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For most lanthanide metals, the stable oxidation state is III [*]. The general electronic structure is $$\ce{[Xe] 4f^{0}^{-14} 5s^2 5p^6 5d^{0}^{-1} 6s^2}\ \ [**].$$

Elements that have the d-electron are La, Ce, Gd, and Lu. Furthermore, the f-subshell is considered relatively stable in states $$f^0, f^7, \text{and} f^{14}.$$

We can conclude that La, Gd, and also Lu easily form $\ce{E^3+}$ ions. Yet, as would be predicted by this easy approach, we would also see $$\ce{Sm+, Tm+ (f^7 \ and \ f^14 ), \\ Pr^5+, Dy^5+ (f^0 \ and \ f^7)}.$$

This is definitely not the case. But the why? alludes me. According to Radiochemistry,

The 4f binding energy is so great that the remaining 4f-electrons are regarded as "core-like". [Italic added]

Could someone explain this further? [***]

Radiochemistry then goes on to assert,

Interelectronic repulsion in related not just to electron pairing, but also to [the] angular momentum of the electrons. [---]

e.g., in $\ce{Pr^2+ (4f^3)->Pr^3+ (4f^2)}$ ionisation removes repulsion between [electrons] of like rotation [---]

$\ce{Pm^2+ (4f^4)->Pm^3+ (4f^3)}$ ionisation removes repulsion between [electrons] of unlike rotation [---] [emphasis added]

What exactly is meant by "electrons of like/unlike rotation"? Plus it mentions some "quarter / three-quarter" rule which I possibly cannot rationalise for a total of 14 electrons.

Any help would be greatly appreciated.


[*] Ce, Pr, and Tb also have the oxidation state IV. Eu and Tm have the additional state II.

[**] Ordering changes with the number of electrons.

[***] I am, of course, aware that

  • each electron would be harder to remove as Culonic attraction increases with positive ionic charge,
  • lathanide contraction is a part of the explanation.

(Somewhat related: Plutonium having more oxidation states than samarium?)


EDIT

Every thanks to Brian for providing awesome title suggestions. These are the four new subquestions:

Just an FYI for future chainposters, there's a sixty-second waiting period after asking two consecutive questions.

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closed as too broad by hBy2Py, Todd Minehardt, M.A.R. ಠ_ಠ, ron, ringo Jun 13 '16 at 3:10

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ There are a lot of different questions in here; I see at least four. I recommend breaking out "Why not $\ce{Sm+}$, etc.?", "What does 'core-like' mean here?", "What is meant by 'electrons of like/unlike rotation'?", and "What's up with this 'quarter / three-quarter' rule?" into separate questions, perhaps edited to contain links among all of them. $\endgroup$ – hBy2Py Jun 12 '16 at 14:15
  • $\begingroup$ @Brian: Are you sure? I thought it would be weird to post so many different questions in a short time. That's why I tried to find the best common ground :D. I will have some time tonight (or tomorrow, worst case scenario), and the corrections shall be made. Maybe not four questions but two or three ;-) $\endgroup$ – Linear Christmas Jun 13 '16 at 6:30
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    $\begingroup$ Yeah, the StackExchange approach favors a larger number of more narrowly scoped Questions. Ideally, any given Question will only contain a single question, about a single topic. Not always possible, but it's the guiding principle. The beauty of the way the site is coded is that when you link to other questions, they show up in the sidebar under "Linked", so it creates pretty strong connections between crosslinked Questions. $\endgroup$ – hBy2Py Jun 13 '16 at 10:51
  • $\begingroup$ @Brian: done and done. Be sure to let me know if anything still needs tweaking. $\endgroup$ – Linear Christmas Jun 14 '16 at 13:17
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    $\begingroup$ On the whole, looks pretty good. Scope of the four subquestions is good, I think. In accord with the downvote it's gotten, the 'one-quarter / three-quarters' question is really unclear. See more comments there. $\endgroup$ – hBy2Py Jun 14 '16 at 13:41