Redox Reaction and Oxidation State

 

Types of Chemical Reactions According to Loss or Gain of Electrons

 

Oxidation and reduction reactions are electrochemical reactions. In electrochemistry oxidation and reduction reactions involve transfer of electrons. The chemical reactions in which chemical energy changes into electrical energy or Vice Versa are called electrochemical reactions.

1. Oxidation    (De-electronation)

2. Reduction   (Electronation)

Difference between Oxidation and Reduction

 

Difference between Oxidation and Reduction

 

Difference between Oxidation and Reduction

Oxidation or De-electronation   

                                                                                                   

1. Oxidation as addition of oxygen

Oxidation is defined as a reaction of addition of oxygen to a substance either other elements or compounds (to produce oxide) e.g. the rusting of iron, burning of magnesium, carbon etc. in air are typical examples of oxidation.

 

2. Oxidation as a removal of Hydrogen

Oxidation is the process of removal of hydrogen from a compound. In organic chemistry, removal of hydrogen from a compound is termed as dehydrogenation.

3. Oxidation as addition of electronegative element

Oxidation is a chemical reaction in which an electronegative element is added into any chemical species (atom, molecule or ion).

4. Oxidation as Removal of electropositive element

Oxidation is a chemical reaction in which an electropositive element is removed from any chemical species (atom, molecule or ion).

5. Oxidation as loss or removal of electrons (Electronic definition)

The most concise definition and broader view of oxidation reaction is in terms of the electron transfers.

 

According to modern electronic concept, the process or a reaction in which a substance (i.e. atom, molecule or ion) loses one or more electrons (which is manifested by an increase in its oxidation number) is called Oxidation. Oxidation involves in producing or increasing the positive charge on the species or decreasing its negative charge

e.g.

The substance that donates electrons is oxidized but it acts as a reducing agent.

6. Oxidation as an increase in oxidation number

In terms of oxidation number concept, the process in which the oxidation state of an element is increased is called Oxidation.

e.g.

(i) In the following redox reaction between zinc powder and dilute hydrochloric acid, zinc has been oxidized to zinc chloride because its oxidation number has been increased from 0 to +2. 

(ii) In the following redox reaction between carbon and oxygen gas, carbon has been oxidized to carbon dioxide because its oxidation number has been increased from 0 to +4. 

            Reduction or Electronation                                

 

1. Reduction as addition of hydrogen

Reduction is the process of addition of hydrogen to a substance. In organic chemistry, addition of hydrogen in a substance is termed as hydrogenation.

2. Reduction as removal of oxygen

Reduction is the process or a reaction (just opposite to oxidation) which involves the removal of oxygen (or electronegative atom) from substances (e.g. oxides).

 

The most of such reactions are simple displacement reactions. In such reactions, oxides are reduced to free elements (usually metal) while other substances (usually an element) are oxidized to their respective oxides.

3. Reduction as addition of an electropositive element

Reduction is a chemical reaction in which an electropositive element is added to any chemical species (atom, molecule or ion). 

4. Reduction as removal of an electronegative element

Reduction is a chemical reaction in which an electronegative element is removed from chemical species (atom, molecule or ion). 

5. Reduction as gain or addition of electrons (Electronic definition)

The most concise definition and broader view of reduction reactions is in terms of the electron transfers.

 

According to modern electronic concept, the process or a reaction in which a substance (i.e. atom, molecule or ion) gains one or more electrons (which is manifested by a decrease in its oxidation number) is called Reduction. Reduction involves in producing or increasing the negative charge on the species or decreasing its positive charge

The substance that gains electrons is reduced but it acts as an oxidizing agent.

e.g.

6. Reduction as a decrease in oxidation number

In terms of oxidation number concept, the process in which the oxidation state of an element is decreased is called Reduction

e.g.

In this reaction, reduction of Br₂ occurs due to decrease in its oxidation number from 0 to –1.

Oxidizing Agent or Oxidant OR Oxidizer        

 

Complete definition

In any redox reaction, the specie that oxidizes the other substance and itself gets reduced is known as oxidizing agent. It is a substance which gains electrons during a reaction from other substance undergoing decrease in oxidation number thereby oxidizing it. Stronger oxidizing agents are found in the lower region of ECS. Thus strongest oxidizing agent is fluorine.

 

Oxidizing agent may be defined as a substance supplying oxygen or electronegative element, removing hydrogen or electropositive element and accepting electrons thereby decreasing oxidation number. Oxidizing agent

1. Gives nascent oxygen

2. accepts hydrogen

3. gains one or more electrons

4. undergoes decrease in oxidation number

5. causes oxidation

6. is reduced

 

1. Oxidizing Agent as donor of oxygen

Oxidizing agent is a substance (element or compound) that releases or supplies oxygen or nascent (or atomic) oxygen either on decomposition or on treatment with other substance.

e.g.

in following reaction, CuO being donor of oxygen is acting as oxidizing agent supplying oxygen to H₂ and thus itself reduces to Cu while H₂ being acceptor of oxygen is acting as reducing agent adding oxygen and thus itself oxidizes to H₂O.

Some other examples of oxidant are illustrated by their decomposition or their reactions with other reagents. 

2. Oxidizing Agent as Acceptor of Hydrogen

Oxidizing agent is a substance that removes or accepts hydrogen from a substance.

3. Oxidizing Agent as Electron Acceptor or Electron recipient  

A substance (in a redox chemical reaction) that accepts or gains or receives one or more electrons from other substance (called the reductant or reducer) is known an oxidizing agent or oxidant. Stated differently, oxidizing agent undergoes decrease in oxidation number. Thus it is a substance that oxidizes the other substance (by removing electrons from it) while itself gets reduced (by accepting electrons from the other substance) to a lower oxidation state.

e.g.

(i) Br₂ molecule accepts electron during reaction and thus it acts as an oxidant.

(ii) Cl₂ molecules decreases its oxidation number from 0 to −1, so it acts as an oxidizing agent.

(iii) zinc reacts with dilute sulphuric acid to form zinc sulphate and hydrogen gas. In this redox reaction, H₂SO₄ acts as an oxidizing agent accepting electrons (undergoing increase in oxidation number) form zinc and thus reduces to H₂ gas while zinc acts as reducing agent by donating electrons (undergoing decreases in oxidation number) and thus oxidizes to Zn²⁺ ions. [SO₄^2– ions being spectator ions, do not appear in net equation].

4. Test for Oxidizing Agent by KI solution

The resulting colour change of colourless to brown by the addition of KI (a reductant) to an oxidizing agent is a test for oxidizing agent.

 

Aqueous potassium iodide (KI) is used to test for the presence of an oxidizing agent. KI is colourless. If a drop of KI is added to a solution containing an oxidizing agent, a brown solution will be formed. The solution turns brown because the iodide ions (I^-) is oxidized to iodine (I₂), by the oxidizing agent. The iodide ion is colourless but aqueous iodine is brown.

Starch-iodide paper can also be used to test for the presence of oxidizing agents. Oxidizing agents change the colour of most starch-iodide paper from white to blue. This is because the iodine produced reacts with the starch to give a blue colour.

