Question

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

Answer 1

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 `"Group 16"`, thus gains 2 electrons (to give an equivalent electronic condiguration to argon, to form:

`S+2e^(-) rarr S^(2-)`

Magnesium, of `"Group 2"`, thus loses 2 electrons (to give an equivalent electronic configuration to neon, to form:

`Mg rarr Mg^(2+) +2e^(-)`

Bromine, of `"Group 17"`, thus gains 1 electron (to give an equivalent electronic configuration to krypton, to form:

`1/2Br_2+e^(-) rarr Br^(-)`

And finally aluminum, loses three electrons to form `Al^(3+)` ion:

`Al rarr Al^(3+) +3e^(-)`

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?