Question

What is the range and magnitude of the Van der Waals force

Answer 1

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?

Answer 2

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

Answer 3

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.

Answer 4

Woops, the first sentence of the second para should read "As for why the dipole field itself falls off like 1/r^3..."

Answer 5

ok that is a real good answer i just need some pictures to visualize