How do you find the oxidation number of the central atom?

1 Answer
Jan 18, 2017

"Oxidation number is the charge left on the central atom........."Oxidation number is the charge left on the central atom.........

Explanation:

Oxidation number is the charge left on the central atom when all the bonding pairs of electrons are removed, with the charge (the electrons) assigned to the MOST electronegative atom.

So clearly we need an example. Let's take SO_4^(2-)SO24. For an S-OSO bond, oxygen is the more electronegative. We break the bond, and get S^(2+)S2+, and O^(2-)O2 (i.e. the 2 electrons that constitute the S-OSO go to OO). In fact oxygen generally takes an oxidation of -IIII in its compounds, and it certainly does so here. We do this for all the S-OSO bonds, and we get S^(VI+)SVI+ and 4xxO^(2-)4×O2.

Given that the sum of the oxidation numbers, of SS and OO, equals the charge on the ion, -22, an assignment of a +VI+VI oxidation number to sulfur (typically we use Roman numerals) is necessarily consistent, as is the formal -22 ionic charge of sulfate ion.

There should be many other examples of oxidation numbers and redox reactions on these boards. Of course, you have to find them. Here is a start.