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`AlCl_3`
Aluminium Al oxidation state = +3
Aluminium atomic number = 13
Electronic construction = `1s^2 2s^2 2p^6 3s^2 3p^1` ` `
It loses 3 electrons from `3s^2` and also `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...
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`AlCl_3`
Aluminium Al oxidation state = +3
Aluminium atomic number = 13
Electronic construction = `1s^2 2s^2 2p^6 3s^2 3p^1` ` `
It loses 3 electrons from `3s^2` and also `3p^1` and form +3 oxidation state.
Chlorine Cl oxidation state = -1
Atomic number = 17
Electronic construction = `1s^2 2s^2 2p^6 3s^2 3p^5`
It gains 1 electron and kind -1 oxidation state.
The 3 electrons lost by Al is welcomed by 3 Cl atom each.
Thus forming `AlCl_3`
Therefore the oxidation says are +3 and -1.
Since its not in alternatives we can take into consideration option D together answer assuming that....
3 electrons lost by Al atom and 3 electrons welcomed by Cl atoms.
Thus choice D +3, -3 is our answer.
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