Oxidation States

Oxidation and reduction

There are three definitions of oxidation and reduction used in different branches of chemistry
Oxidation and reduction can be used to describe any of the following processes

Definitions and Examples of Oxidation & Reduction Table

Oxidation

Reduction

Addition of oxygen

e.g. 2Mg + O₂ → MgO

Loss of oxygen

e.g. 2CuO + C → 2Cu + CO₂

Loss of hydrogen

e.g. CH₃OH $\overset{[O]}{\to}$ CH₂O + H₂O

Gain of hydrogen

e.g. C₂H₄ + H₂ → C₂H₆

Loss of electrons

e.g. Al → Al³⁺ + 3e^–

Gain of electrons

e.g. F₂ + 2e^– → 2F^–

Oxidation Number

The oxidation number or state of an atom is the charge that would exist on an individual atom if the bonding were completely ionic
It is like the electronic ‘status’ of an element
Oxidation numbers are used to...

- tell if oxidation or reduction has taken place
- work out what has been oxidised and/or reduced
- construct half equations and balance redox equations

Atoms and simple ions

The oxidation number is the number of electrons which must be added or removed to become neutral
The oxidation number is always written with the charge before the number

Oxidation Number of Simple Ions Table

Atoms	Na in Na = 0	Neutral already, no need to add any electrons
Cations	Na in Na⁺ = +1	Need to add 1 electron to make Na⁺ neutral
Anions	Cl in Cl^– = –1	Need to take 1 electron away to make Cl^– neutral

Worked example

What are the oxidation states of the elements in the following species?

C
Fe³⁺
Fe²⁺
O^2-
He
Al³⁺

Answers:

0
+3
+2
-2
0
+3

So, in simple ions, the oxidation number of the atom is the charge on the ion:
- Na⁺, K⁺, H⁺ all have an oxidation number of +1
- Mg²⁺, Ca²⁺, Pb²⁺ all have an oxidation number of +2
- Cl^-, Br^-, I^- all have an oxidation number of -1
- O^2-, S^2- all have an oxidation number of -2

Molecules or Compounds

In molecules or compounds, the sum of the oxidation numbers on the atoms is zero

Oxidation Number in Molecules or Compounds Table

Elements	H in H₂ = 0	Both are the same and must add up to zero
Compounds	C in CO₂ = +4	1 x (+4) and 2 x (–2) = 0
Compounds	O in CO₂ = –2	1 x (+4) and 2 x (–2) = 0

Since CO₂ is a neutral molecule, the sum of the oxidation states must be zero
For this, one element must have a positive oxidation number and the other must be negative

How do you determine which is the positive one?

The more electronegative species will have the negative value
Electronegativity increases across a period and decreases down a group
O is further to the right than C in the periodic table so it has the negative value

How do you determine the value of an element's oxidation state?

From its position in the periodic table and/or
The other element(s) present in the formula

Many atoms, such as S, N and Cl can exist in a variety of oxidation states
The oxidation number of these atoms can be calculated by assuming that the oxidation number of the other atom is fixed
Here are six rules to deduce the oxidation number of an element

Oxidation Number Rules Table

Rule	Example
1. The oxidation number of any uncombined element is zero	H₂ Zn O₂
2. Many atoms or ions have fixed oxidation numbers in compounds	Group 1 elements are always +1 Group 2 elements are always +2 Fluorine is always –1 Hydrogen is +1 (except for in metal hydrides like NaH, where it is –1) Oxygen is -2 (except in peroxides, where it is -1 and in F₂O where it is +2)
3. The oxidation number of an element in a mono-atomic ion is always the same as the charge	Zn²⁺ = +2 Fe³⁺ = +3 Cl^– = –1
4. The sum of the oxidation number in a compound is zero	NaCl Na = +1 Cl = –1 Sum of oxidation numbers = 0
5. The sum of oxidation numbers in an ion is equal to the charge on the ion	SO₄^2– S = +6 Four O atoms = 4 x (–2) Sum of oxidation numbers = –2
6. In either a compound or an ion, the more electronegative element is given the negative oxidation number	F₂O Both F atoms = 2 x (–1) O = +2 Sum of oxidation numbers = 0

