GENERAL CHEMISTRY
DETERMINING OXIDATION NUMBERS
If the compound is ionic, first separate it into its component ions. Treat each of the component ions separately, in order 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, (meaning that each H atom has an oxidation number of +1), and the O atom’s oxidation number is –2.]
The rules below are applied as follows. Rule 1 tells us to separate an ionic compound into its component ions before proceeding. Rule 2 is the overall rule that must be applied. It involves the charge on the entire molecule or ion of concern. If necessary, we then apply Rules 3 and 4, in that order, until only one element is left undefined. Then, Rule 2 is used to determine the remaining oxidation number.
Note: Unless an element has an oxidation number equal to zero, its oxidation number is always displayed with both its sign and its magnitude – i.e., the following are legitimate forms for oxidation numbers: 0, +1, +2, –1, –2. However, the following are NOT legitimate forms for oxidation numbers: –, +, 1, 2, etc.
1. Assign known charges on atoms or polyatomic ions first. I.e., always separate ionic species into their component ions before continuing. 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.] Another case: Pb(SO4)2 contains 2 sulfate anions, . Thus the cation must be Pb4+, according to the rule that the sum of all charged ions in an ionic compound must sum to 0.
2. Overall Charge = sum of all oxidation numbers. This means that the sum of all oxidation numbers on 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]
: sum of all ox. nos. = –2 [I.e., 1 × (ox. no. S) + 4 × (ox. no. O) = –2]
Parts 2a through 2c result directly from applying Rule 2 to substances that contain only 1 element:
a) The oxidation number on each atom in a pure element is 0.
H2: ox. no. on each H = 0
S8: ox. no. on each S = 0
Mg: ox. no. on each Mg = 0
b) The oxidation number on each atom in a monatomic ion is equal to its charge.
Mg2+: ox. no. on Mg = +2
: ox. no. on Cl = –1
c) Polyatomic species containing a single element will share the total charge equally.
: ox. no. on each Hg = +1 [I.e., 2 × (ox. no. Hg) = +2]
: ox. no. on each O = –½ [I.e., 2 × (ox. no. O) = –1] [This is the superoxide anion]
3. The oxidation number on H (in compounds) is usually +1, unless it is bound to a metal to make a hydride, MHn, where its charge, (and thus oxidation number), is –1. [The only other exception is in the pure element H2, where it is forced to be 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 on O (in compounds) is usually –2. The only exceptions, (other than the pure elemental forms of O2 or O3, where it is 0), are those where it is forced to be something else:
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; F HAS TO BE -1.
[Note that all of the exceptions given above for oxygen, O, can be determined using Rule 2, after assigning the oxidation numbers of the other elements in the formula. I.e., in MgO2, Mg is in Group 2 and must have a +2 charge, so Rule 2 gives the following:
Total charge = 0 = (ox. no. Mg) + 2(ox. no. O) = +2 + 2(ox. no. O).
Substituting that the oxidation number on Mg is +2 and solving for oxygen gives (ox. no. O) = –1.]
Thus, oxygen’s oxidation number would already be assigned after applying Rules 1 and 2. Thus exceptions 4a) and 4b) are unnecessary. Whenever the oxidation number on oxygen cannot be determined from the other elements present, (if any), then it is always assigned a value of –2.
Rule 4c) arises from the fact that fluorine is more electronegative than oxygen. [F is the most electronegative element and O is the second-most electronegative element.] Thus we assign F to have an oxidation number equal to its normal charge, i.e., –1, since it is in group 17. Applying Rule 2 gives an oxidation number of +2 to O.
Use these 4 rules as follows. Apply Rule 1 – i.e., if the substance is ionic, separate it into its component ions and treat each ion separately. I.e., apply Rules 2 – 4 on each ion, independent of the other. Next, set up the equation in Rule 2, (and solve it, if only 1 element is present). Finally, apply Rules 3 and 4, in that order, until only one element is left unassigned. Then, use your Rule 2 equation to solve for that sole remaining oxidation number.
