Question

In $N{a_2} O$, what is the oxidation state of oxygen?

Answer 1

The oxidation state of an atom (sometimes referred to as the oxidation number) in a chemical compound provides information on the number of electrons it has lost and therefore the degree of the atom's oxidation.

Complete answer:
In basic words, the oxidation number is the number assigned to the components in a chemical combination. The oxidation number is the total number of electrons that atoms in a molecule may share, lose, or gain while establishing chemical interactions with atoms of another element.
The term "oxidation number" is also used to refer to the oxidation state. However, depending on whether we are considering the electronegativity of the atoms or not, these words may have a different meaning. In coordination chemistry, the phrase "oxidation number" is often used.
The charge that an atom seems to have while establishing ionic connections with other heteroatoms is defined as its oxidation number. A negative oxidation state is assigned to an atom with a greater electronegativity (even if it forms a covalent connection).
The definition assigns an atom an oxidation state based on the circumstances in which the atom –
i) Interactions with heteroatoms.
ii) Ionic bonds are always formed by either acquiring or losing electrons, regardless of the type of the bond.
Now in the case of $N{a_2} O$
O is more electronegative than Na, hence, Na is given +1 oxidation state and O shows oxidation state as given below.
Let us assume that the oxidation state of oxygen in $N{a_2} O$ is x.
Therefore: -

$$2\left ({+ 1} \right) + x = 0 \\\ \+ 2 + x = 0 \\\ x = - 2 \\\$$

In $N{a_2} O$ oxidation state of O is $- 2$.

Note:
Since one atom may possess numerous valence electrons and establish multiple bonds, they will all be considered to be ionic and given an oxidation state equal to the number of electrons participating in the bonding. Thus, oxidation number or state is a fictitious instance of atoms establishing an ionic connection.