Class 12

Balbharti Maharashtra State Board 12th Chemistry Important Questions Chapter 10 Halogen Derivatives Important Questions and Answers.

Maharashtra State Board 12th Chemistry Important Questions Chapter 10 Halogen Derivatives

Question 1.
What are halogen derivatives of hydrocarbons?
Answer:
The replacement of hydrogen atom/s in aliphatic or aromatic hydrocarbons by halogen atom/s results in the formation of halogen derivatives of hydrocarbons.

Question 2.
How are halogen derivatives of hydrocarbons classified?
Answer:
Halogen derivatives of alkane are classified as :

1. Monohalogen derivative (or alkyl halide) : It is a halogen derivative of an alkane in which one hydrogen atom is replaced by one halogen atom and it is also called alkyl halide. E.g. C₂H₅Br.
2. Poh halogen derivatives : These are halogen derivatives in which more than one hydrogen atoms of an alkane are substituted by corresponding number of halogen atoms.

They are classified as follows :
(i) Dihalogen derivatives : The compounds formed by the substitution of two hydrogen atoms of an alkane by two halogen atoms are called dihalogen derivatives.

They are further classified as :

• Vicinal dihalides (Two halogen atoms on vicinal or adjacent carbon atoms)
• Geminal dihalides : CH₃ – CHCI₂ (Two halogen atoms on the same carbon atom)

(ii) Trihalogen derivatives : The compounds formed by the substitution of three hydrogen atoms of an alkane by three halogen atoms are called trihalogen derivatives.

(iii) Tetrahalogen derivatives : The compounds formed by the substitution of four hydrogen atoms of an alkane by four halogen atoms are called tetrahalogen derivatives. E.g. CCI₄.

Question 3.
What are alkyl halides? How are they classified?
Answer:
The compound formed by the replacement of one hydrogen atom in an alkane by a halogen atom is called an alkyl halide. The halogen atom is bonded to sp³ hybridised carbon. Alkyl halides are classified into the following three classes depending on the type of the carbon to which halogen atom is bonded.

(1) Primary (1°) alkyl halide : Alkyl halide in which a halogen atom is bonded to a primary carbon atom is called primary alkyl halide.

[Primary (1°) carbon atom i.e., the carbon atom which is attached to only one carbon atom.]

They are represented by the general formula R – ^1°CH₂ – X.

(2) Secondary (2°) alkyl halide : Alkyl halide in which a halogen atom is bonded to a secondary carbon atom is called secondary alkyl halide. [Secondary (2°) carbon i.e., the carbon atom which is attached to two other carbon atoms.]

They are represented by the general formula (R and R’ can be same or different)

(3) Tertiary (3°) alkyl halide : Alkyl halide in which halogen atom is bonded to a tertiary carbon atom is called tertiary alkyl halide. [Tertiary (3°) carbon i.e., the carbon atom which is attached to three other carbon atoms.]

They are represented by the general formula  (R, R’ and R” may be same or different)

Question 4.
Explain the following :
(1) Alkyl halide or haloalkanes
(2) Allylic halides
(3) Benzylic halide
(4) Vinylic halide
(5) Haloalkyne
(6) Aryl halide or haloarenes.
Answer:
(1) Alkyl halide or haloalkanes : In alkyl halides or haloalkanes the halogen atom is bonded to sp3 hybridized carbon which is a part of saturated carbon chain.
Example :

(2) Allylic halides : In allylic halides, halogen atom is bonded to a sp³ hybridized carbon atom next to a carbon-carbon double bond.

Example :

(3) Benzylic halide : In benzylic halides, halogen atom is bonded to a sp³ hybridized carbon atom which is further bonded to an aromatic ring.
Example :

(4) Vinylic halides : In vinylic halides, halogen atom is bonded to a sp² hybridized carbon atom of aliphatic chain. Vinylic halide is a haloalkene.
Example :

(5) Haloalkyne : In haloalkynes, halogen atom is bonded to a sp hybridized carbon atom.
Example :

(6) Aryl halides or haloarenes : In aryl halides, halogen atom is directly bonded to the sp² hybridized carbon atom of aromatic ring.
Example :

Question 5.
Give the IUPAC names of the following :
Answer:

Question 6.
Draw the structures of the following compounds:
Answer:

Question 7.
Write the structure of-
(a) 3-chloro-3-ethylhex-l-ene
(b) 1-Iodo-2, 3-dimethylbutane
(c) 1, 3, 5-tribromobenzene
Answer:

Question 8.
Write structures of
(a) 2-iodo-3-methyl pentane
(b) 3-chiorolleNane
(c) 1-chloro-2, 2-dimethyl propane
(d) 1-chloro-4-ethyl cyclohexane.
Answer:

Question 9.
Write the possible isomers of monochloro derivatives of 2,3-Dimethylbutane and write their IUPAC names.
Answer:
The given parent hydrocarbon has molecular formula, C₆H₁₄. The monochloro derivative of this compound has molecular formula C₆H₁₃CI.
The parent hydrocarbon is,

Hence the structures of isomers of monochioroderivative are.

Question 10.
Write structures and IUPAC names of all possible isomers of C₅H₁₁Br and classify them as l°/2°/3°.
Answer:
C₅H₁₁Br is a monohalogen derivative.

Question 11.
How are following compounds obtained from alcohols :
(1) ethyl chloride C₂H₅CI
(2) isopropyl chloride (CH₃ CHCI – CH₃)
(3) tert-butyl chloride (CH₃)₃ – CI?
Answer:
Alcohols in the presence of Lucas reagent which is a solution of concentrated HCI and ZnCI₂ form alkyl halides. Hydrogen chloride is used with zinc chloride (Grooves’ process) for primary and secondary alcohols.

(3) Tertiary alcohols don’t need ZnCI₂ to react with HCI.

The order of reactivity of alcohols with a given halo acid is 3° >2°> 1°.

Question 12.
How are following compounds prepared from alcohols:
(1) ethyl bromide (C₂H₅Br)
(2) isopropyl bromide (CH₃ – CHBr – CH₃)
(3) tert-butyl bromide (CH₃)₃ C – Br?
Answer:
(1) Ethyl alcohol on heating with conc. hydrobromic acid (48%) forms ethyl bromide.
OR
When ethyl alcohol is treated with a mixture of NaBr and H₂SO₄, ethyl bromide is formed. Here HBr is generated in situ.

(2) Isopropyl alcohol, on reaction with NaBr and dil. H₂SO₄ forms isopropyl bromide.

(3) Tertiary alcohol on reaction with sodium bromide and dil. H₂SO₄ forms tert-butyl bromide.

Question 13.
How is ethyl iodide obtained from ethyl alcohol?
Answer:
When ethyl alcohol is treated with sodium or potassium iodide in 95 % phosphoric acid, ethyl iodide is formed. Here HI is generated in situ.

Question 14.
How will you prepare the following :

(1) Ethyl chloride (chloroethane) from ethyl alcohol using
(i) PCI₃
(ii) PCI₅ and
(iii) SOCI₂.
Answer:
(i) When ethyl alcohol is refluxed with phosphorus trichloride, ethyl chloride is formed.

(ii) When ethyl alcohol is refluxed with phosphorus pentachloride, ethyl chloride is formed.

(iii) When ethyl alcohol is refluxed with thionyl chloride, in the presence of pyridine, ethyl chloride is formed. The by-products obtained are gases. Therefore, this method is preferred for preparation of alkyl chloride.

(2) Isopropyl chloride (2-chloropropane) from isopropyl alcohol using
(i) PCI₃
(ii) PCI₅
(iii) SOCI₂.
Answer:
When isopropyl alcohol is refluxed with phosphorus trichloride, isopropyl chloride is formed.

When isopropyl alcohol is refluxed with phosphorus pentachloride, isopropyl chloride is formed.

When isopropyl alcohol is refluxed with thionyl chloride, in the presence of pyridine, isopropyl chloride is formed. The by-products obtained are gases. Therefore, this method is preferred for the preparation of alkyl chloride.

(3) Ethyl bromide (bromoethane) from ethyl alcohol.
Answer:
When ethyl alcohol is treated with a mixture of red phosphorus and bromine or hydrobromic acid (phosphorus tribromide is generated in situ), ethyl bromide is formed.

(4) Ethyl iodide (do ethane) from ethyl alcohol.
Answer:
When ethyl alcohol is heated with a mixture of red phosphorus and iodine, (phosphorus triiodide is generated in situ), ethyl iodide is formed.

Question 15.
Explain halogenation of methane.
Answer:
Halogenation : A reaction of alkanes with halogens (CI₂, Br₂, I₂) in the presence of appropriate conditions forming a mixture of alkyl halides.
(1) Chlorination:

(2) When excess of chlorine is used, tetrachioro methane, a major product is obtained. When excess of methane is used, chioromethane, a major product is obtained. The order of reactivity of halogens towards alkane is

(3) lodination:

However, iodination reaction is a reversible reaction. HI being a strongest reducing agent reduces methyl iodide back to methane.

(4) Fluorination: A reaction of alkane with fluorine is explosive and also hydrofluoric acid is poisonous and corrosive. Hence, alkyl fluorides are not prepared by halogenation of alkane.

Question 16.
Predict the possible products of the following reaction :
(1) Bromination of propane
(2) Bromination of n-butane
(3) Bromination of 2-Methyl propane.
Answer:
(1) Bromination of propane

(2) Bromination of n-butane

(3) Bromination of 2-Methyl propane

Question 17.
How are following compounds prepared by halogenation of ethane :
(1) Chloroethane
(2) Bromoethane
(3) Iodoethane?
Answer:
(1) Chlorination of ethane : When ethane (excess) is reacted with a limited quantity of chlorine in the presence of diffused sunlight or U.V. light or at high temperature, chloroethane is obtained.

(2) Bromination of ethane : When ethane is heated with Br₂ in the presence of anhydrous AlBr₃, bromoethane is obtained.

(3) Iodination : When ethane is reacted with I₂ in the presence of suitable oxidising agents like-HgO or HIO₃ or dilute HNO₃ iodoethane is obtained.

Question 18.
Direct iodination of alkanes is not possible.
Answer:
(1) Direct iodination of alkanes using iodine is highly reversible.
\(\mathrm{RH}+\mathrm{I}_{2} \rightleftharpoons \mathrm{RI}+\mathrm{HI}\)
(2) Hydroiodic acid HI being strong reducing agent, it reduces RI to alkane RH.
(3) The reaction takes place only in the presence of a suitable oxidizing agent like HgO, HIO₃ or dilute HNO₃ which decomposes HI. Hence, direct iodination of alkanes is not possible.

Question 20.
How are following compounds obtained from alkenes :

Answer:
(1) Ethene on reaction with hydrogen chloride forms

(2) Ethene on reaction with hydrogen bromide forms

(3) Propene on reaction with hydrogen iodide forms

(4) but-2-ene on reaction with hydrogen iodide forms 2-iodobutane.

Question 19.
State and explain Markovnikov’s rule.
Answer:
Markovnikov’s rule : When an unsymmetrical reagent is added to an unsymmetrical alkene, the negative part of the reagent gets attached to that carbon atom of the double bond which carries less number of hydrogen atoms.

Example : Addition of HBr’to unsymmetrical alkene like propene gives two products.

Isopropyl bromide is the major product, since the negative part (Br^–) of HBr is attached to carbon atom of a double bond with less number of hydrogen atoms.

Question 20.
Explain peroxide effect.
OR
Write a note on the Kharasch-Mayo effect.
OR
Explain the addition of HBr to (unsymmetrical alkene) propane in the presence of benzoyl peroxide.
Answer:
The addition of HBr to an unsymmetrical alkene (propane) in the presence of benzoyl peroxide takes place in the opposite orientation to that of Markovnikov’s rule and this is known as Kharasch-Mayo effect or peroxide effect or Anti-Markovnikov addition.

Question 21.
Write the structure of alkyl halide obtained by the action hydrogen bromide on 2-Methyiprop-1-ene in the presence of peroxide.
Answer:
In the presence of peroxide. HBr to 2-Methyl prop-I-cne forms l-Bromo-2-methylpropane.

Question 22.
How are alkyl iodides prepared from alkyl chlorides/bromides?
Answer:
Alkyl iodide is prepared by treating alkyl chloride or alkyl bromide with sodium iodide, in the presence of dry acetone, sodium chloride or sodium bromide precipitates from the solution and can be separated by filtration. This reaction is known as Finkelstein reaction.

Question 23.
How are alkyl fluorides prepared with alkyl chlorides/alkyl bromides?
Answer:
When alkyl chloride or alkyl bromide is heated with metallic fluorides like AgF, CaF₂, CoF₂ or Hg₂F₂, alkyl fluoride is formed.

This reaction is known as Swarts reaction.

Question 24.
Explain the preparation of haloarenes using electrophilic substitution.
Answer:
When arene is treated with chlorine or bromine in dark at ordinary temperature in the presence of lewis acid as a catalyst like Fe, FeCI₃ or anhydrous AlCI₃, aryl chloride or aryl bromide is formed.

When toluene is brominated in dark at ordinary temperature in the presence of iron, a mixture of ortho and para bromo tolerene is obtained.

Ortho and para isomers can be easily separated as there is large difference in melting points of ortho and para isomers.

Question 25.
Write a note on Sandmeyer’s reaction.
Answer:
Aryl halides are most commonly prepared by replacement of nitrogen of diazonium salt. The replacement of diazonium group by -Cl or -Br using cuprous salt is called Sandmeyer’s reaction. When a primary aromatic amine (like aniline) suspended in cold F1C1, is treated with sodium nitrite, a diazonium salt (benzene diazonium chloride) is formed. When diazonium salt is treated with cuprous chloride or cuprous bromide, aryl halide (chlorobenzene or bromobenzene) is formed.

When benzene diazonium salt is mixed with potassium iodide, iodobenzene is formed.

Question 26.
Define the following :
Answer:
(1) Monochromatic light : It consists of rays of single wavelength vibrating in different planes perpendicular to the direction of propagation of the light.

