Oxidation States (Oxidation Numbers)
Oxidation states simplify the process of determining what is being oxidized and what is being reduced in redox reactions. However,
for the purposes of this introduction, it would be useful to review and be familiar with the following concepts:
oxidation and reduction in terms of electron transfer
electron-half-equations
To illustrate this concept, consider the element vanadium, which forms a number of different ions (e.g., V 2 +
and V
3 +
). The 2+
ion will be formed from vanadium metal by oxidizing the metal and removing two electrons:
2 + −
V → V +2 e (1)
The vanadium in the V 2 +
ion has an oxidation state of +2. Removal of another electron gives the V 3 +
ion:
2 + 3 + −
V → V +e (2)
The vanadium in the V 3 +
ion has an oxidation state of +3. Removal of another electron forms the ion VO : +
2
3 + 2 + + −
V + H O → VO +2 H +e (3)
2
The vanadium in the VO 2 +
is now in an oxidation state of +4.
Notice that the oxidation state is not always the same as the charge on the ion (true for the products in Equations 1 and 2), but
not for the ion in Equation 3).
The positive oxidation state is the total number of electrons removed from the elemental state. It is possible to remove a fifth
electron to form another the VO ion with the vanadium in a +5 oxidation state.
+
2
2 + + + −
VO + H O → VO +2 H +e (4)
2 2
Each time the vanadium is oxidized (and loses another electron), its oxidation state increases by 1. If the process is reversed, or
electrons are added, the oxidation state decreases. The ion could be reduced back to elemental vanadium, with an oxidation state of
zero.
If electrons are added to an elemental species, its oxidation number becomes negative. This is impossible for vanadium, but is
common for nonmetals such as sulfur:
− 2 −
S+2 e → S (5)
Here the sulfur has an oxidation state of -2.
Summary
The oxidation state of an atom is equal to the total number of electrons which have been removed from an element (producing a
positive oxidation state) or added to an element (producing a negative oxidation state) to reach its present state.
Oxidation involves an increase in oxidation state
Reduction involves a decrease in oxidation state
Recognizing this simple pattern is the key to understanding the concept of oxidation states. The change in oxidation state of an
element during a reaction determines whether it has been oxidized or reduced without the use of electron-half-equations.
Determining oxidation states
Counting the number of electrons transferred is an inefficient and time-consuming way of determining oxidation states. These rules
provide a simpler method.
1 https://chem.libretexts.org/@go/page/3660
, Rules to determine oxidation states
The oxidation state of an uncombined element is zero. This applies regardless of the structure of the element: Xe, Cl2, S8,
and large structures of carbon or silicon each have an oxidation state of zero.
The sum of the oxidation states of all the atoms or ions in a neutral compound is zero.
The sum of the oxidation states of all the atoms in an ion is equal to the charge on the ion.
The more electronegative element in a substance is assigned a negative oxidation state. The less electronegative element is
assigned a positive oxidation state. Remember that electronegativity is greatest at the top-right of the periodic table and
decreases toward the bottom-left.
Some elements almost always have the same oxidation states in their compounds:
Element Usual oxidation state Exceptions
Group 1 metals Always +1
Group 2 metals Always +2
Oxygen Usually -2 Peroxides and F2O (see below)
Hydrogen Usually +1 Metal hydrides (-1) (see below)
Fluorine Always -1
Chlorine usually -1 Compounds with O or F (see below)
The reasons for the exceptions
Hydrogen in the metal hydrides: Metal hydrides include compounds like sodium hydride, NaH. Here the hydrogen exists as a
hydride ion, H-. The oxidation state of a simple ion like hydride is equal to the charge on the ion—in this case, -1.
Alternatively, the sum of the oxidation states in a neutral compound is zero. Because Group 1 metals always have an oxidation
state of +1 in their compounds, it follows that the hydrogen must have an oxidation state of -1 (+1 -1 = 0).
Oxygen in peroxides: Peroxides include hydrogen peroxide, H2O2. This is an electrically neutral compound, so the sum of the
oxidation states of the hydrogen and oxygen must be zero.
Because each hydrogen has an oxidation state of +1, each oxygen must have an oxidation state of -1 to balance it.
Oxygen in F2O: The deviation here stems from the fact that oxygen is less electronegative than fluorine; the fluorine takes
priority with an oxidation state of -1. Because the compound is neutral, the oxygen has an oxidation state of +2.
Chlorine in compounds with fluorine or oxygen: Because chlorine adopts such a wide variety of oxidation states in these
compounds, it is safer to simply remember that its oxidation state is not -1, and work the correct state out using fluorine or
oxygen as a reference. An example of this situation is given below.
Example 1 : Chromium
What is the oxidation state of chromium in Cr2+?
Solution
For a simple ion such as this, the oxidation state equals the charge on the ion: +2 (by convention, the + sign is always included
to avoid confusion)
What is the oxidation state of chromium in CrCl3?
This is a neutral compound, so the sum of the oxidation states is zero. Chlorine has an oxidation state of -1 (no fluorine or
oxygen atoms are present). Let n equal the oxidation state of chromium:
n + 3(-1) = 0
n = +3
The oxidation state of chromium is +3.
