Class 11 Chemistry

Class 11 Chemistry Organic Chemistry Basic Principles And Techniques Organic Chemistry Classification And Nomenclature

It has been known from ancient times that minerals, plants, and animals are the three major sources of naturally occurring substances. However, very little information was known regarding their chemistry until the beginning of the eighteenth century. In 1675, Lemery classified the natural substances into three classes such as mineral substances, vegetable substances, and animal substances based on the sources from which they were obtained and it was readily accepted.

• In 1784, it was Lavoisier who first showed that all compounds derived from vegetable and animal sources always contained carbon and hydrogen and sometimes oxygen, nitrogen, sulfur, and even phosphorus. So, there is a dose relationship between the vegetable and animal products.

This led to the classification of natural substances into two categories: 

1. Organic compounds: All those substances which were obtained from plants and animals, i.e., the substances which were obtained from living organisms and
2. Inorganic compounds: All those substances that were obtained from non-living sources such as rocks and minerals etc. This classification, however, found justification by the fact that in several cases, the same compound could be derived from both vegetable and animal sources.

A detailed investigation of the structure of the organic compounds revealed that almost all of them essentially contain both carbon and hydrogen atoms (hydrocarbon) and some of them also contain the atoms of a few other elements such as nitrogen, oxygen, phosphorus, halogens etc.

Since these are formed by replacing the hydrogen atoms in the hydrocarbons by these atoms, therefore, these are regarded as derivatives of hydrocarbons.  Organic compounds are, therefore, hydrocarbons and their derivatives and the branch of chemistry that deals with the study of these compounds is known as organic chemistry.

Inorganic chemistry, on the other hand, is defined as the chemistry of all elements other than carbon and their compounds

Tetrahedral Arrangement Of The Four Valencies (Bonds) Of Carbon

In 1859, Kekule proposed that carbon exhibits the normal valency of four units in simple as well as complicated organic molecules and an atom of carbon is joined to other atoms by four covalent bonds which remain in a plane. Also, the angle between any two adjacent bonds is 90°.

• This proposal, however, could not explain some special characteristics of organic compounds, and a change in Kekule’s theory was urgently required.
• In 1874, both can’t Hoff and Le Bel predicted that the four bonds of a carbon atom are directed towards the four corners of a regular tetrahedron, i.e., the angle between any two adjacent bonds is I09°28′ (tetrahedral angle).
• This representation is regarded as a tetrahedral model or space model. It was supported by electron diffraction and spectroscopic studies.
• The tetrahedral arrangement of four bonds of carbon laid the foundation for a fascinating field of ‘stereochemistry It is for this reason, that the first Nobel Prize in chemistry was awarded to van’t Hoffin 1901

Explanation for the existance of a large number of organic compounds:

An important and interesting property of carbon atom is its unique capacity to form bonds with other carbon atoms.This property of forming bonds with atoms of the same element is called catenation. Carbon shows maximum catenation in its group (group 14) in the periodic table.

This is because of the greater strength of the C —C bond as compared to other atoms. For example, the C— C bond is very strong (335 kj- mol^-1 ) in comparison to the Si—Si bond (220 kj-mol^-1 ) or Ge—Ge bond (167 kj – mol^-1). As a result, carbon atoms can link with each other to form either linear chains of various lengths or branched chains and also rings of different sizes.

Again, two or more organic compounds having the same molecular formula may be formed. Compounds having the same formula but different properties are called isomers and the phenomenon is called isomerism.

Catenation and isomerism are the two properties of carbon that are responsible for the existence of a large number of organic compounds. Carbon is also involved in forming multiple bonds with other carbon atoms (C=C, C≡C) and also with oxygen (C—O) and nitrogen atoms (C=N, C≡N). Themultiplebondingis also responsible for the existence of a variety of carbon compounds.

Electronic explanation of the tetra covalency of carbon: Lewis’ theory:

In 1916, G.N. Lewis put forward the electronic concept regarding the formation of bonds. According to him, atoms with similar or almost similar electrochemical nature combine by forming one or more electron pair using odd electron(s) in their valence shells, and in this way, they attain stable electronic configurations (electron octets) of the nearest noble gases.

Carbon exists in group IVA of the second period of the periodic table. In the perspective of its position just between the electropositive and electronegative elements, it can be said that it is neither an electropositive nor an electronegative element. Its ionization enthalpy is sufficiently high and its electron affinity is of moderate magnitude.

Therefore, carbon (ground state electronic configuration: ls²2s²2p_x¹2p_y¹) can neither produce C⁴⁺ ion by losing 4 electrons from its valence shell nor form C^-4- ion by gaining 4 electrons.

Both processes require large amounts of energy which are not ordinarily available during a chemical reaction. Thus, carbon does not usually form electrovalent compounds

Exception:

During the reaction between carbon and highly electropositive sodium, carbon forms the C^4- ion thereby producing the electrovalent compound sodium carbide, Na₄C ).

According to the electronic configuration, carbon should be bivalent, i.e., it should exhibit a valency of two because of the presence of two odd electrons in 2p_x^– and 2p_y^–orbitals.

To account for tetracovalency, it is believed that during the process of bond formation (an energy-releasing process), the two electrons in the 2s -orbital get unpaired, and one of them is promoted to the empty 2pz -orbital. Therefore, the electronic configuration of carbon in the excited state is ls²2s²2p_x¹2p_y¹ 2p_z¹) So, carbon can complete its octet by sharing the four odd electrons in its valence shell

With the 4 electrons of the valence shells of the other 4 atoms. Hence, carbon fulfills its octet by covalency and In the formation of covalent compounds, its valency is always 4

Example: One C -atom forms four electron pairs by combining with four H-atoms and produces one methane (CH₄) molecule with consequent completion of its octet According to Lewis’s theory, 1 electron pair stands for 1 covalent single bond, represented by dash sign

Now, two carbon atoms can form 1, 2, or 3 electron pairs by contributing 1, 2, or 3 electrons respectively from each carbon atom and use them equally to form a carbon-carbon single bond (C—C), carbon-carbon double bond (C= C) or carbon-carbon triple bond (C = C) respectively. For example, ethane (CH₃ —CH₃), ethylene (CH₂=CH₂), and acetylene (HC = CH) contain one carbon-carbon single, double, and triple bond respectively.

Although Lewis’s theory gives a satisfactory explanation to the tetracovalency of carbon, it fails to offer any idea about how these valencies are oriented in three-dimensional space. Moreover, this concept of covalent bond formation by sharing of electrons is purely a qualitative approach that tells nothing about the attractive forces involved in bond formation

Hybridization Of Carbon And Shapes Of Some Simple Organic Molecules

For sp³, sp² and sp -hybridisations of carbon and the shapes ofmethane, ethane, ethylene and acetylene see article no. 4.8 of the chapter “Chemical Bonding and Molecular Structure”. From the knowledge of the state of hybridisation of atoms in an organic molecule, an idea about shapes ofthe molecules can be obtained.

These maybe summarised as follows:

1. Atoms bonded to an sp³ -hybridised carbon atom are tetrahedrally oriented. So, any molecule containing an sp³ -hybridised carbon atom must have threedimensional shape (irrespective of the presence of any sp2 or sp -hybridised carbon atom in it).

2. The two sp² -hybridized C-atoms and the atoms directly attached to them remain in the same plane. So, if any organic molecule contains only sp² -hybridized C-atoms which are arranged in a chain or cyclic pattern, then the whole molecule becomes planar.

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For example: 1,3-butadiene (CH₂=CH—CH=CH₂) and benzene molecules are planar.

If a molecule contains both sp and sp² – hybridized C-atoms, then the molecule will also be planar. For example, the but-l-en-3-yne (CH₂= CH—C= CH) molecule is planar. If sp² or sp -hybridized carbon atoms remain attached to a benzene ring, then the molecule will also be planar. For example, vinylbenzene and ethylbenzene molecules are found to be planar. When an atom is present as a substituent in benzene, even then the molecule becomes planar.

For example: Fluorobenzene, chlorobenzene, bromobenzene etc…  are planar molecules.

3. The two sp -hybridized carbon atoms and the atoms directly attached to them remain on the same line. So, if a

A molecule contains only sp -hybridized carbon atoms which remain bonded one after another, and then the molecule will be linear in shape. But 1,3-diyne, for example, is a linear molecule.

Structure of some familiar organic compounds:

Prediction of state of hybridization of a carbon atom from the nature of bonding:

If a carbon atom is bonded to four atoms by four single bonds (i.e., 4 cr -bonds), then that carbon atom is sp³ -hybridized;

• If a carbon atom is bonded to three atoms by two single bonds [i.e., 2 σ – bonds) and one double bond [i.e., 1 σ and 1 π -bond), then that carbon atom is sp² -hybridized; and
• If a carbon atom is bonded to two atoms by one single bond [i.e., one bond) and one triple bond [i.e., one σ -bond and two π bonds) or by two double bonds [i.e., two car and two n bonds), then that carbon atom is sp -sp-hybridized

Effect Of Hybridisation On Bond Lengths And Bond Strengths

The bond length as well as bond strength of any bond depends upon the size of the hybrid orbitals involved in bond formation.

1. The s -characters of sp3, sp², and sp -hybrid orbitals are 25%, 33.33%, and 50% respectively. As s -the orbital is more closer to the nucleus as compared to p -orbital, more the percentage of s -the character of the hybrid orbital, more it will be attracted by the nucleus, and as a result, its size will

The s -characters of sp³, sp² and sp -hybrid orbitals are 25%, 33.33%, and 50% respectively. As s -orbital is closer to the nucleus as compared to p -orbital, the more the percentage of s -character of the hybrid orbital, the more it will be attracted by the nucleus, and as a result, its size will decrease more.

Therefore, the sizes of sp³, sp², and sp hybrid orbitals follow the order: sp < sp² < sp³. Consequently, the bond formed by an sp³ -hybrid orbital is longer than the bond formed by an sp² hybrid formed orbital by an which-hybrid orbital turn is longer than the bond

Example:

The C_sp³   bond is longer than the C_sp²  —H bond which in turn is longer than the C_sp —H bond. Similarly, the C_sp³ —C_sp³  bond is longer than the C_sp² —C_sp²  bond which turns longer than the C_sp — C_sp bond. For the same reason, the lengths of single, double, and triple bonds follow the order: C —C>C=C>≡ C. The presence and the increase in the number of n -bonds between two carbon atoms are also responsible for the decrease in bond length.

2. Shorter the bond, the greater is its strength i.e., stronger itis. Therefore, a <r -bond formed by sp -hybrid orbitals is stronger than a σ -bond formed by sp² -hybrid orbitals which in turn forms more stronger σ -bond than formed by sp³ -hybrid orbitals. Again, the bond strength increases with an increase in bond multiplicity and so, a triple bond is stronger than a double bond whichin turn is stronger than a single bond

Bond lengths and bond dissociation enthalpies of different bonds:

Different Structural Representations Of Organic Compounds

Lewis Structure Or electron dot Structure

In Lewis structure or electron dot structure, only the electrons of the valence shell of each atom of a molecule is H shown. For example, the electron dot structure of propene is shown at the right side. Since this style of writing is time-consuming and inconvenient, it is not generally followed

Dash formula

The Lewis structure, however, can be simplified by representing each pair ofelectrons involved in forming a covalent bond by a dash(—). In this representation, a single dash represents a single bond, a double dash (=) represents a double bond, and a triple dash (≡) represents a triple bond.

The lone pair of electrons on the heteroatoms e.g., oxygen, nitrogen, halogens, sulfur, etc. May or may not be shown. Thus, ethane, ethylene, acetylene and ethanol can be represented by their structural formulas as given above. Such structural representation is also called complete structural formula or graphic formula.

Condensed structural formula

The complete structural formulas can be further abbreviated by deleting some or all of the covalent bonds and by indicating the number of identical groups attached to an atom by a subscript. Such simplified structural representations of molecules are called condensed structural formulas. Thus, ethane, ethylene, acetylene and ethanol can be represented as: CH₃CH₃ (ethane); H₂C — CH₂ (ethylene); HC=CH (acetylene); CH₃CH₂OH (ethanol)

Similarly, a long chain of carbon atoms such as CH₃CH₂CH₂CH₂CH₂COOH can be represented as CH₃(CH₂)₄COOH.

Bond-line Structural formula

It is a very simple, short and convenient method of representing the structures of organic molecules. In this representation, only the carbon-carbon bonds are shown by lines drawn in a zig-zag fashion.

The line ends and the intersection of lines represent carbon atoms carrying an appropriate number of H -atoms so that its tetravalency is fulfilled (i’.e., the terminals denote CH₃ -groups, and the unsubstituted intersections or line junctions denote CH₂ -groups). The heteroatoms and the H-atoms attached to them are, however, specifically shown. A single bond is represented by a single line ( —), a double bond is represented by two parallel lines (= ) and a triple bond is represented by three parallel lines (≡).

In bond-line notation, the cyclic compounds are represented by an appropriate ring or polygon without showing carbon and hydrogen atoms. Each comer of a polygon represents a carbon atom while each side of the polygon represents a carbon-carbon bond.

For example:

Three-dimensional representation of organic molecules

The three-dimensional structure of organic molecules is quite difficult to represent on paper (two-dimensional). So, certain graphic conventions have been proposed. For example, by using a solid wedge and dashed wedge the threedimensional structure of a molecule from a two-dimensional presentation can be perceived.

In these formulas, the solid wedge (i.e., the thick solid or heavy line) is used to indicate a bond lying above the plane ofthe paper and projecting towards the observer, and the dashed wedge is used to represent a bond lying below the plane ofthe paper and projecting away from the observer. The bonds lying in the plane of the paper are shown by normal or ordinary lines (—).

A three-dimensional representation of a methane molecule, for example, is shown below:

Classification Of Organic Compounds Based On Carbon Skeleton

Due to the existence large number of organic compounds is rather inconvenient to study the chemical nature of these compounds individually. To simplify and systematize the study of organic chemistry, organic compounds have been broadly classified into two categories depending on the nature of their carbon skeleton or structures. The flow sheet given here is the general classification of organic compounds.

Acyclic or open chain or aliphatic compounds:

The compounds containing open chains of carbon atoms with the appropriate number of H -atoms and functional groups are called acyclic or open-chain compounds. These compounds are also called aliphatic compounds because the earlier members of this class were obtained either from animal or vegetable fats (Greek, aliphatic = fat). Due to the variation in the structure of carbon chains, two types of open-chain compounds have been formed.

These are:

• Unbranched or straight-chain compounds and
• Branched chain compounds.

1. Straight chain compounds

The open chain or aliphatic compounds in which no carbon atom (except the two terminal C -atoms) is attached to more than two carbon atoms are called unbranched or straight chain compounds.

Examples:

2. Branched-chain compounds

The open-chain or aliphatic compounds in which at least one carbon atom is attached to three or more carbon atoms are called branched-chain compounds

Examples:

Cyclic or closed chain or ring compounds

Compounds containing one or more closed chains or rings of atoms in their molecules are called cyclic or closed chain or ring compounds. Depending upon the nature of the atoms present in the ring, the cyclic compounds may be divided into the following two classes

1. Carbocydic or homocyclic compounds:

The compounds containing rings that are made up ofonly carbon atoms are called carboxylic or homocyclic compounds.

These are further divided into two sub-classes as follows:

1. Alicyclic compounds:

The carboxylic compounds which show resemblance in properties with the aliphatic compounds are called alicyclic compounds

Examples:

2. Aromatic compounds: For aromatic compounds, see the aromatic hydrocarbon portion

2. Heterocyclic compounds:

Cyclic compounds containing more heteroatoms (atoms other than C and H i.e., O, N, S, etc.) in their rings are called heterocyclic compounds. Depending upon their chemical behaviour, they are further divided into the following two categories

1. Alicyclic heterocyclic compounds:

Aliphatic cyclic compounds containing one or more heteroatoms in their rings are known as alicyclic heterocyclic compounds

Examples:

2. Aromatic heterocyclic compounds:

Aromatic compounds containing one or more heteroatoms in their molecules are called aromatic heterocyclic compounds.

Examples:

Saturated and unsaturated compounds

1. Saturated compounds:

The organic compounds in which the carbon atoms are linked with each other only by single covalent bonds are called saturated compounds.

Examples: Methane (CH₄), ethane (CH₃—CH₃), ethyl alcohol (CH₃ —CH₂ —OH), methylamine (CH₃ —NH₂) etc.

2. Unsaturated compounds:

The organic compounds which contain at least one carbon-carbon double bond or triple bond are called unsaturated compounds

Examples:

Ethylene (H₂C= CH₂), l-butene(C₂H₅CH= CH₂)

1,3-butadiene (H₂C=CH—CH=CH₂) etc.

Organic compounds containing unsaturated groups such as are considered saturated compounds. -CHO, -COOH, -COOR etc. But no C =C or C = C For example, propionic acid (CH₃CH₂COOH) is a saturated compound but acrylic acid (CH₂=CH—COOH) is an unsaturated compound.

Functional Groups & Homologous Series

Functional group Definition

A functional group may be defined as an atom or group of atoms present in an organic compound, which is responsible for its characteristic chemical properties.

Generally, compounds having the same functional group have similar properties while compounds with different functional groups have different chemical properties. For this reason, organic compounds are classified into different classes or families based on the functional groups present In the reactions of organic compounds, the organic groups or radicals generally do not suffer any change but the functional groups actively participate:

Some common functional groups present in various organic compounds along with their classes:

1. Though the functional groups govern the chemical properties, they are also found to influence the physical properties in some cases. For example, alcohols (ROH) due to the presence of —OH group remain associated through intermolecular H -bonding and as a result, the boiling points of alcohols are much higher than that of ethers having similar molecular masses.

2. Organic compounds with 2 or more different functional groups exhibit characteristic properties of all the functional groups present in it. For example, aldol [CH₃CH(OH)CH₂CHO] exhibits the properties of ; both alcohol and aldehyde.

3. Although different compounds having the same functional group show similarities in their chemical properties, their properties are not identical. For example, formaldehyde (HCHO) and acetaldehyde (CH₃CHO) containing the same functional group ( —CHO), do not respond to the same type of reaction. The former participates in the Cannizzaro reaction but not in Aldol condensation reaction while the latter takes part in Aldol condensation reaction but the Cannizzaro reaction

Homologous series

Homologous series definition:

A homologous series is defined as a series or group of similarly constituted organic compounds which have the same functional group and thus similar chemical properties and any two successive members differ in their molecular formula by a CH₂– group. Members of such a series are called homologues.

Some homologous series, their general formulas and structures of compounds upto C₄

Characteristics of homologous series:

1. All the members of a homologous scries can be represented by the same general formula. For example, C_nH_n+1 + j OH is the general formula of alcohols.

2. Same functional group is present in ail members of any homologous series. So, the members of any homologous series have almost identical chemical properties (the phenomenon of such resemblance in properties among the compounds of the same homologous series is called homology). However, with increase in molecular mass, the chemical reactivity ofthe compounds usually decreases.

3. Any two successive members of aparticular series differ in molecular formulary CHHomologous series group or14 massunits.

The physical properties such as density, melting point, boiling point of the members of a homologous series increase gradually with the increase in molecular mass. However, solubility and volatility show a declining trend with rise in molecular mass.

4. The members of a homologous series can be prepared by almost identical methods, known as the general methods of preparation.

Significance of homologous series:

• From the knowledge ofthe method of preparation and the properties of a particular member of a homologous series, the method of preparation and properties of the other members of the same series can easily be predicted.
• Therefore, by dividing the vast number of organic compounds into homologous series followed by the study of the method of preparation, properties, and reactions of a representative member, an overall idea about the whole family can be obtained.
• However, the first member of a series often differs from the other members in the method of preparation and properties.
• For example, the method of preparation and properties of formic acid (HCOOH), the first member ofthe carboxylic acid family, are different from that ofthe other members of the family.

Classification Of Organic Compounds On The Basis of Functional Group

Hydrocarbon

Hydrocarbons are the binary compounds of C and H. Based on their structural formulae, they are given types

1. Saturated hydrocarbons or alkanes or paraffins:

The open chain hydrocarbons in which the carbon atoms (except methane, CH4, in which the single carbon is bonded to four H -atoms) of each molecule are linked mutually by single covalent σ -bonds, and rest of the valencies of C atoms are satisfied by single covalent σ -bonds with H atoms, are known as saturated hydrocarbons or alkanes.

Since these saturated hydrocarbons are quite less reactive due to absence of any functional group, they are also called paraffins (Latin: parum =little and affins = affinity or reactivity). The compounds ofthis class are representedby the general formula: C_n2n+ 2.

Classification of carbon and hydrogen atoms present in saturated hydrocarbons or alkanes:

The C-atoms present in an alkane molecule may be classified into four types primary (1°), secondary (2°), tertiary (3°), and quaternary (4°), C-atom as follows:

• A C-atom attached to only one (or no other) C-atom is called primary C-atom It is designated as 1° carbon.
• A C-atom attached to two other C-atoms is called a secondary C-atom.It is designated as 2°.
• A C-atom attached to three other C-atoms is called a tertiary C-atom.It is designated as 3° carbon.
• A C-atom attached to four other C-atoms is called a quaternary C-atom. It is designated as 4° carbon.

The hydrogen atoms may similarly be classified.

The hydrogen atoms attached to 1°, 2°, and 3° carbon atoms are called primary (1°), secondary (2°) and tertiary (3°) hydrogen atoms respectively. It is to be noted that unlike quaternary carbon atoms quaternary H atom has no existence because a quaternary carbon does not carry any hydrogen atom. The former example clearly illustrates the various types of carbon and hydrogen atoms.

2. Unsaturated hydrocarbons:

The open chain hydrocarbons which contain at least one carbon-carbon double bond  or carbon-carbon triple (—C= C— ) bond in their molecules are called unsaturated hydrocarbons. These unsaturated hydrocarbons are further classified into two types: alkenes and alkynes.

The unsaturated hydrocarbons containing carbon-carbon double bonds are called alkenes, for example, ethylene (CH₂=CH₂), propylene (CH₃CH=CH₂) etc. A double bond is made up of one cr -bond and one n -bond. These hydrocarbons are also called olefins (Greek: olefiant = oil forming) because the lower members of this class react with chlorine to form products. The general formula of alkenes is C_nH_2n where n = 1, 2, 3— etc.

The unsaturated hydrocarbons containing carbon-carbon triple bonds are called alkynes, for example, acetylene (HC = CH), methyl acetylene (CH3C= CH) etc. A triple bond is made up of one σ-bond and two π-bonds. The general formula of alkynes is C_nH_2n-2 where n= 1, 2,3 – etc

Classification of hydrocarbon derivatives based on functional group:

IUPAC Nomenclature Of Aliphatic Organic Compounds

According to the IUPAC system, the name of an organic compound consists of 3 parts:

1. Word root
2. Suffix and
3. Prefix.

1. Word root:

Word root, the basic unit of the name, denotes the number of carbon atoms present in the parent chain (the longest possible continuous chain of carbon atoms including the functional group and multiple bonds) of the organic molecule.

For chains containing up to four carbon atoms, special word roots (based upon the common names of alkanes) such as ‘meth’ for C₁, ‘eth’ for C₂, ‘prop’ for C_3, and ‘but’ for C₄ are used where C₁, C₂, C₃ and C₄ represent the number of carbon atoms in the chain. For chains of five or more carbon atoms, Greek numerals or number roots are used to represent the word roots.

For example:

‘Pent’ is used for the C₅ chain, ‘hex’ is used for C₆ chain etc. The general word root for any carbon chain is ‘alk’

2. Suffix

These are of the following two types

1. Primary suffix:

A primary suffix is always added to the word root to indicate whether the carbon chain is saturated or unsaturated. There are three basic primary suffixes. If the carbon atoms are linked only by single covalent bonds (C —C) ‘-ane’ is used.

If there is at least one double bond (C= C) present in the chain, the primary suffix ‘- ene’ and if there is at least one triple bond (C = C) in the chain, the primary suffix ‘-yne’ is used. Hence, the name of

1. CH₃CH₂CH₁₃ is prop + ane = propane and
2. CH₃CH=CH₂ is prop + ene = propene.

If the parent carbon chain contains 2, 3, or more double or triple bonds, numerical suffixes such as ‘di’ (for two), ‘tri’ (for three), ‘tetra’ (for four), etc.

Are added to the primary suffix. If the primary suffix begins with a consonant then an extra ‘a’ is to be added to the word root. For example, the primary suffix used for two double bonds is diene. Now if it is added to the word root ‘but’ for C₄ chain, then the name of the compound will be butadiene.

2. Secondary suffix:

A secondary suffix is used to indicate the functional group presents an organic molecule and is to be added to the primary suffix while writing the IUPAC name. Secondary suffixes of some important functional groups are listed:

It is to be noted that while adding the secondary suffix to the primary suffix, terminal ‘e’ of the primary suffix [i.e., ‘ane,’ ‘ene’ or ‘yne’) is dropped if the secondary suffix begins with a vowel but is retained if the secondary suffix begins with a consonant.

For example: The name of CH₃CH₂OH is: eth + ane + ol = ethanol and that of CH₃CH₂CN is: prop + ane + nitrile = propanenitrile.

3. Prefix

It is a part of the IUPAC name of a compound that appears before the word root. Prefixes are of two types

1. Primaryprefix: A primary prefix is used to differentiate a cyclic compound from an acyclic compound

For Example:

In case of carbocyclic compounds, the primary prefix ‘cyclo’ is used just before the word root. For example, the cyclic hydrocarbon, maybe named as

2. Secondaryprefix:

In the IUPAC system of nomenclature, some groups are not considered functional groups or secondary suffixes. These are treated as substituents and are called secondary prefixes. These are added just before the word root (or the primary prefix in the case of alicyclic compounds) in alphabetical order.

The secondary prefixes of a few substituents are given in the following table:

Therefore, while writing the IUPAC name of an aliphatic organic compound the various parts are to be added in the following sequence: Secondary prefix + Primary prefix+ Wordroot+ Primary suffix + Secondary suffix. It is not necessary that all the parts may be present in a particular compound. However, the word root and primary suffix must be present.

Example:

In case ofthe compound, \(\stackrel{4}{\mathrm{C}} \mathrm{H}_3 \stackrel{3}{\mathrm{C}} \mathrm{HCl} \stackrel{2}{\mathrm{C}} \mathrm{H}_2 \stackrel{1}{\mathrm{C}} \mathrm{H}_2 \mathrm{OH}\), the word roots ‘but’, the primary suffix ‘ane’, the secondary suffix is ‘ol’ and the secondary prefix is ‘chloro’. As the compound is acyclic, the primary prefix is absent. Therefore, the IUPAC name of the compound is chloro (at position C-3 ) + but + an (e is omitted) + ol (at position C-l )- 3-chlorobutan-l-ol.

IUPAC Nomenclature of some organic compounds

Common & Iupac Nomenclature Of Some Important Classes Of Organic Compounds

Saturated hydrocarbons (Alkanes)

In the IUPAC system, saturated acyclic hydrocarbons are called alkanes. IUPAC names of alkanes are obtained by adding the suffix ‘ane’ to the word root indicating the number of C-atoms present in the chain. The first 4 alkanes (CH₄ to C₄H₁₀) have their special names, i.e., methane, ethane, propane, and butane.

The names of alkanes containing 5 or more C-atoms are obtained by adding prefixes such as ‘pent’ (5), ‘hex’ (6), ‘hept’ (7), ‘Oct’ (8), etc., indicating the number of C-atoms in the molecule to the suffix ‘ane! Although common and IUPAC names of alkanes are the same, the prefix (normal)is added to the common name of alkanes containing 4 or more C-atoms.

IUPAC and common names of first ten members of the alkane family:

Alkyl groups

An organic group produced by the removal of one H -atom from an alkane molecule is called an alkyl group or alkyl radical.

For example: 

1. Removal of one H-atom from methane (CH₄) produces methyl group (-CH₃),
2. An ethyl group ( —C₂H₅) is formed by the removal of any one of six equivalent H -atoms from ethane (C₂H₆)

All the H -atoms of alkanes containing more than two carbon atoms are not always equivalent. So, two or more alkyl groups can be derived from these alkanes.

For example:

In the case of propane (CH₃CH₂CH₃), the removal of one hydrogen atom attached to the terminal carbon yields an unbranched propyl group. But when one H -atom linked to the middle carbon is removed, an isopropyl group is obtained.

The alkyl groups are generally represented by the letter R. So, an alkane is represented as R —H. The monovalent alkyl groups have the general formula: C_nH_2n+1 [n = 1.2,3 etc.]

Nomenclature of alkyl groups

The names of the alkyl groups are derived by replacing the suffix ‘ane’ of the corresponding alkane by the suffix ‘yl’

Classification of alkyl groups

1. Primary (1°) alkyl groups:

The removal of one 1° H -atom from an alkane gives a primary or 1° alkyl group. Ethyl (CH₃CH₂-), isobutyl (Me₂CHCH₂—), neopentyl (Me₃CCH₂—), etc. Are some examples of primary alkyl groups. The primary alkyl groups obtained from simple straight-chain alkanes are called normal alkyl groups

In the common nomenclature system, those are designated as n -alkyl group Imt In the IUPAC system, n Is omitted

For example:

In the common or trivial system of nomenclature, CH,CH₂CH₂– IS written as n -propyl group hut In the IUPAC system, it Is designated as propyl group. The primary alkyl groups In which the second last carbon in the chain is branched to one

The group are named by using the prefix Mso’

For example:

The —CH₂CH(CH₃)₃ group is called the isobutyl group. (As a group, up to isohexyl and as an alkane up to isohexane, the use ofthe ‘iso’ has been accepted by the IUPAC system).

The primary alkyl groups in which the second last carbon in the chain is branched to two —CH₃ groups are named by using the prefix ‘nco’.

For example: The —CH₂C(CH₃ )₃ group is called neopentyl group. (As a group, upto neo hexyl and as an alkane upto neohexane, the use of‘neo’ is accepted by die IUPAC system.)

2. Secondary (2°) alkyl group:

Removal of one 2° H-atom from an alkane forms a secondary (2°) alkyl group. In both trivial & IUPAC system,itis written as sec-alkyl (pronounced as secondary alkyl group),

For example: CH₃ CH₂ CHCH₃ is a 2° alkyl group named as a sec-butyl group.

3. Tertiary (3°) alkyl group:

The removal of one 3° H -atom from an alkane leads to the formation of a tertiary or 3° alkyl group. In both the trivial and IUPAC systems, it is written as ferf-alkyl group or f-alkyl group (pronounced as tertiary alkyl group),

For example:—C(CH₃ )₃ is a tertiary alkyl group whose name is a fert-butyl group or a t-butyl group

Structures and IUPAC names of some alkyl groups

Monovalent radicals derived from unsaturated acyclic hydrocarbons end with ‘-enyl ‘-ynyl,’ ‘-dienyl,’ etc., depending on the nature of the radicals or groups. Positions of double and triple bonds are indicated by numerals where necessary.

The c-atom of any radical containing free valence is always numbered as ‘1’

For example:

CH ≡ C— (ethynyl), \(\stackrel{3}{\mathrm{C}} \mathrm{H} \equiv \stackrel{2}{\mathrm{C}} \stackrel{1}{\mathrm{C}} \mathrm{H}_2\) (prop-2-any)

⇒ \(\stackrel{3}{\mathrm{C}} \mathrm{H}_3 \stackrel{2}{\mathrm{C}} \mathrm{H}=\stackrel{1}{\mathrm{C}} \mathrm{H}-\) (prop-l-enyl) etc. The following trivial names are retained in the IUPAC system: CH₂=CH— (vinyl),

CH₂=CHCH₂— (allyl), etc. The radical (CH₂= ) is called ‘methylene’ and (=CH—)is called ‘methine.

The presence of one or more free valency in radicals derived from parent alkenes is indicated by suffixes like monovalent(-yl), divalent (-diyl), trivalent (-triyl) etc.

For example: CH₃CH< is ethane-1, 1-diyl; (CH₃)₂C< is propane-2,2-diyl etc.

Aryl groups:

The organic groups derived from benzene and other benzene derivatives are termed as aryl groups. Aryl groups are generally represented by Ar. The simplest aryl group is phenyl group (—C₆H₅ ). It can be obtained by removing one hydrogen atom from a molecule of benzene (C₆H₆ )

Trivial or common system of nomenclature of other classes of compounds:

IUPAC nomenclature of different classes of compounds at a glance:

Rules Branched For Iupac Nomenclature Of Chain Alkanes

Longest chain rule

The longest continuous chains of the alkane is to be identified first.It is known as the parent or root chain. The number of carbon atoms in the parent chain determines the word root

It is to be noted that the longest chain may or may not be straight but it must be continuous. All other carbon atoms that are not included in the parent chain are called branched chains side chains or substituents. The branched chain alkane is, therefore, named as a derivative ofthe parent chain

Example:

The parent chain in the compound (I) contains 6 carbon atoms and the CH₃ -group is a side chain or substituent. Therefore, it is to be named as a derivative of hexane. The parent chain in the compound (II) contains 8 C-atoms but is not straight so, it is named as a derivative of octane. CH₃ -and C₂H₆ -groups are the two substituents here. If two chains of equal lengths are possible, then the one with a maximum number of side chains or substituents is to be considered as the parent chain.

Example:

In the compound (III), the parent chain is the horizontal six-carbon chain containing two alkyl substituents ( —CH₃ and C₂H₅) but not the other six-carbon chain containing only one alkyl substituent [(CH₃)₂CH-]

Lowest number rule

The carbon atoms of the parent chain is to be numbered as 1, 2, 3, 4, . etc. from one end in such a way that the carbon atom carrying the substituent gets the lowest possible number.

Example:

In the following compound, the numbering can be done in two different ways. The numbering of the carbon chain as given in the structure (IVA) is correct because the carbon carrying the substituent gets a lower number i.e., 3. However, the numbering of the carbon chain as given in the structure (IVB) is incorrect because the carbon carrying the substituent gets a higher number i.e., 5.

The number indicating the position of the substituent in the parent chain is called its positional number or locant. Thus, the correct locant for the methyl side chain in the above compoundis 3.

When two substituents are present in the chain, then the lowest set of locant rule is applied. It states that when two or more different sets of locants are possible, then that set of locants will be the lowest which (when compared term by term with other sets, each in order of increasing magnitude) has the lowest term at the first point of difference. This rule is used irrespective ofthe nature ofthe substituent.

Example:

When the carbon atoms of the parent chain of the following compound (V) are numbered from the sides, two sets of locants are obtained. Out of the two sets of locants (2,3) and (3,4), the first set is lower and hence preferred because the first term, i.e., 2 in the first set is lower than the first term, i.e., 3 in the second set

Similarly , for the compound (VI):

Out ofthe two sets of officiants (2,2,4) and (2,4,4), the first set is lower and hence preferred as the second term in the first set i.e.,2 is lower than the second term 4in the second set.

According to the latest IUPAC recommendations of nomenclature (1993), the lowest set of locant rules Is preferred even If it violates the lowest sum rule.

For example:

In the case of the following compound, the numbering of the carbon chain from left and right gives two different set of locants with two different sum of locants.

The numbering from the left is correct because the first term CH₃ i.e., 2 in the set (2, 7, 8) is lower than the first term i.e., 3 in the set (3, 4, 9), even though the sum of locants is lower when the numbering is done from the right. Thus, the correct name ofthe alkanes 2, 7, 8-trimethyIdecane.

Name of the branched chain alkanes

When there is one alkyl group in the parent chain, its name is to be prefixed to the name of the parent alkane, and its position on the chain is to be indicated by writing before it the number of the carbon atom carrying the substituent.

The name of the substituent is separated from its positional number or locant by a hyphen (-). The final name of the alkane is to be written as one word, i.e., there will be no gap between the name of the substituent and the parent alkane.

Example:

Alphabetical order of the side chains or substituents

When two or more different alkyl groups (side chains or substituents) are present on the parent chain, such groups prefixed by their positional numbers or locants, are to be arranged in alphabetical order irrespective of their positional numbers and written before the name of the parent alkane.

Example:

In the given compound, between ethyl and methyl groups, ethyl comes first in the alphabetical order and therefore, its name is 3-ethyl-2-methylhexane. When a number appears between two substituent groups then hyphens are used on both sides ofthe number

It is to be noted that while deciding the alphabetical order of various alkyl groups, prefixes such as ‘iso’ or‘neo’ are to be considered as a part of the fundamental name of the alkyl group while the prefixes, ‘second ‘tert’ are not

Examples:

Numbering of different alkyl groups at equivalent positions

If two different alkyl groups or substituents are present at equivalent positions, i.e., at the same position from the two ends of the parent chain, then numbering of the chain is to be done in such a way that the alkyl group which comes first in the alphabetical order gets the lower number.

Example: 

Naming of same alkyl groups at different positions

When the parent chain contains two or more same alkyl groups at different positions (or at the same position), the positional number of each alkyl group is separated by commas, and suitable prefixes such as ‘di’ (for two), ‘tri’ (for three), ‘tetra’(for four) etc., and then they are to be attached to the name of the alkyl group. When two alkyl groups are attached to the same carbon atom, the positional number or locant is to be written twice. It is to be noted that the prefixes like ‘di’, ‘tri’, ‘tetra’, etc. are not considered while deciding the alphabetical order ofthe alkyl groups

Naming of complex substituents/substituted substituents

1. If the alkyl group on the parent chain is complex, i.e., if it has a branched chain, it is named as a substituted alkyl group by numbering the C-atom of this group attached to the parent chain as 1. The name of the complex substituent is generally enclosed in brackets to avoid any confusion with the numbering of the parent chain.

2. When the same complex substituent occurs more than once on the parent chain, it is indicated by multiplying the prefixes such as “bis’ (for two), ‘tris’ (for three), ‘tetrakis’ (for four), ‘pentakis'(for five), etc

Example:

IUPAC nomenclature of bicyclic compounds

The name of a bicyclic compound in the IUPAC system is based on the name ofthe alkane having the same number of carbons as there in the ring system. The name follows the prefix bicyclo and a set of brackets enclosing numbers indicating the number of carbons in each of the three bridges connecting the bridgehead carbons in order of decreasing size.

For example:

The following bicyclic compounds containing nine and eight carbon atom are named bicyclo [4.3.0]nonane and bicyclo[3.2.1]octane respectively.

What is wrong with the following names? Draw the structures they represent and write their correct names. (i)1,1-dimethylhexane (iii)3-methyl-5-ethylheptane (iv) 4, A-dimethyl-3-ethylpentane (v) 3, 4,7-trimethyloctane (vi) 3,3-diethyl-2,A,Atrimethylpentane 36. Give the IUPAC name of the following alkane containing complex substituents:

Rules For Iupac Nomenclature Of Unsaturated Hydrocarbons

For naming the compounds containing multiple (double and triple) bonds, the following additional rules are to be applied:

1. The parent chain must contain multiple bonds (double or triple) regardless of the fact whether it denotes the longest continuous chain or not

For example:

In compound (I), the parent chain contains 5 carbon atoms and not 6 carbon atoms since the latter does not include the double bond

Example:

2. While naming a particular member ofthe alkene or alkyne family, the primary suffix ‘ane’ of the corresponding alkanes to be replaced by ‘ene’ or ‘yne’ respectively.

3. The numbering of the parent chain is to be done in such a way that the first C-atom associated with the multiple bond gets the lowest possible number

Examples:

4. If the parent chain contains a side chain, then also the multiple bond get spriority in numbering.

If the parent chain contains 2 or 3 double or triple bonds, then the primary suffix ‘diene’ (or ‘triene’) or ‘diyne’ (or ‘triyne’) are to be used to represent them. In these cases, terminal ‘a’ is also added to the wordroot Fornumbering, the lowest set oflocants rule is to be followed.

Examples:

5. If the parent chain contains both double and triple bonds, the following points are to be remembered while writing their names:

The unsaturated hydrocarbon is always named as a derivative of alkyne, i.e., the primary suffix ‘ene’ is always to be written before ‘yne’.

In all these cases, the terminal ‘e’ of the one is dropped. The numbering of the parent chain is to be done from that end which is nearer to the double or triple bond, i.e., the lowest set officiants rule is to be followed

Examples:

If the positions of double and triple bonds are identical, i.e., if  the set officiants from both sides ofthe chain is the same, then the double bond is always given preference over the triple bond

Example:

6. If the unsaturated hydrocarbon contains a side chain along with double and triple bonds, then the numbering of the parent chain is to be done in such a way that the multiple bonds get the lowest set of locants. However, if the numbering from both ends of the parent chain gives the same set of locants to the multiple bonds, then the locant for the side chain must be minimum.

Examples:

When there are more than two double bonds in the hydrocarbon and it is impossible to include all of the min the parent chain, then the double bond which is not included in the parent chain is treated as a substituent

IUPAC Nomenclature Of Compounds With Functional Groups, Multiple Bonds And Substituents

The following additional rules are to be followed while naming organic compounds containing one functional group, double and triple bonds and substituents:

1. In these compounds, the longest chain of carbon atoms containing the functional group and the maximum number ofdouble and triple bonds are to be considered as the parent chain.It may or maynot be the longest possible carbon chain.

For example:

In the following compound (I), the parent chain containing the functional group and the double bond has 6 carbon atoms while the longest possible carbon chain has 7 carbon atoms.

Example:

2. The parent chain is to be numbered in such a way that the functional group gets the lowest locant, even if it violets the lowest set oflocants rule for substituents. For example, in the following compound, the lowest locant for the functional group >C= O is 3 and not 5

Example:

3. If the organic compound contains a terminal functional group such as —CHO, —COOH, —COCl, —CONH₂, —COOR, —C =N, etc., The numbering of the parent chain must be started from the functional group, we., it is always given number but the number is usually omitted from the final name ofthe compound

Example:

4. If a compound contains two or more similar functional groups, numerical prefixes [di, tri, tetra etc.) are used to indicate their numbers, and the terminal ‘e’ of the primary suffix (ane, ene or yne) is retained while writing the name.

Examples:

5. If an organic compound contains more than two similar terminal functional groups and all of them are directly attached to the parent chain, then none of them are considered as a part of the parent chain. Special suffixes such as carboxylic acid (for —COOH), carbaldehyde (for—CHO), carbonitrile (for —C=N), carboxamide (for —CONH₂) etc. are used to name these

Examples:

IUPAC Nomenclature Of Compounds With Pentane-2,4-Dione Two Different Functional Groups

When an organic compound contains two or more different functional groups, then one of the functional groups is to be selected as the principal functional group while all other functional groups (also called secondary 7 functional groups) are to be treated as substituents. The choice of principal group is made on the basis ofthe following order of preference:

—COOH>—SO₃H >— COOCO — >— COX (X=halogen) >—CONH₂> —C≡ N>—CHO>C= O> —OH> —SH>—NH₂

All the remaining groups such as halo (fluoro, chiro, bromo, and iodo), nitroso (—NO), nitro (—NO₂), alkoxy (—OR), alkyl (—R), aryl {Example: C₆H₃), etc., are always treated as substituents or simply as prefixes.

Suffixes and prefixes of some Important functional groups [with decreasing priority

Polyfunctional compounds are named as follows:

• The chain containing the principal functional group, the maximum number of secondary functional groups, and multiple (double or triple) bonds, if any, is to be considered as the principal chain in the compound.
• The principal chain is to be numbered in such a way that the principal functional group gets the lowest possible number followed by double bond, triple bond, and substituents.
• The prefixes for the secondary functional groups and other substituents are to be placed in alphabetical order before the word root. If two or more identical secondary functional groups are present, these are to be indicated by using di, tri, tetra, etc. as prefixes
• The principal functional group is to be written after the word root and the compound is to be named as a member of that particular class of compound.

Examples of some functional compounds:

IUPAC nomenclature of other classes of compounds (According to 1993 Recommendations)

Structural Formulas Of Organic Compounds From Their Iupac Names

• To write the structure of an organic compound if its IUPAC name is given, the given steps are to be followed:
• The longest carbon chain (parent chain) is to be selected from the word root ofthe IUPAC name ofthe compound
• The parent chain is to be numbered from either end.
• If the name of the compound contains primary suffix ‘ene’ or ‘yne\itis placed at the indicatedposition along the chain. Iv] Name and position of the functional group (secondary suffix) is to be identified from the IUPAC name andit is to be placed atits rightpositionin the carbon chain.
• The names and positions of other prefixes, if any, are to be identified from the IUPAC name and to be attached at proper positionin the carbon chain.
• Finally, the required number ofH- atoms are added, wherever
  necessary, to satisfy tetra covalency ofeach C-atom

.
Examples 1: Letus write the structural formula of 5-hydroxy- 2-methylhex-3-enoic acid.

Step 1: The wordroot ‘hex! indicates that the parent chain contains 6 carbon atoms: C —C —C—C —C—C

Step 2: Numbering ofthe carbon chain 6 5 is done as indicated: \(\stackrel{6}{\mathrm{C}}-\stackrel{5}{\mathrm{C}}-\stackrel{4}{\mathrm{C}}-\stackrel{3}{\mathrm{C}}-\stackrel{2}{\mathrm{C}}-\stackrel{1}{\mathrm{C}}\)

Step 3: TheIUPACname of the compound has the primary C suffix ‘ene’ at position 3. Therefore, C-3 and C-4 ofthe parent chain are linked by a double bond. \(\stackrel{6}{\mathrm{C}}-\stackrel{5}{\mathrm{C}}-\stackrel{4}{\mathrm{C}}=\stackrel{3}{\mathrm{C}}-\stackrel{2}{\mathrm{C}}-\stackrel{1}{\mathrm{C}}\)

Step 4: The secondary suffixes ‘oic acid’. Therefore, the carbon atom of the —COOH group is indicated as 1. \(\stackrel{6}{\mathrm{C}}-\stackrel{5}{\mathrm{C}}-\stackrel{4}{\mathrm{C}}=\stackrel{3}{\mathrm{C}}-\stackrel{2}{\mathrm{C}}-\stackrel{1}{\mathrm{C}} \mathrm{OOH}\)

Step 5: The prefixes ‘hydroxy’ and ‘methyl’ are attached at the positons 5 and 2 respectively

Step 6: A required number of H- atoms are added to various carbon atoms to get the final structure ofthe compound.

2. IUPAC name: But-2-en-l-o

Step 1: C —C —C—C —C—C

Step 2: \(\stackrel{4}{\mathrm{C}}-\stackrel{3}{\mathrm{C}}-\stackrel{2}{\mathrm{C}}-\stackrel{1}{\mathrm{C}}\)

Step 3: \(\stackrel{4}{\mathrm{C}}-\stackrel{3}{\mathrm{C}}=\stackrel{2}{\mathrm{C}}-\stackrel{1}{\mathrm{C}}\)

Step 4: \(\stackrel{4}{\mathrm{C}}-\stackrel{3}{\mathrm{C}}=\stackrel{2}{\mathrm{C}}-\stackrel{1}{\mathrm{C}}-\mathrm{OH}\)

Step 5: \(\mathrm{H}_3 \stackrel{4}{\mathrm{C}}-\stackrel{3}{\mathrm{C}} \mathrm{H}=\stackrel{2}{\mathrm{C}} \mathrm{H}-\stackrel{1}{\mathrm{C}} \mathrm{H}_2 \mathrm{OH}\)

3. UPAC name: 3-amino-4-methylpentanoic acid

Step 1: C —C —C—C —C—C

Step 2: \(\stackrel{5}{\mathrm{C}}-\stackrel{4}{\mathrm{C}}-\stackrel{3}{\mathrm{C}}-\stackrel{2}{\mathrm{C}}-\stackrel{1}{\mathrm{C}}\)

Step 3: \(\stackrel{5}{\mathrm{C}}-\stackrel{4}{\mathrm{C}}-\stackrel{3}{\mathrm{C}}-\stackrel{2}{\mathrm{C}}-\stackrel{1}{\mathrm{C}} \mathrm{OOH}\)

Step 4: 

Step 5:

4. IUPAC name: 3-ethyl-4, 5-dimethyl hex-l-yn-3-o

Step 1: C —C —C—C —C—C

Step 2: \(\stackrel{1}{\mathrm{C}}-\stackrel{2}{\mathrm{C}}-\stackrel{3}{\mathrm{C}}-\stackrel{4}{\mathrm{C}}-\stackrel{5}{\mathrm{C}}-\stackrel{6}{\mathrm{C}}\)

Step 3: \(\stackrel{1}{\mathrm{C}} \equiv \stackrel{2}{\mathrm{C}}-\stackrel{3}{\mathrm{C}}-\stackrel{4}{\mathrm{C}}-\stackrel{5}{\mathrm{C}}-\stackrel{6}{\mathrm{C}}\)

Step 4: 

Step 5:

Step 6:

Isomerism And Organic Reaction Mechanism Introduction

Organic compounds tend to exist as isomers. There are two or more compounds which have the same molecular formula (or molecular mass) but different physical and chemical properties, i.e., these are isomers of each other. This phenomenon is known as isomerism.

Again, organic compounds being covalent normally participate in molecular reactions. The mechanism of a reaction is the path followed by the substrate and reagent (the reacting species) while forming the products and in fact, it explains how the bonds in the reacting molecules break and new bonds in the product molecules are formed. To understand the mechanisms of organic reactions, some fundamental concepts are to be conceived. In this chapter, the isomerism in organic compounds and some basic concepts of organic reaction mechanisms have been discussed.

Isomerism In Organic Compounds

The property of isomerism in organic compounds is due to different sequence of bonding of their atoms or due to different arrangements of their atoms or groups in space when the sequence of bonding is same

Isomerism in organic compounds Definition:

The phenomenon of the existence of two or more compounds possessing the same molecular formula but different physical and chemical properties is known as isomerism. Such compounds are called isomers

Example:

Two compounds having the same molecular formula, C₂H₅O but with completely different physical and chemical properties are found to exist. One compound is a liquid which boils at 78°C and reacts with metallic sodium to liberate hydrogen gas and the other compound is a gas which boils at -24°C and does not react with sodium. Therefore, these two compounds are entirely different in nature.

The first compound is ethyl alcohol belonging to the class of compounds known as alcohols and the second compound is dimethyl ether belonging to the class of compounds known as ethers. Because of the difference in their structures, they are completely different in their properties.

These two compounds are, therefore, isomers and the phenomenon of the existence of these two compounds of identical molecular formulas but belonging to different families is known as isomerism

CH₃— CH₂— OH(  Ethyl alcohol)

CH₃— O—CH₃( Dimethyl Ether)

Structural Isomerism

The compounds having the same molecular formula but different structures or molecular constitutions, i.e., different atom-to-atom bonding sequences or atomic connectivity are called structural or constitutional isomers and the phenomenon Is known as structural isomerism. Structural isomerism can be further subdivided into five different categories. Besides, tautomerism is also considered structural Isomerism.

1. Chain Isomerism

Chain Isomerism Definition:

Two or more compounds (belonging to the same family) having the same molecular formula but different carbon skeletons are called chain or nuclear isomers and the phenomenon Is called chain or nuclear isomerism

Example:

1. n -butane and isobutane are two chain isomers because they have die same molecular formula (C₄H₁₀) but differ in their carbon skeletons.

2. n-pentane, isopentane and neopentane are the three yJJJj function, group isomerism Chain isomers because they (molecular formula C₅H₁₂) differ in their carbon skeleton

2. Position isomerism

Position isomerism Definition:

When two or more compounds having the same structure of the carbon chain, i.e., the same carbon skeleton, differ in the position of substituent, multiple (double or triple) bond or functional group, these are called position isomers and this phenomenon is called position isomerism

Examples:

1. n -propyl alcohol and isopropyl alcohol are two position isomers. They have the same molecular formula (C₃H₈O) and have identical carbon skeletons. But in n propyl alcohol, hydroxy (-OH) group is at the terminal C-atom of the chain while in isopropyl alcohol the hydroxy (-OH) group is attached to the middle C -atom.

3. Functional group isomerism

Functional, group isomerism Definition:

Two or more compounds having the same molecular formula but different functional groups [i.e., belonging to different families) are called functional isomers and this phenomenon is called functional group isomerism or functional isomerism

1. Alcohols & ethers (C_nH_{2n+ 2}O) exhibit functional group isomerism. Ethyl alcohol and dimethyl ether having the same molecular formula, C₃H₆O represent two functional isomers. The functional group of ether is divalent oxygen (— O —) while the functional group of alcohol is the alcoholic hydroxy (— OH) group.

2. Aldehydes, ketones, unsaturated alcohols and unsaturated ethers exhibit functional group isomerism. The molecular formula, C₃H₆O for example, represents the following two functional isomers:

3. Carboxylic acid and esters (C_nH_2nO₂) exhibit functional group isomerism. For example, the molecular formula C₂H₄O₂ represents the following two functional isomers

CH -COOH (Acetic acid) (Func. gr: —COOH)

H – COOCH(Methyl formate)( Func. gr:—COOCH₃)

4. 1°, 2° and 3° amines (C_nH_3nN) exhibit functional group isomerism. The molecular formula C₃H₉N, for example, represents

The following three functional isomers:

CH₃CH₂CH₂NH₂ Propan-l-amine (1° amine)

CH₃CH₂NHCH₃  N-methylhexanamine (2° amine)

(CH₃)₃N [N,N-dimethylethanolamine (3° amine)]

5. Aromatic alcohols, phenols and ethers exhibit functional group isomerism. For example, the molecular formula, C₇H₈O represents the following three functional isomers,

6. Dienes, alienes & alkynes (C_nH_2n-2) exhibit functional group isomerism. The molecular formula C5H8, for example, represents four functional group isomers

7. Cyanides & isocyanides (C_nH_2n-1N) exhibit functional group isomerism. The molecular formula C₃H₅N, for example, represents two functional group isomers:

CH₃CH₂CN – Propanenitrile

CH₃CH₂NC – Ethyl isocyanide

5. Metamerism

Metamerism Definition:

When two ‘or more compounds having the same molecular formula but different numbers of carbon atoms (or alkyl groups) on either side of the functional groups such as —O —S—, — NH—, —CO — etc. are called metamers and the phenomenon is called metamerism. Metamerism occurs among the members of the same homologous series

Examples: The molecular formula, C₄H₁₀O represents the following metamers of ether family:

Examples of some metamers:

6. Ring-chain isomerism

 Ring-chain isomerism Definition:

Compounds having the same molecular formula but possessing open-chain and closed-chain structures are called ringchain isomers and the phenomenon is called ring-chain isomerism.

Examples:

1. Propene and cyclopropane are ring-chain isomers with the molecular formula C₃H₆

2. Propyne and cyclopropene are ring-chain isomers with the molecular formula

7. Tautomerism

This is a special kind of functional group isomerism involving dynamic equilibrium between the isomers.

Tautomerism Definition:

The functional group isomerism which arises due to the reversible transfer of a group or atom from one polyvalent atom to the other within the same molecule witth necessary rearrangement of linkages and the resulting isomers exist in dynamic equilibrium with each other is called tautomerism. The interconvertible isomers are called tautomers or tautomerides. Tautomerism is also called desmotropism

Conditions for tautomerism:

1. There must be at least one or H -atom present concerning each functional group in the compound.
2. The compound must contain an electronegative atom bonded by a double or triple bond e.g., C=O, N=O, C=NH etc.

Keto-enol tautomerism:

In this type of tautomerism, one form is the keto-form containing a keto group >C=O ) while the other form is the enol-form containing an enolic group >C=C—OH). The term ‘enol’ comes from ‘ene’ of the double bond and ‘ol’ of the hydroxy (—OH) group. Ketoenol tautomerism is possible only for those carbonyl compounds which contain at least one a -H -atom

Example:

Ethyl acetoacetate is an important example of this type. Tautomeric equilibrium generally (if no other factors ! operate) favours the structure in which the H-atom is bonded to the C -atom rather than the more electronegative O -atom, i.e., equilibrium favours the weaker acid. However, in this case, because of conjugation and intramolecular H-bonding the percentage of enol-form is much higher (8%) as compared ! to that of acetone where no such factors operate

Keto and enol-forms are, in reality, two dynamic structural isomers. These isomers are always interconvertible and . one H-atom is shifted from one C-atom to an O-atom and vice-versa to establish the equilibrium.

Factors affecting the percentage of enol content:

1. Intramolecular H-bonding (chelation) stabilises the enol and thus increases the amount of enol-form.
2. Conjugation resulting in resonance stability of the enol-form helps to increase the enol content.
3. Polar protic solvents usually decrease the percentage of enol form because the more polar keto-form becomes relatively more stabilised in this medium

The percentage of the enol-form is greater because of its much higher stability caused by strong intramolecular H-bonding and effective resonance.

1. The following compounds do not exhibit keto-enol tautomerism due to lack of a -H-atom

2. Keto-enol actually involves interconversion of group and a tautomerism —C=C(OH) group.

The sum of bond energies of the is 347.9 kcal-mol-1 and that of the —C=C(OH) group is 336.0 kcal. mol-1 . So, the keto form is thermodynamically more stable than the enol form in the absence of other factors which can stabilise the enol form.

3. Pseudotropism: Tautomerism in which there is practically; no existence of one tautomer is called pseudotropism. For example

Nitro-acinitro tautomerism:

Nitroso-oximino tautomerism:

CH₃—CH₂—N Nitrosoetane(nitroso form) ⇌  CH₃ —CH—N — OH(Ethanal oxime (oximino form)

Stereoisomerism

Stereoisomerism Definition:

The isomers having the same structure formula, i.e., er same atom-to-atom bonding sequence or connectivity which differ in the relative arrangement of atoms or groups in space are called stereoisomers and the phenomenon is called stereoisomerism. It is the specific directional property of covalent bonds in threedimensional space which gives rise to stereoisomerism.

Stereoisomerism is broadly classified into two categories: 

1. Configurational isomerism
2. Conformational isomerism.

The isomerism that arises due to different spatial arrangements of groups or atoms (which are not interconvertible) in the same structural isomer is called configurational isomerism.

Configurational isomerism can be subdivided into two types: 

1. Optical isomerism and
2. Geometrical isomerism.

1. Optical isomerism

Optical isomerism Definition: 

If two molecules having the same atom-to-atom bonding sequence or connectivity (i.e., the same constitution) are mirror images of each other and are non-superimposable, then these are called enantiomers. Enantiomers rotate the plane polarised light to an equal degree but in opposite directions. Because of their effect on plane polarised light, separate enantiomers are said to be optically active and because of this property, they are called optical isomers and the phenomenon is called optical isomerism

Example:

As shown in the figure, lactic acid or 2-hydroxypropanoic acid (CH₃CHOHCOOH) exists as a pair of enantiomers. In (-)-lactic acid, the sequence of occurrence of the groups: OH->COOH→CH₃ appears in a clockwise direction but in (+)-lactic acid, it appears in an anticlockwise direction when viewed along the C — Ii bond axis from the side opposite to that of the H-atom. Both the enantiomers rotate the plane of plane polarised light to the same extent but in opposite directions. Enantiomer which rotates the plane of polarised light towards the right is called dextrorotatory or d- or (+)-lactic acid and which rotates the plane of polarised light towards the left is called laevorotatory or l- or (—)-lactic acid

Compounds which can rotate the plane of plane polarised light are called optically active compounds and those which cannot are called optically inactive compounds. Both the enantiomers of lactic acid are optically active.

Symmetric and asymmetric molecules:

If a molecule has at least one of the following elements of symmetry:

1. Plane of symmetry,
2. Centre of symmetry
3. Alternating axis of symmetry

Then it is a symmetric or achiral molecule. Symmetric molecules are optically inactive. However, if a molecule has none of these symmetry elements, it is called an asymmetric or chiral molecule. Asymmetric molecules are optically active.

A symmetric molecule is superimposable on its mirror image while an asymmetric molecule is not. Therefore, whether a molecule is symmetric or asymmetric may also be verified by constructing models of the molecule & its mirror image and then placing one model on the other. If they are found to be superimposable, then the molecule must be symmetric; if not, then the molecule is asymmetric.

Example:

The two mirror-image forms of lead-acid are asymmetric and so, they are optically active. In our daily lives, we come across many things which are related to mirror images and do not get superimposed,

For example: Our left hand is the mirror image of our right hand but the left hand and the right hand are non-superimposable. So, the glove of the left hand do not fit in the right hand

Plane of symmetry or Sigma plane (σ ):

The plane of symmetry (σ) is defined as an imaginary plane that bisects a molecule in such a way that one half of the molecule is the mirror image of the other half (the plane acting as a mirror). The plane is also called a mirror plane.

Example: Meso-tartaric acid has a plane of symmetry.

Centre of symmetry or Centre of inversion (i):

A centre of symmetry is a point from which lines, when drawn on one side and produced an equal distance on the other side, will meet identical points (i.e., atoms) in the molecule

Asymmetric carbon atom and optical activity:

When a carbon atom of an organic molecule is attached to four different atoms or groups (Cabde), then that carbon atom is called an asymmetric carbon atom or chiral carbon. It is also called a stereogenic centre. Molecules containing only one asymmetric carbon atom are asymmetric and optically active. If a molecule contains more than one asymmetric carbon atom, it may be asymmetric or symmetric, i.e., it may be optically active or inactive.

Example:

(+) tartaric acid having two asymmetric C-atoms is optically active but meso-tartaric acid is optically inactive.

The presence of asymmetric carbon atoms is not essential for exhibiting optical activity:

Some of the substituted allenes (abC=C=Cab) are found to be optically active, even though they contain no asymmetric carbon atom. The reason for their optical activity is the overall asymmetry of their molecules.

Example: The two mirror images of 1,3-dichloro propadiene are overall asymmetric and so, they are optically active

Therefore, the presence of asymmetric C-atom in a molecule is not essential for exhibiting optical activity. If the structure is overall asymmetric, then the molecule will be optically active.

Conditions necessary for optical activity:

1. The compound is non-superimposable on its mirror image, or
2. It contains only one asymmetric C-atom or
3. Plane, centre and alternating axis of symmetry are absent in the molecule.

E. Eliel in his book, has mentioned a molecule which does not contain any element of symmetry, yet it is not optically active. He further stated that principally possibility of the existence of such molecules cannot be discarded outright. Hence, the necessary and sufficient condition for the optical activity of any compound is the non-superimposability ofthe molecule on its mirror image

• Meso-compounds: If a compound having more than one chiral centre is found to be optically inactive, then it is called a meso-compound. Forexample, meso-tartaric acid.
• Enantiomers:  Stereoisomers that are not superimposable on each other but related to each other as mirror images are called enantiomers. Thus (+) & (-)-lactic acid form a pair of enantiomers. Similarly, stereoisomers I & II of 3- bromopentan-2- ol form a pair of enantiomers. Enantiomers are optically active molecules having equal but opposite specific rotations. All other physical and chemical properties of enantiomers are the same.
• Diastereoisomers: Stereoisomers that are not mirror images of each other are called diastereoisomers. Stereoisomers I & III of 3-bromopentan-2-ol are diastereoisomers of each other

Calculation of no. of stereoisomers:

No. of stereoisomers, both optically active and inactive (meso-form) can be obtained from the number of chiral centres present in the molecule

Examples:

1.  3-bromopentan-2-ol (CH₃CHOHCHBrCH₂CH₃) has two dissimilar chiral centres. Therefore, it has 2² = 4 optically active stereoisomers and no optically inactive meso-isomer. Fischer projections of these four stereoisomers are shown below

(I, II) and (III, IV) represent two pairs of enantiomers and they are optically active. Each of (I, III), (I, IV), (II, III) and (II, IV) represents a pair of diastereoisomers.

