Chapter 14, Problem 28QP

(a)

Interpretation Introduction

Interpretation:

In the ion ${\text{H}}_2{\text{PO}}_4^{-}$ , the oxidation number of each element is to be determined.

Explanation of Solution

The given ion is ${\text{H}}_2{\text{PO}}_4^{-}$ . The oxidation number of oxygen and hydrogen is $-2$ and $-1$ which is predicted from the periodic table. The net charge on ${\text{H}}_2{\text{PO}}_4^{-}$ is $-1$ . Using the net charge and chemical formula, the oxidation number of phosphorus is determined as follows:

$\begin{array}{c}{\text{Net charge H}}_2{\text{PO}}_4^{-}\text{=}\left[\begin{array}{l}\text{Total positive }\\ \text{oxidation number}\\ 2\times \text{H}+1\times \text{P}\end{array}\right]+\left[\begin{array}{l}\text{Total negative }\\ \text{oxidation number}\\ 4\times \text{O}\end{array}\right]\\ -1=\left[\begin{array}{l}\text{Total positive }\\ \text{oxidation number}\\ 2\times 1+1\times \text{P}\end{array}\right]+\left[4\times \left(-2\right)\right]\\ \left[\begin{array}{l}\text{Total positive }\\ \text{oxidation number}\\ \text{P}\end{array}\right]=+5\end{array}$

The oxidation number of hydrogen $\left(\text{H}\right)$ , phosphorus $\left(\text{P}\right)$ and oxygen $\left(\text{O}\right)$ in ${\text{H}}_2{\text{PO}}_4^{-}$ is $+1$ , $+5$ and $-2$ .

(b)

Interpretation Introduction

Interpretation:

In the ion ${\text{FeCl}}_6^{3-}$ , the oxidation number of each element is to be determined.

Explanation of Solution

The given ion is ${\text{FeCl}}_6^{3-}$ . The oxidation number of chorine is $-1$ which is predicted from the periodic table. The net charge on ${\text{FeCl}}_6^{3-}$ is $-3$ . Using the net charge and chemical formula, the oxidation number of iron is determined as follows:

$\begin{array}{c}{\text{Net charge FeCl}}_6^{3-}\text{=}\left[\begin{array}{l}\text{Total positive }\\ \text{oxidation number}\\ 1\times \text{Fe}\end{array}\right]+\left[\begin{array}{l}\text{Total negative }\\ \text{oxidation number}\\ 6\times \text{Cl}\end{array}\right]\\ -3=\left[\begin{array}{l}\text{Total positive }\\ \text{oxidation number}\\ 1\times \text{Fe}\end{array}\right]+\left[6\times \left(-1\right)\right]\\ \left[\begin{array}{l}\text{Total positive }\\ \text{oxidation number}\\ \text{Fe}\end{array}\right]=+3\end{array}$

The oxidation number of iron $\left(\text{Fe}\right)$ and chlorine $\left(\text{Cl}\right)$ in ${\text{FeCl}}_6^{3-}$ is $+3$ and $-1$ .

(c)

Interpretation Introduction

Interpretation:

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

Explanation of Solution

The given ion is ${\text{ClO}}_2^{-}$ . The oxidation number of oxygen is $-2$ which is predicted from the periodic table. The net charge on ${\text{ClO}}_2^{-}$ is $-1$ . Using the net charge and chemical formula, the oxidation number of chlorine is determined as follows:

$\begin{array}{c}{\text{Net charge ClO}}_2^{-}\text{=}\left[\begin{array}{l}\text{Total positive }\\ \text{oxidation number}\\ 1\times \text{Cl}\end{array}\right]+\left[\begin{array}{l}\text{Total negative }\\ \text{oxidation number}\\ 2\times \text{O}\end{array}\right]\\ -1=\left[\begin{array}{l}\text{Total positive }\\ \text{oxidation number}\\ 1\times \text{Cl}\end{array}\right]+\left[2\times \left(-2\right)\right]\\ \left[\begin{array}{l}\text{Total positive }\\ \text{oxidation number}\\ \text{Cl}\end{array}\right]=+3\end{array}$

The oxidation number of oxygen $\left(\text{O}\right)$ and chorine $\left(\text{Cl}\right)$ in ${\text{ClO}}_2^{-}$ is $-2$ and $+3$ .

(d)

Interpretation Introduction

Interpretation:

In the ion ${\text{SiF}}_6^{2-}$ , the oxidation number of each element is to be determined.

Explanation of Solution

The given ion is ${\text{SiF}}_6^{2-}$ . The oxidation number of fluorine is $-1$ which is predicted from the periodic table. The net charge on ${\text{SiF}}_6^{2-}$ is $-2$ . Using the net charge and chemical formula, the oxidation number of silicon is determined as follows:

$\begin{array}{c}{\text{Net charge SiF}}_6^{2-}\text{=}\left[\begin{array}{l}\text{Total positive }\\ \text{oxidation number}\\ 1\times \text{Si}\end{array}\right]+\left[\begin{array}{l}\text{Total negative }\\ \text{oxidation number}\\ 6\times \text{F}\end{array}\right]\\ -2=\left[\begin{array}{l}\text{Total positive }\\ \text{oxidation number}\\ 1\times \text{Si}\end{array}\right]+\left[6\times \left(-1\right)\right]\\ \left[\begin{array}{l}\text{Total positive }\\ \text{oxidation number}\\ \text{Si}\end{array}\right]=+4\end{array}$

The oxidation number of silicon $\left(\text{Si}\right)$ and fluorine $\left(\text{F}\right)$ in ${\text{SiF}}_6^{2-}$ is $+4$ and $-1$ .

(e)

Interpretation Introduction

Interpretation:

In the ion ${\text{AsO}}_4^{3-}$ , the oxidation number of each element is to be determined.

Explanation of Solution

The given ion is ${\text{AsO}}_4^{3-}$ . The oxidation number of oxygen is $-2$ which is predicted from the periodic table. The net charge on ${\text{AsO}}_4^{3-}$ is $-3$ . Using the net charge and chemical formula, the oxidation number of arsenic is determined as follows:

$\begin{array}{c}{\text{Net charge AsO}}_4^{3-}\text{=}\left[\begin{array}{l}\text{Total positive }\\ \text{oxidation number}\\ 1\times \text{As}\end{array}\right]+\left[\begin{array}{l}\text{Total negative }\\ \text{oxidation number}\\ 4\times \text{O}\end{array}\right]\\ -3=\left[\begin{array}{l}\text{Total positive }\\ \text{oxidation number}\\ 1\times \text{As}\end{array}\right]+\left[4\times \left(-2\right)\right]\\ \left[\begin{array}{l}\text{Total positive }\\ \text{oxidation number}\\ \text{As}\end{array}\right]=+5\end{array}$

The oxidation number of arsenic $\left(\text{As}\right)$ and oxygen $\left(\text{O}\right)$ in ${\text{AsO}}_4^{3-}$ is $+5$ and $-2$ .