Question

1 mol of $Mn{ { { { O }_{ 4 } }^{ 2- } } }$ in neutral aqueous medium disproportionates to?
[A] $\dfrac{2}{3}mol\text{ }of\text{ }Mn{{O}_{4}}^{-}\text{ }and\text{ }\dfrac{1}{3}\text{ }of\text{ }Mn{{O}_{2}}$
[B] $\dfrac{1}{3}\text{ }mol\text{ }of\text{ }Mn{{O}_{4}}^{-}\text{ }and\text{ }\dfrac{2}{3}\text{ }of\text{ }Mn{{O}_{2}}$
[C] $\dfrac{1}{3}\text{ }mol\text{ }of\text{ }M{{n}_{2}}{{O}_{7}}\text{ }and\ \dfrac{2}{3}\ of\ Mn{{O}_{2}}$
[D] $\dfrac{2}{3}\text{ }mol\text{ }of\text{ }M{{n}_{2}}{{O}_{7}}\text{ }and\,\text{ }\dfrac{1}{3}\text{ }of\text{ }Mn{{O}_{2}}$

Answer 1

Hint: To answer this, firstly understand what a disproportionation reaction is. Here, the same species will undergo oxidation as well as reduction to give us two products where one is oxidized and the other is reduced.

Complete step by step answer:
To answer this, firstly let us discuss the meaning of a disproportionation reaction.
We can define disproportionation reaction as a redox reaction where both oxidation as well as reduction takes place from the same element/species i.e. the same element is oxidised as well as reduced and forms two or more different products.
In simpler words, we can explain it as a reaction where the same element is both reduced and oxidised at the same time to give us a two different products or, we can say that if the compound forms two products, one as a result of oxidation and other as a result of reduction, we can say it undergoes disproportionation.

 Now let us discuss the compound given in the question and see what it disproportionates to-
$3Mn{ { { { O }_{ 4 } }^{ -2 } } }\xrightarrow [ neutral aqueous ]{ medium } Mn{ O }_{ 2 }+2Mn{ { { { O }_{ 4 } }^{ - }+ } }{ O }_{ 2 }$
$Mn{ { { { O }_{ 4 } }^{ -2 } } }\xrightarrow [ neutral aqueous ]{ medium } \dfrac { 1 }{ 3 } Mn{ O }_{ 2 }+\dfrac { 2 }{ 3 } Mn{ { { { O }_{ 4 } }^{ - }+\dfrac { 1 }{ 3 } } }{ O }_{ 2 }$

We can see from the above reaction that in neutral aqueous medium, $Mn{ { { { O }_{ 4 } }^{ -2 } } }$ disproportionates to give $\dfrac{2}{3}mol\text{ }of\text{ }Mn{{O}^{-}}_{4}$ and $\dfrac{1}{3}mol\text{ }of\text{ }Mn{{O}_{2}}$.

Therefore, the correct answer is option [A] $\dfrac{2}{3}mol\text{ }of\text{ }Mn{{O}_{4}}^{-}\text{ }and\text{ }\dfrac{1}{3}\text{ }of\text{ }Mn{{O}_{2}}$
So, the correct answer is “Option A”.

Note: Every disproportionation reaction is a redox reaction but every redox reaction is not a disproportionation reaction. Thus, disproportionation is a special case of a redox reaction. Certain compounds undergo disproportionation in order to gain higher stability which results from oxidation or reduction. Also we should remember that gain of electron is reduction and loss of electron is oxidation.