# 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.

Contents

## Steps

To properly draw the XeF6 Lewis structure, follow these steps:

#1 Draw a rough sketch of the structure
#2 Next, indicate lone pairs on the atoms
#3 Indicate formal charges on the atoms, if necessary

Let’s break down each step in more detail.

### #1 Draw a rough sketch of the 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 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

• 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

### #2 Next, indicate lone pairs on the atoms

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:

### #3 Indicate formal charges on the atoms, if necessary

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.

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.

Therefore, this structure is the stable Lewis structure of XeF6.

Next: SbF5 Lewis structure