$[NiCl_{4}]^{2-}$ is paramagnetic while $[Ni(CO)_{4}]$ is diamagnetic though both are tetrahedral. Why?

(N/A) Though both $[NiCl_{4}]^{2-}$ and $[Ni(CO)_{4}]$ are tetrahedral,their magnetic characters are different. This is due to a difference in the nature of ligands.
In $[NiCl_{4}]^{2-}$,$Ni$ is in the $+2$ oxidation state with a $3d^{8}$ configuration. $Cl^{-}$ is a weak field ligand and it does not cause the pairing of unpaired $3d$ electrons. Hence,$[NiCl_{4}]^{2-}$ is paramagnetic.
In $[Ni(CO)_{4}]$,$Ni$ is in the zero oxidation state i.e.,it has a configuration of $3d^{8}4s^{2}$.
$CO$ is a strong field ligand. Therefore,it causes the pairing of unpaired $3d$ electrons. Also,it causes the $4s$ electrons to shift to the $3d$ orbital,thereby giving rise to $sp^{3}$ hybridization. Since no unpaired electrons are present in this case,$[Ni(CO)_{4}]$ is diamagnetic.