Compare relative stability of O2+ and O2- using bond order?
The electronic configuration of the O2+ ion containing 15 electrons can be written as:
σ1s2< σ*1s2< σ2s2< σ*2s2< σ2p2z < (π2p2x= π2p2y) < (π*2p1x= π*2py) < σ*2pz
The bond order can be found as:
B.O = (Nb-Na)/2
Nb = Number of electrons in the bonding orbitals = 10
Na = Number of electron in the anti-bonding orbitals = 5
B.O = (10-5)/2 = 5/2 = 2.5
The electronic configuration of the O2- ion containing 17 electrons can be written as:
σ1s2< σ*1s2< σ2s2< σ*2s2< σ2p2z < (π2p2x= π2p2y) < (π*2p2x= π*2p1y) < σ*2pz
The bond order can be calculated as:
B.O = (10-7)/2 = 3/2 = 1.5.
The higher the value of bond order, higher is the stability of the bond, so on the basis of above information, we can say that, O2+ ion is more stable than O2- ion.