JEE Main 2024 — Chemistry Physical Chemistry

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)${\mathrm{Cu}}^{+}\rightarrow {\mathrm{Cu}}^{2+}+\mathrm{Cu}$Cu+→Cu2++Cu

1. $2{H}_{2}{\overset{-1}{O}}_{2}\overset{}{\rightarrow }{H}_{2}\overset{-2}{O}+{\overset{0}{O}}_{2}$2H2​O−12​→H2​O−2+O02​

$$3)3{\mathrm{Cl}}_{2}+6{\mathrm{OH}}^{-}\rightarrow 5{\mathrm{Cl}}^{-}+{{\mathrm{ClO}}_{3}}^{-}+3{H}_{2}O$$3)3Cl2​+6OH−→5Cl−+ClO3​−+3H2​O

$$4)2{\mathrm{NO}}_{2}(g)+2{\mathrm{OH}}^{-}(\mathrm{aq})\rightarrow {{\mathrm{NO}}_{2}}^{-}(\mathrm{aq})+{H}_{2}O(l)+{{\mathrm{NO}}_{3}}^{-}$$4)2NO2​(g)+2OH−(aq)→NO2​−(aq)+H2​O(l)+NO3​−

$$5){P}_{4}+3{\mathrm{OH}}^{-}+3{H}_{2}O\rightarrow {\mathrm{PH}}_{3}+3{H}_{2}{{\mathrm{PO}}_{2}}^{-}$$5)P4​+3OH−+3H2​O→PH3​+3H2​PO2​−

Here, the chlorine reactant is in oxidation state $0$0, and if we look at the products, the chlorine in the Cl$^{−}$− ion has an oxidation number of $-1$−1, where it has undergone reduction. Meanwhile, the oxidation number of the chlorine in the ${{{\mathrm{ClO}}_{3}}^{-}}_{}$ClO3​−​ ion is $+5$+5, which means it has been oxidised.

Hence, ${H}_{2}{O}_{2},{\mathrm{Cu}}^{+},{\mathrm{Cl}}_{2},{\mathrm{ClO}}_{3},{P}_{4},{\mathrm{NO}}_{2}$H2​O2​,Cu+,Cl2​,ClO3​,P4​,NO2​ shows disproportionation reaction.