Lewis structure of HBr

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Lewis structure of HBr
Lewis structure of HBr | Image: Root Memory

The Lewis structure of HBr contains a single bond between the hydrogen atom and bromine atom. There are three lone pairs on the bromine atom, and the hydrogen atom does not have any lone pair.

Steps

By using the following steps, you can easily draw the Lewis structure of HBr:

#1 Draw skeleton
#2 Show chemical bond
#3 Mark lone pairs
#4 Calculate formal charge and check stability (if octet is already completed on central atom)

Let’s one by one discuss each step in detail.

#1 Draw skeleton

In this step, first calculate the total number of valence electrons. And then, decide the central atom.

  • Let’s calculate the total number of valence electrons

We know that… hydrogen is a group 1 element and bromine is a group 17 element. Hence, hydrogen has one valence electron and bromine has seven valence electrons.

Now HBr has one hydrogen atom and one bromine atom.

So the total number of valence electrons = valence electrons of hydrogen atom + valence electrons of bromine atom

Therefore, the total number of valence electrons = 1 + 7 = 8

  • Now decide the central atom

We can not assume hydrogen as 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, assume that bromine is the central atom.

So now, put hydrogen and bromine next to each other. And draw the rough skeleton structure for the Lewis structure of HBr something like this:

Skeleton structure for Lewis structure of HBr | Image: Root Memory

#2 Show chemical bond

Place two electrons between the atoms to show a chemical bond as follows:

One chemical bond shown between atoms | Image: Root Memory

#3 Mark lone pairs

As calculated earlier, we have a total of 8 valence electrons. And in the above structure, we have already used two valence electrons. Hence, six valence electrons are remaining.

Two valence electrons represent one lone pair. So six valence electrons = three lone pairs.

Note 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.

Also, make sure that you start marking these lone pairs on outside atoms first. And then, on the central atom.

The outside atom is hydrogen. But hydrogen can not keep more than 2 electrons in its last shell. Hence, don’t mark the lone pairs on hydrogen.

So the central atom (bromine) will get three lone pairs.

Now draw the Lewis structure of HBr something like this:

Lone pairs marked on Lewis structure of HBr | Image: Root Memory

In the above structure, you can see that the octet is completed on the central atom (bromine), and also on the outside atom. Therefore, the octet rule is satisfied.

After completing the octet, one last thing we need to do is, calculate the formal charge and check the stability of the above structure.

#4 Calculate formal charge and check stability

The following formula is used to calculate the formal charges on atoms:

Formal charge = valence electrons – nonbonding electrons – ½ bonding electrons

Collect the data from the above structure and then, write it down below as follows:

  • For hydrogen atom

Valence electrons = 1
Nonbonding electrons = 0
Bonding electrons = 2

Formal charge = 1 – 0 – ½ (2) = 0

  • For bromine atom

Valence electrons = 7
Nonbonding electrons = 6
Bonding electrons = 2

Formal charge = 7 – 6 – ½ (2) = 0

Mention the formal charges of atoms on the structure. So the Lewis structure of HBr looks something like this:

Formal charges are calculated, and got the stable Lewis structure of HBr | Image: Root Memory

In the above structure, you can see that the formal charges of both (hydrogen and bromine) are zero. Therefore, this is the stable Lewis structure of HBr.

And the horizontal line drawn in the above structure represents a pair of bonding valence electrons.

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Deep

Rootmemory.com was founded by Deep Rana, who is a mechanical engineer by profession and a blogger by passion. He has a good conceptual knowledge on different educational topics and he provides the same on this website. He loves to learn something new everyday and believes that the best utilization of free time is developing a new skill.

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