[Kalibasa]

I'm really confused, because in my book they said that in creating metal ions, electrons are subtracted from the p-orbitals, the s-orbitals and then if needed the d-orbitals from the preceding energy level. This would make sense in general, but I thought d-orbitals were usually higher-energy than the s orbital directly above them because of their configuration around the nucleus. For instance, I thought that you would go from 5s to 4d to 5p. Why, then, would the configuration of In 3+ be [Kr]4d¹⁰ instead of [Kr]5s²4d⁸? Why wouldn't you subtract from the d-orbitals first?

[Schrödinger]

You remove electrons from the outermost shells(valence shells), one by one, penetrating into the inner levels. 5p is outermost, then 5s, then 4d and so on.

[tamim83]

In general, remove electrons with the highest value of n first.  For electrons with the same n  (like your In example), remove electrons with the highest value of l first.  So for In ([Kr] 5s² 4d¹⁰ 5p¹), if you want In³⁺ remove the n=5 electrons first. 

[Schrödinger]

@tamim83 : I think the following is the reason for removing electrons in that order...can you tell me if i'm right?

Orbitals with greater n+l have higher energy (Aufbau's rule). Those with higher energy are further away from the nucleus because the system is bound and hence potential and total energy are negative, and they become less negative as we move away from the nucleus.

[Abstractineum]

The energy of the orbitals are as you wrote, 5s<4d<5p and you would theoretically detract electrons from the orbitals in the listed order (starting from the left). However, there are exceptions. In the periodic system, the two first are the Cr and Cu, where lower energy 4s-orbital electrons are promoted to a 3d-orbital. In these cases the resulting electronic configurations are (instead of [Ar] 4s²3d⁴ and [Ar] 4s²3d⁹) [Ar] 4s¹3d⁵ and [Ar] 4s¹3d¹⁰ respectively. Both these generate half/-filled d-orbitals.
  In your case it seems to be the same; the s-electrons are promoted to the higher-energy d orbital which in some way must stabilize the molecule.

[tamim83]

@tamim83 : I think the following is the reason for removing electrons in that order...can you tell me if i'm right?

Orbitals with greater n+l have higher energy (Aufbau's rule). Those with higher energy are further away from the nucleus because the system is bound and hence potential and total energy are negative, and they become less negative as we move away from the nucleus.

That is mostly correct.  However, it really doesn't explain why the 4s electrons are ionized before the 3d electrons.  It would make more sense for the electrons that are not as "tightly bound" to the system to get ionized first.  But, that doesn't happen.  I think it may be due to the fact that, on average, 4s electrons are found further away from the nucleus than 3d electrons (look at the radial distribution functions).  However this sort of contradicts the reason why the 4s orbital is lower in energy (it can penetrate closer to the nucleus- again see the radial distribution functions).  This is one of those things where there is a rule but no easy answer as to why the rule works.  

[Kalibasa]

I actually stumbled on the "answer" in my book about two hours after I asked this, but it isn't really a true explanation, more of a statement of fact. Apparently when you build up, you add electrons in the order I listed: 4s to 3d to 4p. But "once they contain electrons, the 3d orbitals lie lower in energy than the 4s-orbital," so you remove electrons from 4p then 4s then 3d. Basically you add electrons in one order but remove them in another, which is very strange but that's how it is... It's like the last post said, there's a rule that we use but we don't really understand it 

The book is Chemical Principles by Atkins, by the way, which I trust

[tamim83]

Hmm, sounds plausible.  The only problem I have is that, well, orbitals don't really exist until electrons occupy them (the orbitals are the electrons).