IIT JEE Advanced 2007 Paper 1

In Na2O2, oxygen is present as O22−, that is, as peroxide and considering molecular orbital theory for peroxide, there are no unpaired electrons.
O22− = σ1s2σ*1s22s2σ*2s2σ2pz2π2px2
= π2py2π*2px2
= π*2py2
In the structure of O3, there are no unpaired electrons

In N2O, there are no unpaired electrons.
N = N→O
σ1s2σ*1s2σ2s2σ*2s2σ2px2π2px2
= π2py2π*2px2σ*2pz2σ3s2
In KO2, oxygen is present as O2−, that is, superoxide and considering molecular orbital theory for superoxide ion, there is one unpaired electron in π*2py orbital
O2− = σ1s2σ*1s2σ2s2σ*2s2σ2pz2π2px2
= π2py2π*2px2
= π*2py′