Chapter 19, Problem 48QRT

Interpretation Introduction

Interpretation:

Given table has to be filled with the respective values or formulas.

Explanation of Solution

Entry 1:

Name is given as phosphorus.  The formula of phosphorus is $\text{P}$.  Oxidation state of phosphorus in its elemental form is always zero.

Entry 2:

Formula of the compound is given as ${{\text{(NH}}_{\text{4}}\text{)}}_{\text{2}}{\text{HPO}}_{\text{4}}$.  Name of the given formula is ammonium hydrogen phosphate.  The oxidation state of phosphorus is ${{\text{(NH}}_{\text{4}}\text{)}}_{\text{2}}{\text{HPO}}_{\text{4}}$ can be calculated by considering the hydrogen phosphate ion alone.  Hydrogen phosphate ion is given as ${\text{HPO}}_{\text{4}}{}^{2-}$.

    $\begin{array}{l}+1+x+4(-2)=-2\\ x-7=-2\\ x=+5\end{array}$

Therefore, oxidation state of phosphorus in ammonium hydrogen phosphate is $+5$.

Entry 3:

Name is given as Phosphoric acid.  The formula of Phosphoric acid is ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$.  Oxidation state of phosphorus in Phosphoric acid can be calculated as shown below,

    $\begin{array}{l}3(+1)+x+4(-2)=0\\ x-5=0\\ x=+5\end{array}$

Therefore, oxidation state of phosphorus in phosphoric acid is $+5$.

Entry 4:

Name is given as Tetraphosphorus decaoxide.  The formula of Tetraphosphorus decaoxide is ${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}$.  Oxidation state of phosphorus in Tetraphosphorus decaoxide can be calculated as shown below,

    $\begin{array}{l}4x+10(-2)=0\\ 4x-20=0\\ x=+5\end{array}$

Therefore, oxidation state of phosphorus in Tetraphosphorus decaoxide is $+5$.

Entry 5:

Formula of the compound is given as ${\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}$.  Name of the given formula is calcium phosphate.  The oxidation state of phosphorus is ${\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}$ can be calculated as shown below,

    $\begin{array}{l}3(+2)+2x+8(-2)=0\\ 2x=10\\ x=5\end{array}$

Therefore, oxidation state of phosphorus in ${\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}$ is $+5$.

Entry 6:

Name is given as calcium dihydrogen phosphate.  The formula of calcium dihydrogen phosphate is ${\text{Ca(H}}_{\text{2}}{\text{PO}}_{\text{4}}{\text{)}}_{\text{2}}$.  The oxidation state of phosphorus is ${\text{Ca(H}}_{\text{2}}{\text{PO}}_{\text{4}}{\text{)}}_{\text{2}}$ can be calculated as shown below,

    $\begin{array}{l}(+2)+4(+1)+2x+8(-2)=0\\ 2x=10\\ x=5\end{array}$

Therefore, oxidation state of phosphorus in calcium dihydrogen phosphate is $+5$.

The complete table can be given as,

Formula	Name	Oxidation state of phosphorus
${\text{P}}_{\text{4}}$	Phosphorus	Zero
${{\text{(NH}}_{\text{4}}\text{)}}_{\text{2}}{\text{PO}}_{\text{4}}$	Ammonium hydrogen phosphate	$+5$
${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$	Phosphoric acid	$+5$
${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}$	Tetraphosphorus decaoxide	$+5$
${\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}$	Calcium phosphate	$+5$
${\text{Ca(H}}_{\text{2}}{\text{PO}}_{\text{4}}{\text{)}}_{\text{2}}$	calcium dihydrogen phosphate	$+5$