XeOF4 Lewis Structure: Drawings, Hybridization, Shape, Charges, Pair And Detailed Facts

In this post, we will study about the chemistry inside the Xenon Oxytetrafluoride molecule using xeof4 lewis structure, geometry, and hybridization.

The inorganic compound Xenon Oxytetrafluoride is colourless. It is particularly unstable and reactive, just like the other Xenon oxides. It is made by reacting XeF6 with either NaNO3 or silica.

  1. Lewis Structure of XeOF4
  2. Steps to Draw Lewis Structure of XeOF4
  3. Valence Electrons
  4. XeOF4 Hybridization
  5. XeOF4 Lewis structure resonance
  6. Octet Rule

It combines vigorously with water to generate very corrosive and toxic chemicals that are prone to exploding, thus it should be kept away from water at all times.

1. Lewis Structure of XeOF4:

In a graphical form, a Lewis structure depicts the placement of valence electrons around the atoms of a molecule. The atoms are represented by their atomic symbols, whereas the electrons are portrayed as dots. G. N. Lewis was the first to say it in 1916.

These structures aid in the development of a better knowledge of chemical bond formation and the quantity of non-bonding electrons in a molecule.

The octet of all or most atoms is satisfied in the best feasible Lewis structure for a molecule, and the individual formal charge for each atom is equal to or near to zero.

The Lewis structure of XeOF4 is as follows:

In the XeOF4 molecule’s Lewis structure, oxygen with six valence electrons established a double bond with Xenon, completing its octet.

Furthermore, all fluorine atoms with seven valence electrons form a single connection with Xenon at first, but now have eight electrons.

Xenon, on the other hand, has more than eight valence electrons, i.e. 14. In fact, Xenon’s octet can be expanded, meaning it can hold more than eight electrons in its valence shell. This is owing to the fact that empty 5d-subshells are readily available.

2. Steps to Draw Lewis Structure of XeOF4:

Let’s look at how to design the Lewis structure of the XeOF4 molecule step by step:

1st step: The first step in sketching a Lewis structure for a molecule is to compute the total number of valence electrons in that molecule.

Xenon is a noble gas and a group 18 element with 8 valence electrons, oxygen is a group 16 element with 6 electrons in its outermost shell, and fluorine is a group 17 element with 7 valence electrons in the XeOF4 molecule.

The total number of electrons in the XeOF4 molecule is now calculated as follows:

8 valence electrons = Xe

Valence electrons: F = 7 X 4 = 28

6 valence electrons in O

As a result, Total equals 42 valence electrons.

Step 2: Next, select a centre atom. All of a molecule’s other atoms are thought to be related to the centre atom. The centre atom is Xenon, which is the least electronegative atom.

Step 3: A single link is used to connect all of the atoms to the core atom. This is done to figure out how many more electrons are needed to complete the octet of the linked atoms.

Step 4: Each bond is a sign for a shared pair of electrons. As a result, each fluorine atom now possesses a full octet, indicating that it is stable.

The oxygen atom, on the other hand, still needs one more electron to complete its octet. This need can be met by creating a double bond with the Xenon atom once the Lewis structure of the XeOF4 molecule has been formed as follows:

Simple Structure of XeOF4 from wikipedia

Step 5: The extended octet of the xenon atom is defined as the ability of the xenon atom to store more than 8 electrons in its valence shell, as discussed in the preceding section.

Step 6: The computation of formal charge is the final step in the authentication of a derived Lewis structure.

In reality, formal charge is a hypothetical concept in which the net charge on each atom of a molecule should be close to zero.

The formal charge is calculated using the following formula:

Formal Charge (FC) = [Total no. of valence e in Free State] – [Total no. of non-bonding e 1/2 (Total no. of bonding e)]

Step 7: Calculating the formal charge on the XeOF4 molecule

Total number of valence electrons in Free State for a Xenon atom = 8.

The total number of non-bonding electrons is equal to two.

The total number of bonding electrons is equal to 12.

