XeO2 Lewis structure

XeO₂ (xenon dioxide) has one xenon atom and two oxygen atoms.

In XeO₂ Lewis structure, there are two double bonds around the xenon atom, with two oxygen atoms attached to it, and each atom has two lone pairs.

Alternative method: Lewis structure of XeO₂

Rough sketch

• 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 XeO₂ has one xenon atom and two oxygen atoms, so…

Valence electrons of one xenon atom = 8 × 1 = 8
Valence electrons of two oxygen atoms = 6 × 2 = 12

And the total valence electrons = 8 + 12 = 20

Learn how to find: Oxygen valence electrons

• Second, find the total electron pairs

We have a total of 20 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 = 20 ÷ 2 = 10

• 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

Lone pair

Here, we have a total of 10 electron pairs. And two Xe — O bonds are already marked. So we have to only mark the remaining eight 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 are two lone pairs.

Mark the lone pairs on the sketch as follows:

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 – 4 – ½ (4) = +2

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.

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

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:

Final structure

The final structure of XeO₂ consists of a central xenon atom linked to two oxygen atoms. In this configuration, the xenon atom utilizes an expanded valence shell to form two double covalent bonds with the oxygen atoms while retaining two lone pairs. Within this layout, each oxygen atom satisfies the octet rule by maintaining two lone pairs alongside its double bond. This arrangement represents the most stable state for the molecule because it results in a formal charge of zero for every atom involved—xenon contributes eight valence electrons (four in bonds, four in lone pairs), and each oxygen contributes six. Therefore, this specific electronic distribution serves as the definitive and most accurate Lewis representation of xenon dioxide.

Next: CF₂ Lewis structure

External video

• How to Draw the Lewis Dot Structure for XeO2: Xenon dioxide – YouTube • Wayne Breslyn

External links

• https://lambdageeks.com/xeo2-lewis-structure/