Tetrahedral Geometry and Potential energy

Question

Intuitively, I'd think it possible to use potential energy to find the bonding angles in methane, but I'm not getting the right answer. Am I missing something?

In a tetrahedral, the faces are equilateral triangles. Center a sphere having as radius the same side length as the equilateral triangle. These 3 spheres will intersect at two points each corresponding to the 4th vertex of a tetrahedral. Take the center of mass to get the center of the tetrahedron. Now you can construct vectors and use the dot product to demonstrate the bonding angle in tetrahedral geometry is $\approx 109.47^\circ$.

I'd think there's a more generalized technique with potential energy, but it seems there's a flaw in my reasoning.

Let $\hat{u_i}$s be unit position vectors centered at the origin, representing the placement of electrons on the unit sphere.

$V=\frac{-e^2}{4\pi \epsilon_0}\sum_{i\ne j}\frac{1}{|\hat{u_i}-\hat{u_j}|}+\frac{e^2}{4\pi \epsilon_0}\sum_{i=1}^4\frac{1}{|\hat{u_i}|}$

Using the cosine rule:

$V=\frac{-e^2}{4\pi \epsilon_0}\sum_{i\ne j}\frac{1}{\sqrt{2(1-\cos \theta_{ij})}}+\frac{4e^2}{4\pi\epsilon_0}$, where $\cos \theta_{ij} =\hat{u_i}\cdot \hat{u_j}$

From the 3 spheres argument above, we expect the bonding angles to be equal. Heuristically, this seems likely by symmetry (though for some reason the distribution is not symmetric with 5 electrons). The potential energy of any pair of electrons on the sphere of the surface is the same and $\binom{4}{2}=6$. So all the $\theta_{ij}$ are equal we can let $\theta_{ij}=\theta$ for all combinations. Then V can be rewritten as

$V=\frac{-6e^2}{4\pi\epsilon_0}\frac{1}{\sqrt{2(1-\cos \theta)}}+\frac{4e^2}{4\pi \epsilon_0}$

With a trig identity:

$V=\frac{-6e^2}{8 \pi \epsilon_0}\frac{1}{\sin \theta/2}+\frac{4e^2}{4\pi \epsilon_0}$

Now we find the minimum using :

$\frac{dV}{d\theta}=\frac{-6e^2}{8\pi \epsilon_0} \frac{-1}{2}\frac{\cos \theta /2}{\sin ^2 \theta/2}=0\implies \cos \theta/2 =0\implies \theta = (2k+1)\pi$ for integer $k$.

But $\theta$ is never $109.47^\circ$ for any $k$.

There has to be a mistake, but I'm not sure where. Anyone see it?

Answer 1

The conclusion "all $\theta_{ij}$ are equal we can use $\theta_{ij}=\theta$ for all combinations" does not work. A priori, before determining the actual minimum, one does not know that all $\theta_{ij}$ are equal. Thus to determine the minimum one would have to solve a problem involving six angles (but with some complicated additional constraint equations since the six angles are determined by only four points, which also are confined to lie on a sphere). What you computed later can be interpreted as the minimum of the potential energy for only two electrons confined to lie on a sphere, which occurs for $\theta=(2k+1)\pi$ (for integer $k$), when they are opposite to each other.

The approach with six free parameters and constraints is likely not an elegant way to solve the problem.