Lewis Structure of C2Br4 (With 6 Simple Steps to Draw!)

Ready to learn how to draw the lewis structure of C2Br4?

Awesome!

Here, I have explained 6 simple steps to draw the lewis dot structure of C2Br4 (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 C2Br4 contains one double bond between the two Carbon (C) atoms and a single bond between Carbon (C) & Bromine (Br) atoms. The two Carbon atoms (C) are at the center and they are surrounded by 4 Bromine atoms (Br). All the four Bromine atoms have 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 C2Br4).

6 Steps to Draw the Lewis Structure of C2Br4

Step #1: Calculate the total number of valence electrons

Here, the given molecule is C2Br4. In order to draw the lewis structure of C2Br4, first of all you have to find the total number of valence electrons present in the C2Br4 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 C2Br4

• 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 Bromine:

Bromine is a group 17 element on the periodic table. ^[2]

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

Hence in a C2Br4 molecule, 

Valence electrons given by each Carbon (C) atom = 4
Valence electron given by each Bromine (Br) atom = 7
So, total number of Valence electrons in C2Br4 molecule = 4(2) + 7(4) = 36

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). ^[3]

Here in the C2Br4 molecule, if we compare the carbon atom (C) and bromine atom (Br), then carbon is less electronegative than bromine.

So, both the carbon atoms should be placed in the center and the remaining 4 bromine atoms will surround it.

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

Now in the above sketch of C2Br4 molecule, put the two electrons (i.e electron pair) between the carbon-carbon atoms and carbon-bromine atoms to represent a chemical bond between them.

These pairs of electrons present between the Carbon atoms as well as between the Carbon and Bromine atoms form a chemical bond, which bonds these atoms with each other in a C2Br4 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 C2Br4, the outer atoms are bromine atoms.

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

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

Also, only 34 valence electrons of C2Br4 molecule are used in the above structure.

But there are total 36 valence electrons in C2Br4 molecule (as calculated in step #1).

So the number of electrons left to be kept on the central atom = 36 – 34 = 2.

So let’s keep these two electrons (i.e 1 electron pair) on the central atom.

Now, 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 convert the lone pair into double bond or triple bond

In this step, we have to check whether the central atoms (i.e two carbon atoms) have an octet or not. 

In simple words, we have to check whether the central Carbon (C) atoms has 8 electrons or not.

As you can see from the above image, one carbon atom has 8 electrons, while the other carbon atom has only 6 electrons. So this carbon atom does not fulfill the octet rule.

Now, in order to fulfill the octet of this carbon atom, we have to convert the lone pair into a double bond.

Now you can see from the above image that both the central carbon atoms are having 8 electrons. So they fulfill the octet rule and both the carbon atoms are 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 atoms (C) as well as bromine atoms (Br).

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 Bromine:
  Valence electrons = 7 (as it is in group 17)
  Nonbonding electrons = 6
  Bonding electrons = 2

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

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

Hence, there will not be any change in the above structure and the above lewis structure of C2Br4 is the final stable structure only.

Each electron pair (:) in the lewis dot structure of C2Br4 represents the single bond ( | ). So the above lewis dot structure of C2Br4 can also be represented as shown below.

Related lewis structures for your practice:
Lewis Structure of TeBr2
Lewis Structure of AsF5
Lewis Structure of HI
Lewis Structure of PO3-
Lewis Structure of BBr3 

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