IUPAC Nomenclacture of Inorganic Compounds and Concept of Oxidation Number

Oxidation number is the charge on the valency of an element, i.e.

Oxidation state = Valency + Charge

The valency of metal carries a positive charge, e.g. Na+ while the valency of non-metal carries the negative charge, e.g. Cl

The total oxidation state of any compound is zero. This is gotten by adding the oxidation number of all the elements in a compound. For example, zinc chloride ZnCl2

Oxidation number of Zn = +2 and Cl = -1

Oxidation number of Zn + 2 x Oxidation number of Cl

= +2 + (2 x -1)

= +2 – 2 = 0

Some elements exhibit variable oxidation state, e.g. copper, lead, nitrogen and iron.

Example 1: Find the oxidation state of nitrogen in the following: (i) N2O (ii) NO (iii) NH3 (iv) NO2 (v) NO3

Solution:

(1)  The sum of oxidation state of N2O = 0

The oxidation number of oxygen = -2

N2O = 0

2 x N + (-2) = 0

2N – 2 = 0

2N = +2

N = +2/2

N = +1

The oxidation state of nitrogen in N2O is +1 and the name of the compound is dinitrogen (I) oxide.

(2) Oxidation state of nitrogen in NO

NO = 0

N + (-2) = 0

N = 0 + 2

N = +2

The oxidation state of nitrogen in NO is +2. Hence, the name is nitrogen (II) oxide.

(3) Oxidation state of nitrogen in NH3

N + 3H = 0

N + 3 x +1 = 0

N = 0 – 3

N = -3

The oxidation state of nitrogen in NH3 is -3. Hence, the name is ammonia.

(4) Oxidation state of Nitrogen in NO2

N + 2O = 0

N + 2 x -2 = 0

N = 0 + 4

N = +4

The oxidation state of nitrogen in NO2 is +4. Hence, the name is nitrogen (IV) oxide

(5) In a radical or an ion, the sum of the oxidation state is equal to its charge.

Oxidation state of nitrogen in NO3

N + 3O = -1

N – 2 x 3 = -1

N – 6 = -1

N = -1 + 6

N = +5

The oxidation state of nitrogen in NO3- is +5. Hence, the name is Trioxonitrate (V).

Example 2: Find the oxidation number of the manganese in potassium tetraoxomanganate (VII) KMnO4

Solution: Oxidation state of potassium is +1

Oxidation state of oxygen is -2

kMnO4 = 0

K + Mn + 4O = 0

+1 + Mn + 4 x -2 = 0

+1 + Mn – 8 = 0

Mn – 7 = 0

Mn = 0 + 7

Mn = +7

The oxidation number of the manganese in potassium tetraoxomanganate (VII) is +7.

Example 3: Find the oxidation number of the chromium in potassium heptaoxodichromate (VI) K2Cr2O7.

Solution: Oxidation number of K = +1 and Oxidation number of oxygen = -2

K2Cr2O7 = 0

+1 x 2 + Cr x 2 + (-2 x 7) = 0

+2 + 2Cr – 14 = 0

2Cr – 12 = 0

2Cr = 0 + 12

Cr = +12/2

Cr = +6

The oxidation number of chromium in potassium heptaoxodichromate (VI) is +6

The IUPAC names of some salts:

KClO4 = Potassium tetraoxochlorate (VII)

Ca(NO3)2 = Calcium trioxonitrate (V)

CuSO4 = Copper (II) tetraoxosulphate (VI)

Na2CO3 = Sodium trioxocarbonate (IV)

FeSO4 = Iron (II) tetraoxosulphate (VI)

Fe(SO4)3 = Iron (III) tetraoxosulphate (VI)

Example: Interprete the compounds represented by the formulae (i) Ca(NO3)2 (ii) Fe2(SO4)3 (iii) KNO3

Solution:

1. Ca(NO3)2 represents 1 molecule of calcium trioxonitrate (V), containing:
1. 1 atom of Calcium (Ca)
2. 2 atoms of Nitrogen (N)
3. 6 atoms of Oxygen (O)
2. Fe2(SO4)3 represents 1 molecule of iron (III) tetraoxosulphate (VI), containing:
1. 2 atoms of Iron (Fe)
2. 3 atoms of Sulphur (S)
3. 12 atoms of Oxygen (O)
3. KNO3 represents 1 molecules of potassium trioxonitrate (V), containing:
1. 1 atom of Potassium (K)
2. 1 atom of Nitrogen (N)
3. 3 atoms of Oxygen (O)

EXERCISES

Lets see how much you’ve learnt, attach the following answers to the comment below

1. 3NH3 is (a) three moles of ammonium (b) three moles of ammonia (c) six moles of ammonia (d) six moles of ammonium
2. What is the oxidation number of sulphur in H2SO4? (a) +2 (b) +4 (c) +6 (d) +7
3. Find the oxidation number of Manganese atom in KMnO4. (a) +7 (b) -7 (c) +6 (d) -6
4. _________ is the charge on the valency of an element. (a) reduction number (b) mole number (c) oxidation number (d) isotopy
5. What is the IUPAC name of this compound Fe2O3? (a) iron (II) oxide (b) iron monoxide (c) iron (IV) oxide (d) iron (III) oxide