QJEEM24Ch5f2f: JEE Main 2024 Chemistry - Class 12 JEEM24Ch5f2f Solution: Among the following halogens Which can undergo…

To determine which halogens among $\mathrm{F}_2, \mathrm{Cl}_2, \mathrm{Br}_2$ , and $\mathrm{I}_2$ can undergo disproportionation reactions, we need to understand what a disproportionation reaction is.

In a disproportionation reaction, a single substance is simultaneously oxidized and reduced, forming two different products. The halogen X₂ disproportionates in water as follows:

$\mathrm{X}_2 + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{HX} + \mathrm{HXO}$

Here, X represents a halogen.

Let's analyze each halogen to see if disproportionation is possible:

Fluorine ( $\mathrm{F}_2$ ): Fluorine is the most electronegative element and has a very high oxidation potential. It does not undergo disproportionation because it prefers to remain in the $-1$ oxidation state and does not form higher oxidation states easily. Therefore, $\mathrm{F}_2$ cannot undergo a disproportionation reaction.

Chlorine ( $\mathrm{Cl}_2$ ): Chlorine can undergo disproportionation. In water, chlorine disproportionates as follows:

$\mathrm{Cl}_2 + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{HCl} + \mathrm{HOCl}$

Here, chlorine is both reduced to $\mathrm{HCl}$ (chloride, -1 oxidation state) and oxidized to $\mathrm{HOCl}$ (hypochlorite, +1 oxidation state).

Bromine ( $\mathrm{Br}_2$ ): Bromine can also undergo disproportionation. In water, bromine disproportionates as follows:

$\mathrm{Br}_2 + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{HBr} + \mathrm{HOBr}$

Here, bromine is both reduced to $\mathrm{HBr}$ (bromide, -1 oxidation state) and oxidized to $\mathrm{HOBr}$ (hypobromite, +1 oxidation state).

Iodine ( $\mathrm{I}_2$ ): Iodine can also undergo disproportionation. In alkaline solutions, iodine disproportionates as follows:

$\mathrm{I}_2 + \mathrm{OH}^- \rightarrow \mathrm{I}^- + \mathrm{IO}^-$

Here, iodine is both reduced to $\mathrm{I}^-$ (iodide, -1 oxidation state) and oxidized to $\mathrm{IO}^-$ (hypoiodite, +1 oxidation state).

Based on the above analysis, all halogens except $\mathrm{F}_2$ can undergo disproportionation. Therefore, the correct answer is:

Option B

$\mathrm{Cl}_2, \mathrm{Br}_2$ and $\mathrm{I}_2$