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

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

Awesome!

Here, I have explained 6 simple steps to draw the lewis dot structure of a BrCN molecule (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 BrCN has a triple bond between the Carbon-Nitrogen atom and a single bond between Carbon-Bromine atom. The Carbon atom (C) is at the center and it is surrounded by Bromine and Nitrogen atoms. The Bromine atom has 3 lone pairs and the Nitrogen atom has 1 lone pair, while the carbon atom does not have 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 BrCN).

## 6 Steps to Draw the Lewis Structure of BrCN

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

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

• For Bromine:

Bromine is a group 17 element on the periodic table. [1]

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

• For Carbon:

Carbon is a group 14 element on the periodic table. [2]

Hence, the valence electrons present in carbon is 4 (see below image).

• For Nitrogen:

Nitrogen is a group 15 element on the periodic table. [3]

Hence, the valence electrons present in nitrogen is 5 (see below image).

Hence in a BrCN molecule,

Valence electrons given by Bromine (Br) atom = 7
Valence electrons given by Carbon (C) atom = 4
Valence electrons given by Nitrogen (N) atom = 5
So, total number of Valence electrons in BrCN molecule = 7 + 4 + 5 = 16

### Step #2: Select the center atom

While selecting the 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 BrCN molecule, if we compare the bromine atom (Br), carbon atom (C) and nitrogen atom (N), then the carbon is less electronegative.

So, carbon should be placed in the center and the remaining bromine and nitrogen atom will surround it.

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

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

These pairs of electrons present between the Bromine (Br), Carbon (C) and Nitrogen (N) atoms form a chemical bond, which bonds these atoms with each other in a BrCN 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 BrCN, the outer atoms are bromine atom and nitrogen atom.

So now, you have to complete the octet on these outer atoms.

Now, you can see in the above image that both the bromine atom as well as nitrogen atom form an octet.

Also, all the 16 valence electrons of BrCN 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 BrCN.

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 has 8 electrons or not.

As you can see from the above image, the central atom (i.e carbon) has only 4 electrons. So it does not fulfill the octet rule.

Now, in order to fulfill the octet of carbon atom, we have to move the electron pair from the outer atom to form a double bond.

Now the question is, from which atom should we move the electron pair?
From nitrogen? or
From Bromine? or
Both?

Well, the halogens (like fluorine, chlorine, bromine, etc.) generally form a single bond. So here you have to shift the electron pair from the nitrogen atom so that it can form a double (or triple) bond.

Still, the octet of carbon atom is not fulfilled as it has only 6 electrons.

So again moving the electron pair from nitrogen atom only, we will get the following structure.

Now you can see from the above image that the central atom (i.e carbon), is having 8 electrons. So it fulfills the octet rule.

### 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 Bromine atom (Br), Carbon atom (C) as well as Nitrogen atom (N).

For that, you need to remember the formula of formal charge;

Formal charge = Valence electrons – Nonbonding electrons – (Bonding electrons)/2

• For Bromine:
Valence electrons = 7 (as it is in group 17)
Nonbonding electrons = 6
Bonding electrons = 2
• For Carbon:
Valence electron = 4 (as it is in group 14)
Nonbonding electrons = 0
Bonding electrons = 8
• For Nitrogen:
Valence electron = 5 (as it is in group 15)
Nonbonding electrons = 2
Bonding electrons = 6

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

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

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

Related lewis structures for your practice:
Lewis Structure of BeI2
Lewis Structure of CHBr3
Lewis Structure of SiCl2Br2
Lewis Structure of SbF5
Lewis Structure of ClBr3

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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 learning platform that provides students with easily understandable explanations.