Q42 / 180 · AIIMS 2010 · Chemistry

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

  • A
    During the reaction bromine is present in four different oxidation states.
  • Correct answer
    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.

Solution

$$ \begin{aligned} & 2 \mathrm{NaOH}_{\text {(dilute) }}+\mathrm{Br}_2 \\ & \stackrel{\text { cold }}{\longrightarrow} \mathrm{NaBrO}+\mathrm{NaBr}+\mathrm{H}_2 \mathrm{O} \\ & \underset{\text { Sodium hypobromide }}{3 \mathrm{NaBrO}} \stackrel{300 \mathrm{~K}}{\longrightarrow} 3 \mathrm{NaBrO} \end{aligned} $$ On acidification, the final mixture gives bromine. $$ \begin{aligned} & 5 \mathrm{NaBrO}+\mathrm{NaBrO}_3+6 \mathrm{HCl} \longrightarrow \\ & 6 \mathrm{NaCl}+3 \mathrm{Br}_2+3 \mathrm{H}_2 \mathrm{O} \end{aligned} $$ 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.

Hence, the correct answer is option: 'B'.

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