Worked example

State the oxidation number of the atoms in blue in these compounds or ions.

a) P₂O₅

b) SO₄^2-

c) H₂S

d) Al₂Cl₆

e) NH₃

f) ClO₂^-

Answer:

P₂O₅	5 O atoms = 5 x (–2) = –10 Overall charge compound = 0 2 P atoms = +10 P = +5
SO₄^2-	4 O atoms = 4 x (–2) = –8 Overall charge compound = –2 S = +6
H₂S	2 H atoms = 2 x (+1) = +2 Overall charge compound = 0 S = –2
Al₂Cl₆	6 Cl atoms = 6 x (–1) = –6 Overall charge compound = 0 2 Al atoms = +6 Al = +3
NH₃	3 H atoms = 3 x (+1) = +3 Overall charge compound = 0 N = –3
ClO₂^-	2 O atoms = 2 x (–2) = –4 Overall charge compound = –1 Cl = +3

Are oxidation numbers always whole numbers?

The answer is yes and no
When you try and work out the oxidation of sulfur in the tetrathionate ion S₄O₆^2-you get an interesting result!

Oxidation number of S in S₄O₆^2–

How to determine the oxidation number of S in S4O62-

The oxidation number of sulfur in S₄O₆^2-is a fraction

The fact that the oxidation number comes out to +2.5 does not mean it is possible to get half an oxidation number
- This is only a mathematical consequence of four sulfur atoms sharing +10 oxidation number
- The four sulfur atoms are in two different environments and the +2.5 is showing the average oxidation number of these two environments
Single atoms can only have an integer oxidation number, because you cannot have half an electron!

Examiner Tip

Oxidation number and oxidation state are often used interchangeably, though IUPAC does not distinguish between the two terms
Oxidation numbers are represented by Roman numerals according to IUPAC

Naming Transition Metal Compounds

Transition metals are characterized by having variable oxidation numbers.
Oxidation numbers can be used in the names of compounds to indicate which oxidation number a particular element in the compound is in
Where the element has a variable oxidation number, the number is written afterwards in Roman numerals.
This is called the STOCK NOTATION (after the German inorganic chemist Alfred Stock), but is not widely used for non-metals, so SO₂ is sulphur dioxide rather than sulphur(IV) oxide
For example, iron can be both +2 and +3 so Roman numerals are used to distinguish between them
- Fe²⁺ in FeO can be written as iron(II) oxide
- Fe³⁺ in Fe₂O₃ can be written as iron(III) oxide

Worked example

Name these transition metal compounds.

Cu₂O
MnSO₄
Na₂CrO₄
KMnO₄
Na₂Cr₂O₇

Answers:

Copper(I) oxide
- The oxidation number of 1 O atom is -2
- Cu₂O has overall no charge
- So, the oxidation number of Cu is +1
Manganese(II) sulfate
- The charge on the sulfate ion is -2
- So, the charge on Mn and oxidation number is +2
Sodium chromate(VI)
- The oxidation number of 2 Na atoms is +2
- Therefore, CrO₄has an overall -2 charge
- So, the oxidation number of Cr is +6
Potassium manganate(VII)
- The oxidation number of a K atom is +1
- Therefore, MnO₄has an overall -1 charge
- So, the oxidation number of Mn is +7
Sodium dichromate(VI)
- The oxidation number of 2 Na atoms is +2
- Therefore, Cr₂O₇has an overall -2 charge
- So the oxidation number of Cr is +6
  - To distinguish it from CrO₄we use the prefix di in front of the anion

Examiner Tip

The answer to question 2 should strictly speaking be manganese (II) sulfate(VI) since sulfur is an element with a variable oxidation number
However, the sulfate ion is a common ion whose name and formula you should know and you are only required to name transition metal compounds using Stock Notation

Oxidation States (HL IB Chemistry)

Revision Note