To determine which halogens among $$\mathrm{F}_2, \mathrm{Cl}_2, \mathrm{Br}_2$$, and $$\mathrm{I}_2$$ can undergo disproportionation reactions, we need to understand what a disproportionation reaction is.

In a disproportionation reaction, a single substance is simultaneously oxidized and reduced, forming two different products. The halogen X₂ disproportionates in water as follows:

$$ \mathrm{X}_2 + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{HX} + \mathrm{HXO} $$

Here, X represents a halogen.

Let's analyze each halogen to see if disproportionation is possible:

Fluorine ($$\mathrm{F}_2$$): Fluorine is the most electronegative element and has a very high oxidation potential. It does not undergo disproportionation because it prefers to remain in the $$-1$$ oxidation state and does not form higher oxidation states easily. Therefore, $$\mathrm{F}_2$$ cannot undergo a disproportionation reaction.

Chlorine ($$\mathrm{Cl}_2$$): Chlorine can undergo disproportionation. In water, chlorine disproportionates as follows:

$$ \mathrm{Cl}_2 + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{HCl} + \mathrm{HOCl} $$

Here, chlorine is both reduced to $$\mathrm{HCl}$$ (chloride, -1 oxidation state) and oxidized to $$\mathrm{HOCl}$$ (hypochlorite, +1 oxidation state).

Bromine ($$\mathrm{Br}_2$$): Bromine can also undergo disproportionation. In water, bromine disproportionates as follows:

$$ \mathrm{Br}_2 + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{HBr} + \mathrm{HOBr} $$

Here, bromine is both reduced to $$\mathrm{HBr}$$ (bromide, -1 oxidation state) and oxidized to $$\mathrm{HOBr}$$ (hypobromite, +1 oxidation state).

Iodine ($$\mathrm{I}_2$$): Iodine can also undergo disproportionation. In alkaline solutions, iodine disproportionates as follows:

$$ \mathrm{I}_2 + \mathrm{OH}^- \rightarrow \mathrm{I}^- + \mathrm{IO}^- $$

Here, iodine is both reduced to $$\mathrm{I}^-$$ (iodide, -1 oxidation state) and oxidized to $$\mathrm{IO}^-$$ (hypoiodite, +1 oxidation state).

Based on the above analysis, all halogens except $$\mathrm{F}_2$$ can undergo disproportionation. Therefore, the correct answer is:

Option B

$$\mathrm{Cl}_2, \mathrm{Br}_2$$ and $$\mathrm{I}_2$$