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Why does fluorine not show disporportionation reaction? - Chemistry

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प्रश्न

Why does fluorine not show disporportionation reaction?

दीर्घउत्तर
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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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Oxidation Number - Types of Redox Reactions
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अध्याय 8: Redox Reactions - Multiple Choice Questions (Type - I) [पृष्ठ ११०]

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एनसीईआरटी एक्झांप्लर Chemistry [English] Class 11
अध्याय 8 Redox Reactions
Multiple Choice Questions (Type - I) | Q 35 | पृष्ठ ११०

संबंधित प्रश्न

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\[\ce{2K(s) + F2(g) → 2K+F– (s)}\]


Justify that the following reaction is redox reaction:

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


Identify the substance oxidised, reduced, oxidising agent and reducing agent for the following reaction:

\[\ce{HCHO(l) + 2[Ag (NH3)2]+(aq) + 3OH–(aq) → 2Ag(s) + HCOO–(aq) + 4NH3(aq) + 2H2O(l)}\]


Why does the following reaction occur?

\[\ce{XeO^{4-}_6 (aq) + 2F- (aq) + 6H+ (aq) -> XeO3(g) + F_2(g) + 3H_2O(l)}\]

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\[\ce{{(CN)}_{2(g)} + 2OH-_{(aq)} -> CN-_{(aq)} + CNO-_{(aq)} + H_2O_{(l)}}\]


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Al, Cu, Fe, Mg and Zn.


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

(i) Phosphorus is undergoing reduction only.

(ii) Phosphorus is undergoing oxidation only.

(iii) Phosphorus is undergoing oxidation as well as reduction.

(iv) Hydrogen is undergoing neither oxidation nor reduction.


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\[\ce{Mg + Fe^{2+} -> Mg^{2+} + Fe}\]


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\[\ce{H2O2 -> H2O + O2}\]

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