Question

The molecules having the same hybridization, shape and number of lone pairs of electrons are:
A. ${\text{Se}}{{\text{F}}_{\text{4}}}$, ${\text{Xe}}{{\text{O}}_{\text{2}}}{{\text{F}}_{\text{2}}}$
B. ${\text{S}}{{\text{F}}_{\text{4}}}$, ${\text{Xe}}{{\text{F}}_{\text{2}}}$
C. ${\text{XeO}}{{\text{F}}_{\text{4}}}$, ${\text{Te}}{{\text{F}}_{\text{4}}}$
D. ${\text{SeC}}{{\text{l}}_{\text{4}}}$, ${\text{Xe}}{{\text{F}}_{\text{4}}}$

Answer 1

We have to determine the hybridisation and shape of the molecules. The mixing of two atomic orbitals with the same energy levels to form a new degenerate orbital is known as hybridisation. The arrangement of the electrons around the central atom is known as the shape of the molecule. The lone pair of electrons can be determined from the Lewis structures.

Complete step by step solution:
1. Determine hybridisation, shape and number of lone pairs of ${\text{Se}}{{\text{F}}_{\text{4}}}$:
- The valence electrons of selenium are six and fluorine are seven. Thus, valence electrons of ${\text{Se}}{{\text{F}}_{\text{4}}}$
$\Rightarrow \left( {1 \times {\text{Valence electrons of Se}}} \right) + \left( {4 \times {\text{Valence electrons of F}}} \right)$
$\Rightarrow \left( {1 \times 6} \right) + \left( {4 \times 7} \right)$
- Valence electrons of ${\text{Se}}{{\text{F}}_{\text{4}}}$ $= 34$
- The four fluorine atoms form four bonds with the central selenium atom. Thus, eight electrons are involved in bonding. Thus, the remaining electrons are,
$\Rightarrow$ Remaining electrons $= 34 - 8 = {\text{26}}$
- Place the remaining ${\text{26}}$ electrons around the fluorine and selenium atoms such that all the atoms complete their octets.
Thus, the structure of ${\text{Se}}{{\text{F}}_{\text{4}}}$ is as follows:

- From the structure of ${\text{Se}}{{\text{F}}_{\text{4}}}$, we can conclude that there are four bond pairs and one lone pair.
- Thus, the hybridisation of ${\text{Se}}{{\text{F}}_{\text{4}}}$ is $s{p^3}d$ and the shape is trigonal bipyramidal and the number of lone pairs is one.

2. Determine hybridisation, shape and number of lone pairs of ${\text{Xe}}{{\text{O}}_{\text{2}}}{{\text{F}}_{\text{2}}}$:
The valence electrons of xenon are eight, oxygen are six and fluorine are seven. Thus,
Valence electrons of ${\text{Xe}}{{\text{O}}_{\text{2}}}{{\text{F}}_{\text{2}}}$
$\Rightarrow \left( {1 \times {\text{Valence electrons of Xe}}} \right) + \left( {2 \times {\text{Valence electrons of O}}} \right) + \left( {2 \times {\text{Valence electrons of F}}} \right)$
$\Rightarrow \left( {1 \times 8} \right) + \left( {2 \times 6} \right) + \left( {2 \times 7} \right)$
Valence electrons of ${\text{Xe}}{{\text{O}}_{\text{2}}}{{\text{F}}_{\text{2}}}$ $= 34$
- The two fluorine and two oxygen atoms form four bonds with the central xenon atom. Thus, eight electrons are involved in bonding. Thus, the remaining electrons are,
$\Rightarrow$ Remaining electrons $= 34 - 8 = {\text{26}}$
- Place the remaining ${\text{26}}$ electrons around the fluorine, oxygen and xenon atoms such that all the atoms complete their octets.
Thus, the structure of ${\text{Xe}}{{\text{O}}_{\text{2}}}{{\text{F}}_{\text{2}}}$ is as follows:

- From the structure of ${\text{Xe}}{{\text{O}}_{\text{2}}}{{\text{F}}_{\text{2}}}$, we can conclude that there are four bond pairs and one lone pair.
- Thus, the hybridisation of ${\text{Xe}}{{\text{O}}_{\text{2}}}{{\text{F}}_{\text{2}}}$ is $s{p^3}d$ and the shape is trigonal bipyramidal and the number of lone pairs is one.
- Thus, the hybridization, shape and number of lone pairs of electrons of ${\text{Se}}{{\text{F}}_{\text{4}}}$, ${\text{Xe}}{{\text{O}}_{\text{2}}}{{\text{F}}_{\text{2}}}$ are same.
Thus, option (A) is correct.

