Question

The combustion of sodium in excess air yields a higher oxide. What is the oxidation state of the oxygen in the product?
(Enter the magnitude value i.e. if the answer is $- 2$ enter $2$ )

Answer 1

When the sodium metal is reacted with an excess amount of air or oxygen, it produces sodium peroxide. So we know it exists as sodium peroxide. So the oxidation number of oxygen will be $- 1$ . After entering the magnitude value it will change to $1$ . For further explanation refer to the solution.

Complete answer:
In ordinary air, sodium metal reacts to form a sodium hydroxide film, which can rapidly absorb carbon dioxide from the air, forming sodium bicarbonate. The temperature of burning sodium increases rapidly to more than ${800^ \circ }C$ , and under these conditions the fire is extremely difficult to extinguish.
4Na + {O_2} \to 2N{a_2}O \\\ N{a_2}O + \dfrac{1}{2}{O_2} \to N{a_2}{O_2} \\\
It exists as $2Na$ and ${O_2}^{ - 2}$ . So, the oxidation state of $O$ is $- 1$ .
The magnitude of a number (also called its absolute value) is its distance from zero, so. The magnitude of $6$ is $6$ . The magnitude of $- 6$ is also $6$ .
So, the answer is $| - 1| = 1$ .

Note:
Oxidation number, also called oxidation state, is the total number of electrons that an atom either gains or losses in order to form a chemical bond with another atom. The oxidation number of any free element has an oxidation number equal to zero. For monatomic ions, the oxidation number always has the same value as the net charge corresponding to the ion. The oxidation state of a free element (uncombined element) is zero.