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Question: To an acidic solution of an anion, a few drops of \(KMn{{O}_{4}}\) solution are added. Which of the ...

To an acidic solution of an anion, a few drops of KMnO4KMn{{O}_{4}} solution are added. Which of the following if present, will not decolourise the KMnO4KMn{{O}_{4}} solution?
(A) CO32−CO_{3}^{2-}
(B) NO2−NO_{2}^{-}
(C) S2−{{S}^{2-}}
(D) Cl−C{{l}^{-}}

Explanation

Solution

Hint: The colour of potassium permanganate is deep purple. When the oxidation state of manganese in potassium permanganate changes from +7 to +2, we observe decolourisation. This is a reduction reaction. So any ion which will reduce the KMnO4KMn{{O}_{4}} will be our answer.

Complete step by step answer:
Let’s look at the answer to the question:
We know that decolourization is observed when the KMnO4KMn{{O}_{4}} gets reduced.
The process can be represented as:
MnO4− + 5e−→ Mn+2MnO_{4}^{-}\,+\,5{{e}^{-}}\to \,M{{n}^{+2}}
Now, we can see that Mn is getting reduced then the other ion should get oxidized.
So, we will look for the ion which will get oxidised, that is its oxidation number will increase.
*Now, in NO2−NO_{2}^{-} the oxidation state of N is +3 and it can increase its oxidation state to +5. So, will decolourise the KMnO4KMn{{O}_{4}}solution.
*In S2−{{S}^{2-}}the oxidation state of S is -2 and it can increase its oxidation state to maximum +6. So, S2−{{S}^{2-}}will decolourise theKMnO4KMn{{O}_{4}}solution.
*In Cl−C{{l}^{-}}the oxidation state of Cl is -1 and it can increase its oxidation state to maximum +7. So, Cl−C{{l}^{-}} will decolourise the KMnO4KMn{{O}_{4}}solution.
*In CO32−CO_{3}^{2-}the oxidation state of C is +4 which is the maximum oxidation state of C. So, CO32−CO_{3}^{2-} will not decolourise the KMnO4KMn{{O}_{4}} solution. This happens as it will not get oxidized and if oxidation doesn’t take place then reduction of KMnO4KMn{{O}_{4}}will also not take place.
Therefore, addition of CO32−CO_{3}^{2-} will not decolourise the KMnO4KMn{{O}_{4}}solution.
Hence, the answer for the given question is option (A).

Note: Permanganate ion is coloured due to d-d electronic transitions whereas Mn+2M{{n}^{+2}} is colourless as there are no d-d electronic transitions. When electrons gain energy they get excited and when they lose energy, radiations are emitted which provide colour to the metal.