= Solution
Choose x along O–O. The valence orbital order appropriate to O$_2$ is approximately
$$
2\sigma_g(2s)<2\sigma_u^*(2s)<3\sigma_g(2p_x)<1\pi_u(2p_y,2p_z)<1\pi_g^*(2p_y,2p_z)<3\sigma_u^*(2p_x).
$$
Its twelve valence <electrons> occupy
$$
\boxed{(2\sigma_g)^2(2\sigma_u^*)^2(3\sigma_g)^2(1\pi_u)^4(1\pi_{g,y}^*)^1(1\pi_{g,z}^*)^1.}
$$
The two degenerate pi-antibonding orbitals are singly occupied with parallel spins according to <Hund's rules>. Thus \b[$S=1$, <multiplicity> $2S+1=3$], with three possible $M_S$ values. A triplet is a many-electron spin classification, not three <electrons> in one orbital. The resulting ground term is ${}^3\Sigma_g^-$. Exchange stabilization favours this triplet over the singlets formed within the same frontier-orbital manifold.
\Image[/past-exam-of-the-chemistry-course-of-the-year-3-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-2a-orbitals.png]
{title=Original valence-O2 diagram with the two singly occupied pi-antibonding orbitals}
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