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 ion or radical the O.N is equal to the charge on it for example
Na+ = +1, Cl– = –1, O2– = -2
CO32- = –2, NO3- =-1, SO42- =-2
3 The algebraic sum of the O.N of all the atoms in a compound is equal O e.g.
H2O = O, i.e O.N of H + 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 table groups
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 group6 = -2 except Oxygen (O) in peroxides).
VI For group 7 = –1 respectively especially in their binary compounds
Determining 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 refer to the general rules for assigning O.N to an element.
a). ZnCl2: The algebraic sum of the O.N of all the atoms in a compound is equal to 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
If we recall from the rules s for assigning O.N to an element. 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 0 ´ 3) = –1
x + (–2 ´ 3) = –1
x – 6 = –1
x = +6 – 1
= +5
Trioxonitrate (V) ion
d)Ca2+ – here we refer to the rule that says 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 in the
1 It is used in the IUPAC (Internal Union of Pure and Applied chemistry) system of naming compounds e.g. H2SO4: TetraoxosulphateVI acid
2. It is used to know the oxidation state or number of an element in a compound
1. Find the O.N of the following underlined elements
a) Na2SO4 (a = +4, b = =2, c = +6, d = –5)
b) [Al(H2O)6]3+ (a = +3, b = –3, c = +6, d = –6)
c) K2Cr2O7 (a = +5, b = +4, c = –6, d = +6)
d) Mn (a = +6, b = +7, c = +5, d = +3)
2 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
3. Find the oxidation numbers of the following underlined elements.
a) K2Cr2O7
b).KMnO4
c). HNO3
d). S2-
e). Cl-
f). Cr2
OBJECTIVE QUESTIONS
1. 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
2.
THEORY QUESTIONS
1. State two applications of oxidation numbers
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