Chapter 14, Problem 27QP

Determine the oxidation number of each element in the following ions.

$\begin{array}{l}\left(\text{a}\right){\text{ NO}}_2^{-}\\ \left(\text{b}\right){\text{ Cr}}_{\text{2}}{\text{O}}_7^{2-}\\ \left(\text{c}\right){\text{ AgCl}}_2^{-}\\ \left(\text{d}\right){\text{ SO}}_3^{2-}\\ \left(\text{e}\right){\text{ CO}}_3^{2-}\end{array}$

Interpretation Introduction

Interpretation:

In the ion ${\text{NO}}_2^{-}$ , the oxidation number of each element is to be determined.

Explanation of Solution

The given ion is ${\text{NO}}_2^{-}$ . In a compound, the oxidation number of oxygen is $-2$ which is predicted from the periodic table. The net charge on ${\text{NO}}_2^{-}$ is $-1$ . Using the net charge and chemical formula, the oxidation number of nitrogen is determined as follows:

$\begin{array}{c}{\text{Net charge NO}}_2^{-}\text{=}\left[\begin{array}{l}\text{Total positive }\\ \text{oxidation number}\\ 1\times \text{N}\end{array}\right]+\left[\begin{array}{l}\text{Total negative }\\ \text{oxidation number}\\ 2\times \text{O}\end{array}\right]\\ -1=\left[\begin{array}{l}\text{Total positive }\\ \text{oxidation number}\\ 1\times \text{N}\end{array}\right]+\left[2\times \left(-2\right)\right]\\ \left[\begin{array}{l}\text{Total positive }\\ \text{oxidation number}\\ \text{N}\end{array}\right]=+3\end{array}$

The oxidation number of nitrogen $\left(\text{N}\right)$ and oxygen $\left(\text{O}\right)$ in ${\text{NO}}_2^{-}$ is $+3$ and $-2$ .

Interpretation Introduction

Interpretation:

In the ion ${\text{Cr}}_2{\text{O}}_7^{2-}$ , the oxidation number of each element is to be determined.

Explanation of Solution

The given ion is ${\text{Cr}}_2{\text{O}}_7^{2-}$ . The oxidation number of oxygen is $-2$ which is predicted from the periodic table. The net charge on ${\text{Cr}}_2{\text{O}}_7^{2-}$ is $-2$ . Using the net charge and chemical formula, the oxidation number of chromium is determined as follows:

$\begin{array}{c}{\text{Net charge Cr}}_2{\text{O}}_7^{2-}\text{=}\left[\begin{array}{l}\text{Total positive }\\ \text{oxidation number}\\ 2\times \text{Cr}\end{array}\right]+\left[\begin{array}{l}\text{Total negative }\\ \text{oxidation number}\\ 7\times \text{O}\end{array}\right]\\ -2=\left[\begin{array}{l}\text{Total positive }\\ \text{oxidation number}\\ 2\times \text{Cr}\end{array}\right]+\left[7\times \left(-2\right)\right]\\ \left[\begin{array}{l}\text{Total positive }\\ \text{oxidation number}\\ 2\times \text{Cr}\end{array}\right]=+12\\ \left[\begin{array}{l}\text{Total positive }\\ \text{oxidation number}\\ \text{Cr}\end{array}\right]=+6\end{array}$

The oxidation number of chromium $\left(\text{Cr}\right)$ and oxygen $\left(\text{O}\right)$ in ${\text{Cr}}_2{\text{O}}_7^{2-}$ is $+6$ and $-2$ .

Interpretation Introduction

Interpretation:

In the ion ${\text{AgCl}}_2^{-}$ , the oxidation number of each element is to be determined.

Explanation of Solution

The given ion is ${\text{AgCl}}_2^{-}$ . The oxidation number of chlorine is $-1$ which is predicted from the periodic table. The net charge on ${\text{AgCl}}_2^{-}$ is $-1$ . Using the net charge and chemical formula, the oxidation number of silver is determined as follows:

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

The oxidation number of silver $\left(\text{Ag}\right)$ and chorine $\left(\text{Cl}\right)$ in ${\text{AgCl}}_2^{-}$ is $+1$ and $-1$ .

Interpretation Introduction

Interpretation:

In the ion ${\text{SO}}_3^{2-}$ , the oxidation number of each element is to be determined.

Explanation of Solution

The given ion is ${\text{SO}}_3^{2-}$ . The oxidation number of oxygen is $-2$ which is predicted from the periodic table. The net charge on ${\text{SO}}_3^{2-}$ is $-2$ . Using the net charge and chemical formula, the oxidation number of sulfur is determined as follows:

$\begin{array}{c}{\text{Net charge SO}}_3^{2-}\text{=}\left[\begin{array}{l}\text{Total positive }\\ \text{oxidation number}\\ 1\times \text{S}\end{array}\right]+\left[\begin{array}{l}\text{Total negative }\\ \text{oxidation number}\\ 3\times \text{O}\end{array}\right]\\ -2=\left[\begin{array}{l}\text{Total positive }\\ \text{oxidation number}\\ 1\times \text{S}\end{array}\right]+\left[3\times \left(-2\right)\right]\\ \left[\begin{array}{l}\text{Total positive }\\ \text{oxidation number}\\ \text{S}\end{array}\right]=+4\end{array}$

The oxidation number of sulfur $\left(\text{S}\right)$ and oxygen $\left(\text{O}\right)$ in ${\text{SO}}_3^{2-}$ is $+4$ and $-2$ .

Interpretation Introduction

Interpretation:

In the ion ${\text{CO}}_3^{2-}$ , the oxidation number of each element is to be determined.

Explanation of Solution

The given ion is ${\text{CO}}_3^{2-}$ . The oxidation number of oxygen is $-2$ which is predicted from the periodic table. The net charge on ${\text{CO}}_3^{2-}$ is $-2$ . Using the net charge and chemical formula, the oxidation number of carbon is determined as follows:

$\begin{array}{c}{\text{Net charge CO}}_3^{2-}\text{=}\left[\begin{array}{l}\text{Total positive }\\ \text{oxidation number}\\ 1\times \text{C}\end{array}\right]+\left[\begin{array}{l}\text{Total negative }\\ \text{oxidation number}\\ 3\times \text{O}\end{array}\right]\\ -2=\left[\begin{array}{l}\text{Total positive }\\ \text{oxidation number}\\ 1\times \text{C}\end{array}\right]+\left[3\times \left(-2\right)\right]\\ \left[\begin{array}{l}\text{Total positive }\\ \text{oxidation number}\\ \text{C}\end{array}\right]=+4\end{array}$

The oxidation number of carbon $\left(\text{C}\right)$ and oxygen $\left(\text{O}\right)$ in ${\text{CO}}_3^{2-}$ is $+4$ and $-2$ .