Why are salts of strongly electropositive elements less heavily hydrated in aqueous solution than the those of less electropositive elements? Shouldn't it be the other way round as the salts of electropositive elements are mostly ionic and hence strongly attracted by water molecules?
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$\begingroup$ related chemistry.stackexchange.com/questions/31868/… $\endgroup$– Martin - マーチン ♦Jun 25, 2015 at 7:07
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$\begingroup$ What state of aggregation are you referring to? Aqueous solution or solid (crystal)? $\endgroup$– Martin - マーチン ♦Jun 25, 2015 at 7:10
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$\begingroup$ @Martin-マーチン I'm talking about salts in aqueous medium. $\endgroup$– ApoorvJun 25, 2015 at 8:16
1 Answer
The degree of hydration in solution does not directly depend on electropositivity. It depends on the ratio of charge to volume. The higher charged and the smaller a fragment is, the more water molecules hydrate it.
Note that almost every salt in solution can be considered as separated ionic pairs. For those that do not seperate, i.e. have a significantly large covalent bonding proportion, you would be looking at the solvation of the non-dissociated ion pair meaning that you can’t properly compare that hydration to simple salts’ hydration.
Metals get more electropositive the closer they are to the bottom left of the periodic table with cesium being the (non-radioactive) corner. Electropositivity tends to increase with an atom’s size. But the larger the size, the less the degree of hydration is, because the charge per volume of the metal ion will be smaller!
Also, the most electropositive metals are alkaline metals, meaning the only create $\ce{M+}$ ions. However, the stronger the charge, the smaller an ion gets: $\ce{Ga^3+}$ has an ionic radius of $76\,\mathrm{pm}$ — smaller than $\ce{Li+}$ whose ionic radius is $90\,\mathrm{pm}$ although they are two periods apart! So $\ce{Ga^3+}$ will be a lot stronger hydrated than $\ce{Li+}$ because it contains much more charge per volume.