Lewis Structure of XeOF4 (With 5 Simple Steps to Draw!)

Lewis Structure of XeOF4

I’m super excited to teach you the lewis structure of XeOF4 in just 5 simple steps.

Infact, I’ve also given the step-by-step images for drawing the lewis dot structure of XeOF4 molecule.

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

Lewis structure of XeOF4 contains a double bond between Xenon & Oxygen and single bonds between Xenon (Xe) & each Fluorine (F) atom. The Xenon atom (Xe) is at the center and it is surrounded by 1 Oxygen (O) and 4 Fluorine atoms (F). The Xenon atom has 1 lone pair, Oxygen has 2 and Fluorine atoms has 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 XeOF4).

5 Steps to Draw the Lewis Structure of XeOF4

Step #1: Calculate the total number of valence electrons

Here, the given molecule is XeOF4. In order to draw the lewis structure of XeOF4, first of all you have to find the total number of valence electrons present in the XeOF4 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 XeOF4

  • 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).

  • For Fluorine: 

Fluorine is a group 17 element on the periodic table.

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

Hence in a XeOF4 molecule, 

Valence electrons given by Xenon (Xe) atom = 8
Valence electrons given by Oxygen (O) atom = 6
Valence electrons given by each Fluorine (F) atom = 7
So, total number of Valence electrons in XeOF4 molecule = 8 + 6 + 7(4) = 42

Step #2: Select the center atom

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

(Remember: Fluorine is the most electronegative element on the periodic table and the electronegativity decreases as we move right to left in the periodic table as well as top to bottom in the periodic table).

Here in the XeOF4 molecule, if we compare the Xenon atom (Xe), Oxygen atom (O) and Fluorine atom (F), then Xenon is less electronegative.

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

step 1

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

Now in the above sketch of XeOF4 molecule, put the two electrons (i.e electron pair) between each atom to represent a chemical bond between them.

step 2

These pairs of electrons present between the atoms form a chemical bond, which bonds these atoms with each other in a XeOF4 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 XeOF4, the outer atoms are oxygen atom and fluorine atoms.

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

step 3

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

Also, only 40 valence electrons of XeOF4 molecule are used in the above structure.

But there are total 42 valence electrons in XeOF4 molecule (as calculated in step #1).

So the number of electrons left to be kept on the central atom = 42 – 40 = 2.

So let’s keep these two electrons (i.e 1 electron pair) on the central atom.

step 4

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), oxygen atom (O) 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

step 5
  • For Xenon:
    Valence electrons = 8 (as it is in group 18)
    Nonbonding electrons = 2
    Bonding electrons = 10
  • For Oxygen:
    Valence electrons = 6 (as it is in group 16)
    Nonbonding electrons = 6
    Bonding electrons = 2
  • For Fluorine:
    Valence electron = 7 (as it is in group 17)
    Nonbonding electrons = 6
    Bonding electrons = 2
Formal charge=Valence electronsNonbonding electrons(Bonding electrons)/2
Xe=8210/2=+1
O=662/2=-1
F=762/2=0

So you can see above that the formal charge on Xenon is +1 and the formal charge on the oxygen atom is -1.

This indicates that the above lewis structure of XeOF4 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 atom to the positively charged xenon atom to form a bond.

step 6

Now, in the above structure, you can see that the charges are minimized and the above lewis structure of XeOF4 is the final stable structure.

Each electron pair (:) in the lewis dot structure of XeOF4 represents the single bond ( | ). So the above lewis dot structure of XeOF4 can also be represented as shown below.

xeof4 lewis structure

Related lewis structures for your practice:
Lewis Structure of ICl
Lewis Structure of H2SO3
Lewis Structure of HSO4-
Lewis Structure of CCl2F2
Lewis Structure of C2H2Cl2 

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