# Which metal hydroxides are soluble?

Alkali metals (Li, Na, K, Rb, Cs) and the alkaline earth metals (Ca, Sr, Ba - sometimes Mg is included also) form hydroxides that are well-known to be soluble or at least slightly soluble and create strongly basic solutions. Apparently these solutions of hydroxides are also known as alkali.

Since the more well-known d metals (Cr, Fe, Mn, Cu, Zn, Ni, Co, Ti etc.) forms no soluble hydroxide, I used to assume that apart from the 9 metals stated at the top, there is no more metals that corresponds to soluble hydroxides.

However, while searching my (non-English) version of Constants of Inorganic Substances: A Handbook by Lidin, Andreyeva and Molochko, I discovered that Europium (II) hydroxide was written as if it is soluble in a chemical equation in the Europium section. Later, I did search for information on the compound with the keywords ""Europium II hydroxide" soluble" , and I found some results:

europium (II) sulfate is scarcely soluble in water, while europium (II) hydroxide dissolves readily and gives an alkaline reaction

Is there any other research on the compound Europium (II) hydroxide? Is it really soluble?

Which metal hydroxides, apart from those of those group 1 and 2 metals, are soluble? Is there any specific reason why europium (II) is such an anomaly? If there are other such cations, is there a specific reason for that particular cation to form such a soluble hydroxide?

• Thallium hydroxide is one example because the size of $\ce{Tl+}$ is appox. same to size of $\ce{K+}$ and thus thallium hydroxide dissolve in water easily and form a strong basic solution. Sep 30, 2016 at 12:16
• Thallium (I) hydroxide, of course. Mar 11, 2017 at 13:02

Europium (II) is an anomaly because it is, well, "II". Most metals in Period 5 or higher, apart from the alkali and alkaline earth metals, favor higher oxidation states which lead to highly insoluble hydroxides or oxides. Europium is different because the $$+2$$ ion is stabilized by having a spherically symmetric electron configuration $$[\ce{Xe}]\mathrm{4f^7}$$.
So we have a $$+2$$ ion similar in size and spherically symmetric electron configuration to the heavier alkaline earth metals, and $$\ce{Eu(II)}$$ compound tend to resemble their alkaline earth counterparts -- including the solubility of the hydroxide.