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:



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> europium (II) sulfate is scarcely soluble in water, while europium (II) hydroxide dissolves readily and gives an alkaline reaction
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([Concise Encyclopedia Chemistry, page 392](https://books.google.com.vn/books?id=Owuv-c9L_IMC&pg=PA392&lpg=PA392&dq=%22europium+II+hydroxide%22+soluble&source=bl&ots=zXsZ_jiMjd&sig=5tDf5QGAxKwf-RlATAa3zJnFfkc&hl=en&sa=X&redir_esc=y#v=onepage&q=%22europium%20II%20hydroxide%22%20soluble&f=false))


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?




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.


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