Chapter 14, Problem 20QP

Determine the oxidation number of iodine in each of the following compounds.

$\begin{array}{l}\left(\text{a}\right)\text{ NaI}\\ \left(\text{b}\right){\text{ I}}_2\\ \left(\text{c}\right){\text{ IO}}_2\\ \left(\text{d}\right){\text{ I}}_2{\text{O}}_7\\ \left(\text{e}\right){\text{ KIO}}_4\\ \left(\text{f}\right)\text{ Ca}{\left(\text{IO}\right)}_2\end{array}$

Interpretation Introduction

Interpretation:

The oxidation number of iodine in $\text{NaI}$ is to be determined.

Explanation of Solution

The given compound is $\text{NaI}$ . The oxidation number of sodium is $+1$ which is predicted from the periodic table. The net charge on $\text{NaI}$ is zero. Using the net charge and chemical formula, the oxidation number of iodine is determined as follows:

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

The oxidation number of iodine $\left(\text{I}\right)$ in $\text{NaI}$ is $-1$ .

Interpretation Introduction

Interpretation:

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

Explanation of Solution

The given compound is ${\text{I}}_2$ which is present in its pure form. The compounds present in the pure form has zero oxidation number. The oxidation number of iodine in ${\text{I}}_2$ is $0$ .

Interpretation Introduction

Interpretation:

The oxidation number of iodine in ${\text{IO}}_2$ is to be determined

Explanation of Solution

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

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

The oxidation number of iodine $\left(\text{I}\right)$ in ${\text{IO}}_2$ is $+4$ .

Interpretation Introduction

Interpretation:

The oxidation number of iodine in ${\text{I}}_2{\text{O}}_7$ is to be determined

Explanation of Solution

The given compound is ${\text{I}}_2{\text{O}}_7$ . Oxygen is more electronegative than phosphorus. Thus, oxidation number of oxygen is $-2$ . The net charge on ${\text{I}}_2{\text{O}}_7$ is zero. Using the net charge and chemical formula, the oxidation number of iodine is determined as follows:

$\begin{array}{c}{\text{Net charge I}}_2{\text{O}}_7\text{=}\left[\begin{array}{l}\text{Total positive }\\ \text{oxidation number}\\ 2\times \text{I}\end{array}\right]+\left[\begin{array}{l}\text{Total negative }\\ \text{oxidation number}\\ 7\times \text{O}\end{array}\right]\\ 0=\left[\begin{array}{l}\text{Total positive }\\ \text{oxidation number}\\ 2\times \text{I}\end{array}\right]+\left[7\times \left(-2\right)\right]\\ \left[\begin{array}{l}\text{Total positive }\\ \text{oxidation number}\\ 2\times \text{I}\end{array}\right]=+14\\ \left[\begin{array}{c}\text{Total positive }\\ \text{oxidation number}\\ \text{I}\end{array}\right]=+7\end{array}$

The oxidation number of each iodine $\left(\text{I}\right)$ in ${\text{I}}_2{\text{O}}_7$ is $+7$ .

Interpretation Introduction

Interpretation:

The oxidation number of iodine in ${\text{KIO}}_4$ is to be determined.

Explanation of Solution

The given compound is ${\text{KIO}}_4$ . Oxidation number of oxygen is $-2$ and the oxidation number of potassium is $+1$ which is predicted from the periodic table. The net charge on ${\text{KIO}}_4$ is zero. Using the net charge and chemical formula, the oxidation number of iodine is determined as follows:

$\begin{array}{c}{\text{Net charge KIO}}_4\text{=}\left[\begin{array}{l}\text{Total positive }\\ \text{oxidation number}\\ 1\times \text{K}+1\times \text{I}\end{array}\right]+\left[\begin{array}{l}\text{Total negative }\\ \text{oxidation number}\\ 4\times \text{O}\end{array}\right]\\ 0=\left[\begin{array}{l}\text{Total positive }\\ \text{oxidation number}\\ 1\times 1+1\times \text{I}\end{array}\right]+\left[4\times \left(-2\right)\right]\\ \left[\begin{array}{l}\text{Total positive }\\ \text{oxidation number}\\ \text{1+I}\end{array}\right]=+8\\ \left[\begin{array}{c}\text{Total positive }\\ \text{oxidation number}\\ \text{I}\end{array}\right]=+7\end{array}$

The oxidation number of iodine $\left(\text{I}\right)$ in ${\text{KIO}}_4$ is $+7$ .

Interpretation Introduction

Interpretation:

The oxidation number of iodine in $\text{Ca}{\left(\text{IO}\right)}_2$ is to be determined

Explanation of Solution

The given compound is $\text{Ca}{\left(\text{IO}\right)}_2$ . Oxidation number of oxygen is $-2$ and the oxidation number of calcium is $+2$ which is predicted from the periodic table. The net charge on $\text{Ca}{\left(\text{IO}\right)}_2$ is zero. Using the net charge and chemical formula, the oxidation number of iodine is determined as follows:

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

The oxidation number of each iodine $\left(\text{I}\right)$ in $\text{Ca}{\left(\text{IO}\right)}_2$ is $+1$ .