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Rules in assigning oxidation numbers with examples

Answer Text:
Rules in assigning oxidation numbers
1. Oxidation number of an uncombined element is zero (0)
2. The charge on a monoatomic ion is equivalent to the oxidation number of that element;
3. The oxidation number of hydrogen in all compounds is +1 except in metal hydrides where its – 1;
4. The oxidation number of oxygen in all compounds is – 2 except in peroxides where it is – 1 and 0F2 where it is +2.
5. In complex ions the overall charge is equal to the sum of the oxidation states of the constituent elements.
6. In compounds, the sum of oxidation numbers of all constituent atoms is equal to zero.
Worked examples
1. Calculate the oxidation number of nitrogen in:
(i). #NO^(3-)#
Solution:
N + (-2 x 3) = -1
N = -1 + 6
= +5
Note: thus nitric acid with a nitrate ion (#NO^(3-)#) is called nitric (V) acid since the oxidation number of nitrogen in it is +5;
(ii). #NO_2#;
Solution:
N + (-2 x 2) = 0
N = 0 + 4= +4
Note: Thus the gas #NO_2# isreferred to as nitrogen (IV) oxide because the oxidation number of nitrogen in it is +4
(iii). #NO_2#;
Solution:
N + (-2 x 2) = -1
N = -1 + 4 = +3
Note: thus nitrous acid containing nitrite ion is called nitrous (III) acid since the oxidation number of nitrogen in it is +3.
(iii). #AgNO_3#;
Solution:
1 + N + (-2 x 3) = 0
1 + N + (-6) = 0
N = 0 – 1 + 6;
N = +5
2. Determine the oxidation number of manganese in each of the following, and hence give the systematic names of the compounds.
(i). #MnSO_4#
Solution:
Mn + 6 + (-2 x 4) = 0
Mn = 0 – 6 + 8;
Mn = +2
Systematic name:Manganese (II) sulphate;
(ii). #Mn_2O_3#
Solution:
2Mn + (3 x -2) = 0;
2Mn = 0 + 6
Mn = ½ x 6;
Mn = +3;
Systematic name:Manganese (III) oxide;
(iii). #KMnO_4#
Solution:
1 + Mn + (-2 x 4) = 0
Mn = 0 – 1 + 8;
Mn = +7;
Systematic name: Potassium manganate (VII) oxide
(iv). #MnO_3#-;
Solution:
Mn + (-2 x 3) = -1
Mn = -1 + 6;
Mn = +5
Systematic name: Manganese (V) ion;