2.1.5a-c - Oxidation numbers and redox nomenclature

2.1.5a-c - Oxidation numbers and redox nomenclature

Oxidation numbers are a bookkeeping system for tracking electrons in formulae, compounds and ions. In this lesson, the focus is not yet on full redox reactions. You will learn the rules for assigning oxidation numbers, how to use them to write formulae, and how Roman numerals make names such as iron(II), iron(III), chlorate(I) and chlorate(III) unambiguous.

What an oxidation number records

An oxidation number is an assigned number for an atom in a substance. It is useful because it gives chemists a consistent way to describe formulae, names and later redox changes.

Oxidation number

An oxidation number is the charge an atom is assigned for electron-bookkeeping purposes. For a simple ion it is the same numerical value as the ion charge, but in covalent compounds it is an assigned value rather than a real full charge.

The notation matters. Oxidation numbers are usually written with the sign before the number, such as +2, +3 or -1. Ion charges in formulae are written as charges, such as Fe^2+, Fe^3+ or Cl^-.

For example:

speciesoxidation number statement
Fe in uncombined iron0
O in O20
Fe in Fe^2++2
Fe in Fe^3++3
Cl in Cl^--1

The key distinction is that Fe^3+ is an ion charge, while +3 is the oxidation number of iron in that ion.

Rules and exceptions

Oxidation numbers are assigned by rules. Use the fixed rules first, then make the total add up to the charge of the whole formula.

RuleExample
An uncombined element has oxidation number 0.Na, Mg, O2, Cl2 all contain atoms with oxidation number 0.
A simple ion has oxidation number equal to its charge.Na^+ is +1; O^2- is -2; Al^3+ is +3.
Group 1 metals are usually +1; Group 2 metals are usually +2.Na is +1 in NaCl; Mg is +2 in MgO.
Oxygen is usually -2.Oxygen is -2 in MgO, CO2 and SO4^2-.
Oxygen is -1 in peroxides.Oxygen is -1 in H2O2 and Na2O2.
Hydrogen is usually +1.Hydrogen is +1 in HCl and H2O.
Hydrogen is -1 in metal hydrides.Hydrogen is -1 in NaH and CaH2.

Worked example 1: oxygen in a peroxide

In Na2O2, sodium is a Group 1 metal, so each sodium has oxidation number +1.

There are two sodium atoms:

1++1=+21 + +1 = +2

The compound is neutral, so the total from the two oxygen atoms must be -2. Each oxygen atom is therefore:

−22=−1\frac{-2}{2} = -1

This matches the peroxide exception: oxygen is -1 in peroxides.

Worked example 2: hydrogen in a metal hydride

In CaH2, calcium is a Group 2 metal, so calcium has oxidation number +2.

The compound is neutral:

(+2)+2(oxidation number of H)=0(+2) + 2(\text{oxidation number of H}) = 0

So the two hydrogens must total -2, and each hydrogen is -1. This is the metal hydride exception.

The peroxide and metal hydride exceptions are not optional details: oxygen can be -1 in peroxides, and hydrogen can be -1 in metal hydrides.

Calculating unknown oxidation numbers

When an oxidation number is not fixed by a simple rule, use the total.

Sum rule

For a neutral compound:

∑oxidation numbers=0\sum \text{oxidation numbers} = 0

For an ion:

∑oxidation numbers=overall ion charge\sum \text{oxidation numbers} = \text{overall ion charge}

Write a small equation instead of trying to do every step in your head. That makes the sign and the number of atoms visible.

Worked example 1: sulfur in sulfuric acid, H2SO4

Hydrogen is usually +1, and oxygen is usually -2.

Let the oxidation number of sulfur be x.

2(+1)+x+4(−2)=02(+1) + x + 4(-2) = 0 2+x−8=02 + x - 8 = 0 x=+6x = +6

So sulfur has oxidation number +6 in H2SO4.

Worked example 2: nitrogen in nitrate, NO3^-

Oxygen is -2. The whole nitrate ion has charge -1.

Let the oxidation number of nitrogen be x.

x+3(−2)=−1x + 3(-2) = -1 x−6=−1x - 6 = -1 x=+5x = +5

So nitrogen has oxidation number +5 in NO3^-.

