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

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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?

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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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3

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3

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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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Yes .

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