Explain the following order of bond dissociation enthalpies: `F-F lt Cl -Cl lt O = O lt N-=N`

Explanation

Since the size of F-atom is smaller than that of Cl-atom, the F-F bond length is less than the Cl-Cl bond length. Therefore, the F-F bonddissociation enthalpy is expected to be higher than that Cl-Cl bond dissociation enthalpy. But actually the reverse is true. in `F_(2)` molecule, each F-atom contains three lone pairs of electrons. because of their close procimity, these unshared electron pairs exert strong repulsive forces towards each other. similar forces of repulsion caused by the same number of lone pairs is much less in case of `Cl_(2)` because of larger size of Cl. so inspite of smaller bond length, the bond dissociation enthalpy of F-F bond is less than the Cl-Cl bond dissociation enthalpy. in `O_(2)` molecule, the two O-atoms are attached by a double bond. due to the presence of two unshared pairs of electron in each O-atom, the force of repulsion is relatively lower. so, O=O bond dissociation enthalpy is higher than Cl-Cl bond dissociation enthalpy. in `N_(2)` molecule, the two N-atoms are linked togetehr by a triple bond. each N-atom contains only one unshared electron pair which caused minimum force of repulsion. thus, the bond dissociation enthalpy of the `N-=N` bond is the highest. hence, the bond dissociation enthalpies of the bonds present in the given molecules follow the given sequence.