Worked example 3: chlorine in ClO2^-

Oxygen is -2. The whole ion has charge -1.

Let the oxidation number of chlorine be x.

x+2(−2)=−1x + 2(-2) = -1 x−4=−1x - 4 = -1 x=+3x = +3

So chlorine has oxidation number +3 in ClO2^-.

Writing formulae using oxidation numbers

Oxidation numbers help you write formulae because the totals must balance. For simple ions, the oxidation number has the same numerical value as the ion charge, so you can use it to balance positive and negative parts of an ionic formula.

Worked example 1: magnesium nitride

Magnesium is in Group 2, so it forms Mg^2+. Nitride is N^3-.

The lowest whole-number ratio that balances total charge is:

3(Mg2+)=+63(\mathrm{Mg}^{2+}) = +6 2(N3−)=−62(\mathrm{N}^{3-}) = -6

So the formula is:

Mg3N2\mathrm{Mg_3N_2}

Worked example 2: iron(III) oxide

The Roman numeral tells you iron has oxidation number +3, so use Fe^3+. Oxide is O^2-.

The lowest whole-number ratio is:

2(Fe3+)=+62(\mathrm{Fe}^{3+}) = +6 3(O2−)=−63(\mathrm{O}^{2-}) = -6

So the formula is:

Fe2O3\mathrm{Fe_2O_3}

Worked example 3: sodium chlorate(I)

In chlorate(I), the Roman numeral tells you chlorine has oxidation number +1. Oxygen is -2.

For ClO^-:

(+1)+(−2)=−1(+1) + (-2) = -1

So chlorate(I) is the ion ClO^-. Sodium is Na^+, so sodium chlorate(I) is:

NaClO\mathrm{NaClO}

Worked example 4: calcium chlorate(III)

In chlorate(III), chlorine has oxidation number +3. Two oxygen atoms make the total:

(+3)+2(−2)=−1(+3) + 2(-2) = -1

So chlorate(III) is ClO2^-. Calcium is Ca^2+, so two chlorate(III) ions are needed:

Ca(ClO2)2\mathrm{Ca(ClO_2)_2}

Use brackets when a polyatomic ion appears more than once.

Roman-numeral names

Roman numerals are used when an element may have compounds or ions with different oxidation numbers. The Roman numeral gives the magnitude of the oxidation number, without a sign.

NameMeaning
iron(II)iron has oxidation number +2, commonly Fe^2+ in a simple ionic compound
iron(III)iron has oxidation number +3, commonly Fe^3+ in a simple ionic compound
chlorate(I)chlorine has oxidation number +1; the ion is ClO^-
chlorate(III)chlorine has oxidation number +3; the ion is ClO2^-

Do not put a plus sign inside the Roman numeral. Write iron(III), not iron(+III). Use brackets around the Roman numeral.

Worked example 1: naming FeCl2

Chlorine in chloride is -1. There are two chloride ions, giving total -2.

The compound is neutral, so iron must be +2.

The name is:

iron(II) chloride\text{iron(II) chloride}

Worked example 2: naming FeCl3

Three chloride ions give total -3, so iron must be +3.

The name is:

iron(III) chloride\text{iron(III) chloride}

Worked example 3: naming NaClO

Sodium is +1, so the ClO ion must be -1. Oxygen is -2.

For chlorine:

x+(−2)=−1x + (-2) = -1 x=+1x = +1

The ion is chlorate(I), so the compound is:

sodium chlorate(I)\text{sodium chlorate(I)}

Nitrate and sulfate without shown oxidation numbers

When no oxidation number is shown, use these default ions:

NameFormula
nitrateNO3^-
sulfateSO4^2-

So iron(III) sulfate uses Fe^3+ and SO4^2-. The formula is:

Fe2(SO4)3\mathrm{Fe_2(SO_4)_3}

This is an HSW8 skill: good chemical communication is systematic and unambiguous. A name such as iron chloride is incomplete if the formula could be FeCl2 or FeCl3.

Explain It Back

Use this as a self-explanation check after the section above. It is for diagnosing what you can already explain, not for learning new material from scratch.

Before leaving the topic, check that you can move both ways: formula to name, and name to formula. Keep oxidation numbers and ion charges separate in your writing.