Chapter 11, Problem 11.44P

(a)

Interpretation Introduction

Interpretation:

The hybridization change of boron in the following reaction is to be determined.

  ${\text{BF}}_3+\text{NaF}\to {\text{Na}}^{+}{\text{BF}}_4^{-}$

Concept introduction:

The atomic orbital is the wave function that is used to find the probability to find an electron around the nucleus of an atom. It is the space around the nucleus of an atom where the electrons are supposed to be found.

Hybridization is the process of intermixing of atomic orbital of slightly different energies to form hybrid orbitals that have similar energy. These orbital have lower energy and more stability than the atomic orbital.

Hybridization of the central atom can be determined from the number of electron groups around the central atom in the Lewis structure of the molecule. Single bond, double bond, triple bond and lone pair all are considered as single electron group.

Answer to Problem 11.44P

The hybridization of boron changes from $s{p}^2$ to $s{p}^3$.

Explanation of Solution

The Lewis structure of ${\text{BF}}_3$ is as follows:

Boron forms three single bonds with three fluorine atoms so three hybrid orbitals are required and therefore the hybridization of boron in ${\text{BF}}_3$ is $s{p}^2$.

The Lewis structure of ${\text{BF}}_4^{-}$ is as follows:

Boron forms four single bonds with four fluorine atoms so four hybrid orbitals are required and therefore the hybridization of boron in ${\text{BF}}_4^{-}$ is $s{p}^3$.

The hybridization of boron changes from $s{p}^2$ to $s{p}^3$.

Conclusion

Hybridization is determined from the number of electron groups around the central atom in the Lewis structure of the molecule.

(b)

Interpretation Introduction

Interpretation:

The hybridization change for phosphorus in the following reaction is to be determined.

  ${\text{PCl}}_3+{\text{Cl}}_2\to {\text{PCl}}_5$

Concept introduction:

The atomic orbital is the wave function that is used to find the probability to find an electron around the nucleus of an atom. It is the space around the nucleus of an atom where the electrons are supposed to be found.

Hybridization is the process of intermixing of atomic orbital of slightly different energies to form hybrid orbitals that have similar energy. These orbital have lower energy and more stability than the atomic orbital.

Hybridization of the central atom can be determined from the number of electron groups around the central atom in the Lewis structure of the molecule. Single bond, double bond, triple bond and lone pair all are considered as single electron group.

Answer to Problem 11.44P

The hybridization of phosphorus changes from $s{p}^3$ to $s{p}^{3}d$.

Explanation of Solution

The Lewis structure of ${\text{PCl}}_3$ is as follows:

Phosphorus forms three single bonds with three chlorine atoms and has a lone pair on it so four hybrid orbitals are required and therefore the hybridization of phosphorus in ${\text{PCl}}_3$ is $s{p}^3$.

The Lewis structure of ${\text{PCl}}_5$ is as follows:

Phosphorus forms five single bonds with five chlorine atoms so five hybrid orbitals are required and therefore the hybridization of phosphorus in ${\text{PCl}}_5$ is $s{p}^{3}d$.

The hybridization of phosphorus changes from $s{p}^3$ to $s{p}^{3}d$.

Conclusion

Hybridization is determined from the number of electron groups around the central atom in the Lewis structure of the molecule.

(c)

Interpretation Introduction

Interpretation:

The hybridization change for carbon in the following reaction is to be determined.

Concept introduction:

The atomic orbital is the wave function that is used to find the probability to find an electron around the nucleus of an atom. It is the space around the nucleus of an atom where the electrons are supposed to be found.

Hybridization is the process of intermixing of atomic orbital of slightly different energies to form hybrid orbitals that have similar energy. These orbital have lower energy and more stability than the atomic orbital.

Hybridization of the central atom can be determined from the number of electron groups around the central atom in the Lewis structure of the molecule. Single bond, double bond, triple bond and lone pair all are considered as single electron group.

Answer to Problem 11.44P

The hybridization of carbon changes from $sp$ to $s{p}^2$.

