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The beryllium iodide $\ce{BeI2}$ as covalent compound is apolar, and thus, it should be water non-soluble. However, it is. Making a research on the web, I learned it is just in the middle of being ionic. Dos it happen with the $\ce{MgI2}$ of $\ce{MgF2}$ as well? I usually explain the following to my students:

  • Solubitity "general" rule: "Similar dissolves similar".
  • Solubitity is also affected by differences in electronegativity and bonding polarity.
  • Solubility is also affected by the capability to displace charge and solvation (that is, intermolecular forces also apply to some extent).
  • As a practical rule, we can consider ionic compounds those formed by G1 or G2 with G16 or G17, but this rule has exceptions, like the rules of electronic configurations.

Is all stuff correct? Solubility in water of $\ce{BeH2}$ and $\ce{XeF2}$ can be also considered with these rules?

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    $\begingroup$ General rules have the general tendency of falling apart in a generally rapid manner. $\endgroup$
    – Jan
    Nov 19, 2017 at 15:12
  • $\begingroup$ Be is very small, iodine very big, no wonder Be2+ prefers to be bound to water. $\endgroup$
    – Mithoron
    Nov 19, 2017 at 17:19
  • $\begingroup$ Just because something is covalent and has zero (permanent) dipole moment does not mean It's doomed to water insolubility. There is zero permanent dipole in en.wikipedia.org/wiki/Hexamethylenetetramine, yet this is soluble in water up to almost a 1:1 mass ratio. $\endgroup$
    – Oscar Lanzi
    Nov 19, 2017 at 19:23
  • $\begingroup$ I know, similar dissolves the similar thumb rule is not general, as the principle of electronic configurations. I just came to remember the Usanovich theory, so to solvatate something, we have to be able to displace charge...Any way, the rules are to be improved... $\endgroup$
    – riemannium
    Nov 19, 2017 at 19:26

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