Lewis Structure of XeO2F2 (With 5 Simple Steps to Draw!)

Ready to learn how to draw the lewis structure of XeO2F2?

Awesome!

Here, I have explained 5 simple steps to draw the lewis dot structure of XeO2F2 (along with images).

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

Lewis structure of XeO2F2 contains double bonds between the Xenon (Xe) atom & Oxygen (O) atoms and single bonds between the Xenon (Xe) atom and Fluorine (F) atoms. The Xenon atom (Xe) is at the center and it is surrounded by 2 Oxygen atoms (O) and 2 Fluorine atoms (F). The Xenon atom has 1 lone pair.

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 XeO2F2).

5 Steps to Draw the Lewis Structure of XeO2F2

Step #1: Calculate the total number of valence electrons

Here, the given molecule is XeO2F2. In order to draw the lewis structure of XeO2F2, first of all you have to find the total number of valence electrons present in the XeO2F2 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 XeO2F2

• 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 Oxygen:

Oxygen is a group 16 element on the periodic table. ^[2]

Hence, the valence electron present in oxygen is 6 (see below image).

• For Fluorine: 

Fluorine is a group 17 element on the periodic table. ^[3]

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

Hence in a XeO2F2 molecule, 

Valence electrons given by Xenon (Xe) atom = 8
Valence electrons given by each Oxygen (O) atom = 6
Valence electrons given by each Fluorine (F) atom = 7
So, total number of Valence electrons in XeO2F2 molecule = 8 + 6(2) + 7(2) = 34

Step #2: Select the center atom

While selecting the center 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). ^[4]

Here in the XeO2F2 molecule, if we compare the xenon atom (Xe), oxygen atom (O) and fluorine atom (F), then xenon is less electronegative than oxygen and fluorine.

So, xenon should be placed in the center and the remaining 2 oxygen atoms as well as 2 fluorine atoms will surround it.

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

Now in the above sketch of XeO2F2, put the two electrons (i.e electron pair) between each atom to represent a chemical bond between them.

These pairs of electrons form a chemical bond, which bonds these atoms with each other in a XeO2F2 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 XeO2F2 molecule, the outer atoms are oxygen atoms and fluorine atoms.

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

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

Also, only 32 valence electrons of XeO2F2 molecule are used in the above structure.

But there are total 34 valence electrons in XeO2F2 molecule (as calculated in step #1).

So the number of electrons left to be kept on the central atom = 34 – 32 = 2.

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

Now, 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 the XeO2F2 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 = 2
  Bonding electrons = 8

• For Oxygen:
  Valence electron = 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 electrons	–	Nonbonding electrons	–	(Bonding electrons)/2
Xe	=	8	–	2	–	8/2	=	+2
O	=	6	–	6	–	2/2	=	-1
F	=	7	–	6	–	2/2	=	0

Let’s keep these charges on the atoms in the above lewis structure of XeO2F2 molecule.

As we know that xenon can hold more than 8 electrons, we can further reduce the charges as mentioned below.

By doing this, the formal charges on the two oxygen atoms as well as one central xenon atom becomes “zero”, and this gives us the most stable lewis structure of XeO2F2.

Each electron pair (:) in the lewis dot structure of XeO2F2 molecule represents the single bond ( | ). So the above lewis dot structure of XeO2F2 molecule can also be represented as shown below.

Related lewis structures for your practice:
Lewis Structure of XeH4
Lewis Structure of S2Cl2
Lewis Structure of N2O5
Lewis Structure of BeBr2
Lewis Structure of CSe2 

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