Chapter 4, Problem 24E

(a-i)

Interpretation Introduction

Interpretation: The molecular structure of given compound is to be predicted from exercise 82 and 84 in chapter 3.

Concept introduction: Molecular structure is a three-dimensional shape of a molecule.

Bond angle is the angle between atoms of molecule.

To determine: The molecular structure of ${\text{POCl}}_3$ (including bond angles).

Explanation of Solution

The steps are to be followed to determine the molecular structure is:

• Identify the central atom.
• Count its valence electrons.
• Add or subtract electrons for charge.
• Draw the Lewis structure.
• Divide the total number of electrons by 2 to find the number of electron pairs.
• Use this number to predict the shape.

The central atom in ${\text{POCl}}_3$ is phosphorous $\left(\text{P}\right)$. The atomic number of phosphorous is 15 and its electronic configuration is,

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

The valence electron of phosphorous is 5

The atomic number of chlorine is 17 and its electronic configuration is,

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

The valence electron of chlorine is 7

The atomic number of oxygen is 8 and its electronic configuration is,

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

The valence electron of oxygen is 6

The molecule ${\text{POCl}}_3$ is made of three chlorine atoms and one phosphorus and oxygen atom; hence, the total number of valence electrons is,

$\begin{array}{c}\text{P}\text{+}\text{O}+3\text{Cl}\text{=}5\text{+}6+3\times 7\\ \text{=}\text{32}\end{array}$

The Lewis structure of ${\text{POCl}}_3$ is

Figure 1

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

$\begin{array}{c}X=\frac{V+M\pm C}{2}\\ X=\frac{5+3}{2}\\ =4\end{array}$

The shape of compound is tetrahedral and bond angle is ${109.5}^0$

(a-ii)

Interpretation Introduction

Interpretation: The molecular structure of given compound is to be predicted from exercise 82 and 84 in chapter 3.

Concept introduction: Molecular structure is a three-dimensional shape of a molecule.

Bond angle is the angle between atoms of molecule.

To determine: The molecular structure of ${\text{SO}}_4{}^{2-}$ (including bond angles).

Explanation of Solution

The steps are to be followed to determine the molecular structure is:

• Identify the central atom.
• Count its valence electrons.
• Add or subtract electrons for charge.
• Draw the Lewis structure.
• Divide the total number of electrons by 2 to find the number of electron pairs.
• Use this number to predict the shape.

The central atom in ${\text{SO}}_4{}^{2-}$ is sulfur $\left(\text{S}\right)$. The atomic number of sulfur is 16 and its electronic configuration is,

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

The valence electron of sulfur is 6

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

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

The valence electron of oxygen is 6

The molecule ${\text{SO}}_4{}^{2-}$ is made of four oxygen atoms and one sulfur atom; hence, the total number of valence electrons is,

$\begin{array}{c}\text{S+}4\text{O}+2e^{-}\text{=}6\text{+}4\times 6+2\\ \text{=}\text{32}\end{array}$

The Lewis structure of ${\text{SO}}_4{}^{2-}$ is

Figure 2

The formula of number of electron pairs is

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

The number of electron pairs is

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

The shape of compound is tetrahedral and bond angle is ${109.5}^0$

(a-iii)

Interpretation Introduction

Interpretation: The molecular structure of given compound is to be predicted from exercise 82 and 84 in chapter 3.

Concept introduction: Molecular structure is a three-dimensional shape of a molecule.

Bond angle is the angle between atoms of molecule.

To determine: The molecular structure of ${\text{XeO}}_4$ (including bond angles).

Explanation of Solution

The steps are to be followed to determine the molecular structure is:

• Identify the central atom.
• Count its valence electrons.
• Add or subtract electrons for charge.
• Draw the Lewis structure.
• Divide the total number of electrons by 2 to find the number of electron pairs.
• Use this number to predict the shape.

The central atom in ${\text{XeO}}_4$ is xenon $\left(\text{Xe}\right)$. The atomic number of xenon is 54 and its electronic configuration is,

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

The valence electron of xenon is 8

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

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

The valence electron of oxygen is 6

The molecule ${\text{XeO}}_4$ is made of four oxygen atoms and xenon atom; hence, the total number of valence electrons is,

$\begin{array}{c}\text{Xe+}4\text{O}+\text{=}8\text{+}4\times 6+2\\ \text{=}\text{32}\end{array}$

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

Figure 3

The formula of number of electron pairs is

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

The number of electron pairs is

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

The shape of compound is tetrahedral and bond angle is ${109.5}^0$

(a-iv)

Interpretation Introduction

Interpretation: The molecular structure of given compound is to be predicted from exercise 82 and 84 in chapter 3.

