AlBr3 Lewis structure

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AlBr₃ (aluminum bromide) has one aluminum atom and three bromine atoms.

In the AlBr₃ Lewis structure, there are three single bonds around the aluminum atom, with three bromine atoms attached to it, and on each bromine atom, there are three lone pairs.

Steps

Use these steps to correctly draw the AlBr₃ Lewis structure:

#1 First draw a rough sketch
#2 Mark lone pairs on the atoms
#3 Calculate and mark formal charges on the atoms, if required

Let’s discuss each step in more detail.

#1 First draw a rough sketch

• First, determine the total number of valence electrons

In the periodic table, aluminum lies in group 13, and bromine lies in group 17.

Hence, aluminum has three valence electrons and bromine has seven valence electrons.

Since AlBr₃ has one aluminum atom and three bromine atoms, so…

Valence electrons of one aluminum atom = 3 × 1 = 3
Valence electrons of three bromine atoms = 7 × 3 = 21

And the total valence electrons = 3 + 21 = 24

Learn how to find: Aluminum valence electrons and Bromine valence electrons

• Second, find the total electron pairs

We have a total of 24 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 = 24 ÷ 2 = 12

• Third, determine the central atom

We have to place the least electronegative atom at the center.

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

Therefore, place aluminum in the center and bromines on either side.

• And finally, draw the rough sketch

#2 Mark lone pairs on the atoms

Here, we have a total of 12 electron pairs. And three Al — Br bonds are already marked. So we have to only mark the remaining nine electron pairs as lone pairs on the sketch.

Also remember that aluminum is a period 3 element, so it can keep more than 8 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 atoms are bromines.

So for each bromine, there are three lone pairs, and for aluminum, there is zero lone pair because all nine electron pairs are over.

Mark the lone pairs on the sketch as follows:

#3 Calculate and mark formal charges on the atoms, if required

Use the following formula to calculate the formal charges on atoms:

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

For aluminum atom, formal charge = 3 – 0 – ½ (6) = 0

For each bromine atom, formal charge = 7 – 6 – ½ (2) = 0

Here, both aluminum 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 (aluminum) doesn’t form an octet. But, aluminum has an exception that it does not require eight electrons to form an octet. So no need to worry about the octet rule here.

Therefore, this structure is the stable Lewis structure of AlBr₃.

Next: BrCN Lewis structure

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