2
$\begingroup$

I've read that oxygen has a lower electron affinity (as shown the picture below), because it has a smaller atomic radius than sulfur and thus the electrons experience significant electron-electron repulsion.

However, I was wondering... Why is this not the case for sulfur and selenium or chlorine and bromine? Because of sulfur's and chlorine's smaller atomic radius compared to selenium and bromine, the incoming electron should also experience a slight electron-electron repulsion (just like it does with oxygen), so $\ce{Se}$ and $\ce{Br}$ should have a higher electron affinity. enter image description here

$\endgroup$

1 Answer 1

4
$\begingroup$

This is likely due to second period elements' being quite small, so electron-electron repulsion is much more significant than in a third period element. The general trend is that EA is more positive as you move down the periodic table since effective nuclear charge is the same, but the electrons are farther away from the nucleus. Maybe it's easier to consider the second period the anomaly...

$\endgroup$
5
  • 1
    $\begingroup$ Then is the 3rd period bigger because of the d orbitals or something? $\endgroup$ Dec 21, 2016 at 2:57
  • $\begingroup$ Not sure what you mean by bigger. Are you talking about the atoms? They're bigger because the electrons are in higher energy (and thus "bigger") orbitals. $\endgroup$
    – Zhe
    Dec 21, 2016 at 4:20
  • $\begingroup$ Yes, but I still don't get why the 2nd period atoms are the anomaly. Shouldn't the same apply to the third period atoms since they are also smaller than the 4th period atoms? $\endgroup$ Dec 21, 2016 at 20:27
  • $\begingroup$ For example, Chlorine atom is smaller than Bromine, so why doesn't its electron -electron repulsion repel the incoming atom just like it does for Flourine? What I'm saying is, Chlorine should have a lower EA than Bromine, Bromine should have a lower EA than Iodine, etc for the same reason Flourine has a lower EA than Chlorine, which is because of the electron-repulsion. $\endgroup$ Dec 21, 2016 at 20:35
  • $\begingroup$ See, that's the great thing about chemistry. You have two competing factors. One factor (the small size) is much more significant in period 2. In all other cases, it's the fact that the orbital sizes are larger. It's important to consider that these effects come from slightly different reasons. One is the electron-electron repulsion, which is less as the atom increases in size because of the increased distances, but also because of increased polarizability. The other is simply the decrease in stabilization due to larger (higher energy) orbitals. $\endgroup$
    – Zhe
    Dec 21, 2016 at 21:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.