The **Lewis structure of AlBr _{3}** contains three single bonds, with aluminum in the center, and three bromines on either side. There are three lone pairs on each bromine atom, and the aluminum atom does not have any lone pair.

## Steps

By using the following steps, you can easily draw the Lewis structure of AlBr_{3}.

#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… aluminum is a group 13 element and bromine is a group 17 element. Hence, aluminum has **three** valence electrons and bromine has **seven** valence electrons.

Now AlBr_{3} has one aluminum atom and three bromine atoms.

So the total number of valence electrons = valence electrons of aluminum atom + (valence electrons of bromine atom × 3)

Therefore, the **total number of valence electrons** = 3 + 21 = 24

- Now decide the central atom

The atom with the least electronegative value is placed at the center. By looking at the periodic table, we get the electronegativity values for aluminum and bromine as follows:

Electronegativity value of aluminum = 1.61

Electronegativity value of bromine = 2.96

Obviously, aluminum is less electronegative than bromine. Hence, assume that **aluminum is the central atom**.

So now, put aluminum in the center and bromines on either side. And draw the rough skeleton structure for the Lewis structure of AlBr_{3} something like this:

**Also read:** How to draw Lewis structure of N_{2}O_{5} (5 steps)

### #2 Show chemical bond

Place two electrons between the atoms to show a chemical bond. Since aluminum is surrounded by three bromines, use six electrons to show **three chemical bonds** as follows:

**Also read:** How to draw Lewis structure of CHBr_{3} (4 steps)

### #3 Mark lone pairs

As calculated earlier, we have a total of 24 valence electrons. And in the above structure, we have already used six valence electrons. Hence, eighteen valence electrons are remaining.

Two valence electrons represent one lone pair. So eighteen valence electrons = **nine lone pairs**.

Note that both (aluminum and bromine) are the period 3 elements, so they can keep more than 8 electrons in their last shell.

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

The outside atoms are bromines, so each bromine will get three lone pairs. And the central atom (aluminum) will not get any lone pair, because all nine lone pairs are used.

So the Lewis structure of AlBr_{3} looks something like this:

In the above structure, you can see that the octet is completed on outside atoms. But, the central atom (aluminum) doesn’t form an octet.

Now beryllium has an exception in AlBr_{3}, that it does not require eight electrons to complete its octet.

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

**Also read:** How to draw Lewis structure of BrCl (4 steps)

### #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
**aluminum**atom

Valence electrons = 3

Nonbonding electrons = 0

Bonding electrons = 6

Formal charge = 3 – 0 – ½ (6) = 0

- For
**each 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 AlBr_{3} looks something like this:

In the above structure, you can see that the formal charges of both (aluminum and bromine) are zero. Therefore, this is the **stable Lewis structure of AlBr _{3}**.

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

## Related

- Lewis structure of N
_{2}O_{5} - Lewis structure of CHBr
_{3} - Lewis structure of BrCl
- Lewis structure of BrCN
- Lewis structure of AsCl
_{3}

## External links

- AlBr3 Lewis Structure in 5 Steps (With Images) – Pediabay
- How do you determine the Lewis structure of AlBr3? – Quora
- How to draw AlBr3 Lewis Structure? – Science Education and Tutorials
- Albr3 Lewis Structure,Geometry:9 Facts You Should Know – Lambda Geeks
- Aluminum Tribromide, AlBr3 Molecular Geometry & Polarity – Tutor-Homework.com
- Write Lewis formulas for the following: BeF2, AlBr3 – Homework.Study.com
- Draw the Lewis structure for AlBr3 – Chegg
- Lewis structure C2H2Br2? – Answers
- Give the total number of valence electrons for the compound AlBr3. What is its corresponding Lewis structure? – OneClass
- Is AlBr3 a Lewis acid or Lewis base with explanation? – Brainly
- AlBr3 (Aluminum Bromide) Oxidation Number – ChemicalAid
- Draw the Lewis structure for AlBr3 – Numerade

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.