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# Degeneracy

This problem was asked in JEE 2015. It states:
If the degeneracy of hydrogen atom in second excited state ($$n=3$$) is 9 then find the degeneracy of hydride anion i.e. $$\ce{H^-}$$ ion in second excited state.

Note: electronic spin is not to be considered here.

Note by Dipanjan Chowdhury
1 year, 4 months ago

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1s$$^2$$ is the ground state electronic configuration. So, 1s$$^1$$ 2s$$^1$$ is the first excited state and 1s$$^1$$ 2p$$^1$$ is the second excited state. Now in 2p, all 3 orientations are degenerate so the degeneracy is 3. · 1 month, 1 week ago

Yes . · 2 months, 4 weeks ago

Sir isn't the ground state of an atom 1s¹ · 2 months, 4 weeks ago

Before I answer this, let me ask you a question that will be key to understanding what's happening here:
Why is the degeneracy of hydrogen atom 9 when n = 3? · 1 year, 4 months ago

I think because of the one 3s three 3p and five 3d orbitals which are not excited as have no electron and as a whole exhibit degeneracy · 1 year, 4 months ago

Okay, good! The fact that hydrogen has only one electron means that orbitals with the same value of n have the same energies. n = 3 has 9 orbitals, as you said, so 9 is the degeneracy.
But for H-, there are two electrons. The orbitals' energies now depend on l (the azimuthal quantum number that differentiates between s,p, d, f and so on.) like, $$1s < 2s < 2p < 3s < 3p ...$$
Now the ground state of H- is $$1s^2$$.
The 1st excited state is $$1s^1 2s^1$$.
The second excited state is $$1s^1 2s^0 2p^1$$ (Compare with H atom for which the 2nd excited state is $$3s^1$$.)
But there are 3 p orbitals for the electron to 'choose' between when making the transition from first excited state to second excited state.
So the degeneracy is 3. Hope this helps. · 1 year, 4 months ago