N_{3}^{–} (azide) has **three nitrogen** atoms. In the lewis structure of N_{3}^{–}, there are two double bonds around the nitrogen atom, with two other nitrogen atoms attached to it, and on left and right nitrogen atoms, there are two lone pairs.

Also, there is a negative (-1) charge on the left and right nitrogen atoms, and a positive (+1) charge on the center nitrogen atom.

## Steps

Here’s how you can draw the I_{3}^{–} lewis structure step by step.

Step #1: draw sketch

Step #2: mark lone pairs

Step #3: mark charges

Step #4: minimize charges

Step #5: minimize charges again (if there are)

Let’s break down each step in detail.

### #1 Draw Sketch

- First, determine the total number of valence electrons

In the periodic table, nitrogen lies in group 15. Hence, nitrogen has **five** valence electrons.

Since N_{3}^{–} has three nitrogen atoms, so…

Valence electrons of three nitrogen atoms = 5 × 3 = 15

Now the N_{3}^{–} has a negative (-1) charge, so we have to add one more electron.

So the **total valence electrons** = 15 + 1 = 16

- Second, find the total electron pairs

We have a total of 16 valence electrons. And when we divide this value by two, we get the value of total electron pairs.

Total electron pairs = total valence electrons ÷ 2

So the **total electron pairs** = 16 ÷ 2 = 8

- Third, determine the central atom

Here, there are three atoms and all atoms are nitrogen, so we can assume any one as the central atom.

Let’s assume that the **central atom is center nitrogen**.

- And finally, draw the rough sketch

### #2 Mark Lone Pairs

Here, we have a total of 8 electron pairs. And two N — N bonds are already marked. So we have to only mark the remaining six electron pairs as lone pairs on the sketch.

Also remember that nitrogen is a period 2 element, so it can not keep more than 8 electrons in its last shell.

Always start to mark the lone pairs from outside atoms. Here, the outside atoms are left nitrogen and right nitrogen.

So for left nitrogen and right nitrogen, there are **three** lone pairs, and for canter nitrogen, there is **zero** lone pair because all six electron pairs are over.

Mark the lone pairs on the sketch as follows:

### #3 Mark Charges

Use the following formula to calculate the formal charges on atoms:

Formal charge = valence electrons – nonbonding electrons – ½ bonding electrons

For **left nitrogen** and **right nitrogen** atom, formal charge = 5 – 6 – ½ (2) = -2

For **center nitrogen** atom, formal charge = 5 – 0 – ½ (4) = +3

Here, all nitrogen atoms have charges, so mark them on the sketch as follows:

The above structure is not a stable lewis structure because all nitrogen atoms have charges. Therefore, reduce the charges (as below) by converting lone pairs to bonds.

### #4 Minimize Charges

Convert a lone pair of the left nitrogen atom to make a new N — N bond with the center nitrogen atom as follows:

### #5 Minimize Charges Again

Since there are charges on nitrogen atoms, again convert a lone pair of the right nitrogen atom to make a new N — N bond with the center nitrogen atom as follows:

In the above structure, you can see that the central atom (center nitrogen) forms an octet. Hence, the octet rule is satisfied.

Now there are still charges on the atoms.

This is okay, because the structure with a negative charge on the most electronegative atom is the best lewis structure. And in this case, the most electronegative element is nitrogen.

Therefore, this structure is the most stable lewis structure of N_{3}^{–}.

And since the N_{3}^{–} has a negative (-1) charge, mention that charge on the lewis structure by drawing brackets as follows:

**Next:** BH_{3} Lewis Structure