2.  2,3-dibromobutane (CH₃CHBrCHBrCH₃) has 2 similar chiral centres and can be divided into two mirror-image halves. Therefore, it can have 2^(2-1)_ = 2¹ = 2 optically active isomers and 2^(2-2)/2_ = 2⁰ = 1 optically inactive meso-isomer. Fischer projections of these 3 stereoisomers are shown below

V and VI represent a pair of enantiomers and are optically active. VII represents an optically inactive meso-isomer having a plane of symmetry. Each of (V, VII) and (VI, VII) represents a pair of diastereoisomers.

3.    or 3-bromopentane- 2,4-diol has three chiral centres and can be divided into two mirror-image halves by passing a plane through the central carbon atom. Therefore, it can have 2(3-1)-2(3-1)//2 = 4-2 = 2 optically active isomers and 2(3-1)/2 = 21 = 2 optically inactive mesoisomers.

Fischer projections of these 4 stereoisomers are as follows:

 

IX and X represent a pair of enantiomers and are optically active. XI and XII represent two optically inactive meso-isomers and both of them have a plane ofsymmetry (cr -plane). Each of (DC, XI), (X, XI), (IX, XII), (X, XII) and (XI, XII) represents a pair of diastereoisomers.

Fischer projection formula:

Fischer developed a twodimensional plane projection formula for three-dimensional molecules. Fischer projection uses a cross to represent the stereocentre and the four bonds attached to it Stereocentre lies at the centre of the cross but is not explicitly shown. Horizontal bonds point towards the observer (i.e:, bonds inclined upwards), while the verticle bonds are directed away from the observer (i.e., bonds are inclined downwards). As per the IUPAC convention, the number-1 carbon atom is placed at the top of the vertical line. For example,

Racemic modification:

The racemic modification is an equimolecular mixture of a pair of enantiomers. The racemic modification is optically inactive due to external compensation, i.e., optical rotation caused by one enantiomer is compensated by the opposite rotation produced by the other. A racemic modification is indicated by writing (d, l ) or (±) before the name of the compound.

Example: (±)-or(d, l )-2-hydroxypropanoic acid or lactic acid. R/S nomenclature of optical isomers:

2.  Geometrical or cis-trans isomerism

Geometrical Definition:

Isomers which have the same structural formula, i.e., the same atom-to-atom bonding sequence or connectivity but have different spatial arrangements of atoms or groups around the double bond or a ring system are called geometrical isomers and the phenomenon is called geometrical isomerism.

A π-bond prevents free rotation of the carbon atoms of a double bond concerning each other. Due to this restricted rotation, the relative positions of the atoms or groups attached to the doubly bonded carbon atoms get fixed. As a result of this, many substituted alkenes can exist in two distinctly isomeric forms which differ from each other only in the relative positions of the atoms or groups in space around the double bond

The isomer in which the similar atoms or groups lie on the same side of the double bond or a ring system is called the cis-isomer while the isomer in which the similar atoms or groups lie on the opposite sides of the double bond or a ring system is called the trans-isomer. Because of this, geometrical isomerism is also called cis-trans isomerism

The two geometrical or cis-trans isomers are stereoisomers which are not mirror images of each other. Therefore, these are diastereoisomers of each other.

E/Z nomenclature of geometrical isomers:

In this system of nomenclature, each ofthe two atoms or groups attached to each doubly bonded carbon atom are assigned according to their priority based on Cahn-Ingold-Prelog rules or simply CIP rules. If the atoms or groups of higher priority are on the same side of the double bond, the isomer is designated as Z (Zusammen in German means together) and if the groups or atoms of higher priority are on the opposite sides ofthe double bond, the isomer is designated as E (Entgegen in German means opposite).

Example: In l-bromo-2-chloropropene, for C l, Br > H (in priority) and for C -2, Cl > CH₃ (in priority).

Geometrical isomers of three types of olefins:

Like the compounds containing a C-C double bond, compounds containing

1. Carbon-Nitrogen double bond (C= N)

Example: Oxime, Hydrazone etc.],

2. Nitrogen-Nitrogen double bond (N=N)

Example: azo, diazo compounds etc.] and some alicyclic compounds exhibit cis-trans or geometrical isomerism

3. Conformational isomerism

Distinction between cis and trans-isomers:

The cis and trans isomers can be distinguished with the help of certain physical characteristics as follows:

1. Melting point:

The molecules of the trans-isomer of a compound is relatively more symmetrical than those of the os-isomer and hence remain closely packed in the crystal lattice. As a consequence, the melting point of the trans-isomer is usually higher than that of the corresponding cis-isomer

2. Solubility:

In general, a cis-isomer is found to be relatively more soluble in a particular solvent because the molecules of the cis-isomer, being less symmetrical, are weakly held in the crystal lattice than the molecules of relatively more symmetrical trans-isomer

3. Dipole moment:

In general, the cis-isomer of  an alkene is found to be more polar than the irans-isomer (in which there is a possibility of cancellation of moments of two oppositely oriented groups or atoms)

Equivalent And Non-Equivalent H-Atoms

If each of the two or more hydrogen atoms present in an organic compound on being replaced by any other atom or group in turn produces the same compound, then these hydrogen atoms are regarded as equivalent hydrogen atoms.

Example:

All 6 hydrogen atoms in the ethane (CH₃—CH₃) molecule are equivalent and this is because, if any one of these six H-atoms is replaced by an atom (or group) such as Br, the same compound ethyl bromide (CH₃CH₂Br) is obtained.

Again, in a propane (CH₃CH₂CH₃) molecule, the 6 hydrogens of two methyl groups are equivalent because the replacement of any one of them by Clatom, produces the same compound, 1-chloropropane (CH₃CH₂CH₂Cl) .

Also, the 2 hydrogens of the methylene ( —CH₃—) group are equivalent because the replacement of either of them by Cl-atom produces the same compound, 2-chloropropane (CH₃CHClCH₃). However, one methyl hydrogen and one methylene hydrogen are non¬equivalent because each of them when displaced by Cl atom gives two different compounds.

Therefore, propane contains two types of non-equivalent H atoms. In the given compounds, equivalent H -atoms are marked by the same English letter while the non-equivalent H atoms are designated by different English letters

In following compounds, all H-atoms are equivalent:

Determination Of No. Of Covalent Bonds In An Organic Compound From Its Molecular Formula

If the total number of electrons required for the completion of If the total number of electrons required for the completion of the molecule is x and the total number of valence electrons of all the atoms present is y, then the total number of covalent bonds \(\frac{(x-y)}{2}\) (for the formation of a covalent bond, 1 electron pair is required hence, the division by 2 has been effected).

If the molecular formula of a compound is C_aH_bO_c, then the number of electrons required for the completion of octets of several C -atoms = 8a, the number ofelectrons required for the completion of duplets of b number of H-atoms = 2b and the number ofelectrons required for the completion of octets of c number of 0 -atoms = 8c.

Hence, the total number of electrons required for the completion of duplets of H-atoms and octets of C and O -atoms present in the molecule, x = (8a + 2b + 8c). The number of valence electrons for several C -atoms, b number of H-atoms and c number of 0 -atoms are 4a, lb and 6c respectively. Therefore, the total number of valence electrons of C, H and O -atoms present in the molecule, y = (4a + b + 6c). So, the total number of covalent bonds present in the molecule of compound

⇒ \(\mathrm{C}_a \mathrm{H}_b \mathrm{O}_c=\frac{(8 a+2 b+8 c)-(4 a+b+6 c)}{2}\)

Examples:

1. Determine the number of covalent bonds in a compound having the molecular formula, C₂H₄O₂.
Answer:

The total number ofelectrons required for the completion of octets of two C and two O -atoms and duplets of four H atoms present in the molecule = (2×8 + 2×8 + 4×2 ) = 40 and the total number of valence electrons of all these atoms =(2 × 4 + 2 ×6 + 4 × 1) =24.

⇒ \(\mathrm{C}_2 \mathrm{H}_4 \mathrm{O}_2=\frac{40-24}{2}\)

= \(\frac{16}{2}\)

= 8

2. Determine the number of covalent bonds in a compound having the molecular formula, C₂H₂Cl₄. Write the probable structure and name of the compound.
Answer:

The total number ofelectrons required for the completion of octets of two C and four Cl -atoms and duplets of two H atoms present in the molecule =(2× 8 + 4×8 + 2×2) = 52
and the total number of valence electrons of all these atoms

=(2×4 + 4×7 + 2 × 1) = 38.

Thus, the no. of covalent bonds present in the compound having molecular formula

⇒ \(\mathrm{C}_2 \mathrm{H}_2 \mathrm{Cl}_4=\frac{52-38}{2}\)

= \(\frac{14}{2}\)

= 7

The possible structures of the compound are ClCHCHCl₂ (1,1,2,2-tetrachloroethane) or ClCH₂CCl (1,1,1,2- tetrachloroethane)

Double Bond Equivalent (Dbe) Or Index Of Hydrogen Deficiency (Ihd)

For determining the structure of an organic compound, it is necessary to ascertain whether unsaturation is present in that compound or not and if present, the amount of unsaturation is also needs to be known. In an organic compound, the amount of unsaturation is expressed in terms of Double Bond Equivalent (DBE).

It is also called the Index of Hydrogen Deficiency (IHD). If a hydrocarbon contains 2 hydrogen atoms less than the alkane having the same number of carbon atoms, its double bond equivalent is I, i.e., the compound may contain 1 double bond or a ring.

For example:

The compound, C4H8 contains two H -atoms less than the alkane (butane, C₄H₁₀ ) having the same number of C -atoms. So, its double bond equivalent is 1.

Thus, the compound may contain 1 double bond or it may be a cyclic one. That is, the compound may be (CH₃CH₂CH=CH₂) but-l-ene (CH₃CH=CHCH₃) or cyclobutane.

SimUarly, the double bond equivalent 2 indicates the presence of 2 double or but-2-ene + 1 2 bonds or I triple bond or 1 double bond and 1 ring or 2 rings in the compound.

So, the term SODAR (Sum of Double Bonds And Rings) is also frequently used. From the molecular formula of a compound, its Double Bond Equivalent (or Index of Hydrogen Deficiency or Sum of Double bonds And Rings) can easily be calculated

The double bond equivalent (DBE) of a compound \(=\frac{\Sigma n(v-2)}{2}+1\) + j where n js the number of different types of atoms present in the molecule and v is the valency of each type of atom.

It is to be remembered that if the value of the DBE of a compound is less than 4, the compound is not a benzenoid aromatic compound.

Examples: 1. The molecular formula of a compound is C₆H₈. Calculate its double bond equivalent (DBE). State whether the compound is an aromatic compound or not.
Answer:

Double Bond Equivalent =  \(\frac{6(4-2)+8(1-2)}{2}+1\)

= 3

Since the double bond equivalent ofthe compound is less than 4, it is not a benzenoitÿaromatic compound

2. Determine the double bond equivalent of each of the following compounds. On catalytic hydrogenation, each of the compounds consumes 2 moles of hydrogen. What is the number of rings (if present) in each of the compounds:

1. C₈H₈Br₂
2. C₈H₁₀O₂
3. C₅H₆F₂
4. C₈H₉C₁₀

Answer:

1. DBE of compound = \(\frac{8(4-2)+8(1-2)+2(1-2)}{2}\) + 1

= 4

The compound on catalytic hydrogenation consumes 2 moles of hydrogen per mole. So, there are 2 double bonds or I triple bond and (4- 2) or 2 rings present in the compound

2. DBE ofthe compound = \(\frac{8(4-2)+10(1-2)+2(2-2)}{2}\) + 1

= 4

The compound on catalytic hydrogenation consumes 2 moles of hydrogen per mole. So, there are 2 double bonds or 1 triple bond and (4- 2) or 2 rings present in the compound.

3. DBE of the compound = \(\frac{5(4-2)+6(1-2)+2(1-2)}{2}\)+ 1

= 2

Since the compound consumes 2 moles of hydrogen on catalytic hydrogenation, the compound contains 2 double bonds or 1 triple bond. There is no ring present in the compound.

4. DBE ofthe compound

= \(\frac{8(4-2)+9(1-2)+1(1-2)+1(2-2)}{2}\) + 1

= 4

Since the compound on catalytic hydrogenation consumes 2 moles of hydrogen per mole, it contains 2 double bonds or 1 triple bond and 2 rings.

3. Write the structures and the IUPAC names of all the isomeric compounds having molecular formula, C4Hg by determining its double bond equivalent.
Answer:

DBE ofthe compound = \(=\frac{4(4-2)+6(1-2)}{2}\) + 1

= 2

Therefore, the compound contains 2 double bonds or 1 triple bond or 1 double bond and 1 ring or two rings. The following 9 isomers ofthe compound are possible:

1. CH₂=CH = CH=CH₂ (Buta-1,3-diene)

2. CH₃CH₂C ≡ CH (But-l-one)

3. CH₃C ≡ CCH₃ (But-2-yne))

4. CH₂ = C = CHCH₃ (Buta-1,2-diene)

Fundamental Concepts Of Organic Reaction Mechanism

In an organic reaction, the organic molecule (called the substrate) reacts with a suitable attacking species (called the reagent) to form products. The formation of produces) may take place either directly from the reactants (i.e., substrate and reagent) through a transition state or the formation of one or more intermediates. Some by-products may also be formed from the intermediate

The reagents are mostly either positively or negatively charged. A positively charged reagent attacks the site which is rich in electrons while a negatively charged reagent attacks that site which is electron deficient. So, for the reaction to take place in the covalent bond of the substrate, the bond must have some degree of ionic character.

Although the bond of an organic compound is mainly covalent, that bond becomes partially ionic due to some permanent or temporary displacement of the bonding electrons.

The ionic nature of a bond may be attributed to the following reasons:

1. Inductive effect
2. Electromeric effect
3. Resonance
4. Hyperconjugation.

Besides these electronic effects, steric effect or steric hindrance plays a very important role in determining the reactivity of organic compounds

1. Inductive effect

Inductive effect Definition:

The permanent displacement of electrons along a carbon chain which occurs when some atom or group, either more or less electronegative than carbon is attached to the carbon chain is called the inductive effect

Inductive effect When a covalent bond is formed between two atoms having different electronegativities, the bonding electron pair is not shared equally by the two atoms. The electron pair being attracted by the more electronegative atom gets shifted more towards it. Consequently, the more electronegative atom acquires a partial negative charge (i.e., 6- ) and the less electronegative atom acquires a partial positive charge (Le., d+).

Example:

When an electronegative (electron-withdrawing) Clatom (or a group such as — NO₂ ) is attached to the end of a carbon chain (whose carbon atoms are designated as 1, 2, 3,… etc.), the cr -electrons ofthe Cx —Cl bond are attracted by or displaced more towards the Cl-atom. As a result, the Cl-atom acquires a partial negative charge (δ-) and the carbon, C₁, acquires a partial positive charge (δ+). As Cj is now somewhat positively charged, it in turn, attracts the cr -electrons of the C₁-C₂ bond towards it So, C₂ acquires a partial positive charge (SS+) smaller in magnitude than that on C₁. Similarly, C₃ acquire a partial positive charge (δδδ+) even smaller than that on C₂

Similarly, if an element less electronegative than carbon, such as lithium (Li) (or any other electron-releasing group or atom), is attached to the terminal C -atom bf a carbon chain, then a partial positive charge (<5+) is developed on the Li-atom and a partial negative charge (6-) is developed on the Cx -atom.

The small negative charge on C₁, in turn, repels the cr -electrons of the C₁—C₂ bond towards C₂. As a result, C₂ acquires a partial negative charge (δδ-) smaller in magnitude than that of C₁ Similarly, C₃ acquires a partial negative charge (δδδ-) which is even smaller than that of C₂

Inductive effect Characteristics:

• This type of charge dispersal, which diminishes rapidly as the distance from the source increases, almost becomes negligible after the third carbon atom and is ignored.
• It is to be noted that although the inductive effect causes some degree of polarity in covalent bonds, the bond is never cleaved due to the effect.
• The inductive effect is represented by the symbol ( —<—). The arrow always points towards the more electronegative atom or group.

Measurement of inductive effect:

The inductive effect is always transmitted along a chain of carbon atoms. It cannot be expressed by any absolute value. The relative inductive effect of an atom or group is measured by taking H -atom of the R₃C — H molecule as standard. When an atom or group- Z of the C — Z bond of the R₃C —Z molecule attracts the bonding electrons more strongly than hydrogen of the C—H bond in the R₃C— H molecule.

Then according to the definition introduced by Ingold, Z is said to have a negative inductive effect or electron-withdrawing inductive effect or -I effect. On the other hand, if the atom or group Z attracts the bonding electrons of the C — Z bond less strongly than the hydrogen atom of the C — H bond, then it is said to have a positive inductive effect or electron-releasing inductive effect or +1 effect

+I effect:  —NH^– > —O^– > — COO^– > (CH₃)₃C— > (CH₃)₂CH — > CH₃CH₂ — > CH₃ — > D

– I effect: — ⁺NR₃> — ⁺SR₂> — ⁺NH₃ > — NO₂ > -SO₂R > — CN > — COOH > — F > — Cl > — Br > — I > — OR > — OH

Impact of inductive effect and its explanation:

Some important properties of organic compounds such as acidic property, basic property, bond polarity, and chemical reactivity vary remarkably due to inductive effect. Some examples of the influence of + 1 and – I effects on the properties of organic compounds are discussed here

1. Strength of monocarboxylic acids

Rule 1:

The relative strengths of monocarboxylic acids can be explained by the inductive effect ofthe substituent present in the carbon chain. If an electron releasing group or atom is attached directly to the -COOH group or to the carbon chain close to the -COOH group, then the positive inductive effect (+1) of such group increases the electron density on the oxygen atom of the

O— H group and consequently, the shared pair of electrons of the O — H bond is less strongly attracted towards the oxygen atom. As a result, the dissociation of the O —H bond to give H+ ion is less favoured. Thus, a group having a + I effect, when present in a monocarboxylic acid molecule decreases the strength of that acid.

On the other hand, if an electron-withdrawing group or atom is attached to the carbon chain close to the —COOH group, the negative inductive effect (-1) of such group decreases the electron density on the oxygen atom of the O— H group and consequently, the shared pair of electrons of O —H bond are more strongly attracted towards the oxygen atom. As a result, the dissociation of the O—H bond to give H+ ion is facilitated. Thus, a group having – I effect, when present in a monocarboxylic acid molecule, increases the strength of that acid molecule.

Example: Chloroacetic acid is stronger than acetic acid

Rule 2:

The strength of carboxylic acid increases as the extent of the effect ofthe substituent increases.

Example:

-I effect of the halogens follows the order: fluorine > chlorine > bromine > iodine. So, among the halogen-substituted acetic acids, trifluoroacetic acid (FCH₂COOH) is the strongest while iodoacetic acid (ICH₂COOH) is the weakest

Rule 3:

With the increase in the number of electron-attracting substituents, the strength of the acid increases.

Example:

Dichloroacetic acid is stronger than monochloroacetic acid while trichloroacetic acid is stronger than dichloroacetic acid.

Rule 4:

As the distance ofthe electron attracting substituent from the carboxyl group increases, the strength of the acid decreases.

Example:

2-chlorobutanoic acid is a stronger acid than 3- chlorobutanoic acid which in turn is stronger than 4- chlorobutanoic acid

Acid strength can also be explained in terms of the relative stabilities of the acid and its conjugate base. Electron withdrawing groups (EWG) disperse the negative charge on the anion [i.e., conjugate base), thus stabilising it and hence increasing acidity. On the contrary, electron-donating groups EDG intensify the negative charge on the anion, thus destabilise it and hence decrease acidity

2. The Basic Strength Of Amines

The increase in the strength of nitrogenous bases, e.g., amines, is related to the readiness with which they are prepared to take up protons and therefore, to the availability of the unshared pair ofelectrons on nitrogen.

Example:

We might expect the order of basic strength:

NH₃<CH₃NH₂<(CH₃)₂NH<(CH₃)₃N> due to the increasing inductive effect (+1) of successive —CH3 groups making the N -atom more negatively charged, i.e., making the unshared pair of electrons more readily available. However, this sequence of basic strength of amines agrees with the results if measurements of basicity are made in the gas phase or in a solvent in which H -H-bonding does not take place

Basic strength increases (in the gas phase or in a solvent which does not form H-bond with amines). The introduction of electron-withdrawing groups close to the basic centre causes a decrease in F₃Cthe basicity, due to their electron- Tri-trifluoromethylamine (virtually non-basic) withdrawing inductive effect.

An interesting example is the amine, (CF₃)₃N which is found to be virtually non-basic, due to the presence of three powerful electron-withdrawing — CF₃ groups, each of which contains three highly electronegative F -atoms.

The order of basic strength of amines in aqueous medium is:

⇒ \(\left(\mathrm{CH}_3\right)_2 \ddot{\mathrm{N}} \mathrm{H}\left(2^{\circ}\right)>\mathrm{CH}_3 \ddot{\mathrm{N}} \mathrm{H}_2\left(1^{\circ}\right)>\left(\mathrm{CH}_3\right)_3 \ddot{\mathrm{N}}\left(3^{\circ}\right)\)

Due to the combined effect of hydrogen bonding and +1 effect of— CH₃, groups, the conjugate acid of (CH₃)₂ NH i.e.,   (CH₃)₂ ⁺NH₂ is the most stable while the conjugate acid of (CH₃)₃N, i.e., (CH₃)₃ NH is the least stable and for this reason, in the aqueous medium, the above order of basicity  observed

2.  Electromeric effect

The complete transfer of a pair of 7t -electrons of a multiple bond (double bonds such as C=C, C=0 and triple bonds such as C = C and G= N ) to one of the multiple bonded atoms (usually the more electronegative one) in the presence of an attacking reagent is called electromeric effect or E-effect.

The transfer of the electron pair is indicated by a curved arrow. As soon as the reagent is removed, this effect vanishes and the molecule reverts back to its original position. Since this effect occurs by the presence ofthe attacking reagent, it takes place in the direction which facilitates the reaction. The electromeric effect may be represented as follows

Types of electromeric effect:

+ E-effect: If the electron pair of the -bond is transferred to that doubly bonded atom to which the attacking species gets finally attached, then the effect is called +E-effect

– E-effect:

If the electron pair of the π-bond is transferred to that doubly bonded atom to which the attacking species Mous Lewis structures, which differ in the do not get finally attached, the effect is called -E-effect

The distinction between inductive and electromeric effect:

3. Resonance

Resonance Definition:

Various Lewis structures, which differ in the positions of non-bonding or ;r -electrons but not in the relative positions of atoms, are called resonance structures, contributing structures or canonical forms. This concept is known as resonance

There are some molecules or ions which cannot be represented adequately by a single electronic (Lewis) structure as all the properties of such molecules or ions do not correspond to a single Lewis structure. In such cases, it becomes necessary to represent the molecule or ion by writing two or more Lewis structures which differ in the arrangements of valence electrons but the basic structure involving cr -bonds remains the same.

It should be remembered that resonance is not a phenomenon, because there is no real existence of different resonance structures. These structures are all imaginary and are taken into consideration to explain the different physical and chemical properties of molecules or ions. The actual structure of the molecule or the ion lies in between these structures. We say that the actual molecule or ion is a resonance hybrid (weighted average) of all these resonance structures.

Resonance is also known as mesomerism. The various resonance structures are connected by double-headed arrows. They contribute to the actual structure in proportion to their stability. The magnitude of internal energy of the resonance hybrid of a molecule or ion is less than that of any resonating structure. Thus the molecule or ion gets stabilised by resonance

Examples:

1. Benzene molecule can be represented as a resonance ~ hybrid (III) of two Kekule structures, I and II. Neither ofthe two structures can fully explain all the properties of benzene. For example, both structures I and II contain two types of carbon-carbon bonds such as C—C (1.54 Å) and C = C (1.34 Å). But actually, it has been found that all the 6 carbon-carbon bonds in benzene are of equal length (1.39 Å).

This suggests that the actual structure of benzene can neither be represented by I nor by II but by a resonance hybrid of these two structures in which all the six carbon-carbon bonds are of equal length and lie in between carbon-carbon single bond length of 1.54 Å and carbon-carbon double bond length of 1.34 Å. So, benzene is quite often represented by the non-Lewis structure III. The circle inside the ring indicates completely delocalised 6 n -electrons. Since I and II are exactly equivalent, They are of the same stability and make equal contributions to the hybrid

2. Carbonate ion (CO₃^2-) may be represented as a resonance hybrid ofthe following three structures: IV, V and VI. None of the three structures can individually explain all the properties of carbonate ions. For example, in all three structures, the carbon-oxygen single bond (1.43Å) and carbon-oxygen double bond (1.20 Å) are present.

But, it has been found experimentally that all the carbon-oxygen bonds in carbonate ion are equal in length (1.28 A) and this bond length is slightly greater than that of the double bond but less than that of the single bond. All the three carbon-oxygen bonds are equivalent. So, the structural formula of the carbonate ion denotes a state equidistant from the three structures IV, V and VI and it is frequently expressed by the non-Lewis structure VII

Rules for writing meaningful resonance structures:

The following rules are to be followed while writing realistic resonance structures:

• The various resonance structures should differ only in the positions of electrons and not in the positions of atoms, Le., the basic structure involving cr -bonds should remain undisturbed.
• The number of paired and unpaired electrons in each resonance structure must be the same.
• All the atoms involved in the process of resonance must be coplanar (or nearly coplanar).
• All the resonance structures should have nearly the same energy.
• Each resonance structure must be a bona fide Lewis structure, i.e., all atoms in a resonance structure must exhibit proper valencies.
• For example: There must not be any structure with pentavalent carbon, pentavalent nitrogen, bivalent hydrogen and so on.

Resonance energy:

The difference in internal energy between the actual molecule (observed value) and that of the resonance structure having the lowest internal energy or highest stability (obtained by calculation) is called resonance energy. The resonance energy is greater when

• The contributing structures are all equivalent and
• The number of contributing structures of roughly comparable energy is greater.

Calculation of resonance energy:

Resonance energy is not a measurable quantity. It can only be estimated from thermochemical data. If the theoretically calculated internal energy of a gram-mole of the most stable resonance structure is E_C and the experimentally determined internal energy of the actual molecule (resonance hybrid) is E_O, then the resonance energy E_R = E_C-E_O. Resonance energy is expressed in kcal mol^-1 or kj mol^-1.

The greater the resonance energy, more the stability of the compound. The resonance energy becomes maximum when the contributing structures are equivalent, i.e., have equal energy content. Also, the more the number of resonance structures having a large contribution, the greater will be the resonance energy.

In determining the relative stabilities of similar molecules or ions, resonance energy is an important factor among various other factors like bond energy, internal strain etc.

Example:

The resonance energy of benzene can be calculated from the heat of hydrogenation values. The heat of hydrogenation is the quantity of heat evolved when the mol of an unsaturated compound is hydrogenated. Cyclohexene containing 1 double bond has a heat of hydrogenation of 28.6 kcal mol^-1.

We might reasonably expect 1,3,5-cyclohexatriene to have a heat of hydrogenation of about three times as large as cyclohexene, i.e.,3 × 28.6 = 05.8 kcal .mol^-1.

The value for benzene is 49.8 kcal .mol^-1. It is 36 kcal .mol^-1 less than the expected value. So, benzene evolves 36 kcal less energy per mole than predicted. This can only mean that benzene is more stable than hypothetical cyclohexadiene by 36 kcal .mol^-1 energy. This 36 kcal. mol^-1 energy is the resonance energy of benzene.

Relative contributions of resonance structures towards resonance hybrid:

All resonance structures do not contribute equally towards resonance hybrid. Relative contributions of resonance structures towards resonance hybrid depend on their relative stabilities. The more stable the resonance structure, the more will be its contribution to resonance hybrid.

Factors that govern the stability of a resonance structure and its relative contribution towards hybrid are :

Rule 1:

Non-polar resonance structures, being more stable than the dipolar resonance structures, contribute more towards the resonance hybrid.

Example:

In the following alkadiene, the first resonance structure is more stable and thus contributes more than the second dipolar resonance structure

Rule 1:

Resonance structures with a greater number of covalent bonds are more stable and contribute more towards the resonance hybrid.

Example:

In the following acyl cation, the second resonance structure having a greater number of covalent bonds is more stable and more contributing.

Rule 2:

In case of anions, the most stable structure is the one in which the negative charge resides on the most electronegative the one in which the positive charge resides on the least electronegative atom. So, these structures are more contributing.

Example:

Resonance structures II and IV of the following anion and cation are relatively more stable hence more contributing towards their respective resonance hybrids.

Rule 3:

In the case of anions, the most stable structure is the one in which negative charge resides on the most electronegative atom. In the case of cations, the most stable structure is the one in which a positive charge resides on the least electronegative atom. So, these structures are contributing

Example:

Resonance structures II and IV of the following anion and cation are relatively more stable hence more contributing towards their respective resonance hybrids

Rule 4:

Canonical structures in which octets of all the atoms are fulfilled are relatively more stable and therefore, make a larger contribution towards the resonance hybrid.

Example:

The second resonance structure of the following acylium ion is more stable and more contributing because all the atoms have octets of electrons in their valence shells (except H which has a duplet).

Rule 5:

Aromatic resonance structures are more stable and more contributing than the non-aromatic resonance structures having the same number of covalent bonds. Example: The aromatic resonance structure I of benzyl cation is more stable and more contributing than the non-aromatic resonance structure II.

Rule 6:

A resonance structure having two units of charge on the same atom is not stable and hence it has a very poor contribution. Again, structures having like charges on adjacent atoms are highly unstable and hence it has a negligible contribution. On the other hand, a resonance structure having two dissimilar charges close to each other is relatively more stable and more contributing than the structure in which the charges are further apart.

Rule 7:

The resonance energy of a system involving monopolar resonance structures is greater than that involving dipolar resonance structures. So the former type of systems (i.e., molecules or ions) are more stable than the latter type.

Example:

Carboxylate ion is more stable (in fact more stabilised by resonance) than the corresponding carboxylic acid

Some facts about resonance structures:

• Resonance structures are not real.
• Resonance structures are not in equilibrium with each other.
• Resonance structures are not isomers because the two isomers differ in the arrangement of both atoms and electrons, whereas resonance structures differ only in the arrangement ofelectrons

Effect of resonance on the properties of molecules:

The following properties of different molecules or ions can be explained by resonance

1. Bond length

Because of resonance, the single bond present in a molecule or ion may acquire a partial double bond character with a consequent decrease in bond length. Similarly, the double bond may acquire some single bond character with a consequent increase in bond length.

Example: Due to resonance, the C — Cl bond (1.69 Å) of vinyl chloride (CH₂ = CHCl) becomes shorter than C — Cl bond (1.76 Å) of ethyl chloride and its C=C bond (1.38 Å) becomes longer than the C=C bond (1.34 Å) of ethene

2. Dipole moment

As a result of resonance, both the magnitude ofthe charge separated (e) and the distance between two charged centres (d) in any molecule may increase. So, the value of dipole moment (p = e × d) increases.

Example:

Due to resonance, the amount of charge separated and the distance between the centres of charges in nitroethane (CH₂=CHNO₂) is greater as compared to nitroethane (CH₃CH₂NO₂). Consequently, the dipole moment <p) nitroethane is greater than that of nitroethane.

3. Acidity And Basicity Of Organic Compounds

Acidic character of phenol:

The greater the ease with which a compound releases proton (H⁺) in its aqueous solution, the stronger it will be as an acid. That phenol is acidic and is a stronger acid than alcohol can be well explained in terms of resonance. Phenol in its aqueous solution ionises to produce phenoxide ions as follows

Phenol can be represented as a resonance hybrid of the following resonance structures (I -V):

Due to the contribution ofthe resonance structures n, in an IV the O -atom becomes positively charged. Because of this, the polarity of the O—H bond increases and hence the tendency of O—H bond fission (to release a proton) also increases. On the other hand, no such resonance is possible in a molecule of alcohol.

So, the alcoholic O — H bond is relatively less polar and the tendency of bond cleavage resulting in proton release is indeed very small. Hence, phenol is more acidic than alcohol.

Alternative explanation:

This relative acidity may also be explained by considering the phenol-phenoxide ion and Alcohol-alkoxide ion equilibria. Like phenol, phenoxide ion may also be represented as a resonance hybrid of the following (VI – X) resonance structures:

Three (II, III and IV) out of five resonance structures of phenol involve charge separation, but the resonance structures of phenoxide ion involve no charge separation. The negative charge is only delocalised. Because of this, phenoxide ion is more resonance stabilised than phenol.

As a consequence, the equilibrium of phenol-phenoxide ion tends to shift towards the right, i.e., phenol exhibits acidic properties by releasing proton (H⁺) easily. On the other hand, both alcohol and alkoxide ions can be satisfactorily represented by single (localised) structures. Due to the absence of differential stabilisation caused by resonance, alcohol is very reluctant to produce alkoxide ions. So, phenol is a stronger acid than alcohol.

The basic character of aniline:

Aniline is a weak base and its basicity is much weaker than aliphatic amines (RNH₂). This can be explained by resonance. It can be represented as a resonance hybrid ofthe following resonance structures:

1. An unshared pair of electrons on the N-atom of aniline becomes involved in resonance interaction with the ring. As a result, N-atom acquires a partial positive charge. Consequently, aniline exhibits little tendency to take up a proton. So aniline is a weak bases

2. However, in the case of aliphatic amines, similar delocalisation of electrons by resonance is not possible, Naturally the electron density on N-atom is not reduced. In fact, due to the +1 effect of the alkyl group (R-), the electron density on Natom is somewhat increased.

As a consequence, nitrogen can easily donate its electron pair to a proton (H+) to combine with it. Thus, aniline (C₆H₅NH₂) is a weaker base than aliphatic amines (RNH₂). Apart from this, relative basicity can also be explained by considering aniline-anilinium ion and amine ammonium ion equilibrium systems.

 Aniline is a resonance hybrid of five resonance structures (I-V).In the conjugate acid anilinium ion, the lone pair of electrons on the N atom is localised in the N —H bond and so, only two structures (VI and VII) can be drawn for its hybrid. Therefore, aniline is more resonance-stabilised concerning the anilinium ion. As a result of this, protonation of aniline is disfavoured.

None of the aliphatic amine and its conjugate acid can be stabilised by resonance. The conjugate acid is stabilised by the weak +1 effect ofthe -R group. Protonation ofthe aliphatic amine is, therefore, not disfavoured and is is somewhat favoured. Thus, aromatic amines are weaker bases than aliphatic amines.

Finally, in aromatic amine, the amino group is attached to sp² -carbon (more electronegative), whereas in aliphatic amine, it is attached to sp³ -carbon (less electronegative). This factor is also partly responsible for decreased basicity of aromatic amines than aliphatic amines.

Resonance effect or mesomeric effect

Resonance effect or mesomeric effect Definition:

The displacement of non-bonding or electrons from one part of a conjugated system (having alternate single and double bonds) to the other part causing permanent polarity in the system (creating centres of high and low electron density) is called resonance effect (R-effect) or mesomeric effect (M-effect)

There are two types of resonance or mesomeric effect: 

1. +R or +M-effect:

An atom or a group is said to have a +R or +M effect if it involves the transfer of electrons away from the atom or the substituent group attached to a double bond or a conjugated system

+ R or +M groups: —OH, —OR, —SH , — NH₂, —Cl, — Br, —I etc.

Example: +M effect of Cl -atom in vinyl chloride may be shown as follows

2. — R or —M- effect:

An atom or a group is said to have a — R or — M effect if it involves the transfer of electrons towards the atom or the substituent group attached to a double bond or a conjugated system.

R or -M groups: >C= O, —CHO, —COOR, — CN, —NO₂ etc.

Example: -M -effect of —CHO group in acetaldehyde may be shown as follows:

4. Hyperconjugation 

Hyperconjugation Definition:

When a carbon containing at least one H -atom is attached to multiple bonds such as C=C, C ≡ C, C= O, C ≡ N etc., the cr -electrons of the C — H bond become involved in delocalisation with the π electrons of the unsaturated system, i.e., there occurs a σ-π conjugation.

Similarly, σ-p type of conjugation may also take place when a carbon-containing at least one H-atom is attached to a carbon-containing partially filled or vacant p -p-orbital. This special type of resonance or conjugation, giving stability to the species (molecule, free radical or carbocation) is called hyperconjugation.

Hyperconjugation causes a permanent polarity in the molecule and is known as the hyperconjugation effect

Examples:

1. Incaseofpropene (CH₃CH=CH₂)

Although one C—H bond ofthe methyl group is shown to be broken in each hyperconjugative structure, H+ is never free from the rest of the molecule nor does it change its position in the molecule. However, from the point of view of apparent fission of the C —H bond, hyperconjugation is also called no-bond resonance.

2. In the case of ethyl cation, hyperconjugation may be shown as:

3. In Case of ethyl radical:

Conditions for effective hyperconjugation:

For effective hyperconjugation, the p-orbital concerned and the a-C —H bond, i.e., the sp³-s orbital must remain in the same plane. Orbital representations of hyperconjugation in propene and ethyl cation are shown as follows:

Although the stability of a molecule, ion or free radical increases due to hyperconjugation, this stability is less than that contributed by resonance. After the names of the scientists who proposed this theory, hyperconjugation is also called the BakerNathan effect.

Effects of hyperconjugation:

1. Relative stabilities of alkenes:

The stabilities of alkenes can easily be explained by hyperconjugation. The greater the number of a -hydrogen atoms (II-atom present on the carbon atom attached directly to a double bonded carbon), i.e., the greater the number of hyperconjugative structures, the higher the stability of the alkene due to hyperconjugation

Example:

2-Methylpropcne [(CH)₂C=CH₂ having 6 hyperconjugable or-H atoms gives 6 no-bond resonance structures while the isomeric compound, 1 -butene (CH₃CH₂CH = CH₂) having only 2 hyperconjugable a-H atoms gives only 2 no-bond resonance structures.

It thus follows that 2-methyl propyne is thermodynamically more stable than its isomer

Directive influence of alkyl groups:

The ortho- and paradirective influence of alkyl groups can be explained by hyperconjugation.

Example:

Directive influence of the —CH₃ group in toluene, can be explained based on hyperconjugation as follows:

As a result of hyperconjugation, the electron density at ortho- and para-positions increases and as a consequence, the electrophilic substitution reactions in toluene occur mainly at these two positions. It thus follows that the alkyl groups are o, p-directing.

Relative stabilities of carbocations:

Due to hyperconjugation, the C — H bonding electron pair is attracted towards the positively charged C -atom of the carbocation.

This helps in dispersing the positive charge in different parts of the alkyl group, i.c., charge delocalisation resulting in instability of the carbocation.

As the number of a-H atoms increases, the number of no-bond resonance structures of carbocation increases which enhances the extent of charge delocalisadon and consequent stabilisation.

Hence, the order of stability of ethyl, isopropyl and left-butyl cation, due to hyperconjugation is:

Relative stabilities of free radicals:

Because of hyperconjugation, the odd electron of a free radical undergoes delocalisation for which it becomes stabilised.

With the increase in the number of a-H atoms, the number of no-bond resonance structures of a free radical increases and as a result, delocalisation of the odd electron takes place to a greater extent and the stability of free radicals also increases.

Therefore, the stability of ethyl, isopropyl and for-butyl radicals follow the order:

Bond length:

Because of hyperconjugation, the carbon-carbon single bond in propene (CH₃—CH=CH₂) acquires some double bond character and the carbon-carbon double bond acquires some single bond character.

As a result, the C — C bond in propene is found to be a little shorter (1.488 Å) than the normal C —C bond (1.543 Å) in ethane and the C=C bond is found to be a little longer (1.353 Å) than the normal C=C bond (1.334 Å) in ethylene

Electron-releasing power of alkyl groups attached to unsaturated systems or electron-deficient carbon atoms:

This depends on the number of or-H atoms. The methyl ( —CH₃) group having three α-H atoms has the highest hyperconjugative effect while this effect is non-existent with the t-butyl group (Me₃C— ) having no a-H atom. So, electron releasing power of various alkyl groups when attached to a double bond (or an electron-deficient carbon) follows the order:

CH₃ →CH₃CH₃→ (CH₃)₂CH→ (CH₃)₃C—

This order is exactly the reverse of the order of the +I-effect of these alkyl groups

5. Steric hindrance or steric strain

Steric hindrance  Definition:

Steric hindrance or steric strain refers to repulsive interactions between non-bonded atoms or groups which arise when the atoms or groups come very close to each other.

When two non-bonded atoms in a molecule Are closer to each other than the sum of their van der Waals radii, they repel each other due to spatial -crowding.

The repulsion arises primarily due to electron-electron repulsive forces involving the non-bonded atoms. Such repulsive interactions between non-bonded atoms is known as steric hindrance or steric strain. Steric strain is responsible for decreased stability or destabilisation of molecules.

When the sheer bulk of groups at or near a reacting site of a molecule hinders or retards a reaction, it is called steric hindrance. On the other hand, if the constituent atoms or groups of a molecule or ion owing to their bulky nature require more space than what is available for them, i.e., when they are forced too close to one another, then mechanical interference forced too close to one another, mechanical interference ion is then said to be under steric strain. Steric strain makes the species unstable, i.e., its energy increases

Example:

In cis. 2-butene, two -CH₃ groups lying on the same side of the double bond are quite close to each other and so they get involved in steric interaction. In trans-2- butene, two —CH₃ groups lying on the opposite sides of the double bond are far apart from each other so they are not involved in steric interaction. Thus, cis-2-butene is thermodynamically less stable than Frans-2-butene

The heat of hydrogenation of cis-isomer is 28.6 kcal mol-1 and for trans-isomer is 27.6 kcal .mol^-1. This observation agrees with the relative stabilities ofthese isomeric alkenes. Because of steric hindrance, tertbutyl chloride does not hydrolysis by an SN₂ mechanism

Effect on Stability or reactivity:

Steric hindrance is responsible for decreasing the stability and increasing the reactivity of many compounds. Due to steric strain, resonance or delocalisation of electrons may be inhibited (Steric inhibition of resonance or SIR). Again, steric hindrance created at the reaction centre decreases the rate of that reaction or does not even allow the reaction to occur. So, steric hindrance plays a vital role in determining the reactivity of a compound

The acidic character of substituted aromatic acids, phenols and basic character of substituted aromatic amines:

Fission Of Covalent Bond: Generation Of Reaction Intermediates

Formation of product molecule(s) from the reactant molecules involves processes like bond fission, bond formation etc. A chemical equation rarely indicates how the reaction proceeds. The mechanism of an organic reaction is a sequential account of each step of the reaction, describing details of electron movement, energetics during bond breaking and bond formation and the CH rates of conversion of reactants into products (kinetics).

During the breaking and formation of bonds, the transfer of y\H electrons is shown by the use of arrow signs. Curved arrow signs containing two bar  indicate the shifting of a pair of electrons while die transfer of one electron is indicated by curved arrow signs containing one barb or ‘fish hook notation’ [it is to be noted that the symbol  is incorrect]

1. Types of the fission of covalent bonds

Cleavage of covalent bonds can take place in two ways depending upon the nature of the bond involved, the nature of the attacking agent and the conditions of the reaction.

Homolytic fission or homolysis:

If a covalent bond in a molecule undergoes fission in such a way that each bonded atom gets one electron of the shared pair, it is called homolytic fission.

This type of bond cleavage results in the formation of neutral species called free radicals. Homolytic cleavage or fission is usually favoured by conditions such as the non-polar nature of the bond, high temperature and the presence of high energy (UV) radiations.

Example:

Homolytic cleavage of a bond, A— B leads to the formation of free radicals, \(\dot{\mathrm{A}} \text { and } \dot{\mathrm{B}}\) (each containing odd electrons), may be shown as follows

This type of bond fission requires less energy than heterolytic bond fission.

Heterolytic fission or heterolysis:

If a covalent bond undergoes fission in such a way that both the bonding electrons are taken away by one of the bonded atoms, it is called heterolytic bond cleavage. This type of bond cleavage results in the formation of a cation having a sextet of electrons and an anion having an octet of electrons in the valence shells ofthe participating atoms

This type of bond cleavage resulting in the formation of charged species is favoured by the conditions such as the polar nature ofthe covalent bond and the presence of polar solvents Due to heterolytic fission of bond, ions involving charge on carbon are usually formed. According to the nature of the charge, these are of two types—carbocations and carbanions

3. Intermediates formed by the fission of bonds

Under the influence of attacking reagents, suitable bonds in most organic compounds undergo homolytic or heterolytic fission to form short-lived and highly reactive (hence cannot be isolated) chemical species called reaction intermediates.

Some common examples of reaction intermediates are :

Carbocations, carbanions, free radicals, carbenes, arynes etc.

1. Carbocations:

Chemical species having a positively charged carbon atom possessing a sextet of electrons are called carbocations.

Carbocations Formation:

Carbocations are formed by heterolytic fission in which the leaving group is removed along with its shared pair ofelectrons. These are represented by R

Carbocations Nomenclature:

In naming a carbocation, the word ‘cation’ is fission added to the name of the alkyl or aryl group.

For example: ⁺CH₃, CH₃ ⁺CH₂, C₆H₅⁺CH₂ etc. are named methyl cation, ethyl cation, and benomyl cation respectively

Carbocations Classification:

Carbocations are classified as primary (1°), secondary (2°) and tertiary (3°) according to the positive charge is present on a primary, secondary and tertiary carbon atom respectively

Examples: Ethyl cation (CH₃⁺CH₂) is a primary, isopropyl cation [(CH₃)₂⁺CH] is a secondary and terf-butyl cation [(CH₃)₃⁺C] is a tertiary carbocation

Carbocations Structure:

The positively charged C-atom of a carbocation is sp² hybridised. Therefore the structure of a carbocation is trigonal planar and the bond angle is 120°. The vacant p-orbital is perpendicular to the plane

Carbocations Stability:

The stability of carbocations follows the order 3°>2°>1°> methyl. This stability order can be explained based on inductive effect, hyperconjugation and resonance.

Inductive effect: When an alkyl group having +1 effect (electron releasing inductive effect) is attached to a positively charged carbon atom, it reduces the positive charge on the central carbon and by doing this, the alkyl group itself becomes positively polarised, i.e., the positive charge on the central carbon atom is dispersed or delocalised.

As a result of this charge delocalisation, the carbocation is stabilised. Thus, the more the number of the alkyl groups attached to the central positively charged carbon atom greater the stability of the carbocation.

Therefore, the stability of the carbocations ⁺CH₃ ,CH₃⁺CH₂ , (CH₃)₂⁺CH and (CH₃)₃⁺C follows the order:

Hyperconjugation:

The effect of hyperconjugation which depends on the number of a —H atoms also leads to the same order of stability

Resonance:

Carbocations in which a positively charged C-atom is attached to a double bond are stabilised by charge delocalisation involving resonance. Stability due to resonance is greater than that contributed by +1 effect. For example, allyl and benzyl cations are stabilised by resonance.

It is to be rememberedResonance structure electron-releasingof benzyl cation groups, by J H their +1 or +R effect stabilise a carbocation (by dispersing the positive charge) while electron attracting groups by their -I or -R effect destabilise a carbocation (by intensifying the positive charge)

Reactivity:

Carbocations are chemically very reactive species because the positively charged carbons present in them have 6 electrons in their valence shell and hence they have strong tendency to complete their octets. The order of their reactivity is opposite to that of their stability and hence, the order of their reactivity is: methyl cation > primary (1°) > secondary (2°) > tertiary (3°). Carbocations behave as electrophiles.

2. Carbanions:

Chemical species carrying a negative charge on carbon atom possessing eight electrons in its valence shell are called carbanions.

Carbanions Formation:

Carbanions are produced by heterolytic cleavage of covalent bonds in which the bonding electron pair remains with the carbon atom. Carbanions are represented by the symbol, Re

Carbanions Classification:

Like carbocations, carbanions are also classified as primary (1°), secondary (2°) and tertiary

(3°) according as the negative charge is present on a primary, secondary and tertiary carbon atom respectively.

Examples:

Ethyl anion (CH₃CH₂) is a primary, isopropyl anion [(CH₃)₂CH] is a secondary and terf-butyl anion [(CH₃)₃C₆] is a tertiary carbanion.

Carbanions Structure:

In alkyl carbanions, the negatively charged carbon atom is sp³ -sp³-hybridised.

Thus, the negative carbon contains 4 pairs of electrons one of which exists as a lone pair. The structure of simple carbanions is usually pyramidal just like that of ammonia. However, the central carbon in resonance stabilised carbanions are sp² – hybridised and therefore, their structures are planar. For example, the structure of allyl anion is planar

Carbanions Stability:

The stability of the carbanions follows the order:  CH₃ > primary (1° ) > secondary ( 2° ) > tertiary (3° ). This order of stability can be explained on the basis of the following factors.

1. Inductive effect:

When an alkyl group having a +1 effect (electron-releasing inductive effect) is attached to a negatively charged carbon atom, it tends to release electrons towards that carbon.

As a result, the intensity of the negative charge on that carbon is increased and so, the carbanion gets destabilised. Evidently, the greater the number of alkyl groups on the carbon carrying the negative charge, the greater the intensity of the negative charge on the carbon atom and hence the carbanion will be less stable.

Hence, the stability of the carbanions \(\stackrel{\ominus}{\mathrm{C}} \mathrm{H}_3, \mathrm{CH}_3 \stackrel{\ominus}{\mathrm{C}} \mathrm{H}_2,\left(\mathrm{CH}_3\right)_2 \stackrel{\ominus}{\mathrm{C}} \mathrm{H}\) and \(\left(\mathrm{CH}_3\right)_3 \stackrel{\ominus}{\mathrm{C}}\) follows the order as given below:

Electron-attracting groups (Example:  —CN, — NO₂, — Br, —Cl, — F etc.) by their -I effect stabilise a carbanion by dispersing the negative charge. Thus, the greater the number of electron-attracting groups, the more will be the stability of the carbanion.

For example:

2. Resonance:

When the negatively charged carbon of a carbanion remains attached to an unsaturated system or to a benzene ring, tire carbanion gets stabilised by resonance. For example, tire negative charge in each of the acetone anion and benzyl anion is highly delocalised by resonance and consequently, they get stabilised.

Reactivity:

Carbanions are very reactive species because the carbon-bearing negative charge is electron-rich and can easily donate its unshared electron pair to some other group or atom to form a covalent bond.

Hence, carbanions behave as nucleophiles. The order of reactivity of carbanions is reverse ofthe order of stability, i.e., tertiary (3°) > secondary (2°) > primary (1°) >CH₃

3. Free radicals

An atom or a group of atoms possessing an odd (unpaired) electron is called a free radical. Homolytic cleavage of a covalent bond leads to the formation of free radicals

Examples:

Free radicals containing odd electrons on carbon are collectively called alkyl free radicals or simply alkyl radicals. For example, methyl radical (CH3), ferf-butyl radical (Me3C) etc.

Free radicals Classification:

Alkyl free radicals are classified as primary (1°), secondary’ (2°) and tertiary (3°) depending on the nature carbon atom bearing the unpaired electron.

Examples: CH₃CH₂ (ethyl radical) is a primary (1°), (CH₃)₃CH (isopropyl radical) is a secondary (2°) and (CH₃)₃C (for-butyl radical) is a tertiary (3°) alkyl radical.

Free radicals Structure:

The structure of alkyl free radicals may be planar or pyramidal. The carbon atom of the planar free radical is sp² -hybridised while the carbon atom of a pyramidal free radical is sp³ -hybridised

Free radicals Stability:

The stability of radicals can be explained on the basis ofthe following factors: 

Hyperconjugation:

Discussed earlier

Resonance:

Free radicals in which the carbon carrying the odd electron is attached to a double bond or a benzene ring is stabilised by resonance

Example:

Reactivity:

Free radicals are highly unstable and reactive species because they have a strong tendency to gain an additional electron, i.e., to share with some other atom or group to have a complete octet. The order of reactivity of alkyl radicals is reverse of the order of stability, Le., primary (1°) > secondary (2°) > tertiary (3°).

4. Carbenes

A neutral group of atoms which contains a carbon atom with only 6 electrons in its valence shell, out of which two electrons are unshared, are called carbenes.

For example: Methylene (: CH₂), dichlorocarbene (: CCl₂ etc. Because of the strong tendency to achieve an octet, carbenes are highly reactive and unstable. They behave as electrophiles.

There are two types of carbenes:

1. Singlet carbene and
2. Triplet carbene.

The central carbon atom of singlet carbene and most of the triplet carbene is sp² -sp²-hybridised.

In singlet carbene, the two unshared electrons occupy an sp² -hybrid orbital while in triplet carbene, these occupy one sp² -orbital and one p-orbital respectively.

The triplet carbene however may also assume a linear structure where the central carbon atom is sp -hybridised. The second one is more stable than the first

5. Arynos:

Benzenoid aromatic compounds having a carbon-carbon triple bond are known as arynes. The most simple among the arynes is benzyne or 1,2-dehydrobenzene.

Arynes are neutral, unstable and highly reactive intermediates. In an aryne molecule, the two sp² -hybrid orbitals, because of their diverging orientation, overlap to a very small extent to form the additional bond and for this weak overlapping, arynes are highly reactive.

It is to be noted that the triple bond in benzyne is not like the triple bond in acetylene because in acetylene two sp -orbitals overlap to form a cr -bond and two pairs of p -orbitals overlap to form two n -bonds

Various reactive intermediates at a glance:

Stability Order:

Classification Organic Reactionsof The Mechanisms Of

Based on carbon-carbon bond cleavage, mechanisms of organic reactions are divided into two classes—

1. The free radical mechanism,
2. Polar or ionic mechanism.

1. Free radical mechanism:

When a chemical reaction occurs through the formation of free radicals, then the mechanism is called free radical mechanism. This type of mechanism, therefore, applies to heat or light-induced organic reactions in which homolytic bond fission takes place. Many substitution and addition reactions occur by this mechanism.

2. Polar or Ionic mechanism:

When a chemical reaction occurs through the formation of ions, then the mechanism is called a polar or ionic mechanism. This type of mechanism, therefore, applies to organic reactions in which heterolytic bond fission takes place. Many substitution and addition reactions occur also by this mechanism.

Reactions involving polar mechanisms, the reagents generally participate as ions. Depending on nature of the charge on reagents, they can be divided into two classes—

Electrophiles or electrophilic reagents:

An electrophile or electrophilic reagent {Greek: electron loving) is an electron-deficient species which can accept an electron pair from an electron rich species (molecule or anion) to form a covalent bond with it

Nucleophiles or nucleophilic reagents:

A nucleophile or nucleophilic reagent (Greek: nucleus loving) is an electron rich species which can donate an electron-pair to an electron deficient species (molecule or cation) to form covalent bond with it. Nucleophiles tend to attack electrophiles

Negatively charged nucleophiles:

Cl-,Br-, OH-, RO-, RS-, CN~, CH₃C = C- etc.

O Neutral nucleophiles:

H20, ROH, NHg, RNH₂, RgN etc.

Since the nucleophiles are electron-donors, they are considered as Lewis bases

Classification Of Organic Reactions

Since the number of organic compounds is quite large, the number of their chemical reactions is also expected to be numerous.

All the organic reactions in general can be classified into the following four types:

1. Substitution reactions
2. Addition reactions
3. Elimination reactions and
4. Rearrangement reactions.

1. Substitution reactions

Substitution reactions Definition:

Reactions involving replacement or substitution of an atom or group in organic molecules by some other atom or group without any change in the remaining part of the molecules are called substitution reactions

The products formed as a result ofsubstitutions are called substitution products.

Depending upon the nature of the attacking species (nucleophile, electrophile or free radical),

The substitution reactions may further be classified into the following 3 types:

1. Nucleophilic substitution reactions:

Substitution reactions involving nucleophiles as the attacking agents are called nucleophilic substitution reactions.

• These reactions are usually designated as SN {Substitution Nucleophilic) reactions.
• Again, depending upon the number of species (molecule, ion or free radical) participating in the rate-determining step (r.d.s.) of the reaction, the SN reactions are further classified as S_N1 and S_N2 reactions.
• The number of particles taking part in the rate-determining step is called the molecularity of the reaction.
• If the molecularity of a reaction is 1, i.e., if the reaction is unimolecular, then the mechanism of the reaction is said to be S_N1 {Substitution Nucleophilic Unimolecular) and if the molecularity is 2 i.e., if the reaction is bimolecular, then the mechanism of the reaction is called S_N2 {Substitution Nucleophilic Bimolecular).

S_N1 Reaction

The reactions which proceed through S_N1 mechanism are called S_N1 reactions. These are two-step processes. In the first step of such a reaction, the atom or group which is to be replaced (the leaving group) is removed to form a stable carbocation. This is the rate-determining step (i.e., the slowest step) of the reaction in which only one particle (the substrate) participates. In the second step, the nucleophile gets covalently attached to the carbocation to form the substituted compound. The rate of an S_N1 reaction depends only on the concentration of the substrate

Example: The hydrolysis of ferf-butyl bromide by aqueous KOH solution to form ferf-butyl alcohol proceeds via KOH solution to form ferf-butyl alcohol proceeds via

Reaction mechanism:

In the first step of the reaction, the C—Br bond of ferf-butyl bromide undergoes fission to form ferfbutyl cation (a stable carbocation) and bromide ion (the leaving group). This is the slowest or rate-determining step. The rate of the reaction depends on the concentration of the substrate, (CH₃)₃CBr . In the second step [fast), the nucleophile (here OH-) attacks the carbocation to form ferf-butyl alcohol.

The reactions which proceed through S_N2 mechanism are called S_N2 reactions. In this type of reaction, the removal of the leaving group (bond breaking) and attachment of the nucleophile with the substrate (bond formation) take place simultaneously, i.e., the reaction occurs in one step. Therefore, it is the rate-determining step of the reaction.

The rate ofan S_N2 reaction depends on the molar concentration of both the substrate and the nucleophile.

Example: Hydrolysis of CH₃Cl by aqueous KOH solution to produce CH₃OH proceeds via S_N2 mechanism

Reaction mechanism: Because of much higher electronegativity of Cl compared to C, the C-atom of CH₃CI becomes The rate of an S_N1 reaction depends only on the electrophilic centre. The nucleophile, OH” attacks the C atom of CH₃Cl from the opposite side of the Cl -atom i.e., at an angular distance of 180°).

Such an approach of the nucleophile requires the lowest energy as this avoids electrostatic repulsions between the negatively charged nucleophile and the leaving group. A backside attack is also more feasible sterically. In this one-step reaction, the formation of C— O and cleavage of the C —Cl bond take place simultaneously. The reaction, thus, proceeds

through a single transition state in which the carbon is partially bonded to both —OH group and Cl-atom and full-bonded to the three H-atoms. In the transition state, the —OH group possesses a diminished negative charge as it starts sharing its electrons with carbon while chlorine acquires a partial negative charge as it tends to depart with the bonding electron pair.

The C-atoms and the three Hatoms become coplanar (bond angle 120° ) and the plane is perpendicular to the line containing the grouping HO^δ- ——C -—Cl^δ- . Once the transition state is formed, HO₅– further approaches the C-atom to form a frill covalent bond while the Cl₅– atom is eliminated as Cl^– by taking full possession of the electron pair of C—Cl bond.

Comparison of S_N1 And S_N2mechanism 

Alkyl halides may undergo hydrolysis by both SN₁ and SN₂ mechanisms. SN₁ and SN₂ mechanisms of hydrolysis of RX may be compared concerning the following factors:

The terms “transition state” and “intermediate” are not synonymous. Although intermediates are (For example:  Carbocations, carbanions, free radicals etc.), very unstable they have real existence, but the transition states represent hypothetical arrangements of atoms possessing a definite shape and charge distribution. These have no real existence. Each step of every reaction proceeds through a transition state but each reaction may or may not proceed through an intermediate.

For example: 

There is no intermediate involved in a one-step reaction such as SN₂ but the two-step reaction SN₁ proceeds through the formation of an intermediate

Electrophilic substitution reactions:

The substitution reaction in which the attacking reagent is an electrophile are called an electrophilic substitution reaction. These reactions are expressed by the symbol, SE (Substitution Electrophilic)

Example:

Nitration of benzene with mixed acid (concentrated HNO₃ and concentrated H₂SO₄) yields nitrobenzene. It is an electrophilic substitution reaction. Nitronium ion (NO₂) produced by the reaction between HNO₃ and H₂SO₄, acts as an electrophile in this reaction.

⇒ \(\mathrm{HNO}_3+2 \mathrm{H}_2 \mathrm{SO}_4 \rightleftharpoons \stackrel{\oplus}{\mathrm{N}} \mathrm{O}_2+\stackrel{\oplus}{\mathrm{H}_3} \mathrm{O}+2 \mathrm{HSO}_4^{\ominus}\)

Reaction mechanism :

The reaction occurs in two steps. In the first step, NO₂ ion is attacked by the n-electrons of a benzene ring and gets attached to any of the six carbon atoms to form a resonance-stabilised carbocation (complex). This is the rate-determining (slow) step of the reaction. In this step, benzene loses its aromaticity. In the second step, the HSO₄ ion accepts a proton (H⁺) from the tr -complex and results in the formation ofthe stable aromatic compound i.e., nitrobenzene. In this step, the aromaticity of the ring is regained.

Free radical substitution reaction:

The substitution reactions in which the attacking reagent is a free radical are called free radical substitution reactions.

Example:

Chlorination of methane in the presence of heat or diffused sunlight to give methyl chloride and hydrogen chloride occurs by a free radical mechanism

Reaction Mechanism:

Addition reactions

Reactions Definition: 

Reactions In which two reacting molecules combine to give a single product molecule arc called addition reactions.

This type of reaction is typical for compounds containing multiple (double or triple) bonds. Depending upon the nature of the attacking species (electrophiles, nucleophiles or free radicals), addition reactions may be classified into the following three types

1. Nucleophilic addition reactions:

Addition reactions In which the attacking reagent Is u nucleophile arc are called nucleophilic addition reactions. Example: Base-catalysed addition of HCN to acetaldehyde to form acetaldehyde cyanohydrin is an example of a nucleophilic addition reaction

Reaction mechanism:

In this reaction, NaOII acts as a catalyst. The reaction between HCN and NaOII gives rise to the nucleophile, CN” ion

Step 1:

Because of resonance and electromeric effect, the carbonyl carbon atom of acetaldehyde acquires a partial positive charge. The positively polarised carbonyl carbon undergoes nucleophilic attack by CN^– ion to form an alkoxide ion. This Is the rate-determining (slowest) step.

Step 2:

In this step, the strongly basic alkoxide ion almost Immediately takes up a proton from HCN or from the reaction medium (i.e, H₂O ) and finally gets converted to the cyanohydrin compound

2. Electrophilic addition reactions:

Addition reactions in which the attacking reagent is an electrophile are called electrophilic addition reactions.