2 https://chem.libretexts.org/@go/page/3660
Oxidation states simplify the process of determining what is being oxidized and what is being reduced in redox reactions. However,
for the purposes of this introduction, it would be useful to review and be familiar with the following concepts:
oxidation and reduction in terms of electron transfer
electron-half-equations
To illustrate this concept, consider the element vanadium, which forms a number of different ions (e.g., V 2 +
and V
3 +
). The 2+
ion will be formed from vanadium metal by oxidizing the metal and removing two electrons:
2 + −
V → V +2 e (1)
The vanadium in the V 2 +
ion has an oxidation state of +2. Removal of another electron gives the V 3 +
ion:
2 + 3 + −
V → V +e (2)
The vanadium in the V 3 +
ion has an oxidation state of +3. Removal of another electron forms the ion VO : +
2
3 + 2 + + −
V + H O → VO +2 H +e (3)
2
The vanadium in the VO 2 +
is now in an oxidation state of +4.
Notice that the oxidation state is not always the same as the charge on the ion (true for the products in Equations 1 and 2), but
not for the ion in Equation 3).
The positive oxidation state is the total number of electrons removed from the elemental state. It is possible to remove a fifth
electron to form another the VO ion with the vanadium in a +5 oxidation state.
+
2
2 + + + −
VO + H O → VO +2 H +e (4)
2 2
Each time the vanadium is oxidized (and loses another electron), its oxidation state increases by 1. If the process is reversed, or
electrons are added, the oxidation state decreases. The ion could be reduced back to elemental vanadium, with an oxidation state of
zero.
If electrons are added to an elemental species, its oxidation number becomes negative. This is impossible for vanadium, but is
common for nonmetals such as sulfur:
− 2 −
S+2 e → S (5)
Here the sulfur has an oxidation state of -2.
Summary
The oxidation state of an atom is equal to the total number of electrons which have been removed from an element (producing a
positive oxidation state) or added to an element (producing a negative oxidation state) to reach its present state.
Oxidation involves an increase in oxidation state
Reduction involves a decrease in oxidation state
Recognizing this simple pattern is the key to understanding the concept of oxidation states. The change in oxidation state of an
element during a reaction determines whether it has been oxidized or reduced without the use of electron-half-equations.
Determining oxidation states
Counting the number of electrons transferred is an inefficient and time-consuming way of determining oxidation states. These rules
provide a simpler method.
1 https://chem.libretexts.org/@go/page/3660
, Rules to determine oxidation states
The oxidation state of an uncombined element is zero. This applies regardless of the structure of the element: Xe, Cl2, S8,
and large structures of carbon or silicon each have an oxidation state of zero.
The sum of the oxidation states of all the atoms or ions in a neutral compound is zero.
The sum of the oxidation states of all the atoms in an ion is equal to the charge on the ion.
The more electronegative element in a substance is assigned a negative oxidation state. The less electronegative element is
assigned a positive oxidation state. Remember that electronegativity is greatest at the top-right of the periodic table and
decreases toward the bottom-left.
Some elements almost always have the same oxidation states in their compounds:
Element Usual oxidation state Exceptions
Group 1 metals Always +1
Group 2 metals Always +2
Oxygen Usually -2 Peroxides and F2O (see below)
Hydrogen Usually +1 Metal hydrides (-1) (see below)
Fluorine Always -1
Chlorine usually -1 Compounds with O or F (see below)
The reasons for the exceptions
Hydrogen in the metal hydrides: Metal hydrides include compounds like sodium hydride, NaH. Here the hydrogen exists as a
hydride ion, H-. The oxidation state of a simple ion like hydride is equal to the charge on the ion—in this case, -1.
Alternatively, the sum of the oxidation states in a neutral compound is zero. Because Group 1 metals always have an oxidation
state of +1 in their compounds, it follows that the hydrogen must have an oxidation state of -1 (+1 -1 = 0).
Oxygen in peroxides: Peroxides include hydrogen peroxide, H2O2. This is an electrically neutral compound, so the sum of the
oxidation states of the hydrogen and oxygen must be zero.
Because each hydrogen has an oxidation state of +1, each oxygen must have an oxidation state of -1 to balance it.
Oxygen in F2O: The deviation here stems from the fact that oxygen is less electronegative than fluorine; the fluorine takes
priority with an oxidation state of -1. Because the compound is neutral, the oxygen has an oxidation state of +2.
Chlorine in compounds with fluorine or oxygen: Because chlorine adopts such a wide variety of oxidation states in these
compounds, it is safer to simply remember that its oxidation state is not -1, and work the correct state out using fluorine or
oxygen as a reference. An example of this situation is given below.
Example 1 : Chromium
What is the oxidation state of chromium in Cr2+?
Solution
For a simple ion such as this, the oxidation state equals the charge on the ion: +2 (by convention, the + sign is always included
to avoid confusion)
What is the oxidation state of chromium in CrCl3?
This is a neutral compound, so the sum of the oxidation states is zero. Chlorine has an oxidation state of -1 (no fluorine or
oxygen atoms are present). Let n equal the oxidation state of chromium:
n + 3(-1) = 0
n = +3
The oxidation state of chromium is +3.
2 https://chem.libretexts.org/@go/page/3660
