Hybridization of XeF4

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XeF₄ Hybridization

The term ‘Hybridization’ refers to the formation of newly hybridized orbitals by fusing the atomic orbitals. On the other hand, these newly formed hybridized orbitals affect molecular geometry and bonding properties. 

Also, the process of hybridization is the development of the valence bond theory. For exploring this knowledge in advance, we will apply three kinds of hydrocarbon compounds to explain sp3, sp2, and sp hybridization.

As we know, in the case of XeF₄ or xenon tetrafluoride, the hybridization of xeof₄ occurs in the central atom, which is Xenon (Xe). In this case, if we gaze upon the valence shell of Xe, the total amount of electrons is six in the 5p orbital as well as two electrons in the 5s orbital. 

Let’s keep an observation on the 5th orbital; we will find that some of the orbitals such as d orbital and f orbital exist which possess no electrons. 

There are two 5p orbital electrons present. We can say that these orbitals are in an excited state. These orbitals transfer to complete the empty 5d orbitals in the process of making the XeF₄. 

This results in 4 unpaired hybridized electrons which consist of 2 in 5p and 2 in 5d orbitals. So, finally, we get the actual orbital used in XeF₄ development, and it results in sp2d2 hybridization.

But if we consider fluorine, there are four F atoms combined with these four half-filled orbitals. The placement of the fluorine atoms will then be on both sides of the central atom.


Hybridization of XeOF₄ 

Do you know about the hybridization of XeOF₄?

As we discussed earlier, the concept is the same. The s-orbital is used by the central atom as usual and the mixing of the p-orbitals as well as the rest of the d-orbitals together to create the hybrid orbitals.

Thus, in the case of XeOF₄ formation, s orbital will be needed for Xe along with its three p-orbitals as well as 2d-orbitals. So, sp3d2 or d2sp3 will be its hybridization state.


The Brief Details of Xeof₄ Hybridization are Given in the Table Below.

Title of the Chemical

Xenon Tetrafluoride

Chemical Formula

XeF₄

Type of Hybridization 

sp2d2

The angle of the Bond

90o or 180o

Structure

Square Planar


XeF₄ consists of two lone pair electrons. Let’s consider the VSEPR theory, which says that there is a repulsion experienced between the bond pair electrons and lone pair electrons. 

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With the help of the structure, they will obtain a stable state. The sole pairs of Xenon stay in the vertical surface in an octahedral arrangement. This is the reason behind the square planar geometry for the XeF₄ molecule.

The angles of the bond are 90o or 180°. The singular couples stay on the contrary sides of the molecule fundamentally at 180° from each other.


Xeof₄ Structure Hybridization

The geometry of Xeof₄​ is square pyramidal. 

As we have learned before the s-orbital will be used by the central atom, the hybrid orbitals will be mixed via its p-orbitals along with its d-orbitals as much as available. 

Xenon will require its s orbital along with its p-orbitals which are three in number, and 2 of its d-orbitals to form the hybridization state as sp3d2, or d2sp3.


Iodine Pentafluoride Hybridization

By adding the number of σ-bonds designed by the chosen atom (in this case ‘I’) and the lone pair’s number on it, we can simply distinguish the hybridization of it.

If the addition is 2 → hybridization−sp

If the addition is 3 → hybridization−sp2

If the addition is 4 → hybridization−sp3

If the addition is 5 → hybridization−sp3d

If the addition is 6 → hybridization−sp3d2

Reviewing the Lewis structure of IF5

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In this case, 5 sigma bonds and 1 lone pair of electrons is possessed by the IF5. That’s why the sum is 6. So, the hybridization of it is sp3d2.

We can also observe the hybridization process of IF5 in this picture.

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Yes, this is known as iodine pentafluoride. It is a wonderful illustration of a molecule that doesn't satisfy the simple Lewis theory. 

Sometimes you will discover that this is the case for weightier elements frequently, where the electrons from numerous subshells can donate to the bonding.

Well here in iodine, the subshells such as 5p and 5s and also the 4d subshells are pretty close in energy to each other, and all are valence subshells.

As such, iodine utilizes the entire electrons in chemical bonding: these are viz:

2 in the 5s, 

5 in the 5p, and 

up to 10 in the 4d, even if it needs only to use 4 of its 4d electrons for bonding. Therefore, it can have more than 8 electrons to be involved in its bonding process.

“sp3d2 hybridized bonding” method has been used by the molecule. We can explain it simply in brief such as the iodine produces one standard covalent bond to an F-atom, four dative covalent bonds to 4 F-atoms, by keeping one lone pair as of its remaining.


Xef₅ Hybridization

Do you know certain facts regarding Xef₅ hybridization?

Central atom: Xe whose configuration is 5s2 5p6

Number of valence electrons: 8 

Number of BP = 5 

Number of lone LP = 2 

Number of e-pairs = BP + LP = 5 + 2 =7 

Hybridization: sp3d3 

Electron pair geometry: Pentagonal bipyramid 

Molecular geometry:

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FAQ (Frequently Asked Questions)

Q1. What is the Requirement of Hybridization?

Ans: Hybridization is necessary as it allows the formation of the most stable and perfect structures. 

At the end of the hybrid orbitals’ existence, a sufficient amount of electrons is available to finish the necessary bonds irrespective of their number of valence electrons.

Q2. Can You Tell Me about the Factors that Make the Most Stable Hybridization?

Ans: The answer is very simple. 

  • The presence of the ‘s’ character will bring more stability.

  • Yes, the ‘sp’ possesses the highest ‘s’ character about 50%; after that sp2 which has 33.33%, and sp3 which has 25%. 

  • Bonding strength also has a major role; i.e. bond strength = high stability.

Q3. How Do We Calculate Hybridization?

Ans: We can calculate the hybridization of an atom in a molecule by just computing the total number of atoms linked to it. 

Here, just calculate the atoms but not bonds. Also, measure the number of lone pairs connected to it. Finally, calculate these two numbers together.