From the LCAO method: The overlap of the 1s AO on one H atom with the is AO on a second gives one bonding MO (s) at lower energy to the constituent AOs and one antibonding MO (s*) at higher energy. Each MO can hold 2 e-. H2^- has 3e- and so the MO \"scheme\" is s(2e-) s*(1e-) Bond Order = ½[S (bonding e-) - S (antibonding e-)] = ½[2-1] = 0.5 so H2^2- is predicted to be bonding. When irradiated with UV light an e- in the s MO is promoted to the s*: s(2e-) s*(1e-) ? h? ? s(1e-) s*(2e-) ? dissociation The bond order of the excited state is formally ½[1-2] = -0.5 which is repulsive: the species flies apart into H^- and H. (H atom not a period). You have to look at an MO diagram for O2; if you do that, focusing on the p-based orbitals, you\'ll see that there are 6 bonding electrons (2 in the p-sigma orbital, and 4 in the p-pi orbitals), and two electrons in the p-pi* (one in each). Thus there are six bonding electrons and two antibonding electrons, so the net is four bonding electrons. Since it takes two bonding electrons to make a bond, the bond order is 2 for O2. If you take away one of the antibonding electrons, now you have a net of five bonding electrons, which - when divided by 2 - gives you a bond order of 2.5. Because you\'ve taken away one electron from neutral O2, the charge will be +1. Solution From the LCAO method: The overlap of the 1s AO on one H atom with the is AO on a second gives one bonding MO (s) at lower energy to the constituent AOs and one antibonding MO (s*) at higher energy. Each MO can hold 2 e-. H2^- has 3e- and so the MO \"scheme\" is s(2e-) s*(1e-) Bond Order = ½[S (bonding e-) - S (antibonding e-)] = ½[2-1] = 0.5 so H2^2- is predicted to be bonding. When irradiated with UV light an e- in the s MO is promoted to the s*: s(2e-) s*(1e-) ? h? ? s(1e-) s*(2e-) ? dissociation The bond order of the excited state is formally ½[1-2] = -0.5 which is repulsive: the species flies apart into H^- and H. (H atom not a period). You have to look at an MO diagram for O2; if you do that, focusing on the p-based orbitals, you\'ll see that there are 6 bonding electrons (2 in the p-sigma orbital, and 4 in the p-pi orbitals), and two electrons in the p-pi* (one in each). Thus there are six bonding electrons and two antibonding electrons, so the net is four bonding electrons. Since it takes two bonding electrons to make a bond, the bond order is 2 for O2. If you take away one of the antibonding electrons, now you have a net of five bonding electrons, which - when divided by 2 - gives you a bond order of 2.5. Because you\'ve taken away one electron from neutral O2, the charge will be +1..