The molecular orbital schemes for both forms of singlet oxygen ($\mathrm{^1\Delta_g}$ and $\mathrm{^1\Sigma_g^+}$) and triplet oxygen ($\mathrm{^3\Sigma_g^-}$) are typically given as shown in the image below.
Figure 1: Molecular orbital schemes of two types of singlet oxygen and triplet oxygen with the highest energy electrons highlighted in red.
Hund’s rule sufficiently predicts that the triplet $\mathrm{^3\Sigma_g^-}$ state is the most stable, i.e. lowest in energy. When one speaks of singlet oxygen, one typically has the $\mathrm{^1\Delta_g}$ form in mind which is $\pu{0.98eV}$ higher in energy than triplet oxygen according to a group member’s recent literature seminar. This is often explained on this site by the fact that spin pairing puts another electron into the space that is already populated by one electron, causing electron-electron repulsion that must be overcome by adding energy.
Extending this explanation, I am inclined to think that the $\mathrm{^1\Sigma_g^+}$ form of singlet oxygen, whose MO scheme is depicted in the centre of figure 1, should have a lower energy than the $\mathrm{^1\Delta_g}$ form because the spins are occupying spacially different (and orthogonal) orbitals. However, the same seminar talk also included the energy difference of $\pu{0.65eV}$ between $\mathrm{^1\Delta_g}$ and $\mathrm{^1\Sigma_g^+}$ with the $\mathrm{^1\Sigma_g^+}$ being higher in energy.
Why is this so?