$B{{r}^{-}}$ is converted into $B{{r}_{2}}$ ​ by using
A. $C{{l}_{2}}$
B. conc. $HCl$
C. $HBr$
D. ${{H}_{2}}S$

Hint: It is clearly known that the reactivity of elements go down in a group. Similarly, the reactivity decreases as we go down the halogen group. Therefore, any element which is below in the periodic table than the top will be replaced by the element which is above.

Complete Step by Step Answer:
Bromine is a halogen element. It generally exists as a yellow colored liquid. Chlorine is more reactive than bromine but less reactive than fluorine so, it can easily remove bromine from sodium bromide solution and convert bromide ion to bromine gas. Chlorine gas reacts with the salt of bromine that is sodium bromide to produce a salt of chlorine that is sodium chloride along with bromine gas as products. The reaction can be given as follows:
$C{{l}_{2}}+2NaBr\to 2NaCl+B{{r}_{2}}$

Thus in this reaction one mole of Chlorine gas reacts with two moles of the salt of bromine that is sodium bromide to produce two moles of a salt of chlorine that is sodium chloride along with one mole of bromine gas as products.
Thus we can write that $B{{r}^{-}}$ is converted into $B{{r}_{2}}$ ​ by using $C{{l}_{2}}$ .
Thus the correct option is A.

Note: Among the halogens fluorine is most reactive due to small size and high electronegativity. Iodine is least reactive due to large size and lowest electronegativity.