AlI3 Lewis structure

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AlI₃ (aluminum iodide) has one aluminum atom and three iodine atoms.

In AlI₃ Lewis structure, there are three single bonds around the aluminum atom, with three iodine atoms attached to it, and on each iodine atom, there are three lone pairs.

Steps

Here’s how you can easily draw the AlI₃ Lewis structure step by step:

#1 Draw a rough skeleton structure
#2 Mention lone pairs on the atoms
#3 If needed, mention formal charges on the atoms

Now, let’s take a closer look at each step mentioned above.

#1 Draw a rough skeleton structure

• First, determine the total number of valence electrons

In the periodic table, aluminum lies in group 13, and iodine lies in group 17.

Hence, aluminum has three valence electrons and iodine has seven valence electrons.

Since AlI₃ has one aluminum atom and three iodine atoms, so…

Valence electrons of one aluminum atom = 3 × 1 = 3
Valence electrons of three iodine atoms = 7 × 3 = 21

And the total valence electrons = 3 + 21 = 24

Learn how to find: Aluminum valence electrons

• 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 iodine, assume that the central atom is aluminum.

Therefore, place aluminum in the center and iodines on either side.

• And finally, draw the rough sketch

#2 Mention lone pairs on the atoms

Here, we have a total of 12 electron pairs. And three Al — I bonds are already marked. So we have to only mark the remaining nine electron pairs as lone pairs on the sketch.

Also remember that aluminum is a period 3 element, so it can keep more than 8 electrons in its last shell. And iodine is a period 5 element, so it can keep more than 8 electrons in its last shell.

Always start to mark the lone pairs from outside atoms. Here, the outside atoms are iodines.

So for each iodine, 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 If needed, mention formal charges on the atoms

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 iodine atom, formal charge = 7 – 6 – ½ (2) = 0

Here, both aluminum and iodine 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 AlI₃.

Next: PF₂^– Lewis structure

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External links

• https://lambdageeks.com/ali3-lewis-structure/
• https://study.com/learn/lesson/aluminum-iodide-formula-molar-mass-lewis-structure.html
• https://brainly.com/question/13908649