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Interatomic potential is usually described as consisting of two parts: one attractive and one repulsive, where the repulsive one prevails at short distances.

The repulsive part is typically explained in terms of the Pauli repulsion principle, by saying that in order to allow for the electron clouds of the two atoms to overlap, electrons from filled subshells will have to be promoted to higher levels, which costs energy.

However, in the case of two hydrogen atoms, there are no filled subshells. First of all, for two hydrogen atoms where the electrons are in different spin states, there is not going to be any Pauli repulsion, so the repulsion must have a different origin. Of course there is some electrostatic repulsion between the electrons, but at very small distances I guess it's the repulsion between the nuclei that prevails.

Consider now the case where the two electrons have the same spin state. One way to overcome Pauli repulsion is then of course the promotion of one electron from 1s to 2s. Perhaps this promotion should be viewed in terms of wave functions for Helium? However, would it not also be possible to instead alter the spin state of one electron? If they are both in "spin up" state, actually this would release energy. I know such a transition is forbidden under E1 (although allowed under e.g. M1) but since we consider two interacting atoms, I reckon the transition could occur through collision.

If the two electrons are both in "spin down" perhaps its more complicated. Could one electron be promoted to "spin up" (still in the 1s orbital) by collision between the atoms?

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  • $\begingroup$ Note that the two electrons are happy in their individual orbitals around different protons. Trying to overlap the two 'electron clouds' as the two nuclei approach each other is not as simple as trying to put the two electrons in to energy levels around just one proton. $\endgroup$ – Jon Custer Mar 23 '17 at 13:07
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If you're asking about the principal contribution to the repulsive part of the interaction between two hydrogen atoms as they get close together, it's easy: it's almost exclusively driven by the electrostatic repulsion between the two protons. When the protons are closer together than one angstrom, there is very little electron charge between the two protons, and the repulsion between them dominates.

At equilibrium, and in the electronic ground state, the two electrons both go into the $\sigma_g$ molecular orbital, which is concentrated between the two protons. There is some amount of electrostatic repulsion between the two electrons, but this mostly serves to keep some of the charge on the far ends of the protons. As far as the protons go, though, there is this large cloud with between $-e$ and $-2e$ of charge sitting between them, shielding the interactions between them into equilibrium.

The equilibrium is found at the point where, if you push the two protons closer together, they 'squeeze' the electron density out of the middle in such a way that it's no longer able to screen their mutual repulsion.

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  • $\begingroup$ Thanks! Is it true however that repulsion between atoms that do have filled shells (i.e. any atom in its ground state apart from Hydrogen) is mainly due to Pauli repulsion? Could you perhaps also address my question regarding the spin, i.e. whether it could change for one of the electrons due "collision" between the atoms. The question was prompted by a discussion in Kittel's book on solid state physics, where he seems to describe the repulsion between two hydrogen atoms with both electrons in the same spin state as due to the energy required in He to promote one electron from 1s to 2s. $\endgroup$ – Étienne Bézout Mar 23 '17 at 13:22

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