The **Lewis structure of XeO _{2}F_{2}** contains two double bonds and two single bonds, with xenon in the center, and two oxygens and two fluorines on either side. There are two lone pairs on each oxygen atom, three lone pairs on each fluorine atom, and one lone pair on the xenon atom.

## Steps

By using the following steps, you can easily draw the Lewis structure of XeO_{2}F_{2}.

#1 Draw skeleton

#2 Show chemical bond

#3 Mark lone pairs

#4 Calculate formal charge and check stability (if octet is already completed on central atom)

#5 Convert lone pair and calculate formal charge again (if formal charges are not closer to zero)

Let’s one by one discuss each step in detail.

### #1 Draw skeleton

In this step, first calculate the total number of valence electrons. And then, decide the central atom.

- Let’s calculate the total number of valence electrons

We know that… xenon is a group 18 element, oxygen is a group 16 element and fluorine is a group 17 element. Hence, xenon has **eight** valence electrons, oxygen has **six** valence electrons, and fluorine has **seven** valence electrons.

Now XeO_{2}F_{2} has one xenon atom, two oxygen atoms, and two fluorine atoms.

So the total number of valence electrons = valence electrons of xenon atom + (valence electrons of oxygen atom × 2) + (valence electrons of fluorine atom × 2)

Therefore, the **total number of valence electrons** = 8 + 12 + 14 = 34

- Now decide the central atom

The atom with the least electronegative value is placed at the center. By looking at the periodic table, we get the electronegativity values for xenon, oxygen, and fluorine as follows:

Electronegativity value of xenon = 2.6

Electronegativity value of oxygen = 3.44

Electronegativity value of fluorine = 3.98

Obviously, xenon is less electronegative than oxygen and fluorine. Hence, assume that **xenon is the central atom**.

So now, put xenon in the center and two oxygens and two fluorines on either side. And draw the rough skeleton structure for the Lewis structure of XeO_{2}F_{2} something like this:

**Also read:** How to draw Lewis structure of AlBr_{3} (4 steps)

### #2 Show chemical bond

Place two electrons between the atoms to show a chemical bond. Since xenon is surrounded by two oxygens and two fluorines, use eight electrons to show **four chemical bonds** as follows:

### #3 Mark lone pairs

As calculated earlier, we have a total of 34 valence electrons. And in the above structure, we have already used eight valence electrons. Hence, twenty-six valence electrons are remaining.

Two valence electrons represent one lone pair. So twenty-six valence electrons = **thirteen lone pairs**.

Note that xenon is 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.

Also, make sure that you start marking these lone pairs on outside atoms first. And then, on the central atom.

The outside atoms are oxygens and fluorines, so each oxygen and each fluorine will get three lone pairs. And the central atom (xenon) will get one lone pair.

So the Lewis structure of XeO_{2}F_{2} looks something like this:

In the above structure, you can see that the octet is completed on the central atom (xenon), and also on the outside atoms. Therefore, the octet rule is satisfied.

Now calculate the formal charge and check the stability of the above structure.

**Also read:** How to draw Lewis structure of N_{2}O_{5} (5 steps)

### #4 Calculate formal charge and check stability

The following formula is used to calculate the formal charges on atoms:

Formal charge = valence electrons – nonbonding electrons – ½ bonding electrons

Collect the data from the above structure and then, write it down below as follows:

- For
**xenon**atom

Valence electrons = 8

Nonbonding electrons = 2

Bonding electrons = 8

Formal charge = 8 – 2 – ½ (8) = +2

- For
**each oxygen**atom

Valence electrons = 6

Nonbonding electrons = 6

Bonding electrons = 2

Formal charge = 6 – 6 – ½ (2) = -1

- For
**each fluorine**atom

Valence electrons = 7

Nonbonding electrons = 6

Bonding electrons = 2

Formal charge = 7 – 6 – ½ (2) = 0

Mention the formal charges of atoms on the structure. So the Lewis structure of XeO_{2}F_{2} looks something like this:

In the above structure, you can see that the formal charges of atoms are not closer to zero. Therefore, convert lone pair and calculate formal charge again.

**Also read:** How to draw Lewis structure of CHBr_{3} (4 steps)

### #5 Convert lone pair and calculate formal charge again

As mentioned earlier, xenon is a period 5 element, so it can keep more than 8 electrons in its last shell.

So convert one lone pair from each oxygen atom to make a new bond with the xenon atom. And then, the Lewis structure of XeO_{2}F_{2} looks something like this:

Now one last thing we need to do is, calculate the formal charge again and check the stability of the above structure.

Formal charge = valence electrons – nonbonding electrons – ½ bonding electrons

Collect the data from the above structure and then, write it down below as follows:

- For
**xenon**atom

Valence electrons = 8

Nonbonding electrons = 2

Bonding electrons = 12

Formal charge = 8 – 2 – ½ (12) = 0

- For
**each oxygen**atom

Valence electrons = 6

Nonbonding electrons = 4

Bonding electrons = 4

Formal charge = 6 – 4 – ½ (4) = 0

- For
**each fluorine**atom

Valence electrons = 7

Nonbonding electrons = 6

Bonding electrons = 2

Formal charge = 7 – 6 – ½ (2) = 0

Mention the formal charges of atoms on the structure. So the Lewis structure of XeO_{2}F_{2} looks something like this:

In the above structure, you can see that the formal charges of all atoms are zero. Therefore, this is the **stable Lewis structure of XeO _{2}F_{2}**.

And each horizontal line drawn in the above structure represents a pair of bonding valence electrons.

## Related

- Lewis structure of AlBr
_{3} - Lewis structure of N
_{2}O_{5} - Lewis structure of CHBr
_{3} - Lewis structure of BrCl
- Lewis structure of BrCN

## External links

- XeO2F2 Lewis Structure: Drawings, Hybridization, Shape, Charges, Pair and Detailed Facts – Lambda Geeks
- XeO2F2 Lewis Structure, Geometry, Hybridization, and Polarity – Techiescientist
- Chemical Bonding: XeO2F2 Lewis Structure – The Geoexchange
- XeO2F2 Lewis Structure in 5 Steps (With Images) – Pediabay
- XeO2F2 Lewis Structure – Laurence Lavelle
- what is the Lewis structure for XeO2F2 which correctly minimizes formal charges – Studocu
- What is the Lewis bond diagram of XeO2f2? – Quora
- Select the lewis structure for xeo2f2 which correctly minimizes formal charges – Brainly
- Select the Lewis structure for XeO2F2 that correctly minimizes formal charges – Chegg
- Why does the XeO2F2 Lewis structure have the Xe in the middle? – Homework.Study.com
- Draw the Lewis structure of XeO2F2 – Quizlet
- Select the Lewis structure for XeO2F2 which correctly minimizes formal charges – Bartleby
- The real structure of difluoro(dioxo)xenon (XeO2F2) – Chemistry Stack Exchange
- Select the Lewis structure for XeO2F2 which has the most favorable formal charges – Numerade

Deep

Rootmemory.com was founded by Deep Rana, who is a mechanical engineer by profession and a blogger by passion. He has a good conceptual knowledge on different educational topics and he provides the same on this website. He loves to learn something new everyday and believes that the best utilization of free time is developing a new skill.