Compare the melting points of hydrogen fluoride (HF) and hydrogen iodide (HI).

I know the following things:

  1. $\ce{HF}$ melts at $189.6~\mathrm{K}$ and $\ce{HI}$ at $222.35~\mathrm{K}$.

  2. There are a lot of factors that have to be considered while comparing melting point, i.e., lattice enthalpy, packing fraction, intermolecular forces.

  3. Generally, compounds that show hydrogen bonding show it in the solid and liquid phases because in the gaseous phase the molecules have enough energy to break the electrostatic interactions of hydrogen bonding.
    So, HF shows hydrogen bonding in solid state. In fact, it even shows hydrogen bonding in gaseous state, though its not relevant here.

  4. As molecular mass increases, van der Waals interactions between molecules also increase. If we only consider this factor, then melting point of HI must be definitely higher.

  5. The strength of hydrogen bonding of HF is so much that the boiling point of $\ce{HF}>\ce{HI}$ although boiling point of $\ce{SbH3}>\ce{NH3}$, i.e. the hydrogen bonding dominates the van der Waals interactions in HI molecules because of their high molecular mass.

  6. It's the combined effect of all the factors that determine the melting point of a substance.

  7. Lattice energy is inversely related to the internuclear distance, it is also inversely proportional to the size of the ions. So, I expect HF to have more lattice energy.

Every factor that I can think of is in favour of HF except molecular mass. But if molecular mass of iodine is the reason that melting point of HI is higher, than boiling point of HI must have also been higher. Then why is the melting point of HI higher than that of HF?

Why is this contradiction?

Do let me know if I am wrong in the above mentioned facts or if I have any gaps in my understanding of the above points.


2 Answers 2


The increase in dispersive attraction between HI molecules is greater than the loss of dipole interactions. Only in $\ce{H2O}$ do dipole forces dominate (Intermolecular and Surface Forces, Jacob N. Israelachvili). Contribution of dispersion to intermolecular bonding for various molecules:

  • $\ce{CH4-CH4}$ - 100%
  • $\ce{HI-HI}$ - 99%
  • $\ce{HBr-HBr}$ - 96%
  • $\ce{HCl-HI}$ - 96%
  • $\ce{H2O-CH4}$ - 87%
  • $\ce{HCl-HCl}$ - 86%
  • $\ce{CH3Cl-CH3Cl}$ - 68%
  • $\ce{NH3-NH3}$ - 57%
  • $\ce{H2O-H2O}$ - 24%

What makes the contribution of dispersion so great is that it's non-directional and always attractive. Think of atoms as sticky spheres.

  • 1
    $\begingroup$ Why H20 serves as an exception? $\endgroup$ Mar 31, 2021 at 11:43
  • 1
    $\begingroup$ 1:1 ratio of H-bond donors and acceptors means much stronger attractive forces and with H and O being small atoms, the dispersive forces are weak. $\endgroup$
    – gsurfer04
    Apr 1, 2021 at 12:16

Like you said many factors contribute to melting points. It is an interesting trend, that melting points increase moving down while boiling points increase. I'd be interested to know how the data was measured.

One thing you did not take into account is polarizability of the larger halides which will greatly influence inter and intra molecular forces. If I can think of a better answer I'll post.


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