As a result, the formal charge on the Xenon atom is equal to 8 – 2 – 12(12) = 0.

Total number of valence electrons in Free State for an oxygen atom = 6.

The total number of nonbonding electrons is equal to four.

The total number of bonding electrons is equal to four.

Therefore, 6 – 4 – 12(4) = 0 formal charge on nitrogen atom.

Total number of valence electrons in Free State for fluorine atom = 7.

The total number of nonbonding electrons in the system is 6.

The total number of bonding electrons is equal to two.

Therefore, 7 – 6 – 12(2) = 0 formal charge on nitrogen atom.

Step 8: Because the XeOF4 molecule’s net formal charge is zero, the above-derived Lewis structure is the most correct for this molecule.

3. Valence Electrons:

Valence electrons are the electrons in an atom’s outermost orbit that revolve the furthest from its nucleus. Valence electrons are vital to understand because they play a role in the creation of links between atoms by gaining, losing, or exchanging electrons. The valence electrons are also significant in establishing a compound’s form or geometry.

4. XeOF4 Hybridization:

We know that the total of bond pairs and lone pairs determines hybridization. To begin, determine the number of bond pairs and lone pairs in XeOF4 and multiply by that number. We already know that the combination of hybridization and VSEPR (Valence shell electron pair repulsion theory) theory aids in predicting the molecule’s structure and geometry.

Hybridization is the process of combining a certain number of atomic orbitals to produce an equal number of new orbitals with the same shape and energy. It is used to determine the molecule’s shape. Hybrid orbitals are the novel orbitals that result from the process.

The atomic orbitals with distinct properties are combined together during hybridization.

The following are the several sorts of orbitals: sp, sp2, sp3, sp3d, sp3d2, sp3d3.

One of the hybridizations is detailed in detail here for everyone’s benefit.

Hybridization of sp: sp hybridization is the process of combining one s and one p orbital of almost equal energy to produce two identical and degenerate hybrid orbitals.

By forming an angle of 180o, these sp hybrid orbitals are organised linearly. They are made up of 50% s-character and 50% p-character.

The sum of each atom’s valence electrons equals total valence electrons. One Xe atom, one O atom, and four F atoms make up this molecule.

Geometry and Shape Calculation:

The electrons in an atom’s outermost shell are known as valence electrons. Let’s look at an example to see how valence electrons are shown. Consider the following scenario: The atomic number of oxygen atoms is 8. So there are 6 electrons in the outer shell. As a result, oxygen possesses six valence electrons. Xenon contains 8 valence electrons, while fluorine has 7 valence electrons.

The total valence electrons are calculated as follows:

Central atom Xe: 8

4F contribution: 28

O contribution: 6

As a result, the total valence electron is 42, which is equal to the sum of the central atom’s X + O + 4F contributions. When you split it by 8, you get 5 bond pairs and 1 lone pair.

As a result, sp3d2 is the hybridization.

5. XeOF4 Lewis structure resonance:

Resonance does not exist in XeOF4. There are just one Xe-O bond and all Xe-F bonds are single bonds. Only a Xe-O single or triple bond, or a Xe-F double or triple bond, would cause resonance. Yes, the molecule’s VSEPR formula is AX5E.

6. Octet Rule:

This rule, proposed by Walther Kossel and Gilbert N. Lewis, establishes the groundwork for atom bonding.

According to this rule, an atom’s valence shell with eight electrons is the most stable. The rule of eight is another name for this. This also aids our understanding of various elements’ combining capacities and why atoms form chemical bonds.

The electrical arrangement of noble gases with eight valence electrons is taken into consideration by this rule. Helium, on the other hand, is a stable noble gas with two electrons in its valence shell and serves as a model element for the hydrogen atom. As a result, it is an outlier.

Madhusudan DN

Hi....Myself D N Madhusudan, I have completed my Masters in General Chemistry at University of Mysore. Mysore. Apart from this, I like to read, and listen to Music. Let's connect through LinkedIn:https://www.linkedin.com/in/madhusudan-d-n-b83a05156

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