Question

Statement 1: $AgI$ is less soluble in water than $AgF$ due to more polarization of $I^{-}$ ion in comparison to $F^{-}$ ion.
Statement 2: The melting point of $BaC{l}_2$ is higher than the melting point of $BeC{l}_2$ due to the greater ionic nature of $BaC{l}_2$.
Statement 3: The order of hydrated radii is: $A{l}^{3+}\left(aq\right)>M{g}^{2+}\left(aq\right)>N{a}^{+}\left(aq\right)$

• A. T T T
• B. T T F
• C. T F T
• D. F T T

Answer 1

The correct option is A T T T

As $I^{-}$ is larger in size than $F^{-}$, the polarisability of $I^{-}$ is greater than that of $F^{-}$ and $AgF$ shows less covalent character thus more soluble in polar solvents.
Smaller the cation, more is the covalent character. Thus $BaC{l}_2$ is less covalent (more ionic) than $BeC{l}_2$ due to larger size of $B{a}^{2+}$ and will be having higher melting point.
Smaller the cation more will be the extent of hydration. The order for the size is: $A{l}^{3+}<M{g}^{2+}<N{a}^{+}$. Hence, the order of hydrated radii will be just reverse, i.e., $A{l}^{3+}\left(aq\right)>M{g}^{2+}\left(aq\right)>N{a}^{+}\left(aq\right)$