Transition Metal Valence Electron Energies

[brlin]

I was going through a BR practice questions today and ran into this problem, which I narrowed down to 2 answers:

How can the reduced radii of transition metal cations be explained?
A) Transition metals lose their outermost electrons from the 4s-orbital, when becoming a cation
C) Transition metals lose their highest-energy electrons from the 4s-orbital, when becoming a cation

The answer was A. I understand why the outermost electrons are at the 4s orbital, but aren't these outermost electrons also the highest-energy electrons? So isn't it technically both of these answers?

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[ReggieD6868]

I was going through a BR practice questions today and ran into this problem, which I narrowed down to 2 answers:

How can the reduced radii of transition metal cations be explained?
A) Transition metals lose their outermost electrons from the 4s-orbital, when becoming a cation
C) Transition metals lose their highest-energy electrons from the 4s-orbital, when becoming a cation

The answer was A. I understand why the outermost electrons are at the 4s orbital, but aren't these outermost electrons also the highest-energy electrons? So isn't it technically both of these answers?

I understand why the answer choices are confusing. As you said, the 4s shell is the outermost shell in this specific element and electrons in the 4s shell have higher energy than the 1s, 2s,..3d shells etc; however, answer choice C is the "booby trap" because although the information in C is true, it is missing the point in regard to decreasing atomic radii.

In other words, choice A specifically states that losing the outermost election leads to a change in atomic radii size. Whereas choice C does is not exactly on the point.

It's annoying question, but you appeared to narrow down the question well, so you seem to be on track.

I hope that this analysis helps.

 

[brlin]

That makes a lot of sense. Thanks!

 

[ChemTutor]

d orbitals usually have higher energy than s. this is why s orbitals are filled first.
thus, C is false.

 

[Temperature101]

d orbitals usually have higher energy than s. This is why s orbitals are filled first.
Thus, c is false.

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