Chapter 4, Problem 23E

Predict the molecular structure and bond angles for each molecule or ion in Exercises 81 and 87 from Chapter 3

Interpretation Introduction

Interpretation: Interpretation: The molecular structure of given compound is to be predicted from exercise 81 and 87 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 ${\text{CCl}}_4$ (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 ${\text{CCl}}_4$ is carbon $\left(\text{C}\right)$. The atomic number of carbon is 6 and its electronic configuration is,

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

The valence electron of carbon is 4

The atomic number of chlorine $\left(\text{Cl}\right)$ 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 molecule ${\text{CCl}}_4$ is made of four chlorine atoms and one carbon and atom; hence, the total number of valence electrons is,

$\begin{array}{c}\text{4Cl}\text{+}\text{C}\text{=}\text{28}\text{+}\text{4}\\ \text{=}\text{32}\end{array}$

The Lewis structure of ${\text{CCl}}_4$ 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

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

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

Interpretation Introduction

Interpretation: Interpretation: The molecular structure of given compound is to be predicted from exercise 81 and 87 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 ${\text{NCl}}_3$ (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 ${\text{NCl}}_3$ is nitrogen $\left(\text{N}\right)$. 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 chlorine $\left(\text{Cl}\right)$ 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 molecule ${\text{NCl}}_3$ is made of three chlorine atoms and one carbon atom; hence, the total number of valence electrons is,

$\begin{array}{c}3\text{Cl}\text{+}\text{1N}\text{=}\text{21}\text{+}5\\ \text{=}\text{26}\end{array}$

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

Figure 2

The formula of number of electron pairs is

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

The number of electron pairs is

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

The shape of compound is tetrahedral.

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

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

Interpretation Introduction

Interpretation: Interpretation: The molecular structure of given compound is to be predicted from exercise 81 and 87 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 ${\text{SeCl}}_2$ (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 ${\text{SeCl}}_2$ is selenium $\left(\text{Se}\right)$. The atomic number of selenium is 34 and its electronic configuration is,

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

The valence electron of selenium is 6

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 molecule ${\text{SeCl}}_2$ is made of selenium and two chlorine atoms; hence, the total number of valence electrons is,

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

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

Figure 3

The formula of number of electron pairs is

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

The number of electron pairs is

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

The shape of compound is tetrahedral.

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

Hence, the structure becomes bent shape and bond angle is ${104.5}^0$

Interpretation Introduction

Interpretation: Interpretation: The molecular structure of given compound is to be predicted from exercise 81 and 87 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 $\text{ICl}$ (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 $\text{ICl}$ is iodine $\left(\text{I}\right)$. The atomic number of iodine is 53 and its electronic configuration is,

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

The valence electron of iodine is 7

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 molecule $\text{ICl}$ is made of iodine and chlorine atoms; hence, the total number of valence electrons is,

$\begin{array}{c}\text{I+}\text{Cl}\text{=}\text{7+7}\\ \text{=14}\end{array}$

The Lewis structure of $\text{ICl}$ is

Figure 4

The formula of number of electron pairs is

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

The number of electron pairs is

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

The shape of compound is linear and bond angle is ${180}^0$

Interpretation Introduction

Interpretation: Interpretation: The molecular structure of given compound is to be predicted from exercise 81 and 87 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 ${\text{NO}}_2{}^{-}$ (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 ${\text{NO}}_2{}^{-}$ is nitrogen. The atomic number of nitrogen $\left(\text{N}\right)$ 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 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 ${\text{NO}}_2{}^{-}$ is made of two oxygen atoms and one nitrogen atom. Also oxygen atom contains one negative charge; hence, the total number of valence electrons is,

$\begin{array}{c}\text{N}\text{+}2\text{O}+1e\text{=}5\text{+}2\times 6+1\\ \text{=}\text{18}\end{array}$

The Lewis structure of ${\text{NO}}_2{}^{-}$ is

Figure 5

The formula of number of electron pairs is

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

The number of electron pairs is

$\begin{array}{c}X=\frac{V+M\pm C}{2}\\ X=\frac{5+1}{2}\\ =3\end{array}$

The shape of compound is linear

In ${\text{NO}}_2{}^{-}$ 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 ${\text{NO}}_2{}^{-}$.

Hence, the structure becomes V-shaped and bond angle is ${115}^0$.

