Chemistry - Complete Pblock- Question with Solution | TestHub

$$\begin{array}{rl}& 2{\mathrm{NaOH}}_{\text{(dilute) }}+{\mathrm{Br}}_2\\ & \stackrel{\text{ cold }}{⟶}\mathrm{NaBrO}+\mathrm{NaBr}+{\mathrm{H}}_2\mathrm{O}\\ & \underset{\text{ Sodium hypobromide }}{3\mathrm{NaBrO}}\stackrel{300\mathrm{K}}{⟶}3\mathrm{NaBrO}\end{array}$$ On acidification, the final mixture gives bromine. $$\begin{array}{rl}& 5\mathrm{NaBrO}+{\mathrm{NaBrO}}_3+6\mathrm{HCl}⟶\\ & 6\mathrm{NaCl}+3{\mathrm{Br}}_2+3{\mathrm{H}}_2\mathrm{O}\end{array}$$ Thus, during the reaction, bromine is present in four different oxidation states i.e., zero in ${\mathrm{Br}}_2,+1$ in $\mathrm{NaBrO},-1$ in $\mathrm{NaBr}$ and +5 in ${\mathrm{NaBrO}}_3$. The greatest difference between various oxidation states of bromine is 6 and not 5 . On acidification of the final mixture, ${\mathrm{Br}}_2$ is formed and disproportionation of ${\mathrm{Br}}_2$ occurs during the reaction giving ${\mathrm{BrO}}^{-},{\mathrm{Br}}^{-}$ and ${\mathrm{BrO}}_3^{-}$ions.