`AlCl_3`

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.

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