Among the following halogens F2, Cl2, Br2 and I2. Which can undergo disproportionation reactions?

Disproportionation requires the element to simultaneously increase and decrease its oxidation state. For halogens (X₂, OS = 0), they must be able to form both $\mathrm{X^-}$ (OS = $-1$) and $\mathrm{XO^-}$ or higher oxyanions (OS > 0).

Step 1 — $\mathrm{F_2}$: Fluorine is the most electronegative element and has no positive oxidation states. It cannot be oxidised to positive states → cannot disproportionate ✗

Example: $\mathrm{F_2 + H_2O \rightarrow HF + HOF}$ — but this is not disproportionation (F stays at 0 in HOF and goes to -1 in HF, but HOF has O at -1, not F at +1).

Step 2 — $\mathrm{Cl_2}$, $\mathrm{Br_2}$, $\mathrm{I_2}$: These halogens can form positive oxidation states (in oxyanions): $\mathrm{Cl_2 + 2OH^- \rightarrow Cl^- + ClO^- + H_2O}$ Cl: $0 \rightarrow -1$ (reduced) and $0 \rightarrow +1$ (oxidised) → Disproportionation ✓

Similarly $\mathrm{Br_2}$ and $\mathrm{I_2}$ can disproportionate in alkaline conditions.

Answer: Option (4) — $\mathrm{Cl_2}$, $\mathrm{Br_2}$ and $\mathrm{I_2}$