AlBr3 Lewis Structure in 5 Steps (With Images)

So you have seen the above image by now, right?

Let me explain the above image in short.

AlBr3 lewis structure has an Aluminum atom (Al) at the center which is surrounded by three Bromine atoms (Br). There are 3 single bonds between the Aluminum atom (Al) and each Bromine atom (Br).

If you haven’t understood anything from the above image of AlBr3 lewis structure, then just stick with me and you will get the detailed step by step explanation on drawing a lewis structure of AlBr3.

So let’s move to the steps of drawing the lewis structure of AlBr3.

Steps of drawing AlBr3 lewis structure

Step 1: Find the total valence electrons in AlBr3 molecule

In order to find the total valence electrons in an AlBr3 molecule, first of all you should know the valence electrons present in the aluminum atom as well as the bromine atom.
(Valence electrons are the electrons that are present in the outermost orbit of any atom.)

Here, I’ll tell you how you can easily find the valence electrons of aluminum as well as bromine using a periodic table.

Total valence electrons in AlBr3 molecule

→ Valence electrons given by aluminum atom:

Aluminum is a group 13 element on the periodic table. ^[1] Hence the valence electrons present in aluminum is 3.

You can see the 3 valence electrons present in the aluminum atom as shown in the above image.

→ Valence electrons given by bromine atom:

Bromine is a group 17 element on the periodic table. ^[2] Hence the valence electrons present in bromine is 7.

You can see the 7 valence electrons present in the bromine atom as shown in the above image.

Hence, 

Total valence electrons in AlBr3 molecule = valence electrons given by 1 aluminum atom + valence electrons given by 3 bromine atoms = 3 + 7(3) = 24.

Step 2: Select the central atom

For selecting the center atom, you have to remember that the atom which is less electronegative remains at the center.

Now here the given molecule is AlBr3 and it contains aluminum atom (Al) and bromine atoms (Br).

You can see the electronegativity values of aluminum atom (Al) and bromine atom (Br) in the above periodic table.

If we compare the electronegativity values of aluminum (Al) and bromine (Br) then the aluminum atom is less electronegative.

So here the aluminum atom (Al) is the center atom and the bromine atoms (Br) are the outside atoms.

Step 3: Connect each atoms by putting an electron pair between them

Now in the AlBr3 molecule, you have to put the electron pairs between the aluminum atom (Al) and bromine atoms (Br).

This indicates that the aluminum (Al) and bromine (Br) are chemically bonded with each other in a AlBr3 molecule.

Step 4: Make the outer atoms stable

Now in this step, you have to check the stability of the outer atoms.

Here in the sketch of AlBr3 molecule, you can see that the outer atoms are bromine atoms.

These outer bromine atoms are forming an octet and hence they are stable.

Also, in step 1 we have calculated the total number of valence electrons present in the AlBr3 molecule.

The AlBr3 molecule has a total 24 valence electrons and all these valence electrons are used in the above sketch of AlBr3.

Hence there are no remaining electron pairs to be kept on the central atom. 

So now let’s proceed to the next step.

Step 5: Check the stability of lewis structure

Now you have come to the final step in which you have to check the stability of lewis structure of AlBr3.

The stability of lewis structure can be checked by using a concept of formal charge.

In short, now you have to find the formal charge on aluminum (Al) atom as well as bromine (Br) atoms present in the AlBr3 molecule.

For calculating the formal charge, you have to use the following formula;

Formal charge = Valence electrons – (Bonding electrons)/2 – Nonbonding electrons

You can see the number of bonding electrons and nonbonding electrons for each atom of AlBr3 molecule in the image given below.

For Aluminum (Al) atom:
Valence electrons = 3 (because aluminum is in group 13)
Bonding electrons = 6
Nonbonding electrons = 0

For Bromine (Br) atom:
Valence electron = 7 (because bromine is in group 17)
Bonding electrons = 2
Nonbonding electrons = 6

Formal charge	=	Valence electrons	–	(Bonding electrons)/2	–	Nonbonding electrons
Al	=	3	–	6/2	–	0	=	0
Br	=	7	–	2/2	–	6	=	0

From the above calculations of formal charge, you can see that the aluminum (Al) atom as well as bromine (Br) atom has a “zero” formal charge.

This indicates that the above lewis structure of AlBr3 is stable and there is no further change in the above structure of AlBr3.

In the above lewis dot structure of AlBr3, you can also represent each bonding electron pair (:) as a single bond (|). By doing so, you will get the following lewis structure of AlBr3.

I hope you have completely understood all the above steps.

For more practice and better understanding, you can try other lewis structures listed below.

Try (or at least See) these lewis structures for better understanding:

AlF3 Lewis Structure	IBr Lewis Structure
SeCl4 Lewis Structure	HOF Lewis Structure
XeO2F2 Lewis Structure	XeH4 Lewis Structure