Geometry, Hybridization, and Polarity of XeO4 Lewis Structure

The chemical compound XeO4 (Xenon Tetraoxide) is made up of Xenon and Oxygen. It’s made by mixing anhydrous sulphuric acid with barium perxenate. The molar mass of this substance is 195.29 g/mol.

It is unique in that it is a stable noble gas compound containing Xenon in its maximum oxidation state, +8.

However, the combination’s stability is limited to temperatures below 35.9 °C, where it exists as a yellow crystalline solid; above this temperature range, the molecule bursts, decomposing into fundamental elements such as Xenon and Oxygen (O2). It has a melting point of 35.9 degrees Celsius and a boiling point of 0 degrees Celsius.

We will look at the principles of Lewis structure, geometry, hybridization, and polarity for XeO4 molecules in this article.

So, let’s get started.

Electrons of Valence

The valence electrons in an atom’s outermost shell are known as the valence electrons.

These electrons are crucial since they are responsible for the atom’s unique features.

During the chemical bonding process, valence electrons can be exchanged or shared with other atoms.

The Rule of the Octet

The octet rule asserts that atoms have a tendency to bind with other atoms to form a valence shell with eight electrons, which is the electronic configuration of noble gases, which are the most stable elements in the periodic table.

Helium, a noble gas with two electrons in its valence shell, is the only exception to this rule. Gilbert Lewis, an American scientist, came up with this rule.

XeO4 Lewis Structure

Lewis structure, also known as Lewis dot structure or electron dot structure, is a depiction of a molecule’s electron configuration in its valence shell.

The valence electrons are represented by dots in the diagram, which show their distribution in the molecule. It was created by Gilbert N. Lewis and is named after him.

The goal is to come up with the greatest possible electrical configuration for the molecule while keeping the octet rule and formal charge in mind.

XeO4 has the following Lewis structure:

The octet of all the concerned elements is satisfied in the above figure, with all four Oxygen atoms establishing a double bond with the Xenon atom, while Xenon now has more than 8 electrons in the valence shell, allowing it to have an enlarged octet, lowering the formal charge and achieving stability.

Drawing the Lewis Structure of XeO4 in Steps

Step 1: To sketch the Lewis structure of XeO4, we must first figure out how many valence electrons there are in the molecule.

Xenon is a group 18 element with 8 valence electrons and a complete octet, whereas oxygen is a group 16 element with 6 electrons in its outermost shell and hence requires two more electrons to complete its octet.

Now, for XeO4, count the total number of electrons:

Xenon has an electron valence of eight.

6 valence electrons for 4 oxygen atoms = 24 valence electrons

As a result, XeO4 has a total of 32 valence electrons.

Step 2: Now we’ll draw the molecule’s skeletal structure, with all of the atoms connected by a single bond.

This step aids in determining the number of additional electrons needed to complete the octet of all the atoms in the molecule.

The main aim here is to satisfy the octet rule for the Oxygen atom, as Xenon already has a complete octet.

Step 3: Because each bond represents one shared pair of electrons, the aforementioned structure shows that each Oxygen atom is missing one electron to complete their octet.

As a result, they establish a double bond with the Xenon atom by sharing two electrons and completing their octet, resulting in a stable state.

Step 4: This sharing causes the valence shell of the Xenon atom to become overfilled, resulting in a total of 16 electrons. However, because Xenon’s valence electrons are in the fourth energy level, it has access to the 4d sublevel and so can accommodate more than eight electrons.

Step 5: Finally, the Lewis structure of XeO4 looks like this after completing the octet for Oxygen atoms and overfilling the Xenon atom:

XeO4 Molecular Geometry

The Valence Shell Electron Pair Repulsion (VSEPR) Theory posits that inside a molecule, electrons from different atoms always try to organise themselves in such a way that they avoid inter-electronic repulsion, i.e. they try to stay as far apart as possible.

Between lone pairs of electrons, electrons involved in the creation of a covalent bond, and lone pairs and linked electrons, this type of repulsive force exists.

The degree of repulsion exerted by lone pairs is greater than that of electrons involved in the formation of chemical bonds since they are free in space.

The difference in electronegativity of the core atom and other atoms involved in the molecule’s synthesis also influences the extent of repulsion.

The shape of a molecule is also determined by the extent of these electronic repulsion forces, as well as the number of lone pairs of electrons present in the molecule.

The distortion of the bond angle between the central atom and adjacent atoms is also determined by the lone pair of electrons present on the central atom.

We must first identify a central atom to comprehend the molecular geometry of XeO4.

All of the other atoms in a molecule are thought to be related to the centre atom, according to VSEPR Theory. Because all of the oxygen atoms in XeO4 are bound to Xenon, it can be used as the central atom.

There are no lone pairs of electrons left in the molecule, as shown by the Lewis structure.

Furthermore, because all of the oxygen atoms have a double bond with the Xenon atom, the bond angle between all of the atoms must remain constant, resulting in a tetrahedral geometry of the molecule with a bond angle of 109.5° or 109°28′.

The electron geometry of XeO4 will be the same as the molecular geometry, i.e. tetrahedral, because there are no lone pairs of electrons left in the molecule to impact it.

XeO4 hybridization

The term “hybridization” refers to the creation of a new hybrid orbital by combining two or more atomic orbitals, such as s, p, d, f, and so on, as long as their energies are similar.

Linus Pauling proposed the hybridization idea in 1931.

The hybrid orbital gets its name from the atomic orbital that is merged in their development.

The sp2 orbital, for example, is made up of one d, one s, and two p orbitals, whereas the dsp2 orbital is made up of one d, one s, and two p orbitals.

When it comes to XeO4 hybridization, the following is what we have to say:

  1. Xenon possesses eight valence electrons, indicating that it exhibits sp3 hybridization.
  2. Electrons from the 5s and 5p orbitals jump to the unoccupied 5d orbital in the excited state.

The four excited electrons in Xenon’s 5d orbital make pi () bonds with four oxygen atoms, while the other four electrons form sigma () connections with the same four oxygen atoms, one in the 5s orbital and three in the 5p orbital, respectively.

Finally, each Oxygen atom would have four non-bonding electrons.

There are 16 bonding and 16 nonbonding electrons as a result.

Because the core atom of Xenon in the XeO4 molecule has a steric number of 4, it produces Sp3 hybridization with tetrahedral geometry.

XeO4’s Polarity

The distribution of electric charge between two atoms connected by a chemical bond is referred to as polarity.

It usually happens when two opposed charges, positive and negative, coexist on distinct atoms of the same molecule.

The difference in electronegativity between the atoms causes the creation of polar bonds, which gives the bond its polarity.

Partial charges on different atoms are minor electric charges that indicate the formation of a polar bond.

XeO4 is a non-polar molecule because the four oxygen atoms are grouped in a tetrahedral pattern around the Xenon atoms, symmetrically opposing each other, neutralising the charges.

In the molecule, the dipole moment also cancels out, resulting in a net dipole moment of 0 for the molecule.

Conclusion

When the octet rule is followed, why do we have to remove one oxygen and add three double bonds to the xenon in the Lewis structure, XeO4? | Examine.com

  1. XeO4 has a tetrahedral molecular geometry.
  2. The XeO4 hybridization is sp3.
  3. XeO4 has a net dipole moment of 0 D.

I hope you found this post to be useful. Please let me know if I was able to assist you in any way. Thanks!

Read more: Intermolecular Forces of H2S (Strong or Weak)

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.

LEAVE A REPLY

Please enter your comment!
Please enter your name here

Read More

Recent