Question

What is the hybridization state of xenon in  $ XeOF_4 $ ?

A)  $ \text{s}{{\text{p}}^{3}} $ 

B) $ \text{s}{{\text{p}}^{3}}\text{d} $ 

C)  $ \text{s}{{\text{p}}^{3}}{{\text{d}}^{2}} $ 

D) $ \text{s}{{\text{p}}^{3}}{{\text{d}}^{3}} $ 

Answer 1

Hint: We know that Hybridization is given by the sum of bond pairs and lone pairs. First find out the number of bond pairs and lone pairs in  $ XeOF_4 $  and add them.

Complete step by step answer:

We already know that both hybridization and VSEPR (Valence shell electron pair repulsion theory) theory together helps in determining the shape and geometry of the molecule.

Hybridization is the intermixing of a particular number of atomic orbitals to form an equal number of new orbitals which have the same shape and energy. It is used to determine the shape of the molecule. The new orbital that is formed are also known as hybrid orbitals.

During hybridization, the atomic orbitals with different characteristics are mixed with each other.

There are different types of orbitals as follows:

 $sp$

 $ \text{s}{{\text{p}}^{2}} $ 

 $ \text{s}{{\text{p}}^{3}} $ 

 $ \text{s}{{\text{p}}^{3}}\text{d} $ 

 $ \text{s}{{\text{p}}^{3}}{{\text{d}}^{2}} $ 

 $ \text{s}{{\text{p}}^{3}}{{\text{d}}^{3}} $ 

One of the hybridizations is explained below for the understanding of all of you.

sp hybridization: Intermixing of one s and one p orbitals of almost equal energy gives rise to two identical and degenerate hybrid orbitals and it is called sp hybridization.

These sp hybrid orbitals are arranged linearly by making an angle of  $ {{180}^{\text{o}}} $ . They possess 50% s-character and 50% p-character.

Total valence electrons are given by the sum of valence electrons of each atom. In this molecule, there is one Xe atom, one O atom and 4 F atoms.

Calculation of Geometry and Shape:

Valence electrons are the electrons located at the outermost shell of an atom. Let us look at one example to visualize valence electrons. For example: Oxygen atoms have an atomic number 8. So, the outer shell has 6 electrons. So oxygen has 6 valence electrons. Similarly, fluorine has 7 valence electrons and Xenon has 8 valence electrons.

Total valence electrons is given by

Central atom Xe: 8

O contribution: 6

4 F contribution: 28

Therefore, the total valence electron is equal to the sum of the central atom X + O contribution + 4F contribution and is 42. It is divided by 8 which gives 5 as a bond pair and 1 lone pair.

Therefore, the hybridization is  $ \text{s}{{\text{p}}^{3}}{{\text{d}}^{2}} $ .

Shape:

Hence, the correct option of the above question is option (C),  $ \text{s}{{\text{p}}^{3}}{{\text{d}}^{2}} $ .

Note: The oxygen is as far from the lone pair as possible to minimize lone-pair double bond interaction. As per VSEPR theory, the order of repulsion is lp-lp>lp-bp>bp-bp.