Question

What is the \( n \)-factor of \(KMnO_{4}\) in acidic medium?

• A. 2
• B. 3
• C. 5
• D. 7

Hint:

The \( n \)-factor of an oxidizing agent is the total number of electrons gained per molecule or ion. For KMnO\textsubscript{4} in acidic medium, Mn goes from +7 to +2, so \( n \)-factor = 5.

Answer 1

The Correct Option is C

Step 1: In acidic medium, potassium permanganate \( \text{KMnO}_4 \) acts as an oxidizing agent. 
The Mn in \( \text{KMnO}_4 \) has an oxidation state of +7.
Step 2: The reduction half-reaction in acidic medium is: \[ \text{MnO}_4^- \rightarrow \text{Mn}^{2+} \]
Step 3: In this process, the oxidation state of Mn changes from +7 to +2.
Step 4: Therefore, the change in oxidation number is: \[ 7 - 2 = 5 \]
Step 5: Hence, the \( n \)-factor, which is the number of electrons gained or lost per formula unit, is 5.