Chapter 9, Problem 63AE

Draw the Lewis structures, predict the molecular structures, and describe the bonding (in terms of the hybrid orbitals for the central atom) for the following.

a. XeO₃

b. XeO₄

c. XeOF₄

d. XeOF₂

e. XeO₃F₂

Interpretation Introduction

Interpretation: The Lewis structures and the molecular structures of the given compounds; and the bonding of these compounds are to be stated.

Concept introduction: The following steps are to be followed to determine the molecular structure of a given compound,

• The central atom is identified.
• Its valence electrons are determined.
• $1{\text{e}}^{-}$ is added to the total number of valence electrons for each monovalent atom present.
• The total number obtained is divided by $2$, to find the number of electron pairs.
• This further gives us the hybridization of the given compound.

To determine: The Lewis structure and the molecular structure of ${\text{XeO}}_3$

Explanation of Solution

The central atom in ${\text{XeO}}_3$ is xenon $\left(\text{Xe}\right)$. The electronic configuration of xenon is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}{\text{4d}}^{\text{10}}{\text{5s}}^{\text{2}}{\text{5p}}^{\text{6}}$

The valence electrons in xenon are eight.

The electronic configuration of oxygen is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^4$

The valence elctrons in oxygen is six.

The total number of valence electrons are,

$\begin{array}{c}\text{Xe}+3\text{O}=\left(8+6\times 3\right){\text{e}}^{-}\\ =\left(26\right){\text{e}}^{-}\end{array}$

The formula of number of electron pairs is,

$X=\frac{V+M\pm C}{2}$

Where,

• $X$ is number of electron pairs.
• $V$ is valence electrons of central atom.
• $M$ is number of monovalent atoms.
• $C$ is charge on compound.

The number of electron pairs is,

$\begin{array}{c}X=\frac{V+M\pm C}{2}\\ X=\frac{8}{2}\\ =4\end{array}$

This means that the central atom shows ${\text{sp}}^3$ hybridization and should have a trigonal pyramidal geometry.

The Lewis structure of ${\text{XeO}}_3$ is,

Conclusion

The geometry of a given compound can be predicted using the hybridization of the central atom present. The molecular shape can be predicted on the basis of the Lewis structure of the given compound. The lone pairs are depicted in the Lewis structure diagrams.

Interpretation Introduction

Interpretation: The Lewis structures and the molecular structures of the given compounds; and the bonding of these compounds are to be stated.

Concept introduction: The following steps are to be followed to determine the molecular structure of a given compound,

• The central atom is identified.
• Its valence electrons are determined.
• $1{\text{e}}^{-}$ is added to the total number of valence electrons for each monovalent atom present.
• The total number obtained is divided by 2, to find the number of electron pairs.
• This further gives us the hybridization of the given compound.

To determine: The Lewis structure and the molecular structure of ${\text{XeO}}_4$

Answer to Problem 63AE

The molecular structure of ${\text{XeO}}_4$ is tetrahedral.

Explanation of Solution

Explanation

The central atom in ${\text{XeO}}_4$ is xenon $\left(\text{Xe}\right)$. The electronic configuration of xenon is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}{\text{4d}}^{\text{10}}{\text{5s}}^{\text{2}}{\text{5p}}^{\text{6}}$

The valence electrons in xenon are eight.

The electronic configuration of oxygen is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^4$

The valence elctrons in oxygen is six.

The total number of valence electrons are,

$\begin{array}{c}\text{Xe}+4\text{O}=\left(8+6\times 4\right){\text{e}}^{-}\\ =\left(32\right){\text{e}}^{-}\end{array}$

The formula of number of electron pairs is,

$X=\frac{V+M\pm C}{2}$

The number of electron pairs is,

$\begin{array}{c}X=\frac{V+M\pm C}{2}\\ X=\frac{8}{2}\\ =4\end{array}$

This means that the central atom shows ${\text{sp}}^3$ hybridization and should have a tetrahedral geometry.

The Lewis structure of ${\text{XeO}}_4$ is,

Conclusion

The geometry of a given compound can be predicted using the hybridization of the central atom present. The molecular shape can be predicted on the basis of the Lewis structure of the given compound. The lone pairs are depicted in the Lewis structure diagrams.

Interpretation Introduction

Interpretation: The Lewis structures and the molecular structures of the given compounds; and the bonding of these compounds are to be stated.

Concept introduction: The following steps are to be followed to determine the molecular structure of a given compound,

• The central atom is identified.
• Its valence electrons are determined.
• $1{\text{e}}^{-}$ is added to the total number of valence electrons for each monovalent atom present.
• The total number obtained is divided by 2, to find the number of electron pairs.
• This further gives us the hybridization of the given compound.

To determine: The Lewis structure and the molecular structure of ${\text{XeOF}}_4$

Answer to Problem 63AE

Answer

The molecular structure of ${\text{XeOF}}_4$ is square pyramidal.

Explanation of Solution

The central atom in ${\text{XeOF}}_4$ is xenon $\left(\text{Xe}\right)$. The electronic configuration of xenon is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}{\text{4d}}^{\text{10}}{\text{5s}}^{\text{2}}{\text{5p}}^{\text{6}}$

The valence electrons in xenon are eight.

The electronic configuration of oxygen is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^4$

The valence elctrons in oxygen is six.

