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Polarisability and Fajan's Rule

Knowing that ...

Question

Knowing that $N a + > M g^{2 +}$ and $S^{2 -} > C l^{-}$ , predict which compound will be the least soluble in a 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 A $M g S$
We know that, more the covalent character lower will be the solubility. Compound with smaller cation and larger anion are more covalent in nature. Among these, $M g S$ is more covalent and it is the least soluble in polar solvents.

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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^{-}

N a^{+}

M g^{2 +}

S^{2 -}

C l^{-}

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