11th Class Chemistry The s-Block Elements

  • question_answer 34) How would you explain? (i) \[BeO\] is insoluble but \[BeS{{O}_{4}}\] is soluble in water. (ii) \[BaO\] is soluble but\[BaS{{O}_{4}}\] is insoluble in water. (iii) \[LiI\] is more soluble than \[KI\] in ethanol.  

    Answer:

    (i) Due to small size of Beryllium ion,\[BeO\]is covalent and polymeric in nature and hence, is insoluble in water. The hydration energy of\[BeS{{O}_{4}}\] is higher than its lattice energy and hence it is soluble in water. (ii) Both \[BaO\] and \[BaS{{O}_{4}}\] are ionic compounds. The size of \[{{O}^{2-}}\]ion is very small in size in comparison to size of \[SO_{4}^{2-}\]ion while \[B{{a}^{2+}}\] ion is comparatively a large ion. Since, bigger anion stabilizes a bigger cation,\[BaS{{O}_{4}}\] is more stable than\[BaO\]. Hence,\[BaO\] is soluble due to small size of \[{{O}^{2-}}\]ion and high hydration energy while\[BaS{{O}_{4}}\] is insoluble due to low hydration energy. (iii) Covalent character of \[LiI\] is greater than that of \[KI\] hence \[LiI\] is more soluble in ethanol (non-polar solvent) than \[KI\].

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