Oxidation States (DP IB Chemistry): Revision Note

Philippa Platt

Written by: Philippa Platt

Reviewed by: Richard Boole

Updated on

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

Oxidation:

  • Addition of oxygen (e.g. 2Mg + O2 → MgO)

  • Loss of hydrogen (e.g. CH3OH rightwards arrow with open square brackets straight O close square brackets on top CH2O + H2O)

  • Loss of electrons (e.g. Al → Al3+ + 3e)

Reduction:

  • Loss of oxygen (e.g. 2CuO + C → 2Cu + CO2)

  • Gain of hydrogen (e.g. C2H4 + H2 → C2H6)

  • Gain of electrons (e.g. F2 + 2e → 2F)

Oxidation number

  • The oxidation number or state of an atom is the charge it would have if all bonding were completely ionic

    • It represents the electronic 'status' of an element

  • Oxidation numbers are used to:

    • Tell if oxidation or reduction has taken place

    • Identify 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 that must be added or removed to make an atom neutral

    • Oxidation numbers are always written with the sign before the number

  • Examples:

    • Na in Na = 0 (neutral atom)

    • Na in Na+ = +1 (must add one electron to make it neutral)

    • Cl in Cl- = –1 (must remove one electron to make it 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

  • 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 total oxidation number of all atoms must equal zero

  • Example: H2O

    • H has oxidation number +1

      • There are two H atoms, giving a total of 2 x (+1) = +2

    • To balance this, O must be -2 so the total is 0

  • Example: CO2

    • O has oxidation number –2

      • There are two O atoms, giving a total of 2 x (-2) = –4

    • To balance this, C must be +4 so the total is 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

General rule:

  • The more electronegative atom is assigned the negative oxidation number

  • Electronegativity increases across a period and decreases down a group

How to determine an atom’s oxidation number:

  • Use the position in the periodic table

  • Assume the oxidation state of known elements and work out the unknowns

Oxidation number rules

  1. The oxidation number of any uncombined element is 0 (e.g. H2, Zn, O2)

  2. Many elements have fixed oxidation states:

    • Group 1 → +1

    • Group 2 → +2

    • Fluorine → –1

    • Hydrogen → +1 (or –1 in metal hydrides like NaH)

    • Oxygen → –2 (except in peroxides = –1 or in F2O = +2)

  3. The oxidation number of a monoatomic ion equals its charge:

    • Zn2+ = +2

    • Cl- = –1

  4. The sum of oxidation numbers in a neutral compound is 0:

    • NaCl → Na = +1, Cl = –1

  5. The sum of oxidation numbers in a polyatomic ion equals the ion’s charge:

    • SO42- → S = +6, O = –2 × 4

  6. In compounds/ions, the more electronegative element gets the negative oxidation number:

    • F2O → F = –1, O = +2

Worked Example

Determine 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:

a) P2O5

  • 5 O atoms  = 5 x (–2) = –10

  • Overall charge compound = 0

  • 2 P atoms = +10

  • P = +5

b) SO42-

  • 4 O atoms  = 4 x (–2) = –8

  • Overall charge compound = –2

  • S = +6

c) H2S

  • 2 H atoms  = 2 x (+1) = +2

  • Overall charge compound = 0

  • S = –2

d) Al2Cl6

  • 6 Cl atoms  = 6 x (–1) = –6

  • Overall charge compound = 0

  • 2 Al atoms = +6

  • Al = +3

e) NH3

  • 3 H atoms  = 3 x (+1) = +3

  • Overall charge compound = 0

  • N = –3

f) ClO2-

  • 2 O atoms  = 2 x (–2) = –4

  • Overall charge compound = –1

  • Cl = +3

Are oxidation numbers always whole numbers?

  • Oxidation numbers are not always whole numbers

    • They are an average of the oxidation numbers of atoms in a compound

    • This means that they can be fractional

  • Example: the oxidation of sulfur in the tetrathionate ion, S4O62-:

(S x 4) + (O x 6) = -2

(S x 4) + (-2 x 6) = -2

(S x 4) + (-12) = -2

(S x 4) = +10

S = fraction numerator plus 10 over denominator 4 end fraction = +2.5

  • The +2.5 oxidation number 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 must always have whole-number oxidation states, because you cannot have half an electron!

Examiner Tips and Tricks

  • The terms oxidation number and oxidation state can be used interchangeably

    • IUPAC allows either term

  • Oxidation numbers are often shown using Roman numerals in compound names

Naming transition metal compounds

  • Transition metals have variable oxidation numbers

    • Oxidation numbers can be used in the names of compounds

    • They indicate the oxidation number of a particular element in the compound

  • When an 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)

    • It is not widely used for non-metals, e.g. 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, CrOhas 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, MnOhas 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, Cr2Ohas an overall -2 charge

    • So the oxidation number of Cr is +6

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

Examiner Tips and Tricks

  • For question 2, the sulfate technically contains sulfur(VI)

    • So, the name of MnSO4 should be manganese(II) sulfate(VI)

  • However, in exams you are only expected to name the transition metal using Stock notation

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Philippa Platt

Author: Philippa Platt

Expertise: Chemistry Content Creator

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

Richard Boole

Reviewer: Richard Boole

Expertise: Chemistry Content Creator

Richard has taught Chemistry for over 15 years as well as working as a science tutor, examiner, content creator and author. He wasn’t the greatest at exams and only discovered how to revise in his final year at university. That knowledge made him want to help students learn how to revise, challenge them to think about what they actually know and hopefully succeed; so here he is, happily, at SME.