3. Determine hybridisation, shape and number of lone pairs of ${\text{S}}{{\text{F}}_{\text{4}}}$:
- The valence electrons of sulphur are six and fluorine are seven. Thus,
Valence electrons of ${\text{S}}{{\text{F}}_{\text{4}}}$
$\Rightarrow \left( {1 \times {\text{Valence electrons of S}}} \right) + \left( {4 \times {\text{Valence electrons of F}}} \right)$
$\Rightarrow \left( {1 \times 6} \right) + \left( {4 \times 7} \right)$
Valence electrons of ${\text{S}}{{\text{F}}_{\text{4}}}$ $= 34$
- The four fluorine atoms form four bonds with the central selenium atom. Thus, eight electrons are involved in bonding. Thus, the remaining electrons are,
$\Rightarrow$ Remaining electrons $= 34 - 8 = {\text{26}}$
- Place the remaining ${\text{26}}$ electrons around the fluorine and selenium atoms such that all the atoms complete their octets.
Thus, the structure of ${\text{S}}{{\text{F}}_{\text{4}}}$ is as follows:

- From the structure of ${\text{S}}{{\text{F}}_{\text{4}}}$, we can conclude that there are four bond pairs and one lone pair.
- Thus, the hybridisation of ${\text{S}}{{\text{F}}_{\text{4}}}$ is $s{p^3}d$ and the shape is trigonal bipyramidal and the number of lone pairs is one.

4. Determine hybridisation, shape and number of lone pairs of ${\text{Xe}}{{\text{F}}_{\text{2}}}$:
The valence electrons of xenon are eight and fluorine are seven. Thus,
Valence electrons of ${\text{Xe}}{{\text{F}}_{\text{2}}}$
$\Rightarrow \left( {1 \times {\text{Valence electrons of Xe}}} \right) + \left( {2 \times {\text{Valence electrons of F}}} \right)$
$\Rightarrow \left( {1 \times 8} \right) + \left( {2 \times 7} \right)$
- Valence electrons of ${\text{Xe}}{{\text{F}}_{\text{2}}}$ $= {\text{22}}$
- The two fluorine atoms form two bonds with the central xenon atom. Thus, four electrons are involved in bonding. Thus, the remaining electrons are,
$\Rightarrow$ Remaining electrons $= 22 - 4 = {\text{18}}$
- Place the remaining ${\text{18}}$ electrons around the fluorine and xenon atoms such that all the atoms complete their octets.
Thus, the structure of ${\text{Xe}}{{\text{F}}_{\text{2}}}$ is as follows:

- From the structure of ${\text{Xe}}{{\text{F}}_{\text{2}}}$, we can conclude that there are two bond pairs and three lone pairs.
- Thus, the hybridisation of ${\text{Xe}}{{\text{F}}_{\text{2}}}$ is $s{p^3}d$ and the shape is trigonal bipyramidal and the number of lone pairs is three.
- Thus, the hybridization, shape and number of lone pairs of electrons of ${\text{Se}}{{\text{F}}_{\text{4}}}$, ${\text{Xe}}{{\text{F}}_{\text{2}}}$ are not same.

Thus, option (B) is not correct.

5. Determine hybridisation, shape and number of lone pairs of ${\text{XeO}}{{\text{F}}_{\text{4}}}$:
- The valence electrons of xenon are eight, oxygen are six and fluorine are seven. Thus,
Valence electrons of ${\text{XeO}}{{\text{F}}_{\text{4}}}$
$\Rightarrow \left( {1 \times {\text{Valence electrons of Xe}}} \right) + \left( {1 \times {\text{Valence electrons of O}}} \right) + \left( {4 \times {\text{Valence electrons of F}}} \right)$
$\Rightarrow \left( {1 \times 8} \right) + \left( {1 \times 6} \right) + \left( {4 \times 7} \right)$
Valence electrons of ${\text{XeO}}{{\text{F}}_{\text{4}}}$ $= 42$
- The four fluorine atoms and one oxygen atom form five bonds with the central xenon atom. Thus, ten electrons are involved in bonding. Thus, the remaining electrons are,
$\Rightarrow$ Remaining electrons $= 42 - 10 = {\text{32}}$
Place the remaining ${\text{32}}$ electrons around the fluorine, oxygen and xenon atoms such that all the atoms complete their octets.
Thus, the structure of ${\text{XeO}}{{\text{F}}_{\text{4}}}$ is as follows:

- From the structure of ${\text{XeO}}{{\text{F}}_{\text{4}}}$, we can conclude that there are five bond pairs and one lone pair.
- Thus, the hybridisation of ${\text{XeO}}{{\text{F}}_{\text{4}}}$ is $s{p^3}{d^2}$ and the shape is square pyramidal and the number of lone pairs is one.

