When a free radical shares the extra electron rather than stealing it outright, a new bond is formed. For example, two chlorine atoms react to form a Cl2 molecule. The bond allows a pair of electrons to spend time wedged between two nuclei, and their energy is lowered tremendously. Bond formation is exothermic; forming a new bond leads to a lower energy relative the energy of the separate free radical and the molecule it attacks.
O2 has two unpaired electrons- but if you look at its MO diagram, you can see that if you add additional electrons, you must put them into an antibonding orbital. And adding electrons to an antibonding orbital raises the energy of the molecule relative to that of the separated atoms. Pairing O2's unpaired electrons does not lead to a lower energy configuration.
Free radicals can be greatly stabilized if the odd electron is delocalized. Sharing electrons from other atoms mitigates the fact that an odd electron can provide only half the bonding a pair can provide. To see this, draw the structure of NO2, which is produced in large quantities in the atmosphere by reactions that occur in lightning bolts and automobile tailpipes. True, it's reactive- but it can persist for some time before reacting. Several resonance structures mean the odd electron is delocalized and contributes to the overall binding (and stability) of the molecule. Some of the free radicals found in smoke and cigarette tar are extremely stable for this reason.
To summarize, just having an odd electron doesn't make free radicals reactive. The odd electron does not have a mysterious 'urge' to pair. Free radicals are most reactive when
Copyright © 1997-2010 by Fred Senese
Comments & questions to fsenese@frostburg.edu
Last Revised 02/23/18.URL: http://antoine.frostburg.edu/chem/senese/101/bonds/faq/print-why-are-free-radicals-reactive.shtml