Oxidation number and rules for the assignment of oxidation number

In ionic substances, the redox reactions can be explained on the basis of electron transfer. However, the redox reactions of covalent compounds cannot be explained in terms of electron transfer. The term oxidation number explains the phenomenon of oxidation-reduction in covalent and ionic substances.

Oxidation number is defined as the charge, which an atom appears to have when all other atoms are removed from it as ions. The oxidation number is a fictitious charge in case of covalent species. It can have positive, zero or negative values depending upon their state of combination.

During the removal of atoms, the electrons on the atoms are counted as follows:
  • Electrons shared between two like atoms are divided equally between the sharing atoms. For example, the electron pair is equally shared between two hydrogen atoms. Therefore, one electron is counted with each hydrogen atom,

Each hydrogen atom is having one electron while its nucleus has one proton and so there is no net charge on each atom of hydrogen. The oxidation number of hydrogen in hydrogen molecule is zero.

  • Electrons shared between two unlike atoms are counted with more electronegative atom. For example, in hydrogen chloride molecule, chlorine is more electronegative than hydrogen. Therefore, the shared pair is counted towards chlorine atom as shown below:

Chlorine gets one extra electron and acquires a unit negative charge with oxidation number of -1. Hydrogen atom without electron has a unit positive charge and acquires an oxidation number of +1.

Determination of Oxidation Number of an Atom

The rules used for the calculation of oxidation number of an atom in a molecule are:

  • The oxidation number of an element in the free or elementary state is always zero. For example, oxidation numbers of helium in He, hydrogen in H2, oxygen in O2, iron in Fe, bromine in Br2, phosphorus in P4, Sulphur in S8, are zero.
  • The oxidation number of an element in a single (monoatomic) ion is the same as the charge on the ion. For example, oxidation number of K+ is +1, of Ca2+ is + 2, of Al3+ is +3, Similarly, the oxidation numbers of Cl-, SO and PO are -1, -2 and -3 respectively,
  • In binary compounds of metal and non-metal, the oxidation number of metal is always positive while that of the non-metal is negative. For example, in NaCl, the oxidation number of sodium is +1 and that of chlorine is 1.
  • In compounds of non-metallic atoms, the atom with high electronegativity is given negative oxidation number. For example, in HCl, the oxidation number of chlorine is -1 because of its high electronegativity. Atoms having high electronegativity have a tendency to form negative ions (anions) while atoms having low electronegativity have great tendency to form positive ions (cations).
  • In all compounds of hydrogen, the oxidation number of hydrogen is +1 except in hydrides of active metals such as LiH, NaH, KH, MgH2 CaH2, etc., where hydrogen has the oxidation number of -1.

In compounds containing oxygen, the oxidation number of oxygen is -2 except in peroxides such as Na2O2 (-1) and in OF2 (+2).

  • For neutral molecule, the sum of the oxidation numbers of all the atoms is equal to zero. For example, in NH3 the sum of the oxidation numbers of nitrogen atom and hydrogen atoms is equal to zero.
  • For a complex ion, the sum of the oxidation number of all the atoms is equal to charge on the ion. For example, in SO ion, the sum of the oxidation numbers of sulphur atom and 4 oxygen atoms must be equal to -2.

Procedure for Calculation of Oxidation Numbers

  • Write down the formula of the given molecule/ion leaving some space between the atoms.
  • Write oxidation number on the top of each atom with 'x' in case of the atom whose oxidation number has to be calculated.
  • Write down the total oxidation numbers of each element beneath the formula. For this purpose multiply the oxidation numbers of each atom with the number of atoms of that kind in the molecule/ion. Write the product in a bracket.
  • Equate the sum of the oxidation numbers to zero for neutral molecule and equal to charge in case of the ion.
  • Solve for the value of 'x'.

Problems

9. What is the oxidation number of the underlined atoms in each of the following molecules/ions?

(a) ClO (b) BrF3 (c) CH4 (d) C6H12O6 (e) Na2B4O7

(f) Na4[Fe(CN6)].

Solution

Suppose x be the oxidation number of the underlined atom:

10. Calculate the oxidation number of (i) Fe in Fe3O4 (ii) S in Na2S4O6.

Solution

(i) Fe in Fe3O4. Let the oxidation number of Fe be 'x'.

(ii) S in Na2S4O6 . Let the oxidation number of S be x.

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