The \[M{{n}^{3+}}\] ion is unstable in solution and undergoes disproportionation reaction to give\[M{{n}^{2+}}\], \[Mn{{O}_{2}}\]and\[{{H}^{+}}\]ion. Write a balanced ionic equation for the reaction.

Hint: The disproportionation reaction is the type of redox reaction where one reactant undergoes oxidation and reduction simultaneously.
Oxidation is the half of the reaction where the reactant gains electrons and hence the oxidation state decreases from the reactant to the product. The reduction is the half of the reaction where the reactant loses electrons and hence the oxidation state increases from the reactant to the product.

Complete step by step solution:
- The \[M{{n}^{3+}}\] ion is unstable in solution and undergoes disproportionation reaction to give\[M{{n}^{2+}}\], \[Mn{{O}_{2}}\]and\[{{H}^{+}}\]ion.
- Now, let’s see the balance of the net ionic equation.
The reaction as illustrated in the question can be represented as: \[M{{n}^{3+}}(aq)\to M{{n}^{2+}}(aq)+Mn{{O}_{2}}(s)+{{H}^{+}}(aq)\]
- The oxidation half is associated with the increase in the oxidation state of Manganese ion or loss of electrons as follows:
\[\overset{(+3)}{\mathop{Mn}}\,{{\text{ }}^{3+}}(aq)\to \overset{(+4)}{\mathop{Mn}}\,{{O}_{2}}(s)\]
One electron is added in the right side of the equation to balance the number of electrons.
The reaction proceeds in the acidic medium. So, \[{{H}_{2}}O\] is added to the left side of the equation to balance the number of oxygen atoms of \[Mn{{O}_{2}}\]and \[{{H}^{+}}\]is added to the right side of the equation to balance the number of hydrogen atoms of\[{{H}_{2}}O\].
\[\overset{(+3)}{\mathop{Mn}}\,{{\text{ }}^{3+}}(aq)\text{ }+\text{ }2{{H}_{2}}O(l)\to \overset{(+4)}{\mathop{Mn}}\,{{O}_{2}}(s)\text{ }+\text{ }{{e}^{-}}\text{ }+\text{ }4{{H}^{+}}\]
- The reduction half is associated with the decrease in the oxidation state of Manganese ion or gain of electrons as follows:
\[\overset{(+3)}{\mathop{Mn}}\,{{\ }^{3+}}(aq)\text{ + }{{\text{e}}^{-}}\to \overset{(+2)}{\mathop{Mn}}\,{{\text{ }}^{2+}}(aq)\]

One electron is added in the left side of the equation to balance the number of electrons.
The net ionic equation is:
\[\overset{(+3)}{\mathop{2Mn}}\,{{\text{ }}^{3+}}(aq)\text{ }+\text{ }2{{H}_{2}}O(l)\to \overset{(+4)}{\mathop{Mn}}\,{{O}_{2}}(s)\text{ }+\text{ }\overset{(+2)}{\mathop{Mn}}\,{{\text{ }}^{2+}}(aq)\text{ + 4}{{\text{H}}^{+}}\]

Note: \[M{{n}^{3+}}\]ion undergoes disproportionation to form \[M{{n}^{2+}}\] and \[M{{n}^{4+}}\]. The reason is:
Electronic configuration of \[M{{n}^{3+}}\]= \[[Ar]3{{d}^{4}}4{{s}^{0}}\]
Electronic configuration of \[M{{n}^{2+}}\]= \[[Ar]3{{d}^{5}}4{{s}^{0}}\]
 +2 oxidation state is more stable than +3 oxidation state. This is due to the stable half-filled electronic configuration. So, the \[M{{n}^{3+}}\]ion is unstable and disproportionates.