Activation energy of electron transfers

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The discussion centers on the reactivity of elements in groups one and two of the periodic table, highlighting that increased reactivity is linked to decreasing ionization energies as outer electrons are further from the nucleus. The conversation also explores why fluorine is more reactive than oxygen in terms of electron gain, attributing this to fluorine's lower activation energy for gaining an electron, which is influenced by effective nuclear charge and atomic radius. The effective nuclear charge concept explains that heavier metals have more electron shielding, making it easier to lose valence electrons, while fluorine's minimal shielding leads to a strong attraction for additional electrons, resulting in high electron affinity. The complexities of predicting activation energies for electron gain reactions are acknowledged, with a reference to Marcus theory as a framework for understanding electron transfer rates.
Moogie
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Hi

As you progress down groups one and two, the elements become more reactive. This is because the ionisation energies of the elements decreases [outer electron further from the nucleus] so the proportion of colliding molecules that have sufficient energy to overcome the activation energy for the reaction increases.

What about reactions involving electron gain? Why is fluorine more reactive than oxygen? I am presuming it is something to do with the gain of an electron by fluorine having a lower activation energy than electron gain by oxygen. But what factors influence the activation energy of electron gain?

thanks
 
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Both cases can be explained using effective nuclear charge idea. In the case of metals losing electron the heavier the metal, the more electrons screen the outermost shell, making valence electrons easier to remove. In the case of fluorine there are not many electrons shielding the nucleus, so the attraction of additional electron is high (electron affinity is high).

You can also think in terms of radius. Force with which electron is attracted by nucleus is inversely proportional to distance squared. So even ignoring electrons between the nucleus and valence electron, the higher the atom radius, the lower the attraction (hence low ionization energy of heavy alkali metals), the lower the radius, the higher the attraction (hence high electron affinity of fluorine).

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Thanks for your reply. I do understand those general principals regarding Zeff. With the information you have provided I can see how the end product is more stable and lower in energy but I'm not sure what affects the activation energy /transition state of an electron gain reaction .
 
Moogie said:
With the information you have provided I can see how the end product is more stable and lower in energy but I'm not sure what affects the activation energy /transition state of an electron gain reaction .

Activation energies are, as a rule, much more difficult to predict than the overall reaction thermodynamics. In any case, the general descriptive framework for ET rates is that of http://en.wikipedia.org/wiki/Marcus_theory" . Unfortunately the Wikipedia article on the subject isn't that great. But at least you can get the idea.
 
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thanks, will look into it
 
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