Chemistry - Complete Pblock- Question with Solution | TestHub

${\mathrm{I}}_2$cannot liberate$\mathrm{B}{\mathrm{r}}_2$from KBr, since${\mathrm{I}}_2$is weaker oxidant than$\mathrm{B}{\mathrm{r}}_2.{\mathrm{F}}_2$and$\mathrm{C}{\mathrm{l}}_2$, being stronger oxidants than$\mathrm{B}{\mathrm{r}}_2$, can liberate$\mathrm{B}{\mathrm{r}}_2$from KBr.
${\mathrm{F}}_2+2\mathrm{K}\mathrm{B}\mathrm{r}\to 2\mathrm{K}\mathrm{F}+\mathrm{B}{\mathrm{r}}_2$
$\mathrm{C}{\mathrm{l}}_2+2\mathrm{K}\mathrm{B}\mathrm{r}\to 2\mathrm{K}\mathrm{C}\mathrm{l}+\mathrm{B}{\mathrm{r}}_2$
Reaction of conc.${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$with KBr forms HBr. HBr is a strong reducing agent. It reduces${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$to form a mixture of$\mathrm{S}{\mathrm{O}}_2$and$\mathrm{B}{\mathrm{r}}_2$.
$\mathrm{K}\mathrm{B}\mathrm{r}+{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4\to \mathrm{H}\mathrm{B}\mathrm{r}+\mathrm{K}\mathrm{H}\mathrm{S}{\mathrm{O}}_4$
$\mathrm{K}\mathrm{H}\mathrm{S}{\mathrm{O}}_4+\mathrm{K}\mathrm{B}\mathrm{r}\to \mathrm{H}\mathrm{B}\mathrm{r}+{\mathrm{K}}_2\mathrm{S}{\mathrm{O}}_4$
HBr so formed reduces${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$.
${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4+2\mathrm{H}\mathrm{B}\mathrm{r}\to \mathrm{S}{\mathrm{O}}_2+2{\mathrm{H}}_2\mathrm{O}+\mathrm{B}{\mathrm{r}}_2$