Question

When sodium reacts with excess of oxygen, oxidation number of oxygen changes from:

• A. 0 to -1
• B. 0 to -2
• C. -1 to -2
• D. +1 to -1

Answer 1

The correct option is A

0 to -1

$2Na+O_2\left(excess\right)\to N{a}_2{O}_2$

The oxidation state of oxygen is 'zero' in its native state, where as its oxidation state in its peroxide form is $\left(N{a}_2{O}_2\right)$ is -1.

You have to go other way round to calculate the oxidation state of oxygen in the peroxide. $2\times (+1)+x=0$ gives us $x=-2$. Therefore each oxygen has an oxidation state of $\frac{-2}{2}=-1$.