JEE Main · 2022 · 28 Jun 2022 · Shift 2 · Coordination Compounds

Writing the coordination formulas:

(a) $\mathrm{CoCl}{}_3\cdot 4\mathrm{NH}{}_3=[\mathrm{Co}(\mathrm{NH}{}_3){}_4\mathrm{Cl}{}_2]Cl$ — This is an octahedral complex with 4 NH₃ and 2 Cl⁻ ligands. It can exhibit cis-trans (geometrical) isomerism.

(b) $\mathrm{CoCl}{}_3\cdot 5\mathrm{NH}{}_3=[\mathrm{Co}(\mathrm{NH}{}_3){}_5\mathrm{Cl}]Cl2$ — Only one Cl⁻ in coordination sphere; no G.I.

(c) $\mathrm{CoCl}{}_3\cdot 6\mathrm{NH}{}_3=[\mathrm{Co}(\mathrm{NH}{}_3){}_6]Cl3$ — All identical ligands; no G.I.

(d) $\mathrm{CoCl}(\mathrm{NO}{}_3){}_2\cdot 5\mathrm{NH}{}_3$ can be $[\mathrm{Co}(\mathrm{NH}{}_3){}_5\mathrm{Cl}](\mathrm{NO}{}_3{)}_2$ or $[\mathrm{Co}(\mathrm{NH}{}_3){}_5(\mathrm{NO}{}_3)]\mathrm{Cl}(\mathrm{NO}{}_3)$ — both have only one different ligand in coordination sphere; no G.I.

Only complex (a) exhibits geometrical isomerism. Answer = 1