The document discusses molecular orbital theory applied to hydrogen and oxygen, explaining the bond order calculation for H2^- and O2. It states that H2^- is predicted to be bonding with a bond order of 0.5, while excitation of electrons can lead to dissociation. For O2, it details the presence of bonding and antibonding electrons resulting in a bond order of 2, which can change based on electron removal.

1. 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.