Diatomic halogens - bond strength
I see as I write the answer, that many comments are spot-on.

Whether or not $\ce{F2}$ has lone pairs depends on your view of molecular orbitals and valence bond (or Lewis) descriptions of bonding.

Covalent radii for fluorine compounds are usually around 0.64 Å for F, but yet the F-F bond distance is 1.43 Å.

If we go for the Lewis or valence bond view, we'd argue that the large number of lone pairs on a small atom will be highly repulsive.

If I go for the molecular orbital view, I'd point out that in reality, the anti bonding orbitals $\pi^*$ and $\sigma^*$ are higher in energy than their bonding counterparts, particularly in the 2nd row, so the F-F molecule will be weaker than a simple single bond.

There's a really nice article by Peter Politzer in JACS that I like to use when discussing this "anomaly": "Anomalous properties of fluorine" J. Am. Chem. Soc., 1969, 91 (23), pp 6235–6237


  It appears rather that the anomaly involves the fluorine atom, in whatever circumstances, and its interaction with an external electron which enters its outer shell. [...] In any case, it is suggested that because of the exceptionally small size of the fluorine atom (see Table I), the entering electron feels unusually large forces of attraction from the nucleus and repulsion from the electrons already associated with the atom. It appears that the magnitude of the anomaly in the repulsive forces is greater than in the attractive forces.


In other words, regardless of the MO or VB view, the issue is one of electrostatic repulsion.