U1.8 · Valence Electrons and Ionic CompoundsPractice · 价电子与离子化合物

Mg has 2 valence electrons and forms $\ce{Mg^2+}$; N has 5 and gains 3 to form $\ce{N^3-}$. Balancing charges, 3 × (+2) = 2 × (−3), gives $\ce{Mg3N2}$.

X has 3 valence electrons and forms X³⁺; Y has 6 and gains 2 to form Y²⁻. Balancing charges, 2 × (+3) = 3 × (−2), gives X₂Y₃ (like Al₂O₃).

Rb is in group 1 and, like Na and K, forms a +1 ion; S is in group 16 and forms $\ce{S^2-}$. It takes 2 $\ce{Rb+}$: $\ce{Rb2S}$, just like $\ce{Na2S}$. You can predict an unfamiliar element’s ion charge from its group.

Period 4 with 2 + 5 = 7 valence electrons: bromine (element 35), in group 17. Gaining 1 electron gives 4s²4p⁶, like Kr, so it forms $\ce{Br-}$.

The first two values differ by less than a factor of 2, but the third is 5.3 times the second: the third electron would come from the inner n = 2 shell, which costs far too much. So Mg loses 2 electrons to form $\ce{Mg^2+}$, which pairs with 2 $\ce{Cl-}$ as $\ce{MgCl2}$.