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(B) The oxidizing power of halogens decreases down the group: $F_2 > Cl_2 > Br_2 > I_2$.$A$ halogen with higher oxidizing power can displace a halogen

Vedclass · Accepted Answer

Only $I_2$ is liberated

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(B) The oxidizing power of halogens decreases down the group: $F_2 > Cl_2 > Br_2 > I_2$.$A$ halogen with higher oxidizing power can displace a halogen with lower oxidizing power from its salt solution.$Br_2$ is a stronger oxidizing agent than $I_2$,so it can displace $I^-$ ions from $NaI$ to form $I_2$.$2NaI + Br_2 \rightarrow 2NaBr + I_2$However,$Br_2$ is a weaker oxidizing agent than $F_2$ and $Cl_2$,so it cannot displace $F^-$ from $NaF$ or $Cl^-$ from $NaCl$.Therefore,only $I_2$ is liberated.

When $Br_2$ is treated with aqueous solutions of $NaF$,$NaCl$,and $NaI$ separately,what happens?

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