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

Hint: 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 \[C{H_2}C{l_2}\].