Consider the change in oxidation state of bromine corresponding to different $E^{\circ }$ values as shown in the Latimer diagram below: $\mathrm{BrO}{}_4{}^{-}$ $\stackrel{1.82\text{V}}{\to }$ $\mathrm{BrO}{}_3{}^{-}$ $\stackrel{1.5\text{V}}{\to }$ $\mathrm{HBrO}$ $\stackrel{1.595\text{V}}{\to }$ $\mathrm{Br}{}_2$ $\stackrel{1.0652\text{V}}{\to }$ $\mathrm{Br}{}^{-}$ Then the species undergoing disproportionation is

Solution & Explanation

\textbf{Answer:} (d) $\mathrm{HBrO}$ \textbf{Solution:} A species disproportionates when the $E^{\circ }$ to its right (reduction) is greater than the $E^{\circ }$ to its left (the oxidation step), giving a positive cell potential. For $\mathrm{HBrO}$: $E_{\text{cell}}^{\circ }=E^{\circ }(\mathrm{HBrO}/\mathrm{Br}{}_2)-E^{\circ }(\mathrm{BrO}{}_3{}^{-}/\mathrm{HBrO})=1.595-1.5=+0.095\text{V}$. Since $E_{\text{cell}}^{\circ }$ is positive, the disproportionation of $\mathrm{HBrO}$ (Br in the +1 state going to both $\mathrm{Br}{}_2$ ($0$) and $\mathrm{BrO}{}_3{}^{-}$ ($+5$)) is feasible. Hence $\mathrm{HBrO}$ undergoes disproportionation, option (d).