Oxidation-reduction(redox) reactions involve the transfer of electrons from one atom or ion to another. In case of ionic, it is very easy to determine the total number of electrons transferred from one atom or ion. However, in many redox reactions involving covalent compounds, it is is not so easy to determine the direction of transfer of electron and the number of electrons transferred from one reaction to the other simply by looking at chemical equation. For example, in the following redox reaction,

$$2H_2(g) + O_2(g) \rightarrow 2H_2O$$

Both the reactants and the products are covalent compounds. From our knowledge of chemical bonding, we know that during the formation of $H_2O$, an electron pair is shared between each hydrogen and oxygen atom and the electron is not completely transferred from hydrogen to oxygen atom. Yet in $H_2O$ molecule, O atom has higher electron density and thus there is a partial transfer of electronic charge better called electron shift from hydrogen to oxygen. In other word, we can say that hydrogen is oxidized and oxygen is reduced. What is said about the reaction between $H_2$ and $O_2$ may be true for a number of other reaction involving covalent compounds. Two such reactions are mentioned below:-

$$H_2(g) + Cl_2(g) \rightarrow 2HCl(g)$$ $$CH_4(g) + 4 Cl_2(g) \rightarrow CCl_4(l) + 4HCl(g)$$

In these reactions, $H_2$ and $CH_4$ act as reducing agent while $Cl_2$ acts as oxidising agent.

With a view to identify oxidizing and reducing agent by keeping track of number of electrons transferred from one reactant to other both in ionic and covalent compounds, and to help in balancing of equation, the concept of oxidation number was introduced.

OXIDATION NUMBER OF AN ELEMENT MAY BE DEFINED AS CHARGED WITH AN ATOM OF THE ELEMENT HAS IN ITS IRON ORE APPEARS TO HAVE WHEN PRESENT IN THE COMBINED STATE WITH OTHER ATOMS PERIOD OXIDATION NUMBER ARE ALSO CALLED OXIDATION STATES PERIOD

How to find Oxidation Number ?

Step 1

The oxidation number of all the atoms of different elements in their respective elementary state and allotropic forms is taken to be zero. For Example, In

$$N_2, Cl_2, H_2, He, P_4, S_8, O_2, O_3, C$$

The oxidation number of each atom is zero.

Step 2

The oxidation number of mono-atomic ions is the same as charge on it. For Example, oxidation numbers of $N^{+1},MG^{+2}$ and $Al^{+3}$ ions are +1, + 2 and + 3 respectively. While those of $Cl^{-1}$, $S^{-2}$ and $N^{-3}$ ions are -1, -2 and -3 respectively.

Step 3

The oxidation number of hydrogen is +1 when combined when combined with non-metal and is -1 when combined with active metal called metal hydrides such as

$$LiH, KH, MgH_2, CH_2$$

Step 4

The oxidation number of oxygen is -2 in most of its compounds, except in peroxides like $H_2O_2, BaOH_2$. Where it is -1. Other interesting exceptions are found in compound such as $F_2 and O_2F_2$ where the oxidation number of oxygen is +2 and +1 respectively. This is due to the fact that fluorine being the most electronegative element known has always an oxidation number of -1.

Step 5

In compound formed by union of metals with non-metals, the metal atom will have positive oxidation number and a non metal will have negative oxidation number. For Example,

• The oxidation number of alkali metals is always + 1 and those of alkaline earth metal +2.

• The oxidation number of halogen is always -1 in the metal halides such as $KF, AlCl_3, MgBr_2, CdI_2$ .

Step 6

In compounds formed by the union of different elements, the more electronegative atom will have negative oxidation number whereas less electronegative atom will have positive oxidation number. For example,

• “N” is given and oxidation number of -3 when it is bonded to less electronegative atom as in $NH_3$ and and $NI_3$, but is given an oxidation number of +3 when it is bonded to more electronegative atom as $NF_3$.
• Since fluorine is the most electronegative element known as so its oxidation number is always -1 in its compounds, that is oxides, interhalogen compounds.
• In interhalogen compounds of F, Cl, Br, and I ; The more electronegative of two halogens gets the oxidation number of -1. For Example, in $IF_7$, oxidation number of “F” is -1 while of “I” is +7.
• Similarly, in $BrCl_3$, the oxidation number of $Cl$ is -1 while the $Br$ is +3.

Step 7

In neutral compounds, the sum of oxidation number of all the atoms is zero.

Step 8

In complex ions, the sum of oxidation numbers of all the atoms in the ion is equal to the charge on the ion.

Step 9

In polyatomic ions, the algebraic sum of oxidation number of all the atoms in the ions is equal to the charge on the ion. For Example, the sum of oxidation number of all three oxygen atoms and one carbon atom in the carbonate ion $CO_3^{2-}$ is -2.

A term which is more often used for oxidation number is Oxidation State. For Example in $CO_2$, the oxidation state as well as the oxidation number of carbon is + 4 while the oxidation state or oxidation number of oxygen is -2. This means that the oxygen oxidation number denotes the oxidation state of the elements in compound.

Metals invariably have positive oxidation states while non-metals may have positive or negative oxidation states. Transition metals usually display several oxidation States. The highest positive oxidation state for s-block elements is equal to its group number but for the p-block elements it is equal to group number -10. However, the highest negative oxidation state for the p-block elements is equal to eight minus the number of electrons in the valence shell. In other words, the highest positive oxidation State increases across the a period in the periodic table. For example, in the third period, the highest positive oxidation State / number increases from + 12 + 7 as shown below :

Group	1	2	13	14	15	16	17
Element	Na	Mg	Al	Si	P	S	Cl
Compound/s	$NaCl$	$MgSO_4$	$AlF_3$	$SiCl_4$	$PF_5,P_4O_10$	$SF_6,SO_3$	$HClO_4,Cl_2O_7$
Oxidation state of the underlined element in the compound	+1	+2	+3	+4	+5	+6	+7

Sample Problems on Oxidation State

Calculate the oxidation number of:-

1. S in $H_2S$
   • Let the oxidation numer of S in $H_2S$ be ‘x’. Writing the oxidation number each atom above its symbol,
     
   • Sum of oxidation numbers of various atoms in $H_2S=2(+1)+x=2+x$
   • But the sum of the oxidation numbers of various atoms in $H_2S$ is Zero (Step 7)
   • $$x = -2$$
   • Thus, the oxidation number of S in $H_2S$ is -2.
2. C in $CH_2Cl_2$
   • Let the oxidation number of C in $CH_2Cl_2$ be ‘x’. Writing the oxidation number each atom above its symbol,
   • $$C^xH_2^{+1}Cl_2^{-1}$$
   • Sum of the oxidation numbers of various atoms in $CH_2Cl_2 = x + 2 (+1) + 2 (-1) = x$
   • Sum of various oxidation number of various atoms in $CH_2Cl_2$ is zero. Step 7
   • $$x=0$$
   • Thus, the oxidation number of C in $CH_2Cl_2$ is zero.