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.


1 Answer 1


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! (:

  • $\begingroup$ I agree with you, but is not the Iodine free radical more radioactive than Fluorine one? I mean, Iodine can react with any Hydrogen, meanwhile Fluorine is selective. Thanks. $\endgroup$
    – Philip
    Apr 25, 2017 at 5:23
  • $\begingroup$ Radioactivity is from the other chapter of the book and has nothing in common with radicals. And about reactivity ... well I would say it will be completely opposite to your statement, which is kind of vague btw. I highly recommend to you go through general and inorganic chemistry to sort the major principles out. Btw may you please answer my previous question? Will appreciate it. $\endgroup$
    – Olena K.
    Apr 25, 2017 at 16:41

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