= 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 $^1S$, $^1D$ and $^3P$.
In <LS coupling>, $J=|L-S|,\ldots,L+S$. The complete list of <term symbols> is therefore
$$
\boxed{{}^1S_0,\qquad{}^1D_2,\qquad{}^3P_0,\ {}^3P_1,\ {}^3P_2.}
$$
They contain $1+5+(1+3+5)=15=\binom62$ magnetic substates, exhausting the antisymmetric two-electron space. All have even spatial <parity>, since each p orbital is odd.
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