GaI3 Lewis structure

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GaI3 Lewis Structure
GaI3 Lewis structure

GaI3 [gallium(Ⅲ) iodide] has one gallium atom and three iodine atoms.

In GaI3 Lewis structure, there are three single bonds around the gallium atom, with three iodine atoms attached to it, and on each iodine atom, there are three lone pairs.


Here’s how you can easily draw the GaI3 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
Periodic table

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

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

Since GaI3 has one gallium atom and three iodine atoms, so…

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

And the total valence electrons = 3 + 21 = 24

  • 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 gallium is less electronegative than iodine, assume that the central atom is gallium.

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

  • And finally, draw the rough sketch
GaI3 Lewis Structure (Step 1)
Rough sketch of GaI3 Lewis structure

#2 Mention lone pairs on the atoms

Here, we have a total of 12 electron pairs. And three Ga — 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 gallium is a period 4 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 gallium, there is zero lone pair because all nine electron pairs are over.

Mark the lone pairs on the sketch as follows:

GaI3 Lewis Structure (Step 2)
Lone pairs marked, and got the stable Lewis structure of GaI3

#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 gallium atom, formal charge = 3 – 0 – ½ (6) = 0

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

Here, both gallium 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 (gallium) doesn’t form an octet. But, gallium 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 GaI3.

Next: SeO42- Lewis structure

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