Bromine is added to cold dilute aqueous solution of $\mathrm{NaOH}$. The mixture is boiled. Which of the…
- During the reaction bromine is present in four different oxidation states.
- The greatest difference between the various oxidation states of bromine is 5 .
- On acidification of the final mixture bromine is formed.
- Disproportionation of bromine occurs during the reaction.
Solution
\(5 \mathrm{NaBrO}+\mathrm{NaBrO}_3+6 \mathrm{HCl} \longrightarrow 6 \mathrm{NaCl}+3 \mathrm{Br}_2+3 \mathrm{H}_2 \mathrm{O}\) 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.
Asked in: NEET 2018
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