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.

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewest1s\(^2\) 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.

Log in to reply

Yes .

Log in to reply

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

Log in to reply

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?

Log in to reply

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

Log in to reply

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.

Log in to reply

Log in to reply