I_{2} (iodine) has **two iodine** atoms.

In the I_{2} Lewis structure, there is a single bond between the two iodine atoms, and on each iodine atom, there are three lone pairs.

## Steps

Use these steps to correctly draw the I_{2} Lewis structure:

#1 First draw a rough sketch

#2 Mark lone pairs on the atoms

#3 Calculate and mark formal charges on the atoms, if required

Let’s discuss each step in more detail.

### #1 First draw a rough 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_{2} has two iodine atoms, so…

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

So the **total valence electrons** = 14

- Second, find the total electron pairs

We have a total of 14 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** = 14 ÷ 2 = 7

- Third, determine the central atom

Here, there are only two atoms and both atoms are iodine, so we can assume any one as the central atom.

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

- And finally, draw the rough sketch

### #2 Mark lone pairs on the atoms

Here, we have a total of 7 electron pairs. And one I — I bond is already marked. So we have to only mark the remaining six 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 atom is left iodine.

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

Mark the lone pairs on the sketch as follows:

### #3 Calculate and mark formal charges on the atoms, if required

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

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

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

Here, both iodine atoms do not have charges, so no need to mark the charges.

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

Therefore, this structure is the **stable Lewis structure** of I_{2}.

**Next:** Acetone Lewis structure