(2) Plane polarized light : A light having oscillations only in one plane perpendicular to direction of propagation of light is known as plane polarized light.

(3) Optical isomerism : The steroisomerism in which the isomers have different spatial arrangements of groups/atoms around a chiral atom is called optical isomerism.

(4) Optical activity : The property of a substance by which it rotates plane of polarization of incident plane polarized light is known as optical activity.

(5) Optically active compound : The compound which rotate the plane of plane polarized light is called optically active compound.

(6) Enantiomers : The optical isomers which are non-superimposable mirror images of each other are called enantiomers or enantiomorphs or optical antipodes.
Example : 2-chlorobutane, lactic acid

(7) Chiral carbon atom : Carbon atom in a molecule which carries four different groups/atoms is called chiral carbon atom.

Chiral atom in a molecule is marked with asterisk (*)

For example : C in lactic acid

(8) Chiral molecule : When a molecule contains one chiral atom, it acquires a unique property i.e. it is non- superimposable with its mirror image is said to be chiral molecule.

(9) Chirality : The relationship between a chiral molecule and its mirror image is similar to the relationship between left and right hands. Therefore it is called handedness or chirality.

(10) Dextrorotatory substance or r/-Isomer : An optically active substance (or isomer) which rotates the plane of a plane polarized light to the right hand side (RHS) is called dextrorotatory substance (or isomer) and denoted by d or (+) sign.

(11) Laevorotatory substance or /-Isomer : An optically active substance (or isomer) which rotates the plane of a plane polarized light to the left hand side (LHS) is called laevorotatory substance (or isomer) and denoted by / or (-) sign.

(12) Racemic mixture or Racemate : A mixture containing equimolar quantities of dextro (d) and laevo (/) optical
isomers which is optically inactive due to molecules of one enantiomer is cancelled by equal and opposite optical rotation due to molecules of the other enantiomer is called a racemic mixture or racemate. It is represented as (dl) or (+).

Question 27.
Calculate the number of isomers for 2-chlorobutane.
Answer:
The number of optical isomers possible for a compound is 2ⁿ where n = number of asymetric carbon atoms.
As n = 1 for 2-ehlorobutane, 2ⁿ = 2¹ = 2. Hence, it has two optical isomers.

Question 28.
How many optical isomers are possible for C₅H₁₁ CI?
Answer:
The number of optical isomers : 3.

Question 29.
How many optical isomers are possible for glucose?
Answer:
The number of optical isomers : 16.

Question 30.
Draw the structures and indicate the chiral carbon atoms in
(1) Lactic acid
(2) 2-Chlorobutane.
Answer:
(1) In lactic acid structure, the starred carbon atom is chiral carbon atom as it is attached to four different substituents, COOH, OH, CH₃ and H.
(2) In 2-chlorobutane structure, the starred carbon atom is chiral carbon atom as it is attached to four different substituents, -CH₂ – CH₃ (ethyl), CH₃ (methyl), Cl and H.

Question 31.
Identify chira! and achiral molecules.

Question 32.
Complete the following reactions and explain optical activity of the products formed:
(i) Pent-1-ene with HBr
Answer:

(ii) Pent-2-ene with HBr
Answer:

Question 33.
C₆H₁₂ (A) on treatment with HCI produced a compound Y. Which is optically active, what is structure A?
Answer:

Question 34.
A racemic mixture is optically inactive. Explain.
Answer:

• A racemic mixture contains equimolar (or equimolecular) quantities of the dextrorotatory (d-) and laevorotatory (l-) isomers (enantiomers) of a compound.
• The d-enantiomer rotates the plane of plane-polarized light to the right, while the l-enantiomer rotates the same to the left to the same extent.
• The quantities of the d- and l-enantiomers being the same, both the rotations are of the same magnitude, but of opposite directions. Hence, they cancel each other. Hence, a racemic mixture is optically inactive.
• It is represented as dl or ( + ). Example : ( ± ) lactic acid

Question 35.
Explain Fischer projection formula with illustration.
OR
Write a note on Fischer projection formula.
Answer:
Fischer projection formula or cross formula : The three dimensional (3-D) view of a molecule is presented on plane of paper. A Fischer projection formula can be drawn by visualizing the main carbon chain verical in the molecule. Each carbon on the vertical chain is represented by a cross.

Conventionally the horizontal lines of the cross represent bonds projecting up from the carbon and the vertical lines represent the bonds going below the carbon.

Question 36.
Explain Wedge formula with illustration
OR
Write a note on Wedge formula.
Answer:
Wedge formula : When a tetrahedral carbon is imagined to be present in the plane of paper all the four bonds at this carbon cannot lie in the same plane. The bonds in the plane of paper are represented by normal lines, the bonds projecting above the plane of paper are represented by solid wedges (or simply by bold lines) while bonds going below the plane of paper are represented by broken wedges (or simply by broken lines).

Question 37.
Give a laboratory test to confirm the presence of halogen in the original organic compound.
Answer:
Haloalkanes are of neutral type in aqueous medium. On warming with aqueous sodium or potassium hydroxide the covalently bonded halogen in haloalkane is converted to halide ion.

When this reaction mixture is acidified by adding dilute nitric acid and silver nitrate solution is added a precipitate of silver halide is formed which confirms presence of halogen in the original organic compound.

Question 38.
Define the following :
Answer:
(1) Mechanism of a reaction : It is a step by step description of exactly how the reactants are transformed into products in as much details as possible.
(2) Substitution reaction : When a group bonded to a carbon in a substrate is replaced by another group to get a product with no change in state of hybridization of that carbon, the reaction is called substitution reaction.

Question 39.
Describe the action of aqueous KOH (or NaOH) on :
(1) ethyl bromide
(2) isopropyl bromide
(3) tert-butyl chloride
(4) methyl bromide
(5) 2-chlorobutane.
Answer:
(1) Ethyl bromide : When ethyl bromide (bromoethane) is refluxed with aqueous potassium hydroxide, ethyl alcohol is formed. The reaction is called a hydrolysis reaction.

(2) Isopropyl bromide : When isopropyl bromide (2-bromopropane) is boiled with aqueous potassium hydroxide, isopropyl alcohol is formed.

(3) Tert-butyl chloride : When tert-butyl chloride is refluxed with aqueous potassium hydroxide, tert-butyl alcohol is formed.

(4) Methyl bromide : When methyl bromide (bromomethane) is heated with aq. KOH, it is hydrolysed to methyl alcohol (methanol).

(5) 2-chlorobutane : When 2-Chlorobutane is boiled with aqueous KOH, Butan-2-ol is obtained.

Question 40.
Describe the action of sodium ethoxide on
(1) ethyl bromide
(2) methyl bromide :
OR
Write a note on Williamson’s synthesis.
OR
How are ethers prepared from alkyl halides?
Answer:
Williamson’s synthesis : When an alkyl halide (R – X) is heated with sodium alkoxide (R – O – Na), an ether is obtained. In this reaction halide (-X) of alkyl halide is replaced by an alkoxy group (-OR). This reaction is known as Williamson’s synthesis. This method is used to prepare simple (or symmetrical) ethers and mixed (or unsymmetrical) ethers.

Sodium alkoxide is obtained by a reaction of sodium with an alcohol.

(1) Simple (symmetrical) ether : When an alkyl halide and sodium alkoxide having similar alkyl groups are heated, symmetrical ether is obtained.

e.g., When ethyl bromide is heated with sodium ethoxide, diethyl ether is formed.

(2) Mixed (unsymmetrical) ether : When an alkyl halide and sodium alkoxide having different alkyl groups are heated, unsymmetrical ether is obtained.

When methyl bromide is heated sodium ethoxide, ethyl methyl ether is formed.
When ethyl bromide is heated with sodium meihoxide, ethyl methyl ether is formed.

Question 41.
What is the action of silver salt of carboxylic acid on alkyl halide?
Answer:
When an alkyl halide (R – X) is heated with silver salt of carboxylic acid (R -COOAg). an ester is obtained.

Question 42.
Describe the action of alcoholic silver acetate on
(1) methyl bromide
(2) ethyl bromide.
Answer:
(1) Methyl bromide : When methyl bromide is heated with an alcoholic silver acetate, methyl acetate is formed.

(2) Ethyl bromide : When ethyl bromide is heated with an alcoholic silver acetate, ethyl acetate is formed.

Question 43.
What is the action of alcoholic silver propionate on ethyl bromide?
Answer:
When ethyl bromide is heated with an alcoholic silver propionate. ethyl propionate is formed.

Question 44.
Describe the action of excess of ammonia on (I) ethyl bromide (2) n.propyl bromide.
Answer:
(1) Ethyl bromide : When ethyl bromide is boiled under pressure with an excess of alcoholic ammonia, ethylamine (ethanamine) is formed. This is known as ‘ammonolysis of ethyl bromide.

(2) n-propyl bromide : When n-propy1 bromide is boiled under pressure with an excess of ammonia, n-propyl amine (propanamine) is formed.

Question 45.
What is ammonolysis? Give a suitable example for the reaction.
Answer:
When an alkyl halide is boiled under pressure with an excess of alcoholic solution of ammonia (NH₃), corresponding (primary amine) alkyl amine is formed. This reaction is known as ammonolysis of alkyl halide.

(1) Ethyl bromide : When ethyl bromide (bromoethane) is refluxed with aqueous potassium hydroxide, ethyl alcohol is formed. The reaction is called a hydrolysis reaction.

(2) Isopropyl bromide : When isopropyl bromide (2-bromopropane) is boiled with aqueous potassium hydroxide, isopropyl alcohol is formed.

(3) Tert-butyl chloride : When tert-butyl chloride is refluxed with aqueous potassium hydroxide, tert-butyl alcohol is formed.

(4) Methyl bromide : When methyl bromide (bromomethane) is heated with aq. KOH, it is hydrolysed to methyl alcohol (methanol).

(5) 2-chlorobutane : When 2-Chlorobutane is boiled with aqueous KOH, Butan-2-ol is obtained.

Question 46.
Describe the action of aqueous alcoholic potassium cyanide on
(1) ethyl bromide
(2) methyl iodide.
Answer:
Ethyl bromide : When ethyl bromide (bromoethane) is boiled with alcoholic solution of potassium cyanide in aqueous ethanol, ethyl cyanide (ethyl nitrile) is formed.

(2) Methyl iodide : When methyl iodide is boiled with alcoholic solution of potassium cyanide, methyl cyanide is formed.

Question 47.
Describe the action of alcoholic silver cyanide on
(1) ethyl bromide
(2) methyl chloride.
OR
Explain isocyanide reaction of
(1) ethyl bromide
(2) methyl chloride.
Answer:
(1) Ethyl bromide : When ethyl bromide is heated with alcoholic silver cyanide, ethyl isocyanide is formed.

(2) Methyl chloride : When .methyl chloride is heated with alcoholic silver cyanide, methyl isocyanide is formed.

The above reactions (1) and (2) are called isocyanide reaction.

Question 48.
Describe the action of potassium nitrite on
(i) ethyl bromide,
(ii) methyl chloride.
Answer:
(1) Ethyl bromide : When ethyl bromide is treated with potassium nitrite, ethyl nitrite is formed.

(2) Methyl chloride : When methyl chloride is treated with potassium nitrite, methyl nitrite is formed.

Question 49.
Describe the action of silver nitrite on (1) ethyl chloride (2) n-propyl bromide.
Answer:
(1) Ethy chloride : When ethyl chloride is treated with silver nitrite, nitroethane is obtained.

(2) n-Propyt bromide: When n-propyl bromide is treated with silver nitrate, nitropropane is obtained.

Question 50.
How will you convert (the following:

(1) Ethyl bromide to ethanol.
Answer:

(2) Ethyl bromide to propane nitrile.
Answer:

(3) Ethyl bromide to ethyl amine.
Answer:

(4) Ethyl bromide to ethyl acetate.
Answer:

(5) Ethyl bromide to ethyl isocyanide.
Answer:

(6) Ethyl bromide to ethyl methyl ether.
Answer:

(7) Ethyl bromide to n-butane.
Answer:

(8) Ethyl bromide to Ethyl magnesium bromide.
Answer:

Question 51.
Define the following :
Answer:
(1) Nucleophilic bimolecular reaction (SN²) : The substitution reaction in which a nucleophile reacts with the substrate and the rate of the reaction depends on the concentration of the substrate and the nucleophile is called a nucleophilic bimolecular reaction.

Example :

(2) SN¹ reaction : The substitution reaction in which a nucleophile reacts with the substrate and the rate of the reaction depends only on the concentration of the substrate is called nucleophilic unimolecular or first order reaction or SN¹ reaction.

Question 52.
Explain, the mechanism of alkaline hydrolysis (reaction with aqueous KOH) of tert-butyl bromide (2-Bromo-2-methylpropane) with energy profile diagram.
OR
Explain only reaction mechanism for alkaline hydrolysis of tert-butyl bromide.
Answer:
(i) Consider alkaline hydrolysis of tert-butyl bromide (2-Bromo-2 methylpropane) with aqueous NaOH or KOH.

(ii) Kinetics of the reaction : Due to steric hindrance of voluminous three methyl groups around carbon, nucleophile OH- cannot attack carbon atom directly. Hence, the reaction takes place in two steps.

Step I : This involves heterolytic fission of C – Br covalent bond in the substrate forming carbocation and Br^–. This is a slow process.

Step II : This step involves attack of nucleophile OH- or carbocation forming C – OH bond and product tert-butyl alcohol. Since it involves ionic charge neutralisation, it is a fast step.

Rate Determining Step (R.D.S.) : Since the first step is a slow step, it is R.D.S., and therefore the rate of the reaction depends on the concentration of only one reactant, (CH₃) C – Br.
Rate = R = k [(CH₃)₃ C – Br] where k is a rate constant of the reaction.

