Question

Equivalent weight of $Br _{2}$ is 96 in the following disproportionation reaction:
$Br _{2}+ OH ^{-} \rightarrow Br + H _{2} O + X$
The oxidised product is: $( Br =80)$

• A. $BrO ^{-}$
• B. $BrO _{2}^{-}$
• C. $BrO _{3}^{-}$
• D. $BrO _{4}^{-}$

Answer 1

Answer: (C)

Eq. wt. in disproportionation reaction

$=\frac{ M }{ n \text { factor }_{1}}+\frac{ M }{ n \text { factor }_{2}}$

$96=\frac{160}{2}+\frac{160}{ n \text { factor }_{2}}$

$ \Rightarrow 16=\frac{160}{ n \text { factor }_{2}}$

n factor $_{2}=10$

which is possible only if the oxidised product is $BrO _{3}^{-}$.