Question

The compound $MX_6$ is octahedral. Find the number of 90° X - M - X angles in the compound.

Answer 1

The question can be solved by considering a simple relation. If there are X number of atoms surrounding the complex then there are in total $\dfrac{\text{X}}{\text{3}}$ number of $\text{9}{{\text{0}}^{\text{0}}}$ .The total number of ${{180}^{0}}$ can be given by $\dfrac{\text{X}}{2}$ .

Complete step by step solution:

Octahedral geometry describes the shape of a compound. In the octahedral complex, the six atoms or the groups or ligands are symmetrically arranged around the central atom. Each atom surrounding the central atom is located at the imaginary corner of the octahedron and the one atom or central atom is at the centre of the octahedron. Let's consider a complex where the central atom is M and surrounding atoms are X forming a $\text{M}{{\text{X}}_{\text{6}}}$ complex.

All the complexes which possess such an arrangement of atoms surrounded by the six atoms are octahedral geometry. The geometry arises due to ${{\text{d}}^{\text{2}}}\text{s}{{\text{p}}^{\text{3}}}\text{ or s}{{\text{p}}^{\text{3}}}{{\text{d}}^{\text{2}}}$ the hybridisation of the central atom.

The structure associated with the octahedral complexes is as shown below.

There are a total of six $\text{M-X}$ bonds. Each bond is making each side of the octahedron.

The four $\text{M-X}$ are placed in the x and y-axis. They are arranged based on the angle between them. Let's consider the total angle which is lying in a plane is equal to ${{360}^{0}}$ . Thus the bond angle between each $\text{M-X}$ bond in the plane is:

$\text{bond angle between M-X axial bond=}\dfrac{\text{36}{{\text{0}}^{\text{0}}}}{\text{4}}\text{=9}{{\text{0}}^{\text{0}}}$

Thus each $\text{M-X}$ bond in the plane is having an $\text{9}{{\text{0}}^{\text{0}}}$ angle between them resulting in the square planar structure.

Two bonds are at the equatorial position. This bond is in the direction of the z-axis. These bonds are perpendicular to the axial bonds.

The bond angle between these equatorial bonds is:

$\text{bond angle between M-X bond equatorial=}\dfrac{\text{36}{{\text{0}}^{\text{0}}}}{\text{2}}\text{=18}{{\text{0}}^{\text{0}}}$

But since these bonds are perpendicular to the axial bonds the angle between axial and equatorial X $\angle {{X}_{axial}}-M-{{X}_{equatorial}}$ is $\text{9}{{\text{0}}^{\text{0}}}$ .

Thus there are a total of four $\angle X\text{ }-\text{ }M\text{ }-\text{ }X$ angels in the axial plane and two in the equatorial plane.

Therefore there are a total of six $\text{9}{{\text{0}}^{\text{0}}}$ $\angle X\text{ }-\text{ }M\text{ }-\text{ }X$ angles in $\text{M}{{\text{X}}_{\text{6}}}$ the complex.

Note: The bond angle plays an important role in determining structure. For example, $\text{M}{{\text{X}}_{7}}$ complex, which is a bipyramid pentagonal geometry, has five bonds in the axial plane. Thus the bond angle in the axial plane is equal to $=\dfrac{{{360}^{0}}}{5}={{72}^{0}}$ . Similarly can be found for other complexes.