Which one of the following orders is correct for the bond dissociation enthalpy of halogen molecules?

Generally, as we move down the halogen group, the atomic size increases, which leads to longer and weaker bonds, so bond dissociation enthalpy should decrease . However, fluorine ($\text{F}_2$) is anomalous. Because of the extremely small size of the fluorine atom, the lone pairs of electrons on the two adjacent bonded atoms experience strong interelectronic repulsions , making the $\text{F-F}$ bond weaker than expected. As a result, the bond dissociation enthalpy of $\text{F}_2$ is lower than that of both $\text{Cl}_2$ and $\text{Br}_2$. According to the standard values , the bond dissociation enthalpies are roughly $\text{Cl}_2$ ($\approx 243$ kJ/mol) > $\text{Br}_2$ ($\approx 192$ kJ/mol) > $\text{F}_2$ ($\approx 155$ kJ/mol) > $\text{I}_2$ ($\approx 151$ kJ/mol). Therefore, the correct order is $\text{Cl}_2 > \text{Br}_2 > \text{F}_2 > \text{I}_2$.