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

1 Answer
Jan 28, 2017

Answer:

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?