Total Triangular Faces with Specific Atoms in Octahedral Complexes

To find the sum of the total number of triangular faces having one $\text{N}$ atom and two $\text{Cl}$ atoms at their corners in both octahedral complexes, we analyze the geometric arrangement of the ligands on an octahedron.

An octahedron has $6$ vertices and $8$ triangular faces. Let the $6$ vertices be represented by $3$ pairs of trans (opposite) positions: $(V_1, V_2), \quad (V_3, V_4), \quad (V_5, V_6)$

A triangular face is formed by selecting exactly one vertex from each of the three opposite pairs. Therefore, the 8 faces are formed by the combinations:

$$\begin{array}{cccc} 1. & (V_1, V_3, V_5) & 5. & (V_2, V_3, V_5) \\ 2. & (V_1, V_3, V_6) & 6. & (V_2, V_3, V_6) \\ 3. & (V_1, V_4, V_5) & 7. & (V_2, V_4, V_5) \\ 4. & (V_1, V_4, V_6) & 8. & (V_2, V_4, V_6) \end{array}$$

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1. Complex 1: cis-$[\text{Co}(\text{NH}_3)_4\text{Cl}_2]\text{Cl}$

In a cis-isomer, the two $\text{Cl}$ atoms are adjacent ($90^\circ$ apart) to each other, which means they do not occupy opposite vertices.

• Let the two $\text{Cl}$ atoms occupy vertices $V_1$ and $V_3$.
• The four $\text{N}$ atoms (from $\text{NH}_3$) occupy the remaining vertices: $V_2, V_4, V_5, V_6$.

We want to find faces with two $\text{Cl}$ atoms and one $\text{N}$ atom. Such a face must contain both $V_1$ ($\text{Cl}$) and $V_3$ ($\text{Cl}$), along with a third vertex from $\{V_5, V_6\}$ (which are both $\text{N}$ atoms).

The faces containing both $V_1$ and $V_3$ are:

1. $(V_1, V_3, V_5) \longrightarrow (\text{Cl}, \text{Cl}, \text{N})$
2. $(V_1, V_3, V_6) \longrightarrow (\text{Cl}, \text{Cl}, \text{N})$

Thus, for cis-$[\text{Co}(\text{NH}_3)_4\text{Cl}_2]\text{Cl}$, the number of faces with one $\text{N}$ and two $\text{Cl}$ atoms is $2$.

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2. Complex 2: mer-$[\text{Co}(\text{NH}_3)_3\text{Cl}_3]$

In a meridional (mer) isomer, three identical ligands lie on a meridian (a plane passing through the central metal atom). Thus, two of the $\text{Cl}$ atoms are trans (opposite) to each other, and the third $\text{Cl}$ atom is cis (adjacent) to both.

• Let the three $\text{Cl}$ atoms occupy vertices $V_1, V_2$ (which are trans to each other) and $V_3$.
• The three $\text{N}$ atoms occupy vertices $V_4, V_5, V_6$.

Evaluating all $8$ faces:

1. $(V_1, V_3, V_5) \longrightarrow (\text{Cl}, \text{Cl}, \text{N})$ — Valid
2. $(V_1, V_3, V_6) \longrightarrow (\text{Cl}, \text{Cl}, \text{N})$ — Valid
3. $(V_1, V_4, V_5) \longrightarrow (\text{Cl}, \text{N}, \text{N})$
4. $(V_1, V_4, V_6) \longrightarrow (\text{Cl}, \text{N}, \text{N})$
5. $(V_2, V_3, V_5) \longrightarrow (\text{Cl}, \text{Cl}, \text{N})$ — Valid
6. $(V_2, V_3, V_6) \longrightarrow (\text{Cl}, \text{Cl}, \text{N})$ — Valid
7. $(V_2, V_4, V_5) \longrightarrow (\text{Cl}, \text{N}, \text{N})$
8. $(V_2, V_4, V_6) \longrightarrow (\text{Cl}, \text{N}, \text{N})$

Thus, for mer-$[\text{Co}(\text{NH}_3)_3\text{Cl}_3]$, the number of faces with one $\text{N}$ and two $\text{Cl}$ atoms is $4$.

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Total Sum

$\text{Total number of faces} = 2 + 4 = 6$