Electrons present
NO2+= 7+16-1=22
BaO2=56+16=72
AlO2-=13+16+1=30
KO2 = 19+16=35
Among KO2, AlO2-, BaO2 and NO2+, unpaired electron is present in
a) NO2+ and BaO2
b) KO2 and AlO2-
c) KO2 only
d) BaO2 only
Electrons present
NO2+= 7+16-1=22
BaO2=56+16=72
AlO2-=13+16+1=30
KO2 = 19+16=35