SN¹ reaction : The reaction between tert.butyl bromide and hydroxide ion to form tert.butyl alcohol follows a first-order kinetics. The rate of this reaction depends only on the concentration of one substance (tert-butyl bromide) and is independent of the concentration of alkali added. It is an unimolecular first (1st) order Nucleophilic Substitution reaction denoted as SN reaction.

Stereochemistry and mechanism of the reaction : The reaction takes place in two steps and both the steps involve formation of transition states (T.S.).

T.S. -1 for first step :

In this transition state, C – Br bond is partially broken, so that carbon atom carries partial positive charge (+δ) and Br carries partial negative charge (-δ) which further breaks forming carbocation and Br . Tert-butyl cation (carbocation) has a planar structure and the CH₃ – C – CH₃ bond angle is 120°. It is the intermediate of the reaction. It is unstable. In this step, hybridisation of carbon atom changes from sp³ (tetrahedral geometry) to sp (planar geometry).

T.S. – II for second step :

In this transition state, C – OH bond is partially fonned so that carbon atom carries partial positive charge (+ δ) and OH carries partial negative charge ( -δ) which further forms tert-butyl alcohol.

Formation of a racemic mixture : Since OH^– has equal probability of the attack on carbocation from frontside and from backside, the products obtained are equal. In case of optical active alkyl halide, a racemic mixture is obtained.

Question 53.
Discuss SN² mechanism of methyl bromide using aqueous KOH. Draw energy profile diagram.
OR
Discuss the mechanism of alkaline hydrolysis of methyl bromide or Bromomethane.
Answer:
(1) Consider alkaline hydrolysis of methyl bromide (Bromomethane). CH₃Br with aqueous NaOH or KOH.

(2) Stereochemistry and Kinetics of the reaction iR.D.S.) : This hydrolysis reaction takes place only in one step which is a rate determining step i.e. R.D.S. The rate of hydrolysis reaction depends on the Concentration of CH3Br
and 0H which are present in the R.D.S. of the reaction.
Rate = R = k [CH₃Br] (OH]
where k is rate constant of the reaction.

SN² reaction : The reaction between methyl bromide and hydroxide ion to form methanol follows a second order kinetics, since the rate of the reaction depends on the concentrations of two reacting species, namely methyl bromide and hydroxide ion it is bimolecular second order (2nd) Nucleophilic Substitution reaction denoted by SN².

(3) Mechanism of the reaction :
(i) It is a single step mechanism. The reaction takes place in the following steps :

(ii) Backside attack of the nucleophile : Nucleophile, OH^– attacks carbon atom of CH₃Br from back side i. e. from opposite side to that of the leaving group i.e. Br^– to experience minimum steric repulsion and electrostatic repulsion between the incoming nucleophile (OH^–) and leaving Br^–.

(iii) Transition state : When a nucleophile, OH^– approaches carbon atom of CH₃Br, the potential energy of the system increases until a transition state (T.S.) of maximum potential energy is formed in which C – Br bond is partially broken and C – OH bond is partially formed. The negative charge is equally shared by both incoming nucleophile- OH^– and outgoing, leaving group-Br^–. (Thus, the total negative charge is diffused.)

(iv) In CH₃Br, carbon atom is sp³ -hybridized and CH₃Br molecule is tetrahedral. The hybridisation of carbon atom changes to sp² hybridisation. The transition state contains pentacoordinate carbon having three δ (sigma) bonds in one plane making bond angles of 120° with each other i.e., H₁; H₂ and H₃ atoms lie in one plane while two partial covalent bonds containing Br and OH lie collinear and on opposite sides perpendicular to the plane.

(v) Inversion of configuration : The transition state decomposes fast by the complete breaking of the C-Br bond and the new C-OH bond is formed on the other side. The breaking of C-Br bond and the formation of C-OH bond take place simultaneously. The energy required to break the C-Br bond is partly obtained from the energy released when C-OH bond is formed. The formation of product CH₃OH is accompanied by complete or 100% inversion of configuration forming again sp³-hybridized carbon atom giving tetrahedral CH₃OH molecule. But in this structure the positions of H₂ and H₃ atoms in the reactant (CH₃Br) and in product are on the opposite side. This inversion of configuration is called Walden inversion.

Question 54.
Discuss the factors influencing S_N1 and S_N2 mechanism.
Answer:
(1) Nature of substrate : S_N2 : The transition state (T.S.) of S_N2 mechanism + is pentacoordinate, it is crowded. As a result S_N2 mechanism is favoured in primary halides and least favoured in tertiary halides.

S_N1 : A planar carbocation intermediate is formed in S_N1 reaction. Bulky alkyl groups can be easily accommodated in planar carbocation and it has no steric crowding. As a result SN1 mechanism is favoured in tertiary halides and least favoured in primary halides.

The carbocation intermediate is stabilized by + effect of alkyl substituents and also by hyperconjugation y effect of alkyl substituents containing a-hydrogens. As a result, S_N1 mechanism is favoured in tertiary halides and least favoured in primary halides.

Thus, tertiary alkyl halides undergo nucleophilic substitution by S_N1 mechanism while primary halides follow S_N2 mechanism.

(2) Nucleophilicity of the reagent : A strong nucleophile attacks the substrate faster and favours S_N2 mechanism. The rate of S_N1 mechanism is independent of the nature of nucleophile. Nucleophile does not react in the 1st step (slow step) of S_N1. Nucleophile reacts fast after the carbocation intermediate is formed.

(3) Solvent polarity : (1) S_N1 reaction proceeds more rapidly in polar protic solvents than in aprotic solvent. Polar protic solvent decreases the rate of S_N2 reaction. (2) In S_N2 mechanism, rate depends on substrate as well as nucleophile. A polar solvent stabilizes nucleophile by solvation. Thus solvent deactivates the nucleophile by stabilizing it. Hence, aprotic solvents or solvent of low polarity will favour S_N2 mechanism.

Question 55.
How does relative reactivity for alkaline hydrolysis with respect to S_N2 and S_N1 vary in the following alkyl halides :
(1) Bromomethane
(2) Bromoethane
(3) 2-Bromopropane
(4) 2-Bromo-2-methylpropane ?
Answer:
(A) Relative reactivity for S_N2 mechanism decreases in the order of :

(B) Relative reactivity for S_N1 mechanism decreases in the order of :
2-Bromo-2-methylpropane > 2-Bromopropane > Bromoethane > Bromomethane

Question 56.
Explain with reason the relative order of reactivity of l°/2°/3° alkyl halides by S_N1 mechanism.
Answer:
In alkaline hydrolysis of an alkyl halide by S_N1 mechanism, the formation of carbocation as an intermediate product is involved.

The increasing order of a stability of carbocation is,

The stability order for carbocation is 3° > 2° > 1°.
Therefore the increasing order of reactivity by S_N1 mechanism of alkyl halides is
(1°) primary < (2°) secondary < (3°) tertiary

Question 57.
Which one of the following is more easily hydrolysed in S_N1 and S_N2 reaction by aqueous KOH, C₆H₅ CHCIC₆H₅ and C₆H₅CH₂CI?
Answer:
In S_N1 reaction C₆H₅CHCI C₆H₅ will be more easily hydrolysed than C₆H₅CH₂CI
In S_N2 reaction C₆H₅ CH₂CI will be more easily hydrolysed than C₆H₅CHCIC₆H₅.

Question 58.
Choose the member that will react faster than the following pairs by S_N1 mechanism.
(1) l-bromo-2, 2-dimethyl propane or 2-bromopropane.

Answer:
The reactivity of S_N1 reaction depends on the steric hindrance, in 2-bromopropane, a-carbon atom is attached to two methyl groups suffers greater steric hindrance to nucleophilic attack than l-bromo-2, 2-dimethyl propane. Hence, 2-bromopropane react faster by S_N1 mechanism.

(2) 2-Iodo-2-methyl butane or 2-iodio-3-methyl butane.

Answer:
Since, 2-Iodo-2-methyl butane is a tertiary alkyl halide, it undergoes SN^-1 reaction faster than 2-iodo-3-methyl butane.

(3) 1-Chloro propane or 2-chloropropane.

Answer:
Since, 2-chloropropane is a secondary alkyl halide, it undergoes SN^-1 reaction faster than 1-chloropropane.

(4) 2-Iodo-2-methyl butane or tert-butyl chloride.

Answer:
Since, iodine is a better leaving group than chloride 2-iodo-2-methyl butane undergo SN^-1 reaction faster than tert-butyl chloride.

Question 59.
Write a note on elimination reaction.
OR
Explain dehydrohalogenation reaction.
Answer:
When alkyl halide having at least one β-hydrogen is boiled with alcoholic solution of potassium hydroxide (KOH), an alkene is formed due to elimination of hydrogen atom from β-carbon and halogen atom from α-carbon, is called dehydrohalogenation.

Tertiary butyl bromide when heated with alcoholic solution of potassium hydroxide forms isobutylene.

This reaction is called β-elimination (or 1,2-elimination) reaction as it involves elimination of halogen and a β-hydrogen atom.

As hydrogen and halogen is removed in this reaction it is also known as dehydrohalogenation reaction.

Question 60.
Describe the action of alcoholic potassium hydroxide (aic. KOH) on
(1) ethyl bromide
(2) n-propyl bromide
(3) isopropyl bromide
(4) tert-butyl chlorIde.
Answer:
(1) Ethyl bromide : When ethyl bromide (bromoethane) is heated with alcoholic potassium hydroxide (alcoholic alkali). ethene (gas) is formed by the dehydrobrominaion reaction.

(2) n-PropI bromide : When n-propyl bromide is heated with alcoholic potassium hydroxide, propene is formed.

(3) Isopropvl bromide : When isopropyl bromide (2-bromopropane) is boiled with alcoholic potassium hydroxide, propcne is formed.

(4) Tert-hutyl chloride: When ten-butyl chloride (2-chloro-2-methyl propanc) is hcatcd with alcoholic KOH.

Question 61.
Describe the action of alc.KOH on 2-bromobutane.
When 2-bromobutane is boiled with alc.KOH on 2-bromobutane, a mixture of but-l-ene and but-2-ene is formed.
Answer:

Question 62.
Explain Saytzelf’s rule with suitable example.
Answer:
Saytzcff’s rule : In dehydrohalogenation reaction the preferred product is that alkene which has the greater number of alkyl groups attached to the doubly bonded carbon atoms.

Hence the number of alkyl substituents on doubly bonded Carbon atoms increases, the stability of the alkene giving its major products.

Hence the increasing stability of alkenes is.

There are two types of fi hydrogens (β₁ and β₂) therefore two alkenes are expected.

Question 63.
What is a Grignard reagent ?
Answer:
Grignard reagent : An organometallic compound in which the divalent magnesium is directly linked to an alkyl group (R -) and a halogen atom (X), and has general formula R – Mg – X is called Grignard reagent. OR When alkyl halide is treated with magnesium in dry ether as solvent, it gives alkyl magnesium halide. It is known as Grignard reagent.

The carbon-magnesium bond is highly polar and magnesium-halogen bond is in ionic in nature. Grignard reagent is highly reactive. It is an important reagent and used in the preparation of a large number of organic compounds.

Question 64.
How is Grignard reagent prepared ?
Answer:
Grignard reagent is an alkyl magnesium halide, R – Mg – X obtained by the reaction of alkyl halide R – X with magnesium (Mg) in dry ether.

When an alkyl halide like CH₃I is added from a dropping funnel to a flask containing pieces of pure Mg in pure and dry ether (diethyl ether) and a trace of iodine, Grignard reagent, CH₃ – Mg – I is formed.

Ethyl iodide when treated with magnesium in presence of dry ether forms ethyl magnesium iodide.

Question 65.
Write a note on Grignard reagent.
Answer:
(1) Ethyl bromide : When ethyl bromide (bromoethane) is heated with alcoholic potassium hydroxide (alcoholic alkali). ethene (gas) is formed by the dehydrobrominaion reaction.

(2) n-PropI bromide : When n-propyl bromide is heated with alcoholic potassium hydroxide, propene is formed.

(3) Isopropvl bromide : When isopropyl bromide (2-bromopropane) is boiled with alcoholic potassium hydroxide, propcne is formed.

(4) Tert-hutyl chloride: When ten-butyl chloride (2-chloro-2-methyl propanc) is hcatcd with alcoholic KOH.

Question 66.
Describe the action of water on
(1) methyl magnesium iodide
(2) ethyl magnesium iodide.
Answer:
(1) Methyl magnesium iodide : When methyl magnesium iodide is treated with water, methane is obtained

(2) Ethyl magnesium iodide : When ethyl magnesium iodide is treated with water, ethane is obtained.

Question 67.
Describe the action of ammonia on
(1) ethyl magnesium bromide
(2) n-propyl magnesium chloride.
Answer:
(1) Ethyl magnesium bromide : When ethyl magnesium bromide is treated with ammonia, ethane is formed.

(2) n-Propyl magnesium chloride : When n-propyl magnesium chloride is treated with ammonia, propane is formed.

Question 68.
Explain Wurtz reaction. OR Explain the action of sodium with alkyl halides.
Answer:
(1) When an alkyl halide is treated with metallic sodium in dry ether, the corresponding higher alkane is formed. This is called Wurtz reaction or Wurtz coupling reaction.

(2) In this reaction the alkyl radicals from two molecules of the reacting alkyl halide combine or couple to form the higher alkane.

(3) Thus, methyl bromide reacts with sodium in ether to form ethane (C₂H₆), while ethyl bromide under the same conditions forms n-butane (C₄H₁₀).

(4) If a mixture of two different alkyl halides is treated with Na in dry ether, then a mixture of alkanes is obtained called self coupling products. For example, a mixture of CH₃Br and C₂H₅Br gives propane along with C₂H₆ and C₄H₁₀.

Question 69.
Explain the reaction of haloarene with alkyl halide and sodium metal.
Write a note on Wurtz-Fittig reaction.
Answer:
When an alkyl halide and an aryl halide is treated with sodium metal in dry ether the corresponding alkylarene (alkyl benzene) is formed. The reaction is known as Wurtz-Fittig reaction. This reaction allows alkylation of alkyl halides.

