Oxidation Numbers: Assigning oxidation numbers to atoms in a chemical compound - Examples

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) H₂S   b) SO₄^-2

 

α)

Step 1: Find the most electronegative atom

The most electronegative atom in H₂S 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 H₂S when it bonds to nonmetals (like S) is +1.

Since H₂S is neutral the sum of the oxidation numbers must be equal to 0

 

H₂S

(+1) * 2 + x = 0

2 + x = 0 and therefore x=oxidation number of S = -2

Therefore the oxidation number of S in H₂S is equal to -2

 

b)

Step 1: Find the most electronegative atom

The most electronegative atom in SO₄^-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 SO₄^-2 has a charge of -2 the sum of the oxidation numbers of the atoms of SO₄^-2 must be equal to -2. Let us suppose that the oxidation number of S is equal to x. 

 

SO₄^-2

x+ 4 * (-2) = -2

x – 8 = -2 and therefore x = +6

Therefore, the oxidation number of S in SO₄^-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 C₃H₈ and  b) S₄O₆^-2

a)

Step 1: Find the most electronegative atom

The most electronegative atom in C₃H₈ 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.

C₃H₈

 (3 * x) + (8 * 1) = 3x + 8 = 0

⇒ x = -8/3

Therefore, the oxidation number of C in C₃H₈ is -8/3.

 

b)

Step 1: Find the most electronegative atom

The most electronegative atom in S₄O₆^-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.

S₄O₆^-2

 (4 * x) + (6 * (-2)) = -2

⇒ 4x  - 12 = -2

⇒ 4x = 10 ⇒ x = +5/2

Therefore, the oxidation number of S in S₄O₆^-2 is equal to +5/2

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Relevant Posts - Relevant Videos

Oxidation – Reduction (Redox) Reactions – Balancing Redox Reactions

Oxidation - Reduction Reactions (Redox)

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References

1. David W. Oxtoby, H.P. Gillis, Alan Campion, “Principles of Modern Chemistry”, Sixth Edition, Thomson Brooks/Cole, 2008
2. Steven S. Zumdahl, “Chemical Principles”  6^th Edition, Houghton Mifflin Company, 2009
3. Ralph H. Petrucci, “General Chemistry”, 3^rd Edition, Macmillan Publishing Co., 1982

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Key Terms

oxidation, reduction, redox, oxidation number, balancing redox reactions, , covalent compound, , charge of a compound, , oxidation state,

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