# Lewis Structure of HBrO2 (With 6 Simple Steps to Draw!)

Ready to learn how to draw the lewis structure of HBrO2?

Awesome!

Here, I have explained 6 simple steps to draw the lewis dot structure of HBrO2 (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 HBrO2 contains one double bond between the Bromine atom (Br) & one Oxygen atom (O) and the rest other atoms are single bonded with each other. The bromine atom is at the center and it is surrounded by 1 oxygen atom and one O-H group.

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 HBrO2).

## 6 Steps to Draw the Lewis Structure of HBrO2

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

Here, the given molecule is HBrO2. In order to draw the lewis structure of HBrO2, first of all you have to find the total number of valence electrons present in the HBrO2 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 HBrO2

• 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 HBrO2 molecule,

Valence electrons given by 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 HBrO2 molecule = 1 + 7 + 6(2) = 20

### 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 HBrO2 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 atom 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 HBrO2 molecule, put the two electrons (i.e electron pair) between the bromine-oxygen atoms and oxygen-hydrogen atoms to represent a chemical bond between them.

These pairs of electrons present between the Bromine-Oxygen atoms as well as between the Oxygen & Hydrogen atoms form a chemical bond, which bonds these atoms with each other in a HBrO2 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 HBrO2, the outer atoms are hydrogen atom as well as oxygen atoms.

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, only 16 valence electrons of HBrO2 molecule are used in the above structure.

But there are total 20 valence electrons in HBrO2 molecule (as calculated in step #1).

So the number of electrons left to be kept on the central bromine atom = 20 – 16 = 4.

So let’s keep these four electrons (i.e 2 electron pairs) on the central bromine atom.

Now, 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 and the bromine 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 oxygen atoms (O), bromine atom (Br) as well as hydrogen atom (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 electrons = 7 (as it is in group 17)
Nonbonding electrons = 4
Bonding electrons = 4
• 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 charge on bromine is +1 and the formal charge on the oxygen atom is -1.

This indicates that the above lewis structure of HBrO2 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 atom to the positively charged bromine atom to form a bond.

Now, in the above structure, you can see that the charges are minimized and the above lewis structure of HBrO2 is the final stable structure.

Each electron pair (:) in the lewis dot structure of HBrO2 represents the single bond ( | ). So the above lewis dot structure of HBrO2 can also be represented as shown below.

Related lewis structures for your practice:
Lewis Structure of HBrO3
Lewis Structure of HBrO4
Lewis Structure of PO2-
Lewis Structure of TeF5-
Lewis Structure of SeCl6

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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.