Identification of HOMO/LUMO in radicals

In summary, the conversation discusses the identification of HOMO and LUMO orbitals in radical molecules, specifically focusing on molecules with singly occupied orbitals. The majority of cases, such as O2, have two singly occupied antibonding orbitals that are considered the HOMO, with the antibonding sigma*2s being the LUMO. However, in molecules like NO where only one antibonding orbital is singly occupied, there is confusion on which orbital is considered the HOMO and LUMO. The conversation also mentions the asymmetry of energy levels in non-homoatomic bi-atomic radicals and how it can affect the filling-up of orbitals. The cited reference confirms the correctness of the energy diagrams
  • #1
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TL;DR Summary
How to identify HOMO/LUMO in molecules with singly occupied orbitals?
Hello,
as the title says, how do you treat singly occupied orbitals in radical molecules, when trying to identify the HOMO and LUMOs?

In the majority of cases I stumbled upon, like O2, both the antibonding orbitals, pi*2px pi*2py, are singly occupied, so they would be considered the HOMO, with the antibonding sigma*2s being the LUMO. Same goes for other radicals that have one fully occupied orbital (eg pi*2px) and one singly occupied (eg pi*2py).

But in molecules like NO, where only one of the two antibonding orbitals is singly occupied (eg, the pi*2px OR the pi*2py), what is considered the HOMO and which the LUMO? Are both the pi*2px and pi*2py considered HOMO even if only one of them is occupied (and only by 1 e), with the sigma*2s being the LUMO? Or both the pi*2px and pi*2py are considered HOMO and LUMO at the same time (is that even possible?) Even if we call the singly occupied orbitals as SOMO, what would be the corresponding LUMO?
 
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  • #3
Hyperfine said:
Molecular Orbital Diagram of NO
That is for the ground state.
Are the energy diagrams correct in the cited reference? For non-homoatomic bi-atomic radicals, the energy levels of the orbitals of the heavier atom are drawn downwards. That asymmetry is characteristic of such radicals. Whether that affects the filling-up of the orbitals, thus defining which is LUMO and which is HOMO, depends on the atoms involved. So, they are nearly leveled for NO, but quite prominent for HF.
 
  • #5
Hyperfine said:
Sorry to press on, but while this latter configuration agrees with my description, the former doesn't. In the first citation, no attention is paid to the difference in energy levels for the corresponding orbitals for N and O. It's obvious unless the text explains sth about it. I wouldn't like to further comment on that.
 
  • #6
I did not say that the diagrams were to scale. They present the molecular orbitals of the molecule, and that was the question originally posed.
 
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  • #7
Hyperfine said:
I did not say that the diagrams were to scale. They present the molecular orbitals of the molecule, and that was the question originally posed.
Sorry for the late reply, but my question was not how to draw the MO of NO. I wanted to know which one would be identified as HOMO and which as LUMO, considering that there's only 1 electron on the highest level but there's also a free orbital next to it
 

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