Answering some kinetics questions I came across this statement:

Free radicals are more stable in the gas phase than ions.

Is this correct or not? Why? Is there a different trend in liquid or solid phase?

  • 4
    $\begingroup$ I'd assume that the statement refers to the fact that free radicals wouldn't be charged, hence no electrostatic attraction. $\endgroup$
    – MaxW
    Commented Apr 20, 2019 at 2:14
  • 4
    $\begingroup$ The word stable is meaningless without a context. It is misused a lot especially by organic chemists. Stable with respect to what? This should the question in our mind whenever we see this word. Free radicals are extremely reactive but some of them can survive in liquids with a very long lifetime. Our favorite O2 is also a diradical. $\endgroup$
    – ACR
    Commented Apr 20, 2019 at 3:01

1 Answer 1


It is difficult to comment on such a broad statement because often it depends on the specifics. So instead, I will talk about a single example: Will the HCl molecule dissociate into radicals (atoms, in this case) or ions?

$$\tag{1}\ce{H-Cl -> H+ + Cl-}$$

$$\tag{2}\ce{H-Cl -> H. + Cl.}$$

Under many circumstances, neither (1) nor (2) happens, H-Cl is the major species.

In water, where the dipole of water solvates chloride ions and the hydrogen ion "reacts" with water to form hydronium ions and large structures, HCl ionizes completely (strong acid). In other solvents, it ionizes less but would also act as acid if there is an appropriate base.

$$\ce{HCl(aq) + H2O(l) -> H3O+(aq) + Cl-(aq)}$$

In the gas phase, ions of opposite charge have a strong long-range attraction, so reaction (1) is unlikely. Reaction (2) is endothermic (bond dissociation energy is defined as homolytic cleavage into radicals, and it has a positive value). If you raise the temperature, you favor the products in reaction (2), so at higher temperatures, the concentration of radicals would increase.

To directly compare ions and radicals in the gas phase, you would consider the following reaction:

$$\tag{3}\ce{H. (g) + Cl. (g) <=> H+(g) + Cl-(g)}$$

Taking away an electron from hydrogen corresponds to the ionization energy of hydrogen atoms (1313. kJ/mol, i.e. it takes energy to remove the electron from the electrostatic field of the nucleus). Adding an electron to the chlorine atom corresponds to the electron affinity of chlorine atoms (-359 kJ/mol, i.e. energy is released when an additional electron encounters a neutral chlorine). Overall, the process is endothermic. Different from reactions (1) or (2), however, the number of reactant particles is equal to the number of product particles, so in this case, it is not clear if the change in entropy is positive or negative, but it won't dominate the reaction at modest temperature. So at room temperature, this reaction will not happen, the radicals are favored over the ions.

What is stabler in gas phase, free radicals or ions?

For the example I chose, free radicals are more stable than ions.

Is there a different trend in liquid or solid phase?

For the example I chose, it is the opposite in aqueous solution.


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