Explanation of Solution

The Lewis structure of acetylene is as follows:

Carbon forms one triple bond with other carbon atom and one single bond with hydrogen atom so two hybrid orbitals are required and therefore the hybridization of carbon in acetylene is $sp$.

The Lewis structure of ethene is as follows:

Carbon forms two single bonds with two hydrogen atoms and one double bond with another carbon atom so three hybrid orbitals are required and therefore the hybridization of carbon in ethene is $s{p}^2$.

The hybridization of carbon changes from $sp$ to $s{p}^2$.

Conclusion

Hybridization is determined from the number of electron groups around the central atom in the Lewis structure of the molecule.

(d)

Interpretation Introduction

Interpretation:

The hybridization change for silicon in the following reaction is to be determined.

  ${\text{SiF}}_4+2{\text{F}}^{-}\to {\text{SiF}}_6^{2-}$

Concept introduction:

The atomic orbital is the wave function that is used to find the probability to find an electron around the nucleus of an atom. It is the space around the nucleus of an atom where the electrons are supposed to be found.

Hybridization is the process of intermixing of atomic orbital of slightly different energies to form hybrid orbitals that have similar energy. These orbital have lower energy and more stability than the atomic orbital.

Hybridization of the central atom can be determined from the number of electron groups around the central atom in the Lewis structure of the molecule. Single bond, double bond, triple bond and lone pair all are considered as single electron group.

Answer to Problem 11.44P

The hybridization of silicon changes from $s{p}^3$ to $s{p}^3{d}^2$.

Explanation of Solution

The Lewis structure of ${\text{SiF}}_4$ is as follows:

Silicon forms four single bonds with four fluorine atoms so four hybrid orbitals are required and therefore the hybridization of silicon in ${\text{SiF}}_4$ is $s{p}^3$.

The Lewis structure of ${\text{SiF}}_6^{2-}$ is as follows:

Silicon forms six single bonds with six fluorine atoms so six hybrid orbitals are required and therefore the hybridization of silicon in ${\text{SiF}}_6^{2-}$ is $s{p}^3{d}^2$.

The hybridization of silicon changes from $s{p}^3$ to $s{p}^3{d}^2$.

Conclusion

Hybridization is determined from the number of electron groups around the central atom in the Lewis structure of the molecule.

(e)

Interpretation Introduction

Interpretation:

The hybridization of sulphur in the following reaction is to be determined.

  ${\text{SO}}_2+\frac{1}{2}{\text{O}}_2\to {\text{SO}}_3$

Concept introduction:

The atomic orbital is the wave function that is used to find the probability to find an electron around the nucleus of an atom. It is the space around the nucleus of an atom where the electrons are supposed to be found.

Hybridization is the process of intermixing of atomic orbital of slightly different energies to form hybrid orbitals that have similar energy. These orbital have lower energy and more stability than the atomic orbital.

Hybridization of the central atom can be determined from the number of electron groups around the central atom in the Lewis structure of the molecule. Single bond, double bond, triple bond and lone pair all are considered as single electron group.

Answer to Problem 11.44P

There is no change in the hybridization of sulphur.

Explanation of Solution

The Lewis structure of ${\text{SO}}_2$ is as follows:

Sulphur forms one single, one double bond with two oxygen atoms separately. It has a lone pair of electrons on it. So three hybrid orbitals are required and therefore the hybridization of sulphur in ${\text{SO}}_2$ is $s{p}^2$.

The Lewis structure of ${\text{SO}}_3$ is as follows:

Sulphur forms two single bonds and one double bond with three oxygen atoms separately so three hybrid orbitals are required and therefore the hybridization of sulphur in ${\text{SO}}_3$ is $s{p}^2$.

The hybridization of sulphur remains the same in going from ${\text{SO}}_2$ to ${\text{SO}}_3$.

Conclusion

Hybridization is determined from the number of electron groups around the central atom in the Lewis structure of the molecule.