Example: The addition of HBr to propene to form 2-bromopropane as the major product is an example of this type of reaction.

Reaction mechanism: The reaction proceeds through the following steps:

Step 1:

Hydrogen bromide provides an electrophile H+, which attacks the double bond and becomes attached to C-l of propene to form preferably the relatively stable secondary (2°) isopropyl cation. It is the rate-determining (slowest) step of the reaction

Step 2: The carbocation is attacked by the nucleophile Br© ion to form 2-bromopropane predominantly

3. Free radical addition reactions:

Addition reactions in which the attacking reagent is a free radical are called free radical addition reactions.

Example:

The addition of HBr to propene in the presence of peroxides to form 1-bromopropane as the major product is an example of a free radical addition reaction

Reaction mechanism:

The reaction follows the given steps:

1. Initiation:

The organic peroxide undergoes homolytic fission of the O —  bond in the presence of heat or light to form alkoxy free radicals. These free radicals then take up H from HBr to Br

2. Propagation:

Bromine free radical gets attached to C-l of propene to form a relatively more stable secondary (2°) free radical. This alkyl radical then takes up H from HBr to from 1-bromopropane predominantly.

3. Termination:

Two bromine free radicals combine to form bromine molecule.

Elimination reactions

Elimination reactions Definition:

The reactions in which two atoms or groups get eliminated from the substrate molecule leading to the formation of a carbene, a multiple bond (double or triple) or a cyclopropane derivative are called elimination reactions

Depending upon the relative positions of the groups or HBr atoms eliminated, these reactions are classified as α (alpha), β(beta) and γ (gamma) elimination reactions.

α  Elimination reactions:

The elimination reactions in which the loss of two atoms or groups occurs from the same atom of the substrate molecule are called α -elimination reactions. Base-catalysed dehydrochlorination of chloroform to form dichlorocarbene is an example of α – elimination reaction

β -Elimination reactions:

The elimination reactions in which the loss of two atoms or groups occurs from the adjacent positions (α, β) of the substrate molecule to form multiple bonds are called β-elimination or, 1,2-elimination reactions. Based on the mechanism involved, β-elimination reactions are discussed through E2, E1 and E1cB mechanisms

E2 mechanism:

The E2 mechanism is a single-step process.

• Base (BI:-) pulls a proton away from the β -carbon atom and simultaneously a leaving group (Y^–) is removed from the β -β-carbon atom resulting in the formation of a carbon-carbon double bond.
• In the transition state, there are three partial bonds — one between the base and β -H atom, one between β -H atom and β-C atom and one between the a -carbon atom and the leaving group, Y.
• The carbon-carbon single bond also acquires a partial double bond character.
• Since the rate-determining step (in this case, it is the only step) involves a reaction between a substrate and a base, the mechanism is designated as E2 (Elimination bimolecular).
• Rate of the reaction is proportional to the molar concentrations of both the reactant and the base.

Example: When 2-bromopropane is heated with alcoholic KOH, propene is obtained

Reaction mechanism: The mechanism of the reaction is:

E1 Mechanism:

Elimination reactions by the El mechanism form to form dichlorocarbene is an example of takes place in two steps.

• In the first step, the substrate undergoes heterolytic fission of the cr -bond between the a -carbon atom and the leaving group to form a carbocation.
• This is the rate-determining (slowest) step of the action.
• Since the rate-determining step (in this case, it is the first step) involves only the substrate, the mechanism is designated as El (Elimination Unimolecular).
• In the second step, the β-C atom loses a proton to the base to produce an alkene.
• The rate of the reaction is proportional to the molar concentration of the substrate only

Example: When an alcoholic solution of tert-butyl chloride is heated, 2-methylpropene is formed by El mechanism.

Reaction mechanism:

The reaction proceeds via the following steps:

ElcB mechanism:

• The substrates having acidic β- hydrogen and a very poor leaving group undergo an elimination reaction by the ElcB mechanism.
• This type of reaction takes place in two steps.
• In the first step, one acidic hydrogen atom attached to β -the carbon atom of the substrate is abstracted by the base to form a stable carbanion.
• This carbanion is the conjugate base of the substrate molecule. In the second step, the leaving group becomes detached from the a -carbon atom as an anion to form an alkene.
• This step is generally the rate-determining (slow) step ofthe reaction. Since the rate of the reaction depends only on the concentration of the conjugate base, so this mechanism is known as ElcB (Elimination Unimolecular Conjugate Base).

Example:

The reaction of sodium ethoxide with 2,2-dichloro-l, 1,1-trifluoroethane gives rise to l,l-dichloro-2,2- difluoroethene. The reaction follows the ElcB mechanism because two electronegative Cl-atoms and electron-withdrawing —CF₃ group causes the β —H atom to become sufficiently acidic and the F- ion behaves as a very poor leaving group

Reaction mechanism:

The mechanism is as follows:

γ -Elimination reactions:

The reactions in which the loss of two atoms or groups occurs from α and γ-positions of the molecule leading to the formation of three-membered rings are called γ-elimination reactions.

Rearrangement reaction

Rearrangement reaction Definition:

The reaction  involving the shift or migration of an atom or group from a particular position in a molecule or ion to another position under suitable conditions to form a rearranged product is called rearrangement reactions

Atom or group generally shifts with bonding electron pair.

Example: When pinacol is treated with concentrated H₂SO₄, it undergoes dehydrative rearrangement to give pinacolone.

Reaction mechanism:

In the first step, pinacol takes a proton to form its conjugate acid. In the second step, the H₂O molecule is eliminated to form a carbocation. In the third step, a methyl group migrates to the adjacent positive C with its bonding electron pair (1, 2-shift) to form a resonance-stabilized carbocation. In the fourth step, the carbocation (the conjugate acid of pinacolone) loses a proton to yield pinacolone.

Purification And Analysis Of Organic Compounds Introduction

The presence of impurities even in very small amounts may sometimes result in deviation of some properties of organic compounds to a marked degree. Therefore, to characterise an organic compound thoroughly, it is essential to obtain it in the purest form. Again, an organic compound must be in a certain state of purity before it can be analysed qualitatively and quantitatively to arrive at its correct molecular formula. The organic compounds whether isolated from a natural prepared in the laboratory are mostly impure. These are generally contaminated with some other substances. A large number of methods are available for the purification of organic compounds.

Different Methods For The Purification Of Organic Compounds

Some of the important methods which are commonly employed for the purification of organic compounds are as follows:

1. Crystallisation
2. Sublimation,
3. Distillation,
4. Extraction and
5. Chromatography.

1. Crystallisation

Crystallisation Definition: 

Crystals are the purest form of a compound having definite geometrical shapes and the process by which an impure compound is converted into its crystals is known as crystallisation.

This is one of the most commonly used methods for the purification of solid organic compounds. It is based on the difference in solubilities ofthe compound and the impurities in a suitable solvent.

A solvent is said to be the most suitable one which fulfils the conditions such as: 

• The organic solid must dissolve in the solvent on heating and must crystallise out on cooling
• The solvent must not react chemically with the organic compound, and
• The impurities should not be normally dissolved in the solvent or if they dissolve, they should be soluble to such an extent that they remain in the solution, i.e., in the mother liquor.

The various solvents which are commonly used for crystallisation are water, alcohol, ether, chloroform, carbon tetrachloride, benzene, acetone, petroleum ether etc.

Crystallisation Procedure:

• A certain amount of an impure organic compound is added to a minimum amount of a suitable solvent and the mixture is then heated to get a hotsaturated solution ofthe compound. The hot solution is then filtered to remove the insoluble impurities, if present.
• The clear solution is then allowed to cool down undisturbed when the solid organic compound separates in the form of fine crystals. The crystals do not separate even after a long time, the inner surface of the vessel is scratched with the round end of a glass rod to facilitate crystallisation.
• The addition of a few crystals of the pure compound to the solution may also hasten the crystallisation process. The process of inducing crystallisation by adding a few crystals of the pure compound into its saturated solution is called seeding.
• The crystallised compound is then filtered as usual. The crystals on the filter paper are washed with a small amount of solvent to remove the impurities.
• The compound is then pressed in between the folds of filter paper to remove water as far as practicable. It is then dried in an esteem or air oven and finally in a vacuum desiccator.

Fractional crystallisation:

This method is used for the separation of a mixture of two (or more) compounds which have unequal solubilities in a particular solvent.

Fractional crystallisation Procedure:

• A saturated solution of the mixture of compounds is prepared in a suitable solvent by applying heat and the hot solution is then allowed to cool when the less soluble component crystallises out earlier than the more soluble component.
• The crystals are separated by filtration.
• The mother liquor is then concentrated and the hot solution is allowed to cool when the crystals of the more soluble second component are obtained.
• By repeating the process, all the components of the mixture are separated.
• It thus follows that fractional crystallisation is the process of separation of different components of a mixture by repeated crystallisation.

2. Sublimation

Sublimation Definition:

The Process of conversion of a solid into the gaseous state on heating without passing through the intervening liquid state and vice versa on cooling is called sublimation.

• Only those substances, whose vapour pressures become equal to the atmospheric pressure much before their respective melting points, undergo ready sublimation when heated
• This process is very useful for the separation of volatile solids which sublime on heating from the non-volatile impurities.

Sublimation Procedure:

• The impure sample is taken in a china dish covered with perforated filter paper (or porcelain plate).
• An inverted funnel is placed over the dish and its stem is plugged with cotton.
• The dish is then heated gently when vapours of the volatile substance pass through perforations of the filter paper and condense on the cooler walls of the funnel leaving behind non-volatile impurities in the dish

Sublimation Applications:

Benzoic acid, camphor, naphthalene, anthracene, iodine etc. are purified by this method. In the case of other compounds like indigo which are very susceptible to thermal decomposition, sublimation is done under reduced pressure

3. Distillation

Distillation Definition:

The process of conversion of a liquid into its vapours by heating followed by condensation of vapours thus produced by cooling is called distillation

• The process of simple distillation is commonly used for the purification of liquids which do not undergo decomposition, on boiling Le., which are sufficiently stable at their boiling points and which contain non-volatile impurities.
• Organic liquids such as benzene, ethanol, acetone, chloroform, carbon tetrachloride, toluene etc. can be purified by the process of simple distillation

Distillation Procedure:

• The impure organic liquid is taken in a distillation flask which is fitted with a water condenser and a thermometer.
• A receiver is attached to the lower end of the condenser.
• One or two pieces of unglazed porcelain or glass beads are added to prevent bumping of the liquid during distillation.
• The flack is then heated in a water bath or a sand bath (in the case of volatile and inflammable liquid) or directly (in the case of liquids having high boiling points) when the temperature rises gradually.
• The liquid starts boiling when its vapour pressure becomes equal to the atmospheric pressure.
• The vapours then pass through the water condenser and condense to form the liquid which is collected in the receiver
• The non-volatile impurities are left behind in the distillation flask

Distillation:

1. Fractional distillation:

Fractional distillation Definition:

The”distillation process in which a mixture of two or more miscible liquids having boiling points close to each other are separated is called fractional distillation

If the boiling points ofthe two liquids of a mixture are very close to each other, i.e., differ only by 10- 20K, their separation cannot be achieved by a simple distillation method. In such cases, the separation can be achieved by fractional distillation which involves repeated distillation and condensations by using a fractionating column

Fractional distillation Procedure:

• The apparatus used is the same as in the simple distillation process except for a fractionating column.
• When the mixture is heated, the temperature rises slowly and the mature starts boiling. The formed mainly consists of the more volatile liquid with a little of the less volatile liquid.
• As these vapours travel up in the fractionating column, the vapours ofthe less volatile liquid condense more readily than those of the more volatile liquid.
• Therefore, vapours rising become rich in the vapours of more volatile liquid and the liquid flowing down becomes rich in less volatile liquid.
• This process is repeated throughout the length of the fractionating column.
• As a consequence, the vapours which escape from the top of the column into the condenser consist of almost the more volatile liquid.
• Thus, the distillate received contains the more volatile component almost in pure form whereas the liquid left behind in the flask is very rich in the less volatile component.
• This process may sometimes be repeated to achieve complete separation of liquids.

Fractional distillation Applications:

One of the most important applications of fractional distillation is to separate crude petroleum into various fractions like gasoline, kerosene oil, diesel oil etc. 0 Fractional distillation is also used to separate methanol (b.p. 338 K) and acetone (b.p. 329 K) from pyroligeneous acid obtained by destructive distillation of wood.

2. Distillation under reduced pressure (Vacuum distillation):

The distillation process which involves the purification of high boiling liquids (which decompose at or below their boiling points) by reducing the pressure over the liquid surface is called vacuum distillation.

Some liquids which tend to decompose at or below their boiling points cannot be purified by ordinary distillation. Such liquids can be purified by distillation under reduced pressure. A liquid boils when its vapour pressure becomes equal to the external pressure.

Therefore if the pressure acting on It is reduced, the liquid boils at a lower temperature and so, its decomposition does not occur. Glycerol, for example, decomposes at its boiling point (563 K). However, if the external pressure is reduced to 12 mm, it boils at 453 K without decomposition. Some other compounds such as phenylhydrazine, diethyl malonate, ethyl acetoacetate etc. are also purified by this method.

Vacuum distillation Procedure:

• The distillation is carried out in a two-necked flask called Claisen’s flask.”A capillary tube is fitted into one neck ofthe flask and is kept immersed in the liquid to be distilled.
• The other neck of the flask is fitted with a thermometer. The side tube of this neck is connected to a condenser carrying a receiver at the other end.
• The receiver is connected to a vacuum pump and a manometer.
• The flask is usually heated in a sand or oil bath. To prevent bumping, a steady flow of air is maintained by the capillary tube with the help ofthe screw-type cock attached to it The desired pressure is maintained by using the vacuum pump

3. Steam distillation:

Steam distillation is applied for the separation and purification of those organic compounds which

• Are insoluble in water
• Are steam volatile,
• Possess a vapour pressure of 10-15 mm Hg at 100°C and
• Contain non-volatile organic or inorganic impurities.

Steam distillation Principle:

In steam distillation, the liquid boils when the sum of the vapour pressures of the organic liquid (pt) and that of water (p₁) becomes equal to the atmospheric pressure (p), i.e.,

p = p₁ + p₂. Since pt is lower than p, the organic liquid boils at a temperature lower than its normal boiling point and hence its decomposition can be avoided. Thus, the principle of this method is similar to that of distillation under reduced pressure.

 Steam distillation Procedure:

• The impure organic liquid is taken in a roundbottom flask. Steam from a steam generator is passed into the flask which is gently heated.
• The mixture starts boiling when the combined vapour pressure becomes equal to the atmospheric pressure
• At this temperature, the vapours of the liquid with the steam escape from the flask and after getting condensed it is collected in the receiver.
• The distillate contains the desired organic liquid and water which can easily be separated by using a separating funnel

4. Differential extraction

Differential extraction Method:

The method of separation of an organic compound from its aqueous solution by shaking with a suitable solvent is called differential extraction

A solid or liquid organic compound can be recovered from its aqueous solution by shaking the solution in a separating funnel with a suitable organic solvent which is insoluble in water but in which the organic compound is highly soluble. Some commonly employed solvents for extraction are ether, benzene, chloroform, carbon tetrachloride etc.

Differential extraction Procedure:

• The aqueous solution ofthe organic compound is mixed with a small quantity of the organic solvent in a separating funnel
• The funnel is stopped and its contents are shaken vigorously when the organic solvent dissolves out of the organic compound.
• The separating funnel is then allowed to stand for some time when the solvent and water form two separate layers.
• The lower aqueous layer (when the organic solvent used is ether or benzene) is run out by opening the tap of the funnel and the organic layer is collected.
• The whole process is repeated to remove the organic compound completely from the aqueous solution.
• The organic compound is finally recovered from the organic solvent by distilling off the latter.

5. Chromatography

Chromatography Definition:

The technique used for the separation of the components of a mixture in which the separation is achieved by the differential movement of individual components through a stationary phase under the influence of a mobile phase is called chromatography.

Chromatography is the most useful and modern technique extensively used for the separation of mixtures into their components, to purify the compounds and also to test the purity of compounds.

This method was first invented by M. Tswett, a Russian botanist in 1906 for the separation of coloured substances into individual components. The word ‘Chromatography’ was originally derived from the Greek word chroma means colour and graphy means writing.

Types of chromatography:

Depending upon the nature of the stationary phase (either a solid or a tightly held liquid on a solid support) and the mobile phase (either a liquid or a gas),

The various types of chromatographic techniques commonly used are:

1. Column or Adsorption chromatography,
2. Thin Layer Chromatography (TLC),
3. High-performance liquid Chromatography (HPLC),
4. Gas Liquid Chromatography (GLC),

Paper or partition Chromatography. In the first three cases, the mobile phases are liquid and the stationary phases are solid.

In the fourth case, the mobile phase is gas while the stationary phase is liquid and in the fifth case, both the mobile and stationary phases are liquid.

The chromatographic separation is based on the principle that the components of the mixture present in the moving phase move at different rates through the stationary phase and thus get separated. Now, depending on the basic principle, chromatography can be divided into two categories:

Absorption chromatography and Partition chromatography.

Adsorption chromatography Principle:

This category of chromatography is based upon the differential adsorption of the various components of a mixture on a suitable adsorbent such as alumina, cellulose, silica gel, magnesium oxide etc. Since some compounds undergo adsorption better than others, they travel through the column at different rates and thus get separated.

Adsorption chromatography is of two types:

1. Column chromatography and
2. Thin Layer Chromatography (TLC).

1. Column chromatography

It is the simplest of all the chromatographic techniques and is extensively used.

Chromatography Procedure:

• A plug of cotton or glass wool is placed at the bottom of a clean and dry glass column and it is then covered with a layer of acid-washed sand.
• A suitable adsorbent such as alumina, silica gel, magnesium oxide, starch etc. is made into a slurry with a non-polar solvent such as hexane or petroleum ether and the slurry is then added to the column gradually and carefully so that no air bubble is entraped in the column.
• The excess of the solvent above the adsorbent is removed by opening the stop-cock. This constitutes the stationary phase.
• The mixture of compounds (say A, B and C) to be separated is dissolved in a minimum volume of a suitable highly polar solvent in which it is readily soluble. It is then added to the top of the column with the help of a dropper or a microsyringe and allowed to pass slowly through it (if the mixture is liquid, it is added as such).
• As the solution travels down, the different components of the mixture get adsorbed to different extents depending upon their polarity (say, A>B>C) and form a narrowband which is quite close to the top of the column. This band or zone is called a chromatogram.
• A suitable solvent called eluent is then made to run through the column. The polarity of the solvent is gradually increased to elute the adsorbed materials. The eluent acts as the mobile phase.
• As the solvent moves down the column, the components A, B and C present in the chromatogram begin to separate. The eluent dissolves out the different components selectively.
• The component which is strongly adsorbed on the stationary phase moves slowly down the column, whereas the component that is weakly adsorbed moves at a faster rate. Therefore, the three components [A, B and C) form three bands at different places in the column.
• As the addition of eluent is continued, the adsorbed components present in the bands are dissolved by the solvent and are then collected in the form of different fractions in separate conical flasks.
• The eluent from each flask is then removed by evaporation or distillation to get the various components in pure form.
• The process of separation of different components of the mixture from the adsorbent and their recovery with the help of a suitable solvent is called elution.

2. Thin Layer Chromatography (TLC):

It is another type of adsorption chromatography which involves the separation of the components of a mixture over a thin layer of adsorbent. This technique is particularly useful in rapid analysis ofthe purity of samples.

Chromatography Procedure:

• A thin layer (0.2 mm thick) of adsorbent such as silica gel or alumina is spread over a plastic or glass plate of suitable size (5 cm  × 20 cm).
• This thin layer of adsorbent acts as the stationary phase. This plate is called a thin-layer chromatography plate or TLC plate or chromatophore.
• Two pencil lines are drawn across the width of the plate at distances about 1 cm from each end. The lower line is called the base line or starting line and the upper line is called the finish line or solvent front.
• A drop of the solution of the mixture to be separated is placed on the starting line with the help of a capillary.
• The plate is then dried and placed in a vertical position in a jar called a developing chamber containing a suitable solvent or a mixture of solvents. It acts as the mobile phase. The height of the solvent in the jar should be such that its upper surface does not touch the sample spot.
• The chamber is then closed and kept undisturbed for half an hour
• As the solvent slowly rises by the capillary action, the components of the mixture also move up along the plate to different distances depending upon their degree of adsorption and thus separation takes place.
• When the solvent front reaches the finish line, the plate is removed from the jar and then dried. The spots of coloured components, due to their original colour, is visible on the TLC plate.

The spots of the colourless components which are invisible to the eye can be detected:

• By placing the plate under a UV lamp because certain organic compounds produce a fluorescence effect in UV light
• By placing the plate in a covered jar containing a few crystals of iodine because certain organic compounds which absorb iodine turn brown and
• By spraying the plate with the solution of a suitable chemical reagent
• For example – Ninhydrin in case of amino acids; 2,4-dinitrophenylhydrazine in case of carbonyl compounds
• Different components developed on the TLC plate are identified through their retention factors (or retardation factor), i.e., R j values.

It may be defined as:

⇒ \(R_f=\frac{\text { Distance travelled by the compound from the baseline }}{\text { Distance travelled by the solvent from the baseline }}\)

• If the two components of the mixture, for example, are A and B, then according to the given their  R₂ values will be a/l and  b/l respectively.
• The two compounds can be identified by comparing these R_f values with the R_f values of pure compounds. Since the solvent front on the TLC plate always moves faster than the compounds, R_f values are always less than 1.
• Each spot is finally eluted separately with suitable solvents and collected

Partition chromatography

Unlike adsorption chromatography (column chromatography or TLC) which represents solid-liquid chromatography, partition chromatography is a liquid-liquid chromatography, i.e., both the stationary phase and mobile phase are liquids.

Partition chromatography Principle:

Partition chromatography is based on continuous differential partitioning (distribution) of components of a mixture between the stationary and the mobile phases Paper chromatography is a common example of partition chromatography.

In paper chromatography, a special type of paper known as chromatographic paper is used. Although the paper is made up of cellulose, the stationary phase in paper chromatography is not the cellulose but water which is adsorbed or chemically bound to cellulose. The mobile phase is usually a mixture of two or three liquids with water as one ofthe components.

Partition chromatography Procedure:

• A suitable strip of chromatographic paper (20cm × 5cm, Whatman filler paper) is taken and a starting line (baseline) is drawn across the width of the paper at about 1 to 2 cm from the bottom.
• The mixture to be separated is dissolved in a minimum amount of a suitable solvent and applied as a spot on the starting line with the help of a fine capillary or micro syringe.
• The spotted chromatographic paper is then suspended in a suitable solvent or a mixture of solvents. The position of the paper should be such that the spot on the starting line remains above the surface of the solvent. Thus, the solvent acts as the mobile phase.
• The solvent rises up the paper strip by capillary action and flows over the sample spot. The different components of the mixture depending upon their solubility (or partitioning between) in the stationary and mobile phases, travel through different distances.
• When the solvent reaches the finish line the paper strip is taken out and dried in air. The paper strip so developed is called a chromatogram.
• The spots of the separated coloured compounds are visible at different distances from their initial position on the starting line.
• The spots of the separated colourless compounds are observed either by placing the paper strip under UV light or by using an appropriate spray reagent as discussed in TLC.
• The components of the mixture are then identified by determining their R j values as discussed in TLC.
• The process can also be performed by folding the chromatographic paper into a cylinder

Criteria of purity of organic compounds

Several methods for the purification of organic compounds have already been discussed. The next important step is to test their purity, i.e., to know whether a particular compound has been purified or not. A pure organic solid has a definite and sharp melting point. An impure solid melts over a range of temperatures and even the presence of traces

Qualitative Analysis Of Organic Compounds

Since organic compounds are either hydrocarbons or their derivatives, the elements which occur in them are carbon (always present), hydrogen (nearly always present), oxygen (generally present), nitrogen, halogens, sulphur (less commonly present), phosphorus and metals (rarely present). All these elements can be detected by suitable methods.

1. Detection of carbon and hydrogen

Detection of carbon and hydrogen Principle:

A small amount of a pure and dry organic compound is strongly heated with dry cupric oxide in a hard glass test tube when carbon present in it is oxidised to carbon dioxide and hydrogen is oxidised to water.

Liberated CO₂ turns lime water milky and the liberated water vapours (H₂O) which get condensed in the bulb of the delivery tube, turn white anhydrous CuSO₄into blue hydrated copper sulphate (CuSO₄-5H₂O

Ca(OH)₂ (Lime Water)+ CO₂→CaCO₃↓ (Milky)+ H₂O

CuSO₄ Anhydrous (white)+ 5H₂O→CuSO₄.5H₂O(Blue)

If the organic compound is a volatile liquid or gas, then the vapours of the compound are passed through heated copper oxide and the presence of C02 and H₂O in the liberated gas can be proved in the same way.

If the organic compound contains sulphur in addition to carbon and hydrogen, then sulphur is oxidised to SO₂ which also turns lime water milky due to the formation of calcium sulphite. In that case, the resulting gases are first passed through an acidified solution of potassium dichromate which absorbs SO₂ and then through lime water.

2. Detection of nitrogen

1. Sodalime test

A very small amount of an organic compound is strongly heated with soda lime (NaOH + CaO) In a test tube. The evolution of ammonia having a typical smell Indicates the presence of nitrogen in the compound.

Example:

CH₃CONH₂ (Acetamide)+ [NaOH + CaO]→ CH₃COONa (Sodium acetate)+ NH₃↑

 Sodalime test Limitation:

Organic compounds containing nitrogen as nitro ( — NO₂) or azo (— N =N— ) groups do not evolve NH₃upon heating with soda lime, i.e., do not give this test.

2. Lassaigne’s test

Preparation of Lassaigne’s extract or sodium extract:

A pea-sized freshly cut dry sodium metal is heated gently in a fusion tube till it melts to a shining globule. A small amount ofthe organic compound is added and the tube is heated strongly for 2-3 minutes till it becomes red hot. The hot tube is then plunged into 10-15 mL of distilled water taken in a mortar. The mixture is then ground thoroughly by a pestle and filtered. The filtrate is known as sodium extract or Lassaigne’s extract.

Test for nitrogen:

The Lassaigne extract is usually alkaline in nature because the excess of metallic sodium reacts with water to form sodium hydroxide. A small amount of freshly prepared FeSO₄ solution is added to a part of sodium extract and the contents are boiled when a light green precipitate of Fe(OH)₂ is obtained.

The mixture is then cooled and acidified with dil.H₂SO₄. The immediate appearance of blue or green colouration or deep blue coloured precipitate of Prussian blue, Fe₄[Fe(CN)₆)₃ indicates the presence of nitrogen.

Chemistry of Lassaigne’s test:

1. When organic compound is fused with metallic Na, carbon and nitrogen present in it combine with Na to form NaCN

2. When the sodium extract is heated with FeSO₄ solution, Fe(OH)₂ is obtained which reacts with sodium cyanide to form sodium ferrocyanide. This is also obtained when FeSO₄ reacts with NaCN.

FeSO₄ + 2NaOH →Fe(OH)₂ + Na₂SO₄

6NaCN + Fe(OH)₂ →Na₄ [Fe(CN)₆ ] (Sodium ferrocyanide) + 2NaOH

FeSO₄ + 6NaCN→Na₂ SO₄ + Na₄ [Fe(CN)₆ ]

Ferrous (Fe²⁺) ions present in hot alkaline solution undergo oxidation by O₂ to give ferric (Fe³⁺) ions. These ferric ions then react with sodium ferrocyanide to produce ferric ferrocyanide (prussian blue)

4Fe(OH)₂ + O₂  + 2H₂O→4 Fe(OH)₃

2Fe(OH)₃ + 3H₂ SO₄ →Fe₂(SO₄ )₃ + 6H₂O

3Na₄ [Fe(CN)₆] + 2Fe₂(SO₄)₃→ Fe₄[Fe(CN)₆]₃  (Prussian Blue)+ 6Na₂SO₄

1. In this test, it is desirable not to add FeCl₃ solution because yellow FeCl₃ causes Prussian blue to appear greenish. For the same reason, the alkaline extract should not be acidified with hydrochloric acid (which produces ferric chloride).

2. When the compound under investigation contains both nitrogen and sulphur, it may combine with sodium during fusion to form sodium thiocyanate (sulphocyanide). This gives blood red colouration with ferric chloride due to the formation of ferric thiocyanate. Thus, Prussian blue is not obtained.

⇒ \(\mathrm{Fe}^{3+}+3 \mathrm{NaSCN} \rightarrow \mathrm{Fe}(\mathrm{SCN})_3+3 \mathrm{Na}^{+}\)

3. If fusion is carried out with excess of sodium, the resulting thiocyanate decomposes to give sodium cyanide and sodium sulphide. So, in that case no blood red colouration is visualised.

NaSCN + 2Na→NaCN + Na₂S

However, in that case, black precipitate of FeS is obtained when FeS04 solution is added to sodium extract. The black precipitate dissolves on the addition of dil. H₂SO₄ and the test of nitrogen is then performed with that clear solution.

FeSO₄ + Na₂S→FeS↓ (Black) + Na₂SO₄

FeS + H₂SO₄ →H₂S↑+ FeSQ₄

Lassaigne’s test Limitations:

• Volatile organic compounds if present, escape before reacting when fused with metallic sodium.
• Some nitro compounds may lead to explosion during the fusion process.

3. Detection of sulphur

1. Lassaigne’s test:

Lassaigne’s extract is prepared as described in the case of nitrogen Sulphur present in the compound (which does not contain nitrogen) reacts with sodium metal to form sodium sulphide:

The following tests are then performed with the extract to detect the presence of sulphur

Lead acetate test:

One part ofthe extract is acidified with acetic acid and then a lead acetate solution is added to it. The formation of a black precipitate of lead sulphide (PbS) confirms the presence of sulphur in the compound

Na₂S + Pb(CH₃COO)₂→PbS↓(Black) + 2CH₃COONa

Sodium nitroprusside test:

A few drops of sodium nitroprusside solution are added to another part of the extract. The appearance of a violet or purple colouration confirms the presence of sulphur in the compound.

Na₂S + Na₂[Fe(CN)₅NO](Sodium nitroprusside)→ Sodium sulphonitroprusside Na₄[Fe(CN)₅NOS](Violet)

2. Oxidation test:

The organic compound is fused with a mixture of potassium nitrate and sodium carbonate and as a result, sulphur, if present, gets oxidised to sodium sulphate.

The fused mass is extracted with distilled water and filtered. The filtrate is acidified with dil. HCl and then barium chloride solution is added to it. The formation of a white precipitate insoluble in hydrochloric acid indicates the presence of sulphur in the compound

Na₂SO₄ + BaCl₂→2NaCl + BaSO₄↓ (White)

4. Detection of halogens

1. Beilstein’s test:

A clean and stout Cu-wire flattened at one end is heated in the oxidising flame of a Bunsen burner until it imparts any green or bluish-green colour to the flame. The hot end of the Cu-wire is then touched with the organic compound under investigation and is once again introduced into the flame. The reappearance of green or bluish-green flame due to the formation of volatile copperhalide indicates the presence of halogens in the compound.

Beilstein’s test Limitations:

Many halogen-free compounds,

For example: 

Certain derivatives of pyridine and quinoline, purines, acid amides, urea, thiourea, cyano compounds etc. give this test presumably owing to the formation of volatile copper cyanides.

Therefore, this test is not always trustworthy. It does not indicate which halogen (Cl, Br or I) is present in the organic compound.

The test is not given by fluoro compounds since copper fluoride is non-volatile.

2.  Lassaigne’s test:

Lassaigne’s extract is prepared as described in the case of nitrogen. During fusion, Na combines with the halogen present in the compound to form sodium halide

The extract is then boiled with dilute HNO₃, cooled and a few drops of AgN03 solution are added to it White or yellow precipitation confirms the presence of halogen.

NaX + AgNO₃ → AgX↓ + NaNO₃ [X = Cl,Br,I]

1. Formation of a curdy white precipitate soluble in ammonium hydroxide solution indicates the presence of chlorine in the organic compound

NaCl + AgNO₃→AgCU(White) + NaNO₃

AgCl + 2NH₄OH→[Ag(NH₃)₂]Cl(Soluble) + 2H₂O

The formation of a pale yellow precipitate partially soluble in ammonium hydroxide solution indicates the presence of bromine in the organic compound.

NaBr + AgNO₃→ AgBr₄-(Pale yellow) + NaNO₃

AgBr + 2NH₄OH → [Ag(NH₃)₂]Br(Soluble) + 2H₂O

Formation of a yellow precipitate insoluble in NH4OH solution indicates the presence of iodine in the organic compound.

Nal + AgNO₃→AgI ↓ (Yellow) + NaNO₃

Function Ol nitric acid:

If the organic compound contains nitrogen and sulphur along with halogens, Lassaigne’s extract contains sodium sulphide (Na₂S) and sodium cyanide (NaCN) along with sodium halide (NaX) . These will form a precipitate with silver nitrate solution and hence will interfere with the test.

Na₂S + 2AgNO₃→Ag₂S↓  (Silver sulphide (Black) + 2NaNO₃

NaCN + AgNO₃→ AgCN↓   (Silver cyanide) (white)+ NaNO₃

For this reason, before the addition of AgNO₃ solution to the sodium extract for the detection of halogens, the extract is boiled with dilute HNO₃ which decomposes sodium sulphide and sodium cyanide to vapours of H₂S and HCN respectively

Alternatively, sulphide and cyanide ions can be removed by adding 5% nickel (II) nitrate solution which reÿct? with these ions forming precipitates of nickel (II) sulphide and nickel (II) cyanide

3. Chlorine water test for bromine and iodine:

• A portion of Lassaigne’s extract is boiled with dilute nitric acid or dilute sulphuric acid to decompose NaCN and Na₂S.
• The solution is then cooled, and acidified with dil. H₂SO₄ and a few drops of carbon disulphide or carbon tetrachloride solution are added to it.
• The resulting mixture is then shaken with a few drops of freshly prepared chlorine water and allowed to stand undisturbed for some time.

An orange or brown colouration in the carbon disulphide or carbon tetrachloride layer confirms the presence of bromine, whereas a violet colouration in the layer confirms the presence of iodine in the compound

2NaBr + Cl₂→2NaCl + Br₂ (turns CS₂ or CCl₄ layer orange)

2NaI + Cl₂→2NaCl + I₂ (turns CS₂ or CCl₄ layer violet)

5. Detection of phosphorus

1. The organic compound under investigation is fused with sodium peroxide (an oxidising agent) when phosphorus is oxidised to sodium phosphate

2. The fused mass is then extracted with water, filtered and the filtrate is boiled with concentrated nitric acid. The mixture is then cooled and an excess of ammonium molybdate solution is added to it

3. The appearance of a yellow precipitate or colouration due to the formation of ammonium phosphomolybdate, (NH₄ )₃ PO₄ -12MoO₃, confirms the presence of phosphorus in the compound.

Quantitative Analysis Of Organic Compounds

After detecting the presence of various elements in a particular organic compound, the next step is to determine their respective percentages. This is known as quantitative analysis.

1. Estimation of carbon and hydrogen

Both carbon and hydrogen present in an organic compound are estimated by Liebig’s method.

Liebig’s method

Liebig’s method Principle:

A pure and dry organic compound of known mass is heated strongly with pure and dry copper oxide (CuO) in an atmosphere of air or oxygen-free from carbon dioxide. Both carbon and hydrogen present in the organic compound undergo complete oxidation and get converted into carbon dioxide and water respectively

COthus produced is absorbed in a previously weighed potash bulb containing a strong KOH solution while water produced is absorbed in a previously weighed U-tube containing anhydrous CaCl₂. The U-tube and the bulb are weighed again and from the difference between the two weights, the amount of CO₂ and HaO are determined.

Liebig’s method Procedure:

• The apparatus for the estimation of C and H is The apparatus consists of the following units:
  • Combustion tube, 
  • U-tube containing anhydrous CaCl₂ and
  • a bulb containing strong KOH solution.
• The tube is heated l strongly for 2-3 hours till the whole of the organic compound is burnt up.
• After combustion is over, the absorption units (the Utube and the potash bulb) are disconnected and are weighed separately

Results and calculations:

Let the mass of organic compound taken = w g. The increase in the mass of the potash bulb, i.e., the mass of CO₂ formed = x g.

The increase in the mass of the U-tube, i.e., the mass of water formed =y g.

Percentage ot carbon: 1 mol of CO₂ (44g) contains 1 gram atom of carbon (12 g).

∴ x g of CO₂ contains = \(\frac{12x}{44}\) g of C

Now, \(\frac{12x}{44}\) g carbon is present in w g organic compound.

∴ The percentage of carbon in the compound

= \(\frac{2 y \times 100}{18 w}=\frac{2}{18} \times \frac{\text { Mass of } \mathrm{H}_2 \mathrm{O} \text { formed }}{\text { Mass of the compound taken }} \times 100\)

Percentage ot hydrogen:  1 mol of water (18g) contains 2 gram-atom hydrogen (2 g).

∴  y g of H₂ O contains = \(\frac{2y}{18}\) g of H

Now, \(\frac{2y}{18}\) hydrogen is present in w g organic compound.

∴ The percentage of hydrogen in the compound

Estimation of C and H by Liebig’s method is suitable for organic compounds containing C, H and O only, but it requires some modifications for compounds containing nitrogen, halogens and sulphur.

1. Compounds containing nitrogen:

During combustion, N present in the organic compound undergoes oxidation to give its oxides (NO, NOz etc.) which are

Also absorbed in an alkali solution along with CO₂. Oxides of nitrogen are decomposed back to nitrogen by placing a copper gauze roll near the exit of the tube. The N₂ so produced is not absorbed by the alkali solution.

2. Compounds containing halogens:

During combustion, halogens present in the organic compound get converted into volatile copper halides which partly decompose to give free halogens. These halogens and volatile copper halides get dissolved in an alkali solution. This can be prevented by placing a silver gauze roll near the exit of the combustion tube. Halogens combine with silver to give non-volatile silver halides.

2Ag + X₂ →2AgX; 2Ag + CuX₂ → 2AgX + Cu

3. Compounds containing only sulphur or sulphur and halogen:

During combustion, elemental sulphur present in the organic compound is oxidised to SO₂ which is absorbed in the potash bulb.

This can be prevented by placing a layer of fused lead chromate near the exit of the tube. SO₂ combines with lead chromate to produce non-volatile lead sulphate.

Lead chromate also reacts with copper halides and free halogens to form lead halides which remain in the combustion tube

Liebig’s method Precautions:

• All the joints ofthe combustion must be air-tight.
• The combustion must be free from CO₂ and water vapour.
• The airflow is controlled in such a way that only 2-3 bubbles are generated per second. A faster flow of air may lead to the formation of carbon monoxide.
• If carbon is deposited on the surface of the combustion mbe, then oxygen instead of air is to be passed at the final stage ofthe process

Example 1: 0.90g of an organic compound on complete combustion yields 2.20 g of carbon dioxide and 0.60 g of water. Calculate the percentages of carbon and hydrogen in the compound.
Answer:

Mass of the organic compound = 0.90 g

Mass of CO₂ formed = 2.2 g & mass of H₂O formed = 0.6 g

Percentage of carbon: 44 g CO₂ contains = 12 g carbon

0.90 g organic compound contains = \(\frac{12 \times 2.20}{44}\) g of carbon

0.20 g organic compound contains = \(\frac{12 \times 2.20}{44}\) g carbon

Percentage of carbon = \(\frac{12 \times 2.20 \times 100}{44 \times 0.90}\)

= 66.67

Percentage of hydrogen:  18 g of H₂O contains = 2g of hydrogen

0.60 g of H₂O contains = \(\frac{2 \times 0.60}{18}\) of hydrogen

0.90 g compound contains = \(\frac{2 \times 0.60}{18}\) g hydrogen

Percentage of hydrogen = \(\frac{2 \times 0.60 \times 100}{18 \times 0.90}\)

= 7.41

2. Estimation ot nitrogen

The following two methods are commonly used for the estimation of nitrogen in an organic compound

1. Duma’s method
2. Kjeldahl’s method.

1. Dumas method

Dumas method Principle:

A weighed amount of the organic compound is heated with an excess of copper oxide in an atmosphere of CO₂. Carbon and hydrogen present in the compound are oxidised to CO₂ and H₂O respectively while nitrogen is set free as nitrogen gas (N₂). If any oxide of nitrogen is formed during this process, it is reduced back to nitrogen by passing over hot reduced copper gauze

When the gaseous mixture thus obtained is passed through a 40% KOH solution taken in a Schiff’s nitrometer, all gases except nitrogen are absorbed by KOH. The volume of nitrogen collected over the KOH solution is noted and from this, the percentage of nitrogen in the compound is calculated

Dumas method Procedure:

• The apparatus used for the estimation of nitrogen by the Dumas method ). It consists of:
  • CO₂ generator combustion tube (a glass tube of diameter 15m/r and length 90 cm) and Schiff’s nitrometer. 
  • The combustion tube is packed with [a] coarse CuO which prevents backward diffusion of gases produced,
  • an accurately weighed amount of organic compound of approximately 0.2g mixed with excess of CuO.
  • Coarse CuO and
  • A reduced copper gauze can reduce any oxides of nitrogen back to N₂.
• The Schiff’s nitrometer (connected to the combustion tube) consists of a graduated tube with a reservoir and a tap at the upper end. It has a mercury seal at the bottom to check the backward flow ofthe KOH solution.
• CO₂ generated by heating NaHCO₃ or MgCO₃ is dried by passing through concentrated HS04 and then passed through the combustion tube to displace the air present in the tube.
• The combustion tube is then heated in the furnace. When the combustion is complete, a rapid stream of CO₂ is passed through the tube to sweep away the last traces of N₂.
• The volume of nitrogen collected over the KOH solution in the nitrometer tube is recorded after careful levelling (by making the level of KOH solution both in the tube and reservoir the same).
• The room temperature is recorded and the vapour pressure of the KOH solution at the experimental temperature is noted in the table.

Results and calculations:

Let the mass of the organic compound taken = w g, the volume of N₂ gas collected = V mL, the atmospheric pressure = P mm, the room temperature = t°C and the vapour pressure of KOH solution at t°C =f mm. Hence, the pressure of dry N₂ gas = (P- f) mm.

By applying the gas equation = \(\frac{P_1 V_1}{T_1}=\frac{P_2 V_2}{T_2}\)

Volume of N₂ at STP, = \(V_2=\frac{P_1 V_1 T_2}{P_2 T_1}=\frac{(P-f) \times V \times 273}{760(273+t)}\)

Percentage of nitrogen: 22400 mL of N₂ gas at STP will weigh = 28 g.

2 mL of N₂ gas at STP will weigh = \(\frac{28 \times V_2}{22400} \mathrm{~g}\)

Percentage of nitrogen:

= \(\frac{\text { Mass of nitrogen }}{\text { Mass of the compound }} \times 100\)

= \(\frac{28}{22400} \times \frac{V_2}{w} \times 100\)

The amount of nitrogen estimated by this method is 0.2% higher than the actual quantity. This is because a fraction of the small amount of air (nitrogen) entrapped in copper oxide gets deposited in the nitrometer

Example:

In Dumasg of an method organic for compound, the estimation gave 31.7 of nitrogen, mL of nitrogen at 14°C and 758 mm pressure. Calculate the percentage of nitrogen in the compound (vapour pressure of water at 14°C = 12 mm.
Answer:

If the volume ofnitrogen at STP is V mL, then

V = \(\frac{31.7(758-12)}{(273+14)} \times \frac{273}{760}\)

= 29. 6 mL

Now, at STP 22400 mL of nitrogen will weigh = 28 g

∴ 29.6 mL nitrogen at STP will weigh = \(\) = 0.037%

∴  Percentage of nitrogen = \(=\frac{0.037 \times 100}{0.1877}\)

= 19.71

2. Kjeldahl’s method

Kjeldahl’s method Principle:

A known mass of a nitrogenous organic compound is digested (heated strongly) with concentrated H₂SO₄ in the presence of a little potassium sulphate (to increase the boiling point of H₂SO₄) and a little CuSO₄ (a catalyst) when the nitrogen present in the compound is quantitatively converted into ammonium sulphate. The resulting mixture is heated with excess of NaOH solution and the ammonia evolved is passed into a known but excess volume of a standard acid (HCl or H₂SO₄ ).

The excess acid left after the neutralisation of ammonia is estimated by titration with some standard alkali

2NaOH + H₂SO₄  →Na₂SO₄ + 2H₂O

Kjeldahl’s Method Procedure

•  The organic compound (0.5-5 g) is heated with a cone. H₂SO₄ in a Kjeldahl’s flask which is a long-necked round-bottomed flask with a loose stopper. A small amount of potassium sulphate and a few drops of mercury (or a little CuS ) are added.
• The reaction mixture is heated for 2-3 hours when carbon and hydrogen present in the compound are oxidised to CO₂ and H₂O respectively while nitrogen is quantitatively converted to ammonium sulphate.
• CO₂ and water vapours escape through the loose stopper while ammonium sulphate remains in the flask.
• The Kjeldahl’s flask is then cooled and the contents of the flask are transferred to a round-bottomed flask.
• The mixture is diluted with water and excess caustic soda solution (40%) is added.
• The flask is then connected to a Liebig condenser through Kjeldahl’s tap. The lower end of the condenser is dipped in a known volume of standard acid (N/10 HCl or H₂SO₄ ) taken in a conical flask.
• The flask is then heated. The evolved ammonia gas is absorbed in the acid solution.

Results and calculations:

Let the mass of the organic compound = w g, volume of acid taken = V₁ mL, normality of acid solution x(N) and volume of x(N) alkali required to neutralise unused acid = V₂ mL.

Now, V₂ mL  × (N) alkali = V₂ mL × (N) acid.

∴ The quantity of acid used for neutralising ammonia = VjmL x(N) acid solution- V2 mL x(N) acid solution = (V₁– V₂) mL x(N) acid solution = (V₁– V₂)mL x(N) ammonia solution.

Now, 1000 mL (N) ammonia solution contains 17 g of NH₃ or 14 g of nitrogen.

Therefore, (V₁–  V₂)mL × (N)

Ammonia solution contains = \(\frac{14 \times\left(V_1-V_2\right) x}{1000}\) g of nitrogen

This amount of nitrogen is present in w g compound

%  of nitrogen = \(\mathrm{n}=\frac{14 \times\left(V_1-V_2\right) x \times 100}{1000 \times w}=\frac{1.4\left(V_1-V_2\right) x}{w}\)

= \(\frac{1.4 \times \text { Vol. of acid used } \times \text { Normality of the acid used }}{\text { Mass of the compound taken }}\)

Kjeldahl’s method  Limitations:

• Nitrogen present in pyridine, quinoline, diazo compound, azo compound, and nitro compound, cannot be converted into ammonium salts by this method.
• So this method does not apply to such compounds. Thus although Dumas’s method applies to all compounds, Kjeldahl’s method has limited applications.

The percentage of nitrogen estimated in this method is not correct.

Kjeldahl’s method Utility:

This experiment can be performed quite easily and quickly. So in those cases where correct estimation of nitrogen content is not necessary

For example: Fertiliser, soil, food¬ stuff etc.), this method has its application.

This method is not troublesome and hence the possibility of error can be minimised by repeating the experiment several times.

Example:  0.257 g of a nitrogenous organic compound was analysed by Kjeldahl’s method and ammonia evolved was absorbed in 50 mL of (N/10) H₂SO₄. The unused acid required 23.2 mL of (N/10) NaOH solution for complete neutralisation. Determine the percentage of nitrogen in the compound.
Answer:

Acid taken = 50 mL (N/10) H₂SO₄ solution

= 5 mL(N) H₂SO₄ solution

Alkali required to neutralise the excess acid

= 23.2 mL (N/10) NaOH = 2.32 mL (N) NaOH

Now, 2.32 mL(N) NaOH = 2.32 mL(N) H₂SO₄ solution

The acid required to neutralise the ammonia evolved

= 5 mL (N) H₂SO₄– 2.32 mL (N) H₂SO₄ solution

= 2.68 mL (N) H₂SO₄ s 2.68 mL (N) NH₃

[v KmL(N) H₂SO₄S VmL(N) ammonia]

1000 mL (N) ammonia solution contains 14g nitrogen

So, 2.68mL (N) ammonia solution contains \(\frac{14 \times 2.68}{1000}\) g nitrogen

So, 0.257g compound contains = \(\frac{14 \times 2.68}{1000}\) g nitrogen

Percentage of nitrogen = \(=\frac{14 \times 2.68 \times 100}{1000 \times 0.257}\)

= 14.6

3. Estimation of halogens: Carius method

Halogens Carius method Principle:

An organic compound containing halogen of known mass is heated with fuming nitric acid and a few crystals of silver nitrate. The halogen present in the compound is converted into the corresponding silver halide (AgX). From the mass of the compound taken and that of the silver halide formed, the percentage of halogen in the compound can be calculated.

Carius method Procedure:

• About 5 mL of fuming nitric acid and 2 g of silver nitrate crystals are placed in a hard glass tube (Carius tube) of about 50 cm in length and 2 cm in diameter.
• A small amount of accurately weighed organic compound is taken in a small tube and the tube is placed carefully into the Carius tube in such a way that nitric acid does not enter the tube.
• The Carius tube is then sealed. Now, it is placed in an outer iron jacket and heated in a furnace at 550-560 K for about six hours
• As a result, C, H and S present in the compound are oxidised to CO₂, H₂O and H₂SO₄ respectively. The halogen present in the compound gets converted into silver halide which is precipitated.
• The tube is cooled and when the sealed capillary end is heated in a burner’s flame, a small hole is formed through which the gases escape.
• The capillary end is now cut off and the precipitate of silver halide is filtered, washed, dried and weighed

Results & calculations:

Let, the mass of organic compound taken = w g and the mass of silver halide (AgX) formed—x g. Now, 1 mol of AgX contains 1 gram-atom of X (X = Cl, Br or I ), i.e., (108 + atomic mass of X) g of AgX contain (atomic mass of X)g of X

∴ xg AgX contain = \(\frac{\text { atomic mass of } X}{(108+\text { atomic mass of } X)}\) X JCg of X

This amount of halogen (X) is present in w g compound.

Percentage of halogen in the compound

= \(\frac{\text { atomic mass of } \mathrm{X}}{(108+\text { atomic mass of } \mathrm{X})} \times \frac{x}{w} \times 100\)

1. Percentage of chlorine (atomic mass = 35.5 )

= \(\frac{35.5}{(108+35.5)} \times \frac{x}{w} \times 100\)

= \(\frac{35.5}{143.5} \times \frac{x}{w} \times 100\)

2. Percentage of bromine (atomic mass = 80 )

= \(\frac{80}{(108+80)} \times \frac{x}{w} \times 100\)

= \(\frac{80}{188} \times \frac{x}{w} \times 100\)

3. Percentage of iodine (atomic mass = 127 )

= \(\frac{127}{(108+127)} \times \frac{x}{w} \times 100\)

= \(\frac{127}{235} \times \frac{x}{w} \times 100\)

Example: In the Carius method, 0.256 g of an organic compound containing bromine gave 0.306 g of AgBr. Find out the percentage of bromine in the compound.
Answer:

The mass of the organic compound taken = 0.256 g and the mass of AgBr formed = 0.306 g.

Now, 1 mol of AgBr = 1 gram-atom of Br

or, (108 + 80) g or 188 g of AgBr = 80 gofBr

i.e., 188 g of AgBr contains = 80 g of bromine

0.306 g of Ag Br contain \(\frac{80}{188} \times 0.306\) g of bromine

This amount of bromine is present in 0.256g compound

Percentage6 of bromine = \(\frac{80}{188} \times \frac{0.306}{0.256} \times 100\)

= 50.9

4. Estimation of sulphur: Carius method

Sulphur Carius method Principle:

When an organic compound containing sulphur is heated with fuming nitric acid in a sealed tube (Carius tube), sulphur is quantitatively oxidised to sulphuric acid. It is then precipitated as barium sulphate by adding a barium chloride solution. The precipitate is then filtered, washed, dried and weighed. From the amount of BaSO₄ formed, the percentage of sulphur is calculated.

Resultsandcalculations: Let the mass of organic compound

= w g and the mass of barium sulphate formed = xg

Now, 1 mol of BaSO₄ contains 1 gram-atom of S, i.e.,

(137 + 32 + 64) g or, 233 g BaO₄ contain 32 g sulphur.

∴ x g of BaSO₄ contain \(\frac{32}{233}\) × x g of sulphur

So, this amount of sulphur is present in the w g compound.

∴ Percentage of sulphur = \(\frac{32}{233} \times \frac{x}{w} \times 100\)

Example: In sulphur estimation by the Carius method, 0.79 g of an organic compound gave 1.164 g of barium sulphate. Calculate the percentage of sulphur in the compound.
Answer:

The mass of the organic compound taken = 0.79 g and the mass of BaSO₄ obtained = 1.164 g.

Now, 1 mol of BaSO₄ = 1 gram-atom of S or,

233 g of BaSO₄= 32 g of i.e.,

233 g of BaSO₄ contain = 32 g ofsulphur.

1.164 g of BaSO₄ contain \(\frac{32}{233} \times 1.164\) × 1.164 g of sulphur

So, 0.79 g compound contain \(\frac{32}{233} \times 1.164\) × 1.164 g of sulphur

Percentage of sulphur =  \(\frac{32}{233} \times \frac{1.164}{0.79} \times 100\)

= 20.24

5. Estimation of phosphorus: Carius method

Phosphorus Carius method principle:

An organic compound (containing P) of known mass is heated with fuming nitric acid in a sealed tube (Carius tube). Phosphorus present in the compound is oxidised to phosphoric acid which is precipitated as ammonium phosphomolybdate by heating it with concentrated nitric acid and then adding ammonium molybdate.

Ammonium phosphomolybdate (Yellow)

The precipitate of ammonium phosphomolybdate thus obtained is filtered, washed, dried and weighed.

Alternative method:

The phosphoric acid is precipitated as magnesium ammonium phosphate (MgNH₄PO₄) by the addition of magnesia mixture (a solution containing MgCl₂, NH₄Cl and NH₄OH ).

The precipitate is filtered, washed, dried and then ignited to give magnesium pyrophosphate (Mg₂P₂O)

From the mass of magnesium pyrophosphate, the percentage of phosphorus in the compound can be easily calculated.

Results and calculations:

Let the mass of the organic compound taken = wg and the mass of ammonium phosphomolybdate formed = xg.

1 mol (NH₄)₃PO₄-12MoO₃ contains 1 gram-atom of P.

or, 3(14 + 4) + 31 + 4 × 16 + 12(96 + 3 × 16) = 1877g of

(NH₄)₃PO₄. 12MOO₃ = 31 g of P, i.e., 1877 g of

(NH₄)₃PO₄ . 12MoO₃ contain 31 g of phosphorus

xg of (NH₄)₃PO₄ -12MoO₃ contain \(\)

xg of phosphorus. This amount of phosphorus is present in w g of the compound.

Percentage of phosphorus = \(\frac{31}{1877} \times \frac{x}{w} \times 100\)

= \(\frac{31}{1877} \times \frac{\text { Mass of ammonium phosphomolybdate }}{\text { Mass of the compound taken }} \times 100\)

Alternative calculation: Let the mass of Mg2P20? formed

= xg. Now, 1 mol of (Mg₂P₂O) contains 2 gram-atom of P

or, (24 × 2 + 31 × 2 + 16 × 7) = 222 g of (Mg₂P₂O)

i.e., 222g of (Mg₂P₂O) contains 62 g of phosphorus.

x g of (Mg₂P₂O)contain \(\frac{62}{222} \times x\) g of phosphorus

w g organic compound contain \(\frac{62}{222} \times x\) xg phosphorus

Percentage of phosphorus = \(\frac{62}{222} \times \frac{x}{w} \times 100\)

= \(\frac{62}{222} \times \frac{\text { Mass of } \mathrm{Mg}_2 \mathrm{P}_2 \mathrm{O}_7 \text { formed }}{\text { Mass of the compound taken }} \times 100\)

Example: 0.35 g of an organic compound containing phosphorus gave 0.65 g magnesium pyrophosphate, Mg₂P₂O in the Carius method. Calculate the percentage of phosphorus in the given compound.
Answer:

The mass of the organic compound taken = 0.35 g and the mass of Mg₂P₂O formed = 0.65 g.

Now, 1 mol of Mg₂P₂O contains 2 gram-atom of P.

Or, (2 ×  24 + 2 × 31 + 16 × 7) g of Mg₂P₂O= 2 × 31 g of P

i.e., 222 g of Mg₂P₂O contains 62 g of phosphorus

0.65 g Mg₂P₂Ocontain \(\frac{62}{222} \times 0.65\) x 0.65 g of phosphorus

This amount of P is present in a 0.35 g compound.

Percentage of phosphorus = \(\frac{62}{222} \times \frac{0.65}{0.35} \times 100\)

6. Estimation of oxygen

Percentage of oxygen = 100 – (sum of the percentages of all other elements). However, the oxygen content of a compound can be estimated directly as follows:

Oxygen Principle:

An organic compound of known mass is decomposed by heating in a stream of nitrogen gas. The mixture of the gaseous products including oxygen is passed over red-hot coke when all the oxygen combines with carbon to form carbon monoxide. The mixture is then passed through warm iodine pentoxide (I₂O₅ ) when CO undergoes oxidation to CO₂ and iodine is liberated.

From the knowledge of the mass of iodine liberated or C02 produced, the percentage of oxygen in the compound can be calculated easily

Results and calculations: Let, the mass of the organic compound taken = wg and the mass of CO₂ formed =xg. We find that each mole of oxygen liberated from the compound will produce 2 mol of CO₂

x g. of CO₂ is obtained from \(\frac{32}{88} \times x=\frac{16}{44} \times\) = x x g of O₂

This amount of oxygen is present in w g ofthe compound.

Percentage of oxygen = \(\frac{16}{44} \times \frac{x}{w} \times 100\)

1. Percentage of carbon (Liebig’s method)

= \(\frac{12}{44} \times \frac{\text { Mass of } \mathrm{CO}_2 \text { formed }}{\text { Mass of the compound taken }} \times 100\)

2. Percentage of carbon (Liebig’s method)

= \(\frac{2}{18} \times \frac{\text { Mass of } \mathrm{H}_2 \mathrm{O} \text { formed }}{\text { Mass of the compound taken }} \times 100\)

3. Percentage of nitrogen (Dumas method)

= \(\frac{28}{22400} \times \frac{\text { Vol. of } \mathrm{N}_2 \text { gas at STP }}{\text { Mass of the compound taken }} \times 100\)

4. Percentage of nitrogen (KjeldahTs method)

= \(\frac{28}{22400} \times \frac{\text { Vol. of } \mathrm{N}_2 \text { gas at STP }}{\text { Mass of the compound taken }} \times 100\)

5. Percentage of halogen (Carius method)

= \(\frac{\text { At. mass of } X}{(108+\text { At. mass of } X)} \times \frac{\text { Mass of AgX }}{\text { Mass of the compound taken }} \times 100\)

X = Cl(35.5) , Br(80) or 1(127)

6. Percentage of sulphur (Carius method)

= \(\frac{32}{233} \times \frac{\text { Mass of } \mathrm{BaSO}_4 \text { formed }}{\text { Mass of the compound taken }} \times 100\)

7. Percentage of phosphorus (Carius method)

\(\frac{62}{222} \times \frac{\text { Mass of } \mathrm{Mg}_2 \mathrm{P}_2 \mathrm{O}_7 \text { formed }}{\text { Mass of the compound taken }} \times 100\)

8. Percentage of oxygen

= \(\frac{16}{44} \times \frac{\text { Mass of } \mathrm{CO}_2 \text { formed }}{\text { Mass of compound taken }} \times 100\).

Class 11 Chemistry Environmental Chemistry Introduction

The word ‘environment’ derived from the French word ‘environment can be defined at an assembly of physical, chemical and biological factors, which act upon an organism or an ecological community to determine its form and mode of survival.

Environment mainly consists of three major components:

1. Biotic or living: 
   • For example – All living creatures)
2. Abiotic or non-living:
   • For example –  Lithosphere, hydrosphere and atmosphere) and
3. Energy components: 
   • For example – Solar energy, thermochemical energy, nuclear energy etc.)

Environmental Chemistry

Environmental Chemistry Definition

The branch of science, which deals with the sources of the chemical components of the environment, the chemical reaction occurring among them, the products formed in the reaction and their impact on the living world is called environmental chemistry.

The study of environmental chemistry is important because

• It makes us aware of the adverse effects of various chemical constituents on the environment.
• It gives us an idea about the toxic effects of various chemical substances and their by-products which are extensively used to fulfill our requirements.
• It gives us an idea about the sources of various toxic chemicals, their adverse effects and the antidotes to combat their toxicity.

1. Some terms used in environmental chemistry

1. Pollutant:

A pollutant is a solid, liquid or gaseous substance (produced either by natural sources or by human activity) which is present in the environment to such an extent that it causes harmful or detrimental effects on living organisms (plants, animals and human beings) or nonliving components.

Example:

Air contains trace amounts of CO (0.1 ppm). If, for any reason, the amount of CO increases to 40 ppm or more, then it is regarded as pollutant

Pollutants are of two kinds :

1. Primary pollutant:  The pollutants which are emitted from any source, directly escape into the environment without sustaining any change are called primary pollutants.
2. Example: S, NO, NO₂, CO, CO₂, hydrocarbons etc.
3. Secondary pollutant: There are some pollutants which do not appear in the environment directly from their source. They are produced as harmful substances by Interaction with pollutants), already present in the environment. Tills type of pollutants are called secondary pollutants.
4. Example: Peroxyacyl nitrate (PAN), dimethyl mercury [(CH₃)₂Hg]

2. Biodegradable pollutants

The pollutants which are decomposed by bacteria or germs are known as biodegradable pollutants.

Examples: Household garbage, cow dung and other biomass etc.

3. Non-biodegradable pollutants:

The pollutants which are not decomposed by bacteria or germs or decomposed very slowly are known as non-biode gradable pollutants.

Example:

Mercury, DDT, Gammaxene etc. The presence of these substances even in trace amounts is injurious to human beings and other animals.

4. Contaminant:

A contaminant is a substance which does not occur in nature under normal conditions but is introduced into the environment either accidentally or through indiscriminate human use.It may or may not be harmful to the living organisms or non-living components. The contaminant is considered as a pollutant when it has some harmful effect

Examples:

• In Delhi pyrosulphuric acid (H₂S₂O₇) leakage from a defective tank killed many persons and caused skin and breathing problems to many others. As pyrosulphuric acid does not occur in the atmosphere, therefore it is a contaminant. Again, because of its dangerous effect, it is also regarded as a pollutant.
•  In Kerala, in 1953, 108 people died after consuming wheat flour contaminated with parathion (an agricultural pesticide).

5. Source:

The source ofany contaminantis a chemical substance or the place from whereitis produced.

