The **Lewis structure of XeO _{4}** contains four double bonds, with xenon in the center, and four oxygens on either side. There are two lone pairs on each oxygen atom, and the xenon atom does not have any lone pair.

## Steps

By using the following steps, you can easily draw the Lewis structure of XeO_{4}.

#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 and oxygen is a group 16 element. Hence, xenon has **eight** valence electrons and oxygen has **six** valence electrons.

Now XeO_{4} has one xenon atom and four oxygen atoms.

So the total number of valence electrons = valence electrons of xenon atom + (valence electrons of oxygen atom × 4)

Therefore, the **total number of valence electrons** = 8 + 24 = 32

- 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 and oxygen as follows:

Electronegativity value of xenon = 2.6

Electronegativity value of oxygen = 3.44

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

So now, put xenon in the center and oxygens on either side. And draw the rough skeleton structure for the Lewis structure of XeO_{4} something like this:

**Also read:** How to draw Lewis structure of CO_{3}^{2-} (5 steps)

### #2 Show chemical bond

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

### #3 Mark lone pairs

As calculated earlier, we have a total of 32 valence electrons. And in the above structure, we have already used eight valence electrons. Hence, twenty-four valence electrons are remaining.

Two valence electrons represent one lone pair. So twenty-four valence electrons = **twelve lone pairs**.

Note that xenon is a period 5 element, so it can keep more than 8 electrons in its last shell. And oxygen is a period 2 element, so it can not keep more than 8 electrons in its 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, so each oxygen will get three lone pairs. And the central atom (xenon) will not get any lone pair, because all twelve lone pairs are used.

So the Lewis structure of XeO_{4} 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 SO_{4}^{2-} (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 = 0

Bonding electrons = 8

Formal charge = 8 – 0 – ½ (8) = +4

- For
**each oxygen**atom

Valence electrons = 6

Nonbonding electrons = 6

Bonding electrons = 2

Formal charge = 6 – 6 – ½ (2) = -1

Mention the formal charges of atoms on the structure. So the Lewis structure of XeO_{4} 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 SiS_{2} (5 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_{4} 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 = 0

Bonding electrons = 16

Formal charge = 8 – 0 – ½ (16) = 0

- For
**each oxygen**atom

Valence electrons = 6

Nonbonding electrons = 4

Bonding electrons = 4

Formal charge = 6 – 4 – ½ (4) = 0

Mention the formal charges of atoms on the structure. So the Lewis structure of XeO_{4} looks something like this:

In the above structure, you can see that the formal charges of both (xenon and oxygen) are zero. Therefore, this is the **stable Lewis structure of XeO**** _{4}**.

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

## Related

- Lewis structure of CO
_{3}^{2-} - Lewis structure of SO
_{4}^{2-} - Lewis structure of SiS
_{2} - Lewis structure of PBr
_{5} - Lewis structure of OCS

## External links

- What is the Lewis Structure of XeO4? – Quora
- XeO4 Lewis Structure in 5 Steps (With Images) – Pediabay
- XeO4 lewis structure, Molecular geometry, Polar or nonpolar, Hybridization – Topblogtenz
- XeO4 Lewis Structure, Geometry, Hybridization, and Polarity – Techiescientist
- XeO4 Lewis Structure: Drawings, Hybridization, Shape, Charges, Pair, And Detailed Facts – Lambda Geeks
- XeO4 Lewis Structure (Xenon Tetroxide) – Pinterest
- Draw the Lewis structure for XeO4 – Homework.Study.com
- Preferred Lewis structure for sulphate anion and XeO4 – Chemistry Stack Exchange
- What is the total number of valence electrons in the lewis structure of xeo4? – Brainly
- XeO4 Geometry and Hybridization – Chemistry Steps
- What is the total number of valence electrons in the Lewis structure of XeO4? – Numerade
- XeO4 (Xenon Tetroxide) Oxidation Number – ChemicalAid
- How many equivalent Lewis structures are necessary to describe the bonding in XeO4? – Answers

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.