Is the bond dissociation enthalpy of Cl-F greater than that of Cl-Cl?
From this source, it can be seen that the $\ce {Cl-F}$ bond (BDE $\approx\pu{251 kJ/mol}$) is indeed stronger than the $\ce {Cl-Cl}$ bond (BDE $\approx\pu{243 kJ/mol}$). This can be explained by the partial ionic character of the $\ce {Cl-F}$ bond. The Pauling electronegativity value of fluorine ($\chi = 3.98$) is appreciably higher than that of chlorine ($\chi = 3.16$). Thus, there are significant partial positive and negative charges on the fluorine and chlorine atoms in the $\ce {ClF}$ molecule. Although the difference in orbital energies and size would mean that the covalent interaction between the chlorine $\ce {3p}$ and fluorine $\ce {2p}$ orbital would be less strong compared to that between 2 chlorine $\ce {3p}$ orbitals, the ionic character of the bond overcompensates for this and in fact, results in the $\ce {Cl-F}$ bond being stronger than the $\ce {Cl-Cl}$ bond.