HO 8 -OXIDATION NUMBERS
Oxidation-reduction reactions (redox reactions) are reactions in which
electrons are lost by an atom or ion in one reactant and gained by an
atom or ion in another reactant. Although electrons are gained and lost in
these reactions, the balanced equation for a redox reaction does not
show the electrons that are being transferred. In order to tell whether
a redox reaction has occurred or not, we need a way to keep track of
electrons. The best way to do so is by assigning oxidation numbers to the
atoms or ions involved in a chemical reaction.
Oxidation numbers are hypothetical numbers assigned to an individual
atom or ion present in a substance using a set of rules. Oxidation
numbers (or oxidation states as they are also called) can be positive,
negative, or zero. It is VERY IMPORTANT to remember that oxidation
numbers are always reported for one individual atom or ion and not for
groups of atoms or ions.
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HO 8 -OXIDATION NUMBERS
Oxidation Number Rules
1. The oxidation number for an atom in its elemental form is always
zero.
o A substance is elemental if both of the following are true:
 only one kind of atom is present
 charge = 0
o Examples:
 S8: The oxidation number of S = 0
 Fe: The oxidation number of Fe = 0
2. The oxidation number of a monoatomic ion = charge of the
monatomic ion.
o Examples:
 Oxidation number of S2- is -2.
 Oxidation number of Al3+ is +3.
3. The oxidation number of all Group 1A metals = +1 (unless
elemental).
4. The oxidation number of all Group 2A metals = +2 (unless
elemental).
5. Hydrogen (H) has two possible oxidation numbers:
o +1 when bonded to a nonmetal
o -1 when bonded to a metal
6. Oxygen (O) has two possilbe oxidation numbers:
o -1 in peroxides (O22-)....pretty uncommon
o -2 in all other compounds...most common
7. The oxidation number of fluorine (F) is always -1.
8. The sum of the oxidation numbers of all atoms (or ions) in a
neutral compound = 0.
9. The sum of the oxidation numbers of all atoms in a polyatomic ion =
charge on the polyatomic ion.
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HO 8 -OXIDATION NUMBERS
When assigning oxidation numbers to the elements in a substance, take a
systematic approach. Ask yourself the following questions:
1.
2.
3.
4.
5.
Is the substance elemental?
Is the substance ionic?
If the substance is ionic, are there any monoatomic ions present?
Which elements have specific rules?
Which element(s) do(es) not have rules?
o Use rule 8 or 9 from above to calculate these.
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HO 8 -OXIDATION NUMBERS
Example: Determine the oxidation number of each element in Na2SO4.
1. Is the substance elemental?
o No, there are three elements present.
2. Is the substance ionic?
o Yes, because metal + non-metal = ionic.
3. If the substance is ionic, are there any monoatomic ions present?
o Yes, the sodium ion (Na+) is monoatomic.
o Therefore, the oxidation number of Na is +1.
4. Which elements have specific rules?
o Oxygen has a rule (-2 in most compounds).
o Oxidation number of O = -2.
5. Which element(s) do(es) not have rules?
o S does not have a rule.
o Use rule 8 (and a little simple algebra) to find the oxidation
number of S.
 Let S = oxidation number of sulfur.
 According to rule 8:
 (# Na) (Oxid. # of Na) + (# S)(Oxid. # S) + (#
Oxygens) (Oxid. # of O) = 0
 So: 2(+1) + S + 4(-2) = 0
 Solve for S using algebra. S = +6.
 Therefore, the oxidation number of S = +6
Example: Determine the oxidation number of each element in K2C2O4.
1. Is the substance elemental?
1. No, there are three elements present.
2. Is the substance ionic?
o Yes, because metal + non-metal = ionic.
3. If the substance is ionic, are there any monoatomic ions present?
o Yes, the potassium ion (K+) is monoatomic.
o Therefore, the oxidation number of K is +1.
4. Which elements have specific rules?
o Oxygen has a rule (-2 in most compounds).
o Oxidation number of O = -2.
5. Which element(s) do(es) not have rules?
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HO 8 -OXIDATION NUMBERS
o
o
C does not have a rule.
Use rule 8 (and a little simple algebra) to find the oxidation
number of C.
 Let C = oxidation number of carbon.
 According to rule 8:
 (# K) (Oxid. # of K) + (# C)(Oxid. # C) + (#
Oxygens) (Oxid. # of O) = 0
 So: 2(+1) + 2C + 4(-2) = 0
 Solve for C using algebra. C = +3.
 Therefore, the oxidation number of C = +3
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HO 8 -OXIDATION NUMBERS
Practice Problems
Practice problems:
Determine the oxidation number of each element in the following
compounds:
1.
2.
3.
4.
5.
6.
Ba(NO3)2
NF3
(NH4)2SO4
NH3
Na2S
HClO4
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