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Assigning oxidation numbers to atoms in a chemical compound

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 ionwere 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 3Oxidation 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
author

DESH DEEPAK A P S CHAUHAN

M.Sc. CHEMISTRY!

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