AP EAPCET 07-Jul-2022 Shift 2 Paper

Whenever one of the halogens is involved in oxidising a species in solution, the halogen is reduced to a halide ion associated with water molecules. The following reactions illustrate this process $\mathrm{F}_2(g) + 2e^{-} \rightarrow 2\mathrm{F}^{-}(aq)$ $\Rightarrow \mathrm{Cl}_2(g) + 2e^{-} \rightarrow 2\mathrm{Cl}^{-}(aq)$ $\mathrm{Br}_2(l) + 2e^{-} \rightarrow 2\mathrm{Br}^{-}(aq)$ $\mathrm{I}_2(g) + 2e^{-} \rightarrow 2\mathrm{I}^{-}(aq)$ Smaller the size of ion, higher is the energy released when the ion is hydrated. Since, the size of ions increases in the order $\mathrm{F}^{-} < \mathrm{Cl}^{-} < \mathrm{Br}^{-} < \mathrm{I}^{-}$. The order of hydration energy is $\mathrm{F}^{-}(aq) > \mathrm{Cl}^{-}(aq) > \mathrm{Br}^{-}(aq) > \mathrm{I}^{-}(aq)$ Because of very high hydration enthalpy of the fluoride ion, $\mathrm{F}_2$ gets very easily reduced followed by $\mathrm{Cl}_2$, then $\mathrm{Br}_2$ and $\mathrm{I}_2$. Therefore, the correct order of oxidising power of halogens is $\mathrm{F}_2 > \mathrm{Cl}_2 > \mathrm{Br}_2 > \mathrm{I}_2$.