Concept introduction: Molecular structure is a three-dimensional shape of a molecule.

Bond angle is the angle between atoms of molecule.

To determine: The molecular structure of ${\text{PO}}_4{}^{3-}$ (including bond angles).

Explanation of Solution

The steps are to be followed to determine the molecular structure is:

• Identify the central atom.
• Count its valence electrons.
• Add or subtract electrons for charge.
• Draw the Lewis structure.
• Divide the total number of electrons by 2 to find the number of electron pairs.
• Use this number to predict the shape.

The central atom in ${\text{PO}}_4{}^{3-}$ is phosphorous $\left(\text{P}\right)$. The atomic number of phosphorous is 15 and its electronic configuration is,

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

The valence electron of phosphorous is 5

The atomic number of oxygen is 8 and its electronic configuration is,

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

The valence electron of oxygen is 6

The molecule ${\text{PO}}_4{}^{3-}$ is made of four oxygen atoms and phosphorous atom; hence, the total number of valence electrons is,

$\begin{array}{c}\text{P+}4\text{O}+3e\text{=}5\text{+}4\times 6+3\\ \text{=}\text{32}\end{array}$

The Lewis structure of ${\text{PO}}_4{}^{3-}$ is

Figure 4

The formula of number of electron pairs is

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

The number of electron pairs is

$\begin{array}{c}X=\frac{V+M\pm C}{2}\\ X=\frac{5+3}{2}\\ =4\end{array}$

The shape of compound is tetrahedral and bond angle is ${109.5}^0$

(a-v)

Interpretation Introduction

Interpretation: The molecular structure of given compound is to be predicted from exercise 82 and 84 in chapter 3.

Concept introduction: Molecular structure is a three-dimensional shape of a molecule.

Bond angle is the angle between atoms of molecule.

To determine: The molecular structure of ${\text{ClO}}_4{}^{-}$ (including bond angles).

Explanation of Solution

The steps are to be followed to determine the molecular structure is:

• Identify the central atom.
• Count its valence electrons.
• Add or subtract electrons for charge.
• Draw the Lewis structure.
• Divide the total number of electrons by 2 to find the number of electron pairs.
• Use this number to predict the shape.

The central atom in ${\text{ClO}}_4{}^{-}$ is chlorine $\left(\text{Cl}\right)$. The atomic number of chlorine is 17 and its electronic configuration is,

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

The valence electron of chlorine is 7

The atomic number of oxygen is 8 and its electronic configuration is,

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

The valence electron of oxygen is 6

The molecule ${\text{ClO}}_4{}^{-}$ is made of four oxygen atoms and chlorine atom; hence, the total number of valence electrons is,

$\begin{array}{c}\text{Cl+}4\text{O}+e\text{=}7\text{+}4\times 6+1\\ \text{=}\text{32}\end{array}$

The Lewis structure of ${\text{ClO}}_4{}^{-}$ is

Figure 5

The formula of number of electron pairs is

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

The number of electron pairs is

$\begin{array}{c}X=\frac{V+M\pm C}{2}\\ X=\frac{7+1}{2}\\ =4\end{array}$

The shape of compound is tetrahedral and bond angle is ${109.5}^0$

(b-i)

Interpretation Introduction

Interpretation: The molecular structure of given compound is to be predicted from exercise 82 and 84 in chapter 3.

Concept introduction: Molecular structure is a three-dimensional shape of a molecule.

Bond angle is the angle between atoms of molecule.

To determine: The molecular structure of ${\text{NF}}_3$ (including bond angles).

Explanation of Solution

The steps are to be followed to determine the molecular structure is:

• Identify the central atom.
• Count its valence electrons.
• Add or subtract electrons for charge.
• Draw the Lewis structure.
• Divide the total number of electrons by 2 to find the number of electron pairs.
• Use this number to predict the shape.

The central atom in ${\text{NF}}_3$ is nitrogen (N). The atomic number of nitrogen is 7 and its electronic configuration is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^3$

The valence electron of nitrogen is 5

The atomic number of fluorine is 9 and its electronic configuration is,

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

The valence electron of fluorine is 7

The molecule ${\text{NF}}_3$ is made of three fluorine atoms and one nitrogen atom; hence, the total number of valence electrons is,

$\begin{array}{c}3\text{F}\text{+}\text{1N}\text{=}\text{3}\times \text{7}\text{+}5\\ \text{=}\text{26}\end{array}$

The Lewis structure of ${\text{NF}}_3$ is

Figure 6

The formula of number of electron pairs is

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

The number of electron pairs is

$\begin{array}{c}X=\frac{V+M\pm C}{2}\\ X=\frac{5+3}{2}\\ =4\end{array}$

The shape of compound is trigonal planner

In ${\text{NF}}_3$ compound nitrogen contains one lone pair and each oxygen atom contains three lone pairs The repulsion between the lone pairs and bond pairs causes the rearrangement of shape of ${\text{NF}}_3$.

