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प्रश्न
When an oxide of manganese (A) is fused with KOH in the presence of an oxidising agent and dissolved in water, it gives a dark green solution of compound (B). Compound (B) disproportionates in neutral or acidic solution to give purple compound (C). An alkaline solution of compound (C) oxidises potassium iodide solution to a compound (D) and compound (A) is also formed. Identify compounds A to D and also explain the reactions involved.
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उत्तर
The compounds (A), (B), (C) and (D) are given as under:
A = \[\ce{MnO2}\]
B = \[\ce{K2MnO4}\]
C = \[\ce{KMnO4}\]
D = \[\ce{KlO3}\]
The reactions are explained as under:
\[\ce{\underset{(A)}{2MnO2} + 4KOH + O2 -> \underset{(B)}{2K2MnO4} + 2H2O}\]
\[\ce{3MnO^{2-}4 + 4H^+ -> \underset{(C)}{2MnO^{-}4} + MnO2 + 2H2O}\]
\[\ce{2MnO^{-}4 + H2O + Kl -> \underset{(A)}{2MnO2} + 2OH^{-} + \underset{(D)}{KlO3}}\]
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संबंधित प्रश्न
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| Column I (Catalyst) | Column II (Process) |
| (i) \[\ce{Ni}\] in the presence of hydrogen | (a) Zieglar Natta catalyst |
| (ii) \[\ce{Cu2C12}\] | (b) Contact process |
| (iii) \[\ce{V2O5}\] | (c) Vegetable oil to ghee |
| (iv) Finely divided iron | (d) Sandmeyer reaction |
| (v) \[\ce{TiCl4 + Al (CH3)3}\] | (e) Haber's Process |
| (f) Decomposition of KCIO3 |
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- Why Manganese has lower melting point than Chromium?
- Why do transition metals of 3d series have lower melting points as compared to 4d series?
- In the third transition series, identify and name the metal with the highest melting point.
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[Ti(H2O)]3+ is coloured while [Sc(H2O)6]3+ is colourless.
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Ce4+ is a strong oxidising agent.
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\[\ce{Fe^{3+}_{ (aq)} + e^- -> Fe^{2+}_{ (aq)}}\], E0 = +0.77 V
\[\ce{MnO^{-4}_{ (aq)} + 8H^+ + 5e^- -> Mn^{2+}_{ (aq)} + 4H2O_{(l)}}\], E0 = +1.51 V
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Transition metals possess a great tendency to form complex compounds.
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29, 59, 74, 95, 102, 104
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