 

5) Examples of Oxidizing Agent

Following are the examples of Oxidant:

Reducing Agent or Reductant or Reducer

 

Complete Definition

In any redox reaction, the specie that reduces the other substance and itself gets oxidized is known as Reducing Agent or Reductant. It is a substance which loses electrons and give to the other for reduction undergoing oxidation. Strongest reducing agent are located on upper region of ECS. Thus strongest reducing agent is Li.

 

A reducing agent may be defined as a substance supplying hydrogen or electropositive element, removing oxygen or electronegative element and donating electrons thereby increasing oxidation number. Reducing agent

1. accepts nascent oxygen

2. loses hydrogen

3. loses one or more electrons

4. oxidation number of atom increases

5. causes reduction

6. is oxidized

 

1. Reducing Agent as Acceptor of Oxygen

Reducing agent is a substance (element or component) that accepts oxygen or nascent (or atomic) oxygen (released by oxidizing agent).

e.g.

in following reaction, C being acceptor of oxygen is acting as reducing agent receiving oxygen from ZnO and thus itself oxidizes to CO while ZnO being donor of oxygen is acting as oxidizing agent supplying oxygen and thus itself reduces to Zn.

2.   Reducing Agent as Donor of Hydrogen

Reducing agent is a substance (element or compound) that releases nascent (or atomic) hydrogen either on decomposition or on treatment with other substance.

3) Reducing Agent as Electron Donor

A substance that donates or loses one or more electrons is called a reducing agent or reductant. Thus it is a substance that reduces the other substance (by supplying electrons to it) while itself gets oxidized (by losing electrons). Stated differently, reducing agent undergoes increase in oxidation number during a reaction.

e.g.

(i) Zn atom loses electrons during reaction & thus it is a reducing agent.

(ii) Na atom loses electrons during reaction & thus it is a reducing agent.

(iii)  Magnesium reacts  with dilute sulphuric acid to form magnesium sulphate and hydrogen gas. In this redox reaction, H₂SO₄ acts as an oxidizing agent accepting electrons (undergoing increase in oxidation number) from magnesium and thus reduces to H₂ gas while magnesium acts as reducing agent by donating electrons (undergoing decreases in oxidation number) and thus oxidizes to Mg²⁺ ions. [SO₄^2– ions being spectator ions, do not appear in net equation].

4)   Identification Test for Reducing Agent by acidified potassium dichromate

The resulting colour change of orange to green by the addition of acidified potassium dichromate (an oxidant) to a reducing agent can be used as a test for reducing agent.

 

Acidified potassium dichromate (VI) can be used to test for the presence of a reducing agent. Acidified potassium dichromate (VI) is made by adding dilute sulphuric acid to aqueous potassium dichromate (VI). The colour of Acidified potassium dichromate (VI) solution changes from orange to green in the presence of a reducing agent. The half reaction is shown here:

In this reaction, the dichromate (VI) ion (Cr₂O₇^2–) is reduced to the chromium (III) ion (Cr³⁺). Cr₂O₇^2– loses oxygen and the oxidation state of chromium decreases from +6 to +3.

5) Examples of Reducing Agent

Following are the examples of Reductant:

1. Non-reactive non-metals …… e.g. C, H₂, S

2. All metals …………………………. e.g.  Li, K, Na, Al, Mg, Zn, Cu etc.

3. Few acids (binary acids) ……. e.g. HCl, HBr, HI, H₂S, HCOOH, ascorbic acid, H3PO3, H₂C₂O₄ etc.

4. Ionic Hydrides …………………... e.g. NaH, CaH₂ etc.

5. Complex Hydrides ……….. e.g. LiAlH₄, NaBH₄

6. Miscellaneous ……………… e.g. SO₂, CO, H₂O₂, NH₃, iodides (KI), ferrous salts (FeSO₄, Mohr’s salt), stannous salts, cyanides salts, Na₂S₂O₃, Na₂SO₃, hydrazine etc.

7. Organic reducing agent …… e.g. Formaldehyde, glucose etc.

8. Metal alloys … e.g. Sodium amalgam (Na-Hg), Zinc amalgam (Zn-Hg), Sodium-lead alloy (Na + Pb)

Difference between Oxidizing and Reducing Agent

 

 

Criteria of deciding oxidizing and reducing nature of compounds

 

A substance acts only as an oxidizing if the oxidation number of one of its element (central atom) is in its highest oxidation state and as a reducing agent if the oxidation number of one of its element is in its lowest oxidation state. However, if the oxidation number of one of the elements of a substance is in its intermediate oxidation state, it can act both as an oxidizing as well as a reducing agent.

 For example;

1. The oxidation number of N in HNO₃ is maximum i.e.+5, therefore, it can act only as an oxidizing agent by accepting one or more electrons e.g.

Here, the oxidation number of N decreases from +5 in HNO₃ to +4 in NO₂ and hence it acts as an oxidizing agent. 

2. The oxidation number of S in H₂S is least i.e. -2 and hence it can act only as a reducing agent by losing one or more electrons. Hence the oxidation number of S increases from -2 in H2S to 0 in elemental Sulphur and hence it acts as a reducing agent.

3. The oxidation number of N in HNO₂ is intermediate (+3), it is neither maximum (+5) nor minimum (-3), therefore, it can act both as an oxidizing as well as a reducing agent. E.g.

Here, the oxidation number of N increases from +3 in HNO₂ to +5 in HNO₃, therefore, it acts a reducing agent.

Here, the oxidation number of N decreases from +3 in HNO₂ to +2 in NO, therefore, it acts as a an oxidizing agent.

 

Factors Affecting Strength of Oxidizing and Reducing Agent

1. A substance can act as oxidizing agent if the oxidation number of one of its element is maximum.

e.g.

HNO₃ in which O.N of N is +5 which is its maximum oxidation state is a strong oxidizing agent

 

2. The more the electronegativity of central element and the more is its oxidation number, the more is the oxidizing power.

e.g.

KClO₄, KBrO₄, HClO₄, KMnO₄, K₂Cr₂O₇, HNO₃, H₂SO₄ etc.

 

3. Oxyanions are stronger oxidizing agents in acidic solution than in basic or neutral solution.

 

4. A substance can act as reducing agent if the oxidation number of one of its element is minimum.

e.g. SnCl₂ (O.N of Sn =+2 which is least), FeSO₄ (O.N of Fe =+2), Na₂S₂O₃ (O.N of S =+2), H₂S (O.N of S = −2), H₂C₂O₄ (O.N of C =+3) etc.

 

5. Anions of electronegative elements like I-, Br⁻, N³⁻ are powerful reducing agents.

6. A substance that can act as both reducing as well as oxidizing agent if oxidation number of one its element is in between its maximum and the minimum oxidation number value.

e.g.

HNO₂ (O.N of N =+3 which is intermediate of +5 and 0).

 

Oxidation number and Acid Strength

The greater the oxidation number of the central element, the greater is the acid strength.

