XeF2 Lewis structure

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XeF2 Lewis Structure
XeF2 Lewis structure

XeF2 (xenon difluoride) has one xenon atom and two fluorine atoms.

In the XeF2 Lewis structure, there are two single bonds around the xenon atom, with two fluorine atoms attached to it, and each atom has three lone pairs.


Use these steps to correctly draw the XeF2 Lewis structure:

#1 First draw a rough sketch
#2 Mark lone pairs on the atoms
#3 Calculate and mark formal charges on the atoms, if required

Let’s discuss each step in more detail.

#1 First draw a rough sketch

  • First, determine the total number of valence electrons
Periodic table

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 XeF2 has one xenon atom and two fluorine atoms, so…

Valence electrons of one xenon atom = 8 × 1 = 8
Valence electrons of two fluorine atoms = 7 × 2 = 14

And the total valence electrons = 8 + 14 = 22

  • Second, find the total electron pairs

We have a total of 22 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 = 22 ÷ 2 = 11

  • 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
XeF2 Lewis Structure (Step 1)
Rough sketch of XeF2 Lewis structure

#2 Mark lone pairs on the atoms

Here, we have a total of 11 electron pairs. And two Xe — F bonds are already marked. So we have to only mark the remaining nine 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 atom, there are three lone pairs.

Mark the lone pairs on the sketch as follows:

XeF2 Lewis Structure (Step 2)
Lone pairs marked, and got the stable Lewis structure of XeF2

#3 Calculate and mark formal charges on the atoms, if required

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 – 6 – ½ (4) = 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.

In the above structure, you can see that the central atom (xenon) forms an octet. And the outside atoms (fluorines) also form an octet. Hence, the octet rule is satisfied.

Therefore, this structure is the stable Lewis structure of XeF2.

Next: XeF4 Lewis structure

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