ICl4- 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.

ICl4- lewis structure has an Iodine atom (I) at the center which is surrounded by four Chlorine atoms (Cl). There are 4 single bonds between the Iodine atom (I) and each Chlorine atom (Cl). There are 2 lone pairs on the Iodine atom (I) and 3 lone pairs on all the four Chlorine atoms (Cl).

If you haven’t understood anything from the above image of ICl4- lewis structure, then just stick with me and you will get the detailed step by step explanation on drawing a lewis structure of ICl4- ion.

So let’s move to the steps of drawing the lewis structure of ICl4- ion.

Steps of drawing ICl4- lewis structure

Step 1: Find the total valence electrons in ICl4- ion

In order to find the total valence electrons in ICl4- ion, first of all you should know the valence electrons present in 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 ICl4- ion

→ 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 ICl4- ion = valence electrons given by 1 iodine atom + valence electrons given by 1 chlorine atom + 1 more electron is added due to 1 negative charge = 7 + 7(4) + 1 = 36.

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 ion is ICl4- ion and it contains iodine atom (I) and chlorine atoms (Cl).

You can see the electronegativity values of iodine atom (I) and chlorine atom (Cl) in the above periodic table.

If we compare the electronegativity values of iodine (I) and chlorine (Cl) then the iodine atom is less electronegative.

So here the iodine atom (I) is the center atom and the chlorine atoms (Cl) are the outside atoms.

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

Now in the ICl4 molecule, you have to put the electron pairs between the iodine atom (I) and chlorine atoms (Cl).

This indicates that the iodine (I) and chlorine (Cl) are chemically bonded with each other in a ICl4 molecule.

Step 4: Make the outer atoms stable

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

Here in the sketch of ICl4 molecule, you can see that the outer atoms are chlorine atoms.

These outer chlorine atoms are forming an octet and hence they are stable.

Also, in step 1 we have calculated the total number of valence electrons present in the ICl4- ion.

The ICl4- ion has a total 36 valence electrons and out of these, only 32 valence electrons are used in the above sketch.

So the number of electrons which are left = 36 – 32 = 4.

You have to put these 4 electrons on the central iodine atom in the above sketch of ICl4 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 ICl4.

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 ICl4 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 ICl4 molecule in the image given below.

For Iodine (I) atom:
Valence electron = 7 (because iodine is in group 17)
Bonding electrons = 8
Nonbonding electrons = 4

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

Formal charge	=	Valence electrons	–	(Bonding electrons)/2	–	Nonbonding electrons
I	=	7	–	8/2	–	4	=	-1
Cl	=	7	–	2/2	–	6	=	0

From the above calculations of formal charge, you can see that the iodine (I) atom has -1 charge, while the chlorine atom has 0 charge.

So let’s keep these charges on the respective atoms in the ICl4 molecule.

This overall -1 charge on the ICl4 molecule is represented in the image given below.

In the above lewis dot structure of ICl4- ion, you can also represent each bonding electron pair (:) as a single bond (|). By doing so, you will get the following lewis structure of ICl4- ion.

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:

SiF4 lewis structure	CH3F lewis structure
NH4+ lewis structure	NH2- lewis structure
AlCl3 lewis structure	BeH2 lewis structure