Condition for optical isomerism

A coordination compound shows optical isomerism if it is chiral, i.e. its mirror image is non-superimposable. This usually happens in certain octahedral complexes with suitable ligand arrangement, especially with bidentate ligands.

Check each option

Option A: $[Cu(NH_3)_4]^{2+}$

• This is typically a square planar or distorted geometry for $Cu^{2+}$.
• Such a complex has symmetry and does not show optical isomerism.
• Not optically active.

Option B: $[ZnCl_4]^{2-}$

• This is a tetrahedral complex.
• Since all four ligands are identical ($Cl^-$), the complex is symmetric.
• It does not show optical isomerism.
• Not optically active.

Option C: $[Cr(C_2O_4)_3]^{3-}$

• Oxalate, $C_2O_4^{2-}$, is a bidentate ligand.
• The complex is octahedral, with three bidentate ligands.
• Complexes of the type $[M(AA)_3]$ show optical isomerism due to the formation of $\Delta$ and $\Lambda$ forms.
• Therefore, this compound does show optical isomerism.

Option D: $[Co(CN)_6]^{3-}$

• This is an octahedral complex with six identical monodentate ligands.
• It is highly symmetric and cannot be chiral.
• Not optically active.

Conclusion

The only compound that shows optical isomerism is: $[Cr(C_2O_4)_3]^{3-}$

Comparison with stored correct answer

• Derived answer: C
• Stored correct answer: C
• They match.