# HBrO Lewis structure

HBrO (hypobromous acid) has one hydrogen atom, one bromine atom, and one oxygen atom.

In HBrO Lewis structure, there are two single bonds around the oxygen atom, with one bromine atom and one hydrogen atom attached to it. The bromine atom has three lone pairs, and the oxygen atom has two lone pairs.

Contents

## Steps

Here’s how you can easily draw the HBrO Lewis structure step by step:

#1 Draw a rough skeleton structure
#2 Mention lone pairs on the atoms
#3 If needed, mention formal charges on the atoms

Now, let’s take a closer look at each step mentioned above.

### #1 Draw a rough skeleton 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 HBrO has one hydrogen atom, one bromine atom, and one oxygen atom, so…

Valence electrons of one hydrogen atom = 1 × 1 = 1
Valence electrons of one bromine atom = 7 × 1 = 7
Valence electrons of one oxygen atom = 6 × 1 = 6

And the total valence electrons = 1 + 7 + 6 = 14

• Second, find the total electron pairs

We have a total of 14 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 = 14 ÷ 2 = 7

• 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, the central atom should be bromine, right?

But if we place bromine in the center and hydrogen and oxygen outside, 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.

Hence, here we have to assume that the central atom is oxygen.

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

• And finally, draw the rough sketch

### #2 Mention lone pairs on the atoms

Here, we have a total of 7 electron pairs. And two bonds are already marked. So we have to only mark the remaining five 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 bromine and hydrogen. But no need to mark on hydrogen, because hydrogen already has two electrons.

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

Mark the lone pairs on the sketch as follows:

### #3 If needed, mention formal charges on the atoms

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 – 6 – ½ (2) = 0

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

Here, the atoms do not have charges, so no need to mark the charges.

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

Therefore, this structure is the stable Lewis structure of HBrO.

Next: IO2 Lewis structure