Oxidation number
Definition: The oxidation number of an element is the
charge that the atom of the element would
have if complete transfer of electron occurs
Tips from the rules:
· the oxidation number for atom and molecule is zero
Atom or molecule | Zn | C | Mg | Ag | H2 | O2 | Cl2 |
Oxidation number | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
· the oxidation number for monoatomic ion is equal to its charge
Monoatomic ion | H+ | Fe2+ | Mn2+ | Fe3+ | Cr3- | O2- | N3- |
Oxidation number | +1 | +2 | +2 | +3 | -1 | -2 | -3 |
· the sum of oxidation numbers of all elements in the compound is zero
Compound | H2SO4 | KMnO4 | SO2 | Na2SO4 | K2Cr2O7 |
Oxidation number | 0 | 0 | 0 | 0 | 0 |
· the sum of oxidation number of all elements in polyatomic ions is equal to the charge of the ions
Polyatomic ion | NH4+ | SO42- | MnO4- | Cr2O72- | S2O32- | Fe(CN)63- |
Oxidation number | +1 | -2 | -1 | -2 | -2 | -3 |
Calculate the oxidation numbers for the underlined elements.
(i) SO2 [compound]
[refer to rules above]
Element | S | O | Sum of oxidation numbers |
Number of element | 1 | 2 | 0 |
Oxidation number | x | -2 |
[let the unknown oxidation number is equal to x]
1(x) + 2(-2) = 0
x - 4 = 0
x = +4
Thus;
Oxidation number of S in SO2 = +4
(ii) H2SO4
Element | H | S | O | Sum of oxidation numbers |
Number of element | 2 | 1 | 4 | 0 |
Oxidation number | +1 | x | -2 |
2(+1) + 1(x) + 4(-2) = 0
2 + x - 8 = 0
x = +6
Thus;
Oxidation number of S in H2SO4 = +6
(iii) Na2S2O3
Element | Na | S | O | Sum of oxidation numbers |
Number of element | 2 | 2 | 3 | 0 |
Oxidation number | +1 | x | -2 |
2(+1) + 2(x) + 3(-2) = 0
2 + 2x - 6 = 0
x = 4
2
= +2
Thus;
Oxidation number of S in Na2S2O3 = +2
(iv) NH3
Element | N | H | Sum of oxidation numbers |
Number of element | 1 | 3 | 0 |
Oxidation number | x | +1 |
1(x) + 3(+1) = 0
x = -3
Thus;
Oxidation number of N in NH3 = -3
(v) S2O32- [polyatomic ion]
[refer to rules above]
Element | S | O | Sum of oxidation numbers |
Number of element | 2 | 3 | -2 |
Oxidation number | x | -2 |
2(x) + 3(-2) = -2
2x - 6 = -2
x = (6 – 2) /2
x = +2
Thus;
Oxidation number of S in S2O32- = +2
(vi) NH4+ : ammonium ion [polyatomic ion]
Element | N | H | Sum of oxidation numbers |
Number of element | 1 | 4 | +1 |
Oxidation number | x | +1 |
x(1) + 4(+1) = +1
x + 4 = +1
x = 1 - 4
x = -3
Thus;
Oxidation number of N in NH4+ = -3
Exercises;
Calculate the oxidation numbers for the underlined elements.
i. KMnO4 potassium manganate(VII)
ii. MnO4-
iii. K2Cr2O7 Potassium dichromate(VI)
iv. Cr2O72-
Answer:
i. KMnO4
Element | K | Mn | O | Sum of oxidation numbers |
Number of element | 1 | 1 | 4 | 0 |
Oxidation number | +1 | x | -2 |
1(+1) + 1(x) + 4(-2) = 0
1 + x – 8 = 0
x = +7
Thus;
Oxidation number of Mn in KMnO4 = +7
ii. MnO4-
Element | Mn | O | Sum of oxidation numbers |
Number of element | 1 | 4 | -1 |
Oxidation number | x | -2 |
1(x) + 4(-2) = -1
x – 8 = -1
x = +7
Thus;
Oxidation number of Mn in KMnO4 = +7
iii. K2Cr2O7
Element | K | Cr | O | Sum of oxidation numbers |
Number of element | 2 | 2 | 7 | 0 |
Oxidation number | +1 | x | -2 |
2(+1) + 2(x) + 7(-2) = 0
2 + 2x – 14 = 0
2x = 12
2 x = 12 / 2
x = +6
Thus;
Oxidation number of Cr in Cr2O72- = +6
iv. Cr2O72-
Element | Cr | O | Sum of oxidation numbers |
Number of element | 2 | 7 | -2 |
Oxidation number | x | -2 |
2(+1) + 2(x) + 7(-2) = 0
2 + 2x – 14 = 0
2x = 12
2 x = 12 / 2
x = +6
Thus;
Oxidation number of Cr in Cr2O72- = +6
Prepared by;
alias huri
