The reaction of iron with sulfur to produce Iron(II) sulfide is inconvenient (for me)

Question

I am taking a lot of iron properties this semester, but I have found some contradicting piece of information about the reaction of iron with nonmetals illustrated with two cases
\begin{align}\ce{ **First Case** }\\\end{align} If we react the iron with chlorine, only FeCl3 will result as following:

$$\begin{align}\ce{2Fe +3Cl2 &->[\space \Delta ]2FeCl3}\tag{1}\\\end{align}$$

No FeCl2 has produced because iron prefers to have oxidation state at +3 as sublevel 3d will be half-filled.

One case that FeCl2 will result when H2 exists as $$\begin{align}\ce{Fe +2HCl &->FeCl2 +H2}\tag{2}\\\end{align}$$ In this situatuon if any FeCl3 formed, H2 will reduce it to FeCl2 \begin{align}\ce{ **Second Case** }\\\end{align} Although, Iron favor +3 oxidation state (as illustrated in 1), If we react iron with sulfur; it will actually favor +2 oxidation state as follows: $$\begin{align}\ce{Fe +S &->[\space \Delta ]FeS}\tag{3}\\\end{align}$$ I tried to find a convenient way that explains how this happen (as the way illustrated at 2), but

I failed to do so. I am really confused:

Does the iron actually favor +3 state or not?

If yes, Why 3 happens? Why there isn't any Fe2S3 in the products?

If no, Why 1 happens?

Is this has to do anything with some hidden conditions or unobvious reaction mechanism?

Answer 1

Not an answer per se (I lack sufficient reputation to comment): are these reactions in aqueous solution, or dry gases? Just a suggestion, but by omitting the charges on these species and not writing them as half-reactions you are obscuring the electron exchange in each. Thus in your first example (iron reacting with chlorine gas), you take metallic Fe, whose oxidation half-reaction can be written as $$\ce{Fe^0 -> Fe^3+ + 3e-}$$ The equivalent for reduction of chlorine would be $$\ce{2e- + Cl2_{(g)} -> 2Cl-}$$ Normalizing half reactions to electrons transferred ($6$), adding and simplifying stoichiometry gives

$$\ce{2Fe^0 + 3Cl2 -> 2FeCl3}$$

Thus chlorine is the electron acceptor (strong oxidizer) and iron the donor. In your reaction (2), iron is again a donor (this time of only $2$ electrons), hydrogen is the acceptor, with chloride ion just along for the ride:

$$\ce{Fe^0 -> Fe^2+ + 2e-}$$ $$\ce{e- + H+ -> \tfrac{1}{2}H2} $$

In contrast to these simple reactions, those involving sulfur are exceeding complex because of the diversity of species that can form. In your reaction to form iron monosulfide, one possible way for Fe to be oxidized but native sulfur (zero-valent S actually present as stable ring, $\mathrm{S}_8$) to be both reduced and oxidized is via a disproportionation reaction, with water reduction to hydrogen consuming additional electrons in an anoxic environment, thus yielding both FeS and iron sulfate: $$\ce{Fe^0 + \tfrac{1}{8} S8 + 2H2O -> \tfrac{1}{2}FeSO4 + \tfrac{1}{2}FeS + 2H2}$$

Amended answer: If you're heating metallic iron and S to significant temperatures, the elemental sulfur will form a gas, $\mathrm{S}_{n}(g)$ ($n$ indicates which T-dependent allotrope is present). To synthesize FeS however, I think you would need an oxygen-free system. If $\mathrm{O_2}$ is initially present, consumption could be accomplished by sacrificing some of the reactants via in situ oxidation of the metal, or production of $\mathrm{SO_2}$. Then this can proceed directly, using the remaining $\mathrm{S}_{n}(g)$, e.g.: $$ \tfrac{1}{2}\mathrm{S_2}(g) + \mathrm{Fe}(s) = \mathrm{FeS}(s) $$ I believe the ratio of S/Fe of the product will increase with temperature, and at sufficiently high temperature reach 2 ($\mathrm{FeS_2}$, pyrite).