Animated Solution for Chemistry - Coordination Compounds: Total number of $cis$ $\text{N}-\text{Mn}-\text{Cl}$ bond angles (that is, $\text{Mn}-\text{N}$ and $\text{Mn}-\text{Cl}$ bonds in $cis$ positions) present in a molecule of $cis-[\text{Mn}(\text{en}{)}_2{\text{Cl}}_2]$ complex is ____ ($\text{en}={\text{NH}}_2{\text{CH}}_2{\text{CH}}_2{\text{NH}}_2$)

The world of coordination chemistry is essentially a playground of 3D geometry. When we talk about complexes like $cis-[\text{Mn}(\text{en}{)}_2{\text{Cl}}_2]$, we aren't just looking at a chemical formula; we are looking at a microscopic architectural structure. The challenge—and the fun—lies in visualizing this structure in our minds and counting the specific angles within it.

In this problem, we are asked to find the total number of cis $\text{N}-\text{Mn}-\text{Cl}$ bond angles. Let's break down exactly what that means and how to systematically count them without getting lost in the 3D space.

First, let's look at the central metal and its ligands. We have a Manganese ($\text{Mn}$) atom surrounded by two chloride (${\text{Cl}}^{-}$) ligands and two ethylenediamine ('en') ligands.

The 'en' ligand is a bidentate ligand. This means it has two nitrogen donor atoms that can bind to the metal. A crucial geometric constraint of the 'en' ligand is its size: the carbon chain connecting the two nitrogen atoms is relatively short. Because of this, an 'en' ligand can only bridge two adjacent positions on the octahedron. It simply cannot stretch across the metal to connect two opposite (trans) positions.

The prefix cis in the name $cis-[\text{Mn}(\text{en}{)}_2{\text{Cl}}_2]$ tells us the relative positioning of the two identical chloride ligands. They must be placed adjacent to each other, forming a $90^{\circ }$ angle with the central Manganese atom.

An octahedron has six vertices. If you pick any single vertex, it has exactly one position directly opposite to it (at $180^{\circ }$, the trans position) and four positions adjacent to it (at $90^{\circ }$, the cis positions).

This gives us a powerful shortcut. If we want to find which nitrogen atoms are cis to a specific chlorine atom, it is much easier to first identify what is trans to that chlorine. Once we know the single trans neighbor, we know that the other four positions must be cis.

Let's draw our octahedron and place the two chlorine atoms in cis positions. We'll call them ${\text{Cl}}_{(a)}$ and ${\text{Cl}}_{(b)}$. Next, we wrap the two 'en' ligands around the remaining four positions.

Let's focus our attention entirely on ${\text{Cl}}_{(a)}$. If we look across the octahedron, directly opposite to ${\text{Cl}}_{(a)}$, we will find one of the nitrogen atoms from an 'en' ligand. Let's call this ${\text{N}}_{(3)}$.

Since ${\text{Cl}}_{(a)}$ is trans to ${\text{N}}_{(3)}$, it cannot form a cis angle with it. However, this means ${\text{Cl}}_{(a)}$ must be cis to all the other ligands in the complex. The other ligands are ${\text{Cl}}_{(b)}$ and three nitrogen atoms: ${\text{N}}_{(1)}$, ${\text{N}}_{(2)}$, and ${\text{N}}_{(4)}$.

Therefore, ${\text{Cl}}_{(a)}$ forms exactly 3 cis $\text{N}-\text{Mn}-\text{Cl}$ bond angles.

Now, we shift our focus to the second chlorine atom, ${\text{Cl}}_{(b)}$. We apply the exact same logic.

Looking directly opposite ${\text{Cl}}_{(b)}$, we find a different nitrogen atom, let's say ${\text{N}}_{(2)}$. Because ${\text{Cl}}_{(b)}$ is trans to ${\text{N}}_{(2)}$, it must be cis to the remaining three nitrogen atoms: ${\text{N}}_{(1)}$, ${\text{N}}_{(3)}$, and ${\text{N}}_{(4)}$.

Just like the first chlorine, ${\text{Cl}}_{(b)}$ also forms exactly 3 cis $\text{N}-\text{Mn}-\text{Cl}$ bond angles.

To find the total number of cis $\text{N}-\text{Mn}-\text{Cl}$ bond angles in the entire molecule, we simply add the contributions from both chlorine atoms.

Total angles = (Angles from ${\text{Cl}}_{(a)}$) + (Angles from ${\text{Cl}}_{(b)}$)

Total angles = $3+3=6$

By understanding the geometric constraints of the ligands and using the relationship between cis and trans positions, we turned a potentially confusing 3D visualization problem into a simple counting exercise.