Rank the following molecules in order of increasing London dispersion forces: \(\mathrm{CCl}_{4}, \mathrm{CI}_{4}\), and \(\mathrm{CBr}_{4}\).

Intermolecular forces are the attractions between molecules. These forces are crucial because they determine many physical properties such as boiling points and melting points. One of these forces is the London dispersion force, which is a type of van der Waals force. It is the weakest among intermolecular forces, but it is present in all molecules, regardless of whether they are polar or nonpolar. London dispersion forces occur due to temporary fluctuations in electron distribution within molecules. These fluctuations create brief, temporary dipoles that attract neighboring molecules.

The size of a molecule significantly influences its London dispersion forces. Larger molecules have more surface area, allowing more temporary dipoles to form, which in turn increases the attraction between molecules. In our exercise example, iodine (I) in \(\text{CI_4}\) has the largest atomic radius compared to bromine (Br) in \(\text{CBr_4}\) and chlorine (Cl) in \(\text{CCl_4}\). This means that \(\text{CI_4}\) has the largest molecule size, and therefore stronger London dispersion forces.

The number of electrons in a molecule also plays a vital role in determining the strength of London dispersion forces. More electrons mean greater electron cloud distortion possibilities, which enhance the temporary dipoles in the molecule. In the provided molecules, chlorine (Cl) has 17 electrons, bromine (Br) has 35 electrons, and iodine (I) has 53 electrons. Consequently, \(\text{CCl_4}\) will have the weakest London dispersion forces, followed by \(\text{CBr_4}\), and \(\text{CI_4}\) will have the strongest. This relationship between electron count and London dispersion forces directly explains why the molecules are ranked as \(\text{CCl_4}\) < \(\text{CBr_4}\) < \(\text{CI_4}\) in terms of increasing London dispersion forces.