= 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 $ {}^3P$ term. Since four <electrons> exceed half filling of the six-state $p$ subshell, its lowest <atomic fine-structure level> has the largest $J=L+S=2$:
$$
\boxed{1s^2\,2s^2\,2p^4\;{}^3P_2,\qquad S=1,\ L=1,\ J=2.}
$$
Its zero-field magnetic degeneracy is $2J+1=5$. The experimentally tabulated neutral-oxygen ground level has precisely this term and $J$.<Energy Levels of Neutral Oxygen by National Institute of Standards and Technology>[first configuration block, 2s2 2p4, triplet P, J=2 at zero energy]{ref}
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