Question 70.
Describe the action of aryl halide on sodium metal.
Answer:
Aryl halide reacts with sodium metal in dry ether, biphenyl is formed. This reaction is known as Fittig reaction.

Question 71.
Identify the product A of following reaction.

Answer:

Question 72.
Explain the following substitution reactions of chlorobenzene :
(1) Halogenation
(2) Nitration
(3) Sulphonation.
Answer:
(1) Halogenation : When chlorobenzene is reacted with chlorine in presence of anhydrous ferric chloride, a mixture of ortho and para-dichlorobenzene (major product) is formed.

(2) Nitration : When chlorobenzene is heated with nitrating mixture (cone, nitric acid -I- cone, sulphuric acid) a mixture of l-chloro-4-nitro benzene (major product) and l-chloro-2-nitrobenzene is formed.

(3) Sulphonation : When chlorobenzene is heated with concentrated sulphuric acid, a mixture of 4-chlorobenzene sulphonic acid (major product) and 2-chlorobenzene sulphonic acid is formed.

Question 73.
Describe the action of the following on chlorobenzene :
(1) Methyl chloride in the presence of anhydrous AICI₃
(2) Acetyl chloride in the presence of anhydrous AICI₃.
Answer:
(1) Methyl chloride in the presence of anhydrous AICI₃ : When chlorobenzene is treated with methyl chloride in the presence of anhydrous AICI₃, a mixture of l-chloro-4-methyl benzene (major product) and l-chloro-2-methyl benzene is formed. Since, the alkyl group is introduced in the benzene ring, the reaction is termed as Friedel Craft’s alkylation.

(2) Acetyl chloride in the presence of anhydrous AICI₃ : When chlorobenzene is reacted with acetyl chloride in the presence of anhydrous AICI₃, a mixture of 2-chloro acetophenone and 4-chloro acetophenone (major product) is formed. Since, the acetyl group is introduced in the benzene ring, the reaction is termed as Friedel Craft’s acylation.

Question 74.
Write a note on Friedel Craft’s reaction.
Answer:

(1) Methyl chloride in the presence of anhydrous AICI₃ : When chlorobenzene is treated with methyl chloride in the presence of anhydrous AICI₃, a mixture of l-chloro-4-methyl benzene (major product) and l-chloro-2-methyl benzene is formed. Since, the alkyl group is introduced in the benzene ring, the reaction is termed as Friedel Craft’s alkylation.

(2) Acetyl chloride in the presence of anhydrous AICI₃ : When chlorobenzene is reacted with acetyl chloride in the presence of anhydrous AICI₃, a mixture of 2-chloro acetophenone and 4-chloro acetophenone (major product) is formed. Since, the acetyl group is introduced in the benzene ring, the reaction is termed as Friedel Craft’s acylation.

Question 75.
Convert 1-chlorobutane into the following compounds :

Answer:
(1) 1-Chlorobutane to butan-l-ol :

(2) 1-Chlorobutane to 1-iodobutane :

(3) 1-Chlorobutane to n-butyl cyanide (CH₃ – CH₂ – CH₂ – CH₂ – CN) :

(4) l-Chlorobutane to

Question 76.
Predict the expected product of substitution reactions :
(1) Isobutyl chloride + sodium ethoxide.
Answer:

(2) n-butyl chloride + sodium.
Answer:

(3) 1-chloropropane + aq. potassium hydroxide.
Answer:

(4) Aniline + NaNO₂/HCl.
Answer:

Question 77.
Write the products:

Question 78.
Identify A and B in the following :

Answer:

Answer:

Answer:

Answer:

Answer:

Answer:

Question 79.
State the uses of the following compounds :
(1) Dichloromethane (CH₂CI₂)
(2) Trichloromethane or Chloroform (CHCI₃)
(3) Tetrachloromethane or carbon tetrachloride (CCI₄)
(4) Iodoform (CHI₃)
(5) Freons
(6) DDT (p, p’-Dichlorodiphenyl trichloroethane).
Answer:
(1) Dichloromethane (CH₂CI₂) :

• Dichloromethane dissolves wide range of organic compounds, hence it is used as solvent for many chemical reactions.
• It is used as a solvent as a paint remover and degreaser.
• It is used as propellant in aerosols and as a fumigant pesticide for grains and strawberries.
• It is used to decaffinate tea or coffee.

(2) Trichloromethane or Chloroform (CHCI₃) :

• Chloroform in the production of chlorofluoromethane, freon refrigerant R-22.
• It is used as solvent in pharmaceuticals, pesticides, gums, fats, resins and dye industry.
• It is a good source of dichlorocarbene species.

(3) Tetrachloromethane or carbon tetrachloride (CCI₄) :

• Carbon tetrachloride is used in the manufacture of refrigerants.
• It is used as a dry cleaning agent and as a pesticide for stored grains.
• It is very useful solvent for oils, fats and resins. It serves as a source of chlorine.

(4) Iodoform (CHI₃) :

• Iodoform is used as antiseptic, dressing of wounds and sores.
• On small scale it is used as disinfectant.

(5) Freons :

• Freons are widely used as propellants in aerosol, products of food, cosmetics and pharmaceutical industries.
• Freons containing bromine in their molecules are used as fire extinguishers.
• They are used in aerosol insecticides, solvent for cleaning clothes and metallic surfaces.
• It is used as foaming agents in the preparation of foamed plastics and in production of certain fluorocarbons.
• It is used as refrigerants and air conditioning purposes.

Question 80.
State the environmental effects of the following compounds :
(1) Dichloromethane (CH₂CI₂)
(2) Trichloromethane or chloroform (CHCI₃)
(3) Tetrachloromethane or carbon tetrachloride (CCI₄)
(4) Iodoform (CHI₃)
(5) Freon.s (CCI₂F₂, CCI₃F, CHCIF₂)

Answer:
(1) Dichloromethane (CH₂CI₂) :

• Higher levels of dichloromethane in air causes nausea, numbness in fingers and toes, dizziness.
• Lower levels of dichloromethane causes impaired vision and hearing.
• Direct contact with eyes can damage cornea.

(2) Trichloromethane or chloroform (CHCI₃) :

• When chloroform is exposed to air in the presence of sunlight, it slowly oxidised to phosgene, a poisonous compound, therefore it is stored in dark, amber coloured bottles.
• Chloroform vapour when inhaled for a short time causes dizziness, headache and fatigue and if inhaled for a long time affects central nervous system.

(3) Tetrachloromethane or carbon tetrachloride (CCI₄) :

• Exposure to carbon tetrachloride causes eye irritation, damages nerve cells, vomiting sensation, dizziness, unconciousness or death. Long exposure to chloroform may affect liver.
• When mixed with air it causes depletion of the ozone layer, which affects human skin leading to cancer.

(4) Iodoform (CHI₃) : Iodoform has a strong smell. It causes irritation to skin and eyes. It may cause respiratory irritation or breathing difficulty, dizziness, nausea, depression of central nervous system, visual disturbance.

(5) Freons (CCI₂F₂, CCI₃F, CHCIF₂) :

• Freon as refrigerant causes ozone depletion.
• Freons have low toxicity and low biological activity.
• Freons from propane group are more toxic in nature.
• Regular large inhalation of freon results in breathing problems, organ damage, loss of consciousness.

(6) DDT :

• DDT is not readily metabolised by animals.
• It is deposited and stored in fatty tissues.
• Exposure to high doses of DDT may cause vomiting, tremors or shakiness.
• Laboratory animal studies showed adverse effect of DDT on liver and reproduction.
• DDT is a pressistent organic pollutant, readily absorbed in soils and tends to accumulate in the ecosystem.
• When dissolved in oil or other lipid, it is readily absorbed by the skin. It is resistant to metabolism.
• There is a ban on use of DDT due to all these adverse effects.

Question 81.
What is the chemical name of freon?
Answer:
The chemical name of freon is Dichlorodifluoromethane.

Question 82.
What is the chemical name of DDT ?
Answer: The chemical name of DDT is p, p’-Dichlorodiphenyltrichloroethane.

Activity :
(1) Collect detailed information about Freons and their uses.
(2) Collect information about DDT as a persistent pesticide.
Reference books :
(1) Organic chemistry by Morrison, Boyd, Bhattacharjee, 7th edition, Pearson.
(2) Organic chemistry by Finar, Vol 1, 6th edition, Pearson

Multiple Choise Questions

Question 83.
Select and write the most appropriate answer from the given alternatives for each sub-question :

Question 1.

Answer:
(b) CH₃ – CH₂ – CH₂ – I

Question 2.
The rate of S_N2 reaction depends on the concentra¬tion of
(a) only the substrate
(b) only the reagent
(c) both the substrate and the reagent
(d) neither the substrate nor the reagent
Answer:
(c) both the substrate and the reagent

Question 3.
In S_N2 reaction, the hydrolysis of alkyl halide shows
(a) the retention of configuration
(b) the inversion of configuration
(c) both retention and inversion of configuration
(d) no change in the configuration
Answer:
(b) the inversion of configuration

Question 4.
The one step exothermic reaction is
(a) S_N1
(b) S_N2
(C) S_N
(d) S2_N
Answer:
(b) S_N2

Question 5.
Which of the following is correct about S_N2 mechanism?
(a) Two step reaction
(b) Complete inversion of configuration
(c) Formation of carbonium ion
(d) Favoured by polar solvent
Answer:
(b) Complete inversion of configuration

Question 6.
Which of the following is not a nucleophile?
(a) Ammonia
(b) Ammonium ion
(c) Primary amine
(d) Secondary amine
Answer:
(d) Secondary amine

Question 7.
Which of the following undergoes nucleophilic substitution exclusively by S_N2 mechanism ?
(a) ethyl chloride
(b) isopropyl chloride
(c) chlorobenzene
(d) benzyl chloride
Answer:
(d) benzyl chloride

Question 8.
Which of the following is most reactive towards nucleophilic substitution reaction ?
(a) CH₂ = CH – CI
(b) CH₃CH = CHCI
(c) C₆H₅CI
(d) CICH₂ – CH = CH₂
Answer:
(d) CICH₂ – CH = CH₂

Question 9.
The stability order of carbocation is
(a) 2° > 3° > 1°
(b) 3° > 2° > 1°
(c) 3° > 1° > 2°
(d) 1° > 3° > 2°
Answer:
(b) 3° > 2° > 1°

Question 10.

(a) ethane
(b) propane
(c) n-butane
(d) n-pentane
Answer:
(c) n-butane

Question 11.
Which of the following characteristic properties of the enantiomers is correct?
(a) The enantiomers possess same physical and chemical properties
(b) The enantiomers are optically active compounds
(c) The enantiomers have different optical rotations
(d) All of these
Answer:
(d) All of these

Question 12.
The optically inactive compound is
(a) glucose
(b) lactic acid
(c) isopropyl alcohol
(d) 2-bromo butane
Answer:
(c) isopropyl alcohol

Question 13.
A compound with the molecular formula CH₂OH(CHOH)₃CH₂OH has optically active forms
(a) 3
(b) 4
(c) 6
(d) 8
Answer:
(d) 8

Question 14.
A racemic mixture consists of
(a) equal amount of d and l isomers
(b) unequal amounts of d and / isomers
(c) unknown amounts of d and / isomers
(d) only d isomers
Answer:
(a) equal amount of d and l isomers

Question 15.
Which of the following compounds is not optically active ?
(a) Lactic acid
(b) Secondary butyl chloride
(c) n-propyl iodide
(d) Glucose
Answer:
(c) n-propyl iodide

Question 16.
Which of the following compounds shows optical activity ?
(a) n-butyl chloride
(b) isobutyl chloride
(c) sec-butyl chloride
(d) t-butyl chloride
Answer:
(c) sec-butyl chloride

Question 17.
The major product of the following reaction is

Answer:
(c)

Question 18.
The above reaction is known as

(a) Wurtz-Fittig reaction
(b) Friedel Craft’s reaction
(c) Sandmeyer’s reaction
(d) Swarts reaction
Answer:
(b) Friedel Craft’s reaction

Question 19.
Iodoform is used as
(a) an anaesthetic
(b) an antiseptic
(c) an analgesic
(d) an antibiotic
Answer:
(b) an antiseptic

Question 20.
p, p’-dichlorodiphenyl trichloroethane is used as
(a) insecticide
(b) anaesthetic
(c) antiseptic
(d) refrigerant
Answer:
(a) insecticide

Question 21.
The order of reactivity in nucleophilic substitution reaction is
(a) CH₃F < CH₃C1 < CH₃I < CH₃Br
(b) CH₃F < CH₃C1 < CH₃Br < CH₃I
(c) CH₃F < CH₃Br < CH₃C1 < CH₃I
(d) CH₃I < CH₃Br < CH₃C1 < CH₃F
Answer:
(b) CH₃F < CH₃C1 < CH₃Br < CH₃I

Question 22.
Racemate is
(a) optically active
(b) optically dextro rotatory
(c) optically inactive
(d) optically laevorotatory
Answer:
(c) optically inactive

Question 23.
The number of asymmetric carbon atoms in glucose are
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(c) 4

Question 24.
The geometry of carbon lum ion is
(a) Tetrahedral
(b) planar
(c) linear
(d) pyramidal
Answer:
(b) planar

Question 25.
In its nucleophilic substitution reaction, aryl halide resembles
(a) Vinyl chloride
(b) allyl chloride
(e) Benzyl chloride
(d) ethyl chloride
Answer:
(a) Vinyl chloride

Question 26.
The weakest C-Cl bond is present in

Answer:
(d)

Question 27.
Which alkyl halide among the following com¬pounds has the highest boiling point ?
(a) (CH₃)₃CCI
(b) CH₃CH₂CH₂CH₂CI
(c) CH₃CH₂CH₂C1
(d) CH₃CH(CH₃)CH₂CI
Answer:
(b) CH₃CH₂CH₂CH₂CI

Question 28.
It is difficult to break C-Cl bond in CH₂ = CH – CI due to
(a) Hyper conjugation
(b) Resonance
(c) Electromeric effect
(d) Inductive effect
Answer:
(b) Resonance

Question 29.
Which one of the following when heated with metallic sodium will not give the corresponding alkane ?

