Why does chlorine oxidise iron to iron(III), not iron(II)? I was asked to predict the product of the reaction between iron metal and chlorine gas:

$$\ce{Fe +Cl2-&gt;} ?$$

The product here is supposed to be $\ce{FeCl3}$. But how would we know if the product is $\ce{FeCl2}$ or $\ce{FeCl3}$? Why should it be $\ce{FeCl3}$ and not $\ce{FeCl2}$?
The enthalpy of formation ($\Delta H_\mathrm{f}^\circ$) of $\ce{FeCl3}$ is $\pu{-399.49 kJ mol-1}$, while the $\Delta H_\mathrm{f}^\circ$ of $\ce{FeCl2}$ is $\pu{-341.79 kJ mol-1}$. This means that $\ce{FeCl3}$ is $\pu{57.7 kJ mol-1}$ more stable than $\ce{FeCl2}$, a considerable amount. This means that it is more thermodynamically favorable for $\ce{FeCl3}$ to form than $\ce{FeCl2}$, likely due to the larger lattice energy.

Furthermore, in the +2 oxidation state, one electron remains paired in the $\mathrm{3d}$ orbital. When $\ce{Fe}$ is in the +3 oxidation state, however, it has a half filled $\mathrm{3d}$ orbital, a state which is known to be particularly stable, which you can read about further here.