Knowing that $N a^{+} > M g^{2 +}$ and $S^{2 -} > C l^{-}$ , predict which compound will be least soluble in polar solvent?

M g S

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N a_{2} S

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M g C l_{2}

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N a C l

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Solution

The correct option is D $M g S$
Higher the lattice energy lower will be the solubility. Out of the four combination possible, the lattice energy of $M g S$ (Bi-bivalent ionic solid) is higher than those of $N a_{2} S$ , $M g C l_{2}$ (uni-bivalent or bi-univalent ionic solids) and $N a C l$ (uni-univalent ionic solids) and hence $M g S$ is the least soluble.

So, the correct answer is option A.

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Similar questions

N a + > M g^{2 +}

S^{2 -} > C l^{-}

N a + > M g^{2 +}

S^{2 -} > C l^{-}

N a^{+}

M g^{2 +}

S^{2 -}

C l^{-}

N a^{+}

M g^{2 +}

S^{2 -}

C l^{-}

M g C l_{2} \to M g^{2 +} + 2 C l^{-}

P b B r_{2} \to P b^{2 +} + 2 B r^{-}

N a C l \to N a^{+} + C l^{-}

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