Lewis structure of XeOF4

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The Lewis structure of XeOF₄ contains one double bond and four single bonds, with xenon in the center, and oxygen and four fluorines on either side. There are three lone pairs on each fluorine atom, two lone pairs on the oxygen atom, and one lone pair on the xenon atom.

Steps

By using the following steps, you can easily draw the Lewis structure of XeOF₄.

#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)
#5 Convert lone pair and calculate formal charge again (if formal charges are not closer to zero)

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, oxygen is a group 16 element, and fluorine is a group 17 element. Hence, xenon has eight valence electrons, oxygen has six valence electrons, and fluorine has seven valence electrons.

Now XeOF₄ has one xenon atom, one oxygen atom, and four fluorine atoms.

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

Therefore, the total number of valence electrons = 8 + 6 + 28 = 42

• 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, oxygen, and fluorine as follows:

Electronegativity value of xenon = 2.6
Electronegativity value of oxygen = 3.44
Electronegativity value of fluorine = 3.98

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

So now, put xenon in the center and oxygen and four fluorines on either side. And draw the rough skeleton structure for the Lewis structure of XeOF₄ something like this:

Also read: How to draw Lewis structure of BrCl₃ (4 steps)

#2 Show chemical bond

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

#3 Mark lone pairs

As calculated earlier, we have a total of 42 valence electrons. And in the above structure, we have already used ten valence electrons. Hence, thirty-two valence electrons are remaining.

Two valence electrons represent one lone pair. So thirty-two valence electrons = sixteen lone pairs.

Note that xenon is a period 5 element, so it can keep more than 8 electrons in its last shell. And both (oxygen and fluorine) are the period 2 elements, so they can not keep more than 8 electrons in their 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 oxygen and four fluorines, so oxygen and four fluorines will get three lone pairs. And the central atom (xenon) will get one lone pair.

So the Lewis structure of XeOF₄ 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.

Now calculate the formal charge and check the stability of the above structure.

Also read: How to draw Lewis structure of CSe₂ (5 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 = 2
Bonding electrons = 10

Formal charge = 8 – 2 – ½ (10) = +1

• For oxygen atom

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

Formal charge = 6 – 6 – ½ (2) = -1

• 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 XeOF₄ looks something like this:

In the above structure, you can see that the formal charges of atoms are not closer to zero. Therefore, convert lone pair and calculate formal charge again.

Also read: How to draw Lewis structure of BeBr₂ (4 steps)

#5 Convert lone pair and calculate formal charge again

As mentioned earlier, xenon is a period 5 element, so it can keep more than 8 electrons in its last shell.

So convert one lone pair from the oxygen atom to make a new bond with the xenon atom. And then, the Lewis structure of XeOF₄ looks something like this:

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

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 = 2
Bonding electrons = 12

Formal charge = 8 – 2 – ½ (12) = 0

• For oxygen atom

Valence electrons = 6
Nonbonding electrons = 4
Bonding electrons = 4

Formal charge = 6 – 4 – ½ (4) = 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 XeOF₄ looks something like this:

In the above structure, you can see that the formal charges of all atoms are zero. Therefore, this is the stable Lewis structure of XeOF₄.

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

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Related

• Lewis structure of BrCl₃
• Lewis structure of CSe₂
• Lewis structure of BeBr₂
• Lewis structure of XeO₂F₂
• Lewis structure of AlBr₃

External video

• XeOF4 Lewis Structure – How to Draw the Lewis Structure for XeOF4 – YouTube • Wayne Breslyn

External links

• XeOF4 Lewis Structure, Geometry, Hybridization, and Polarity – Techiescientist
• XeOF4 Lewis Structure in 5 Steps (With Images) – Pediabay
• Lewis Dot of Xenon Oxytetrafluoride XeOF4 – Kent’s Chemistry
• What’s the XeOF4 Lewis structure? – Quora
• XeOF4 lewis structure, molecular geometry, bond angle, hybridization – Topblogtenz
• What is the formal charge on the central atom in the most reasonable Lewis structure for XeOF4? – Homework.Study.com
• What is the Lewis structure of XeOF4? – Answers
• Write Lewis structure for XeOF4. Calculate the formal charge on each atom, show your work – Chegg
• Draw a Lewis structure for XeOF4. Identify the number of lone pairs, single bonds, and double bonds on the central Xe atom – OneClass