Oxidation Numbers: Assigning oxidation numbers to atoms in chemical compounds  Examples
In a previous post entitled “Oxidation – Reduction (Redox) Reactions – Balancing Redox Reactions” the general steps for assigning oxidation numbers to atoms in a molecule or ion were presented. Assignment of oxidation numbers is a basic step for balancing redox reactions.
A few solved examples are given below:
Example #1
Calculate the oxidation number of the S atom in the following compounds: α) H_{2}S b) SO_{4}^{2}
α)
Step 1: Find the most electronegative atom
The most electronegative atom in H_{2}S is the S atom
Step 2: The oxidation number of the most electronegative atom is equal to the charge of its ion
Since the charge of sulfur’s ion is unknown let’s suppose that is equal to x
Step 3: Oxidation numbers that are not known have to be calculated.
The oxidation number of H_{2}S when it bonds to nonmetals (like S) is +1.
Since H_{2}S is neutral the sum of the oxidation numbers must be equal to 0
H_{2}S
(+1) * 2 + x = 0
2 + x = 0 and therefore x=oxidation number of S = 2
Therefore the oxidation number of S in H_{2}S is equal to 2
b)
Step 1: Find the most electronegative atom
The most electronegative atom in SO_{4}^{2} is the O atom
Step 2: The oxidation number of the most electronegative atom is equal to the charge of its ion
The charge of oxygen ion is 2 in covalent compounds
Step 3: Oxidation numbers that are not known have to be calculated.
Since SO_{4}^{2} has a charge of 2 the sum of the oxidation numbers of the atoms of SO_{4}^{2} must be equal to 2. Let us suppose that the oxidation number of S is equal to x.
SO_{4}^{2}
x+ 4 * (2) = 2
x – 8 = 2 and therefore x = +6
Therefore, the oxidation number of S in SO_{4}^{2} is +6.
The above two examples show that an element can have more than one oxidation number (oxidation state) depending upon the other elements to which is attached.
Example #2
Calculate the oxidation number of the elements: a) propane C_{3}H_{8} and b) S_{4}O_{6}^{2}
a)
Step 1: Find the most electronegative atom
The most electronegative atom in C_{3}H_{8} is the C atom
Step 2: The oxidation number of the most electronegative atom is equal to the charge of its ion
The oxidation number of oxygen x is unknown and it has to be calculated. The oxidation number of hydrogen is +1 in covalent compounds.
Step 3: Oxidation numbers that are not known have to be calculated.
C_{3}H_{8}
(3 * x) + (8 * 1) = 3x + 8 = 0
⇒ x = 8/3
Therefore, the oxidation number of C in C_{3}H_{8} is 8/3.
b)
Step 1: Find the most electronegative atom
The most electronegative atom in S_{4}O_{6}^{2} is the O atom
Step 2: The oxidation number of the most electronegative atom is equal to the charge of its ion
The oxidation number of oxygen is equal to 2 in covalent compounds. Let us suppose that x is the oxidation number of the S atom
Step 3: Oxidation numbers that are not known have to be calculated.
S_{4}O_{6}^{2}
(4 * x) + (6 * (2)) = 2
⇒ 4x  12 = 2
⇒ 4x = 10 ⇒ x = +5/2
Therefore, the oxidation number of S in S_{4}O_{6}^{2} is equal to +5/2
Relevant Posts  Relevant Videos
Oxidation – Reduction (Redox) Reactions – Balancing Redox Reactions
Oxidation  Reduction Reactions (Redox)
References

David W. Oxtoby, H.P. Gillis, Alan Campion, “Principles of Modern Chemistry”, Sixth Edition, Thomson Brooks/Cole, 2008

Steven S. Zumdahl, “Chemical Principles” 6^{th} Edition, Houghton Mifflin Company, 2009

Ralph H. Petrucci, “General Chemistry”, 3^{rd} Edition, Macmillan Publishing Co., 1982
Key Terms
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