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प्रश्न
Why does fluorine not show disporportionation reaction?
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उत्तर
Disproportionate is defined as the reaction in which one compound of intermediate oxidation state converts to two compounds, one of higher and one of lower oxidation states So, to occur such type of redox reaction, the element should exist in at least three oxidation states. So that element present in the intermediate state and it can change to both higher and lower oxidation state during disproportionate reaction. Fluorine is the most electronegative element and a strong oxidizing agent and is the smallest in size of all the halogens. It does not show a positive oxidation state (shows only −1 oxidation state) and hence, does not undergo disproportionate reaction.
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संबंधित प्रश्न
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\[\ce{Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)}\]
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\[\ce{4 NH3(g) + 5 O2(g) → 4NO(g) + 6H2O(g)}\]
Suggest a list of the substances where carbon can exhibit oxidation states from –4 to +4 and nitrogen from –3 to +5.
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\[\ce{2AgBr (s) + C6H6O2(aq) → 2Ag(s) + 2HBr (aq) + C6H4O2(aq)}\]
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\[\ce{HCHO(l) + 2[Ag (NH3)2]+(aq) + 3OH–(aq) → 2Ag(s) + HCOO–(aq) + 4NH3(aq) + 2H2O(l)}\]
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\[\ce{N2H4(l) + 2H2O2(l) → N2(g) + 4H2O(l)}\]
Identify the substance oxidised, reduced, oxidising agent and reducing agent for the following reaction:
\[\ce{Pb(s) + PbO2(s) + 2H2SO4(aq) → 2PbSO4(s) + 2H2O(l)}\]
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\[\ce{P4 + 3OH- + 3H2O -> PH3 + 3H2PO^{-}2}\]
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\[\ce{Fe + Cd^{2+} -> Cd + Fe^{2+}}\]
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