My chemistry teacher said that pure $\ce{Fe(OH)2}$ is a "white" crystalline solid in anaerobic environments, and acquires the usual green tinge, the colour of $\ce{Fe(II)}$, when in contact with air. Wikipedia also agrees with this:

It is produced when iron(II) salts, from a compound such as iron(II) sulfate, are treated with hydroxide ions. Iron(II) hydroxide is a white solid, but even traces of oxygen impart a greenish tinge.

The above paragraph does not have any references. However, it suggests that the green colour is due to substances collectively known as "green rust", which is composed of hydroxides of $\ce{Fe(II) / Fe(III)}$ in different proportions. The following photo shows different colours of $\ce{Fe(OH)2}$.

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I was puzzled over the mechanism behind this colour change, as $\ce{Fe(III)}$ has a yellow tinge, not green. My teacher suggested the possibility that the oxidation of $\ce{Fe(II)}$ to $\ce{Fe(III)}$ induces changes to the lattice structure, causing a shift in its spectrum which happens to coincide with the colour of the ferrous ion.

Is this correct? Are there any reliable sources that support or contradict with this explanation?


1 Answer 1


All your tubes show a strong air oxidation, as the stuff deposited in the higher part of the tubes above the liquid is a brown powder, made of $\ce{Fe(OH)3}$ and due to a complete oxidation of $\ce{Fe(OH)2}$. It can also be seen in the liquid of the second tube.

But this oxidation occurs in two steps : first formation of green-black $\ce{Fe3O4}$ and then into brown $\ce{Fe(OH)3}$ The nearly black deposit in the first tube at the left-hand side of the picture is due to a partial oxidation of $\ce{Fe(OH)2}$. It is made of magnetite $\ce{Fe3O4}$, containing a stoichiometric amount of ferrous and ferric ions ($\ce{Fe3O4}$ looks like $\ce{ 1 FeO + 1 Fe2O3})$. When $\ce{Fe(OH)2}$ starts to become oxidized, as in the two last tubes on the right-hand side, it produces a few $\ce{Fe3O4}$ in much $\ce{Fe(OH)2}$, and this mixture is dark green, due to the dark green color of the diluted magnetite. At the end of the oxidation process, like in the second tube, $\ce{Fe3O4}$ gets more thoroughly oxidized, and the deposit becomes brown, but not so dark, and it is made of pure $\ce{Fe(OH)3}$. The successive reactions are : $$\ce{6 Fe(OH)2 + O2 -> 2 Fe3O4 + 6 H2O}$$ $$\ce{4 Fe3O4 + O2 + 18 H2O -> 12 Fe(OH)3}$$ As shown in the equations, the white precipitate of $\ce{Fe(OH)2}$ is rather sensitive to the tiniest amount of $\ce{O2}$, as one molecule of $\ce{O2}$ is able to oxidize a huge amount of $\ce{Fe(OH)2}$ (6 molecules $\ce{Fe(OH)2}$ - $6$ is a lot with respect to $1$) not magnetite. And even magnetite is still more sensitive to $\ce{O2}$. as one molecule $\ce{O2}$ is able to oxidize $12$ $\ce{Fe}$ atoms of magnetite, so as to produce $\ce{12}$ molecules $\ce{Fe(OH)3}$. $\ce{Fe(OH)2}$ is easily transformed in magnetite by oxidation. But dark magnetite is still more easily destroyed by $\ce{O2}$ into brown, but relatively pale $\ce{Fe(OH)3}$.


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