# Oxidation Number

Oxidation number (Ox. No.)

It is the number of electrons of an atom of that element which has lost or gained in going from the free element in its natural state to its new state in that particular compound”.

It is quite clear that oxidation state (number) implies the effective charge associated with an element. Hence it may be positive or negative. Positive oxidation number (Ox. No.) implies the number of electrons which must be added to a cation to give a neutral atom. Similarly, negative oxidation number means the number of electrons which must be removed from an anion to give a neutral atom.

The following arbitrary rules are observed in determining the oxidation number:

1. The oxidation number of element in the uncombined or free state is zero.

E.g.

$Cl^0_2, Na^0_2, Mg^0_2, C^0_2, N^0_2, O^0_3, P^0_4, S^0_8$etc.

2. The sum of oxidation numbers of all the atoms in the formula for a neutral compound is zero.

E. g. $NH_3, CO, H_2O$ etc.

3. The Ox. No. of hydrogen in its compounds is +1, except in ionic metal hydrides, (e. g. , NaH, $CaH_2$ etc.) where its Ox. No. is -1.

4. The Ox. No. of oxygen in its compounds is -2, except in peroxides (e.g., $H_2O_2, Na_2O_2, H_2SO_5, H_2S_2O_8$) etc. where its Ox. No. is -1. The Ox. No. of oxygen in super  oxides is -1/2 e. g., in $NaO_2, KO_2 , Ca(O_2)_2, Al(O_2)_3$ etc. The Ox. No. of oxygen in $OF_2$ is +2.

5. The Ox. No. of fluorine is -1. The Ox. No. of other halogens is -1 but if oxygen is present in the compound then it may be +1, +3, +5 or +7.

6. Oxidation No. of metal in amalgam and metal carbonyl $[M[CO]_X]$ is always zero.

E. g.

$Ni(CO)_4$

7. The Ox. No. of alkali metal ions $(Na^+ , Mg{++}$ etc.) is +1 and that of alkaline earth metal ions $(Be^{++} , Mg^{++}$etc.) is +2.

8. The oxidation number of an ion is the same as its charge, e. g., the oxidation numbers of $Na^+, Mg^{++}, SO^+_4 \text{and} PO^{3-}_4$ are +1, +2, -2 and -3, respectively.

9. If in any compound molecules like $NH_3, PH_3, CO, H_2O$ are taken as $()_x$ put their values equal to zero.

E.g.

$[Fe(H_2O)_5NO]SO_4 \\ x + 0 + 1- 2 = 0 \\ x = \pm 1$

10. There is no direct rule to calculate Ox. No. of N and C both. Therefore indirect method is used which is based on the following facts:

(a)    Each covalent bond contributes one unit for Ox. No.

(b)   Covalently bonded atoms with less electronegativity acquire positive Ox. No. whereas other with more electronegativity acquire negative Ox. No.

(c)    In case of coordinate bond, give +2 value for Ox. No. to atom from which coordinate bond is directed to a more electronegative atom.

If coordinate bond is directed from more electronegative to less electronegative atom then neglect contribution of coordinate bond.

On the basis of oxidation number, oxidation is a process in which there is an increase in the oxidation number while reduction is a process in which there is a decrease in the oxidation number.

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