# Why does chlorine have a higher boiling point than hydrogen chloride?

Chlorine has a boiling point of $238~\mathrm{K}$ while hydrogen chloride has a boiling point of $188~\mathrm{K}$.

Hydrogen chloride has dipole-dipole forces so I would expect it to have greater inter-molecular forces and thus a higher boiling point. However, since this is not the case, this must mean that chlorine has a greater intermolecular forces due to London dispersion forces.

But why? I thought London dispersion forces were miniscule and only had a notable effect in large molecules. Shouldn't the dipole-dipole forces be way stronger than the London dispersion forces?

• You also have to consider that a diatomic chlorine molecule is almost twice as heavy as a hydrogen chloride molecule. – A.K. Jan 9 '16 at 5:22
• @A.K. I don't think molecular mass is directly correlated to boiling point, even if it is often mentioned. – Nicolau Saker Neto Jan 9 '16 at 13:07
• @NicolauSakerNeto Molar mass does affect boiling point, though its not the only factor. Heavy water melts and boils at higher temperatures than regular water. – A.K. Jan 9 '16 at 14:51
• @A.K. There always seems to be a better alternate explanation though. For example, a higher boiling point in water isotopologues may be better explained by a reduced zero-point energy in the intermolecular hydrogen bonds. This seems to be supported by the fact that $\ce{D2 ^{16}O}$ and $\ce{H2 ^{18}O}$ have essentially the same mass, but their boiling points are 101.42 ºC and 100.15 ºC respectively; the increased effect of hydrogen replacement is consistent with a larger decrease in bond vibration ZPE. Furthermore, $\ce{T2}$ and $\ce{H2}$ boil at 25.04 K and 20.39 K, a rather small difference. – Nicolau Saker Neto Jan 9 '16 at 21:52

This is because $\ce{Cl2}$ has close to double the mass and is also a larger molecule compared to $\ce{HCl}$. You were correct in saying that London dispersion forces are weaker but because of $\ce{Cl2}$s size they overcome the dipole-dipole forces in $\ce{HCl}$.