The three more energy electron layers for Indium(49) are, sorted from lower to higher energy:

[Kr] $5\text{s}^24\text{d}^{10}5\text{p}^1$

From this configuration, it could be said that is easy to subtract the $5\text{p}^1$ electron (oxidation state +1), next start subtracting electrons from the 4d layer. However, the more common oxidation state of Indium is +3 (as in $\ce{In2Se3}$). It seems electrons are subtracted from layer 5s before to the ones of 4d. This could break the rule of filling the lower energy layer first.

How to explain the +3 oxidation state of Indium taken into account electron configurations?

  • $\begingroup$ Indium is hardly the first element which exhibits similar behavior. $\endgroup$ Jan 11, 2021 at 11:52
  • $\begingroup$ @IvanNeretin: yes. Question and answer can be extend to some others. Could be Copper +1 is the first one. $\endgroup$ Jan 11, 2021 at 11:59
  • $\begingroup$ chemistry.stackexchange.com/questions/27662/… $\endgroup$
    – Mithoron
    Jan 11, 2021 at 15:20

1 Answer 1


When electrons are removed from an atom, those in the outermost orbital is removed first. The p orbitals are most away from the nucleus, then the s orbitals, and finally the d orbitals. In fact, the closeness of d orbitals to the nucleus accounts for higher energy of the 4d orbitals than the 5s orbital. Because electrons are closer, there is a stronger repulsion between them, increasing the energy. In the case of indium, electrons in the 5s orbital are farther away from the nucleus and are removed first.

I've also found a relevant discussion here from a chemistry community at UCLA.


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