2.1.5 (a) Oxidation Numbers

Syllabus

(a) rules for assigning and calculating oxidation number for atoms in elements, compounds and ions

{Learners will be expected to know oxidation numbers of 0 in Peroxides and H in Metal Hydrides.}

What does this mean?

Atoms, molecules and mono-atomic ions

Elements have Oxidation number = 0 (even in molecules like Cl_{2 ,} O_{2 ,} O_{3 ,} P₄or in Metal lattices or Covalent Lattices.

So, Na atoms have an Oxidation Number 0 (element)

And, C atoms in a diamond have an Ox. No. 0 (element)

The oxidation number in a simple ion is genuinely simple:

Na⁺ ions have an Oxidation Number +1

Fe²⁺ ions have an Oxidation Number +2

Fe³⁺ ions have an Oxidation Number +3

Cl^- ions have an Oxidation Number -1

O^2- ions have an Oxidation Number -2

N^3- ions have an Oxidation Number -3

Not all Oxidation numbers are so obvious.

What happens when the atom is part of a molecule or polyatomic ion?

The rules

1. Any atom in any elemental form has Oxidation Number = 0 eg O₂, P₄, C₆₀ all have Ox. No. =0

2. Oxidation Numbers of neutral molecules add up to zero eg CO₂ -- Assume O = -2

Therefore C=+4 to completely cancel

3. Oxidation Numbers of ions add up to the overall charge eg SO₄^2- ---- Assume O = -2

4 x O = -2 x 4 = -8

Therefore S = +6 to leave -2

4. Oxidation Numbers of F in a compound is always -1

5. Oxidation Numbers of O in a compound is usually -2 (except in Peroxides)

6. Oxidation Numbers of H in a compound is usually +1 (except in Metal Hydrides)

7. Oxidation Numbers of Cl in a compound is usually -1 (except where it is bonded to F or O)

8. Oxidation Numbers of Metal ions in Grps 1,2 or 3 in compounds are always +1, +2,+3

Using the Rules

So, the Oxidation number of Nitrogen in the following is:

a) NO₂ O = -2 so there is a total of -4 negative charges.

But there is no overall charge so the Nitrogen must be cancelling out all 4 negatives.

So, the Oxidation number of N = +4

b) N₂O O = -2 so there is a total of -2 negative charges.

But there is no overall charge so the Nitrogen must be cancelling out all 2 negatives.

So, the Oxidation number of each N = +1

c) NO₃^- O = -2 so there is a total of -6 negative charges.

But there is an overall charge so the Nitrogen must be cancelling out only 5 negatives.

So, the Oxidation number of N = +5

Exceptions to the O= -2, H= +1 rules

H - Metal Hydrides

These are compounds where metals are bonded to Hydrogen only.

Eg NaH or CaH₂

In both cases we assume that the metal ions have the charges we expect for elements in their groups (Rule 8)

Ox. No. (Na) = +1 so the Oxidation Number of H must be -1 (to cancel)

Ox. No. (Ca) = +2 so the Oxidation Number of both H's must be -1 (to cancel)

O - Peroxides/ Superoxides

In practice, this is always Hydrogen Peroxide (H₂O₂)

In this case we assume that the H atoms are both +1, as usual.

Which means that both O atoms must be -1 to cancel.

This will be the case for any other Peroxide ion.

Exam-style Questions

1. (a) In terms of electrons, give the meaning of the term oxidation.

..................................................................................................................................................................(1)

(b) Nitrogen monoxide, NO, is formed when Silver metal reduces Nitrate ions, NO₃^- in acid solution.

(i) Deduce the oxidation state of nitrogen in NO and in NO₃^-

NO..................................................................................................................................................................

NO₃^- .....................................................................................................................................................................

(ii) Write a half-equation for the reduction of NO₃^- ions in acid solution to form Nitrogen Monoxide and water.

.......................................................................................................................................................................

(iii) Write a half-equation for the oxidation of silver metal to Ag+(aq) ions.

.......................................................................................................................................................................

(iv) Hence, deduce an overall equation for the reaction between Silver metal and Nitrate ions in acid solution.

..................................................................................................................................................................(5)

(Total 6 marks)

2. (a) In terms of electron transfer, what does the reducing agent do in a redox reaction?

..................................................................................................................................................................(1)

(b) What is the oxidation state of an atom in an uncombined element?

..................................................................................................................................................................(1)

(c) Deduce the oxidation state of nitrogen in each of the following compounds.

(i) NCl₃ ...........................................................................................................................................................

(ii) Mg₃N₂ ........................................................................................................................................................

(iii) NH₂OH ....................................................................................................................................................(3)

Answers

1. (a) removal/loss of electrons 1

(b) (i) +2 1

+5 1

(ii) NO₃– + 4H+ + 3e– → NO +2H₂O 1

(iii) Ag → Ag+ + e– 1

(iv) NO₃– + 4H+ + 3Ag → NO + 2H₂O + 3 Ag+ 1

[6]

1. (a) A reducing agent gives electrons (1)

Not electron pairs

1

(b) Zero (1) 1

(c) (i) (+)3(1)

(ii) –3(1)

(iii) –1(1)

Allow answers in roman numerals

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