XeO4 Lewis Structure in 5 Steps (With Images)

XeO4 Lewis Structure

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

Let me explain the above image in short.

XeO4 lewis structure has a Xenon atom (Xe) at the center which is surrounded by four Oxygen atoms (O). There is a double bond between the Xenon (Xe) & all the four Oxygen (O) atoms. There are 2 lone pairs on the Oxygen atoms (O).

If you haven’t understood anything from the above image of XeO4 lewis structure, then just stick with me and you will get the detailed step by step explanation on drawing a lewis structure of XeO4.

So let’s move to the steps of drawing the lewis structure of XeO4 molecule.

Steps of drawing XeO4 lewis structure

Step 1: Find the total valence electrons in XeO4

In order to find the total valence electrons in XeO4 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 XeO4 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 XeO4 molecule = valence electrons given by 1 xenon atom + valence electrons given by 4 oxygen atoms = 8 + 6(4) = 32.

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 XeO4 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.

XeO4 step 1

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

Now in the XeO4 molecule, you have to put the electron pairs between the xenon atom (Xe) and oxygen atoms (O).

XeO4 step 2

This indicates that the xenon (Xe) and oxygen (O) are chemically bonded with each other in a XeO4 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 XeO4 molecule, you can see that the outer atoms are oxygen atoms.

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

XeO4 step 3

Also, in step 1 we have calculated the total number of valence electrons present in the XeO4 molecule.

The XeO4 molecule has a total 32 valence electrons and all these valence electrons are used in the above sketch.

Hence there are no remaining electron pairs to be kept on the central atom.

So now let’s proceed to the next step.

Step 5: 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 XeO4 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 XeO4 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 XeO4 molecule in the image given below.

XeO4 step 4

For Xenon (Xe) atom:
Valence electrons = 8 (because xenon is in group 18)
Bonding electrons = 8
Nonbonding electrons = 0

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)/2 Nonbonding electrons
Xe = 8 8/2 0 = +4
O = 6 2/2 6 = -1

From the above calculations of formal charge, you can see that the xenon (Xe) atom has +4 charge while all the oxygen atoms have -1 charge.

So let’s keep these charges on the respective atoms of the XeO4 molecule.

XeO4 step 5

The above image shows that the lewis structure of XeO4 is not stable.

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

XeO4 step 6

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).

XeO4 step 7

In the above lewis dot structure of XeO4 molecule, you can also represent each bonding electron pair (:) as a single bond (|). By doing so, you will get the following lewis structure of XeO4.

lewis structure of XeO4

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:

SF3+ Lewis Structure XeO3 Lewis Structure
H2CO3 Lewis Structure SBr2 Lewis Structure
HOCl Lewis Structure C6H6 (Benzene) Lewis Structure

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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