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Every definition I see of Van der Waals interactions claim that they are short-range, nonspecific interactions between two chemical species. What does "nonspecific" mean in this context?

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I would assume that it means that all elements and molecules have them. I may be wrong though – Quantum MOCHACCINO Feb 24 '14 at 11:31
up vote 10 down vote accepted

The definition of the van der Waals force that I like (and apparently so too do the folks who have contributed to the Wikipedia article on the topic) is much more *specific about what is and is not a van der Waals force. This definition originates in the International Union of Pure & Applied Chemistry Compendium of Chemical Terminology

The van der Waals force is the sum of the attractive or repulsive forces between molecules (or between parts of the same molecule) other than those due to covalent bonds, the hydrogen bonds, or the electrostatic interaction of ions with one another or with neutral molecules or charged molecules.

Thus, a van der Waals force is a catch-all term to include all other very weak interactions between atoms and molecules:

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Your answer is not full, it does not answer why van der Waals force is "nonspecific" forsce. – saldenisov Feb 24 '14 at 12:05
It's nonspecific in that it is any force that is not one of those specific three forces. – Ben Norris Feb 24 '14 at 19:43
Don't forget chalcogen bonding ($\sigma$-hole bonding), $\pi$-hole bonding (snicker), pnictogen bonding, $n\rightarrow\pi$* interactions, hydrogen bonding... the list goes on and on. – LordStryker Feb 25 '14 at 17:21

There are 3 forces that could be named as van der Waals' force (see wikipedia article)

What is common between them? The common between them is that these interactions are electrostatic interactions between dipoles of different nature (permanent dipoles, induced dipoles). And electrostatic interaction is not specific: there is no space specificity, or chemical specificity.

An electrostatic interaction between dipoles according to Coulomb law decays as $$\sim 1/r^6$$, that is why van der Waals' forces (Keesom force,Debye force,London dispersion) force are short-range.

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I vaguely remember London and Debye forces proportional to $\frac{1}{r^6}$. – Klaus Warzecha Feb 24 '14 at 12:06
See here for an excellent description of the distance dependence of van der Waals interactions. – Nicolau Saker Neto Feb 24 '14 at 12:08
My mistake, I've corrected. – saldenisov Feb 24 '14 at 12:11

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