XeO3 Lewis Structure in 5 Steps (With Images)

XeO3 Lewis Structure

So you have seen the above image by now, right?

Let me explain the above image in short.

XeO3 lewis structure has a Xenon atom (Xe) at the center which is surrounded by three Oxygen atoms (O). There is a double bond between the Xenon (Xe) & all the three Oxygen (O) atoms. There is 1 lone pair on Xenon atom (Xe) and 2 lone pairs on the Oxygen atoms (O).

If you haven’t understood anything from the above image of XeO3 lewis structure, then just stick with me and you will get the detailed step by step explanation on drawing a lewis structure of XeO3.

So let’s move to the steps of drawing the lewis structure of XeO3 molecule.

Steps of drawing XeO3 lewis structure

Step 1: Find the total valence electrons in XeO3

In order to find the total valence electrons in XeO3 molecule, first of all you should know the valence electrons present in xenon atom as well as oxygen atom.
(Valence electrons are the electrons that are present in the outermost orbit of any atom.)

Here, I’ll tell you how you can easily find the valence electrons of xenon as well as oxygen using a periodic table.

Total valence electrons in XeO3 molecule

→ Valence electrons given by xenon atom:

Xenon is a group 18 element on the periodic table. [1] Hence the valence electrons present in xenon is 8.

You can see the 8 valence electrons present in the xenon atom as shown in the above image.

→ Valence electrons given by oxygen atom:

Oxygen is group 16 element on the periodic table. [2] Hence the valence electrons present in oxygen is 6.

You can see the 6 valence electrons present in the oxygen atom as shown in the above image.

Hence, 

Total valence electrons in XeO3 molecule = valence electrons given by 1 xenon atom + valence electrons given by 3 oxygen atoms = 8 + 6(3) = 26.

Step 2: Select the central atom

For selecting the center atom, you have to remember that the atom which is less electronegative remains at the center.

Now here the given molecule is XeO3 and it contains xenon atom (Xe) and oxygen atoms (O).

You can see the electronegativity values of xenon atom (Xe) and oxygen atom (O) in the above periodic table.

If we compare the electronegativity values of xenon (Xe) and oxygen (O) then the xenon atom is less electronegative.

So here the xenon atom (Xe) is the center atom and the oxygen atoms (O) are the outside atoms.

XeO3 step 1

Step 3: Connect each atoms by putting an electron pair between them

Now in the XeO3 molecule, you have to put the electron pairs between the xenon atom (Xe) and oxygen atoms (O).

XeO3 step 2

This indicates that the xenon (Xe) and oxygen (O) are chemically bonded with each other in a XeO3 molecule.

Step 4: Make the outer atoms stable

Now in this step, you have to check the stability of the outer atoms.

Here in the sketch of XeO3 molecule, you can see that the outer atoms are oxygen atoms.

These outer oxygen atoms are forming an octet and hence they are stable.

XeO3 step 3

Also, in step 1 we have calculated the total number of valence electrons present in the XeO3 molecule.

The XeO3 molecule has a total 26 valence electrons and out of these, only 24 valence electrons are used in the above sketch.

So the number of electrons which are left = 26 – 24 = 2.

You have to put these 2 electrons on the central xenon atom in the above sketch of XeO3 molecule.

XeO3 step 4

Now let’s proceed to the next step.

Step 5: Check the octet on the central atom

In this step, you have to check whether the central xenon atom (Xe) is stable or not.

In order to check the stability of the central xenon (Xe) atom, we have to check whether it is forming an octet or not.

XeO3 step 5

You can see from the above picture that the xenon atom is forming an octet. That means it has 8 electrons.

And hence the central xenon atom is stable.

Now let’s proceed to the final step to check whether the lewis structure of XeO3 is stable or not.

Step 6: Check the stability of lewis structure

Now you have come to the final step in which you have to check the stability of lewis structure of XeO3 molecule.

The stability of lewis structure can be checked by using a concept of formal charge.

In short, now you have to find the formal charge on xenon (Xe) atom as well as oxygen (O) atoms present in the XeO3 molecule.

For calculating the formal charge, you have to use the following formula;

Formal charge = Valence electrons – (Bonding electrons)/2 – Nonbonding electrons

You can see the number of bonding electrons and nonbonding electrons for each atom of XeO3 molecule in the image given below.

XeO3 step 6

For Xenon (Xe) atom:
Valence electrons = 8 (because xenon is in group 18)
Bonding electrons = 6
Nonbonding electrons = 2

For Oxygen (O) atom:
Valence electrons = 6 (because oxygen is in group 16)
Bonding electrons = 2
Nonbonding electrons = 6

Formal charge=Valence electrons(Bonding electrons)/2Nonbonding electrons
Xe=86/22=+3
O=62/26=-1

From the above calculations of formal charge, you can see that the xenon (Xe) atom has +3 charge while all the oxygen atoms have -1 charge.

This indicated that the lewis structure of XeO3 is not stable.

So we have to minimize these charges by shifting the electron pairs from the oxygen atoms to the xenon atom.

XeO3 step 7

After shifting the electron pairs from the oxygen atoms to the xenon atom, the charges on xenon and oxygen atoms become zero. And this is a more stable lewis structure. (see below image).

XeO3 step 8

In the above lewis dot structure of XeO3 molecule, you can also represent each bonding electron pair (:) as a single bond (|). By doing so, you will get the following lewis structure of XeO3.

lewis structure of XeO3

I hope you have completely understood all the above steps.

For more practice and better understanding, you can try other lewis structures listed below.

Try (or at least See) these lewis structures for better understanding:

H2CO3 Lewis StructureSBr2 Lewis Structure
HOCl Lewis StructureC6H6 (Benzene) Lewis Structure
NBr3 Lewis StructureSeF4 Lewis Structure
About author

Jay is an educator and has helped more than 100,000 students in their studies by providing simple and easy explanations on different science-related topics. He is a founder of Pediabay and is passionate about helping students through his easily digestible explanations.

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