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Oxidation States (DP IB Chemistry)

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Philippa Platt

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DP Chemistry: SLDP Chemistry: HL

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 rightwards arrow with open square brackets straight O close square brackets on top 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?

  1. C

  2. Fe3+

  3. Fe2+

  4. O2-

  5. He

  6. Al3+

 

Answers:

  1. 0

  2. +3

  3. +2

  4. -2

  5. 0

  6. +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

    • O2-, S2- 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

O in CO2 = –2

  • 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

SO42–

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) SO42-

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

SO42-

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 S4O62- you get an interesting result!

Oxidation number of S in S4O62–

How to determine the oxidation number of S in S4O62-

The oxidation number of sulfur in S4O62- 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 Tips and Tricks

  • 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.

  1. Cu2O

  2. MnSO4

  3. Na2CrO4

  4. KMnO4

  5. Na2Cr2O7

 

Answers:

  1. 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

  2. Manganese(II) sulfate

    • The charge on the sulfate ion is -2

    • So, the charge on Mn and oxidation number is +2

  3. Sodium chromate(VI)

    • The oxidation number of 2 Na atoms is +2

    • Therefore, CrO4 has an overall -2 charge

    • So, the oxidation number of Cr is +6

  4. Potassium manganate(VII)

    • The oxidation number of a K atom is +1

    • Therefore, MnO4 has an overall -1 charge

    • So, the oxidation number of Mn is +7

  5. Sodium dichromate(VI)

    • The oxidation number of 2 Na atoms is +2

    • Therefore, Cr2O7 has an overall -2 charge

    • So the oxidation number of Cr is +6

      • To distinguish it from CrO4 we use the prefix di in front of the anion

Examiner Tips and Tricks

  • 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

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    Philippa Platt

    Author: Philippa Platt

    Expertise: Chemistry

    Philippa has worked as a GCSE and A level chemistry teacher and tutor for over thirteen years. She studied chemistry and sport science at Loughborough University graduating in 2007 having also completed her PGCE in science. Throughout her time as a teacher she was incharge of a boarding house for five years and coached many teams in a variety of sports. When not producing resources with the chemistry team, Philippa enjoys being active outside with her young family and is a very keen gardener