Consider the following compounds :
KO₂, H₂O₂ and H₂SO₄
The oxidation state of the underlined elements in them are, respectively,

We are asked to find the oxidation numbers of the underlined elements in the three compounds KO₂, H₂O₂ and H₂SO₄.

1. KO₂ – the underlined element is potassium (K).
In KO₂ the oxygen atoms form a superoxide ion O₂⁻. The total charge of the O₂ unit is –1, which means each oxygen atom has an oxidation state of –½. The overall charge of the neutral compound is 0, so the oxidation state of K must balance the –1 contributed by the O₂ unit:
$x$_K+(-1)=0
Thus $x$_K=+1. Therefore potassium is in the +1 oxidation state.

2. H₂O₂ – the underlined element is oxygen.
Hydrogen is always +1. Let the oxidation number of each oxygen be x.
$2(+1)+2x=0$
Solving gives $2+2x=0\; \Rightarrow \; 2x=-2\; \Rightarrow \; x=-1$. Hence each oxygen in hydrogen peroxide is –1.

3. H₂SO₄ – the underlined element is sulfur.
Hydrogen is +1, oxygen is –2. Let the oxidation number of sulfur be y.
$2(+1)+y+4(-2)=0$
That is $2+y-8=0\; \Rightarrow \; y=+6$. So sulfur carries a +6 oxidation state.

Collecting the results, the oxidation states of the underlined elements are +1, -1, and +6, respectively.