# Geometry and chemistry

There's a really obvious geometric reason why the cosine of the bond angle in graphite is $-1/2$: the stuff consists of sheets shaped like honeycombs.

There's also a really obvious geometric reason why the cosine of the bond angle in methane is $-1/3$: four hydrogen atoms are arranged in a symmetric way about a center, with all six angles between them equal: tetrahedral symmetry.

Is there a geometric reason why the cosine of the bond angle in water is $-1/4$?

(I suspect chemists might dismiss this question as being about magic and voodoo and mystical and the like. Mathematicians don't mind magic and voodoo and mysticism, but maybe they object to vagueness or chemistry. So we'll see if this question is tolerated here.)

(Full disclosure: once upon a time I posted nearly this same question somewhere else.....)

• "I suspect chemists might dismiss this question as being about magic and voodoo and mystical and the like." - I, for one, don't. On the other hand, you're probably aware of the two lone electron pairs in water contributing to the deviation from the ideal tetrahedral angle, so I'm not entirely sure that the reasons are entirely geometric. – J. M. is not a mathematician Nov 14 '11 at 17:58
• Maybe people are less accustomed to thinking about the $-1/3$ situation than about the honeycomb, so here it is: Picture four unit vectors pointing out from the origin in $\mathbb{R}^3$, arranged so that all of the six angles between two of them are equal. What are the components of the four vectors in the direction of one of them? The component of one of them is $1$. The components of the other three are equal to each other, and the components of all four must of add up to $0$. – Michael Hardy Nov 14 '11 at 18:07
• Michael: a probably more geometrically intuitive way to see the point you're presenting is to consider the embedding of the tetrahedron into a cube... – J. M. is not a mathematician Nov 15 '11 at 0:43

That there's nothing else hidden behind this numerical coincidence is evidenced by very precise Hartree-Fock calculations for water molecules, which yield a bond angle of $106.339^\circ$, quite a bit off from $\arccos(-\frac14)$. These calculations should pick up on any geometric aspects of the problem; the difference between this value and the actual value is due to the electronic correlations that are neglected in the Hartree-Fock approximation.
• If you're right about the value of the bond angle, then I think you're right about your other comments. However, why is it that at en.wikipedia.org/wiki/Water_%28data_page%29 we see the bond angle given as $104.4775^\circ$? Obvious answer: because some idiot edited the article that way. But by superficial inspection that number has persisted even after some debate about it on the article's talk page a few months ago. (On my home computer, I can't view the article that you cite; I'll try it from a university computer later.) – Michael Hardy Nov 14 '11 at 19:55
• @Michael: That's a misunderstanding. The number I quoted is the precise value (to the given accuracy) in the Hartree-Fock approximation. That is, if you neglect correlation between the electrons and write the wavefunction as an antisymmetrized product of individual orbitals and then use a very good basis set such that the discretization error is negligible, you get this value. The actual value, which you can get either by measurement or by calculations that take correlation into account, is $104.5^\circ$. My argument was that the Hartree-Fock value should reflect any geometric symmetries. – joriki Nov 14 '11 at 20:04
• For instance, in the two examples you cited, with cosines $-1/2$ and $-1/3$, a very cheap Hartree-Fock calculation would already yield those angles, because they arise from geometric symmetry and not from details of the electronic wavefunction. – joriki Nov 14 '11 at 20:05
• OK, so $104.5^\circ$ is $\arccos(-1/4)$ rounded to the nearest half-degree. Any idea where the number in the Wikipedia article comes from? – Michael Hardy Nov 14 '11 at 21:19
• If there's no geometric reason for this, the next question is whether we can exploit the coincidence somehow. $\qquad$ – Michael Hardy Jul 22 '16 at 17:41