JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Coordination Compounds

$[\mathrm{PtC}{\mathrm{l}}_2(\mathrm{N}{\mathrm{H}}_3{)}_2]$ is square planar, and the two chlorides can be placed next to each other or opposite each other, giving cis and trans forms: $2$ isomers.
$[\mathrm{Ni}(\mathrm{CO}{)}_4]$ is tetrahedral, where all four positions are mutually adjacent and no two can be distinguished as cis or trans; with four identical ligands there is in any case only one arrangement: $0$ isomers.
$[\mathrm{Ru}({\mathrm{H}}_2\mathrm{O}{)}_3\mathrm{C}{\mathrm{l}}_3]$ is octahedral of the $\mathrm{M}{\mathrm{a}}_3{\mathrm{b}}_3$ type, so the three like ligands either occupy one triangular face or lie on a meridian, giving the facial and meridional forms: $2$ isomers.
$[\mathrm{CoC}{\mathrm{l}}_2(\mathrm{N}{\mathrm{H}}_3{)}_4{]}^{+}$ is octahedral of the $\mathrm{M}{\mathrm{a}}_4{\mathrm{b}}_2$ type, with the two chlorides adjacent or opposite: $2$ isomers.
The counts are $2,0,2,2$.