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(D) The solubility $(S)$ is calculated from $K_{sp}$ as follows:$1$. $AgBr \rightleftharpoons Ag^+ + Br^-$; $K_{sp} = S^2 \implies S = \sqrt{K_{sp}} =

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$Zn(OH)_2 > AgBr > Hg_2Cl_2$

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(D) The solubility $(S)$ is calculated from $K_{sp}$ as follows:$1$. $AgBr ightleftharpoons Ag^+ + Br^-$; $K_{sp} = S^2 \implies S = \sqrt{K_{sp}} = \sqrt{5.0 	imes 10^{-13}} \approx 7.07 	imes 10^{-7} \ M$$2$. $Zn(OH)_2 ightleftharpoons Zn^{2+} + 2OH^-$; $K_{sp} = 4S^3 \implies S = \sqrt[3]{K_{sp}/4} = \sqrt[3]{1.0 	imes 10^{-15} / 4} = \sqrt[3]{0.25 	imes 10^{-15}} \approx 0.63 	imes 10^{-5} = 6.3 	imes 10^{-6} \ M$$3$. $Hg_2Cl_2 ightleftharpoons Hg_2^{2+} + 2Cl^-$; $K_{sp} = 4S^3 \implies S = \sqrt[3]{K_{sp}/4} = \sqrt[3]{1.3 	imes 10^{-18} / 4} = \sqrt[3]{0.325 	imes 10^{-18}} \approx 0.688 	imes 10^{-6} = 6.88 	imes 10^{-7} \ M$Comparing the values: $Zn(OH)_2 (6.3 	imes 10^{-6}) > AgBr (7.07 	imes 10^{-7}) > Hg_2Cl_2 (6.88 	imes 10^{-7})$.Thus, the correct order is $Zn(OH)_2 > AgBr > Hg_2Cl_2$.

Salt	$K_{sp}$ at $298 K$
$AgBr$	$5.0 \times 10^{-13}$
$Zn(OH)_2$	$1.0 \times 10^{-15}$
$Hg_2Cl_2$	$1.3 \times 10^{-18}$

The correct order of solubility of the given salts in water at $298 K$ is:
Salt $K_{sp}$ at $298 K$
$AgBr$ $5.0 \times 10^{-13}$
$Zn(OH)_2$ $1.0 \times 10^{-15}$
$Hg_2Cl_2$ $1.3 \times 10^{-18}$

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The correct order of solubility of the given salts in water at $298 K$ is:Salt$K_{sp}$ at $298 K$$AgBr$$5.0 \times 10^{-13}$$Zn(OH)_2$$1.0 \times 10^{-15}$$Hg_2Cl_2$$1.3 \times 10^{-18}$