Lewis Structure of XeH4 (With 5 Simple Steps to Draw!)

I’m super excited to teach you the lewis structure of XeH4 in just 5 simple steps.

Infact, I’ve also given the step-by-step images for drawing the lewis dot structure of XeH4 molecule.

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

Lewis structure of XeH4 contains four single bonds between the Xenon (Xe) atom and each Hydrogen (H) atom. The Xenon atom (Xe) is at the center and it is surrounded by 4 Hydrogen atoms (H). The Xenon atom has 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 XeH4).

5 Steps to Draw the Lewis Structure of XeH4

Step #1: Calculate the total number of valence electrons

Here, the given molecule is XeH4. In order to draw the lewis structure of XeH4, first of all you have to find the total number of valence electrons present in the XeH4 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 XeH4

• 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 Hydrogen: 

Hydrogen is a group 1 element on the periodic table. ^[2]

Hence, the valence electron present in hydrogen is 1 (see below image).

Hence in a XeH4 molecule, 

Valence electrons given by Xenon (Xe) atom = 8
Valence electrons given by each Hydrogen (H) atom = 7
So, total number of Valence electrons in XeH4 molecule = 8 + 1(4) = 12

Step #2: Select the center atom (H is always outside)

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

Here in the XeH4 molecule, if we compare the xenon atom (Xe) and hydrogen atom (H), then hydrogen is less electronegative than xenon. But as per the rule, we have to keep hydrogen outside.

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

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

Now in the above sketch of XeH4 molecule, put the two electrons (i.e electron pair) between each xenon atom and hydrogen atom to represent a chemical bond between them.

These pairs of electrons present between the Xenon (Xe) and Hydrogen (H) atoms form a chemical bond, which bonds the xenon and hydrogen atoms with each other in a XeH4 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 XeH4, the outer atoms are hydrogen atoms.

So now, you have to check whether these hydrogen atoms are forming a duplet or not! (because hydrogen requires only 2 electrons to have a complete outer shell).

You can see in the above image that all the hydrogen atoms form a duplet.

Also, only 8 valence electrons of XeH4 molecule are used in the above structure.

But there are total 12 valence electrons in XeH4 molecule (as calculated in step #1).

So the number of electrons left to be kept on the central atom = 12 – 8 = 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 hydrogen atom (H).

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 = 4
  Bonding electrons = 8
• For Hydrogen:
  Valence electron = 1 (as it is in group 1)
  Nonbonding electrons = 0
  Bonding electrons = 2

Formal charge	=	Valence electrons	–	Nonbonding electrons	–	(Bonding electrons)/2
Xe	=	8	–	2	–	8/2	=	0
H	=	1	–	0	–	2/2	=	0

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

Hence, there will not be any change in the above structure and the above lewis structure of XeH4 is the final stable structure only.

Each electron pair (:) in the lewis dot structure of XeH4 represents the single bond ( | ). So the above lewis dot structure of XeH4 can also be represented as shown below.

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

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