Example:

A source of the pollutants like CO, NO etc., is the gas 2 emitted from petrol or diesel automobile engines

6. Sink:

If any medium continuously reacts with a pollutant for a long period and causes destruction to it, then it is said to be the sink of that particular pollutant.

Examples:

1.  Sea water acts as a sink of CO₂ present in the atmosphere

⇒ H₂O (sink) +CO₂ (Pollutant ) H₂CO₃

2. An automobile wall or memorial acts as a sink of sulfuric acid, present as an atmospheric pollutant.

⇒ (Sink )CaCO₃+ (polluntant)H₂SO₄→CaSO₂↓ + H₂O + C₂O

7. Receptor or target:

If any plant or animal body or any biotic component is adversely affected by a pollutant, then that particular body or component is called a receptor or the target of that pollutant.

Examples:

•  Smoke or smog causes a burning sensation in our respirator tract and eyes.In this case, man is the receptor or target of smoke or smog.
• The aquatic animals are the receptors of the oil layer, floating on the surface of seawater.

8. Pathways Of pollutants:

The mechanism by which any pollutant is liberated from its source, spreads in the environment and ultimately gets destroyed, is called the pathway of that pollutant.

Example:

Nitric oxide, a pollutant, liberated from petrol or diesel engine reacts with oxygen present in the air to form nitrogen dioxide. Nitrogen dioxide reacts with rainwater and falls on the surface of the earth as nitric acid.

2NO + O₂ → 2NO₂; 2HNO₂+O₂→2HNO₃

2NO₂ + H₂O→HNO₃ + HNO₂

9. Threshold Limit Value (TLV):

Anypollutantin in the environment is considered to be harmful to living organisms if its concentration exceeds a particular limit. This particular limit (of concentration) is called ‘threshold limit value’ (TLV) ofthat particular pollutant.  Atmosphere

Examples: Threshold limit values for CO and CO₂ are. 40 ppm and 5000 ppm respectively. Ho However, the TLV for phosgene (COCl₂) is only 0.1 ppm.

Threshold limit value for factory workers: It is the maximum limit of a pollutant present in the environment maximum limit of a pollutant present in the environment hours per day without suffering from health hazards Examples: Threshold limit values for SO₂ and CO₂ are 5 mg^-3. m^-3 and 5000 mg m^-3 respectively

Air Pollution

When one or more undesirable chemical substances produced by natural phenomena or uncontrolled human activity get mixed with the air to bring about health hazards to human or any other living being and affect their life processes, then air pollution is said to have occurred. The substances which cause air pollution are called air pollutants

Examples:

Sulphur dioxide (SO₂), carbon monoxide (CO), carbon dioxide (CO₂), nitric oxide (NO), nitrogen dioxide (NO₂), hydrogen sulphide (H₂S), hydrocarbons (C_xH_y), ammonia (NH₃), fly ash, dust particles, smoke, fumes etc., are some air pollutants

Causes of air pollution

Natural causes of air pollution:

• Emission of SO₂, CO, H₂S gases etc., due to volcanic eruption.
• Gases liberated due to the decomposition of living bodies.
• Dust storm, forest fires and the fall of a meteorite.
• Spreading of virus, bacteria, fungus, pollen-grain of flowers in air etc

Causes of air pollution due to human activities:

• Sulphur dioxide gas, carbon monoxide gas and fly ash produced from thermal power plant.
• Gases such as sulphur dioxide, sulphur trioxide, oxides of nitrogen, hydrogen chloride, chlorine etc., from acid manufacturing factories.
• Ammonia gas liberated from fertiliser factories and cold storages.
• Sulphur dioxide, carbon monoxide and different metallic oxides obtained during extraction of metals.
• Fine and bulky solid particles produced in cement and asbestos factories.
• Carbon monoxide, nitrogen dioxide, sulphur dioxide and other such gaseous substances evolved from petroleum refineries.
• Dust and sand in the region of coal mines.
• CO, SO₂ and oxides of nitrogen released from motor vehicles.
• Extensive deforestation affects the balance of oxygen and carbon dioxide and increases the quantity of carbon dioxide.
• Highly poisonous gas evolved from destructive weapons used in warfare.0 Emission of radioactive rays due to accidents in nuclear reactors, nuclear power plants etc.
• The extensive use offossil fuel resultsin evolution of gaseous pollutants.

Class 11 Physics	Class 12 Maths	Class 11 Chemistry
NEET Foundation	Class 12 Physics	NEET Physics

Major Air Pollutants

Major air pollutants are divided into two classes:

1. Inorganic and organic gaseous substances and
2. Particulates.

Inorganic and organic gaseous substances

1. Carbon monoxide (CO):

Natural sources:

• During volcanic eruptions,
• Due to decomposition of dead plants and animals in marshy land,
• During forest fires,
• During the extraction of petroleum and natural gases,
• During lightning, carbon monoxide is produced.

Man-made sources:

• Incomplete combustion of fuel in automobile engines,
• Incomplete combustion of carbonaceous matter in industrial furnaces,
• Incomplete combustion of agriculture and slash materials,
• Heating of carbon dioxide at a high temperature in the presence of coke in the blast furnace produces carbon monoxide. Sulphur dioxide (SO₂)

Harmful effects:

This colourless, odourless gas is very harmful

Natural sources:

For human beings and animals. It has a greater affinity towards haemoglobin than of oxygen. So, it readily combines with haemoglobin to give a stable compound, carboxyhaemoglobin

(HbO₂ + CO ⇌ HbCO + O₂ ).

• Consequently, oxygen carrying capacity of haemoglobin decreases which ultimately results in death.
• The presence of CO even in small amounts may result in nausea, headache, dizziness etc.
• Cigarette smoke contains a considerable amount of CO which ultimately increases CO content in the smoker’s blood.
• Thus, chain smokers often suffer from mental imbalance, headaches, heart attacks etc. Acute oxygen starvation in the body due to poisoning by CO is called anoxia or asphyxiation.

Sink of carbon monoxide:

• Some bacteria present in the soil absorb CO and convert it to CO₂.
• Inorganic radicals like HO and HOO, atomic oxygen and ozone oxidise CO to CO₂

Control of CO pollution: 

1. One of the main sources of CO pollution is the use of internal combustion engines in automobiles. These engines emit a mixture of CO, NO_x, hydrocarbons and particulates.
   • In order to control the CO emission suitable modifications have to be done in internal combustion engines
   • The carburettor is to be adjusted so as to give a proper air fuel ratio
   • Catalytic converters can be fitted into the exhaust pipe which may bring about oxidation of CO to CO₂ .
   • Internal combustion engines should be modified and improved. Some automobile makers have improved automobile engines by the use of an extra combustion chamber so that fuel undergoes complete combustion and the exhaust gas does not contain CO.
2. CNG, LNG, LPG and LHG (liquefied hydrogen gas) should be used instead of gasoline.
3. Alternative power sources such as fuel cells, solar energy etc., be used.

2. Oxides of sulphur (SO_x):

Two oxides of sulphur, sulphur dioxide (SO₂) and sulphur trioxide (SO₃) are represented by the general formula SO_x. Both of them are colourless gases having suffocating odour.

Sulphur Dioxide (SO₂)

Sulphur Dioxide (SO₂)

Natural sources of  SO₂:

SO₂ gas is liberated in the atmosphere due to volcanic eruptions. 67% of the total SO₂ content of the atmosphere is contributed from this source.

Sources created by human activities:

1. Combustion of coal in thermal plants produces SO₂ gas in abundant quantities.
2. SO₂ gas is evolved during roasting of sulphide ores
3. For example:  FeS, CuFeS₂, Cu₂S, ZnS, PbS etc.) in the extraction of metals.
4. Smoke emitted from oil refineries and automobile engines contains a large quantity of SO₂

Harmful effects of SO₂ :

SO₂ is a gas having extremely pungent and suffocating odour. It adversely affects the respiratory system and causes a burning sensation to eyes.

• If inhaled, it may lead to breathing problems, asthma and bronchitis.
• The presence of excess SO₂ in the atmosphere may result in lung cancer.
• SO₂ also causes allergies.
• In the presence of metallic oxide or other catalysts,  SO₂ reacts with oxygen present in the air and water vapour to form H₂SO₄.
• This sulphuric acid is much more harmful since it corrodes building materials such as marble or limestone, metals like iron, steel, aluminium etc.

The presence of SO₂ hampers the production of chlorophyll and the leaves turn colourless. This is known as chlorosis.

Control of SO₂ pollution:

• Removal of sulphur from fuel i.e., desulphurisation of fuel by using chemical scrubbers
• Use of fuel containing less amount of sulphur
• For example: Natural gas.
• Removal of SO₂ from gaseous fuels.
• Production of electricity in nuclear power plants instead of thermal powerplants

Sulphur Trioxide (SO₃)

1. Sources of SO₃

A significant amount of SO₂, emitted from natural sources or man-made processes is oxidised to sulphur trioxide (SO₃)

1. SO₃ molecules get activated by absorbing radiations having a wavelength of 300-400 run and react with aerial

SO₂ is also oxidised to SO₃ by ozone, hydrogen peroxide and oxides of nitrogen (NO_x) present in the atmosphere

 Harmful effects of SO₃:

SO₃ reacts with water vapour present in the atmosphere to yield sulphuric acid (H₂SO₄). Produced H₂SO₄ combines with water droplets present in air to form an aerosol and ultimately come down as acid rain.

Historical buildings For example:

Tajmahal, Victoria Memorial, monuments, sculptures etc., made up of limestone, marble etc., are adversely affected by

Acid rain*: CaCO₃+ H₂SO₄→ CaSO₄+H₂O + CO₂

Apart from these, metallic (aluminium and iron) structures, bridges, etc., are severely damaged by acid rain. Acid rain increases the acidity(i.e., decreases the pH level of ponds, lakes etc., and in consequence, aquatic plants and animals get dangerously affected.

Acid rain also decreases the stability and glossiness of paint and varnishes.

Sink of SO_x:

Edifices, historical memorials, monuments, and sculptures made of marble, limestone and natural sources of ammonia act as a sink of SO_x.

2. Oxides of nitrogen (NO_x)

Nitrogen forms five oxides—N₂O, NO,N₂O₃, NO₂ and N₂O₅. Out of these, NO and NO₂ act as chief air pollutants. NO and NO₂ are designated

Natural sources of NO_x:

1. During lightning discharge nitrogen and oxygen combine together to form nitric oxide (NO) and this nitric oxide on reaction with excess oxygen, produces nitrogen dioxide (NO₂ ).

N₂ + O₂ → 2NO ; 2NO + O₂ →2NO₂

2. Decomposition of ammonium salts in the soil by some bacteria leads to the formation of oxides of nitrogen (NOx), mainly NO.

Sources of NO_x created by human activities:

In thermal power plants and different industries, atmospheric N₂ and O₂ combine to produce large amounts of nitric oxide (NO). This NO reacts with aerial oxygen to give NO₂.

• The high temperature produced during the burning of fuels in petrol and diesel engines also favours the formation of NO by mutual interaction of atmospheric N₂ and O₂
• NO thus formed is subsequently converted to NO₂ by aerial O₂
• An abundant quantity of NO_x gas escapes into the atmosphere from industries producing organic acid.
• Atomic explosions also add to NO_x in the air.

Harmful effects of NO_x:

1. NO₂ gas is relatively toxic but its adverse effect depends on its concentration in the air and the extent of the reaction.

• Higher concentration of NO₂ in the air produces diseases like inflammation of the lungs, bronchitis, pneumonia etc.
• If NO₂ of higher concentration is inhaled, pulmonary oedema and haemorrhage of the lungs may occur.

2. The most harmful effect of NO_x in the atmosphere is the formation of photochemical smog.

• In the presence of sunlight, hydrogen, nitrogen dioxide and oxygen in the atmosphere, on reaction, gives a mixture of ozone, peroxyacyl nitrate (PAN) and aldehyde.
• All these substances unitedly form photochemical smog. Photochemical smog irritates the eyes and nose and induces sneezing, cough and difficulty in breathing (dyspnoea). If photochemical smog persists for a long period, it may cause death.

3. NO_x drastically reduces the rate of the photosynthesis of plants. Leaves and fruits start shedding due to the presence of NO_x in the air.

4. NO and NO₂ present in the atmosphere react with ozone to form NO₂ and NO₃ respectively. From the latter (z’.e., NO₂ and NO₃), N₂O₅ is obtained. N₂O₅ thus produced, on reaction with rainwater, forms nitric acid

NO +O₃→NO₂+ O₂ ; NO₂+ O₃→NO₃ + O₂

NO₂ + NO₃→N₂O₅; N₂O₅+ H₂O→2HNO

Sink of NOx:

In the atmosphere, the maximum stability of NO and NO₂ are of four days and three days respectively. After that, they are converted to nitric acid (HNO₃). This transformation may occur in two ways following path 1 and path 2 . Ozone plays a major role in the transformation occurred by Path 2

Nitric acid thus produced comes down during the rain. A part of it falls on waterbodies

For example: Ponds, rivers, lakes etc

The remaining part, on reaction with different basic Compounds belonging to hydrocarbon series (organic substances {e.g., ammonia, lime etc.) present in the soil, Is pollutants): The main air-polluting hydrocarbons are converted into nitrate salts. Here, water and the different basic materials in the soil work as the sink of NO_x.

Control of pollution caused by NO_x:

When the gas released from automobile engines is passed through a catalytic converter in the presence of Pt-catalyst, oxides of nitrogen (NO_x) are reduced to produce mainly N₂ and a small amount of NO₃.

Production of NO_x may be reduced by conducting the combustion at a lower temperature in the presence of excess air.

Before releasing the gas mixture containing NO_x, produced in the factories, it is passed over the metallic oxide catalyst

For example:  Cr₂O₃, ZnO, CuO etc.)

Heated to 500°C, thus nitric oxide (NO) is reduced to N₂ and O₂: 2NO ⇌ N₂ + O₂.

In the chemical absorption process, NO_x can be removed from the waste gas mixture. In this case,

1. Acidic For example H₂SO₄
2. Basic substances For example: Ca(OH)₂ or Mg(OH)₂ are used

3. Hydrogen sulphide (H₂S)

Natural sources:

• H₂S gas is liberated from volcanoes.
• Due to the decomposition of proteinaceous compounds containing sulphur, H₂S has evolved. Thus, rotten fish or eggs smell like H₂S gas.

Sources created by human activities:

• In oil refineries, during the production of paper and paints (containing sulphur) appreciable amount of H₂S is produced. Harmful effects of H₂S .’H₂S is a poisonous gas.
• It has severe harmful effects on man. It causes headache, nausea, irritation to the eyes, throat and nose, loss of appetite and diarrhoea.
• When taken in higher doses, it may lead to respiratory problems like bronchial pneumonia or even death.
• The reaction of H₂S with essential proteins is the primary cause of its toxic effects on human bodies. H₂S binds with iron in the mitochondrial cytochrome enzymes, thus preventing cellular respiration.

Sink of H₂S : H₂S undergoes slow oxidation to SO₂ in the atmosphere

2H₂ S + 3O₂→  2SO₂ + 2H₂O

H₂S + O₃ → SO₂ + H₂O

It also combines with various metallic salts to form insoluble metallic sulphides

4. Compounds belonging to the hydrocarbon series (organic pollutants)

The main air-polluting hydrocarbons are methane, benzene etc. Beiides thiete,
acetylene, ethylene, propylene, 1,3 butadiene etc., act as air pollutants. Among the gaseous air pollutants, the one which Is present In maximum quantity In air Is methane

Natural sources:

• In paddy fields and other muddy marshy lands, bacterial decomposition of plants, almost In the absence of air, produces plenty of methane.
• Putrefaction of animal excreta produces an abundant quantity of methane.
• The anaerobic decomposition of animal bodies produces methane.
• Forest fires and the evaporation of hydrocarbons of the terpenoid class from plants In forests are also important sources of hydrocarbon.

 Sources created by human activities:

• Hydrocarbons are produced due to the incomplete combustion or evaporation of the liquid fuels used in vehicles.
• Hydrocarbons also escape into the atmosphere as a result of incomplete combustion of fuels used for different purposes

For example: Cooking, lighting etc

• Liquid substances such as benzene, toluene etc., used as solvents in different chemical industrial units get easily evaporated and pollute the atmosphere
• Gaseous hydrocarbon, 1,3-butadiene used in the preparation of rubber and other polymers causes air pollution.
• During destructive distillation of coal, some air pollutants [e.g., benzopyrene) escape into the atmosphere.

Harmful effects of hydrocarbons: 

• Methane is a greenhouse gas.
• Due to photochemical reactions with oxygen and oxides of nitrogen, hydrocarbons form photochemical oxidants and photochemical smog, which are responsible for irritating eyes, nose and lungs and also cause breathing problems.
• Polycyclic aromatic hydrocarbons (PAH)
  • For example: Benzopyrene are carcinogenic and its presence in the atmosphere in large amounts may cause cancer.

Sink of hydrocarbon:

Hydrocarbons are sufficiently stable. However, they undergo slow oxidation or photochemical reactions involving several steps and ultimately form products like CO₂ or other water-soluble compounds which are then washed away by rain.

Control of pollution caused by hydrocarbons:

Since automobiles are the main sources of hydrocarbon pollution, so such pollution can be controlled by following those steps which are taken to control CO pollution.

Particulates

Generally, finely divided solid and liquid particles, suspended in air are referred to as particulates. The diameter of the particulates varies from 0.0002μ to 500μ (1 micron or 1 μ= 10^-4cm).

The particles of this dimension, being dispersed in air, form aerosol. Depending on the diameter, density of the particles and the intensity of air current, these particles exist in the atmosphere for a period ranging from a few seconds to a few months.

Particulates can be of two types:

• Viable (living microorganisms such as bacteria, viruses, fungi etc.) and
• Non-viable (non-living matters such as mist, smoke, dust etc.)

1. Suspended particulate matter

Smoke, soot, dust particles, metallic oxides and chlorides, fly ash, asbestos dust, acid mist(H₂SO₄, HNO₃)etc.

Natural sources of particulates:

Small particles on the surface of the earth are scattered into the atmosphere by air currents, cyclones, volcanic eruptions etc.

Sources created human activities:

• Soot:  These are fine carbon particles formed by incomplete combustion of carbonaceous fossil fuels.
• Metal particles: These are generated in the metal extraction involving processes like grinding, calcination, smelting of ores etc.
• Particles of metal oxides: During the combustion of fuels containing metal oxides, fine particles of metal oxides are generated.
• For example: Coke containing FeS₂, on combustion, produces fine particles of Fe₃O₄.

3FeS₂ + 8 O₂ → Fe₃O₄+ 6SO₂

• PbCI₂ and PbBr₂:  Compounds like tetraethyl lead (TEL), dichloroethane, dibromoethane etc., are used in petrol as anti-knocking agents. The Pb-compound during combustion of petrol produces PbO which later turns into volatile PbCl₂ and PbBr₂. These further escape into the air and exist as small particles.
• Inorganic silicates: Fine particles of silicates are scattered into the atmosphere from cement industries.
• Asbestos particles: Asbestos is a fibre-like silicate mineral. During the formation of the asbestos sheet, fine particles of asbestos arÿÿatteredinto the atmosphere.
• Sulphuric and nitric acid mist: The SO_x and NO_x  vapours present in the atmosphere react with water vapour and form their corresponding acid mists
• Particles of metal oxides: During the combustion of fuels containing metal oxides, fine particles of metal oxides are generated.
  • For example: Coke containing FeS₂, on combustion, produces fine particles of Fe₃O₄
• Organic particulates: These are mainly the particles of alkane, alkene and aromatic hydrocarbons
  • For example: PAH.
• The formation of such particles is associated with the combustion of petrol and petroleum refining. PAH particles get attached to the soots floating in the air easily and create severe health hazards.

2. Harmful effects Of particulates

The harmful effects of particulates depend on the particle size and the nature of the substances from which these particles originate. Generally, finer particles are more harmful. This is because, the particles with a diameter greater than 5μ get trapped in the nostril but particles having a diameter less than 5μ enter the lungs through the nostril.

The magnitude of the surface area of extremely fine particles being exceedingly high, carcinogenic particulates

For example:

Polycyclic aromatic hydrocarbons, asbestos etc.) can easily find their shelter on these particles and cause cancer, asthma, tuberculosis and different lung diseases. Apart from this, finely divided suspended particles in the air, enter the body through the eyes, ears, nose etc.,

Leads to different types of diseases:

• Smoke released from automobiles contains lead particles which adversely affect the child’s brains and cause nerve diseases.
• The normal production and development of red blood corpuscles (R.B.C.) are disturbed in the presence of lead particles. If an excess amount of lead particles are inhaled regularly, haemoglobin disintegrates and is eliminated through urine
•  Accumulation of cadmium particles in the body during respiration even in trace amounts, may cause breathing trouble and heart disease.
• Beryllium compounds
  • For example: BeCl₂, BeSO, ) affect the lung and create berylliosis disease.
• Workers of coal mines and cotton mills are prone to be attacked by diseases like black lung and white lung.
•  Workers of cement factories are susceptible to five attacks of silicosis due to the intake of SiO₂ particles during respiration.
• Inhalation of asbestos particles results in asbestosis and it leads to cancer in the case of workers of asbestos factories.
• The cause of arsenicosis is due to inhalation of arsenic compounds. In this disease, the skin becomes rough and wounds appear on the skin.

The harmful effects of particulates also extend to the plants present in the atmosphere react with the Water kingdom. Accumulation of dust and other particles on the vapours and their corresponding acid mists. leaves of plants close to the stomata. As a result, transpiration as well as photosynthesis of plants get severely affected. Naturally, the growth of plants and the production of crops also get depreciated

Ozone Layer

In the lower region of the stratosphere (the region at the height of 15 km to 35 km from the ground), there exists a layer of ozone gas. This layer is known as the ozone layer or ozonosphere.

Almost 90% of total ozone gas exists in this layer. In the absence of the ozonosphere, the existence of living beings would have been at stake. It is this ozone layer in the stratosphere which absorbs the harmful ultraviolet radiations coming from the sun. Thus it prevents most of the ultraviolet rays from reaching the Earth’s surface

1. Formation of the ozone layer

Oxygen molecules (O₂) present in the stratosphere absorb ultraviolet radiations coming from the sun and decompose to form oxygen atoms (O). This atomic oxygen combines with oxygen molecules to form ozone molecule

O*₃ + M → O₃ + *M [* Asterisk sign indicates excited state]

M  denotes a neutral colliding species

For example:

O₂ or N₂ with which O₂  on collision, releases its excess energy] On the other hand, the ozone molecule ( O₃  ) also absorbs UVradiation and gets converted into oxygen molecule ( O₂ )

In the ozone layer of the stratosphere, these two opposite processes (i.e., the formation of ozone molecule and the decomposition of ozone molecules) occur in a cyclic order and finally attain a state of equilibrium. Due to the existence of such an equilibrium state, the quantity of ozone in the stratosphere remains fixed.

2. Role of ozone layer in the environment

The ozone sphere works as a protective layer for the fitting world. Acting as a protective umbrella, the ozone layer absorbs most of the harmful UV radiation emitted by the sun because these rays are utilised in the production and decomposition of ozone.

In the absence of the ozone layer, the earth’s surface and the adjacent air would be so heated by the UV-radiation that the existence of the plants and the animals including bacteria in land and water would have been impossible.’

Ultraviolet radiation:

• Causes cancer in human skin
• Damages the cornea of the eyes and develops premature cataracts
• Decreases immunity against diseases and fertility in living beings. Genetic diseases are also the consequence of the harmful effects of ultraviolet radiation

3. Depletion of ozone layer: Ozone hole

In 1982, the British scientist Jo Foreman first observed that the ozone layer of the stratosphere above the Antarctic (the south polar regions) was gradually getting thinner. Extensive research in the subsequent years has revealed that the ozone layer is gradually becoming thinner not only in Antarctica but almost everywhere in the stratosphere. The phenomenon of thinning of the ozone layer of the stratosphere is known as depletion of the ozone layer or the ozone hole.

Depletion of the ozone layer signifies that the equilibrium between! the two contradictory’ Processes i.e., the formation and the decomposition of ozone has been disturbed somehow. Due to the influx of several foreign substances into the atmosphere, the rate of decomposition of ozone has far exceeded than the rate of its formation. This has resulted in the thinning of the ozone layer almost every where in the stratosphere

Ozone layer before formation of hole:

Ozone layer after formation of hole:

Causes of depletion of ozone layer (formation of ozone hole):

1. Scientists believe that some chemical substances belonging to the class of chlorofluorocarbons (Freons or CFCs) are mainly responsible for the formation of ozone hole. Chlorofluorocarbons (CFCs) are the different chloro and fluoro derivatives of methane and ethane.

Some examples of CFCs are given below:

The above-mentioned chemicals were extensively used as

• Refrigerants
• Propellants in aerosols,
• Foaming agent in plastic production,
• Ingredients of fire extinguisher,
• Solvents for various purposes etc. In the troposphere, these gases are non-corrosive, non-toxic, non-inflammable and chemically inert.

As a result, these gases after being liberated from the field of their applications, gradually reach the upper stratosphere after a long period of time.

In the stratosphere, they absorb the UV radiation coming from the sun and decompose to produce highly active chlorine atoms which subsequently with O₃ of the ozone layer to liberate oxygen and chlorine monoxide free radical (CIO) which, on further reaction with ozone, forms O₂ molecules and active chlorine atoms. The chlorine atoms again combine with O₃ molecules to form O₂ molecules. In this way, a cyclic process continues, which eventually causes depletion of the ozone layer.

It has been experimentally found that a single Cl -atom is capable of decomposing millions of O₃ molecules.

2. Experiments head revealed that halons widely damage the ozone layer. Halons are halocarbons. Most of them contain bromine as halogen. These are mainly used as fire extinguishers

For example: Halon—1211 (CF₂BrCl, bromochlorodifluoromethane),

Halon: 1301 (CF₃Br, bromotrifluoromethane) etc.

Halons are stable in the troposphere. But in the stratosphere, they absorb UV-radiations and decompose to produce active bromine atoms. These active bromine atoms combine with O₃ of the ozone layer to liberate oxygen and bromine monoxide free radical (BrO).

The produced BrO again reacts with ozone to evolve O₂ molecules and active bromine atoms. Bromine atoms thus obtained, in reaction with O₃, give O₂ molecule. In this way, the entire process proceeds continuously in a cyclic manner which leads to the decay of the ozone layer.

3. Extensive studies have unquestionably proved that increase in quantity of die oxides of nitrogen in the stratosphere adversely affect the ozone layer. The main source of these oxides are the supersonic aeroplanes which emit plenty of NO gas while flying through the stratosphere. Like Cl and Br-atoms, NO molecule brings about catalytic decomposition of O₃ into O₂

NO + O₃→NO₂ + O₂ ; NO₂ + O → NO + O₂

The reaction of O₃ with NO yields NO₂ yields. This NO₂ combines with the oxygen atom (which ”highways produced in the stratosphere due to the decomposition of O₃ and O₂ under the influence of UV-radiation) to regenerate NO. This explains why the O₃ molecules undergo continuous decomposition although the quantity of NO is not diminished.

4. Effect of ozone hole on the environment

Effect on climate:

If the ozone layer in the stratosphere is destroyed, the UV radiation emitted by the sun, instead of being absorbed by this region, will be incident on the earth’s surface. Consequently, the temperature of the earth’s surface will increase. Owing to this rise in temperature the earth will be continuously heated and the ice in the polar regions will melt, resulting in a rise in
the water level of the sea.

Effect on mankind:

In the absence of ozone layer in the stratosphere, the UV radiation will directly reach the earth’s surface. This radiation is extremely harmful to human beings. It causes skin cancer and premature cataract in the eyes. Exposure to the UV-radiation damages the immune system which thereby increases susceptibility to viral infections. Moreover, this radiation motivates the photochemical reactions which increases the tendency of smog formation. This in turn creates severe respiratory problems such as bronchitis, tracheal irritation, lung diseases etc.

Effect on plants, animals and other living organisms:

The incidence of UV radiation on the earth’s surface will hinder the process of photosynthesis. As a result, the production of crops will decrease. UV radiation would naturally increase the earth’s average surface temperature. Water bodies will dry up and water from the soil will evaporate. Consequently, agriculture will be greatly affected and the production of crops will fall drastically.

Also, UV radiation precludes photosynthesis. Therefore plants and aquatic phytoplanktons will die. Thus, marine life which depends on phytoplankton will also perish. Therefore, UV radiation disrupts the entire ecological system and composition of the sea

Greenhouse And Greenhouse Effect

1. What is a greenhouse?

Greenhouse means a glass room or glass chamber. Plants of the tropical region cannot adapt themselves to the climatic conditions of the cold countries. So for sustaining plant life (particularly for plants of tropical regions in cold countries), this type of chamber made of glass is used. Sun rays enter the glass chamber through the transparent glass roof and walls and due to this the soil gets heated.

The heated soil inside the chamber radiates infrared rays of longer wavelengths which cannot pass through the glass. The glass absorbs a part of these rays and the rest are reflected to the soil inside the chamber. As a result, the temperature inside the chamber always remains higher than that of the outside temperature. Thus, proper growth of these plants becomes possible.

In fact, some gaseous substances present in the earth’s atmosphere such as CO₂ water vapour etc., together act like a glass of the greenhouse and keep the atmosphere adjacent to the earth’s surface warmer and create a favourable environment for the living world.

2. Greenhouse effect and its importance

Greenhouse Effect Definition:

The natural process by which CO₂  water vapour and some other gases are present in the atmosphere, prevents the return of the radiation emitted by the earth’s surface to outer space, thereby keeping the surface of the earth and the adjacent environment suitable for the effect.

‘ Some gases like CO₂, water vapour etc., present in the atmosphere allow sun rays of smaller wavelengths to be incident on the earth’s surface but prevent the rays of longer wavelength (infrared rays) emitted from the hot earth’s surface, from returning to the outer space. Those gases absorb a significant portion of the reflected radiation of longer wavelength and arc heated.

The rest of the infrared rays fall on the earth’s surface and remain in the adjacent atmosphere to keep the surface of the earth and the adjacent atmosphere warm and make it favourable for the existence of the living world. due to indiscriminate deforestation, the quantity of CO₂ absorbed by the plants is gradually decreasing.

Greenhouse gases like CO₂, water vapour etc., help to keep the atmosphere warm to a certain level of temperature (average as 5°C) which is essential for the existence of life on earth. If those gases were not present in the atmosphere then the average temperature of the earth’s surface and that of the surrounding atmosphere would have dropped to about -30°C and eventually, survival of life on earth would have been impossible

3. Greenhouse gases, their sources and contribution towards the greenhouse effect

The gases that absorb a significant portion of radiation of longer wavelengths (infrared rays) emitted by the hot earth’s surface and reflect the rest to the earth’s surface, to keep the adjacent environment warm, are referred to as greenhouse gases. Some greenhouse gases are—carbon dioxide (CO₂), methane (CH₄), chlorofluorocarbon (CFCs), ozone (O₃), nitrous oxide (N20), water vapour (H₂O) etc.

1. Carbon dioxide:

Out of all the greenhouse gases, CO₂ is present in the largest amount in the atmosphere. Naturally, CO₂ plays the most vital role in absorbing the radiation emitted by the earth’s surface. The contribution of CO₂ towards the greenhouse effect is approximately 50%.

During respiration, plants and animals take in O₂ and give up CO₂  while plants accept CO₂ for the preparation of their food. In this way, the CO₂  -level in the environment is maintained. But at present, due to the progressive increase in the quantity of CO₂  in air, the equilibrium of CO₂  gas in the atmosphere has been disturbed. The possible reasons for the continuous

Increase in the percentage of COz gas are as follows:

1. 1.  Due to the indiscriminate use of fossil fuels in factories, motor vehicles etc., the quantity of CO₂ released in the atmosphere is not being used up completely by different natural processes.
2. Consequently, the concentration of CO in the atmosphere keeps on increasing.
3. In industrial regions, particularly during the manufacture of cement, a large &ftU>unt of CO₂ gas is released into the atmosphere.
4. During the process of photosynthesis, plants absorb CO₂ from the air. But due to indiscriminate deforestation, the quantity of CO₂ absorbed by the plants is gradually decreasing.

2.  Methane:

The role of methane gas in preventing the outflow of the heat emitted from the earth’s surface is worth mentioning. Due to the bacterial decomposition plants in paddy fields and other marshy lands, putrefaction of dung and other excreta and anaerobic decomposition of dead animals, methane gas is produced. Besides these, different waste organic compounds, oil mines etc., are the other sources of methane gas.

The capacity of each methane molecule to prevent the outflow of heat is 25 times a much as that of a molecule of CO₂  gas. But the quantity of methane gas in the atmosphere being less than that of CO₂ gas, its contribution to the greenhouse effect is about 16-20%.

3. Chlorofluorocarbons (CFCs):

CFCs are widely used as refrigerants, propellants in aerosol sprays, fire extinguishing agents, solvents for cleaning electronic types of equipment and foaming agents. These compounds destroy the ozone layer in the stratosphere and act as greenhouse gases in the troposphere.

The capacity of the chlorofluorocarbon molecules to prevent the release of heat emitted from the earth’s surface is 15000-20000 times greater than that of CO₂ molecules. These compounds are extremely stable. So, they can exist in the atmosphere for a long time. The contribution of these compounds towards the greenhouse effectis found to be 13-18%.

4. Tropospheric ozone:

Ozone gas present in the troposphere acts as a greenhouse gas. The combustion of fossil fuels in automobile engines, thermal power plants and different chemical industries gives rise to a profuse quantity of oxides of nitrogen (NO_x). The combination of ; different hydrocarbons and oxygen present in the atmosphere with these oxides results in the formation of ozone gas. The contribution of ozone gas to the greenhouse effect is about 7-8%.

5. Nitrous oxide:

Extensive combustion of fossil fuels, motor vehicles, and bacterial decomposition of nitrogenous chemical fertilisers in agricultural lands generates nitrous oxide. Again, its heat retention capacity per molecule is 200 times greater than that of CO₂ per molecule. The contribution of this gas towards the greenhouse effect is about 4-5%.

4. Global warming

For tire last few centuries, the average temperature of Tire Earth has been gradually increasing.

For example: During the period from 1800-1900 AD, the average temperature of the earth has increased by nearly 0.4°G. Again, in the following century i.e., 1900-2000 AD., this increase in temperature has been almost of 1°C. So it cannot be denied that the natural environment is gradually becoming warmer.

The phenomenon of this progressive rise in temperature all over the world is called global warming. The reason for this global warming can be attributed to the increased concentration of greenhouse gases in the atmosphere, caused by various human activities

Harmful effects of global warming: 

1. Because of global warming 2100 AD, the polar ice caps will melt, thereby releasing an enormous amount of water. Then the situation will be almost similar to what it was 1,30,000 years ago, when the surface water level of the sea was 6 years ago, when the surface water level of the sea was 6 coastal regions like Holland, America, New Orleans, Florida, Bangladesh etc., will be inundated and will go under water forever.

2. Global warming is a great threat to human health. Respiratory problems occur frequently due to human health.

Respiratory problems occur frequently due great concern. Global warning would initiate a favourable temperature for the breeding of microorganisms resulting in the epidemic spread of dreadful diseases such as dengue, malaria, encephalitis etc. If the CO₂ content in the air becomes twice that of the present value, then many species will become extinct from the earth

5. Consequences of the greenhouse effect

Scientists have predicted about the effect of increased concentration of greenhouse gases such as

• The temperature of the earth’s surface and the troposphere will go to increase day by day and by the middle of this century, the temperature of the earth will be increased by at least of 2°-4°C.
• Due to the increase in earth’s temperature, the polar caps, accumulated in tyre polar regions (Greenland and Antarctica) will melt and this will cause an increase in the water level of the sea. As a consequence, vast coastal regions like India, Bangladesh, Myanmar, Maldives etc., will sink, causing colossal devastation. If the populated area be inundated in this way, the resettlement of the affected people will pose a great problem to many countries
• Owing to the increased greenhouse effect, droughts will be more frequent during summer in the countries of the mid¬ latitudes in the northern hemisphere. As a result, crop production in the fertile lands of North America and the previous Soviet Russia will be reduced.
• More devastating cyclones, supercyclones, tornados or hurricanes will occur with an increase in temperature.
• Increase in temperature may lead to the destruction of forests due to forest fires.
• Due to the inability to sustain at high temperatures, living beings will die. Consequently, the ecosystem will be severely affected. In the marshy lands, due to increased decomposition of plants, methane will be liberated.

In a word, it can be said that by the middle of this century, man will have to face severe natural calamities.

6. Measures to check global warming

Global warming cannot be eradicated or reduced overnight. A concerted effort is necessary to attain this goal. In overnight. A concerted effort is necessary to attain this goal.

In measures are mentioned below:

• The addition of CO₂ to the atmosphere should be minimised by reducing the use of fossil fuels such as wood, coal, petroleum etc.
• Unlawful cutting of trees should be stopped and the forests should be saved from destruction.
• Afforestation ought to be encouraged so that plants absorb more CO₂ (for the preparation of their food). Koiget !r ’Boron
• Use of unconventional forms of energy
• For example: Solar energy, wind energy, tidal energy etc.) is to be increased.
• Use of CFCs is to be prohibited

Smog Or Classical Smog

In December 1952, the city of London was covered with a dense layer of fog continuously for five days. The people, irrespective of age and sex, fell ill and 4000 people ultimately lost their lives.

Subsequently, it was found that the fog contained poisonous gases emitted from automobile engines and factories and the constituent which was present in the largest quantity, was found to be sulphur dioxide gas (SO₂). A British physician named it smog (smog = smoke + fog). As the horrible effect of such smog was first observed in London, it was called London smog

Several accidents caused by such smog (of course of less alarming proportions), occurred in different cities. Smog is frequently observed in big cities

For example:

Delhi, Mumbai Kolkata) of India, during the winter season. Mixture of particulates with gaseous oxides of Mixture of particulates with gaseous oxides of the presence of SO₂ and carbon (soot) particles, classical smog possesses a reducing character. Thus it is also called reducing smog

1. Formation of smog

During winter, particularly after evening or early in the During winter, particularly after evening or early in the earth’s surface becomes heavier If it is suddenly cooled down earth’s surface becomes heavier ifit is suddenly cooled down cannot go upwards and remains confined in that layer. Impure cannot go upwards and remains confined in that layer. Impure finely divided particles liberated from local factories motor vehicles, mix with that confined air to create finely divided particles liberated from local factories motor vehicles, mix with that confined air to create and and

Harmful effects of smog:

• Smog irritates the nose, eyes and throat, resulting in sneezing and coughing.
• It affects the respiratory system, causing bronchitis, asthma, heart disease etc.
• It lowers visibility level, posing great problems while driving cars. So, accidents are likely to happen.
• It also has adverse effects on electronic systems and plants

2. Photochemical smog or Los Angeles smog

This type of smog was first observed in the city of Los Angeles in America, in the year 1950. So, it is called Los Angeles smog. Highly poisonous substances like nitrogen dioxide(NO₂) and ozone (O₃). peroxyacyl nitrate (PAN), smog was formed due to chemical reactions in the presence of bright sunlight, it is commonly known as photochemical smog. Generally, during the mid-days of the summer when the sun shines brightly, (i.e., the intensity of solar radiation is very high) this kind of smog is observed

In big cities, where there is considerable vehicular traffic on the roads throughout the whole day and night, the atmosphere contains the oxides of nitrogen particularly, nitrogen dioxide (NO₂) in the largest proportions.

Apart from these, hydrocarbons (produced by evaporation or incomplete combustion of liquid fuels) and other gaseous substances

For example: SO₂> CO₂  are present.

In the presence of bright substances like ozone (O₃), peroxyacyl nitrate (RCO₃NO₂), aldehyde and ketone. These gaseous substances and the particulates mix together in the air to form photochemical smog.

It formation of smog is to be noted that, it is not real smog, because it contains particulates mix in air to form photochemical smog. The formation of smog is to be noted that, it is not a real smog, because it contains Hence it is also known as oxidising smog

Mechanism for the formation of photochemical smog:

In the presence of sunlight, nitrogen dioxide (NO₂) molecule decomposes into nitric oxide (NO) molecules and atomic oxygen

In the reaction of hydrocarbons with this atomic oxygen, at first highly reactive free radical RCO is produced and this radical again combines with oxygen molecule to give peroxyacetyl radical

Peroxyacyl radical is highly reactive. It combines with hydrocarbon,O₂ and XO₂ to form a mixture of aldehyde, ketone, ozone and peroxyacyl nitrate respectively. This peroxyacylnitrate is extremely harmful for eyes.

Harmful effects of photochemical smog:

• Presence of large amounts of ozone (O₃), peroxyacetyl nitrate (PAN), acetaldehyde, ketone etc., causes
  • Irritation of eyes, nose, and throat but it sill effect on the eyes is much more intense
  • Congestion of nostrils, sneezing and cough
  • Respiratory problems and chest pain.
• By brown colour, it reduces visibility and hence the car drivers and pilots face extreme difficulties.
• PAN and other oxidising materials damage plant cells and produce white spots on leaves.
• PAN hinders the process of photosynthesis.
• Rubber goods lose elastic properties and become brittle.

To control or suppress the formation of photochemical smog, the following methods can be adopted.

• Certain chemical compounds, which are capable of generating free radicals, are sprayed into the atmosphere. The free radicals readily combine with the free radicals responsible for the formation of photochemical smog (such as O^•, R^•, R^•O, ^•H etc.) and hence nullify their effects.
• Efficient catalytic converters are being developed for installation in automobiles so that emission of nitrogen oxides and hydrocarbons can be prevented or minimized

Certain plants such as Pinus, Juniperus, Pyrus, Vritis etc., can directly assimilate oxides of nitrogen for their metabolic activity. So their plantation could be helpful.

Comparison between ordinary smog and photochemical smog:

Acid Rain

Ordinary water is slightly acidic (pH = 5.6) because a portion of carbon dioxide gas present in the air gets dissolved in water and forms carbonic acid

CO₂ + H₂O→H₂CO₃ .

But if rainwater contains an excess amount of dissolved acid, then it is called acid rain. Acid rain is mainly a mixture of H₂SO₄, HNO₃ and HCl. The pH of such rainwater generally lies within the range of 5.6 to 3.5. The proportion of the above acids in the rainwater of different localities depends upon the quantities of sulphur dioxide (SO₂), nitrogen oxides (NO_x) and hydrogen chloride (HCl) present in the air of that particular locality

1. Origin of Acid Rain

Huge quantities of the oxides of sulphur, nitrogen and carbon (SO_x, NO_x, CO_x etc.) are released In the air due to natural processes as well as tyre human activities. These oxides combine with oxygen, ozone and water vapour present in air to give different acids. These acids float in the air in the form of fine particles as an aerosol. Moreover, HCl gas is also liberated in sufficient quantity from the factories. These acids come down to the earth through dew, snow and rainfall

H₂SO₄ has the highest contribution (60-65%) to acid rain followed by HNO₃ having 30-35% contribution.

2. Harmful effects of acid rain

Effect on soil and plants:

Acid rain increases the acidity of the soil, changes the solubility of different metals and metallic oxides present in the soil. Thus, living creatures and bacteria living inside the soil are severely affected or die. As a result, the fertility of the soil decreases and the production of crops Is drastically reduced. Due to increased acidity of the soil, leaf pigments get spoiled, the process of photosynthesis and as a consequence the growth of plants and their immunity drastically fall, i.e., agricultural productivity is reduced.

Effect on aquatic plants and animals:

Due to acid rain, the pH of different water bodies decreases significantly. As a result, the production of spawn of fish is reduced. The biological processes of fishes are affected. An increase in acidity results in the elimination of many species of algae and zooplankton, aquatic insects, fishes etc. That is, polluted water disrupts the aquatic food chains and consequently disturbs the ecosystem. In countries like America, Sweden etc., in a large number of lakes, virtually no fish exists due to acid rain.

Effect on mankind:

Acid rain dissolves different metallic substances. These dissolved substances enter human bodies through water and result in severe health hazards. Acid rain has profound ill effects on human skin, hair and cells. H₂SO₄ and HNO₃ present in acid rain enter the human body and adversely affect the nervous, respiratory and digestive systems.

 Effect on architecture and edifice:

Because of acid rain buildings, monuments of historical importance

For example:

Tajmahal, Victoria Memorial, British Parliament House), states, sculptures etc., made of marble;, limestone, dolomite, mortar and slate suffer irreparable damages

Marble, limestone etc., react with H₂SO₄ to form an insoluble layer of CaSO₄ and lose its glossiness

CaCO₃+ H₂SO₄ → CaSO₄ ↓+ CO₂↑ + H₂O

Few years ago, scars developed on the surface of Tajmahal Few years ago, scars developed on the surface ofTajmahal Few years ago, scars developed on the surface of Tajmahal

Slone cancer:

• The scars that are developed on the surface of architectural edifices, memorials, sculptures etc., are termed stone leprosy or stone cance
• Metallic surfaces,
  • For example: Aluminium, steel or iron structures, bridges etc., exposed to acid rain, suffer steady.
• Textile materials, leather and paper products are also not spared from the ill effects of acid rain

3. Measures to check acid rain

Only the drastic reduction in the quantities of SOx and NOx in the environment can eliminate the apprehension rather than the threat of acid rain.

The following measures can be adopted to check acid rain:

Use of fossil fuels is to be decreased as far as practicable. Fuels of low sulphur content should be used so that the emission of SO₂ can be controlled.

Suitable technological devices should be developed for the removal of those gases

For example: SO_x, NO_x etc.)

Released from the thermal plants, factories, furnaces for metal extraction and various other sources

Vehicles involving engines with improved technology must be launched so that the emission of NO_x can be controlled

Water Pollution

In nature, water is an indispensable component. 97% of the In the nature, water is an indispensable component—97% of the water, which is practically of no use to human beings.

• The remaining 3% is sweet water, of which 2% remains in condensed form in polar regions and in various permanent glaciers.
• The remaining 1% of sweet water, which is accumulated in rivers, fountains, lakes, ponds, under the soil etc., is consumed for different useful purposes.
• Due to the polar nature of water, a large number of inorganic salts get dissolved in it. As a result, these salts become easily available to living beings.
• These salts are extremely important to aquatic life.
• The various gaseous pollutants such as CO₂, SO_x, NO_x etc., present in the atmosphere and organic and inorganic pollutants on the earth’s surface are swept by rainwater and mix with the rivers, lakes, seas etc.
• Due to different chemical reactions in water, these pollutants decompose to give unpolluted water. As a result, the extent of environmental pollution decreases.

Water pollution:

When the water of different water bodies gets contaminated with one or more chemical substances, evolved either by natural phenomena or indiscriminate human activities and tend to cause health hazards to man and other living beings or adversely affect the processes of their livelihood, then the water is said to be polluted.

1. Domestic wastes

Solid waste of various materials of domestic use

For example: Discarded paper, plastics, torn cloth, vegetable refuse, remains of food etc.),

Excreta of man and domestic animals are mostly left in open places. With time, these discarded materials are carried by wind or rainwater to the nearby water bodies. This contaminates the water and makes it unfit for use.

Water pollution caused by domestic wastes:

1. Domestic wastes mostly contain organic compounds. These organic compounds are decomposed by bacteria in the presence of dissolved oxygen (DO). This process is called biodegradation.

In the process of biodegradation, carbon, hydrogen, nitrogen, phosphorus etc., present in the organic compounds are oxidised to CO₂, H₂O, nitrate, phosphate and other salts. During the process the quantity of dissolved oxygen gradually decreases. Naturally, aquatic plants, fi hes and other aquatic organisms do not get sufficient oxygen for respiration.

Consequently, aquatic living beings face serious problems. DO is considered to be an important parameter in predicting the quality of water. For aquatic plants and animals, the value of DO must not be less than of 4-6mg L^-1 . With the increasing value of DO, the quality of the water gradually improves. Lowering the value of DO indicates that the water is getting polluted.

Biochemical Oxygen Demand or BOD:

Biochemical oxygen demand (BOD) may be defined as the number of milligrams of oxygen required for biodegradation (i.e., biochemical degradation) of organic matter present per litre of polluted water.

• For the determination of the BOD of a sample of water, the sample of water, kept at 20°C, is saturated with oxygen & is subjected to biodegradation (i.e., oxidation) of organic compounds for 5 days by the bacteria present in that water.
• The statement, the BOD of a sample of water is 60 or 60 mg.L^-1   means for the biochemical decomposition of organic matter present per litre of water, 60mg of oxygen is required. BOD of the sample of water, if expressed in ppm (parts per million), also gives the same value.
• Greater the value of BOD of water, the higher will be the extent of pollution of that water because if the water contains a large amount of organic matter, the quantity of oxygen required will be high.
• If the value of BOD of a certain sample of water is greater than 5 ppm then, the water is considered to be impure.

Chemical Oxygen Demand or COD :

• Water may sometimes contain organic or inorganic pollutants which are not decomposed by bacteria. These are called nonbiodegradable pollutants.
• So, the value of the BOD of any sample of polluted water does not truly reflect the extent of pollution of that sample of water. Thus, for the determination of the total quantity of biodegradable and non-biodegradable pollutants in a sample of water, the sample of water is oxidised by a strong oxidising agent (K₂Cr₂O₇+ H₂SO₄) in the laboratory.
• In this case, the oxidising agents supply the necessary oxygen required for the complete oxidation of the pollutants.
• The total amount of oxygen required for the complete oxidation of biodegradable and non-biodegradable pollutants is called Chemical Oxygen Demand (COD). Naturally, the value of COD of any sample of water is always greater than that of BOD.

2. If dissolved oxygen is deficient in water the oxidation of organic pollutants does not get completed. As a consequence of incomplete oxidation, methane (CH₄), hydrogen sulphide (H₂S), phosphine (PH₃), different amino compounds etc., are formed which creates an extremely off ensive odour.

3. Waste materials, sewage from dispensaries, hospitals and domestic wastes carrying pathogenic microorganisms are drained into the water bodies which may result in various diseases such as cholera, typhoid, paratyphoid, dysentery, hepatitis, polio, gastroenteritis, jaundice etc.

4. Waste materials like plastics do not undergo bacterial decomposition in the presence of oxygen, i.e., they are nonbiodegradable. They remain unaffected even if they areleft in water for years. Thus, they decrease the depth of water as well as increase the extent of water pollution under the influence of their constituent chemical compounds

2. Industrial wastes

• Industries release wastes, due to the production of organic and inorganic materials. Factories producing or using mineral acids like HCl, H₂SO₄, HNO₃, H₃PO₄etc., and alkalis
  • For example: NaOH, KOH, and NH₃ give up profuse quantities of waste materials or effluents which are thrown directly into the water of rivers, lakes, ponds etc.
• These acids or alkalis get mixed with water and increase the acidity or alkalinity of water.
• Again, the industrial wastes of different factories, containing metallic elements
  • For example: Pb, Hg, Cd, Zn, Cr, Mn, As, Be etc.)  mixes with different water bodies
• These metals have profound ill effects on aquatic plants and animals in particular.
• Direct use of this polluted water entails attack by several diseases.
• Besides, these metals, accumulated in human bodies through food chains, cause a wide range of diseases

1. Cadmium (cd):

Refining of zinc, copper and lead, electro¬ plating industries, iron and steel factories, Ni-Cd battery factories etc., release cadmium as waste material into rivers and other water reservoirs.

Harmful effects:

Cadmium, introduced into the body through water pollution causes vomiting, irritation of the lungs, malfunctioning of the liver and kidney high blood pressure, anaemia, disorder of bone marrow etc.

Ital-llai disease:

• In 1970, a disease caused by pollution due to cadmium occurred in Toyama Japan and came to be known as Itnl-Itai. The water containing cadmium discharged from a zinc extraction factory situated in that locality was used for irrigation.
• As a result, cadmium was incorporated into rice because of the cultivation of paddy with this polluted water.
• Cadmium was thus introduced into human bodies through this rice when consumed as food. Thus the disease Itai-Itai originated. Pain in bone and joints, weakening or brittleness Of the bone etc., are the symptoms of this disease.

2. Mercury (Hg):

The water discharged as industrial waste from the factories producing caustic soda, chlorine, pesticides etc., is die source of mercury pollution.

Harmful effects:

Mercury is highly toxic. It causes stomach pain, dropsy (oedema), headache etc. Moreover affects the nervous system and kidneys, decreases the reproductive power of males, and babies are found to be born with deformity.

Minamata disease:

In 1953-69, die disease caused by pollution due to mercury appeared in the Minamata area on the sea-coast of Japan and came to be known as Minamata disease. In that area, more than 100 people died of this disease and thousands of people became crippled. In this coastal region, waste materials contaminated with mercury from a polyvinyl factory were regularly discharged into sea¬ water.

Mercury present in the effluent was converted into highly poisonous methyl mercury by different reactions. This poisonous compound was introduced into human bodies through sea fishes and led to the outbreak of Minamata disease. The primary symptoms of this disease are the lack of sensation in muscles, lips, tongue etc., which culminate in blindness and loss of memory

3. Lead (Pb):

The waste materials discharged from factories For example:  extraction and refining of lead, paints, varnishes, alloys, batteries and ship-building etc.), containing lead, pollute the water of rivers, lakes and other sources. Apart from these, the anti-knocking compound [Pb(C₂H₅)₄] used in gasoline and petrol is a potential source of lead

Harmful effects:

If water contaminated with lead enters the body, lead gets accumulates in the body. Most of the lead is ultimately deposited in the bone. Lead poisoning gives rise to symptoms such as loss of appetite, vomiting, constipation, anaemia, insomnia, headache etc., and also affects the digestive system.

4. Manganese (Mn):

Effluents containing Mn from ferromanganese producing industries, welding factories and MnO₂ as waste materials released from dry batteries, mix with water as pollutants.

Harmful effects:

If Mn-containing water is consumed for a prolonged time, it causes nervous disorder.

5. Coball (Co):

Industrial discharge from ceramic, paint or dye industries results in the Co-pollution of water.

Harmful effects: If cobalt-contaminated water is consumed, symptoms such as lowering of blood pressure, diarrhoea, deformation of bones etc., are developed.

6. Arsenic (As):

The main sources of arsenic poisoning are pesticides, chemical wastes, pharmaceutical industries, mining by-products etc. In the tube well water in some places, arsenic compounds are present.

Harmful effects:

Water polluted by arsenic disturbs blood circulation in the skin and black or ash spots appear on the skin of the throat, neck and back. The skin of the hands and legs becomes rough and spots like moles appear on the skin. Continuous use of water containing arsenic for a longer time causes cirrhosis of the liver, cancer in the lungs and urinary track

7. Arsenic pollution:

According to the recommendations of the World Health Organisation (WHO), water containing 0.01 mg of arsenic per litre is quite safe for drinking. The limit of arsenic in water that human bodies can sustain is 0.05 mg L^-1. But the average arsenic content in the tube well water of some places in the vast region of Bangladesh including some districts of Gangetic West Bengal

For example:

North and South 24-Parganas, Nadia, Murshidabad, Maldah etc.) is 0.25mg. L^-1 . As a result of the indiscriminate use of this water, nearly ten lakhs of people have been victimised in West Bengal. Out of these, at least two lakhs of people have been suffering from acute skin diseases

8. Black-foot disease:

Consumption of arsenic contaminated for a long period also causes severe damage to. lower limbs and formation of black lumps on palm and foots. This is known as ‘black-foot disease.

3. Fertilisers used in agriculture

Chemical fertilisers or nutrients are extensively used for increasing the agricultural production. Mainly urea or organic fertilizers and ammonium sulphate, ammonium nitrate, monocalcium phosphate etc., are used as inorganic fertilisers. A certain portion of these fertilisers remains unutilised and being carried by the rainwater, falls into the nearby, lakes etc., and thus causes water pollution

Water containing nitrate ions cannot be used as potable water because nitrate ion cannot be removed by the usual process of purification of water. Consumption of such water affects the haemoglobin of babies, causing the disease called ‘blue baby syndrome’. Moreover, nitrate ions are converted to carcinogenic nitrosamines inside the body

Eutrophication: 

• Inorganic fertilizers or nutrients
  • For example: Nitrates, phosphates, sulphates etc.)
• Left unutilised in agricultural production, mix with water as waste materials and act as pollutants of the water.
• But, these materials enrich that water with nutrients and help in the rapid growth population of the aquatic plants.
• This higher rate of growth is found to be remarkably high in the case of algae.
• This phenomenon of enrichment of water mixed with fertilizers, causing rapid growth of the population of aquatic plants is called eutrophication.

The ill effects of this over-nutrition i.e., eutrophication may be summarised as:

• The aquatic plants, because of their rapid growth require abundant quantity of oxygen and cause depletion of dissolved oxygen (DO), thereby threatening the survival of aquatic life.
• When the quantity of dissolved oxygen decreases, the anaerobic bacteria grow abnormally and these bacteria react with those waste materials to form different gaseous substances such as methane, ammonia, hydrogen sulphide etc. As aresult, foul smell is emitted.
• With time when the plants die, the remains of the dead plants get deposited at the bottom of lakes, ponds etc., and become shallow.
• In extreme instances of eutrophication, when the population of plants explodes, they exhaust almost the whole of the dissolved oxygen. Consequently, fishes, insects and other aquatic animals die due to the absence of oxygen.

4. Pesticides used in agriculture

A wide range of synthetic organic chemicals are used for the better production and preservation of crops. For example, insecticides, fungicides, herbicides etc., are applied to the field to kill insects, fungi, herbs etc. These chemicals are collectively known as pesticides. Pesticides, when used in agricultural fields, are carried by flowing water. Thus, they enter the hydrosphere and cause pollution of water. Again, when pesticides are sprayed in the field, a part of them get mixed with the atmosphere which come down along with rain water and mixes with the water of the rivers, lakes etc. Water pollution is also caused by the wastes discharged from the factories producing pesticides.

Different classes of pesticides and their harmful effects:

Biomagnification:

There are some pollutants which do no There are some pollutants which do no aldrin, heptachlor etc. These compounds exist for years together, keeping their poisonous effect Intact. These are called permanent organic pollutants.

They are not soluble in water but soluble in fats and oils. So they dissolve in body fats and go on accumulating. These highly toxic substances accumulated in living bodies are transmitted to the bodies of other living beings through food chains. These persistent organic pollutants (POP) exist at highly toxic levels in the bodies of living beings owing to repeated consumption of polluted food

5. Detergent

Detergent is widely used as a cleaning agent in household work and in industry. The effluent released after its use mixes with the nearby ponds, rivers etc., and causes water pollution. Two chief constituents of detergent are—

• Surface active agent: For example: Alkylbenzene sulphonate (ABS).
• Builder of filler: For example sodium Yripolyphosphate [Na₅P₃O₁₀],if>Both these constituents are responsible for water pollution

Water pollution caused by surface active agents:

1. Surface active agents decrease the surface tension of water and consequently help in the formation of foam emulsion and oily substances with water. These surface active agents are non-biodegradable and thereby entail water pollution.

2. Foam created by detergents forms a layer on the surface of water and thus prevents water from coming in contact with air and sun rays. Consequently, water cannot absorb oxygen from air and the dissolved oxygen level (DO) in water decreases.

3. Furthermore, sun rays being obstructed, the aquatic plants at the bottom cannot release oxygen by the process of photosynthesis. For this reason, also, the dissolved oxygen level gradually gets diminished. This results in the deficiency of oxygen required for the respiration of aquatic plants and animals.

4.. Surface active agents form a layer on some organic pollutants

For example Phenolic compounds

So, phenolic compounds present in water can no longer come in contact with bacteria and hence the biodegradation of organic pollutants becomes inhibited. Consequently, the extent of pollution in water increases.

Pollution caused by builders or fillers

Detergent contains phosphate salts known as builders or fillers. Phosphate ions produced from them form water-soluble complexes by combination with the basic radicals Ca⁺², Mg⁺² etc. Iff These complex phosphate salts serve as nutrients for algae and aquatic plants, consequently affecting their rapid population growth (Eutrophication). Plenty of oxygen is required for their respiration. This results in rapid decrease in the level of dissolved oxygen (DO) and the survival of aquatic animals becomes extremely difficult.

6. Radioactive substances

Radioactive substances, during mining and refining as well as from nuclear power plants, may be carried into water. Radioactive discharges from medical and scientific institutions using radioactive isotopes may also lead to water pollution.

Harmful effects:

The presence of radioactive substances in trace amounts may cause nervous debility, physical deformity, miscarriage, sterility, cancer, blindness etc. The harmful influence of this radioactivity continues from one generation to another.

7. Thermal pollution

In hydroelectric power plants, generally, the water from rivers or lakes is converted into superheated steam which is used to rotate the turbine. Only a negligible fraction of heat carried by steam is transformed into electrical energy and the rest returns to rivers or lakes with the help of water. This process continues, in cyclic order.

As a result, the temperature of water of the river or the lake rises considerably and the dissolved oxygen (DO) level decreases, causing great harm to the aquatic animals, particularly the fishes. In thermal nuclear power plants and many other industries, water is used as a coolant, which is discharged at a high temperature to rivers or lakes resulting in a rise in the temperature of the water. This increased temperature accelerates the faster assimilation of the waste materials, causing the depletion of dissolved oxygen (DO).

8. Oil-slicks on sea-water

Mineral oils and by-products of oil spread into seawater for several reasons. Consequently, a layer of floating oil (oil slicks) on sea-water is formed and the transfer of atmospheric biochemical level oxygen dissolved of into dissolved oxygen decomposes sea-water(DO)oxygen levels is prevented reduced

• Again, from which further water oxygen in and Naturally required hence reduces this brings about a shortfall of oxygen required for respiration aquatic plants and animals and their survival becomes extremely difficult
• Moreover, oil-slicks on seawater do not allow sun rays to Moreover, oil-slicks on seawater do not allow sun rays to photochemical reactions of aquatic plants are hindered and their growth is remarkably inhibited thereby.
• At sea, the oil layer causes the death of birds. The oil floating on the sea penetrates through the feathers and wings of birds and thus their insulation and buoyancy are adversely affected. Consequently, their body temperature decreases and ultimately they die. This phenomenon is called hypothermia

9. Controls of water pollution

• Septic tanks should be used in every house.
• Bathing and washing of clothes in water bodies like ponds, lakes, rivers etc., should be controlled.
• Application of chemical fertilisers and pesticides must be done within a safe limit.
• Water from sewage systems has to be treated properly.
• Effluents from the industries should not be released directly to the water bodies before proper treatment.
• Oil leakages from oil-loaded ships must be stopped to avoid water pollution.

Soil

Soil is a constantly changing mixture of materials composed of organic and inorganic substances water air microorganisms etc., which allows plants to grow. Various kinds of organic and inorganic materials, mixed with ground are extremely difficult. rocks, give rise to the formation of soil.

 Chemical composition of soil:

Soil is a complex substance. Its various constituents are:

1. Minerals:

Soil contains different kinds of minerals. The chemical nature of the rock from which the soil originates determines the variety and quantity of minerals in it. The particles present in the soil are basically silicate minerals. The chief constituent elements of soil are silicon, calcium, sodium, potassium, magnesium, iron, aluminium, oxygen etc. These elements are present in the form of silica (SiO₂), silicate (KAISi308, NaAlSi308), epidote [4CaO, 3(AlFe)₂O₃, 6SiO₂, HO₂] etc.

