= Solution
The oxygen atom has \b[$1s^2\,2s^2\,2p^4$]. Each oxygen contributes six valence electrons, so the molecular diagram contains twelve. Taking the internuclear direction as $z$, the relevant ordering for oxygen is
$$
\sigma_g(2s)<\sigma_u^*(2s)<\sigma_g(2p_z)
<\pi_u(2p_x)=\pi_u(2p_y)
<\pi_g^*(2p_x)=\pi_g^*(2p_y)<\sigma_u^*(2p_z).
$$
The valence occupation is
$$
\boxed{[\sigma_g(2s)]^2[\sigma_u^*(2s)]^2
[\sigma_g(2p_z)]^2[\pi_u(2p_x)]^2[\pi_u(2p_y)]^2
[\pi_g^*(2p_x)]^1[\pi_g^*(2p_y)]^1}.
$$
The final two electrons occupy the degenerate antibonding orbitals singly with parallel spin, following <Hund's rules>. Thus oxygen has two unpaired electrons and is paramagnetic. The $1s$ core orbitals are deliberately omitted from the valence-only <molecular orbital diagram>.
\Image[/past-exam-of-the-mphil-in-micro-and-nanotechnology-enterprise-course-of-the-university-of-cambridge/2023/paper-1-additional-oxygen-mo.png]
{title=Valence molecular-orbital occupation of ground-state oxygen}
{description=Original qualitative energy-level diagram with both oxygen atomic-orbital sets and all valence molecular orbitals. Energies and separations are illustrative; arrows indicate electron occupation. The two singly occupied antibonding pi orbitals explain the triplet ground state.}
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