Timeline for Which orbitals of the hydrogen atom are degenerate for n=3?
Current License: CC BY-SA 3.0
14 events
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| Dec 14, 2019 at 3:59 | comment | added | DavePhD | @paracetamol only for hydrogen | |
| Dec 14, 2019 at 3:02 | comment | added | ibuprofen | is this only valid for Hydrogen? Or for any element is it valid? | |
| Sep 1, 2016 at 15:40 | comment | added | porphyrin | Its not an ideal question, I agree, and it is a pity that external examiners did not point this out (assuming its a year 1 exam question), but presumably at this level the examiners thought that generic orbital diagrams are acceptable. | |
| Sep 1, 2016 at 15:02 | history | edited | DavePhD | CC BY-SA 3.0 |
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| Feb 16, 2015 at 19:43 | comment | added | DavePhD | But for 3s, you wouldn't be able to see the nodes in such drawings, in the sense that there is still an outer spherical equal-probability surface. | |
| Feb 16, 2015 at 19:09 | comment | added | Geoff Hutchison | Same thing with a 3s.. there would be two radial nodes in a 3s. | |
| Feb 16, 2015 at 14:23 | comment | added | DavePhD | I don't like it either, on the one hand the question does specifically says "n=3", but II looks like 2p, not 3p because 3p would have a radial node winter.group.shef.ac.uk/orbitron/AOs/3p | |
| Feb 16, 2015 at 14:13 | comment | added | Geoff Hutchison | I'm not thrilled about this question, since II and III could easily be a 2p and 1s orbital, respectively. There's nothing to indicate that they're really a 3p and 3s orbital. So my first instinct was choice "B" because I identified the "p" and "s" of a different n than the clearly 3d orbitals. | |
| Feb 16, 2015 at 13:36 | history | edited | DavePhD | CC BY-SA 3.0 |
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| Feb 9, 2015 at 20:54 | vote | accept | Dan | ||
| Feb 9, 2015 at 20:49 | comment | added | DavePhD | If n=3, then l could be l=0 (s), l=1 (p) or l=2 (d). So when n=3, the degenerate orbitals (according to the non-relativistic Schrodinger equation) are 3s, the three 3p orbitals, and the five 3d orbitals. | |
| Feb 9, 2015 at 20:39 | comment | added | Dan | What I meant to say was "Doesn't degenerate mean there are multiple places pairs of orbitals can be within the same subshell?" I'm still confused why the answer is all of them. I thought the s orbitals were never degenerate because there is only one place and like you said it doesn't make sense to say one orbital is degenerate. So why is it degenerate? what orbital with n = 3 has the same energy as it? And how do you know? | |
| Feb 9, 2015 at 20:39 | history | edited | DavePhD | CC BY-SA 3.0 |
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| Feb 9, 2015 at 20:32 | history | answered | DavePhD | CC BY-SA 3.0 |