XeF3- Lewis structure

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XeF₃^– has one xenon atom and three fluorine atoms.

In the XeF₃^– Lewis structure, there are three single bonds around the xenon atom, with three fluorine atoms attached to it, and each atom has three lone pairs.

Also, there is a negative (-1) charge on the xenon atom.

Steps

Here’s how you can easily draw the XeF₃^– 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, xenon lies in group 18, and fluorine lies in group 17.

Hence, xenon has eight valence electrons and fluorine has seven valence electrons.

Since XeF₃^– has one xenon atom and three fluorine atoms, so…

Valence electrons of one xenon atom = 8 × 1 = 8
Valence electrons of three fluorine atoms = 7 × 3 = 21

Now the XeF₃^– has a negative (-1) charge, so we have to add one more electron.

So the total valence electrons = 8 + 21 + 1 = 30

Learn how to find: Fluorine valence electrons

• Second, find the total electron pairs

We have a total of 30 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 = 30 ÷ 2 = 15

• Third, determine the central atom

We have to place the least electronegative atom at the center.

Since xenon is less electronegative than fluorine, assume that the central atom is xenon.

Therefore, place xenon in the center and fluorines on either side.

• And finally, draw the rough sketch

#2 Mention lone pairs on the atoms

Here, we have a total of 15 electron pairs. And three Xe — F bonds are already marked. So we have to only mark the remaining twelve electron pairs as lone pairs on the sketch.

Also remember that xenon is a period 5 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.

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

So for each fluorine, there are three lone pairs, and for xenon, there are three lone pairs.

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 xenon atom, formal charge = 8 – 6 – ½ (6) = -1

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

Here, the xenon atom has a charge, so mark it on the sketch as follows:

In the above structure, you can see that the central atom (xenon) forms an octet. And the outside atoms (fluorines) also form an octet. Hence, the octet rule is satisfied.

Now there is still a negative (-1) charge on the xenon atom.

This is not okay, right? Because the structure with a negative charge on the most electronegative atom is the best Lewis structure. And in this case, the most electronegative element is fluorine.

But if we convert a lone pair of the xenon atom to make a new Xe — F bond with the fluorine atom, and calculate the formal charge, then we do not get the formal charges on atoms closer to zero.

And the structure with the formal charges on atoms closer to zero is the best Lewis structure.

Therefore, this structure is the most stable Lewis structure of XeF₃^–.

And since the XeF₃^– has a negative (-1) charge, mention that charge on the Lewis structure by drawing brackets as follows:

Next: XeF5⁺ Lewis structure

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