Hence, the structure becomes trigonal pyramidal and bond angle is $<{109.5}^0$.

(b-ii)

Interpretation Introduction

Interpretation: The molecular structure of given compound is to be predicted from exercise 82 and 84 in chapter 3.

Concept introduction: Molecular structure is a three-dimensional shape of a molecule.

Bond angle is the angle between atoms of molecule.

To determine: The molecular structure of ${\text{SO}}_3{}^{2-}$ (including bond angles).

Explanation of Solution

The steps are to be followed to determine the molecular structure is:

• Identify the central atom.
• Count its valence electrons.
• Add or subtract electrons for charge.
• Draw the Lewis structure.
• Divide the total number of electrons by 2 to find the number of electron pairs.
• Use this number to predict the shape.

The central atom in ${\text{SO}}_3{}^{2-}$ is sulfur $\left(\text{S}\right)$. The atomic number of sulfur is 16 and its electronic configuration is,

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

The valence electron of sulfur is 6

The atomic number of oxygen is 8 and its electronic configuration is,

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

The valence electron of oxygen is 6

The molecule ${\text{SO}}_3{}^{2-}$ is made of three oxygen atoms and one sulfur atom. Also oxygen atom contains two negative charge; hence, the total number of valence electrons is,

$\begin{array}{c}\text{S+}3\text{O}+2e^{-}\text{=}6\text{+}3\times 6+2\\ \text{=}26\end{array}$

The Lewis structure of ${\text{SO}}_3{}^{2-}$ is

Figure 7

The formula of number of electron pairs is

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

The number of electron pairs is

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

The shape of compound is trigonal planner

In ${\text{SO}}_3{}^{2-}$ compound sulfur contains one lone pair and each oxygen atom contains three lone pairs The repulsion between the lone pairs and bond pairs causes the rearrangement of shape of ${\text{SO}}_3{}^{2-}$.

Hence, the structure becomes trigonal pyramidal and bond angle is $<{109.5}^0$.

(c-iii)

Interpretation Introduction

Interpretation: The molecular structure of given compound is to be predicted from exercise 82 and 84 in chapter 3.

Concept introduction: Molecular structure is a three-dimensional shape of a molecule.

Bond angle is the angle between atoms of molecule.

To determine: The molecular structure of ${\text{PO}}_3{}^{3-}$ (including bond angles).

Explanation of Solution

The steps are to be followed to determine the molecular structure is:

• Identify the central atom.
• Count its valence electrons.
• Add or subtract electrons for charge.
• Draw the Lewis structure.
• Divide the total number of electrons by 2 to find the number of electron pairs.
• Use this number to predict the shape.

The central atom in ${\text{PO}}_3{}^{3-}$ is phosphorous $\left(\text{P}\right)$. The atomic number of phosphorous is 15 and its electronic configuration is,

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

The valence electron of phosphorous is 5

The atomic number of oxygen is 8 and its electronic configuration is,

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

The valence electron of oxygen is 6

The molecule ${\text{PO}}_3{}^{3-}$ is made of three oxygen atoms and phosphorous atom. Also oxygen atoms contain three negative charges; hence, the total number of valence electrons is,

$\begin{array}{c}\text{P+}3\text{O}+3e\text{=}5\text{+}3\times 6+3\\ \text{=}26\end{array}$

The Lewis structure of ${\text{PO}}_3{}^{3-}$ is

Figure 8

The shape of compound is trigonal planner

In ${\text{PO}}_3{}^{3-}$ compound phosphorous contains one lone pair whereas each oxygen atom contains three lone pairs The repulsion between the lone pairs and bond pairs causes the rearrangement of shape of ${\text{PO}}_3{}^{3-}$.

Hence, the structure becomes trigonal pyramidal and bond angle is $<{109.5}^0$.

(d-iv)

Interpretation Introduction

Interpretation: The molecular structure of given compound is to be predicted from exercise 82 and 84 in chapter 3.

Concept introduction: Molecular structure is a three-dimensional shape of a molecule.

Bond angle is the angle between atoms of molecule.

To determine: The molecular structure of ${\text{ClO}}_3{}^{-}$ (including bond angles).

