You’ve heard of atomic number, mass number and even valency of elements. But what is Oxidation number? Have no idea? Don’t worry! We are here to help you with understanding what this number is. Let’s begin exploring about oxidation number.
Oxidation number, also known as oxidation state, is used for determining how many electrons an atom has. However, oxidation numbers do not necessarily mean real charges on molecules. Therefore, we can determine oxidation numbers for atoms of any element irrespective of covalent or ionic bonding.
Let us go through the article to understand oxidation numbers in details! One of the less known examples of electron transfer is when hydrogen and oxygen combine to form water. All of us know the reaction,
Even though it is not a very common reaction but we can notice that the H atom in the neutral state (zero) in H2 converts to positive state after the formation of H2O. Similarly, the oxygen atom in O2 is present in its zero state converts to di-negative state in H2O. Thus, we can clearly see that there is an electron transfer taking place in hydrogen and oxygen in the reaction. As a result of H2 and O2 are oxidized and reduced, respectively.
Browse more Topics Under Redox Reactions
- Balance Redox Reactions
- Classical Idea of Redox Reactions
- Oxidation Number
- Redox Reactions and Electrode Potential
- Types of Redox Reactions
- Redox Reactions as the Basis of Titrations
- Redox Reactions – Electron Transfer Reactions
However, when we study in depth about charge transfer we will understand that the reaction is only partial. Therefore, it can be described as electron shift rather than electron transfer or electron loss where complete loss and gain of electrons occurs in Hydrogen and oxygen, respectively. This reason is also applicable to other reaction examples as well. Refer to the example to understand this class of reactions:
It is important to keep track of the electron shifts in the reactions that involve the covalent compound formations. The more practical method is the use of oxidation number to keep track of the electron shifts happening in the chemical reactions. In the oxidation number method, we consider that a complete transfer of an electron from less electronegative to more electronegative atom takes place. Refer to the example below to see how to show charge on each atom as part of the reaction in atomic number.
It is important to understand that assumptions on electron transfer are made just for keeping track or recordkeeping purpose only. Later on, it will become clear that electron transfer is nothing but the simple descriptions of redox reactions.
Definition of Oxidation Number
Oxidation number refers to the oxidation state of a particular element in a compound by following a set of rules. The rules were set on the basis of the electron pair of a covalent compound and the electronegativity of an element. However, it is difficult to make out which is more electronegative element than the other in any given compound or ion.
Thus, it led to the formulation of a certain set of rules for the determination of an element in a particular compound or ion. Moreover, if in a molecule/compound/ion two or more elements are present then the average of all the atoms of the particular element is taken to find the oxidation number of the given element. We will further understand this later with an example.
Rules for Determination of Oxidation Number
When the elements are present in free or its elemental state then the oxidation number of the particular element will be zero. Therefore, each atom in H2, O2, Na, Cl2, O3, P4, S8, Mg, etc in their free form has oxidation number zero.
Ions having one atom bear oxidation number equal to the charge present on the ion. For instance, Na+ ion has the oxidation number +1. Similarly, Mg2+, Fe3+ ion, Cl– ion, O2– ion will have charge +2, +3, –1, –2, respectively. This rule will apply to all ions. All alkali metals in the compound form will have oxidation number +1. Similarly, all alkaline earth metals have an oxidation number of +2. However, aluminium in all its compound form has oxidation number +3.
Oxygen in most of its compounds state has an oxidation number of -2. However, there are exceptions in case of peroxides, superoxides, and oxygen bonded with fluorine. Superoxides and Peroxides are compound of oxygen in which atoms of oxygen are linked directly to one another. In the case of peroxides such as H2O2, Na2O2 has oxidation number -1 whereas in the case of superoxides KO2, RbO2 each oxygen atom will have the oxidation number of –(½).
The second exception is a rare case. Oxygen forms bond with fluorine forming compounds such as oxygen difluoride and dioxygen difluoride. The oxidation number of oxygen difluoride (OF2) is +2. The oxidation number of dioxygen difluoride (O2F2) is +1. The changes in oxidation number occur due to the bonding state of oxygen.
Usually, hydrogen bears the oxidation number +1. However, there are exceptions in the case when hydrogen bonds to metals in the case of binary compounds or compounds that contain two elements. For instance, NaH, CaH2, LiH. In all the examples the oxidation state of hydrogen is -1.
Generally, fluorine bears oxidation number -1. Similarly, other members of the halogen group such as Br, I, Cl also have an oxidation state of -1 if they are present as halide ions in the compound form. However, chlorine, bromine, and iodine in combination with oxygen have positive oxidation numbers. For example, oxoanions and oxoacids.
A very important thing to understand while determining oxidation numbers is that the algebraic sum of the oxidation number of every element present in the compound will always be zero. However, in case of the polyatomic ion, the algebraic sum of the oxidation number of all the atoms will be same/ equal to the charge on the ion.
For example in case of carbonate ion (CO3)2– the oxidation numbers of one carbon atom and 3 oxygen atoms will be equal to -2. Applying this rule oxidation numbers of any element can be calculated in a molecule or an ion. For now, we know that metallic elements will have positive oxidation state whereas nonmetallic element will have negative oxidation state.
Important Points to Remember
Transition elements consist of several positive oxidation states. The maximum oxidation number of a particular element is the same as the group number in case of the first two groups. It can be also calculated as group number minus 10 with respect to the long form of periodic table.
We can also say it in this way that p block elements except fluorine and oxygen will have highest oxidation number same as their group number and the lowest oxidation state will be equal to group number minus 8. In case of transition elements, the lowest oxidation number calculation is possible by a number of electrons present in ns whereas highest oxidation state calculation is equal to unpaired electrons in ns and (n–l)d.
Therefore, the highest number of the oxidation state of the atom of any element will gradually increase across the period in the periodic table. For example, the maximum value of oxidation number in the third period will range from 1 to 7. Oxidation state and oxidation number are used interchangeably.
For example oxidation state of carbon in CO2 is 4 and for oxygen is -2. This means that oxidation numbers of the elements are equal to the oxidation state of an element in this compound.
In certain cases, the oxidation numbers or states in different compounds are written in notation given by German chemist, Alfred Stock. This notation is known as Stock notation. According to this notation, the Roman numeral in parenthesis after the metal’s symbol represents the oxidation number in a molecular formula of a compound.
For example, auric chloride and aurous chloride are written as Au(III)Cl3 and Au(I)Cl. Similarly, other compounds like stannous chloride and stannic chloride can be written as Sn(II)Cl2 and Sn(IV)Cl4.
Solved Questions for You
Q1. Find oxidation number of Cr2 in the formula K2Cr2O7
Solution: K2= +2; O7= (-2 × 7); Cr2= 2 ×x
Therefore, 2+ 2x-14=0. Thus x=+6 (Oxidation state of chromium)
Q2. Find oxidation state of N in the formula NH4NO3
Solution: There are two different nitrogen atoms in the compound. Therefore we need to do the calculation separately
In NH+4, or X +4 (+1) = +1
Therefore x= -3 (Oxidation state in NH+4). The oxidation state of N in NO–3
y+3 × (-2)= -1
y= +6-1 =5