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Is there any range of distance between the nuclei of the atoms (in Angstrom, say between 4 to 12 Angstrom) within which there will be an occurrence of Van der Waals force (attraction) between them?
For example, say within a distance of 4 A between two atoms of same type, there will van der Waals attraction present. Can such a certainty be obtained? If two atoms are very far apart, then there will be not Van der Waals attraction between them, this is known. So my question is how close should the nuclei of two atoms be so that one can say with certainty that there will be Van der Waals attraction?

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Van der Waals force is due to interaction of two dipoles. It is never strictly zero, but it diminish very fast with distance.

It is common to look at VdW interaction at Van der Waals radius. The largest radius should be around 2.60 for Fr making the total distance between nuclei more than 4 A. Heavier elements will have larger radius, but heavier elements are not stable. Are we really interested in VdW force between two atoms that will decay in 10^-15 sec?

Also, by "4-12A" do you actually mean distance between nuclei or the surfaces of atoms (VdW radius A + VdW radius B + "4-12 Angstrom")? In the first case iodine has VdW = 2.15A, so nuclei interact on distances more than 4A. This is the distance between atoms of neighboring I2 molecules in the crystal. In the second case you will see some interaction, but it will be quite small.

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  • $\begingroup$ I meant nuclei of atom .... If it is surface of atom, how does it change $\endgroup$ – girl101 Jul 19 '16 at 12:36
  • $\begingroup$ okay so I can understand that each type of atom has their own VdW radius. If two atoms intersect within the radius then there will be force, else no force. Am I correct? $\endgroup$ – girl101 Jul 19 '16 at 12:39

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