Aluminium Al oxidation state = +3
Aluminium atomic number = 13
Electronic Configuration = `1s^2 2s^2 2p^6 3s^2 3p^1` ` `
It loses 3 electrons from `3s^2` and `3p^1` and form +3 oxidation state.
Chlorine Cl oxidation state = -1
Atomic number = 17
Electronic Configuration = `1s^2 2s^2 2p^6 3s^2 3p^5`
It gains 1 electron and form -1 oxidation state.
The 3 electrons lost by Al is accepted by 3 Cl atom each.
Thus forming `AlCl_3`
Therefore the oxidation states are +3 and -1.
Since its not in options we can consider option D as answer assuming that....
3 electrons lost by Al atom and 3 electrons accepted by Cl atoms.
Thus option D +3, -3 is our answer.