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While going through my book, I came across a problem which gave some options about the order of energy of 3s, 3p and 3d orbitals of hydrogen atom. The correct answer was given as:

3s, 3p and 3d orbitals all have the same energy


I also read somewhere that the order of energy of different orbitals increases as the value of $n+l$ increases where $n$ is the principal quantum number and $l$ is the azimuthal quantum number. So my guess was $3s<3p<3d$ .

What is the reason for this?

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    $\begingroup$ The former is true for hydrogen atom, the latter for other atoms. $\endgroup$ Commented Apr 8, 2021 at 8:51
  • $\begingroup$ 1s. 2s. 2p. 3s. 3p. 3d. 4s. 4p. 4d. 4f. 5s. 5p. 5d. 5f. 6s. 6p. 6d. 7s. 7p. 8s. $\endgroup$
    – user100905
    Commented Sep 24, 2021 at 22:30
  • $\begingroup$ moeller diagram $\endgroup$
    – user100905
    Commented Sep 24, 2021 at 22:33

1 Answer 1

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3s, 3p and 3d orbitals all have the same energy in hydrogen atoms ( or monoelectron atomic ions ) according to the nonrelativistic Schroedinger model.

Fine measurements and relativistic models of hydrogen atoms shows there is slight difference in their energies, related to the fine structure of spectra and Lamb shift ( Wikipedia Hyperphysics )

For multielectron atoms, s, p, d orbitals have different energies, as there come in the play 2 major effects, which affect electron energies:

  1. Mutual electron repulsion
  2. Screening/shielding of the nucleus charge by other electrons.
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    $\begingroup$ oh so as the 2 effects don't take place in the case of mono electron species, we can say that they'll have the same energy. If for a moment we imagine a hypothetical case where these 2 effects don't exist at all, would the energy be same in multi electron species just like it was in the case of Hydrogen? $\endgroup$ Commented Apr 8, 2021 at 9:59
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    $\begingroup$ Yes, we can say they have the same energy as the first approximation.// Yes, they would have had approximately the same energy too, but what is the point of such highly hypothetical thought ? $\endgroup$
    – Poutnik
    Commented Apr 8, 2021 at 10:03

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