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प्रश्न
Refer to the periodic table given in your book and now answer the following question:
Select three metals that can show disproportionation reaction.
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उत्तर
In disproportionation reactions, one of the reacting substances always contains an element that can exist in at least three oxidation states.
Mn, Cu, and Ga can show disproportionation reactions as these elements can exist in three or more oxidation states.
संबंधित प्रश्न
Justify that the following reaction is redox reaction:
\[\ce{CuO(s) + H2(g) → Cu(s) + H2O(g)}\]
Justify that the following reaction is redox reaction:
\[\ce{4 NH3(g) + 5 O2(g) → 4NO(g) + 6H2O(g)}\]
Fluorine reacts with ice and results in the change: \[\ce{H2O(s) + F2(g) → HF(g) + HOF(g)}\]
Justify that this reaction is a redox reaction.
Identify the substance oxidised, reduced, oxidising agent and reducing agent for the following reaction:
\[\ce{2AgBr (s) + C6H6O2(aq) → 2Ag(s) + 2HBr (aq) + C6H4O2(aq)}\]
Identify the substance oxidised, reduced, oxidising agent and reducing agent for the following reaction:
\[\ce{HCHO(l) + 2[Ag (NH3)2]+(aq) + 3OH–(aq) → 2Ag(s) + HCOO–(aq) + 4NH3(aq) + 2H2O(l)}\]
Consider the reactions:
\[\ce{2S_2O_3^{(2-)}(aq) + l_2(S) -> S_4O_6^{(2-)}(aq) + 2l-(aq)}\]
\[\ce{S_2O_3^{(2-)}(aq) + 2Br_2(l) + 5H_2O(l) -> 2SO_4^{2-} (aq) + 4Br-(aq) + 10H+ (aq)}\]
Why does the same reductant, thiosulphate react differently with iodine and bromine?
Why does the following reaction occur?
\[\ce{XeO^{4-}_6 (aq) + 2F- (aq) + 6H+ (aq) -> XeO3(g) + F_2(g) + 3H_2O(l)}\]
What conclusion about the compound Na4XeO6 (of which `"XeO"_6^(4+)` is a part) can be drawn from the reaction.
What sorts of informations can you draw from the following reaction ?
\[\ce{{(CN)}_{2(g)} + 2OH-_{(aq)} -> CN-_{(aq)} + CNO-_{(aq)} + H_2O_{(l)}}\]
Arrange the following metals in the order in which they displace each other from the solution of their salts.
Al, Cu, Fe, Mg and Zn.
Identify disproportionation reaction
Identify the correct statements with reference to the given reaction.
\[\ce{P4 + 3OH- + 3H2O -> PH3 + 3H2PO^{-}2}\]
(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.
\[\ce{MnO^{2-}4}\] undergoes disproportionation reaction in acidic medium but \[\ce{MnO^{-}4}\] does not. Give reason.
Assertion (A): Redox couple is the combination of oxidised and reduced form of a substance involved in an oxidation or reduction half cell.
Reason (R): In the representation \[\ce{E^Θ_{Fe^{3+}/Fe^{2+}}}\] and \[\ce{E^Θ_{Cu^{2+}/Cu}, Fe^{3+}/Fe^{2+}}\] and \[\ce{Cu^{2+}/Cu}\] are redox couples.
Write redox couples involved in the reactions given.
\[\ce{Cu + Zn^{2+} ->Cu^{2+} + Zn}\]
Find out the oxidation number of chlorine in the following compounds and arrange them in increasing order of oxidation number of chlorine.
\[\ce{NaClO4, NaClO3, NaClO, KClO2, Cl2O7, ClO3, Cl2O, NaCl, Cl2 , ClO2}\].
Which oxidation state is not present in any of the above compounds?
Which of the following statement is CORRECT for the decomposition reaction of KClO3?
\[\ce{2KClO3 → 2KCl +3O2}\]
The species given below that does NOT show a disproportionation reaction is ______.
\[\ce{H2O2 -> H2O + O2}\]
This represents ______.
In an experiment O3 undergo decomposition as \[\ce{O3 -> O2 + O}\] by the radiations of wavelength 310 Å. The total energy falling on the O3 gas molecules is 2.4 × 1026 eV and quantum yield of the reaction is 0.2.
The volume strength of the H2O2 solution which is obtained from reaction of 1 l H2O and nascent oxygen [O] obtained from the above reactions is (Assuming no change in volume of H2O)
\[\ce{H2O + O -> H2O2}\]
[Given: Na (Avogadro's No.) = 6 × 1023]
