# 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.
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.