Explanation of Solution

The steps are to be followed to determine the molecular structure is:

• Identify the central atom.
• Count its valence electrons.
• Add or subtract electrons for charge.
• Draw the Lewis structure.
• Divide the total number of electrons by 2 to find the number of electron pairs.
• Use this number to predict the shape.

The central atom in ${\text{ClO}}_3{}^{-}$ is chlorine $\left(\text{Cl}\right)$. . The atomic number of chlorine is 17 and its electronic configuration is,

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

The valence electron of chlorine is 7

The atomic number of oxygen is 8 and its electronic configuration is,

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

The valence electron of oxygen is 6

The molecule ${\text{ClO}}_3{}^{-}$ is made of three oxygen atoms and chlorine atom. Also oxygen atoms contain three negative charges; hence, the total number of valence electrons is,

$\begin{array}{c}\text{Cl+}3\text{O}+e\text{=}7\text{+}3\times 6+1\\ \text{=}26\end{array}$

The Lewis structure of ${\text{ClO}}_3{}^{-}$ is

Figure 9

The shape of compound is trigonal planner

In ${\text{ClO}}_3{}^{-}$ compound chlorine contains one lone pair whereas each oxygen atom contains three lone pairs The repulsion between the lone pairs and bond pairs causes the rearrangement of shape of ${\text{ClO}}_3{}^{-}$.

Hence, the structure becomes trigonal pyramidal and bond angle is $<{109.5}^0$.

(c)

(c-i)

Interpretation Introduction

Interpretation: The molecular structure of given compound is to be predicted from exercise 82 and 84 in chapter 3.

Concept introduction: Molecular structure is a three-dimensional shape of a molecule.

Bond angle is the angle between atoms of molecule.

To determine: The molecular structure of ${\text{ClO}}_2{}^{-}$ (including bond angles).

Explanation of Solution

The central atom in ${\text{ClO}}_2{}^{-}$ is chlorine $\left(\text{Cl}\right)$. The atomic number of chlorine is 17 and its electronic configuration is,

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

The valence electron of chlorine is 7

The atomic number of oxygen is 8 and its electronic configuration is,

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

The valence electron of oxygen is 6

The molecule ${\text{ClO}}_2{}^{-}$ is made of two oxygen atoms and chlorine atom. Also oxygen atoms contain one negative charges; hence, the total number of valence electrons is,

$\begin{array}{c}\text{Cl+}2\text{O}+e\text{=}7\text{+}2\times 6+1\\ \text{=}20\end{array}$

The Lewis structure of ${\text{ClO}}_2{}^{-}$ is

Figure 10

The shape of compound is linear.

In ${\text{ClO}}_3{}^{-}$ compound chlorine contains two lone pair whereas each oxygen atom contains three lone pairs The repulsion between the lone pairs and bond pairs causes the rearrangement of shape of ${\text{ClO}}_3{}^{-}$.

Hence, the structure becomes bent shaped and bond angle is $<{109.5}^0$.

(c-ii)

Interpretation Introduction

Interpretation: The molecular structure of given compound is to be predicted from exercise 82 and 84 in chapter 3.

Concept introduction: Molecular structure is a three-dimensional shape of a molecule.

Bond angle is the angle between atoms of molecule.

To determine: The molecular structure of ${\text{SCl}}_2$ (including bond angles).

Explanation of Solution

The central atom in ${\text{SCl}}_2$ is sulfur $\left(\text{S}\right)$. The atomic number of chlorine is 17 and its electronic configuration is,

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

The valence electron of chlorine is 7

The atomic number of sulfur is 16 and its electronic configuration is,

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

The valence electron of sulfur is 6

The molecule ${\text{SCl}}_2$ is made of two chlorine atoms and sulfur atom; hence, the total number of valence electrons is,

$\begin{array}{c}\text{2Cl+}\text{S}\text{=}2\times 7\text{+}6\\ \text{=}20\end{array}$

The Lewis structure of ${\text{SCl}}_2$ is

Figure 11

The shape of compound is linear.

In ${\text{SCl}}_2$ compound sulfur contains two lone pair whereas each chlorine atom contains three lone pairs The repulsion between the lone pairs and bond pairs causes the rearrangement of shape of ${\text{SCl}}_2$.

Hence, the structure becomes bent shaped and bond angle is $<{109.5}^0$.

(d-iii)

Interpretation Introduction

Interpretation: The molecular structure of given compound is to be predicted from exercise 82 and 84 in chapter 3.

Concept introduction: Molecular structure is a three-dimensional shape of a molecule.

Bond angle is the angle between atoms of molecule.

To determine: The molecular structure of ${\text{PCl}}_2{}^{-}$ (including bond angles).

