I have just started inorganic chemistry in school and I have just learnt about metal aqua ions and how they react in water. I learnt that $\ce{[Fe(H2O)6]^2+}$ reacts in water by losing a $\ce{H+}$ acting as a Bronsted-Lowry acid producing $\ce{H+}$ ions. Due to this reaction we can test for metals such as $\ce{Fe^2+}$ by reacting it with $\ce{NaOH}$. This is because the $\ce{OH-}$ ions react with the $\ce{H+}$ ions produced pushing the equilibria to the right forming a insoluble precipitate. My questions is, why does the reaction stop there? The reaction stops once $\ce{[Fe(OH)2(H2O)4]}$ forms which is the insoluble precipitate but why doesn't it continue to form $\ce{[Fe(H2O)3(OH)3]-}$ and become soluble again?

  • 1
    $\begingroup$ Actually precipitate isn't [Fe(OH)2(H2O)4]. Precipitation starts with it, but it's rather complex condensation/polymerisation process. $\endgroup$
    – Mithoron
    Feb 6, 2019 at 19:50

2 Answers 2


This is a pretty interesting question. First of all, you are right, if you ever happened to see a yellow $\ce{Fe^3+}$ solution that was no pure aqua-complex but more likely $\ce{[Fe(H2O)5(OH)]^2-}$. The real $\ce{[Fe(H2O)6]^3+}$ for example, like it is formed in conc. $\ce{HNO3}$ or $\ce{HClO4}$ is actually somewhat purple. We call these compounds cationic acids. So the aqua-complexes will act as acids and become deprotonated in water. For your $\ce{Fe^2+}$ example it's pretty much the same. I often read in student reports that, if for example a hexaqua complex is treated with a base the $\ce{OH^-}$ would exchange for one water ligand. Many then concluded that $\ce{OH^-}$ had to be the stronger ligand in the spectrochemical series. This is, of course, wrong and not related to each other here. What is far more logical, at least to me is, that by coordination the water gets closer towards the highly charged metal center. But that means there is a repulsion from the $\ce{H}$ in $\ce{H2O}$ and the metal center since both have a positive (formal) charge. This makes it much easier to become deprotonated by a base.

But why does this stop at some points? I asked myself the same question for the example of $\ce{BeO}$ reacting with $\ce{OH^-}$ to form $\ce{[Be(OH)4]^2-}$. Why is Beryllium oxide amphoteric forming a berrylate while $\ce{MgO}$ to $\ce{BaO}$ do not show this ability? It took me quite a while to find an answer to this. There is an equation by Cartledge that tries to describe this phenomenon:

$$\ce{Φ = z/r}$$

Where Φ is the ionic potential, z the charge and r the radius. If $\ce{Φ^{1/2}}$ is smaller than 2.2 the ion is considered basic, above 3.2 it's considered to be acidic. In the middle, it's considered amphoteric. What does that mean? It seems like there is a ratio of radius to charge. Small, highly charged ions seem to polarize the surrounding water molecules so much that they quickly lose a proton.

In your case, we are not specifically talking about oxides but I guess that something similar should be happening. The high charge on the metal and the fact that $\ce{Fe}$ is already in the middle of the d-row where atomic radii decrease along the row should be enough to actually cause even more deprotonation, this time perhaps not of surrounding water molecules like in the case of $\ce{BeO}$ but the ones that are already ligated to the metal center. At some point it will require additional help which happens when the pH is increased. And in fact for $\ce{Fe^2+}$ a $\ce{Na4[Fe(OH)6]}$ seems to exist.

I guess the comparison with the $\ce{BeO}$ may perhaps be a bit far fetched but it would be an actual example where you can see how the charge to radius ratio acts on the way soluble hydroxo-complexes are formed.

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    $\begingroup$ Can you give a reference, other than textbook problems that often make things up, for $\ce{Na4Fe(OH)6}$? $\endgroup$ Feb 6, 2019 at 23:16
  • $\begingroup$ Good question. It is mentioned in the Riedel - Anorganische Chemie and the new edition of the former Holleman Wiberg to exist. But I will check the Gmelin on iron later. $\endgroup$ Feb 7, 2019 at 7:46

Since you said you have just started inorganic chemistry in school, I give you some simpler insight:

Simply, a complex metal ion has a metal ion at its center with a number of other molecules (e.g., $\ce{H2O}$) or ions (e.g., $\ce{CN-}$) surrounding it. When you start learning complex ions (or molecules) you must learn why complex ions of the type $\ce{[M(H2O)6]^n+}$ (hexaaqua ions) are acidic. Examples are, $\ce{[Fe(H2O)6]^2+}$, $\ce{[Fe(H2O)6]^3+}$, $\ce{[Cu(H2O)6]^2+}$, $\ce{[Co(H2O)6]^2+}$, $\ce{[Ni(H2O)6]^2+}$, $\ce{[V(H2O)6]^2+}$, etc. The first thing you need to know is the solutions of hexaaqua ions with equal concentrations have varying range of acidity (pH of solution), which depend a lot on type of metal's charge, ionic radius, etc. See the following graph extracted from the web page, Complex Metal Ions - The Acidity of the Hexaaqua Ions.

pKa Vs Ionic Radius

Let's take the hexaaquairon(III), $\ce{[Fe(H2O)6]^3+}$ as typical complex ion. The structure of the ion (octahedral) is given in the diagram A below and an explanation for its acidity is given in the diagram B (From Complex Metal Ions - The Acidity of the Hexaaqua Ions):

Hexaaqua Iron(III)

This explanation is given by visualizing the 3+ charge of the ion is located entirely on $\ce{Fe}$ in the center. When one of lone pair on the $\ce{O}$ of each $\ce{H2O}$ form a coordinate bond with the iron, it can be visualize that these lone pairs move closer towards $\ce{Fe}$. That has an effect on the electrons in all $\ce{O-H}$ bonds within the complex (imagine all $\ce{O}$ atoms have partial positive charges due to this electron movement). The electrons of $\ce{O-H}$ bonds, in turn, get pulled towards the oxygen even more than usual because of this. That leaves each $\ce{H}$ nucleus more exposed than normal (compared to that in $\ce{H2O}$ molecules in the solution). The overall effect is that each of $\ce{H}$ atoms is more positive than it is in ordinary water molecules. Thus, the 3+ charge is no longer located entirely on $\ce{Fe}$ but spread out over the whole ion--much of it on the hydrogen atoms of coordinated water.

The $\ce{H}$ atoms attached to $\ce{H2O}$ ligands are sufficiently positive that they can act as Brønsted acid and can be pulled off in a acid-base reaction involving $\ce{H2O}$ molecules in the solution according to the following sequence of equations illustrating the first three stages of the process: $$\ce{[Fe(H2O)6]^3+ + H2O <=> [Fe(H2O)5(OH)]^2+ + H3O+} \\\ce{Fe(H2O)5(OH)]^2+ + H2O <=> [Fe(H2O)4(OH)4]^+ + H3O+} \\\ce{Fe(H2O)4(OH)2]^+ + H2O <=> [Fe(H2O)3(OH)3] + H3O+}$$

As you see, the process end up with a complex with no charge on it after third stage. This complex is described as a Neutral Complex. Because it has no charge, thus, it doesn't dissolve in water to any extent, and will precipitate as it is formed. See below diagram for color illustration of the reaction of $\ce{[NaOH}$ with $\ce{[Fe(H2O)6]^2+}$ and $\ce{[Fe(H2O)6]^3+}$, respectively.

Fe(II) & Fe(III)


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