Question

What are the oxidation numbers of oxygen? Give an example.

Answer 1

In order to answer this question, to know the oxidation numbers of oxygen, we should first know what is the oxidation number of any element, and we will explain the oxidation number briefly and we will also give an example and explain it.

Complete answer:
Due to its high electronegativity, oxygen usually has a negative two charges. For example in the compound, calcium oxide, $CaO$ , calcium has an oxidation number of +2 and the oxygen has -2 charge. In all peroxides (oxygen-oxygen linkage), oxygen has an oxidation state of −1. For example, consider ${H_2}{O_2}$ , here H is less electronegative so it will acquire a charge of +1. In all superoxides ( $K{O_2},Cs{O_2},Rb{O_2}$ ), oxygen has an oxidation state of $- \dfrac{1}{2}$ . The more electronegative atom acquires a negative charge while less electronegative atom acquire a positive charge. Depending on this hypothesis oxygen has 5 oxidation states.
The following rules may be used to assign an oxidation number to a given element or compound. The oxidation number of any free element is zero. The oxidation number of monatomic ions is often equal to the net charge associated with the ion. The oxidation state of the hydrogen atom $(H)$ is +1.
Oxidation number or state of an atom/ion is the number of electrons an atom/ion that the molecule has either gained or lost compared to the neutral atom.

Note:
Rules for Assigning Oxidation Numbers The convention is that the cation is written first in a formula, followed by the anion. The oxidation number of a free element is always 0. The oxidation number of a monatomic ion equals the charge of the ion. The usual oxidation number of hydrogen is +1.