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.

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## Comments

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TopNewestBefore 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?

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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

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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.

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Sir isn't the ground state of an atom 1s¹

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Yes .

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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.

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3

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3

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