Orbital angular momentum of O2

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Discussion Overview

The discussion revolves around the orbital angular momentum of the O2 molecule, particularly in relation to its paramagnetic behavior and the implications of unpaired electrons in degenerate pi* orbitals. Participants explore theoretical aspects of angular momentum in molecular and atomic contexts.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant notes that the paramagnetic behavior of O2 is attributed to two unpaired electrons in a degenerate pi* orbital, questioning the consideration of orbital angular momentum.
  • Another participant explains that in the ground state ^3Σg^-, the total orbital angular momentum of the occupied antibonding pi orbitals is zero, while in the excited state ^1Δg, the orbital angular momentum is 2 hbar.
  • A participant questions whether the combined angular momentum of degenerate states always adds to zero, referencing atomic configurations.
  • Another participant counters that according to Hund's rules, both spin (S) and orbital (L) angular momentum tend to be maximal in the ground state of an atom with degenerate shells.
  • One participant provides an example using nitrogen's electronic configuration, suggesting that maximizing S results in L being zero.
  • A later reply agrees that for main group elements with p orbitals, L cannot exceed 1 if S is to be maximal, and notes that Hund's rules are more complex for d or f shell elements.

Areas of Agreement / Disagreement

Participants express differing views on the relationship between combined angular momentum and degeneracy, with some asserting that it can be zero while others reference Hund's rules suggesting otherwise. The discussion remains unresolved regarding the implications of these principles in the context of O2.

Contextual Notes

Participants reference specific states and configurations, but the discussion does not resolve the implications of these configurations on angular momentum in molecular versus atomic contexts.

Pete99
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Hi,

I have read several texts where it says that the paramagnetic behavior of O2 molecule is due to two unpaired electrons in a degenerate pi* orbital.

I have not read, however, any comment about the orbital angular momentum of these pi* states.

Is there any reason why the orbital angular momentum is not considered? Is it zero? Why?

Thanks for any help!
 
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Thats an interesting question. In the ground state ^3\Sigma_g^-, the orbital angular momenta of the occupied antibonding pi orbitals combine to total orbital angular momentum 0 (hence the \Sigma). Classically, the two unpaired electron orbit around the molecular axis in different directions. In the lowest excited state (singlett oxygen)
^1\Delta_g the total spin of the electrons is zero but the orbital angular momentum is 2 hbar. Hence the paramagentism of singlet oxygen is nearly as high as that of ground state oxygen although spins are paired.
 
Thanks for your answer!

Is it always the case that the combined angular momentum of a set of degenerate states adds to zero? I know that it is zero for an atom (for example px, py, pz).
 
Pete99 said:
Thanks for your answer!

Is it always the case that the combined angular momentum of a set of degenerate states adds to zero? I know that it is zero for an atom (for example px, py, pz).

Quite on the contrary! According to Hunds rules, both S and L tend to be maximal in the ground state of an atom with degenerate shell.
 
Maybe I did not make myself clear, or I am totally wrong, but in an atom such as nitrogen in the ground state, the electronic configuration would be 1s2 2s2 2p3

To maximize S we would have the three electrons with parallel spins in the three degenerate p states, but then L would be zero, isn't it?
 
Yes, for sure. As long as you only consider atoms or ions of main group elements with maximally p orbitals L cannot be greater as 1 if S is to be maximal. The Hund rules become more interesting for atoms and ions with open d or f shells, like lanthanides.
 

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