Examples:
FeSO4: This is an ionic substance that contains the sulfate anion, . Since the charges on the ions must sum to 0, the charge on Fe, (and its oxidation number), must be +2. To determine the values for the rest of the elements, we must set up the Rule 2 Equation:
–2 = 1 ´ (ox. no. S) + 4 ´ (ox. no. O).
This equation contains 2 unknowns, so we must continue through the rules. Since there are no H atoms, we go to Rule 4 and set the oxidation number on O to be –2. This leaves 1 unknown – the one for S. Using our result for oxygen, we get
–2 = (ox. no. S) + 4(–2)
Solving, we get ox. no. S = +6.
CH4O: This is not ionic, so we skip Rule 1. Rule 2 gives the following equation:
0 = 1 ´ (ox. no. C) + 4 ´ (ox. no. H) + 1 ´ (ox. no. O)
Applying Rule 3, we set the ox. no. on H to +1. Since 2 unknowns remain, we apply Rule 4 and set the ox. no. on O to –2. This leaves 1 unknown only, that for C. We get
0 = (ox. no. C) + 4 ´ (+1) + 1 ´ (–2)
Solving, we get (ox. no. C) = –2.
N2O4: Not ionic – we skip Rule 1. Rule 2 yields
0 = 2 ´ (ox. no. N) + 4 ´ (ox. no. O)
Since there are no H atoms, we skip to Rule 4 and set the ox. no. on O to –2. We get
0 = 2 ´ (ox. no. N) + 4 ´ (–2) = 2(ox. no. N) –8.
Solving, we get (ox. no. N) = +4.
Cr2O7–2: We skip Rule 1, since the ions have already been separated – this is the resulting anion. Rule 2:
–2 = 2 ´ (ox. no. Cr) + 7 ´ (ox. no. O)
The ox. no. on O is set to –2, (Rule 4). Substituting, we get:
–2 = 2 ´ (ox. no. Cr) + 7 ´ (–2) = 2 (ox. no. Cr) – 14
Solving, we get (ox. no. Cr) = +6.
Fe(CH3COO)3: First, application of Rule 1 yields the presence of the acetate anion, CH3COO–. In order to balance the charge, the charge on Fe must be +3, (and so must be its oxidation number). To determine the oxidation numbers on the rest of the elements, we apply Rule 2 to the acetate anion:
–1 = 2 ´ (ox. no. C) + 3 ´ (ox. no. H) + 2 ´ (ox. no. O)
There are 3 unknowns. Applying Rules 3 and 4 gives oxidations of +1 on H and –2 on O. The Rule 2 equation becomes the following:
–1 = 2 ´ (ox. no. C) + 3 ´ (+1) + 2 ´ (–2) = 2 ´ (ox. no. C) +3 – 4 = 2 ´ (ox. no. C) –1
This gives
0 = 2 (ox. no. C)
And thus we get (ox. no. C) = 0.
NH4NO2: Applying Rule 1, we note that this compound contains the ammonium ion, NH4+, and the nitrite anion, NO2–. We will apply the remaining rules to each ion, separately:
NH4+: The Rule 2 equation is given by
+1 = 1 ´ (ox. no. N) + 4 ´ (ox. no. H)
Rule 3 gives (ox. no. H) = +1, and the equation becomes
+1 = 1 ´ (ox. no. N) + 4 ´ (+1) = (ox. no. N) + 4
Solving, we get (ox. no. N) = –3
NO2–: The Rule 2 equation is given by
–1 = 1 ´ (ox. no. N) + 2 ´ (ox. no. O)
Rule 4 gives (ox. no. O) = –2, and the equation becomes
–1 = 1 ´ (ox. no. N) + 2 ´ (–2) = (ox. no. N) – 4
Solving, we get (ox. no. N) = +3
Note that the oxidation number on N is –3 in NH4+ and is +3 in NO2–. NH4NO3 is a substance where different oxidation numbers occur on atoms from the same element. This can happen only when the atoms occur in different ions – i.e., the oxidation number on an element in the cation can be different from the oxidation number on the same element in the anion. This phenomenon can never occur among the atoms from identical elements within the same cation or within the same anion or within a non-ionic compound. [For example, the oxidation numbers on the N atoms in N2O4 must be identical to one another.]
– 3 – Raynor: 9/11/2010