Question

When ${ Br }_{ 2 }$ is treated with aqueous solutions of NaF, NaCl, NaI separately:
A. ${ F }_{ 2 },{ Cl }_{ 2 }{ and I }_{ 2 }$ are liberated
B. Only ${ F }_{ 2 }{ ,Cl }_{ 2 }$ are liberated
C. Only ${ I }_{ 2 }$ is liberated
D. Only ${ Cl }_{ 2 }$ is liberated

Answer 1

Hint: The more reactive halogen can displace the less reactive halogen from a compound. In the halogen series, the reactivity of the halogens decreases down the group.

Complete step-by-step answer:
The reactivity of halogens is in the order of ${ F }_{ 2 }{ >Cl }_{ 2 }{ { >Br }_{ 2 }>I }_{ 2 }$

So, when ${ Br }_{ 2 }$ is treated with aqueous solutions of NaF and NaCl, Br being less reactive than F and Cl, is not capable of carrying out a displacement reaction.
When NaI is treated with ${ Br }_{ 2 }$, halogen displacement becomes possible as Br is more reactive than I.
The following reaction will take place:
${ NaI+Br }_{ 2 }{ \rightarrow NaBr+I }_{ 2 }$
As we can see that ${ I }_{ 2 }$ is being liberated in this reaction, the correct option is C.

Additional Information:
1) All the halogens are quite reactive and react with a variety of metals and nonmetals.
2) The reactivity of halogen is due to the presence of ${ 7 }$ electrons in the valence shell.
3) Hence, in order to complete their octet, they readily accept an electron.
4) The reactivity of halogen decreases down the group (note: exceptions are present).

Hence, the trend for reactivity is as follows:
${ F }_{ 2 }{ >Cl }_{ 2 }{ { >Br }_{ 2 }>I }_{ 2 }$

Note: The possibility to make a mistake is that reactivity decreases down the group, not increases. Here, ${ Br }_{ 2 }$ is a stronger oxidizing agent displaces weaker oxidizing agents ${ I }_{ 2 }$ from its salt.