= Solution
Let $\sigma_L,\sigma_R$ be localized bonding orbitals of the two new C–O bonds, and let $\sigma_L^*,\sigma_R^*$ be their antibonding partners. Their <symmetry> combinations are
$$
\sigma_\pm=\frac{\sigma_L\pm\sigma_R}{\sqrt2},\qquad
\sigma_\pm^*=\frac{\sigma_L^*\pm\sigma_R^*}{\sqrt2}.
$$
The two remaining active orbitals are out-of-ring <oxygen> lone-pair combinations, $n_\pm=(p_{y,L}\pm p_{y,R})/\sqrt2$. The opposite-phase combination $n_-$ is the HOMO in the supplied ordering and $n_+$ is HOMO−1. Both are filled, as are the two C–O bonding combinations, giving
$$
\boxed{(\sigma_+)^2(\sigma_-)^2(n_+)^2(n_-)^2(\sigma_+^*)^0(\sigma_-^*)^0.}
$$
The original active-orbital sketches show these phase relations, while the correlation diagram gives a qualitative energy ordering. Precise splittings between unrelated <symmetry> blocks are not determined by the fragment <ionization energies>. The retained O–O <sigma bond>, the C–C <sigma bond> and the other framework orbitals lie outside this six-orbital active comparison; the eight-electron count does not describe all valence <electrons> of the product.
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