Explanation of Solution

Explanation:

The central atom in ${\text{PCl}}_2{}^{-}$ is sulfur $\left(\text{S}\right)$. The atomic number of chlorine is 17 and its electronic configuration is,

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

The valence electron of chlorine is 7

The atomic number of phosphorous is 15 and its electronic configuration is,

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

The valence electron of sulfur is 5

The molecule ${\text{PCl}}_2{}^{-}$ is made of two chlorine atoms and phosphorous atom. Also chlorine atom contains single negative charge; hence, the total number of valence electrons is,

$\begin{array}{c}\text{2Cl+}\text{P}+e\text{=}2\times 7\text{+5}+1\\ \text{=}20\end{array}$

The Lewis structure of ${\text{PCl}}_2{}^{-}$ is

Figure 12

The shape of compound is linear.

In ${\text{PCl}}_2{}^{-}$ compound phosphorous contains two lone pair whereas each chlorine atom contains three lone pairs The repulsion between the lone pairs and bond pairs causes the rearrangement of shape of ${\text{PCl}}_2{}^{-}$.

Hence, the structure becomes bent shaped and bond angle is $<{109.5}^0$.

(d-i)

Interpretation Introduction

Interpretation: The molecular structure of given compound is to be predicted from exercise 82 and 84 in chapter 3.

Concept introduction: Molecular structure is a three-dimensional shape of a molecule.

Bond angle is the angle between atoms of molecule.

To determine: The molecular structure of $\left({\text{O}}_3\right)$ (including bond angles).

Explanation of Solution

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

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

The valence electron of oxygen is 6

The molecule $\left({\text{O}}_3\right)$ is made of three oxygen atoms.; hence, the total number of valence electrons is,

$\begin{array}{c}\text{3O}\text{=}3\times 6\\ \text{=}\text{18}\end{array}$

The Lewis structure of $\left({\text{O}}_3\right)$ is

Figure 13

Hence, the shape of compound is V-shaped and bond angle is ${120}^0$

(ii)

To determine: The molecular structure of ${\text{SO}}_3$ (including bond angles).

Explanation:

The atomic number of sulfur $\left(\text{S}\right)$ is 16 and its electronic configuration is,

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

The valence electron of sulfur is 6

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

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

The valence electron of oxygen is 6

The molecule ${\text{SO}}_2$ is made of two oxygen atoms and one sulfur atom; hence, the total number of valence electrons is,

$\begin{array}{c}\text{S}\text{+}2\text{O}\text{=}6\text{+}2\times 6\\ \text{=}18\end{array}$

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

Figure 14

Hence, the shape of compound is V-shaped and bond angle is ${120}^0$

(d-ii)

Interpretation Introduction

Interpretation: The molecular structure of given compound is to be predicted from exercise 82 and 84 in chapter 3.

Concept introduction: Molecular structure is a three-dimensional shape of a molecule.

Bond angle is the angle between atoms of molecule.

To determine: The molecular structure of ${\text{SO}}_{\text{2}}$ (including bond angles).

Explanation of Solution

The atomic number of sulfur $\left(\text{S}\right)$ is 16 and its electronic configuration is,

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

The valence electron of sulfur is 6

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

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

The valence electron of oxygen is 6

The molecule ${\text{SO}}_2$ is made of two oxygen atoms and one sulfur atom; hence, the total number of valence electrons is,

$\begin{array}{c}\text{S}\text{+}2\text{O}\text{=}6\text{+}2\times 6\\ \text{=}18\end{array}$

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

Figure 14

Hence, the shape of compound is V-shaped and bond angle is ${120}^0$

(d-iii)

Interpretation Introduction

Interpretation: The molecular structure of given compound is to be predicted from exercise 82 and 84 in chapter 3.

Concept introduction: Molecular structure is a three-dimensional shape of a molecule.

Bond angle is the angle between atoms of molecule.

To determine: The molecular structure of ${\text{SO}}_3$ (including bond angles).

Explanation of Solution

Explanation:

The atomic number of sulfur $\left(\text{S}\right)$ is 16 and its electronic configuration is,

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

The valence electron of sulfur is 6

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

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

The valence electron of oxygen is 6

The molecule ${\text{SO}}_3$ is made of two oxygen atoms and one sulfur atom; hence, the total number of valence electrons is,

$\begin{array}{c}\text{S}\text{+}2\text{O}\text{=}6\text{+}2\times 6\\ \text{=}18\end{array}$

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

Figure 15

Hence, the shape of compound is trigonal planner and bond angle is ${120}^0$.

Conclusion

The molecular structure describes the relative position of atoms in a molecule