Structure, Geometry, Hybridization, and Polarisation of XeOF4 According to Lewis

The inorganic compound xenon octafluoride has no discernible colour. Similar to other xenon oxides, it is highly reactive and unstable. Preparation involves reacting XeF6 with NaNO3 or silica.

It forms extremely corrosive and toxic chemicals when exposed to water, which can cause explosions, hence it must be kept away from liquids at all times.

I will use the lewis structure, geometry, hybridization, and polarity of Xenon oxytetrafluoride to describe its internal chemistry.

In that case, let’s launch into action.

Electrons in Valence Shells

An atom’s valence electrons are those that orbit the nucleus in the outermost orbits. Valence electrons play a key role in bond formation between atoms by gaining, losing, or exchanging electrons with their counterparts.

A compound’s structure or geometry is also heavily influenced by its valence electrons.

Octet Principle

This rule, given by Walther Kossel and Gilbert N. Lewis, provides the basis for chemical bonding.

This rule states that the optimal number of electrons for an atom’s valence shell is eight. The rule of eight is another name for this. This also sheds light on the atoms that comprise chemical bonds and their combining capacities.

The electrical arrangement of noble gases with eight valence electrons is, in fact, taken into consideration by this rule. However, despite its status as a noble gas, Helium is chemically stable with only two electrons in its valence shell and serves as a prototypical element for hydrogen. As a result, it stands out as an unusual case.

In the Lewis Structure of XeOF4

The Lewis structure of a molecule is a graphical representation of the distribution of valence electrons around the atoms in the molecule. Dots stand in for electrons, whereas atomic symbols stand in for the atoms. In 1916, G. N. Lewis presented this.

The development of chemical bonds and the amount of unpaired electrons in a molecule can both be better understood with the aid of these structures.

In the optimal Lewis structure for a molecule, all or almost all of the atoms have their octets filled, and the formal charges of the individual atoms are either zero or close to zero.

XeOF4 has the following Lewis structure:

By forming a double bond with Xenon, oxygen, which has six valence electrons, completed Xenon’s octet (as seen in the Lewis structure for XeOF4 shown above).

It should be noted that all the fluorine atoms with seven valence electrons initially form a single connection with Xenon, increasing their total number of electrons from seven to eight.

However, Xenon is seen to have a greater number of valence electrons than eight, specifically 14. Xenon can accommodate more than eight electrons in its valence shell because it can increase the size of its octet. The reason for this is the abundance of free 5d-subshells.

How to Represent the XeOF4 Lewis Structure

Let’s have a look at how the Lewis structure of XeOF4 is depicted, step by step:

The first thing to do when sketching a Lewis structure for a molecule is to figure out how many valence electrons it has.

The XeOF4 molecule consists of the noble gas Xenon, which is in group 18 and has 8 valence electrons, oxygen, which is in group 16 and has 6 electrons in its outermost shell, and fluorine, which is in group 17 and has 7 valence electrons.

Getting an estimate for the total number of electrons in a XeOF4 molecule now:

There are 8 electrons in the shell of Xe.

Valence electrons for F = 74 = 28

O has a valence electron configuration of 6.

So, the sum of the valence electrons is 42.

The next step is to select an atomic core. All the other atoms in a molecule are thought to have connections with the central atom. Since Xenon has the lowest electronegative charge, it is used as the focal atom.

Third, a single link is formed between all the atoms and the central atom. The purpose of this exercise is to calculate how many additional electrons are needed to fill out the octet of the associated atoms.

Four, every bond is a metaphor for one pair of electrons that are shared. In this way, each fluorine atom possesses an entire octet and can no longer lose an electron, rendering it stable.

The oxygen atom, however, needs one extra electron to reach full octet stability. Since the Lewis structure of XeOF4 looks like this, a double bond with the Xenon atom can be formed to meet this criterion.


In Step 5, we reviewed the concept of the expanded octet of the xenon atom, which refers to the element’s ability to accommodate more than 8 electrons in the valence shell.

Step 6: Formal charge computation is the last step in verifying a derived Lewis structure.

The net charge on a single atom within a molecule, according to the concept of formal charge, should be very close to zero.

It can be shown that the formula for determining formal charge is as follows:

An object’s formal charge (FC) is defined as follows: FC = [Total number of e– valence objects in Free State]. It can be written as follows: – [Total number of non-bonding e–– 1/2 (Total number of bonding e–)]

The formal charge of the XeOF4 molecule was determined in Step 7.

Total number of valence electrons in the ground state for a xenon atom = 8

Two electrons are not involved in a covalent bond.

There are a total of 12 bonding electrons.

Accordingly, the formal charge of a Xenon atom is = 8 – 2 – 1/2. (12)

= 0

There are six valence electrons in the ground state of the oxygen atom.

