9
$\begingroup$

This is a question presented in a IIT-JEE 2015 paper I exam. It says,

. Not considering the electronic spin, the degeneracy of the second excited state (n = 3) of H atom is 9, while the degeneracy of the second excited state of H- is-

I saw a solution which says it's 3.

I think by 9 degenerate energy levels the question meant 9 possible orbitals for M shell. But I am not sure how's that 3 for for H-.

Please don't delete this question as I haven't found it anywhere.

Improve this question
$\endgroup$

2 Answers 2

Reset to default
21
$\begingroup$

I think it is important to understand that for hydrogen atom (or any other one-electron system) all orbitals from the same shell have same energy. For instance, $E_\mathrm{2s} = E_\mathrm{2p}$, $E_\mathrm{3s} = E_\mathrm{3p} = E_\mathrm{3d}$, etc. Thus,

  • The first excited state of hydrogen atom would be one in which either $\mathrm{2s}$ or one of the three $\mathrm{2p}$ orbitals is occupied and it will be 4-fold degenerate: $\mathrm{1s^0 2s^1}$, $\mathrm{1s^0 2p_x^1}$, $\mathrm{1s^0 2p_y^1}$, $\mathrm{1s^0 2p_z^1}$.
  • Analogously, the second excited state of hydrogen atom would be one in which either $\mathrm{3s}$ or one of the three $\mathrm{3p}$ or one of the five $\mathrm{3d}$ orbitals is occupied and it will be 9-fold degenerate: $\mathrm{1s^0 2s^0 2p_x^0 2p_y^0 2p_z^0 3s^1}$, $\mathrm{1s^0 2s^0 2p_x^0 2p_y^0 2p_z^0 3s^0 3p_x^1}$, $\mathrm{1s^0 2s^0 2p_x^0 2p_y^0 2p_z^0 3s^0 3p_y^1}$, $\mathrm{1s^0 2s^0 2p_x^0 2p_y^0 2p_z^0 3s^0 3p_z^1}$, plus 5 configurations with only $\mathrm{3d}$-orbitals occupied.

For hydride the situation is different: it is not a one-electron system, so different orbitals from the same shell do not have same energy anymore. For instance, $E_\mathrm{2s} < E_\mathrm{2p}$, $E_\mathrm{3s} < E_\mathrm{3p} < E_\mathrm{3d}$, etc. Thus,

  • The first excited state of the hydride would be one in which one electron populates $\mathrm{1s}$ orbital and another $\mathrm{2s}$ one, i.e. a non-degenerate $\mathrm{1s^1 2s^1}$ state.
  • The second excited state of the hydride would be one in which one electron populates $\mathrm{1s}$ orbital and another one of the three $\mathrm{2p}$ ones, i.e. a 3-fold degenerate state: $\mathrm{1s^1 2s^0 2p_x^1}$, $\mathrm{1s^1 2s^0 2p_y^1}$, $\mathrm{1s^1 2s^0 2p_z^1}$.
Improve this answer
$\endgroup$
4
  • 3
    $\begingroup$ So one of the electrons remains in $\ce1s^1$ at all times? Is this a rule? This question was intended for students who only know how to work around with hydrogen-like species (i.e single electron species). PS I am aware about this particular examination, its syllabi as well as the difficult level of its questions. @Wildcat $\endgroup$
    – McSuperbX1
    May 12, 2019 at 13:41
  • $\begingroup$ I am confused about the same that why one of electrons remains in 1s1 all the time? $\endgroup$
    – Khushi Ladha
    Jul 22, 2020 at 12:32
  • $\begingroup$ For hydrogen atom (or any other one-electron system) all orbitals from the same shell have same energy, why is it so? @wildcat $\endgroup$
    – Khushi Ladha
    Jul 22, 2020 at 12:34
  • 2
    $\begingroup$ @KhushiLadha The energy difference between 1s and 2s is more than that of 2s and 2p. Thus, in the 2nd excited state, the electron jumps from 2s to 2p. $\endgroup$
    – Apurvium
    Aug 5, 2021 at 6:11
1
$\begingroup$

In a single electron system like H atom or He+ all orbitals with same principal quantum number (n) have same energy Therefore the degeneracy refers to the 3 subshells 3s 3p 3d which have same energy

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.