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  1. Which of the following orbitals are degenerate in the hydrogen atom with $n = 3$? enter image description here
    A. II and III only;
    B. I and IV only;
    C. I, II, and IV only;
    D. II, III, and IV only;
    E. all.

The answer says its E. All of them.

First of all isn't there only 1 electron in hydrogen? And how could the $\mathrm{s}$ orbital be degenerate? Doesn't degenerate mean there are multiple places pairs of orbitals can be?

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First of all isn't there only 1 electron in hydrogen?

yes

And how could the s orbital be degenerate? Doesn't degenerate mean there are multiple places pairs of orbitals can be?

"degenerate" means having the same energy. "Degenerate" refers to a set of orbitals. It doesn't make sense to say one orbital is degenerate.

Solving the non-relativistic Schrodinger equation, all the orbitals for a given "n" are degenerate. Energy only depends upon n.

More complete consideration including relativity, spin and quantum electrodynamics shows that they are not all degenerate however.

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  • $\begingroup$ 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? $\endgroup$ – Dan Feb 9 '15 at 20:39
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    $\begingroup$ 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. $\endgroup$ – DavePhD Feb 9 '15 at 20:49
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    $\begingroup$ 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. $\endgroup$ – Geoff Hutchison Feb 16 '15 at 14:13
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    $\begingroup$ 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 $\endgroup$ – DavePhD Feb 16 '15 at 14:23
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    $\begingroup$ 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. $\endgroup$ – DavePhD Feb 16 '15 at 19:43
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The one electron of hydrogen, when excited, reaches 3rd energy level and degeneracy is determined by $(n + l)$, where $n$ - energy level, $l = 0$ for $\mathrm{s}$, $1$ for $\mathrm{p}$, $2$ for $\mathrm{d}$, $3$ for $\mathrm{f}$.

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First of all yes there is only 1 e- present in atom of H. But here the atom is "hydrogen like" which can be He+, Li+2 Be3+, etc. The question was twisted at the point n=3, when it said n=3 then the atom can be "hydrogen like" as I mentioned above.

Now n=3 makes the above orbitals with 3s, 3p, 3d and 3f. Now (l) quantum number for 3s=0 for 3p=1, for 3d=2 and for 3f=3.

Now, electron can jump in case of 3s=0 to 1, 3p=0 to 3, 3d= 0 to 5 and 3d= 0 to 7

Thus, we can say that from above orbitals, there will be degeneration in all of them.

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  • $\begingroup$ I can make sense of your last two sentences. Are you saying that all orbitals in the third layer are degenerate? $\endgroup$ – M.A.R. Sep 1 '16 at 18:24
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    $\begingroup$ There are 3f orbitals? That's kind of new to me. $\endgroup$ – Martin - マーチン Aug 4 '17 at 7:59
  • $\begingroup$ @Mithoron Just for your information, your edit (i -> I) dequeued the item from the VLQ queue, and the "recommend deletion" never got executed. That's a bummer, isn't it ;) $\endgroup$ – Martin - マーチン Aug 4 '17 at 8:02
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Ok so the best way to think of this is to compare two atoms. Take for example {Ne} which has 10 electrons so this atom would have a number of n=3. The problem asks what orbitals would be degenerate for n=3 and we know that H has a number of n=1 so this would be just 1s^1 the question is a trick question to make you think that all orbitals above the n=1 orbital are degenerate so really even though it gives you hydrogen its just to throw you off. Picture a n=3 atom for this problem that has no element attached to it in that case all orbitals would be equal energy therefore it would be option E because all orbitals s,p,and d would be degenerate. Hope that helps took my a while to conceptualize this.

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  • $\begingroup$ Welcome to chemistry. Feel free to take a tour. Visit the help center for any unanswered questions about the site. I don’t get your answer. But neon’s configuration is $\mathrm{1s^2\ 2s^2\ 2p^8}$, so there is no $n = 3$ in there … $\endgroup$ – Jan Dec 14 '15 at 2:14

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