# Lewis Structure of CIF3 (With 6 Simple Steps to Draw!)

Ready to learn how to draw the lewis structure of CIF3?

Awesome!

Here, I have explained 6 simple steps to draw the lewis dot structure of CIF3 (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 CIF3 contains a single bond between the Carbon-Fluorine atoms and Carbon-Iodine atoms. The Carbon atom (C) is at the center and it is surrounded by 3 Fluorine atoms (F) and 1 Iodine atom (I). The Carbon atom does not have a lone pair while the iodine & fluorine 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 CIF3).

## 6 Steps to Draw the Lewis Structure of CIF3

### Step #1: Calculate the total number of valence electrons

Here, the given molecule is CIF3. In order to draw the lewis structure of CIF3, first of all you have to find the total number of valence electrons present in the CIF3 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 CIF3

• 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).

• For Fluorine:

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

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

Hence in a CIF3 molecule,

Valence electrons given by Carbon (C) atom = 4
Valence electrons given by each Fluorine (F) atom = 7
Valence electrons given by Iodine (I) atom = 7
So, total number of Valence electrons in CIF3 molecule = 4 + 7 + 7(3) = 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). [4]

Here in the CIF3 molecule, if we compare the carbon atom (C), iodine atom (I) and fluorine atom (F), then carbon is less electronegative.

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

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

Now in the above sketch of a CIF3 molecule, put the two electrons (i.e electron pair) between each carbon-iodine and carbon-fluorine atoms to represent a chemical bond between them.

These pairs of electrons present between the Carbon (C), Iodine (I) and fluorine (F) atoms form a chemical bond, which bonds these atoms in a CIF3 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 CIF3, the outer atoms are iodine atoms and fluorine atoms.

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

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

Also, all the 32 valence electrons of CIF3 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 CIF3.

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), iodine atom (I) as well as fluorine atoms (F).

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 Fluorine:
Valence electron = 7 (as it is in group 17)
Nonbonding electrons = 6
Bonding electrons = 2
• For Iodine:
Valence electron = 7 (as it is in group 17)
Nonbonding electrons = 6
Bonding electrons = 2

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

Hence, there will not be any change in the above structure and the above lewis structure of CIF3 is the final stable structure only.

Each electron pair (:) in the lewis dot structure of CIF3 represents the single bond ( | ). So the above lewis dot structure of CIF3 can also be represented as shown below.

Related lewis structures for your practice:
Lewis Structure of ClCN
Lewis Structure of CH2S
Lewis Structure of BrF4-
Lewis Structure of AsF6-
Lewis Structure of SCl6

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