XeO4 Lewis structure

XeO4 Lewis Structure
XeO4 Lewis structure | Image: Learnool

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.

Rough sketch

  • First, determine the total number of valence electrons
Periodic table | Image: Learnool

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
XeO4 Lewis Structure (Step 1)
Rough sketch of XeO4 Lewis structure | Image: Learnool

Lone pair

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:

XeO4 Lewis Structure (Step 2)
Lone pairs marked on XeO4 Lewis structure | Image: Learnool

Formal charge

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:

XeO4 Lewis Structure (Step 3)
Formal charges marked on XeO4 Lewis structure | Image: Learnool

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.

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

XeO4 Lewis Structure (Step 4)
Lone pair of left oxygen is converted, but still there are charges | Image: Learnool

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:

XeO4 Lewis Structure (Step 5)
Lone pair of right oxygen is converted, but still there are charges | Image: Learnool

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:

XeO4 Lewis Structure (Step 6)
Lone pair of top oxygen is converted, but still there are charges | Image: Learnool

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:

XeO4 Lewis Structure (Step 7)
Lone pair of bottom oxygen is converted, and got the stable Lewis structure of XeO4 | Image: Learnool

Final structure

The final structure of XeO4 includes a central xenon atom connected to four oxygen atoms through double covalent bonds. In this configuration, the xenon atom serves as an exception to the octet rule, utilizing an expanded valence shell to accommodate sixteen electrons through four double bonds. Each oxygen atom satisfies its octet by retaining two lone pairs alongside its double bond with the central atom. This arrangement is the most stable because it results in formal charges of zero for all atoms involved, representing the most energetically favorable state for the molecule. Consequently, this specific electronic distribution serves as the definitive and most accurate Lewis representation of xenon tetroxide.

Next: CNO Lewis structure

External video

External links

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.

Leave a Comment