# XeO2F2 Lewis structure

XeO2F2 (xenon dioxydifluoride) has one xenon atom, two oxygen atoms, and two fluorine atoms.

In the XeO2F2 Lewis structure, there are two single bonds and two double bonds around the xenon atom, with two fluorine atoms and two oxygen atoms attached to it. Each fluorine atom has three lone pairs, each oxygen atom has two lone pairs, and the xenon atom has one lone pair.

Contents

## Steps

Use these steps to correctly draw the XeO2F2 Lewis structure:

#1 First draw a rough sketch
#2 Mark lone pairs on the atoms
#3 Calculate and mark formal charges on the atoms, if required
#4 Convert lone pairs of the atoms, and minimize formal charges
#5 Repeat step 4 if needed, until all charges are minimized, to get a stable Lewis structure

Let’s discuss each step in more detail.

### #1 First draw a rough sketch

• 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 XeO2F2 has one xenon atom, two oxygen atoms, and two fluorine atoms, so…

Valence electrons of one xenon atom = 8 × 1 = 8
Valence electrons of two oxygen atoms = 6 × 2 = 12
Valence electrons of two fluorine atoms = 7 × 2 = 14

And the total valence electrons = 8 + 12 + 14 = 34

• Second, find the total electron pairs

We have a total of 34 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 = 34 ÷ 2 = 17

• 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 Mark lone pairs on the atoms

Here, we have a total of 17 electron pairs. And four bonds are already marked. So we have to only mark the remaining thirteen 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 oxygens and fluorines.

So for each oxygen and 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 Calculate and mark formal charges on the atoms, if required

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 (8) = +2

For each 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 Convert lone pairs of the atoms, and minimize formal charges

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

### #5 Repeating step 4 to get a stable Lewis structure

Since there are charges on xenon and oxygen atoms, again 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 (oxygens 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 XeO2F2.

Next: XeH4 Lewis structure