The electronic configuration of fluorine is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^5$

The valence elctrons in fluorine is seven.

The total number of valence electrons are,

$\begin{array}{c}\text{Xe}+\text{O}\text{+}\text{4F}=\left(8+6+7\times 4\right){\text{e}}^{-}\\ =\left(42\right){\text{e}}^{-}\end{array}$

The formula of number of electron pairs is,

$X=\frac{V+M\pm C}{2}$

The number of electron pairs is,

$\begin{array}{c}X=\frac{V+M\pm C}{2}\\ X=\frac{8+4}{2}\\ =6\end{array}$

This means that the central atom shows ${\text{sp}}^3{\text{d}}^{\text{2}}$ hybridization and should have a square pyramidal geometry.

The Lewis structure of ${\text{XeOF}}_4$ is,

Conclusion

The geometry of a given compound can be predicted using the hybridization of the central atom present. The molecular shape can be predicted on the basis of the Lewis structure of the given compound. The lone pairs are depicted in the Lewis structure diagrams.

Interpretation Introduction

Interpretation: The Lewis structures and the molecular structures of the given compounds; and the bonding of these compounds are to be stated.

Concept introduction: The following steps are to be followed to determine the molecular structure of a given compound,

• The central atom is identified.
• Its valence electrons are determined.
• $1{\text{e}}^{-}$ is added to the total number of valence electrons for each monovalent atom present.
• The total number obtained is divided by 2, to find the number of electron pairs.
• This further gives us the hybridization of the given compound.

To determine: The Lewis structure and the molecular structure of ${\text{XeOF}}_2$

Answer to Problem 63AE

Answer

The molecular structure of ${\text{XeOF}}_2$ is T-shape.

Explanation of Solution

The central atom in ${\text{XeOF}}_2$ is xenon $\left(\text{Xe}\right)$. The electronic configuration of xenon is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}{\text{4d}}^{\text{10}}{\text{5s}}^{\text{2}}{\text{5p}}^{\text{6}}$

The valence electrons in xenon are eight.

The electronic configuration of oxygen is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^4$

The valence elctrons in oxygen is six.

The electronic configuration of fluorine is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^5$

The valence elctrons in fluorine is seven.

The total number of valence electrons are,

$\begin{array}{c}\text{Xe}+\text{O}\text{+}2\text{F}=\left(8+6+7\times 2\right){\text{e}}^{-}\\ =\left(28\right){\text{e}}^{-}\end{array}$

The formula of number of electron pairs is,

$X=\frac{V+M\pm C}{2}$

The number of electron pairs is,

$\begin{array}{c}X=\frac{V+M\pm C}{2}\\ X=\frac{8+2}{2}\\ =5\end{array}$

This means that the central atom shows ${\text{sp}}^3{\text{d}}^{\text{2}}$ hybridization and should have T-shape geometry.

The Lewis structure of ${\text{XeOF}}_2$ is,

Conclusion

The geometry of a given compound can be predicted using the hybridization of the central atom present. The molecular shape can be predicted on the basis of the Lewis structure of the given compound. The lone pairs are depicted in the Lewis structure diagrams.

Interpretation Introduction

Interpretation: The Lewis structures and the molecular structures of the given compounds; and the bonding of these compounds are to be stated.

Concept introduction: The following steps are to be followed to determine the molecular structure of a given compound,

• The central atom is identified.
• Its valence electrons are determined.
• $1{\text{e}}^{-}$ is added to the total number of valence electrons for each monovalent atom present.
• The total number obtained is divided by 2, to find the number of electron pairs.
• This further gives us the hybridization of the given compound.

To determine: The Lewis structure and the molecular structure of ${\text{XeO}}_3{\text{F}}_2$

Answer to Problem 63AE

Answer

The molecular structure of ${\text{XeO}}_3{\text{F}}_2$ is trigonal bipyramidal.

Explanation of Solution

The central atom in ${\text{XeO}}_3{\text{F}}_2$ is xenon $\left(\text{Xe}\right)$. The electronic configuration of xenon is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}{\text{4d}}^{\text{10}}{\text{5s}}^{\text{2}}{\text{5p}}^{\text{6}}$

The valence electrons in xenon are eight.

The electronic configuration of oxygen is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^4$

The valence elctrons in oxygen is six.

The electronic configuration of fluorine is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^5$

The valence elctrons in fluorine is seven.

The total number of valence electrons are,

$\begin{array}{c}\text{Xe}+3\text{O}\text{+}2\text{F}=\left(8+\left(6\times 3\right)+7\times 2\right){\text{e}}^{-}\\ =\left(40\right){\text{e}}^{-}\end{array}$

The formula of number of electron pairs is,

$X=\frac{V+M\pm C}{2}$

The number of electron pairs is,

$\begin{array}{c}X=\frac{V+M\pm C}{2}\\ X=\frac{8+2}{2}\\ =5\end{array}$

This means that the central atom shows ${\text{sp}}^3\text{d}$ hybridization and should have trigonal bipyramidal geometry.

The Lewis structure of ${\text{XeO}}_3{\text{F}}_2$ is,

Conclusion

The geometry of a given compound can be predicted using the hybridization of the central atom present. The molecular shape can be predicted on the basis of the Lewis structure of the given compound. The lone pairs are depicted in the Lewis structure diagrams.