Question

Which of the following cannot be explained on the basis of Fajan's Rule?

• A. $A{g}_{2}S$ is much less soluble than $A{g}_{2}O$
• B. $Fe(OH{)}_3$ is much less soluble than $Fe(OH{)}_2$
• C. $BaC{O}_3$ is much less soluble than $MgC{O}_3$.
• D. Melting boint of $AlC{l}_3$ is much less than that of $NaCl$

Answer 1

The correct option is C $BaC{O}_3$ is much less soluble than $MgC{O}_3$.

(a) $A{g}_{2}S$ has a lower solubility than $A{g}_{2}O$ in water because $A{g}_{2}S$ is more covalent due to bigger size of $S^{2-}$.
(b) Smaller size of anion favours the formation of ionic bond. Out of $Ca{F}_2$ and $Ca{I}_2$, $Ca{F}_2$ is more ionic, as the size of $F^{-}$ ion is smaller than that of $I^{-}$ ion.
(c) The order of solubility of carbonates is: $BeC{O}_3>MgC{O}_3>CaC{O}_3>SrC{O}_3>BaC{O}_3$
This can only be explained on the basis of lattice energy and hydration energy.
(d) Melting point of $AlC{l}_3$ is much lower than that of $NaCl$ because the charge on $A{l}^{3+}$ is more than $N{a}^{+}$ and also size for $A{l}^{3+}$ is less than $N{a}^{+}$. So $AlC{l}_3$ is more covalent than $NaCl$.