Question

Using the following Latimer diagram for bromine, pH = 0;

$Br{O}_4^{-}$ $\stackrel{1.82V}{\to }$ $Br{O}_3^{-}\stackrel{1.50V}{\to }$ $HBrO\stackrel{1.595V}{\to }$ $B{r}_2\stackrel{1.06552}{\to }B{r}^{-}$

the species undergoing disproportionation is:

• A. $Br{O}_4^{-}$
• B. $Br{O}_3^{-}$
• C. $HBrO$
• D. $B{r}_2$

Answer 1

The correct option is C $HBrO$

$Br{O}_4^{-}\stackrel{1.82V}{\to }Br{O}_3^{-}\stackrel{1.50V}{\to }HBrO\stackrel{1.595V}{\to }B{r}_2$

Assigning the oxidation state of Br in each molecule using oxidation number of $O=-2,H=+1$

$\stackrel{+7}{Br}{O}_4^{-}\stackrel{1.82V}{\to }\stackrel{+5}{Br}{O}_3^{-}\stackrel{1.50V}{\to }H\stackrel{+1}{Br}O\stackrel{1.595V}{\to }\stackrel{0}{B{r}_2}$

Since the oxidation potential of HBrO is less than that of the reduction potential i.e left side potential is less than the right side potential of the conversion. Thus HBrO is the species that will undergo disproportionation.

Hence, the correct option is $C$