# What physical forces give rise to the peculiar bond angle of hydrogen peroxide?

In one of the Periodic Table videos, Prof. Poliakoff shows a model of a hydrogen peroxide molecule and claims that the H-O bonds will always be at right angles to each other. I have a rudimentary intuition for how molecules arrange themselves based on mutual attraction and repulsion - I've solved some Foldit puzzles - but the H2O2 defies that kind of common sense explanation.

I get that there are forces pushing the hydrogens apart and forces pulling them together, but how does that fragile equilibrium of forces result in an exact right angle?

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I strongly suspect that this will be better answered at chemistry.stackexchange.com. The physics answer is that the only forces present are electrostatic interactions. The chemistry description is likely to be more useful. – Colin McFaul Jul 19 '12 at 2:37
I consider the question sufficiently on-topic to stay here if you want (bonds are quantum mechanics and all that), but I am with @Colin in thinking that you might get better answers on Chemistry. If you want it migrated just flag or comment and we'll ask the Chem mods. – dmckee Jul 19 '12 at 3:28
The offer to ask the Chemistry mods if they want this remains open...but I won't migrate it to a beta site without your say-so. – dmckee Jul 21 '12 at 15:30

The dihedral angle is only 90º in solid hydrogen peroxide. In the liquid or gas it's nearer 120º.

Steric factors would obviously prefer the dihedral angle to be 180º, i.e. the molecule would be flat, but the interaction of the two O-H bonds would prefer a 90º angle. You end up with a compromise of about 120º. In the crystal you get many body interactions that complicate matters.

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Why do the two OH bonds want 90 degrees? How do you get intuition for this? – Ron Maimon Jul 19 '12 at 7:27
I'm not sure there is an intuitive answer. The Wikipedia article claims it's an interaction between the lowest unoccupied molecular ($\pi^*$) orbitals on the two OH bonds, but this strikes me as arm waving. Experience suggests that arm waving arguments are dangerous because you can rationalise pretty much anything if you try hard enough. – John Rennie Jul 19 '12 at 7:42
That's why bad arm-waving arguments are bad. But good arm-waving is indispensible. You can tell it works when you get correct predictions for things you don't know in advance. The protons are stuck on to a slight bump in electron density on one side of each O, and there is a restriction of orthogonality in the region where the electrons overlap on the two O's. The orthogonality might be correlating the e's, but it also might be electron electron interaction. I don't know. Nobody simulated this? It's simple quantum chemistry today. – Ron Maimon Jul 19 '12 at 16:32
As you say, it's easy to calculate a wavefunction for hydrogen peroxide. I was doing similar calculations in the early eighties! When I Googled for background on the question I did find several papers calculating the energy as a function of bond angle, and they end up with a minimum at about 112°. But these numerical calculations don't give you a good feel for why that angle is the optimum. You can probably get software for doing these calculations free - if I can find the time it might be interesting to try the calculation for myself. – John Rennie Jul 19 '12 at 16:38
I agree that if you use canned software you won't get any intuition, but if it's a form of numerical exact diagonalization (perhaps feasable for these small atoms, not sure), then you can look at the states directly and try to figure out if it's e-e repulsion or exclusion forcing the angles. – Ron Maimon Jul 19 '12 at 17:15

According to J. Chem. Theory Comput. 1 (2005) 394, it seems it's mainly the combination of hyperconjugation and repulsion between the lone pairs that favours a minimum at around 120°.

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