MCQ
$B{r^ - }$ is converted into $B{r_2}$ by using
- ✓$C{l_2}$
- BConc. $HCl$
- C$HBr$
- D${H_2}S$
✓
Answer
Correct option: A.
$C{l_2}$
a
(a) $C{l_2} + 2B{r^ - }\, \to \,2C{l^ - } + B{r_2}$
(a) $C{l_2} + 2B{r^ - }\, \to \,2C{l^ - } + B{r_2}$
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