Compare relative stability of O2+ and O2- using bond order?

The electronic configuration of the O_{2}^{+} ion containing 15 electrons can be written as:

σ1*s ^{2}*< σ*1

*s*< σ2

^{2}*s*< σ*2

^{2}*s*< σ2

^{2}*p*

^{2}*= π2*

_{z}< (π2p^{2}_{x}*p*

^{2}*) < (π*2*

_{y}*p*

^{1}*= π*2*

_{x}*p*

*) < σ*2*

_{y}*p*

_{z}The bond order can be found as:

B.O = (N_{b}-N_{a})/2

N_{b }= Number of electrons in the bonding orbitals = 10

N_{a }= Number of electron in the anti-bonding orbitals = 5

B.O = (10-5)/2 = 5/2 = 2.5

The electronic configuration of the O_{2}^{-} ion containing 17 electrons can be written as:

σ1*s ^{2}*< σ*1

*s*< σ2

^{2}*s*< σ*2

^{2}*s*< σ2

^{2}*p*

^{2}*= π2*

_{z}< (π2p^{2}_{x}*p*

^{2}*) < (π*2p*

_{y}

^{2}*= π*2*

_{x}*p*

^{1}*) < σ*2*

_{y}*p*

_{z}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, O_{2}^{+} ion is more stable than O_{2}^{-} ion.

**
**