Lewis Structure of C2H4Br2 (With 6 Simple Steps to Draw!)

I’m super excited to teach you the lewis structure of C2H4Br2 in just 6 simple steps.

Infact, I’ve also given the step-by-step images for drawing the lewis dot structure of C2H4Br2 molecule.

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

Lewis structure of C2H4Br2 (1, 2-dibromoethane) contains a single bond between the two Carbon (C) atoms, Carbon-Hydrogen atoms and Carbon-Bromine atoms. The Bromine atom has 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 C2H4Br2).

6 Steps to Draw the Lewis Structure of C2H4Br2

Step #1: Calculate the total number of valence electrons

Here, the given molecule is C2H4Br2 (1, 2-dibromoethane). In order to draw the lewis structure of C2H4Br2, first of all you have to find the total number of valence electrons present in the C2H4Br2 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 C2H4Br2

• 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 Hydrogen: 

Hydrogen is a group 1 element on the periodic table. ^[2]

Hence, the valence electron present in hydrogen is 1 (see below image).

• For Bromine:

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

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

Hence in a C2H4Br2 molecule, 

Valence electrons given by each Carbon (C) atom = 4
Valence electron given by each Hydrogen (H) atom = 1
Valence electrons given by each Bromine (Br) atom = 7
So, total number of Valence electrons in C2H4Br2 molecule = 4(2) + 1(4) + 7(2) = 26

Step #2: Select the center atom (H is always outside)

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 C2H4Br2 molecule, if we compare the carbon atom (C) and bromine atom (Br), then the carbon is less electronegative than bromine.

So in a C2H4Br2 molecule (1, 2-dibromoethane), carbon is placed in the center and the bromine atoms will surround it.

Also as per the rule, we have to keep hydrogen outside.

(Note: Here the given compound is 1, 2-dibromoethane, which indicates that the bromine atom is attached to the 1st carbon as well as 2nd carbon.
If the compound is 1, 1-dibromoethane, then both the bromine atoms will be attached to the same carbon.) 

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

Now in the above sketch of C2H4Br2 molecule, put the two electrons (i.e electron pair) between the atoms to represent a chemical bond between them.

These pairs of electrons present between the Carbon-Carbon atoms, Carbon-Hydrogen atoms as well as between the Carbon-Bromine atoms form a chemical bond, which bonds these atoms with each other in a C2H4Br2 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 C2H4Br2, the outer atoms are hydrogen atoms and bromine atom.

So now, you have to check whether these hydrogen atoms are forming a duplet or not! (because hydrogen requires only 2 electrons to have a complete outer shell).

You also have to see whether the bromine atoms are forming an octet or not! (because bromine requires 8 electrons to have a complete outer shell).

You can see in the above image that both the hydrogen atoms form a duplet. And the bromine atom also forms an octet.

Also, all the 26 valence electrons of C2H4Br2 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 C2H4Br2.

Let’s move to the next step.

Step #5: Check whether the central atom has octet or not

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, the central atoms (i.e carbon), has 8 electrons. So it fulfills 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), bromine atom (Br) as well as hydrogen atoms (H).

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 Hydrogen:
  Valence electron = 1 (as it is in group 1)
  Nonbonding electrons = 0
  Bonding electrons = 2
• 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
H	=	1	–	0	–	2/2	=	0
Br	=	7	–	6	–	2/2	=	0

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

Hence, there will not be any change in the above structure and the above lewis structure of C2H4Br2 is the final stable structure only.

Each electron pair (:) in the lewis dot structure of C2H4Br2 represents the single bond ( | ). So the above lewis dot structure of C2H4Br2 can also be represented as shown below.

Related lewis structures for your practice:
Lewis structure of NO2
Lewis structure of NH3
Lewis structure of HCN
Lewis structure of H2O
Lewis structure of N2

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