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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 $\ce{FeCl2}$?

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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.

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    $\begingroup$ 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? $\endgroup$
    – Ivan Neretin
    Commented Oct 13, 2015 at 9:26
  • $\begingroup$ This does not really answer it - it doesn't really matter which is more thermodynamically stable. $\endgroup$
    – Mithoron
    Commented Oct 13, 2015 at 10:53
  • $\begingroup$ It is thermodynamically stable for diamond to spontaneously turn to graphite at room temperature. This does not mean we observe it to do so. $\endgroup$
    – Ali Caglayan
    Commented Oct 13, 2015 at 16:26

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