Among the following halogens

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

Which can undergo disproportionation reactions?

<p>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.</p> <p>In a disproportionation reaction, a single substance is simultaneously oxidized and reduced, forming two different products. The halogen X₂ disproportionates in water as follows:</p> <p> <p>$\mathrm{X}_2 + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{HX} + \mathrm{HXO}$</p> </p> <p>Here, X represents a halogen.</p> <p>Let's analyze each halogen to see if disproportionation is possible:</p> <p><strong>Fluorine ($\mathrm{F}_2$):</strong> 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.</p> <p><strong>Chlorine ($\mathrm{Cl}_2$):</strong> Chlorine can undergo disproportionation. In water, chlorine disproportionates as follows:</p> <p> <p>$\mathrm{Cl}_2 + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{HCl} + \mathrm{HOCl}$</p> </p> <p>Here, chlorine is both reduced to $\mathrm{HCl}$ (chloride, -1 oxidation state) and oxidized to $\mathrm{HOCl}$ (hypochlorite, +1 oxidation state).</p> <p><strong>Bromine ($\mathrm{Br}_2$):</strong> Bromine can also undergo disproportionation. In water, bromine disproportionates as follows:</p> <p> <p>$\mathrm{Br}_2 + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{HBr} + \mathrm{HOBr}$</p> </p> <p>Here, bromine is both reduced to $\mathrm{HBr}$ (bromide, -1 oxidation state) and oxidized to $\mathrm{HOBr}$ (hypobromite, +1 oxidation state).</p> <p><strong>Iodine ($\mathrm{I}_2$):</strong> Iodine can also undergo disproportionation. In alkaline solutions, iodine disproportionates as follows:</p> <p> <p>$\mathrm{I}_2 + \mathrm{OH}^- \rightarrow \mathrm{I}^- + \mathrm{IO}^-$</p> </p> <p>Here, iodine is both reduced to $\mathrm{I}^-$ (iodide, -1 oxidation state) and oxidized to $\mathrm{IO}^-$ (hypoiodite, +1 oxidation state).</p> <p>Based on the above analysis, all halogens except $\mathrm{F}_2$ can undergo disproportionation. Therefore, the correct answer is:</p> <p>Option B</p> <p><strong>$\mathrm{Cl}_2, \mathrm{Br}_2$ and $\mathrm{I}_2$</strong></p>