Chapter 14, Problem 19QP

Indicate the oxidation number of phosphorus in each of the following compounds.

$\begin{array}{l}\left(\text{a}\right){\text{ AlPO}}_{\text{4}}\\ \left(\text{b}\right){\text{ PF}}_5\\ \left(\text{c}\right){\text{ H}}_3{\text{PO}}_4\\ \left(\text{d}\right){\text{ H}}_3{\text{PO}}_2\\ \left(\text{e}\right){\text{ PH}}_3\\ \left(\text{f}\right){\text{ H}}_{\text{3}}{\text{PO}}_3\end{array}$

Interpretation Introduction

Interpretation:

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

Explanation of Solution

The given compound is ${\text{AlPO}}_4$ . Oxygen is more electronegative than phosphorus. Thus, the oxidation number of oxygen is $-2$ . The oxidation number of aluminum is $+3$ . This is predicted from the periodic table. The net charge on ${\text{AlPO}}_4$ is zero. Using the net charge and chemical formula, the oxidation number of phosphorus is determined as follows:

$\begin{array}{c}{\text{Net charge AlPO}}_4=\left[\begin{array}{l}\text{Total positive }\\ \text{oxidation number}\\ 1\times \text{Al}+1\times \text{P}\end{array}\right]+\left[\begin{array}{l}\text{Total negative }\\ \text{oxidation number}\\ 4\times \text{O}\end{array}\right]\\ 0=\left[\begin{array}{l}\text{Total positive }\\ \text{oxidation number}\\ 1\times 3+1\times \text{P}\end{array}\right]+\left[4\times \left(-2\right)\right]\\ \left[\begin{array}{l}\text{Total positive }\\ \text{oxidation number}\\ \text{3+P}\end{array}\right]=+8\\ \left[\begin{array}{c}\text{Total positive }\\ \text{oxidation number}\\ \text{P}\end{array}\right]=+5\end{array}$

The oxidation number of phosphorus $\left(\text{P}\right)$ in ${\text{AlPO}}_4$ is $+5$ .

Interpretation Introduction

Interpretation:

The oxidation number of phosphorus in ${\text{PF}}_5$ is to be determined.

Explanation of Solution

The given compound is ${\text{PF}}_5$ . Fluorine is more electronegative than phosphorus. Thus, the oxidation number of fluorine is $-1$ . The net charge on ${\text{PF}}_5$ is zero. Using the net charge and chemical formula, the oxidation number of phosphorus is determined as follows:

$\begin{array}{c}{\text{Net charge PF}}_5=\left[\begin{array}{l}\text{Total positive }\\ \text{oxidation number}\\ 1\times \text{P}\end{array}\right]+\left[\begin{array}{l}\text{Total negative }\\ \text{oxidation number}\\ 5\times \text{F}\end{array}\right]\\ 0=\left[\begin{array}{l}\text{Total positive }\\ \text{oxidation number}\\ 1\times \text{P}\end{array}\right]+\left[5\times \left(-1\right)\right]\\ \left[\begin{array}{l}\text{Total positive }\\ \text{oxidation number}\\ \text{P}\end{array}\right]=+5\end{array}$

The oxidation number of phosphorus $\left(\text{P}\right)$ in ${\text{PF}}_5$ is $+5$ .

Interpretation Introduction

Interpretation:

The oxidation number of phosphorus in ${\text{H}}_3{\text{PO}}_4$ is to be determined.

Explanation of Solution

The given compound is ${\text{H}}_3{\text{PO}}_4$ . Oxygen is more electronegative than phosphorus. Thus, the oxidation number of oxygen is $-2$ . The oxidation number of hydrogen is $+1$ , which is predicted from the periodic table. The net charge on ${\text{H}}_3{\text{PO}}_4$ is zero. Using the net charge and chemical formula, the oxidation number of phosphorus is determined as follows:

