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JEE Main 202430 JanuaryEvening Shift
Exam
JEE Main
Year
2024
Date
30 January
Shift
Evening Shift
Subject
Chemistry
Chapter
Redox Reactions

JEE Main 2024 · 30 January · Evening Shift · Chemistry · Redox Reactions

Total number of species from the following which can undergo...

Total number of species from the following which can undergo disproportionation reaction ___________.
H2O2,ClO3−,P4,Cl2,Ag,Cu+1,F2,NO2,K+{H}_{2}{O}_{2}, {ClO}_{3}^{-}, {P}_{4}, {Cl}_{2}, Ag, {Cu}^{+ 1}, {F}_{2}, {NO}_{2}, {K}^{+}

Correct answer

0

Step-by-step solution

The compounds or ions or elements in which central atom has intermediate oxidation state undergo a disproportionation reaction.

Disproportionation reactions are the reactions in which the same element/compound gets oxidised and reduced simultaneously.

1)Cu+→Cu2++Cu{Cu}^{+} \to {Cu}^{2 +} + Cu

2) 2H2O−12→H2O−2+O022{H}_{2}\overset{- 1}{O}2\overset{}{\to}{H}_{2}\overset{- 2}{O} + \overset{0}{O}2

3)3Cl2+6OH−→5Cl−+ClO3−+3H2O3 ) 3{Cl}_{2} + 6{OH}^{-} \to 5{Cl}^{-} + {{ClO}_{3}}^{-} + 3{H}_{2}O

4)2NO2(g)+2OH−(aq)→NO2−(aq)+H2O(l)+NO3−4 ) 2{NO}_{2} ( g ) + 2{OH}^{-} ( aq ) \to {{NO}_{2}}^{-} ( aq ) + {H}_{2}O ( l ) + {{NO}_{3}}^{-}

5)P4+3OH−+3H2O→PH3+3H2PO2−5 ) {P}_{4} + 3{OH}^{-} + 3{H}_{2}O \to {PH}_{3} + 3{H}_{2}{{PO}_{2}}^{-}

Here, the chlorine reactant is in oxidation state 00, and if we look at the products, the chlorine in the Cl− ion has an oxidation number of −1- 1, where it has undergone reduction. Meanwhile, the oxidation number of the chlorine in the ClO3−{{ClO}_{3}}^{-} ion is +5+ 5, which means it has been oxidised.

Hence, H2O2,Cu+,Cl2,ClO3,P4,NO2{H}_{2}{O}_{2}, {Cu}^{+}, {Cl}_{2}, {ClO}_{3}, {P}_{4}, {NO}_{2} shows disproportionation reaction.

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