Question

Optical isomerism is shown by:

• A. \(\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5\left(\mathrm{NO}_2\right)\right]^{+}\)
• B. \(\left[\mathrm{Co}(\mathrm{en})_3\right]^{3+}\)
• C. \(\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right]^{+}\)
• D. \(\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_3 \mathrm{Cl}_3\right]\)

Hint:

For optical isomerism:
- Requires chirality (no plane of symmetry).
- Common in octahedral \(\mathrm{M}(\mathrm{AA})_3\) or cis-\(\mathrm{MA}_2(\mathrm{AA})_2\).
- Check for symmetry planes in the structure.

Answer 1

The Correct Option is B

Optical isomerism occurs when a molecule is non-superimposable on its mirror image, requiring no plane of symmetry (chiral). For coordination compounds, this often occurs in octahedral complexes with specific ligand arrangements.
Step 1: Analyze each complex
All complexes are octahedral (coordination number 6). - (A) \(\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5\left(\mathrm{NO}_2\right)\right]^{+}\): Type \(\mathrm{MA}_5\mathrm{B}\). Has a plane of symmetry (e.g., through \(\mathrm{NO}_2\), Co, and trans \(\mathrm{NH}_3\)). Not chiral.
- (B) \(\left[\mathrm{Co}(\mathrm{en})_3\right]^{3+}\): Type \(\mathrm{M}(\mathrm{AA})_3\), where en (ethylenediamine) is a bidentate ligand. This forms a propeller-like structure with no plane of symmetry, exhibiting optical isomerism (enantiomers: \(\Delta\) and \(\Lambda\)). Chiral.
- (C) \(\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right]^{+}\): Type \(\mathrm{MA}_4\mathrm{B}_2\). Has cis and trans isomers. The trans isomer has a plane of symmetry; the cis isomer has a plane of symmetry through the Cl-Co-Cl plane. Not chiral.
- (D) \(\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_3 \mathrm{Cl}_3\right]\): Type \(\mathrm{MA}_3\mathrm{B}_3\). Has facial (fac) and meridional (mer) isomers, both with planes of symmetry. Not chiral.
Step 2: Conclusion
Only \(\left[\mathrm{Co}(\mathrm{en})_3\right]^{3+}\) shows optical isomerism, matching option (B).