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Oxidation Number
The oxidation number of an element can be said to be an index number indicating the amount of oxidizing or reducing that an atom of the element in its free state requires to get converted to the state it is found in the compound under study.
Oxidation number also expresses the combining power of elements in a compound, and thus can be used to predict the formulae of the products of chemical reductions.
For free and uncombined elements, the oxidation number is zero. Hence,
the oxidation number of hydrogen, H_{2}, chlorine, Cl_{2},
copper, Cu and sodium, Na is zero.
 The oxidation number of hydrogen (H) is +1 in all compounds except in hydrides. In hydrides, example,
sodium hydride, NaH, it is –1.
 Except in peroxides and superoxides, the oxidation number of oxygen in all compounds is –2.
In peroxides, e.g. H_{2}O_{2}, it is –1, and in superoxides, e.g. KO_{2}, it is  1/2.
 The oxidation number of a radical or ion
is equal to its electrical charge. Example,
Cu^{2+}, the oxidation number is +2; Cl^{}, the oxidation number is –1.
For a charged group of atoms chemically combined, the sum of oxidation numbers of all
the atoms is equal to the charge carried by the radical.
Example, MnO_{4}^{}, the sum of the oxidation numbers of
manganese, Mn and the 4 oxygen atoms gives
1. The 4 atoms of oxygen have a total oxidation number of 8.
Therefore, Mn  8 = 1; Mn = +7
Therefore, the oxidation number of Mn in MnO_{4}^{}
is +7. Notice that the sign
(+ or ) must be assigned to the oxidation number.
 For neutral molecules, the oxidation numbers of their atoms must sum up to zero. Example, SO_{2}, H_{2}S and H_{2}O_{2}.
To find the oxidation number of sulphur in SO_{2}
SO_{2} = 0
S 2(2) = 0
S  4 = 0
S = +4
Notice that there are two oxygen atoms, each having oxidation number of
2. Therefore, total oxidation number of oxygen present is 2(2) = 4
Therefore, the oxidation number or state of sulphur in SO_{2} is +4
The knowledge of oxidation numbers can be used in balancing redox reactions.
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