# Iron-II to iron-III change in complex formation

We have an iron dust sample, which is treated with hydrogen chloride in aqueous solution. The solution enters in a reaction with a sodium hydroxide powder, from which we obtain a precipitate. Could we solubilize the precipitate with excess of ammonia? Note that, at the end of the reaction, qualitative chemical analysis identified the ion $$\ce{Fe^{3+}}$$, but not $$\ce{Fe^{2+}}$$.

First reaction is, obviously: $$\ce{Fe + 2HCl -> FeCl2 + H2 ^}$$

Then, we have: $$\ce{FeCl2 + 2NaOH -> 2NaCl + Fe(OH)2}$$

We observe that we have, now, only $$\ce{Fe^{2+}}$$ in the aqueous solution, not $$\ce{Fe^{3+}}$$. Precipitate solubilization means coordinative compounds formation. If for iron we have a similar reaction to the one with copper ion: $$\ce{Cu(OH)2 + 4NH3 -> [Cu(NH3)](OH)2}$$

We would have only $$\ce{Fe^{2+}}$$. So, is there any redox process happening in background? Or a physical process (non-chemical)?

First iron(II) hydroxide $$\ce{Fe(OH)2}$$ is a precipitate made in aqueous solution, and it does not react with ammonia. Second, the iron(II) hydroxide $$\ce{Fe(OH)2}$$ is a green substance which is extremely sensitive to the oxygen of the air. In a couple of minutes, it gets brown, due to the formation of iron(III) hydroxide $$\ce{Fe(OH)3}$$ according to $$\ce{4 Fe(OH)2 + O2 + 2 H2O -> 4 Fe(OH)3}$$ If water is absent, the reaction may produce another compound of iron (III) $$\ce{4 Fe(OH)2 + O2 -> 4 FeOOH + 2 H2O}$$