XeF4 Lewis Structure

XeF4 Lewis Structure

XeF4 (xenon tetrafluoride) has one xenon atom and four fluorine atoms. In the lewis structure of XeF4, there are four single bonds around the xenon atom, with four fluorine atoms attached to it. Each fluorine atom has three lone pairs, and the xenon atom has two lone pairs.


Here’s how you can draw the XeF4 lewis structure step by step.

Step #1: draw sketch
Step #2: mark lone pairs
Step #3: mark charges (if there are)

Let’s break down each step in detail.

#1 Draw Sketch

  • First, determine the total number of valence electrons

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

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

Since XeF4 has one xenon atom and four fluorine atoms, so…

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

And the total valence electrons = 8 + 28 = 36

  • Second, find the total electron pairs

We have a total of 36 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 = 36 ÷ 2 = 18

  • Third, determine the central atom

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

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

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

  • And finally, draw the rough sketch
XeF4 Lewis Structure (Step 1)

#2 Mark Lone Pairs

Here, we have a total of 18 electron pairs. And four Xe — F bonds are already marked. So we have to only mark the remaining fourteen 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 fluorine is a period 2 element, so it can not keep more than 8 electrons in its last shell.

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

So for each fluorine, there are three lone pairs, and for xenon, there are two lone pairs.

Mark the lone pairs on the sketch as follows:

XeF4 Lewis Structure (Step 2)

#3 Mark Charges

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

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

And in the above structure, you can see that the central atom (xenon) forms an octet. Hence, the octet rule is satisfied.

Therefore, this structure is the stable lewis structure of XeF4.

Next: PO43- Lewis Structure

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