6. Determine hybridisation, shape and number of lone pairs of ${\text{Te}}{{\text{F}}_{\text{4}}}$:
- The valence electrons of tellurium are six and fluorine are seven. Thus,
Valence electrons of ${\text{Te}}{{\text{F}}_{\text{4}}}$
$\Rightarrow \left( {1 \times {\text{Valence electrons of Te}}} \right) + \left( {4 \times {\text{Valence electrons of F}}} \right)$
$\Rightarrow \left( {1 \times 6} \right) + \left( {4 \times 7} \right)$
- Valence electrons of ${\text{Te}}{{\text{F}}_{\text{4}}}$ $= 34$
The four fluorine atoms form four bonds with the central tellurium atom. Thus, eight electrons are involved in bonding. Thus, the remaining electrons are,
$\Rightarrow$ Remaining electrons $= 34 - 8 = {\text{26}}$
Place the remaining ${\text{26}}$ electrons around the fluorine and tellurium atoms such that all the atoms complete their octets.
Thus, the structure of ${\text{Te}}{{\text{F}}_{\text{4}}}$ is as follows:

From the structure of ${\text{Te}}{{\text{F}}_{\text{4}}}$, we can conclude that there are four bond pairs and one lone pair.
Thus, the hybridisation of ${\text{Te}}{{\text{F}}_{\text{4}}}$ is $s{p^3}d$ and the shape is trigonal bipyramidal and the number of lone pairs is one.
Thus, the hybridization, shape and number of lone pairs of electrons of ${\text{XeO}}{{\text{F}}_{\text{4}}}$, ${\text{Te}}{{\text{F}}_{\text{4}}}$ are not same.
Thus, option (C) is not correct.

7. Determine hybridisation, shape and number of lone pairs of ${\text{SeC}}{{\text{l}}_{\text{4}}}$:
- The valence electrons of selenium are six and chlorine are seven. Thus,
- Valence electrons of ${\text{SeC}}{{\text{l}}_{\text{4}}}$
$\Rightarrow \left( {1 \times {\text{Valence electrons of Se}}} \right) + \left( {4 \times {\text{Valence electrons of Cl}}} \right)$
$\Rightarrow \left( {1 \times 6} \right) + \left( {4 \times 7} \right)$
Valence electrons of ${\text{SeC}}{{\text{l}}_{\text{4}}}$ $= 34$
The four chlorine atoms form four bonds with the central selenium atom. Thus, eight electrons are involved in bonding. Thus, the remaining electrons are,
$\Rightarrow$ Remaining electrons $= 34 - 8 = {\text{26}}$
- Place the remaining ${\text{26}}$ electrons around the chlorine and selenium atoms such that all the atoms complete their octets.
Thus, the structure of ${\text{SeC}}{{\text{l}}_{\text{4}}}$ is as follows:

- From the structure of ${\text{SeC}}{{\text{l}}_{\text{4}}}$, we can conclude that there are four bond pairs and one lone pair.
Thus, the hybridisation of ${\text{SeC}}{{\text{l}}_{\text{4}}}$ is $s{p^3}d$ and the shape is trigonal bipyramidal and the number of lone pairs is one.

8. Determine hybridisation, shape and number of lone pairs of ${\text{Xe}}{{\text{F}}_{\text{4}}}$:
- The valence electrons of xenon are eight, oxygen are six and fluorine are seven. Thus,
Valence electrons of ${\text{Xe}}{{\text{F}}_{\text{4}}}$
$\Rightarrow \left( {1 \times {\text{Valence electrons of Xe}}} \right) + \left( {4 \times {\text{Valence electrons of F}}} \right)$
$\Rightarrow \left( {1 \times 8} \right) + \left( {4 \times 7} \right)$
Valence electrons of ${\text{Xe}}{{\text{F}}_{\text{4}}}$ $= 36$
- The four fluorine atoms form four bonds with the central xenon atom. Thus, eight electrons are involved in bonding. Thus, the remaining electrons are,
$\Rightarrow$ Remaining electrons $= 36 - 8 = {\text{28}}$
Place the remaining ${\text{28}}$ electrons around the fluorine and xenon atoms such that all the atoms complete their octets.
Thus, the structure of ${\text{Xe}}{{\text{F}}_{\text{4}}}$ is as follows:

- From the structure of ${\text{Xe}}{{\text{F}}_{\text{4}}}$, we can conclude that there are five bond pairs and two lone pairs.
- Thus, the hybridisation of ${\text{Xe}}{{\text{F}}_{\text{4}}}$ is $s{p^3}{d^2}$ and the shape is octahedral and the number of lone pairs is one.
Thus, the hybridization, shape and number of lone pairs of electrons of ${\text{SeC}}{{\text{l}}_{\text{4}}}$, ${\text{Xe}}{{\text{F}}_{\text{4}}}$ are not same.
Thus, option (D) is not correct.

Thus, the hybridization, shape and number of lone pairs of electrons of ${\text{Se}}{{\text{F}}_{\text{4}}}$, ${\text{Xe}}{{\text{O}}_{\text{2}}}{{\text{F}}_{\text{2}}}$ are same.

**Thus, the correct option is (A) ${\text{Se}}{{\text{F}}_{\text{4}}}$, ${\text{Xe}}{{\text{O}}_{\text{2}}}{{\text{F}}_{\text{2}}}$.

Note:**
To determine the shape of the molecule we must know the number of bond pairs and the number of lone pairs around the central atom. These can be determined by drawing the Lewis structures.