Hybridization Of XeO₂F₂

The valence bond theory predicts the valency of an atom to be equal to the number of unpaired electrons or partially filled orbits, present in its ground state since the bonds are formed by overlapping of orbitals having unpaired electrons of two atoms which may be different or same. 

However, the electronic configuration of the ground state of certain atoms shows the valency which may or may not be equal to their actual valency in their compounds. For example, Be-atom has no unpaired electron (or partially filled orbital) in its ground state. This configuration shows that Be cannot form any covalent bond. However, it is found to be divalent in its compounds such as BeH2,BeCl2, etc. The very simple picture of the overlap of half-filled atomic orbitals cannot be used to account for all molecular structures. For example, consider the formation of methane (CH4) from carbon. Four C-H bonds in case of methane are formed by the overlap of four partially filled orbitals of carbon with the four partially filled orbitals of four hydrogen atoms.      

In the excited state of carbon, s and p orbitals have different energies. Consequently, four bonds of carbon must be of two types. Three of the bonds should be of one type (s-p bonds) while the fourth bond should be of a different type (s-s bond). However, experimental evidence indicates that for all four bonds in the case of CH4, the concept of Hybridization is used. The concept of hybridization is also useful to explain the observed values of bond angles. 

Hybridization is the concept of intermixing of the orbitals of an atom having nearly the same energy to give exactly equivalent orbital with the same energy, identical shapes, and symmetrical orientations in space. 

Hybridization of XeO2F2 

Determining the hybridization of XeO2F2 is easy. We just have to know the number of valence electrons and use the basic hybridization formula i.e. 

Number of electrons = ½ [V+N-C+A].

where, 

V = The number of valence electrons that is present in the central atom (xenon).

N = The number of monovalent (fluorine) atoms bonded to the central atom. 

C = The charge of cation. 

And, A = The charge of anion.

The basic information is given below.

Name of the Molecule: Xenon Dioxide Difluoride

Molecular Formula: XeO2F2

Hybridization Type: sp3d

Bond Angle: 91°, 105°, and 174°

Geometry: Trigonal Bipyramidal or See-Saw

In Xenon Dioxide Difluoride, xenon is the central atom having 8 valence electrons. The fluorine atom is the monovalent surrounding atom and the oxygen atom is the divalent surrounding atom. We will be taking the 8 valence electrons of Xenon adding 2 monovalent fluorine atoms. At the end, the whole sum will have to be divided by 2.

Taking the values then,

Number of electrons = ½ [8+2-0+0] = 5

Therefore, the final number that we get is 5 which further suggests sp3d hybridization. In Xenon Dioxide Difluoride, there has to be 5 sp3d hybrid orbitals. We can find 5 electron pairs around the central atom where it will be containing 4 bond pairs and 1 lone pair.

Tips to Predict The Type Of Hybridization of Central Atom in A Molecule or ION

Step 1: Add the number of valence electrons of all the atoms present in the given molecule/ion.

Step 2: In the case of a cation, subtract the number of electrons equal to the charge on the cation and in the case of an anion, add the number of electrons equal to the charge on the anion. 

Step 3: (i) If the result obtained in step 2 is less than 8, divide it by 2 and find the sum of the quotient and remainder. 

(ii) If the result obtained in step 2 lies between 9 and 56, divide it by 8 and find the first quotient. Divide the remainder (if any) by 2 and find the second quotient. Add all the quotients and the final remainder. 

Solved Examples

Example 1) Which of the following represents the given mode of hybridization

sp2-sp2-sp-sp from left to right. 

1. CH2 = CH - C = N

2. CH2 = C = C = CH2

3. CH2 = CH-CH = CH2

4. CH2 = C + CH - CH3

Solution 1) CH2 = CH - C = N is the correct representation of the mode of hybridization.  

Example 2) The bond angle between two hybrid orbital is 100 degree. Calculate the percentage of s-character of the hybrid orbital.

Solution 2) The percentage of s-character can be calculated as follows.

Decrease in angle = 120° - 109°28' = 10°32' ≈ 21°/2

Also, decrease in s-character = 33.3 - 25 = 8.3 or 25/3

Actual decrease in bond angle = 109°28' - 100° = 9°28' = 19/2

Hence, expected decrease in s-character = \[\frac{25}{3}\] ×  \[\frac{2}{21}\] × \[\frac{19}{2}\] =  7.54%

Thus, s-character when the bond angle between two hybrid orbital = 25 - 7.54 = 17.46%