AlBr3 (aluminum bromide) has one aluminum atom and three bromine atoms. In the lewis structure of AlBr3, 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.
Here’s how you can draw the AlBr3 lewis structure step by step.
Step #1: draw sketch
Step #2: mark lone pairs
Step #3: mark charges (if there are)
Let’s break down each step in detail.
#1 Draw Sketch
- First, determine the total number of valence electrons
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
- 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
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.
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 Mark Charges
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