The **Lewis structure of XeF _{5}^{+}** contains five single bonds, with xenon in the center, and five fluorines on either side. There are three lone pairs on each fluorine atom, and one lone pair on the xenon atom.

Plus, there is a positive (+1) charge on the xenon atom.

## Steps

By using the following steps, you can easily draw the Lewis structure of XeF_{5}^{+}.

#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)

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 fluorine is a group 17 element. Hence, xenon has **eight** valence electrons and fluorine has **seven** valence electrons.

Now XeF_{5}^{+} has one xenon atom and five fluorine atoms.

So the total number of valence electrons = valence electrons of xenon atom + (valence electrons of fluorine atom × 5)

And XeF_{5}^{+} has a positive (+1) charge, so we have to subtract one electron.

Therefore, the **total number of valence electrons** = 8 + 35 – 1 = 42

- 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 fluorine as follows:

Electronegativity value of xenon = 2.6

Electronegativity value of fluorine = 3.98

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

So now, put xenon in the center and fluorines on either side. And draw the rough skeleton structure for the Lewis structure of XeF_{5}^{+} something like this:

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

### #2 Show chemical bond

Place two electrons between the atoms to show a chemical bond. Since xenon is surrounded by three fluorines, use ten electrons to show **five chemical bonds** as follows:

**Also read:** How to draw Lewis structure of CH_{3}COO^{–} (5 steps)

### #3 Mark lone pairs

As calculated earlier, we have a total of 42 valence electrons. And in the above structure, we have already used ten valence electrons. Hence, thirty-two valence electrons are remaining.

Two valence electrons represent one lone pair. So thirty-two valence electrons = **sixteen lone pairs**.

Note 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.

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

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

So the Lewis structure of XeF_{5}^{+} 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.

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

**Also read:** How to draw NO_{2}^{–} Lewis structure (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 = 10

Formal charge = 8 – 2 – ½ (10) = +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 XeF_{5}^{+} looks something like this:

In the above structure, you can see that the formal charges of atoms are closer to zero. Therefore, this is the **most stable Lewis structure of XeF _{5}^{+}**.

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

Now XeF_{5}^{+} is an ion having a positive (+1) charge, so draw brackets around the above Lewis structure and mention that charge on the top right corner. And then, the Lewis structure of XeF_{5}^{+} looks something like this:

## Related

- Lewis structure of ClBr
_{3} - Lewis structure of CH
_{3}COO^{–} - NO
_{2}^{–}Lewis structure - Lewis structure of SnCl
_{2} - Lewis structure of CH
_{2}NH

## External links

- XeF5+ Lewis Structure & Characteristics: 13 Complete Facts – Lambda Geeks
- How many right angles are there in a XeF+5 ion? – Socratic
- What is the Lewis structure of XeF5+, with the central atom of Xe – Chegg
- Draw the Lewis structure of XeF5+ – Numerade
- Draw the Lewis structure for XeF5+ – Bartleby
- What’s the hybridisation of XeF5+? – Quora
- Draw the Lewis structure and 3D drawing for XeF5+ – Course Hero
- What is the Lewis structure for XeF5+ – Quizlet

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.