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A violet compound of manganese (A) decomposes on heating to liberate oxygen and compounds (B) and (C ) of manganese are formed. Compound (C ) reacts with KOH in the presence of potassium nitrate to give compound (B). On heating compound (C ) with conc. `H_(2)SO_(4)` and NaCl, chlorine gas is liberate and a compound (D) of manganese alongwith other products is formed. Identify compounds A to D and also explain the reaction involved. |
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Answer» Since, compound (C ) on treating with conc. `H_(2)SO_(4)` and NaCl gives `Cl_(2)` gas, so it is manganese dioxide `(MnO_(2))`. It is obtained alongwith `MnO_(4)^(2-)` when `KMnO_(4)` (violet) is heated. Thus, (A) `=KMnO_(4)" (B)"=K_(2)MnO_(4)` (C ) `=MnO_(2)" (D)"=MnCl_(2)` `underset([A])(KMnO_(4))overset(Delta)rarrunderset([B])(K_(2)MnO_(4))+underset([C])(MnO_(2))+O_(2)` `2MnO_(2)+4KOH + O_(2) rarr2K_(2)MnO_(4)+2H_(2)O` `MnO_(2) + 4NaCl+4H_(2)SO_(4)rarrunderset([D])(MnCl_(2))+4NaHSO_(4) + 2H_(2)O + Cl_(2)` |
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