How many geometrical isomers (including any possible optical isomers) can exist for the octahedral complex $[M (N H_{3})_{2} C l_{2} (H_{2} O)_{2}]$ where the two $N H_{3}$ , two $C l$ , and two $H_{2} O$ ligands are each identical?

medium 2 marks Class 12 · Chemistry · Coordination Compounds

Answer

3 geometrical isomers

The correct answer is 3 geometrical isomers — Therefore the total number of geometrical isomers is 3.

Step-by-step solution

Identify the ligand sets: two $N H_{3}$ (A), two $C l$ (B), two $H_{2} O$ (C).
For an octahedral complex of type $M A_{2} B_{2} C_{2}$ , three distinct arrangements are possible.
Isomer 1: all three pairs are trans to each other (all‑trans).
Isomer 2: one pair is cis while the other two pairs remain trans (cis‑trans).
Isomer 3: all three pairs are cis, giving a ‘cis‑cis‑cis’ arrangement.
Each arrangement possesses a plane of symmetry, so none are optically active.
Therefore the total number of geometrical isomers is 3.

Common mistake: Counting only cis and trans (2) and overlooking the third mixed arrangement

Concept tested: Geometrical isomerism in $MA_{2}B_{2}C_{2}$ octahedral complexes