GENERAL CHEMISTRY
DETERMINING OXIDATION NUMBERS (from Dr. Raynor)
If the compound is ionic, first separate it into its component ions. Treat each of the component ions separately,
using the rules given below, to assign oxidation numbers to each of the elements in each ion. [Note: the oxidation
number is for each individual atom in the compound, not for groups of atoms – i.e., each H atom in H2O has an
oxidation number of +1.]
The rules below are applied as follows: Rule 1 is the overall rule that must be applied to the charge on the entire
molecule or ion involved. Use Rules 2-4, in order, until only one element is left undefined. Then, use Rule 1 to
find the last oxidation number.
1.
The sum of all oxidation numbers for all atoms in a compound or ion must equal the total charge on that
compound or ion.
H2O: sum of all ox. nos. = 0 [i.e., 2 × (ox. no. H) + 1 × (ox. no. O) = 0]
SO24 : sum of all ox. nos. = –2 [i.e., 1 × (ox. no. S) + 4 × (ox. no. O) = –2]
Parts 1a through 1c result directly from applying Rule 1 to substances containing only 1 element:
a)
The oxidation number of each atom in a pure element is 0.
H2: ox. no. on H = 0
S8: ox. no. on S = 0
Mg: ox. no. on Mg = 0
b)
The oxidation number of each atom in a monatomic ion is equal to its charge.
Mg2+: ox. no. on Mg = +2
Cl : ox. no. on Cl = –1
c)
Polyatomic species containing a single element will share the total charge equally.
Hg 22 : ox. no. on each Hg = +1 [2 × (ox. no.Hg) = +2]
O 2 1 : ox. no. on each O = –½ [2 × (ox. no.O) = –1] [This is the superoxide anion]
2.
Assign known charges on atoms or polyatomic ions first. For example, Na in salts always has a charge of
+1 and so has an oxidation number of +1. Similarly, if the anion is a chloride, it always has a –1 charge,
and thus has an oxidation number of –1. [i.e., FeCl3: the ox. no. of Cl is –1. Applying Rule 1 we find that
the oxidation number on Fe is +3.]
3.
The oxidation number of H (in compounds) is always +1, unless it is bound to a metal to make a hydride,
MHn, where its oxidation number is –1. [The only other exception is in the pure element H2, where it is 0 –
see Rule 1a.]
HCl: ox. no. on H = +1 [Thus, ox. no. on Cl = –1]
CH4: ox. no. on H = +1 [Thus, ox. no. on C = –4]
MgH2: ox. no. on H = –1 [Thus, ox. no. on Mg = +2]
4.
The oxidation number of O (in compounds) is usually –2. The only exceptions, (other than the pure
elemental forms of O2 or O3, where it is 0), are as follows:
a)
The ox. no. of O in peroxides, (compounds containing the O2–2 ion), is –1: e.g. H2O2, Na2O2, MgO2,
etc.
b)
The ox. no. of O in superoxides, (compounds containing the O2– ion), is –½: e.g. KO2, RbO2, etc.
c)
The ox. no. of O in OF2 is +2.
–1–
R. Lalancette: 2/16/2016
[Note that all of these exceptions for O can be determined using Rule 1, after assigning the oxidation
numbers of the other elements in the formula. I.e., in MgO2, Mg must have a +2 charge, (it is in group 2),
so Rule 1 gives the following: total charge = 0 = (ox. no. Mg) + 2(ox. no. O) = +2 + 2(ox. no. O). Solving
gives (ox. no. O) = –1 = peroxide.]
Use these rules as follows. Set up the equation for Rule 1. Then apply Rules 2-4 in the order given until only one
element is left unassigned. Then, use your Rule 1 equation to solve for the single remaining oxidation number.
Finally, unless an element has a zero oxidation number, its oxidation number is always displayed with its sign –
i.e., we can have oxidation numbers of 0, +1, +2, –1, –2, etc. However, the following are NOT legitimate forms
for oxidation numbers: –, +, 1, 2, etc.
Examples:
FeSO4:
Rule 1: 0 = [1  (ox. no. Fe)] + [1  (ox. no. S)] + [4  (ox. no. O)].
The charge on Fe, and thus its oxidation number, is +2, since the charge on the sulfate anion is
–2. (Rule 2). We will next set the oxidation number of O to be –2, (Rule 3). This leaves 1
unknown – the one for S. Substituting our results for Fe and O, we get
0 = +2 + (ox. no. S) + 4(–2)
Solving, we get ox. no. S = +6.
Note – the same result for S occurs if we first separate the ions from one another. We then have
that the total charge on the sulfate anion, SO24 , is –2. The Rule 1 equation thus becomes
–2 = 1  (ox. no. S) + 4  (ox. no. O)
Substituting –2 for the oxidation number on oxygen gives
–2 = (ox. no. S) + (–8)
Solving, we again get (ox. no. S) = +6, (the same result).
CH4O:
Rule 1: 0 = [1  (ox. no. C)] + [4  (ox. no. H)] + [1  (ox. no. O)]
The ox. no. on H is set to +1, (Rule 2), and that on O is set to –2, (Rule 3). This gives
0 = (ox. no. C) + (4 – 2)
Solving, we get (ox. no. C) = –2.
N2O4:
Rule 1: 0 = [2  (ox. no. N)] + [4  (ox. no. O)]
The ox. no. on O is set to –2. (Rule 3) Substituting, we get
0 = [2(ox. no. N)] –8.
Solving, we get (ox. no. N) = +4.
Cr2O7–2:
Rule 1: –2 = [2  (ox. no. Cr)] + [7  (ox. no. O)] = [2 (ox. no. Cr)] –14
The ox. no. on O is set to –2. Substituting, we get:
2 (ox. no. Cr) = 14 – 2 = 12
Solving, we get (ox. no. Cr) = +6.
Fe(CH3COO)3: Rule 1: 0 = [1  (ox. no. Fe)] + [6  (ox. no. C)] + [9  (ox. no. H)] + [6  (ox. no. O)]
The ox. no. on Fe is +3, since the charge on Fe is +3, (due to the fact that the charge on each
acetate anion is –1 and there are 3 of them). Next, choose the ox. no. on H to be +1, (Rule 2),
and that on O to be –2, (Rule 3), and solve for the ox. no. on C, the last unassigned element:
0 = 1 (+3) + 6 (ox. no. C) + 9 (+1) + 6 (–2)
We get 6 (ox. no. C) = 0. Thus (ox. no. C) = 0.
–2–
R. Lalancette: 2/16/2016