# Orbital angular momentum of O2

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

DrDu
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.

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).

DrDu

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?

DrDu