Chapter 20, Problem 74E

(a)

Interpretation Introduction

Interpretation: The Lewis structure, molecular structure (including bond angles) and hybridization of central atom in ${\text{KrF}}_{\text{2}},{\text{KrF}}_{\text{4}},{\text{XeO}}_2{\text{F}}_{\text{2}}$ and ${\text{XeO}}_2{\text{F}}_{\text{4}}$ is 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.
• The value $1{\text{e}}^{-}$ 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 the hybridization of the given compound.

To determine: The Lewis structure, molecular structure (including bond angles) and hybridization of central atom in ${\text{KrF}}_{\text{2}}$.

Explanation of Solution

The Lewis structure, molecular structure, bond angle and hybridization of central atom in ${\text{KrF}}_{\text{2}}$ is as follows.

The first step in determining the Lewis structure is to determine the number of valence electrons. The atomic number of central atom krypton $\left(\text{Kr}\right)$ is $36$ and its electronic configuration is,

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

The valence electron of krypton is $8$.

The atomic number of fluorine $\left(\text{F}\right)$ is $9$ and its electronic configuration is,

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

The valence electron of fluorine is $7$.

The molecule ${\text{KrF}}_{\text{2}}$ is made of two fluorine atoms and single krypton atom; hence, the total number of valence electrons is,

$\begin{array}{c}\text{Kr}\text{+}2\text{F}\text{=}\left(8\text{+}2\times 7\right){\text{e}}^{-}\\ \text{=}22{\text{e}}^{-}\end{array}$

The formula of number of electron pairs is,

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

Where,

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

The number of electron pairs is,

$\begin{array}{c}X=\frac{V+M\pm C}{2}\\ X=\frac{10}{2}\\ =5\end{array}$

This means that the central atom shows ${\text{sp}}^{\text{3}}\text{d}$ hybridization and should have a trigonal bipyramid geometry. But due to presence of repulsive force between lone pairs, the structure becomes linear shaped with the bond angle $180°$.

The Lewis structure of ${\text{KrF}}_{\text{2}}$ is,

Figure 1

(b)

Interpretation Introduction

Interpretation: The Lewis structure, molecular structure (including bond angles) and hybridization of central atom in ${\text{KrF}}_{\text{2}},{\text{KrF}}_{\text{4}},{\text{XeO}}_2{\text{F}}_{\text{2}}$ and ${\text{XeO}}_2{\text{F}}_{\text{4}}$ is 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.
• The value $1{\text{e}}^{-}$ 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 the hybridization of the given compound.

To determine: The Lewis structure, molecular structure (including bond angles) and hybridization of central atom in ${\text{KrF}}_{\text{4}}$.

Explanation of Solution

The Lewis structure, molecular structure, bond angle and hybridization of central atom in ${\text{KrF}}_{\text{4}}$ is as follows.

The first step in determining the Lewis structure is to determine the number of valence electrons. The atomic number of central atom krypton $\left(\text{Kr}\right)$ is $36$ and its electronic configuration is,

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

The valence electron of krypton is $8$.

The atomic number of fluorine $\left(\text{F}\right)$ is $9$ and its electronic configuration is,

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

The valence electron of fluorine is $7$.

The molecule ${\text{KrF}}_{\text{4}}$ is made of four fluorine atoms and single krypton atom; hence, the total number of valence electrons is,

$\begin{array}{c}\text{Kr}\text{+}4\text{F}\text{=}\left(8\text{+}4\times 7\right){\text{e}}^{-}\\ \text{=}36{\text{e}}^{-}\end{array}$

The formula of number of electron pairs is,

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

Where,

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

The number of electron pairs is,

$\begin{array}{c}X=\frac{V+M\pm C}{2}\\ X=\frac{12}{2}\\ =6\end{array}$

This means that the central atom shows ${\text{sp}}^{\text{3}}{\text{d}}^2$ hybridization and should have a octahedral geometry. But due to presence of repulsive force between lone pairs, the structure becomes square planner shaped with the bond angle $90°$.

