Are there any conditions that'll improve the interaction? Just like, I assume, hydrogen bonds are strongest when the difference in electronegativity is biggest.


2 Answers 2


An electric dipole is formed in a molecule as the vectorial sum of all the partial charges on the atoms multiplied by their separation. Thus if, these charges are large and well separated and not symmetrically situated about the molecule the dipole will be large. For example chloro-benzene has a far larger dipole than toluene. However, 1,4-dichlorbenzene has zero dipole as the charges cancel out. (Molecular point groups can be used to decide if a molecule has a dipole, but not its magnitude. Only molecules in groups $C_1, C_s,C_n,C_{nv}, n>1$ have dipoles). Hydrogen bonds have no special significance with respect to dipoles.

Dipoles interact through space, and the strength of their interaction depends on the distance between the dipoles, relative orientation and relative permittivity (dielectric constant) of the intervening medium. Each of these have a dramatic effect for any given pair of dipoles.

The interaction energy E for dipole vector $\mu_1, \mu_2$ at separation R and where $\bar R$ is the vector joining the centres of the dipoles is

$$E = \frac{1}{4\pi\epsilon_0\epsilon} \left( \frac{\mu_1\cdot\mu_2}{R^3} - 3\frac{(\mu_1\cdot \bar R)(\mu_2\cdot \bar R)}{R^5} \right) $$

The dot product between two vectors e.g. $\mu_1\cdot \bar R = |\mu_1||\bar R|\cos(\theta_1)$ where $\theta_1$ is the angle between the vectors, see figure.

If the vectors are now represented as unit vectors e.g. $\bar \mu_1=\bar u_1 d_1$ where d is the magnitude of the dipole, and $\bar R = \bar u_R R$ then the energy becomes $$E = \frac{K}{4\pi\epsilon_0\epsilon}\frac{d_1d_2}{R^3} $$ where $$K= \bar u_1\cdot \bar u_2 -3(\bar u_1 \cdot \bar u_R)(\bar u_2 \cdot \bar u_R) $$

The figure shows the geometry.


If the dot product $\bar \mu_1 \cdot \bar \mu_2 =0 $ and one of the two dot products with $\bar u_R $ are $0$ then the energy is zero, and this can occur when dipoles are at 90 degrees and one points to the centre of the other. Physically this makes sense as each end of a dipole sees equal positive and negative charges on the other dipole.


To some degree, hydrogen bonding can be thought of as a subset or type of dipole-dipole interaction. This Wikipedia article defines dipole-dipole interactions as:

Dipole-dipole interactions are electrostatic interactions between permanent dipoles in molecules. These interactions tend to align the molecules to increase attraction (reducing potential energy).

The same article states, regarding hydrogen bonding:

The hydrogen bond is often described as a strong electrostatic dipole-dipole interaction. However, it also has some features of covalent bonding.

So, the assumption/example in your question is generally correct: just as a large difference in electronegativity of two atoms within a hydrogen bonding molecule results in stronger hydrogen bonds, the same tends to hold for non-hydrogen bonding dipole-dipole interactions. And while there are other forces affecting dipole-dipole interactions (i.e. solvent effects, molecular geometry/steric effects, etc.) it is the strength of the permanent dipole moment resulting primarily from the bond length and the difference in electronegativity between the atoms at the permanent dipole location that determine how strong the intermolecular dipole-dipole interactions will be.


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