Oxidation Numbers and Naming

Learning Objectives (OCR Spec 2.1.5)

By the end of this lesson you should be able to:

• Define oxidation and reduction in terms of electron transfer and oxidation number
• Assign oxidation numbers to atoms in elements, compounds and ions using a set of rules
• Use Roman numerals to name compounds where an element can have multiple oxidation states
• Recognise disproportionation as a reaction in which the same element is both oxidised and reduced

The Concept of Oxidation Number

An oxidation number (or oxidation state) is a hypothetical charge that an atom in a compound would have if all of its bonds were fully ionic. It is a bookkeeping device for tracking electron transfer in redox reactions.

Key definitions:

• Oxidation = loss of electrons = increase in oxidation number
• Reduction = gain of electrons = decrease in oxidation number

The mnemonic is OIL RIG - Oxidation Is Loss, Reduction Is Gain (of electrons).

Rules for Assigning Oxidation Numbers

Apply these rules in order:

1. Uncombined elements have an oxidation number of 0. (e.g. Na, O2, S8, Cl2, Fe)
2. Monatomic ions equal their charge. (Na+ = +1; O^2- = -2)
3. Sum rule: Oxidation numbers sum to 0 in a neutral compound and to the overall charge in a polyatomic ion
4. Group 1 metals are always +1; Group 2 always +2; Aluminium always +3
5. Hydrogen is +1 except in metal hydrides (NaH, CaH2) where it is -1
6. Oxygen is -2 except:
   • In peroxides (H2O2, Na2O2) where it is -1
   • In superoxides (KO2) where it is -1/2
   • When bonded to fluorine (OF2) where it is +2
7. Fluorine is always -1 (the most electronegative element)
8. Other halogens are -1 except when bonded to O or F (where they become positive)

Worked Examples

Example 1: Sulfur in H2SO4

Let oxidation state of S = x. 2(+1) + x + 4(-2) = 0 2 + x - 8 = 0 x = +6

Example 2: Manganese in MnO4-

Let oxidation state of Mn = x. x + 4(-2) = -1 x - 8 = -1 x = +7

Example 3: Chromium in Cr2O7^2-

Let oxidation state of Cr = x. 2x + 7(-2) = -2 2x - 14 = -2 2x = +12 x = +6

Example 4: Nitrogen in NH4NO3

This compound contains two nitrogens in different oxidation states:

• In NH4+: x + 4(+1) = +1, so x = -3
• In NO3-: x + 3(-2) = -1, so x = +5

Roman Numeral Nomenclature

Transition metals and other elements often exhibit multiple oxidation states. To avoid ambiguity, IUPAC uses Roman numerals (in brackets, no space) to indicate the oxidation state of an element within a compound name.

Compound	Name
FeCl2	Iron(II) chloride
FeCl3	Iron(III) chloride
Cu2O	Copper(I) oxide
CuO	Copper(II) oxide
KMnO4	Potassium manganate(VII)
K2Cr2O7	Potassium dichromate(VI)
HNO2	Nitrous acid / nitric(III) acid
HNO3	Nitric acid / nitric(V) acid
H2SO3	Sulfurous acid / sulfuric(IV) acid
H2SO4	Sulfuric acid / sulfuric(VI) acid
NaClO	Sodium chlorate(I)
NaClO3	Sodium chlorate(V)
NaClO4	Sodium chlorate(VII)

Note: OCR uses the modern "(VI) acid" nomenclature. Always state the oxidation number of the named element, not the overall charge.

Changes in Oxidation Number during Reactions

Consider the reaction of magnesium with dilute hydrochloric acid:

Mg(s) + 2HCl(aq) -> MgCl2(aq) + H2(g)

Atom	Before	After	Change
Mg	0	+2	+2 (oxidised)
H	+1	0	-1 (reduced)
Cl	-1	-1	unchanged (spectator)

Magnesium has lost 2 electrons; each hydrogen has gained 1 electron. Two hydrogens are reduced per magnesium, balancing the electron transfer.

Disproportionation

Disproportionation is a redox reaction in which the same element is simultaneously oxidised and reduced. Two classic examples:

Chlorine and Cold Dilute NaOH

Cl2 + 2NaOH -> NaCl + NaClO + H2O

Chlorine begins at 0. In NaCl it becomes -1 (reduced). In NaClO it becomes +1 (oxidised). Same element, both processes, one reaction.

Copper(I) in Aqueous Solution

2Cu+ -> Cu + Cu^2+

Cu(+1) disproportionates to Cu(0) and Cu(+2).