High e-e repulsion = high electron affinity?

8/19/2018 4:13:09 PM
One of the questions in the Chemistry section is about "Oxygen is the most electronegative element of group 6A. However it also has the lowest electron affinity in this group. explain why?"
The answer to this question is that oxygen's valance electrons experience more electron-electron repulsion than other group 6A elements, and further explains that small atoms (ex: oxygen, fluorine) have several electrons crowding around a small nucleus and thus have greater electron-e repulsion, which is why it's more difficult to add an electron to oxygen than to a larger atom with lower repulsion forces.

The explanation makes perfect sense to me, but my understanding is that electron affinity increases when it's harder to add more electron to an element. Here since it's harder to add another electron to oxygen, wouldn't the electron affinity of oxygen be HIGHER, which also makes sense in consideration of the trend of electron affinity (increases going up and to the right)? Can someone please explain to me why there's such discrepancy in the question "lowest electron affinity of oxygen" and the "more electron-electron repulsion of oxygen?"

Thank you so much

8/22/2018 5:27:31 PM
I'm not sure mathematically, but conceptually, it makes sense to me that if it's easier to add an electron, then the affinity would be higher (the atom is more likely to pick up an electron). So if there is more repulsion, that would make the atom less likely to pick up an electron and thus a smaller affinity ("like") for electrons due to that repulsion.

Sorry if this didn't answer your question but I hope it helps.

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