ICl_{2}^{–} (iodine dichloride) has **one iodine** atom and **two chlorine** atoms. In the lewis structure of ICl_{2}^{–}, there are two single bonds around the iodine atom, with two chlorine atoms attached to it, and each atom has three lone pairs.

Also, there is a negative (-1) charge on the iodine atom.

## Steps

Here’s how you can draw the ICl_{2}^{–} 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, both iodine and chlorine lie in group 17.

Hence, both iodine and chlorine have **seven** valence electrons.

Since ICl_{2}^{–} has one iodine atom and two chlorine atoms, so…

Valence electrons of one iodine atom = 7 × 1 = 7

Valence electrons of two chlorine atoms = 7 × 2 = 14

Now the ICl_{2}^{–} has a negative (-1) charge, so we have to add one more electron.

So the **total valence electrons** = 7 + 14 + 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

We have to place the least electronegative atom at the center.

Since iodine is less electronegative than chlorine, assume that the **central atom is iodine**.

Therefore, place iodine in the center and chlorines on either side.

- And finally, draw the rough sketch

### #2 Mark Lone Pairs

Here, we have a total of 11 electron pairs. And two I — Cl 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. And chlorine is a period 3 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 chlorines.

So for each atom, 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 **iodine** atom, formal charge = 7 – 6 – ½ (4) = -1

For **each chlorine** atom, formal charge = 7 – 6 – ½ (2) = 0

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

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

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

This is not okay, right? 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 chlorine.

But if we convert a lone pair of the iodine atom to make a new I — Cl bond with the chlorine atom, and calculate the formal charge, then we do not get the formal charges on atoms closer to zero.

And the structure with the formal charges on atoms closer to zero is the best lewis structure.

Therefore, this structure is the most stable lewis structure of ICl_{2}^{–}.

And since the ICl_{2}^{–} has a negative (-1) charge, mention that charge on the lewis structure by drawing brackets as follows:

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