The correct order of bond dissociation enthalpy of halogens is:

The bond dissociation enthalpy of halogens depends on the size of the atoms and the effectiveness of orbital overlap.

For $$F_2$$, despite its small size, the bond dissociation enthalpy is anomalously low (159 kJ/mol) because the lone pairs on the two small fluorine atoms are very close together, leading to strong electron-electron repulsion that weakens the bond.

$$Cl_2$$ has the highest bond dissociation enthalpy (242 kJ/mol) among the halogens because chlorine atoms are large enough to avoid significant lone pair repulsion while still being small enough for effective orbital overlap.

$$Br_2$$ has a bond dissociation enthalpy of 193 kJ/mol, lower than $$Cl_2$$ because of the larger atomic size leading to less effective overlap.

$$I_2$$ has the lowest bond dissociation enthalpy (151 kJ/mol) due to its very large atomic size and poor orbital overlap.

So the order is: $$Cl_2 (242) > Br_2 (193) > F_2 (159) > I_2 (151)$$, which corresponds to option (4).

Therefore, the correct answer is option (4): $$Cl_2 > Br_2 > F_2 > I_2$$.