# Why does chlorine gas have a higher boiling point than hydrogen iodide

Hydrogen iodide, $$\ce{HI}$$, is a dipolar molecule much larger than chlorine, $$\ce{Cl2}$$.

The melting point of $$\ce{HI}$$ $$(222.35\ \mathrm K)$$ is definitely higher than that of $$\ce{Cl2}$$ $$(171.6\ \mathrm K)$$. However, the boiling point of $$\ce{HI}$$ $$(237.79\ \mathrm K)$$ is less than that of $$\ce{Cl2}$$ $$(239.11\ \mathrm K)$$.

This is very confusing since $$\ce{HI}$$ has more London dispersion forces than $$\ce{Cl2}$$. It also has a dipole, which chlorine does not.

A similar question to this looked at how silicon tetrafluoride $$(\ce{SiF4})$$ had a higher melting point than sulfur tetrafluoride $$(\ce{SF4})$$, and the answer concluded that the exception happened because symmetrical molecules fit into lattices better than asymmetrical molecules. However, I am not sure if this applies to the situation between $$\ce{Cl2}$$ and $$\ce{HI}$$ because they both only contain two atoms. This situation is also in the opposite order, where the molecule with greater London dispersion forces and a dipole has a higher melting point but a lower boiling point.

With this in mind, can anyone explain why this exception between $$\ce{Cl2}$$ and $$\ce{HI}$$ exists?