Oxygen Ground State

Oxygen, with \bgroup\color{black}$Z=8$\egroup has the 1S and 2S levels filled giving \bgroup\color{black}$j=0$\egroup as a base. It has four valence 2P electrons which we will treat as two valence 2P holes. Hund's first rule , maximum total \bgroup\color{black}$s$\egroup, tells us to couple the two hole spins to \bgroup\color{black}$s=1$\egroup. This is the symmetric spin state so we'll need to make the space state antisymmetric. Hund's second rule, maximum \bgroup\color{black}$\ell$\egroup, doesn't play a role because only the \bgroup\color{black}$\ell=1$\egroup state is antisymmetric. Since the shell is more than half full we couple to the the highest \bgroup\color{black}$j=\ell+s=2$\egroup. So the ground state is \bgroup\color{black}$^3P_{2}$\egroup.
\bgroup\color{black}$m_\ell$\egroup e
1 \bgroup\color{black}$\uparrow\downarrow$\egroup
0 \bgroup\color{black}$\uparrow$\egroup
-1 \bgroup\color{black}$\uparrow$\egroup
\bgroup\color{black}$s=\sum m_s=1$\egroup
\bgroup\color{black}$\ell=\sum m_\ell=1$\egroup

Jim Branson 2013-04-22