Fluorine boils at -188.1 °C and oxygen boils at -183 °C, but shouldn't $\ce{F2}$ boil after $\ce{O2}$?

Despite being electronegative elements, both are nonpolar molecules and posses dispersion forces as the only mean of intermolecular interactions. Yet, $\ce{F2}$ has more electrons than $\ce{O2}$, and should have stronger dispersion forces. Hence, $\ce{F2}$ should boil after $\ce{O2}$, but it doesn't. Why?

  • 1
    $\begingroup$ Obviously van der Waals forces are stronger for O2, perhaps because it is more polarizable than F2. $\endgroup$
    – Poutnik
    Apr 3, 2022 at 10:42
  • $\begingroup$ @Poutnik I didn't know that. Can you explain why? $\endgroup$
    – John Hon
    Apr 3, 2022 at 10:55
  • $\begingroup$ It is quite well known polarizability of fluorine and fluoride is very low, due its high electronegativity and small size. $\endgroup$
    – Poutnik
    Apr 3, 2022 at 10:57
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    $\begingroup$ although there are no stable O4 molecules in liquid oxygen, O2 molecules do tend to associate in pairs with antiparallel spins, forming transient O4 units. from tetraoxygen $\endgroup$
    – Poutnik
    Apr 3, 2022 at 11:16
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    $\begingroup$ Bonding in there is completely different - the two molecules are hardly trivial. There's no point in comparing them like they were two elements of homologous series. $\endgroup$
    – Mithoron
    Apr 3, 2022 at 12:21

1 Answer 1


It would be tempting to argue that fluorine is so electronegative and holds its electrons so tightly that their polarizability is reduced, thus so are the dispersion forces in $\ce{F2}$. But upon further review, this does not stand up. We would expect nitrogen, being less electronegative than oxygen, to offer more polarizability still, yet molecular nitrogen boils at a temperature lower than both oxygen and fluorine (-196 °C). The real question is why $\ce{O2}$ boils higher than both neighboring diatomic molecules, $\ce{N2}$ and $\ce{F2}$.

Putting two and two together

What really distinguishes oxygen from its neighbors is its existence as a diradical, which arises from the degeneracy of its partially filled π molecular orbitals. This creates the possibility of interaction between unpaired electrons from different molecules.

Such an interaction is described, in terms of the magnetic properties of liquid oxygen, in this answer. Pairs of oxygen molecules tend to have "sticky collisions" in which they are indeed temporarily dimerized through bonding between their unpaired electrons. This counts as an attractive interaction that selectively occurs upon condensation and it is just enough to favor condensation at a slightly higher temperature than both neighboring, non-radical diatomics.

When one (and one) is enough

Another comparison is possible with the monoxides of carbon and nitrogen, versus $\ce{O2}$ and each other. Both $\ce{CO}$ and $\ce{NO}$ are weakly dipolar as well as having dispersion forces, but $\ce{NO}$ has the potential to dimerize like $\ce{O2}$ whereas $\ce{CO}$ does not. Here is how boiling points compare in this series:

Oxide b.p./°C Wikipedia page
$\ce{CO}$ −192 en.wikipedia.org/wiki/Carbon_monoxide
$\ce{NO}$ −152 en.wikipedia.org/wiki/Nitric_oxide
$\ce{O2}$ −183 en.wikipedia.org/wiki/Oxygen

Nitric oxide combines a weak dipole with the dimerization capability between molecules with an unpaired electron, forming a dimer. The tendency of nitric oxide to dimerize, though weak, is stronger than that in $\ce{O2}$, so this interaction produces a larger effect on boiling point with the nitrogen oxide.


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