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When halogens have a low Bond dissociation energy, BDE, their free radicals are stable.
$\ce{Cl-Cl}$ (BDE: $\pu{58 kcal/mol}$) and $\ce{F-F}$ (BDE: $\pu{38 kcal/mol}$). $\ce{Cl}$ free radical is less stable than $\ce{F}$ free radical. But why that doesn't happen with iodine ($\ce{I-I}$, BDE: $\pu{36 kcal/mol}$). Iodine free radicals are unstable.
Where have you found the information that the iodine free radicals are unstable? It does not sound right for me.
BDEs of chlorine, bromine and iodine decrease down the group as the size of the atom increases: 243, 192, and 151 kJ/mol. The BDE of fluorine, is however, lower (158 kJ/mol) than those of chlorine and bromine. Here you were right! But the free radical stability depends also on other factors. It increases with increasing the atom size and with decreasing the electronegativity. Additionally the stability of the free radical decreases as the orbital is held closer to the nucleus. Thus, as we can expect, the iodide radical will be the most stable, and the fluoride - the least./
Btw correlation does not mean causality! (:
