Past exam of the advanced quantum physics course of the University of Cambridge / 2015 / Paper 1 / 1 / c / Solution

Closed $1s^2$ and $2s^2$ subshells have $L=S=0$. In the open $2p^4$ subshell, three electrons can have parallel spins in $m_{\ell }=-1,0,1$, and the fourth must have the opposite spin, giving maximum $S=1$. Treating the shell as two $p$ holes gives the same allowed terms as $p^2$; at this spin, antisymmetry permits $L=1$, rather than $L=2$. One can also see $M_L=1$ by placing the fourth electron in $m_{\ell }=1$.

Hund's rules therefore select the ${}^{3}P$ term. Since four electrons exceed half filling of the six-state $p$ subshell, its lowest atomic fine-structure level has the largest $J=L+S=2$:Its zero-field magnetic degeneracy is $2J+1=5$. The experimentally tabulated neutral-oxygen ground level has precisely this term and $J$.^[ref]