The geometry of $\mathrm{Ni}(\mathrm{CO}{)}_4$  and ${\left[{\mathrm{NiCl}}_4\right]}^{-2}$ are

Atomic number of Ni is 28 and its electronic configuration is:
$\left[\mathrm{Ar}\right]3{\mathrm{d}}^{8}4{\mathrm{s}}^2$
Oxidation number of Ni in $\mathrm{Ni}{\left(\mathrm{CO}\right)}_4$ is zero. So, sp³ orbitals of Ni is utilized in bonding with CO ligands in $\mathrm{Ni}{\left(\mathrm{CO}\right)}_4$. Therefore, geometry of $\mathrm{Ni}{\left(\mathrm{CO}\right)}_4$is tetrahedral.

Oxidation state of Ni is ${\left[{\mathrm{NiCl}}_4\right]}^{-2}$ is +2 and its electronic configuration becomes
$\left[\mathrm{Ar}\right]3{\mathrm{d}}^8$

As Cl is a weak ligand, therefore pairing of 3d electrons won't take place. So, sp³ orbitals of Ni is utilized in bonding with Cl ligands in ${\left[{\mathrm{NiCl}}_4\right]}^{-2}$. Therefore, geometry of ${\left[{\mathrm{NiCl}}_4\right]}^{-2}$ is tetrahedral.