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

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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

Relevant Posts - Relevant Videos

Oxidation – Reduction (Redox) Reactions – Balancing Redox Reactions

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”  6th Edition, Houghton Mifflin Company, 2009
  3. Ralph H. Petrucci, “General Chemistry”, 3rd Edition, Macmillan Publishing Co., 1982

Key Terms

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

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