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

Question

$(i)$ $"sodium chloride;"$
$(ii)$ $"lithium fluoride;"$
$(iii)$ $"potassium bromide;"$
$(iv)$ $"rubidium bromide."$

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Answer 1 · 2017-09-13T16:34:58

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

And so we gots $Na^+Cl^-$ , i.e. 10 electrons associated with the $Na^+$ cation, and 18 electrons associated with the $Cl^+$ anion.

And for $Li^+F^-$ , i.e. 2 electrons associated with the $Li^+$ cation, and 10 electrons associated with the $Cl^-$ anion.

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

And for $Rb^+Br^-$ , i.e. 36 electrons associated with the $Rb^+$ cation, and 36 electrons associated with the $Br^-$ anion.