2. Air:

The particles of soil leave enough space in between, which is occupied by air. The air present in the soil contains carbon dioxide, oxygen, nitrogen and water vapour. But the quantity of O₂ present in the soil is less than that of O₂ present in air while the quantity of CO₂ in the soil is comparatively greater than the corresponding amount in air.

3. Water:

Water content in the soil always varies from place to place. The constitution of soil determines its water tension capacity. If the amount of organic compounds present in the soil is increased then the water-retention capacity of soil will also increase. This water serves as a solvent for mineral and organic matter. Moreover, water retained in the soil plays a vital role in maintaining the structural arrangement of the soil.

4. Organic compounds:

Generally, organic substances are produced from the remnants of dead plant and animal bodies. Besides this, the waste material of living beings is also a potential source of organic substances. The organic matter liberated due to the bacterial decomposition of the remains of plant and animal bodies mix with soil to form humus. This humus is a very significant part of the soil.

Some of its qualitative features are:

• It holds particles of soil in a state of aggregation.
• It increases the water retention capacity of the soil.
• The movement of water and air inside the soil is enhanced by it.
• It serves as a source of food for the microorganisms present in the soil.

Effect of soil on the environment:

• The role of soil in the existence of life is very significant.
• Without soil, the evolution of plants and animals on land would not have been possible. If igneous rocks were heated by the scorching rays of the sun, then the environment would have been inimical to life.
• It is the layer of wet soil that makes the existence of life possible.
• Plants produce organic food by taking the required water and salt;

For example:

Sodium chloride from the soil. The animal kingdom survives by taking this organic food, otherwise, the existence of animal life would have been endangered. So, soil is a component of immense importance for the living world i.e., the flora and fauna

Soil Pollution

Soil pollution is caused when industrial wastes, radioactive pollutants, domestic and municipal wastes, and agricultural pollutants, are either thrown or dumped into the soil. These reduce the overall quality of soil and are harmful for living beings.

1. Various sources of soil pollution:

1. Pollution caused by industrial waste:

Plastic and paint industries, coal and mining industries, metallurgical units and the industries for the production of sugar, leather, cotton, pesticides, glass, cement etc., discharge a large amount of their discarded waste, which causes soil pollution.

• Moreover, the waste released by thermal and nuclear power plants also pollutes the soil
•  Industrial waste containing heavy metals is retained by the soil and these metals, on being absorbed by plants, enter into the human bodies through the food grains and display their toxic effect.
• Industrial effluents such as mercury, lead, zinc, arsenic etc., even destroy many bacteria which are useful for soil.

2. Pollution caused by municipal waste disposal:

Waste material discharged by municipalities

• For example:  Plastics, accumulated garbage in dustb, broken glasses, waste cloth pieces, paper, ash and other discarded materials) are dumped on the roadside or at some particular places.
• These waste materials not only pollute the soil but also act as breeding centres for pathogenic germs.

3. Pollution caused by fertiliser:

• Nowadays for agricultural production, chemical fertilisers are widely used. As a result of this indiscriminate use of fertilisers
• The microorganisms which produce humus and nutrients are set to face odds threatening their very existence.
• Use of excessive potassium-containing fertilisers decreases the vitamin-C content (ascorbic acid) in vegetables and fruits, indispensable for our health.
• Due to the application of superphosphate. Assimilation of copper, zinc etc., by the plants becomes very difficult.

4. Pollutionbypesticides:

• The chemical substances used to protect plants from the harmful effects of germs, insects, weeds, fungi etc., are called pesticides.
• Pesticides of various types are used to increase the production of crops. Pesticides are of three types, viz., insecticides, fungicides and herbicides.

The harmful effects of these pesticides are as follows:

• Use of excessive insecticides causes pollution of the soil. Insecticides belonging to the class of organophosphates weaken the muscles of animals. Chlorinated pesticides affect the nervous system and cause stomach cancer.
• Fungicides which destroy fungi are mainly compounds of mercury and copper. Extensive application of these compounds brings about severe soil pollution and disturbs the natural ecosystem.
• Herbicides destroy the unwanted weeds or herbs etc., grown in the soil. When these substances are used in excess amounts, they pollute the soil. In females, these interfere with reproduction and harm the foetus or result in the birth of crippled newborn babies.

5. Pollution caused by acid rain:

• In industrial areas due to air pollution, air becomes enriched with oxides of sulphur and nitrogen (SO₂ and NO₂ Which ultimately result in acid rain.
• Acid rain increases the acidity of the soil and adversely affects the cultivation of crops.

Pollution caused by radioactive substances:

Radio¬ active waste materials, emitted from atomic reactors, as a result of experimental studies on atom bombs and nuclear experiments, are added to the soil. The radioactive emissions from the waste pollute the whole environment including the land mass.

2. Controls of soil pollution

• Use of sanitary landfills i.e., where untreated waste is buried in layers and covered with earth
• Wastes of glass, plastics, and paper should be recycled for further uses.
• Proper and scientific sanitation in every household is necessary to prevent soil pollution.
• Organic pesticides should be used instead of chemical pesticides.
• Proper afforestation can control soil erosion significantly.
• Proper treatment of industrial effluents can lower the amount of pollutants causing soil pollution.
• Radioactive waste material should be treated cautiously to avoid soil pollution.

Green Chemistry

Chemistry is undeniably an important part of our lives since it leads to the formulation and fabrication of medicines, materials, polymers, paints, coatings, electronics etc. Chemists also address fundamental problems like global warming, ozone layer depletion, soil and water pollution, efficient food production via photosynthesis etc.

However, processes on an industrial scale not only produce the desired material, but also large quantities of undesired and toxic chemicals in the form of solids, liquids and gases and have become the biggest challenge that chemists need to face. Hence, there has been a considerable effort to shift to synthetic methods which would minimise environmental pollution. This is where the concept of green chemistry steps in

The U.S. Environmental Protection Agency (USEPA) defines green chemistry as the design of chemical products and processes that reduce the generation of hazardous substances. The use and production of these chemicals and processes may involve reduced waste products, non-toxic components and improved efficiency. Green chemistry is a highly effective approach to pollution prevention because it applies innovative scientific solutions to real-world environmental situations. Green chemistry is also known as sustainable chemistry.

A key difference between environmental chemistry and green chemistry: Environmental chemistry deals with the study of chemical pollutants in the environment whereas green chemistry is concerned with the design of chemicals and processes that minimise toxicity to the environment. This is the key difference between environmental chemistry and green chemistry

1. Applications of green chemistry

The term green chemistry was coined by P. T. Anastas who elucidated the principles of green chemistry in his book ‘Green Chemistry:

1. Prevention: Prevention of waste is better than treatment or cleanup of waste materials.
2. Atom economy: Synthetic methods should be designed such as to maximise the incorporation of all materials into the product.
3. Less hazardous chemical synthesis: Synthetic methods should be designed to minimise toxicity to humans and the environment.
4. Designing safer chemicals: Chemical products must be designed in a manner that their toxicity is reduced, without affecting their desired functions.
5. Safer solvents & Auxiliaries: Use of auxiliary substances (e.g. solvents or separating agents) should be minimised whenever practicable and innocuous when used.
6. Design for energy efficiency: Energy requirements for chemical processes should be minimised and alternative routes for conducting synthesis at ambient temperature and pressure should be probed
7. Use of renewable feedstocks: Green chemistry encourages the use of renewable resources (raw materials or feedstocks) whenever possible.
8. Reduce derivatives: Derivatisation (such as the use of protecting/ de-protecting groups, modification of physical/ chemical processes etc.) leads to the use of additional reagents and the possibility of generating wastes. Such practices should be avoided unless imperative
9. Catalysis: Catalytic reagents are superior to stoichiometric reagents.
10. Design for degradation: Chemical products should be designed so that they decompose into benign products that do not accumulate in the environment.
11. Real-time analysis for pollution prevention: Analytical methodologies should be devised and optimised for real-time, in-process monitoring of chemical processes before the formation of toxic substances.
12. Inherently safer chemistry for accident prevention: Substances and the form of substances must be cautiously chosen to avoid the risk of accidents due to accidental release, explosions, fires etc.

2. Contribution of green chemistry

A few classic chemical processes where green chemistry has proved beneficial are outlined below.

1. Synthesis of Ibuprofen:

Ibuprofen is the active ingredient of several analgesic and anti-inflammatory drugs. The initial synthesis of ibuprofen consisted of a six-step process with a very poor atom economy. However, recent advances has made possible the synthesis of ibuprofen with an atom economy of more than 90%. This synthesis produces less waste and is a three-step process.

2. Use of dense-phase carbon dioxide:

Dense-phase carbon dioxide is used in both homogeneous and heterogeneous catalysis. Its use allows us to replace organic solvents with chemically inert and environmentally non-toxic carbon dioxide. It is used in the food industry as a reusable solvent to ensure minimal nutrient loss and better preservation of the food products. Dense-phase carbon dioxide may also be used to enhance the quality of cement and to reduce the industrial waste of coal plants.

3. Use of liquid carbon dioxide in dry cleaning:

Carbon dioxide is a new environment-friendly alternative for dry’ cleaning. Liquid carbon dioxide effectively removes stains. At the same time, it is less harmful than perchloroethylene, the solvent which is used by 80% dry cleaners.

4. Use of carbon dioxide as a refrigerant:

Chlorofluorocarbons (CFCs) have been extensively used as refrigerants. However, CFCs are now known to be the prince reason for the ‘ozone hole’ in the stratosphere. Carbon dioxide is now used as a refrigerant and has zero ODP (ozone depletion potential) and minimal CWP (global warming potential).

5. Catalytic hydrogenation of diethyl amine:

A greener approach to the catalytic hydrogenation of diethyl amine furnishes a herbicide with the least environmental toxicity.

6. Antifouling agent Sea-Nine:

Sea-Nine (The Dow Chemical Company) is a rapidly biodegradable settlement inhibitor. It is a highly effective antifoulant against bacterial slime, algae, hydrozoids, etc., and is free from heavy metals. Sea-Nine is a good alternative to organotin compounds which cause aquatic toxicity

7. Paper industry and laundry:

Chlorine has long been used for producing good quality paper from wood (by removing all lignin). However, the use of chlorine leads to the formation of chlorinated hydrocarbons which are known to be potential. H₂O₂ is now being used as an alternative in the presence of some activators. The use of H₂O₂ produces lesser environmental concerns.lt is also used in laundry leads lesser use of water.

8. Pyrocool foam:

Pyrocool is used in portable fire extinguishers. It uses a non-toxic foam that cools and extinguishes fire without causing risk to human life. It is also free from volatile organic compounds (VOCs), CFCs and carcinogenic chemicals.

9. Synthesis of antibiotics: Antibiotic drugs like ampicillin and amoxicillin can be synthesised by biochemical methods using environment-friendly enzymes.

10. Sonochemistry: Sonochemistry deals with the study of chemical reactions induced by sound waves.

11. Single-step synthesis of ethanal: A single-step synthesis of ethanal from ethyl alcohol has been studied. This method employs water-soluble ionic catalysts and is environmentally friendly.

12. Fuel cell: A new variety of fuel cells have been fabricated which can be used as batteries in cell phones. Such fuel cells are based on the combustibility of ethanol.

Class 11 Chemistry Environmental Chemistry Long Question And Answers

Question 1. What do you mean by reducing smog and oxidising smog?
Answer:

• Ordinary smog contains sulphur dioxide (SO₂), very fine carbon particles and some other reducing agents.
• As a result of this ordinary smog exhibits reducing property.
• So smog of this type is called reducing smog. Photochemical smog by the presence of oxidising substances such as ozone, NO₂ peroxyacyl nitrate etc. shows oxidising properties.
• As a result of this photochemical smog is sometimes called oxidising smog.

Question 2. What is the Montreal Protocol?
Answer:

• Nowadays, the depletion of the ozone layer in the stratosphere by various greenhouse gases has been a matter of great concern to the scientists of the whole world.
• As a result of this, the decision to prohibit the use of the chief greenhouse gas i.e., chlorofluorocarbon (CFC) was adopted in 1993 in a convention of scientists, arranged in Montreal Canada.
• This is known as the Montreal Protocol. India also signed this agreement

Question 3. What is the Bhopal gas tragedy? Mention the after-effects of the Bhopal gas tragedy.
Answer:

• Bhopal gas tragedy was a gas leak incident in India. This incident is considered as the world’s worst industrial disaster.
• It occurred at the midnight of 2-3 December 1984 at the Union Carbide India Limited (UCIL) pesticide plant in Bhopal, Madhya Pradesh.
• Over 5,00,000 people were exposed to the poisonous methyl isocyanate (MIC) gas.
• Among those people around 3000 people died from the immediate effect of the gas leakage.
• The initial effects of exposure were coughing, vomiting, severe eye irritation and suffocation.
• The people of the affected area still suffering from the after-effects of the tragedy

Question 4. In the stratosphere, ozone is useful but in the troposphere, it is harmful to us— explain.
Answer:

The layer of ozone gas present in the stratosphere which extends from a height of 15 km to 75 km from the sea level of the earth’s surface is known as the ozone layer or ozonosphere.

• The harmful ultraviolet rays (UV rays) coming from the sun is mostly absorbed in the ozone layer because these rays are utilised in the production and dissociation of ozone gas.
• The absence of this ozone layer would allow the ultraviolet rays coming from the sun to reach the earth’s surface entirely.
• This would have heated the earth’s surface and the adjacent air to such an extent that the existence of the living world in land and water would have been jeopardised.
• But, ozone gas present in the troposphere acts as a greenhouse gas.
• Ozone gas contributes nearly 7-8% to the creation of the greenhouse effect.
• Due to greenhouse effects, the surface temperature of the earth will rise and this eventually will melt the polar caps accumulated in polar regions which will cause colossal devastation by tidal waves, cyclones, super cyclones.
• Thus ozone, in the stratosphere is useful but in the troposphere is harmful

Question 5. The extensive depletion of the ozone layer occurs in the months of September to October. Explain this phenomenon.
Answer:

In Antarctica, during the months (March to August) just before the advent of spring season (September-October), the temperature drops below -90°C.

• As a result, the water vapour in the atmosphere condenses to form polar stratospheric clouds.
• Different oxides of nitrogen which are floating in the atmosphere produce nitric acid (HNO₃) in contact with the crystals of ice in the cloud.
• In this condition, chlorine derived from the chlorofluorocarbon (CFC) compounds does not find any opportunity to become inert by reaction with the oxides of nitrogen because chlorine is not capable of reacting with the nitric acid.
• As a consequence of this phenomenon, during the few months from March to August (when the sky in Antarctica remains covered with darkness), chlorine keeps on accumulating in the stratosphere.
• Then, with the arrival of spring, the chlorine present in the atmosphere becomes very reactive in the presence of sunlight and triggers the process of breaking of ozone (O₃) molecules in the ozone layer.
• Thus, extensive depletion of the ozone layer takes place in September to October

Question 6. Name the greenhouse gases and mention their sources due to human activities.
Answer:

Question 7. What is an atomic power plant? What Is Chernobyl Disaster? What is the cause of this accident?
Answer:

The power plant produces electricity by the nuclear fission reaction of radioactive elements such as uranium. plutonium etc. is called atomic power plant

Chornobyl is a city in Ukraine in the former Soviet Union. On the 26th of April, 1986, the accident that occurred with horrifying consequences and destroyed the environment has remained alive in our memory as the Chornobyl disaster.

As a result of this tragic accident, the radioactive emission spread over an area of about 3000 sq km and nearly ten crores of people had to be rehabilitated. Uranium was used as fuel in the plant for the generation of electricity. On the day of the accident, due to the lack of proper safety measures, uranium fuel in the atomic reactor, being exceedingly heated, caused the explosion

Question 8. Carbon monoxide gas is more dangerous than carbon dioxide gas. Why?
Answer: 

The colourless, odourless carbon monoxide gas is severely harmful for human beings and animals. It has a greater affinity towards haemoglobin than that of oxygen. So, it readily displaces oxygen from oxyhaemoglobin (HbO₂) to form the more stable compound carboxyhaemoglobin (HbCO) to give a stable compound, carboxyhaemoglobin

HbO₂ + CO ⇌  HbCO + O₂

In blood, when the concentration of carboxy haemoglobin reaches 3-4%, the oxygen-carrying capacity of the blood is greatly reduced. In other words, the body becomes oxygen-starved. results in headache, nervousness, cardiovascular disorder, weak eye-sight etc., On the other hand, CO₂ does not combine with haemoglobin and hence is less harmful as a pollutant. CO₂. is mainly responsible for the greenhouse effect and global warming

Question 9. What are the harmful effects of photochemical smog and how can they be controlled?
Answer:

Photochemical smog can be controlled in the given ways: 

By using efficient catalytic converters in the automobiles which will check the release of both NO₂  and certain hydrocarbons known as primary precursors.

• This will automatically check the formation of secondary precursors. Such as Ozone and PAN.
• By spraying certain compounds into the atmosphere which will control hydrocarbons, NO₂, and PAN.
• Certain plants like pinus, Pyrus, Vitis Quercus etc., are capable of causing the metabolism of the oxides of nitrogen. Hence their plantation could be helpful.

Question 10. What are herbicides? Explain giving
Answer:

Herbicides:

• These are the chemicals employed to control weeds. The common herbicides are sodium chlorate (NaClO₃) and sodium arsenite (Na₃AsO₃).
• These herbicides are no longer preferred because they are toxic towards mammals.
• At present organic herbicides like triazines are used as weed controllers and have no adverse effect on human beings

Question 11. A large number of fish are suddenly found floating dead on a lake. There is no evidence of toxic dumping but you find an abundance of phytoplankton. Suggest a reason for the fish kill
Answer:

• The presence of excess phosphate and nitrate compounds increases the growth of phytoplankton (organic pollutants such as leaves, grass, trash etc.).
• A large population of bacteria decomposes this organic pollutant.
• During this process, they consume the dissolved oxygen of water which is of course very much essential for the life of sea animals, particularly fish.
• When the level of dissolved oxygen falls below 6 ppm, the fish cannot survive.
• Hence a large number of fish are found floating dead on the lake

Question 12. How can domestic waste be used as manure?
Answer:

• Domestic wastes consist of two types of materials, biodegradable such as leaves, rotten food, vegetable refuse etc., and non-biodegradable portion which consists of plastic, glass, metal scrap etc.
• The biodegradable waste should be deposited in the landfills.
• Then this waste gets converted into time

Question 13.  For your agricultural field or garden, you have developed a compost-producing pit.  Discuss the process in the light of bad odour, flies and recycling
Answer:

• The compost-producing pit should be developed at a suitable place to protect ourselves from bad odour and flies.
• It should be covered properly to prevent the entry of flies and the emission of foul odour.
• The waste materials like plastics, glass, newspapers etc must be handed over to the vendors.
• These are finally sent to the recycling industry without creating a pollution problem

Class 11 Chemistry Environmental Chemistry Short Question And Answers

Question 1. How are NO and NO₂ produced in the atmosphere?!
Answer:

Due to lightning discharge in the upper atmosphere, nitrogen and oxygen combine to produce nitric oxide. This nitric oxide (NO) reacts with aerial oxygen to give nitrogen dioxide.  (NO₂) Because of the bacterial decomposition of ammonium salts in the soil, NO is produced. Besides these, combustion of fossil fuels also serves as a potential source of NO and NO₂

Question 2. In the presence of carbon monoxide, haemoglobin loses its oxygen-carrying capacity causing oxygen starvation of body cells—explain
Answer:

Carbon monoxide has a strong affinity for haemoglobin This gas combines with haemoglobin to form highly stable carboxyhaemoglobin

Consequently, the availability of oxygen in the body cells decreases, because haemoglobin fails to carry the oxygen necessary for the life process to continue. So carbon monoxide present in excess may sometimes cause death

Question 3. Many spray bottles from which a perfume is sprayed contain a very harmful substance. By what name is it commonly known? Why is it harmful?
Answer:

The harmful constituent is chlorofluorocarbons (CFCs) or Freons. When they diffuse into the upper atmosphere, they absorb ultraviolet radiation which ruptures carbon-chlorine bonds to give chlorine atoms. The chlorine atoms thus produced cause destruction of the ozone layer which shields the earth from the harmful effects of ultraviolet radiation of the sun. This is the reason for which CFCs are very harmful.

Question 4. What is hypothermia?
Answer:

Mineral oil and by-products of oil get dispersed in water for various reasons, thereby contaminating it. When a bird comes in contact with this polluted water, the oil floating on the surface of the water penetrates the feathers and wings of the bird. This in turn annihilates the possibility of the bird’s flight. In addition, the temperature of the bird’s body drops considerably resulting in its death. This phenomenon is called hypothermia

Question 5. What were the components of London Smog? What was its nature?
Answer:

Finely divided particles, water vapour, SOx and quantity of NO It was reducing in nature because the non-metallic oxide SO₂ was mainly present in it as a reducing agent

Question 6. It was reducing in nature because the non-metallic oxide, SO₂ was mainly present in it as a reducing agent
Answer:

O₃, NO₂ Peroxyacyl nitrate (PAN), aldehyde, ketone, hydrocarbons and CO. It was oxidising in character, because O₃, NO₂ peroxyacyl nitrate etc., were present in it as oxidising agents. peroxyacetyl nitrate etc., were present in it as oxidising agents

Question 7. What is the temperature range of the atmosphere? What do you mean by inversion temperature in different regions of the atmosphere?
Answer:

-56°C to 1200°C . When we traverse from one region of the atmosphere to the next adjoining region, the trend of temperature changes successively from higher to lower or vice versa. This is called inversion temperature

Question 8. What is marine pollution? What is siltation?
Answer:

The pollution of seawater due to the discharge of wastes from different sources into it thereby making it harmful for human health and aquatic life is called marine pollution. Mixing of soil and rock particles into water is called siltation. The soil particles produce turbidity in water thereby hindering free movement of aquatic organisms.

Question 9. What is the Polar Vortex? What is its effect?
Answer:

A tight whirlpool of wind formed in the stratosphere surrounding Antarctica is called the Polar Vortex. It is so rigid that it cuts off Antarctica from the surrounding ozone-rich air of the non-polar regions. Thus, as long as the polar vortex surrounds Antarctica, the ozone hole remains unfilled

Question 10. What is an ozone umbrella? Why is it called so?
Answer:

Ozone layer present in the stratosphere is called the ozone umbrella. Like an umbrella, the ozone layer prevents harmful. UV radiation from reaching the earth. Thus, the ozone layer is also called the ozone umbrella.

Question 11. In hospitals, patients with CO poisoning are kept in high-pressure chambers containing oxygen at 2 to 2.5 atm pressure—why?
Answer:

Under high pressure of oxygen, CO of carboxyhaemoglobin (HbCO) is replaced by O₂ and thus transport of O₂ to different parts of the body starts.

HbCO + O₂ ⇌  HbO₂+CO

Question 12. What is the role of CO₂ in creating the greenhouse effect?
Answer:

A part of the infrared rays of longer wavelength, emitted by the earth’s surface on being heated by the sun rays is absorbed by CO₂ gas molecules and the return to the earth’s surface along with the surrounding atmosphere gets heated

Question 13. What is Polar Stratospheric Clouds (PSCs)?
Answer:

In Antarctica, the climatic conditions are quite different. In winter, there is no sunlight and the temperature is very low. The low temperature causes the formation of special types of clouds over Antarctica which are called Polar Stratospheric Clouds (PSCs).

Question 14. In which regions atmosphere, temperature increases with altitude and in which regions it decreases?
Answer:

• In the stratosphere and thermosphere temperature increases with altitude, while in the troposphere and mesosphere, temperature decreases with altitude
• In which season the depletion of the ozone layer in Antarctica takes place and when is it replenished
• During spring (in September and October) ozone layer depletion occurs in Antarctica, while after spring

The gaseous and particulate pollutants are:

1. Gaseous air pollutants: These are mainly oxides of sulphur (SO₂, SO₃), oxides of nitrogen (NO, NO₂) and oxides of carbon (CO, CO₂), H₂S, hydrocarbons, ozone and other oxidants.
2. Particulate pollutants: Particles in the form of smog, dust, mist, smoke etc., belong to this category

Class 11 Chemistry Environmental Chemistry Multiple Choice Question

Question 1. The ozone layer forms naturally by—

1. The interaction of CFC with oxygen
2. The interaction of UV radiation with oxygen
3. The interaction of IR radiation with oxygen
4. The interaction of oxygen and water vapour

Answer: 2. The interaction of UV radiation with oxygen

The ozone layer forms naturally by the interaction of UV radiation with oxygen

Question 2. Among the following, the one which is not a “greenhouse gas” is

1. N₂O
2. CO
3. CH ₄
4. O

Answer: 4. O

O₂ is not a gas responsible for the rise in temperature of the earth. So, O₂ is not a ‘greenhouse gas

Question 3. Metal ion responsible for the Minamata disease is

1. CO
2. Hg²⁺
3. Cu
4. Zn²⁺

Answer: 2. Hg²⁺

Hg²⁺ ion is responsible for the Minamata disease.

Question 4. What is DDT among the following

1. A fertilizer
2. Biodegradable pollutant
3. Non-biodegradable pollutant
4. Greenhouse gas

Answer: 3.  Non-biodegradable pollutant

DDT is a non-biodegradable pollutant

Question 5. The gas leaked from a storage tank of the Union Carbide plant in Bhopal gas tragedy was O₂

1. Phosgene
2. Methylisocyanate
3. Methylamine
4. Ammonia

Answer: 2. Methylisocyanate

Methylisocyanate gas was leaked from a storage tank of the Union Carbide plant in the Bhopal gas tragedy.

Question 6. The concentration of fluoride, lead, nitrate and iron in a water sample from an underground lake was found to be 1000 ppb, 40 ppb, 100 ppm and 0.2 ppm respectively. The water is unsuitable for drinking due to high concentration

1. Fluoride
2. Lead
3. Nitrate
4. Iron

Answer: 3. Nitrate

In the sample of water, the concentration of fluoride, lead and iron are in permissible limit but the concentration of nitrate ion is much higher than its permissible limit. Thus the water is not suitable for drinking.

Question 7. A water sample has ppm level concentration of the following anions, F^– = 10, SO₄^2-= 10, NO₃^– = 50. The anion/ anions that make/makes the water sample unsuitable for drinking is/are—

1. Only F^–
2. Only SO₄^2-
3. Only NO₃^–
4. Both SO₄^2- And NO₃^–

Answer: 1. Only F^–

In drinking water, if the concentration of SO₄^2- is more than 500 ppm, it shows a laxative effect and it is not suitable for drinking. If the concentration of SO₄^2-  is less than 500 ppm, it is consumable.

In drinking water, if the concentration of NO₃  ion is more than 50 ppm it causes methemoglobinemia disease. This is not suitable for drinking. If the concentration of F“ ion in drinking water is more than 1 ppm it damages teeth and bones. Thus it is not suitable for drinking

Question 8. The recommended concentration of fluoride ion in drinking water is upto 1 ppm as fluoride ion is required to make teeth enamel harder

1. [3Ca₃(PO₄)₂ Ca(OH)₂]
2. [3Ca₃(PO₄)₂ .CaF₂]
3. [3{Ca(OH)₂} . CaF₂]
4. [3(CaF₂) . Ca(OH)₂]

Answer: 1. [3Ca₃(PO₄)₂ Ca(OH)₂]

Generally, tooth enamel is hydroxyapatite [3Ca₃(PO₄)₂ Ca(OH)₂].  Fluoride ion (F^–) reacts with hydroxyapatite to form a more rigid solid compound fluorapatite.

Question 9. Which one of the following statements is not true 

1. Oxides of sulphur, nitrogen and carbon are the most widespread air pollutants
2. PH of drinking water should be between 5.5-9.5
3. A concentration of DO below 6 ppm is good for the growth of fish
4. Clean water would have a BOD value of less than 5 ppm

Answer: 3.  A concentration of DO below 6 ppm is good for the growth of fish

Fish growth is facilitated if the DO value is less than 6 ppm. A decrease in the Do value means an increase in water pollution

Question 10. Which one of the following statements regarding photochemical smog is not correct

1. Photochemical smog is formed through a photochemical reaction involving solar energy
2. Photochemical smog does not irritate the eyes and throat
3. Carbon monoxide does not play any role in photochemical smog formation
4. Photochemical smog is an oxidising agent in character

Answer: 2. Photochemical smog does not irritate eyes and throat

Question 11. Which one of the following is not a common component of photochemical smog

1. Ozone
2. Acrolein
3. Peroxyacetyl nitrate
4. Chlorofluorocarbons

Answer: 4. Chlorofluorocarbons

Question 12. Which of the following is a sink for CO

1. Microorganisms present in the soil
2. Oceans
3. Plants
4. Haemoglobin

Answer: 1.  Microorganisms present in the soil

Question 13. Which oxide of nitrogen is not a common pollutant introduced into the atmosphere both due to natural and human activity

1. NO
2. N₂O₅
3. N₂O
4. NO₂

Answer: 2.  N₂O₅

Question 14. Living in the atmosphere of CO is dangerous because it

1. Combines with O₂ present inside to form CO₂
2. Reduces organic matter of tissues
3. Combines with haemoglobin and makes it incapable of absorbing oxygen
4. Dries up the blood

Answer: 3. Combines with haemoglobin and makes it incapable of absorbing oxygen

Question 15. Which of the following is not a greenhouse gas

1. Hydrogen
2. Carbon dioxide
3. Methane
4. Nitrous oxide or NO₂

Answer: 1.  Hydrogen

Carbon dioxide, methane, water vapour, nitrous oxide, CFCs and ozone are greenhouse gases

Question 16. Which of the following has the highest concentration of PAN

1. Smoke
2. Ozone
3. Photochemical smog
4. Reducing smog

Answer: 3.  Photochemical smog

The main component of photochemical smog is peroxyacetyl nitrate, (PAN). The other components are ozone, nitric oxide, acrolein and formaldehyde

Question 17. Which of the following is not a greenhouse gas?

1. Carbon dioxide
2. Water vapours
3. Methane
4. Oxygen

Answer: 4.  Oxygen

Carbon dioxide, water vapours and methane are greenhouse gases

Question 18. Which air pollutants do not evolve from motor vehicles

1. Formaldehyde
2. Carbon dioxide
3. Fly ash
4. Sulphur dioxide

Answer: 3. fly ash

Question 19. The top layer of the atmosphere is

1. Stratosphere
2. Troposphere
3. Exosphere
4. Ionosphere

Answer: 3. Exosphere

Question 20. Which of the following is not an air pollutant

1. NO
2. CO
3. O₃
4. C_xH_y

Answer: 2. CO

Question 21. Which of the following has the highest affinity towards haemoglobin 

1. CO
2. NO
3. O₂
4. CO₂

Answer: 1. CO

Question 22. Which gas is not present in the ozone layer

1. O₂
2. O₃
3. N₂
4. CO

Answer: 4. CO

Question 23. Fluoride pollution mainly affects

1. Teeth
2. Brain
3. Kidney
4. Heart

Answer: 1.  Teeth

Question 24. Which metal is mainly responsible for the decline of the Roman Empire

1. Copper
2. Lead
3. Arsenic
4. Zinc

Answer: 2.  Lead

Question 25. Which of the given pollutants does not affect the lungs

1. CO
2. SO₂
3. CO
4. NO

Answer: 3. CO

Question 26. Which of the following statements is not true

1. Ozone gas has no role in the greenhouse effect
2. Ozone gas oxidises sulphur dioxide of the atmosphere to sulphur trioxide
3. Gradual thinning of the ozone layer leads to the formation of an ozone hole
4. Oxygen molecule in the stratosphere forms ozone molecules in the presence of ultraviolet radiation

Answer: 1.  Ozone gas has no role in the greenhouse effect

Question 27. Although nitrogen and oxygen are the major constituents of air, they do not react with each other to produce oxides of nitrogen because

1. As the reaction is exothermic, a high temperature is required
2. A catalyst is required for the initiation of the reaction
3. Oxides of nitrogen are unstable
4. Nitrogen and oxygen do not take part in the reaction

Answer: 1. As the reaction is exothermic, a high temperature is required

Question 28. Which of the following is a secondary air pollutant

1. CO
2. CH
3. PAN
4. NO

Answer: 3. PAN

Question 29. Which of the given fuels used in motor vehicles is not environment friendly

1. Dye
2. Petrol
3. LPG
4. CNG

Answer: 4. LPG

Question 30. A major source of methane in India

1. Fruit garden
2. Sugarcane field
3. Paddy fields
4. Wheat fields

Answer: 3.  Paddy fields

Question 31. Cause of stenosis diseases

1. Fly ash
2. Cement particles
3. Cotton fibre
4. Lead particles

Answer: 3.  Cotton fibre

Question 32. The poisonous substance used in the paper is

1. Cadmium
2. Lead
3. Manganese
4. Mercury

Answer: 2.  Lead

Question 33. Which reaction is carried out in the catalytic converter of motor vehicles to eliminate NO_x from smoke

1. Oxidation
2. Reduction
3. Both oxidation-reduction
4. None

Answer: 2. Reduction

Question 34. Which of the following causes depletion in the ozone layer directly

1. SO₂
2. CFCs
3. H₂O
4. NO

Answer: 1. SO₂

Question 35. The incomplete combustion of gasoline produces

1. CO₂
2. CO
3. SO₂
4. NO₂

Answer: 2. CO

Question 36. Value of pH in drinking water—

1. Between 5.3 to 6.5
2. <5.5
3. Between 5.5 to 9.5
4. 9.5

Answer: 3. Between 5.5 to 9.5

Question 37. The radiation has a specific biological effect but is unable to cause Ionisation Is

1. UV-radiation
2. β -ray
3. X-ray
4. y-ray

Answer: 1.UV-radiatlon

Question 38. Which of the following compounds increases the BOD value of the water supply

1. CO₂
2. O
3. H₂O
4. CHOH

Answer: 4.  CHOH

Question 39. Which is most harmful for a human being

1. UV- A
2. UV – BOD
3. UV – CO₂
4. UV- DDT

Answer: 2. UV – BOD

Question 40. DDT and BHC are

1. Antibiotic
2. Chemical fertilizer
3. Non-biodegradable pollutant
4. Biodegradable compound

Answer: 3.  Non-biodegradable pollutant

Question 41. Which component of motor vehicle smoke causes nerve and mental diseases

1. Hg
2. SO₂
3. Pb
4. NO

Answer: 3. Pb

Question 42. Which mercury compound is the most toxic in nature

1. CH₃Hg⁺
2. HgCl₂
3. Hg₂Cl₂
4. Hg metal

Answer: 1. CH₃Hg⁺

Question 43. Example of herbicide

1. DDT
2. Triazines
3. Methylmercury
4. PCBS

Answer: 2.  Triazines

Question 44. Nitrogen oxide is not a major air pollutant

1. NO₂
2. N₂O
3. NO
4. N₂O₅

Answer: 4. N₂O₅

Question 45. pH of natural rainwater

1. 6.5
2. 3.5
3. 4.6
4. 5.6

Answer: 4. 4.6

Question 46. Which water pollutant metal causes sterility disease

1. Cu
2. Hg
3. Cd
4. Mn

Answer: 4.  Mn

Question 47. The minimum permissible level of sound pollution is

1. 75 dB
2. 65 dB
3. 55 dB
4. 50 dB

Answer: 1. 75 dB

Question 48. In acid rain which of the following are present

1. H₂CO₃
2. HNO₃
3. CH₃COOH
4. H₂SO₄

Answer: 1, 2, and 4

Question 49. If fertilizer containing phosphate is dissolved in water

1. The amount of dissolved oxygen decreases
2. Calcium phosphate precipitates
3. Growth of fish increases
4. The growth of aquatic plants increases

Answer: 1 and 4

Question 50. Result of global warming

1. The temperature of the earth’s surface will increase
2. Glaciers of the Himalayan region will melt
3. Demand of biochemical oxygen will increase
4. Eutrophication

Answer: 1 and 2

Question 51. Which are responsible for photochemical smog

1. Oxides of nitrogen
2. Hydrocarbons
3. Carbon monoxide
4. Nobel gases

Answer: 1, 2, and 3

Question 52. Which gases absorb IR radiation

1. O₂
2. NO₂
3. CO
4. CFC

Answer: 3 and 4

Question 53.  Depletion in the ozone layer is caused by

1.  So₂
2. Halons
3. NO
4. C_xH_y

Answer: 2 and 3

Question 54. Which of the following states are responsible for environmental pollution

1. pH value in rainwater is 5.6
2. Eutrophication
3. The BOD value in the water sample is 15 ppm
4. The amount of CO₂ in the atmosphere is 0.03%

Answer: 2, and 3

Question 55. Which processes occur in the troposphere

1. Photosynthesis
2. Combustion
3. Greenhouse effect
4. Acid rain

Answer: 1, 2, 3 and 4

Question 56. Which statements are true

1. Mainly the effects of HNO₃ are more in acid rain
2. NO is more toxic than NO₂
3. Ozone gas is responsible for the greenhouse effect
4. IR radiation cannot pass through CO₂ gas but gets absorbed by it

Answer: 3 and 4

Question 57. Which radical causes depletion in the Ozone layer

1. CH₃
2. F
3. Cl
4. Br

Answer: 3 and 4

Question 58. Which greenhouse gases are produced in the agriculture field

1. CH₄
2. NH
3. Nobel
4. SO

Answer: 1 and 4

Question 59. Which are the following statements are incorrect

1. SO₂ does not affect the larynx
2. SO₂ is a more harmful pollutant than SO₃
3. In the case of living cell NO₂ is more toxic than NO
4. There is no role of NO_x in photochemical smog

Answer: 1, 2 and 4

Question 60. Diseases caused by the harmful effects of SO₂ 

1. Digestion problem
2. Breathing problem
3. Bronchitis
4. Asthma

Answer: 2, 3 and 4

Question 61. Which of the following processes are responsible for the formation of CO₂ in the atmosphere

1. Respiration
2. Combustion of fossil fuel
3. Decay of animals
4. Production of cement in factories

Answer: 2 and 4

Question 62. Which of the following react to produce PAN

1. NO
2. O₂
3. Hydrocarbon
4. CO

Answer: 1, 2 and 3

Question 63. Which constituents of phytochemicals responsible for eye irritation

1. Ozone
2. PAN
3. Hydrocarbon
4. O₂

Answer: 1 and 2

Question 64. The main constituents of London smog are

1. Oxides of sulphur
2. O₂
3. O₃
4. Oxides of nitrogen

Answer: 1 and 4

Question 65. Which of the following is responsible for the depletion of the ozone layer in the stratosphere

1. So
2. CFCl
3. CF
4. CF Br

Answer: 2 and 4

Question 66. Which of the following are primary pollutants

1. PAN
2. SO
3. NOz
4. Me₂Hg

Answer: 2 and 3

Question 67. Contribution of CO₂ and CH₄ in greenhouse effects

1. The contribution of CO₂ is 50%
2. The contribution of CH₄ is 16%
3. The contribution of CO₂ is 19%
4. The contribution of CH₄ is 19%

Answer: 1 and 4

Question 68. In which region there is a greater possibility of formation of photochemical smog

1. Region where a large number of automobiles are used
2. Region where sulphur-containing coal is used
3. Marshy land region
4. Orest region

Answer:  1. Region where large number of automobiles are used

Question 69.  Acid rain is a dilute aqueous solution of which of the following pairs of acids

1. H₂SO₄ and HCl
2. H₂CO₃ and HCl
3. H₂SO₄ and HNO₃
4. H₂CO₃ and HCl

Answer: 3 . H₂SO₄ and HNO

Question 70. Which of the following metallic air pollutants is present in the gas emitted by motor vehicles

1. Iron
2. Lead
3. Copper
4. Mercury

Answer: 2.  Lead

Question 71.  Which of the following is not a Greenhouse gas

1. CFCs
2. Ammonia
3. Carbon dioxide
4. Methane

Answer: 2. Ammonia

Question 72. Which compound is responsible for hole formation in the stratosphere of the ozone layer

1. C₆F₆
2. C₆H₄Cl₂
3. CCl₂F₂
4. C₆H₆

Answer: 3. CCl₂F₂

Question 73. Which of the following gases emitted by motor vehicles is responsible for the formation of photochemical smog

1. SO₂
2. CO
3. NO
4. CO₂

Answer: 3. NO

Class 11 Chemistry Environmental Chemistry Very Short Question And Answers

Question 1. What are the chief air pollutants?
Answer: 

SO₂, SO₃, CO, CO₂, NO₂, NO, O₃, hydrocarbons, fine particles of solid or liquid suspended in air.

Question 2. What are the main pollutants emitted from thermal power plants?
Answer: CO, CO₂, NO, NO₂, fly ash etc.

Question 3. Name the sink of CO.
Answer: A special type of bacteria present in the soil, which converts CO into C0

Question 4.  Name two sinks of CO₂
Answer: Sea water (which dissolves CO₂) and plants (which use CO₂ for photosynthesis).

Question 5. Which aromatic compound is present in the air as particulate?
Answer: Polycyclic aromatic hydrocarbons (PAH) like benzopyrene

Question 6. What is PCB?
Answer:  PCB is polychlorinated biphenyl. It is highly toxic

Question 7. What is the role of particulates in the formation of clouds?
Answer:  Particulates act as nuclei in the formation of clouds.

Question 8.  What 31-6 are the main compounds responsible for causing damage to the ozone layer?
Answer:  Freons and nitric Oxide (NO) are the main compounds fo cause damage to the ozone layer

Question 9. What is the role of the builder in synthetic detergents?
Answer: It removes hardness-producing ions {viz., Ca²⁺, Mg²⁺) from water

Question 10. Which of the atmospheric layers contains the maximum ozone gas?
Answer: Stratosphere

Question 11. Mention the range of temperature of the atmosphere.
Answer: From -92°C to +1200°C

Question 12. What is the main source of carbon monoxide in the atmosphere?
Answer: Coal, Petrol and incomplete combustion of other fossil fuels,

Question 13. Which one is more harmful to the human body— CO or C°2?
Answer: CO

Question 14. What are the main pollutants produced by forest fires?
Answer: CO , CO₂ , NO , NO₂

Question 15. What are the major pollutants emitted by thermal power plants?
Answer: CO, CO₂, NO, NO₂,flash

Question 16. What are the ads present in acid rain?
Answer:  H₂SO₄, HNO₃ and HC

Question 17. What is the size of the particulates?
Answer: From 0.0002 μ to 500p

Question 18. What Is the main chemical responsible for the Bhopal gas tragedy?
Answer: Methyl isocyanate (MIC)

Question 19. By which disease do the workers of asbestos factories suffer?
Answer: Asbestosis

Question 20. Give one example of a fire extinguisher made by Pyrocool technology.
Answer: Pyrocoolfoam

Question 21. Which acid contributes most to the formation of acid rain?
Answer: Sulphuric acid (H₂SO₄)

Question 22. Mention the H limit of acid rain
Answer: From 5.6 to 3.5

Question 23. Give thefullform of’PAN’.
Answer: Peroxyacyl nitrate

Question 24. Which unit is used to measure the columnar density of O₃ gas in Earth’s atmosphere?
Answer: Dobson (Du) unit.

Question 25. Which gas has the maximum contribution to the greenhouse effect?
Answer: Carbon dioxide (CO₂)

Question 26. Among the air pollutants gas is responsible for the damage caused to the TajMahal
Answer: Sulphur dioxide (SO₂)

Question 27. Mention two diseases originating from water pollution
Answer: Cholera, typhoid

Question 28. What is the cause? Bhopal gas tragedy
Answer: Methyl isocyanate gas

Question 29. Name a nitrogen-fixing bacteria
Answer: Rhizobium

Question 30. Name the main air pollutant that is present in automobile exhausts.
Answer: Carbon monoxide (CO)

Class 11 Chemistry Environmental Chemistry Fill In The Blanks

Question 1. The total mass of gaseous substances in the atmosphere is nearly ____________________
Answer: 5.5 × 10¹⁵ ton

Question 2. The amount of CO₂ in the atmosphere is approximately is more harmful ___________________
Answer: 2. 0.031%

Question 3. CO₂ when mixed with blood, forms ___________________
Answer: Carboxyhaemoglobin,

Question 4. The word, ‘CFC’ means___________________
Answer: Chlorofluorocarbon

Question 5. The word, ‘PAN’ stands for___________________
Answer: Peroxyacyl nitrate

Question 6. The formation of ozone hole increases the tendency of human beings to be attacked by ___________________
Answer: Cancer

Question 7. One remarkable phenomenon happened in the troposphere is___________________
Answer: Green House effect

Question 8. Between NO₂ and NO__________________
Answer: NO₂

Question 9. The word, ‘PAH’ denotes __________________
Answer: Polyaromatic hydrocarbon

Question 10. The lung disease caused by silica is __________________
Answer: Silicosis

Question 11. Among the following gases _____________ is a greenhouse gas (NO₂, N₂O, SO₂ , NO)
Answer: NO

Question 12. The causes of Minamata is____________ containing effluent.
Answer: Mercury

Class 11 Chemistry Environmental Chemistry Warm-Up Question And Answers

Question 1. What Is fly ash?
Answer:

Fly ash is emitted from thermal power plants due to the combustion of coal. It consists of fine particles of SiO₂

Al₂O₃, CaO, Fe₂O₃, NO₂, SO₂, P₂O₃

Question 2. What are the sinks for CO₂ and CO gases?
Answer:

Sea water (CO₂ is soluble here) and some special bacteria (which absorb CO and convert it into CO₂

Question 3. Mention three hydrocarbons which function as air pollutants.
Answer:

1,3-butadiene, 1,2-benzopyrene, 1,2-benzanthracene

Question 4. Name three primary and three secondary air pollutants.
Answer:

1. Primary air pollutants: CO, SO₂, NO₂;
2. Secondary air pollutants: O₃, PAN, formaldehyde

Question 5. Name some hydrocarbons that are present in the atmosphere as organic particulates.
Answer:

Methane, benzene, benzopyrene

Question 6. Name the compounds responsible for ozone hole formation.
Answer:

Chlorofluorocarbons, halons etc

Question 7. Why the temperature of the stratosphere increase with the increase in height?
Answer:

The ozone layer in the stratosphere absorbs the harmful ultraviolet radiation coming from the sun and converts it into heat. Thus the temperature of this layer increases.

Question 8. What are halons? State their uses
Answer:

Halons arc halocarbons. They used as fire extinguishers

Question 9. Why is the tropospheric ozone harmful?
Answer:

Ozone gas present In the troposphere acts as a greenhouse gas.

Question 10. Mention the season and time of the day when London smog is generally observed.
Answer:

During winter, particularly after evening or early in the morning London smog is generally observed

Question 11. Mention the season and time of the day when Los Angeles smog is generally observed
Answer:

During the mid-days of the summer season when the sun shines brightly this kind of smog is observed.

Question 12. Which region is most susceptible to the formation of photochemical smog?
Answer:

Photochemical smog is mostly observed in big cities, where there is considerable vehicular traffic on the roads throughout the whole day and night

Question 13. Why is photochemical smog called Los Angeles smog?
Answer:

This type of smog was first discovered in the city of Los Angeles in America in the year 1950. So it is called Los Angeles smog.

Question 14. ‘There is “a ‘ tendency of environmental degradation of Tajmahal Explain
Answer:

SO₂ released from the industries situated around Tajmahal reacts photochemically with atmospheric O₂ and water vapour to form H₂SO₄. The H₂SO₄ hence produced reacts with white marble and damages it

Question 15. Give two examples of chlorinated organic pesticides.
Answer:

p, p’ -dichlorodiphenyltrichloroethane (DDT) and benzene hexachloride (BHC)

Question 16. What is loam soil?
Answer:

The soil containing almost equal amounts of sand, slit and clay along with humus, 34% of air and 64% of water is called loam soil

Question 17. Give two examples of each insecticide and herbicide
Answer:

Insecticide: DDT, BHC

Herbicide: 2,4-dichloro phenoxy acetic acid, dioxin

Question 18.   Write one effect of the depletion of the ozone layer and one measure for the prevention of ozone layer depletion.
Answer:

We have to reduce the use of compounds made by CFCs and halons

Question 19. Explain tropospheric pollution in 100 words.
Answer:

Tropospheric pollution occurs due to the presence of undesirable poisonous gases and solid particles in the air.

Question 20. What is anoxia or asphyxiation?
Answer:

Acute oxygen starvation in the body due to poisoning by carbon monoxide is called anoxia or asphyxiation

Question 21. What is humification
Answer:

The process of decomposition of organic matter (roots, leaves etc.) in the soil by microorganisms to produce humus is called humification.

Question 22. Why does the population of fish get hindered by clouds? thermally polluted water
Answer:

Thermal pollution increases the temperature which in turn decreases the DO level of the water. Thus, it affects the fishes badly and their growth gets retarded

Question 23.  Name four natural sources of air pollution.
Answer:

Volcanic eruptions, forest fires, lightning, decomposition of dead plant and animal bodies in marshyland.

Class 11 Chemistry S Block Elements Group 1 Elements (Alkali Metals) Introduction

Class 11 S-Block Notes All the alkali metals have one valence electron ( ns¹ ) outside the noble gas core. The loosely held s -electronin the outermost valence shell makes them the most electropositive metals.

• To get the stable electronic configurations of noble gases, they readily lose the valence electron to generate the monovalent (M⁺) ions. Hence, they are never found in a free state but in the combined state of nature.
• Since the last electron enters ns -orbital, these are called s -block elements.
• Since all these elements have similar valence shells or outer electronic configurations, all the alkali metals exhibit a striking resemblance in their physical and chemical properties and they are placed in a definite group (Gr-1).
• Lithium shows some abnormal behavior as its electronic configuration is slightly different from the rest of the members of Gr-1 and also because of its extremely small atomic and ionic radii.

Again, lithium shows some similarities with magnesium present in the group- 2 of the third period

Electronic configuration of alkali metals: 

Occurrence Of Alkali Metals

• Since the alkali metals are highly reactive, they do not exist in a free state. In nature, they mostly occur as compounds like halides, oxides, silicates, borates, and nitrates.
• According to abundance, lithium is placed at the 35th position. It mainly occurs in nature in the tire form of silicates,
• For example: Spodumene: LiAl(SiO₃)₂ and Lepidolite: Li₂Al₂(SiO₃)₃(F, OH)₂
• Sodium and potassium are respectively placed at 7th and 8th position in order of their abundance. Sea water is a major source of NaCl and KCl.
• Sodium is abundantly present in the form of rock salt (NaCl). Other important minerals are Chile salt petre: NaNO₃, borax: Na₂B₄O₇.10H₂O, mirabilite: Na₂SO₄, and trona: Na₂CO₃-NaHCO₃-2H₂O.
• Important ores of potassium are sylvite: KCl, carnallite: KCl-MgCl₂-6H₂O, and feldspar: (K₂O-Al₂O₃-6SiO₂).
• Rubidium and cesium are much less abundant than lithium. Radioactive francium does not occur appreciably in nature. It is obtained from the radioactive decay of actinium
• ²²⁷Ac ₈₉ → ²²³Fr₈₇  +⁴He₂  Its longest-lived isotope ²²³Fr₈₇  has a half¬life period of only 21 minutes.
• Since most of the compounds of alkali metals are water soluble, they are found in adequate amounts in seawater.

General Trends In Atomic And Physical Flv Properties Of Alkali Metals

The alkali metals show regular trends in their physical and chemical properties with an increase in atomic number. Some important atomic and physical properties of alkali metals are given in the following table:

Atomic and physical properties of alkali metals:

Class 11 S-Block Notes

General trends in different atomic and physical properties of alkali metals and their explanations:

1. Atomic and ionic radii 

The atomic and ionic radii of alkali metals are the largest in their respective periods and these values further increase on moving down the group from Li to Cs.

Atomic and ionic radii  Explanation:

On moving from left to right in a period, the number of electronic shells remains the same but the nuclear charge increases with each succeeding element Thus, the valence shell electrons experience a greater pull towards the nucleus and this results in successive decreases in atomic and ionic radii with an increase in atomic number

• Thus, the atomic and ionic radii of alkali metals are the largest in their respective periods.
• On moving down the group, a new electronic shell is ) added to each element and the nuclear charge increases with an increase in atomic number.
• The addition of an electronic shell tends to increase the size of the atom but the increase in nuclear charge tends to decrease the atomic radii by attracting the electron cloud inward. Thus, the two factors oppose each other.
• However, the increase in the number of shells increases the screening effect of the inner electrons on the outermost s -electron and as the screening effect is, quite large, it overcomes the contractive effect of the increased nuclear charge.
• The net result is an increase in atomic and ionic radii down the group from Li to Cs.

2. Ionization enthalpy

The first ionization enthalpies of alkali metals are the lowest in their respective periods. Explanation: Since the alkali metal atoms are largest in their respective periods, their outermost electrons being far away from the nucleus experience less force of attraction and hence, can be removed easily.

1. Ionization enthalpy of group-1 alkali metals decreases down the group.

Explanation:

Since the alkali metal atoms are largest in their respective periods, their outermost electrons being far away from the nucleus experience less force of attraction and hence, can be removed easily.

2. Ionisation enthalpy7 of group-1 alkali metals decreases down the group

Explanation:

On moving down the group from Li to Cs, the distance of the valence s-electron from the nucleus progressively increases due to the addition of a new shell with each succeeding element With an increase in the number of inner shells, the screening effects progressively increase and as a result, the effective nuclear charge experienced by the valence electron progressively decreases and hence, the ionization enthalpies decrease down the group.

3. The second ionization enthalpies of alkali metals are very high.

Explanation:

The monovalent cation formed by the removal of an electron from the alkali metal atom has a very stable noble gas configuration,

For example – 

1. Li⁺: 1s² or [He], Na⁺: 1s²2s²2p⁶ 
2. [Ne], k⁺: 1s²2s²2p⁶3s²3p⁶ or [Ar] etc.

Removal of another electron from the monovalent ion having a stable noble gas configuration is very difficult and requires a huge amount of energy. For this reason, the second ionization enthalpies of alkali metals are very high.

3. Hydration of ions, hydrated radii, and hydration enthalpy

The salts of alkali metals are generally ionic and are soluble in water because the cations get hydrated in water to form hydrated cations: M⁺ + aq —> [M(aq)]⁺.

1. The degree of hydration of ions and the hydrated radii decrease as we move down the group

Explanation:

The smaller the cation, the greater its degree of hydration. Since ionic radii increase down the group, the degree of hydration decreases, and consequently, the radii of die-hydrated ions decrease from Li⁺ to Cs⁺.

2. The order of mobilities of die alkali metal ions in aqueous solution is: Li⁺ < Na⁺ < K⁺ < Rb⁺ < Cs⁺

Explanation:

Smaller ions are more easily hydrated. As Li+ is the smallest ion among the given ions, it is most easily hydrated and has the least ionic mobility in an aqueous solution whereas Cs+ is the largest and is least hydrated. So its mobility is die highest.

3. Ionic conductance of the hydrated ions increases from [Li(m7)]+ to [Cs(aq)]+.

Explanation:

The ionic conductance of these hydrated ions increases from [Li(aq)]⁺ to [Cs(ag)]⁺ because die size decreases and mobility increases in this order. Hydration of ions is an exothermic process. The energy released when 1 gram-mol of an ion undergoes hydration is called hydration energy or hydration enthalpy,

4. Hydration enthalpy of alkali metal ions decreases from Li⁺ to Cs⁺.

Class 11 Physics	Class 12 Maths	Class 11 Chemistry
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Class 11 S-Block Notes

Explanation:

The hydration enthalpy of an ion depends upon the ratio of charge to radius (q: r). Since the radii of alkali metal ions increase down the group, the hydration enthalpies decrease from Li⁺ to Cs⁺. Li⁺ ion has the maximum degree of hydration and for this reason, most of the lithium salts are found to be hydrated

For example: LiCl-2H₂O, LiClO₄-3H₂O etc.

4. Oxidation State

Alkali metals exhibit a +1 oxidation state in their compounds and it remains restricted in a +1 state only.

Oxidation State Explanation: 

Alkali metals have low ionization enthalpies and by losing their valence s -electrons they acquire the stable electronic configurations of the nearest noble gases. Thus, they have a strong tendency to form M⁺ ions and exhibit a +1 oxidation state in their compounds.

The second ionization enthalpies required to pull out another electron from M⁺ ions having very unstable noble gas electronic configuration are very high indeed and are not available under the conditions of chemical bond formation. Hence,v the alkali metals do not form M²⁺ ions, i.e. their oxidation state remains restricted to +1 state.

5. Metallic character

The elements of this group are typical metals that are soft (can be easily cut with a knife) and light. When freshly cut, they are silvery white but on exposure to air, they turn tarnished. The metallic character, which refers to the level of reactivity of a metal, increases on moving down the group.

Metallic character Explanation:

As the ionization enthalpy decreases down the group,  the tendency to lose the valence electron increases, and consequently, the metallic character increases.

6. Photoelectric effect

Alkali metals (except Li) exhibit a photoelectric effect. The emission of electrons from the surface of a metal exposed to electromagnetic radiations of suitable wavelength is called the photoelectric effect.

Photoelectric effect Explanation:

Due to low ionization enthalpies, the alkali metals exhibit a photoelectric effect. It is to be noted that lithium having the highest ionisation enthalpy does not exhibit a photoelectric effect. Cesium having the lowest ionisation enthalpy possesses the highest tendency to exhibit a photoelectric effect.

Potassium anti-cesium, rather than lithium is used in photoelectric cells:

The ionization enthalpies of potassium and cesium are much lower than that of lithium. For this reason, these two metals on exposure to light easily emit electrons from their surface but lithium does not. Hence, potassium and cesium rather than lithium are used in photoelectric cells.

7. Electronegativity

The alkali metals have low electronegativity which further decreases down the group.

Electronegativity Explanation:

The alkali metals having ns¹ electronic configuration preferably show electron releasing tendency rather than electron accepting. Thus, they have low electronegativities. Since the atomic sizes increase down the group, the tendency of atoms to hold their valence electrons decreases down the group, and consequently, electronegativity decreases down the group.

Class 11 S-Block Notes

8. Conductivity

Alkali metals are good conductors of heat and electricity.

Conductivity Explanation:

Due to the presence of loosely bound valence electrons (ns¹) which are free to move throughout the metal structure, the alkali metals are good conductors of heat and electricity.

9. Melting and boiling points 

Melting and boiling points: Melting and boiling points of alkali metals are low and decrease down tire group.

Melting and boiling point Explanation:

The cohesive energy that binds the atoms in the crystal lattices of these metals is relatively low (weak metallic bonding) due to the presence of only one valence electron (ns¹) which can take part in bonding. Hence, their melting and boiling points are low. These further decrease down the group as the strength of the metallic bonds and cohesive energy decrease with increasing atomic size.

10. Nature of bonds formed

Alkali metals form ionic compounds and the ionic character of compounds increases down the group from Li to Cs.

Nature of bonds formed Explanation:

For low ionization enthalpies, alkali metals readily form monovalent cations by losing their valence electrons. As ionization enthalpies decrease down the group, the ionic character of the compounds increases down the group.

11. Density

The densities of alkali metals are quite low and increase down the group from Li to Cs.

Density Explanation:

Due to their large atomic size and weak metallic bond, alkali metals have low density. Both the atomic volume and the atomic mass increase down the group but the corresponding increase in atomic mass is not balanced by the increase in atomic volume. As a result, the densities of alkali metals increase down the group. However, the density of K is less than that of Na because the atomic size and atomic volume of potassium are quite higher than that of sodium. As a result, the ratio of mass/ volume decreases.

Li is the lightest metal having a density of 0.53 g. cm^-3. It cannot be preserved in kerosene because it floats over it. Generally kept wrapped in paraffin wax.

12. Flame coloration

Alkali metals or their salts on heating in the flame of the bunsen burner, impart characteristic colors and they can be easily identified from the color of the flame

Class 11 S-Block Notes

Flame coloration Explanation:

Ionization enthalpies of alkali metals are not much higher. Thus, when an alkali metal or its salt (especially chloride due to its more volatile nature) is heated in a Bunsen burner flame, the electrons in the valence shell get excited and jump to higher energy levels by absorbing energy. When the excited electron drops back to its ground state, the emitted radiation falls in the visible region and as a result, alkali metals or their salts impart color to the flame.

Alkali metals can be detected by flame tests and can be estimated by flame photometry or atomic absorption spectroscopy.

13. Softness

Alkali metals are soft (can be cut easily with die help of a knife) and their softness increases down the group.

Softness Explanation:

The softness of alkali metals is due to their low cohesive energy and weak metallic bonding. Further, on moving down the group, the strength of metallic bonding decreases due to an increase in atomic size and as a result, the softness of the metals increases down the group.

Chemical Properties Of Alkali Metals

Alkali metals are highly reactive. Such reactivity may be attributed to their large atomic size, low ionization enthalpies, and low heats of atomization. –

Action of air and moisture

Alkali metals, being highly reactive, react readily with atmospheric oxygen to form oxides. These oxides further react with moisture to form hydroxides which in turn produce carbonates by reacting with atmospheric CO₂

Metal oxides also react with CO₂ to form carbonates.

These metals lose their glossiness and become tarnished due to the formation of carbonate layers on their surface. To protect from atmospheric oxygen and moisture, these metals are always stored in inert hydrocarbon solvents such as kerosene, petroleum ether, etc.

Reaction with oxygen

When the alkali metals are heated with oxygen or excess air, they form different types of oxides depending upon the nature of the metal involved. Lithium mainly forms monoxide (Li₂O), sodium forms peroxide (Na₂O₂), and the other alkali metals (K, Rb, and Cs) mainly form superoxides having the general formula MO₂. The temperature required for the reaction decreases down the group from Li to Cs.

4Li + O₂ → 2Li₂ O₂ (Lithium monoxide)

2Na + O₂ → Na₂O₂ (Sodium peroxide)

M +O₂ → MO₂ (Superoxide) [here, M = K, Rb, Cs]

 Oxygen Explanation

A smaller cation can stabilize a smaller anion while a larger cation can stabilize a larger anion. If both the ions are similar in size, the coordination number will be high and this results in higher lattice energy.

A cation having a weak positive electric field can stabilize an anion having a weak negative electric field. Li⁺  ion and oxide ion (O^2-) have small ionic radii and high charge densities. Hence, these small ions combine to form a very’ stable lattice of Li₂O.

Sodium forms peroxide (Na₂O₂) but potassium forms superoxide (KO₂), even though the peroxide ion,  is larger in size than the superoxide ion, This can be explained in terms of charge densities. Due to the bigger size

Na+ ion has a weaker positive field around It and therefore it can stabilise peroxide Ion which also has a weaker negative field around it. Thus, Na⁺ forms peroxide. K⁺ ion is still bigger in size and the magnitude of the positive field around it is much weaker so it can’t stabilize superoxide ion which also has a much weaker negative field around it. Hence, K forms superoxide.

