Which of the following compound shows optical isomerism?

The correct answer is Option (3) → \([Cr(C_2O_4)_3]^{3-}\)

An entity shows optical isomerism when it is chiral: its mirror image cannot be superimposed on it, so it exists as two enantiomers (d and l) that rotate plane-polarised light in opposite directions. A molecule with a plane of symmetry is not chiral.

Optical isomerism is common in octahedral complexes with didentate ligands. In \( \ce{[Cr(C2O4)3]^{3-}} \) three oxalate ions (each bonding through two O atoms) span the six positions round \( \ce{Cr^{3+}} \) like the blades of a propeller. The left-handed and right-handed propellers are mirror images that do not superimpose, as for \( \ce{[Co(en)3]^{3+}} \). So the trioxalatochromate(III) ion shows optical isomerism.

Why the other options are incorrect

• Option (1) \([Cu(NH_3)_4]^{2+}\) — Incorrect. This ion is square planar; the plane of the four N atoms is a plane of symmetry, so it is not chiral.

• Option (2) \([ZnCl_4]^{2-}\) — Incorrect. A tetrahedral ion with four identical ligands is highly symmetrical and has no enantiomers.

• Option (4) \([Co(CN)_6]^{3-}\) — Incorrect. An octahedron with six identical unidentate ligands is superimposable on its mirror image; it shows neither geometrical nor optical isomerism.

NCERT (2026-27 edition):
Optical Isomerism – “Optical isomers are mirror images that cannot be superimposed on one another. These are called as enantiomers. The molecules or ions that cannot be superimposed are called chiral.”

Optical Isomerism – “… Optical isomerism is common in octahedral complexes involving didentate ligands (Fig. 5.6).”