Lewis structure of AlF3

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Lewis structure of AlF3
Lewis structure of AlF3

The Lewis structure of AlF3 contains three single bonds, with aluminum in the center, and three fluorines on either side. There are three lone pairs on each fluorine atom, and the aluminum atom does not have any lone pair.

Steps

By using the following steps, you can easily draw the Lewis structure of AlF3.

#1 Draw skeleton
#2 Show chemical bond
#3 Mark lone pairs
#4 Calculate formal charge and check stability (if octet is already completed on central atom)

Let’s one by one discuss each step in detail.

#1 Draw skeleton

In this step, first calculate the total number of valence electrons. And then, decide the central atom.

  • Let’s calculate the total number of valence electrons

We know that… aluminum is a group 13 element and fluorine is a group 17 element. Hence, aluminum has three valence electrons and fluorine has seven valence electrons.

Now AlF3 has one aluminum atom and three fluorine atoms.

So the total number of valence electrons = valence electrons of aluminum atom + (valence electrons of fluorine atom × 3)

Therefore, the total number of valence electrons = 3 + 21 = 24

  • Now decide the central atom

The atom with the least electronegative value is placed at the center. By looking at the periodic table, we get the electronegativity values for aluminum and fluorine as follows:

Electronegativity value of aluminum = 1.61
Electronegativity value of fluorine = 3.98

Obviously, aluminum is less electronegative than fluorine. Hence, assume that aluminum is the central atom.

So now, put aluminum in the center and fluorines on either side. And draw the rough skeleton structure for the Lewis structure of AlF3 something like this:

Skeleton structure for Lewis structure of AlF3

#2 Show chemical bond

Place two electrons between the atoms to show a chemical bond. Since aluminum is surrounded by three fluorines, use six electrons to show three chemical bonds as follows:

Three chemical bonds shown between atoms

#3 Mark lone pairs

As calculated earlier, we have a total of 24 valence electrons. And in the above structure, we have already used six valence electrons. Hence, eighteen valence electrons are remaining.

Two valence electrons represent one lone pair. So eighteen valence electrons = nine lone pairs.

Note that aluminum is a period 3 element, so it can keep more than 8 electrons in its last shell. And fluorine is a period 2 element, so it can not keep more than 8 electrons in its last shell.

Also, make sure that you start marking these lone pairs on outside atoms first. And then, on the central atom.

The outside atoms are fluorines, so each fluorine will get three lone pairs. And the central atom (aluminum) will not get any lone pair, because all nine lone pairs are used.

So the Lewis structure of AlF3 looks something like this:

Lone pairs marked on Lewis structure of AlF3

In the above structure, you can see that the octet is completed on outside atoms. But, the central atom (aluminum) doesn’t form an octet.

Now aluminum has an exception in AlF3, that it does not require eight electrons to complete its octet.

So one last thing we need to do is, calculate the formal charge and check the stability of the above structure.

#4 Calculate formal charge and check stability

The following formula is used to calculate the formal charges on atoms:

Formal charge = valence electrons – nonbonding electrons – ½ bonding electrons

Collect the data from the above structure and then, write it down below as follows:

  • For aluminum atom

Valence electrons = 3
Nonbonding electrons = 0
Bonding electrons = 6

Formal charge = 3 – 0 – ½ (6) = 0

  • For each fluorine atom

Valence electrons = 7
Nonbonding electrons = 6
Bonding electrons = 2

Formal charge = 7 – 6 – ½ (2) = 0

Mention the formal charges of atoms on the structure. So the Lewis structure of AlF3 looks something like this:

Formal charges are calculated, and got the stable Lewis structure of AlF3

In the above structure, you can see that the formal charges of both (aluminum and fluorine) are zero. Therefore, this is the stable Lewis structure of AlF3.

And each horizontal line drawn in the above structure represents a pair of bonding valence electrons.

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External links

Deep

Rootmemory.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.

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