What is the range and magnitude of the Van der Waals force
Stacey Warren - Expert brainly.com
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If you mean dispersion (London) forces, they fall off like 1/r^6, and the magnitude is directly proporional to the number of electrons in the interacting molecules and the polarizability of the electron orbitals. Does that help?
yeah the london dispersion forces one of the Van der Waals forces,
what i really want to know is how come it is 1/r^6, not 1/r^2.
i cant visualize where the 1/r^6 is coming from
It's a dipole-induced dipole force. The field created by the instantaneous dipole on one molecules falls of as 1/r^3, that being the nature of an electric dipole field. That means the strength of the induced dipole on the other molecule goes like 1/r^3. Then the field of the induced dipole falls off like 1/r^3 on its way back to the original molecule, so to speak. So that gives you two factors of 1/r^3, for a total of 1/r^6.
As for why the dipole field itself falls off like 1/r^2: because the Coulomb field itself falls off like 1/r^2, and the only way you can tell a dipole field exists is if you are close enough to "see" both charges (otherwise it looks like a net charge of zero, from far enough away). So that gives you an extra factor of d/r, where d = effective charge separation, because as d/r -> 0 you are too far away to "see" the charge separation.