XeH4 Lewis Structure in 5 Steps (With Images)

So you have seen the above image by now, right?

Let me explain the above image in short.

XeH4 lewis structure has a Xenon atom (Xe) at the center which is surrounded by four Hydrogen atoms (H). There are 4 single bonds between the Xenon atom (Xe) and each Hydrogen atom (H). There are 2 lone pairs on the Xenon atom (Xe).

If you haven’t understood anything from the above image of XeH4 lewis structure, then just stick with me and you will get the detailed step by step explanation on drawing a lewis structure of XeH4.

So let’s move to the steps of drawing the lewis structure of XeH4.

Steps of drawing XeH4 lewis structure

Step 1: Find the total valence electrons in XeH4 molecule

In order to find the total valence electrons in XeH4 molecule, first of all you should know the valence electrons present in xenon atom as well as hydrogen atom.
(Valence electrons are the electrons that are present in the outermost orbit of any atom.)

Here, I’ll tell you how you can easily find the valence electrons of xenon as well as hydrogen using a periodic table.

Total valence electrons in XeH4 molecule

→ Valence electrons given by xenon atom:

Xenon is a group 18 element on the periodic table. ^[1] Hence the valence electrons present in xenon is 8.

You can see the 8 valence electrons present in the xenon atom as shown in the above image.

→ Valence electrons given by hydrogen atom:

Hydrogen is group 1 element on the periodic table. ^[2] Hence the valence electron present in hydrogen is 1.

You can see that only 1 valence electron is present in the hydrogen atom as shown in the above image.

Hence, 

Total valence electrons in XeH4 molecule = valence electrons given by 1 xenon atom + valence electrons given by 4 hydrogen atoms = 8 + 1(4) = 12.

Step 2: Select the central atom

For selecting the center atom, you have to remember that the atom which is less electronegative remains at the center.

(Remember: If hydrogen is present in the given molecule, then always put hydrogen outside.)

Now here the given molecule is XeH4 and it contains xenon atom (Xe) and hydrogen atoms (H).

You can see the electronegativity values of xenon atom (Xe) and hydrogen atom (H) in the above periodic table.

If we compare the electronegativity values of xenon (Xe) and hydrogen (H) then the hydrogen atom is less electronegative. But as per the rule we have to keep hydrogen outside.

So here the xenon atom (Xe) is the center atom and the hydrogen atoms (H) are the outside atoms.

Step 3: Connect each atoms by putting an electron pair between them

Now in the XeH4 molecule, you have to put the electron pairs between the xenon atom (Xe) and hydrogen atoms (H).

This indicates that the xenon (Xe) and hydrogen (H) are chemically bonded with each other in a XeH4 molecule.

Step 4: Make the outer atoms stable. Place the remaining valence electrons pair on the central atom.

Now in this step, you have to check the stability of the outer atoms.

Here in the sketch of XeH4 molecule, you can see that the outer atoms are hydrogen atoms.

These outer hydrogen atoms are forming a duplet and hence they are stable.

Also, in step 1 we have calculated the total number of valence electrons present in the XeH4 molecule.

The XeH4 molecule has a total 12 valence electrons and out of these, only 8 valence electrons are used in the above sketch.

So the number of electrons which are left = 12 – 8 = 4.

You have to put these 4 electrons on the central xenon atom in the above sketch of XeH4 molecule.

Now let’s proceed to the next step.

Step 5: Check the stability of lewis structure

Now you have come to the final step in which you have to check the stability of lewis structure of XeH4.

The stability of lewis structure can be checked by using a concept of formal charge.

In short, now you have to find the formal charge on xenon (Xe) atom as well as hydrogen (H) atoms present in the XeH4 molecule.

For calculating the formal charge, you have to use the following formula;

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

You can see the number of bonding electrons and nonbonding electrons for each atom of XeH4 molecule in the image given below.

For Xenon (Xe) atom:
Valence electrons = 8 (because xenon is in group 18)
Bonding electrons = 8
Nonbonding electrons = 4

For Hydrogen (H) atom:
Valence electron = 1 (because hydrogen is in group 1)
Bonding electrons = 2
Nonbonding electrons = 0

Formal charge	=	Valence electrons	–	(Bonding electrons)/2	–	Nonbonding electrons
Xe	=	8	–	8/2	–	4	=	0
H	=	1	–	2/2	–	0	=	0

From the above calculations of formal charge, you can see that the xenon (Xe) atom as well as hydrogen (H) atom has a “zero” formal charge.

This indicates that the above lewis structure of XeH4 is stable and there is no further change in the above structure of XeH4.

In the above lewis dot structure of XeH4, you can also represent each bonding electron pair (:) as a single bond (|). By doing so, you will get the following lewis structure of XeH4.

I hope you have completely understood all the above steps.

For more practice and better understanding, you can try other lewis structures listed below.

Try (or at least See) these lewis structures for better understanding:

S2Cl2 Lewis Structure	N2O5 Lewis Structure
BeBr2 Lewis Structure	CSe2 Lewis Structure
BrCl Lewis Structure	AsCl3 Lewis Structure