Orbital angular momentum of O2

  • Thread starter Pete99
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  • #1
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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!
 

Answers and Replies

  • #2
DrDu
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Thats an interesting question. In the ground state [tex] ^3\Sigma_g^-[/tex], the orbital angular momenta of the occupied antibonding pi orbitals combine to total orbital angular momentum 0 (hence the [tex]\Sigma[/tex]). Classically, the two unpaired electron orbit around the molecular axis in different directions. In the lowest excited state (singlett oxygen)
[tex] ^1\Delta_g [/tex] 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.
 
  • #3
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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).
 
  • #4
DrDu
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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.
 
  • #5
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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?
 
  • #6
DrDu
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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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