Download Chapter 7: Chemical Formulas and Chemical Compounds

Chapter 7-2:
Oxidation Numbers
Coach Kelsoe
Chemistry
Pages 232-235
Section 7-2 Objectives
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List the rules for assigning oxidation
numbers.
Give the oxidation number for each element
in the formula of a chemical compound.
Name binary molecular compounds using
oxidation numbers and the Stock system.
Oxidation Numbers


The charges on the ions composing an ionic
compound reflect the electron distribution of
the compound.
In order to indicate the general distribution of
electrons among the bonded atoms in a
molecular compound or a polyatomic ion,
oxidation numbers, or oxidation states
are assigned to the atoms composing the
compound or ion.
Oxidation Numbers
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Oxidation numbers do not have an exact
physical meaning. However they do come in
handy when naming compounds, writing
formulas, and balancing chemical equations.
As a general rule in assigning oxidation
numbers, shared electrons are assumed to
belong to the more-electronegative atom in
each bond.
Here are the rules for assigning oxidation
numbers.
Rules for Assigning Oxidation
Numbers
1.
2.
The atoms in a pure element have an
oxidation number of zero. For example, the
oxygens in O2 and the sulfur in S8.
The more-electronegative element in a
binary molecular compound is assigned the
number equal to the negative charge it
would have as an anion. The lesselectronegative atom is assigned the
number equal to the positive charge it
would have as a cation.
Rules for Assigning Oxidation
Numbers
3.
4.
Fluorine has an oxidation number of -1 in all
of its compounds because it is the most
electronegative element.
Oxygen has an oxidation number of -2 in
almost all compounds. Exceptions include
when it is in peroxides, such as H2O2, in
which its oxidation number is -1, and when
it is in compounds with halogens, such as
OF2, in which its oxidation number is +2.
Rules for Assigning Oxidation
Numbers
5.
6.
Hydrogen has an oxidation number of +1 in
all compounds containing elements that are
more-electronegative than it; it has an
oxidation number of -1 in compounds with
metals.
The algebraic sum of the oxidation numbers
of all atoms in a neutral compound is equal
to zero.
Here’s How It Works…
Least
electronegative
XY
Most
electronegative
O = +2
O = -2
H = +1
H = -1
Rules for Assigning Oxidation
Numbers
7.
8.
The algebraic sum of the oxidation numbers
of all atoms in a polyatomic ion is equal to
the charge of the ion.
Although rules 1-7 apply to covalently
bonded atoms, oxidation numbers can also
be assigned to atoms in ionic compounds.
For example, the ions Na+, Ca2+, and Clhave oxidation numbers of +1, +2, and -1,
respectively.
Oxidation Numbers

Let’s examine hydrogen fluoride and water.

In HF, the bond is polar with a partial negative
charge on the fluorine and a partial positive
charge on the hydrogen. If HF were an ionic
compound in which an electron was fully
transferred to the fluorine ion, H would have a 1+
charge and F would have a 1- charge. Thus the
oxidation numbers of H and F in hydrogen
fluoride are +1 and -1 respectively.
Oxidation Numbers

Let’s examine hydrogen fluoride and water.


In a water molecule, the oxygen atom is more
electronegative than the hydrogen atoms. If H2O were an
ionic compound, the oxygen would have a charge of 2and the hydrogen ions would each have a charge of 1+.
The oxidation numbers of H and O in water are therefore
+1 and -2 respectively.
Because the sum of the oxidation numbers of the
atoms in a compound must satisfy rule 6 or 7 of the
guidelines, it is often possible to assign oxidation
numbers when they are not known.
Sample Problem 1

Assign oxidation numbers to each atom in the
following compounds or ions:
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UF6
H2SO4
ClO3-
Solutions to Sample Problem 1

UF6

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H2SO4
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F has an O.N. of -1 and there are 6, so to make the sum
be zero, U must have an oxidation number of +6.
H has an O.N. of +1 giving the total H’s a +2, and O has
an O.N. of -2 giving the total O’s a -8. So S must have a
O.N. of +6.
ClO3
O has an O.N. of -2, there are 3 giving the total O’s a -6.
To keep the -1 charge, Cl must have a +5 charge.
Sample Problem 2

Assign oxidation numbers to each atom in the
following compounds or ions:
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CF4
N2O5
HClO3
Solutions to Sample Problem 2

CF4


N2O5
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F has an O.N. of -1 and there are 4, so to make the sum
be zero, U must have an oxidation number of +4.
O has an O.N. of -2 and there are 5, so to make the sum
be zero, N must have an oxidation number of +5.
HClO3

H has an O.N. of +1 giving the total H’s a +1, and O has
an O.N. of -2 giving the total O’s a -6. So Cl must have a
O.N. of +5.
Using Oxidation Numbers for
Formulas and Names
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Many nonmetals can have more than one
oxidation number (See page 903). These
numbers can sometimes be used in the same
manner as ionic charges to determine
formulas.
Oxidation numbers alone, however, cannot be
used to predict the existence of a compound.
For example, sulfur (+4, +6) and oxygen can
form either SO2 or SO3. This is not enough!
Using Oxidation Numbers for
Formulas and Names
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We used Roman numerals to denote ionic
charges in the Stock system of naming ionic
compounds.
The Stock system is actually based on
oxidation numbers, and it can be used as an
alternative to the prefix system for naming
binary molecular compounds.
For example, SO2 is called sulfur dioxide in
the prefix system and sulfur(IV) oxide in the
Stock system.
Using Oxidation Numbers for
Formulas and Names


The international body that governs
nomenclature has endorsed the Stock system,
which is more practical for complicated
compounds.
Prefix-based names and Stock-system names
are still used interchangeably for many simple
compounds, however.
Examples of Names in Both
Systems
Prefix system
Stock system
PCl3
Phosphorus trichloride
Phosphorus(III) chloride
PCl5
Phosphorus pentachloride Phosphorus(V) chloride
N2O
Dinitrogen monoxide
Nitrogen(I) oxide
NO
Nitrogen monoxide
Nitrogen(II) oxide
PbO2
Lead dioxide
Lead(IV) oxide
Mo2O3
Dimolybdenum trioxide
Molybdenum(III) oxide