Question

Bromine is added to cold dilute aqueous solution of $NaOH$. The mixture is boiled. Which of the following statements is not true?

• A. During the reaction bromine is present in four different oxidation states
• B. The greatest difference between the various oxidation states of bromine is $5$
• C. On acidification of the final mixture bromine is formed
• D. Disproportionation of bromine occurs during the reaction

Answer 1

Answer: (B)

The greatest difference between the various oxidation states of bromine is $5$

On acidification the final mixture gives bromine:

$5NaBrO+NaBr{O}_3+6HCl⟶6NaCl+3{Br}_2+3H_{2}O$

Thus, during the reaction, bromine is present in four different oxidation states ie., zero in ${Br}_2$ $,+1$ in $NaBrO$ $,-1$ in $NaBr$ and $,+5$ in $NaBr{O}_3$.

The greatest difference between various oxidation states of bromine is $6$ and not $5$. On acidification of the final mixture, ${Br}_2$ is formed and disproportionation of ${Br}_2$ occurs during the reaction giving $Br{O}^{-}$, ${Br}^{-}$ and $Br{O}_3^{-}$ ions.

Hence, option $B$ is correct.