According to valence bond theory, two O-atoms in superoxide ion (O₂^–) are attached by a 2- 2-electron bond (common covalent bond) and a 2-electron bond. As the unpaired electron is present in the 2-electron bond, the superoxide ion is paramagnetic and all tire superoxides are colored (LiO₂ and NaO₂ are yellow, KO₂ & CsO₂ are orange, RbO₂ is brown).

According to MO theory, there is an unpaired electron in one of the n antibonding orbitals and because of this, the superoxide ion is paramagnetic. The electronic configuration of O₂^–

Class 11 S-Block Notes

O₂^–  ion present in common oxides [For example,  Li₂O, NaO₂etc.) and the 02~ ion in peroxides (For example,  Na₂O₂), contain no unpaired electron and for this reason, these are diamagnetic and colorless.

Reaction with water

The alkali metals having high negative reduction potentials (E°) can act as a better-reducing agent than hydrogen. Hence, they react with water to form water-soluble hydroxides and liberate hydrogen gas.

2M + 2H₂O → 2MOH + H₂T [M = alkali metal]

Reactivity with water increases down the group as the electropositive character of the metals increases clown the group. Lithium decomposes water slowly. Sodium reacts with water vigorously. K, Rb, and Cs react with water explosively and the evolved hydrogen gas catches fire.

Alkali metals also react with compounds containing acidic H-atoms

For example:  halogen halides (HX), alcohols (ROH), acetylene HC=CH, etc., to form their corresponding salts and H₂

2Na + 2HX → 2NaX  (Sodium halide) + H₂ ↑

Li + 2C₂H₅OH → 2 C₂ H₅ OLi (Lithium ethoxide) + H₂ ↑

2Na + 2HC=CH →  2NaC = CH (Sodium acetylide)+ H₂ ↑

The standard electrode potential of Li is most negative while that of sodium is least negative i.e., in the reaction with water, Li releases a greater amount of heat than sodium. Despite that, Li reacts less vigorously with water than Na. j This can be explained concerning chemical kinetics. Na has a low melting point and the heat of the reaction is sufficient j to melt it. Molten metal spreads out and exposes a relatively large surface to water and as a result, it reacts with water 1 readily and violently. On the other hand, the melting point of Li is much higher and the heat of the reaction is not sufficient to melt it. Hence, its surface area does not increase and it reacts slowly with water.

Reaction with dihydrogen

All alkali metals react with dihydrogen at about 673 K (Li at 1073 K) to form colorless, crystalline hydrides (MH). These hydrides have a high melting point.

1. The reactivity of the alkali metals towards dihydrogen decreases down the group.

Explanation:

As the size of the metal cation increases down the group, the lattice energy of the hydrides decreases down the group. Consequently, the reactivity of the alkali metals towards hydrogen decreases down the group,

2. The ionic character of the alkali metal hydrides increases from Li to Cs.

Explanation:

Since the ionization enthalpy of alkali metals decreases down the group, the tendency to form cations as well as the ionic character of the hydrides increases.

Reaction with halogens

All alkali metals react vigorously with halogens to form crystalline halide compounds having the general formula MX. Lithium halides are covalent due to the very high polarising power of the Li⁺ ion. Halides of other alkali metals are ionic in nature’

⇒ \(2 \mathrm{M}+\mathrm{X}_2 \rightarrow 2 \mathrm{M}^{+} \mathrm{X}^{-}\) ( X = F, Cl, Br or I)

The reactivity of the alkali metals towards a particular halogen increases down the group.  Due to a decrease in ionization enthalpies or an increase in the electropositive character of the metals down the group, the reactivity increases down the group.

The reactivity of halogens towards a particular alkali metal decreases in the order:

F₂> Cl₂ > Br₂ > I₂

Reducing nature

The alkali metals act as strong reducing agents because of their low ionization enthalpies. Since the ionization enthalpies decrease on moving down the group, therefore, in the free state the reducing power also increases in the same order, i.e., Li  < Na < K < Rb < Cs.

Class 11 S-Block Notes

The tendency of a metal to lose an electron in solution is measured by its standard electrode potential (E°). The alkali metals have low values (higher negative values) of E° and so they have a strong tendency to lose electrons and can act as strong reducing agents. Lithium, although, has the highest ionization enthalpy, is the strongest reducing agent in solution (E° = -3.04 V). On the other hand, Na is the weakest reducing agent and the reducing character increases from Na to Cs, i.e., Na < K < Rb < Cs.

Explanation of anomalous behavior of lithium

The anomalous behavior of lithium can be explained because the ionization enthalpy is the property of an isolated atom in the gaseous state while the standard electrode potential is concerned when the metal atom goes into solution.

The ionization enthalpy involves the change:

M(g) → M⁺+(g) + e, while the standard electrode potential involves the change: M(s) → M⁺(aq) + e.

The latter change occurs in three steps as follows:

1. M(s)→ M(g) – sublimation enthalpy
2. M(g)→ M+(g) + e – ionisation enthalpy
3. M⁺ (g) + H₉O → M⁺(aq) + hydration enthalpy

The overall tendency for the change depends on the net effect of these three steps. Among the alkali metal cations, Li⁺ ion has the maximum tendency to get hydrated due to its very small size. The high hydration enthalpy compensates the energy required in the first two steps to a large extent and the overall energy required to convert M(s) to M⁺ (aq) is minimum for lithium.

Thus, small size and high hydration enthalpy are responsible for the strong reducing character of lithium.

The solution in liquid ammonia

Alkali metals dissolve in liquid ammonia to give highly conducting deep blue solutions which are highly reducing and paramagnetic. As the concentration increases (> 3M), the color of the solution changes to copper-bronze. These concentrated solutions are diamagnetic.

Solution in liquid ammonia Explanation:

1. When an alkali metal is dissolved in liquid v ammonia, ammoniated cations, and ammoniated electrons are formed as shown below:

M + (x+ y)NH₃ →  [M(NH₃)_x]⁺ (Ammoniated cation) + [e(NH₃)_y]^–(Ammoniated electron)

2. The blue color of these solutions is due to the excitation of the free ammoniated electrons to higher energy levels by absorbing energy corresponding to the red region of visible light.  The transmitted light is blue which imparts a blue color to the solutions.

3. With the increase in the concentration of the alkali metal, the formation of clusters of metal ions starts and because of this, at a much higher concentration (> 3M) the solutions possess metallic luster and attain the color of copper-bronze.

4. These blue solutions are highly conducting because of the presence of ammoniated electrons and ammoniated cations but the conductivity decreases with increasing concentration as the ammoniated cations get attached to the free unpaired electrons.

5. These blue solutions are paramagnetic due to the presence of unpaired electrons. However, tire paramagnetism decreases with increasing concentration due to the association of ammoniated electrons to yield diamagnetic species.

6. The free ammoniated electrons make these solutions very powerful reducing agents.

7. These solutions when kept, form metal amides and release H₂. However, these solutions can be stored in anhydrous conditions in the absence of impurities like Fe, Pt, Zn, etc.

M⁺(am) + e(am) + NH₃(l)→ MNH₂(am) + ½H₂(g)

Where ‘am’ stands for ‘solution in ammonia

2M + 2NH₃→ 2MNH₂ (metal amide)+ H₂

Class 11 S-Block Notes

Extraction of alkali metals

Alkali metals cannot be extracted by applying common processes used for the extraction of other metals.

Alkali metals Explanation:

• The alkali metals are strong reducing agents. Hence, they cannot be extracted by reduction of their oxides or other compounds.
• Since they are highly electropositive, the method of displacing them from their salts by any other element is not possible.
• The aqueous solution of their salts cannot be used for extraction by electrolytic method because hydrogen, instead of the alkali metal is discharged at the cathode (discharge potentials of alkali metals are much higher).
• However, by using Hg as a cathode, the alkali metals can be deposited but in that case, the alkali metals readily combine with mercury to form amalgams from which the recovery of metals becomes quite difficult.
• The electrolysis of their fused salts (usually chlorides) is the only successful method for their extraction, Another metal . ‘ chloride is generally added to lower its fusion temperature

General Characteristics Of The Compounds Of Alkali Metals

The compounds of alkali metals are predominantly ionic. Some of the general characteristics of these compounds are described below.

1. Oxides and hydroxides

1. Typical oxides or monoxides of alkali metals

For example: Li₂O and Na₂O are white ionic solids and basic. These oxides react with water to form strong alkalis (MOH).

Example:  Na₂O + HO → 2NaOH

2. All peroxides are strong oxidizing agents. They react with water or acid to give hydrogen peroxide (H₂O₂) and the corresponding metal hydroxide. Na₂O₂ is widely used as an oxidizing agent in inorganic chemistry.

M₂O₂ + 2H₂O →  2MOH + H₂O₂

Example: Na₂O₂ + 2H₂O₂→ 2NaOH + H₂O₂

3. Superoxides are stronger oxidizing agents than peroxides and react with water or acid to give both H₂O₂ and O₂ along with metal hydroxide.

2MO₂ + 2H₂O→  2MOIH + H₂O₂+ O₂

Example: 2KO₂ + 2H₂O₂ →2KOH + H₂O₂ + O₂

The alkali metal hydroxides (MOH) are all white crystalline solids and corrosive. They are the strongest of all bases and readily dissolve in water. Due to excess hydration, a large amount of heat is released. These hydroxides are thermally stable except Li OH. The basic strength of alkali metal hydroxides increases on moving down the group from Li to Cs.

Explanation:

The ionization enthalpies of alkali metals decrease on moving down the group and this causes a weakening of the bond between the alkali metal and the hydroxyl group (M —OH). This results in an increase in the concentration of hydroxyl ions in the solution, i.e.,  the basic character of the solution increases on moving down the group.

Thus, the basic strength of the hydroxides follows the order:

CsOH > RbOH > KOH > NaOH > LiOH

2. Halides

The alkali metal halides can be prepared by combining metals directly with halogens or by reacting appropriate oxides, hydroxides or carbonates with aqueous halogen acids (HX).

2M + X₂  →  2MX ; M₂ O + 2HX↓ 2MX + H₂ O

MOH + HX→ MX + H₂ O

M₂ CO₃+ 2HX→ 2MX + HO₂ + CO₂

The enthalpy of formation (ΔH°_f) of alkali metal halides is highly negative. For a given metal, AHj values decrease from fluoride to iodide. These halides are colorless crystalline solids having high melting and boiling points.

1. The melting point of halides of a particular alkali metal decreases as:

Fluoride > Chloride > Bromide > Iodide.

Explanation:

For a particular alkali metal ion, the lattice enthalpies decrease as the size of the halide ion increases.

Lattice enthalpies of NaP, NaCl, Nalir, and Nal are 893, 770,730, and 685 kJ. mol^-1 respectively, A.s the lattice enthalpy decreases, the energy required to break the crystal lattice decreases, and consequently, the melting points decrease. Thus, the melting points of NaF, NaCl, NaBr, and Nal are found to be 1201K, 10IMK, 1028 K, and 944 K respectively.

2. For a particular halide ion, the melting point of IJX is less than that of

NaX and thereafter the melting points decrease on moving down the group from Na to Cs.

Class 11 S-Block Notes

Explanation:

The melting point of LiCl (887K) is less than that of NaCl (I084K), because LiCl is covalent (for smaller atomic size of Li compared to that of Na), but NaCl is ionic. Thereafter, the order of melting point is:

NaCl(1084K)>KCl(1039K)>RbCI(988K)>CsCl(925K) This is observed because the lattice enthalpies decrease as the size of the alkali metal atom increases.

3. Solubilities of the alkali metal halides (except fluorides) decrease on moving down the group since the decrease in hydration enthalpy is more than the corresponding decrease in the lattice enthalpy.

For example, the difference in lattice enthalpy between NaCl and KCl is 67kJ. mol^-1 whereas the difference in hydration enthalpy between Na⁺ and K⁺ ion is 76 kj -mol^-1 Thus, KCl is relatively less soluble in water compared to NaCl.

Explanation:

The solubility of a salt in water depends on its lattice enthalpy as well as its hydration enthalpy. In general, if hydration enthalpy > lattice enthalpy, the salt dissolves in water but if the hydration enthalpy < lattice enthalpy, the salt does not dissolve.

Further, the extent of hydration depends on the ionic size. The smaller the size of the ion, the more it will get hydrated and the greater will be its hydration enthalpy. LiF, for example, is almost insoluble in water because of its higher lattice enthalpy (-1005 kJ . mol^-1 ).

On the other hand, the low solubility of Csl in water is due to smaller hydration enthalpies of the two large ions [-276(Cs⁺)-305(I^–) = -581 kJ.-mol^-1]. o Due to the smaller size and relatively higher electronegativity of Li, lithium halides except LiF are predo¬minantly covalent and hence, are soluble in organic solvents such as acetone, alcohol, ethyl acetate etc.

In contrast, sodium chloride, being ionic, is insoluble in organic solvents.

3. Soils of oxoacids

Alkali metals react with c to acids such as carbonic acid (H₂CO₃), nitric acid (HNO₃), sulphuric acid (H₈SO₄), etc., to form corresponding salts and release H₂. Due to the high polarising power and lattice energy of small Li ions, lithium salts behave abnormally.

4. Nature of carbonates and bicarbonates

All alkali metals form carbonates of the type M₂CO₃. Since the alkali metals are highly electropositive, their carbonates are remarkably stable up to l000°C above which they first melt and then decompose to form oxides. These salts are readily soluble in water. As electropositive character increases down the group, the stability of carbonates increases in the same order:

Cs₂CO₃ > Rb₂CO₃ > K₂CO₃ > Na₂CO₃ > Li₂CO₃

Li₂CO₃ is insoluble in water and unstable towards heat. It decomposes readily to give Li₂O and CO₂.

Explanation:

1. The very small L ion exerts a strong polarising power on the large carbonate (CO₃^2-) ion and distorts the electron cloud of its nearby oxygen atom.

This results in the weakening of the C—O bond and the strengthening of the Li — O bond. This eventually facilitates the decomposition of Li₂CO₃ leading to the formation of Li₂O and CO₂. %

2. The crystal lattice formed by a smaller Li⁺ ion with a smaller O₂ ion is more stable than that 2  formed by a larger CO₃ ion and a smaller Li+ ion. This also favors the decomposition of Li₂CO₃

Class 11 S-Block Notes

3. The aqueous solution of carbonates is alkaline. This is because carbonates being the salts of strong bases and weak acids (H₂CO₃) undergo hydrolysis.

M₂CO₃ + 2H₂O  ⇌  2MOH (strong base) + H₂CO₃ (weak acid)

4. Bicarbonates or hydrogen carbonates (MHCO₃) of the alkali metals except LiHCO₃ are obtained in the solid state. These bicarbonates are soluble in water and stable towards heat. On strong heating, all the bicarbonates undergo decomposition to yield carbonates with the evolution of carbon dioxide.

2MHCO₃ (heat)→ M₂CO₃ + CO₂ + H₂O

As the electropositive character of the metals increases down the group from Li to Cs, the stability of the bicarbonates increases in the same order.

5. Nature of nitrates

The alkali metal nitrates (MNO₃) are prepared by the action of HNO₃ on the corresponding carbonates or hydroxides. They are ionic crystalline solids having low melting points and are highly soluble in water. On strong heating, they (except LiNO₃ ) decompose into nitrites and at higher temperatures oxides.

For example:

LiNO₃ decomposes readily on heating to give

6. Nature of sulphates

The alkali metals form sulfates of the type M₂SO₄. All the sulfates except Li₂SO₄ are soluble in water. The sulfates when fused with charcoal, form sulphides.

M₂SO₄.+ 4C→  M₂ S + 4CO

Sulfates of alkali metals form double salts with the sulfates of trivalent metals like Fe, Al, Cr, etc. These double salts crystallize with a large number of water molecules to form alum. A typical example is potassium aluminum

[K₂SO₄→ Al₂(SO₄)₃ -24H₂O].

Lithium sulfate (Li₂SO₄) is not known to form alum.

Anomalous Behaviour Of Lithium (Li) And Similarity Between Li And Mg

Although lithium, the first element of group 1, exhibits most of the characteristic properties of this group, yet it differs from other members of this group in several respects.

Reasons for anomalous behavior of lithium

1. Both Li- atom and Li⁺ ion have very small sizes.
2. Much higher polarising power of very small Li⁺ ion results in increased Points of difference between lithium and other alkali metals covalent character of its compounds.
3. Lithium has the lowest electropositive character, the highest ionization enthalpy, and the highest electronegativity compared to the rest of the members.
4. Non-availability of d -d-orbital in its valence (outermost) shell.
5. Strong intermetallic bonding (cohesive force) due to its small size. On the other hand, lithium shows a diagonal relationship with magnesium

Class 11 S-Block Notes

Points of difference between lithium and other alkali metals:

Reasons for the similarities between lithium & magnesium:

Lithium exhibits a diagonal relationship with the 3rd-period group-2 element, magnesium. Reasons for the similarities between lithium and magnesium

• The atomic as well as ionic radii of Li and Mg are almost the same (Li⁺ = 76 pm and Mg²⁺ = 72 pm).
• Both lithium and magnesium have almost similar electronegativities (Li = 0.98 and Mg = 1.2).

Similarities between lithium and magnesium:

Uses Of Alkali Metals

Class 11 S-Block Notes

Preparation, Properties, And Uses Of Some Important Compounds Of Sodium

1. Sodium carbonate (washing soda), (Na₂CO₃-10H₂O)

1. Manufacture: Ammonia-soda or Solvay process

Sodium carbonate is commonly known as washing soda. It is generally manufactured by the Solvay process or ammonia-soda process.

Principle: When carbon dioxide is passed through an aqueous solution of NaCl (brine, 28% NaCl solution) saturated with ammonia, sodium bicarbonate is formed.

NH₃ + CO₂ + H₂O > NH₄HCO₃

NH₄HCO₃ + NaCl  ⇌ NaHCO₃ + NH₄Cl

Due to the common ion effect of Na⁺ ion, sodium bicarbonate so formed gets precipitated. Such removal of solid NaHCO₃ shifts the reaction more and more towards the right. This results in a greater yield of NaHCO₃.  In this way, a nearly two-thirds portion of NaCl is converted into NaHCO₃. The precipitated NaHCO₃ is then filtered off, dried, and heated at 150°C to get sodium carbonate.

Evolved CO₂ is reused to saturate the ammoniated brine.

Class 11 S-Block Notes

Raw materials:

1. Brine solution (28% aqueous solution of NaCl),
2. Limestone or calcium carbonate (CaCO₃) it is the source of CO₂ and
3. Ammonia.

2. Description of the process

Preparation of ammoniated brine: 

1. This process is carried out in the absorption tower made of iron

2.  From an overhead tank, brine is allowed to trickle down slowly along the tower and ammonia gas from the ammonia recovery tower which is mixed with a small amount of CO₂ is allowed to pass through a tube situated near the bottom of the tower. As a result, the brine solution gets saturated with ammonia while calcium chloride and magnesium chloride are present as impurities in commercial.

3. Sodium chloride gets precipitated as their corresponding insoluble carbonates.

2NH₃ + CO₃ + H₂O → (NH₄)₂CO₃

CaCl₂ + (NH₄)CO₃ → 2NH₄Cl + CaCO₃↓

MgCl₂ + (NH₄)2CO₃ → 2NH₄Cl + MgCO₃↓

4. The ammoniated brine is then filtered to remove the precipitated calcium and magnesium carbonates and the filtrate thus obtained is passed into the carbonation tower.

Class 11 S-Block Notes

Carbonation of ammoniated brine:

1. This operation is carried out in a long cast iron tower (carbonation or Solvay tower). The tower is fitted with several horizontal plates.

2. The ammoniated brine solution is trickled down from the top of the tower while CO₂ gas from the lime kiln is introduced into the tower under high pressure through a pipe fitted at the base of the tower.

3. In this way, CO₂ comes in contact with the descending stream of ammoniated brine and they react with each other to form ammonium bicarbonate which subsequently combines with NaCl to produce sodium bicarbonate and ammonium chloride.

NH₃ + CO₂ + H₂O →  NH₄HCO₃

NaCl + NH₄HCO₃ →  NaHCO₃↓ + NH₄Cl

Separation of sodium bicarbonate:

1. The solution coming out of the carbonation tower contains crystals of NaHCO₃. These are separated by passing the solution through vacuum filters.

2. The separated sodium bicar¬bonate is washed with water to remove any sodium or ammonium chloride that may adhere to it and then dried.

3. The filtrate containing NH₄Cl and a small amount of NH₄HCO₃ is taken to the ammonia recovery tower where it comes in contact with Ca(OH)₂.

Calcination:

When the dry NaHCO₃ is heated strongly in a furnace at 180°C, it decomposes to form anhydrous Na₂CO₃. It is called soda ash. It is nearly 99.5 % pure.

Evolved CO₂ is reused in the carbonation tower or absorption tower

Recovery of ammonia:

The filtrate from the carbonation tower which contains ammonium chloride and a little ammonium bicarbonate is made to flow down the ammonia recovery tower. NH₄HCO₃ is decomposed by the heat of steam and NH₄Cl reacts with calcium hydroxide to form ammonia, carbon dioxide, and CaCl₂. The mixture of NH₃ and CO₂ is used for tire saturation of brine while calcium chloride is obtained as a by-product.

1.

2.

Potassium carbonate (K₂CO₃) cannot be prepared by the Soh’ay process. This is because unlike sodium bicarbonate, potassium bicarbonate (KHCO₃) which is fairly soluble in water does not get precipitated when CO₂ is passed through the ammoniated solution of KCl.

 Properties of Sodium carbonate: 

1. State:

Sodium carbonate is available either as anhydrous salt or as hydrated salt. The hydrated salts are white crystalline substances and are mainly of two types

1. Decahydrate (Na₂CO₃-10H₂O) and
2. Monohydrate (Na₂CO₂-H₂O).

The decahydrate is also called washing soda. The anhydrous salt commonly known as soda ash, is a white powder. When sodium carbonate is crystallized from water, the decahydrate is obtained as white transparent crystals. These crystals are efflorescent. When exposed to air for a long time, crystals of decahydrate partially lose their water of crystallization and is converted to monohydrate, a powdery substance which is known as crystal carbonate.

Class 11 S-Block Notes

⇒ \(\mathrm{Na}_2 \mathrm{CO}_3 \cdot 10 \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{Na}_2 \mathrm{CO}_3 \cdot \mathrm{H}_2 \mathrm{O}+9 \mathrm{H}_2 \mathrm{O}\)

2. Action of heat:

When the decahydrate is heated up to 100°C, it slowly loses nine molecules of water of crystallization and gets converted into monohydrate. When the monohydrate is heated above 100°C, the anhydrous salt (Na₂CO₃) is produced as a white powder which melts at high temperatures but never undergoes decomposition.

1.

2.

Anhydrous Na₂CO₃ or soda ash melts at higher temperatures (melting point 852°C) but does not decompose. It turns to monohydrate when kept in the air.

3. Hydrolysis:

It dissolves in water with the evolution of a considerable amount of heat. Being a salt of a weak acid (HCO₃) and a strong base (NaOH), it undergoes hydrolysis in water to give an alkaline solution.

⇒\(\mathrm{Na}_2 \mathrm{CO}_3+2 \mathrm{H}_2 \mathrm{O} \rightleftharpoons 2\left[\mathrm{Na}^{+}+\mathrm{OH}^{-}\right]+\mathrm{H}_2 \mathrm{CO}_3\)

4. Reaction with acid:

At ordinary temperature, Na₂CO₃ reacts with dilute mineral acids to form the corresponding sodium salts and water along with the evolution of CO₂

Na₂CO₃ + 2HCl→2NaCl + CO₂↑ + H₂O

Na₂CO₃ + 2CH₃COOH→2CH₃COONa + CO₂↑ + H₂O

Reaction with slaked lime: When a solution of Na₂CO₃ is heated with slaked lime (milk of lime) at 80°C, sodium hydroxide with insoluble calcium carbonate is obtained.

Na₂CO₃  + Ca(OH)₂ → CaCO₃↓+ 2NaOH

Class 11 S-Block Notes

Uses of sodium carbonate:

• Sodium carbonate is mainly used for softening hard water and for washing clothes.
• It is used in fire extinguishers.
• It is largely used in the manufacture of soap, glass, borax, and caustic soda.
• It is used in the paper, paint, and textile industries.
• A mixture of Na₂CO₃ and K₂CO₃ is used as a fusion mixture.
• It is used as an important laboratory reagent both in qualitative and quantitative analysis.

2. Sodium bicarbonate or sodium hydrogen carbonate (baking soda), NaHCO₃

Preparation of sodium bicarbonate:

Sodium hydrogen carbonate is obtained as the intermediate product in the Solvay process of manufacturing sodium carbonate.

It can also be prepared by passing CO₂ through a saturated solution of sodium carbonate. Being less soluble, the white crystals of sodium hydrogen carbonate can be filtered out and dried at room temperature.

Na₂CO₃ + H₂O+ CO₂ ⇌ 2NaHCO₃↓

Properties of sodium bicarbonate:

1. State: It is a white crystalline solid and is sparingly soluble in cold water. It is also stable in air.

2. Hydrolysis: Being a salt of weak acid (H₂CO₃) and strong base (NaOH), it hydrolyses to give a faintly alkaline solution.

NaHCO₃ + H₂O ⇌  [Na⁺ + OH^–] + H₂CO₃

3. Action of heat: On heating, it decomposes to form CO₂, water, and sodium carbonate.

4. Reaction with acids: At ordinary temperature, it reacts with mineral acids to form CO₂, water, and the sodium salt of the acid:

NaHCO₃ + HCl→ NaCl + CO₂ ↑ + H₂O

Uses of sodium bicarbonate:

• It is used as an antacid (known as soda bi-carb). It is also used as a mild antiseptic for skin infections.
• It is the chief ingredient of ‘baking powder’ which is used in preparing breads, biscuits, cakes etc.
• It is used in the preparation of soft drinks like soda- water, lemonades, etc.
• It is also used in fire extinguishers.

Class 11 S-Block Notes

3. Sodium hydroxide (caustic soda), NaOH

Manufacture of sodium hydroxide:

Sodium hydroxide is industrially prepared by the electrolysis of an aqueous solution of NaCl (brine) in a specially designed cell called the Castner-Kellner cell or mercury cathode cell.

Sodium hydroxide Principle:

When a brine solution is electrolyzed in a cell using a mercury cathode and graphite anode, metallic sodium discharged at the cathode combines with mercury to form sodium amalgam. Now, electrolysis of slightly alkaline water in the cell using sodium amalgam as anode and iron rod as cathode produces NaOH. The reaction between sodium amalgam and water also produces NaOH.

Sodium hydroxide  Procedure:

1.  The cell consists of a large rectangular iron tank divided into three compartments by two slate partitions which do not touch the bottom of the tank but remain suspended in mercury placed in the grooves

2. The graphite anodes are fixed in the two outer compartments and the cathode which consists of several iron rods is fitted in the central compartment.

3. The layer of mercury at the bottom serves as an intermediate electrode as a cathode in the outer compartment and as an anode in the central compartment by induction.

4. The brine solution is taken in the two outer compartments and a very dilute NaOH solution is taken in the central compartment.

5. The mercury layer is made to flow from one compartment to another by rocking the cell with the help of an eccentric wheel.  On passing electric current, the following reactions take place in the outer and central compartments.

6. In the outer compartment, NaCl undergoes electrolysis. Cl₂ gas formed at the anode comes out from the outlet tube while sodium liberated at the cathode combines with mercury to form sodium amalgam.

NaCl → Na⁺+ Cl^– ; H₂ O → H⁺ + OH^–

• At cathode: Na⁺ + e →Na; Na + Hg→ Na/Hg
• At anode,: Cl^–→ Cl + e; Cl + Cl→ Cl₂ ↑

7. In the central compartment, sodium amalgam (Na/Hg) acts as an anode by induction.

⇒ \(\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}^{+}+\mathrm{OH}^{-}, \mathrm{NaOH} \rightarrow \mathrm{Na}^{+}+\mathrm{OH}^{-}\)

• At cathode: \(\mathrm{H}^{+}+e \longrightarrow \mathrm{H} ; \mathrm{H}+\mathrm{H} \longrightarrow \mathrm{H}_2 \uparrow\)
• At anode: \(\mathrm{Na} / \mathrm{Hg} \longrightarrow \mathrm{Na}^{+}+e+\mathrm{Hg}\)

Net reaction: \(2 \mathrm{Na} / \mathrm{Hg}+2 \mathrm{H}_2 \mathrm{O} \rightarrow 2\left(\mathrm{Na}^{+}+\mathrm{OH}^{+}\right)+2 \mathrm{Hg}+\mathrm{H}_2 \uparrow\)

The concentration of NaOH in the central compartment gradually increases with the progress of electrolysis and when it becomes 20%, the solution is withdrawn, evaporated and converted into pellets or flakes of NaOH.

Class 11 S-Block Notes

Properties of sodium hydroxide:

1. State: It is a white, crystalline hygroscopic solid having a melting point of 318°C.

2. Solubility: It dissolves in water with the evolution of heat, giving a strong alkaline solution. It also dissolves in alcohol.

3. Hygroscopic and corrosive nature:

The crystals of NaOH are deliquescent (hygroscopic). When exposed to air they absorb moisture from air and dissolve in the absorbed water. Moist caustic soda generally absorbs CO₂ from air to form sodium carbonate which forms a coating over the surface of the material. As Na₂CO₃ is non-hygroscopic, wet sodium hydroxide becomes dry again.

⇒ \(2 \mathrm{NaOH}+\mathrm{CO}_2 \rightarrow \mathrm{Na}_2 \mathrm{CO}_3+\mathrm{H}_2 \mathrm{O}\)

It is corrosive. When its concentrated solution comes in contact with the skin it produces a burning sensation. It breaks down the proteins of the skin and because of this property, it is commonly called caustic soda,

4. Reaction with acids, acidic oxides, and amphoteric oxides: Being a strong alkali, it reacts with acids, acidic oxides, and amphoteric oxides to form corresponding salts.

NaOH + HCl→ NaCl +H₂O

2NaOH + SO₂→ Na₂SO₃ +H₂O

Al₂O₃ + 2NaOH →2NaAlO₂ (Sodium aluminate) + H₂O

ZnO + 2NaOH → Na₂ZnO₂ (Sodium zincate) + H₂O

Uses of sodium hydroxide:

It is used

• In the manufacture of soap, paper, artificial silk, dyes, and several chemicals
• In the refining of [etroleum and vegetable oil,
• In the purification of bauxite,
• As a cleaning agent for greasy machines and metal NItoots,
• As a laboratory reagent etc.

4. Sodium chloride (common salt), NaCI

Preparation of sodium chloride:

• The main source of sodium chloride Is seawater which contains 2.7-2,9% of the salt by mass. In tropical countries like India, common salt is generally obtained by the evaporation of seawater.
• Crude sodium chloride obtained by this process contains calcium sulfate (CaSO.), sodium sulfate (Na₂SO₄), calcium chloride (CaCl₂), magnesium chloride (MgCl₂), etc as impurities.
• Since MgCI₂ and CaCl₂ are deliquescent (absorb moisture from the air), impure common salt gets wet in the rainy season.
• For purification, a saturated solution of crude NaCI is prepared and the insoluble impurities are removed by filtration.
• The filtrate is then saturated with hydrogen chloride gas and crystals of pure NaCI separate out due to the common ion effect.
• Chlorides of Ca and Hg being more soluble remain dissolved in the solution. NaCI can also be prepared from rock salt.

Properties of sodium chloride:

• NaCI is a white crystalline solid that melts at 1081K.
• 36 g of NaCI is soluble in 100g of water at 373K. However, solubility does not increase much with an increase in temperature.

Uses of sodium chloride:

• It is used as common salt or table salt for domestic purposes.
• It is used in the manufacture of sodium, caustic soda (NaOH), chlorine, washing soda, sodium peroxide, sodium sulfate, etc.
• It is used in soap industry, in softening hard water, in freezing mixtures, and for regenerating ion exchange resins.

Biological Importance Of Sodium And Potassium

Sodium and potassium ions are the most common cations present in biological fluids. A person weighing 70kg contains about 90g of Na and 170g of K along with 5g of Fe and 0.06 g of Cu.

The daily requirement of Na and K for the human body is about 2 g each.

1. The Na⁺ ions are mainly found outside the cells, in blood plasma, and in the interstitial fluid that surrounds the cells. These ions take part in the transmission of nerve signals, in regulating the flow of water across the cell membranes, and in the transportation of various amino acids and sugars into the cells.
2.  K⁺ ions are the most abundant cations in the cell fluids, where they activate a variety of enzymes, and promote the oxidation of glucose into ATP (adenosine triphosphate), and Na⁺ ions promote the transmission of nerve signals.
3. The Na⁺ and K⁺ ions differ considerably in concentration on the opposite sides of the cell membrane. In blood plasma, for example, the concentrations of Na+ and K+ ions are 143 million-L^-1 and 5 million-L^-1 respectively.
4. Within the blood cells, however, the concentrations of these ions are 10 millimol-L^-1 and 105 millimol-L^-1 respectively.
5. The activities in a nerve cell depend upon the sodium-potassium ion gradient. These ionic gradients are maintained by an ion transport mechanism that operates the active inclusion of K⁺ ions and active exclusion of Na⁺ ions across the cell membrane.
6. The transportation of ions requires energy which is obtained by hydrolysis of ATP. The hydrolysis of one ATP molecule to ADP provides enough energy to move three Na⁺ ions out of the cell two K⁺ ions and one H+ ion back into the cell.

Class 11 Chemistry S Block Elements Group-2 Elements (Alkaline Earth Metals) Introduction

The outermost shell of these elements contains two electrons and the penultimate shell contains eight electrons except forthe first member beryllium which contains two electrons.0 Since the last electron enters the ns orbital, these are also called s-block elements.

Their outermost electronic configuration may be represented as ns², where n- 2 to 7. Due to their similarity in electronic configuration, they are placed in the same group (Group- 2) of the periodic table and closely resemble each other in physical and chemical properties. Two valence electrons are always involved together giving rise to uniform bivalency of the elements.

Class 11 S-Block Notes

Beryllium shows some abnormal properties as its electronic configuration is slightly different from the rest of the members. The main reason is that both the beryllium atom and Be²⁺ ion are extremely small. Beryllium also shows some similarities with aluminum of group 13. Like alkali metals, the compounds of these metals are also predominantly ionic. The electronic configurations of alkaline earth metals are given in the following table

Electronic configuration!) of alkaline earth metals

Occurrence Of Alkaline Earth Metals

Due to low ionization enthalpies and high electropositive character, the alkaline earth metals are chemically very reactive and hence, do not occur in the free state but are widely distributed in nature as silicates, carbonates, sulfates, and phosphates.

• Relative abundance of Be, Mg, Ca, Sr, Ba, and Ra in the earth’s crust is 2, 27640, 46600, 384, 390, and 1.3 x 10-6 ppm respectively. 0 Beryllium, the fifty-first most abundant element by mass in the earth’s crust, is found as silicate minerals like beryl (Be₃Al₂Si₆O₁₈) and phenacite (Be₂SiO₄).
• Magnesium, the sixth most abundant element is found as carbonate, sulphate, and silicate. Its two important minerals are magnesite (MgCO₃) and dolomite [MgCO₃.CaCO₃].
• It is also found in seawater at 0.13% as MgCl₂ and MgSO₄.
• Calcium, the fifth most abundant element by mass found in the earth’s crust, occurs mainly as CaCO₃ in the form of limestone, marble and chalk. Its other important minerals are fluorspar (CaF2), fluorapatite, [3Ca₃(PO₄)2-CaF₃], gypsum (CaSO₄-2H_2O) anhydride, (CaSO₄).
• Strontium and barium are respectively the fifteenth and sixteenth most abundant element. Strontium occurs principally as the mineral celestite (SrSO₄) and strontianite (SrCO₃) while barium occurs mainly as the mineral barytes (BaSO₄).
• Radium is radioactive and extremely scarce. It occurs in very small amounts (1 gin 7 ton) in pitchblende as the decay product of uranium.

General Trends In Atomic And Physical Properties Of Alkaline Earth Metals

The alkaline earth metals show regular trends in their physical and chemical properties with an increase in atomic number. Some important atomic and physical properties of alkaline earth metals are given in the

Atomic and physical properties of alkaline earth metals:

General trends in different atomic and physical properties of alkali metals and their explanations

1. Atomic and ionic radii

The atomic and ionic radii of alkaline earth metals are fairly large but smaller than those of the corresponding alkali metals and these increase on moving down a group.

Atomic and ionic radii Explanation:

The electrons of alkaline earth metals having a higher nuclear charge are more strongly attracted towards the nucleus.  On moving down the group, the atomic as well as ionic radii increase. The addition of new shells and the increasing screening effect jointly overcome the effect of increasing nuclear charge down the group

2. Ionization enthalpy

1. The first and second ionization enthalpies of alkaline earth metals are quite low and decrease down the group from Be to Ra

Explanation:

The low ionization enthalpies of alkaline earth metals are due to their smaller nuclear charge and larger atomic size (compared to the other succeeding elements of the same period) which result in weaker forces of attraction between valence electrons (ns2) and nucleus. On moving down the group, atomic size increases and the screening effect of the inner shell electrons also increases

Class 11 S-Block Notes

Since the alkali metal atoms are largest in their respective periods, their outermost electrons being far away from the nucleus experience less force of attraction and hence, can be removed easily. These two effects jointly overcome the effect of increasing nuclear charge down the group. Thus, first and second ionization enthalpies decrease down the group.

2. The first ionization enthalpies of alkaline earth metals are higher than those of the corresponding alkali metals but their second ionization enthalpies are lower than those of the corresponding alkali metals.

Explanation:

The alkaline earth metals have higher values of first ionization enthalpy than those of the corresponding alkali metals because they have smaller size and higher nuclear charge which result in stronger forces of attraction between the valence electrons and the nucleus.

The second ionization enthalpy values of alkaline earth metals are much lower than those of the corresponding alkali metals because the loss of the second electron from an alkaline earth metal cation (M⁺) leads to the attainment of a stable noble gas configuration (ns²np⁶) while the loss of the second electron from an alkali metal cation (M⁺) causes loss ofits stable noble gas configuration

Example:

3. Electropositive or metallic character

The alkaline earth metals are highly electropositive and possess high metallic character. However, they are less electropositive than the alkali metals. Their electropositive or metallic character increases on moving down the group.

Electropositive Explanation:

• Due to their relatively low ionization enthalpies, alkaline earth metals have a strong tendency to lose both valence electrons to form dipositive ions. Thus, they exhibit high electropositive or metallic character.
• As their atoms have smaller sizes and higher ionization enthalpies compared to those of the corresponding alkali metals, their tendency to lose valence electrons is less than that of alkali metals. Hence, alkaline earth metals have less electropositive or metallic character as compared to tine alkali metals.
• On moving down the group from Be to Ra, ionization enthalpies decrease due to an increase in atomic radii. Therefore, the tendency to lose electrons increases and so does the electropositive character

Class 11 S-Block Notes

4. Hydration enthalpy

Hydration enthalpies of alkaline earth metal ions are much greater than that of the alkali metal ions & decrease down the group from Be²⁺ to Ba²⁺

Be²⁺ > Mg²⁺ > Ca²⁺ > Sr2+ > Ba²⁺

Hydration enthalpy Explanation:

Due to the smaller size of alkaline earth metal ions, their hydration enthalpies are much greater than those of the alkali metal ions. Therefore, the compounds of alkaline earth metals are found to be more extensively hydrated than those of alkali metals. Magnesium chloride and calcium chloride, for example, exist as hexahydrates (MgCl₂-6H₂O and CaCl₂-6H₂O) while sodium chloride and potassium chloride do not form such hydrates.

The ionic conductance of hydrated alkaline earth metal ions increases from [Be(H₂O)_x]²⁺ to [Ba(HO)_2x]²⁺ due to a decrease in the extent of hydration. The hydration enthalpy of an ion is directly proportional to its charge/radius ratio {q/r). On moving down a group, the radii of the alkaline earth metals increase. As a result, the hydration enthalpies of these metals decrease.

5. Oxidation State

Alkaline earth metals exhibit an oxidation in their compounds. Although the second date of +2 in their compounds. Although the second ionization enthalpy of these elements is nearly double that of the first ionization enthalpy, yet they exist as divalent ions (M²⁺) in most of their compounds.

Oxidation State Explanation:

1. The divalent ions (M²⁺) of alkaline earth metals have stable noble gas configurations. Thus, M²⁺ ion is more stable than M+ ion.

M ([Noble gas] ns²) → M²⁺ [Noble gas] + 2e

2. Due to greater charge and smaller size, the divalent cations lead to the formation of very stable lattices, and hence, a huge amount of energy is released. The high lattice enthalpy easily compensates for the high second ionization enthalpy.

3. Divalent cations for their smaller size get hydrated in water to a greater extent and the energy thus released (hydration enthalpy) is large enough to compensate for the second ionization enthalpy

The ΔH_i(3) values of alkaline earth metals are very high because the electron now has to be removed from the stable noble gas configuration. For this reason, the alkaline earth metals do not exhibit an oxidation state of more than +2.

6. Melting & boiling points

Alkaline earth metals have higher melting & boiling points than that of alkali metals. However, on moving down the group, no regular trend is observed.

Melting & boiling points Explanation: 

• Due to their smaller size, the atoms of alkaline earth metals form a more close-packed crystal lattice. Moreover, alkaline earth metals have two electrons in their valence shell whereas alkali metals have only one.
• The larger number of valence electrons leads to the formation of stronger metallic bonds.
• No regular trend in melting and boiling point is observed down the group because the atoms adopt different crystal structures.

7. Nature of bonds formed:

Like alkali metals, alkaline earth metals predominantly form ionic compounds. However, these are less ionic than the corresponding alkali metal compounds. Beryllium, the first member of this group, is an exception as its compounds are covalent. Magnesium also tends to form covalent compounds to some extent. On moving down the group, the tendency to form ionic compounds increases.

Nature of bonds formed Explanation:

Alkaline earth metals form ionic compounds because they have low ionization enthalpies. Their compounds, however, are less ionic because their ionization enthalpies are higher than those of the corres¬ ponding alkali metals. Due to its much smaller size and much higher ionization enthalpy, beryllium forms compounds that are predominantly covalent. Down the group, the tendency to form ionic compounds increases because ionization enthalpy decreases.

Class 11 S-Block Notes

8. Density and hardness

The alkaline earth metals are denser and harder than the corresponding alkali metals. However, on moving down the group, no regular trend is observed. It initially decreases from Be to Ca and then increases from Ca to Ba.

Density and hardness Explanation:

The extent of cohesive energy determines the density and hardness of metals and this depends on the number of electrons involved in metallic bonding and the size of the atom. In alkali metals, one electron per atom (the valence electron) is involved in metallic bonding while in alkaline earth metals, two electrons per atom (the valence electrons) are involved. Moreover, the atoms of alkaline earth metals are heavier and smaller in size.

Therefore, the extent of cohesive energy is relatively higher in the case of alkaline earth metals & consequently, the atoms in alkaline earth metals are packed more closely in their lattices. Cohesive energy decreases from Be to Ca due to a gradual increase in size while it is found to increase from Ca to Ba due to the formation of different crystal lattices.

9. Conductivity

The Gr-2 metals are good conductors of heat and electricity.

Conductivity Explanation:

Due to the presence of two loosely bound valence electrons (per atom) which can move freely throughout the crystal lattice, the alkaline earth metals are good conductors of heat and electricity.

10. Flame coloration

When the alkaline earth metals and their salts, except beryllium and magnesium, are heated in the flame of a bunsen burner, they impart characteristic color to the flame.

These colors are as follows:

• Ca: Brickred
• Ba: Apple green
• Sr & Ra: Crimson red

Flame coloration Explanation:

• When the alkaline earth metals or their salts are put into a flame, the electrons of their valence shell absorb energy and get excited to higher energy levels.
• When they drop back to the ground state, the absorbed energy is emitted in the form of visible light having characteristic wavelengths.
• Depending upon the wavelength of light emitted, different colors are imparted to the burner flame.
• Due to their smaller size, the valence electrons in Be and Mg are too strongly bound to get excited by the energy available from the flame. Therefore, they do not impart any color to the flame.

Alkaline earth metals (except Be and Mg) can easily be identified by flame test in qualitative analysis. Further, they can be estimated by flame photometry or atomic absorption spectroscopy.

11. Magnetic property

The alkaline earth metals and their salts are diamagnetic.

Magnetic property Explanation:

Since the divalent ions (M²⁺) of alkaline earth metals have noble gas configurations with no unpaired electrons, their salts are diamagnetic. The metals are also diamagnetic as all the orbitals are filled up with paired electrons.

Chemical Properties Of Alkaline Earth Metals (Group-2 Metals)

Due to their low ionization enthalpies and high electropositive character, alkaline earth metals have a strong tendency to lose their valence electrons. Therefore, they are highly reactive and do not exist in the free state in nature.

1. Reducing nature

The alkaline earth metals are strong reducing agents. However, they are weaker reducing agents than alkali metals. Again, like alkali metals, their reducing strength increases down the group.

Class 11 S-Block Notes

Reducing nature Explanation:

The alkaline earth metals except Be, have a fairly strong tendency to lose two valence electrons to form dipositive ions (M→ M²+  2e) i.e. they possess low ionization enthalpies and hence, they are strong reducing agents.

This is indicated by their high negative values of reduction potentials (E°). Their reducing strength, however, is less than the alkali metals as their atomization enthalpies and ionization enthalpies are relatively higher. Reducing strength increases on moving down the group as their ionization enthalpies decrease & electrode potentials become progressively more negative from Be to Ba.

2. Action of air

• Being fairly reactive, the alkaline earth metals are oxidized by the oxygen of the air and get tarnished due to the formation of a fine layer of oxide on their surface. With increasing atomic numbers, the effect of air on the metals gradually increases.
• Be and Mg being less reactive are not much affected by air. Ca and Sr get easily tarnished in air while Ba readily burns when exposed to air. Hence, Ca, Ba, and Sr are usually stored in paraffin.

3. Reaction with oxygen

Alkaline earth metals burn in oxygen to form oxides. Be, Mg, and Ca form monoxides while Sr and Ba form peroxides when they react with oxygen. This is because larger cation stabilizes a larger anion and hence the tendency to form peroxide increases as the size of the metal ion increases

(M = Be, Mg or Ca)

(M = Ba, Sr)

4. Reaction with water

• Alkaline earth metals except beryllium react with water to form the corresponding hydroxides along with the liberation of H₂ gas.
• Beryllium having the lowest negative standard electrodepotential (E° of Be²⁺/Be = -1.97V) among all

The group-2 metals is the least electropositive and hence, do not react with water or steam even at red hot conditions.

Ca, Sr, and Ba have relatively higher negative standard electrode potentials similar to those of the corresponding Gr-1 metals and hence, react even with cold water.

Mg has an intermediate value of E° and does not react with cold water but decomposes in boiling water.

M + 2H₂O→M(OH)₂ + H₂↑ (M = Mg, Ca; Sr or Ba)

Thus, the reactivity of the alkaline earth metals towards water increases on moving down the group. However, they are less reactive towards water as compared to the corresponding alkali metals.

5. Reaction with nitrogen

1. All alkaline earth metals burn in nitrogen to form nitrides of the type M3N2. However, Li forms Li3N.

3M + N₂ →M₂N₂ (M = Be, Mg, Ca, Sr and Ba)

2.  The ease of formation of nitrides decreases from Be to Ba. Since N₂ molecule is very stable, it requires very high energy to form nitride ions (N^3-). This large amount of energy is supplied from the lattice enthalpy evolved when crystalline solids containing ions with high charges (M²⁺ and N^3-) are formed.

Be₃N₂ is volatile because it is covalent. Other nitrides of this group are not volatile as they are ionic crystalline solids.

Class 11 S-Block Notes

6. Reaction with halogens

The alkaline earth metals directly combine with halogens at higher temperatures to form halides having the general formula, MX₂

Halides can also be obtained by the action of halogen acids on metals, their oxides, hydroxides, or carbonates

M + 2HX→ MX₂ + H₂; MO + 2HX→MX₂ + H₂O

M(OH)₂ + 2HX→MX₂ + 2H₂O

MCO₃ + 2HX→MX₂ + CO₂ + H₂O

BeCl₂ is, however, conveniently prepared by heating BeO with Cl₂ in the presence of charcoal at 1073K

7. Reaction with hydrogen:

All the elements of group-2 except Be, form metal hydrides of the general formula MH₂ when heated with hydrogen

Beryllium hydride can be prepared indirectly by reducing beryllium chloride with lithium aluminum hydride.

2BeCl₂ + LiAlH₄→ 2BeH₂ + LiCl + AlCl₃

Both beryllium hydride (BeH₂) and magnesium hydride (MgH₂) are covalent compounds. In these molecules, both Be and Mg have four electrons in their valence shell. Therefore, these molecules are electron deficient. To make up for their electron deficiency, these two compounds exist as polymers, (BeH₂)_n and (MgH₂)_n in which each Be or Mg -atom forms four three-centre two-electron (3c-2e) bonds or hydrogen bridge bonds or banana bonds.

Class 11 S-Block Notes

The structure of polymeric beryllium hydride is shown below:

CaH₂, SrH_2, and BaH₂ are ionic compounds in which a hydride ion (H^–) exists as an anion. Calcium hydride (CaH₂ ) which is also called hydrolith is used for the production of H₂ by the action of HaO on it.

8. Reaction with carbon

When the alkaline earth metals except for Be, are heated with carbon in an electric furnace or when their oxides are heated with carbon, carbides of the type MC₂ are obtained. These carbides are also called acetylides (containing discrete C₂ ions) as on hydrolysis they form acetylene.

(M = Mg, Ca, Sr or Ba)

At much higher temperatures ( ~ 1700°C), beryllium reacts with carbon to form Be₂C. This carbide is called methanide (containing discrete C^4- ion) as on hydrolysis it produces methane. On heating, MgC₂ forms Mg₂C₃, which is called allylide (containing discrete C₃^4- ion) as hydrolysis yields allylene (methyl acetylene).

CaC₂ + 2H₂O → HC≡ CH + Ca(OH)₂

Be₂C + 4H₂O →  2Be(OH)₂ + CH₄

Mg₂C₃+ 4H₂O →  CH₃C ≡ CH + 2Mg(OH)₂

When calcium carbide (CaC₂), an important chemical intermediate, is heated in an electric furnace with atmospheric nitrogen at 1375K, it produces calcium cyanamide (CaNCN).

The mixture of CaNCN and carbon is called nitrolim. It is used as a slow-acting nitrogen fertilizer as it undergoes very slow hydrolysis and evolves NH3 gas for a long period.

CaC₂ + 3H₂O → CaCO₃ + 2NH₃

9. Reaction with acids

The alkaline earth metals react with dilute acids to form the corresponding salt with the liberation of H₂ gas.

M + H₂SO₄ → MSO₄ + H₂↑ (M = Be, Mg, Ca, Sr or Ba)

Beryllium is the only group-2 metal which reacts with alkali to form H₂ and beryllate salt.

Be + 2NaOH + 2H₂O → Na₂ [Be(OH)₄] (Sodium beryllate)+ H₂ ↑

This is observed due to the diagonal relationship between aluminum and beryllium.

10. Solutions in liquid ammonia

Like alkali metals, alkaline earth metals dissolve in ammonia to give deep blue-colored solutions containing ammoniated cations and ammoniated electrons.

M + (x+ 2y)NH₃ → [M(NH₃)_x]²⁺ + 2[e(NH₃)_y ]^–

Class 11 S-Block Notes

Evaporation of ammonia from these solutions leads to the formation of hexammoniates M(NH₃)₆ which slowly decompose to yield the corresponding metal amides, M(NH₂)₂ and H₂

11. Tendency to form complexes

The group-2 elements tend to form stable complexes and it is found to be greater than that of alkali metals because their ions have smaller size and higher charge. The tendency to form complexes decreases down the group and this is due to the decrease in ion-dipole interaction with increasing size of the metal ion. Be and Mg have the maximum tendency to form complexes.

Examples of two stable complexes of Be and Mg are \(\left[\mathrm{BeF}_4\right]^{2-}\&\left[\mathrm{Mg}\left(\mathrm{NH}_3\right)_6\right]^{2+}\) respectively

1. Complexation of Ca²⁺ by EDTA and polyphosphates plays an important role in the removal of the metal in water softening.
2.  In chlorophyll, the complex formed by the combination of Mg and the tetrapyrrole system (porphyrin) is very crucial in photosynthesis.

12. Extraction of alkaline earth metals

Like alkali metals, alkaline earth metals are also very reactive and strong reducing agents. So they cannot be extracted by ordinary chemical reduction methods. These metals also cannot be prepared by electrolysis of aqueous solutions of their salts because in that case, hydrogen is discharged at the cathode instead of the metal which has a much higher discharge potential. However, electrolysis can be carried out using a Hg-cathode, but in that case, recovery of the metal from amalgam becomes difficult. These metals are best isolated by electrolysis of their fused salts, usually chlorides.

General Characteristics Of The Compounds Of Alkaline Earth Metals

Compounds of group-2 elements are predominantly ionic but are less ionic than the corresponding compounds of group 1 elements and this is due to their increased nuclear charge and smaller size. The general characteristics of some of the compounds of alkaline earth metals are discussed below.

1. Oxides of alkaline earth metals

1. Crystal structure:

Except for BeO (covalent solid), the oxides of the remaining alkaline earth metals are crystalline ionic solids and possess a rock-salt (NaCl) structure with coordination number 6. BeO though covalent, is an extremely hard solid because of its polymeric nature. BeO possesses a covalent lattice with coordination number 4. Both BeO and MgO have several properties that make them useful as refractory materials (for lining furnaces).

These properties are:

• They have high melting points (BeO-2500°C and MgO – 2800°C),
• They have very low vapor pressures,
• They are good conductors of heat, O they are chemically inert and
• They can act as electrical insulators.

2. Stability:

Due to much higher lattice enthalpies, the oxides are very stable towards heat. The lattice enthalpies decrease with an increase in the size of the metal ion.

3. Basic character:

Beryllium oxide, BeO reacts with both acids and alkalis, i.e., it is amphoteric while the oxides of other group-2 metals are basic.

BeO + 2HCl→BeCl₂ + H₂O

BeO + 2NaOH → Na₂ BeO₂ (Sodium beryllate) + H₂O

The basic strength of the oxides increases on moving down the group.

BeO (Amphoteric) < MgO (Weakly basic) < CaO (Basic ) < SrO, BaO( Strongly basic)

4. Reaction with water:

All these oxides except BeO and MgO, react with water to form sparingly soluble hydroxides. These reactions are exothermic.

MO + H₂O→M(OH)₂ + heat, M = Ca, Sr or Ba

2. Hydroxides

1. Basic character:

All the alkaline earth metal hydroxides are basic except Be(OH)₂ which is amphotericin nature. Their basic strength increases on moving from Be(OH)₂ to Ba(OH)₂

The alkaline earth metal hydroxides are, however, less basic than the alkali metal hydroxides.

Basic character Explanation:

Due to low ionization enthalpies of the alkaline earth metals, the M — O bond present in their hydroxides is weak and breaks up easily to give OH^– ions. For this reason, their hydroxides exhibit basic character. On moving down the group, the tendency of the M — OH bond to break heterolytically increases because ionisation enthalpies decrease and consequently, the basic character of the hydroxides increases.

Class 11 S-Block Notes

Due to larger ionic sizes and lower ionization enthalpies of alkali metals, the M — OH bonds in their hydroxides are still weaker than those in alkaline earth metal hydroxides. Thus, the alkali metal hydroxides are more basic than the alkaline earth metal hydroxides.

2. Solubility in water:

Hydroxides of alkaline earth metals are less soluble in water than the hydroxides of alkali metals. Again, the solubility of these hydroxides increases markedly on moving down the group

Solubility in water Explanation:

On moving down the group, both the lattice enthalpy and the hydration enthalpy decrease with increasing ionic size. However, the lattice enthalpy decreases more rapidly than the hydration enthalpy, and consequently, their solubility increases down the group.

3. Thermal stability:

The alkaline earth metal hydroxides decompose on heating to give the metal oxide and water.

Thermal stability Explanation:

Thermal stability of these hydroxides increases down the group as the polarising power of the M2+ ion and the lattice enthalpy of the oxide formed decreases with increasing ionic size down the group.

3. Halides

• Due to the high polarising power of the Be²⁺ ion, beryllium halides have a covalent nature having low melting points.
• All other alkaline earth metal halides are ionic and their ionic character increases as the size of the metal ion (M²⁺) increases down the group. These ionic halides are non-volatile solids having high melting points.
• Due to its covalent nature, beryllium halides are sparingly soluble in water but readily soluble in organic solvents. The halides of other group-2 alkaline earth metals are readily soluble in water.
• Except for BeCl₂, all other anhydrous halides of the alkaline earth metals are hygroscopic in nature and form hydrates.
• For example: MgCl₂-6H₂O, CaCl₂-6H₂O, SrCl₂-2H₂O and BaCl₂-2H₂O
• The tendency to form hydrate decreases down the group. Thus, anhydrous calcium chloride is used as a dehydrating agent in the laboratory.
• The dehydration of the hydrated chlorides, bromides, and iodides of Ca, Sr, and Ba can be achieved by heating. However, the corresponding hydrated halides of Be and Mg on heating suffer hydrolysis.
• BeF₂ is highly soluble in water due to the much higher hydration enthalpy of the very small Be²⁺ ion.
• All other fluorides (MgF₂ > CaF₂, SrF_2, and BaF₂ ) are almost insoluble in water because their lattice enthalpies are higher than their hydration enthalpies.
• Except for BeCl₂ and MgCl₂, all other alkaline earth metal chlorides impart characteristic color to a flame.
• For example: CaCl₂: is brick red, SrCl₂:  Is crimson red, BaCl₂: Is grassy green, etc.

Structure of BeCl₂ In the solid state, beryllium chloride has a polymeric chain structure with chlorine bridges as given below:

Which are bonded by two covalent bonds while the other two by coordinate bonds. The Be -atoms in (BeCl₂)n sp³ -hybridized.

In the vapor phase, BeCl₂ exists as a chlorine-bridged dimer which dissociates into linear triatomic monomer at about 1200K. In the dimer, Be is sp² -hybridized while in the monomeric is sp² -hybridized

• CaF₂ an industrially important compound, is the main source of both F₂ and HF

CaF₂ + H₂SO₄4-→ 2HF + CaSO₄

• CaF₂ is also used for making prisms and cell windows for spectrophotometers, an important instrument used in the spectroscopic analysis of compounds.
• In cold countries, CaCl₂ is Used for treating ice on roads, because 30% eutectic mixture of CaCl₂ +H₂O freezes at  – 55%C.

4. Salts of Oxoacids

Trends in the properties of some salts of group-2 elements are discussed below

1. Carbonates:

Solubility in water:

The carbonates of the alkaline earth metals are practically insoluble in water. Their solubilities decrease on moving down the group. BeCO₃ is sparingly soluble in water while BaCO₃ is insoluble in water.

Solubility in water Explanation:

On moving down the group, lattice enthalpy of carbonates remains almost unchanged” (the size of the metal ion is much smaller compared to CO₃^2-  ion) but hydration enthalpies of cations (M²⁺) decrease. Consequently, the solubilities of carbonates decrease down the group.

The extremely low solubility of alkaline earth metal carbonates in water is very important in the precipitation of Ba²⁺, Sr^2+, and Ca²⁺ ions as their carbonates, in Gr-IV qualitative analysis of basic radicals.

Thermal stability: Carbonates of gr.-2 metals decompose on heating to give metal oxide and carbon dioxide

Thermal stability of the carbonates increases down the group with increasing cationic size

Thermal stability Explanation:

The trend can be explained in terms of the stability of the monomeric is sp –hybridized. of the resulting metal oxides. With increasing stability of  Cl the metal oxide, the carbonate becomes more unstable BeCl₂ [monomer] towards heat. The stability of metal oxides decreases down the group due to a decrease in lattice enthalpy with increasing cationic (M²⁺) size. Hence, the stability of the carbonates towards heat increases down the group.

2. Bicarbonates:

The bicarbonates of alkaline earth metals do not exist in the solid state but are found in solutions. When such solutions are heated, bicarbonates decompose to form carbonates with the evolution of CO₂

3. Sulfates

1. Solubilityin water:

The sulfates of alkaline earth metals are relatively less soluble in water than the corresponding sulfates of alkali metals. Further, their solubilities decrease down the group. For example, BeSO₄ and MgSO₄ are more soluble in water, CaSO₄ is less soluble in water and SrSO₄, BaSO_4, and RaSO₄ are practically insoluble in water.

Solubilityin water Explanation:

On moving down the group, the hydration enthalpies decrease with increasing cationic size but the lattice enthalpy remains almost unchanged because the anion, SO₄ is much larger as compared to the metal ions (M²⁺). For this reason, the solubilities of sulfates decrease down the group.

2. Thermal stability: The alkaline earth metal sulfates white solids which dissociate on heating to give the metal oxides and sulfur trioxide.

The thermal stability of the sulfates increases down the group due to an increase in ionic character. This is due to a decrease in the polarising power of the metal ions with increasing ionic size. It becomes evident from the dissociation temperatures of the sulfates given below

4. Nitrates

Nitrates of alkaline earth metals are prepared by heating corresponding metal carbonate with dilute HNO₃

MCO₃  + 2HNO₃→ M(NO₃) + H₂ O + CO₂ (M = Be, Mg, Ca, Sr or Ba)

Magnesium nitrate crystallizes as Mg(NO₃)₂. 6H₂O whereas barium nitrate crystallizes as an anhydrous salt. This again shows that the tendency to form hydrates decreases with increasing ionic size and decreasing hydration enthalpy as we move down the group. Upon heating, all nitrates decompose to give the corresponding oxides with the evolution of NO₂ and O₂.

The nitrates of all these metals are soluble in water. Beryllium is unusual for the fact that it forms a basic nitrate in addition to the normal salt.

Anomalous Behaviour Of Be And Similarities Between Be And Al

Beryllium, the first member of group 2, shows some anomalous behavior, i.e.,it differs from the rest of the members of its family.

Reasons for anomalous behavior of beryllium:

• The extremely small size of the Be atom and Be²⁺ ion,
• Much higher polarising power of Be²⁺ ion,
• Higher ionization enthalpy and electronegativity as compared to the other members, absence of vacant d -d-orbitals in its valence shell.
• Again, beryllium resembles its diagonally placed element aluminum, the second typical element of group 13 and period 3, in several properties

Class 11 S-Block Notes

Difference between beryllium and other alkaline earth metals

Reasons for the similarities between beryllium and aluminum:

• They have approximately the same polarising power. The polarising power of Be²⁺ is 0.064, while that of Al³⁺is 0.060.0
• The standard electrode potentials of Be and A1 are much closer i.e. Be²⁺/Be = -1.85V and Al³⁺+/Al = -1.66V.0
• The electronegativity of both the elements i.e., beryllium and aluminum are the same (1.5in the Pauling scale).

Similarities between beryllium and aluminium:

Uses Of Alkaline Earth Metals

Preparation, Properties, And Uses of Some Important Compounds Of Calcium

1. Calcium oxide (quicklime), CaO

Preparation of calcium oxide:

Calcium oxide or quicklime is prepared commercially by heating limestone at about 1250K temperature in lime kilns.

