Lewis Structure of CH3I (With 6 Simple Steps to Draw!)

I’m super excited to teach you the lewis structure of CH3I in just 6 simple steps.

Infact, I’ve also given the step-by-step images for drawing the lewis dot structure of CH3I molecule.

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

Lewis structure of CH3I contains a single bond between the Carbon (C) & Hydrogen (H) atoms as well as between the Carbon (C) & Iodine (I) atom. The Carbon atom (C) is at the center and it is surrounded by three Hydrogen (H) and one Iodine atom (I). The Iodine atom has 3 lone pairs.

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 CH3I).

6 Steps to Draw the Lewis Structure of CH3I

Step #1: Calculate the total number of valence electrons

Here, the given molecule is CH3I. In order to draw the lewis structure of CH3I, first of all you have to find the total number of valence electrons present in the CH3I 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 CH3I

• 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 Hydrogen: 

Hydrogen is a group 1 element on the periodic table. ^[2]

Hence, the valence electron present in hydrogen is 1 (see below image).

• For Iodine:

Iodine is a group 17 element on the periodic table. ^[3]

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

Hence in a CH3I molecule, 

Valence electrons given by Carbon (C) atom = 4
Valence electron given by each Hydrogen (H) atom = 1
Valence electrons given by Iodine (I) atom = 7
So, total number of Valence electrons in CH3I molecule = 4 + 1(3) + 7 = 14

Step #2: Select the center atom (H is always outside)

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). ^[4]

Here in the CH3I molecule, if we compare the carbon atom (C) and iodine atom (I), then the carbon is less electronegative than iodine.

So, carbon should be placed in the center and the iodine atom will surround it.

Also as per the rule, we have to keep hydrogen outside.

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

Now in the above sketch of CH3I molecule, put the two electrons (i.e electron pair) between the carbon-hydrogen atoms and carbon-iodine atoms to represent a chemical bond between them.

These pairs of electrons present between the Carbon & iodine atom as well as between the Carbon & Hydrogen atoms form a chemical bond, which bonds these atoms with each other in a CH3I 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 CH3I, the outer atoms are hydrogen atoms and iodine atom.

So now, you have to check whether these hydrogen atoms are forming a duplet or not! (because hydrogen requires only 2 electrons to have a complete outer shell).

You also have to see whether the iodine atom is forming an octet or not! (because iodine requires 8 electrons to have a complete outer shell).

You can see in the above image that all the hydrogen atoms form a duplet. And the iodine atom also forms an octet.

Also, all the 14 valence electrons of CH3I 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 CH3I.

Let’s move to the next step.

Step #5: Check whether the central atom has octet or not

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), is having 8 electrons. So it fulfills the octet rule and the carbon atom is stable.

Step #6: Final step – 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), iodine (I) atom as well as hydrogen atoms (H).

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
• For Hydrogen:
  Valence electron = 1 (as it is in group 1)
  Nonbonding electrons = 0
  Bonding electrons = 2

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

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

Hence, there will not be any change in the above structure and the above lewis structure of CH3I is the final stable structure only.

Each electron pair (:) in the lewis dot structure of CH3I represents the single bond ( | ). So the above lewis dot structure of CH3I can also be represented as shown below.

Related lewis structures for your practice:
Lewis Structure of BrO-
Lewis Structure of SeOF2
Lewis Structure of SBr6
Lewis Structure of IO3-
Lewis Structure of HOFO 

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