Question

Compare the relative stability of the given species and indicate their magnetic properties: O₂, O₂⁺, O₂⁻(superoxide), O₂²⁻

Answer 1

a) O2 = (σ1s)2(σ*1s)2 (σ2s)2(σ*2s)2(σ2z)2(π2px2= π2py2) (π*2px1= π*2py1)

Bond order will be (Nb-Na) /2 = (10-6)/2= 2

It is paramagnetic as it contains two unpaired electrons.

b) O2+ = = (σ1s)2(σ*1s)2 (σ2s)2(σ*2s)2(σ2z)2(π2px2= π2py2) (π*2px1)

The bond order will be  2.5

It is paramagnetic as it contains one unpaired electron.

c) O2+ = = (σ1s)2(σ*1s)2 (σ2s)2(σ*2s)2(σ2z)2(π2px2= π2py2) (π*2px2= π*2py1)

The bond order will be 1.5

It is paramagnetic as it contains one unpaired electron.