CaCO₃ ⇌ CaO + CO₂ ↑- 425 kcal

The reaction is endothermic and reversible. To get a good yield of quicklime, the forward reaction is facilitated by removing CO₂ as soon as it is formed. Again, the temperature in the kiln is not allowed to rise above 1270K because above that temperature, silica (SiO₂ ) present as

impurity in limestone combines with CaO to form calcium silicate (CaSiO₃).

CaO + SiO₂ >127°K>CaSiO₃

Properties of calcium oxide:

1. State: It is a white amorphous solid having a melting point of of2870 K.

2. The action of heat:

Calcium oxide does not melt even when heated in an oxy-hydrogen flame (2270K). On strong heating, it becomes incandescent and emits bright white light (known as limelight). It melts only when heated in an electric furnace at 2850K.

3. The action of air:

When dry lumps of CaO are exposed to moist air, they absorb moisture and CO₂ from the air to form calcium hydroxide and calcium carbonate respectively. As a result, heat is evolved and the lumps of CaO are converted into powder. Calcium hydroxide thus produced also reacts with CO₂ of air to form calcium carbonate. Therefore, calcium carbonate is the final product we get. However, once the outer surfaces of the lumps of CaO become fully covered with CaCO₃, the core material is not further acted upon by moist air.

CaO + H₂O-+Ca(OH)₂; CaO + CO₂→CaCO₃

Ca(OH)₂ + CO₂→ CaCO₃ + H₂O

4. Reaction with water:

Calcium oxide possesses a high affinity towards water. Many organic liquids and moist gases are dried with CaO. It reacts vigorously with water to form Calcium hydroxide:

CaO + H₂O→ Ca(OH)₂

When a limited amount of water is sprinkled on the lumps of calcium oxide, a vigorous reaction starts. For its highly exothermic nature, the water added gets immediately transformed into steam with a hissing sound. As a result, the lumps of CaO swell up, crack, and finally crumble to a fine, dry, white powder of calcium hydroxide. Such powdered calcium hydroxide is known as slaked lime and the above process is called the slaking of lime

Quicklime when slaked with caustic soda (NaOH), produces a solid called soda lime (CaO + NaOH).

5. Reaction with acids and acidic oxides:

Calcium oxide is a basic oxide and hence reacts with acids to form the corresponding calcium salts and water.

Examples:

CaO + 2HCl→CaCl₂+ H₂O

CaO + H₂SO₄→CaSO₂ + H₂O

CaO + 2HNO₂→ Ca(NO₃)₂ + H₂O

In the reaction with sulphuric acid, insoluble calcium sulfate is produced and it forms a protective coating on the lumps of CaO. Consequently, the reaction proceeds only to a small extent and then stops.

CaO reacts with acidic oxides to form calcium salts.

Class 11 S-Block Notes

Examples:

CaO + CO₂ → CaCO₃ ; CaO + SO₂→ CaSO₃

6CaO + P₄O₁₀ →\(\rightarrow{\Delta}\) 2Ca₃(PO₄)₂

It reacts with silica at a much higher temperature to form calcium silicate: CaO + SiO₂→ CaSiO₃

6. Reaction with ammonium salts:

Being a strong base, CaO displaces ammonia forming ammonium salts. The reaction occurs rapidly on gentle heating. This reaction may be used for preparing NH3 in the laboratory.

2NH₄Cl + CaO→ 2NH₃↑ + CaCl₂ + H₂O

7. Reaction with chlorine gas:

When calcium oxide is heated in the presence of dry Cl₂ gas above 573K, calcium chloride is obtained with the evolution of oxygen gas.

8. Reaction with carbon:

When calcium oxide is heated with coke in an electric furnace at about 2273K, it forms calcium carbide and carbon monoxide. This reaction is used in the industrial preparation of calcium carbide.

Uses of calcium oxide:

• Quicldime is an important primary material for manufacturing cement and glass.
• It is used to prepare caustic soda from sodium carbonate.
• It is used in the purification of sugar.
• It finds application in the manufacture of dyes.
• It is largely used in the preparation of slaked lime which has many industrial uses.
•  It is used as a flux in metallurgy to remove siliceous impurities.
• It is used to dry several gases and alcohols.
• It is used in the preparation of calcium carbide and soda lime.
• It is used in softening of hard water and in tanning industries.

2. Calcium hydroxide or (slaked lime) Ca(OH₂)

Preparation of calcium hydroxide:

Calcium hydroxide or slaked lime is prepared by sprinkling a limited amount of water on the lumps of calcium oxide. The process is known as slaking of lime. The reaction is highly exothermic.

CaO + H₂O→Ca(OH)₂

It can also be prepared by treating a concentrated aqueous calcium chloride solution with a solution of caustic soda.

CaCl₂ + 2NaOH→Ca(OH)₂↓+ 2NaCl

Calcium hydroxide Physical properties:

Calcium hydroxide or slaked lime is available as a white amorphous powder. It is soluble in water to a very small extent. The clear dilute aqueous solution of calcium hydroxide is known as lime water. When a large amount of calcium hydroxide is added to water, a white suspension (like milk) of the substance in water is obtained. This is known as milk of lime.

Calcium hydroxide Chemical properties:

1. The action of air: When calcium hydroxide is exposed to air, it slowly absorbs CO₂ from the air and is converted into water-insoluble calcium carbonate.

Ca(OH)₂ + CO₂→ CaCO₃ ↓+ H₂O

It is to be noted that the formation of a white scum on the surface of clear lime water, exposed to air, is due to the formation of insoluble CaCO₃.

2. The action of heat: When calcium hydroxide is heated above 723K, it undergoes complete dehydration to yield CaO.

3. Reaction with carbon dioxide:

When carbon dioxide is passed through clear lime water, calcium carbonate is formed. The resultant insoluble calcium carbonate remains suspended in water as fine particles. As a result, clear lime water becomes milky (turbid) in appearance.

Ca(OH)₂ + CO₂ →CaCO₃↓ + H₂O

When excess of CO₂ gas is passed through this milky suspension, the water-insoluble particles of CaCO₃ further react with CO₂  in the presence of water to form water-soluble colorless calcium bicarbonate, Ca(HCO₃)₂. As a result, the turbidity of the solution disappears and it becomes transparent (clear) again.

CaCO₃ + CO₂ + H₂O→ Ca(HCO₃)₂(soluble)

When tiie clear solution of calcium bicarbonate is heated, the solution again becomes turbid due to the decomposition of Ca(HCO₃) into insoluble CnCO₃.

4. Reaction with sulfur dioxide:

When SO₂ gas is passed through clear lime water, water-insoluble, white calcium sulfite (CaSO₃) is formed and the clear solution becomes turbid. When an excess of SO₂ gas is passed through this turbid solution, SO₂ reacts with CaSO₃ in die presence of water to form water-soluble, colourless calcium bisulfite, Ca(HSO₃)₂. Thus, the turbid solution becomes clear again. When the resultant clear solution is heated, calcium bisulfite decomposes to give back insoluble calcium sulfite along with SO₂ and water. So, the clear solution becomes turbid again.

Ca(OH)2 + SO₂→ CaSO₃↓ + H₂O

CaSO₃ + SO₂ + H₂O→Ca(HSO₃)₂ (soluble)

Class 11 S-Block Notes

5. Reaction with acid:

Calcium hydroxide being a quite strong base reacts with acids and acidic oxides to form the corresponding salts and water.

Examples:

Ca(OH)₂ + 2HCl→CaCl₂ + 2H₂O

Ca(OH)₂ + H₂SO₄→ CaSO₄ ↓+ 2H₂O

Ca(OH)₂ + CO₂ →CaCO₃↓- + H₂O

Ca(OH)₂ + SO₂→ CaSO₃ + H₂O

3Ca(OH)₂+ P₂O₅→ Ca₃(PO₄)₂ + 3H₂O

In its reaction with sulphuric acid, insoluble white calcium sulfate is formed and it forms a protective coating on solid Ca(OH)₂ and stops the reaction.

Reaction with ammonium salts: Being a stronger base than ammonia, calcium hydroxide displaces ammonia from its salts when heated with an ammonium salt.

Examples: 2NH₄Cl + Ca(OH)₂→ 2NH₃T↑ + CaCl₂ + 2H₂O

This reaction is generally used for the preparation of ammonia in the laboratory.

6. Reaction with chlorine:

At about 313K, chlorine gas reacts with slightly moist slaked lime to form a dry, white, and powdery substance with a pungent smell. This powder having bleaching and disinfecting properties is commonly called bleaching powder. Regarding the formation and . composition of bleaching powder, the following two views have been proposed.

1. According to Odling (1861), bleaching powder Is a mixture of calcium hypochlorite, Ca(OCl)₂, and calcium chloride, CaCl_2, and Is called calcium chlorohypochlorite, Ca(OCl). Its formation can be shown as follows

2Ca(OH)₂+ 2Cl₂ → Ca(OCl)₂ + CaCl(Bleaching powder) +2 H₂O

Or, 2Ca(OH)₂ + 2CI₂ → 2Ca(OCl)Cl + 2H₂O

Or, Ca(OH)₂ + Cl₂ → Ca(OCl)Cl(Bleaching powder) + H₂O

2. According to Bunn, Clork, and Ghifford (1935), bleaching powder is supposed to be a mixture of

Ca(OCl)₂, CaCl₂ & Ca(OH)₂. Its formation can be represented as

2Cl₂ + 3Ca(OH)₂→ Ca(OCl)₂-Ca(OH)₂-CaCl₂-2H₂O (Bleaching powder)

Wlien Cl₂ gas is passed through excess of cold lime water, calcium chloride, and calcium hypochlorite are formed.

2Ca(OH)₂ + 2CI₂→ CaCl₂ + Ca(OCl)₂ + 2H₂O

When excess chlorine is passed through hot lime water calcium chloride and calcium chlorate are formed.

6Ca(OH)₂ + 6Cl₂→ 5CaCl₂ + Ca(ClO₃)₂ + 6H₂O

Uses of calcium hydroxide

• Calcium hydroxide is used in the manufacture of caustic soda, bleaching powder, superphosphate of lime (a chemical fertilizer), etc.
• It is used in the preparation of mortar, a building material.
• Slaked lime is mixed with three to four times of the weight of sand.
• The mixture is made into a thick paste with a gradual addition of water.
• The paste is called mortar. As the paste becomes dry, it hardens due to the formation of

CaCO₃:  Ca(OH)₂ + CO₂→ CaCO₃ + H₂O

• It is used in the manufacture of glass and in the manufacture of calcium hydrogen sulfate, Ca(HSO₄)₂ which is used in the paper industry.
• It is used in the recovery of ammonia from NH₄Cl (a by-product of the Solvay process), in coal gas purification, in tanneries for removing hair from hides, and for softening hard water. 0 Milk of lime is used for white washing due to its disinfectant properties.
• Lime water is a laboratory reagent for the detection of carbon dioxide.

3. Calcium carbonate (limestone), CaCO₃

Natural occurrence: In nature, calcium carbonate occurs in large quantities as limestone, marble, calcite, chalk, etc. It occurs also in the mineral, dolomite which is the double carbonate of calcium and magnesium (CaCO₃-MgCO₃). Besides these minerals, CaCO₃ occurs in abundance in corals, eggshells, outer covering of oysters, snail’s conch, and in teeth and bones of man and animals.

Preparation of calcium carbonate: Calcium carbonate may

Be prepared by passing CO₂ through lime water or by adding a solution of Na₂CO₃ to a solution of CaCl.

1 Ca(0H)₂ + CO₂ → CaCO₃↓+ H₂O

CaCl₂ + Na₂ CO₃→ CaCO₃↓+ 2NaCl

The precipitate of CaCO₃ is known as precipitated chalk.

Calcium carbonate Physical properties:

1. It is a white solid.
2. It is a stable compound which is almost insoluble in water.

Calcium carbonate Chemical properties:

1. The action of heat:

When heated at a higher temperature ( ~ 1270K), calcium carbonate decomposes to give calcium oxide (quicklime) & carbon dioxide. The reaction is reversible and endothermic. So, it proceeds towards completion when the reaction is carried out in an open vessel.

CaCO₃  ⇌  CaO + CO₂ ↑-heat

2. Reaction with dilute acids: It reacts with dilute acids to form the corresponding calcium salts.

CaCO₃  + 2HCl → CaCl₂ + CO₂↑ + H₂O

CaCO₃ + H₂SO₄ → CaSO₄ + CO₂↑ + H₂O

CaCO₃ + 2HNO₃ → Ca(NO₃)₂ + CO₂↑+ H₂O

3. Reaction with carbon dioxide:

When CO₂  gas is passed through a fine suspension of calcium carbonate in water, the latter slowly dissolves to produce calcium bicarbonate and thus, a clear solution is obtained.

CaCO₃ + CO₂ + H₂O→ Ca(HCO₃)₂

When the resultant clear solution is heated, calcium bicarbonate decomposes back to insoluble CaCO₃. Thus, the clear solution becomes turbid again. ‘

Uses of calcium carbonate:

• Calcium carbonate is largely used in the manufacture of quicklime, cement, and glass.
• Along with MgCO_3, it is used as a flux in the extraction of metals such as iron.
• It is used as a building material in the form of marble and the construction of statues.
• Specially precipitated>calcium carbonate is used in the manufacture of high-quality paper.
• Precipitated chalk is used in toothpowder and toothpaste, cosmetics, and also in some medicines (antacids).

Class 11 S-Block Notes

4. Plaster of Paris (hemihydrate of calcium sulfate), (CaSO₃)₂-H₂O

Preparation of Plaster of Paris:

Plaster of Paris is prepared by heating gypsum, (CaSO₄-2H₂O) at about 383-393K in a rotating burner.

Properties of Plaster of Paris:

• It is a white powder.
• On mixing with an adequate amount of water, it forms a concentrated mixture which solidifies in 5 to 15 minutes due to rehydration. This is called the setting of Plaster of Paris.

• When Plaster of Paris is heated at about 473K, it forms anhydrous CaSO₄. It is called dead burnt plaster and it does not solidify with water. For this reason, during the preparation of Plaster of Paris from gypsum, the temperature should not be allowed to rise above 393K.

Uses of Plaster of Paris:

• Plaster of Paris is largely used in the building industry.
• It is used in surgical bandages for plastering fractured bones.
• It is used for making casts of statues, molds in pottery work, ornamental castings, and blackboard chalks.
• It is also used in dentistry.

5. Cement

Cement is a mixture of finely powdered calcium silicates and aluminates along with small quantities of gypsum.

The raw materials used for cement are:

1. Limestone (CaCO₃) 
2. Clay containing silica (SiO₂) and
3. Alumina (Al₂O₃) and
4. Gypsum (CaSO₄ – 2H₂O).

Cement Composition:

Different types of cement have different compositions.  The composition of Portland cement is given below:

For cement to have good quality, the ratio of silica to alumina should be between 2.5 to 4 and the ratio of CaO to the total oxides of silicon, aluminum, and iron should be close to 2.

Cement Preparation:

For the manufacture of cement, limestone, and clay are fused by strong heating to form cement clinker. This is mixed with the gypsum and ground to a very thin powder.

Cement Settings:

When cement is mixed with water, it forms a plastic mass. After some time it becomes solid. This change is due to the three-dimensional linking between — Si—O — Si — and —Si—O—Al — chains. This transition from plastic to solid is called setting.

Fly ash is a waste product of the steel industry produced mainly due to the burning of coal and carbon compounds. It has similar properties to that of cement. Sometimes fly ash is used with the cement to reduce the cost without compromising on the quality.

Class 11 S-Block Notes

Cement Uses: 

• Cement is the most important construction material.
• It is used in the construction of tunnels, roads, bridges, etc.
• It is used in concrete and reinforced concrete. These are made by mixing cement with sand, pebbles, and water.
• Mixing with sand it is used for plastering.

Biological Importance Of Magnesium (Mg) And Calcium (Ca)

• Magnesium and calcium ions found in biological fluids play an important role in biological processes. Mg²⁺ ions are concentrated in cells while C²⁺ ions are concentrated in body fluids, outside the cell.
• It is known that the energy is stored in the form of ATP.  The formations of phosphate linkages are catalyzed by Mg²⁺ ions. Also, the hydrolysis of phosphate linkages, (Which is accompanied by the release of energy, is also catalyzed by Mg²⁺ ions.
• Mg²⁺ ions are present in chlorophyll-a, the green pigment of plants, which absorbs light and is essential for photosynthesis.
• Both these ions are also essential for the transmission of impulses along nerve fibers. About 99% of calcium in the body is present in bones and teeth as apatite, Ca₃(PO₄)₂. In the enamel of teeth, it is present as fluorapatite [3Ca₃(PO₄)₂.CaF₂].
• Ca²⁺ ions also play an important role in blood clotting and are necessary to trigger the contraction of muscles and to maintain regular heartbeats.
• The concentration of Ca²⁺ ions in blood plasma (about 100mg L^-1) is maintained by two hormones namely calcitonin and parathyroid.
• The calcium ions in bones exchange readily with those in blood plasma. About 400 mg of Ca²⁺ enters and leaves our bones every day.
• In normal adults, there is a balance between this exchange. However, in aged people, especially women, sometimes there occurs a net loss of calcium in the bone leading to a disease called osteoporosis.
• An adult human body contains 1200 g of Ca and 25 g of Mg compared to only 59 g of Fe and 0.06 g of Cu. The daily requirement of calcium and magnesium in the human body has been estimated to be about 200-300 mg. The sources of Mg in our food are nuts, green vegetables, wheat, coffee, etc. while that of Ca are milk, paneer, and different milk products.

Class 11 Chemistry S Block Elements Long Questions And Answers

Question 1. How can anhydrous magnesium chloride be prepared from magnesium chloride hexahydrate?
Answer:

Anhydrous magnesium chloride cannot be prepared by simply heating MgCl₂-6H₂O because it gets hydrolyzed by its water of crystallization.

However, when hydrated MgCl₂ is heated at 650K in the presence of HCl, its hydrolysis is prevented; and it loses its water of crystallisation to form anhydrous MgCl₂

However, when hydrated MgCl₂ is heated at 650K in the presence of HCI, its hydrolysis is prevented, and it loses its water of crystallization to form anhydrous MgCl₂.

Question 2. Which out of BeCl₂ and CaCl₂ would give an acidic solution when dissolved in water?
Answer: 

Being a covalent compound and a good Lewis acid, BeCl₂ forms a hydrated salt, Be(H₂O₂)4Cl₂. The hydrated ion undergoes hydrolysis in solution producing H₃O⁺. This occurs because the Be — O bond is very strong and so in the hydrated ion this weakens the O —H bond. Hence, there is a strong tendency to lose protons. For this reason, the aqueous solution of BeCl₂ is acidicin nature.

[Be(H₂O)₄]²⁺ + H₂O ⇌ [Be(H₂O)₃OH]⁺ + H₃O⁺

On the other hand, CaCl₂ is a strongly ionic compound and does not behave as a Lewis acid (the size of Ca is relatively large and its octet is filled up). Moreover, since it is a salt of strong acid and strong base, it does not undergo hydrolysis and therefore, its aqueous solution is neutral.

Question 3 Explain The below Observation

1.

2.

Answer:

1. The smaller Li⁺ ion exerts strong polarising power and distorts the electron cloud of the nearby oxygen atom of the OH^– ion. This results in the formation of a strong Li —O bond and the weakening of the O —H bond. This ultimately facilitates the decomposition of LiOH into Li₂O and H₂O. The polarising power of the large Na+ ion is much lower and thus, NaOH remains unaffected by heating.

Class 11 S-Block Notes

2. The smaller Li⁺ ion exerts a strong polarising power on highly polarisable H^– ion and as a result, the two atoms remain strongly attached by a covalent bond. On the other hand, due to the low polarising power of Na⁺ ion, NaH is essentially ionic & so it dissociates on heating to yield metallic sodium and dihydrogen.

Question 4. Discuss the roles of Na₂O₂, KO_2, and LiOH in the purification of air.
Answer:

Sodium peroxide (Na₂O₂) is used to purify the air in submarines and confined spaces as it removes carbon dioxide (CO₂) and produces O₂.

Potassium superoxide (KO₂) is used to purify air in space capsules, submarines, and breathing masks because it can absorb carbon dioxide (CO₂) thereby removing it and producing O₂ Both functions are important life support systems.

Lithium hydroxide (LiOH) is used for the absorption of carbon dioxide (CO₂) in space capsules and submarines

Question 5. Lithium forms monoxide while sodium forms) peroxide in the presence of excess oxygen why
Answer:

Larger cations can be stabilised by larger anions because if both the ions are comparable in size, the coordination number will be high and this gives rise to a high lattice enthalpy. Lithium-ion, Li⁺ as well as oxide ion, O^2- , have small ionic radii and high charge densities.

Hence these small ions combine and form a very stable lattice of lithium monoxide (Li₂O). Similarly, the formation of sodium peroxide (Na₂O₂) can be explained based on the stable lattice formed by the packing of relatively large Na⁺ ion and peroxide ion O₂

Question 6. A freshly cut piece of sodium metal appears shiny but its metallic lustre soon gets tarnished when exposed to air. Give reason
Answer:

When a freshly cut piece of metallic sodium is exposed to moist air, it readily reacts with oxygen to form sodium monoxide (Na₂O). The resultant sodium monoxide and also the metal itself readily react with the moisture of the air to form sodium hydroxide (NaOH).

In the subsequent step, both Na₂O and NaOH combine with CO₂ of air to form sodium carbonate (Na₂CO₃). Thus, a coating of sodium carbonate is formed on the surface of the metal and as a result of this, the metallic luster is tarnished.

4Na + O₂→ 2Na₂O; Na₂O + H₂O →2NaOH

2Na + 2H₂O→2NaOH + H₂↑; Na₂O+ CO₂→Na₂CO₃

2NaOH + CO₂ →Na₂CO₃ + H₂O

Question 7. What happens when each of the following compounds is heated? 

1. Li₂CO₃ 
2. Na₂CO₃
3. LiNO₃
4. KNO₃

Answer:

1. Lithium carbonate (Li₂CO₃ ) decomposes readily on heating to give lithium monoxide (Li₂O) and carbon dioxide (CO₂).

2. Na₂CO₃ does not decompose on heating.

(3) Lithium nitrate (LiNO₃) decomposes on heating to form lithium monoxide (Li₂O), nitrogen dioxide (NO₂), and dioxygen (O₂).

4.  Potassium nitrate (KNO₃) decomposes on heating to give potassium nitrite (KNO₂) and dioxygen

Question 8. What happens when

1. HCl gas is passed through a concentrated solution of common NaCl-containing impurities like Na₂SO₄, CaSO₄, CaCl₂, MgCl_2, etc.
2. caustic soda beads are exposed to air for a long time.
3. How will you convert Na₂CO₃ into NaHCO₃ and vice versa?

Answer:

1. When HCl gas is passed through a concentrated solution of common NaCl with impurities, crystals of pure NaCl separate out because of the common ion effect,  Caustic soda (NaOH) is a deliquescent substance and becomes wet on exposure to air.

On long exposure, the solid beads dissolve in the absorbed water Moist caustic soda then absorbs carbon dioxide (CO₂) from air to form sodium carbonate (Na₂CO₃) which forms a coating over the surface of the material. As sodium carbonate (Na₂CO₃) is not a deliquescent substance the wet sodium hydroxide becomes dry again.

2NaOH + CO₃→ Na₂CO₃+ H₂O

2.  Sodium bicarbonate (NaHCO₃) can be obtained by passing carbon dioxide (CO₂) through a saturated solution of sodium carbonate. Sodium bicarbonate, being less soluble gets separated from the solution as a white crystalline substance.

Na₂ CO₃ + CO₂+ H₂O→2NaHCO₃

When sodium bicarbonate (NaHCO₃) is heated, it decomposes to give sodium carbonate (Na₂CO₃) and carbon dioxide (CO₂)

Question 9. Why are the group-2 metals harder and have higher melting and boiling points than group-1 metals?
Answer:

The magnitude of the cohesive energy determines the hardness as well as melting and boiling points of the metals and it depends on the number of electrons involved in metallic bonding. In the case of group 1 metals, one electron per atom (valence electron, ns² ) is involved in metallic bonding while in group 2 metals, two electrons per atom (valence electrons, ns²) are involved in metallic bonding. Moreover, atoms of group 2 metals are smaller in size than those of group 1 metals.

Consequently, stronger metallic bonding exists in group 2 metals which results in higher cohesive energy and close packing of the atoms. This accounts for the greater hardness and higher melting and boiling points of group 2 metals as compared to group 1 metals.

Class 11 S-Block Notes

Question 10.

1. Give some common tests used for the detection of calcium compounds. 

2. A white solid When heated liberates a colorless gas that does not support combustion. The residue is dissolved in water to form (B) which can be used for whitewashing. When excess CO₂ gas is passed through the solution of (B), it gives a compound (C) which on heating forms (A). Identify (A), (B) and (C). Give the reactions
Answer:

The following tests may be performed for the detection of Ca -compounds:

1. Calcium salts give a brick-red color in the flame test,
2. When ammonium oxalate solution is added to a solution of a calcium salt, a white precipitate of crystalline calcium oxalate is obtained which is insoluble in acetic acid but soluble in mineral acids
3. In addition of a solution of sodium carbonate to a neutral (or ammoniacal) solution of a Ca-salt, white calcium carbonate is precipitated, which is soluble in acids.
4. The observations suggest that the compound (A) is limestone, i.e., CaCO₃, the compound (B) is calcium hydroxide & the compound (C) is calcium bicarbonate.

The corresponding reactions are as:

Question 11. Compare the alkali metals and alkaline earth metals concerning their

1. Basicity of oxides
2. Solubility of hydroxides and
3. Solubility of nitrates.

Answer:

1. Basicity of oxides:

The ionization enthalpy of alkali metals is less than the corresponding alkaline earth metals. So the alkali metal oxides are more basic than the corresponding alkaline earth metal oxides.

2. Solubility of hydroxides:

Due to of small size and higher ionic charge, the lattice enthalpies of alkaline earth metal hydroxides are much higher than those of alkali metal hydroxides and hence, the solubility of alkali metal hydroxides is much higher than that of alkaline earth metal hydroxides.

3. Solubility of nitrates:

Nitrates of alkali and alkaline earth metals are soluble in water. However, the solubility of alkali metal nitrates increases down the group because their lattice enthalpies decrease more rapidly than their hydration enthalpies. Nitrates of alkaline earth metals follow the reverse trend i.e., their solubility decreases down the group and this is because their hydration enthalpies decrease more rapidly than their lattice enthalpies.

Question 12. What are the properties that make oxides of MgO and  BeO useful for lining furnaces?
Answer:

The given properties make MgO and BeO useful for lining furnaces

• They have high melting points [melting point of Beo is 2500°C (approx) and MgO is 2800°C (approx)].
• They are very good conductors of heat.
• They have very low vapor pressure.
• They are chemically inert.
• They are insulators

Question 13. The halides of alkali metals are soluble in water except for LiF. Why?
Answer:

The solubility of a salt in water depends on its lattice enthalpy as well as its hydration enthalpy. A salt dissolves in water when its hydration enthalpy exceeds its lattice enthalpy value. The lattice enthalpy of LiF is very high and its hydration enthalpy value does not exceed this value. As a result, LiF is insoluble in water

However, the other halides of alkali metals possess higher hydration enthalpy values compared to their corresponding lattice enthalpies. Hence, they are soluble in water

Question 14. Why Is LiCO₃ decomposed at a lower temperature whereas Na₃ CO₃ at a higher temperature
Answer: 

Li+ ion being smaller in size forms a more stable lattice with the smaller oxide ion (O^2-) than the larger carbonate ion (CO₃ ) and consequently, Li₂CO₃  decomposes into Li₂ O at a much lower temperature. The high polarising power of very small Li⁺ ions also facilitates the decomposition of Li₂CO₃.

On the other hand, the larger Na+ ion forms a more stable lattice with the larger CO₃ ion than with the smaller O^2- ion. Therefore, Na₂ CO₃ is quite stable and decomposes only at very high temperatures.

Question 15. Compare the solubility and thermal stability of the given compounds of the alkali metals with those of the alkaline earth metals. 

1. Nitrates
2. Carbonates Sulphates.

Answer:

1. Nitrates:

The nitrates of alkali metals as well as alkaline earth metals are highly soluble in wOn heating, nitrates of both alkali and alkaline earth metals undergo decomposition.

Nitrates of alkali metals decompose to form metallic nitrite and oxygen. On the other hand, nitrates of alkaline earth metals decompose to form the corresponding oxides with the evolution of NO₂ and O₂

Due to the diagonal relationship, lithium nitrates behave similarly to magnesium nitrate

2. Carbonates:

1. Except for Li₂CO₃, other carbonates of alkali metals readily dissolve in water. However, carbonates of alkaline earth metals are practically insoluble in water. Their solubilities decrease on moving down the group. So, BeCO₃ is sparingly soluble in water while BaCO₃ is insoluble in water.

2. Carbonates of alkali metals except Li₂CO₃ are stable and do not decompose on heating, but carbonates of alkaline earth metals decompose on heating to give metal oxide and carbon dioxide.

The thermal stability of the carbonates increases down the group. Like MgCOg, Li₂CO₃ decomposes on heating.

3. Sulphates:

1. Except Li₂SO₄, the remaining sulfates of alkali metals are water-soluble. The sulfates of alkaline earth metals are relatively less soluble in water than the corresponding sulfates of alkali metals. Further, their solubilities decrease down the group,

2. The sulfates of alkali metals are stable compounds and do not decompose on heating. On the other hand, alkaline earth metals dissociate on heating to give metal oxides and sulfur trioxide.

The thermal stability of the sulfates increases down the group. Li₂SO₄ dissociates on heating just like MgSO₄

Class 11 S-Block Notes

Question 16. Starting with sodium chloride how would you proceed to prepare

1. Sodium metal
2. Sodium peroxide

Answer:.

1. Metallic sodium can be obtained by the electrolysis of a mixture of sodium chloride (40%) and calcium chloride (60 %) in a fused state. The function of calcium chloride is to lower the reaction temperature from 807°C (m.p. of NaCl) to about 577°C. The molten sodium metal thus obtained is liberated at the cathode

Overall reaction: Nacl → Na⁺ + Cl^–

At cathode: Na⁺ + e→ Na;

At anode: Cl^– → Cl + e; Cl + Cl → Cl₂ ↑

2.  Sodium chloride is first converted to sodium by electrolytic reduction. The metal is then heated more than air. The initially formed sodium oxide reacts with excess O₂ to form Na₂O₂.

Question 17. What happens when  

1. Magnesium is burnt in the air 
2. Calcium nitrate is heated?

Answer:

1. When magnesium bums in the air, magnesium oxide (MgO) and magnesium nitride (Mg₃N₂) are obtained as products

2. On heating calcium nitrate, it decomposes to form CaO, NO_2, and O₂ reaction

Question 18. The hydroxides and carbonates of sodium and potassium are easily soluble in water while the corresponding salts of magnesium and calcium are sparingly soluble in water. Explain.
Answer:

Alkali metals form monovalent cations (such as Na⁺, and K⁺) while alkaline earth metals form divalent cations (such as Mg²⁺, and Ca²⁺). Due to the increase in charge of the cations, the lattice energy of the corresponding salt increases. For this reason, hydroxides and carbonates of sodium and potassium have lower lattice enthalpy values than the hydroxides and carbonates of magnesium and calcium.

As hydration enthalpies of the hydroxides and carbonates of sodium and potassium are greater than their lattice enthalpies, these salts readily dissolve in water. However, in the case of the hydroxides and carbonates of calcium and magnesium, the lattice enthalpy values is greater than that of hydration enthalpy and consequently, these salts are less soluble in water.

Question 19. Why are lithium salts commonly hydrated and those of the other alkali ions usually anhydrous?
Answer:

As Li⁺ is the smallest ion among all the alkali metal ions, it can polarise water molecules more easily than the other alkali metal ions. So, numerous water molecules get attached to lithium salts as water of crystallization. However, this is not observed in the case of other alkali metal ions. Thus, lithium salts are commonly hydrated

Example: LiCl-2H₂O  and those of the other alkali ions are usually anhydrous.

Class 11 S-Block Notes

Question 20. Why is LiF almost insoluble in water whereas LiCl is soluble not only in water but also in acetone?
Answer:

The lattice enthalpy of ionic LiF (formed by small Li⁺ion and F^– ion) is higher than its hydration enthalpy. On the other hand, the lattice enthalpy of LiCl containing small Li⁺ ion and large Cl^– ion is considerably lower than its hydration enthalpy.

Thus, LiF is almost insoluble in water while LiCl is soluble. Furthermore, Li⁺ ions can polarise bigger Cl^– ions more easily than smaller F^– ions. As a result, LiCl has more covalent character than LiF and so, it is also soluble in the organic solvent acetone.

Question 21. What happens when

1. Sodium metal is dropped in water?
2. Sodium metal is heated in a free supply of air
3. Sodium peroxide dissolves in water?

Answer:

Sodium hydroxide is formed. H₂ gas is evolved which catches fire due to the exothermicity of the reaction.

2Na(s) + 2H₂O(l)→ 2NaOH(aq) + H₂(g)

2. Sodium peroxide is formed by heating sodium metal in a free supply of air.

2Na(s) + O₂ (g)→ 2Na₂O₂(s)

3. H₂O₂ is formed when sodium peroxide dissolves in water

Na₂O₂(s) + 2H₂O(l)→2NaOH(aq) + H₂O₂(Z)

Question 22. Comment on each of the given observations:

1. Lithium is the only alkali metal to form a nitride directly.
2.  M²⁺+(aq) + 2e → M(s) (where M = Ca, Sr, or Ba) is nearly constant.

Answer:

1. The lattice energy of lithium nitride (Li₃N) which consists of a small cation (Li⁺) and a small anion (N₃-) is much higher and this energy compensates for the high bond dissociation energy of the N=N bond and the energy to form N₃– ion. Larger alkali metal ions cannot compensate for these energy requirements. Hence, Li⁺ is the only alkali metal that forms nitride directly.

2. M²⁺ (aq) + 2e → M(s) Where M = Ca, Sr, Ba

Standard electrode potential, \(E_{\mathrm{M}^{2+} / \mathrm{M}}^0\) depends on 3

1. Enthalpy of vaporisation
2. Ionization enthalpy and
3. Enthalpy of hydration.

As the combined effect of these factors is almost the same for Ca, Sr and Ba, their E° values are nearly constant.

Class 11 S-Block Notes

Question 23. State as to why

1.  A solution of Na₂CO₃ is alkaline?
2. Alkali metals are prepared by electrolysis of their fused chlorides.
3. Sodium is found to be more useful than potassium.?

Answer:

1. Na₂CO₃ is a salt of weak acid H₂CO₃ and strong base NaOH. In an aqueous solution, it ionises to form Na⁺ and CO^2-₃ – ions.

Na₂CO₃ ⇌  2Na⁺  (aq) + CO^2-₃ (aq)

The formed CO^2-₃ ions hydrolyse in an aqueous solution to produce acetic acid and OH^– ion.

CO^2-₃ (aq) + 2H₂O(Z) ⇌ H₂CO₃ ⇌+ 2OH^– ( aq)

As H₂CO₃ is a weak acid, it remains mostly unionized. Consequently, the concentration of OH^– ions increases in the solution thereby making the solution alkaline.

2.

• As alkali metals are strong reducing agents, they cannot be extracted by chemical reduction from their oxides or other compounds,
• As alkali metals are highly electropositive, these metals cannot be displaced from their salts with the help of other elements,
• Alkali metals cannotbe obtained even by the electrolysis of aqueous solutions of their salts.

In this case, H₂, instead of the alkali metal is liberated at the cathode because the discharge potential of alkali metals is higher than that of hydrogen.

Thus, to prepare alkali metals, electrolysis of their fused chlorides is carried out. For example,

NaCl →Na⁺ + Cl^–

During electrolysis, at the cathode,

2Na⁺ + 2e →2Na; and at the anode, 2Cl→ Cl₂ + 2e

3. Sodium is found to be more useful because Na is not as reactive as K. For this reason, reactions of sodium with different substances can be controlled and usage of sodium is far more safe than potassium. Thus, sodium is more useful than potassium

Question 24. Write balanced equations for reactions between 

1. Na₂O₂ & Water,
2. KO₂ & water,
3. Na₂ O& CO₂

Answer: The balanced equations of the given reactions are.

1.

2. 2K O₂ (S) + 2H₂ O(l) → 2KOH(aq) + H₂ O₂ (aq) + O₂ (g)

3. Na₂ O + CO₂→ Na₂ CO₂

Question 25. How would you explain the given observations? 

1. BeO is almost insoluble but BeSO₄ is soluble in water.
2. BaO is soluble but BaSO4 is insoluble in water.
3. Lil is more soluble than KI in ethanol.

Answer:

1. O^2- is smaller in size than SO₄^2-. Consequently, a small Be²⁺ ion is tightly packed with a small O^2- ion, and thus, the lattice enthalpy of BeO is greater than its hydration enthalpy. So BeO is insoluble in water. On the other hand, a small Be²⁺ ion is loosely packed with a large SO^2-. ion and thus, the lattice enthalpy of BeSO₄ is less than its hydration enthalpy. So, BeSO₄ is soluble in water

2. Large Be²⁺ ion is tightly packed with large SO₄^2-.ion and thus, the lattice enthalpy of BaSO₄ is greater than its hydration enthalpy. So, BaSO₄ is insoluble in water On the contrary, a large Ba²⁺ ion is loosely packed with a small Oion, and thus lattice enthalpy of BaO is less than its hydration enthalpy. So, BaO is water soluble.

KI is predominantly ionic. On the other hand, due to the high polarising power of very small Li+ ions, Lil is predominantly covalent. For this reason, Lil is more soluble than KI in the organic solvent ethanol.

Question 27. How will you distinguish between:

1. Mg and Ca
2. Na₂SO₄ and BaSO₄
3. Na₂CO₃ and NaHCO₃,
4. LiNO₃ and KNO₃

Answer:

1. Calcium, when heated, imparts brick red color to the flame but magnesium does not.

2. Sodium sulfate (Na₂SO₄) is soluble in water but barium sulfate (BaSO₄) is insoluble.

3. Na₂CO₃ is stable to heat but NaHCO₃ decomposes on heating to produce CO₂ gas which turns limewater milky

4. When LiNO₃ is heated, it decomposes to yield reddish-brown vapors of NO₂. However, when KNO₃ is heated, it decomposes to yield colorless O₂ gas

Class 11 Chemistry S Block Elements Short Question And Answers

Question 1. Li+ ion is the smallest one among the ions of group- 1 elements. It would, therefore, be expected to have much higher ionic mobility and hence the solutions of lithium salts would be expected to have higher conductivity than the solutions of cesium salts. However, in reality, the reverse is observed. Explain
Answer: 

Due to high charge density, very small Li⁺ ion gets much more hydrated compared to the large Cs⁺ ion. Thus, the size of the hydrated lithium-ion is much larger than that of the hydrated cesium ion. For this reason, the mobility of Li⁺ ion is much lower than that of Cs⁺ ion and consequently, the solutions of lithium salts have much lower conductivity than the solutions of cesium salts.

Question 2. The E° value for Cl₂/Cl^– is +1.36, for I₂/I^– is + 0.53V, for Ag⁺/Ag is + 0.79 V, for Na⁺/Na is -2.71V and for Li⁺/Li is -3.04V. Arrange the following atoms and ions in decreasing order of their reducing strength: I^–, Ag, Cl^–, Li, Na
Answer:

The lower the value of standard reduction potential, the greater the tendency of the reduced form to be oxidized, i.e., the reduced form will serve as a stronger reductant. Therefore, the decreasing order of reducing strength of the given atoms and ions is

Li > Na > I^–> Ag > Cl^–

Class 11 S-Block Notes

Question 3. The alkali metals are obtained not by the electrolysis of the aqueous solutions of their salts but by the electrolysis of their molten salts. Explain.
Answer:

The solutions of alkali metal salts contain metal cations, anions, H⁺ ions, and OH^– ions. The discharge potential of H⁺ ions is lower than that of metal cations. So, on electrolysis of tire solutions of alkali metal salts, hydrogen is discharged at the cathode rather than the metal. However, when the molten salts of alkali metals are electrolysed, the metal cation being the only cation present, gets discharged at the cathode

Question 4. The alkali metals are paramagnetic but their salts are diamagnetic—why?
Answer:

Due to the presence of anupaired valence electron the alkali metals are paramagnetic. During salt formation, this unpaired electron of the outermost shell leaves the metal atom and becomes attached to a non-metal atom. As a result, the cation and the anion thus obtained contain no unpaired electron. Hence, the alkali metal salts are diamagnetic.

Question 5. Beryllium & magnesium do not give color to flame whereas other alkaline earth metals do so. Why?
Answer: 

Due to their smaller size, valence electrons of Be and Mg are more tightly held by the nucleus. Therefore, they need a large amount of energy for the excitation of their valence electrons to higher energy levels. Since such a large amount of energy is not available from Hansen flame, these two metals do not impart any color to the flame.

Question 6. E° for M²⁺(aq) + e →M(s) (where, M = Ca, Sr or Ba) Is nearly constant. Comment.
Answer:

The value of standard electrode potential (E°) of any M²⁺/M electrode depends upon three factors

1. Enthalpy of vaporisation
2. Ionization enthalpy, and
3. Enthalpy of hydration.

Since the combined effect of these factors is approximately the same for Ca, Sr, and Ba, their standard electrode potential (E°) values are nearly constant.

Question 7. Both alkaline earth metals and their salts are diamagnetic. Explain.
Answer:

The alkaline earth metals are diamagnetic as all the orbitals are filled with paired electrons. The ions of alkaline earth metals, M²⁺ have stable noble gas configurations in which all the orbitals are doubly occupied. Also, there are no unpaired electrons in the anions. Hence, the salts of alkaline earth metals are also diamagnetic.

Question 8. Beryllium salts can never have more than 4 molecules of water of crystallization, i.e., it can never achieve a coordination number > 4 while other metal ions tend to have a coordination number of 6, for example: [Ca(H₂O₆)²⁺. Explain.
Answer: 

Beryllium does not exhibit coordination numbers more than 4 because, in its valence shell of Be²⁺ ion, there are only four available orbitals (one s and three p) present. The remaining members of the group can have a coordination number of six by using their d -d-orbitals along with s -and p -orbitals.

Question 9. Anhydrous is used as a drying agent — why?
Answer:

Anhydrone or magnesium perchlorate, Mg(ClO₄)₂ is  used as a drying agent because, due to its smaller size and higher charge, Mg has a greater tendency to complexation with water molecules by forming Mg(ClO₄)₂ -6H₂O

Question 10. Li salts are commonly hydrated while other alkali metal salts are usually anhydrous. Explain.
Answer:

Due to the smallest size among all alkali metal ions, the Li⁺ ion can interact with water molecules more easily than the other alkali metal ions. Hence the salts of lithium are commonly hydrated. On the other hand, other alkali metal ions being larger have little tendency to get hydrated. Therefore, their salts are generally anhydrous. For example, lithium chloride crystallises as LiCl. 2H₂O₂ but sodium chloride crystallizes as NaCl.

Question 11. Although the abundance of Na and K in the earth’s crust are comparable, sodium is nearly 30 times more abundant than potassium in seawater—why?
Answer:

These metals were leached from the aluminosilicate rocks by weathering. The potassium salts having large anions are less soluble than the sodium salts because of higher lattice energy. Moreover, potassium is preferentially absorbed by the plants. For this reason, sodium is more abundant than potassium in seawater.

Question 12. When caustic soda solution is kept in a glass bottle, the inner surface of the bottle becomes opaque. Explain
Answer:

Caustic soda (NaOH) being strongly basic reacts with acidic silica (SiO₂) present in glass to form sodium silicate (Na₂SiO₃)

SiO₂ + 2NaOH → Na₂SiO₃ + H₂O

As a result, the inner surface of the bottle becomes opaque.

Question 13. Why are alkali metals stored in kerosene?
Answer:

When the highly reactive alkali metals are exposed to air,  they readily react with oxygen, moisture, and carbon dioxide of air to form oxides, hydroxides and carbonates respectively. To prevent these reactions, alkali metals are normally stored in kerosene, an inert liquid.

Question 14. The second ionization enthalpies of alkaline earth metals are much lower than those of the corresponding alkali metals. Explain.
Answer:

The loss of a second electron from an alkali metal cation (M⁺) causes a loss of stable noble gas configuration while the loss of a second electron from an alkaline earth metal cation leads to the attainment of a very stable noble gas configuration. This explains why the second ionization enthalpies of alkaline earth metals are much lower than those of the corresponding alkali metals.

Class 11 S-Block Notes

Question 15. MgO is used as a refractory material— Explain why
Answer:

Due to greater charge on both the cation (Mg²⁺) and die anion (O^2-), MgO possesses higher lattice energy and for tills, it has very high melting point and does not decompose on heating. For this reason, it is used as a refractory material

Question 16. BaSO₄ is insoluble in water whereas BeSO₄ is soluble in water—1-explain with reason._
Answer: 

The lattice enthalpy of BaSO₄ is much higher than its hydration enthalpy and hence it is insoluble in water. On the other hand, the hydration enthalpy of BeSO₄ is much higher than its lattice enthalpy because of the loose packing of the small Be²⁺ ion with the relatively large sulfate ion. Hence, it becomes soluble in water.

Question 17. BeCl₂ fumes in moist air but BaCl₂ does not. Explain
Answer:

BeCl₂ being a salt of a weak base, Be(OH)_2, and a strong acid, HCl undergoes hydrolysis in moist air to form HCl; which fumes in air. BaCl₂ on the other hand, being a salt of a strong base, Ba(OH)₂ and a strong acid, HCl does not undergo hydrolysis to form HCl and hence does not fume in moist air

BeCl₂ + 2H₂O→Be(OH)₂ + 2HCl↑

BaCl₂ +H₂O→ Ba(0H)₂ + 2HCl

Question 18. Mg₃N₂ when reacts with water, gives off NH₂ but HCl is not evolved when MgCl₂ reacts with water at room temperature. Give reasons.
Answer:

Mg₃N₂ is a salt of the strong base, Mg(OH)₂ and the weak acid, NH₃

Hence it gets hydrolyzed to give NH₃.

Mg₃N₂ + 6H₂O→ 3Mg(OH)₂ + 2NH₃T↑

MgCl₂,  On the other hand, is a salt of the strong base, Mg(OH)_2, and the strong acid, HCl. Hence, it does not undergo hydrolysis

Question 19. A piece of burning magnesium ribbon continues to bum in sulphur dioxide.
Answer:

A piece of magnesium ribbon continues to bum in SO₂ since it reacts to form MgO and S. This reaction is such exothermic that heat evolved keeps the magnesium ribbon burning.

This reaction is such exothermic that heat evolved keeps the magnesium ribbon burning

Question 20. Ba²⁺ ions are poisonous, still, they are provided to patients before taking stomach X-rays. Explain
Answer:

A barium meal (suspension of BaSO₄ in water) is generally given to patients when X-ray photographs of the alimentary canal are required. The salt provides a coating of the alimentary canal and hence X-ray photograph can be taken since it is quite transparent to the X-ray otherwise. BaSO₄ is almost insoluble in water and hence it does not pass from the digestive system to the circulatory system and can therefore be safely used for the purpose.

Question 21. Can sodium hydride be dissolved in water? Justify.
Answer:

Sodium hydride cannot be dissolved in water because it gets hydrolyzed with brisk effervescence of hydrogen gas.

NaH + H₂ O→  H₂ + NaOH

Question 22. Why does sodium impart a yellow color in the flame?
Answer:

The ionization enthalpy of Na is relatively low. Therefore when this metal or its salt is heated in Bunsen flame, its valence shell electron is excited to higher energies by absorption of energy. When the excited electron returns to its initial position in the ground state, it liberates energy in the form of light in the yellow region of the electromagnetic spectrum. That’s why sodium imparts yellow color to the flame.

Class 11 S-Block Notes

Question 23. Wind makes lithium exhibit uncommon properties compared to the rest of the alkali metals.
Answer:

The unusual properties of lithium as compared to other alkali metals is since

• Li – atom and L⁺  ion are exceptionally small in size and
• Li⁺ ion has the highest polarising power (i.e„ charge/size ratio).

Question 24. What is the common oxidation state exhibited by the alkali metals and why?
Answer:

The alkali metals easily lose their valence electrons (ns¹) to acquire a stable octet, (i.e., the stable electronic configuration of the nearest noble gas) and because of this, the common oxidation state exhibited by the alkali metals is +1.

Question 25. What is the difference between baking soda and baking powder?
Answer:

Both baking soda and baking powder are leavening agents. Baking soda is pure sodium bicarbonate. When baking soda is combined with moisture and an acidic ingredient (for example,  Yoghurt, buttermilk) the resulting chemical reaction produces CO₂ gas bubbles that cause baked goods to rise. Baking powder contains NaHCO₃, but it includes an acidifying agent (cream of tartar) already, and also a drying agent (usually starch).

Baking powder is available as single-acting baking powder and as double-acting baking powder. Single-acting baking powders are activated by moisture, so we must bake recipes that include this product immediately after mixing. Double-acting powder reacts in two phases and can stand for a while before baking.

Question 26. Though table salt is not deliquescent it gets wet ! in the rainy season— Explain.
Answer:

Pure NaCl is not deliquescent but table salt contains impurities like MgCl₂ and CaCl₂ These impurities being deliquescent absorb moisture from air in the rainy season. As a result, table salt gets wet

Question 27. What precautions should be taken while handling beryllium compounds and why?
Answer:

Contact of Be compounds with the skin dermatitis, and inhaling dust or smoke of Be-compounds causes a disease called berylliosis which is rather similar to success- Therefore, beryllium compounds should be handled with care

Question 28. Explain why the elements of group 2 form M²⁺ Ions, but not M3+ ions.
Answer:

Loss of third electron from group-2 metal atoms causes loss of stable noble gas configuration and for this reason group- 2 elements form M²⁺ ions but not M³⁺ ions. fifl Arrange Be(OH)₂, Ba(OH)₂ & Ca(OH)₂in order of increasing solubility in water and explain the order. Answer: Among the alkaline earth metal hydroxides having a common anion, the cationic radius influences the lattice enthalpy. Since the lattice enthalpy decreases much more than the hydration enthalpy with increasing cationic size, the solubility increases on moving down the group.

Question 29. The reaction between marble and dilute H₂SO₄ is not used to prepare CO₂ gas—why?
Answer:

Marble (CaCO₃) reacts with dilute H₂SO₄ to form insoluble CaSO₄ which deposits on the surface of marble and prevents further reaction. So, the evolution of CO₂ ceases after some time. Thus the reaction between marble and dilute H₂SO₄ prepares CO₂ gas

Question 30. Name the important compound of Li used in organic synthesis. How the compound is prepared?
Answer:

The compound is lithium aluminum hydride (LiAlH₄). It is a useful reducing agent and is used in organic synthesis. It is prepared by the reaction of lithium hydride and aluminum chloride in a dry ether solution.

4LiH + AlCl₃ → LiAlH₄ + 3LiCl

Question 31. What is the oxidation state of K in KO₂ and why is this compound paramagnetic?
Answer:

The superoxide ion is represented as O^–₂ . It has one unit of negative charge. Since the compound is neutral, therefore, the oxidation state of K is +1. The structure of Or is 6J0 O^–₂    is  Since it has one unpaired electron in π∗_2p MO, therefore, the compound is paramagnetic.

Question 32. The crystalline salts of alkaline earth metals contain more water of crystallization than the corresponding alkali metals. Explain.
Answer:

In the salts of alkaline earth metals, the metal ions have a smaller size and higher charge compared to the corresponding metal ions of the alkali metals of the same period. Thus, alkaline earth metals have a greater tendency to get hydrated & form crystalline salts compared to alkali metals. Thus, NaCl is completely anhydrous whereas MgCl₂ exists as MgCl₂-6H₂O

Question 33. Lithium salts are more stable if the anion present in the salt is small. Explain.
Answer:

Small anions have more ionic character and hence the salts of lithium containing those ions have more lattice enthalpy. Large anions, on the other hand, are highly polarisable and hence they impart covalent character to the salt. Thus, lithium salts are more stable with small anions than that with large anions

Question 34. Alkali metals become opaque when they are kept open in the air Why?
Answer:

As the alkali metals are highly reactive, they readily react 20; with oxygen to form oxides. These oxides undergo a reaction with the water vapor present in the air to produce hydroxides. The formed hydroxides immediately react with CO₂ of air to produce carbonate compounds. These carbonate compounds form layers on the surface of alkali metals. Consequently, they become opaque

Class 11 S-Block Notes

Question 35. BaSO₄ is insoluble in water, but BcSO₄ is soluble in water-Explain.
Answer:

The lattice enthalpy of BaSO₄ is much higher than its hydration enthalpy and hence, it is insoluble in water. On the other hand, the hydration enthalpy of BeSO₄ is much higher than its lattice enthalpy because of the loose packing of the small Be²⁺ ion with the relatively large sulfate ion. Hence, it becomes soluble in water.

Question 36. An aqueous solution of Be(NO₃)₂ is strongly acidic. Explain.
Answer:

In the hydrated ion, [Be(H₂O)₄]²⁺, water molecules are extensively polarised, ultimately leading to the weakening of the O —H bond.

Hydrolysis takes place and the solution becomes distinctly acidic:

(H₂O)₃Be₂ +—OH₂ + H₂O→(H₂O)₃ 3 Be⁺ —OH + H₃O⁺

Question 37. The hydroxides and carbonates of Na and K are readily soluble in water while the corresponding salts of Mg and Ca are sparingly soluble. Explain.
Answer:

Due to smaller size and higher ionic charge, the lattice enthalpies of alkaline earth metals are much higher than those of alkali metals and hence the solubility of alkali metal hydroxides and carbonates is much higher than those of alkaline earth metal hydroxides and carbonates;

Question 38. BeO is insoluble but BeSO₄ is soluble in water. Explain.
Answer:

The higher lattice enthalpy of BeO formed by the combination of a small cation and small anion is more than its hydration enthalpy but the lattice enthalpy of BeSO₄ formed by the combination of a small cation and large anion is less than its hydration enthalpy;

Question 39. BaO is soluble but BaSO₄ is insoluble in water — why?
Answer:

The lattice enthalpy of BaO formed by the combination of a large cation and a small anion is less than its hydration enthalpy but the lattice enthalpy of BaSO₄ formed by the combination of a large cation and a large anion is more than its hydration enthalpy;

Question 40. Lil is more soluble than KIin ethanol. Explain.
Answer:

Due to the much higher polarising power of very small Li⁺ ion, Lil is predominantly covalent but due to the low polarising power of relatively large K⁺ ion, KI is predominantly ionic and for this reason, Lil is more soluble in the organic solvent ethanol;

Question 41. How can fused calcium chloride be prepared? Give two important uses of it.
Answer:

When CaCl₂ 2H₂O is heated above 533K, anhydrous CaCl₂ forms. This melts at 1046K. When the molten salt is cooled, it solidifies as white lumps of crystalline mass which is known as fused calcium chloride;

Question 42. Hydrated magnesium chloride is heated in the presence of ammonium chloride (NH₄Cl).
Answer:

Magnesium chloride when heated with NH₄Cl forms an additional compound (MgCl₂-NH₄Cl₆H₂O) which on heating forms anhydrous MgCl₂

Question 43. Why are alkali metals not found in nature?
Answer:

Alkali metals are highly electropositive and extremely reactive elements. Thus, they easily react with atmospheric oxygen and carbon dioxide. These metals have a high tendency to lose electrons to form cations because of their low ionization potential. For this reason, they readily react with highly electronegative elements or radicals to form compounds. So, alkali metals are not found in a free state in nature

Question 44. Explain why is sodium less reactive than potassium.
Answer:

Due to its small size, the ionization enthalpy of sodium (496kL-mol^-1) is greater than potassium (419kJ. mol^-1) and the standard electrode potential value of potassium (-2.925V) is more negative than that of sodium (-2.714V). Thus, sodium is less reactive than potassium.

Question 45. Basicity of oxides
Answer:.

Basicity of oxides: The ionization enthalpy of alkali metals is less than that of the corresponding alkaline earth metals, i.e., the electropositive character of alkali metals is greater than that of alkaline earth metals. Thus, the basicity of oxides of alkali metals is more than the oxides of alkaline earth

Class 11 Chemistry S Block Elements Multiple Choice Questions

Question 1. NO₂ is not obtained on heating—

1. AgNO₃
2. KNO₃
3. Cu(NO₃)₂
4. Pb(NO₃)₂

Answer: 2. KNO₃

KNO₃ on heating decomposes to form potassium nitrite (KNO₂) and oxygen gas

Class 11 S-Block Notes

Question 2. Which one of the following has the lowest ionization

1. ls²2s²2p⁶
2. ls²s²2p⁶3s¹
3. ls²2s²2p⁵
4. ls²2s²2p³

Answer: 2. ls²s²2p⁶3s¹

Since the electronic configuration ls²s²2p⁶3s¹ is that of an alkali metal, its ionization potential value is the lowest.

Question 3. Which of the following represents the composition of carnallite mineral—

1. K₃O-Al₂O₃.6SiO₂
2. KNO₃
3. K₂SO4.MgSO₄-MgCl₂-6H₂O
4. KCl-MgCl₂-6H₂O

Answer: 4. KCl-MgCl₂-6H₂O

Carnallite:  KCl-MgCl₂-6H₂O

Question 4. Chlorine gas reacts with red-hot calcium oxide to give—

1. Bleaching powder and dichlorine monoxide
2. Bleaching powder and water
3. Calcium chloride and chlorine dioxide
4. Calcium chloride and oxygen

Answer: 4. Calcium chloride and oxygen

2CaO + 2Cl₂ → 2CaCl₂ + O₂

Question 5. The decreasing order of the basic characters of K₂O, BaO, CaO, and MgO is

1. K₂O > BaO > CaO > MgO
2. K₂O > CaO > BaO > MgO
3. MgO > BaO > CaO > K₂O
4. MgO>CaO>BaO>K₂O

Answer:   4. MgO>CaO>BaO>K₂O

The oxides of alkali metals are highly basic. The basic character of the oxides of alkaline earth metals increases on moving down the group.

Question 6. Match the flame colors of the alkaline earth metal salts in the Bunsen burner.

Answer: 4. 1 -A, 2-B,3-C

1. 1 -A, 2-C,3-B
2. 1 -C, 2-A,3-B
3. 1 -B, 2-C,3-A
4. 1 -A, 2-B,3-C

Question 7. The correct order of solubility in water is

1. CaSO₄>BaSO₄>BeSO₄>MgSO₄>SrSO₄
2. BeSO₄ > MgSO₄> CaSO₄> SrSO₄ > BaSO₄
3. BaSO₄>SrSO₄>CaSO₄>MgSO₄>BeSO₄
4. BeSO₄> CaSO₄ > MgSO₄ > SrSO₄> BaSO₄

Answer: 2. BeSO₄ > MgSO₄> CaSO₄> SrSO₄ > BaSO₄

Question 8. When BaCl₂ is added to an aqueous solution, a white precipitate is obtained. The anion among CO^2-₃ SO^2-₃ and SO^2-₄ that was present in the solution can be

1. CO^2-₃ But any of the other two
2. SO^2-₃ But not any of the other two
3. SO^2-₄ But not any of the other two
4. Any of them

Answer: 4. Any of them

BaSO₃, BaCO₃ and BaSO₄ are insoluble in water, thus they are precipitated out in an aqueous solution.

Question 9. Which of the following is least thermally stable

1. MgCO₃
2. CaCO₃
3. SrCO₃
4. BeCO₃

Answer: 4. BeCO₃

Question 10. Which one of the following orders presents the correct sequence of the increasing basic nature of the given oxides

1. MgO < K₂O < Al₂O₃ < Na₂O
2. Na₂O<K₂O<MgO<Al₂O₃
3. K₂O<Na₂O<Al₂O₃<MgO
4. Al₂O₃ < MgO < Na₂O < K₂ O

Answer: 4.Al₂O₃ < MgO < Na₂O < K₂ O

The basicity of the metallic oxides increases with the increase in electronegativity values of the metals

Class 11 S-Block Notes

Question 11. Which of the following on thermal decomposition yields a basic as well as an acidic oxide

1. KClO
2. CaCO₃
3. NH₄NO₃
4. NaNO₃

Answer: 2.CaCO₃

CaCO₃ at high temperature decomposes to form CaO and CO₂

Question 12. Which one of the following alkaline earth metal sulfates has its hydration enthalpy greater than its lattice

1. BaSO₄
2. SrSO₄
3. CaSO₄
4. BeSO₄

Answer: 4. BeSO₄

The hydration enthalpy of BeSO₄ is greater than Its lattice energy because of the very small size of Be ²⁺

Question 13. The main oxides formed on combustion of, Na and Kin excess of air respectively are

1. LiO₂, Na₂O and KOH
2. LiO₂, Na₂O₂ and K₂O
3. Li₂O₂, Na₂O₂ and KO
4. Li₂O₂, Na₂O₂ and KO₂

Answer: 4. Li₂O₂, Na₂O₂ and KO₂

4Li + O₂ →2Li₂O

2Na + O→Na₂O₂

K + O→KO

Question 14. Although lithium and magnesium resemble each other in properties due to a diagonal relationship, the following statement is not correct

1. Both of them form a nitride compound
2. When nitrates of both Li and Mg are heated, NO₂ and O₂ are obtained
3. Both of them form basic carbonate salt
4. Both of them form soluble bicarbonate salt

Answer: 3. Both of them form basic carbonate salt

Magnesium forms basic bicarbonate salt [3MgCO₃.Mg(OH)₂ .3H₂O] whereas lithium forms carbonate salt [Li₂CO₃]. I cannot form any basic bicarbonates

Question 15. Match Column – 1 Column – 2 for the compositions of substances and select the correct answer using the code given below

1. 1 -C, 2-D, 3-  A, 4- B
2. 1 -B, 2-C, 3-  D, 4- A
3. 1 -A, 2-B, 3-  C, 4- E
4. 1 -D, 2-C, 3-  A, 4- B

Answer: 2. 1 -B, 2-C, 3-  D, 4- A

Plaster of Paris: CaSO₄– ½H₂O

Epsomite: MgSO₄-7H₂O

Kieserite: MgSO₄-H₂O

Gypsum: CaSO₄-2H₂O

Question 16. Which one of the following is present as an active ingredient in bleaching powder for bleaching action

1. CaOCl₂
2. Ca(OCl)₂
3. CaO₂Cl
4. CaCl₂

Answer: 2.Ca(OCl)₂

Bleaching powder is – a mixture of calcium chlorohypochlorite [Ca(OCl)₂] and basic calcium chloride [CaCl₂-Ca(OH)₂-2H₂O]. The compound present as an active ingredient in bleaching powder for bleaching action is Ca(OCl)₂

Question 17. Which of the following has the lowest melting point

1. CaCl₂
2. CaBr₂
3. Cal₂
4. CaF₂

Answer: 3.Cal₂

With the increase in the size of the halogens (from toI), the covalent character of the corresponding compounds (from CaF₂ to Cal₂) increases. Therefore, the melting points of the compounds decrease

Class 11 S-Block Notes

Question 18. In the case of replacement reaction, the reaction will be most favorable if M happens to be

1. K
2. Rb
3. Li
4. Na

Answer: 2. Rb

For a particular alkyl halide, the reactivity of the alkali metal fluorides increases gradually from Li to Cs because their ionization enthalpies decrease and their electronegativities increase. Therefore, for the given reaction, if M = Rb, then the reaction becomes most favorable.

