Which of the following is/are correct order of solubility in water?

L i I < L i B r < L i C l < L i F

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B e (O H)_{2} < M g (O H)_{2} < C a (O H)_{2} < S r (O H)_{2} < B a (O H)_{2}

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N a H C O_{3} < K H C O_{3} < R b H C O_{3} < C s H C O_{3}

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H g F_{2} > H g C l_{2} > H g B r_{2} > H g I_{2}

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Solution

The correct option is D $H g F_{2} > H g C l_{2} > H g B r_{2} > H g I_{2}$
a) $L i F$ is the least soluble in water because it has highest lattice enthalpy.
b) This trend can be explain on the basis of decreasing lattice energy down the group.
c) Solubility of bicarbonates increases down the group, because smaller the cation greater is the covalent character, hence lesser is the solubility in water.
d) Here solubility decreases gradually from $H g F_{2}$ to $H g I_{2}$ because as the size of the anion increases from $F^{-}$ to $I^{-}$ the covalent character also increases in the compound and hence solubility decreases in water.

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