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.

#Chemistry

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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?

Ameya Daigavane - 5 years, 2 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

Dipanjan Chowdhury - 5 years, 2 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.

Ameya Daigavane - 5 years, 2 months ago

@Ameya Daigavane – Absolutely clear, Sir. Thank You . :D I was unaware of this sort of splitting.

Dipanjan Chowdhury - 5 years, 2 months ago

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

Abhinav Rajyakirti - 4 years, 1 month ago

Yes .

Dipanjan Chowdhury - 4 years, 1 month ago

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.

Sai Krishna Attaluri - 4 years ago

Qwerty Asdfghjkl - 3 years, 1 month ago

Pratham Pahuja - 2 years, 11 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$