In peroxides, the oxidation state of
O is -1 and they give
H2O2, with dilute acids, and have peroxide linkage. In
KO31 +1+(X×2)=0 x=−‌ (thus, it is a superoxide, not a peroxide.)
In
BaO2,‌‌+2+(x×2)=0 x=−1 Thus, it is a perioxide. Only it gives
H2O2 when reacts with dilute acids and has peroxide linkage as
| Ba2+[O−O]2− |
| peroxide linkage |
In
MnO2 and
NO2,Mn and
N exhibit variable oxidation states, thus, the oxidation state of
O in these is
−2. Hence, these are not peroxides. Thus, it is clear, that among the given molecules only
BaO2 is a peroxide.