excited state electronic configuration?

[thestormpetrel]

This is one of the questions on one of the local MCAT books:
The electronic configuration of an element is 1s^2,2s^2,2p^6,3d^5,4s^1, this represents its:
a) ground state b)excited state c)cation form d) excited state

since, instead of 3s coming after 2p there was 3d, i chosed b as the answer. But, the answer key says it's the ground state and there is this really vague explanation about how the unpaired s orbital shows the last orbital, which is the ground state of the atom.

But, I still don't understand how it's the ground state. Am I wrong or the book?

Thanks!

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[gettheleadout]

It's a typo. The configuration given is for chromium, which steals an electron from 4s to get 3d half-filled. That's what they wanted you to recognize. Leaving out the 3s electrons in the configuration is an error that makes the question needlessly confusing.

 

[ninjiliu]

Yes, and the reason the electron falls into the 3d orbital is because it's more stable to have a half filled d orbitals than to have just 4 electrons. Not sure why but go with it.

 

[sps27]

The trick question here is and just for grins and chuckles - if Cr were to loose an electron, which one will it loose, 3d or 4s? And why so?

 

[Helixkam]

Even though the 3d orbital gets filled first i believe it is the 4s orbital that loses the electron first, i kind of remember there is a trick here but I think i may have it backwards? Whats the right answer 😛?

 

[gettheleadout]

The trick question here is and just for grins and chuckles - if Cr were to loose an electron, which one will it loose, 3d or 4s? And why so?

It would lose the 4s electron first when ionized.

 

[sps27]

It would lose the 4s electron first when ionized.

Yes, that is correct, it does lose the 4s electron first. So is 4s more energetic than 3d and if so, why does it get filled first? I don't know why it gets filled first but there must be some logic to that Aufbau's principle. But as far as losing, if I have to guess I would say that losing an entire quantum number (4) is more energetically beneficial than losing an electron from 3'rd shell. So perhaps that is why it loses the 4s.

 

[FeinMS]

4s is less energetic than 3d. However since 4s is more outer than 3d (Note n=4 vs n=3), electron is removed from 4s.