Why is an energy cost associated with splitting the HOMO spins in singlet oxygen?

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?

• I used to think of it in terms of Hund's second rule; higher angular momentum = lower energy. But I'm not sure about the validity of (1) the extension to molecules; and (2) the extension to comparing excited states. Technically, Hund's rules are only used to find the ground state. On a side note, I'd suggest actually putting the term symbols into the title; this is one rare case where it's really helpful to have MathJax in the title. – orthocresol Oct 3 '17 at 11:43
• I need to clarify if I understood this correctly, You are asking: Why is $\mathrm{^1\Delta_g}$ lower in energy than $\mathrm{^1\Sigma_g^+}$? – Martin - マーチン Oct 5 '17 at 5:24
• @Martin-マーチン Yes. – Jan Oct 5 '17 at 16:14
• @orthocresol you are correct. I think the extension to molecules and excited states is valid (with caveats, but that is just being technical) – getafix Jan 1 '18 at 7:01
• @getafix - both the Mulliken and Herzberg references in my answer give examples where the ordering of excited states predicted through Hund's rules can be wrong. They're a good rule of thumb, but that implies exceptions can and do exist. – Geoff Hutchison Jan 1 '18 at 19:13

Short story: The comment by orthocresol is correct, based on Hund's rules. Given the same spin multiplicity (i.e., the two singlet states), the state with higher orbital angular momentum ($\mathrm{^1\Delta_g}$) wins.

Loosely speaking, one can imagine that comparing the two singlet states, the higher angular momentum implies that the electrons are a bit further out, and suffer less electron-electron repulsion.

Longer story...

The "unified atom" model can apply to simple diatomics - that is, many of the same considerations for the electronic structure can be used for multi-electron atoms. There's discussion in both Mulliken (Rev. Mod. Phys. 1932, 4, 1. and Herzberg's Spectra of Diatomic Molecules).

The vibrational-electronic spectra of $\ce{O2}$ was determined fairly readily: (Figure from Chem. Rev., 2003, 103 (5), pp 1685–1758, adapted from Herzberg).

We can assign the lines as discussed by Mulliken in 1932:

The $\mathrm{^1\Sigma_g^+}$, is 1.62 volts above the $\mathrm{^3\Sigma_g^-}$ according to theoretical considerations,[118] the $\mathrm{^1\Delta_g}$, is probably about halfway between $\mathrm{^1\Sigma_g^+}$, and $\mathrm{^3\Sigma_g^-}$ (cf. Fig 48).

Reference 118 is to Hund and Hückel:

• F. Hund, Zeits. f. Physik 63, 726 (1930).
• E. Huckel, Zeits. f. Physik 60, 442-3 (1930).

Both Mulliken and Herzberg discuss the Hund's rules in regards to many diatomics - that the triplet will lie lower than the singlet because of higher S, but then the two singlet states will differ based on angular momentum (here L).

• Incidentally, the $\mathrm{^1\Delta_g}$ state is doubly-degenerate, but that doesn't affect the ordering of the states. – Geoff Hutchison Dec 28 '17 at 20:26