Lewis Structure of CI4 (With 6 Simple Steps to Draw!)

I’m super excited to teach you the lewis structure of CI4 in just 6 simple steps.

Infact, I’ve also given the step-by-step images for drawing the lewis dot structure of CI4 molecule.

So, if you are ready to go with these 6 simple steps, then let’s dive right into it!

Lewis structure of CI4 contains four single bonds between the Carbon (C) atom and each Iodine (I) atom. The Carbon atom (C) is at the center and it is surrounded by 4 Iodine atoms (I). The Carbon atom does not have a lone pair while all four iodine atoms have three lone pairs each.

Let’s draw and understand this lewis dot structure step by step.

(Note: Take a pen and paper with you and try to draw this lewis structure along with me. I am sure you will definitely learn how to draw lewis structure of CI4).

6 Steps to Draw the Lewis Structure of CI4

Step #1: Calculate the total number of valence electrons

Here, the given molecule is CI4. In order to draw the lewis structure of CI4, first of all you have to find the total number of valence electrons present in the CI4 molecule.
(Valence electrons are the number of electrons present in the outermost shell of an atom).

So, let’s calculate this first.

Calculation of valence electrons in CI4

• For Carbon:

Carbon is a group 14 element on the periodic table. ^[1]

Hence, the valence electrons present in carbon is 4 (see below image).

• For Iodine:

Iodine is a group 17 element on the periodic table. ^[2]

Hence, the valence electrons present in iodine is 7 (see below image).

Hence in a CI4 molecule, 

Valence electrons given by Carbon (C) atom = 4
Valence electrons given by each Iodine (I) atom = 7
So, total number of Valence electrons in CI4 molecule = 4 + 7(4) = 32

Step #2: Select the center atom

While selecting the center atom, always put the least electronegative atom at the center.

(Remember: Fluorine is the most electronegative element on the periodic table and the electronegativity decreases as we move right to left in the periodic table as well as top to bottom in the periodic table). ^[3]

Here in the CI4 molecule, if we compare the carbon atom (C) and iodine atom (I), then carbon is less electronegative than iodine.

So, carbon should be placed in the center and the remaining 4 iodine atoms will surround it.

Step #3: Put two electrons between the atoms to represent a chemical bond

Now in the above sketch of CI4 molecule, put the two electrons (i.e electron pair) between each carbon atom and iodine atom to represent a chemical bond between them.

These pairs of electrons present between the Carbon (C) and Iodine (I) atoms form a chemical bond, which bonds the carbon and iodine atoms with each other in a CI4 molecule.

Step #4: Complete the octet (or duplet) on outside atoms. If the valence electrons are left, then put the valence electrons pair on the central atom

Don’t worry, I’ll explain!

In the Lewis structure of CI4, the outer atoms are iodine atoms.

So now, you have to complete the octet on these iodine atoms (because iodine requires 8 electrons to have a complete outer shell).

Now, you can see in the above image that all the iodine atoms form an octet.

Also, all the 32 valence electrons of CI4 molecule (as calculated in step #1) are used in the above structure. So there are no remaining electron pairs.

Hence there is no change in the above sketch of CI4.

Let’s move to the next step.

Step #5: Check whether the central atom has octet or not. If it does not have an octet, then move the electron pair from the outer atom to form a double bond or triple bond

In this step, we have to check whether the central atom (i.e carbon) has an octet or not. 

In simple words, we have to check whether the central Carbon (C) atom is having 8 electrons or not.

As you can see from the above image, the central atom (i.e carbon), has 8 electrons. So it fulfills the octet rule and the carbon atom is stable.

Step #6: Check the stability of lewis structure by calculating the formal charge on each atom

Now, you have come to the final step and here you have to check the formal charge on carbon atom (C) as well as each iodine atom (I).

For that, you need to remember the formula of formal charge;

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

• For Carbon:
  Valence electrons = 4 (as it is in group 14)
  Nonbonding electrons = 0
  Bonding electrons = 8
• For Iodine:
  Valence electron = 7 (as it is in group 17)
  Nonbonding electrons = 6
  Bonding electrons = 2

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

So you can see above that the formal charges on carbon as well as iodine are “zero”.

Hence, there will not be any change in the above structure and the above lewis structure of CI4 is the final stable structure only.

Each electron pair (:) in the lewis dot structure of CI4 represents the single bond ( | ). So the above lewis dot structure of CI4 can also be represented as shown below.

Related lewis structures for your practice:
Lewis Structure of BI3
Lewis Structure of CH3I
Lewis Structure of BrO-
Lewis Structure of SeOF2
Lewis Structure of SBr6 

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