JEE Main · 2021 · 18 Mar 2021 · Shift 1 · Coordination Compounds

$[\mathrm{NiB}{\mathrm{r}}_2(\mathrm{PP}{\mathrm{h}}_3{)}_2]$ is square planar, and in its trans isomer the two bromides lie at $180^{\circ }$ to each other with the two $\mathrm{PP}{\mathrm{h}}_3$ groups likewise opposite. $[\mathrm{Co}(\mathrm{N}{\mathrm{H}}_3{)}_3(\mathrm{N}{\mathrm{O}}_2{)}_3]$ is octahedral, and the meridional isomer places each set of three like ligands along a meridian of the octahedron, so within a set two ligands are mutually trans and the third is cis to both; the facial isomer, by contrast, has all three mutually cis on one triangular face. The correct pair is the square planar nickel complex with bromide opposite bromide and the octahedral cobalt complex with its three nitro groups spread along a meridian.