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

Lewis Structure of HBrO

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

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

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

Lewis structure of HBrO contains a single bond between the Hydrogen (H) & Oxygen (O) atom as well as between the Oxygen (O) and Bromine (Br) atom. The Oxygen atom (O) is at the center and it is surrounded by Hydrogen and Bromine atom. The 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 HBrO).

6 Steps to Draw the Lewis Structure of HBrO

Step #1: Calculate the total number of valence electrons

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

  • For Hydrogen: 

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

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

  • For Oxygen:

Oxygen is a group 16 element on the periodic table. [3]

Hence, the valence electron present in oxygen is 6 (see below image).

Hence in a HBrO molecule, 

Valence electron given by Hydrogen (H) atom = 1
Valence electrons given by Oxygen (O) atom = 6
Valence electrons given by Bromine (Br) atom = 7
So, total number of Valence electrons in HBrO molecule = 1 + 7 + 6 = 14

Step #2: Make the rough sketch

From the chemical formula (i.e HOBr), you can get the idea that an oxygen atom is at the center which is surrounded by Hydrogen (H) and Bromine (Br) on the outer sides.

So let’s draw a rough sketch for the HOBr molecule.

step 1

(Note: If you put a bromine atom at the center then, you will not get the most stable structure. Hence the Oxygen atom is kept in the center. You can check the stability of the lewis structure by calculating the formal charges as mentioned in the step #6).

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

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

step 2

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

Hydrogen already has a duplet (see below image).

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

step 3

Now, you can see in the above image that the bromine atom forms an octet.

Also, only 10 valence electrons of HBrO molecule are used in the above structure.

But there are total 14 valence electrons in HBrO molecule (as calculated in step #1).

So the number of electrons left to be kept on the central atom = 14 – 10 = 4.

So let’s keep these four electrons (i.e 2 electron pairs) 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 oxygen) has an octet or not.

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

step 5

As you can see from the above image, the central atom (i.e oxygen), has 8 electrons. So it fulfills the octet rule and the oxygen 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 hydrogen atom (H), 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 6
  • For Hydrogen:
    Valence electrons = 1 (as it is in group 1)
    Nonbonding electrons = 0
    Bonding electrons = 2
  • 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
H=102/2=0
O=644/2=0
Br=762/2=0

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

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

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

hbro lewis structure

Related lewis structures for your practice:
Lewis Structure of IO2-
Lewis Structure of CI4
Lewis Structure of BI3
Lewis Structure of CH3I
Lewis Structure of BrO- 


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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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