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

Lewis Structure of ICN

Ready to learn how to draw the lewis structure of ICN?

Awesome!

Here, I have explained 6 simple steps to draw the lewis dot structure of ICN molecule (along with images).

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

Lewis structure of ICN has a triple bond between the Carbon-Nitrogen atom and a single bond between Carbon-Iodine atom. The Carbon atom (C) is at the center and it is surrounded by Iodine and Nitrogen atoms. The Iodine atom has 3 lone pairs and the Nitrogen atom has 1 lone pair, while the carbon atom does not have 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 ICN).

6 Steps to Draw the Lewis Structure of ICN

Step #1: Calculate the total number of valence electrons

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

  • For Iodine:

Iodine is a group 17 element on the periodic table.

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

  • For Carbon:

Carbon is a group 14 element on the periodic table.

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

  • For Nitrogen:

Nitrogen is a group 15 element on the periodic table.

Hence, the valence electrons present in nitrogen is 5 (see below image).

Hence in a ICN molecule, 

Valence electrons given by Iodine (I) atom = 7
Valence electrons given by Carbon (C) atom = 4
Valence electrons given by Nitrogen (N) atom = 5
So, total number of Valence electrons in ICN molecule = 7 + 4 + 5 = 16

Step #2: Select the center atom

While selecting the 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). [1]

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

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

step 1

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

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

step 2

These pairs of electrons present between the Iodine (I), Carbon (C) and Nitrogen (N) atoms form a chemical bond, which bonds these atoms with each other in a ICN 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 ICN, the outer atoms are iodine atom and nitrogen atom.

So now, you have to complete the octet on these outer atoms.

step 3

Now, you can see in the above image that both the iodine atom as well as nitrogen atom form an octet.

Also, all the 16 valence electrons of ICN 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 ICN.

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 has 8 electrons or not.

step 4

As you can see from the above image, the central atom (i.e carbon) has only 4 electrons. So it does not fulfill the octet rule.

Now, in order to fulfill the octet of carbon atom, we have to move the electron pair from the outer atom to form a double bond.

step 5

Still, the octet of carbon atom is not fulfilled as it has only 6 electrons.

So again moving the electron pair from nitrogen atom only, we will get the following structure.

step 6

Now you can see from the above image that the central atom (i.e carbon), is having 8 electrons. So it fulfills the octet rule.

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 Iodine atom (I), Carbon atom (C) as well as Nitrogen atom (N).

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

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

step 7
  • For Iodine:
    Valence electrons = 7 (as it is in group 17)
    Nonbonding electrons = 6
    Bonding electrons = 2
  • For Carbon:
    Valence electron = 4 (as it is in group 14)
    Nonbonding electrons = 0
    Bonding electrons = 8
  • For Nitrogen:
    Valence electron = 5 (as it is in group 15)
    Nonbonding electrons = 2
    Bonding electrons = 6
Formal charge=Valence electronsNonbonding electrons(Bonding electrons)/2
I=762/2=0
C=408/2=0
N=526/2=0

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

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

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

icn lewis structure

Related lewis structures for your practice:
Lewis Structure of P2H4
Lewis Structure of SI6
Lewis Structure of CBr2F2
Lewis Structure of SiH3-
Lewis Structure of AsBr3 


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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. With a desire to make learning accessible for everyone, he founded Knords Learning, an online learning platform that provides students with easily understandable explanations.

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