Question

If $N{a^ + }$ ion is larger than $M{g^{2 + }}$ ion, and ${S^{2 - }}$ ion is larger than $C{l^ - }$ ion, which of the following will be less soluble in water?
(A) $NaCl$
(B) $N{a_2}S$
(C) $MgC{l_2}$
(D) $MgS$

Answer 1

Hint: To check whether the compound is soluble in water or not we must know that Lattice energy inversely varies with solubility in water. Hence, we will first check whether the given salts are formed of univalent or divalent ions. If the salt is composed of divalent ions, lattice energy will be high in that particular salt.

Complete Step by Step Solution:
The given data is:
Size of $N{a^ + }$ > Size of $M{g^{2 + }}$
Size of ${S^{2 - }}$ > Size of $C{l^ - }$

Let us consider all the four options given write their nature as:
$NaCl \to N{a^ + } + C{l^ - } \to $ Pair of univalent ions
$N{a_2}S \to 2N{a^ + } + {S^{2 - }} \to $ Pair of univalent and divalent ions
$MgC{l_2} \to M{g^{2 + }} + 2C{l^ - } \to $ Pair of univalent and divalent ions
$MgS \to M{g^{2 + }} + {S^{2 - }} \to $ Pair of divalent ions

Now, we know that the Lattice Energy of Bivalent ions is higher than that of univalent ions therefore, the Lattice Energy of the given salts in ascending order is given as:
$NaCl < N{a_2}S \sim MgC{l_2} < MgS$
And, Lattice energy inversely varies with solubility in water. Therefore, Solubility in the descending order is:
$NaCl > N{a_2}S \sim MgC{l_2} > MgS$
Thus, $MgS$ is less soluble in the water than all the other given salts.
Hence, the correct option is (D) $MgS$.

Note: Since this is a conceptual-based problem hence, it is essential that the options given in the question are analysed very carefully to give an accurate solution. Also, remember to identify the nature of ions i.e., univalent/divalent, and use the concept of lattice energy to find the solubility of salts in water.