Lewis Structure of XeCl4 (With 5 Simple Steps to Draw!)

Lewis Structure of XeCl4

Ready to learn how to draw the lewis structure of XeCl4?

Awesome!

Here, I have explained 5 simple steps to draw the lewis dot structure of XeCl4 (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 XeCl4 contains four single bonds between the Xenon (Xe) atom and each Chlorine (Cl) atom. The Xenon atom (Xe) is at the center and it is surrounded by 4 Chlorine atoms (Cl). The Xenon atom has 2 lone pairs and all the Chlorine 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 XeCl4).

5 Steps to Draw the Lewis Structure of XeCl4

Step #1: Calculate the total number of valence electrons

Here, the given molecule is XeCl4. In order to draw the lewis structure of XeCl4, first of all you have to find the total number of valence electrons present in the XeCl4 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 XeCl4

  • 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 Chlorine:

Chlorine is a group 17 element on the periodic table.

Hence, the valence electron present in chlorine is 7 (see below image).

Hence in a XeCl4 molecule, 

Valence electrons given by Xenon (Xe) atom = 8
Valence electrons given by each Chlorine (Cl) atom = 7
So, total number of Valence electrons in XeCl4 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 XeCl4 molecule, if we compare the xenon atom (Xe) and chlorine atom (Cl), then the xenon is less electronegative than chlorine.

So, xenon should be placed in the center and the remaining 4 chlorine 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 XeCl4 molecule, put the two electrons (i.e electron pair) between each xenon atom and chlorine atom to represent a chemical bond between them.

step 2

These pairs of electrons present between the Xenon (Xe) and Chlorine (Cl) atoms form a chemical bond, which bonds the xenon and chlorine atoms with each other in a XeCl4 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 XeCl4, the outer atoms are chlorine atoms.

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

step 3

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

Also, only 34 valence electrons of XeCl4 molecule are used in the above structure.

But there are total 36 valence electrons in XeCl4 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.

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) as well as each chlorine atom (Cl).

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 electron = 8 (as it is in group 18)
    Nonbonding electrons = 4
    Bonding electrons = 8
  • For Chlorine:
    Valence electron = 7 (as it is in group 17)
    Nonbonding electrons = 6
    Bonding electrons = 2
Formal charge=Valence electronsNonbonding electrons(Bonding electrons)/2
Xe=848/2=0
Cl=762/2=0

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

Hence, there will not be any change in the above structure and the above lewis structure of XeCl4 is the final stable structure only.

Each electron pair (:) in the lewis dot structure of XeCl4 represents the single bond ( | ). So the above lewis dot structure of XeCl4 can also be represented as shown below.

xecl4 lewis structure

Related lewis structures for your practice:
Lewis Structure of AlBr3
Lewis Structure of AlF3
Lewis Structure of IBr
Lewis Structure of SeCl4
Lewis Structure of HOF 

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