Lewis Structure of XeF4 (With 5 Simple Steps to Draw!)

I’m super excited to teach you the lewis structure of XeF4 in just 5 simple steps.

Infact, I’ve also given the step-by-step images for drawing the lewis dot structure of XeF4 molecule.

So, if you are ready to go with these 5 simple steps, then let’s dive right into it!

Lewis structure of XeF4 contains four single bonds between the Xenon (Xe) atom and each Fluorine (F) atom. The Xenon atom (Xe) is at the center and it is surrounded by 4 Fluorine atoms (F). The Xenon atom has 2 lone pairs and all the Fluorine atoms have 3 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 XeF4).

5 Steps to Draw the Lewis Structure of XeF4

Step #1: Calculate the total number of valence electrons

Here, the given molecule is XeF4 (xenon tetrafluoride). In order to draw the lewis structure of XeF4, first of all you have to find the total number of valence electrons present in the XeF4 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 XeF4

• For Xenon: 

Xenon is a group 18 element on the periodic table. ^[1]

Hence, the valence electron present in xenon is also 8 (see below image).

• For Fluorine: 

Fluorine is a group 17 element on the periodic table. ^[2]

Hence, the valence electrons present in fluorine is 7 (see below image).

Hence in a XeF4 molecule, 

Valence electrons given by Xenon (Xe) atom = 8
Valence electrons given by each Fluorine (F) atom = 7
So, total number of Valence electrons in XeF4 molecule = 8 + 7(4) = 36

Step #2: Select the center atom

While selecting the atom, always put the least electronegative atom at the center. 

Here in the XeF4 molecule, if we compare the xenon atom (Xe) and fluorine atom (F), then the xenon is less electronegative than fluorine.

So, xenon should be placed in the center and the remaining 4 fluorine atoms will surround it.

Step #3: Put two electrons between the atoms to represent a chemical bond

Now in the above sketch of XeF4 molecule, put the two electrons (i.e electron pair) between each xenon atom and fluorine atom to represent a chemical bond between them.

These pairs of electrons present between the Xenon (Xe) and Fluorine (F) atoms form a chemical bond, which bonds the xenon and fluorine atoms with each other in a XeF4 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 XeF4, the outer atoms are fluorine atoms.

So now, you have to complete the octet on these fluorine atoms (because fluorine requires 8 electrons to have a complete outer shell).

Now, you can see in the above image that all the fluorine atoms form an octet.

Also, only 34 valence electrons of XeF4 molecule are used in the above structure.

But there are total 36 valence electrons in XeF4 molecule (as calculated in step #1).

So the number of electrons left to be kept on the central atom = 36 – 32 = 4.

So let’s keep these four electrons (i.e 2 electron pairs) on the central atom.

Now, let’s move to the next step.

Step #5: Final step – Check the stability of lewis structure by calculating the formal charge on each atom

Now, you have come to the final step and here you have to check the formal charge on xenon atom (Xe) as well as each fluorine atom (F).

For that, you need to remember the formula of formal charge;

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

• For Xenon:
  Valence electron = 8 (as it is in group 18)
  Nonbonding electrons = 4
  Bonding electrons = 8
• For Fluorine:
  Valence electron = 7 (as it is in group 17)
  Nonbonding electrons = 6
  Bonding electrons = 2

Formal charge	=	Valence electrons	–	Nonbonding electrons	–	(Bonding electrons)/2
Xe	=	8	–	4	–	8/2	=	0
F	=	7	–	6	–	2/2	=	0

So you can see above that the formal charges on xenon as well as fluorine are “zero”.

Hence, there will not be any change in the above structure and the above lewis structure of XeF4 is the final stable structure only.

Each electron pair (:) in the lewis dot structure of XeF4 represents the single bond ( | ). So the above lewis dot structure of XeF4 can also be represented as shown below.

Related lewis structures for your practice:
Lewis structure of PO43-
Lewis structure of I3-
Lewis structure of CN-
Lewis structure of PF3
Lewis structure of PCl5

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