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If two hydrogen atoms are far apart, they have no effect on one another. But as they are bought closer together, they begin to excerpt an effect. The two nuclei, having the same positive charge, repel one another, and the two electron clouds also repel one another. However, most important of all is the attraction between the nucleus of one atom and the electron cloud of the other atom. As the atoms approach, the electron clouds are pulled toward the region between the nuclei (d). - Chemical Principles, Dickerson, Fourth Edition

First of all, how is it that the electron clouds do not repel each other (as they approach and do not yet have a binding molecular orbital)? Shouldn't (d) look more like this if the first attractive force is from London dispersion force (is it?):

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Is the attraction between the nucleus and the cloud only so much stronger because the nucleus is stationary and therefore the electrons repel each other much less than the nucleus attracts them due to the constant change in their residence probability?

In principle, two Xe atoms, for example, should first attract each other in this way, shouldn't they? The only difference is that they do not form a bond. I know that this behaviour is described by the Lennard Jones potential but I would just like an explanation for all this.

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    $\begingroup$ Consider that at low temperatures, noble gases do liquefy, i.e., condense because there is attraction between atoms. All, except for He, even form solids at low enough temperature, with atoms regularly aligned. (Willem Hendrik Keesom did solidify He, only under pressure of 2500 kPa.) $\endgroup$ Commented Dec 19, 2021 at 22:18
  • $\begingroup$ See: journal.csj.jp/doi/10.1246/bcsj.57.3062 $\endgroup$ Commented Dec 20, 2021 at 3:39
  • $\begingroup$ You make this far more complex than you need to. The issue is that there are several different forces involved, some attractive, some repulsive and the overall result is a combination of all of them. $\endgroup$
    – matt_black
    Commented Dec 20, 2021 at 10:18
  • $\begingroup$ See physics.stackexchange.com/q/684215/313612. $\endgroup$
    – Ed V
    Commented Dec 20, 2021 at 17:59

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I've written a lengthy answer to a similar question here. But in essence, I think you have the right idea with this part of your question:

First of all, how is it that the electron clouds do not repel each other (as they approach and do not yet have a binding molecular orbital)? (...) Is the attraction between the nucleus and the cloud only so much stronger because the nucleus is stationary and therefore the electrons repel each other much less than the nucleus attracts them due to the constant change in their residence probability?

The electrons on one atom feel more of the stationary nuclear charge of the other atom than its dispersed electron cloud. That's why you see the overall "inwards" polarization of the electron densities.

However, the text book you cite gets one detail slightly wrong:

However, most important of all is the attraction between the nucleus of one atom and the electron cloud of the other atom.

It's not the electron cloud of the other atom that a nucleus is attracted to, but its own shifted electron cloud.

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