In the molecular orbital model, which species is more likely to gain an electron: N2^2+ or O2^2+?

In the molecular orbital (MO) model, the likelihood of a species to gain an electron depends on its electron affinity and the energy of its molecular orbitals. Nitrogen (N2) and Oxygen (O2) are both diatomic molecules, and when they are ionized as N2^2+ or O2^2+, they lose electrons.

To determine which one is more likely to gain an electron, we look at the highest occupied molecular orbital (HOMO). The species with the higher energy HOMO can provide a lower-energy (more stable) vacant molecular orbital for an incoming electron.

For N2^2+, two electrons are removed from the neutral N2 molecule, which has an electron configuration (σ(2s))^2(σ*(2s))^2(π(2p))^4(σ(2p))^2. When two electrons are removed to make N2^2+, they come from the highest occupied molecular orbitals which are the π(2p) orbitals in this case. Thus, N2^2+ would have the configuration (σ(2s))^2(σ*(2s))^2(π(2p))^2(σ(2p))^2.

For O2^2+, the neutral O2 molecule has a different ordering of energy levels: (σ(2s))^2(σ*(2s))^2(σ(2p))^2(π(2p))^4(π*(2p))^2. When you remove two electrons to form O2^2+, they are also removed from the highest occupied orbitals, which are the π*(2p) orbitals. Thus, after removal, the O2^2+ would have the configuration (σ(2s))^2(σ*(2s))^2(σ(2p))^2(π(2p))^4.

It is evident that N2^2+ has a pair of electrons in the σ(2p) orbital, making it's HOMO the σ(2p) orbital, which is a bonding orbital and is lower in energy. On the other hand, O2^2+ has its HOMO as the π(2p) orbital, which is a non-bonding orbital and is higher in energy as compared to the σ(2p) orbital in N2^2+.

Hence, since O2^2+ has the higher energy of its π(2p) molecular orbital, it is more likely to gain an electron because it can offer a more stable (lower-energy) arrangement for the incoming electron compared to N2^2+.