
In acid solution, the reaction $ MnO_4^ - \to M{n^{ + 2}} $ involves.
Answer
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Hint :The solutions which contain more hydrogen ions than water are called acidic solutions. When an acidic substance is added to a solution, the concentration of hydronium ions increases in the solution. This happens because when an acid gets dissolved in water, the balance between the hydrogen ions and the hydroxide ions gets shifted. And the solution will have more hydrogen ions than hydroxide ions in the solution and this makes the solution acidic.
Complete Step By Step Answer:
In the reaction $ MnO_4^ - \to M{n^{ + 2}} $ , we can observe that the charge on manganese ion reduces to $ + 2 $ from $ + 7 $ .This is called reduction reaction. $ Mn{O_4} $ is purple in color because manganese is in $ + 7 $ oxidation state and the permanganate ion acts as a very strong oxidizing agent. The geometry of this ion is tetrahedral geometry. These ions are stable in neutral and in slightly alkaline medium. As there is reduction of manganese from $ + 7 $ to $ + 2 $ , there is reduction of $ 5 $ electrons. The reaction is represented in the following way:
$ MnO_4^ - + 5{e^ - } \to M{n^{ + 2}} $
Note :
The purple color of the permanganate ion is not because of the d-d transition. It is because of the charge transition from oxygen to manganese. This transition makes it purple in color irrespective of the unpaired electron in $ M{n^{ + 7}} $ . When a photon of light is absorbed, charge transfer takes place and it results in the purple color.
Complete Step By Step Answer:
In the reaction $ MnO_4^ - \to M{n^{ + 2}} $ , we can observe that the charge on manganese ion reduces to $ + 2 $ from $ + 7 $ .This is called reduction reaction. $ Mn{O_4} $ is purple in color because manganese is in $ + 7 $ oxidation state and the permanganate ion acts as a very strong oxidizing agent. The geometry of this ion is tetrahedral geometry. These ions are stable in neutral and in slightly alkaline medium. As there is reduction of manganese from $ + 7 $ to $ + 2 $ , there is reduction of $ 5 $ electrons. The reaction is represented in the following way:
$ MnO_4^ - + 5{e^ - } \to M{n^{ + 2}} $
Note :
The purple color of the permanganate ion is not because of the d-d transition. It is because of the charge transition from oxygen to manganese. This transition makes it purple in color irrespective of the unpaired electron in $ M{n^{ + 7}} $ . When a photon of light is absorbed, charge transfer takes place and it results in the purple color.
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