}^{2-}$ is square planar because CN is a strong field ligand. In the presence of this ligand, electrons pair up in 3d orbitals and it is undergoing $ds{p}^2$ hybridization. So it is in square planar geometry while $\left[Ni{\left(CO\right)}_4\right]$ undergoes $s{p}^3$ hybridization and so it has a tetrahedral geometry.