# How would you write the following elements as an ion: S, Mg, Br, Al?

Jan 28, 2017

Well, metals tend to lose their valence electrons.........to give the electronic configuration of the preceding Noble gas.

#### Explanation:

And non-metals tend to gain valence electrons, so as to achieve the electronic configuration of the next Noble gas.

Sulfur, of $\text{Group 16}$, thus gains 2 electrons (to give an equivalent electronic condiguration to argon, to form:

$S + 2 {e}^{-} \rightarrow {S}^{2 -}$

Magnesium, of $\text{Group 2}$, thus loses 2 electrons (to give an equivalent electronic configuration to neon, to form:

$M g \rightarrow M {g}^{2 +} + 2 {e}^{-}$

Bromine, of $\text{Group 17}$, thus gains 1 electron (to give an equivalent electronic configuration to krypton, to form:

$\frac{1}{2} B {r}_{2} + {e}^{-} \rightarrow B {r}^{-}$

And finally aluminum, loses three electrons to form $A {l}^{3 +}$ ion:

$A l \rightarrow A {l}^{3 +} + 3 {e}^{-}$

In all these instances, the metal, electron-rich, tends to be reducing. And the non-metal, electron poor on account of its high nuclear charge, tends to be oxidizing. Capisce? How did I use the Periodic Table in this answer?