BrF_{5} (bromine pentafluoride) has **one bromine** atom and **five fluorine** atoms.

In the BrF_{5} Lewis structure, there are five single bonds around the bromine atom, with five fluorine atoms attached to it. Each fluorine atom has three lone pairs, and the bromine atom has one lone pair.

## Steps

Use these steps to correctly draw the BrF_{5} Lewis structure:

#1 First draw a rough sketch

#2 Mark lone pairs on the atoms

#3 Calculate and mark formal charges on the atoms, if required

Let’s discuss each step in more detail.

### #1 First draw a rough sketch

- First, determine the total number of valence electrons

In the periodic table, both bromine and fluorine lie in group 17.

Hence, both bromine and fluorine have **seven** valence electrons.

Since BrF_{5} has one bromine atom and five fluorine atoms, so…

Valence electrons of one bromine atom = 7 × 1 = 7

Valence electrons of five fluorine atoms = 7 × 5 = 35

And the **total valence electrons** = 7 + 35 = 42

Learn how to find: Bromine valence electrons and Fluorine valence electrons

- Second, find the total electron pairs

We have a total of 42 valence electrons. And when we divide this value by two, we get the value of total electron pairs.

Total electron pairs = total valence electrons ÷ 2

So the **total electron pairs** = 42 ÷ 2 = 21

- Third, determine the central atom

We have to place the least electronegative atom at the center.

Since bromine is less electronegative than fluorine, assume that the **central atom is bromine**.

Therefore, place bromine in the center and fluorines on either side.

- And finally, draw the rough sketch

### #2 Mark lone pairs on the atoms

Here, we have a total of 21 electron pairs. And five Br — F bonds are already marked. So we have to only mark the remaining sixteen electron pairs as lone pairs on the sketch.

Also remember that bromine is a period 4 element, so it can keep more than 8 electrons in its last shell. And fluorine is a period 2 element, so it can not keep more than 8 electrons in its last shell.

Always start to mark the lone pairs from outside atoms. Here, the outside atoms are fluorines.

So for each fluorine, there are **three** lone pairs, and for bromine, there is **one** lone pair.

Mark the lone pairs on the sketch as follows:

### #3 Calculate and mark formal charges on the atoms, if required

Use the following formula to calculate the formal charges on atoms:

Formal charge = valence electrons – nonbonding electrons – ½ bonding electrons

For **bromine** atom, formal charge = 7 – 2 – ½ (10) = 0

For **each fluorine** atom, formal charge = 7 – 6 – ½ (2) = 0

Here, both bromine and fluorine atoms do not have charges, so no need to mark the charges.

In the above structure, you can see that the central atom (bromine) forms an octet. And the outside atoms (fluorines) also form an octet. Hence, the octet rule is satisfied.

Therefore, this structure is the **stable Lewis structure** of BrF_{5}.

**Next:** NCl_{3} Lewis structure

## External links

- https://topblogtenz.com/bromine-pentafluoride-brf5-lewis-dot-structure-and-molecular-geometry/
- https://techiescientist.com/brf5-lewis-structure/
- https://www.thegeoexchange.org/chemistry/bonding/Lewis-Structures/BrF5-lewis-structure.html
- https://lambdageeks.com/brf5-lewis-structure/
- https://geometryofmolecules.com/brf5-lewis-structure-molecular-structure-hybridization-bond-angle-and-shape/
- https://sciedutut.com/brf5-lewis-structure/