Concept of Oxidation Number

OXIDATION NUMBER

A less obvious example of electron transfer is realised when hydrogen combines with oxygen to form water by the reaction:

2H₂(g) + O₂ (g) \(\to\)  2H₂O (l)                     (8.18)

Though not simple in its approach, yet we can visualise the H atom as going from a neutral (zero) state in H₂ to a positive state in H₂O, the O atom goes from a zero state in O₂ to a dinegative state in H₂O. It is assumed that there is an electron transfer from H to O and consequently H₂ is oxidised and O₂ is reduced.

However, as we shall see later, the charge transfer is only partial and is perhaps better described as an electron shift rather than a complete loss of electron by H and gain by O. What has been said here with respect to equation (8.18) may be true for a good number of other reactions involving covalent compounds. Two such examples of this class of the reactions are:

H₂(s) + Cl₂(g) \(\to\) 2HCl(g)                         (8.19)

and,

CH ₄(g) + 4Cl₂(g) \(\to\) CCl₄(l) + 4HCl(g)    (8.20)

In order to keep track of electron shifts in chemical reactions involving formation of covalent compounds, a more practical method of using oxidation number has been developed. In this method, it is always assumed that there is a complete transfer of electron from a less electronegative atom to a more electonegative atom. For example, we rewrite equations (8.18 to 8.20) to show charge on each of the atoms forming part of the reaction :

0           0                 +1 –2

2H₂(g) + O2(g) \(\to\)  2H₂O (l)                     (8.21)

0            0                    +1 –1

H₂ (s) + Cl₂(g) \(\to\) 2HCl(g)                        (8.22)

–4 +1             0          +4 –1       +1 –1

CH₄(g) + 4Cl₂(g) \(\to\) CCl₄(l) +4HCl(g)      (8.23)

It may be emphasised that the assumption of electron transfer is made for book-keeping purpose only and it will become obvious at a later stage in this unit that it leads to the simple description of redox reactions.

Oxidation number denotes the oxidation state of an element in a compound ascertained according to a set of rules formulated on the basis that electron pair in a covalent bond belongs entirely to more electronegative element.

It is not always possible to remember or make out easily in a compound/ion, which element is more electronegative than the other. Therefore, a set of rules has been formulated to determine the oxidation number of an element in a compound/ion. If two or more than two atoms of an element are present in the molecule/ion such as Na₂S₂O₃/Cr₂O , the oxidation number of the atom of that element will then be the average of the oxidation number of all the atoms of that element. We may at this stage, state the rules for the calculation of oxidation number. These rules are:

1. In elements, in the free or the uncombined state, each atom bears an oxidation number of zero. Evidently each atom in H₂, O₂, Cl₂, O₃, P₄, S₈, Na, Mg, Al has the oxidation number zero.

2. For ions composed of only one atom, the oxidation number is equal to the charge on the ion. Thus Na⁺ ion has an oxidation number of +1, Mg²⁺ ion, +2, Fe³⁺ ion, +3, Cl^– ion, –1, O^2– ion, –2; and so on. In their compounds all alkali metals have oxidation number of +1, and all alkaline earth metals have an oxidation number of +2. Aluminium is regarded to have an oxidation number of +3 in all its compounds.

3. The oxidation number of oxygen in most compounds is –2. However, we come across two kinds of exceptions here. One arises in the case of peroxides and superoxides, the compounds of oxygen in which oxygen atoms are directly linked to each other. While in peroxides (e.g., H₂O₂, Na₂O₂), each oxygen atom is assigned an oxidation oxygen atom is assigned an oxidation an oxidation number of +2 and +1 number of –1, in superoxides (e.g., KO₂, RbO₂) each oxygen atom is assigned an oxidation number of –(½). The second exception appears rarely, i.e. when oxygen is bonded to fluorine. In such compounds e.g., oxygen difluoride (OF₂) and dioxygen difluoride (O₂F₂), the oxygen is assigned, respectively. The number assigned to oxygen will depend upon the bonding state of oxygen but this number would now be a positive figure only.

4. The oxidation number of hydrogen is +1, except when it is bonded to metals in binary compounds (that is compounds containing two elements). For example, in LiH, NaH, and CaH₂, its oxidation number is –1.

5. In all its compounds, fluorine has an oxidation number of –1. Other halogens (Cl, Br, and I) also have an oxidation number of –1, when they occur as halide ions in their compounds. Chlorine, bromine and iodine when combined with oxygen, for example in oxoacids and oxoanions, have positive oxidation numbers.

