Given that Na^{+} ion and S^{2-} ion are larg | vedclass

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(A) The solubility of ionic compounds is governed by the lattice energy and hydration energy. Compounds with high lattice energy are generally less so

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$MgS$

Vedclass · Answer

(A) The solubility of ionic compounds is governed by the lattice energy and hydration energy. Compounds with high lattice energy are generally less soluble in water. According to Fajan's rule,the covalent character increases with the increase in the size of the anion and the charge density of the cation. $MgS$ has a high degree of covalent character due to the presence of a small $Mg^{2+}$ cation and a large $S^{2-}$ anion,which leads to high lattice energy and low solubility in water. Therefore,the order of solubility is $NaCl > MgCl_2 > Na_2S > MgS$. Thus,$MgS$ is the least soluble.

Given that $Na^{+}$ ion and $S^{2-}$ ion are larger than $Cl^{-}$ ion,which of the following will be least soluble in water?

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