Condition for facial–meridional isomerism

Facial–meridional ( $\mathrm{fac}$–$\mathrm{mer}$) isomerism is shown by octahedral complexes of the type: $[MA_3B_3]$ where three identical ligands and another three identical ligands are present.

• fac: the three identical ligands occupy one face of the octahedron.
• mer: the three identical ligands lie along a meridian, with two trans and one cis.

Check each option

Option A: $[\mathrm{Co}(\mathrm{en})_2\mathrm{Cl}_2]^+$

Here, $\mathrm{en}$ is a bidentate ligand. The complex is of type: $[M(\mathrm{AA})_2B_2]$ Such complexes can show cis–trans isomerism, but not $\mathrm{fac}$–$\mathrm{mer}$ isomerism.

So, A is incorrect.

Option B: $[\mathrm{Co}(\mathrm{NH}_3)_3\mathrm{Cl}_3]$

This is of the form: $[MA_3B_3]$ Therefore, it can show facial–meridional isomerism.

So, B is correct.

Option C: $[\mathrm{Co}(\mathrm{NH}_3)_4\mathrm{Cl}_2]^+$

This is of the type: $[MA_4B_2]$ Such complexes show cis–trans isomerism, not $\mathrm{fac}$–$\mathrm{mer}$.

So, C is incorrect.

Option D: $[\mathrm{Co}(\mathrm{en})_3]^{3+}$

This is of the form: $[M(\mathrm{AA})_3]$ It shows optical isomerism ($\Delta$ and $\Lambda$ forms), but not $\mathrm{fac}$–$\mathrm{mer}$ isomerism.

So, D is incorrect.

Final Answer

The complex showing facial–meridional isomerism is: $\boxed{[\mathrm{Co}(\mathrm{NH}_3)_3\mathrm{Cl}_3]}$ i.e. Option B.

Comparison with stored correct answer

Stored correct answer = B

My derived answer = B

Hence, they agree.