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If Na^(+) ion is larger than Mg^(2+) ion...

If `Na^(+)` ion is larger than `Mg^(2+)` ion and `S^(2-)` ion is larger than `Cl^(-)` ion, which of the following will be least soluble in water?

A

Sodium chloride

B

Soldium sulphide

C

Magnesium chloride

D

Magnesium sulphide

Text Solution

Verified by Experts

The correct Answer is:
D
Magnesium sulphide. Higher the lattice energy lower solubility. Out of the four combinations possible, the lattice energy of MgS (bi-bivalent ionic solid) is higher than those of `Na_(2) S, M_(2)Cl_(2)` (unibivalent or biunivalent ionic solids) and `NaCl` (univi-univalent ionic solids) and hence MgS is the least soluble.
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