# Why group 13 elements don't show +2 as their major oxidation state?

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?

• These electrons excite only when they form covalent bonds. They won't be exciting in free states. – Pritt says Reinstate Monica May 28 '17 at 14:40
• @PrittBalagopal They will lose 2 electrons only in their excited states. – Arishta May 29 '17 at 4:25

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.