$\mathrm{N}\mathrm{i}\left(28\right):\left[\mathrm{A}\mathrm{r}\right]3{\mathrm{d}}^{8}4{\mathrm{s}}^2$
As CO is a strong field ligand so configuration would be

n=0(no unpaired electron)
For, four ‘CO’-ligands hybridisation would be $\mathrm{s}{\mathrm{p}}^3$ and thus the complex would be diamagnetic and of tetrahedral geometry as follows.