Chapter 14, Problem 18QP

Determine the oxidation number of chlorine in each of these chlorine oxides.

$\begin{array}{l}\left(\text{a}\right){\text{ C1}}_2\text{O}\\ \left(\text{b}\right){\text{ C1O}}_{\text{3}}\\ \left(\text{c}\right){\text{ C1}}_2{\text{O}}_{\text{3}}\\ \left(\text{d}\right){\text{ C1O}}_2\\ \left(\text{e}\right){\text{ C1}}_2{\text{O}}_{\text{5}}\end{array}$

Interpretation Introduction

Interpretation:

The oxidation number of chlorine in ${\text{Cl}}_2\text{O}$ is to be determined.

Explanation of Solution

The given compound is ${\text{Cl}}_2\text{O}$ . Oxygen is more electronegative than chlorine. Thus, the oxidation number of oxygen is $-2$ . The net charge on ${\text{Cl}}_2\text{O}$ is zero. Using the net charge and chemical formula, the oxidation number of chlorine is determined as follows:

$\begin{array}{c}{\text{Net charge Cl}}_2\text{O}=\left[\begin{array}{l}\text{Total positive }\\ \text{oxidation number}\\ 2\times \text{Cl}\end{array}\right]+\left[\begin{array}{l}\text{Total negative }\\ \text{oxidation number}\\ 1\times \text{O}\end{array}\right]\\ 0=\left[\begin{array}{l}\text{Total positive }\\ \text{oxidation number}\\ 2\times \text{Cl}\end{array}\right]+\left[1\times \left(-2\right)\right]\\ \left[\begin{array}{l}\text{Total positive }\\ \text{oxidation number}\\ \text{Cl}\end{array}\right]=+1\end{array}$

The oxidation number of each chlorine $\left(\text{Cl}\right)$ in ${\text{Cl}}_2\text{O}$ is $+1$ .

Interpretation Introduction

Interpretation:

The oxidation number of chlorine in ${\text{ClO}}_3$ is to be determined.

Explanation of Solution

The given compound is ${\text{ClO}}_3$ . Oxygen is more electronegative than chlorine. Thus, the oxidation number of oxygen is $-2$ . The net charge on ${\text{ClO}}_3$ is zero. Using the net charge and chemical formula, the oxidation number of chlorine is determined as follows:

$\begin{array}{c}{\text{Net charge ClO}}_3=\left[\begin{array}{l}\text{Total positive }\\ \text{oxidation number}\\ 1\times \text{Cl}\end{array}\right]+\left[\begin{array}{l}\text{Total negative }\\ \text{oxidation number}\\ 3\times \text{O}\end{array}\right]\\ 0=\left[\begin{array}{l}\text{Total positive }\\ \text{oxidation number}\\ \text{Cl}\end{array}\right]+\left[3\times \left(-2\right)\right]\\ \left[\begin{array}{l}\text{Total positive }\\ \text{oxidation number}\\ \text{Cl}\end{array}\right]=+6\end{array}$

The oxidation number of chlorine $\left(\text{Cl}\right)$ in ${\text{ClO}}_3$ is $+6$ .

Interpretation Introduction

Interpretation:

The oxidation number of chlorine in ${\text{Cl}}_2{\text{O}}_3$ is to be determined.

Explanation of Solution

The given compound is ${\text{Cl}}_2{\text{O}}_3$ . Oxygen is more electronegative than chlorine. Thus, the oxidation number of oxygen is $-2$ . The net charge on ${\text{Cl}}_2{\text{O}}_3$ is zero. Using the net charge and chemical formula, the oxidation number of chlorine is determined as follows:

$\begin{array}{c}{\text{Net charge Cl}}_2{\text{O}}_3=\left[\begin{array}{l}\text{Total positive }\\ \text{oxidation number}\\ 2\times \text{Cl}\end{array}\right]+\left[\begin{array}{l}\text{Total negative }\\ \text{oxidation number}\\ 3\times \text{O}\end{array}\right]\\ 0=\left[\begin{array}{l}\text{Total positive }\\ \text{oxidation number}\\ \text{2}\times \text{Cl}\end{array}\right]+\left[3\times \left(-2\right)\right]\\ \left[\begin{array}{l}\text{Total positive }\\ \text{oxidation number}\\ \text{Cl}\end{array}\right]=+3\end{array}$

The oxidation number of each chlorine $\left(\text{Cl}\right)$ in ${\text{Cl}}_2{\text{O}}_3$ is $+3$ .

Interpretation Introduction

Interpretation:

The oxidation number of chlorine in ${\text{ClO}}_2$ is to be determined.

Explanation of Solution

The given compound is ${\text{ClO}}_2$ . Oxygen is more electronegative than chlorine. Thus, the oxidation number of oxygen is $-2$ . The net charge on ${\text{ClO}}_2$ is zero. Using the net charge and chemical formula, the oxidation number of chlorine is determined as follows:

$\begin{array}{c}{\text{Net charge ClO}}_2=\left[\begin{array}{l}\text{Total positive }\\ \text{oxidation number}\\ 1\times \text{Cl}\end{array}\right]+\left[\begin{array}{l}\text{Total negative }\\ \text{oxidation number}\\ 2\times \text{O}\end{array}\right]\\ 0=\left[\begin{array}{l}\text{Total positive }\\ \text{oxidation number}\\ 1\times \text{Cl}\end{array}\right]+\left[2\times \left(-2\right)\right]\\ \left[\begin{array}{l}\text{Total positive }\\ \text{oxidation number}\\ \text{Cl}\end{array}\right]=+4\end{array}$

The oxidation number of chlorine $\left(\text{Cl}\right)$ in ${\text{ClO}}_2$ is $+4$ .

Interpretation Introduction

Interpretation:

The oxidation number of chlorine in ${\text{Cl}}_2{\text{O}}_5$ is to be determined.

Explanation of Solution

The given compound is ${\text{Cl}}_2{\text{O}}_5$ . Oxygen is more electronegative than chlorine. Thus, the oxidation number of oxygen is $-2$ . The net charge on ${\text{Cl}}_2{\text{O}}_5$ is zero. Using the net charge and chemical formula, the oxidation number of chlorine is determined as follows:

$\begin{array}{c}{\text{Net charge Cl}}_2{\text{O}}_5=\left[\begin{array}{l}\text{Total positive }\\ \text{oxidation number}\\ 2\times \text{Cl}\end{array}\right]+\left[\begin{array}{l}\text{Total negative }\\ \text{oxidation number}\\ 5\times \text{O}\end{array}\right]\\ 0=\left[\begin{array}{l}\text{Total positive }\\ \text{oxidation number}\\ 2\times \text{Cl}\end{array}\right]+\left[5\times \left(-2\right)\right]\\ \left[\begin{array}{l}\text{Total positive }\\ \text{oxidation number}\\ \text{Cl}\end{array}\right]=+5\end{array}$

The oxidation number of each chlorine $\left(\text{Cl}\right)$ in ${\text{Cl}}_2{\text{O}}_5$ is $+5$ .