AlI3 Lewis structure

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.

Alternative method: Lewis structure of AlI₃

Contents

Rough sketch

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

AlI3 Lewis Structure (Step 1) — Rough sketch of AlI₃ Lewis structure | Image: Learnool

Lone pair

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:

AlI3 Lewis Structure (Step 2) — Lone pairs marked, and got the stable Lewis structure of AlI₃ | Image: Learnool

Formal charge

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.

Final structure

The final structure of AlI₃ contains a central aluminum atom connected to three iodine atoms through single covalent bonds. In this configuration, the aluminum atom serves as an exception to the octet rule, possessing an incomplete valence shell with only six electrons across three bonding pairs. Each iodine atom fulfills its octet by maintaining three lone pairs of its own alongside the single shared bond. This arrangement is the most stable because it results in formal charges of zero for all atoms involved, representing the most energetically favorable state for the molecule. Consequently, this specific electronic distribution serves as the definitive and most accurate Lewis representation of aluminum iodide.

Next: PF₂^– Lewis structure

External video

How to Draw the Lewis Dot Structure for AlI3: Aluminum iodide – YouTube • Wayne Breslyn

External links

Deep

Learnool.com was founded by Deep Rana, who is a mechanical engineer by profession and a blogger by passion. He has a good conceptual knowledge on different educational topics and he provides the same on this website. He loves to learn something new everyday and believes that the best utilization of free time is developing a new skill.