Do London dispersion forces affect boiling point?

Answer 1

London dispersion forces affect boiling point. For example, larger atoms have stronger London dispersion forces affecting them, thus holding them together stronger, increasing the energy required to pull them apart and thus the boiling temperature. So, for example, the boiling points of ideal gazes increase with size: helium: -269 degrees C, neon: -246 degrees C, argon: -186 degrees C, etc, up to radon: -62 degrees C (note that the temperatures are below zero, so -62 is actually greater than -269 degrees C).

Similar principle applies to molecules, such as CHCl3 and CCl4. While one might expect CHCl3 to have greater boiling temperature than CCl4 because it is polar, and has a permanent dipole, while CCl4 is symmetric and does not contain a permanent dipole, both molecules have approximately equal boiling points with CCl4 slightly GREATER than CHCl3. This is because CCl4 has more electrons around it because of the extra chloride atom, so that the induced dipoles are strong, and London dispersion forces holding the molecules together are also strong.

For more information about London Dispersion Force, check http://en.wikipedia.org/wiki/Van_der_Waals_force

Answer 2

London and dipole forces are intermolecular forces or forces that hold separate molecules together. The stronger the forces the more energy that is required to separate the molecules which is what happens when you change phase from a liquid to a gas. To supply more energy, higher temperatures are required. So, stronger forces cause higher boiling points. London forces tend to increase with increasing molecular weight but if dipole forces are present they will dominate the interaction since dipole forces are stronger than London forces.