 

HClO₄ (+7) > HClO₃ (+75)  > HClO₂ (+3) > HClO (+1)

Some Oxidizing agent and Reducing Agent 

 

Types of Chemical Reactions According to Electron Transfer

 

There are three types of chemical reactions based on oxidation and reduction:

 

1. Non- Redox Reaction

2. Redox Reaction or Oxidation-Reduction Reaction (ORR)

3. Auto Redox Reaction or Self Oxidation-Reduction Reaction (ARR)

 

Non- Redox Reaction

 

Definition of Non-Redox Reactions

The chemical reaction in which there is no electron transfer i.e. no substance is oxidized or reduced not undergoing change in oxidation number is called non-redox reaction.

 

General Examples of Non-Redox Reactions

1. Neutralization

2. Hydrolysis

3. Precipitation reactions

4. Acid displacement reactions

5. Base Displacement Reactions

6. Some decomposition reactions

7. All double decomposition reactions

8. Some molecular addition reactions

Oxidation-Reduction Reactions (ORR) or Redox Reactions

                                                                       

Definition of Oxidation-Reduction (Redox) Reaction (ORR)

Redox reactions are also called electron-transfer reactions since electrons are transferred from the reductant to oxidant.

 

Oxidation and reduction always occur simultaneously during a chemical reaction. The chemical reactions in which oxidation and reduction occur simultaneously are called oxidation-reduction reactions (ORR) or redox reactions. In other words, these are the reactions in which increase and decrease in oxidation number of same or different atoms occurs.

These reactions comprising of simultaneous oxidation and reduction.

 

In terms of electron transfer, a redox reaction is defined as the process in which electrons are transferred from one substance (reducing agent) to another (oxidizing agent).

             

Explanation

All oxidation and reduction reactions are complimentary of one another and occur simultaneously, one cannot take place without the other. No single oxidation and no single reduction process are known. Oxidation-reduction reactions involve two opposing but complementary processes. These processes can never occur singly i.e. every oxidation must necessarily be accompanied by its opposing process reduction and vice versa. The simultaneously oxidation and reduction reactions are generally termed as redox reactions.  The substance which brings reduction is known as reducing agent while a substance which brings oxidation is known as oxidizing agent. 

 

Daily life examples of Redox reactions

The reactions taking place in batteries are redox reactions. Redox reactions take place in the batteries such that electrons transferred can pass through some external circuit so that they produce electric current

Digestion and metabolism of food which takes place in our body in order to supply us the energy required to perform work is also takes place through a series of redox reactions. 

 

Ordinary bleach oxidize the substances that stain fabric, this makes them colourless and easier to remove from fabric.

 

redox couple

A redox couple is defined as having together the oxidized and reduced forms of a substance taking part in an oxidation or reduction half reaction. Represented as Zn²⁺/Zn and Cu²⁺/Cu.

Example No. 1 of Redox Reaction

The addition reaction between H₂ gas and Br₂ to form hydrogen bromide is an example of redox reaction. In this reaction H₂ has been oxidized because its oxidation number has been increased so H₂ is a reducing agent while Br₂ has been reduced because its oxidation number has been decreased so Br₂ is an oxidizing agent. Thus it is a Redox Reaction during which oxidation and reduction takes place simultaneously.

  

General Examples of Redox Reactions                         

Following are the examples of Redox reactions:

 

Types of Redox Reactions

Redox reactions are divided into two main types.

(i)  Inter molecular Redox Reactions

(ii) Intra molecular Redox Reactions

(iii)Disproportionation

(iv) Comproportionatin reaction (reverse of Disproportionation; HClO + Cl^‒ → Cl₂ + OH^‒)

 

Inter molecular Redox Reactions

 

In such redox reactions, one molecule of reactant is oxidized whereas molecule of other reactant is reduced. In this case, one substance is oxidized and another is reduced. In following reaction, HCl is oxidized while MnO₂ is reduced. 

(ii) Intra molecular Redox Reactions

 

In such redox reactions, one atom of a molecule is oxidized and other atom of same molecule is reduced. In this case, one element of the compound is reduced while another element of the same compound is oxidized.

Examples

In the decomposition of KClO₃, its Cl is reduced to KCl and O is oxidized to O₂. 

In the decomposition of (NH₄)₂Cr₂O₇, its Cr is reduced to Cr₂O₃ and N is oxidized is oxidized to N₂. 

 

Disproportionation Reaction/Auto-Redox reaction/Self-Redox reactions

 

Definition

It is an important and special type of redox reaction in which a single substance (specie) undergoes simultaneous oxidation and reduction i.e. it occurs when a same element is both oxidized and reduced simultaneously (i.e. in the meantime). A specie undergoing auto-redox reaction is said to be disproportionate.

disproportionation, also called disputation reaction, is basically a  redox reaction involving simultaneous reduction and oxidation of atoms of the same element of a substance of intermediate oxidation state from one oxidation state to two different oxidation states forming two compounds, one with higher and one with lower oxidation states. So a species is simultaneously reduced and oxidized to form two different products.

Example

Reason

The requirement for disproportionation reaction to occur is, the element undergoing disproportionation should exhibit minimum three different oxidation states and the element must be less stable in a particular oxidation state from which it can be both oxidized as well as reduced to relatively more stable oxidation states.

 

Examples of Auto-Redox Reactions

1. Decomposition or Disproportionation of potassium chlorate to potassium perchlorate and potassium chloride

2. Decomposition of nitrogen (III) oxide into nitric oxide and nitrogen dioxide

3. Decomposition of hydrogen peroxide into water and oxygen

Decomposition reaction of hydrogen peroxide into water and oxygen involves disproportionation of oxygen. In this auto-redox reaction, the relatively less stable oxygen of peroxide in the -1 oxidation state disproportionates into relatively more stable compounds i.e. water and dioxygen changing its oxidation state to the -2 oxidation state in water and zero oxidation state in oxygen gas at the same time.     

                     

4. Dissolution of chlorine gas in water (Reaction of chlorine gas with water)

 

5. Photolysis of Mercurous chloride into mercuric chloride and mercury

Upon UV-irradiation, Mercurous chloride or mercury(I) chloride undergoes disproportionation. under UV light to give mercury and mercuric chloride. The Hg₂²⁺ ion is oxidized to Hg²⁺ and reduced to Hg.

 

6. Dissolution of nitrogen dioxide in water

When nitrogen dioxide in which oxidation state of nitrogen is +5 reacts with water (Ostwald process), it undergoes disproportionation reaction resulting in the formation of both nitric acid and nitrous acid (or nitric oxide; O.S of N = +2) wherein nitrogen has oxidation states +5 and +3 respectively. In this reaction, nitrogen of NO₂ with +4 oxidation state is simultaneously oxidized to nitric acid (+5 oxidation state) and reduced to nitrous acid or NO (with oxidation state +3 or +2). Thus, it is a disproportionation reaction.

