I2 Lewis Structure in 6 Steps (With Images)

I2 lewis structure

So you have seen the above image by now, right?

Let me explain the above image in short.

I2 lewis structure has two Iodine atoms (I) which contain a single bond between them. There are 3 lone pairs on both the Iodine atoms (I).

If you haven’t understood anything from the above image of I2 (iodine) lewis structure, then just stick with me and you will get the detailed step by step explanation on drawing a lewis structure of I2.

So let’s move to the steps of drawing the lewis structure of I2.

Steps of drawing I2 lewis structure

Step 1: Find the total valence electrons in I2 molecule

In order to find the total valence electrons in I2 (iodine) molecule, first of all you should know the valence electrons present in a single iodine atom.
(Valence electrons are the electrons that are present in the outermost orbit of any atom.)

Here, I’ll tell you how you can easily find the valence electrons of iodine using a periodic table.

Total valence electrons in I2 molecule

→ Valence electrons given by iodine atom:

Iodine is a group 17 element on the periodic table. [1] Hence the valence electrons present in iodine is 7.

You can see the 7 valence electrons present in the iodine atom as shown in the above image.


Total valence electrons in I2 molecule = 7(2) = 14.

Step 2: Select the central atom

For selecting the center atom, you have to remember that the atom which is less electronegative remains at the center.

Now here the given molecule is I2 (iodine). Both the atoms are same, so you can select any of the atoms as a center atom.

I2 step 1

Let’s assume the right side iodine as a central atom.

Step 3: Connect each atoms by putting an electron pair between them

Now in the I2 molecule, you have to put the electron pairs between both the iodine atoms (I).

I2 step 2

This indicates that both the iodine (I) atoms are chemically bonded with each other in a I2 molecule.

Step 4: Make the outer atoms stable. Place the remaining valence electrons pair on the central atom.

Now in this step, you have to check the stability of the outer atom.

Here in the sketch of I2 molecule, we have assumed the right side iodine atom as a center atom. So the left side iodine is the outer atom.

Hence you have to make the left side iodine stable.

You can see in the below image that the left side iodine atom is forming an octet and hence it is stable.

I2 step 3

Also, in step 1 we have calculated the total number of valence electrons present in the I2 molecule.

The I2 molecule has a total 14 valence electrons and out of these, only 8 valence electrons are used in the above sketch.

So the number of electrons which are left = 14 – 8 = 6.

You have to put these 6 electrons on the right side iodine atom in the above sketch of I2 molecule.

I2 step 4

Now let’s proceed to the next step.

Step 5: Check the octet on the central atom. If it does not have octet, then shift the lone pair to form a double bond or triple bond.

In this step, you have to check whether the central (i.e right side) iodine atom (I) is stable or not.

In order to check the stability of this iodine (I) atom, we have to check whether it is forming an octet or not.

I2 step 5

You can see from the above picture that both the iodine atoms are forming an octet. That means they have 8 electrons.

And hence the iodine atoms are stable.

Now let’s proceed to the final step to check whether the lewis structure of I2 is stable or not.

Step 6: Check the stability of lewis structure

Now you have come to the final step in which you have to check the stability of lewis structure of I2.

The stability of lewis structure can be checked by using a concept of formal charge.

In short, now you have to find the formal charge on both the iodine (I) atoms present in the I2 molecule.

For calculating the formal charge, you have to use the following formula;

Formal charge = Valence electrons – (Bonding electrons)/2 – Nonbonding electrons

You can see the number of bonding electrons and nonbonding electrons in the image given below.

I2 step 6

For Iodine (I) atom:
Valence electron = 7 (because iodine is in group 17)
Bonding electrons = 2
Nonbonding electrons = 6

Formal charge=Valence electrons(Bonding electrons)/2Nonbonding electrons

From the above calculations of formal charge, you can see that both the iodine (I) atoms have a “zero” formal charge.

This indicates that the above lewis structure of I2 is stable and there is no further change in the above structure of I2.

In the above lewis dot structure of I2, you can also represent each bonding electron pair (:) as a single bond (|). By doing so, you will get the following lewis structure of I2.

Lewis structure of I2

I hope you have completely understood all the above steps.

For more practice and better understanding, you can try other lewis structures listed below.

Try (or at least See) these lewis structures for better understanding:

NOF lewis structureClF5 lewis structure
IF4- lewis structurePBr3 lewis structure
CO2 lewis structureSO2 lewis structure
About author

Jay is an educator and has helped more than 100,000 students in their studies by providing simple and easy explanations on different science-related topics. He is a founder of Pediabay and is passionate about helping students through his easily digestible explanations.

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