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

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

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

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

Lewis structure of BrF contains one single bond between the Bromine (Br) and Fluorine (F) atom. And both the Bromine and Fluorine atoms have three lone pairs on it.

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 BrF).

6 Steps to Draw the Lewis Structure of BrF

Step #1: Calculate the total number of valence electrons

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

• 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 Fluorine: 

Fluorine is a group 17 element on the periodic table.

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

Hence in a BrF molecule, 

Valence electrons given by Bromine (Br) atom = 7
Valence electrons given by Fluorine (F) atom = 7
So, total number of Valence electrons in BrF molecule = 7 + 7 = 14

Step #2: Select the center atom

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

But here in the BrF 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. (You should assume the less electronegative atom as a central atom.)

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

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

These pair of electrons present between the Bromine (Br) and Fluorine (F) atoms form a chemical bond, which bonds both these atoms with each other in a BrF molecule.

Step #4: Complete the octet (or duplet) on outside atom. 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 BrF, we have just assumed the bromine atom as a central atom and so the fluorine atom is an outer atom.

So now, we have to complete the octet on the fluorine atom.

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

Also, only 8 valence electrons of BrF molecule are used in the above structure.

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

So the number of electrons left to be kept on the central atom = 14 – 8 = 6.

So let’s keep these six electrons (i.e 3 electron pairs) on the central atom (i.e 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 atom) has an octet or not. 

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

As you can see from the above image, the central atom (i.e bromine atom) has 8 electrons. So it fulfills the octet rule and this bromine atom is also 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 the bromine atom (Br) as well as fluorine atom (F).

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

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

• For Bromine:
  Valence electron = 7 (as it is in group 17)
  Nonbonding electrons = 6
  Bonding electrons = 2
• For Fluorine:
  Valence electron = 7 (as it is in group 17)
  Nonbonding electrons = 6
  Bonding electrons = 2

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

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

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

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

Related lewis structures for your practice:
Lewis Structure of AlH3
Lewis Structure of MgF2
Lewis Structure of SbF3
Lewis Structure of Cl3-
Lewis Structure of PCl2- 

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