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Exercises · 10.29

Q.How would you explain the following observations?

(i) BeO is almost insoluble but BeSO4 is soluble in water,
(ii) BaO is soluble but BaSO4 is insoluble in water,
(iii) LiI is more soluble than KI in ethanol.
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(i) BeO almost insoluble but BeSO4_4 soluble

BeO consists of two small, similarly-sized ions (Be2+^{2+} and O2−^{2-}), which gives it an exceptionally high lattice enthalpy that hydration enthalpy cannot overcome, so BeO is essentially insoluble (this also mirrors why alkaline earth oxide/hydroxide solubility rises down the group — Be being the smallest cation gives the least soluble oxide/hydroxide). BeSO4_4, however, pairs the very small Be2+^{2+} with the much larger sulphate ion; when a small cation is paired with a large anion, the lattice enthalpy is comparatively lower (dominated by the large anion, similar to a large-cation/small-anion mismatch) and the very high hydration enthalpy of the small Be2+^{2+} ion is enough to overcome it, making BeSO4_4 soluble.

(ii) BaO soluble but BaSO4_4 insoluble

With the small oxide ion, lattice enthalpy is controlled mainly by the (comparatively small) anion, and as we saw for the hydroxides (Example 4), solubility of the Group 2 oxide/hydroxide series increases down the group, so the large Ba2+^{2+} pairs favourably enough with the small O2−^{2-} for BaO to be soluble. In BaSO4_4, however, the anion itself is large, so lattice enthalpy stays high regardless of cation size, while Ba2+^{2+}'s hydration enthalpy (already low, being the largest Group 2 cation) is too small to compensate — so BaSO4_4 is insoluble (this is exactly the carbonate/sulphate trend of Example 5, where solubility falls down the group).

(iii) LiI more soluble than KI in ethanol …

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