There are four electrons that aren’t part of a bond.

The total number of bonding electrons is 4.

To conclude, the formal charge of a nitrogen atom is 6 – 4 – 1/2. (4)

= 0

For the atom of fluorine, the sum of the valence electrons in its free state is seven.

The sum of the unpaired electrons in the atom is six.

Bonding electron count = 2

Thus, the formal charge of a nitrogen atom is calculated as 7 – 6 – 1/2. (2)

= 0

Eighth, as XeOF4 has no net formal charge, the calculated Lewis structure is the most reliable one for this molecule.

Structure of XeOF4 Molecules

The Valence Shell Electron Pair Repulsion Theory is used to find the molecular geometry of a molecule.

This hypothesis proposes that the number of bond pairs and the number of lone pairs of electrons play crucial roles in determining the overall shape of a molecule.

This is because lone pair-lone pair repulsion is greatest because the electrons in these pairs have the most freedom to move away from one another, while repulsion between bond pair-bond pair is minimal since the electrons in these pairs are bonded in a fixed configuration.

The bond angles between atoms in a molecule are also affected by the inter-electronic repulsion between them.

Due to its Lewis structure, we know that xenon is the core atom in the XeOF4 molecule.

This molecule should take on an octahedral structure, but it doesn’t seem to be doing that. All the atoms in this molecule, however, have unpaired electrons. As a result, this molecule does not have the predicted form.

Due to the repulsive forces between lone pairs and bond pairs of distinct atoms, all angles here should be smaller than 90 degrees.

As an additional step, we count the amount of lone pairs and bond pairs surrounding the core atom (Xenon). When the oxygen double bond is counted as one bond pair, there are a total of 5 bond pairs and 1 lone pair of electrons. In addition, it has 6 electron groups (4 fluorine, 2 oxygen).

This chart is used to approximate the molecular geometry of a molecule in accordance with the VSEPR theory, taking into account the aforementioned parameters.

According to the aforementioned table, the XeOF4 molecule, which contains 5 bond pairs and 1 lone pair linked to the central atom, should have a square pyramidal shape. Furthermore, the electron geometry is an octahedron. All of the bond angles are under 90 degrees.

View the article on a compound with a similar form, BrF5, to learn more about its Lewis Structure, geometry, hybridization, and polarity.

Hybridization of XeOF4

Hybridization is the process of combining orbitals of similar energy to create a new orbital whose name is derived from those of its component orbitals.

If an orbital is designated as sp3d, for instance, it is composed of one s-orbital, three p-orbitals, and one d-orbital. Linus Pauling first proposed this idea back in 1931.

A molecule’s hybridization can be predicted using its steric number, as shown in the following table.

Table for Hybridization

In addition, below is the formula to use for determining steric number:

The number of sigma () bonds on the centre atom plus the number of lone pairs on that atom is the steric number.

Now, the steric number of XeOF4 molecules can be determined using the aforementioned formula.

The sigma bond count for a Xe atom is 5.

One lone pair of electrons exists on the Xe atom.

Therefore, the XeOF4 molecule has a steric number of 6.

Taking into account the aforementioned data, the XeOF4 molecule has an sp3d2 hybridization state.

It is also known that Xenon possesses 8 valence electrons and an sp3 hybridization state. As it is, the formula reads:

Electrons in the p subshell of XeOF4 become excited and go to the empty d- subshell.

The fluorine atoms accept four of these electrons, which form covalent bonds. The other two excited electrons join the oxygen atom in sigma and pi bonds.

Polarity of XeOF4

When talking about a molecule’s polarity, we’re talking about the way its charge is split up because of the disparity in how its atoms are distributed.

The more electronegative species tends to draw the shared pair of electrons towards itself, resulting in an unbalanced charge distribution. The polarity of molecules is also affected by their symmetry.

In XeOF4, all of the atoms are linked to the central atom in an asymmetrical fashion, giving the molecule a square pyramidal shape. The centre atom also has an unpaired pair of electrons.

Further, both the oxygen and fluorine atoms that surround the xenon atom are more electronegative than the xenon itself. The polar nature of XeOF4 explains this property.


XeOF4’s Lewis structure.

Molecularly speaking, XeOF4 has a square pyramidal shape.

For the XeOF4 molecule, the hybridization state is sp3d2.

The molecule XeOF4 is polar because of its nature.

Learning Success!!

Read more: To what end does ice float on water?

Misha Khatri
Misha Khatri is an emeritus professor in the University of Notre Dame's Department of Chemistry and Biochemistry. He graduated from Northern Illinois University with a BSc in Chemistry and Mathematics and a PhD in Physical Analytical Chemistry from the University of Utah.


Please enter your comment!
Please enter your name here

Read More