$Ni \,in\, [Ni(Cl) ]^{2-}$ exist as $Ni ^{2+}$ ion.
$\because Cl$ is a weak field ligand (high spin). It will not go for pairing.
Hence, configuration of $\left[ Ni ( Cl )_{4}\right]^{2-}$ is
Tetrahedral with two unpaired electrons (i.e., paramagnetic) with $s p^{3}$ hybridisation.
(2) In $\left[ CO \left( C _{2} O _{4}\right)_{3}\right]^{3-},\left( C _{2} O _{4}\right)_{3}^{2-}$ is a bidentate ligand thus, give octahedral structure.
(3) In $\left[ Ni ( CN )_{4}\right]^{2-}$, $Ni$ exist as $Ni ^{2+}$ ion.
$\because CN ^{-}$ is a strong field ligand (low spin), the unpaired electrons of 3rd orbital go for pairing to give $\left[ Ni ( CN )_{4}\right]^{2-}$ as follows:
Hence, the structure of $\left[ Ni ( CN )_{4}\right]^{2-}$ is square planar with $d s p^{2}$ hybridisation having diamagnetic nature.
In $\left[ Ni ( CO )_{4}\right]$, Ni has zero oxidation state.
i.e.
When CO approaches to Ni-metal, electron of Ni, go for rearrangement to give tetrahedral structure with no unpaired electrons (diamagnetic nature) and $s p^{3}$ hybridisation.
