AlI3 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.

AlI3 lewis structure has an Aluminum atom (Al) at the center which is surrounded by three Iodine atoms (I). There are 3 single bonds between the Aluminum atom (Al) and each Iodine atom (I).

If you haven’t understood anything from the above image of AlI3 lewis structure, then just stick with me and you will get the detailed step by step explanation on drawing a lewis structure of AlI3.

So let’s move to the steps of drawing the lewis structure of AlI3.

Steps of drawing AlI3 lewis structure

Step 1: Find the total valence electrons in AlI3 molecule

In order to find the total valence electrons in an AlI3 molecule, first of all you should know the valence electrons present in aluminum atom as well as iodine 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 iodine using a periodic table.

Total valence electrons in AlI3 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 iodine atom:

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

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

Hence, 

Total valence electrons in AlI3 molecule = valence electrons given by 1 aluminum atom + valence electrons given by 3 iodine 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 AlI3 and it contains aluminum atom (Al) and iodine atoms (I).

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

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

So here the aluminum atom (Al) is the center atom and the iodine atoms (I) are the outside atoms.

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

Now in the AlI3 molecule, you have to put the electron pairs between the aluminum atom (Al) and iodine atoms (I).

This indicates that the aluminum (Al) and iodine (I) are chemically bonded with each other in a AlI3 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 AlI3 molecule, you can see that the outer atoms are iodine atoms.

These outer iodine 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 AlI3 molecule.

The AlI3 molecule has a total 24 valence electrons and all these valence electrons are used in the above sketch of AlI3.

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 AlI3.

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 iodine (I) atoms present in the AlI3 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 AlI3 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 Iodine (I) atom:
Valence electron = 7 (because iodine 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
I	=	7	–	2/2	–	6	=	0

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

This indicates that the above lewis structure of AlI3 is stable and there is no further change in the above structure of AlI3.

In the above lewis dot structure of AlI3, you can also represent each bonding electron pair (:) as a single bond (|). By doing so, you will get the following lewis structure of AlI3.

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:

SI4 Lewis Structure	GaCl3 Lewis Structure
NSF Lewis Structure	C2H4Br2 Lewis Structure
CO2 lewis structure	SO2 lewis structure