XeF4 Lewis structure

XeF₄ (xenon tetrafluoride) has one xenon atom and four fluorine atoms.

In the XeF₄ Lewis structure, there are four single bonds around the xenon atom, with four fluorine atoms attached to it. Each fluorine atom has three lone pairs, and the xenon atom has two lone pairs.

Alternative method: Lewis structure of XeF₄

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 XeF₄ has one xenon atom and four fluorine atoms, so…

Valence electrons of one xenon atom = 8 × 1 = 8
Valence electrons of four fluorine atoms = 7 × 4 = 28

And the total valence electrons = 8 + 28 = 36

Learn how to find: Fluorine valence electrons

• Second, find the total electron pairs

We have a total of 36 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 = 36 ÷ 2 = 18

• 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 18 electron pairs. And four Xe — F bonds are already marked. So we have to only mark the remaining fourteen 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 two 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 – 4 – ½ (8) = 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 XeF₄ comprises a central xenon atom connected to four fluorine atoms through single covalent bonds. Within this layout, the xenon atom utilizes an expanded octet to accommodate twelve valence electrons, consisting of four bonding pairs and two 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 every atom, 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 tetrafluoride.

Next: PO₄^3- Lewis structure

External video

• XeF4 Lewis Structure – How to Draw the Lewis Structure for XeF4 – YouTube • Wayne Breslyn

External links

• https://www.chemistryscl.com/general/XeF4-lewis-structure/
• https://techiescientist.com/xef4-lewis-structure/
• https://geometryofmolecules.com/xef4-xenon-tetrafluoride-molecular-geometry-lewis-structure-polarity/
• https://www.thegeoexchange.org/chemistry/bonding/Lewis-Structures/XeF4-lewis-structure.html
• https://topblogtenz.com/xef4-lewis-dot-structure-molecular-geometry-bond-angle-hybridization/
• https://lambdageeks.com/xef4-lewis-structure/
• https://socratic.org/questions/how-do-you-draw-the-lewis-structure-for-xef-4