Oxidation number define Oxidation state of an element in a compound calculated according to the set of rules formulated on the basis that electron is a covalent bond entirely belongs to more electronegative element
Or
It is the charge over atom in a molecule species which when such molecule is assumed to be hundred percent ionic
Rules for assigning oxidation numbers
(i) Oxidation number of an element in free atomic state (Na,Ag,Fe) or its polyatomic state($H_2, Cl_2, S_8,O_3, P_4$ is always zero
(ii)Oxidation number of F is always -1 in all its compound .Other halogens like CL 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
(iii) 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.
(iv) oxidation number of oxygen in most of the compound is -2 however it is different in peroxides and superoxide the compounds of oxygen in which oxygen atoms are directly linked to each other. In peroxides like $H_2O_2$ and $N_2O_2$ to the oxidation number of oxygen is -1, In superoxide($KO_2,RbO_2$) oxidation number is -1/2 for oxygen ,In Oxygen difluoride ($OF_2$) the oxidation numbers is plus two because F is more electronegative than O. In dioxygen
difluoride ($O_2F_2$),the oxidation number of oxygen is +1
(v) The algebraic sum of the oxidation numbers of all the atoms in a polyatomic ion is equal to the charge on the ion.
(vi)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.
(vi)An increase in oxidation number corresponds to the oxidation of element in a given compound that acts as a reducing agent ,Similarly decreasing oxidation number corresponds to the he reduction of elements in the compound and it behaves as an oxidising agent
Procedure for Calculating Oxidation Numbers
To determine the oxidation numbers of elements in a molecule or ion, follow these steps
Write the formula of the given molecule or ion, leaving space between the atoms.
Assign the known oxidation numbers above each atom. For any atom whose oxidation number is unknown, label it as "x."
Below the formula, multiply the oxidation number of each atom by the number of atoms of that type, and write the total in parentheses.
For neutral molecules, set the sum of oxidation numbers equal to zero. For ions, set the sum equal to the ion's charge.
Solve for "x" to find the unknown oxidation number.
Examples
1. Find the Oxidation of Bromine in $BrF_3$? Answer
we can write as
$Br F_3$
x -1
x -3
x-3=0, x =+3
2. Find the Oxidation Number of Carbon in $CH_4$ Answer
we can write as
$CH_4$
x +1
x +4
x+4=0, x=-4
3.Find the Oxidation Number of Boran in $Na_2B_4O_7$ Answer
we can write as
$Na_2B_4O_7$
+1 x -2
+2 +4x -14
2+4x-14=0, x=3
4.Find the Oxidation of Sulphur in $S_2O_7^{2-}$ Answer
we can write as
$S_2O_7^{2-}$
x -2
2x -14
2x-14=-2
x=6
The Paradox of Fractional Oxidation Number
Sometimes, we come across with certain compounds in which the oxidation number of a particular element in the compound is in fraction.
Examples are:
$C_3O_2$ [where oxidation number of carbon is (4/3)],
$Br_3O_8$ [where oxidation number of bromine is (16/3)]
$Na_2S_4O_6$ (where oxidation number of sulphur is 2.5).
Actually this fractional oxidation state is the average oxidation state of the element under examination and the structural parameters reveal that the element for whom fractional oxidation state is realised is present in different
oxidation states. Example
$C_3O_2$ is written as
O=C=C=C=O
-2 +2 0 +2 -1
So central carbon atom is at oxidation state 0, and other two carbon are at +2
