In the ground-state, is it always the case that the orbitals with the highest energy donate the electrons when the atom is ionized?

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    $\begingroup$ "Donate" is quite an ambiguous term. Think of "donating 20$ to a guy with a knife". You hit an atom with X-ray, it may pretty well lose an electron from any orbital, including the lowermost one, so what? $\endgroup$ – Ivan Neretin Mar 6 '16 at 10:53
  • $\begingroup$ Which is, of course, the whole basis of Auger spectroscopy- electron from a deep level is kicked out, a higher electron drops in to the empty orbital, releasing enough energy to kick a different electron out at a characteristic energy. $\endgroup$ – Jon Custer Mar 6 '16 at 14:22

No, sometimes you should consider ions like $\ce{Mn^2+}$. Its electronic configuration is $\ce{1s^2 2s^2 2p^6 3s^2 3p^6 3d^5}$. Although $\ce{3d>4s}$, it loses the electrons in $\ce{4s}$ first.

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    $\begingroup$ When 3d electrons exist in the same atom as 4s electrons, the 4s electrons are higher energy. $\endgroup$ – SendersReagent Mar 6 '16 at 11:13
  • $\begingroup$ @DGS, always 4s orbitals have greater orbital energy (in absolute value) than 3d orbitals. $\endgroup$ – user1420303 Mar 6 '16 at 12:17
  • $\begingroup$ @user1420303 Then why don't they fill first as you go along the periodic table? $\endgroup$ – SendersReagent Mar 6 '16 at 12:18
  • $\begingroup$ @DGS Sorry, I tried to say that they are always <<lower>> in absolute value. $\endgroup$ – user1420303 Mar 6 '16 at 12:23
  • $\begingroup$ @DGS I am unsure about what you mean by energy of orbitals. When I speak about orbital energy I am refering to the autovalue of the Fock equation for this orbital. Orbitals have not energy itself, I mean, the energy as described by physics in mechanics. In my context the filling order it is not difficult explained by this together with the fact that the sum of the orbital energies is not the molecule/atom energy. $\endgroup$ – user1420303 Mar 6 '16 at 12:36

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