JEE Advanced · 2026 · 17 May 2026 · Paper 2 · Coordination Compounds

For an octahedral complex, we need to count triangular faces with one N and two Cl atoms at corners.

**For cis-$[\text{Co}({\text{NH}}_3{)}_4{\text{Cl}}_2]\text{Cl}$:**
The two Cl atoms are adjacent (cis). Looking at the octahedron, there are 2 triangular faces that have one N atom and both Cl atoms at their corners.

**For mer-$[\text{Co}({\text{NH}}_3{)}_3{\text{Cl}}_3]$:**
The three Cl atoms are in meridional positions (forming a plane through the metal). Each pair of Cl atoms (there are 3 such pairs) can form triangular faces with N atoms. There are 4 triangular faces with one N and two Cl atoms.

Total = 2 + 4 = 6