Read the passage carefully and answer the Questions
\(KMnO _4\) is prepared by the fusion of \(MnO _2\) with an alkali metal hydroxide and an oxidizing agent like \(KNO _3\) to give a dark-green manganate ion which disproportionate to give permanganate as follows.
\(\begin{array}{l}2 MnO _2+4 KOH + O _2 \rightarrow 2 K_2 MnO _4+2 H _2 O \\ 3 K_2 MnO _4+4 H ^{+} \rightarrow 2 KMnO _4+ MnO _2+2 H _2 O \end{array}\)
On heating \(K _2 MnO _4\) decomposes at 513 K to give \(K _2 MnO _4\). Permanganate ion is tetrahedral and diamagnetic
Acidified \(KMnO _4\) acts a strong oxidizing agent which oxidizes oxalic acid, ferrous ions, nitrite ion and iodides.
Reaction of iodide ion \(\left( I ^{-}\right)\)with \(KMnO _4\) will.....

1. A liberate \(I_2\) in acidic solution
2. B produce \(IO _3^{-}\)in acidic solution
3. C liberate \(I_2\) in neutral or mild alkaline solution
4. D produce \(I _2 O _4\) in neutral or mild alkaline solution