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

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

1 Answer
anor277
Sep 13, 2017

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

Explanation:

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

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

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

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

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