Why is the F-F bond in fluorine a p-p sigma bond?

Won't the lone pairs hybridize to $\mathrm{sp^3}$ making it an $\mathrm{sp^3}$-$\mathrm{sp^3}$ $\sigma$ bond? If it still is $\mathrm{p^\sigma}$-$\mathrm{p^\sigma}$ bond what is the spatial location of the lone pairs?




Hybridisation is a mathematical concept chemists use to better understand spatial distribution of bonded atoms. It should *only* be used if we cannot explain the structure will without resorting to hybridisation.


The energy of a 2s orbital is *lower* than that of a 2p orbital in atoms with a fully populated 1s orbital. Therefore, there is a desire to keep its energy low. A bond between two p orbitals (of different atoms) will also lower the energy of *one of these p orbitals* (and create an antibonding σ\* unpopulated one, too).


Furthermore, p orbitals extend into space in a directed manner: Along their coordinate axis, while s-orbitals are spherical in shape. The p orbital is already in a nice direction for bonding; any mixing with s orbitals will make it extend into space slightly less — potentially unfavourable for orbital overlap.


Finally, the lone pairs on each fluorine atom will be a set of two p type orbitals, each carrying a lone pair perpendicular to the bond axis, and an s type orbital. Experimentally, one should be able to detect two different energy levels for the two different lone pairs. (However, I didn’t go looking for papers and I can’t remember any lecturer mentioning any off the top of my head.)


So the $\ce{F-F}$ bond is indeed a $\mathrm{p^\unicode[Times]{x3c3}\ce{-}p^\unicode[Times]{x3c3}}$ bond because it makes the overall system energetically favourable.


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