JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Coordination Compounds

Geometrical isomerism needs at least two distinguishable arrangements of the ligands about the metal. In $[\mathrm{Co}(\mathrm{N}{\mathrm{H}}_3{)}_3(\mathrm{N}{\mathrm{O}}_2{)}_3]$ the three like ligands can occupy one triangular face or lie along a meridian, giving facial and meridional forms; $[\mathrm{CoC}{\mathrm{l}}_2(\mathrm{en}{)}_2]$ and $[\mathrm{Co}(\mathrm{N}{\mathrm{H}}_3{)}_4\mathrm{C}{\mathrm{l}}_2{]}^{+}$ are both of the $\mathrm{M}{\mathrm{A}}_4{\mathrm{B}}_2$ type and give cis and trans forms. $[\mathrm{Co}(\mathrm{CN}{)}_5(\mathrm{NC}){]}^{3-}$ is of the type $\mathrm{M}{\mathrm{A}}_5\mathrm{B}$, and in an octahedron all six sites are equivalent, so wherever the single isocyanide sits the same structure results. The complex ion showing no geometrical isomerism is $[\mathrm{Co}(\mathrm{CN}{)}_5(\mathrm{NC}){]}^{3-}$.