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In kO2 oxidation state of each oxygen is -1/2.
let oxidation state of k in KO2 be x
hence, oxidation state of k is +1
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K i.e. pottSium is a member of group 1 so it readily reacts with oxygen. As we move down the group the tendency of forming peroxides and super oxides increase. KO2 is pottasium super oxide ko2 =k+ + O2( 1-) . O2 (1-) is super oxide in which o has a charge of -1/2. ko2 =+1×1 + x (oxidation state of oxygen )×2=0 (as ko2 is electrically neutral) On solving 1+2x=0 => x= -1/2
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