# XeOF4 Lewis structure

XeOF4 (xenon oxytetrafluoride) has one xenon atom, one oxygen atom, and four fluorine atoms.

In the XeOF4 Lewis structure, there is one double bond and four single bonds around the xenon atom, with one oxygen atom and four fluorine atoms attached to it. Each fluorine atom has three lone pairs, the oxygen atom has two lone pairs, and the xenon atom has one lone pair.

Contents

## Steps

To properly draw the XeOF4 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
#4 Minimize formal charges by converting lone pairs of the atoms
#5 Repeat step 4 if necessary, until all charges are minimized

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, oxygen lies in group 16, and fluorine lies in group 17.

Hence, xenon has eight valence electrons, oxygen has six valence electrons, and fluorine has seven valence electrons.

Since XeOF4 has one xenon atom, one oxygen atom, and four fluorine atoms, so…

Valence electrons of one xenon atom = 8 × 1 = 8
Valence electrons of one oxygen atom = 6 × 1 = 6
Valence electrons of four fluorine atoms = 7 × 4 = 28

And the total valence electrons = 8 + 6 + 28 = 42

• Second, find the total electron pairs

We have a total of 42 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 = 42 ÷ 2 = 21

• Third, determine the central atom

We have to place the least electronegative atom at the center.

Since xenon is less electronegative than oxygen and fluorine, assume that the central atom is xenon.

Therefore, place xenon in the center and oxygen and fluorine on either side.

• And finally, draw the rough sketch

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

Here, we have a total of 21 electron pairs. And five bonds are already marked. So we have to only mark the remaining sixteen 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 both (oxygen and fluorine) are the period 2 elements, so they can not keep more than 8 electrons in their last shell.

Always start to mark the lone pairs from outside atoms. Here, the outside atoms are oxygen and fluorines.

So for each fluorine and oxygen, 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 (10) = +1

For oxygen atom, formal charge = 6 – 6 – ½ (2) = -1

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

Here, both xenon and oxygen atoms have charges, so mark them on the sketch as follows:

The above structure is not a stable Lewis structure because both xenon and oxygen atoms have charges. Therefore, reduce the charges (as below) by converting lone pairs to bonds.

### #4 Minimize formal charges by converting lone pairs of the atoms

Convert a lone pair of the oxygen atom to make a new Xe — O bond with the xenon atom as follows:

In the above structure, you can see that the central atom (xenon) forms an octet. And the outside atoms (oxygen and fluorines) also form an octet. Hence, the octet rule is satisfied.

Also, the above structure is more stable than the previous structures. Therefore, this structure is the stable Lewis structure of XeOF4.

Next: KrF2 Lewis structure