I’m super excited to teach you the lewis structure of AlBr3 in just 5 simple steps.
Infact, I’ve also given the step-by-step images for drawing the lewis dot structure of AlBr3 molecule.
So, if you are ready to go with these 5 simple steps, then let’s dive right into it!
Lewis structure of AlBr3 contains three single bonds between the Aluminum (Al) atom and each Bromine (Br) atom. The Aluminum atom (Al) is at the center and it is surrounded by 3 Bromine atoms (Br). The Aluminum atom does not have a lone pair while all three bromine atoms have three lone pairs each.
Let’s draw and understand this lewis dot structure step by step.
(Note: Take a pen and paper with you and try to draw this lewis structure along with me. I am sure you will definitely learn how to draw lewis structure of AlBr3).
5 Steps to Draw the Lewis Structure of AlBr3
Step #1: Calculate the total number of valence electrons
Here, the given molecule is AlBr3. In order to draw the lewis structure of AlBr3, first of all you have to find the total number of valence electrons present in the AlBr3 molecule.
(Valence electrons are the number of electrons present in the outermost shell of an atom).
So, let’s calculate this first.
Calculation of valence electrons in AlBr3
- For Aluminum:
Aluminum is a group 13 element on the periodic table.
Hence, the valence electrons present in aluminum is 3 (see below image).
- For Bromine:
Bromine is a group 17 element on the periodic table.
Hence, the valence electrons present in bromine is 7 (see below image).
Hence in a AlBr3 molecule,
Valence electrons given by Aluminum (Al) atom = 3
Valence electrons given by each Bromine (Br) atom = 7
So, total number of Valence electrons in AlBr3 molecule = 3 + 7(3) = 24
Step #2: Select the center atom
While selecting the center atom, always put the least electronegative atom at the center.
(Remember: Fluorine is the most electronegative element on the periodic table and the electronegativity decreases as we move right to left in the periodic table as well as top to bottom in the periodic table).
Here in the AlBr3 molecule, if we compare the aluminum atom (Al) and bromine atom (Br), then aluminum is less electronegative than bromine.
So, aluminum should be placed in the center and the remaining 3 bromine atoms will surround it.
Step #3: Put two electrons between the atoms to represent a chemical bond
Now in the above sketch of AlBr3 molecule, put the two electrons (i.e electron pair) between each aluminum atom and bromine atom to represent a chemical bond between them.
These pairs of electrons present between the Aluminum (Al) and Bromine (Br) atoms form a chemical bond, which bonds the aluminum and bromine atoms with each other in a AlBr3 molecule.
Step #4: Complete the octet (or duplet) on outside atoms. If the valence electrons are left, then put the valence electrons pair on the central atom
Don’t worry, I’ll explain!
In the Lewis structure of AlBr3, the outer atoms are bromine atoms.
So now, you have to complete the octet on these bromine atoms (because bromine requires 8 electrons to have a complete outer shell).
Now, you can see in the above image that all the bromine atoms form an octet.
Also, all the 24 valence electrons of AlBr3 molecule (as calculated in step #1) are used in the above structure. So there are no remaining electron pairs.
Hence there is no change in the above sketch of AlBr3.
Let’s move to the next step.
Step #5: Check the stability of lewis structure by calculating the formal charge on each atom
Now, you have come to the final step and here you have to check the formal charge on aluminum atom (Al) as well as each bromine atom (Br).
For that, you need to remember the formula of formal charge;
Formal charge = Valence electrons – Nonbonding electrons – (Bonding electrons)/2
- For Aluminum:
Valence electrons = 3 (as it is in group 13)
Nonbonding electrons = 0
Bonding electrons = 6
- For Bromine:
Valence electron = 7 (as it is in group 17)
Nonbonding electrons = 6
Bonding electrons = 2
|Formal charge||=||Valence electrons||–||Nonbonding electrons||–||(Bonding electrons)/2|
So you can see above that the formal charges on aluminum as well as bromine are “zero”.
Hence, there will not be any change in the above structure and the above lewis structure of AlBr3 is the final stable structure only.
Each electron pair (:) in the lewis dot structure of AlBr3 represents the single bond ( | ). So the above lewis dot structure of AlBr3 can also be represented as shown below.