How do we assign oxidation numbers to constituent atoms in polyatomic ions?

1 Answer
Nov 1, 2015

Answer:

Perhaps this is best served by an example. The complex ion, #Cr_2O_7^(2-)#, has a formal charge of #-2#; the oxidation number of chromium in this ion is #VI^+#.

Explanation:

The sum of all the oxidation numbers (of #Cr, VI^+#, and #O, -II#) leads to the charge on the ion. I could go to a simpler example and get the same result: i.e. ammonium ion, #NH_4^+#. The oxidation number of #N# is here #-III#, and the oxidation number of #H# is #I^+#. Of course, I am using the charge on the ion to get the oxidation state of its constituents, and vice versa, but this formalism allows one to quantify electron transfer, and the transfer of charge across species.

Sometimes (of course), the charge on the ion is also the oxidation number, i.e. #Cr^(3+)#, #Fe^(3+)#, #Mn^(2+)#; but these species are also formalisms. For instance #Fe^(3+)(aq)# is probably #Fe(OH_2)_6^(3+)#. If your unsatisfied with the explanations and definitions I have offered, state your objections and I (and others) will have another go.