Answer:
(c)

Question 30.
The most reactive alkyl halide towards SN2 reac¬tion is
(a) CH₃X
(b) R₃CX
(C) R₂CHX
(d) RCH₂X
Answer:
(a) CH₃X

Question 31.
The number of electrons surrounding the carbon- ium ion is
(a) 6
(b) 8
(c) 10
(d) 7
Answer:
(a) 6

Question 32.
The lowest stability of carbocation among the compounds

Answer:
(a)

Question 33.
Carbon atom in methyl carbocation contains how many pairs of electrons?
(a) 8
(b) 4
(c) 3
(d) 5
Answer:
(b) 4

Question 34.
The optically inactive compound is
(a) Glucose
(b) Lactic acid
(c) 2-Chlorobutane
(d) 2-Chloropropane
Answer:
(d) 2-Chloropropane

Question 35.
The hydrogen halide which does not obey Markownikv rule in presence of peroxide is
(a) HC1
(b) HBr
(c) HF
(d) HI
Answer:
(b) HBr

Question 36.
Which one of the following is NOT used to prepare alkyl halide from an alcohol ?
(a) SOCl₂
(b) PC1₃
(c) HC1 + ZnCl₂
(d) NaCl
Answer:
(d) NaCl

Question 37.
The total number of electrons present in the central carbon atom of a free radical is
(a) 7
(b) 8
(c) 9
(d) 6
Answer:
(a) 7

Question 38.
In which of the following pairs both are nucleophiles ?
(a) BF₃, AICI₃
(b) NO⁺₂, Cl^–
(c) CN^–, NH₃
(d) Br⁺, BC1₃
Answer:
(c) CN^–, NH₃

Question 39.
Which one of the following alkane is NOT formed in Wurtz reaction ?
(a) Methane
(b) Ethane
(c) Propane
(d) Butane
Answer:
(a) Methane

Question 40.
Which of the following groups has highest priority according to R, S convention?
(a) CH₂OH
(b) COOH
(c) COCH₃
(d) COOCH₃
Answer:
(d) COOCH₃

Question 41.
The halogen atom in aryl halides is
(a) o- and p-di reefing
(b) m-directing
(c) o, m and p-di reefing
(d) only m-directing
Answer:
(a) o- and p-di reefing

Question 42.
Chlorobenzene can be obtained by benzene diazonium chloride by
(a) Friedel Craft’s reaction
(b) Wurtz reaction
(c) Gatterman’s reaction
(d) Fittig reaction
Answer:
(c) Gatterman’s reaction

Question 43.
Which of the following carbocations is least stable ?


Answer:
(c)

Question 44.
But-l-ene on reaction with HCI in the presence of sodium peroxide yields
(a) n-butyl chloride
(b) isobutyl chloride
(c) secondary butyl chloride
(d) tertiary butyl chloride
Answer:
(c) secondary butyl chloride

Question 45.
Carbon tetrachloride is used as
(a) anaesthetic
(b) antiseptic
(c) dry cleaning agent
(d) fire extinguisher
Answer:
(c) dry cleaning agent

Question 46.
Identify the product D in the following sequence of reactions :

(a) 2, 2-dimethyl butane
(b) 2, 3-dimethyl butane
(C) hexane
(d) 2, 4-dimethylpentane
Answer:
(b) 2, 3-dimethyl butane

Question 47.
The preparation of alkyl fluoride from alkyl chlor ide, in presence of metallic fluorides is known as
(a) Williamson’s reaction
(b) Finkeistein reaction
(c) Swarts reaction
(d) Wurlz reaction
Answer:
(c) Swarts reaction

Question 48.
UPAC name of the following compound is

(a) 3-Bromo-3, 4-dimetbyiheptane
(b) 3,4-dimethyl-3-bromoheptane
(c) 5-Bromo-4,5-dimethylheptane
(d) 4,5-dimethyl-5-bromoheptane
Answer:
(a) 3-Bromo-3, 4-dimetbyiheptane

Balbharti Maharashtra State Board 12th Chemistry Textbook Solutions Chapter 2 Solutions Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Chemistry Solutions Chapter 2 Solutions

1. Choose the most correct answer.

Question i.
The vapour pressure of a solution containing 2 moles of a solute in 2 moles of water (vapour pressure of pure water = 24 mm Hg) is
(a) 24 mm Hg
(b) 32 mm Hg
(c) 48 mm Hg
(d) 12 mm Hg
Answer:
(d) 12 mm Hg

Question ii.
The colligative property of a solution is
(a) vapour pressure
(b) boiling point
(c) osmotic pressure
(d) freezing point
Answer:
(c) osmotic pressure

Question iii.
In calculating osmotic pressure the concentration of solute is expressed in
(a) molarity
(b) molality
(c) mole fraction
(d) mass per cent
Answer:
(a) molarity

Question iv.
Ebullioscopic constant is the boiling point elevation when the concentration of solution is
(a) 1 m
(b) 1 M
(c) 1 mass%
(d) 1 mole fraction of solute
Answer:
(a) 1 m

Question v.
Cryoscopic constant depends on
(a) nature of solvent
(b) nature of solute
(c) nature of solution
(d) number of solvent molecules
Answer:
(a) nature of solvent

Question vi.
Identify the correct statement
(a) vapour pressure of solution is higher than that of pure solvent.
(b) boiling point of solvent is lower than that of solution
(c) osmotic pressure of solution is lower than that of solvent
(d) osmosis is a colligative property.
Answer:
(b) boiling point of solvent is lower than that of solution

Question vii.
A living cell contains a solution which is isotonic with 0.3 M sugar solution. What osmotic pressure develops when the cell is placed in 0.1 M KCl solution at body temperature ?
(a) 5.08 atm
(b) 2.54 atm
(c) 4.92 atm
(d) 2.46 atm
Answer:
(c) 4.92 atm

Question viii.
The osmotic pressure of blood is 7.65 atm at 310 K. An aqueous solution of glucose isotonic with blood has the percentage (by volume)
(a) 5.41%
(b) 3.54%
(c) 4.53%
(d) 53.4%
Answer:
(a) 5.41%

Question ix.
Vapour pressure of a solution is
(a) directly proportional to the mole fraction of the solute
(b) inversely proportional to the mole fraction of the solute
(c) inversely proportional to the mole fraction of the solvent
(d) directly proportional to the mole fraction of the solvent
Answer:
(d) directly proportional to the mole fraction of the solvent

Question x.
Pressure cooker reduces cooking time for food because
(a) boiling point of water involved in cooking is increased
(b) heat is more evenly distributed in the cooking space
(c) the higher pressure inside the cooker crushes the food material
(d) cooking involves chemical changes helped by a rise in temperature
Answer:
(a) boiling point of water involved in cooking is increased

Question xi.
Henry’s law constant for a gas CH₃Br is 0.159 mol dm^-3 atm at 250°C. What is the solubility of CH₃Br in water at 25 °C and a partial pressure of 0.164 atm?
(a) 0.0159 mol L^-1
(b) 0.164 mol L^-1
(c) 0.026 M
(d) 0.042 M
Answer:
(c) 0.026 M

Question xii.
Which of the following statement is NOT correct for 0.1 M urea solution and 0.05 M sucrose solution ?
(a) osmotic pressure exhibited by urea solution is higher than that exhibited by sucrose solution
(b) urea solution is hypertonic to sucrose solution
(c) they are isotonic solutions
(d) sucrose solution is hypotonic to urea solution
Answer:
(c) they are isotonic solutions

2. Answer the following in one or two sentences

Question i.
What is osmotic pressure ?
Answer:
(1) Definition : The osmotic pressure is defined as the excess mechanical pressure required to be applied to a solution separated by a semipermeable membrane from pure solvent or a dilute solution to prevent the osmosis or free passage of the solvent molecules at a given temperature. The osmotic pressure is a colligative property.

Osmosis and osmotic pressure

(2) Explanation : Consider an inverted thistle funnel on the mouth of which a semipermeable membrane is firmly fastened. It is filled with the experimental solution and immersed in a solvent like water. As a result, solvent molecules pass through the membrane into the solution in the funnel causing rising of level in the arm of thistle funnel. This increases the hydrostatic pressure. At a certain stage this rising level stops indicating an equilibrium between the rates of flow of solvent molecules from solvent to solution and from solution to solvent. The hydrostatic pressure at this stage represents osmotic pressure of the solution in the thistle funnel.

Question ii.
A solution concentration is expressed in molarity and not in molality while considering osmotic pressure. Why ?
Answer:

1. While calculating osmotic pressure by equation, π = CRT, the concentration is expressed in molarity but not in molality.
2. This is because the measurements of osmotic pressure are made at a certain constant temperature.
3. Molarity depends upon temperature but molality is independent of temperature.
4. Hence in osmotic pressure measurements, concentration is expressed in molarity.

Question iii.
Write the equation relating boiling point elevation to the concentration of solution.
Answer:
The elevation in the boiling point of a solution is directly proportional to the molal concentration (expressed in mol kg^-1) of the solution.
Hence, if ΔT_b is the elevation in the boiling point of a solution of molal concentration m then,
ΔT_b ∝ m
∴ ΔT_b = K_b m
where K_b is a proportionality constant.
If m = 1 molal,
ΔT_b = K_b
K_b is called the ebullioscopic constant or molal elevation constant. K_b is characteristic of the solvent.

Question iv.
A 0.1 m solution of K₂SO₄ in water has freezing point of -0.43 °C. What is the value of van’t Hoff factor if K_f for water is 1.86 K kg mol^-1?
Answer:
Given : m = 0.1 m, ΔT_f = 0 – (-0.43) = 0.43 °C
K_f = 1.86 K kg mol^-1, i = ?
ΔT_f = i × K_f × m
∴ i = \(\frac{\Delta T_{\mathrm{f}}}{K_{\mathrm{f}} \times m}=\frac{0.43}{1.86 \times 0.1}\) = 2.312
van’t Hoff factor = i = 2.312

Question v.
What is van’t Hoff factor?
Answer:
Definition of the van’t Hoff factor, i : It is defined as a ratio of the observed colligative property of the solution to the theoretically calculated colligative property of the solution without considering molecular change.

The van’t Hoff factor can be represented as,

This colligative property may be the lowering of vapour pressure of a solution, the osmotic pressure, the elevation in the boiling point or the depression in the freezing point of the solution. Hence,

• When the solute neither undergoes dissociation or association in the solution, then, i = 1
• When the solute undergoes dissociation in the solution, then, i > 1
• When the solute undergoes association in the solution, then i < 1

From the value of the van’t Hoff factor, the degree of dissociation of electrolytes, degree of association of nonelectrolytes can be obtained.

van’t Hoff factor gives the important information about the solute molecules in the solution and chemical bonding in them.

Question vi.
How is van’t Hoff factor related to degree of ionization?
Answer:
Consider 1 dm³ of a solution containing m moles of an electrolyte AxBy. The electrolyte on dissociation gives x number of A^y+ ions and y number of B^x- ions. Let α be the degree of dissociation.

At equilibrium,
AxBy ⇌ xA^y+ + yB^x-
For 1 mole of electrolyte : 1 – α,  xα,  yα
and For ‘m’ moles of an electrolyte : m(1 – α), mxα, myα are the number of particles.
Total number of moles at equilibrium, will be,
Total moles = m(1 – α) + mxα + myα
= m[(1 – α) + xα + yα]
= m[1 + xα + yα – α]
= m[1 + α(x + y – 1)]

The van’t Hoff factor i will be,

If total number of ions from one mole of electrolyte is denoted by n, then (x + y) = n
∴ i = 1 + α(n – 1)
∴ α(n – 1) = i – 1
∴ α = \(\frac{i-1}{n-1}\) ……(1)
This is a relation between van’t Hoff factor i and degree of dissociation of an electrolyte.

Question vii.
Which of the following solutions will have higher freezing point depression and why ?
a. 0.1 m NaCl b. 0.05 m Al₂(SO₄)₃
Answer:
(1) Freezing point depression is a colligative property, hence depends on the number of particles in the solution.
(2) More the number of particles in the solution, higher is the depression in freezing point.
(3) The number of particles (ions) from electrolytes are,

(4) Therefore Al₂(SO₄)₃ solution will have higher freezing point depression.

Question viii.
State Raoult’s law for a solution containing a nonvolatile solute.
Answer:
Statement of Raoult’s law : The law states that the vapour pressure of a solvent over the solution of a nonvolatile solute is equal to the vapour pressure of the pure solvent multiplied by mole fraction of the solvent at constant temperature.

Question ix.
What is the effect on the boiling point of water if 1 mole of methyl alcohol is added to 1 dm³ of water? Why?
Answer:

• The boiling point of water (or any liquid) depends on its vapour pressure.
• Higher the vapour pressure, lower is the boiling point.
• When 1 mole of volatile methyl alcohol is added to 1 dm³ of water, its vapour pressure is increased decreasing the boiling point of water.

Question x.
Which of the four colligative properties is most often used for molecular mass determination? Why?
Answer:

1. Since osmotic pressure has large values, it can be measured more precisely.
2. The osmotic pressure can be measured at a suitable constant temperature.
3. The molecular masses can be measured more accurately.
4. Therefore, it is more useful to determine molecular masses of expensive substances by osmotic pressure.

3. Answer the following.

Question i.
How vapour pressure lowering is related to a rise in boiling point of solution?
Answer:
(1) The boiling point of a liquid is the temperature at which the vapour pressure of the liquid becomes equal to the external pressure, generally 1 atm (101.3 × 10³ Nm^-2).

(2) When a liquid is heated, its vapour pressure rises till it becomes equal to the external pressure.
If the liquid has a low vapour pressure, it has a higher boiling point.

