Arrange the following compounds as decreasing order of their solubility:
$B e S O_{4}, M g S O_{4}, C a S O_{4}, S r S O_{4}, B a S O_{4}$

B e S O_{4} > M g S O_{4} > C a S O_{4} > S r S O_{4} > B a S O_{4}

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B e S O_{4} > M g S O_{4} > B a S O_{4} > C a S O_{4} > S r S O_{4}

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B e S O_{4} > B a S O_{4} > C a S O_{4} > S r S O_{4} > M g S O_{4}

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B a S O_{4} > C a S O_{4} > S r S O_{4} > M g S O_{4} > B e S O_{4}

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Solution

The correct option is A $B e S O_{4} > M g S O_{4} > C a S O_{4} > S r S O_{4} > B a S O_{4}$
The solubility of sulphates in water decreases down the group. The magnitude of the lattice enthalpy remains almost constant as the sulphate ion is so big that small increase in the size of cation from Be to Ba does not make any difference. The hydration enthalpy decreases from $B e^{2 +}$ to $B a^{2 +}$ as the size of the cation increases down the group.
Hence the solubility of sulphates of alkaline earth metal decreases down the group mainly due to decreasing hydration enthalpy from $B e^{2 +}$ to $B a^{2 +}$ . The high solubility of $B e S O_{4}$ and $M g S O_{4}$ is due to the high hydration enthalpy because of the smaller size of $B e^{2 +}$ and $M g^{2 +}$ ions. The reason for this is because the sulphates of alkaline earth metals decompose on heating giving their corresponding oxides and $S O_{3}$ .
The decreasing order of solubility would be-
$B e S O_{4} > M g S O_{4} > C a S O_{4} > S r S O_{4} > B a S O_{4} > R a S O_{4}$
Hence, the correct option is $A$ .

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