CBSE Class 12 · 2019 · d and f Block ElementsShort Answer

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When MnO₂ is fused with KOH in the presence of KNO₃ as an oxidising agent, it gives a dark green compound (A). Compound (A) disproportionates in acidic solution to give a purple compound (B). An alkaline solution of compound (B) oxidises KI to compound (C), whereas an acidified solution of compound (B) oxidises KI to (D). Identify (A), (B), (C) and (D).

Answer: (A) K₂MnO₄, (B) KMnO₄, (C) KIO₃, (D) I₂.

Solution

Fusion:

2Mn O_{2} + 4KOH + O_{2} KN O_{3}, Δ 2 K_{2} Mn O_{4} + 2 H_{2} O

→ (A) = K

_{2}

MnO

_{4}

(dark green).
Disproportionation in acid:

3Mn O_{4}^{2 -} + 4 H^{+} \to 2Mn O_{4}^{-} + Mn O_{2} + 2 H_{2} O

→ (B) = KMnO

_{4}

(purple).
Alkaline KMnO

_{4}

+ KI:

2Mn O_{4}^{-} + H_{2} O + I^{-} \to 2Mn O_{2} + 2O H^{-} + I O_{3}^{-}

→ (C) = KIO

_{3}

.
Acidified KMnO

_{4}

+ KI:

2Mn O_{4}^{-} + 10 I^{-} + 16 H^{+} \to 2M n^{2 +} + 5 I_{2} + 8 H_{2} O

→ (D) = I

_{2}

CBSE Class 12 Board · 2019 · Subjective

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CBSE Class 12 · 2019 · d and f Block ElementsShort Answer

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