# Do molecules with polar bond, but with no dipole moment experiences a greater effect from the london dispersion forces?

Do molecules that have polar bonds and a structure that makes the polarity of those bonds cancel each other out, experience a greater effect from the London dispersion force, than molecules with a similar size, bond length, shape, and mass, but with non-polar bonds?

Yes, an example of a molecule would be CO$$_2$$, which has no dipole moment but does have two polar bonds. Although CO$$_2$$ has no dipole moment, its charge distribution gives rise to a quadrupole moment. This quadrupole moment provides a handle for the molecule to interact with other particles (with other quadrupoles, but also with dipoles, induced dipoles or electric charges). As you might know, dipole-dipole interaction scales as $$r^{-3}$$ where $$r$$ is the distance between the particles and Van der Waals interaction (induced dipole-induced dipole interaction) scales as $$r^{-6}$$. It can be shown that quadrupole-quadrupole interaction scales as $$r^{-5}$$ and therefore represents a stronger interaction than the induced dipole-induced dipole interaction.

• What you say about distance is correct but there is also a scaling or proportionality term that is different for each type of interaction, this can radically change the size of the interaction at a given fixed distance. – porphyrin Nov 30 '18 at 9:38
• Agreed, but for similar sized molecules (as stated in the question) I expect the polarizability to be smaller than the quadrupole moment and, moreover, the quadrupole interaction adds to the the Van der Waals interaction. – Paul Nov 30 '18 at 12:04

Basically, you need dipole polarizability($$\alpha$$) and first ionization energy (I) of each atom, and distance between them (S). You can approximate dispersion interaction for atoms as $$E_{disp}=-1.5\frac{I_1*I_2}{I_1+I_2}*\frac{\alpha_1*\alpha_2}{S^6}$$ As you can see, I and $$\alpha$$ are properties of matter, but, for example, double bonds (which are shorter) can change dispersion interaction.

As for molecules, london dispersion forces are "main" forces for nonpolar molecules. Contribution of LDF to intramolecular energy of two methane molecules is 100%; of two water molecules - only 24%. If molecule is not small and highly polar, LDF contribution will be about 90-100%. In order to have more detailed result, use table of electronegativity as well as 3D-structure of given molecule and determine it's polarity. Actually, perfectly polar bonds as well as perfectly cancelled ionic bonds will give the same 100% LDF contribution, but there is nothing truly perfect in our world.

Also, if two molecules have perfectly "similar size, bond length, shape, and mass", they are probably two identical molecules.