\( \text{ClO}_2^-, \text{F}_2, \text{MnO}_4^- \text{ and } \text{Cr}_2\text{O}_7^{2-} \)
\( \text{Cr}_2\text{O}_7^{2-}, \text{MnO}_4^-, \text{ClO}_2^- \text{ and } \text{Cl}_2 \)
\( \text{ClO}_4^-, \text{MnO}_4^-, \text{ClO}_2^- \text{ and } \text{F}_2 \)
\( \text{MnO}_4^{2-}, \text{ClO}_2^-, \text{Cl}_2 \text{ and } \text{Mn}^{3+} \)
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The Correct Option isD
Solution and Explanation
Step 1: Understanding the Concept:
A disproportionation reaction is a redox reaction in which the same element undergoes both oxidation and reduction simultaneously. This is only possible if the element is in an intermediate oxidation state and can move to both a higher and a lower oxidation state. Step 2: Detailed Explanation:
- Species in their highest oxidation state (like \( \text{Mn}^{+7} \) in \( \text{MnO}_4^- \), \( \text{Cr}^{+6} \) in \( \text{Cr}_2\text{O}_7^{2-} \), \( \text{Cl}^{+7} \) in \( \text{ClO}_4^- \)) cannot be further oxidized, so they cannot disproportionate.
- \( \text{F}_2 \) is the most electronegative element and can only have an oxidation state of 0 or -1. It cannot be oxidized, so it doesn't disproportionate.
- **Let's check Option (D):**
1. **\( \text{MnO}_4^{2-} \):** Mn is in \( +6 \). It can disproportionate to \( \text{MnO}_4^- \text{ (+7)} \) and \( \text{MnO}_2 \text{ (+4)} \) in acidic medium.
2. **\( \text{ClO}_2^- \):** Cl is in \( +3 \). It can go up to \( \text{ClO}_3^- \text{ (+5)} \) and down to \( \text{Cl}^- \text{ (-1)} \).
3. **\( \text{Cl}_2 \):** Cl is in 0. In alkaline solution, it disproportionates to \( \text{Cl}^- \text{ (-1)} \) and \( \text{ClO}^- \text{ (+1)} \).
4. **\( \text{Mn}^{3+} \):** In aqueous solution, it disproportionates to \( \text{Mn}^{2+} \) and \( \text{MnO}_2 \text{ (+4)} \). Step 3: Final Answer:
The set in option (D) contains species that all undergo disproportionation.
