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Its well-known that crystal field splitting occurs because the electrons in d orbitals repel incoming ligands and the d orbitals are no longer degenerate.

My question is now: is the repulsion (with the incoming ligands) between the electrons in the d orbitals or the orbitals itself? If its the former,then does Sc3+ not undergo d orbital splitting ie it exists as 5 degenerate 3d orbitals?

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    $\begingroup$ It is between the electrons and the ligands. "empty" orbitals per se don't exist as they more or less only dictate the probability distribution of an electron around a nucleus. However, if the electrons of the sublevels are excited to the d-states, it will be excited into ligand-split d-orbitals, that is correct. $\endgroup$
    – Mäßige
    Commented Jul 24, 2023 at 7:38
  • $\begingroup$ An orbital has 3 major related meanings: 1/ a mathematical wave function being a solution of the wave equation 2/ An electron quantum state belonging to such a function 3/ 3D region of probable occurrence of such an electron. $\endgroup$
    – Poutnik
    Commented Aug 23, 2023 at 9:32

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The answer to this question is quite obvious, an orbital has no physical meaning, it's a mathematical object, an electron is a physical charge particle. An occupied orbital is therefore the state of the electron, an empty orbital is a virtual state. Virtual orbitals are useless, have no meaning and are used for convenience to understand things like transitions, the available states and so on. Sc$^{3+}$ has only unoccupied virtual $d$ orbitals, which cannot be split by a crystal field.

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  • $\begingroup$ so im guessing that this image in docbrown.info/page07/transition01Sc.htm is wrong? since it shows that Sc3+ has d-orbital splitting? $\endgroup$ Commented Jul 25, 2023 at 16:51
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    $\begingroup$ @whitingsheep Well, energies the electrons would achieve IF reaching quantum states belonging to empty 3d orbitals MAY split, e.g. in magnetic field - Zeeman effect. $\endgroup$
    – Poutnik
    Commented Aug 23, 2023 at 9:53
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    $\begingroup$ "useless" does not mean "don't exist", and if they have "no meaning" how are they "used to understand things"? Surely the things they are used to understand give them meaning $\endgroup$
    – Ian Bush
    Commented Dec 21, 2023 at 14:03

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