Intermolecular Interactions of Apolar Solutes

The apolar particles like methane possess no dipole moment, but there are still interactions between them, including the short-range repulsive force and the long-range attractive force. The attractive force is known as van der Waals interactions.

The interaction energy between two isolated apolar particles such as argon can be measured by molecular beam experiments. It varies with the separation as shown in Fig. 6. The interaction energy is zero at infinite distance. As the separation is reduced, the energy decreases, passing through a minimum at a distance of approximately $3.8$ Å for argon. The energy then rapidly increases as the separation decreases.

Figure 6: The interaction energy between two molecules.

The curve in Fig.6 comes from a balance between attractive and repulsive forces. The attraction between two neutral particles without permanent dipole moments results from the dispersion force, of which London used quantum mechanics to explain its physical origin [17]. An instantaneous dipole can arise from quantum fluctuation in the electron clouds. The instantaneous dipole in a molecule can in turn induce a fluctuating dipole in neighboring atoms, giving rise to an attractive inductive effect.

The van der Waals force goes as an inverse-sixth power. It can be understood as follows. As shown in Fig. 7, there is an instantaneous dipole $\vec{p}$ at the origin pointing along $z$ axis. The electric potential $U_{d}(r,\theta)$ at distance $r$ and angle $\theta$ is given by

Figure 7: The interaction energy between dipole-induced dipole moments.

$$U_{d}(r,\theta)=\frac{p\cos\theta}{4\pi\epsilon_{0}r^{2}}.$$

The electric field is then given by

$$E_{r}=-\frac{\partial U_{d}}{\partial r}=\frac{2p\cos\theta}{4\pi\epsilon_{0}r^{3}},$$

$$E_{\theta}=-\frac{1}{r}\frac{\partial U_{d}}{\partial\theta}=\frac{p\sin\theta}{4\pi\epsilon_{0}r^{3}},$$

$$E_{\phi}=-\frac{1}{r\sin\theta}\frac{\partial U_{d}}{\partial\phi}=0.$$

As a result,

$$\vec{E}_{d}=\frac{p}{4\pi\epsilon_{0}r^{3}}(2\cos\theta\hat{r}+\sin\theta\hat{\theta}).$$

The induced dipole of the molecule at $\vec{r}$ is given by

$$\vec{p}_{i}=\alpha\vec{E_{d}}$$

where $\alpha$ is the polarizability of the molecule. The interaction energy $U$ between the induced dipole and the dipole at the origin is

$$U=-\vec{p}_{i}\cdot\vec{E}_{d}=-\alpha E_{d}^{2}\propto\frac{1}{r^{6}}.$$

Therefore, the van der Waals interaction energy varies as $1/r^{6}$.

When the separation is shorter than $3$ Å, a small decrease of distance will causes a large increase in the energy, which has a quantum mechanical origin. It can be understood by Pauli's exclusion principle which prohibits two electrons in a system occupy the same quantum state. It happens only for two electrons of the same spin. The effect of exchanging is to reduce the electrostatic repulsion between pairs of electrons by forbidding them to occupy the same region of space (i.e. the internuclear region). Therefore, the short-range repulsive forces are often called exchange forces. The reduced electron density in the internuclear region leads to repulsion between the incompletely shield nuclei. Unfortunately, there is no simple theoretical formulation of the repulsion force in the classical force field.