Aside from the general trend of the boiling points of the hydrides, I noticed how the group 15 hydrides somehow "overtook" the group 17 ones in boiling point, which felt weird.

chart of boiling pressures of hydrides

Notably, H-Br has an electronegativity difference of 0.7, which is enough to be considered polar, which to my understanding means HBr will exist with a permanent dipole. The As-H bond has an even smaller difference, so it can't experience these tiny permanent dipole attractions.

That probably means there's a big difference in London dispersion forces, but both molecules have 36 electrons, so they'd be similarly capable of forming instantaneous dipoles. Their molecule shape might affect their ability to induce dipoles, but I don't know where to start with that.

So if both AsH3 and HBr experience similar London forces, but HBr also experiences weak permanent dipole-dipole interactions, why does AsH3 ($\mathrm{-63^\circ C}$) have a slightly higher melting point than HBr ($\mathrm{-68^\circ C}$)? (The same could be said for SbH3 and HI, which has a slightly more significant difference in boiling point.)

  • 1
    $\begingroup$ What's the point of comparing apples with oranges? The molecules are simply different. But if you wonder about polarisability, AsH3 should have higher, simply because it's bigger. $\endgroup$
    – Mithoron
    Dec 12, 2023 at 22:25

1 Answer 1


The question behind the question is why do the hydrogen halides fall behind the other hydrogen compounds shown when we get to heavier periods?

A clue to the answer may be gleaned by completing each isoelectronic series with the noble gas: for example, $\ce{Kr}$ or $\ce{H0Kr}$ in period 4 and $\ce{Xe}$ or $\ce{H0Xe}$ in Period 5. In each case the noble gases are well below the other compounds in their respective series; in Period 4 for instance krypton weighs in at $-153°\mathrm{C}$ which is well below germane and the other Period 4 molecules.

As we move to bigger atoms with more electrons the predominant forces are dispersion forces and dipole-diploe interactions. The dipole-dipole interactions are absent for both the noble gases and the tetrahedral molecules produced by the Group 14 elements. Meanwhile the dispersion forces diminish as we move to later groups because more of the valence electrons are localized onto one atom in Group 17 and especially Group 18 (the noble gases) than in earlier groups.

As the dipole-dipole interactions fade in the heavier halides and they must rely more on dispersion forces, the lower dispersion forces compared with the more electron-delocalized polyhydrogen compounds (such as $\ce{HBr}$ versus $\ce{H2Te}$ and $\ce{H3As}$) lead to a drop is boiling point. This drop from Groups 15 and 16 to Group 17 is just the beginning of the huge drop going into the noble gases, where dipole-dipole interactions are completely gone and dispersion forces take a hit with all electrons stuck on one atom per gas particle.


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