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)

$2NaO{H}_{(dilute)}+B{r}_2\stackrel{\text{cold}}{\to }NaBrO+NaBr+H_{2}O$

$\underset{\text{Sodium hypobromide}}{3NaBrO}\stackrel{300K}{\to }NaBrO$

On acidification, the final mixture gives bromine.

$5NaBrO+NaBr{O}_3+6HCl\to 6NaCl+3B{r}_2+3H_{2}O$

Thus, during the reaction, bromine is present in four different oxidation states i.e., zero in $B{r}_2,+1$ 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, $B{r}_2$ is formed and disproportionation of $B{r}_2$ occurs during the reaction giving $Br{O}^{-},B{r}^{-}$ and $Br{O}_3^{-}$ ions.