How do I know the likely oxidation states of metals, and non-metals?

1 Answer
Nov 21, 2016

Answer:

Well main group metal ions typically form an ion whose charge is equal to their Group number............i.e. isoelectronic with the last Noble Gas.

Explanation:

........And non-metals typically form an ion isoelectronic with the next Noble Gas.

A Group 17 element typically forms an ion with a single negative charge:

#1/2X_2 + e^(-) rarr X^-# #X=F, Cl, Br.......#

And a Group 16 element typically forms an ion with a double negative charge:

#1/2O_2(g) + 2e^(-) rarr O^(2-)#

In each case the element has formed an ion isoelectronic with the next (or last) Noble Gas. And we can go even farther than this, and consider Group 15.

#P(s) + 3e^(-) rarr P^(3-)#

And we can look at oxidation of the alkali metals (Group I):

#M(s) rarr M^(+) + e^-# #M=Li, Na, K, etc.#

And of the alkaline earths (Group 2):

#M(s) rarr M^(2+) + 2e^-# #M=Ca, Ba, Sr, etc.#

The point is that the Group number reflects electronic structure, i.e. the number of electrons present in the valence shell. Group I and Group II metals have 1 and 2 valence electrons respectively. As atomic number increases across a Period, nuclear charge increases accordingly, and non-metals, to the RIGHT of the Period as we face it, tend to be oxidizing, and ADD electrons to their valence shell.