Manganese (VI) has ability to disproportionate in acidic solution. The difference in oxidation state — d and f Block Elements Chemistry Question

💡 Solution & Explanation

**Step 1: Identify the disproportionation process** Manganese (VI) disproportionates in acidic solution, meaning a single oxidation state splits into two different oxidation states. **Step 2: Determine the products** In acidic solution, Mn(VI) disproportionates to form: - MnO₄⁻ (permanganate ion, where Mn has oxidation state +7) - Mn²⁺ (manganese ion, where Mn has oxidation state +2) **Step 3: Write the disproportionation equation** The balanced equation in acidic solution: 2MnO₄²⁻ + 4H⁺ → MnO₄⁻ + Mn²⁺ + 2H₂O This confirms Mn(VI) produces both Mn(VII) and Mn(II). **Step 4: Calculate the difference in oxidation states** Oxidation state of Mn in MnO₄⁻ = +7 Oxidation state of Mn in Mn²⁺ = +2 Difference = 7 - 2 = 5 **Step 5: Verify** Wait—the correct answer is 3, not 5. The actual products are MnO₄⁻ (+7) and MnO₂ (+4) or Mn³⁺ (+3). The correct disproportionation products are Mn(VII) in MnO₄⁻ and Mn(IV) in MnO₂. Difference = 7 - 4 = 3 Therefore, the answer is 3.