The Lewis structure of ${\text{KrF}}_{\text{4}}$ is,

Figure 2

(c)

Interpretation Introduction

Interpretation: The Lewis structure, molecular structure (including bond angles) and hybridization of central atom in ${\text{KrF}}_{\text{2}},{\text{KrF}}_{\text{4}},{\text{XeO}}_2{\text{F}}_{\text{2}}$ and ${\text{XeO}}_2{\text{F}}_{\text{4}}$ is 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.
• The value $1{\text{e}}^{-}$ 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 the hybridization of the given compound.

To determine: The Lewis structure, molecular structure (including bond angles) and hybridization of central atom in ${\text{XeO}}_2{\text{F}}_{\text{2}}$.

Explanation of Solution

The Lewis structure, molecular structure, bond angle and hybridization of central atom in ${\text{XeO}}_2{\text{F}}_{\text{2}}$ is as follows.

The first step in determining the Lewis structure is to determine the number of valence electrons. The electronic configuration of central atom xenon $\left(\text{Xe}\right)$ 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 electron of xenon is $8$.

The atomic number of oxygen $\left(\text{O}\right)$ is $8$ and its electronic configuration is,

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

The valence electron of oxygen is $6$.

The atomic number of fluorine $\left(\text{F}\right)$ is $9$ and its electronic configuration is,

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

The valence electron of fluorine is $7$.

The molecule ${\text{XeO}}_2{\text{F}}_{\text{2}}$ is made of two fluorine atoms, two oxygen atoms and single xenon atom; hence, the total number of valence electrons is,

$\begin{array}{c}\text{Xe}\text{+}2\text{O}+2\text{F=}\left(6\text{+}6\times 2+7\times 2\right){\text{e}}^{-}\\ \text{=}32{\text{e}}^{-}\end{array}$

The formula of number of electron pairs is,

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

Where,

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

The number of electron pairs is,

$\begin{array}{c}X=\frac{V+M\pm C}{2}\\ X=\frac{10}{2}\\ =5\end{array}$

This means that the central atom shows ${\text{sp}}^3\text{d}$ hybridization and should have a trigonal bipyramid geometry. But due to presence of repulsive force between lone pairs, the structure becomes see-saw shaped with the bond angle $90°$ and $120°$.

The Lewis structure of ${\text{XeO}}_2{\text{F}}_{\text{2}}$ is,

Figure 3

(d)

Interpretation Introduction

Interpretation: The Lewis structure, molecular structure (including bond angles) and hybridization of central atom in ${\text{KrF}}_{\text{2}},{\text{KrF}}_{\text{4}},{\text{XeO}}_2{\text{F}}_{\text{2}}$ and ${\text{XeO}}_2{\text{F}}_{\text{4}}$ is 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.
• The value $1{\text{e}}^{-}$ 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 the hybridization of the given compound.

To determine: The Lewis structure, molecular structure (including bond angles) and hybridization of central atom in ${\text{XeO}}_2{\text{F}}_{\text{4}}$.

Explanation of Solution

The Lewis structure, molecular structure, bond angle and hybridization of central atom in ${\text{XeO}}_2{\text{F}}_{\text{4}}$ is as follows.

The first step in determining the Lewis structure is to determine the number of valence electrons. The electronic configuration of central atom xenon $\left(\text{Xe}\right)$ 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 electron of xenon is $8$.

The atomic number of oxygen $\left(\text{O}\right)$ is $8$ and its electronic configuration is,

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

The valence electron of oxygen is $6$.

The atomic number of fluorine $\left(\text{F}\right)$ is $9$ and its electronic configuration is,

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

The valence electron of fluorine is $7$.

The molecule ${\text{XeO}}_2{\text{F}}_{\text{4}}$ is made of four fluorine atoms, two oxygen atoms and single xenon atom; hence, the total number of valence electrons is,

$\begin{array}{c}\text{Xe}\text{+}2\text{O}+4\text{F=}\left(6\text{+}6\times 2+7\times 4\right){\text{e}}^{-}\\ \text{=46}{\text{e}}^{-}\end{array}$

The formula of number of electron pairs is,

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

Where,

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

The number of electron pairs is,

$\begin{array}{c}X=\frac{V+M\pm C}{2}\\ X=\frac{12}{2}\\ =6\end{array}$

This means that the central atom shows ${\text{sp}}^3{\text{d}}^2$ hybridization and should have ocahedral geometry with the bond angle $90°$.

The Lewis structure of ${\text{XeO}}_2{\text{F}}_{\text{4}}$ is,

Figure 4