7. Decomposition of Cuprous chloride into cupric chloride and copper

Decomposition of Cuprous chloride into cupric chloride and copper involves disproportionation of copper. When cuprous chloride in which oxidation state of copper is +1 is heated it is decomposed and simultaneously oxidized to copper chloride changing the oxidation state of copper from +1 to +2 and reduced to elemental copper changing the oxidation state of copper from +1 to 0. Thus, this is a disproportionation reaction.

8. Dissolution of metal superoxides with water

This reaction can serve as a convenient source of oxygen in masks of self-contained breathing apparatus worn by fire fighters. The source of oxygen is the reaction between KO₂ and exhaled water vapours. The KOH so formed serves to remove CO₂ from the exhaled breath.

9. disproportionation of Phosphorus to phosphine and hypophosphite in alkaline medium.

Phosphorus disproportionates to phosphine and hypophosphite in alkaline medium. In this case, one P atom is reduced to -3 oxidation number (in PH₃) and three P atoms get oxidized to +1 (in NaH₂PO₂).

10. Auto-redox Reactions of chlorine gas with dilute or conc Alkalis (sodium hydroxide & lime water)

Chlorine undergoes auto-redox reaction with water, sodium hydroxide (cold and hot) and lime water (cold, hot and dry) in which it reduces to chloride (Cl^-) ion (in HCl or NaCl or CaCl₂) as well as oxidizes itself to Cl⁺¹ (in hypochlorite; ClO^1-) or Cl⁺⁵ (in chlorate; ClO₃^1-).

11. Cannizaro’s reaction/Auto-redox Reactions of formaldehyde with conc. Alkalis

The self-addition oxidation reduction and disproportionation Reaction in which two molecules of aldehyde lacking a-hydrogen are disproportinated into carboxylic acid (which form salt with alkali) and alcohol is known as Cannizaro’s Reaction.

 

Aldehydes lacking a-hydrogen like formaldehyde and benzaldehyde (do not show aldol condensation) undergo self-redox reaction in presence of aqueous alkali, two molecules of such aldehydes disproportionate and simultaneously oxidize and reduce one another into acid and alcohol respectively.

 

For example

formaldehyde on heating with conc. Solution of strong alkali like NaOH undergoes self-oxidation reduction reaction in one molecule of formaldehyde is reduced to methanol and the other is oxidized to formic acid that forms salt with alkali.

Comproportionation reaction /synproportionation (opposite of the disproportionation)

Comproportionation reaction is the opposite of disproportionation reaction. In this reaction, two reactants with the same element in different oxidation states combine to form the same element in the intermediate oxidation state.

 

The reverse of disproportionation, such as when a compound in an intermediate oxidation state is formed from precursors of lower and higher oxidation states, is called comproportionation,

Example

Ag²⁺_(aq) + Ag_(s) → 2Ag⁺_(aq)

Oxidation Number (O.N) OR Oxidation State (Oxi. No.)

 

Definition of Oxidation Number

In covalent bond formation the electrons are not transferred as in ionic bond formation, but partial transfer of electronic charge takes place, known as electron shift. The oxidation number method always assumes that there is a complete transfer of electrons from a less electronegative atom to a more electronegative atom.

 

Oxidation Number is a fictitious charges assigned to the atom of an element in a covalently bonded molecule by arbitrary conventions that results when the electrons in a covalent bond are assigned to the more electronegative atom (assuming the bonding were ionic) making certain the law of charge conservation is strictly obeyed. Oxidation Number is a purely a hypothetical number without any theoretical justification and it does not correspond (coincide with) to the real (actual) charge on the atoms, except in the special case of simple ionic compounds. Oxidation number of an atom in a molecule or ion is the hypothetical or real charge present on an atom due to electronegativity difference.

 

the number of charges an atom would have in a molecule of a compound or polyatomic ion if bonding electrons were transferred completely in the direction indicated by the difference in electronegativity. Thus oxidation number reflects the number of electrons transferred in a covalent molecule or polyatomic ion. It is the number of electrons lost or gained by an atom of an element during its change from free state into a particular compound.

OR

the apparent charge (i.e. not real), either positive or negative or zero, on an atom of element in a molecule of a compound or in a polyatomic ion (radical) that results when the electrons in a covalent bond are assigned to the more electronegative atom is called oxidation number or oxidation state. Its value may be positive, negative or zero even fractional value ranges -4 to +7 (+8) depending upon the charge of combined atoms in the molecule or ion. (In Ni (CO)₄ oxidation number of Ni is zero).

OR

It is the fictitious charge that an atom appears to have in a given species when the bonding electrons are counted towards more electronegative atom i.e. it is the hypothetical charge an atom would possess in a compound if the bonding were completely ionic.

 

Oxidation number as the degree of Oxidation

Oxidation number is the number with positive or negative sign which indicates the extent to which an element has been oxidized or reduced i.e. it shows the number of electrons which an atom has lost or gained as a result of bonding. The oxidation state is a “measure (or indicator) of the degree of oxidation” of an atom in a chemical compound. (In writing oxidation numbers, we will write the sign before the number to distinguish them from actual electronic charges, which we write with number first). It is the fundamental key to understanding redox reactions, reaction mechanisms, catalysis, etc.

 

Basis of Assigning Oxidation Number

The oxidation number for an element in a covalent compound is by taking the oxidation number to be equal to the charge that the element would carry, if all the bonds in the compound were regarded as ionic instead of covalent. In doing this, a shared pair or electrons between two atoms is assigned to the atom with the greater electronegativity. Or, if the two atoms are alike, the shared pair is split between the two, one electron being assigned to each atom. The resulting charges on the various atoms when the bonding electrons are so assigned are the oxidation numbers of the atoms.

Covalency

It is the number of hydrogen atoms which can combine with a given atom. It is equal to the number of single bonds which an atom can form. It is also equal to the number of electrons an atom can share.

 

Oxidation State

It is the oxidation number per atom.

 

Examples of Oxidation Number

1. Oxidation number of Mn in KMnO₄ is +7.

2. Oxidation number of Cr in K₂Cr₂O₇ is +6.

3. Oxidation number of Ni in Ni(CO)₄ is 0.

4. Oxidation number of O in OF₂ is +2.

5. Oxidation number of O in KO₂ is –½.

6. Oxidation number of O in H₂O₂ is –1.

Difference between Valency and Oxidation number

Valency is a different term than oxidation number though sometimes the valency and the oxidation number of an element are same in a compound.

 

1. oxidation number is just the apparent charge (not necessarily actual) over the atom when the electrons are counted according to the arbitrary rules i.e. oxidation number is the number with positive or negative sign which indicates the extent to which an element has been oxidized or reduced. While valency is mere a number without positive or negative sign which expresses the combining or displacing tendency of an atom of an element and valency of an element is given by the number of electrons it actually loses or gains or shares during the formation of a compound.

 

2. The oxidation no. of an atom may be in fraction, whereas the valency is always in whole number. The oxidation number of an atom in a compound may be zero but valency of an element cannot be zero (except noble gases).

 

3. the oxidation state of an element may vary in its different compounds whereas in most of the cases, the valency of an element is constant.

