Rules for assigning oxidation states

the oxidation state of an element is always zero

the sum of the oxidation states of the elements in a neutral molecule e.g. CO₂ is always zero
the oxidation state of an element existing as a simple ion is the charge on that ion e.g. sodium has an oxidation state of +1 in Na⁺ and bromine has an oxidation state of -1 in Br^–

Note … the sign goes before the number, this is an oxidation state not a charge; and elements have an oxidation state (not ions or compounds) so we say that bromine has an oxidation state of -1 in the bromide ion, not that bromide or the bromide ion has an oxidation state of -1

the sum of the oxidation states of elements in a polyatomic ion must equal the charge on the ion
fluorine always has an oxidation state of -1; oxygen usually has an oxidation state of -2 unless it is in a compound with fluorine (the rules are based on electronegativity); hydrogen almost always has an oxidation state of +1; chlorine always has an oxidation state of -1 unless it is in a compound with fluorine or oxygen

Exceptions to the rules ….

the superoxide ion, O₂^–, and peroxides such as H₂O₂ (each oxygen has an oxidation state -½)
the hydride ion, H^– (hydrogen has an oxidation state of -1)

If the oxidation state of an element increases during a chemical reaction, then that element has been oxidised.
If the oxidation state of an element decreases during a chemical reaction, then that element has been reduced.

Practice questions

What are the oxidation states of the elements in each of these substances?

(a) P₄ (b) N^3- (c) CO₂ (d) NO₃^– (e) MnO₄^–

In the compound [ICl₂]⁺[SbCl₆]^– the oxidation state of chlorine is -1. What are the oxidation numbers of iodine and antimony?

(a) Determine the initial and final oxidation states of both iodine and manganese in the following equation

2I^– + MnO₂ + 4H⁺ ⇾ Mn²⁺ + 2H₂O + I₂

(b) Explain in terms of electrons why manganese is said to be reduced.

Molybdenum can form an oxyanion with the formula [Mo₃₆O₁₁₂(H₂O)₁₆]^8-. Calculate the oxidation state of molybdenum in this ion.

Chromium is produced from its ore by the following series of reactions

i. 4FeCr₂O₄ + 8Na₂CO₃ + 7O₂ ⇾ 8Na₂CrO₄ + 2Fe₂O₃ + 8CO₂

ii. 2Na₂CrO₄+ H₂SO₄ ⇾ Na₂Cr₂O₇ + Na₂SO₄ + H₂O

iii. Na₂Cr₂O₇ + 2C ⇾ Cr₂O₃ + CO + Na₂CO₃

iv. Cr₂O₃ + 2Al ⇾ 2Cr + Al₂O₃

Complete the following table.

Equation	Oxidation state of Cr in the reactant	Oxidation state of Cr in the product	Oxidised / reduced / neither?
i
ii
iii
iv

The salt sodium amide, NaNH₂, can be made in a redox reaction between molten sodium and gaseous ammonia. Write a balanced equation for the reaction and state what has been oxidised and what has been reduced.

Which process is NOT oxidation?

A. NH₄⁺ ⇾ NO₂^–

B. NO₃^– ⇾ N₂

C. N₂ ⇾ NO

D. NO₂^– ⇾ NO₃^–

Answers

(a) element (0)

(b) simple ion (-3)

(d) nitrogen +5; oxygen -2

(e) manganese +7; oxygen -2

I is +3; Sb is +5
(a) Mn changes oxidation state from +4 to +2; iodine from -1 to 0

(b) Mn gains 2 electrons during the reaction

Equation	Oxidation state of Cr in the reactant	Oxidation state of Cr in the product	Oxidised / reduced / neither?
i	+3	+6	Oxidised
ii	+6	+6	Neither
iii	+6	+3	Reduced
iv	+3	0	Reduced

6. Na + NH₃ ⇾ NaNH₂ + ½ H₂; Na is oxidised from 0 to +1; H is reduced from +1 to 0

7. B (only one where oxygen is lost rather than gained)

Naming inorganic compounds and ions using oxidation states

We can use the oxidation states of elements in inorganic compounds to help name the compounds systematically – this is really useful because transition metal elements, halogens and the elements of Group 5 and 6 typically form compounds where they can have a number of different oxidation states. It also means that we can deduce the formula of such compounds from their systematic names.

E.g. ClO^– is the chlorate (I) ion because chlorine has an oxidation state of +1 in this ion and so the compound NaClO must be sodium chlorate (I)

BrO₃^– is the bromate (V) ion because bromine has an oxidation state of +5 and Mg(BrO₃)₂ is magnesium bromate (V)

KMnO₄ is potassium manganate (VII) : the number refers to the oxidation state of manganese in the manganate ion (which must have an overall charge of 1- as potassium is a simple ion with a 1+ charge)

Copper (II) oxide is CuO and copper (I) oxide is Cu₂O : oxygen has an oxidation state of -2, copper is a transition metal and forms compounds with variable oxidation states