Question

Does $CC{l_4}$ have a higher boiling point than $C{H_2}C{l_2}$? I thought London forces were weaker than dipole-dipole forces?

Answer 1

We need to know that comparing $CC{l_4}$ and $CHC{l_3}$, $CC{l_4}$ has a nicer / more symmetrical shape. Therefore, we can expect its packing to be more compact. This would mean that in a sample of $CC{l_4}$, there would arguably be a greater surface area of contact between two molecules of $CC{l_4}$.

Complete answer:
We need to know that the Tetrachloromethane ($CC{l_4}$) consists of non-polar molecules interacting via dispersion forces, whereas trichloromethane ($CHC{l_3}$) consists of polar molecules interacting via permanent dipole-permanent dipole (pd-pd) interactions.
To answer this question simply, $CC{l_4}$ has a higher boiling point than $CHC{l_3}$ because dispersion forces in $CC{l_4}$ is extensive enough to be stronger than pd-pd interactions in $CHC{l_3}$.
Some of the factors that affect overall strength of intermolecular forces are listed below:
Strength of each intermolecular interaction (I.e. what textbooks say about one hydrogen bond > one pd-pd interaction > one dispersion force)
Extensiveness of intermolecular interactions.
Thermodynamic changes such as entropy (Explained in detail in some of the other responses)Etc
A greater surface area of contact would then allow for the formation of more extensive intermolecular interactions.
So in $CC{l_4}$, even though the strength of each intermolecular interaction is weaker compared to $CHC{l_3}$, the extensiveness of intermolecular interaction in $CC{l_4}$ far exceeds that in $CHC{l_3}$ such that the overall strength of intermolecular interactions in $CC{l_4}$ is stronger than that in $CHC{l_3}$.

Note:
We have to know that the $C{H_2}C{l_2}$ boils at a lower temperature because its increase in entropy is larger, and $CC{l_4}$ boils at a higher temperature because its increase in entropy is lower. But also, it requires less thermal energy to boil $C{H_2}C{l_2}$.