Arrange solubility of sulphate of alkali metals in increasing order.

{L i}_{2} {S O}_{4} < {N a}_{2} {S O}_{4} < K_{2} {S O}_{4} < {R b S O}_{4}

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{L i}_{2} {S O}_{4} < {N a}_{2} {S O}_{4} < {R b S O}_{4} < K_{2} {S O}_{4}

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{N a}_{2} {S O}_{4} < {L i}_{2} {S O}_{4} < K_{2} {S O}_{4} < {R b S O}_{4}

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None of these

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Solution

The correct option is A ${L i}_{2} {S O}_{4} < {N a}_{2} {S O}_{4} < K_{2} {S O}_{4} < {R b S O}_{4}$
According to Fajan's rule, large anion can be polarised by the charge on small cation to induce covalent nature.
For the given sulphates of the first group, $L i_{2} S O_{4}$ will be most covalent and hence less solubility.
As we go down the group, the ionic nature increase and so the solubility.
Hence, the correct answer is option A.

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K_{2} {S O}_{4} . {A l}_{2} {({S O}_{4})}_{3} .24 H_{2} O

{({N H}_{4})}_{2} {S O}_{4} . {A l}_{2} {({S O}_{4})}_{3} .24 H_{2} O

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