Chapter SH, Problem 18.5APP

To determine

To calculate: The oxidation number of phosphorus in the phosphate ion.

Answer to Problem 18.5APP

The oxidation number of phosphorous in ${\text{(PO}}_4{}^{3-})$ is $5+$ .

Explanation of Solution

Given data:

The symbol for phosphate ion: ${\text{(PO}}_4{}^{3-})$

The oxidation number of phosphate ion is $3-$ .

The oxidation number of oxygen is $2-$ .

As per the problem,

Let the oxidation number of phosphorus be $x+$ .

The oxidation number of phosphate ion is $3-$ .

The oxidation number of oxygen is $2-$ .

There are four oxygen atoms in the ${\text{(PO}}_4{}^{3-})$ .

Therefore, the oxidation number of phosphate ion can be written as

  $\begin{array}{l}(x+)+(4\times 2-)=3-\\ x=(3-)-(8-)\\ x=5+\end{array}$

Conclusion:

The oxidation number of phosphorous in ${\text{(PO}}_4{}^{3-})$ is $5+$ .