In DFT and Hartree-Fock bonding orbitals have negative energies while antibonding orbitals have positive energies. Why is this the case? It seems this is due to the zero-point energy being defined for bare nuclei and electrons at infinite separation. See:




Is this correct and are there any textbooks / peer reviewed references that state/explain this?

Does this all just derive from the binding energy?


1 Answer 1


The zero point of orbital energies is usually defined as the vacuum level, i.e. the energy of a hypothetical orbital that is infinitely diffuse and does not come close to any nuclei or other electron. Note that this is not quite the same thing as "bare nuclei and electrons at infinite separation" as mentioned in the linked answer. The latter is the zero point of the total energy, not the zero point of the orbital energy. Orbital energies are the energies of single electron wavefunctions, so referencing orbital energies against the energy of a multiparticle system does not make sense.

However, antibonding orbitals do not necessarily have positive energies! Very small molecules do frequently have positive antibonding orbital energies, and most textbooks mention small molecules only, which may lead to the false impression that this holds for larger molecules as well. Calculate the antibonding orbitals of moderately sized and electron deficient molecules, such as 2,4,6-trinitrotoluene, and you'll probably find that quite a few of the antibonding orbitals have negative energies, particularly if you use DFT instead of HF. For DFT calculations of large conjugated molecules, negative LUMO energies are probably the rule rather than the exception.

Similarly, bonding orbitals do not necessarily have negative energies, although exceptions are considerably rarer. A trivial example is a molecule surrounded by a sufficient number of negative ions. Negative ions raise both the bonding and the antibonding orbital energies due to their Coulomb potential, so with a sufficient number of negative ions, the bonding orbital of the molecule will be raised above vacuum level; it will then spontaneously ionize, but under favorable conditions the spontaneous ionization may be slow enough such that the unionized molecule is experimentally observable.

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    $\begingroup$ I would add that if one uses a complete basis set, the LUMO energy goes to zero since one can always add an electron at no cost to the non-interacting vacuum. $\endgroup$ Commented Nov 30, 2022 at 14:44

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