# BrF2- Lewis structure

BrF2 has one bromine atom and two fluorine atoms.

In BrF2 Lewis structure, there are two single bonds around the bromine atom, with two fluorine atoms attached to it, and each atom has three lone pairs.

Also, there is a negative (-1) charge on the bromine atom.

Contents

## Steps

To properly draw the BrF2 Lewis structure, follow these steps:

#1 Draw a rough sketch of the structure
#2 Next, indicate lone pairs on the atoms
#3 Indicate formal charges on the atoms, if necessary

Let’s break down each step in more detail.

### #1 Draw a rough sketch of the structure

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 BrF2 has one bromine atom and two fluorine atoms, so…

Valence electrons of one bromine atom = 7 × 1 = 7
Valence electrons of two fluorine atoms = 7 × 2 = 14

Now the BrF2 has a negative (-1) charge, so we have to add one more electron.

So the total valence electrons = 7 + 14 + 1 = 22

• Second, find the total electron pairs

We have a total of 22 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 = 22 ÷ 2 = 11

• 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 Next, indicate lone pairs on the atoms

Here, we have a total of 11 electron pairs. And two Br — F bonds are already marked. So we have to only mark the remaining nine 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 atom, there are three lone pairs.

Mark the lone pairs on the sketch as follows:

### #3 Indicate formal charges on the atoms, if necessary

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 – 6 – ½ (4) = -1

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

Here, the bromine atom has a charge, so mark it on the sketch as follows:

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.

Now there is still a negative (-1) charge on the bromine atom.

This is not okay, right? Because the structure with a negative charge on the most electronegative atom is the best Lewis structure. And in this case, the most electronegative element is fluorine.

But if we convert a lone pair of the bromine atom to make a new Br — F bond with the fluorine atom, and calculate the formal charge, then we do not get the formal charges on atoms closer to zero.

And the structure with the formal charges on atoms closer to zero is the best Lewis structure.

Therefore, this structure is the most stable Lewis structure of BrF2.

And since the BrF2 has a negative (-1) charge, mention that charge on the Lewis structure by drawing brackets as follows: