Consider that the coordinating atoms of the ligands in cis -[Co(NH...

In an octahedral complex, the six coordinating atoms are located at the vertices of an octahedron, which has 8 triangular faces. Each face is formed by 3 vertices, and any two adjacent (cis) vertices form an edge that is shared by exactly two faces.

For the $cis$-[Co(NH$_3$)$_4$Cl$_2$]$^+$ complex:
The two Cl atoms are at adjacent (cis) positions, forming exactly one Cl-Cl edge. This edge is shared by two triangular faces. Since the remaining four vertices are occupied by N atoms, the third vertex of each of these two faces must be an N atom. Therefore, there are 2 faces with exactly two Cl atoms and one N atom.

For the $mer$-[Co(NH$_3$)$_3$Cl$_3$] complex:
The three Cl atoms are in a meridional arrangement, meaning two Cl atoms are trans to each other and the third is cis to both. This arrangement forms exactly two Cl-Cl edges. Since the three Cl atoms do not occupy the vertices of the same face (unlike the $fac$ isomer), no face contains three Cl atoms. Each of the two Cl-Cl edges is shared by two faces, yielding $2 \times 2 = 4$ distinct faces that contain exactly two Cl atoms. The third vertex in all these 4 faces is an N atom. Therefore, there are 4 faces with exactly two Cl atoms and one N atom.

The sum of the total number of such faces in both complexes is $2 + 4 = 6$.

Answer: $6$