 

4. Valency and oxidation states of carbon in its different compounds give a good example to differentiate the two concepts. In CH₄, CH₃Cl, CH₂Cl₂, CHCl₃ and CCl₄, the valency of carbon is always four (due to sharing of four electrons) but its oxidation numbers is -4, -2, 0, +2 and +4 respectively.

 

Oxidation State Vs Valency

Some Important Points on Oxidation Number

 

1.   Oxidation number may be fractional.

 

2.   Oxidation number is positive in metallic elements

 

3.   Oxidation number is positive or negative in non-metallic elements

 

4.   Oxidation number is represented in Roman numbers in parenthesis (brackets) after the symbol of the metal in compounds

Stock Notation

Representation of oxidation state of element by Roman numerals within parenthesis is known as stock notation i.e. Expressing the oxidation state of a metal by Roman numerals like I, II, III etc. within parenthesis is called stock notation.

e.g.

Fe(II) SO₄ or FeSO₄ = Iron(II) sulphates (or ferrous sulphate)  

Fe(III) or FeCl₃     = iron(III) chloride (or ferric chloride)

Au(III) Cl₃ or AuCl₃ = Gold(III) chloride (or auric chloride)

Sn(II) Cl₂ or Sn(II) Cl₂ = Tin(II) chloride (or stannous chloride)

Hg(II)Cl₂ or Hg(II)Cl₂ = Mercury(II) chloride (or mercuric chloride)

Na₂CrO₄  = sodium chromate(VI)

 

5.The oxidation number of metals in amalgams and metal carbonyls i.e. Ni(CO)₄, Fe(CO)₃,       Cr(CO)₆ etc. is zero.

 

6.   In allotropic forms like diamond, graphite etc. oxidation number is 0.

 

7.   In case of coordinate bond, it gives +2 value of oxidation number to less electronegative atom and -2 values to more electronegative atom when coordinate bond is directed form less       electronegative atom to more electronegative atom.

 

8. If coordinate bond is directed from more electronegative to less electronegative atom then its  contribution be zero for both the atoms.

 

9. Oxidation number of O in compounds of fluorine is positive as F is the most electronegative element.

 

10. Electronegativity values of no two elements are same

            P > H, C > H, S > C, Cl > N

 

11. Oxidation state of same element can be different in same or different compounds

O.N of S in H₂S = ‒ 2

O.N of S in H₂SO₃ = +4

O.N of S in H₂SO₄ = +6

 

Details of Important Points of Oxidation Number                 

1.  Positive and Negative Oxidation Numbers

2.  Fractional Oxidation Number

3.  Range of Oxidation Number

4.  Maximum Oxidation Numbers

5. Oxidation Number measures covalent and ionic character

7.  Oxidation number and group number

 

1. Positive and Negative Oxidation Numbers

In general metallic elements (present at the farther left and middle of the periodic table) have only positive oxidation numbers (e.g. Li, Na, K, Mg, Ca, Ba, Al, Pb, Sn, Fe, Cu, etc.) whereas non-metallic elements may have either positive or negative oxidation numbers. But usually non-metals such as F, O, N and other halogens (Cl, Br, and I) have negative oxidation numbers. 

 

2. Fractional Oxidation Number

Elements as such do not have any fractional oxidation numbers. In reality no element can have a fractional oxidation state as electrons cannot be transferred in fraction.

 

When two or more atoms of an element are present in different oxidation states, then calculated oxidation number in a compound or ion may come out as fractional due to average of all the different oxidation states.

 

Fractional oxidation number is the average oxidation number. Fractional oxidation number of a particular element can be calculated only if we know about the structure of the compound in which it is present.

e.g.

(i) In tetrathionate (S₄O₆²⁻) ion, the oxidation number of end S atoms is +5 each and that of the middle S atoms is 0 each. The total oxidation number of 4 S atoms is 5+0+0+5=+10 and the average oxidation number is 10 ÷ 4 = 2.5.

(ii) In sodium tetrathionate (Na₂S₄O₆), the oxidation number of both S⁺ is equal to 0 (pure covalent bond) and other two terminal sulphur atoms have oxidation number = +5.

having the structure

(iii) in C₃O₂ the oxidation number of C is +4/3 or + 1.33.  

3. Range of Oxidation Number

Oxidation number of an atom in a molecule may be positive, negative or any value ranges –4 to zero to +7 (or +8 in Os⁺⁸O₄, Ru⁺⁸O₄ etc. or even fractional value

e.g. Fe₃O₄ (Fe = +2.6), C₃O₄ (C = +2.6), C₃O₂ (C = +1.33). Oxidation number of an atom in a molecule may have zero value. e.g. Ni(CO)₄ (Ni = 0), Fe(CO)₅ (Fe=0), C₆H₁₂O₆ (C=0), C₁₂H₂₂O₁₁ (C =0), CH₂O (C =0) etc.

 

The highest known oxidation state is +8 in the tetroxides of ruthenium, xenon, osmium, iridium, hassium, and some complexes involving plutonium; the lowest known oxidation state is −4 for some elements in the carbon group.

 

4. Variable Oxidation Numbers

The transition metals of group B of the periodic table usually have several possible oxidation states except IIIB group (Sc, Y, La and Ac = +3) and IIB group (Zn, Cd = +2). The variable oxidation state of d-block elements is due to involvement of unpaired electrons of d-subshell. Similarly p-block elements exhibit variable oxidation states which is due to inert pair effect.

5. Maximum Oxidation Numbers

Maximum oxidation number is always positive and maximum oxidation numbers of an atom in a molecule is equal to its group number. e.g. maximum oxidation number of Cl of group VIIA is +7 and that of Cr of group VIB is +6.

 

▶ Os, Ru, Xe show maximum oxidation number i.e. +8

▶ In 3d-series of transition metals, Mn shows maximum oxidation number of +7.

▶ Maximum oxidation number of element = Group number in periodic table

▶ Minimum oxidation number of element = Group number – 8 (applicable only for non-metals).

 

Maximum or highest oxidation state is not stable. (Thus compounds containing central atom with its highest oxidation state are unstable and tend to decompose to reduce oxidation number

e.g. KMnO₄ (Mn = +7), AgNO₃ (N = +5), Mn₂O₇ (Mn = +7), HClO₄ (Cl = +7) etc.

 

▶ For p-block elements (except F and O), the highest oxidation number is equal to their group number and lowest oxidation number is equal to the group number minus eight.

 

▶ In transition elements the lowest oxidation number is equal to the number of ns electrons and highest oxidation number is equal to number of ‘ns’ and (n–1)d unpaired electrons. Maximum oxidation number of transition elements is given by:

Maximum oxidation number of atom = Number of ‘s’ electrons + Number of unpaired ‘d’ electrons

e.g.                                                                                                       

Maximum oxidation number of Mn = 2 + 5 = +7                           

Maximum oxidation number of Fe  = 2 + 4 = +6                           

6. Oxidation Number measures covalent and ionic character

A high oxidation number usually indicates significant (more) covalent character in the bonding of that compound

e.g. Mn₂O₇ (Mn=+7) and MnO₄ (Mn=+7) ion have more covalent character.

