London dispersion forces and distance relationship
London dispersion forces (or van der Waals) can be rationalized (classically) as a relationship between induced dipole moments.

Some electrostatic (Coulombic) relationships:


Two Point Charges: $V = \frac{q_1 q_2}{4\pi\epsilon_0 r}$ (i.e., $r^{-1}$)
Point Charge and Permanent Dipole: $V = \frac{\mu_1 q_2}{4\pi\epsilon_0 r^2}$ (i.e., $r^{-2}$)
Two Permanent Dipoles: $V = \frac{\mu_1 \mu_2}{4\pi\epsilon_0 r^3}(1 - 3 \cos^2 \theta)$ (i.e., $r^{-3}$)


The last assumes that the polar molecules are stationary with angle $\theta$ between the dipole moment vectors. If the molecules rotate then it falls off to $1/r^6$.

In general, the interaction of a $2^n$-pole (dipole, quadrupole, octupole, etc.) with $2^m$-pole is $V \propto \frac{1}{r^{n+m-1}}$.

So why the $1/r^6$ for dispersion? Here's a quick justification. Let's take the system of a permanent dipole that's creating a induced dipole moment in another molecule:

$\mu_2^* = \alpha_2 E$

where $\alpha_2$ is the polarizability in the second molecule and E is the electric field generated by the polar molecule:

$E = \mu \sqrt{1 + 3\cos^2 \theta} / 4\pi\epsilon_0r^3$

The potential energy is given by:

$V = \alpha_2 E^2 / 2$

$V \propto r^{-6}$

Since there are multiple dispersion interactions (two dipole moments averaged over all rotations, permanent dipole-induced dipole, hydrogen bonding, and induced dipole-induced dipole moment) the general potential energy is classically approximated as:

$V = \frac{C}{r^6}$