Vapour pressure curve showing elevation in boiling point

(3) When a nonvolatile solute is added to a solvent, its vapour pressure decreases, hence the boiling point increases.
This is explained by graphical representation of the vapour pressure and the boiling point of the pure solvent and the solution.

If T₀ and T are the boiling points of a pure solvent and a solution, then the elevation in the boiling point is given by,
ΔT_b = T – T₀
The curve AB, represents the variation in the vapour pressure of a pure solvent with temperature while curve CD represents the variation in the vapour pressure of the solution.

(4) This elevation in the boiling point is proportional to the lowering of the vapour pressure, i.e., P₀ – P, where P₀ and P are the vapour pressures of the pure solvent and the solution.
[ΔT_b ∝ (P₀ – P) or ΔT_b ∝ ΔP]

Question ii.
What are isotonic and hypertonic solutions?
Answer:
(1) Isotonic solutions : The solutions having the same osmotic pressure at a given temperature are called isotonic solutions.

Explanation : If two solutions of substances A and B contain n_A and n_B moles dissolved in volume V (in dm³) of the solutions, then their concentrations are,
C_A = \(\frac{n_{\mathrm{A}}}{V}\) (in mol dm^-3) and
C_B = \(\frac{n_{\mathrm{B}}}{V}\) (in mol dm^-3)

If the absolute temperature of both the solutions is T, then by the van’t Hoff equation,
π_A = C_ART and π_B = C_BRT, where π_A and π_B are their osmotic pressures.
For the isotonic solutions,
π_A = π_B
∴ C_A = C_B
∴ \(\frac{n_{\mathrm{A}}}{V}=\frac{n_{\mathrm{B}}}{V}\)
∴ n_A = n_B
Hence, equal volumes of the isotonic solutions at the same temperature will contain equal number of moles (hence, equal number of molecules) of the substances.

(2) Hypertonic solutions : When two solutions have different osmotic pressures, then the solution having higher osmotic pressure is said to be a hypertonic solution with respect to the other solution.

Explanation : Consider two solutions of substances A and B having osmotic pressures π_A and π_B. If π_B is greater than π_A, then the solution B is a hypertonic solution with respect to the solution A.
Hence, if C_A and C_B are their concentrations, then C_B > C_A. Hence, for equal volume of the solutions, n_B > n_A.

Question iii.
A solvent and its solution containing a nonvolatile solute are separated by a semipermable membrane. Does the flow of solvent occur in both directions? Comment giving reason.
Answer:

1. When a solvent and a solution containing a non-volatile solute are separated by a semipermeable membrane, there arises a flow of solvent molecules from solvent to solution as well as from solution to solvent.
2. Due to higher vapour pressure of solvent than solution, the rate of flow of solvent molecules from solvent to solution is higher.
3. As more and more solvent passes into solution due to osmosis, the solvent content increases, and the rate of backward flow increases.
4. At a certain stage an equilibrium is reached where both the opposing rates become equal attaining an equilibrium.

Question iv.
The osmotic pressure of CaCl₂ and urea solutions of the same concentration at the same temperature are respectively 0.605 atm and 0.245 atm. Calculate van’t Hoff factor for CaCl₂.
Answer:
Given : π_Cacl2 = 0.605 atm;
π_Urea = 0.245 atm
For urea solution, van’t Hoff factor, i = 1
π_Cacl2 = i × (CRT)_Cacl2
π_Urea = (CRT)_Urea

van’t Hoff factor = i = 2.47

Question v.
Explain reverse osmosis.
Answer:
Reverse osmosis : The phenomenon of the passage of solvent like water under high pressure from the concentrated aqueous solution like seawater into pure water through a semipermeable membrane is called reverse osmosis.

The osmotic pressure of seawater is about 30 atmospheres. Hence when pressure more than 30 atmospheres is applied on the solution side, regular osmosis stops and reverse osmosis starts. Hence pure water from seawater enters the other side of pure water.

Purification of seawater by reverse osmosis

For this purpose of suitable semipermeable membrane is required which can withstand high pressure conditions over a long period.
This method is used successfully in Florida since 1981 producing more than 10 million litres of pure water per day.

Question vi.
How molar mass of a solute is determined by osmotic pressure measurement?
Answer:
Consider V dm³ (litres) of a solution containing W₂ mass of a solute of molar mass M₂ at a temperature T.
Number of moles of solute, n₂ = \(\frac{W_{2}}{M_{2}}\)
The osmotic pressure π is given by,
π = \(\frac{W_{2} R T}{M_{2} V}\)
∴ M₂ = \(\frac{W_{2} R T}{\pi V}\)
By measuring osmotic pressure of a solution, the molar mass of a solute can be calculated.
Since osmotic pressure can be measured more precisely, it is widely used to measure molar masses of the substances.

Question vii.
Why vapour pressure of a solvent is lowered by dissolving a nonvolatile solute into it?
Answer:
Lowering of vapour pressure of a solution :
When a nonvolatile solute is added to a pure solvent, the surface area is covered by the solute molecule decreasing the rate of evaporation, hence its vapour pressure decreases. This decrease in vapour pressure is called lowering of vapour pressure.

If P₀ is the vapour pressure of a pure solvent (liquid) and P is the vapour pressure of the solution, where P < P₀, then, (P₀ – P) is the lowering of the vapour pressure.

Question viii.
Using Raoult’s law, how will you show that ∆P = \(\boldsymbol{P}_{1}^{0}\)x₂ ? Where x₂ is the mole fraction of solute in the solution and \(\boldsymbol{P}_{1}^{0}\) vapour pressure of pure solvent.
Answer:
If x₁ and x₂ are the mole fractions of solvent and solute respectively, then
x₁ + x₂
By Raoult’s law,
P = x₁ × P₀
where P⁰ is the vapour pressure of a pure solvent and P is the vapour pressure of the solution at given temperature.

Question ix.
While considering boiling point elevation and freezing point depression a solution concentration is expressed in molality and not in molarity. Why?
Answer:

• Boiling point elevation and freezing point depression involve temperature changes, (ΔT_b and ΔT_f).
• Since molarity depends on temperature but molality is independent of temperature we use molality and not molarity in considering boiling point elevation and freezing point depression.

Question 4.
Derive the relationship between degree of dissociation of an electrolyte and van’t Hoff factor.
Answer:
Consider 1 dm³ of a solution containing m moles of an electrolyte AxBy. The electrolyte on dissociation gives number of A^y+ ions and y number of B^x- ions. Let α be the degree of dissociation.

At equilibrium,
AxBy ⇌ xA^y+ + yB^x-
For 1 mole of electrolyte : 1 – α, xα, yα and
For ‘m’ moles of an electrolyte : m(1 – α), mxα, myα are the number of particles.
Total number of moles at equilibrium, will be,
Total moles = m(1 – α) + mxα + myα
= m[(1 – α) + xα + yα]
= m[1 – xα + yα – α]
= m[1 + α(x + y – 1)]
The van’t Hoff factor i will be,

If total number of ions from one mole of electrolyte is denoted by n, then (x + y) = n
∴ i = 1 + α(n – 1)
∴ α(n – 1) = i – 1
∴ α = \(\frac{i-1}{n-1}\) ……..(1)
This is a relation between van’t Hoff factor i and degree of dissociation of an electrolyte.

Question 5.
What is effect of temperature on solubility of solids in water? Give examples.
Answer:
The solubility of a solid solute depends upon temperature.

Variation of solubilities of some ionic solids with temperature

• Generally rise in temperature increases the solubility. This is due to expansion of holes or empty spaces in the liquid solvent. Generally 10 °C rise in temperature, increases the solubility of solids two fold.
• Dissolution process may be endothermic or exothermic.
• The solubility of the substances like NaBr, NaCl, KCl, etc. changes slightly with the increase in temperature.
• The solubility of the salts like NaNO₃, KNO₃, KBr, etc. increases appreciably with the increase in temperature.
• The solubility of Na₂SO₄ first increases and after 30 °C decreases with the increase in temperature.

This variation in solubility with temperature can be used to separate the salts from the mixture by fractional crystallisation.

Question 6.
Obtain the relationship between freezing point depression of a solution containing nonvolatile nonelectrolyte and its molar mass.
Answer:
The freezing point depression, ΔT_f of a solution is directly proportional to molality (m) of the solution.
∴ ΔT_f ∝ m
∴ ΔT_f = K_f m
where Kf is a molal depression constant.
The molality of a solution is given by,

If W₁ grams of a solvent contain W₂ grams of a solute of the molar mass M₂, then the molality m of the solution is given by,

If the weights are expressed in kg then,
ΔT_f = K_f × \(\frac{W_{2}}{W_{1} M_{2}}\)
The unit of K_f is K kg mol^-1
Hence, from the measurement of the depression in the freezing point of the solution, the molar mass of the substance can be determined.

Question 7.
Explain with diagram the boiling point elevation in terms of vapour pressure lowering.
Answer:
(1) The boiling point of a liquid is the temperature at which the vapour pressure of the liquid becomes equal to the external pressure, generally 1 atm (101.3 × 10³ Nm^-2).

(2) When a liquid is heated, its vapour pressure rises till it becomes equal to the external pressure.
If the liquid has a low vapour pressure, it has a higher boiling point.

Vapour pressure curve showing elevation in boiling point

(3) When a nonvolatile solute is added to a solvent, its vapour pressure decreases, hence the boiling point increases.
This is explained by graphical representation of the vapour pressure and the boiling point of the pure solvent and the solution.

If T₀ and T are the boiling points of a pure solvent and a solution, then the elevation in the boiling point is given by,
ΔT_b = T – T₀
The curve AB, represents the variation in the vapour pressure of a pure solvent with temperature while curve CD represents the variation in the vapour pressure of the solution.

(4) This elevation in the boiling point is proportional to the lowering of the vapour pressure, i.e., P₀ – P, where P₀ and P are the vapour pressures of the pure solvent and the solution.
[ΔT_b ∝ (P₀ – P) or ΔT_b ∝ ΔP]

Question 8.
Fish generally needs O₂ concentration in water at least 3.8 mg/L for survival. What partial pressure of O₂ above the water is needed for the survival of fish? Given the solubility of O₂ in water at 0 °C and 1 atm partial pressure is 2.2 × 10^-3 mol/L (0.054 atm)
Answer:
Given : Required concentration of O₂
= 3.8 mg/L
= \(\frac{3.8 \times 10^{-3}}{32} \mathrm{~mol} \mathrm{~L}^{-1}\)
Solubility of O₂ = 2.2 × 10^-3 mol L^-1
P = 1 atm
Partial pressure of O₂ needed = Po₂ = ?

Pressure needed = Po₂ = 0.05397 atm.

Question 9.
The vapour pressure of water at 20 °C is 17 mm Hg. What is the vapour pressure of solution containing 2.8 g urea in 50 g of water? (16.17 mm Hg)
Answer:
Given : Vapour pressure of pure solvent (water) = P0
= 17 mm Hg
Weight of solvent = W₁ = 50 g
Weight of solute (urea) = 2.8 g
Molecular weight of a solvent = M₁ = 18
Molecular weight of a solute (urea) = M₂
= 60 g mol^-1
\(\frac{P_{0}-P}{P_{0}}=\frac{W_{2} \times M_{1}}{W_{1} \times M_{2}}\)
∴ \(\frac{17-P}{17}=\frac{2.8 \times 18}{50 \times 60}\) = 0.0168
∴ 17 – P = 17 × 0.0168
17 – P = 0.2856
∴ P= 17 – 0.2856
= 16.7144 mm Hg
Vapour pressure of solution = 16.7144 mm Hg

Question 10.
A 5% aqueous solution (by mass) of cane sugar (molar mass 342 g/mol) has freezing point of 271K. Calculate the freezing point of 5% aqueous glucose solution.
Answer:
Given : W₂ = 5 g cane sugar; W₁ = 100 – 5 = 95 g
M₂ = 342 g mol^-1; T_f1 = 271 K;
ΔT_f1 = 273 – 271 = 2 K; T_f = ?
W₂ = 5 g glucose, W’₁ = 100 – 5 = 95 g,
M’₂ = 180 g mol-1, ΔT_f2 = ?

= 12.996 K kg mol^-1
≅ 13 K kg mol^-1

∴ Freezing point of solution = T_f
= 273 – 3.801 = 269.2 K
Freezing point of solution = 269.2 K.

Question 11.
A solution of citric acid C₆H₈O₇ in 50 g of acetic acid has a boiling point elevation of 1.76 K. If K_b for acetic acid is 3.07 K kg mol^-1, what is the molality of solution?
Answer:
Given : W₁ = 50 g acetic acid
ΔT_b = 1.76 K
K_b = 3.07 K kg mol^-1
m = ?
ΔT_b = K_b × m
∴ m = \(\frac{\Delta T_{\mathrm{b}}}{K_{\mathrm{b}}}\)
= \(\frac{1.76}{3.07}\)
= 0.5733 m
Molality of solution = 0.5733 m

Question 12.
An aqueous solution of a certain organic compound has a density of 1.063 gmL^-1, an osmotic pressure of 12.16 atm at 25°C and a freezing point of -1.03°C. What is the molar mass of the compound? (334 g/mol)

Question 13.
A mixture of benzene and toluene contains 30% by mass of toluene. At 30°C, vapour pressure of pure toluene is 36.7 mm Hg and that of pure benzene is 118.2 mm Hg. Assuming that the two liquids form ideal solutions, calculate the total pressure and partial pressure of each constituent above the solution at 30°C.
Answer:
Given : 30% by mass of toluene (T) and 70% by mass of benzene (B).
W_T = 30 g; W_B = 70g
\(P_{\mathrm{T}}^{0}\) = 36.7 mm Hg; \(P_{\mathrm{B}}^{0}\) = 118.2 mm Hg
M_T = 92 g mol^-1; MB = 78 g mol^-1
P_T = ?, P_B = ?, P_soln = ?

