When \(Br_2\) is treated with aqueous solution of \(NaF, NaCl, NaI\) separately

Only \(I_2\) is liberated

The correct answer is option 4. Only \(I_2\) is liberated.

Bromine is a halogen, and halogens are very reactive elements. They can displace other halogens from their compounds. The order of reactivity of halogens is:

\[
\begin{align*}
F_2 &> Cl_2 > Br_2 > I_2
\end{align*}
\]

This means that fluorine is the most reactive halogen and iodine is the least reactive halogen.
When bromine gas is treated with aqueous solutions of sodium fluoride (\(NaF\)), sodium chloride (\(NaCl\)), and sodium iodide (\(NaI\)) separately, the following reactions will occur:

\[
\begin{align*}
2NaF + Br_2 &\rightarrow \text{No reaction} \\
2NaCl + Br_2 &\rightarrow \text{No reaction} \\
2NaI + Br_2 &\rightarrow 2NaBr + I_2
\end{align*}
\]

As you can see, bromine can only displace iodine from its compound. This is because bromine is more reactive than iodine.

Therefore, the correct answer is (4) Only \(I_2\) is liberated.