Lewis Structure of AlI3 (With 5 Simple Steps to Draw!)

Ready to learn how to draw the lewis structure of AlI3?

Awesome!

Here, I have explained 5 simple steps to draw the lewis dot structure of AlI3 (along with images).

So, if you are ready to go with these 5 simple steps, then let’s dive right into it!

Lewis structure of AlI3 contains three single bonds between the Aluminum (Al) atom and each Iodine (I) atom. The Aluminum atom (Al) is at the center and it is surrounded by 3 Iodine atoms (I). The Aluminum atom does not have a lone pair while all three iodine 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 AlI3).

5 Steps to Draw the Lewis Structure of AlI3

Step #1: Calculate the total number of valence electrons

Here, the given molecule is AlI3. In order to draw the lewis structure of AlI3, first of all you have to find the total number of valence electrons present in the AlI3 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 AlI3

• 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 Iodine:

Iodine is a group 17 element on the periodic table.

Hence, the valence electrons present in iodine is 7 (see below image).

Hence in a AlI3 molecule, 

Valence electrons given by Aluminum (Al) atom = 3
Valence electrons given by each Iodine (I) atom = 7
So, total number of Valence electrons in AlI3 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). ^[1]

Here in the AlI3 molecule, if we compare the aluminum atom (Al) and iodine atom (I), then aluminum is less electronegative than iodine.

So, aluminum should be placed in the center and the remaining 3 iodine atoms will surround it.

Step #3: Put two electrons between the atoms to represent a chemical bond

Now in the above sketch of AlI3 molecule, put the two electrons (i.e electron pair) between each aluminum atom and iodine atom to represent a chemical bond between them.

These pairs of electrons present between the Aluminum (Al) and Iodine (I) atoms form a chemical bond, which bonds the aluminum and iodine atoms with each other in a AlI3 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 AlI3, the outer atoms are iodine atoms.

So now, you have to complete the octet on these iodine atoms (because iodine requires 8 electrons to have a complete outer shell).

Now, you can see in the above image that all the iodine atoms form an octet.

Also, all the 24 valence electrons of AlI3 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 AlI3.

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 iodine atom (I).

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 Iodine:
  Valence electron = 7 (as it is in group 17)
  Nonbonding electrons = 6
  Bonding electrons = 2

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

So you can see above that the formal charges on aluminum as well as iodine are “zero”.

Hence, there will not be any change in the above structure and the above lewis structure of AlI3 is the final stable structure only.

Each electron pair (:) in the lewis dot structure of AlI3 represents the single bond ( | ). So the above lewis dot structure of AlI3 can also be represented as shown below.

Related lewis structures for your practice:
Lewis Structure of PF2-
Lewis Structure of SI4
Lewis Structure of GaCl3
Lewis Structure of NSF
Lewis Structure of C2H4Br2

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