What is oxidation number?

1 Answer
Apr 3, 2016

Answer:

The charge left on an atom in a molecule or ion, when all the bonding electrons are broken, with the charge assigned to the most electronegative atom.

Explanation:

Oxidation number and oxidation state are fictitious entities. They are a method of electron counting, and have no real physical significance. The use of oxidation numbers can be very helpful when balancing chemical equations, which I suppose is why it is still included on a chemistry syllabus.

Let's take a very simple example: the oxidation of iron metal by concentrated hydrochloric acid. Note, that there is nothing fundamental in what I am doing. The method of redox equations will NOT help predict reactivity; it is a way of ACCOUNTING for OBSERVED reactivity. Chemistry is still an experimental science, and the experiment is always done first.

If iron filings are placed in hydrochloric acid, the filings will foam and, eventually, will go up into solution. We say that the iron is oxidized up to #Fe^(2+)#; it has lost electrons:

#Fe(s) + rarr Fe^(2+) + 2e^-#; #(i)# OXIDATION

Iron metal is zerovalent: it is neutral with an oxidation number of #0#; (all of the ELEMENTS have oxidation number #0#). But where did the electrons go? We conceive that the electrons simultaneously acted to reduce the hydrogen of the acid to give dihydrogen gas:

#2HCl(aq) + 2e^(-)rarr H_2(g)uarr + 2Cl^-#; #(ii)#. REDUCTION

In the final redox equation we add (i) and (ii) so that electrons (which are after all conceptual entities) do not appear in the final redox equation:

#Fe(s) + 2HCl(aq) rarr Fe^(2+) + 2Cl^(-) + H_2(g)uarr#

Equally, I could have written:

#Fe(s) + 2HCl(aq) rarr FeCl_2(aq) + H_2(g)uarr#

Importantly, in all the reactions given, both MASS and CHARGE were balanced, as they must be in any chemical equation.

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