Average bond enthalpy: the energy required for 1 mole of a specific gaseous bonds to undergo homolytic fission

Why specifically homolytic fission? Is it to allow comparison, but if that were the case, couldn't heterolytic also be used?

  • 2
    $\begingroup$ I would think homolytic because homolytic fission returns the two atoms as they would be in neutral state, prior to forming the covalent bond. Heterolytic fission could result in ionic species. Ionic species didn't originally form the bond (you assume) so you use homolytic fission. $\endgroup$ Mar 26, 2020 at 12:09
  • $\begingroup$ @arevmelikyan This makes sense. So a greater energy would be required for heterloytic fission? $\endgroup$
    – XXb8
    Mar 26, 2020 at 12:12
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    $\begingroup$ Not necessarily, its just that using homolytic fission would provide a better indicator of the average bond enthalpy, as I have said. But yes, usually it would require more energy to separate two oppositely charged ions. $\endgroup$ Mar 26, 2020 at 12:34
  • $\begingroup$ Try to think the other way round. If you would like to determine the strength of a bond like H-H, would you start form one H+ and one H- ion, or from 2 H· atoms ? $\endgroup$
    – Maurice
    Mar 26, 2020 at 15:31
  • $\begingroup$ I would guess that it's because that's how the the bond dissociation energy is defined: goldbook.iupac.org/terms/view/B00699 $\endgroup$
    – Buck Thorn
    Mar 26, 2020 at 20:35


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