It is exceptional for being a stable compound of a noble gas comprising of Xenon in its highest oxidation state i.e. +8. However, the stability is only impounded to a temperature below −35.9 °C where it is found as a yellow crystalline solid but, above this temperature range, the compound explodes leading to its decomposition into basic elements viz. Xenon and Oxygen (O2). It has a melting point of −35.9 °C and a boiling point of 0 ºC. In this article, we will study the fundamentals of lewis structure, geometry, hybridization, and polarity for XeO4 molecules. So, let us start it.
Valence Electrons
The electrons present in the outermost shell of an atom are referred to as its valence electrons. These electrons are important as they are responsible for the characteristic properties of the atom. The valence electrons can be transferred or shared with other atoms during the process of chemical bonding.
Octet Rule
The octet rule states that the atoms have a tendency to bond with other atoms to have eight electrons in their valence shell, which is the electronic configuration of noble gases which are considered as the most stable elements in the periodic table. The only exception to this rule is Helium which is a noble gas and has two electrons in its valence shell. This rule was given by American chemist Gilbert Lewis.
Lewis Structure of XeO4
Lewis structure, also sometimes referred to as Lewis dot structure or electron dot structure is the representation of the arrangement of electrons of a molecule in its valence shell. Dots in the diagram symbolize the valence electrons indicating their distribution in the molecule. It is named after its inventor Gilbert N. Lewis. The aim here is to devise the best possible electronic configuration for the molecule, satisfying the octet rule and formal charge. The Lewis structure for XeO4 is:
As per the above diagram, the octet of all the involved elements is satisfied with all four Oxygen atoms forming a double bond with the Xenon atom while Xenon now having more than 8 electrons in the valence shell is allowed to have an expanded octet, therefore, reducing the formal charge and attaining stability.
Steps to Draw the Lewis Structure of XeO4
Step 1: To draw the Lewis structure of XeO4 we will first have to determine the number of valence electrons in the molecule. Xenon is a group 18 element and therefore, has 8 valence electrons and a complete octet, Oxygen on the other hand belongs to group 16 and comprises 6 electrons in its outermost shell and therefore, needs two more electrons to satisfy its octet. Now, counting the total number of electrons for XeO4: Xenon = 8 Valence electron Oxygen = 6 valence electrons for 4 Oxygen atoms, 6 X 4 = 24 Therefore, total number of valence electrons in XeO4 = 32 Step 2: Now, we will draw the skeletal structure of the molecule with all the atoms joined through a single bond. This step helps to recognize the number of electrons further required to complete the octet of all the atoms present in the molecule. As Xenon already has a complete octet the main focus here is to satisfy the octet rule for the Oxygen atom. Step 3: As every single bond represents one shared pair of electrons, it is clear from the above structure that all the Oxygen atoms are in short of one electron each to complete their octet. Therefore, they form a double bond with the Xenon atom by sharing two electrons and acquiring a stable state by completing their octet. Step 4: This sharing results in overfilling of the valence shell of the Xenon atom, which now consists of 16 electrons, but as Xenon has its valence electrons in the fourth energy level it has the access to 4d sublevel and hence, is capable of accommodating more than eight electrons. Step 5: Finally, after completing the octet for Oxygen atoms and overfilling of Xenon atom the Lewis structure of XeO4 looks like this:
Molecular Geometry of XeO4
The Valence Shell Electron Pair Repulsion (VSEPR) Theory states that inside a molecule the electrons of different atoms always tend to arrange themselves in a manner to avoid the inter-electronic repulsion i.e. they have a tendency to remain placed as far as possible inside a molecule. This kind of repulsion force exists between lone pairs of electrons, between the electrons participating in the formation of a covalent bond as well as between lone pairs and bonded electrons. However, being free in space the amount of repulsion exerted by lone pairs is greater than that of electrons involved in the formation of chemical bonds. Further, the extent of repulsion is also regulated by the difference in electronegativity of the central atom and other atoms participating in the formation of the molecule. The extent of these electronic repulsion forces along with the lone pairs of electrons present in the molecule also further determines the shape of a molecule. Also, the lone pair of electrons present upon the central atom determines the distortion of the bond angle between the central atom and other atoms. To understand the molecular geometry of XeO4, we will first have to choose a central atom. According to VSEPR Theory, it is assumed that all the other atoms in a molecule are connected with the central atom. As in the case of XeO4 all the oxygen atoms are bonded with Xenon, it can be chosen as the central atom. It is clear from the Lewis structure that there are no lone pairs of electrons left in the molecule. Also, all the oxygen atoms are attached with the Xenon atom through a double bond, therefore, the bond angle between all the atoms must remain the same resulting in tetrahedral geometry of the molecule with a bond angle of 109.5° or 109°28′.
As there are no lone pairs of electrons left in the molecule to influence it, the electron geometry of XeO4 will be the same as molecular geometry i.e. tetrahedral.
Hybridization of XeO4
The process of hybridization refers to the formation of a new hybrid orbital, by incorporating two or more atomic orbitals viz. s, p, d, f, etc, together, provided the orbital should be similar in their energies. The theory of hybridization was given by Linus Pauling in 1931. The name of the hybrid orbital is derived from the atomic orbital which is meld in their formation. For example, sp2 indicates that one s and two p orbital combined in its formation while dsp2 orbital is made up of one d, one s, and two p orbital. Talking about the hybridization of XeO4: 1. The number of the valence electrons in Xenon is 8 i.e. it has the hybridization of sp3. 2. In the excited state the electrons from 5s and 5p orbital jump to occupy the vacant 5d orbital.
- The four excited electrons in the 5d orbital of Xenon form pi (π) bond with four Oxygen atoms while the other four electrons with one present in 5s and three in 5p orbital, respectively, form sigma (σ) bonds with the same four oxygen atoms.
- Finally, there would be four non-bonding electrons on each Oxygen atom.
- Therefore, there are 16 bonding and 16 nonbonding electrons.
- The steric number of Xenon central atom in the XeO4 molecule is 4, thus, it forms Sp3 hybridization with tetrahedral geometry.
Polarity of XeO4
Polarity refers to the distribution of electric charge between two atoms joined through a chemical bond. It occurs usually owing to the presence of two opposite charges viz. positive and negative, on different atoms of the same molecule. The polarity in a bond arises due to the difference in electronegativity between the atoms which results in the formation of polar bonds. The partial charges on dissimilar atoms are the small electric charges, which signify the occurrence of a polar bond. XeO4 is a non-polar molecule as the four oxygen atoms are arranged around the Xenon atoms in a tetrahedral fashion, symmetrically opposite to each other, therefore, canceling the charges. The dipole moment also cancels out in the molecule and hence, the net dipole moment for the molecule becomes 0.
Conclusion
The Lewis structure of XeO4 is:
The molecular geometry of XeO4 is tetrahedral.
The hybridization of XeO4 is sp3.
The net dipole moment of XeO4 is 0 D. I hope you guys found my article informative. Leave your comments if I helped you in any way. Thanks!
