**Oxidation Number**

Chemists have devised a useful
“accountancy” tool to help keep track of electrons in compounds and reactions.
This is particularly important in redox reactions where some atoms lose (are
oxidised) and others gain (are reduced) electrons.

Each atom in a molecule is
assigned an oxidation number (sometimes called oxidation state). This is the
positive or negative charge the atom would have if the molecule was ionic.

**Rules for working out oxidation
numbers (O.N.)**

The rules should be used in this
order – the higher the rule, the higher its priority.

1. An atom in its elemental form (e.g. Fe, Cl2, graphite
etc) has O.N. = 0

2. The sum of the O.N. of all the atoms in a molecule
equals zero.

3. The sum of the O.N. of all the atoms in an ion equals
the charge of the ion.

4. The O.N. of fluorine is -1 (except in F2 where it is 0
[rule 1]).

5. The O.N. of group 1 elements is +1.

6. The O.N. of group 2 elements is +2

7. The O.N. of oxygen is -2, except peroxides where it is
-1.

8. The O.N. of halogens is usually -1.

9. The O.N. of hydrogen is +1 when bonded to non-metals
and -1 when bonded to metals.

**Example 1**

Find out the oxidation numbers of AlF3

F has O.N. = -1 (rule 4)

Al must have O.N. = +3

So that overall charge is zero:

(+3) + (3×-1) = 0

**Example 2**

Find the oxidation number of NH3

H is attached to a non-metal

So O.N. = +1 (rule 9) N must be -3

so that overall charge is zero:

(-3) + (3×+1) = 0

**Example 3**

Find the oxidation number ofNH4+

H is attached to a non-metal

so O.N. = +1 (rule 9) Ion has
charge of +1.

N must be -3

so that overall charge is +1:

(-3) + (4×+1) = +1

** **

** **

**Example 4**

Find the oxidation number of NH2-

H is attached to a non-metal

so O.N. = +1 (rule 9) Ion has
charge of -1.

N must be -3

so that overall charge is -1:

(-3) + (2×+1) = -1

**Example 5**

Find the oxidation number of Cl2

Elemental chlorine

so O.N. = 0 (rule 1)

**Example 6**

Find the oxidation number of ClF3

F has O.N. = -1 (rule 4)

Cl must have O.N. = +3

so that overall charge is zero:

(+3) + (3×-1) = 0

(Note that rule 4 has higher
priority than rule 8)

**Example 7**

Find the oxidation number of HNO3

O has O.N. = -2 (rule 7)

H is attached to a non-metal

so O.N. = +1 (rule 9)

N must have O.N. = +5

so that overall charge is zero:

(+1) + (+5) + (3×-2) = 0

**Example 8**

Find the oxidation number of NO3-

O has O.N. = -2 (rule 7) Ion has
charge of -1.

N must have O.N. = +5

so that overall charge is -1:

(+5) + (3×-2) = -1

**Example 9**

Find the oxidation number of N2O4

O has O.N. = -2 (rule 7)

N must have O.N. = +4

so that overall charge is zero:

(2×+4) + (2×-2) = 0

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** **

**Example 10**

Find the oxidation number of NO2-

O has O.N. = -2 (rule 7) Ion has
charge of -1.

N must have O.N. = +3

so that overall charge is -1:

(+3) + (2×-2) = -1

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