Q.Explain why the bond dissociation enthalpy is anomalously lower than that of , even though fluorine is the smaller atom.
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Start your 14-day free trial to unlock the full solution →Ordinarily, a shorter bond between two smaller atoms is expected to be stronger (bond dissociation enthalpy generally correlates with bond length for a given bond order). By this logic, since fluorine is smaller than chlorine, one would predict 's bond to be stronger than 's. Experimentally, however, the opposite is observed: the F-F bond dissociation enthalpy is only about , distinctly lower than 's . The explanation lies in fluorine's very small atomic size: in , the two fluorine atoms (and, more specifically, their three non-bonding lone pairs each) are forced very close together across the short F-F bond, so the lone pairs on one fluorine atom experience significant electrostatic and exchange repulsion from the lone pairs on the other -- weakening the net bonding interaction. In the larger molecule, the two chlorine atoms (and their lone pairs) sit considerab …
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