- #1
OneMan98
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Hi guys, my first post/question. Excuse my stupidity if you think this is an easy question, I'm really having a difficult time with reviewing this part in general chemistry.
Which has a more negative electron affinity, Na+ or Na? Na+ or Cl?
2. The attempt at a solution
My thinking was to write down the ground-state electron configurations.
Na+: [Ne]
Na: [Ne]3s^1
Cl: [Ne]3s^2 3p^5
Electron affinity is described as the energy change that occurs when an electron is added to an isolated atom in its gaseous state.
I assumed for the first part of the question that Na had a greater negative electron affinity because it would become paired in the s orbital, and since Na+ already has a noble gas configuration, wouldn't adding an electron make the Electron affinity positive?
For the second part I naturally assumed it was Cl for similar reasons, and since the textbook says halogens have the greatest Electron affinities.
Answer for both questions: Na+
I don't really understand how/why, so if someone could please explain in a manner that's simple yet concise, that would be awesome.
Thanks Physics forum! I look forward to being an active member on here and learning a lot.
Homework Statement
Which has a more negative electron affinity, Na+ or Na? Na+ or Cl?
2. The attempt at a solution
My thinking was to write down the ground-state electron configurations.
Na+: [Ne]
Na: [Ne]3s^1
Cl: [Ne]3s^2 3p^5
Electron affinity is described as the energy change that occurs when an electron is added to an isolated atom in its gaseous state.
I assumed for the first part of the question that Na had a greater negative electron affinity because it would become paired in the s orbital, and since Na+ already has a noble gas configuration, wouldn't adding an electron make the Electron affinity positive?
For the second part I naturally assumed it was Cl for similar reasons, and since the textbook says halogens have the greatest Electron affinities.
Answer for both questions: Na+
I don't really understand how/why, so if someone could please explain in a manner that's simple yet concise, that would be awesome.
Thanks Physics forum! I look forward to being an active member on here and learning a lot.