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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?

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