Question

Which of the following change is not correct about the oxidizing property of $KMnO_4$ in acidic medium?

• A. $S^{2-} \rightarrow S$
• B. $Mn^{2+} \rightarrow MnO_2$
• C. $SO_3^{2-} \rightarrow SO_4^{2-}$
• D. $C_2O_4^{2-} \rightarrow CO_2$

Hint:

Remember the reduction products of $MnO_4^-$:

• Acidic: $Mn^{2+}$ (Colourless)
• Neutral/Faintly Alkaline: $MnO_2$ (Brown ppt)
• Strongly Alkaline: $MnO_4^{2-}$ (Green)

Answer 1

The Correct Option is B

Step 1: Understanding Oxidation with $KMnO_4$:
Potassium Permanganate ($KMnO_4$) is a strong oxidizing agent. In acidic medium, the permanganate ion ($MnO_4^-$) itself gets reduced to the Manganese(II) ion ($Mn^{2+}$). Reaction: $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$.
Step 2: Analyzing the Options:

• (A) $S^{2- \rightarrow S$:} Sulphide ion is oxidized to elemental Sulphur. Correct. ($H_2S \rightarrow S$)
• (B) $Mn^{2+ \rightarrow MnO_2$:} This is an oxidation of manganese from $+2$ to $+4$. However, $KMnO_4$ acts as an oxidizing agent and gets \reduced to $Mn^{2+}$ in acidic medium. It does not oxidize $Mn^{2+}$ further to $MnO_2$ in standard acidic conditions (that is the reverse of what happens to permanganate). In neutral/alkaline medium, $MnO_4^-$ reduces to $MnO_2$. But the question specifies "acidic medium" and asks about the oxidizing \property (what it does to others). $KMnO_4$ cannot oxidize $Mn^{2+}$ because it reduces \to $Mn^{2+}$. This reaction represents the product turning into an intermediate, or is chemically incorrect in this context. Thus, this change is not correct.
• (C) $SO_3^{2- \rightarrow SO_4^{2-}$:} Sulphite is oxidized to Sulphate. Correct.
• (D) $C_2O_4^{2- \rightarrow CO_2$:} Oxalate is oxidized to Carbon dioxide. Correct.

Step 3: Conclusion:
Option (B) is incorrect because $Mn^{2+}$ is the reduction product of $KMnO_4$ in acid, not a substrate it oxidizes to $MnO_2$.