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Question: In acidic medium \(MnO_{4}^{2-}\): A) Disproportionates to \(Mn{{O}_{2}}\) and \(MnO_{4}^{-}\) B) ...

In acidic medium MnO42−MnO_{4}^{2-}:

A) Disproportionates to MnO2Mn{{O}_{2}} and MnO4−MnO_{4}^{-}

B) Is oxidized to MnO4−MnO_{4}^{-}

C) Is reduced to MnO2Mn{{O}_{2}}

D) Is reduced to Mn2+M{{n}^{2+}}

Explanation

Solution

In the manganate ion, Mn is having an oxidation state of +6.

Oxidation reaction is the reaction in which the oxidation number is increased from reactant to the product side and in reduction the oxidation number of an atom gets decreased moving from reactant to product side.

Complete answer:

In the question, it is asked which chemical phenomenon will take place when manganate ions react with acids or if manganate ions are acidified.

We know that we are only concerned about the Mn atom in the manganate ion.Manganese is a transition element and it belongs to the first series of the transition metals.

Manganese have an atomic number 25 and hence there are 25 electrons present in the Mn atom.

Let's write the electronic configuration of the Mn atom,

E.C of Mn=[Ar]4s23d5E.C\,of\,Mn=\left[ Ar \right]4{{s}^{2}}3{{d}^{5}}

So there are seven electrons in the outermost orbitals.

One of the salient features of the transition metal is that the transition metals show various oxidation states.

Mn here can extend their oxidation state or oxidation number till +7.

Mn show oxidation states like +7,+4,+3+,+2 and o

Here lets calculate the oxidation state of Mn in MnO42−MnO_{4}^{2-}

Put the value of oxidation number of Mn as x and then calculate:

Oxidation number = x+4(−2)=−2x+4(-2)=-2

Oxidation number of Mn in MnO42−MnO_{4}^{2-}=x=−2+8=6x=-2+8=6

So in MnO42−MnO_{4}^{2-}, Mn is having +6 oxidation state and hence it can increase its oxidation state i.e. it could undergo oxidation reaction and can also undergo reduction reaction by decreasing the oxidation number.

Hence the MnO42−MnO_{4}^{2-} undergoes a disproportionation reaction in an acidic medium, as a disproportionation reaction is a type of redox reaction.

And the reaction involved can be written as,

3MnO42−+4H+→2MnO4−+MnO2+2H2O3MnO_{4}^{2-}+4{{H}^{+}}\to 2Mn{{O}_{4}}^{-}+Mn{{O}_{2}}+2{{H}_{2}}O

In MnO42−→MnO4−MnO_{4}^{2-}\to Mn{{O}_{4}}^{-}, the oxidation number Mn is increasing from +6 to +7 a hence oxidation takes place.

In MnO42−→MnO2MnO_{4}^{2-}\to Mn{{O}_{2}}, the oxidation number of Mn is decreasing, +6 to +4, hence reduction takes place.

Therefore, the answer is option A.

Note:

If the question was asked for the case of permanganate ion (MnO41−MnO_{4}^{1-}) then it could only undergo reduction reaction, since the oxidation state in MnO41−MnO_{4}^{1-} is +7 and it cannot extend its oxidation state further. So only undergo reduction reactions.