XeF2 Lewis structure

XeF₂ (xenon difluoride) has one xenon atom and two fluorine atoms.

In the XeF₂ Lewis structure, there are two single bonds around the xenon atom, with two fluorine atoms attached to it, and each atom has three lone pairs.

Alternative method: Lewis structure of XeF₂

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 XeF₂ has one xenon atom and two fluorine atoms, so…

Valence electrons of one xenon atom = 8 × 1 = 8
Valence electrons of two fluorine atoms = 7 × 2 = 14

And the total valence electrons = 8 + 14 = 22

Learn how to find: Fluorine valence electrons

• Second, find the total electron pairs

We have a total of 22 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 = 22 ÷ 2 = 11

• 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 11 electron pairs. And two Xe — F bonds are already marked. So we have to only mark the remaining nine 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 atom, 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 (4) = 0

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

Here, both xenon and fluorine atoms do not have charges, so no need to mark the charges.

Final structure

The final structure of XeF₂ involves a central xenon atom connected to two fluorine atoms through single covalent bonds. In this configuration, the xenon atom serves as an exception to the octet rule, utilizing an expanded valence shell to accommodate ten electrons, which include two bonding pairs and three lone pairs. Each fluorine 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, representing the most energetically favorable state for the molecule. Consequently, this specific electronic distribution serves as the definitive and most accurate Lewis representation of xenon difluoride.

Next: XeF₄ Lewis structure

External video

• XeF2 Lewis Structure – How to Draw the Lewis Structure for XeF2 – YouTube • Wayne Breslyn

External links

• https://geometryofmolecules.com/xef2-lewis-structure-polarity-hybridization-and-shape/
• https://lambdageeks.com/xef2-lewis-structure/
• https://www.thegeoexchange.org/chemistry/bonding/Lewis-Structures/XeF2-lewis-structure.html
• https://techiescientist.com/xef2-lewis-structure/
• https://topblogtenz.com/xef2-lewis-structure-molecular-geometry-bond-angle-shape/
• https://www.sciencecoverage.com/2022/02/xef2-lewis-structure.html