XeF6 Lewis structure

XeF₆ (xenon hexafluoride) has one xenon atom and six fluorine atoms.

In the XeF₆ Lewis structure, there are six single bonds around the xenon atom, with six fluorine atoms attached to it. Each fluorine atom has three lone pairs, and the xenon atom has one lone pair.

Alternative method: Lewis structure of XeF₆

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 XeF₆ has one xenon atom and six fluorine atoms, so…

Valence electrons of one xenon atom = 8 × 1 = 8
Valence electrons of six fluorine atoms = 7 × 6 = 42

And the total valence electrons = 8 + 42 = 50

Learn how to find: Fluorine valence electrons

• Second, find the total electron pairs

We have a total of 50 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 = 50 ÷ 2 = 25

• 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 25 electron pairs. And six Xe — F bonds are already marked. So we have to only mark the remaining nineteen 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 is one lone pair.

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 – 2 – ½ (12) = 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 XeF₆ features a central xenon atom linked to six fluorine atoms through single covalent bonds. In this arrangement, the xenon atom utilizes an expanded valence shell to accommodate fourteen electrons, which consist of six bonding pairs and one lone pair. Within this layout, each of the six fluorine atoms successfully satisfies the octet rule by retaining three lone pairs alongside its single shared bond. This configuration represents the most stable state for the molecule because it results in a formal charge of zero for every atom involved, despite the significant steric crowding around the central atom. Consequently, this specific electronic distribution serves as the definitive and most accurate Lewis representation of xenon hexafluoride.

Next: SbF₅ Lewis structure

External video

• XeF6 Lewis Structure – How to Draw the Lewis Structure for XeF6 – YouTube • Wayne Breslyn

External links

• https://www.thegeoexchange.org/chemistry/bonding/Lewis-Structures/XeF6-lewis-structure.html
• https://lambdageeks.com/xef6-lewis-structure/