1. Identify the type of reaction

The reaction is: $2\mathrm{Cu}^{2+}+4\mathrm{X}^- \rightarrow \mathrm{Cu}_2\mathrm{X}_2(s)+\mathrm{X}_2$

Here, halide ion $\mathrm{X}^-$ is doing two things:

• reducing $\mathrm{Cu}^{2+}$ to $\mathrm{Cu}^+$
• itself getting oxidized to $\mathrm{X}_2$

Also, $\mathrm{Cu}^+$ precipitates as $\mathrm{Cu}_2\mathrm{X}_2$.

2. Write the relevant half-reactions

Reduction of copper: $\mathrm{Cu}^{2+} + e^- \rightarrow \mathrm{Cu}^+$

Oxidation of halide: $2\mathrm{X}^- \rightarrow \mathrm{X}_2 + 2e^-$

Combining with precipitation of $\mathrm{Cu}^+$ as $\mathrm{Cu}_2\mathrm{X}_2$ gives the overall reaction.

3. Use standard electrode potentials

For copper: $\mathrm{Cu}^{2+} + e^- \rightarrow \mathrm{Cu}^+, \quad E^\circ = +0.15\,\text{V}$

For halogens:

• $\mathrm{Cl}_2 + 2e^- \rightarrow 2\mathrm{Cl}^-, \quad E^\circ = +1.36\,\text{V}$
• $\mathrm{Br}_2 + 2e^- \rightarrow 2\mathrm{Br}^-, \quad E^\circ = +1.09\,\text{V}$
• $\mathrm{I}_2 + 2e^- \rightarrow 2\mathrm{I}^-, \quad E^\circ = +0.54\,\text{V}$

For the halide to be oxidized, the corresponding oxidation potential is the negative of these values. So oxidation is easiest for $\mathrm{I}^-$, then $\mathrm{Br}^-$, and hardest for $\mathrm{Cl}^-$.

4. Check feasibility without precipitation effect

Cell potential for $2\mathrm{Cu}^{2+} + 2\mathrm{X}^- \rightarrow 2\mathrm{Cu}^+ + \mathrm{X}_2$

is: $E^\circ_{\text{cell}} = E^\circ(\mathrm{Cu}^{2+}/\mathrm{Cu}^+) - E^\circ(\mathrm{X}_2/\mathrm{X}^-)$

• For $\mathrm{Cl}^-$: $E^\circ = 0.15 - 1.36 = -1.21\,\text{V}$ Not feasible.

• For $\mathrm{Br}^-$: $E^\circ = 0.15 - 1.09 = -0.94\,\text{V}$ Not feasible.

• For $\mathrm{I}^-$: $E^\circ = 0.15 - 0.54 = -0.39\,\text{V}$ Still not feasible by standard potentials alone.

5. Include the precipitation effect

In the given reaction, $\mathrm{Cu}^+$ is removed from solution as insoluble $\mathrm{Cu}_2\mathrm{I}_2$ (i.e. $2\mathrm{CuI}$), which drives the reaction forward.

This effect is strong only for iodide because $\mathrm{CuI}$ is highly stable and very insoluble. Hence iodide can reduce $\mathrm{Cu}^{2+}$ to $\mathrm{Cu}^+$ with simultaneous formation of: $2\mathrm{Cu}^{2+} + 4\mathrm{I}^- \rightarrow 2\mathrm{CuI}(s) + \mathrm{I}_2$

Since $\mathrm{Cu}_2\mathrm{I}_2 = 2\mathrm{CuI}$, this matches the given form.

For chloride and bromide, this reaction does not occur under normal conditions because the driving force is insufficient.

6. Evaluate options

• A: All halogens — Incorrect
• B: Only Iodine — Correct
• C: Only Bromine — Incorrect
• D: Only Chlorine — Incorrect

Therefore, the correct answer is: $\boxed{\text{Only Iodine}}$