Which of the following compounds shows optical isomerism?

For a coordination compound to exhibit optical isomerism it must be chiral, i.e. the spatial arrangement of the ligands should lack a plane of symmetry, a centre of symmetry and an improper rotation axis. In practice this happens most often in:

• Octahedral complexes containing three bidentate ligands such as $$en$$ or $$C_2O_4^{2-}$$ (Δ- and Λ-forms).
• Some octahedral complexes of the type $$[MABCD_2]$$ or $$[MA_2BCDE]$$ when the arrangement of monodentate ligands removes all symmetry elements.
• Rarely in tetrahedral complexes when the four groups are all different.

Now examine each option:

Option A: $$[\text{Cu}(\text{NH}_3)_4]^{2+}$$ is square-planar (Cu(II), $$d^9$$). The square plane contains a mirror plane passing through the metal and two opposite ligands, so the complex is achiral. No optical isomerism.

Option B: $$[\text{ZnCl}_4]^{2-}$$ is tetrahedral with four identical $$Cl^-$$ ligands. All four vertices are the same, giving several symmetry elements (mirror planes, an inversion centre), hence it is achiral. No optical isomerism.

Option C: $$[\text{Cr}(\text{C}_2\text{O}_4)_3]^{3-}$$ is octahedral with three bidentate oxalate ($$C_2O_4^{2-}$$) ligands. The chelating nature of oxalate produces two non-superimposable mirror images designated as the Δ (right-handed) and Λ (left-handed) forms. There is no plane or centre of symmetry, so the complex is chiral and shows optical isomerism.

Option D: $$[\text{Co}(\text{CN})_6]^{3-}$$ is a regular octahedron with six identical $$CN^-$$ ligands arranged symmetrically. The complex has both an inversion centre and several mirror planes, so it is achiral. No optical isomerism.

Hence, the only compound that can exist as a pair of optical isomers is

Option C which is: $$[\text{Cr}(\text{C}_2\text{O}_4)_3]^{3-}$$