# HBr Lewis structure

HBr (hydrogen bromide) has one hydrogen atom and one bromine atom.

In the HBr Lewis structure, there is a single bond between the hydrogen and bromine atom, and on the bromine atom, there are three lone pairs.

Contents

## Steps

Here’s how you can easily draw the HBr 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, and bromine lies in group 17.

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

Since HBr has one hydrogen atom and one bromine atom, so…

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

And the total valence electrons = 1 + 7 = 8

• Second, find the total electron pairs

We have a total of 8 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 = 8 ÷ 2 = 4

• 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.

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

• And finally, draw the rough sketch

### #2 Mention lone pairs on the atoms

Here, we have a total of 4 electron pairs. And one H — Br bond is already marked. So we have to only mark the remaining three 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. And bromine is a period 4 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 atom is hydrogen. But no need to mark on hydrogen, because hydrogen already has two electrons.

So for bromine, there are three lone pairs.

Mark the lone pair 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

Here, both hydrogen and bromine 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 atom (hydrogen) also forms a duet. Hence, the octet rule and duet rule are satisfied.

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

Next: N2H4 Lewis structure