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?

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    $\begingroup$ It certainly won't. What you can do is to use the hybridization model to describe the bonding in F-F. The bond is surely not made only by pure p-p overlap. $\endgroup$ – EJC Nov 28 '15 at 13:38
  • $\begingroup$ @Marko Yes it is … $\endgroup$ – Jan Nov 29 '15 at 12:57
  • $\begingroup$ There is contribution from 2sigma and 1pi(u) MOs to the F-F bond: image.slidesharecdn.com/… $\endgroup$ – EJC Nov 29 '15 at 13:35

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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    $\begingroup$ There are so many papers trying to explain that bond, there is no ultimate truth, and I don't think there is a general consensus. I would expect at least some contribution from the s orbitals, as the in-axis lone pair will definitely have to be polarised away from the bonding pair. (In pictorial terms of MO theory.) If I recall correctly, F2 is used as a point in case example for the charge shift bond. $\endgroup$ – Martin - マーチン Sep 8 '19 at 20:42

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