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

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

Awesome!

Here, I have explained 6 simple steps to draw the lewis dot structure of HBr (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 HBr (or Hydrogen bromide) contains one single bond between the Hydrogen (H) and Bromine (Br) atom. The Bromine atom 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 HBr).

6 Steps to Draw the Lewis Structure of HBr

Step #1: Calculate the total number of valence electrons

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

• 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).

Hence in a HBr molecule, 

Valence electrons given by Bromine (Br) atom = 7
Valence electron given by Hydrogen (H) atom = 1
So, total number of Valence electrons in HBr molecule = 1 + 7 = 8

Step #2: Select the center atom (H is always outside)

While selecting the atom, you have to put the least electronegative atom at the center. 

But here in the HBr molecule, there are only two atoms. So you can consider any of the atoms as a center atom.

So, let’s assume that the bromine atom is a central atom. (Because hydrogen is always placed outside in any lewis structure.)

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

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

These pair of electrons present between the Bromine (Br) and Hydrogen (H) atoms form a chemical bond, which bonds the bromine and hydrogen atoms with each other in a HBr 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 HBr, the outer atom is hydrogen atom.

So now, you have to check whether this hydrogen atom is forming a duplet or not! (because hydrogen requires only 2 electrons to have a complete outer shell).

You can see in the above image that the hydrogen atom forms a duplet.

Also, only 2 valence electrons of HBr molecule are used in the above structure.

But there are total 8 valence electrons in HBr molecule (as calculated in step #1).

So the number of electrons left to be kept on the atom (i.e bromine atom) = 8 – 2 = 6.

So let’s keep these six electrons (i.e 3 electron pairs) on the 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 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 Bromine:
  Valence electrons = 7 (as it is in group 17)
  Nonbonding electrons = 6
  Bonding electrons = 2
• For Hydrogen:
  Valence electron = 1 (as it is in group 1)
  Nonbonding electrons = 0
  Bonding electrons = 2

Formal charge	=	Valence electrons	–	Nonbonding electrons	–	(Bonding electrons)/2
Br	=	7	–	6	–	2/2	=	0
H	=	1	–	0	–	2/2	=	0

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

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

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

Related lewis structures for your practice:
Lewis structure of N2H4
Lewis structure of CH3NH2
Lewis structure of SiO2
Lewis structure of SiH4
Lewis structure of ClO4-

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