We say that increasing the energy of the $\ce {HOMO}$ of molecules increases their reactivity towards electrophiles while decreasing the energy of the $\ce {LUMO}$ increases their reactivity towards nucleophiles. However, this seems to be built on the implicit assumption that all molecules will always have a $\ce {HOMO}$ that is lower in energy relative to all $\ce {LUMO}$s in every single other molecule. How can that be justified?

  • $\begingroup$ Have a look at en.wikipedia.org/wiki/Molecular_orbital_diagram, especially the bit mentioning the Aufbau principle, and then have a think ... $\endgroup$ – Ian Bush May 11 '19 at 12:31
  • $\begingroup$ @IanBush Please bear in mind that I am not referring to the HOMO and LUMO of a single molecule. An example to consider specifically: Why is the C-Pb $\sigma^*$ MO necessarily lower in energy than the C-F $\sigma^*$? $\endgroup$ – Tan Yong Boon May 11 '19 at 12:38
  • $\begingroup$ Haven't thought about this in many years, but in the combined system (when they are interacting) if the HOMO were higher in energy than the LUMO it wouldn't remain as the HOMO for very long (symmetry and very small hopping integral exclusions apply) - in which case your argument above still applies, and you need not make the assumption that you think is required. Hopefully somebody who has looked at this in the last 3 decades can point out where I am wrong. $\endgroup$ – Ian Bush May 11 '19 at 13:15
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    $\begingroup$ While HOMO and LUMO have an absolute meaning for each (neutral) molecule, "electronegative" and "nucleophilic" are relative concepts, they are categories used to classify molecules. So if I understand your question correctly, you are right. It is a built in assumption (based on a comparison of different molecules) that a nucleophile will be nucleophilic in the presence of just the right kind of partner, an electrophile. $\endgroup$ – Buck Thorn May 11 '19 at 13:27
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    $\begingroup$ Not every molecule, but every one that you would reasonably regard as a possible reaction partner. $\endgroup$ – Karl May 11 '19 at 21:00

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