
The hybridization of xenon in $Xe{F_2}$ is:
A) $s{p^3}$
B) $s{p^2}$
C) $s{p^3}d$
D) $s{p^2}d$
Answer
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Hint: Hybridization is a concept in which the atomic orbitals of different atoms contribute to form newly hybridized orbitals. The participating atoms in the molecule share their electrons to form new hybrid orbitals. Valence electrons are the maximum number of electrons in which an atom can contribute to the hybridization of a molecule.
Complete step by step answer:
1) Hybridization of a molecule is basically an idea of the contribution of each atomic orbital in the forming of a new atomic orbital.
2) So, when we talk about the contribution of orbitals, there is always the contribution of the electrons. The maximum electrons that can be contributed by an atom in bond formation are its number of valence electrons.
3) The central atom in the molecular structure $Xe{F_2}$ is $Xe$ which has $8$ valence electrons.
4) The $2$ valence electrons of the total $8$ valence electrons will contribute to the formation of bonds with $2Fe$ atoms forming $2$ bond pairs $Xe - F$.
5) The remaining unpaired electrons are $6$ of the total $8$ valence electrons. Two unpaired electrons will contribute to the $1$ lone pair of an electron, which makes the $6$ unpaired electrons into a $3$ lone pair of electrons.
6) Therefore, the molecular structure $Xe{F_2}$ will have $2$ bond pairs and $3$ lone pairs.
${\text{Co - ordination no}}{\text{. of }}Xe{F_2} = {\text{Total no}}{\text{. of bond pairs + Total no}}{\text{. of lone pairs = 2 + 3 = 5 }}$
For the coordination number, $5$ the hybridization is $s{p^3}d$ which shows the option C as a correct choice.
Note:
The coordination number $5$ stands for the contribution of $1 - s$ orbital, $3 - p$ orbital and $1 - d$ orbital which gives the hybridization of the molecule $s{p^3}d$. While calculating the hybridization of a molecule the valence electrons of the central atom are considered and their sharing of electrons will be as per bonding with other atoms or groups present in the molecule which are denoted as bond pairs and unpaired electrons of the central atom are denoted as lone pairs. Two unpaired electrons are calculated as one lone pair of electrons which is a very important point to be noted.
Complete step by step answer:
1) Hybridization of a molecule is basically an idea of the contribution of each atomic orbital in the forming of a new atomic orbital.
2) So, when we talk about the contribution of orbitals, there is always the contribution of the electrons. The maximum electrons that can be contributed by an atom in bond formation are its number of valence electrons.
3) The central atom in the molecular structure $Xe{F_2}$ is $Xe$ which has $8$ valence electrons.
4) The $2$ valence electrons of the total $8$ valence electrons will contribute to the formation of bonds with $2Fe$ atoms forming $2$ bond pairs $Xe - F$.
5) The remaining unpaired electrons are $6$ of the total $8$ valence electrons. Two unpaired electrons will contribute to the $1$ lone pair of an electron, which makes the $6$ unpaired electrons into a $3$ lone pair of electrons.
6) Therefore, the molecular structure $Xe{F_2}$ will have $2$ bond pairs and $3$ lone pairs.
${\text{Co - ordination no}}{\text{. of }}Xe{F_2} = {\text{Total no}}{\text{. of bond pairs + Total no}}{\text{. of lone pairs = 2 + 3 = 5 }}$
For the coordination number, $5$ the hybridization is $s{p^3}d$ which shows the option C as a correct choice.
Note:
The coordination number $5$ stands for the contribution of $1 - s$ orbital, $3 - p$ orbital and $1 - d$ orbital which gives the hybridization of the molecule $s{p^3}d$. While calculating the hybridization of a molecule the valence electrons of the central atom are considered and their sharing of electrons will be as per bonding with other atoms or groups present in the molecule which are denoted as bond pairs and unpaired electrons of the central atom are denoted as lone pairs. Two unpaired electrons are calculated as one lone pair of electrons which is a very important point to be noted.
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