Chapter 14, Problem 14.118P

Interpretation Introduction

Interpretation:

Term disproportionation reaction has to be explained and also the reaction that fits into the description of disproportionation reaction has to be identified.

Explanation of Solution

Disproportionation reaction is a type of reaction where a substance can act as a reducing agent as well as an oxidizing agent since an atom in the substance reacts to give atoms having and lower oxidation states.

Reason for correct option:

In option (b), Chlorine in $Cl{O}_2$ gets reduced from $+4$ to $+3$ (in $HCl{O}_2$) as well as gets oxidized from $+4$ to $+5$ (in $HCl{O}_3.$).  Hence $Cl{O}_2$ disproportionate.

  $\text{2}\stackrel{+4}{\text{Cl}}{\text{O}}_{\text{2}}\left(g\right){\text{ + H}}_{\text{2}}\text{O}\left(l\right)\stackrel{}{\to }\text{ H}\stackrel{+5}{\text{Cl}}{\text{O}}_{\text{3}}\left(\text{aq}\right)\text{ + H}\stackrel{+3}{\text{Cl}}{\text{O}}_{\text{2}}\left(\text{aq}\right)$

In option (c), $C{l}_2$ gets reduced from $0$ to $-1$ (in $NaCl$) as well as gets oxidized from $0$ to $+1$ (in $NaClO$).  Hence $C{l}_2$ disproportionate.

  ${\stackrel{0}{\text{Cl}}}_{\text{2}}\left(g\right)\text{ + 2NaOH}\left(\text{aq}\right)\stackrel{}{\to }\text{ Na}\stackrel{-1}{\text{Cl}}\left(\text{aq}\right)\text{ + Na}\stackrel{+1}{\text{Cl}}\text{O}\left(\text{aq}\right){\text{ + H}}_{\text{2}}\text{O}\left(l\right)$

In option (d), Nitrogen atom of nitrite $\left(N{O}_2{}^{-}\right)$ gets reduced from $+3$ to $0$ and Nitrogen atom of ammonium $\left(N{H}_4{}^{+}\right)$ gets oxidized from $-3$ to $0$.  Hence $N{H}_{4}N{O}_3$ disproportionates.

  $\stackrel{-3}{\text{N}}{\text{H}}_{\text{4}}\stackrel{+3}{\text{N}}{\text{O}}_{\text{2}}\left(s\right)\stackrel{}{\to }{\stackrel{0}{\text{N}}}_{\text{2}}\left(g\right){\text{ + 2H}}_{\text{2}}\text{O}\left(g\right)$

In option (e), Magnesium in $Mn{O}_4{}^{2-}$ gets reduced from $+6\text{ to +4}$ (in $Mn{O}_2$) as well as gets oxidized from $+6\text{ to +7}$ (in $Mn{O}_4{}^{-}$).  Hence $Mn{O}_4{}^{2-}$ disproportionates.

  $\text{3}\stackrel{+6}{\text{Mn}}{\text{O}}_{\text{4}}{}^{2-}\left(\text{aq}\right){\text{ + 2H}}_{\text{2}}\text{O}\left(l\right)\stackrel{}{\to }\text{ 2}\stackrel{+7}{\text{Mn}}{\text{O}}_{\text{4}}{}^{-}\left(\text{aq}\right)\text{ + }\stackrel{+4}{\text{Mn}}{\text{O}}_{\text{2}}\left(s\right){\text{ + 4OH}}^{-}\left(aq\right)$

In option (f), Gold in $AuCl$ gets reduced from $+1\text{ to 0}$ (in $Au$) as well as gets oxidized from $+1\text{ to +3}$ (in $AuC{l}_3$).  Hence $AuCl$ disproportionates.

  $\text{3}\stackrel{+1}{\text{Au}}\text{Cl}\left(s\right)\stackrel{}{\to }\stackrel{+3}{\text{Au}}{\text{Cl}}_{\text{3}}\left(s\right)\text{ + 2}\stackrel{0}{\text{Au}}\left(s\right)$

Therefore, correct options are (b), (c), (d), (e) and (f).

Reason for in-correct option:

In option (a), Iodine atom in $I_2$ gets reduced from $0\text{ to}-\text{1/3}$ (in $K{I}_3$). Iodine atom in $KI$ gets oxidized from $-1\text{ to}-\text{1/3}$ (in $K{I}_3$).

  ${\stackrel{0}{\text{I}}}_{\text{2}}\left(s\right)\text{ + }\stackrel{-\text{1}}{\text{KI}}\left(\text{aq}\right)\stackrel{}{\to }\stackrel{-\text{1/3}}{{\text{KI}}_{\text{3}}}\left(\text{aq}\right)$

Here, oxidizing and reducing agents are different substance. Hence, it is not a disproportionation reaction.

Therefore, incorrect option is (a).