XeH4 Lewis structure

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XeH₄ (xenon tetrahydride) has one xenon atom and four hydrogen atoms.

In the XeH₄ Lewis structure, there are four single bonds around the xenon atom, with four hydrogen atoms attached to it, and on the xenon atom, there are two lone pairs.

Steps

Here’s how you can easily draw the XeH₄ Lewis structure step by step:

#1 Draw a rough skeleton structure
#2 Mention lone pairs on the atoms
#3 If needed, mention formal charges on the atoms

Now, let’s take a closer look at each step mentioned above.

#1 Draw a rough skeleton structure

• First, determine the total number of valence electrons

In the periodic table, xenon lies in group 18, and hydrogen lies in group 1.

Hence, xenon has eight valence electrons and hydrogen has one valence electron.

Since XeH₄ has one xenon atom and four hydrogen atoms, so…

Valence electrons of one xenon atom = 8 × 1 = 8
Valence electrons of four hydrogen atoms = 1 × 4 = 4

And the total valence electrons = 8 + 4 = 12

Learn how to find: Hydrogen valence electrons

• Second, find the total electron pairs

We have a total of 12 valence electrons. And when we divide this value by two, we get the value of total electron pairs.

Total electron pairs = total valence electrons ÷ 2

So the total electron pairs = 12 ÷ 2 = 6

• Third, determine the central atom

Here hydrogen can not be the central atom. Because the central atom is bonded with at least two other atoms, and hydrogen has only one electron in its last shell, so it can not make more than one bond.

Hence, here we have to assume that the central atom is xenon.

Therefore, place xenon in the center and hydrogens on either side.

• And finally, draw the rough sketch

#2 Mention lone pairs on the atoms

Here, we have a total of 6 electron pairs. And four Xe — H bonds are already marked. So we have to only mark the remaining two electron pairs as lone pairs on the sketch.

Also remember that xenon is a period 5 element, so it can keep more than 8 electrons in its last shell. And hydrogen is a period 1 element, so it can not keep more than 2 electrons in its last shell.

Always start to mark the lone pairs from outside atoms. Here, the outside atoms are hydrogens. But no need to mark on hydrogen, because each hydrogen has already two electrons.

So for xenon, there are two lone pairs.

Mark the lone pair on the sketch as follows:

#3 If needed, mention formal charges on the atoms

Use the following formula to calculate the formal charges on atoms:

Formal charge = valence electrons – nonbonding electrons – ½ bonding electrons

For xenon atom, formal charge = 8 – 4 – ½ (8) = 0

For each hydrogen atom, formal charge = 1 – 0 – ½ (2) = 0

Here, both xenon and hydrogen atoms do not have charges, so no need to mark the charges.

In the above structure, you can see that the central atom (xenon) forms an octet. And the outside atoms (hydrogens) also form a duet. Hence, the octet rule and duet rule are satisfied.

Therefore, this structure is the stable Lewis structure of XeH₄.

Next: SiCl₂Br₂ Lewis structure

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External links

• https://www.thegeoexchange.org/chemistry/bonding/Lewis-Structures/XeH4-lewis-structure.html
• https://oneclass.com/homework-help/chemistry/5669768-xeh4-lewis-structure.en.html