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.
$LiF<NaF<KF<RbF<CsF$