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

Lewis Structure of NOBr

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

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

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

Lewis structure of NOBr contains a double bond between the Nitrogen (N) & Oxygen (O) atom and a single bond between the Nitrogen (N) and Bromine (Br) atom. The Nitrogen atom (N) is at the center and it is surrounded by Oxygen and Bromine atoms. The Nitrogen has 1 lone pair, Oxygen has 2 lone pairs and the Bromine 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 NOBr).

6 Steps to Draw the Lewis Structure of NOBr

Step #1: Calculate the total number of valence electrons

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

  • 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 Oxygen:

Oxygen is a group 16 element on the periodic table. [2]

Hence, the valence electron present in oxygen is 6 (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 NOBr molecule, 

Valence electrons given by Nitrogen (N) atom = 5
Valence electrons given by Oxygen (O) atom = 6
Valence electrons given by Bromine (Br) atom = 7
So, total number of Valence electrons in NOBr molecule = 5 + 6 + 7 = 18

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 NOBr molecule, if we compare the nitrogen atom (N), oxygen atom (O) and bromine (Br) atom, then the nitrogen is less electronegative than oxygen and bromine.

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

step 1

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

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

step 2

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

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

step 3

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

Also, only 16 valence electrons of NOBr molecule are used in the above structure.

But there are total 18 valence electrons in NOBr molecule (as calculated in step #1).

So the number of electrons left to be kept on the central atom = 18 – 16 = 2.

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

step 4

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 has 8 electrons or not.

step 5

As you can see from the above image, the central atom (i.e nitrogen) has only 6 electrons. So it does not fulfill the octet rule.

Now, in order to fulfill the octet of nitrogen atom, we have to move the electron pair from the outer atom (i.e oxygen atom) to form a double bond.

step 6

Now you can see from the above image that the central atom (i.e nitrogen), 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 nitrogen atom (N), oxygen atom (O) as well as bromine atom (Br).

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

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

step 7
  • For Nitrogen:
    Valence electrons = 5 (as it is in group 15)
    Nonbonding electrons = 2
    Bonding electrons = 6
  • For Oxygen:
    Valence electron = 6 (as it is in group 16)
    Nonbonding electrons = 4
    Bonding electrons = 4
  • For Bromine:
    Valence electron = 7 (as it is in group 17)
    Nonbonding electrons = 6
    Bonding electrons = 2
Formal charge=Valence electronsNonbonding electrons(Bonding electrons)/2

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

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

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

nobr lewis structure

Related lewis structures for your practice:
Lewis Structure of ClF2-
Lewis Structure of ClF4-
Lewis Structure of CIF3
Lewis Structure of ClCN
Lewis Structure of CH2S 

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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.

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