# Why is the oxidation state of oxygen in nitrite not -4?

Feb 18, 2016

Here's what's going on here.

#### Explanation:

You may be confusing oxygen's oxidation state with its total oxidation state contribution. For oxygen in ${\text{NO}}_{2}^{-}$, it contributes $- 2$ per oxygen, so two oxygens gives $- 4$ total contribution.

You're dealing with the nitrite anion, ${\text{NO}}_{2}^{-}$, so right from the start you know that the oxidation numbers of the atoms that make up this ion must add up to give $- 1$, the charge of the anion.

Now, oxidation numbers are assigned by taking into account the electronegativity of the atoms that form a covalent bond.

More specifically, you assign oxidation numbers by assuming that the more electronegative atom "takes" the bonding electrons shared to form the covalent bond.

In most covalent compounds, oxygen has an oxidation state of $\textcolor{b l u e}{- 2}$. Exceptions to this rule include peroxides and compounds in which oxygen is bonded to fluorine.

Now, it's very important to realize that oxidation numbers are assigned per atom. Since you have $2$ atoms of oxygen and $1$ atom of nitrogen in the nitrite anion, you can say that

$O {N}_{\text{nitrogen" + 2 xx ON_"oxygen}} = - 1$

You know that oxygen has an oxidation state of $\textcolor{b l u e}{- 2}$, so you have

$O {N}_{\text{nitrogen}} + \stackrel{- 4}{\overbrace{2 \times \left(\textcolor{b l u e}{- 2}\right)}} = - 1$

This will of course result in a $\textcolor{b l u e}{+ 3}$ oxidation state for the nitrogen atom.

Therefore, the oxidation states of nitrogen and oxygen in the nitrite anion are $\textcolor{b l u e}{+ 3}$ and $\textcolor{b l u e}{- 2}$, respectively.

$\stackrel{\textcolor{b l u e}{+ 3}}{{\text{N")stackrel(color(blue)(-2))("O}}_{2}^{-}}$