# How do we account for the volatility of the hydrogen halide series?

Jun 4, 2017

$H F$ is MORE involatile than its lower group congeners.......

#### Explanation:

Which is the same thing as saying that $H C l$, and $H B r$, and $H I$, have LOWER boiling points than that of $H F$.

AS a chemist, as a physical scientist, it is your responsibility to assess data not remember them, and we list the NORMAL boiling points as follows:

WHY are the boiling points of $H F$ and ${H}_{2} O$ so disproportionately high? The answer is intermolecular hydrogen bonding. That is the electronegative fluorine atom polarizes electron density towards itself to give a resultant molecular dipole, i.e. a separation of charge: ""^(-delta)F-H^(delta+).

In the condensed phase, these dipoles line up to give a strong contributor to intermolecular force: ""^(-delta)F-H^(delta+)*F-H*F-H etc.

And thus $H F$ is the LEAST VOLATILE hydrogen halide.

Note that hydrogen bonding also occurs for $H C l$, and $H I$, and also ${H}_{2} S$, but because the dipole is of smaller magnitude, hydrogen-bonding does not make such a contribution to intermolecular force. The boiling points here follow the order of dispersion force, which increase with the number of electrons, and thus with the atomic number of the halogen/chalcogen.