Orbital angular momentum of O2

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The discussion centers on the orbital angular momentum of the O2 molecule, specifically regarding its paramagnetic behavior attributed to two unpaired electrons in degenerate pi* orbitals. It is noted that in the ground state (^3Σg^-), the total orbital angular momentum is zero due to the combination of the occupied antibonding pi orbitals. In contrast, the lowest excited state (^1Δg) has an orbital angular momentum of 2ħ, despite paired spins, indicating significant paramagnetism. The conversation also touches on the implications of Hund's rules, clarifying that while maximizing spin (S) can lead to zero orbital angular momentum (L) in some cases, this is not universally applicable across all elements. Overall, the relationship between orbital angular momentum and electron configuration in molecular and atomic systems is complex and varies with the specific states considered.
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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