 Solution

The closed shells have $L=S=0$, so only the two equivalent p electrons matter. Their addition of angular momentum allows $L=0,1,2$ and $S=0,1$, subject to the Pauli exclusion principle. Interchanging two $l=1$ orbital states has sign $(-1{)}^{2-L}=(-1{)}^L$: even L is spatially symmetric and needs the antisymmetric spin singlet; odd L is spatially antisymmetric and needs the symmetric spin triplet. Thus the allowed LS terms of a p-squared configuration are ${}^{1}S$, ${}^{1}D$ and ${}^{3}P$.

In LS coupling, $J=|L-S|,\dots ,L+S$. The complete list of term symbols is thereforeThey contain $1+5+(1+3+5)=15=\left(\genfrac{}{}{0ex}{}{6}{2}\right)$ magnetic substates, exhausting the antisymmetric two-electron space. All have even spatial parity, since each p orbital is odd.