Animated Solution for Chemistry - Coordination Compounds: Complexes ($M{L}_5$) of metals Ni and Fe have ideal square pyramidal and trigonal bipyramidal geometries, respectively. The sum of the $90^{\circ }$, $120^{\circ }$ and $180^{\circ }$ L-M-L angles in the two complexes is ............... .

When dealing with coordination complexes of the type $M{L}_5$, the central metal atom is surrounded by five ligands. Depending on the metal and the nature of the ligands, these complexes can adopt different geometries to minimize repulsion. In this problem, we are given two specific geometries: Square Pyramidal for the Nickel complex and Trigonal Bipyramidal for the Iron complex.

Our goal is to count the exact number of $90^{\circ }$, $120^{\circ }$, and $180^{\circ }$ L-M-L bond angles in both of these structures and find their sum. Let's break them down one by one.

Imagine a square base with the metal atom sitting exactly at the center. Four ligands occupy the corners of this square. The fifth ligand sits directly above the metal atom, forming the apex of the pyramid.

First, let's look for the $180^{\circ }$ angles. These occur when two ligands are exactly opposite to each other, forming a straight line through the central metal. In the square base, the diagonals connect opposite corners. Since a square has two diagonals, there are exactly $2$ such $180^{\circ }$ angles.

Next, we count the $90^{\circ }$ angles. Within the square base itself, adjacent ligands are separated by $90^{\circ }$. Since there are four sides to the square, that gives us $4$ angles. Now, consider the axial ligand at the top. It is perpendicular to the entire square base, meaning it forms a $90^{\circ }$ angle with each of the four equatorial ligands. This adds another $4$ angles.

Total $90^{\circ }$ angles = $4+4=8$.

There are no $120^{\circ }$ angles in a square pyramidal geometry.

Now, let's shift our focus to the trigonal bipyramidal geometry. Here, three ligands form an equilateral triangle around the metal in the equatorial plane. The remaining two ligands are positioned axially—one directly above and one directly below the plane.

How many $180^{\circ }$ angles are there? The only two ligands that are perfectly opposite each other are the top and bottom axial ligands. Thus, there is exactly $1$ such $180^{\circ }$ angle.

What about the $120^{\circ }$ angles? These are found exclusively in the equatorial plane. The three ligands form a perfect equilateral triangle, dividing the $360^{\circ }$ plane into three equal parts. This gives us exactly $3$ angles of $120^{\circ }$.

Finally, let's count the $90^{\circ }$ angles. The top axial ligand is perpendicular to the equatorial plane, so it forms a $90^{\circ }$ angle with all three equatorial ligands. Similarly, the bottom axial ligand also forms a $90^{\circ }$ angle with the same three equatorial ligands.

Total $90^{\circ }$ angles = $3+3=6$.

We have successfully mapped out all the required angles from both geometries. Now, we simply need to sum them up as requested by the problem.

From the Square Pyramidal complex: - $180^{\circ }$ angles: $2$ - $90^{\circ }$ angles: $8$

From the Trigonal Bipyramidal complex: - $180^{\circ }$ angles: $1$ - $120^{\circ }$ angles: $3$ - $90^{\circ }$ angles: $6$

Total Sum = $2+8+1+3+6=20$.

The beauty of this problem lies entirely in spatial visualization. By mentally constructing these molecules, counting the angles becomes a straightforward and highly rewarding exercise!