# XeO4 Lewis structure

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XeO4 (xenon tetroxide) has one xenon atom and four oxygen atoms.

In the XeO4 Lewis structure, there are four double bonds around the xenon atom, with four oxygen atoms attached to it, and on each oxygen atom, there are two lone pairs.

Contents

## Steps

Here’s how you can easily draw the XeO4 Lewis structure step by step:

#1 Draw a rough skeleton structure
#2 Mention lone pairs on the atoms
#3 If needed, mention formal charges on the atoms
#4 Minimize formal charges by converting lone pairs of the atoms, and try to get a stable Lewis structure
#5 Repeat step 4 again if needed, until all charges are minimized

Now, let’s take a closer look at each step mentioned above.

### #1 Draw a rough skeleton structure

• First, determine the total number of valence electrons

In the periodic table, xenon lies in group 18, and oxygen lies in group 16.

Hence, xenon has eight valence electrons and oxygen has six valence electrons.

Since XeO4 has one xenon atom and four oxygen atoms, so…

Valence electrons of one xenon atom = 8 × 1 = 8
Valence electrons of four oxygen atoms = 6 × 4 = 24

And the total valence electrons = 8 + 24 = 32

• Second, find the total electron pairs

We have a total of 32 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 = 32 ÷ 2 = 16

• Third, determine the central atom

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

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

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

• And finally, draw the rough sketch

### #2 Mention lone pairs on the atoms

Here, we have a total of 16 electron pairs. And four Xe — O 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 oxygen is a period 2 element, so it can not keep more than 8 electrons in its last shell.

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

So for each oxygen, there are three lone pairs, and for xenon, there is zero lone pair because all twelve electron pairs are over.

Mark the lone pairs on the sketch as follows:

### #3 If needed, mention formal charges on the atoms

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 – 0 – ½ (8) = +4

For each oxygen atom, formal charge = 6 – 6 – ½ (2) = -1

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 Minimize formal charges by converting lone pairs of the atoms

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

### #5 Since there are charges on atoms, repeat step 4 again

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:

### #6 Minimize charges again

There are still charges on xenon and oxygen atoms, so again convert a lone pair of the oxygen atom to make a new Xe — O bond with the xenon atom as follows:

### #7 Minimize charges again

There are still charges on xenon and oxygen atoms, so 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) 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 XeO4.

Next: CNO Lewis structure