Q.What is the oxidation state of K in KO2?

When burnt in excess dry oxygen, the alkali metals give three different classes of binary oxide depending on the size and hence polarising power of the cation, because the lattice must be able to accommodate the oxide ion ($O^{2-}$), peroxide ion ($O_2^{2-}$) or superoxide ion ($O_2^-$) stably. The very small, highly polarising $Li^+$ can only stabilise the small $O^{2-}$ ion, so lithium gives the normal oxide, $4Li + O_2 \rightarrow 2Li_2O$; sodium, being slightly larger, forms mainly the peroxide, $2Na + O_2 \rightarrow Na_2O_2$; while the large, weakly polarising $K^+$, $Rb^+$ and $Cs^+$ can stabilise the still larger superoxide ion, giving compounds such as $K + O_2 \rightarrow KO_2$. In the superoxide ion $O_2^-$, the extra electron leaves one electron unpaired (bond order 1.5 between the two oxygen atoms), so $KO_2$ has an unpaired electron and is paramagnetic. Assigning oxidation states: the superoxide ion as a whole carries charge $-1$ across its two oxygen atoms (average oxidation number of O $= -\tfrac12$), so in $KO_2$ potassium is $+1$; similarly in $Na_2O_2$ sodium is $+1$ and the peroxide oxygens are $-1$ each. These oxides react vigorously with water: $Na_2O_2 + 2H_2O \rightarrow 2NaOH + H_2O_2$, $2KO_2 + 2H_2O \rightarrow 2KOH + H_2O_2 + O_2$, and $Na_2O + CO_2 \rightarrow Na_2CO_3$.