I_{3}^{–} (triiodide) has **three iodine** atoms. In the lewis structure of I_{3}^{–}, there are two single bonds around the iodine atom, with two other iodine atoms attached to it, and on each iodine atom, there are three lone pairs.

Also, there is a negative (-1) charge on the center iodine 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 (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, iodine lies in group 17. Hence, iodine has **seven** valence electrons.

Since I_{3}^{–} has three iodine atoms, so…

Valence electrons of three iodine atoms = 7 × 3 = 21

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

So the **total valence electrons** = 21 + 1 = 22

- Second, find the total electron pairs

We have a total of 22 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** = 22 ÷ 2 = 11

- Third, determine the central atom

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

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

- And finally, draw the rough sketch

### #2 Mark Lone Pairs

Here, we have a total of 11 electron pairs. And two I — I bonds are already marked. So we have to only mark the remaining nine electron pairs as lone pairs on the sketch.

Also remember that iodine is a period 5 element, so it can 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 iodine and right iodine.

So for each iodine, there are **three** lone pairs.

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 iodine** and **right iodine** atom, formal charge = 7 – 6 – ½ (2) = 0

For **center iodine** atom, formal charge = 7 – 6 – ½ (4) = -1

Here, the center iodine atom has a charge, so mark it on the sketch as follows:

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

Now there is still a negative (-1) charge on the center iodine atom.

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 iodine.

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

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

**Next:** CN^{–} Lewis Structure