Question

Lattice energies of $Be{F}_2,Mg{F}_2,Ca{F}_{2}andBa{F}_2$ are $-2906,-2610,-2459,and-2367kJmo{l}^{-1}$, respectiely. Hydration energies of $B{e}^{2+},M{g}^{2+},C{a}^{2+},B{a}^{2+},and{F}^{-}$ are $-2494,-1921,-1577,-1305,and-457kJmo{l}^{-1}$, respectively. Which of the flourides is not soluble in water.

• A. $Be{F}_2$
• B. $Mg{F}_2$
• C. $Ca{F}_2$
• D. $Ba{F}_2$

Answer 1

The correct option is C $Ca{F}_2$

For a compound to be soluble,
$\Delta H_{lattice}+\Delta H_{hydration}<0$
(a) $Be{F}_2$:
Lattice enrgy =$+2906kJ$
Hydration energy =$-2494+2\times (-457)$
=$-502kJ$

(b) $Mg{F}_2$
Lattice enrgy =$+2610kJ$
Hydration energy =$-1921+2\times (-457)$
=$-225kJ$

(c) $Ca{F}_2$
Lattice enrgy =$+2459kJ$
Hydration energy =$-1577+2\times (-457)$
=$+32kJ$

(d) $Ba{F}_2$
Lattice enrgy =$+2367kJ$
Hydration energy =$-1305+2\times (-457)$
=$-148kJ$