Question 19. Which one of the alkali metals forms only the normal oxide M₂O on heating in air

1. Li
2. Na
3. Rb
4. K

Answer: 1. Li

Question 20. The ease of adsorption of the hydrated alkali metal ions on the ion-exchange resins follows the order

1. K⁺ < Na+ ⁺ Rb⁺ < Li⁺
2. Na⁺ < Li⁺ < K⁺ < Rb⁺
3. Li⁺ < K⁺ < Na⁺ < Rb⁺
4. Rb⁺ < K+< Na+< Li⁺

Answer: 4. Li⁺ < K⁺ < Na⁺ < Rb⁺ 

The ease of adsorption of the hydrated alkali metal ions on the ion-exchange resins decreases with the increase in the size of the alkali metal. The correct order of the sizes of the alkali metals is: Li⁺ < Na⁺ < K⁺ < Rb⁺ .4.

The ease of adsorption of the hydrated alkali metal ions on the ion-exchange resins decreases with the increase in the size of the alkali metal. The correct order of the sizes of the alkali metals is: Li⁺ < Na⁺ < K⁺ < Rb⁺

Question 21. Be²⁺is isoelectronic with which of the following ions

1. Na⁺
2. Mg²⁺
3. H⁺
4. Li⁺

Answer: 4. Li⁺

No. ofelectrons in Be = 4

.-. No. of electrons in Be²⁺ = 2

Again, no. of electrons in Li = 3

No. ofelectrons in Li+ = 2

Therefore, Be²⁺ is isoelectronic with Li⁺

Question 22. On heating which of the following releases CO₂ most easily

1. K₂CO₃
2. Na₂CO₃
3. MgCO₃
4. CaCO₃

Answer: 3.

The thermal stability of the given compounds follows the order: of K₂CO₃ > Na₂CO₂ > CaCO₂ > MgCO₂.

Thus, MgCO₃ Can release CO₂ most easily on heating

Question 23. In contrast with beryllium, which one of the following statements is incorrect

1. It forms Be₂C
2. Its salts rarely hydrolyzed
3. Its hydride is electron deficient and polymeric
4. It is rendered passive by nitric acid

Answer: 2. Its salts rarely hydrolyzed

Beryllium forms covalent compounds, thus hydrolysis of beryllium compounds occurs readily.

Question 24. The suspension of slaked lime in water is

1. Quicklime
2. Milk of lime
3. Aqueous solution of slaked lime
4. Lime water

Answer: 2. Milk of lime

Suspension of slaked lime Is known as ‘milk ofIimei

Class 11 S-Block Notes

Question 25. Which of the following statements is false

1. Mg ²⁺ ions form a complex with ATP
2. Ca²⁺ ions are important in blood clotting
3. Ca²⁺ ions are not important in maintaining the regular beating of the heart
4. Mg²⁺ ions are important in the green parts of plants

Answer: 3. Ca²⁺ ions are not important in maintaining the regular beating of the heart

Ca²⁺ plays an important role in the regular beating of the heart

Question 26. The product obtained as a result of a reaction of nitrogen

1. Ca(CN)₂
2. CaCN₃
3. Ca₂CN
4. CaCN

Answer: The given options are not correct

Question 27. Which one of the following takes up CO₂ and releases

1. K₂O
2. CaO
3. KO₂
4. KOH

Answer: 3.KO₂

2KO₂ + CO₂ → K₂ CO₃ + \(\frac{3}{2}\) O₂

Question 28. Ionic mobility of which of the following alkali metal ions is lowest when the aqueous solution of these salts is put under an electric field

1. K
2. Rb
3. Li
4. Na

Answer: 3. Li

Question 29. Magnesium reacts with an element (X) to form an ionic compound. If the ground state electronic configuration of (X) is ls²2s²2p³ the simplest formula for this compound is

1. Mg₃X₂
2. Mg₂X₃
3. Mg₂X
4. Mg X₂

Answer: 1. Mg₃X₂

Electronic configuration of X: ls²2s²2p³

Valency of X = 3

Here, the valency of Mg = 2

Thus the compound will be Mg₃X₂

Question 30. Which of the following oxides is most acidic nature

1. CaO
2. MgO
3. BaO
4. BeO

Answer: 4. BeO

On moving down the group basic character of the gr-2 On moving down the group basic character of the gr-2

BeO < MgO < CaO < SrO < BaO

Thus, BeO is acidic among all the given oxides.

Class 11 S-Block Notes

Question 31. Among CaH₂, BeH₂, BaH₂ the order of ionic character

1. BaH₂< BeH₂ < CaH₂
2. BeH₂ < CaH₂ < BaH₂
3. BeH₂<BaH₂<CaH₂
4.  CaH₂ < BeH₂, < BaH₂

Answer: 2.BeH₂ < CaH₂ < BaH₂

Question 32. Which of the following is not hygroscopic

1. CsCl
2. MgCl₂
3. CaCl₂
4. LiCI

Answer: 1. CsCl is not hygroscopic in nature while MgCl,  CaCI_4, and LiCl are hygroscopic in nature

Question 33. Which is the correct order of solubility in water—

1. Ba(OH)₂ <Mg(OH)₂
2. BaCO₃ >CaCO₃
3. CaSO₄ <MgSO₄
4. Ca(OH)2^Mg(OH)₂

Answer: 3. CaSO₄ <MgSO₄

Question 34. Bleaching powder does not contain—

1. CaCl₂
2. Ca(OH)₂
3. Ca(OCl)₂
4. Ca(ClO₃ )₂

Answer: 4.Ca(ClO₃)₂

Bleaching powder is a mixture of calcium hypochlorite, Ca(OCI)_2, and the basic chloride CaCl₂, H₂O with some slaked lime, Ca(OH)₂.

Question 35. Which of the following statements is incorrect—

1. Li⁺ has a minimum degree of hydration
2. The oxidation state of K in KO₂ is +1
3. Na is used to make a Na/Pb alloy
4. MgSO₄ is readily soluble in water

Answer: 1.  Li⁺ has a minimum degree of hydration

The hydration enthalpies of alkali metal ions decrease with an increase in ionic sizes. Hence, the orderis Li⁺ > Na⁺ > K⁺ > Rb⁺ > Cs⁺. Therefore Li⁺ has a maximum degree of hydration

Question 36. Which of the following has the highest hydration energy

1. MgCl₂
2. CaCl₂
3. BaCl₂
4. SrCl₂

Answer: 1. MgCl₂

Smaller-sized and highly charged metal ions have higher hydration energy. Therefore, the order of hydration energy will be

Be²⁺ > Mg²⁺ > Ca²⁺ > Sr²⁺ > Ba²⁺

Question 37. Which is the correct order of solubility in water

1. Ba(OH)₂< Mg(OH)₂
2. BaCO₃ >CaCO₃
3. CaSO₄ < MgSO₄
4. Ca(OH)₂ ≅ Mg(OH)₂

Answer: 3.CaSO₄ < MgSO₄

Class 11 S-Block Notes

Question 38. What is the product of the reaction between the dilute solution of Ba(OH)₂ and H₂O₂ + ClO₂ 

1. HOCl
2. Ba(OCl)₂
3. Ba(ClO₃)₂
4. Ba(ClO₂)₂

Answer: 4.Ba(ClO₂)₂ 

Ba(OH)₂  +H₂ O₂  + 2ClO₂  → Ba(ClO₂ )₂  + 2H₂ O + O₂ 

Question 39. Which of the following compounds transforms baking soda into baking powder

1. KHCO₃
2. NaHCO₃
3. KHC₄H₄O₆
4. KCl

Answer: 3.KHC₄H₄O₃

Question 40. Which of the following pair of compounds does not undergo any chemical change on heating

1. MgCO₃,KHCO₃
2. CS₂CO₃,KNO₂
3. Na₂CO₃,NaNO₃
4. Li₂CO₃, KNO₂

Answer: 2.  CS₂CO₃, KNO₂

Question 41. Ca(OH)₂ (anhydrous)  CaCl₂ +B In this reaction, B is

1. CaOCl₂
2. Ca(ClO₃)₂
3. Ca(OH)₂
4. Ca(ClO₂)₂

Answer: 2. Ca(ClO₃)₂

Question 42. The blue color of potassium solution in liquid ammonia is due to the presence of

1. Solvated electron
2. Potassium amide
3. Impurities present in potassium
4. Potassium oxide

Answer: 1. Solvated electron

Question 43. The alkali metal that does not participate in the reaction MI + I₂ → MI₃ is

1. Na
2. K
3. Rb
4. Cs

Answer: 1. Na

Question 44. The lithium compound which is soluble in water is

1. Li₂CO₃
2. LiNO₃
3. LiF
4. Li₃PO₄

Answer: 2. LiNO₃

Question 45. An alkaline earth metal when heated along with nitrogen forms X. On hydrolysis, X forms an insoluble basic compound and a gas which turns CuSO₄ solution deep blue. The metal is

1. Be
2. Ca
3. Mg
4. Ba

Answer: 3. Mg

Class 11 S-Block Notes

Question 46. Which of the following compounds exists in only an aqueous solution

1. Li₃PO₄
2. NaHCO₃
3. [Li(NH₃)₄]I
4. LiHCO₃

Answer: 4.LiHCO₃

Question 47. Which of the following acids does not liberate CO₂ on reacting with sodium carbonate

1. Dilute HCl
2. Dilute H₃BO₃
3. Dilute H₃PO₄
4. Dilute H₂PO₄

Answer: 2. Dilute H₃BO₃

Question 48. Hydrated BeCl₂ acts as a/an

1. Lewis base
2. Arrhenius base
3. Arrhenius acid
4. Lewis acid

Answer: 4. Lewis acid

Question 49. Which of the following compounds or pair of compounds is responsible for turning yellow sodium peroxide white in the presence of air

1. Na
2. NaOH and Na₂CO
3. H₂O₂
4. NaOH and H₂O₂

Answer: 2.NaOH and Na₂CO

Question 50. The chemical formula ofmicrocosmic saltis

1. NaHPO₄.2H₂O
2. NH₄₃HPO₄.2H₂O
3. Na(NH₄)HPO₄.4H₂O
4. (NH₄)3HPO₄.4H₂O

Answer: 3.Na(NH₄)HPO₄.4H₂O

Question 51. The hydride which does not form as a result of a direct reaction between the metal and hydrogen is

1. CaH₂
2. MgH₂
3. BeH₂
4. NaH

Answer: 3.BeH₂

Question 52. An amphoteric oxide dissolves in HCl to form a salt. The salt does not impart any characteristic color to the flame but fumes in moist air. The oxide is

1. BaO₂
2. MgO
3. BeO
4. CaO

Answer: 3. BeO

Question 53. The difference in the number of water molecules in gypsum and Plaster of Paris is—

1. 5/2
2. 2
3. 1/2
4. 3/2

Answer: 4. 3/2

Class 11 S-Block Notes

Question 54. Compound A imparts a brick-red color to the flame and decomposes on heating to produce oxygen and a brown gas. A is

1. Mg(NO₃)₂
2. Ba(NO₃)2
3. Ca(NO₃)₂
4. Sr(NO₃)₂

Answer: 2. Ba(NO₃)₂

Question 55. The sodium salt of an unknown anion when heated with MgCl₂ forms a white precipitate. The anion is

1. SO^2-₄
2. CO^2-₃
3. HCO^–₃
4. NO^–₃

Answer: 3.HCO^–₃

Question 56. The salt which is added to table salt to keep it dry and free-flowing

1. KCl
2. Ca(PO₄)₂
3. KI
4. Na₃PO₄

Answer: 2.Ca(PO₄)₂

Question 57. The compound which acts as an oxidizing as well as a reducing agent

1. NaNO₃
2. Na₂O
3. Na₂O₂
4. KNO₃

Answer: 3.

Question 58. The compound which is used to extinguish fire caused by combustion of alkali metals is—

1. CC1₄
2. Sand
3. Water
4. Kerosene

Answer: 1.  CC1₄

Question 59. An aqueous solution of sodium sulphate is electrolyzed using inert electrodes. The products formed at the cathode and anode respectively are

1. O₂,H₂
2. H₂,O₂
3. O₂, Na
4. O₂,SO₂

Answer: 1.O₂, H₂

Question 60. Sodium when heated at 300°C in air forms X which absorbs CO₂ to form Na₂CO₃ and a compound Y. The compound Y is 

1. H₂
2. O₂
3. H₂O₂
4. O₃

Answer: 2.O₂

Question 61. Excess Na⁺ ions in the human body cause

1. Diabetes
2. Anaemia
3. Low blood pressure
4. High blood pressure

Answer: 4. High blood pressure

Class 11 S-Block Notes

Question 62. The metal carbide which on hydrolysis produces allylene or propyne is

1. Be
2. Ca
3. Al
4. Mg

Answer: 4. Mg

Question 63. A metal M readily forms water-soluble sulfate MSO₄, water-insoluble hydroxide M(OH)_2, and oxide MO. The oxide remains inert on heating. The hydroxide is soluble in NaOH. M is

1. Be
2. Mg
3. Ca
4. Sr

Answer: 1. Be

Question 64. When compound A is heated, it produces a colourless gas, and the residue obtained is dissolved in water to form compound B. The compound C is formed when excess CO₂ is passed through the aqueous solution of B. The compound C can be separated in a solid state from its solution. C in its solid state when heated forms the compound A. Ais

1. CaCO₃
2. Na₂CO₃
3. K₂CO₃
4. CaSO₄.2H₂O

Answer: 1.CaCO₃

Question 65. When metals A and B are heated in air, A only forms oxide but B forms both oxide and nitride. A and B are

1. Cs, K
2. Mg, Ca
3. Li, Na
4. K, Mg

Answer: 4. K, Mg

Question 66. The alkali metal which emits a light of the longest wavelength in the flame testis

1. Na
2. K
3. Cs
4. Li

Answer: 2. K

Question 67. The compound which does not form a double salt is

1. Li₂SO₄
2. Na₂SO₄
3. K₂SO₄
4. Rb₂SO₄

Answer: 1. Li₂SO₄

Question 68. The decreasing order of stability of the chloride salts of alkali metals is

1. LiCl > KCl > NaCl > CsCl
2. CsCl > KCl > NaCl > LiCl
3. NaCl > KCl > LiCl > CsCl
4. KCl > CsCl > NaCl > LiCl

Answer: 4. KCl > CsCl > NaCl > LiCl

Question 69. The affinity of sodium for water is used for drying

1. Alcohol
2. Ammonia
3. Benzene
4. Phenol

Answer: 3. Benzene

Class 11 S-Block Notes

Question 70. The ions present in an anhydrous mixture of potassium fluoride and hydrofluoric acid are

1. K⁺,H⁺ F^–
2. (KF)⁺+ (HF)^–
3. KH⁺, F^–
4. K⁺,HF₂^–

Answer: 4. K⁺,HF₂^–

Question 71. The correct order of covalent character in the following compounds is

1. LiCl < NaCl < BeCl₂
2. BeCl₂ > LiCl > NaCl
3. NaCl < LiCl < BeCl₂
4. BeCl₂ < NaCl < LiCl

Answer: 3. NaCl < LiCl < BeCl₂

Question 72. A chemical compound A is used for the recovery of ammonia during the preparation of washing soda. When CO₂ is passed through the aqueous solution of A, the solution turns turbid. A is used for whitewashing because of its disinfecting properties. The chemical formula of A is

1. Ca(HCO₃)₂
2. CaO
3. Ca(OH)₂
4. CaCO₃

Answer: 3. Ca(OH)₂

Question 73. The compound used for drying neutral or basic gases is—

1. Calcium carbonate
2. Sodium carbonate
3. Sodium bicarbonate
4. Calcium oxide

Answer: 4. Calcium oxide

Question 74. Which of the following compounds does not contain calcium carbonate

1. Dolomite
2. Marble statue
3. Burnt gypsum
4. Snail shell

Answer: 3. Burnt gypsum

Question 75. The compound which on thermal decomposition produces a basic and an acidic oxide is

1. KClO₃
2. Na₂CO₃
3. NaNO₃
4. CaCO₃

Answer: 4.CaCO₃

Question 76. Which of the following oxides does not react with water

1. BeO
2. CaO
3. MgO
4. SrO

Answer: 1. BeO

Question 77. Which of the following carbonates is soluble in water

1. SrCO₃
2. BaCO₃
3. Al₂(CO₃)₃
4. Rb₃CO₃

Answer: 3.Al₂(CO₃)₃

Class 11 S-Block Notes

Question 78. The compound whose aqueous solution is called ‘baryta water’ is

1. BaSO₄
2. BaO
3. BaCO₃
4. Ba(OH)₂

Answer: 4. Ba(OH)₂

Question 79. The alkali metal that emits a light of the shortest wavelength in the flame test is

1. Na
2. K
3. Cs
4. Li

Answer: 3. Cs

Question 80. The correct order of ionic mobility of the following ions in their aqueous solution is

1. Na⁺> K⁺> Rb⁺> Cs⁺
2. K⁺> Na⁺ > Rb⁺> Cs⁺
3. Cs⁺ > Rb⁺ > K⁺ > Na⁺
4. Rb⁺ > K⁺ > Cs⁺ > Na⁺

Answer: 3. Cs⁺ > Rb⁺ > K⁺ > Na⁺

Question 81. Which of the following compounds is not used for storing or immersing metallic sodium

1. Benzene
2. Kerosene
3. Ethanol
4. Toluene

Answer: 3. Ethanol

Question 82. Which of the following compounds is paramagnetic

1. KO₂
2. SiO₂
3. TiO₂
4. BaO₂

Answer: 1.KO₂

Question 83. KO₂ is used in oxygen cylinders that are used for submarines and spacecraft because—

1. It increases the amount of oxygen by absorbing CO₂
2. It eliminates water vapour
3. It absorbs CO₂
4. It forms O₃

Answer: 1.It increases the amount of oxygen by absorbing CO₂

Question 84. Which of the following compounds is the most stable

1. LiF
2. LiCl
3. LiBr
4. Lil

Answer: 1. LiF

Question 85. The alkali metal for which the photoelectric effect is maximum is

1. Cs
2. Na
3. K
4. Li

Answer: 1. Cs

Question 86. The melting points of alkali metals are low. Which of the alkali metals melts when room temperature becomes more than 30 °C 

1. K
2. Na
3. Cs
4. Rb

Answer: 3. Cs

Class 11 S-Block Notes

Question 87. Which of the following alkali metal hydroxides is the most basic in nature

1. CsOH
2. KOH
3. Li OH
4. RbOH

Answer: 1.CsOH

Question 88. The metal whose carbonate salt is the most stable is

1. Na
2. Mg
3. A1
4. Si

Answer: 1. Na

Question 89. Which of the following compounds is used for manufacturing soft soaps

1. KOH
2. NaOH
3. LiOH
4. Mg(OH)₂

Answer: 1.

Question 90. The atomic number of a radioactive alkali metal is

1. 55
2. 87
3. 19
4. 37

Answer: 2. 87

Question 91. Which of the following is used for the manufacture of high? temperature thermometers

1. An alloy of Li and Na
2. An alloy of Na and Cs
3. An alloy of Na and K
4. An alloy of K and Rb

Answer: 3. An alloy of Na and K

Question 92. The mixture of MgCl₂ and MgO is known as

1. Sorel’s cement
2. Portland cement
3. Alum
4. Magnesium oxychloride

Answer: 1. Sorel’s cement

Question 93. Which of the following alkaline earth metals does not form its corresponding hydride by directly reacting with hydrogen

1. Mg
2. Sr
3. Be
4. Ba

Answer: 3. Be

Question 94. Which of the following alkaline earth metal hydroxides is soluble in NaOH solution

1. Ba(OH)₂
2. Ca(OH)₂
3. Mg(OH)₂
4. Be(OH)₂

Answer: 4. Be(OH)₂

Question 95. The chloride salt which is soluble in ethanol is

1. BeCl₂
2. CaCl₂
3. SrCl₂
4. MgCl₂

Answer: 1.BeCl₂

Class 11 S-Block Notes

Question 96. Which of the following is produced when one mole of magnesium nitride reacts with an excess of water

1. One mole of ammonia
2. One mole of nitric acid
3. Two moles of ammonia
4. Two moles of nitric acid

Answer: 3. Two moles of ammonia

Question 97. Which of the following methods is used for the preparation of calcium

1. Reduction of CaO by carbon
2. Reduction of CaO by hydrogen
3. Electrolysis of a mixture of anhydrous CaCl₂ and KCl
4. Electrolysis of molten Ca(OH)₂

Answer: 3. Electrolysis of a mixture of anhydrous CaCl₂ and KCl

Question 98. Which alkaline earth metal ion plays a vital role in the contraction of muscles

1. Ba²⁺
2. Sr²⁺
3. Mg²⁺
4. Ca²⁺

Answer: 4. Ca²⁺

Question 99. Which of the following correctly indicates the formula of halides of alkaline earth metals

1. BeCl₂.2H₂O
2. BeCl₂.4H₂O
3. CaCl₂.6H₂O
4. SrCl₂.4H₂O

Answer: 1.BeCl₂.2H₂O

Question 100. The compounds of sodium which are used in the textile Industry are

1. Na₂CO₃
2. NaHCO₃
3. NaOH
4. NaCl

Answer: 1. Na₂CO₃

Question 101. Which of die following pairs of elements have similar properties

1. Be, Cs
2. K, Cs
3. Sr, Rb
4. Be, Al

Answer: 2. K, Cs

Question 102. The chlorides which are soluble in pyridine are

1. LiCl
2. CsCl
3. NaCl
4. BeCl₂

Answer: 1 and 4

Question 103. The gases in which magnesium burns are

1. CO₂
2. N₂O
3. N₂
4. SO₂

Answer: 1,2,3 and 4

Question 104. Which of the following oxides have rock salt structure with coordination number 6:6 

1. MgO
2. CaO
3. SrO
4. B₂O₃

Answer: 1,2 and 3

Question 105. Which of the following pairs of compounds cannot exist in aqueous solution

1. NaH₂PO₄ and Na₂HCO₃
2. Na₂CO₃ and NaHCO₃
3. NaOH and NaH₂ PO₄
4. NaHCO₃ and NaOH

Answer: 3 and 4

Question 106. The compounds which on heating do not form oxides are

1. NaNO₃
2. CsOH
3. LiOH
4. SrCO₃

Answer: 1 and 2

Question 107. Which of the following pairs of elements will give superoxides and peroxides respectively when heated with excess air

1. K, Br
2. Nay Rb
3. K, Rb d
4. Na, Ba

Answer: 3 and 4

Class 11 S-Block Notes

Question 108. Which of the following do not respond to flame test

1. Be
2. Mg
3. KO₂
4. Sr

Answer: 1 and 2

Question 109. Which of the following compounds are not paramagnetic nature

1. K₂O₂
2. NO₂
3. KO₂
4. K₂O

Answer: 1 and 4

Question 110. Which of the following is incorrect

1. Soda ash: Na₂CO₃
2. Pearl ash: Cu₂CO₃
3. Bone ash: K₂CO₃
4. Baking soda: NaHCO₃

Answer: 2 and 3

Question 111. Which ions of water are replaced by sodium ions when hard water is passed through zeolite (hydrated sodium aluminum silicate)

1. H⁺
2. Mg⁺
3. Ca²⁺
4. SOl₄^2-

Answer: 2 and 3

Question 112. The compounds which are soluble in organic solvents are

1. CaCl₂
2. BaCl₂
3. BeCl₂
4. AlCl₃

Answer: 3 and 4

Question 113. The compounds formed when potassium superoxide reacts with water are

1. KOH
2. H₂O₂
3. K₂O₂
4. O₂

Answer: 1, 2 and 4

Question 114. Which of the following hydrated salts undergo hydrolysis on heating

1. BaCl₂.2H₂O
2. MgCl₂.6H₂O
3. SrCl₂.2H₂O
4. CaCl₂.6H₂O

Answer: 2 and 4

Class 11 S-Block Notes

Question 115. Which of the following compounds are extensively used as drying agents

1. Anhydrous CaCl₂
2. Mg(ClO₄)2
3. BeC
4. Ca(OH)₂

Answer: 1 and 2

Question 116. Each of the following compounds reacts with water but which of these liberates the same gas

1. Na
2. Na₂O₂
3. KO₂
4. NaH

Answer: 1 and 4

Question 117. The polarisability of LiCl is higher than that of NaCl. Concerning this, which of the following statements are true

1. The melting point of LiCl is less than that of NaCl
2. LiCl is sparingly soluble in organic solvents
3. LiCl dissociates to a greater extent in water than NaCl
4. The conductivity of molten LiCl is less than that of NaCl

Answer: 1 and 4

Question 118. Which of the following statements are correct

1. Electronegativity of alkali metals decreases with an increase in atomic number
2. Lithium is the lightest metal
3. Alkali metals are strong reducing agents –
4. The electronegativity of alkali metals ranges from 1.0 to 0.7

Answer: 1, 3, 4

Question 119. Which of the following statements are incorrect about the hydrates of alkali metals

1. Conduct electricity in their molten states
2. These compounds act as oxidizing agents
3. These compounds dissolve in water to liberate
4. These compounds are covalent

Answer: 2 and 4

Question 120. Which of the following statements are correct about the ionic solids KI and CaO lattice enthalpy of CaO is greater than that of K₃

1. I am soluble in benzene
2. The melting point of CaO is high
3. The melting point of KI is high

Answer: 1, 2, 3

Question 121. The compounds which do not form NO₂ on undergoing thermal decomposition are

1. LiNO₃
2. NaNO₃
3. KNO₃
4. RbNO₃

Answer: 2, 3, and 4

Question 122. Metals are identified by their standard reduction potential, enthalpy of fusion, and atomic size. The alkali metals are identified by their

1. High boiling point
2. High negative standard reduction potential
3. High density
4. Greater atomic size

Answer: 2 and 4

Class 11 S-Block Notes

Question 123. Which of the following sulfates easily dissolve in water

1. BeSO₄
2. MgSO₄
3. BaSO₄
4. SrSO₄

Answer: 1 and 2

Question 124. The properties of beryllium nitride which are different from the nitrides of other alkaline earth metals are

1. Its volatility
2. Its covalent nature
3. Unable to undergo hydrolysis
4. Its ionic nature

Answer: 1 and 2

Question 125. Which of the following options is correct for RbO₂

1. It is a peroxide
2. It is diamagnetic
3. It is a superoxide
4. It is paramagnetic

Answer: 3 and 4

Question 126. Which of the following statements is correct for the alkaline earth metals

1. Hydration enthalpy of Sr²⁺is less than that of Ba²⁺
2. CaCO₄ undergoes decomposition at a higher temperature than BaCO₃
3. Ba(OH)₂ is a stronger base than Mg(OH)₂
4. SrSO₄  is more soluble in water than CaSO₄

Answer: 2 and 3

Question 127.  Which of the following statements are correct about the alloy formed by sodium and potassium

1. It is used in Lassaigne’s test
2. It is liquid at ordinary temperature
3. It is used in specially designed thermometers
4. It is used as a coolant in nuclear reactors

Answer: 2 and 3

Question 128. Which of the following statements is incorrect

1. BeCl₂ molecule is linear in the gaseous state
2. Calcium hydride is known as hydrolith
3. Carbides of both beryllium and calcium react with water to form acetylene
4. Oxides of both Be and Ca are amphoteric

Answer: 3 and 4

Question 129. Which of the following is less stable thermally

1. LiF
2. KCl
3. RbF
4. CsF

Answer: 2. KCl

The thermal stability of a compound increases with the increasing value of enthalpy of formation. Among the given compounds the value of enthalpy of formation is minimum (-428kJ.mol^-1) for KCl. Hence, it has the lowest thermal stability among the given compounds

Question 130. Which of the following pairs is responsible for developing an electric potential across the membrane of living cells

1. Ca²⁺and Na⁺
2. Na⁺ and K⁺
3. K⁺ and Ba²⁺
4. Mg²⁺ and Ca²⁺

Answer: 3. K⁺ and Ba²⁺

Class 11 S-Block Notes

Question 131. Which one of the following chlorides is a soluble organic solvent

1. CaCl₂
2. NaCl
3. MgCl
4. BeCl₂

Answer: 4.MgCl

Question 132. Which of the following alkaline earth metal carbonates is thermally least stable

1. BaCO₃
2. CaCO₃
3. SrCO₃
4. BeCO₃

Answer: 4.SrCO₃

Question 133. The cause of the different colors of the flame in the flame test is

1. Lowionisation potential
2. Low melting point
3. Malleability
4. The presence of one electron in the outermost orbit

Answer: 1. Lowionisation potential

Question 134. Which of the following alkaline earth metal sulfate is most soluble in water

1. CaSO₄
2. SrSO₄
3. BaSO₄
4. MgSO₄

Answer: 4. BaSO₄

Question 135. What will be the order of reducing powder of the following elements Li,Na,  K, Rb, Cs

1. Cs > Rb > K > Na > Li
2. Rb > Cs > K > Na > Li
3. K > Rb > Cs > Na > Li
4. Na > Li> K >Cs > Rb

Answer: 1. Cs > Rb > K > Na > Li

The order of reducing free states is:

Cs > Rb > K > Na > Li

Question 136. Which one of the following elements shows a diagonal relationship with magnesium

1. Na
2. Li
3. Be
4. Ca

Answer: 2. Li

Question 137. Sodium is preserved in which of the following liquids

1. Water
2. Ethanol
3. Kerosene oil
4. Methanol

Answer: 3. Kerosene oil

Question 138. Which of the alkali metals has having least melting point?

1. Na
2. K
3. Rb
4. Cs

Answer:  4. Cs

As the atomic size of the metal increases, the strength of metallic bonding decreases and consequently, the melting point decreases. Since the size of the cesium atom is the largest, it has the least melting point

Question 139. Which one of the following alkali metals gives hydrated salts? 

1. Li
2. Na
3. K
4. Cs

Answer: 1.  Li

 Li; Explanation:

Among all the alkali metal ions, Li+ is the smallest and thus it has the highest charge density. As a result, Li⁺ attracts water molecules more strongly than any other alkali metal cation and forms hydrated salts.

Class 11 S-Block Notes

Question 140. Which one of the alkaline earth metal carbonates is thermally the most stable? 

1. MgCO₃
2. CaCO₃
3. SrCO₃
4. BaCO₃
   

Answer:  4. BaCO₃

BaCO₃; Explanation:

Among all the alkaline earth metals, the Ba²⁺ ion is the largest. Again, CO₃^2- ion is also quite large. Thus, among the given carbonates, Ba²⁺ and CO₃^2-  ions are most tightly packed in the crystal lattice of BaCO₃. Consequently, the lattice enthalpy of BaCO₃ is the highest and so, is most thermally stable.

Class 11 Chemistry S Block Elements Very Short Answer Type

Question 1. Name the lightest and the heaviest metal.
Answer:   Lithium is the lightest (density: 0.53g-cm^-3) and osmium is the heaviest (density: 22.6g-cm^-3 ) metal.

Question 2. Which one between water and pyrene (CCl₄), can be used to extinguish fire caused by metallic sodium?
Answer:  Pyrene (CCl₄) can be used.

Question 3. Which among the alkali metal ions has the lowest mobility In aqueous solution?
Answer: Li⁺ ions have the lowest mobility because they remain highly hydrated in aqueous solution.

Question 4. Which of the alkali metal cations has the highest polarising power?
Answer: Due to its very small size, Li⁺ ion has the highest polarising power among the alkali metal ions

Question 5. The density of alkali metals is very low Why?
Answer: Due to large atomic size and weak metallic bonding, the densities of alkali metals are very low.

Question 6. Which out of LIF of Lil is more covalent?
Answer:  Lil is more covalent.

Question 7. Which alkali metal carbonate easily liberates CO₂ on heating?
Answer: Li₂CO₃ easily liberates CO₂ on heating.

Question 8. Which alkali metal chloride imparts violet color to the bunsen burner flame?
Answer: Potassium chloride (KCl)

Question 9. Which alkali metal hydride is used as a source of hydrogen for filling up meteorological balloons?
Answer: Lithium hydride (LiH).

Question 10. Name the products formed in the following reaction. Explain Lil + KF→? + ?
Answer: LiF and KI. The larger K⁺ ion stabilizes the larger I^– ion while the smaller Li⁺ ion stabilizes the smaller F- ion.

Question 11.  What is water glass?
Answer: Sodium silicate (NaOSiO₃).

Question 12. Which alkali metal combines with nitrogen to form the corresponding nitride?
Answer: Lithium (Li).

Question 13. What are the raw materials used for the manufacture of sodium carbonate by the Solvay process?
Answer: The raw materials are NaCl, CaCO and NH₃.

Question 14. Which out of the Na₂CO₃ solution & NaHCO₃ solution, changes the color of phenolphthalein into pink?
Answer: Na₂CO₃ solution.

Question 15. What is the main ingredient of baking powder?
Answer: Sodium bicarbonate (NaHCO₃).

Question 16. Which compound is used to treat the flue gases from coal and oil-fired power stations and to remove SO₂ and H₂SO₄ responsible for acid rain?
Answer: Na₂CO₃.

Question 17. What are the ingredients of the fusion mixture that is used in dry tests in inorganic analysis?
Answer: K₂CO₃ and Na₂CO₃

Question 18. Which alkaline earth metal is the most abundant in the earth’s crust?
Answer: Calcium (Ca).

Question 19. Which alkaline earth metal is radioactive?
Answer: Radium (Ra).

Question 20. Name the metal of group-2 which is used to prepare Grignard reagent.
Answer: Magnesium (Mg).

Question 21. Name the element of group 2 that resembles lithium in characteristics.
Answer: Magnesium (Mg).

Question 22. Which alkaline earth metals do not impart any color to the flame of a Bunsen burner?
Answer: Be anil Mg.

Question 23. Name an alkaline earth metal compound that can be used as a portable source of hydrogen.
Answer: Calcium hydride (CaH₂).

Question 24. Which Gr-2 metal forms covalent compounds?
Answer: Beryllium (Be).

Question 25. Which Gr-2 metal burns readily when exposed to air?
Answer: Barium (B a).

Question 26. Which alkaline earth metal reacts with alkali to form hydrogen gas?
Answer: Beryllium (Be).

Question 27. What is the composition of the alloy, electron?
Answer: 95% of Mg+ 5% of Zn.

Question 28. Which reagent is used to analyze Ca²⁺ and Mg²⁺ ions quantitatively?
Answer: EDTA (Ethylenediaminetetraacetic acid).

Question 29. What is the medicinal name of the aqueous solution of Mg(OH)₂?
Answer: Milk of magnesia.

Class 11 S-Block Notes

Question 30. What is anhydrous?
Answer: Magnesium perchlorate, Mg(ClO₄)₂ is known as anhydrous.

Question 31. How will you distinguish between BeSO₄ & BaSO₄?
Answer: BeSO₄ is water soluble but BaSO₄ is insoluble in water.

Question 32. Which out of MgCO₃, SrCO_3, and BaCO₄, possesses the highest thermal stability?
Answer: BaCO₄ has the highest thermal stability.

Question 33. Distinguish between Be(OH)₂ and Ba(OH)₂
Answer: Be(OH)₂ is soluble in caustic soda solution while Ba(OH)2 is insoluble in it.

Question 34. Why is BeCl₂ soluble in organic solvents?
Answer: BeCl₂ is covalent.

Question 35. Which Gr-2 metal carbonate is unstable in the air?
Answer: BeCO₃.

Question 36. Which alkaline earth metal sulfate is useful in diagnosing stomach ulcers by X-ray?
Answer: BaSO₄ (in ‘barium meal’ X-ray).

Question 37. BeO is covalent but still, it has a much higher melting point— mention the reason.
Answer: This is due to its polymeric structure.

Question 38. What is the difference between lime water and milk of lime?
Answer: Mg²⁺

Question 39. Write a reaction by which BeCl₂ can be prepared.
Answer:

Question 40. Explain why the compounds of beryllium are much more covalent than the other Gr-2 metal compounds.
Answer:

Due to the very small size and high charge of Be²⁺ ion, it has much higher polarising power and because of this, the compounds of Be are much more covalent than the other group-2 metal compounds.

Question 41. Beryllium compounds are extremely toxic why?
Answer:

Beryllium compounds are extremely toxic because of their very high solubility and their ability to form complexes with enzymes in the body.  Also, beryllium displaces magnesium from some enzymes

Question 42. What is the composition of soda lime used for the preparation of hydrocarbons in the laboratory?
Answer: NaOH and CEO
.
Question 43. ED Name two acidic oxides which react similarly with calcium hydroxide [Ca(OH)₂].
Answer: CO₂ and SO₂.

Question 44. Which alkaline earth metal oxide is used as a flux in metallurgy to remove siliceous impurities?
Answer: Calcium oxide (CaO)

Question 45. What is the commercial name of the disinfectant powder obtained when Cl2 reacts with slightly moist slaked lime at 40°C?
Answer: Bleaching powder.

Question 46. What is Plaster of Paris?
Answer: Hemihydrate of calcium sulphate, (CaSO₄)₂-H₂O is called Plaster of Paris

Question 47. What type of impurities in gypsum should be avoided in preparing Plaster of Paris from it?
Answer: Carbonaceous impurities are to be avoided.

Question 45. Why KNO₃ is used instead of NaNO₃ in gunpowder?
Answer: NaNO₃ is deliquescent. Hence, KNO₃ is preferred over NaNO₃ for the preparation of gunpowder.

Question 46. What do you mean by ‘black ash
Answer: A mixture of sodium carbonate (Na₂CO₃) and calcium sulfide (CaS) is called black

Question 47. What happens when magnesium is heated with | acetylene at 875K?
Answer: Magnesium on heating with acetylene at 875K forms magnesium carbide (Mg₃C₂)

Question 48. Which calcium salt causes the formation of the kidney? stones?
Answer:

Calcium oxalate, \(\mathrm{Ca}^2+\stackrel{\ominus}{\mathrm{O}}_2 \mathrm{C}-\mathrm{C} \stackrel{\ominus}{\mathrm{O}}_2\) causes formation of kidney stones

Question 49. Alkali metals are good reducing agents—Why?
Answer: 

The smaller the ionization enthalpy, the greater the reducing strength. Since alkali metals have lower ionization enthalpies, they are good reducing agents.

Question 50. Explain why the alkali metals cannot be obtained by the reduction method.
Answer: Alkali metals are strong reducing agents and it is difficult to reduce their oxide by any other reducing agent

Question 51. Which alkali metal ion has the maximum polarising j power and why?
Answer:  Li⁺ ion has the highest polarising power among all the alkali metal ions as the value of charge to size ratio of the smallest Li+ ion is the highest.

Question 52. The fire caused by sodium in the laboratory cannot be extinguished by spraying water Why?
Answer:

Sodium reacts vigorously with water producing H₂ gas which catches fire by the heat evolved in the reaction. So water cannot be used for extinguishing sodium-fire

Class 11 S-Block Notes

Question 53. Why does Li not exist with Na or K in their minerals?
Answer:

Lithium forms independent minerals and does not exist with Na or K because the Li⁺ ion is too small to replace the more abundant Na+ or K⁺ ions in their minerals

Question 54. How can you prepare propyne from magnesium carbide?
Answer:

When magnesium carbide (MgC₂) is heated, Mg₂C₃ (magnesium allylide) is obtained. This on hydrolysis, produces propyne (CH₃C=CH)

Question 55. Why do halides of Be dissolve in organic solvents while those of Ba do not?
Answer:

Halides of Be are covalent. Hence, they dissolve in organic solvents while of Ba are ionic. Hence, they do not dissolve in organic solvents.

Question 56. Write the composition of gunpowder.
Answer: Gunpowder is an explosive mixture containing KNO₃ along with charcoal and sulfur

Question 57. Why is Na₂S₂O₃ used in photography?
Answer: In photography, Na₂S₂O₃ is used to dissolve the unexposed AgBr.

Question 58. The affinity of sodium towards water is used in drying benzene. Explain.
Answer: Sodium does not react with benzene. Hence it can be used effectively for drying benzene.

Question 59. Write with a balanced equation the reaction for the manufacture of sodium bicarbonate from sodium carbonate.
Answer:

Sodium bicarbonate (NaHCO₃) Is manufactured by passing CO2 through a saturated solution of sodium carbonate (Na₂CO₃).

Na₂CO₃ + CO₂ + H₂O ⇌ 2NaHCO₃↓

Question 60.  Name a pair of elements that exhibits a diagonal relationship.
Answer:  Lithium (Li) and magnesium (Mg) CO₂

Question 62. Name an alkaline earth metal.
Answer: Calcium (ca)

Question 63. Write the balanced equation for the reaction when water is added to calcium carbide.
Answer: CaC₂ + 2M₂O →MC=CH + Ca(OH)₂

Question 64. Write the balanced equation(s) for the reaction when excess carbon dioxide is passed through brine saturated with ammonia
Answer:

When carbon dioxide is passed through an aqueous solution of NaCI (brine, 28% NaCl solution) saturated with ammonia, sodium bicarbonate is formed.

Question 65. Which alkali metal ion has the highest polarising power?
Answer: Li⁺

Question 66. What are the common oxidation states exhibited by group-1 and group-2- 2 metals respectively?
Answer: +2 And +1 respectively

Question 67. Name the alkali metals which form superoxides when heated with excess oxygen.
Answer: K, Rb and cs

Question 68. Which one is the lightest and which one is the heaviest of all the metals?
Answer: Lithium (Li) and Osmium(os) respectively

Question 69. What is the composition of the white powder obtained when metallic magnesium is burnt in the air?
Answer: A mixture of MgO and Mg₃N₂

Question 70. Name two metals of group 2 which do not impart any color to the flame.
Answer: Mg and Be

Question 71. Which alkali metal cannot be stored in kerosene?
Answer: Li

Question 72. Which alkali metal is used as a scavenger in metallurgy to remove O₂ and N₂ gases?
Answer: Li

Question 73. Which Gr-2 metal carbide reacts with water to produce methane?
Answer: Be₂ C

Question 74. In an aqueous solution, which alkali metal ion has the lowest mobility?
Answer: Li⁺

Question 76. Which is the most abundant alkaline earth metal in the earth’s crust?
Answer: Calcium

Question 77. Name one group-2 metal.
Answer: Chlorophyll

Question 78. Which alkali metal ion forms a stable complex with 18- crown-6
Answer: Potassium(K)

Question 79. Which alkaline earth metal is largely used as a lightweight construction metal?
Answer: Mg

Question 80. Which alkaline earth metal forms an organometallic compound known as Grignard reagent?
Answer: Mg

Class 11 S-Block Notes

Question 81. Which alkali and alkaline earth metals are radioactive?
Answer: Fr and Ra

Question 82.  The salts of which alkali metals are commonly hydrated?
Answer: Li

Question 83. Which alkali metal acts as the strongest reducing agent in aqueous solution?
Answer: Li

Question 84. Which is the least stable alkali metal carbonate?
Answer: Li₂CO₃

Question 85. What is baryta water?
Answer: Ba(OH ₂ solution

Question 86. Which alkaline earth metal hydroxide is most soluble in water?
Answer: Ba(OH)₂

Question 87. Which alkaline earth metal imparts crimson color to flame?
Answer: Srcl₂

Question 88. BeO is covalent and still has high melting points — why?
Answer: It is polymeric

Question 89. What is anhydrous?
Answer: Magnesium perchlorate Mg(ClO₄)₂

Question 90. Which alkaline earth metal carbonate can be kept only in an atmosphere of CO₂?
Answer: BeCO₃

Question 91. Which alkaline earth metal chloride is used as a desiccant in the laboratory?
Answer: CaCl₂

Question 92. The basic strength of which alkali metal hydroxide is the highest?
Answer: CsOH

Question 93. Which alkaline earth metal hydroxide is amphoteric?
Answer: Be(OH)₂

Question 94. How will you distinguish between Ba(OH)₂ and Be(OH)₂?
Answer: Be(OH)₂ dissolves in alkali but Ba(OH)₂ does not

Question 95. What are the raw materials used for the manufacture of washing soda by the Solvay process?
Answer: NaCl, CaCO₃ and NH₃

Question 96. What is soda ash?
Answer: . Anhydrous Na₂CO₃

Question 97. Which alkaline earth metal hydroxide and alkali metal carbonate are used for softening of hard water?
Answer: Ca(OH)₂ and Na₂CO₃;

Question 98. What is the formula of Plaster of Paris?
Answer: 2CaSO₄ .H₂O

Question 99. What is dead burnt plaster?
Answer: Anhydrous CaSO₄;

Class 11 S-Block Notes

Question 100. Which alkaline earth metal carbonate is used as an ingredient of chewing gum?
Answer: CaCO₃

Question 101. What is a fusion mixture?
Answer: A mixture of Na₂CO₃ and K₂CO₃

Question 102. What is the most abundant source of sodium chloride?
Answer: Seawater

Question 103. For which of its chief properties is Plaster of Pariswide used?
Answer: On mixing with water, it forms a plastic mass which sets into a hard mass within 5 to 15 minutes;

Question 104. What is used for making blackboard chalks?
Answer: CaCO₃

Question 105. Which compound is generally used for the detection of CO₂ in the laboratory?
Answer: Lime water [Ca(OH)₂]

Question 106. What is the diagonal relationship? Give one example
Answer:

See ‘General Discussion on s -block elements

Question 107. Why is LiCl soluble in organic solvents?
Answer:

As the polarising power of Li⁺ is very high, LiCl is covalent. Hence, it is soluble in an organic solvent

Question 108. Why are fumes seen when barium halides are kepi In open air?
Answer:

In the open air, barium halide undergoes hydrolysis by water vapor (moisture), and fumes of halogen acid (except HF) evolve

Question 109. Which of the alkaline earth metal hydroxides are amphotericin in nature?
Answer:

Beryllium hydroxide [Be(OH)₂] is amphoteric in nature

Question 110. What is hydrolysis?
Answer:

Calcium hydride [CaH₂] is known as hydrolytic

Question 111. Find out the oxidation state of sodium in Na₂O₂.
Answer:

Let the oxidation state of Na be x. The oxidation state of oxygen in the peroxides is -1

2x + 2(-1) = 0; 2x = 2,x= +1.

Question 112. Explain why can alkali and alkaline earth metals not be obtained by chemical reduction methods.
Answer:

Alkali and alkaline earth metals are strong, reducing agents and it is difficult to reduce their oxides or chlorides by any other reducing agent.

Question 113. Which two alkaline earth metals cannot be identified by flame test?
Answer:

Beryllium (Be) and magnesium (Mg) cannot be identified by flame test

Question 114. Beryllium chloride hydrate loses no water over P₄O₁₀ — why?
Answer:

Due to very small size and stronger hydrating tendency of Be²⁺ ion, it is not possible for P₄O₁₀ to abstract water molecules from beryllium chloride hydrate, [Be(H₂O)4]Cl₂;

Question 115. Magnesium occurs in nature largely as MgCO₃but beryllium never occurs as BeCO₃. Explain.
Answer:

The strong polarising power of Be²⁺ ion makes BeCO₃ unstable;

Class 11 S-Block Notes

Question 116. The crystalline salts of alkaline earth metals contain more water of crystallization than the corresponding alkali metal salts—why?
Answer:

Due to their smaller size and higher charge, the alkaline earth metal ions have a greater tendency to coordinate with water molecules as compared to alkali metal ions

Question 117. Out of Na and Mg which one has a higher second ionization enthalpy? Why?
Answer:

The second ionization enthalpy of Na is higher than that of Mg;

Question 118. The chloride of a metal is soluble in an organic solvent. The chloride can be CaCl₂, NaCl, MgCl₂, BeCl₂
Answer: BeCl₂

Question 119. Why does common salt become wet in the rainy season?
Answer:

Due to the absorption of aerial moisture by deliquescent impurities like MgCl₂, CaCl_2, etc;

Question 120. With the help of a drop of an indicator solution, how would you know whether a solution consists of Na₂CO₃ or NaHCO₃?
Answer:

A drop of phenolphthalein will change the colorless solution of Na₂CO₃ to purple but the colorless solution of NaHCO₃ will remain unchanged;

Class 11 Chemistry S Block Elements Fill In The Blanks

Question 1. _____________ has no d-orbital in its valence shell.
Answer: Lithium

Question 2. _____________ reacts with nitrogen to give nitrides.
Answer: Lithium

Question 3. _____________ imparts golden-yellow color to the flame.
Answer: Sodium

Question 4. Potassium, in reaction with dioxygen, produces _____________
Answer: KO₂

Question 5. The outermost electronic configuration of the radioactive alkali metal is _____________
Answer: 7s¹

Question 6. The bicarbonate salt_____________ of does not exist in solid state.
Answer: Lithium

Question 7. Ionic conductance of Li+ ion in aqueous solution is lowest _____________ is highest
Answer: Hydration

Question 8. _____________ion has maximum polarising power
Answer: Li⁺

Question 9. _____________ is the most abundant alkali metal in Earth’s
Answer: Sodium

Question 10. The basicity of the alkali metal hydroxides _____________ down the group.
Answer: Increases

Question 11._____________ is used as a source of oxygen in submarines, space shuttles, and oxygen masks.
Answer: KO₂

Class 11 S-Block Notes

Question 12. The alkali metals combine with mercury to give_____________
Answer: Amalgams

Question 13. The alkali metals exist as_____________ lattices having cordination number 8.
Answer: Body Center cubic

Question 14. Lil _____________ is more soluble than KI in ethanol.
Answer: More

Question 15. K₂CO₃ cannot be produced by the Solvay process because _____________ does not get precipitated in the aqueous solution.
Answer: KHCO₃

Question 16. _____________ and _____________ do not respond to flame test.
Answer: Be,  Mg

Question 17. _____________ ion exhibits maximum does not respond tendency to complexes.
Answer: Be²⁺

Question 18. Only _____________ can displace hydrogen from dilute HNO₃
Answer: Mg

Question 19. Common salt gets wet due to the presence of _____________ as impurity.
Answer: MgCl₂

Question 20. Hydration enthalpy of Mg²⁺ is _____________ Ca²⁺. as than that of
Answer: Greater

Question 21. Hydrolysis of calcium carbide produces _____________
Answer: Acetylene

Question 22. The commercial name of _____________is hydrolith
Answer: CaH₂,

Class 11 S-Block Notes

Question 23. BeCO₃ is stable only in an atmosphere of _____________
Answer: CO₂

Question 24. Lime water is a transparent aqueous solution of _____________
Answer: Ca(OH)₂

Question 25. The second ionisation enthalpy of the alkaline earth metals is _____________ than their first ionization enthalpy.
Answer: Greater

Question 26. _____________ is used to prepare Grignard reagents
Answer: Greater

Question 27. The melting point of the alkaline earth metals is _____________ than that of alkali metals
Answer: Mg

Question 28. Between Ca and Na, _____________ is used to dehydrat alcohols
Answer: Ca

Question 29. Among the alkaline earth metals, _____________ is abundant in the earth’s crust.
Answer: Ca

Question 30. Temperature of the mixture of _____________  -54°C. is most and ice is about
Answer: CaCl²⁺

Class 11 S-Block Notes

Class 11 Chemistry S Block Elements Warm-Up Exercise Question And Answers

Question 1. Name a radioactive alkali metal and write its atomic number.
Answer:  Francium, atomic number =87

Question 2. Mention the similarity shown in the electronic configurations of the alkali metals.
Answer: All alkali metals have similar valence shell electronic configuration of ns¹

Question 3. Why alkali metals are called s -s-block elements?
Answer: Alkali metals are called s -s-block elements because the last electron enters the ns-orbital

Class 11 S-Block Notes

Question 4. Which of the alkali metals exhibits abnormal behavior?
Answer: Lithium

Question 5. Which alkali metal is most abundant in the earth’s crust?
Answer: Sodium (Na)

Question 6. What is trona?
Answer:

Trona ( Na₂ CO₃ , NaHCO₃  .  2H₂O) is an important mineral of sodium

Question 7. Arrange lithium, sodium, and potassium according to their abundance in the earth’s crust.
Answer:

According to their abundance in nature, the elements are arranged as lithium < potassium < sodium.

Question 8. Give an example of a double salt formed by an alkali metal and alkaline earth metal.
Answer:

The double salt formed by an alkali metal and an alkaline earth metal is carnallite (KCl-MgCl₂-6H₂O)

Question 9. What type of crystal lattice is formed by the alkali metals?
Answer:

Alkali metals form a body-centered cubic lattice

Question 10. Arrange LiF, NaF, KF, RbF, and CsF in increasing order of their lattice energies.
Answer:

The increasing order of lattice energies of the given compounds is: CsF < RbF < KF < NaF < LiF

Question 11. Why are alkali metals paramagnetic?
Answer:

Due to the presence of unpaired electrons in the valence shell of alkali metals, they are paramagnetic.

Question 12. Which alkali metal is generally used in photoelectric cells?
Answer:

Cesium is generally used in photoelectric cells

Question 13. Which alkali metals form superoxides when heated in excess air?
Answer:

The alkali metals that form superoxides when they are heated in excess air are potassium (K), rubidium (Rb), and cesium (Cs).

Class 11 S-Block Notes

Question 14. Differentiate between Na₂CO₃ and NaHCO₃
Answer:

When NaHCO₃ is heated, it liberates CO₂ which turns lime-water milky. On the other hand, when Na₂CO₃ is heated, it does not undergo decomposition

Question 15. Arrange MCI, MBr, MF, and MI (where M= alkali metal) according to increasing covalent character.
Answer:

The covalent character of metallic chlorides increases with an increase in the size of the anion. Therefore, the order of the metallic chlorides according to increasing covalent character is MF < MCI < MBr < MI

Question 16. Explain why the peroxides and superoxides of the alkali metals act as strong oxidizing agents.
Answer:

In reaction with water, peroxides produce MOH along with H₂O₂ and superoxides produce MOH and O, along with H₂O₂. H₂O₂ is a strong oxidizing agent. Thus, peroxides and superoxides of the alkali metals act as strong oxidizing agents

M₂O₂ + 2H₂O → 2MOH + H₂O₂

2MO₂ + 2H₂O→ 2MOH + H₂O₂ + O

Question 17. Give a simple test to distinguish between KNO₃ & LiNO₃.
Answer:

Colourless O₂ gas evolves on heating KNO₃. But heating, LiNO₃ dissociates into colorless gas and brown NO₂ gas

Question 18. Explain why a solution of Na₂CO₃ is alkaline in nature whereas a solution of Na₂SO₄ is neutral.
Answer:

For Na₂CO₃, Na₂SO₄ is a salt of strong base (NaOH) and strong acid (H₂SO₄). So, the nature of the solution of Na₂SO₄ is neutral.

Question 19. Among the sulfate salts of lithium, sodium, potassium, and rubidium, which salt does not form double salt?
Answer:

Lithium sulfate (Li₂, SO₄) does not form any double salt

Question 20. What happens when sodium sulfate is fused with charcoal? Give equation.
Answer:

When sodium sulfate is fused with charcoal, it reduces to sodium sulfide, and carbonÿis oxidized to carbon monoxide:

Na₂SO₄ + 4→ Na₂S + 4CO↑

Class 11 S-Block Notes

Question 21. Which alkali metal bicarbonate has no existence in the solid state?
Answer:

The alkali metal bicarbonate which has no existence in a solid state is lithium bicarbonate (LiHCO₃)

Question 22. Mention the property for which lithium is used to separate N2 gas from a gas mixture.
Answer:

The alkali metal bicarbonate which has no existence in a solid state is lithium bicarbonate (LiHCO₃)

Question 23. Give a simple test to distinguish between Li₂CO₃ & Na₂CO₃. 
Answer:

On heating, Li₂CO₃ decomposes to CO₂ which turns lime water milky. On the other hand, Na₂ CO₃ does not decompose on heating

Class 11 S-Block Notes

Question 24. Write down the name of the alkali metal compound which i

1. Effective in the treatment of manic depressive psychosis
2. Used in baking powder

Answer:

1. Li₂CO₃
2. NaHCO₃

Question 25. Name an alkali metal that is used in Lassaigne s test for detection of nitrogen, sulphur and halogen in the organic compounds.
Answer: Sodium

Question 26. Explain why sodium carbonate is used in fire extinguishers.
Answer:

In fire extinguishers, Na₂CO₃ is used as it reacts with dil. H₂SO₄ to produce CO₂

Question 27. Why docs sodium hydrogen carbonate called baking soda?
Answer:

NaHCO₃ is an important ingredient in baking powder. Thus, it is also called baking soda.

Question 28. Why a standard solution of sodium hydroxide (NaOH) cannot be prepared?
Answer:

Sodium hydroxide, being a hygroscopic substance absorbs moisture from the atmosphere. It also absorbs CO₂ from the air and forms Na₂CO₃. Thus, sodium hydroxide cannot be accurately weighed and so a standard solution of NaOH cannotbe prepared.

Question 29. How the group-2 elements are commonly known? Why are they so called?
Answer:

Group-2 elements are commonly known as ‘alkaline earth metals’. These are so-called because their oxides are basic and are found in the earth’s crust.

Class 11 S-Block Notes

Question 30. Write the general electronic configuration of group-2 elements.
Answer:

Electronic configuration: [Inert gas] ns²

Question 31. Why group-2 elements are called s -s-block elements?
Answer:

Group-2 elements are called s -s-block elements because the last electron enters the s -s-orbital

Question 32. Which element of group 2 shows abnormal behavior?
Answer:

Beryllium (Be) shows abnormal behavior

Question 33. Which group-2 element has a slightly different electronic configuration than the rest of the elements?
Answer:

Beryllium has 2 electrons in its penultimate shell while the rest have 8 electrons in their penultimate shells.

Question 34. Explain why the atomic and ionic radii of Mg is less than those of Na and Ca.
Answer:

The electrons of Mg having a higher nuclear charge are more strongly attracted towards the nucleus. Thus, the atomic and ionic radii of Mg are less than Na. Again on moving down the group (from Mg to Ca), the atomic and ionic radii increase due to the addition of new shells and the increasing screening effect joindy overcomes the effect of increasing nuclear charge. Thus, the atomic and ionic radii of Mg is less than Ca

Question 35. Arrange Mg²⁺, Ba²⁺, Sr²⁺, Be²^+, and Cav according to decreasing order of their hydration enthalpies. Explain your answer.
Answer:

The correct order is  → Be²⁺ > Mg²⁺ > Ca²⁺ > Sr²⁺ > Ba²⁺

Class 11 S-Block Notes

Question 36. Alkaline earth metals predominantly form ionic compounds. However, the first member of the group forms covalent compounds—Explain why.
Answer:

Due to relatively high electronegativity, the first member of each of these groups tends to form covalent compounds

Question 37. A white residue is obtained when metallic Mg is burnt in air. This residue when heated with water emits an ammoniacal smell. Explain these observations.
Answer:

On burning metallic magnesium in the air, magnesium oxide (MgO) and magnesium nitride (Mg₃N₂) are formed. So, the white residue obtained is a mixture of MgO & Mg₃N₂.

This residue when heated with water, results in the hydrolysis of Mg₃N₂ which emits an ammoniacal smell.

Question 38. Why calcium is better than sodium in eliminating a small amount of water from alcohol?
Answer:

Both Na and Ca react with water to form the corresponding hydroxide. However, sodium rapidly reacts with alcohol to form sodium ethoxide (NaOC₂H₅) but calcium reacts quite slowly with alcohol

2C₂H₅OH + 2Na → 2C₂H₅ONa + H₂↑

Thus, for eliminating a small amount of water from alcohol, calcium is better than sodium

Question 39. Why Mg-ribbon continues to burn in the presence of SO₂ gas
Answer:

During the combustion of Mg-ribbon, the amount of heat generated leads to the decomposition of sulfur dioxide into sulfur and oxygen. This oxygen is responsible for the continued burning of Mg-ribbon.

2Mg + SO₂→ 2MgO + S

Question 40. Except Be(OH)₂, all other alkaline earth metal hydroxides are basic and their basic strength increases down the group.
Answer:

Be has high ionization enthalpy, for which Be(OH)₂ is amphoteric.

Question 41. Why is Mg(OH)₂ less basic than NaOH?
Answer:

Due to the larger ionic size and lower ionization enthalpy of Na, the Na — OH bond in NaOH is weaker than the Mg — OH bondin Mg(OH)₂. Thus, NaOH is more basic than Mg(OH)₂

Question 42. Sparingly soluble carbonate salts of alkaline earth metals become easily soluble in water in the presence of CO₂. Why?
Answer:

Sparingly soluble carbonate salts of alkaline earth metals are converted into soluble bicarbonate salts in the presence of CO₂ So, these salts become easily soluble in water.

CaCO₃(s) + CO₂(g) + H₂O(l)→ Ca(HCO₃)₂ (aq)

Question 43. The anhydrous chloride salt of which alkaline earth metal is used as a drying agent in the laboratory?
Answer:

Anhydrous chloride salt of calcium metal, i.e., calcium chloride (CaCl₂) is used as a drying agent in laboratory

Class 11 S-Block Notes

Question 44. Why BeCO₃ is kept in an atmosphere of CO₂?
Answer:

BeCO₃ being highly unstable easily decomposes to give off CO₂ when kept in an open atmosphere.

Question 45. Name a bivalent element whose oxide is soluble in excess NaOH solution.
Answer:

Beryllium; BeO + 2NaOH→ Na₂BeO₂ + H₂O

Question 46. Why CaF₂ is considered the most important fluoride salt among all the fluoride salts of alkaline earth metals?
Answer:

CaF₂ is the most important source for the preparation of fluorine

Question 47. Write down some important points of difference between beryllium and magnesium.
Answer:

• Beryllium is harder than magnesium,
• Compounds of Be are largely covalent, whereas most of the compounds of Mg are ionic,
• Be does not exhibit a coordination number of more than 4, while Mg exhibits a coordination number of 6

Question 48. Be usually forms covalent compounds but other elements of group ionic compounds. Why?
Answer:

Be usually forms covalent compounds due to its high ionisation enthalpy and small size. However, due to comparatively low ionization enthalpy and large size, other elements of group-2 form ionic compounds

Question 49. Which compounds of the alkaline earth metals are used as refractory substances?
Answer:

The oxides of alkaline earth metals (MO) have high melting points and so are used as refractory substances

Question 50. Carbonaceous impurities in gypsum and any fuel are avoided during the preparation of Plaster of Paris. Explain.
Answer:

Carbonaceous impurities in gypsum and any fuel are avoided during the preparation of Plaster of Paris because carbon will reduce CaSO₄ to CaS.

CaSO₄ + 4C →  CaS + 4CO ↑