Interpretation Introduction

Interpretation: Interpretation: The molecular structure of given compound is to be predicted from exercise 81 and 87 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 ${\text{NO}}_3{}^{-}$ (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 ${\text{NO}}_3{}^{-}$ is nitrogen $\left(\text{N}\right)$. The atomic number of nitrogen $\left(\text{N}\right)$ 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 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 ${\text{NO}}_2{}^{-}$ is made of three oxygen atoms and one nitrogen atom. Also oxygen atom contains one negative charge; hence, the total number of valence electrons is,

$\begin{array}{c}\text{N}\text{+}2\text{O}+1e\text{=}5\text{+}2\times 6+1\\ \text{=}\text{18}\end{array}$

The Lewis structure of ${\text{NO}}_3{}^{-}$ is

Figure 6

The formula of number of electron pairs is

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

The number of electron pairs is

$\begin{array}{c}X=\frac{V+M\pm C}{2}\\ X=\frac{5+1}{2}\\ =3\end{array}$

The shape of compound is linear

In ${\text{NO}}_3{}^{-}$ 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 ${\text{NO}}_3{}^{-}$.

Hence, the structure is trigonal planner and bond angle is ${120}^0$

Interpretation Introduction

Interpretation: Interpretation: The molecular structure of given compound is to be predicted from exercise 81 and 87 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 ${\text{N}}_2{\text{O}}_4$ (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 ${\text{N}}_2{\text{O}}_4$ is nitrogen $\left(\text{N}\right)$. The atomic number of nitrogen $\left(\text{N}\right)$ 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 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 ${\text{N}}_2{\text{O}}_4$ is made of four oxygen atoms and two nitrogen atom. Also oxygen atom contains one negative charge; hence, the total number of valence electrons is,

$\begin{array}{c}\text{2N}\text{+}4\text{O}+1e\text{=}2\times 5\text{+}4\times 6\\ \text{=}34\end{array}$

The Lewis structure of ${\text{N}}_2{\text{O}}_4$ is

Figure 7

The shape of compound is trigonal planner and bond angle is ${120}^0$

In ${\text{N}}_2{\text{O}}_4$ compound nitrogen contains no 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 ${\text{N}}_2{\text{O}}_4$.

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

Interpretation Introduction

Interpretation: Interpretation: The molecular structure of given compound is to be predicted from exercise 81 and 87 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 ${\text{OCN}}^{-}$ (including bond angles).

Explanation of Solution

The central atom in ${\text{OCN}}^{-}$ is carbon $\left(\text{C}\right)$. The atomic number of carbon is 6 and its electronic configuration is,

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

The valence electron of carbon is 4

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 atomic number of nitrogen $\left(\text{N}\right)$ 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 molecule ${\text{OCN}}^{-}$ is made of one oxygen, carbon and nitrogen atoms. Also nitrogen atom contains one negative charge; hence, the total number of valence electrons is,

$\begin{array}{c}\text{N+}\text{O}+\text{C}+1e\text{=}5\text{+}4+6+1\\ \text{=}16\end{array}$

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

Figure 8

The shape of compound is linear and bond angle is ${180}^0$

Interpretation Introduction

Interpretation: Interpretation: The molecular structure of given compound is to be predicted from exercise 81 and 87 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 ${\text{SCN}}^{-}$ (including bond angles).

Explanation of Solution

The central atom in ${\text{SCN}}^{-}$ is carbon $\left(\text{C}\right)$. The atomic number of carbon is 6 and its electronic configuration is,

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

The valence electron of carbon is 4

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

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

The valence electron of sulfur is 6

The atomic number of nitrogen $\left(\text{N}\right)$ 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 molecule ${\text{SCN}}^{-}$ is made of one sulfur, carbon and nitrogen atoms. Also nitrogen atom contains one negative charge; hence, the total number of valence electrons is,

$\begin{array}{c}\text{S+}\text{C}+\text{N}+1e\text{=}6\text{+}4+5+1\\ \text{=}16\end{array}$

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

Figure 9

The shape of compound is linear and bond angle is ${180}^0$

Interpretation Introduction

Interpretation: The molecular structure and bond angle of given compound is to be predicted from exercise 81 and 87 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 ${\text{N}}_3{}^{-}$ (including bond angles).

Explanation of Solution

The atomic number of nitrogen $\left(\text{N}\right)$ 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 molecule ${\text{N}}_3{}^{-}$ is made of three nitrogen atoms. Also nitrogen atom contains one negative charge; hence, the total number of valence electrons is,

$\begin{array}{c}3\text{N}+1e\text{=}3\times 5+1\\ \text{=}16\end{array}$

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

Figure 10

The shape of compound is linear and bond angle is ${180}^0$

Conclusion

The molecular structure describes the relative position of atoms in a molecule. The molecular structures and bond angle of the given molecules are determined