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)

$2NaOH_{(dilute)} + Br_2 \xrightarrow{\text{cold}} NaBrO + NaBr + H_2O$

$\underset{\text{Sodium hypobromide}}{3NaBrO } \xrightarrow{300\, K} NaBrO$

On acidification, the final mixture gives bromine.

$5NaBrO + NaBrO_3 + 6HCl \rightarrow 6NaCl + 3Br_2 + 3H_2O$

Thus, during the reaction, bromine is present in four different oxidation states i.e., zero in $Br_2, + 1$ in $NaBrO, -1$ in $NaBr$ and $+5$ in $NaBrO_3$.

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 $BrO^{-},\, Br^-$ and $BrO_3^{-}$ ions.