COMEDK 2025 Morning Solved Paper

Concept:A compound must have different ligands arranged so that both non‑superimposable mirror images (optical) and distinct spatial arrangements (geometrical) are possible.Explanation:Option A: $\left[\mathrm{Cr}(\mathrm{ox})_3\right]^{3-}$ has three identical bidentate oxalate ligands.It shows optical isomerism ($\Delta$/$\Lambda$) but no geometrical isomerism because all three ligands are equivalent.Option B: $\left[\mathrm{Co}(\mathrm{en})_2 \mathrm{Cl}_2\right]^{+}$ has two bidentate ethylenediamine (en) and two monodentate Cl$^{-}$ ions.Geometrical isomers: cis (Cl adjacent) and trans (Cl opposite).The cis form is chiral and shows optical isomerism ($\Delta$/$\Lambda$ enantiomers).Thus both geometrical and optical isomerism are present.Option C: $[\mathrm{Co}(\mathrm{CN})_6]^{3-}$ has six identical monodentate CN$^{-}$ ligands.Only one arrangement exists; no geometrical or optical isomerism.Option D: $\left[\mathrm{Co}\left(\mathrm{NO}_3\right)_3\left(\mathrm{NH}_3\right)_3\right]$ has three monodentate NO$_3^{-}$ and three monodentate NH$_3$ ligands.It shows geometrical isomers (fac and mer), but neither isomer is chiral (both are superimposable on their mirror images).So no optical isomerism.Only option B fulfills both conditions.Answer: Option B: $\left[\mathrm{Co}(\mathrm{en})_2 \mathrm{Cl}_2\right]^{+}$