Match List I with List II:

Choose the correct answer from the  options given below:

A-I, B-III, C-IV, D-II

The correct answer is option 2. A-I, B-III, C-IV, D-II.

A. \(XeF_6\)I. \(sp^3d^3\): Distorted octahedral :

We can easily determine the hybridization of xenon hexafluoride by using the common formula which is:

\(\text{Hybridization}\, \ = \frac{1}{2}[V + M – C +A]\)

Here,

\(V \) = number of valence electrons,

\(M\) = monovalent

\(C\) = positive charge

\(A\) = negative charge

Let us put the values according to the formula.

\(\text{Hybridization = } \frac{1}{2}[8+6-0+0]

\(\text{Hybridization = } \frac{1}{2}[14]\)

\(\text{Hybridization = } 7\)

The hybridization number is equal to 7. Now we can say that hybridization is \(sp^3d^3\).

During the formation of \(XeF_6\), xenon has 8 electrons in its valance shell and it forms six bonds with the fluorine atoms. Further, the molecules will have one lone pair and 6 bond pairs. Now if we take the steric number then it will be 7. This can be interpreted as \(sp^3d^3\) hybridization.

After hybridization \(XeF_6\) molecular geometry will be distorted octahedral. What happens here is that the fluorine atoms are placed in the vertices of the octahedron while the lone pairs move in the space to avoid or reduce the repulsion.

B. \(XeO_3\)III. \(sp^3\): Pyramidal

The central atom is Xenon, \(Xe\) and its atomic number is \(54\), so its electronic configuration is

\([_{36}Kr] 4d^{10}5s^25p^6\)

So, \(Xe\) has 8 valence elecetrons.

The central atom is Xenon, \(Xe\) and its atomic number is \(54\), so its electronic configuration is

\([_{36}Kr] 4d^{10}5s^25p^6\)

So, \(Xe\) has 8 valence elecetrons.

\(\text{Hybridization = } \frac{1}{2}[8+0-0+0]

\(\text{Hybridization = } \frac{1}{2}[8]\)

\(\text{Hybridization = } 4\)

The hybridization number of \(XeO_3\) is 4. Thus, the hybridization of \(Xe\) in \(XeO_3\) is \(sp^3\). Due to the presence of one lone pair the structure is pyramidal.

C. \(XeOF_4\)IV. \(sp^3d^2\): Square pyramidal

The central atom is Xenon, \(Xe\) and its atomic number is \(54\), so its electronic configuration is

\([_{36}Kr] 4d^{10}5s^25p^6\)

So, \(Xe\) has 8 valence elecetrons.

\(\text{Hybridization = } \frac{1}{2}[8+4-0+0]\)

\(\text{Hybridization = } \frac{1}{2}[12]\)

\(\text{Hybridization = } 6\)

The hybridization number of \(XeOF_4\) is 6. Thus, the hybridization of \(Xe\) in \(XeOF_4\) is \(sp^3d^2\). Due to the presence of one lone pair the structure is square pyramidal.

D. \(XeF_4\)II. \sp^3d^2\): Square Planar

Both the VSEPR theory and the concept of hybridization are applied to predict the molecular geometries of xenon compounds.

According to the VSEPR theory, the shape of the molecule is predicted by the total number of electron pairs (lone pairs + bond pairs) in the valence shell of the central Xe atom.

To calculate the total number of electron pairs:

\(\frac{\text{valence electrons of the central atom + numver of bonded atoms}}{2}\)

With the above formula,

\(\frac{8 + 4}{2} = 6\)

Hence, there are 6 electron pairs. Since there are 4 fluorine atoms joined to xenon. So, there will be a 4 bond pair of electrons.

Now for calculating the number of lone pairs in the compound: -

\(\text{ Total number of electron pairs –number of bond pairs}\)

\(\text{Lone pairs = }6 – 4 = 2\)

Hence, in the compound, there are 2 lone pairs.

Depending on the number of \(Xe−F\) covalent bonds to be formed, the requisite number of electrons of the of the \(5p\)−orbital valence shell of \(Xe\) get unpaired and promoted to the vacant \(5d\) −orbitals followed by hybridization.

Since, there are 6 electron pairs, the hybridization of the compound will be \(sp^3d^2\).

So, the hybridization is \(sp^3d^2\) and it has 2 lone pairs, the shape of \(XeF_4\) is square planar.