# HBrO4 Lewis structure

HBrO4 (perbromic acid) has one hydrogen atom, one bromine atom, and four oxygen atoms.

In HBrO4 Lewis structure, there are three double bonds and one single bond around the bromine atom, with four oxygen atoms attached to it. The oxygen atom with a single bond is attached with one hydrogen atom, and on each oxygen atom, there are two lone pairs.

Contents

## Steps

To properly draw the HBrO4 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
#4 Minimize formal charges by converting lone pairs of the atoms
#5 Repeat step 4 if necessary, until all charges are minimized

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, hydrogen lies in group 1, bromine lies in group 17, and oxygen lies in group 16.

Hence, hydrogen has one valence electron, bromine has seven valence electrons, and oxygen has six valence electrons.

Since HBrO4 has one hydrogen atom, one bromine atom, and four oxygen atoms, so…

Valence electrons of one hydrogen atom = 1 × 1 = 1
Valence electrons of one bromine atom = 7 × 1 = 7
Valence electrons of four oxygen atoms = 6 × 4 = 24

And the total valence electrons = 1 + 7 + 24 = 32

• Second, find the total electron pairs

We have a total of 32 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 = 32 ÷ 2 = 16

• Third, determine the central atom

Here hydrogen can not be the central atom. Because the central atom is bonded with at least two other atoms, and hydrogen has only one electron in its last shell, so it can not make more than one bond.

Now we have to choose the central atom from bromine and oxygen. Place the least electronegative atom at the center.

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

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

• And finally, draw the rough sketch

### #2 Next, indicate lone pairs on the atoms

Here, we have a total of 16 electron pairs. And five bonds are already marked. So we have to only mark the remaining eleven electron pairs as lone pairs on the sketch.

Also remember that hydrogen is a period 1 element, so it can not keep more than 2 electrons in its last shell. Bromine is a period 4 element, so it can keep more than 8 electrons in its last shell. And oxygen 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 oxygens and hydrogen. But no need to mark on hydrogen, because hydrogen already has two electrons.

So for top oxygen, bottom oxygen, and left oxygen, there are three lone pairs. For right oxygen, there are two lone pairs, and for bromine, there is zero lone pair because all eleven electron pairs are over.

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 hydrogen atom, formal charge = 1 – 0 – ½ (2) = 0

For bromine atom, formal charge = 7 – 0 – ½ (8) = +3

For top oxygen, bottom oxygen, and left oxygen atom, formal charge = 6 – 6 – ½ (2) = -1

For right oxygen atom, formal charge = 6 – 4 – ½ (4) = 0

Here, both bromine and oxygen atoms have charges, so mark them on the sketch as follows:

The above structure is not a stable Lewis structure because both bromine and oxygen atoms have charges. Therefore, reduce the charges (as below) by converting lone pairs to bonds.

### #4 Minimize formal charges by converting lone pairs of the atoms

Convert a lone pair of the left oxygen atom to make a new Br — O bond with the bromine atom as follows:

### #5 Repeat step 4 (minimize charges again)

Since there are charges on bromine and oxygen atoms, again convert a lone pair of the bottom oxygen atom to make a new Br — O bond with the bromine atom as follows:

### #6 Minimize charges again

There are still charges on bromine and oxygen atoms, so again convert a lone pair of the top oxygen atom to make a new Br — O bond with the bromine atom as follows:

In the above structure, you can see that the central atom (bromine) forms an octet. The outside atoms (oxygens) also form an octet, and hydrogen forms a duet. Hence, the octet rule and duet rule are satisfied.

Also, the above structure is more stable than the previous structures. Therefore, this structure is the stable Lewis structure of HBrO4.

Next: PO2 Lewis structure