# What is the angle between electron groups in the tetrahedral geometry?

Jun 24, 2017

This can be geometrically worked out, but we expect an angle of ${109.5}^{\circ}$. Let's verify that by drawing a tetrahedron in a cube.

Here,

• the center of the cube is the central atom in a tetrahedral molecule.
• the upper left rear, upper right front, bottom left front, and bottom right rear points are the surrounding atoms in a tetrahedral molecule.

The base diagonal is $\sqrt{2}$ from the geometry of a $45 - 45 - 90$ triangle of leg lengths $1$. From the diagram, the diagonal from the bottom left front to the upper right rear is from the Pythagorean Theorem:

${1}^{2} + {\left(\sqrt{2}\right)}^{2} = {c}^{2}$

$\implies c = \sqrt{3}$

We want half that diagonal, as that is the length towards the central atom in a tetrahedral molecule, so $\frac{c}{2} = \frac{\sqrt{3}}{2}$.

Finally, we have an $S S S$ triangle to solve to determine the tetrahedral angle. By symmetry, the side lengths are:

• upper-left: $x = \frac{\sqrt{3}}{2}$
• upper-right: $y = \frac{\sqrt{3}}{2}$
• base diagonal: $z = \sqrt{2}$

The law of cosines is:

${z}^{2} = {x}^{2} + {y}^{2} - 2 x y \cos Z$

We can use this to find the tetrahedral angle, $Z$, corresponding to base diagonal length $z = \sqrt{2}$:

$\frac{{z}^{2} - {x}^{2} - {y}^{2}}{- 2 x y} = \cos Z$

$\implies \textcolor{b l u e}{Z} = \arccos \left(\frac{{z}^{2} - {x}^{2} - {y}^{2}}{- 2 x y}\right)$

$= \arccos \left(\frac{{\left(\sqrt{2}\right)}^{2} - {\left(\frac{\sqrt{3}}{2}\right)}^{2} - {\left(\frac{\sqrt{3}}{2}\right)}^{2}}{- 2 \left(\frac{\sqrt{3}}{2}\right) \left(\frac{\sqrt{3}}{2}\right)}\right)$

$= \arccos \left(\frac{2 - \frac{3}{4} - \frac{3}{4}}{- \frac{3}{2}}\right)$

$= \textcolor{b l u e}{{109.5}^{\circ}}$