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MnOX42− undergoes disproportionation reaction in acidic medium but MnOX4− does not. Give reason.

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प्रश्न

\[\ce{MnO^{2-}4}\] undergoes disproportionation reaction in acidic medium but \[\ce{MnO^{-}4}\] does not. Give reason.

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उत्तर

In \[\ce{MnO^{2-}4}\], Mn is in highest oxidation state that is +7 hence here manganese cannot undergo oxidation that is why disproportionate is not possible whereas in \[\ce{MnO^{2-}4}\] manganese is in +6 oxidation state which can be oxidized as well as reduced.

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अध्याय 8: Redox Reactions - Multiple Choice Questions (Type - I) [पृष्ठ १०७]

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एनसीईआरटी एक्झांप्लर Chemistry Exemplar [English] Class 11
अध्याय 8 Redox Reactions
Multiple Choice Questions (Type - I) | Q 18 | पृष्ठ १०७

संबंधित प्रश्न

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\[\ce{4 NH3(g) + 5 O2(g) → 4NO(g) + 6H2O(g)}\]


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Justify that this reaction is a redox reaction.


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  1. \[\ce{H3PO2(aq) + 4 AgNO3(aq) + 2 H2O(l) → H3PO4(aq) + 4Ag(s) + 4HNO3(aq)}\]
  2. \[\ce{H3PO2(aq) + 2CuSO4(aq) + 2 H2O(l) → H3PO4(aq) + 2Cu(s) + H2SO4(aq)}\]
  3. \[\ce{C6H5CHO(l) + 2[Ag (NH3)2]+(aq) + 3OH–(aq) → C6H5COO–(aq) + 2Ag(s) + 4NH3 (aq) + 2 H2O(l)}\]
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\[\ce{{(CN)}_{2(g)} + 2OH-_{(aq)} -> CN-_{(aq)} + CNO-_{(aq)} + H_2O_{(l)}}\]


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\[\ce{P4 + 3OH- + 3H2O -> PH3 + 3H2PO^{-}2}\]

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(ii) Phosphorus is undergoing oxidation only.

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