s relative to p orbital energy question

[HopefulMDclass2020]

I was just working AAMC practice test 9, and had a issue with one of their answers.

The relevant passage information for this question is, "The researchers suggest that two effects involving metal orbitals influence the reaction rate. A full valence s subshell hinders reaction, and valence s and d orbitals of similar energy form sd hybrid orbitals, enhancing the reactions."

Question : Ta reacts slower than Nb because:
A. The valence s orbitals of Ta have a much higher energy than do its valence d orbitals.
B. The valence s orbitals of Ta have a much lower energy than do its valence d orbitals.

I choose B because Ta has valence s electrons in the 6th shell and valence p in the 5th shell (I thought p was n-1). The answer was A because the p orbitals have a higher energy level. Can someone help shed light of this question.

Thanks!

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[Quinoline]

I think the question is asking to compare between s and d orbitals, not s and p. d orbitals are the ones that are n-1. So Ta has higher energy level s orbitals than d.

 

[chemist16]

I think OP should clarify the question and verify whether we're dealing with d or p orbitals here. The answer can't be A as written because 5d is greater than 6s in energy. It's greater because electrons in an s orbital are stabilized by the s-orbital's penetration to the nucleus. In other words, s orbitals don't have nodes at the nucleus.

 

[HopefulMDclass2020]

I am sorry, I meant D orbitals. And how are the 5d at a greater energy level than the 6s. I was under the impression that the higher the principal quantum number, the higher the energy.

Example, when we ionize a transition metal, it will loss electrons from the highest energy level and thus loss electrons from the s orbital before losing electrons from the d orbital.

If the question is still not clear, I can provide more passage information that would help.

 

[chemist16]

I am sorry, I meant D orbitals. And how are the 5d at a greater energy level than the 6s. I was under the impression that the higher the principal quantum number, the higher the energy.

Example, when we ionize a transition metal, it will loss electrons from the highest energy level and thus loss electrons from the s orbital before losing electrons from the d orbital.

If the question is still not clear, I can provide more passage information that would help.

Sorry, it's been awhile and I never took Gen Chem. Yes, you're right - for transition metals, the d orbitals are slightly lower in energy than the s orbitals. So then the question and answer are completely clear - you've answered your own question. A cannot be correct because it says that s orbitals have higher energy than d orbitals.

Your rationale is incorrect because there are no valence p electrons for Ta (no electrons in the 6p subshell). Basically what happens when you have a metallic reaction is that your s electrons get bumped up into the d subshell (or bumped down, depending on how you look at it) because the s-orbitals are in fact more stable than the d-orbitals in terms of bonding for two reasons: 1) s orbitals have greater penetration to the nucleus and 2) 6s has a greater radial extension than 5d. So the higher the energy gap between the 6s and 5d subshells, the harder it is to bump electrons between them.