Question

The solubility of fluorides of alkali metals in water is

• A. $ ~\text{LiF }>\text{ NaF }>\text{ KF }>\text{ RbF }>\text{ CsF} $
• B. $ \text{CsF }>\text{ RbF }>\text{ KF }>\text{ NaF }>\text{ LiF} $
• C. $ \text{LiF }>\text{ KF }>\text{ NaF }>\text{ CsF }>\text{ RbF} $
• D. $ \text{CsF }>\text{ KF }>\text{ NaF }>\text{ RbF }>\text{ LiF} $

Answer 1

Answer: (B)

$LiI > NaI > KI > RbI > Cal$ (solubility)
Lattice energy $\frac{1}{ r _{+}+ r _{-}}$
Since, $r _{-}> r _{+}$the sum will not change too much as $r _{+}$increases. As a result, the lattice energy will not change significantly as $r _{+}$increases. Thus decreases in lattice energy is not as fast as decreases in hydration energy. Thus, more decreases in hydration energy result in a decrease in solubility. Thus, solubility decreases as
$LiI > NaI > KI > RbI > Cal$
With a small anion as $\left( F ^{-}\right)$the lattice energy decreases more rapidly along a series of salts with increasing cation size and thus solubility increases.
$LiF < NaF < KF < RbF < CsF$