# Coulombic energy of repulsion and exchange energy

In this Wikipedia link about the excited states of oxygen, this diagram is shown. The left most molecular orbital diagram is the first excited state of oxygen, and the MO diagram in the middle is the second excited state of oxygen. It seems to imply that the energy of pairing is less than the exchange energy, and thus the first excited state exhibits the two electrons with opposite spins in the same MO, while the second excited state exhibits the two electrons with opposite spins in separate MOs.

However, in Miessler, Fischer & Tarr, Inorganic Chemistry, 5th Edition, this diagram is depicted, implying that the energy of pairing is greater and more unfavorable than the exchange energy.

These two diagrams seem to have different orders for the first and second excited states. Which energy energy state, having both electrons with opposite spin and in the same orbital, or having both electrons with opposite spin and in separate orbitals, has higher energy?

• @porphyrin are you saying that the Wikipedia article is incorrect? Aug 19 '19 at 16:02
• No, apologies, from memory I made a silly mistake, I stupidly confused singlet delta with singlet sigma. The singlet delta is the first excited state at 0.98 eV the singlet sigma at 1.6 eV above the triplet sigma ground state so the labelling in the Wikipedia article is correct. The second diagram does seem to imply that the singlet sigma is lower in energy than the delta but I don't follow why there are three lines in each, orbitals, energy levels? It does not seem to make sense. Aug 19 '19 at 18:27
• @porphyrin I think the three lines refer to an arbitrary degenerate $np_x$, $np_y$, $np_z$ orbitals; not necessarily the antibonding MOs in $O_2$. Aug 19 '19 at 18:42