BrCl4- Lewis structure

BrCl4 has one bromine atom and four chlorine atoms.

In BrCl4 Lewis structure, there are four single bonds around the bromine atom, with four chlorine atoms attached to it. Each chlorine atom has three lone pairs, and the bromine atom has two lone pairs.

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

Contents

Steps

Use these steps to correctly draw the BrCl4 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 chlorine lie in group 17.

Hence, both bromine and chlorine have seven valence electrons.

Since BrCl4 has one bromine atom and four chlorine atoms, so…

Valence electrons of one bromine atom = 7 × 1 = 7
Valence electrons of four chlorine atoms = 7 × 4 = 28

Now the BrCl4 has a negative (-1) charge, so we have to add one more electron.

So the total valence electrons = 7 + 28 + 1 = 36

• Second, find the total electron pairs

We have a total of 36 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 = 36 ÷ 2 = 18

• Third, determine the central atom

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

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

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

• And finally, draw the rough sketch

#2 Mark lone pairs on the atoms

Here, we have a total of 18 electron pairs. And four Br — Cl bonds are already marked. So we have to only mark the remaining fourteen 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 chlorine is a period 3 element, so it can keep more than 8 electrons in its last shell.

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

So for each chlorine, there are three lone pairs, and for bromine, there are two lone pairs.

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 – 4 – ½ (8) = -1

For each chlorine 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 (chlorines) 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 chlorine.

But if we convert a lone pair of the bromine atom to make a new Br — Cl bond with the chlorine 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 BrCl4.

And since the BrCl4 has a negative (-1) charge, mention that charge on the Lewis structure by drawing brackets as follows:

Next: XeO2 Lewis structure