XeOF4 Lewis structure

XeOF₄ (xenon oxytetrafluoride) has one xenon atom, one oxygen atom, and four fluorine atoms.

In the XeOF₄ Lewis structure, there is one double bond and four single bonds around the xenon atom, with one oxygen atom and four fluorine atoms attached to it. Each fluorine atom has three lone pairs, the oxygen atom has two lone pairs, and the xenon atom has one lone pair.

Alternative method: Lewis structure of XeOF₄

Rough sketch

• First, determine the total number of valence electrons

In the periodic table, xenon lies in group 18, oxygen lies in group 16, and fluorine lies in group 17.

Hence, xenon has eight valence electrons, oxygen has six valence electrons, and fluorine has seven valence electrons.

Since XeOF₄ has one xenon atom, one oxygen atom, and four fluorine atoms, so…

Valence electrons of one xenon atom = 8 × 1 = 8
Valence electrons of one oxygen atom = 6 × 1 = 6
Valence electrons of four fluorine atoms = 7 × 4 = 28

And the total valence electrons = 8 + 6 + 28 = 42

Learn how to find: Oxygen valence electrons and Fluorine valence electrons

• Second, find the total electron pairs

We have a total of 42 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 = 42 ÷ 2 = 21

• Third, determine the central atom

We have to place the least electronegative atom at the center.

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

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

• And finally, draw the rough sketch

Lone pair

Here, we have a total of 21 electron pairs. And five bonds are already marked. So we have to only mark the remaining sixteen 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 both (oxygen and fluorine) are the period 2 elements, so they can not keep more than 8 electrons in their last shell.

Always start to mark the lone pairs from outside atoms. Here, the outside atoms are oxygen and fluorines.

So for each fluorine and oxygen, there are three lone pairs, and for xenon, there is one lone pair.

Mark the lone pairs on the sketch as follows:

Formal charge

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 – 2 – ½ (10) = +1

For oxygen atom, formal charge = 6 – 6 – ½ (2) = -1

For each fluorine atom, formal charge = 7 – 6 – ½ (2) = 0

Here, both xenon and oxygen atoms have charges, so mark them on the sketch as follows:

The above structure is not a stable Lewis structure because both xenon and oxygen atoms have charges. Therefore, reduce the charges (as below) by converting lone pairs to bonds.

Convert a lone pair of the oxygen atom to make a new Xe — O bond with the xenon atom as follows:

Final structure

The final structure of XeOF₄ contains a central xenon atom linked to one oxygen atom and four fluorine atoms. In this arrangement, the xenon atom utilizes an expanded valence shell to accommodate fourteen electrons, which consist of one double bond to the oxygen, four single bonds to the fluorine atoms, and one lone pair. Within this layout, the oxygen atom fulfills its octet by forming a double bond and maintaining two lone pairs, while each fluorine atom also satisfies the octet rule by retaining three lone pairs alongside its single shared bond. This configuration represents the most stable state for the molecule because it results in a formal charge of zero for every atom involved. Accordingly, this specific electronic distribution serves as the definitive and most accurate Lewis representation of xenon oxytetrafluoride.

Next: KrF₂ Lewis structure

External video

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

External links

• https://techiescientist.com/xeof4-lewis-structure/
• https://lambdageeks.com/xeof4-lewis-structure/
• https://oneclass.com/homework-help/chemistry/7035246-xeof4-lewis-structure.en.html