aque1408
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Why is the first electron affinity for atoms that normally form negative ions exothermic and the second electron affinity endothermic?

For the second electron affinity, apparently it is endothermic due to the energy taken in to overcome the repulsive forces between the negative ion and the electron. Would there still not be a negative ion and an electron in the first electron affinity for repulsive forces to form?
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username5211340
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No, because in the first electron affinity you're adding an electron to a neutral atom, when an atom gains that electron it becomes an Anion and is therefore negatively charged. If you want to add another electron to it, you need to overcome the repulsion because the ion and electron have the same charges.
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aque1408
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(Original post by Chemastronomical)
No, because in the first electron affinity you're adding an electron to a neutral atom, when an atom gains that electron it becomes an Anion and is therefore negatively charged. If you want to add another electron to it, you need to overcome the repulsion because the ion and electron have the same charges.
Oh, makes a lot more sense now. Thanks!
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aque1408
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(Original post by Chemastronomical)
No, because in the first electron affinity you're adding an electron to a neutral atom, when an atom gains that electron it becomes an Anion and is therefore negatively charged. If you want to add another electron to it, you need to overcome the repulsion because the ion and electron have the same charges.
Looking over it again, I think I am still confused.

I have read that the first electron affinity is only exothermic for atoms that normally form negative ions. Why does it only apply to these types of atoms? I assume it is something to do with a more stable electronic configuration but I don't understand why that would result in less repulsion?
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username5211340
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(Original post by aque1408)
Looking over it again, I think I am still confused.

I have read that the first electron affinity is only exothermic for atoms that normally form negative ions. Why does it only apply to these types of atoms? I assume it is something to do with a more stable electronic configuration but I don't understand why that would result in less repulsion?
As I mentioned above, the initial electron affinity is adding an electron to a neutral atom, so there isn’t much repulsion to overcome, when the electron is added to the shell, energy is given out therefore it’s Exothermic. Now, you have a negatively charged ion because the atom has one more electron than it normally should have. Think of it as a ball of negative charge, you want to add two negative things together so of course there’s going to be repulsion. The energy needed to overcome that repulsion is a lot that’s why you’re putting energy into it.
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aque1408
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(Original post by Chemastronomical)
As I mentioned above, the initial electron affinity is adding an electron to a neutral atom, so there isn’t much repulsion to overcome, when the electron is added to the shell, energy is given out therefore it’s Exothermic. Now, you have a negatively charged ion because the atom has one more electron than it normally should have. Think of it as a ball of negative charge, you want to add two negative things together so of course there’s going to be repulsion. The energy needed to overcome that repulsion is a lot that’s why you’re putting energy into it.
Sorry if I'm just missing out something obvious.

I think I understand the idea of what you're saying but from Golby notes (where I got this information): https://chemrevise.files.wordpress.c...namics-aqa.pdf it mentions the first electron affinity is ONLY exothermic for atoms that are naturally negative (e.g. Cl- or O2-). If this is true, why would the first electron affinity for atoms that naturally form positive ions (e.g. Na+) not be exothermic as you are still adding an electron to a neutral atom, so there would not be much repulsion to overcome?
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username5211340
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(Original post by aque1408)
Sorry if I'm just missing out something obvious.

I think I understand the idea of what you're saying but from Golby notes (where I got this information): https://chemrevise.files.wordpress.c...namics-aqa.pdf it mentions the first electron affinity is ONLY exothermic for atoms that are naturally negative (e.g. Cl- or O2-). If this is true, why would the first electron affinity for atoms that naturally form positive ions (e.g. Na+) not be exothermic as you are still adding an electron to a neutral atom, so there would not be much repulsion to overcome?
The notes seem to be weird, just note that ionisation energies are for Cations and electron affinities are for Anions. Adding electrons to something refers to electron affinity. Keep in mind those notes are just condensed versions from a textbook (perhaps CGP) so don’t take it as fact as they could have errors. You should use a textbook to learn the content.
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aque1408
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(Original post by Chemastronomical)
The notes seem to be weird, just note that ionisation energies are for Cations and electron affinities are for Anions. Adding electrons to something refers to electron affinity. Keep in mind those notes are just condensed versions from a textbook (perhaps CGP) so don’t take it as fact as they could have errors. You should use a textbook to learn the content.
Ok, thanks for sticking with me and the question!
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jack_harrison
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(Original post by aque1408)
Sorry if I'm just missing out something obvious.

I think I understand the idea of what you're saying but from Golby notes (where I got this information): https://chemrevise.files.wordpress.c...namics-aqa.pdf it mentions the first electron affinity is ONLY exothermic for atoms that are naturally negative (e.g. Cl- or O2-). If this is true, why would the first electron affinity for atoms that naturally form positive ions (e.g. Na+) not be exothermic as you are still adding an electron to a neutral atom, so there would not be much repulsion to overcome?
This question came up in one of my first year tutorials as a real head scratcher. It was a question about how an atom can be attractive to an electron despite being neutral. The important thing to note here is that the orbital shapes you see in textbooks showing the shapes are not the whole picture. Those diagrams are simplified to contain only around 95% of the total orbitals, which technically stretch out infinitely like a gravitational field technically does. So at any point in space, there is more positive charge than negative between that point and the nucleus (since some electron density is further away from the nucleus than that point), resulting in a net attraction.

Elements that generally tend to form negative ions are usually very electronegative, so their nuclei are highly charged and their outer shell is held closely to it. The closer you can get to the nucleus, the stronger the attraction, the more negative the enthalpy change. The only deviations you get in electron affinity trends are due to electrons being packed so close together that their repulsion slightly negates the attraction mentioned above. You can see the trend in first electron affinities of the halogens to see that fluorine is a case of this.

Hope this helps
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aque1408
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(Original post by jack_harrison)
This question came up in one of my first year tutorials as a real head scratcher. It was a question about how an atom can be attractive to an electron despite being neutral. The important thing to note here is that the orbital shapes you see in textbooks showing the shapes are not the whole picture. Those diagrams are simplified to contain only around 95% of the total orbitals, which technically stretch out infinitely like a gravitational field technically does. So at any point in space, there is more positive charge than negative between that point and the nucleus (since some electron density is further away from the nucleus than that point), resulting in a net attraction.

Elements that generally tend to form negative ions are usually very electronegative, so their nuclei are highly charged and their outer shell is held closely to it. The closer you can get to the nucleus, the stronger the attraction, the more negative the enthalpy change. The only deviations you get in electron affinity trends are due to electrons being packed so close together that their repulsion slightly negates the attraction mentioned above. You can see the trend in first electron affinities of the halogens to see that fluorine is a case of this.

Hope this helps
Thanks for the explanation, blows my mind how every-time I seem to learn a topic in more depth in Chemistry I realise how little I knew before (I know this is the case for pretty much everything but it seems so prominent in Chemistry)
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(Original post by aque1408)
Thanks for the explanation, blows my mind how every-time I seem to learn a topic in more depth in Chemistry I realise how little I knew before (I know this is the case for pretty much everything but it seems so prominent in Chemistry)
Yeah because fundamentally chemistry needs to be described by quantum mechanics which has a lot of maths, but the maths is extremely complicated for systems with more than one electron (in fact there's no neat mathematical solution for multi-electron systems, you can only calculate things numerically). So in order to do anything and rationalise trends (which is what chemistry is about) you really need to simplify it down, and so at various different levels of education you are fed with different simplifications to the true picture.
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