Explain the bond dissociation energy?

Fluorine has a very small size. It's the smallest atom in the 2nd period. As a result of this, there is large repulsion between the nuclei of fluorine. So, despite the smaller bond length, it's easier to break the bonds of an F2 molecule. So this is why fluorine has a smaller bond dissociation enthalpy. Haha, sorry, I missed out another thing. It's not just nuclear repulsion but repulsion of electrons in a p orbital that has a definite shape and is comparatively small. Having a large electron density in a small orbital causes more repulsion. Hope this helps. :D

the bond dissociation energy in fluorine is low because of its very small size due to which there is strong electron-electron repulsion, moreover it has 2p orbital contrary to the other halogens who have d-orbital also. It is because of this low bond dissociation enthalpy that fluorine is very reactive.

F(a):F(b)→z (by MO convention) the singly occupied 2pz AO of F(a)(↑) overlaps with the corresponding singly occupied 2pz AO of F(b)(↓) to give the F:F(↓↑) σ bond. The bond is contracted due to the unique high electronegativity of F. In order to achieve this overlap the F(a) 2px e⁻ pair interacts with the the F(b) 2px e⁻ pair causing repulsion that weakens the σ bond; the same is true for the 2py e⁻ pair AOs. The situation is a trade-off. For the other halogens, X:X, the same σ bonding takes place but now because the bond is not contrated and little lp-lp repulsion takes place. (The d AOs are core e⁻s and make no significant contribution to X-X bonding). The anomalously weak F-F bond is the primary (but not the only) reason for the reactivity of F2. As has been mention the e⁻ affinity for Cl is > than that for F. Now I have fooled you with the science. Let me just say that I don't believe this is the correct reason but I am saving my explanation for another day.

F(a):F(b)→z (by MO convention) the singly occupied 2pz AO of F(a)(↑) overlaps with the corresponding singly occupied 2pz AO of F(b)(↓) to give the F:F(↓↑) σ bond. The bond is contracted due to the unique high electronegativity of F. In order to achieve this overlap the F(a) 2px e⁻ pair interacts with the the F(b) 2px e⁻ pair causing repulsion that weakens the σ bond; the same is true for the 2py e⁻ pair AOs. The situation is a trade-off. For the other halogens, X:X, the same σ bonding takes place but now because the bond is not contrated and little lp-lp repulsion takes place. (The d AOs are core e⁻s and make no significant contribution to X-X bonding). The anomalously weak F-F bond is the primary (but not the only) reason for the reactivity of F2. As has been mention the e⁻ affinity for Cl is > than that for F. Now I have fooled you with the science. Let me just say that I don't believe this is the correct reason but I am saving my explanation for another day.

the bond dissociation energy in fluorine is low because of its very small size due to which there is strong electron-electron repulsion, moreover it has 2p orbital contrary to the other halogens who have d-orbital also. It is because of this low bond dissociation enthalpy that fluorine is very reactive.