\(A \xrightarrow{4KOH,\ O_2} \underset{(\mathrm{Green})}{2B} + 2H_2O\)
\(3B \xrightarrow{4HCl} \underset{(\mathrm{Purple})}{2C} + MnO_2 + 2H_2O\)
\(2C \xrightarrow{H_2O.KI} \underset{(\mathrm{Purple})}{2A} + 2KOH + D\)
In the above sequence of reactions, A and D, respectively, are :
\(3B \xrightarrow{4HCl} \underset{(\mathrm{Purple})}{2C} + MnO_2 + 2H_2O\)
\(2C \xrightarrow{H_2O.KI} \underset{(\mathrm{Purple})}{2A} + 2KOH + D\)
In the above sequence of reactions, A and D, respectively, are :
Show Hint
- \(\mathrm{KI}\) and \(\mathrm{KMnO_4}\)
- \(\mathrm{KI}\) and \(\mathrm{K_2MnO_4}\)
- \(\mathrm{KIO_3}\) and \(\mathrm{MnO_2}\)
- \(\mathrm{MnO_2}\) and \(\mathrm{KIO_3}\)
The Correct Option is D
Solution and Explanation
\(\mathrm{3K_2MnO_4 + 4HCl \rightarrow 2KMnO_4 + MnO_2 + 2H_2O + 4KCl}\) (disproportionation in acid): C = \(\mathrm{KMnO_4}\) (purple).
In neutral/weakly alkaline medium, permanganate oxidises iodide to iodate: \(\mathrm{2KMnO_4 + H_2O + KI \rightarrow 2MnO_2 + 2KOH + KIO_3}\). So A = \(\mathrm{MnO_2}\) and D = \(\mathrm{KIO_3}\).
Answer: (4) \(\mathrm{MnO_2}\) and \(\mathrm{KIO_3}\)
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