Question

Why does the London dispersion/ van der Waals force exist? Surely the quantum jitteriness of the electron probability cloud will just as often induce a dipole that causes the atoms to repel as attract?

Answer 1

No, they won't. Think about this. The state where a momentary dipole has induced a dipole on something else has a lower energy than states where the dipoles repel, or there are no dipoles at all.

It is a basic characteristic of nature that states of lower energy occur more frequently than those of higher energy. So states in which these dipoles exist and attract each other are slightly more probable.

Answer 2

Also, quantum mechanics has nothing to do with this. Dispersion forces are fully described by classical mechanics.

Answer 3

Sorry, there was a misapprehension on my part- I thought it was completely random. So a dipole is more statistically favourable (or with a lower potential energy) than a non-dipole because of Coulomb attraction between opposites?

Answer 4

Correct. If the atom is all alone, then it will be unpolarized on average. Momentary dipoles will appear all the time, of course, but they will point in random directions, and average out to zero.

But as soon as the atom comes near another atom, the symmetry is broken, and you can no longer expect the dipoles to appear in random directions. They will appear more often in the direction that lowers the energy of the entire system. That inevitably leads to correlated dipoles, and a net attraction -- i.e. London forces.

The essence of the origin of these forces, however, is that states of a system with lower energy occur more frequently than those of higher energy. It's not often explained that way in beginning chemistry classes, because this is a statement from statistical mechanics, which is a college junior subject.