XeF3- Lewis structure

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.

Alternative method: Lewis structure of XeF₃^–

Rough sketch

• 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

Lone pair

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:

Formal charge

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:

Final structure

The final structure of XeF₃^– consists of a central xenon atom connected to three fluorine atoms through single covalent bonds. In this configuration, the xenon atom utilizes an expanded valence shell to accommodate twelve electrons, which include three bonding pairs and three lone pairs. Each fluorine atom fulfills its octet by maintaining three lone pairs alongside its single shared bond. This arrangement is the most stable as it results in a -1 formal charge on the central xenon atom, while each fluorine atom retains a formal charge of zero. Accordingly, this specific electronic distribution serves as the definitive and most accurate Lewis representation of XeF₃^–.

To properly represent this as a polyatomic ion, the entire Lewis structure is enclosed within square brackets. The overall charge of 1- is then written as a superscript outside the brackets at the top right, reflecting the presence of an additional electron beyond the total valence count of the neutral atoms.

Next: XeF5⁺ Lewis structure

External video

• How to Draw the Lewis Structure for XeF3- – YouTube • Wayne Breslyn