These are the things I understand:

1) Group-13 elements show +1 and +3 as their major oxidation states(OS).

2) The general configuration of G-13 elements is $ns^2$ $np^1$.

3) In the excited state, their electronic configuration is $ns^1$ $np_x^1$ $np_y^1$ .

On losing two electrons from their excited state, their configuration will become $ns^1$ $np_x^0$ $np_y^0$

I don't understand why will this configuration be unstable?

  • $\begingroup$ These electrons excite only when they form covalent bonds. They won't be exciting in free states. $\endgroup$ May 28, 2017 at 14:40
  • 1
    $\begingroup$ @PrittBalagopal They will lose 2 electrons only in their excited states. $\endgroup$
    – Arishta
    May 29, 2017 at 4:25

1 Answer 1


The one p electron is energetically much higher than the two s electrons (and any d electrons, if present). So it is rather easily removed to give a monocation or the $\mathrm{+I}$ oxidation state.

Thereafter, to remove a second electron you need to rip out one of the paired s electrons. This is not going to be easy and requires a lot of energy. But once you have arrived there, at a hypothetical $\mathrm{+II}$ cation, you again have one single lone electron in a relatively (compared to the core orbitals) orbital that should, again, be easily removed. Obviously that can happen leading you quickly to the $\mathrm{+III}$ state.

Indeed, it would be rather easy to see that two $\mathrm{+II}$ atoms might come together and transfer a single electron from one to the other—a disproportionation, giving $\mathrm{+I}$ and $\mathrm{+III}$ from two $\mathrm{+II}$. This would be predicted to be an exothermic process.

This only applies to isolated atoms; where bonds are formed other oxidation states are accessible. For example, boron exhibits the $\mathrm{+II}$ oxidation state in bis(pinacolato)diboron as can be deduced easily.


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