The correct option is B −1 and −12
In hydrogen peroxide (H2O2) peroxide linkage (−O−O−) linkage is present. Again electronegativity of oxygen is more than hydrogen so , hydrogen carries positive charge and oxygen carries negative charge.
Let the oxidation number of oxygen in H2O2 be x.
So , 2x+2=02x=−2x=−1
Therefore, the oxidation state of each oxygen is −1.
In KO2 the K atom carries +1 charge as it has only one electron in its valence shell so , O2 should have −1 charge i.e. it acts as a superoxide.
In KO2, let oxidation number of oxygen be x.
+1+2x=0x=−12
Therefore, the oxidation number of oxygen in KO2 is −12