 

Compounds with lower oxidation states have more ionic character

e.g. MnO₂ (Mn=+4) and Mn₂O₃ (Mn=+3) have significant ionic character.

 

7. Oxidation number and group number

Oxidation number is directly related to the group number to which the element belongs.

E.g. the oxidation number of group IA is +1 and that of IIA is +2. Similarly zero group shows zero oxidation state.

 

Oxidation number of p-block elements is the number of electrons in the valence shell or deficiency of electrons in the valence shell.

 

8. Oxidation number of two or more atoms of same elements may be different

If a compound contains two or more atoms of the same element, all of them may or may not have same oxidation number

e.g.

(i) In Na₂S₂O₃, one S-atom has oxidation number = -2 while the other has oxidation number = +6.

 

(ii) In bleaching powder; CaOCl₂ or Ca(OCl)Cl, oxidation number of one Cl = ‒1 while oxidation       number of other Cl = +1.

 

(iii) In Fe₃O₄ or FeO.Fe₂O₃, oxidaiton number of one Fe = +2 while that of each of the other two = +3.

 

(iv) In NH₄NO₃, oxidation number of N of NH₄⁺ = ‒3 while that of N in NO₃^‒ = +5.

 

How to get Oxidation Number

1)  Given a compound, write its Lewis dot structure.

 

2)  For each separate bond decide which element is most electronegative (EN).

 

3) Give the most electronegative element all the electrons of that bond. If the atoms in the bond are the same, give each element half of the electrons.

 

4) When all electrons have been assigned subtract the number of electrons on each atom from the valence of each element to get the oxidation state (number).

Significance of Oxidation number

Oxidation number provides a measure of whether the atom in a molecule is neutral, electron rich or electron-poor. It guides us to identify elements that are oxidized (oxidation number increases) and reduced (oxidation number decreases) at a glance by comparing its oxidation number before and after the reaction.

 

Oxidation numbers are used:

 

1. In nomenclature (naming) of compounds.

 

2. In classifying types of reactions (as redox, non-redox or auto-redox).

 

3. In balancing of equations of redox reactions.

 

4. In examining trends in chemical reactivity across the periodic table.

 

5. In exploring the systematic chemistry of elements.

 

6. In identifying redox (oxidation-reduction) reactions.

 

7. In determining the Equivalent weights

 

8. in comparing the strength of acid and base

(a) Strength of acids increases with increase in oxidation number.

(b) Strength of base decreases with increase in oxidation number.

 

9. In determining the oxidizing and reducing nature of compounds

(a) If any compound is in maximum oxidation state, then it will act as oxidant only.

(b) If any compound is in minimum oxidation state, then it will act as reductant only.

(c) If the oxidation state is intermediate, then compound can act as both reductant as well as oxidant.

 

10. To determine possible molecular formula of any compound

Suppose that there are three atoms A, B, C and their oxidation number are +6, ‒1, ‒2 respectively. Then the molecular formula of compound formed by them will be AB₄C because

+6 = (‒1 4)+( ‒2)

or +6 = ‒6

 

details

in comparing the strength of acid and base

 

Rules for Finding Oxidation Number

 

1. Oxidation Number of Free Elements is zero

The oxidation number of an atom in its elemental form or uncombined state is always zero i.e. the oxidation number of an element in a free atomic state (Na, H, Cl, O, P etc.) or in its poly-atomic state (graphite, H₂, O₂, P₄, S₈ etc.) or alloy form (Na/Hg) is always zero. (Oxidation number is zero for any elemental substance, which occurs in diatomic, triatomic or polyatomic forms or allotropic forms (diamond, graphite) or alloy form (Na/Hg). (Free state = most stable state, uncombined state).

e.g. each atom in Na, Mg, C, O₂, N₂, H₂, Br₂, F₂, I₂, O₃, P₄, S₈, has an oxidation number of zero.

 

(i) Oxidation state of atoms present in homoatomic molecules is zero. e.g. H₂, O₂, N₂, P₄, S₈ = zero

(ii) Oxidation state of an element in any of its allotropic form is zero.

C_diamond = 0, C_graphite = 0, S_monoclinic = 0, S_rhombic = 0

(iii) Oxidation state of all the components of any an alloy are 0 e.g. Na⁰/Hg⁰

(iv) In complex compounds, oxidation state of some neutral ligands is zero. e.g. CO, NO, H₂O, NH₃

2.   Oxidation Number of Monoatomic Ion equal to its charge

The oxidation number of atom in monoatomic ion (composed of only one atom) is equal to its charge.

e.g.

oxidation number of Na in Na¹⁺ is +1, that of Ba in Ba^{2 +} ion is +2, that of Al in Al³⁺ ion is +3, that of Ca in Ca²⁺ is +2, that of Cl in Cl^– ion is – 1, that of O in O^2–  ion is – 2, that of P in P^3–  ion is – 3 and so on.

 

3. Oxidation Number of atoms in Polyatomic Ion

The oxidation number an atom in a polyatomic ion is usually equal to its oxidation number that it would have if it were a monoatomic ion.

For example;

in hydroxide ion (OH^–), the oxygen atom has an oxidation number of –2 as if it were a monatomic oxide (O^2–) ion and the hydrogen atom has an oxidation number of +1 as if it were simple H⁺ ion.

[In oxyanions, the oxidation number of central atom is always positive which is usually equal to its highest oxidation number e.g. in CO₃^2– ion, the oxidation number of carbon is +4 which is its highest oxidation state].

 

4. Sum of Oxidation Numbers of all atoms in Polyatomic Ion equals net ionic charge

In polyatomic ions (or compound radical), the sum of oxidation numbers of its all atoms is equal to overall (net) charge of the ion.

e.g.

in the ammonium ion (NH₄⁺), the oxidation number of each H is +1 and that of N is –3. Thus the sum of the oxidation numbers is –3 + 4(+1) = +1, which is equal to net charge of the ion. 

5. Sum of Oxidation Numbers of all atoms in a molecule is always zero

In a neutral species (molecule of a compound), the sum of the oxidation numbers of all elements is always zero to comply with law of charge conservation. 

e.g.

[This rule is particularly useful for finding the oxidation number of an atom in difficult cases by assigning oxidation numbers to the ‘Easy’ atoms first and then find the oxidation number of the ‘Difficult‘ atom by subtraction]. 

6. Oxidation Number of Atoms in Binary Polar Compounds

In binary polar compounds (those with two different elements), more electronegative element has negative oxidation number (equal to its charge in simple ionic compounds of the element) while less electronegative element has positive oxidation number. 

e.g.

7. Oxidation Number of Atoms in Ternary Compounds

In ternary polar compounds (those with three or more different elements), only more electronegative element has negative oxidation number (equal to its normal oxidation number) while all other elements have positive oxidation numbers.

exceptions

In few organic compounds and in few inorganic compounds where two more electronegative elements are present, more than one element has negative oxidation number while only one element has positive oxidation number. 

 

8. Oxidation Number of Fluorine is always –1

The oxidation number of Fluorine in its compounds is always –1.

