Identify the incorrect statement from the following:

The statement “Oxygen exhibits only −2 oxidation state.” is incorrect because oxygen can show several oxidation numbers besides −2.

1. $O0$ in its elemental form (O₂). In diatomic oxygen each atom shares electrons equally, resulting in an oxidation state of 0.

2. $O-1$ in peroxides such as hydrogen peroxide (H₂O₂) and barium peroxide (BaO₂). Here each oxygen atom is bonded to another oxygen atom, giving an overall −2 charge split equally, i.e., each oxygen is −1.

3. $O-½$ in superoxides like potassium superoxide (KO₂). The O₂ unit carries a −1 charge, so each oxygen has an average oxidation state of −½.

4. Positive oxidation states are also possible when oxygen is bonded to more electronegative elements:

• $O+1$ in dioxygen difluoride (O₂F₂) where each oxygen is bonded to two fluorine atoms.
• $O+2$ in oxygen difluoride (OF₂) where oxygen is less electronegative than fluorine.

Therefore, oxygen does not exhibit only the −2 oxidation state; it can have 0, −1, −½, +1, and +2 oxidation numbers depending on the chemical environment. This makes the given statement false.