Consider the following compounds: $\underline{\text{K}}{\text{O}}_2$, ${\text{H}}_2{\underline{\text{O}}}_2$ and ${\text{H}}_2\underline{\text{S}}{\text{O}}_4$. The oxidation states of the underlined elements in them are, respectively:

1. In ${\text{KO}}_2$ (Potassium superoxide), K is an alkali metal, so its oxidation state is always $+1$.2. In ${\text{H}}_2{\text{O}}_2$ (Hydrogen peroxide), H is $+1$. Let O be $x$. $2(+1)+2x=0⟹x=-1$.3. In ${\text{H}}_2{\text{SO}}_4$ (Sulfuric acid), H is $+1$ and O is $-2$. Let S be $y$. $2(+1)+y+4(-2)=0⟹y=+6$.The sequence is $+1,-1,+6$.