(Due to restriction of negative oxidation number, F cannot form oxyacids or oxyanions for that it has to assign positive oxidation number).

9. Oxidation Number of Other Halogens is usually –1

The oxidation number of other halogens (Cl, Br and I) in binary compounds where they occur as halide ion  (X^‒) is usually –1. e.g.

The major exception is in compounds or ions of Cl, Br and I where they are bonded to oxygen atom e.g. in oxyacids, oxysalts and oxyanions of halogens like H⁺¹Cl⁺⁷O₄^–8, Na⁺¹Cl⁺⁵O₃^–6, Cl⁺¹O^1– etc. 

10. Oxidation Number of Hydrogen is mostly +1

 

11. Oxidation Number of Oxygen

The oxidation number of oxygen in most of its compounds is usually –2  (e.g. MgO^–2, Na₂O^–2, P₂O₅^–2, NO^–2, Cl₂O^–2, SO₂^–2 etc.). However in

12.    Oxidation Number of Elements in Groups of the Periodic Table

The oxidation number of each element of Group IA (Li, Na, K, Rb, Cs), IIA (Be, Mg, Ca, Sr, Ba), IIIA (B,Al), IVA (C, Si, Ge, Sn, Pb), VA (N, P, As, Sb, Bi) and VIA (O, S, Se, Te), in their compounds is +1, +2, +3, –4/+4, –3/+5, –2, –1, +1 to +6 respectively. (The most common oxidation state of group IIIA is +3 but it can also show +1 oxidation state due to inert pair effect).

Some helping rules for calculating oxidation number.

 

Q1. Determine the oxidation number of central atom in following

(i) S in Na₂S₂O₃                              

(ii) Mn in MnO₄⁻                                         

(iii) Cr in Cr₂O₇²⁻

(iv) Cl in ClO₃                                 

(v) Cr in Cr₂(SO₄)₃                       

(vi) P in Ca(H₂PO₄)₂

(vii) S in H₂SO₄                                             

(viii) Ni in Ni(CO)₄                      

(ix) Fe in Fe₃O₄

(x) C in C₃O₄                                    

(xi) Fe in Fe(CO)₃                         

(xii) Cr in Cr(CO)₆

Solution

(i) Finding out Oxidation number of S in Na₂S₂O₃

Oxidation number of Na = +1 

Oxidation number of O = −2

Na₂S₂O₃ is a neutral compound in which sum of the oxidation numbers of all atoms is zero. 

 

2(Na) + 2(S) + 3(O) = 0

2(+1) + 2S + 3(−2) = 0

+2 + 2S + (−6) = 0

2S + −4 = 0

2S  = +4

S  = +4/2 = +2

 

In Na₂S₂O₃, one S-atom (S*) has oxidation number = −2 while the other S-atom (S**) has oxidation number = +6. The total oxidation number of both S atoms = −2 + +6 = +4

The average oxidation number of S = Total Oxidation number/2 = +4/2 = +2 

(ii) Finding out Oxidation number of Mn in MnO₄⁻             

Oxidation number of O = −2

MnO₄⁻ is a polyatomic ion in which sum of the oxidation numbers of all elements is equal to its net ionic charge.

Mn + 4(O) = −1

Mn + 4(−2) = −1

Mn −8 = −1

Mn = −1 + 8

Mn = +7

 

(iii) Finding out Oxidation number of Cr in Cr₂O₇²⁻

Oxidation number of O = −2

Cr₂O₇²⁻ is a polyatomic ion in which sum of the oxidation numbers of all elements is equal to its net ionic charge.

2Cr + 7(O) = −2

2Cr + 7(−2) = −2

2Cr −14 = −2

2Cr = −2 + 14

2Cr = +12

Cr = +12/2 = +6

 

(iv) Finding out Oxidation number of Cl in ClO₃

Oxidation number of O = −2

 

ClO₃ is a neutral compound in which sum of the oxidation numbers of all atoms is zero. 

 

Cl + 3(O) = 0 

⇒ Cl + 3(−2) = 0 

⇒ Cl −6 = 0 

⇒ Cl = +6

(v) Finding out Oxidation number of Cr in Cr₂(SO₄)₃

Oxidation number of SO₄ = −2

Cr₂(SO₄)₃ is a neutral compound in which sum of the oxidation numbers of all elements is equal to zero.

 

2Cr + 3(SO₄) = 0

2Cr  + 3(−2) = 0

2Cr  −6 = 0

2Cr  = +6

Cr  = +6/2 = +3

 

Alternate method

In sulphate (SO₄²⁻) ion, oxidation number of S is +6.

 

2Cr + 3(S) + 12(O) = 0

2Cr  + 3(+6) + 12(−2) = 0

2Cr  + 18 + (−24) = 0

2Cr  + 18 −24 = 0

2Cr  −6 = 0

2Cr  = +6

Cr  = +6/2 = +3

 

(vi) Finding out Oxidation number of P in Ca(H₂PO₄)₂

Oxidation number of Ca = +2

Oxidation number of H = +1

Oxidation number of O = −2

 

Ca(H₂PO₄)₂ is a neutral compound in which sum of the oxidation numbers of all elements is equal to zero.

Ca + 4(H) + 2P + 8(O) = 0

+2 + 4(+1) + 2P + 8(−2) = 0

+2 + 4 + 2P + (−16) = 0

+2 + 4 + 2P −16 = 0

2P −10 = 0

2P = +10

P = +10/2 = +5

 

(vii) Finding out Oxidation number of S in H₂SO₄ 

Oxidation number of H = +1 

Oxidation number of O = –2

H₂SO₄ is a neutral compound in which sum of the oxidation numbers of all atoms is zero.  

(viii) Finding out Oxidation number of Ni in Ni(CO)₄

The oxidation number of metals in amalgams and metal carbonyls i.e. Ni(CO)₄, etc. is zero. In fact, in this transition metal complex, CO (called carbonyl) is a neutral ligand with overall charge of zero, hence giving zero oxidation number to central atom Ni. Some other neutral ligand is NO, H₂O, NH₃, PH₃, en, etc.

The oxidation number may be calculated as 

(ix) Finding out Oxidation number of Fe in Fe₃O₄

Fe₃O₄ is a mixed oxide consisting of two oxides ferrous oxide; FeO (O.N of Fe =+2)and ferric oxide, Fe₂O₃ (O.N of Fe =+3). The average oxidation of Fe is +8/3 or +2.66.

 

Oxidation number of Fe in FeO = +2

Oxidation number of Fe in Fe₂O₃ = +3

Average oxidation number = 2+ 2(+3)/2 = +8/3

 

The oxidation number may be calculated as

 

3(Fe) + 4(O) = 0

3Fe + 4(−2) = 0

3Fe + (−8) = 0

3Fe = +8

Fe = +8/3 or + 2.66

 

(x) Finding out Oxidation number of C in C₃O₄

3C + 4(O) = 0

3C + 4(−2) = 0

3C + (−8) = 0

3C= +8

C= +8/3 or + 2.66

 

(xi) Finding out Oxidation number of Fe in Fe(CO)₃

The oxidation number of metals in amalgams and metal carbonyls i.e. Fe(CO)₃ , Ni(CO)₄, Cr(CO)₆ etc. is zero. In fact, in this transition metal complex, CO (called carbonyl) is a neutral ligand with overall charge of zero, hence giving zero oxidation number to central atom Fe. Some other neutral ligand is NO, H₂O, NH₃, PH₃, en, etc.