6. The algebraic sum of the oxidation number of all the atoms in a compound must be zero. In polyatomic ion, the algebraic sum of all the oxidation numbers of atoms of the ion must equal the charge on the ion. Thus, the sum of oxidation number of three oxygen atoms and one carbon atom in the carbonate ion, (CO )^2– must equal –2.

By the application of above rules, we can find out the oxidation number of the desired element in a molecule or in an ion. It is clear that the metallic elements have positive oxidation number and nonmetallic elements have positive or negative oxidation number. The atoms of transition elements usually display several positive oxidation states. The highest oxidation number of a representative element is the group number for the first two groups and the group number minus 10 (following the long form of periodic table) for the other groups. Thus, it implies that the highest value of oxidation number exhibited by an atom of an element generally increases across the period in the periodic table. In the third period, the highest value of oxidation number changes from 1 to 7 as indicated below in the compounds of the elements.

A term that is often used interchangeably with the oxidation number is the oxidation state. Thus in CO₂, the oxidation state of carbon is +4, that is also its oxidation number and similarly the oxidation state as well as oxidation number of oxygen is – 2. This implies that the oxidation number denotes the oxidation state of an element in a compound.

The oxidation number/state of a metal in a compound is sometimes presented according to the notation given by German chemist, Alfred Stock. It is popularly known as Stock notation. According to this, the oxidation number is expressed by putting a Roman numeral representing the oxidation number in parenthesis after the symbol of the metal in the molecular formula. Thus aurous chloride and auric chloride are written as Au(I)Cl and Au(III)Cl₃. Similarly, stannous chloride and stannic chloride are written as Sn(II)Cl₂ and Sn(IV)Cl₄. This change in oxidation number implies change in oxidation state, which in turn helps to identify whether the species is present in oxidised form or reduced form. Thus, Hg₂(I)Cl₂ is the reduced form of Hg(II) Cl₂.

Problems on Oxidation Number

PROBLEM 3

Using Stock notation, represent the following compounds :HAuCl₄, Tl₂O, FeO, Fe₂O₃, CuI, CuO, MnO and MnO₂.

SOLUTION

By applying various rules of calculating the oxidation number of the desired element in a compound, the oxidation number of each metallic element in its compound is as follows:

HAuCl₄  \(\to\)     Au has 3

Tl₂O       \(\to\)    Tl has 1

FeO       \(\to\)     Fe has 2

Fe₂O₃    \(\to\)     Fe has 3

CuI         \(\to\)    Cu has 1

CuO       \(\to\)    Cu has 2

MnO      \(\to\)     Mn has 2

MnO₂     \(\to\)    Mn has 4

Therefore, these compounds may be represented as:

HAu(III)Cl₄, Tl₂(I)O, Fe(II)O, Fe₂(III)O₃, Cu(I)I, Cu(II)O, Mn(II)O, Mn(IV)O₂.

The idea of oxidation number has been invariably applied to define oxidation, reduction, oxidising agent (oxidant), reducing agent (reductant) and the redox reaction. To summarise, we may say that:

Oxidation: An increase in the oxidation number of the element in the given substance.

Reduction: A decrease in the oxidation number of the element in the given substance.

Oxidising agent: A reagent which can increase the oxidation number of an element in a given substance. These reagents are called as oxidants also.

Reducing agent: A reagent which lowers the oxidation number of an element in a given substance. These reagents are also called as reductants.

Redox reactions: Reactions which involve change in oxidation number of the interacting species.

PROBLEM 4

Justify that the reaction:

2Cu₂O(s) + Cu₂S(s) \(\to\) 6Cu(s) + SO₂(g) is a redox reaction. Identify the species oxidised/reduced, which acts as an oxidant and which acts as a reductant.

SOLUTION

Let us assign oxidation number to each of the species in the reaction under examination. This results into:

+1 –2       +1 –2              0        +4 –2

2Cu₂O(s) + Cu₂S(s) \(\to\) 6Cu(s) + SO₂

We therefore, conclude that in this reaction copper is reduced from +1 state to zero oxidation state and sulphur is oxidised from –2 state to +4 state. The above reaction is thus a redox reaction.

Further, Cu₂O helps sulphur in Cu₂S to increase its oxidation number, therefore, Cu(I) is an oxidant; and sulphur of Cu₂S helps copper both in Cu₂S itself and Cu₂O to decrease its oxidation number; therefore, sulphur of Cu₂S is reductant.