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In the case of $CaC{l}_2$In an aqueous medium, $CaC{l}_2$is dissociated as $C{a}^{2+}$and $C{l}^{-}$ions which means it produces one ion of $C{a}^{2+}$and $C{l}^{-}$two ions of in the solution, therefore, the number of anions is twice the number of cation

$CaC{l}_2\left(s\right)→C{a}^{2+}\left(aq\right)+2C{l}^{-}\left(aq\right)$

In scene (B) four cations are having +2 charge and eight anions have a -1 charge Hence, scene (B) is the best representation of this case.

In the case of $L{i}_{2}S{O}_4:$In an aqueous medium, $L{i}_{2}S{O}_4$is dissociated as $L{i}^{+}$and $S{O_4}^{-2}$ions which means it produces two ions of $L{i}^{+}$and one ion of $S{O_4}^{-2}$in the solution therefore the number of cation is twice the number of the anion.

$L{i}_{2}S{O}_4\left(s\right)→2L{i}^{+}\left(aq\right)+S{O_4}^{-2}\left(aq\right)$

In scene (C) four anions are having -2 charge and eight cations have a +1 charge.

Hence, scene (C) is the best representation of this case.

In the case of$N{H}_{4}Br:$ In an aqueous medium, $N{H}_{4}Br$is dissociated as$N{H_4}^{+}$ and$B{r}^{-}$ ions which means it produces one ion of$N{H_4}^{+}$ and$B{r}^{-}$ one ion of in the solution therefore the number of cations is equal to number of the anions

$N{H}_{4}Br\left(s\right)→N{H_4}^{+}\left(aq\right)+B{r}^{-}\left(aq\right)$

In scene (A) four anions are having -1 charge and four cations have a +1 charge.

Hence, scene (A) is the best representation of this case.