 

The oxidation number may be calculated as

 

Fe + 3(CO) = 0

Fe + 3(0) = 0

Fe  + 0 = 0

Fe = 0

                               

(xii) Finding out Oxidation number of Cr in Cr(CO)₆

The oxidation number of metals in amalgams and metal carbonyls i.e. Fe(CO)₃ , Ni(CO)₄, Cr(CO)₆ etc. is zero. In fact, in this transition metal complex, CO (called carbonyl) is a neutral ligand with overall charge of zero, hence giving zero oxidation number to central atom Cr. Some other neutral ligand is NO, H₂O, NH₃, PH₃, en, etc.

 

The oxidation number may be calculated as

 

Cr+ 6(CO) = 0

Cr + 6(0) = 0

Cr = 0

Q2. Calculate the oxidation number of central element in following:

(i) C in C₂H₆O                        

(ii) C in C₄H₈O₂                              

(iii) C in CH₃Cl

(iv) Cr in CrO₂Cl₂                               

(v) C in COCl₂                                   

(vi) N in NOCl

(vii) U in UO₂Cl₂                  

(viii)Bi in BiOCl                            

(ix) S in SO₂Cl₂

(x) S in SOCl₂                        

(xi) S in HSO₃F                               

(xii) C in HCNS

Solution

Note

In few organic compounds and in few inorganic compounds where two more electronegative elements are present, more than one element has negative oxidation number while only one element has positive oxidation number. 

 

 

Q4. Calculate the oxidation number of central element in following:

(i) P in POCl₃                                        

(ii) CNO⁻                                          

(iii) Cl in CaOCl₂ (Bleaching powder)

(iv) S in Na₂S₄O₆                             

(v) N in NH₄NO₃             

(vi) Br in BrO₃⁻

Solution

(i) Finding out Oxidation number of P in POCl₃

Oxidation number of Cl = −1   (since Cl is more electronegative than P) 

Oxidation number of O = −2    (Normal oxidation number of O)

POCl₃ is a neutral compound in which sum of the oxidation numbers of all atoms is zero.  

(ii) Finding out Oxidation number of C in CNO⁻     

Oxidation number of N = −2      (since N is more electronegative than C) 

Oxidation number of O = −2 (Normal oxidation number of O)

CNO⁻ is a polyatomic ion in which sum of the oxidation numbers of all elements is equal to its net ionic charge.

(iii) Finding out Oxidation number of Cl in CaOCl₂ (Bleaching powder)

CaOCl₂ (Bleaching powder) is a mixed salt containing two anions namely chloride (Cl⁻) ion with −1 oxidation number of Cl and hypochlorite (OCl⁻) ion with +1 oxidation state of Cl.

(iv) Finding out Oxidation number of S in Na₂S₄O₆

(v) Finding out Oxidation number of N in NH₄NO₃

NH₄NO₃ (ammonium nitrate) consists of two nitrogen containing ionic species namely ammonium ion (NH₄⁺) in which oxidation number of N is −3 and nitrate ion (NO₃⁻) in which oxidation number of N is +5. The total oxidation number of both nitrogen is +2 and the average oxidation number is +1.

 

Oxidation number of N in NH₄⁺

N + 4(H) = +1

N + 4(+1) = +1

N + 4 = +1

N = +1 − 4

N = − 3

 

Oxidation number of N in NO₃⁻

N + 3(O) = −1

N + 3(−2) = −1

N − 6 = −1

N = −1 + 6

N = +5

 

Average Oxidation number of N

2(N) + 4(H) + 3(O) = 0

2N + 4(+1) + 3(−2) = 0

2N + (+4) + (−6) = 0

2N + −2 = 0

2N = +2

N = +2/2 = +1

 

(vi) Finding out Oxidation number of Br in BrO₃⁻

Oxidation number of O = −2 (Normal oxidation number of O)

BrO₃⁻ is a polyatomic ion in which sum of the oxidation numbers of all elements is equal to its net ionic charge.

Br + 3(O) = −1

Br + 3(−2) = −1

Br −6 = −1

Br = −1+6

Br = +5

Q5. Calculate the oxidation number of central element in following:

(i)NO₂                                          

(ii) OF₂                                

(iii) NCl₃

(iv) CO                                       

(v) BN                                  

(vi) SO₃    

          

Solution 

 

 

Q6. Calculate

(i) Oxidation number of S in tetrathionate (S₄O₆²⁻) ion

(ii) Oxidation number of C in C₃O₂ (carbon suboxide)

(iii) Oxidation number of U in UO₂(NO₃)₂

(iv)Oxidation number of Cr in CrO₅ (Perchromate; Blue)

(v) Oxidation number of Cl in NaClO₄

(vi) Oxidation number of Cr in SrCr₂O₇

(vii) Oxidation number of B in Na₂B₄O₇

(vii) Oxidation number of Br in Br₃O₈

Solution

(i) Finding out Oxidation number of S in tetrathionate (S₄O₆²⁻) ion

The oxidation number of end S atoms is +5 each and other two middle sulphur  atoms shown by S^* (pure covalent bond) is 0 each. The total oxidation number of 4 S atoms is 5+0+0+5=10 and the average oxidation number is 10 ÷ 4 = 2.5.

 

(ii) Finding out Oxidation number of C in C₃O₂

In C₃O₂ (carbon suboxide), the central C-atom has oxidation number = 0 (pure or non-polar covalent bond) while the other two C-atoms have oxidation number = +2 each.

The total oxidation number of all the three carbon atoms = +2 + 0 +2 = +4

The average oxidation number of C = total oxidation number/3 = +4/3 or + 1.33

(iii) Oxidation number of U in UO₂(NO₃)₂

Oxidation number of O = −2

Oxidation number of NO₃⁻ = −1

 

U + 2(O) + 2(NO₃) = 0

U + 2(−2) + 2(−1) = 0

U + (−4) + (−2) = 0

U −6 = 0

U = +6

 

(iv) Oxidation number of Cr in CrO₅ (Perchromate; Blue)

The oxidation number of Cr in CrO₅ cannot be calculated by the orthodox method of computing oxidation number as it is found to be +10 which is greater than the group number of Cr i.e. VIB and an element cannot exceed its maximum oxidation number than its group number, so this oxidation number is wrong.

 

Cr + 5(O) = 0

Cr + 5(−2) = 0

Cr −10 = 0

Cr = +10 (wrong as it is greater than group number of Cr i.e. VIB)

 

The oxidation number of Cr is calculated from its structure

(vii) Oxidation number of Br in Br₃O₈

3Br + 8(O) = 0

3Br + 8(−2) = 0

3Br −16 = 0

3Br = +16

Br = +16/3 or +5.33