Total number of moles = n_Total = n_T + n_B
= 0.326 + 0.8974
= 1.2234 mol

Mole fractions :
x_T = \(\frac{n_{\mathrm{T}}}{n_{\text {Total }}}=\frac{0.326}{1.2234}\) = 0.2665
x_B = 1 – 0.2665 = 0.7335
Psoln = x_T + \(P_{\mathrm{T}}^{0}\) + x_B × \(P_{\mathrm{B}}^{0}\)
= 0.2665 × 36.7 + 0.7335 × 118.2
= 9.780 + 86.7
= 96.48 mm Hg

Partial pressures :
P_T = x_T × P_soln
= 0.2665 × 96.48
= 25.71 mm Hg
P_B = x_B × P_soln
= 0.7335 × 96.48
= 70.77 mm Hg
Total pressure P_soln = 96.48 mm Hg
Partial pressures : P_Toluene = 25.71 mm Hg
P_Benzene = 70.77 mm Hg

Question 14.
At 25 °C a 0.1 molal solution of CH₃COOH is 1.35% dissociated in an aqueous solution. Calculate freezing point and osmotic pressure of the solution assuming molality and molarity to be identical.
Answer:
Given : T = 273 + 25 = 298 K
C = 0.1 m ≅ 0.1 M; K_f = 1.86 K kg mol^-1
Per cent dissociation = 1.35
Freezing point = t_f = ?
π = ?
α = \(\frac{1.35}{100}\) = 0.0135

i = 1 – α + α + α = 1 + α = 1 + 0.0135 = 1.0135
(i) ΔT_f = i × K_f × m
= 1.0135 × 1.86 × 0.1
= 0.1885 °C
∴ Freezing point of solution = 0 – 0.1885
= – 0.1885 °C

(ii) n = iCRT
= 1.035 × 0.1 × 0.08206 × 298
= 2.53 atm

(i) Freezing point of solution = – 0.1885 °C
(ii) Osmotic pressure = π = 2.53 atm

Question 15.
A 0.15 m aqueous solution of KCl freezes at -0.510 °C. Calculate i and osmotic pressure at 0 °C. Assume volume of solution equal to that of water.
Answer:
Given : c = 0.15 m KCl ≅ 0.15 M KCl
ΔT_f = 0 – T_f = 0 – (-0.510) = 0.510 °C
T = 273 K; K_f = 1.86 K kg mol^-1
i = ?; π = ?
ΔT_f = i × K_f × m
∴ i = \(\frac{\Delta T_{\mathrm{f}}}{K_{\mathrm{f}} \times m}\)
= \(\frac{0.510}{1.86 \times 0.15}\)
= 1.828
π = iCRT
= 1.828 × 0.15 × 0.08206 × 273
= 6.143 atm
i = 1.828, Osmotic pressure = π = 6.143 atm

12th Chemistry Digest Chapter 2 Solutions Intext Questions and Answers

Can you tell ? (Textbook Page No. 29)

Question 1.
Why naphthalene dissolves in benzene but not in water ?
Answer:
Since naphthalene is a covalent nonpolar substance, it is soluble in a nonpolar solvent like benzene but insoluble in polar solvent like water.

Question 2.
Anhydrous sodium sulphate dissolves in water with the evolution of heat. What is the effect of temperature on its solubility ?
Answer:
Since the dissolution of anhydrous sodium sulphate in water is an exothermic process due to evolution of heat, according to Le Chatelier’s principle its solubility decreases with the increase in temperature.

(Textbook Page No. 42)

Question 1.
If 1.25 m sucrose solution has ΔT_f of 2.32 °C, what will be the expected value of ΔT_f for 1.25 m CaCl₂ solution?
Answer:
Sucrose being nonelectrolyte, it has i = 1 but for CaCl₂,
(CaCl₂ → Ca²⁺ + 2Cl^–) the value of i = 3.
Hence
ΔT_f = i × 2.32
= 3 × 2.32
= 6.92 °C
∴ ΔT_f = 6.92 °C.

(Textbook Page No. 44)

Question 1.
Which of the following solutions will have maximum boiling point elevation and which have minimum freezing point depression assuming the complete dissociation? (a) 0.1m KCl (b) 0.05 m NaCl (c) 1 m AlPO₄ (d) 0.1 m MgSO₄.
Solution :
Boiling point elevation and freezing point depression are colligative properties that depend on number of particles in solution. The solution having more number of particles will have large boiling point elevation and that having less number of particles would show minimum freezing point depression.

AlPO₄ solution contains highest moles and hence highest number particles and in turn, the maximum ΔT_b. NaCl solution has minimum moles and particles, it has minimum ΔT_f.

Question 2.
Arrange the following solutions in order of increasing osmotic pressure. Assume complete ionization. (a) 0.5 m Li₂SO₄ (b) 0.5 m KCl (c) 0.5 m Al₂(SO₄)₃ (d) 0.1 m BaCl₂.
Answer:
Consider 1 dm³ of each solution.

Osmotic pressure being a colligative property, it depends on number of particles in the solution.
Therefore, increasing order of osmotic pressure is,

Balbharti Maharashtra State Board Class 12 Secretarial Practice Solutions Chapter 12 Stock Exchange Textbook Exercise Questions and Answers.

Maharashtra State Board Class 12 Secretarial Practice Solutions Chapter 12 Stock Exchange

1A. Select the correct answer from the options given below and rewrite the statements.

Question 1.
A stock exchange is where stock brokers and traders can buy and sell ______________
(a) Gold
(b) Securities
(c) Goods
Answer:
(b) Securities

Question 2.
The ______________ is the first Stock Exchange to be recognized by the Indian Government under the Securities Contracts (Regulation) Act.
(a) BSE
(b) NSE
(c) OTCEI
Answer:
(a) BSE

Question 3.
______________ is a dealer in Stock Exchange who carries on trading of securities in his own name.
(a) Jobber
(b) Broker
(c) Bull
Answer:
(a) Jobber

Question 4.
A ______________ who expects fall in price of securities.
(a) bull
(b) bear
(c) Jobber
Answer:
(b) bear

Question 5.
The practice of buying and selling within the same trading day before the close of the market on that day is called ______________
(а) insider trading
(b) day trading
(c) auction
Answer:
(b) day trading

1B. Match the pairs.

Question 1.

Group ‘A’	Group ‘B’
(a) SEBI	(1) Expects the price of shares to rise in the future.
(b) Day Trading	(2) Expects the price of shares to fall in the future.
(c) Bull	(3) Buying and selling of securities during the same trading day.
(d) Bear	(4) To protect the interest of investors in the securities market.
(e) BSE	(5) Buying and selling of securities to particular investors.
	(6) One of the oldest stock exchanges in India.
	(7) To protect the interest of companies in the securities market.
	(8) Buying and selling of securities within a week.
	(9) Newest Stock Exchange in India.
	(10) One who invests in new issues of securities.

Answer:

Group ‘A’	Group ‘B’
(a) SEBI	(4) To protect the interest of investors in the securities market.
(b) Day Trading	(3) Buying and selling of securities during the same trading day.
(c) Bull	(1) Expects the price of shares to rise in the future.
(d) Bear	(2) Expects the price of shares to fall in the future.
(e) BSE	(6) One of the oldest stock exchanges in India.

1C. Write a word or a term or a phrase that can substitute each of the following statements.

Question 1.
A specific place where the trading of securities is arranged in an organized method.
Answer:
Stock Exchange

Question 2.
The first Stock Exchange to be recognized by the Indian Government under the Securities Contracts Regulation Act.
Answer:
BSE

Question 3.
A dealer in the Stock Exchange who carries on the trading of securities in his own name.
Answer:
Jobber

Question 4.
A speculator who expects the price of shares rises in the future.
Answer:
Bull

1D. State whether the following statements are True or False.

Question 1.
A Stock Exchange is a reliable barometer to measure the economic condition of a country.
Answer:
True

Question 2.
Bombay Stock Exchange is the oldest Stock Exchange in India.
Answer:
True

Question 3.
A broker is a dealer in the Stock Exchange who carries on the trading securities in his own name.
Answer:
False

Question 4.
A Bear is a speculator who expects the prices of shares to rise in the future.
Answer:
False

1E. Complete the sentences.

Question 1.
The oldest Stock Exchange in India is the ______________
Answer:
BSE

Question 2.
A speculator who expects fall in prices of share ______________
Answer:
Bear

Question 3.
A person who buys or sells shares on behalf of his clients is called as ______________
Answer:
broker

Question 4.
The largest and most modern stock exchange in India is the ______________
Answer:
National Stock Exchange

1F. Select the correct option from the bracket.

Question 1.

Group ‘A’	Group ‘B’
(a) Regulator of Capital Market	(1) …………………..
(b) ……………………	(2) Nifty
(c) Jobber	(3) …………………..
(d) …………………..	(4) Oldest Stock Exchange in the world

(London Stock Exchange, Index of NSE, SEBI, Trades in securities in his own name)
Answer:

Group ‘A’	Group ‘B’
(a) Regulator of Capital Market	(1) SEBI
(b) Index of NSE	(2) Nifty
(c) Jobber	(3) Trades securities in his own name
(d) London Stock Exchange	(4) Oldest Stock Exchange in the world

1G. Answer in one sentence.

Question 1.
What is Stock Exchange?
Answer:
A Stock Exchange is a place or a platform where investors-individuals, institutions, or organizations meet to purchase or sell securities.

Question 2.
Who is a Broker?
Answer:
He is a licensed member of a stock exchange who transacts business on the behalf of his clients, being an agent between investors and jobbers.

Question 3.
Who is Jobber?
Answer:
A Jobber is a professional speculator in the stock exchange who carries on the trading securities in his name.

Question 4.
Who is a Bull?
Answer:
Bull (Tejiwala) is an optimistic speculator who expects the price of a share to rise in the future and buys with the hope of selling at a high price to earn profit.

Question 5.
Who is Bear?
Answer:
A bear is a speculator who expects the prices of shares to fall in the future and sells his securities at the prevailing prices to avoid loss.

Question 6.
Who is Lame Duck?
Answer:
A lame-duck is a bear broker whose expectations have gone wrong and makes a loss in his dealings.

Question 7.
What is a trading ring?
Answer:
The trading of shares that takes place during trading hours on the floor of the stock exchange is called the Trade Ring.

Question 8.
What is Sensex?
Answer:
Sensex is the index of the BSE which represents the increase or decrease in prices of stocks of a selected group of companies.

Question 9.
What is Rally?
Answer:
If the Sensex or Nifty moves in an upward direction over a period of 14 to 20 trading sessions, it is called as a rally.

Question 10.
What is Crash?
Answer:
If the Sensex or Nifty moves in a downward direction, it is called a crash.

1H. Correct the underlined word/s and rewrite the following sentences.

Question 1.
One of the functions of SEBI is to protect the interest of issuers of securities in the securities market.
Answer:
One of the functions of SEBI is to protect the interest of investors in securities in the securities market.

Question 2.
A Broker cannot directly deal with investors.
Answer:
A Jobber cannot directly deal with investors.

Question 3.
A Bear expects prices of shares to rise in the future.
Answer:
A Bull expects prices of shares to rise in the future.

Question 4.
A Bull buys new issues of securities from the primary market.
Answer:
A stag buys new issues of securities from the primary market.

Question 5.
A stock market is an important constituent of the money market.
Answer:
A stock market is an important constituent of the capital market.

2. Explain the following terms/concepts.

Question 1.
Stock Exchange
Answer:
Stock Exchange is a specific place where various types of securities are purchased and sold. The term securities include equity shares, preference shares, debentures, government securities, and bonds, etc. including units of mutual funds. They act as intermediaries between investors and borrowers, to provide safety and stability to the investors, stock exchanges in India are regulated by SEBI.

Question 2.
Broker
Answer:
He is a member of the stock exchange and is licensed by the stock exchange to buy or sell shares on his client’s behalf. He is an agent between the investors and Jobber and earns his income in the form of commission or brokerage.

Question 3.
Jobber
Answer:
A Jobber is a professional speculator in the stock exchange who carries on the trading securities in his own name. He buys securities as an owner and sells them at a higher price, and cannot deal with investors directly.

Question 4.
Bull
Answer:
A Bull is a speculator who is optimistic, expects the price of a share to rise in the future, and buys with the hope of selling them at higher prices to earn profit. A bull’s action leads to higher prices for securities as there is an excess purchase over sales.

Question 5.
Bear
Answer:
Bear (Mandiwala) is a pessimistic speculator who expects a fall in the price of a security, so he sells his securities at prevailing prices to avoid loss as he anticipates further fall in prices. His action leads to lowering prices as there is an excess of sales over the purchase.

Question 6.
Contract Note
Answer:
It is a note given by a broker to his client in a specific form, validating the transaction. Its copy comes immediately to both after the transaction within 24 hours.

3. Study the following case/situation and express your opinion.

1. Mr. Y is a practicing Company Secretary offering advisory services to companies, institutions, etc. on corporate laws including the Companies Act. He has received few queries from his clients, please assist Mr. Y in answering them.

Question (a).
BDI bank wants to offer DP services. Whom should they approach for registering as DP?
Answer:
If BDI bank wants to offer DP services, they should approach the concerned Depository for registering themselves.

Question (b).
KM Financial wants to offer Debenture Trustee services. Where should they apply for getting registered?
Answer:
If KM Financial wants to offer Debenture Trustee services then they should be registered with SEBI to act as Debenture Trustee.

Question (c).
TT Ltd. Co. wants to issue an IPO. Should it get itself registered with SEBI?
Answer:
The entire IPO process is regulated by SEBI, TT Ltd. Co should register with The U. S. Securities and Exchange Commission (SEC) which ensures that the company has made disclosures in detail thus TT Ltd will get the green signal to issue IPO.

2. Mr. P has recently got his B.Sc. degree. He has enrolled in a course in the securities market. As a new student of this subject, he has few queries as follows:

Question (a).
Does a company need to be listed on a stock exchange’s ability to sell its securities through the stock exchange?
Answer:
Yes, a company needs to be listed on Stock Exchanges to sell its securities through the Stock Exchange.

