Question

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

Answer 1

`"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-)`. For an `S-O` bond, oxygen is the more electronegative. We break the bond, and get `S^(2+)`, and `O^(2-)` (i.e. the 2 electrons that constitute the `S-O` go to `O`). In fact oxygen generally takes an oxidation of `-II` in its compounds, and it certainly does so here. We do this for all the `S-O` bonds, and we get `S^(VI+)` and `4xxO^(2-)`.

Given that the sum of the oxidation numbers, of `S` and `O`, equals the charge on the ion, `-2`, an assignment of a `+VI` oxidation number to sulfur (typically we use Roman numerals) is necessarily consistent, as is the formal `-2` 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.