Match the condition for $KMnO X_{4}$ in Column I with the species formed or evolved in Column II. Column I: (P) acidic medium (Q) neutral medium (R) strongly alkaline medium (S) solid heated at 513 K Column II: (1) $Mn X^{2 +}$ (2) $MnO X_{2}$ (3) $MnO X_{4} X^{2 -}$ as the reduction product (4) $O X_{2}$ evolved

Answer: A

Acidic:

MnO X_{4} X^{-} \to Mn X^{2 +}

. Neutral:

MnO X_{4} X^{-} \to MnO X_{2}

. Strongly alkaline:

MnO X_{4} X^{-} + e^{-} \to MnO X_{4} X^{2 -}

. Heating:

2 KMnO X_{4} \to K X_{2} MnO X_{4} + MnO X_{2} + O X_{2}

. So P-1, Q-2, R-3, S-4.