# Lewis Structure of NBr3 (With 6 Simple Steps to Draw!)

Ready to learn how to draw the lewis structure of NBr3?

Awesome!

Here, I have explained 6 simple steps to draw the lewis dot structure of NBr3 (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 NBr3 contains three single bonds between the Nitrogen (N) atom and each Bromine (Br) atom. The Nitrogen atom (N) is at the center and it is surrounded by 3 Bromine atoms (Br). The Nitrogen atom has 1 lone pair and all the three 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 NBr3).

## 6 Steps to Draw the Lewis Structure of NBr3

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

Here, the given molecule is NBr3 (nitrogen tribromide). In order to draw the lewis structure of NBr3, first of all you have to find the total number of valence electrons present in the NBr3 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 NBr3

• For Nitrogen:

Nitrogen is a group 15 element on the periodic table. [1]

Hence, the valence electrons present in nitrogen is 5 (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 NBr3 molecule,

Valence electrons given by Nitrogen (N) atom = 5
Valence electrons given by each Bromine (Br) atom = 7
So, total number of Valence electrons in NBr3 molecule = 5 + 7(3) = 26

### 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). [3]

Here in the NBr3 molecule, if we compare the nitrogen atom (N) and bromine atom (Br), then the nitrogen is less electronegative than bromine.

So, nitrogen should be placed in the center and the remaining 3 bromine atoms will surround it.

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

Now in the above sketch of NBr3 molecule, put the two electrons (i.e electron pair) between each nitrogen atom and bromine atom to represent a chemical bond between them.

These pairs of electrons present between the Nitrogen (N) and Bromine (Br) atoms form a chemical bond, which bonds the nitrogen and bromine atoms with each other in a NBr3 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 NBr3, 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 24 valence electrons of NBr3 molecule are used in the above structure.

But there are total 26 valence electrons in NBr3 molecule (as calculated in step #1).

So the number of electrons left to be kept on the central atom = 26 – 24 = 2.

So let’s keep these two electrons (i.e 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 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 nitrogen) has an octet or not.

In simple words, we have to check whether the central Nitrogen (N) atom is having 8 electrons or not.

As you can see from the above image, the central atom (i.e nitrogen), has 8 electrons. So it fulfills the octet rule and the nitrogen atom is 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 nitrogen atom (N) as well as each bromine atom (Br).

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

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

• For Nitrogen:
Valence electrons = 5 (as it is in group 15)
Nonbonding electrons = 2
Bonding electrons = 6
• For Bromine:
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 nitrogen as well as bromine are “zero”.

Hence, there will not be any change in the above structure and the above lewis structure of NBr3 is the final stable structure only.

Each electron pair (:) in the lewis dot structure of NBr3 represents the single bond ( | ). So the above lewis dot structure of NBr3 can also be represented as shown below.

Related lewis structures for your practice:
Lewis Structure of SeF4
Lewis Structure of H3PO4
Lewis Structure of H2Se
Lewis Structure of SOCl2
Lewis Structure of ICl2-

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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.