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Imagine two atoms, and only consider the Van der Waals force. The electron cloud will jitter due to its quantum mechanical nature- some of these jitters forming dipoles, some not. However, on average they form dipoles. Why? This question is equivalent to asking (although it is useful here to ask in another way): why is the potential energy of a dipole lower than a non-dipole?

What determines which electrons are going to move to the middle of the atoms to form a dipole, and which ones move to the outside of an atom (I assume that a dipole is formed like this $e^-..ion...........e^-..ion$ for the centre of positive charge shifted rightwards of the centre of negative charge): given that the jitters are instantaneous, and photons travel only at the speed of light, how are dipoles formed quickly enough before the next jitter? Am I falling into a rut because I am not considering the wave nature of electrons?

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    $\begingroup$ Hi Alyosha! Your two recent questions are heavy with anthropomorphism (atoms or electrons “decide to do X”, etc.). I don't know how to phrase it exactly, and when I manage I'll try to make an answer to these questions, but I believe this view negatively impacts your understanding of these questions. $\endgroup$
    – F'x
    Oct 7, 2012 at 20:37
  • $\begingroup$ Thanks for that. Sometimes anthropomorphisms help in that they remove confusion when asking a question quickly by stressing the fact that I want to know WHY rather than be given a superficial explanation, so I suppose I've gotten into bad habits. I've tried to edit my posts as I agree that here they detracted heavily from the questions' posts. $\endgroup$
    – Meow
    Oct 7, 2012 at 21:00

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on average they form dipoles.

Not quite. On average they form the nice (electrically neutral) electron orbitals we know and love. However, when they happen to be in a dipole state, there is an attraction between the two atoms, and that is evidently more than enough to overcome the randomly occurring opposing dipole states. That's just mathematics.

why is the potential energy of a dipole lower than a non-dipole

It isn't. But quantum mechanics allows electrons to visit higher energy states, with a probability dependent on the energy difference.

What determines which electrons are going to move to the middle of the atoms to form a dipole, and which ones move to the outside of an atom

Nothing. The "movement" of electrons is just a random statistical fluctuation in the location of the electrons' centers of charge. Quantum-mechanical events like this are inherently nondeterministic. And strictly speaking, you can't tell electrons apart anyway. There is no "which electron," at one point in time you have one electron here and another there, the next they may have switched places. Or not, you can never know.

(I assume that a dipole is formed like this $e−..ion...........e−..ion$ for the centre of positive charge shifted rightwards of the centre of negative charge)

Don't think of those as being electrons, just the statistical average of the negative charge distribution.

given that the jitters are instantaneous, and photons travel only at the speed of light, how are dipoles formed quickly enough before the next jitter

First, They're not quite instantaneous (that would have them basically everywhere possible at once), and not quite jitters.

In summary, don't think of electrons as particles with a distinct position moving around, just think of them as waves, with a peak that occasionally forms, and that peak causes the electric arrangement of two nearby atoms to, on average, be mildly attractive.

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  • $\begingroup$ I still don't see why the particles will attract, because: 1- Given that the average state of the electrons is the neutral and non-attractive AND an attractive dipole is no more energetically favourable than a repulsive opposing dipole, can I conclude that the electron peaks are arranged so the atoms are attracting for as much time as they are repulsing? 2- If (1) is true, then it seems to be that the electron peaks are arranged in an attractive way attract more strongly than the repulsive arrangement repulses. I suppose all my former misunderstandings boil down to the above. $\endgroup$
    – Meow
    Oct 8, 2012 at 16:13
  • $\begingroup$ The atoms will only be repelled when the peaks of both electrons are between the two atoms. Anything else and they will be attracted. $\endgroup$
    – Kevin
    Oct 8, 2012 at 16:22
  • $\begingroup$ Also- I take it that this dycem-cc.com/imagedownloads/vdw.gif model of inducing a dipole is an erroneous simplification, then? $\endgroup$
    – Meow
    Oct 8, 2012 at 18:00
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''What determines which electrons are going to move to the middle of the atoms to form a dipole,...?''

Electrons are indistinguishable particles, so the question doesn't make sense. Electrons are excitations of the electron field (a quantum version of like wavelets on a lake), and their formation somewhere is just by concentrating the field at some place.

In some molecules such as benzene, some electrons don't even clump at all near particular locations but are strongly delocalized.

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  • $\begingroup$ I was sloppy with my wording, yes. $\endgroup$
    – Meow
    May 26, 2013 at 13:55

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