# 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:

$$\ce{Fe +Cl2->} ?$$

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.