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From what I understand, adding $\ce{NaOH}$ to a solution containing transition metal ions gives transition metal hydroxide precipitates. Why is $\ce{NaOH}$ soluble but transition metal hydroxides insoluble? I have a feeling it has something to do with ligands and complexes formed, which I am familiar with.

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  • $\begingroup$ Why is Al(OH)3 (and some more hydroxides from main groups) insoluble? $\endgroup$ – Georg Mar 22 '15 at 10:55
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The rule of thumb for hydroxide salts is actually that all hydroxides are insoluble save those of the alkali metals and the heavier alkaline earth metals (namely $\ce{Ca^{2+}}$, $\ce{Sr}^{2+}$, and $\ce{Ba}^{2+}$). There are many resources you can look up to check the solubility rules of certain ions, this one, for example.

Before researching this, I suspected that the solubility of the salts with a certain anion might be determined by the cation's control over its valence electrons, which is measurable through ionization energy. Sure enough, when you compare the first ionization energies of various metals, you find that those with the lowest first ionization energies are those that are soluble hydroxides.

Alkali Metals

The salts of all alkali metals are completely soluble because they have relatively little control over the outermost electron orbiting them, in addition to this being the only electron they must lose in order to attain an octet structure.

Alkaline Earth Metals

For the alkaline earth metals, the complete outermost $\mathrm{s}$ orbital must be lost in order to attain an octet structure. Because the orbital is complete, and among other reasons, the first ionization will require more energy than the alkali metal in the same period. In addition to this, the second electron must then also be lost. This means that overall, it requires much more energy to ionize this group than the alkali metals. This explains the insolubility of $\ce{Be(OH)2}$ and $\ce{Mg(OH)2}$.

The solubility of the hydroxides of the metals lower in the alkaline earth metal group can be explained by the shielding effect, which effectively says that the ability of a positive nucleus to attract electrons decreases as the amount of orbitals surrounding the nucleus decrease.

Transition Metals

The insolubility of the hydroxides of the transition metals can be explained both by the increasingly large ionization energies that result from successively removing electrons (or negatively charged hydroxide molecules in this case) from the metal cation, and the greater effective nuclear charge on the electrons across a period.

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  • $\begingroup$ there is quite a set of transition metal oxides soluble in water. $\endgroup$ – permeakra Jan 20 '16 at 17:44
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Solubility depends on complex interactions between many factors, but ionization and shielding aren't amongst them. IE and shielding have to do with how easily or likely you form the ionic compound at first place, and not how an ionic compound behaves after you make it.

In general solubility depends on...... i) lattice enthalpy of the ionic compound ii) solution enthalpy of the ionic compound iii) entropy (disorderedness) changes of both the solvent and the solute

In general group 1 metal compounds have low lattice enthalpies, because the ionic bonds formed by 1+ ions are relatively weak.

Also, for dissolving ionic compounds in general, solvent molecules will experience a decrease in entropy as solvation shells of eg water molecules surround the ions in a particular orientation in order to form ion-dipole attractions. This effect is not too significant for group 1 ionic compounds.

I'm not saying the above gives a complete story, but both factors above can both help explain the relatively high solubility of group-1 ionic compounds in general.

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  • $\begingroup$ Welcome to Chemistry.SE! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. For more information in general have a look at the help center. $\endgroup$ – Martin - マーチン Jan 20 '16 at 4:49

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