Lewis structure of XeF4 - Root Memory (2024)

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The Lewis structure of XeF₄ contains four single bonds, with xenon in the center, and four fluorines on either side. There are three lone pairs on each fluorine atom, and two lone pairs on the xenon atom.

Steps

By using the following steps, you can easily draw the Lewis structure of XeF₄:

#1 Draw skeleton
#2 Show chemical bond
#3 Mark lone pairs
#4 Calculate formal charge and check stability (if octet is already completed on central atom)

Let’s one by one discuss each step in detail.

#1 Draw skeleton

In this step, first calculate the total number of valence electrons. And then, decide the central atom.

• Let’s calculate the total number of valence electrons

We know that… xenon is a group 18 element and fluorine is a group 17 element. Hence, xenon has eight valence electrons and fluorine has seven valence electrons.

Now XeF₄ has one xenon atom and four fluorine atoms.

So the total number of valence electrons = valence electrons of xenon atom + (valence electrons of fluorine atom × 4)

Therefore, the total number of valence electrons = 8 + 28 = 36

• Now decide the central atom

The atom with the least electronegative value is placed at the center. By looking at the periodic table, we get the electronegativity values for xenon and fluorine as follows:

Electronegativity value of xenon = 2.6
Electronegativity value of fluorine = 3.98

Obviously, xenon is less electronegative than fluorine. Hence, assume that xenon is the central atom.

So now, put xenon in the center and fluorines on either side. And draw the rough skeleton structure for the Lewis structure of XeF₄ something like this:

Also read: How to draw Lewis structure of F2 (4 steps)

#2 Show chemical bond

Place two electrons between the atoms to show a chemical bond. Since xenon is surrounded by four fluorines, use eight electrons to show four chemical bonds as follows:

Also read: How to draw Lewis structure of PO43- (5 steps)

#3 Mark lone pairs

As calculated earlier, we have a total of 36 valence electrons. And in the above structure, we have already used eight valence electrons. Hence, twenty-eight valence electrons are remaining.

Two valence electrons represent one lone pair. So twenty-eight valence electrons = fourteen lone pairs.

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

Also, make sure that you start marking these lone pairs on outside atoms first. And then, on the central atom.

The outside atoms are fluorines, so each fluorine will get three lone pairs. And the central atom (xenon) will get two lone pairs.

So the Lewis structure of XeF₄ looks something like this:

In the above structure, you can see that the octet is completed on the central atom (xenon), and also on the outside atoms. Therefore, the octet rule is satisfied.

After completing the octet, one last thing we need to do is, calculate the formal charge and check the stability of the above structure.

Also read: How to draw Lewis structure of HCl (4 steps)

#4 Calculate formal charge and check stability

The following formula is used to calculate the formal charges on atoms:

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

Collect the data from the above structure and then, write it down below as follows:

• For xenon atom

Valence electrons = 8
Nonbonding electrons = 4
Bonding electrons = 8

Formal charge = 8 – 4 – ½ (8) = 0

• For each fluorine atom

Valence electrons = 7
Nonbonding electrons = 6
Bonding electrons = 2

Formal charge = 7 – 6 – ½ (2) = 0

Mention the formal charges of atoms on the structure. So the Lewis structure of XeF₄ looks something like this:

In the above structure, you can see that the formal charges of both (xenon and fluorine) are zero. Therefore, this is the stable Lewis structure of XeF₄.

And each horizontal line drawn in the above structure represents a pair of bonding valence electrons.

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Related

• Lewis structure of F2
• Lewis structure of PO43-
• Lewis structure of HCl
• Lewis structure of ClO2–
• Lewis structure of H2O2

External links

• Drawing the Lewis Structure for XeF4 – The University of Maryland
• XeF4 Lewis Structure: Drawings, Hybridization, Shape, Charges, Pair, and Detailed Facts – Lambda Geeks
• XeF4 (Xenon tetrafluoride) Lewis Structure – Chemistry School
• XeF4 Lewis structure, Molecular geometry, Bond angle, Shape – Topblogtenz
• How do you draw the Lewis structure for “XeF”_4? – Socratic
• XeF4 Lewis Structure, Molecular Geometry, Hybridization, and MO Diagram – Techiescientist
• XeF4 Lewis Structure: 4 Simple Steps – What’s Insight
• Xef4(Xenon Tetrafluoride) Molecular Geometry, Lewis Structure and Polarity – Geometry of Molecules
• Draw the Lewis structure for XeF4 – Chegg
• How can the Lewis structure for XeF4 be determined? – Quora
• Drawing the Lewis Structure for XeF4 – The Geoexchange
• How to draw XeF4 Lewis Structure? – Science Education and Tutorials
• Lewis Dot of Xenon Tetrafluoride XeF4 – Kent’s Chemistry
• Is Xef4 Polar or NonPolar? – ChemWhite
• How to Draw the Lewis Structure for XeF4 – Pinterest
• Lewis structure of XeF4 – AceOrganicChem
• XeF4 Lewis Structure in 5 Steps (With Images) – Pediabay
• Draw the Lewis structure for XeF4 – Homework.Study.com
• Xenon Tetrafluoride, XeF4 Molecular Geometry & Polarity – Tutor-Homework.com
• The Lewis structure for XeF4 is Xe with four fluorine atoms bonded to it – Numerade
• draw the lewis structure for xenon tetrafluoride (xef4). how many valence electrons surround the central xe atom in this structure? – Brainly
• XeF4 Geometry and Hybridization – Chemistry Steps