$\begin{array}{c}{\text{Net charge H}}_3{\text{PO}}_4=\left[\begin{array}{l}\text{Total positive }\\ \text{oxidation number}\\ 3\times \text{H}+1\times \text{P}\end{array}\right]+\left[\begin{array}{l}\text{Total negative }\\ \text{oxidation number}\\ 4\times \text{O}\end{array}\right]\\ 0=\left[\begin{array}{l}\text{Total positive }\\ \text{oxidation number}\\ 3\times 1+1\times \text{P}\end{array}\right]+\left[4\times \left(-2\right)\right]\\ \left[\begin{array}{l}\text{Total positive }\\ \text{oxidation number}\\ \text{3+P}\end{array}\right]=+8\\ \left[\begin{array}{c}\text{Total positive }\\ \text{oxidation number}\\ \text{P}\end{array}\right]=+5\end{array}$

The oxidation number of phosphorus $\left(\text{P}\right)$ in ${\text{H}}_3{\text{PO}}_4$ is $+4$ .

Interpretation Introduction

Interpretation:

The oxidation number of phosphorus in ${\text{H}}_3{\text{PO}}_2$ is to be determined.

Explanation of Solution

The given compound is ${\text{H}}_3{\text{PO}}_2$ . Oxygen is more electronegative than phosphorus. Thus, the oxidation number of oxygen is $-2$ . The oxidation number of hydrogen is $+1$ , which is predicted from the periodic table. The net charge on ${\text{H}}_3{\text{PO}}_2$ is zero. Using the net charge and chemical formula, the oxidation number of phosphorus is determined as follows:

$\begin{array}{c}{\text{Net charge H}}_3{\text{PO}}_2\text{=}\left[\begin{array}{l}\text{Total positive }\\ \text{oxidation number}\\ 3\times \text{H}+1\times \text{P}\end{array}\right]+\left[\begin{array}{l}\text{Total negative }\\ \text{oxidation number}\\ 2\times \text{O}\end{array}\right]\\ 0=\left[\begin{array}{l}\text{Total positive }\\ \text{oxidation number}\\ 3\times 1+1\times \text{P}\end{array}\right]+\left[2\times \left(-2\right)\right]\\ \left[\begin{array}{l}\text{Total positive }\\ \text{oxidation number}\\ \text{3+P}\end{array}\right]=+4\\ \left[\begin{array}{c}\text{Total positive }\\ \text{oxidation number}\\ \text{P}\end{array}\right]=+1\end{array}$

The oxidation number of phosphorus $\left(\text{P}\right)$ in ${\text{H}}_3{\text{PO}}_2$ is $+1$ .

Interpretation Introduction

Interpretation:

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

Explanation of Solution

The given compound is ${\text{PH}}_3$ . The oxidation number of hydrogen is $+1$ which is predicted from the periodic table. The net charge on ${\text{PH}}_3$ is zero. Using the net charge and chemical formula, the oxidation number of phosphorus is determined as follows:

$\begin{array}{c}{\text{Net charge PH}}_3\text{=}\left[\begin{array}{l}\text{Total positive }\\ \text{oxidation number}\\ 1\times \text{P}\end{array}\right]+\left[\begin{array}{l}\text{Total negative }\\ \text{oxidation number}\\ 3\times \text{H}\end{array}\right]\\ 0=\left[\begin{array}{l}\text{Total positive }\\ \text{oxidation number}\\ 1\times \text{P}\end{array}\right]+\left[3\times \left(+1\right)\right]\\ \left[\begin{array}{l}\text{Total positive }\\ \text{oxidation number}\\ \text{P}\end{array}\right]=-3\end{array}$

The oxidation number of phosphorus $\left(\text{P}\right)$ in ${\text{PH}}_3$ is $-3$ .

Interpretation Introduction

Interpretation:

The oxidation number of phosphorus in ${\text{H}}_3{\text{PO}}_3$ is to be determined

Explanation of Solution

The given compound is ${\text{H}}_3{\text{PO}}_3$ . Oxygen is more electronegative than phosphorus. Thus, the oxidation number of oxygen is $-2$ . The oxidation number of hydrogen is $+1$ , which is predicted from the periodic table. The net charge on ${\text{H}}_3{\text{PO}}_3$ is zero. Using the net charge and chemical formula, the oxidation number of phosphorus is determined as follows:

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

The oxidation number of phosphorus $\left(\text{P}\right)$ in ${\text{H}}_3{\text{PO}}_3$ is $+3$ .