Oxidation number (O.N) of an element is the charge on an atom of the element whether it is by itself or bonded to another atom. It indicates the number of electrons the atom has gained or lose at that moment. That is, it is the charge an element will have if electrons were transferred to or from it. It is usually zero (0) for an element in the uncombined state. It is also referred to as the oxidation state of the element
NOTE: - The sign or charge for O.N is written before the number (–2) but it is written after the number for an ionic charge i.e O2–
Rules for calculating oxidation number
The following rules are applied when assigning an oxidation number or calculating the oxidation number of an element thus
1. The O.N of oxygen is always equal to -2 except in peroxides (–1)
2. O.N of hydrogen is always equal to plus one (+1) except again in metallic hydrides (–1).
3. The O.N for an element in the elemental (or ground) state O.N = O (zero) e.g.
Na = O, Cl2 = O, O2 = O etc.
4. For an for a simple ion is equal to the charge on it for example
Na+ = +1, Cl– = –1, O2– = -2
5. Oxidation number of a radical is equal to the charge on it e.g CO32- = –2, NO3- =-1, SO42- =-2
3 The O.N of a compound is equal O (because it is the sum of the e.g.
H2O = O,
i.e (O.N of H × no of H atoms) + (O.N of O) =
(+1 x 2) + ( -2 x1)
2 -2 = 0
NaOH = (+1 x1) + (-2 x 1) + (1 x 1)
+1 - 2+ 1
+2-2=0
Rules for specific groups in the PT.
I For group 1A elements (comprising Li,Na,K, e.t.c) their O.N = +1
II For group 2A elements (comprising Be, Mg, Ca e.t.c) their O.N = +2 in all compounds
III For group 3 elements (B, Al, e.t.c) their O.N = +3 especially in their binary compounds.
IV For group 5 = -3
V For group 6 = -2 except Oxygen (O) in peroxides).
VI For group 7 = –1 respectively especially in their binary compounds
Determination of the oxidation number of an element
1 Find the O.N of the underlined elements in the following
a). ZnCl2 b). SO3 c). NO3- d). Ca2+
Solution
To determine the O.N of the underlined elements, we must follow the general rules for calculating O.N of an element.
a). ZnCl2: The O.N of a compound is zero, i.e. ZnCl2 = O.
Since Cl is a group 7(A) element and ZnCl2 is a binary compound then the O.N of Cl is –1, therefore, the O.N of Zn is
(O.N of Zn) + (O.N of Cl ´ 2) = 0x + (–1 ´ 2) = 0
x – 2 = 0
x = +2
b) SO3
Solution
(O.N of S) + (O.N of 0 ´ 3) = 0x + (–2 ´ 3) = 0
x – 6 = 0
x = +6
Trioxosulphate(IV) ion
c) NO3-
Solution
The O.N of a radical is equal to the charge on it, hence
NO3- = –1
that is,
(O.N of N) + (O.N of 0x3) = –1
x + (–2x3) = –1
x – 6 = –1
x = +6 – 1
= +5
Trioxonitrate (V) ion
d)Ca2+ The O.N of an ion is the charge on it, i.e., Ca2+ = +2 Calcium = ion
Uses of oxidation number
Oxidation number is used for the
1. It is used in the IUPAC (International Union of Pure and Applied chemistry) system of naming compounds e.g. H2SO4: Tetraoxosulphate(VI) acid
2. It is used to know the oxidation state or number of an element in a compound
OBJECTIVE QUSETION
1. Which species undergoes reduction in the reaction represented by the equation below?
H2S(aq)+2FeCl3(aq)→S(s) + 2HCl +3FeCl2
(a) Fe3+
(b). H2S
(c). Cl–
(d) S
2. Oxidation is a reaction which involves the following except
(a). Loss of electrons
(b). Increase in oxidation number
(c). Gain of oxygen
(d). addition of hydrogen
The O.N of the following underlined elements are
3. Na2SO4
(a) +4,
(b) -2,
(c) +6
(d) –5
4. Al (H2O)6]3+
(a) +3
(b) –3
(c) +6
(d) –6
5. K2Cr2O7
(a) +5
(b) +4
(c) –6
(d) +6
6. Mn
(a) +6
(b) +7
(c) +5
(d) +3
THEORY QUESTION
2. Find the oxidation numbers of the following underlined elements.
(a) K2Cr2O7 (b) KMnO4 (c) HNO3
(d). S2- (e). Cl- (f). Cr2
2.a State two applications of oxidation numbers
b. What is the oxidation state of manganese in each of the following species?
i. MnO2 ii MnO4- iii. MnCl2
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