# Lewis Structure of HBrO4 (With 6 Simple Steps to Draw!)

Ready to learn how to draw the lewis structure of HBrO4?

Awesome!

Here, I have explained 6 simple steps to draw the lewis dot structure of HBrO4 (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 HBrO4 contains the Bromine (Br) atom at the center which is surrounded by three Oxygen atoms (O) and one O-H group. The Bromine atom is double bonded with 3 Oxygen atoms and it is single bonded with O-H group. All the Oxygen atoms have 2 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 HBrO4).

## 6 Steps to Draw the Lewis Structure of HBrO4

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

Here, the given molecule is HBrO4. In order to draw the lewis structure of HBrO4, first of all you have to find the total number of valence electrons present in the HBrO4 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 HBrO4

• For Hydrogen:

Hydrogen is a group 1 element on the periodic table.

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

• For Bromine:

Bromine is a group 17 element on the periodic table.

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

• For Oxygen:

Oxygen is a group 16 element on the periodic table.

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

Hence in a HBrO4 molecule,

Valence electrons given by each Hydrogen (H) atom = 1
Valence electrons given by Bromine (Br) atom = 7
Valence electrons given by each Oxygen (O) atom = 6
So, total number of Valence electrons in HBrO4 molecule = 1 + 7 + 6(4) = 32

### 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). [1]

Here in the HBrO4 molecule, if we compare the bromine atom (Br), oxygen atom (O) and hydrogen atom (H), then hydrogen is less electronegative than bromine and oxygen. But as per the rule, we have to keep hydrogen outside.

So, bromine (which is less electronegative than oxygen) should be placed in the center and the remaining oxygen atoms as well as OH group will surround it.

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

Now in the above sketch of HBrO4, put the two electrons (i.e electron pair) between each bromine atom, oxygen atom and hydrogen atom to represent a chemical bond between them.

These pairs of electrons present between the Bromine (Br), Oxygen (O) and Hydrogen (H) atoms form a chemical bond, which bonds these atoms with each other in a HBrO4 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 HBrO4, the outer atoms are hydrogen atom as well as oxygen atom.

Hydrogen already has a duplet (see below image).

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

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

Also, all the 32 valence electrons of HBrO4 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 HBrO4.

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 atom (i.e bromine) has an octet or not.

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

As you can see from the above image, the central atom (i.e bromine), 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), oxygen atoms (O) 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 Hydrogen:
Valence electron = 1 (as it is in group 1)
Nonbonding electrons = 0
Bonding electrons = 2
• For Bromine:
Valence electron = 7 (as it is in group 17)
Nonbonding electrons = 0
Bonding electrons = 8
• For Oxygen:
Valence electron = 6 (as it is in group 16)
Nonbonding electrons = 6
Bonding electrons = 2
• For Oxygen (of O-H group):
Valence electron = 6 (as it is in group 16)
Nonbonding electrons = 4
Bonding electrons = 4

So you can see above that the formal charges on bromine is +3 and the formal charge on the three oxygen atoms is -1.

This indicates that the above lewis structure of HBrO4 is not stable and so we have to minimize the charges to get a more stable lewis structure.

This can be done by shifting the lone pair from negatively charged oxygen atoms to the positively charged bromine atom to form a double bond.

Now, in the above structure, you can see that the charges are minimized and the above lewis structure of HBrO4 is the final stable structure.

Each electron pair (:) in the lewis dot structure of HBrO4 represents the single bond ( | ). So the above lewis dot structure of HBrO4 can also be represented as shown below.

Related lewis structures for your practice:
Lewis Structure of PO2-
Lewis Structure of TeF5-
Lewis Structure of SeCl6
Lewis Structure of SeBr2
Lewis Structure of HCP

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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.