For which salt are there EQUAL numbers of electrons with respect to the anion, and the cation?

$\left(i\right)$ $\text{sodium chloride;}$ $\left(i i\right)$ $\text{lithium fluoride;}$ $\left(i i i\right)$ $\text{potassium bromide;}$ $\left(i v\right)$ $\text{rubidium bromide.}$

Sep 13, 2017

Well, you have a series of binary PAIRS of ions here....and so I think it is $\text{option 4}$

Explanation:

And so we gots $N {a}^{+} C {l}^{-}$, i.e. 10 electrons associated with the $N {a}^{+}$ cation, and 18 electrons associated with the $C {l}^{+}$ anion.

And for $L {i}^{+} {F}^{-}$, i.e. 2 electrons associated with the $L {i}^{+}$ cation, and 10 electrons associated with the $C {l}^{-}$ anion.

And for ${K}^{+} B {r}^{-}$, i.e. 18 electrons associated with the ${K}^{+}$ cation, and 36 electrons associated with the $B {r}^{-}$ anion.

And for $R {b}^{+} B {r}^{-}$, i.e. 36 electrons associated with the $R {b}^{+}$ cation, and 36 electrons associated with the $B {r}^{-}$ anion.