Why is the boiling point of fluorine lower than that of oxygen?

[Fluorine boils at -188.1 °C](https://www.rsc.org/periodic-table/element/9/fluorine) and [oxygen boils at -183 °C](https://www.rsc.org/periodic-table/element/8/oxygen), 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?




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](https://en.wikipedia.org/wiki/Nitrogen)). 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](https://chemistry.stackexchange.com/a/134713/17175). 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](https://en.wikipedia.org/wiki/Carbon_monoxide) |
| $\ce{NO}$ | −152 | [en.wikipedia.org/wiki/Nitric\_oxide](https://en.wikipedia.org/wiki/Nitric_oxide) |
| $\ce{O2}$ | −183 | [en.wikipedia.org/wiki/Oxygen](https://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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