ICl Lewis Structure in 5 Steps (With Images)

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

Let me explain the above image in short.

ICl lewis structure has one Iodine atom (I) and one Chlorine atom (Cl) which contain a single bond between them. There are 3 lone pairs on both the Iodine atom (I) as well as Chlorine atom (Cl).

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

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

Steps of drawing ICl lewis structure

Step 1: Find the total valence electrons in ICl molecule

In order to find the total valence electrons in an ICl molecule, first of all you should know the valence electrons present in the iodine atom as well as chlorine 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 as well as chlorine using a periodic table.

Total valence electrons in ICl 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.

→ Valence electrons given by chlorine atom:

Chlorine is group 17 element on the periodic table. ^[2] Hence the valence electrons present in chlorine is 7.

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

Hence, 

Total valence electrons in ICl molecule = valence electrons given by 1 iodine atom + valence electrons given by 1 chlorine atom = 7 + 7 = 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 ICl. It has only two atoms, so you can select any of the atoms as a center atom.

Let’s assume the iodine atom as a central atom.
(You should assume the less electronegative atom as a center atom).

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

Now in the ICl molecule, you have to put the electron pairs between the iodine atom (I) and chlorine atom (Cl).

This indicates that the iodine (I) and chlorine (Cl) are chemically bonded with each other in a ICl 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 ICl molecule, we have assumed the iodine atom as a center atom. So the chlorine is the outer atom.

Hence you have to make the chlorine atom stable.

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

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

The ICl 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 iodine atom in the above sketch of ICl molecule.

Now let’s proceed to the next step.

Step 5: 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 ICl.

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 iodine (I) atom as well as chlorine (Cl) atoms present in the ICl 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 for each atom of ICl molecule in the image given below.

For Iodine (I) atom:
Valence electrons = 7 (because iodine is in group 17)
Bonding electrons = 6
Nonbonding electrons = 4

For Chlorine (Cl) atom:
Valence electrons = 7 (because chlorine is in group 17)
Bonding electrons = 2
Nonbonding electrons = 6

Formal charge	=	Valence electrons	–	(Bonding electrons)/2	–	Nonbonding electrons
I	=	7	–	6/2	–	4	=	0
Cl	=	7	–	2/2	–	6	=	0

From the above calculations of formal charge, you can see that the iodine (I) atom as well as chlorine (Cl) atom has a “zero” formal charge.

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

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

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:

HSO4- Lewis Structure	CCl2F2 Lewis Structure
C2H2Cl2 Lewis Structure	NH2OH Lewis Structure
HClO3 Lewis Structure	SF5- Lewis Structure