Oxidation numbers

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Oxidation numbers
Oxidation numbers
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Oxidation numbers are used to
describe the distribution of
electrons among bonded atoms.
Covalent bonds involve sharing of
electrons, but oxidation numbers
show what the distribution would be
if the electrons were completely
transferred.
Guidelines for Assigning Oxidation Numbers
( see p. 180 for a complete list)
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The oxidation of any free (uncombined) element
is zero.
The oxidation number of a monatomic ion is
equal to the charge of the ion.
• e.g. The oxidation number of K+ is +1.
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The oxidation number of each hydrogen atom is
+1, unless it is combined with a metal, then it
has a state of -1.
The oxidation number of fluorine is always -1.
The oxidation number of each oxygen atom in
most of its compounds is -2.
Guidelines for Assigning Oxidation Numbers
( see p. 180 for a complete list)
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The algebraic sum of the oxidation
numbers for all the atoms in a
compound is zero.
The algebraic sum of the oxidation
numbers for all the atoms in a
polyatomic ion is equal to the charge
on that ion.
Example: Determine the oxidation numbers
for each atom in KMnO4
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This compound is made up of a K+ cation and an
MnO4- anion.
The K+ is a monatomic ion with a charge of +1,
so its oxidation number is +1.
Assume that each O atom has an oxidation
number of -2.
The MnO4- has a total charge of -1. There are 4
O atoms, each with an oxidation number of -2
The oxidation number of the Mn may be found by
the equation:
• Mn + 4(-2) = -1
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Therefore, the oxidation number of Mn in this
compound is +7.
Example: Determine the oxidation numbers
for each atom in Co(NO2)2
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The anion is the nitrite ion, NO2-.
Assume that each O atom in the nitrite ion has an
oxidation number of -2.
The NO2- has a total charge of -1. There are 2 O
atoms, each with an oxidation number of -2
The oxidation number of the N may be found by
the equation:
• N + 2(-2) = -1
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Therefore, the oxidation number of N in this
compound is +3.
Since there are two nitrate ions, each with a
charge of -1, the charge on the Co must be +2.
As an ion, its oxidation number is equal to its
charge.
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