# 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 $$\ce{FeCl2}$$?

## 1 Answer

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.

• That's true, but I think there is more to the question. See, when you encounter some simple reaction, do you literally reach for the reference book to check the enthalpies? I don't think so. Then how do we really predict things like this? I imagine an ordered row of redox potentials (without numbers, just the order) and think in terms like "OK, chlorine is strong, it will drive iron all the way to +3..." But then again, maybe I am idealizing it, maybe we just remember all simple reactions and use the reference book for the not-so-simple ones? – Ivan Neretin Oct 13 '15 at 9:26
• This does not really answer it - it doesn't really matter which is more thermodynamically stable. – Mithoron Oct 13 '15 at 10:53
• It is thermodynamically stable for diamond to spontaneously turn to graphite at room temperature. This does not mean we observe it to do so. – Ali Caglayan Oct 13 '15 at 16:26