XeCl_{3}^{–} (xenon trichloride) has **one xenon** atom and **three chlorine** atoms.

In the XeCl_{3}^{–} Lewis structure, there are three single bonds around the xenon atom, with three chlorine atoms attached to it, and each atom has three lone pairs.

Also, there is a negative (-1) charge on the xenon atom.

## Steps

Use these steps to correctly draw the XeCl_{3}^{–} 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

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, and chlorine lies in group 17.

Hence, xenon has **eight** valence electrons and chlorine has **seven** valence electrons.

Since XeCl_{3}^{–} has one xenon atom and three chlorine atoms, so…

Valence electrons of one xenon atom = 8 × 1 = 8

Valence electrons of three chlorine atoms = 7 × 3 = 21

Now the XeCl_{3}^{–} has a negative (-1) charge, so we have to add one more electron.

So the **total valence electrons** = 8 + 21 + 1 = 30

Learn how to find: Chlorine valence electrons

- Second, find the total electron pairs

We have a total of 30 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** = 30 ÷ 2 = 15

- Third, determine the central atom

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

Since xenon is less electronegative than chlorine, assume that the **central atom is xenon**.

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

- And finally, draw the rough sketch

### #2 Mark lone pairs on the atoms

Here, we have a total of 15 electron pairs. And three Xe — Cl bonds are already marked. So we have to only mark the remaining twelve 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 chlorine is a period 3 element, so it can keep more than 8 electrons in its last shell.

Always start to mark the lone pairs from outside atoms. Here, the outside atoms are chlorines.

So for each chlorine, there are **three** lone pairs, and for xenon, there are **three** lone pairs.

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 – 6 – ½ (6) = -1

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

Here, the xenon atom has a charge, so mark it on the sketch as follows:

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

Now there is still a negative (-1) charge on the xenon atom.

This is not okay, right? Because the structure with a negative charge on the most electronegative atom is the best Lewis structure. And in this case, the most electronegative element is chlorine.

But if we convert a lone pair of the xenon atom to make a new Xe — Cl bond with the chlorine atom, and calculate the formal charge, then we do not get the formal charges on atoms closer to zero.

And the structure with the formal charges on atoms closer to zero is the best Lewis structure.

Therefore, this structure is the **most stable Lewis structure** of XeCl_{3}^{–}.

And since the XeCl_{3}^{–} has a negative (-1) charge, mention that charge on the Lewis structure by drawing brackets as follows:

**Next:** POF_{3} Lewis structure

Deep

Learnool.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.