Molecular orbitals of the oxocarbons (CO)n, n = 2-6. Why does (CO)4 have a triplet ground state?

Cyclobutane-1,2,3,4-tetrone has been both predicted and found to have a triplet ground state, in which a b(2g) σ MO and an a(2u) π MO are each singly occupied. The nearly identical energies of these two orbitals of (CO)(4) can be attributed to the fact that both of these MOs are formed from a bonding combination of C-O π* orbitals in four CO molecules. The intrinsically stronger bonding between neighboring carbons in the b(2g) σ MO compared to the a(2u) π MO is balanced by the fact that the non-nearest-neighbor, C-C interactions in (CO)(4) are antibonding in b(2g), but bonding in a(2u). Crossing between an antibonding, b(1g) combination of carbon lone-pair orbitals in four CO molecules and the b(2g) and a(2u) bonding combinations of π* MOs is responsible for the occupation of the b(2g) and a(2u) MOs in (CO)(4). A similar orbital crossing occurs on going from two CO molecules to (CO)(2), and this crossing is responsible for the triplet ground state that is predicted for (CO)(2). However, such an orbital crossing does not occur on formation of (CO)(2n+1) from 2n + 1 CO molecules, which is why (CO)(3) and (CO)(5) are both calculated to have singlet ground states. Orbital crossings, involving an antibonding, b(1), combination of lone-pair MOs, occur in forming all (CO)(2n) molecules from 2n CO molecules. Nevertheless, (CO)(6) is predicted to have a singlet ground state, in which the b(2u) σ MO is doubly occupied and the a(2u) π MO is left empty. The main reason for the difference between the ground states of (CO)(4) and (CO)(6) is that interactions between 2p AOs on non-nearest-neighbor carbons, which stabilize the a(2u) π MO in (CO)(4), are much weaker in (CO)(6), due to the much larger distances between non-nearest-neighbor carbons in (CO)(6) than in (CO)(4).