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

We determine the oxidation states of the specified elements in each compound by using the rules that the sum of oxidation numbers in a neutral molecule equals zero and that hydrogen is usually +1 while oxygen is –2, except in peroxides where oxygen is –1.

1. KO₂

In potassium superoxide the formula can be written as K⁺ (O₂)²⁻. The O₂ unit carries a –2 charge, meaning each oxygen atom has an oxidation state of –1 (characteristic of a peroxide). Since the overall compound is neutral, the oxidation state of potassium must balance the –2 from the peroxide:

Thus potassium (K) has an oxidation state of +1.

2. H₂O₂

Hydrogen is assigned +1. Let the oxidation state of each oxygen be x. The molecule is neutral, so:

Solving gives 2 + 2x = 0 → x = –1. Therefore each oxygen in hydrogen peroxide carries an oxidation state of –1, confirming the peroxide character.

3. H₂SO₄

Hydrogen is +1 (two atoms give +2) and each oxygen is –2 (four atoms give –8). Let the oxidation state of sulfur be y. For a neutral molecule:

Which simplifies to +2 + y – 8 = 0 → y = +6. Hence sulfur in sulfuric acid has an oxidation state of +6.

Putting the three results together, the oxidation states of the underlined elements are +1, –1, and +6.