Lattice dissociation energies of $B e F_{2}, M g F_{2}, C a F_{2}$ and $B a F_{2}$ 2906, 2610, 2459 and 2367 kJ/mol Respectively. Hydration energies $B e^{2 +}, M g^{2 +}, C a^{2 +}, B a^{2 +}$ and $F^{-} - 2494, - 1921, - 1577, - 1305$ and -457 kJ mol respectively. Which of the fluorides is not soluble in water?

B e F_{2}

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M g F_{2}

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C a F_{2}

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B a F_{2}

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Solution

The correct option is D $B a F_{2}$
For an ionic compound to be soluble in water,
$Δ H_{h y d r a t i o n} + Δ H_{l a t t i c e} < 0$
A) $B e F_{2}$

$Δ H_{h y d r a t i o n} + Δ H_{l a t t i c e} = (- 2494 + (2 \times - 457) + 2906 = - 502$

B) $M g F_{2}$
$Δ H_{h y d r a t i o n} + Δ H_{l a t t i c e} = (- 1921 + (2 \times - 457) + 2610 = - 225$

C) $C a F_{2}$
$Δ H_{h y d r a t i o n} + Δ H_{l a t t i c e} = (- 1577 + (2 \times - 457) + 2459 = - 32$

D) $B a F_{2}$
$Δ H_{h y d r a t i o n} + Δ H_{l a t t i c e} = (- 1305 + (2 \times - 457) + 2267 = 48$
Hence $B a F_{2}$ is insoluble in water.

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