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(C) The ability of halogens to act as oxidizing agents decreases down the group $(F_2 > Cl_2 > Br_2 > I_2)$.$Br_2$ can only oxidize the iodide ion $(I

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Only $I_2$ is liberated

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(C) The ability of halogens to act as oxidizing agents decreases down the group $(F_2 > Cl_2 > Br_2 > I_2)$.$Br_2$ can only oxidize the iodide ion $(I^-)$ to iodine $(I_2)$ because $I^-$ is a stronger reducing agent than $Br^-$,$Cl^-$,and $F^-$.$Br_2$ cannot oxidize $Cl^-$ or $F^-$ because $Br_2$ is a weaker oxidizing agent than $Cl_2$ and $F_2$.Therefore,the reaction is: $2NaI + Br_2 	o 2NaBr + I_2$.

When $Br_2$ is treated with aqueous solutions of $NaF$,$NaCl$,and $NaI$ separately,which of the following occurs?

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