I’m super excited to teach you the lewis structure of XeO4 in just 5 simple steps.
Infact, I’ve also given the step-by-step images for drawing the lewis dot structure of XeO4 molecule.
So, if you are ready to go with these 5 simple steps, then let’s dive right into it!
Lewis structure of XeO4 contains four double bonds between the Xenon (Xe) atom and Oxygen (O) atoms. The Xenon atom (Xe) is at the center and it is surrounded by 4 Oxygen atoms (O). All the four Oxygen atoms (O) have 2 lone pairs.
Let’s draw and understand this lewis dot structure step by step.
(Note: Take a pen and paper with you and try to draw this lewis structure along with me. I am sure you will definitely learn how to draw lewis structure of XeO4).
5 Steps to Draw the Lewis Structure of XeO4
Step #1: Calculate the total number of valence electrons
Here, the given molecule is XeO4. In order to draw the lewis structure of XeO4, first of all you have to find the total number of valence electrons present in the XeO4 molecule.
(Valence electrons are the number of electrons present in the outermost shell of an atom).
So, let’s calculate this first.
Calculation of valence electrons in XeO4
- For Xenon:
Xenon is a group 18 element on the periodic table. 
Hence, the valence electron present in xenon is also 8 (see below image).
- For Oxygen:
Oxygen is a group 16 element on the periodic table. 
Hence, the valence electron present in oxygen is 6 (see below image).
Hence in a XeO4 molecule,
Valence electrons given by Xenon (Xe) atom = 8
Valence electrons given by each Oxygen (O) atom = 6
So, total number of Valence electrons in XeO4 molecule = 8 + 6(4) = 32
Step #2: Select the center atom
While selecting the atom, always put the least electronegative atom at the center.
Here in the XeO4 molecule, if we compare the xenon atom (Xe) and oxygen atom (O), then the xenon is less electronegative than oxygen.
So, xenon should be placed in the center and the 4 oxygen atoms will surround it.
Step #3: Put two electrons between the atoms to represent a chemical bond
Now in the above sketch of XeO4, put the two electrons (i.e electron pair) between each xenon atom and oxygen atom to represent a chemical bond between them.
These pairs of electrons present between the Xenon (Xe) and Oxygen (O) atoms form a chemical bond, which bonds the xenon and oxygen atoms with each other in a XeO4 molecule.
Step #4: Complete the octet (or duplet) on outside atoms. If the valence electrons are left, then put the valence electrons pair on the central atom
Don’t worry, I’ll explain!
In the Lewis structure of XeO4, the outer atoms are oxygen atoms.
So now, you have to complete the octet on these oxygen atoms (because oxygen requires 8 electrons to have a complete outer shell).
Now, you can see in the above image that all the oxygen atoms form an octet.
Also, all the 32 valence electrons of XeO4 molecule (as calculated in step #1) are used in the above structure. So there are no remaining electron pairs.
Hence there is no change in the above sketch of XeO4.
Let’s move to the next step.
Step #5: Check the formal charge
You can see from the above image that the central atom (i.e xenon), is having 8 electrons. So it fulfills the octet rule.
But, in order to get the most stable lewis structure, we have to check the formal charge on each atom of XeO4 molecule.
For that, you need to remember the formula of formal charge;
Formal charge = Valence electrons – Nonbonding electrons – (Bonding electrons)/2
- For Xenon:
Valence electrons = 8 (as it is in group 18)
Nonbonding electrons = 0
Bonding electrons = 8
- For Oxygen:
Valence electron = 6 (as it is in group 16)
Nonbonding electrons = 6
Bonding electrons = 2
|Formal charge||=||Valence electrons||–||Nonbonding electrons||–||(Bonding electrons)/2|
So you can see above that the formal charge on xenon is +4 and the formal charge on all the four oxygen atoms is -1.
This indicates that the above lewis structure of XeO4 is not stable and so we have to minimize the charges to get a more stable lewis structure.
This can be done by shifting the lone pair from negatively charged oxygen atoms to the positively charged xenon atom to form a double bond.
Now, in the above structure, you can see that the charges are minimized and the above lewis structure of XeO4 is the final stable structure.
Each electron pair (:) in the lewis dot structure of XeO4 represents the single bond ( | ). So the above lewis dot structure of XeO4 can also be represented as shown below.
Related lewis structures for your practice:
Lewis Structure of SF3+
Lewis Structure of XeO3
Lewis Structure of H2CO3
Lewis Structure of SBr2
Lewis Structure of HOCl