# How do I use Lewis structures to determine the oxidation numbers of "N" in compounds like "N"_2"O", "NO", "N"_2"O"_3, "N"_2"O"_4, and "N"_2"O"_5?

Oct 14, 2016

Warning! Long answer. You count the valence electrons around $\text{N}$ according to a set of rules and then assign the oxidation number.

#### Explanation:

The Rules

1. Lone pair electrons (LPs) belong entirely to the atom on which they reside.
2. Shared electrons (bonding pair electrons or BEs) between identical atoms are shared equally.
3. Shared electrons between different atoms belong entirely to the more electronegative atom.
4. Oxidation number (ON) is the difference between the valence electrons in the isolated atom (VE) and the valence electrons in the bound atom (LP + BE).

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} O N = V E - L P - B E \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

I will calculate the oxidation numbers for only one Lewis structure of each oxide.

An isolated $\text{N}$ atom has five valence electrons ($V E = 5$).

Dinitrogen monoxide

The Lewis structure of $\text{N"_2"O}$ is

(a) The left hand $\text{N}$ atom

LP = 2; BE = 3

$O N = V E - L P - B E = 5 - 2 - 3 = 0$

(b) The central $\text{N}$ atom

LP = 0; BE = 3

$O N = 5 - 0 - 3 = + 2$

Each $\text{N}$ atom has a different formal charge.

The average oxidation number on $\text{N} = \frac{0 + 2}{2} = + \frac{1}{2}$

Nitrogen monoxide

The Lewis structure of $\text{NO}$ is

LP = 3; BE = 0

$O N = V E - L P - B E = 5 - 3 - 0 = + 2$

Dinitrogen trioxide

The Lewis structure of ${\text{N"_2"O}}_{3}$ is

(a) The left hand $\text{N}$ atom

LP = 0; BE = 1

$O N = 5 - 0 - 1 = + 4$

(b) The central $\text{N}$ atom

LP = 2; BE = 1

$O N = 5 - 2 - 1 = + 2$

Each $\text{N}$ atom has a different formal charge.

The average formal charge on $\text{N} = \frac{+ 4 + 2}{2} = + \frac{6}{2} = + 3$

Dinitrogen Tetroxide

The Lewis structure of ${\text{N"_2"O}}_{4}$ is

(a) The left-hand $\text{N}$ atom

LP = 0; BE = 1

$O N = 5 - 0 - 1 = + 4$

(b) The right-hand $\text{N}$ atom

LP = 0; BE = 1

$O N = 5 - 0 - 1 = + 4$

Each $\text{N}$ atom has the same formal charge.

The average oxidation number on $\text{N} = \frac{+ 4 + 4}{2} = + \frac{8}{2} = + 4$

Dinitrogen pentoxide

The Lewis structure of ${\text{N"_2"O}}_{5}$ is

(a) The left-hand $\text{N}$ atom

LP = 0; BE = 0

$O N = 5 - 0 - 0 = + 5$

(b) The right-hand $\text{N}$ atom

LP = 0; BE = 0

$O N = 5 - 0 - 0 = + 5$

Each $\text{N}$ atom has the same formal charge.

The average oxidation number on $\text{N} = \frac{+ 5 + 5}{2} = + \frac{10}{2} = + 5$