The question said "Write the condensed electron configuration for $\ce{Zn^{2+}}$I got [Ar] 4s$^2+$ 3d$^8$. Why was this the incorrect answer? Zinc lost 2 electrons and surely those electrons should go from the d orbitals?


2 Answers 2


By the time the $\ce{3d}$ orbitals are filled with electrons, the $\ce{4s}$ orbitals have become higher in energy, making it easier to remove them. Here are energy level curves for both the neutral and +1 forms of the $\ce{3d}$ atoms, with electron energy on the $y$-axis:

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you have done only wrong thing and that is sufficient to make your answer wrong . Electron will not be removed from 3d subshell but it will be removed from 4s subshell. The reason is whenever we fill electrons we see the value of (n+l){Principal quantum number + azimuthal quantum number}, the subshell having smaller value of (n+l) gets the electroon first, butt when we have to remove an electron,it is removed froom the subshell having greatest value of n(Principal quantum number).

  • $\begingroup$ I don't get it, how do you go from (n+l) to the electron removing from the subshell having greatest value of n? $\endgroup$
    – Hamze
    Oct 4, 2016 at 3:10
  • $\begingroup$ see,by aufbau principal the electrons are filled in subshells in order of their increasing energy and order of increasing energy can be described with the help of n+l,now as far as use of n is concerned we see that in removing the electrons concept of shells is taken into account(not subshell) and quantum number associated with shells is 'n'. $\endgroup$ Oct 4, 2016 at 3:19
  • $\begingroup$ see (n+l) and aufbau principal on wikipedia . $\endgroup$ Oct 4, 2016 at 3:20

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