# Why is $\arccos(-\frac 13)$ the optimal angle between bonds in a methane ($\rm CH_4$) molecule?

Background: In a CH4 molecule, there are 4 C-H bonds that repel each other. Essentially the mathematical problem is how to distribute 4 points on a unit sphere where the points have maximal mutual distance - or, how to distribute 4 position vectors such that the endpoint distances between them are maximized.

This question Angle between lines joining tetrahedron center to vertices shows that the angles between the points/vectors forming the vertices of a regular tetrahedron is $\arccos(-\frac13)$.

I have been told by chemistry teachers that that is the shape which maximizes mutual distance between the points. However, that link was not relevant to me because I am trying to figure out a proof that the tetrahedral model maximizes distance and is the only model which maximizes it.

Therefore, the very similar question Calculations of angles between bonds in CH₄ (Methane) molecule was also irrelevant, because the answer started off with "Note that a regular tetrahedron can be inscribed in alternating vertices of a cube.", and thus assuming that the tetrahedral shape was optimal without any mathematical basis.

So my question is, is $\arccos(-\frac13)$ the optimal angle, and if so, how can I approach this question to prove it? I have limited knowledge in vector calculus but I am willing to learn if this is some multivariable optimization problem.

Thanks!

• To maximize distances you put the atoms as far apart as possible. It doesn't seem to me that this is what you want. Instead you need a model of the attraction which causes the carbon and the hydrogens to stick together to form a molecule. Such a model, as I understand it, is a Schrödinger equation in $3\cdot (6+4)=30$ dimensions, assuming only electrostatic attraction is involved, Born-Oppenheimer and so forth. It seems challenging to prove that the ground state for such a Schrödinger equation must be symmetric and involve tetrahedral angles, but maybe I'm wrong about this. Mar 23, 2016 at 16:01
• As is pointed out at the related question on this site math.stackexchange.com/questions/1663014/… , this is known as the Tammes problem - see e. g. en.wikipedia.org/wiki/Tammes_problem . This is not an answer but it might give you some idea where to look. Mar 23, 2016 at 16:11
• What do you want to maximize, exactly? The sum of the distances between every pair of points? The sum of the square of those distances? Other? Mar 23, 2016 at 16:30
• I apologize for a foolish error. The Schrödinger equation I was referring to is actually for assumed positions of the nuclei. In books (as I recall) the true positions of the nuclei are then obtained by a minimization process which requires solving the equation at each set of positions considered. As far as I know this is done numerically and is tough. But maybe some theorem exists whereby maximizing an objective function such as the sum of the distances between the hydrogens divided by their mutual distance from the carbon (assumed the same for all) is enough to give the H's positions. Mar 23, 2016 at 17:02
• Henry Cohn and Abhinav Kumar, Universally Optimal Distributions of Points on Spheres, J.A.M.S. 20 (2007) 99-147 -- Theorem 1.2 gives a proof for 4 points. Mar 23, 2016 at 17:56

It might be helpful to look at a 2D model. You'll need a nickel and two pennies.

Put the coins flat on a table however you like. Then let your friend move the coins according to some rules.

1. If the pennies are less than 2 diameters apart or more than 4 diameters apart, they can move the pennies to a more reasonable distance apart.
2. If the nickel can be moved to a spot either closer or more equidistant to the pennies, they can move the nickel.

Every time your friend can move the coins, you need to pay them $100. What would the rules be for 3 pennies and a nickel? Alternately, make a computer model. You'll quickly get to models like the Thomson problem. But notice that the 2010 proof for$5$points was incredibly difficult, and$7\$ points is considered unproven.

The 4 point case involves a very strong attractor, the tetrahedron. Add a few more atoms and it's a vastly harder problem. A molecule doesn't care if there is an optimal solution -- if the atoms find a local minimum, they will likely stay in that weird configuration.

Methane doesn't have any other local minima to worry about.

Water has two hydrogens and two free electrons, making the model more interesting.

• "A molecule doesn't care if there is an optimal solution -- if the atoms find a local minimum, they will likely stay in that weird configuration." - for the chemistry buffs, here is a concrete example: although the cyclohexane molecule and its derivatives will often prefer to be in its "chair" conformation, every so often, one will see substituted cyclohexanes in the "twist-boat" conformation, which corresponds to a local optimum of its potential energy surface. Mar 28, 2017 at 12:27