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So the 3d orbital comes after the 4s orbital on the periodic table because it has a higher energy level, but then why are the electrons removed from the 4s orbital before the 3d orbital? For the 3d orbital to be filled after the 4s orbital, it must have greater energy, thereby it should be farther away from the nucleus, and if it is farther away from the nucleus than it should then be the outermost orbital, so electrons should be removed from it first right?

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  • $\begingroup$ Energies of different orbitals depend on electron configuration. See e.g. graylark.com/eve/orbital-energies-table.html $\endgroup$
    – Poutnik
    Commented Feb 22, 2022 at 23:09
  • $\begingroup$ According to the image and the Aufbau principle, the 3d orbital has greater energy than the 4s orbital but that supports the idea that it (the 3d orbital) is farther away from the nucleus, it is the outermost shell and thereby electrons should be removed from it first? $\endgroup$
    – Berry
    Commented Feb 22, 2022 at 23:21
  • $\begingroup$ According to the image , the 4s orbital has greater energy than the 3d orbital? $\endgroup$ Commented Feb 23, 2022 at 1:46
  • $\begingroup$ My bad, but my textbook states that the 3d orbital would have greater energy than the 4s orbital, while this image proves otherwise? $\endgroup$
    – Berry
    Commented Feb 23, 2022 at 2:00
  • $\begingroup$ For the 4th period, 3d orbitals have lower ( more negative ) energy since Sc. The important thing is 4s orbital is filled. For K and Ca it is the opposite. You can see how 3d energy at Sc is falling down. $\endgroup$
    – Poutnik
    Commented Feb 23, 2022 at 5:09

2 Answers 2

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Filling of electrons follows the aufbau principle. So according to this rule $3d$ orbitals are filled after the filling of $4s$ orbitals. This is because $3d$ orbitals have slightly more energy than $4s$ orbitals.

Why are the electrons removed from the $4s$ orbital before the $3d$ orbital ?

The electrons are removed from the $4s$ orbital before the $3d$ orbital. This is because the effective nuclear charge $(Z_{eff})$ is more for $3d$ electrons than $4s$ electrons. As $Z_{eff}$ is more for $3d$ electrons they are tightly bound to nucleus than that of $4s$ electrons. So that's why we remove $4s$ electrons before $3d$ electrons.
This can be simply illustrated by Slater's rule.

  • For example consider iron ($Fe$)
    By using Slater's rule it was found that $Z_{eff}$ for $3d$ electrons is $6.25$ whereas for $4s$ electrons it was just $3.75$ .

Similarly this can also be used to illustrate for other atoms.

  • NOTE : Slater's rule will give you an approximate value for $Z_{eff}$ but not exact values.
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  • $\begingroup$ Do do you think the order of filling is according to energy level. And removal, is based on effective nuclear charge? Or would you say filling is also based on effective nuclear charge? By the way, Slaters rule on Iron indeed shows 4s as lower effective nuclear charge than 3d. But for scandium the effective nuclear charge of 4s and 3d is the same. omnicalculator.com/chemistry/effective-charge (Though as you say slaters rule only gives approximate values). Are you aware of a computational calculation of effective nuclear charge that works better? $\endgroup$
    – barlop
    Commented Jan 31 at 17:59
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There is a subtle effect when you compare transition metal cations with neutral atoms: the ions have greater effective nuclear charge. This overwhelms electron-electron repulsion and tends to force orbitals into the same shell order they would have with only a single electron (meaning $\color{blue}{3}d$ drops below $\color{blue}{4}s$ in the ion). See this answer for a discussion of this effect.

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