Answer
a) $CCl_4$
b) $H_2O$
c) $H_2O$
d) $CCl_4$
e) $H_2O$
f) $H_2O$
g) $CCl_4$
Work Step by Step
a) $KrF_2$ is nonpolar due to its tetrahedral shape; therefore, it dissolves best in the nonpolar $CCl_4$.
b) $SF_2$ is polar due to its bent shape; therefore, it dissolves best in the polar $H_2O$.
c) $SO_2$ is polar due to its bent shape; therefore, it dissolves best in the polar $H_2O$.
d) $CO_2$ is nonpolar due to its linear shape; therefore, it dissolves best in the nonpolar $CCl_4$.
e) $MgF_2$ is polar due to its bent shape; therefore, it dissolves best in the polar $H_2O$.
f) $CH_2O$ is polar due to its asymetric polar $C-O$ bond; therefore, it dissolves best in the polar $H_2O$.
g) $CH_2=CH_2$ is nonpolar due to it containing only nonpolar $C-H$ bonds; therefore, it dissolves best in the nonpolar $CCl_4$.