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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.

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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: $\mathrm{1s^1 2s^0 2p_x^1}$, $\mathrm{1s^1 2s^0 2p_y^1}$, $\mathrm{1s^1 2s^0 2p_z^1}$.
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  • $\begingroup$ 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 at 13:41
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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

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