Chapter 9, Problem 9.81QP

Calculate the bond length for each of the following single bonds, using covalent radii (Table 9.4):

1. a C─H
2. b S─Cl
3. c Br─Cl
4. d Si─O

Interpretation Introduction

Interpretation:

The bond lengths of the given bonds has to be calculated using covalent radii.

Concept introduction:

Bond length: bond order and bond length is inversely proportional to each other.

Bond length is increase means the bond order decreases.

Answer to Problem 9.81QP

The length of the $\text{C-H}$ bond is $\text{107}\text{pm}$

Explanation of Solution

The length of the $\text{C-H}$ bond is calculated via covalent radii

$\begin{array}{l}{\text{r}}_{\text{C}}\text{=}\text{76pm}\\ \\ {\text{r}}_{\text{H}}\text{=}\text{31pm}\\ \\ {\text{d}}_{\text{C-H}}\text{=}{\text{r}}_{\text{C}}\text{+}{\text{r}}_{\text{H}}\\ \\ \text{=}\text{76}\text{pm}\text{+31}\text{pm}\\ \text{=107}\text{pm}\end{array}$

The both covalent radii of the Carbon and Hydrogen are added then we get the length of the $\text{C-H}$ bond.

Interpretation Introduction

Interpretation:

The bond lengths of the given bonds has to be calculated using covalent radii.

Concept introduction:

Bond length: bond order and bond length is inversely proportional to each other.

Bond length is increase means the bond order decreases.

Answer to Problem 9.81QP

The length of the $\text{S-Cl}$ bond is $207\text{pm}$

Explanation of Solution

The length of the $\text{S-Cl}$ bond is calculated via covalent radii

$\begin{array}{l}{\text{r}}_{\text{S}}\text{=}\text{105pm}\\ \\ {\text{r}}_{\text{Cl}}\text{=}\text{102pm}\\ \\ {\text{d}}_{\text{S-Cl}}\text{=}{\text{r}}_{\text{S}}\text{+}{\text{r}}_{\text{Cl}}\\ \\ \text{=}\text{105}\text{pm}\text{+102}\text{pm}\\ \text{=207}\text{pm}\end{array}$

The both covalent radii of the Sulfur and Chlorine are added then we get the length of the $\text{S-Cl}$ bond.

Interpretation Introduction

Interpretation:

The bond lengths of the given bonds has to be calculated using covalent radii.

Concept introduction:

Bond length: bond order and bond length is inversely proportional to each other.

Bond length is increase means the bond order decreases.

Answer to Problem 9.81QP

The length of the $\text{Br-Cl}$ bond is $222\text{pm}$

Explanation of Solution

The length of the $\text{Br-Cl}$ bond is calculated via covalent radii

$\begin{array}{l}{\text{r}}_{\text{Br}}\text{=}\text{120}\text{pm}\\ \\ {\text{r}}_{\text{Cl}}\text{=}\text{102pm}\\ \\ {\text{d}}_{\text{Br-Cl}}\text{=}{\text{r}}_{\text{Br}}\text{+}{\text{r}}_{\text{Cl}}\\ \\ \text{=}\text{120}\text{pm}\text{+102}\text{pm}\\ \text{=222}\text{pm}\end{array}$

The both covalent radii of the Bromine and Chlorine are added then we get the length of the $\text{Br-Cl}$ bond.

Interpretation Introduction

Interpretation:

The bond lengths of the given bonds has to be calculated using covalent radii.

Concept introduction:

Bond length: bond order and bond length is inversely proportional to each other.

Bond length is increase means the bond order decreases.

Answer to Problem 9.81QP

The length of the $\text{Si-O}$ bond is $\text{177}\text{pm}$

Explanation of Solution

The length of the $\text{Si-O}$ bond is calculated via covalent radii

$\begin{array}{l}{\text{r}}_{\text{C}}\text{=}\text{76pm}\\ \\ {\text{r}}_{\text{H}}\text{=}\text{31pm}\\ \\ {\text{d}}_{\text{C-H}}\text{=}{\text{r}}_{\text{C}}\text{+}{\text{r}}_{\text{H}}\\ \\ \text{=}\text{76}\text{pm}\text{+31}\text{pm}\\ \text{=107}\text{pm}\end{array}$

The both covalent radii of the Silicon and Oxygen are added then we get the length of the $\text{Si-O}$ bond.