XeOF4 Lewis Structure, Geometry, Hybridization, and Polarity

Xenon Oxytetrafluoride is a colorless inorganic compound. Similar to other oxides of Xenon it is also very unstable and highly reactive. It can be prepared by reacting XeF6 with NaNO3 or silica.

It vigorously reacts with water to form extremely corrosive and dangerous compounds that tend to explode and hence, it should always be stored away from water.

In this article, I will explain to you the chemistry inside the Xenon Oxytetrafluoride compound with the help of its lewis structure, geometry, hybridization, and its polarity.

So, let’s get started.

Contents show

Valence Electrons

The electrons of an atom that revolve farthest from its nucleus in its outermost orbit are known as valence electrons. It is important to understand the concept of valence electrons, as these participate in the formation of bonds between different atoms by acquiring, losing, or sharing electrons.

Also, the valence electrons play an important role in determining the shape or geometry of a compound.

Octet Rule

Given by Walther Kossel and Gilbert N. Lewis, this rule lays the foundation of bonding between different atoms.

As per this rule, an atom becomes most stable when it has eight electrons in its valence shell. This is also known as the rule of eight. This also helps us understand the combining capacity of various elements and why do they atoms form chemical bonds.

Actually, this rule takes into account the electronic configuration of noble gases that have eight valence electrons. However, Helium even after being a noble gas is stable with two electrons in its valence shell and also acts as a model element for the hydrogen atom. Therefore, it is an exception.

Lewis Structure of XeOF4

A Lewis structure represents the location of valence electrons around the atoms of a molecule in a pictorial form. The electrons are shown as dots while the atoms are represented by their atomic symbols. This was first given by G. N. Lewis in 1916.

These structures help develop a better understanding of chemical bond formation in a molecule and also the number of non-bonding electrons.

In the best possible Lewis structure for a molecule, the octet of all or most atoms is satisfied and the individual formal charge for each atom is equal to or close to zero.

The Lewis structure for XeOF4:

Looking at the above Lewis structure for XeOF4 molecule, oxygen that had six valence electrons formed a double bond with Xenon, hence completing its octet.

Also, all the fluorine atoms having seven valence electrons initially form a single bond with Xenon and now have eight electrons.

However, Xenon is seen to have more than eight valence electrons i.e. 14. Actually, Xenon has the ability to expand its octet i.e. it can house more than eight electrons in its valence shell. This is due to the availability of empty 5d-subshells.

Steps to Draw Lewis Structure of XeOF4

Let us now see the step by step process for drawing the Lewis structure of the XeOF4 molecule:

Step 1: While drawing the Lewis structure for a molecule the first step is to calculate the total number of valence electrons that are present in that molecule.

In the case of the XeOF4 molecule, Xenon is a noble gas and a group 18th element has 8 valence electrons, Oxygen is a group 16 element and had 6 electrons in its outermost shell while fluorine is group 17 element having 7 valence electrons.

Now calculating the total number of electrons in XeOF4 molecule:

Xe = 8 valence electrons

XeOF4 Molecular Geometry

The molecular geometry of a compound is determined by using the Valence Shell Electron Pair Repulsion Theory.

As per this theory, the numbers of bond pairs, as well as lone pairs of electrons, are majorly involved in devising the shape of a molecule.

This is owing to the repulsion forces that exist between these electrons which are highest among lone pair-lone pairs as they have enough free space to move away, while the repulsion forces are least between bond pair-bond pairs as they are attached in a particular position.

Also, the inter-electronic repulsion influence the bond angles between different atoms in a molecule.

In the case of the XeOF4 molecule, we already know through the Lewis structure that xenon is the central atom.

Normally, the shape of this molecule should be octahedral. However, we also notice that all the atoms of this molecule contain one or more lone pairs of electrons. Therefore, the shape of this molecule is distorted and differs from the normally expected shape.

Here, all the angles should be less than 90° due to the repulsion forces that exist between the lone pairs as well as bond pairs of different atoms.

Also, calculating the number of bond pairs and lone pairs on the central atom i.e. Xenon. It has 5 bond pairs as a double bond of oxygen will be considered a single bond pair, and one lone pair of electrons. Also, it has 6 electron groups viz. four fluorine and two with oxygen.

Taking the above values into consideration the following chart is used to estimate the molecular geometry of a compound as per the VSEPR theory:

As per the above chart, the XeOF4 molecule which has 5 bond pairs and one lone pair attached to the central atom should have square pyramidal geometry. Also, it has octahedral electron geometry. The bond angles are less than 90°.

Check out the article on a similarly shaped compound BrF5 Lewis Structure, geometry, hybridization, and polarity.

XeOF4 Hybridization

The process of hybridization involves mixing the orbitals of similar energy to form a new orbital the name of which is derived from its constituent orbitals.

For example, if an orbital is called sp3d, this means that one s, three p, and one d-orbital are involved in its formation. This concept was introduced by Linus Pauling in 1931.

The hybridization of a molecule is determined by using the steric number for that molecule as per the table given below:

Also, the formula for calculating steric number is as follows:

Steric No. = Number of sigma (σ) bond on central atom + lone pair on the central atom

Now, using the above formula to calculate the steric number for XeOF4 molecules

As the central atom for the molecule is Xenon,

Number of sigma bond on Xe atom = 5

No. of lone pair of electrons on Xe atom = 1

Therefore, Steric number for XeOF4 molecule = 6

Talking the above-given table into consideration the hybridization state for the XeOF4 molecule is sp3d2.

• Also, we know that Xenon in itself has sp3 hybridization state and contains 8 valence electrons. It is written as:

• In the case of XeOF4, the electrons in the p subshell get excited and jump into the vacant d- subshell.

• Four of these electrons bond with the fluorine atoms. The remaining two excited electrons bond with the oxygen atom forming a sigma and a pi bond, respectively.

• Therefore, the final hybridization state for the XeOF4 molecule is sp3d2.

Polarity of XeOF4

The polarity of a molecule refers to the separation of charge owing to the unequal distribution amongst different atoms.

It mostly depends upon the electronegativity of atoms as the more electronegative species tends to pull the shared pair of electrons towards itself, hence, creating an imbalance in charge distribution. The symmetry of the molecules also influences their polarity.

In the XeOF4 molecule, the geometry is square pyramidal and all the atoms are bonded with the central atom in an asymmetrical manner. Also, there is a lone pair of electrons attached to the central atom.

Further, all the atoms attached to the Xenon atom viz. oxygen and fluorine, are more electronegative than it. Therefore, XeOF4 is a polar molecule.

Conclusion

The Lewis structure for XeOF4.

The molecular geometry of the XeOF4 molecule is square pyramidal.

The hybridization state for the XeOF4 molecule is sp3d2.

XeOF4 is a polar molecule.

Happy learning!!

XeOF4 Lewis Structure, Geometry, Hybridization, and Polarity - Techiescientist (2024)