CIF3 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.

CIF3 lewis structure has a Carbon atom (C) at the center which is surrounded by one Iodine atom (I) and three Fluorine atoms (F). There are single bonds between the Carbon-Iodine atoms and Carbon-Fluorine atoms. There are 3 lone pairs on the Iodine atom (I) as well as Fluorine atoms (F).

If you haven’t understood anything from the above image of CIF3 lewis structure, then just stick with me and you will get the detailed step by step explanation on drawing a lewis structure of CIF3.

So let’s move to the steps of drawing the lewis structure of CIF3.

Steps of drawing CIF3 lewis structure

Step 1: Find the total valence electrons in CIF3 molecule

In order to find the total valence electrons in a CIF3 molecule, first of all you should know the valence electrons present in carbon atom, iodine atom as well as fluorine 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 carbon, fluorine as well as iodine using a periodic table.

Total valence electrons in CIF3 molecule

→ Valence electrons given by carbon atom:

Carbon is group 14 element on the periodic table. ^[1] 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 iodine atom:

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

Fluorine is group 17 element on the periodic table. ^[3] Hence the valence electron present in fluorine is 7.

You can see the 7 valence electrons present in the fluorine atom as shown in the above image.

Hence, 

Total valence electrons in CIF3 molecule = valence electrons given by 1 carbon atom + valence electrons given by 3 fluorine atoms + valence electrons given by 1 iodine atom = 4 + 7(3) + 7 = 32.

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 CIF3 and it contains carbon atom (C), fluorine atoms (F) and iodine atom (I).

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

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

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

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

Now in the CIF3 molecule, you have to put the electron pairs between the carbon atom (C), fluorine atoms (F) and iodine atom (I).

This indicates that the carbon (C), fluorine (F) and iodine (I) are chemically bonded with each other in a CIF3 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 CIF3 molecule, you can see that the outer atoms are iodine atoms and fluorine atoms.

These outer iodine and fluorine 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 CIF3 molecule.

The CIF3 molecule has a total 32 valence electrons and all these valence electrons are used in the above sketch of CIF3.

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

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.

You can see from the above picture that the carbon atom is forming an octet. That means it has 8 electrons.

And hence the central carbon atom is stable.

Now let’s proceed to the final step to check whether the lewis structure of CIF3 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 CIF3.

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 carbon (C) atom, fluorine (F) atoms as well as iodine (I) atom present in the CIF3 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 CIF3 molecule in the image given below.

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

For Fluorine (F) atom:
Valence electrons = 7 (because fluorine is in group 17)
Bonding electrons = 2
Nonbonding electrons = 6

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

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

This indicates that the above lewis structure of CIF3 is stable and there is no further change in the above structure of CIF3.

In the above lewis dot structure of CIF3, you can also represent each bonding electron pair (:) as a single bond (|). By doing so, you will get the following lewis structure of CIF3.

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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