ICN Lewis Structure in 6 Steps (With Images)

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

Let me explain the above image in short.

ICN lewis structure has a Carbon atom (C) at the center which is surrounded by Iodine atom (I) and Nitrogen atom (N). There is 1 single bond between the Carbon (C) & Iodine (I) atom and 1 triple bond between the Carbon (C) & Nitrogen (N).

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

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

Steps of drawing ICN lewis structure

Step 1: Find the total valence electrons in ICN molecule

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

Total valence electrons in ICN 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 carbon atom:

Carbon is group 14 element on the periodic table. ^[2] Hence the valence electrons present in carbon is 4.

You can see the 4 valence electrons present in the carbon atom as shown in the above image.

→ Valence electrons given by nitrogen atom:

Nitrogen is a group 15 element on the periodic table. ^[3] Hence the valence electrons present in nitrogen is 5.

You can see the 5 valence electrons present in the nitrogen atom as shown in the above image.

Hence, 

Total valence electrons in ICN molecule = valence electrons given by 1 iodine atom + valence electrons given by 1 carbon atom + valence electrons given by 1 nitrogen atom = 7 + 4 + 5 = 16.

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 ICN and it contains iodine atom (I), carbon atom (C) and nitrogen atom (N).

You can see the electronegativity values of iodine atom (I), carbon atom (C) and nitrogen atom (N) in the above periodic table.

If we compare the electronegativity values of iodine atom (I), carbon atom (C) and nitrogen atom (N) then the carbon atom is less electronegative.

So here the carbon atom (C) is the center atom and the iodine atom (I) and nitrogen atom (N) are the outside atoms.

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

Now in the ICN molecule, you have to put the electron pairs between the iodine atom (I), carbon atom (C) and nitrogen atom (N).

This indicates that the iodine atom (I), carbon atom (C) and nitrogen atom (N) are chemically bonded with each other in a ICN 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 an ICN molecule, you can see that the outer atoms are iodine atom and nitrogen atom.

These outer iodine and nitrogen 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 ICN molecule.

The ICN molecule has a total 16 valence electrons and all these valence electrons are used in the above sketch.

Hence there are no remaining electron pairs to be kept on the central atom.

So now let’s proceed to the next step.

Step 5: Check the octet on the central atom. If it does not have octet, then shift the lone pair to form a double bond or triple bond.

In this step, you have to check whether the central carbon atom (C) is stable or not.

In order to check the stability of the central carbon (C) atom, we have to check whether it is forming an octet or not.

Unfortunately, the carbon atom is not forming an octet here. Carbon has only 4 electrons and it is unstable.

Now to make this carbon atom stable, you have to shift the electron pair from the outer nitrogen atom so that the carbon atom can have 8 electrons (i.e octet).

(Note: Here you have 2 choices. You can shift electron pair either from iodine or from nitrogen. But halogens generally form a single bond. So here you should shift the electron pair from nitrogen.)

But after shifting one electron pair, the carbon atom is still not forming an octet as it has only 6 electrons.

So again we have to shift one more electron pair from the nitrogen atom only.

After shifting this electron pair, the central carbon atom will get 2 more electrons and thus its total electrons will become 8.

You can see from the above picture that the carbon atom is forming an octet.

And hence the carbon atom is stable.

Now let’s proceed to the final step to check whether the lewis structure of ICN is stable or not.

Step 6: 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 ICN molecule.

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), carbon (C) atom as well as nitrogen (N) atoms present in the ICN 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 ICN molecule in the image given below.

For Iodine (I) atom:
Valence electrons = 7 (because iodine is in group 17)
Bonding electrons = 2
Nonbonding electrons = 6

For Carbon (C) atom:
Valence electrons = 4 (because carbon is in group 14)
Bonding electrons = 8
Nonbonding electrons = 0

For Nitrogen (N) atom:
Valence electrons = 5 (because nitrogen is in group 15)
Bonding electrons = 6
Nonbonding electrons = 2

From the above calculations of formal charge, you can see that the iodine (I) atom, carbon (C) atom as well as nitrogen (N) atom have a “zero” formal charge.

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

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

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:

About author

Jay Rana

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. He is a founder of Pediabay and is passionate about helping students through his easily digestible explanations.

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