Question

In neutral medium, permanganate reacts with iodide to give:

• A. \(\mathrm{MnO_2}\), \(\mathrm{I_2}\)
• B. \(\mathrm{Mn^{2+}}\), \(\mathrm{I_2}\)
• C. \(\mathrm{Mn^{4+}}\), \(\mathrm{IO^-}\)
• D. \(\mathrm{Mn^{2+}}\), \(\mathrm{IO_3^-}\)

Hint:

In neutral medium, \(\mathrm{MnO_4^-}\) gives \(\mathrm{MnO_2}\); in acidic, it gives \(\mathrm{Mn^{2+}}\); in basic, it gives \(\mathrm{MnO_4^{2-}}\).

Answer 1

The Correct Option is A

- In neutral medium, permanganate ion \(\mathrm{(MnO_4^-)}\) is reduced to \(\mathrm{MnO_2}\) (manganese dioxide). - Iodide \(\mathrm{(I^-)}\) is oxidized to iodine \(\mathrm{(I_2)}\). - The balanced redox reaction in neutral medium is: \[ 2\mathrm{MnO_4^-} + 2\mathrm{H_2O} + 10\mathrm{I^-} \rightarrow 2\mathrm{MnO_2} + 5\mathrm{I_2} + 4\mathrm{OH^-} \] - Therefore, the products are \(\mathrm{MnO_2}\) and \(\mathrm{I_2}\).