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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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4 Answers 4

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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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    $\begingroup$ 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
    Commented Feb 9, 2015 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
    Commented Feb 16, 2015 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
    Commented Feb 16, 2015 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
    Commented Feb 16, 2015 at 19:43
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    $\begingroup$ @paracetamol only for hydrogen $\endgroup$
    – DavePhD
    Commented Dec 14, 2019 at 3:59
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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.
    Commented Sep 1, 2016 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 ♦
    Commented Aug 4, 2017 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 ♦
    Commented Aug 4, 2017 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
    Commented Dec 14, 2015 at 2:14

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