# Why is it energetically favourable for molecular bonds to form from a QM point of view?

For example, if you have two hydrogen atoms and an oxygen atom, they are all electrically neutral and don't attract each other. But then if they manage to get "close enough" somehow they snap together releasing energy, and to get them apart again requires energy be input into the system.

This says that the bound state is a 'lower energy' state than the unbound state, and this is my question:

From a Quantum Mechanical point of view, where does this energy come from that is released when the bond is formed?

I've done some reading and it has something to do with filling electron shells and electron probability wavefunctions spreading between atoms, and something about the Virial Theorom where the kinetic energy of the electron is reduced as it's range of area is increased.

But that doesn't make sense to me, so I was wondering if someone could explain where the bond energy released comes from in terms of the electron wavefunctions? Assume high school physics/freshman maths.

I don't think it helps understanding much to bring the virial theorem in, but the thing about the range is increased is pretty much the key.

First, why the range is increased – well, there used to be one proton to swirl around and now there are two, easy.

Why this is energetically favourable is another matter. I'm afraid you'll have to look at the Schrödinger equation here: $$\bigl(E - V(\mathbf{x})\bigr) \cdot \psi(\mathbf{x}) \propto \tfrac{\partial^2}{\partial \mathbf{x}^2} \psi(\mathbf{x}).$$ Read this as: "it takes energy to bend the wave function". But the smaller you constrain your electron's range, the more you need to "bend" the wave function.

To make this concrete:

1. The wave function is a smooth function on all space.
2. It needs to become zero as you approach infinity.
3. It must be normalised: $\int_{\mathbb{R}^3}\mathrm{d}\mathbf{x}\:|\psi(\mathbf{x})|^2 = 1$. From a Born-probabilistic view this just means: the probability to find the electron anywhere in all space is $1$. (After all, it can hardly escape the universe, can it?)

To fullfill those points, you obviously can't use a wave function that's $0$ everywhere: it wouldn't integrate up to $1$. You need to have some kind of bulge around the proton(s). The integral is something similar to the "volume" of this bulge, if it were something like an ocean wave. So the wider you spread the bulge (i.e. the region in which the electron is typically found), the more shallow it can be. And a more shallow bulge requires less bending of the wave function! Hence it is a state with lower energy, i.e. energy is released as you increase the range of the electron.

• the other answer by @xuanji talks about a larger Δx decreasing the Δp in accordance with the uncertainty principle, and lower Δp means lower Kinetic Energy, is that what you mean by 'energy is released as you increase the range of the electron'? – matao Oct 4 '13 at 7:14
• @matao: you can see it both ways. Talking explicitly about the wavefunction is often considered somehow ugly, but IMO this is the clearest, least ambiguous way of discussing those issues. Talking about $\Delta x$ and $\Delta p$ values is equivalent to those arguments, but such clear distinctions (your classical intuition of kinetic and potential energy doesn't really help you much here) are troublesome in QM. – leftaroundabout Oct 4 '13 at 7:43
• I agree, talking in terms of the wavefunction is the best way to be clear about these things, I just find it hard sometimes to let go of my 'classical' approach while I still don't have the full QM toolbox to discuss things properly. So, back to your answer, it's something like: the amount of energy of the electron WF is proportional to the total curvature of the wavefunction? So spreading the WF out has less overall 'curvature', therefore less energy? – matao Oct 4 '13 at 8:51
• do you mind defining the terms in your Schrodinger equation as stated? I can take X as 'the position of the particle', but I can't see an equals sign, and I don't know what E and V stand for? thanks! – matao Oct 15 '13 at 7:41
• It's just the ordinary time-independent equation for a single non-relativistic particle, with the potential taken over to the left, and the prefactor $\frac{-\hbar^2}{2m}$ compacted into the proportionality sign $\propto$. – leftaroundabout Oct 15 '13 at 9:36

The energy changes come from changes in a) the electrostatic energy between the electrons and the protons as well as b) the reduced kinetic energy of the electron.

The electrostatic energy changes because the geometrical configuration of the proton and electrons are now different, while the electron KE changes because the electron is more delocalized (larger $\Delta x$) and hence has smaller momentum (smaller $\Delta p$, which implies smaller minimum $p$); this is the same reason why electrons can't approach arbitrarily close to the nucleus in an atom.

For $H_2$ at least, (a) is actually energetically unfavourable! It takes energy to concentrate the electron wavefunction into the internuclear region, and that energy comes from the reduced KE of the electron. (http://www.users.csbsju.edu/~frioux/26rio897.pdf)

• both of these answers are good! yours explains the energy source well in terms of the uncertainty principle. Can you comment at all on this energetic favourability with respect to how the orbitals fill up? ie the angular lobes. – matao Oct 4 '13 at 7:30
• Your last paragraph doesn't make much sense this way. If you invest enegy in constraining the electron "into the internuclear region", what you're doing is increase the kinetic energy, nothing else. – leftaroundabout Oct 4 '13 at 7:47
• Why would it increase the kinetic energy? The electron wavefunction is now spread over two protons and the internuclear region, doesn't that make it less localized? – xuanji Oct 4 '13 at 9:45
• Exactly, but you were talking about energy required to squeeze it into the internuclear region, which you don't need to do. – leftaroundabout Oct 5 '13 at 13:47