![BrF Lewis Structure](https://pediabay.com/wp-content/uploads/2022/06/BrF-lewis-structure.jpg)
So you have seen the above image by now, right?
Let me explain the above image in short.
BrF lewis structure has one Bromine atom (Br) and one Fluorine atom (F) which contain a single bond between them. There are 3 lone pairs on both the Bromine atom (Br) as well as Fluorine atom (F).
If you haven’t understood anything from the above image of BrF lewis structure, then just stick with me and you will get the detailed step by step explanation on drawing a lewis structure of BrF.
So let’s move to the steps of drawing the lewis structure of BrF.
Steps of drawing BrF lewis structure
Step 1: Find the total valence electrons in BrF molecule
In order to find the total valence electrons in a BrF molecule, first of all you should know the valence electrons present in the bromine atom as well as fluorine atom.
(Valence electrons are the electrons that are present in the outermost orbit of any atom.)
Here, I’ll tell you how you can easily find the valence electrons of bromine as well as fluorine using a periodic table.
Total valence electrons in BrF molecule
→ Valence electrons given by bromine atom:
![](https://pediabay.com/wp-content/uploads/2022/03/2-7.jpg)
Bromine is a group 17 element on the periodic table. [1] Hence the valence electrons present in bromine is 7.
![](https://pediabay.com/wp-content/uploads/2022/03/3-7.jpg)
You can see the 7 valence electrons present in the bromine atom as shown in the above image.
→ Valence electrons given by fluorine atom:
![](https://pediabay.com/wp-content/uploads/2022/03/4-2.jpg)
Fluorine is group 17 element on the periodic table. [2] Hence the valence electron present in fluorine is 7.
![](https://pediabay.com/wp-content/uploads/2022/03/5-1.jpg)
You can see the 7 valence electrons present in the fluorine atom as shown in the above image.
Hence,
Total valence electrons in BrF molecule = valence electrons given by 1 bromine atom + valence electrons given by 1 fluorine atom = 7 + 7 = 14.
Step 2: Select the central atom
For selecting the center atom, you have to remember that the atom which is less electronegative remains at the center.
Now here the given molecule is BrF. It has only two atoms, so you can select any of the atoms as a center atom.
![BrF step 1](https://pediabay.com/wp-content/uploads/2022/06/BrF-step-1.jpg)
Let’s assume the bromine atom as a central atom.
(You should assume the less electronegative atom as a center atom).
Step 3: Connect each atoms by putting an electron pair between them
Now in the BrF molecule, you have to put the electron pairs between the bromine atom (Br) and fluorine atom (F).
![BrF step 2](https://pediabay.com/wp-content/uploads/2022/06/BrF-step-2.jpg)
This indicates that the bromine (Br) and fluorine (F) are chemically bonded with each other in a BrF molecule.
Step 4: Make the outer atoms stable. Place the remaining valence electrons pair on the central atom.
Now in this step, you have to check the stability of the outer atom.
Here in the sketch of BrF molecule, we have assumed the bromine atom as a center atom. So the fluorine is the outer atom.
Hence you have to make the fluorine atom stable.
You can see in the below image that the fluorine atom is forming an octet and hence it is stable.
![BrF step 3](https://pediabay.com/wp-content/uploads/2022/06/BrF-step-3.jpg)
Also, in step 1 we have calculated the total number of valence electrons present in the BrF molecule.
The BrF molecule has a total 14 valence electrons and out of these, only 8 valence electrons are used in the above sketch.
So the number of electrons which are left = 14 – 8 = 6.
You have to put these 6 electrons on the bromine atom in the above sketch of BrF molecule.
![BrF step 4](https://pediabay.com/wp-content/uploads/2022/06/BrF-step-4.jpg)
Now let’s proceed to the next step.
Step 5: Check the stability of lewis structure
Now you have come to the final step in which you have to check the stability of lewis structure of BrF.
The stability of lewis structure can be checked by using a concept of formal charge.
In short, now you have to find the formal charge on bromine (Br) atom as well as fluorine (F) atoms present in the BrF molecule.
For calculating the formal charge, you have to use the following formula;
Formal charge = Valence electrons – (Bonding electrons)/2 – Nonbonding electrons
You can see the number of bonding electrons and nonbonding electrons for each atom of BrF molecule in the image given below.
![BrF step 5](https://pediabay.com/wp-content/uploads/2022/06/BrF-step-6.jpg)
For Bromine (Br) atom:
Valence electrons = 7 (because bromine is in group 17)
Bonding electrons = 6
Nonbonding electrons = 4
For Fluorine (F) atom:
Valence electrons = 7 (because fluorine is in group 17)
Bonding electrons = 2
Nonbonding electrons = 6
Formal charge | = | Valence electrons | – | (Bonding electrons)/2 | – | Nonbonding electrons | ||
Br | = | 7 | – | 6/2 | – | 4 | = | 0 |
F | = | 7 | – | 2/2 | – | 6 | = | 0 |
From the above calculations of formal charge, you can see that the bromine (Br) atom as well as fluorine (F) atom has a “zero” formal charge.
This indicates that the above lewis structure of BrF is stable and there is no further change in the above structure of BrF.
In the above lewis dot structure of BrF, you can also represent each bonding electron pair (:) as a single bond (|). By doing so, you will get the following lewis structure of BrF.
![lewis structure of BrF](https://pediabay.com/wp-content/uploads/2022/06/lewis-structure-of-BrF.jpg)
I hope you have completely understood all the above steps.
For more practice and better understanding, you can try other lewis structures listed below.
Try (or at least See) these lewis structures for better understanding:
AlH3 Lewis Structure | MgF2 Lewis Structure |
SbF3 Lewis Structure | Cl3- Lewis Structure |
PCl2- Lewis Structure | AsO2- Lewis Structure |
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. He is a founder of Pediabay and is passionate about helping students through his easily digestible explanations.
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