XeOF4 Lewis Structure in 5 Steps (With Images)

So you have seen the above image by now, right?

Let me explain the above image in short.

XeOF4 lewis structure has a Xenon atom (Xe) at the center which is surrounded by four Fluorine atoms (F) and one Oxygen atom (O). There are single bonds between the Xenon atom (Xe) & each Fluorine atom (F) and a double bond between Xenon atom (Xe) and Oxygen atom (O).

If you haven’t understood anything from the above image of XeOF4 lewis structure, then just stick with me and you will get the detailed step by step explanation on drawing a lewis structure of XeOF4.

So let’s move to the steps of drawing the lewis structure of XeOF4.

Steps of drawing XeOF4 lewis structure

Step 1: Find the total valence electrons in XeOF4 molecule

In order to find the total valence electrons in a XeOF4 molecule, first of all you should know the valence electrons present in xenon atom, oxygen atom as well as fluorine atom.
(Valence electrons are the electrons that are present in the outermost orbit of any atom.)

Here, I’ll tell you how you can easily find the valence electrons of xenon, oxygen as well as fluorine using a periodic table.

Total valence electrons in XeOF4 molecule

→ Valence electrons given by xenon atom:

Xenon is a group 18 element on the periodic table. ^[1] Hence the valence electrons present in xenon is 8.

You can see the 8 valence electrons present in the xenon atom as shown in the above image.

→ Valence electrons given by oxygen atom:

Oxygen is group 16 element on the periodic table. ^[2] Hence the valence electrons present in oxygen is 6.

You can see the 6 valence electrons present in the oxygen atom as shown in the above image.

→ Valence electrons given by fluorine atom:

Fluorine is group 17 element on the periodic table. ^[3] Hence the valence electron present in fluorine is 7.

You can see the 7 valence electrons present in the fluorine atom as shown in the above image.

Hence, 

Total valence electrons in XeOF4 molecule = valence electrons given by 1 xenon atom + valence electrons given by 1 oxygen atom + valence electrons given by 4 fluorine atoms = 8 + 6 + 7(4) = 42.

Step 2: Select the central atom

For selecting the center atom, you have to remember that the atom which is less electronegative remains at the center.

Now here the given molecule is XeOF4 and it contains xenon atom (Xe), oxygen atom (O) and fluorine atoms (F).

You can see the electronegativity values of xenon atom (Xe), oxygen atom (O) and fluorine atom (F) in the above periodic table.

If we compare the electronegativity values of xenon (Xe), oxygen (O) and fluorine (F) then the xenon atom is less electronegative.

So here the xenon atom (Xe) is the center atom and the oxygen atom (O) and fluorine atoms (F) are the outside atoms.

Step 3: Connect each atoms by putting an electron pair between them

Now in the XeOF4 molecule, you have to put the electron pairs between the xenon atom (Xe), oxygen atom (O) and fluorine atoms (F).

This indicates that the xenon (Xe), oxygen (O) and fluorine (F) are chemically bonded with each other in a XeOF4 molecule.

Step 4: Make the outer atoms stable. Place the remaining valence electrons pair on the central atom.

Now in this step, you have to check the stability of the outer atoms.

Here in the sketch of XeOF4 molecule, you can see that the outer atoms are oxygen atom and fluorine atoms.

These outer atoms are forming an octet and hence they are stable.

Also, in step 1 we have calculated the total number of valence electrons present in the XeOF4 molecule.

The XeOF4 molecule has a total 42 valence electrons and out of these, only 40 valence electrons are used in the above sketch.

So the number of electrons which are left = 42 – 40 = 2.

You have to put these 2 electrons on the central xenon atom in the above sketch of XeOF4 molecule.

Now let’s proceed to the next step.

Step 5: Check the stability of lewis structure

Now you have come to the final step in which you have to check the stability of lewis structure of XeOF4.

The stability of lewis structure can be checked by using a concept of formal charge.

In short, now you have to find the formal charge on xenon (Xe) atom, oxygen (O) atom as well as fluorine (F) atoms present in the XeOF4 molecule.

For calculating the formal charge, you have to use the following formula;

Formal charge = Valence electrons – (Bonding electrons)/2 – Nonbonding electrons

You can see the number of bonding electrons and nonbonding electrons for each atom of XeOF4 molecule in the image given below.

For Xenon (Xe) atom:
Valence electrons = 8 (because xenon is in group 18)
Bonding electrons = 10
Nonbonding electrons = 2

For Oxygen (O) atom:
Valence electrons = 6 (because oxygen is in group 16)
Bonding electrons = 2
Nonbonding electrons = 6

For Fluorine (F) atom:
Valence electrons = 7 (because fluorine is in group 17)
Bonding electrons = 2
Nonbonding electrons = 6

From the above calculations of formal charge, you can see that the xenon (Xe) atom has +1 charge and the oxygen (O) atom has -1 charge.

Because of this reason, the above obtained lewis structure of XeOF4 is not stable. 

So we have to minimize these charges by shifting the electron pairs towards the xenon atom.

After shifting the electron pair from oxygen atom to xenon atom, the lewis structure of XeOF4 becomes more stable.

In the above lewis dot structure of XeOF4, you can also represent each bonding electron pair (:) as a single bond (|). By doing so, you will get the following lewis structure of XeOF4.

I hope you have completely understood all the above steps.

For more practice and better understanding, you can try other lewis structures listed below.

Try (or at least See) these lewis structures for better understanding:

About author

Jay Rana

Jay is an educator and has helped more than 100,000 students in their studies by providing simple and easy explanations on different science-related topics. He is a founder of Pediabay and is passionate about helping students through his easily digestible explanations.

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