$\left[\mathrm{C}\mathrm{o}\right(\mathrm{O}\mathrm{x}{)}_3{]}^{3-}$
$\mathrm{C}{\mathrm{o}}^{3+}=\left[\mathrm{A}\mathrm{r}\right]4{\mathrm{s}}^{0}3{\mathrm{d}}^6$

$\left[\mathrm{N}\mathrm{i}\right(\mathrm{C}\mathrm{N}{)}_4{]}^{2-}$ and $\left[\mathrm{N}\mathrm{i}\right(\mathrm{C}\mathrm{O}{)}_4]$ have zero magnetic moment as have no unpaired electron.
$[\mathrm{N}\mathrm{i}\mathrm{C}{\mathrm{l}}_4{]}^{2-}$ and $[\mathrm{P}\mathrm{t}\mathrm{C}{\mathrm{l}}_4{]}^{2-}$ have different shapes that is tetrahedral and square planar respectively.
In brown-ring complex $\left[\mathrm{F}\mathrm{e}\right({\mathrm{H}}_2\mathrm{O}{)}_5\mathrm{N}\mathrm{O}]\mathrm{S}{\mathrm{O}}_4$ oxidation state of Fe is +1 as NO is taken as +1 here.