Manganese (VI) has ability to disproportionate in acidic solution. The difference in oxidation states of two ions it forms in acidic solution is ____________.

Solution

Chemistry ma’am · 4 steps

Manganate(VI) splits both ways in acid: one part rises to +7, the rest drops to +4.

1. Disproportionation. In disproportionation, one species is oxidised and reduced at the same time. So Mn(VI) must give one product above +6 and one below +6.

   3MnO₄²⁻ + 4H⁺ → 2MnO₄⁻ + MnO₂ + 2H₂O

2. Find the states. In MnO₄⁻, x + 4(-2) = -1, so x = +7. In MnO₂, x + 2(-2) = 0, so x = +4.
3. Check electrons. Two Mn atoms each lose 1 electron going from +6 to +7. One Mn atom gains 2 electrons going from +6 to +4. The electrons balance.
4. So. The difference in oxidation states is 7 - 4 = 3.

The common slip: 5. Taking the reduced product as Mn²⁺ gives 7 - 2 = 5. Disproportionation of manganate gives MnO₂ (+4), not Mn²⁺.

The idea to keep. In disproportionation, the products always sit on either side of the starting oxidation state. Manganate(VI) in acid gives permanganate (+7) and MnO₂ (+4).

Answer from the NTA final answer key. Solution by PrepWiser.