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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?
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.
