## 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 sp^{3}, sp^{2}, and sp hybridization.

As we know, in the case of XeF_{4} 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 XeF_{4}.

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 XeF_{4} development, and it results in sp^{2}d^{2} 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 XeOF_{4}

Do you know about the hybridization of XeOF_{4}? 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 XeOF_{4} formation, s orbital will be needed for Xe along with its three p-orbitals as well as 2d-orbitals. So, sp^{3}d^{2} or d^{2}sp^{3} will be its hybridization state.

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

XeF_{4} 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 90^{o} or 180°. The singular couples stay on the contrary sides of the molecule fundamentally at 180° from each other.

Xeof_{4} Structure Hybridization

The geometry of Xeof

_{4}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 sp

^{3}d^{2}, or d^{2}sp^{3}.

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−sp

^{2}If the addition is 4 → hybridization−sp

^{3}If the addition is 5 → hybridization−sp

^{3}dIf the addition is 6 → hybridization−sp

^{3}d^{2}

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 sp^{3}d^{2}. 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.

“sp^{3}d^{2} 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.

Xef_{5} Hybridization

Do you know certain facts regarding Xef_{5} hybridization?

Central atom: Xe whose configuration is 5s^{2} 5p^{6}

Number of valence electrons: 8

Number of BP = 5

Number of lone LP = 2

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

**Hybridization:** sp^{3}d^{3}

**Electron pair geometry:** Pentagonal bipyramid

**Molecular geometry:**

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Characteristics of Hybridization:

Ions with similar energies or configurations are hybridized.

Hybridization takes place if the orbitals belong to the same atom.

The hybrid orbitals can make stronger or more stable bonds than the pure atom orbitals.

All the hybrid orbitals are equivalent in energy and shape.

Hybridization leads to the production of equivalent orbitals. This gives the orbitals its maximum symmetry.

Hybridization explains the behavior of the molecule in an orbit.

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## FAQs on Hybridization of XeF4

**1. What is the Requirement of Hybridization?**

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. Hybridization explains the formation of bonds in atoms, like Carbon, at the orbital level. With the help of Vedantu, students can also learn about the 'hybridization level' to understand the concept of hybridization in depth.

**2. Can You Tell Me About the Factors that Make the Most Stable Hybridization?**

**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 sp

^{2}which has 33.33%, and sp3 which has 25%.

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

**3. How Do We Calculate Hybridization?**

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 the bonds. Also, measure the number of lone pairs connected to it. Finally, calculate these two numbers together. Below are the steps to calculate hybridization by an easy way:

Observe the atom.

Count all the other atoms connected to that chosen atom.

Then, count the lone pairs.

Add both of these numbers.

Then, if this number is:

4, the hybridization is sp

^{3}.

3, the hybridization is sp

^{2}.

2, the hybridization is sp.

Also, if the number comes as '1', then it is a hydrogen atom.

**4. List the points of difference between sp, sp ^{2}, and sp^{3} hybridization. **

**Difference between sp, sp ^{2}, and sp^{3} hybridization are :**

In sp, s characteristics percent is 50% and p characteristics percent is 50%. In sp

^{2}, s characteristic percent is 33.33% and p characteristic percent is 66.66%. In sp^{3}, s characteristic percent is 25% and p characteristic percent is 75%.

In terms of orbital arrangement, sp is linear, sp

^{2}is trigonal planar and sp^{3}is tetrahedral.

sp configuration gives 2 unhybridized p orbitals. sp

^{2}gives one unhybridized p orbital. sp^{3}does not result in any unhybridized p orbital.

In sp configuration, the angle between orbitals is 180°. In sp

^{2}, the angle between orbitals is 120°. In sp^{3}, the angle between orbitals is 109.5°.

sp is the simplest orbital hybridization state whereas sp

^{2}is more complex than sp. sp^{3}is the most complex between the three.

**5. What conditions are necessary for the Hybridization of XeF _{4}.**

**The following conditions are necessary for the Hybridization of XeF _{4} :**

All the orbitals participating in hybridization should have only a small amount of difference in enthalpies.

The orbital that undergoes hybridization belongs to the valence shell of the atom.

Hybridization takes place between orbitals that are completely filled, half-filled and empty.

It is not necessary for all the orbitals of the valence shell to undergo hybridization in XeF

_{4}.Promotion of electrons is not a necessary condition for the hybridization but the presence of at least 1 electron must be there.