Question (b).
What is the term used for referring to a stock exchange’s ability to reflect the economic conditions of a country?
Answer:
A Stock Exchange is the “Economic Barometer” and acts as an economic mirror that reflects the economic conditions of a country, eg. Boom, recession period.

Question (c).
Which term refers to the functions of the stock exchange as a provider of a ready market for sale and purchase of security?
Answer:
The “Liquidity” function is the main function of the Stock Exchange as it provides a ready market for the sale and purchase of securities.

4. Distinguish between the following.

Question 1.
Jobber and Broker
Answer:

Points	Jobber	Broker
1. Meaning	A Jobber is one who buys and sells securities in his own name.	A broker is an agent who deals in buying and selling securities on behalf of his client.
2. Nature of Trading	A jobber carries out trading activities only with the broker.	A broker carries out trading activities with the jobber on behalf of his investors.
3. Restrictions on Dealings	A jobber is prohibited to directly buy or sell securities in the stock exchange. Also, he cannot directly deal with the investors.	A broker acts as a link between the jobber and the investors. He trades i.e. buys and sells securities on behalf of his investors.
4. Agent	A jobber is a special mercantile agent.	A broker is a general mercantile agent.
5. Form of Consideration	A jobber gets consideration in the form of profit. The positive difference between sale and purchase of securities.	A broker gets consideration in the form of commission or brokerage. The rate/amount of brokerage is fixed by stock exchanges.
6. Amount of Consideration	The amount of Consideration payable to Jobber is determined by the competition of jobbers.	The rate or amount of brokerage of a broker is fixed as per stock exchange provisions.

5. Answer in brief.

Question 1.
State the functions of SEBI.
Answer:
The various functions of SEBI are

• To protect the interest of investors in the securities stock market.
• To promote the development of securities markets.
• To regulate the business in stock exchanges and any other securities market.
• To register and regulate the working of stockbrokers, sub-brokers, share transfer agents, bankers to an issue, trustee of trust deeds, registrars to an issue, merchants bankers, underwriters, and such other intermediaries who may be associated with the securities market.
• To register and regulate the working of the Depositories, Depository Participants, Custodians of securities, foreign institutional investors, credit rating agencies.
• To register and regulate the working of venture capital funds and collective investment schemes including mutual funds.
  • To promote and regulate self-regulatory organizations.
  • To prohibit fraudulent and unfair trade practices relating to securities markets.
• To promote investors’ education and training of intermediaries of the securities market.
• To prohibit insider trading in securities.
• To conduct research and carry out publications.

Question 2.
State any four features of the Stock Exchange.
OR
What are the features of the Stock Exchange?
Answer:
According to the Securities Contracts (Regulation) Act 1956, the term stock exchange is defined as, “An association, organization or body of individuals, whether incorporated or not, established for the purpose of assisting, regulating and controlling of business in buying, selling and dealing in securities.”

Husband and Dockerary have defined stock exchange as “Stock exchanges are the privately organized market which is used to facilitate trading in securities.”

The important features of a stock exchange are as follows:
(i) Market for Securities:
The stock exchange is a place where all types of corporate securities, as well as securities of government and semi-government bodies, are traded.

(ii) Second Hand Securities:
Securities traded in the Stock exchange are those securities that are already issued by the companies. In other words, second-hand securities are bought and sold among investors in a stock exchange.

(iii) Listed Securities:
Only securities that are listed with the stock exchange can be traded on a stock exchange. Listing of securities helps in protecting the interest of investors as companies have to strictly comply with the rules laid down by the stock exchange.

(iv) Organised and Regulated Market:
All Listed Companies have to comply with the guidelines of SEBI. Companies will also have to function as per the rules and regulations laid down by the Stock exchange.

(v) Specific Location:
The stock exchange is a specific physical place where securities are traded. It is a marketplace where brokers and intermediaries meet to conduct dealings in securities. Today, all trading is done electronically on a stock exchange.

(vi) Trading only through Members:
Securities in a Stock exchange can be traded only by the members of the exchange on their own behalf or through authorized brokers.

6. Justify the following statements.

Question 1.
The Securities and Exchange Board of India SEBI is the regulator for the securities market in India.
Answer:

• The Securities and Exchange Board of India was set up on 12th April 1988. The main purpose of setting up SEBI was to develop and regulate stock exchanges in India.
• The objectives of SEBI are to protect the interest of the investors and regulate the securities market in India.
• To bring professionalism in the working of intermediaries in the capital markets, i.e., brokers, mutual funds, stock exchanges, Demat- depositories, etc. is also a feature of SEBI.
• The role of SEBI also includes creating a good financial climate, so that companies can raise long-term funds through the issue of securities – shares and debentures.
• The main function of SEBI is to register and regulate the working of stockbrokers, sub-brokers, share transfer agents, bankers to an issue, trustee of trust deeds, registrars to an issue, merchant bankers, underwriters, and such other intermediaries who may be associated with securities market.
• Thus, it is rightly said that SEBI is the regulator of the securities market in India.

Question 2.
Stock exchanges work for the growth of the Indian economy.
Answer:

• The stock exchange is a specific place where the trading of securities is arranged in an organized method.
• The stock exchanges help in the process of rapid economic development by speeding up the process of capital formation as well as resource mobilization in India.
• It helps in raising medium-term capital as well as long-term capital for the development and expansion of the companies in the Indian economy.
• New industries and commercial enterprises can easily acquire capital funds for economic growth.
• It reflects a healthy financial and investment conducive atmosphere in the economy. It stimulates investment in the productive sector which accelerates the process of economic development of the nation.
• Thus, it is rightly said that the stock exchanges work for the growth of the Indian economy.

7. Answer the following questions.

Question 1.
Explain the functions of the Stock Exchange.
Answer:
Definition Of Stock Exchange: According to the Securities Contracts (Regulation) Act of 1956, the term ‘stock exchange’ is defined as “An association, organization or body of individuals, whether incorporated or not established for the purpose of assisting, regulating and controlling of business in buying, selling and dealing in securities.”

Husband and Dockerary have defined stock exchange as: “Stock exchanges are privately organized markets which are used to facilitate trading in securities.”

Stock Exchange performs various important functions discussed as follows:
(i) Mobilisation of Savings:
Stock markets are organized and regulated markets that protect the interests of the investors. It obtains surplus funds (savings) from individual households private and public sector units etc. and channelizes them in the proper direction. It thus provides a ready market for buying and selling securities.

(ii) Capital Formation:
Investors in securities are attracted due to good returns on investments and capital appreciation. The stock exchanges encourage investors to invest in the primary and secondary stock markets for investing in stock markets, investors need to save money. Savings lead to investment in shares and other securities. Such investments lead to capital formation.

(iii) Pricing of Securities:
The price of the securities are sold in the stock markets is based on demand and supply forces listed securities get prestige and reputation. When the prices of the shares go up constantly, their security value increases. The valuation of securities is useful to investors, the government, and creditors. The investors thus can gauge their investment worth and the creditors too can estimate the creditworthiness of a company.

(iv) Economic Barometer:
A stock exchange is a reliable barometer to measure the economic condition of a country. They encourage investors to invest and help companies to generate long-term funds thus promoting industrial development. The rise or fall in the share prices indicates the boom or recession cycle of the economy. The stock exchange is the pulse of the economy and the mirror that reflects the country’s economic status.

(v) Protecting Interest of Investors:
In the stock markets, only the listed securities are traded. The stock exchanges protect the interests of the investors through the strict enforcement of their rules and regulations. The securities Control (Regulation) Act 1956, provides rules for the functioning, licensing, and controlling speculations of stock exchanges. The SEBI also plays an important role in monitoring stock exchanges thus protect the interests of the investors by regulating intermediaries, monitoring speculation, and making the investors aware of their rights through IEPF, etc.

(vi) Liquidity:
The stock exchange facilitates liquidity by providing a ready market for the sale and purchase of securities. It provides marketability along with liquidity to investments in corporate enterprises. Because of stock exchange investors can convert a long-term investment into short-term and medium-term as it provides a two-way outlet by transforming money into an investment and vice versa without much delay.

(vii) Better Allocation of Capital:
The stock exchange regulates and controls the flow of investment from unproductive to productive, uneconomic to economic, unprofitable to profitable enterprises. Thus, savings of the people are channelized into industry yielding good returns, and under utilization of capital is avoided.

(viii) Contributes to Economic Growth:
The stock exchange help in the process of rapid economic development by speeding up the process of capital formation as well as resource mobilization. It helps in raising medium as well as long-term capital for the development and expansion of the companies. The resource of the economy flows from one company to another. This leads to capital formation as well as economic growth.

(ix) Providing Scope for Speculation:
Stock Exchanges’ like any other market provides a mechanism for evaluating the prices of securities through the basic law of demand and supply. Stock Exchange prices help to check the real worth of the securities in the market and thus permit healthy speculation of securities.

(x) Promotes the Habit of Savings and Investment: The stock market offers attractive opportunities for investment in various securities by obtaining funds from surplus units such as households, individuals, public sector units, central government, etc, and channelizing these funds for productive purposes.

Balbharti Yuvakbharati English 12th Digest Chapter 2.8 Small Towns and Rivers Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 12 English Yuvakbharati Solutions Chapter 2.8 Small Towns and Rivers

12th English Digest Chapter 2.8 Small Towns and Rivers Textbook Questions and Answers

Question (i)
Most of the civilizations have flourished on the banks of the rivers. Discuss the reasons in the class.
Points:
(a) Availability of water
(b) Fertile soil for agriculture
(c) Fishing ground
(d) Transport

Question (ii)
Write down the names of the famous cities that are situated on the banks of the rivers given below.
Answer:

River	City
Ganga	Varanasi
Yamuna	Mathura
Godavari	Rajahmundry
Varada	Sagara
Krishna	Vijayawada
Tapi	Surat

Question (iii)
Write down the names of the rivers on the banks of which following cities have prospered:
Answer:

City	River
London	Thames
Cairo	Nile
New York	Hudson
Paris	Siene

Question (iv)
Divide your class into groups and discuss the changes that might have taken place when the cities grow on the banks of the rivers.
Points: Dwellings are built – Population grows – Fishing flourishes – Outsiders come to settle – Trading takes place – Prosperity increases – Increased population, trade, traffic and fishing begin to harm the river system – River system shows damage – River starts to decay

Question (v)
Share your views in the class on the topic ‘Conservation of Rivers and Development of the Cities.’
(Students may list their own points of views.)

(A1)

Question 1.
State the importance of Nature in the lives of the people from the North-eastern part of India as expressed in the poem with reference to:
(a) Flowers
(b) River
(c) Bamboo
(d) East
Answer:
(a) Flowers – Tuberoses are woven into wreaths to be placed on the body of a departed as a mark of love and respect.
Sole: Most of the textual ‘Wh’ questions are converted to Activity Format.
(b) River: The people believe the river has a soul. They respect their rivers and even revere it as divine as its waters are immortal.
(c) Bamboo: The poet says ‘in the cool bamboo’. The colour is a cool green. The place where the bamboo grows is also cool.
(d) East: The direction of sunrise is very important for the people of Arunachal Pradesh. They ensure the dead are placed pointing west so that their soul directly enters the golden house of the sun. They believe that finally souls must attain the sun’s abode in the east.

(A2)

Question (i)
The poet has described her small town in Arunachal Pradesh. Pick out from the extract, the lines that describe the poet’s town.
Answer:
1. ‘My hometown lies calmly amidst the trees’,
2. ‘it is always the same In summer and winter With the dust flying Or the wind howling down the gorge’

Question (ii)
Make a list of natural elements mentioned in the extract.
Answer:
trees, winter, summer, wind, tuberoses, life, land, river, fish, stars

Question (iii)
‘The river has a soul.’ Elaborate the concept in your words as the poet has explained in the extract.
Answer:
When the poet says ‘The river has a soul’ she personifies the river. The river flows with great force – ‘like a torrent of grief. The river flowing with great force can be like a person pouring out grief in a storm of emotion. The river also seems to be holding its breath, maybe because it is choking with filth. There are no fishes. It is not clear and sparkling. So the poet says -‘I think it holds its breath seeking a land of fish and stars’.

Question (iv)
The poet is convinced with the thought of immortality of water. Pick out the relevant lines from the extract.
Answer:
‘The river has a soul’.

(v) The poet has used some unconventional expressions. Illustrate them in your words.

Question (a)
‘Torrent of grief.
Answer:
The poet describes the river flowing in summer with great speed. Just like someone becomes emotional in great sadness, the force of the water seems to be like the outpouring of sorrow of the river.

Question (vi)
The poet has connected the need to preserve Nature with the belief of particular community and her childhood memories. Write down the measures you would take to convince the people regarding the need to conserve the Nature.
Answer:
We have to make people at large realize that we are a part of the nature not apart from the nature. Saving nature is to save ourselves, To bring about this realization I can address my steps to two sections. The first would be the children. Saving water would be the first thing to teach children. Story-telling, poems, songs, games, cartoons and can easily bring the conservation ideas to young minds. I would take my ideas to schools, parks, malls and try to spread this to the young ones.

The other section is of course the adult public across economic and social levels. The well- off people must not be careless if they can afford to pay bills they can waste resources.

I will do everything possible. I will use social media to spread various messages/ mottos.

(A3)

Question (i)
Write down the expressions related to ‘the seasons’ from the extract.
Answer:
1. Summer or winter
2. ‘in the summer’

Question (ii)
Match column ‘A’ with column ‘B’.

A	B
1. cool 2. happy 3. dreadful 4. dry	(a) silence (b) earth (c) bamboo (d) pictures

Answer:

1. cool-bamboo
2. happy – pictures
3. dreadful – silence
4. dry-earth

(A4)

Question (i)
Read the expression ‘a sad wreath of tuberoses’.
‘Is the wreath sad?’ Explain the figure of speech.
Answer:
The figure of speech is transferred epithet. The sad mourners have placed the wreath of tuberoses on the dead. The emotion of sadness has been transferred to the flowers for effect.

Question (ii)
List and explain the metaphorical expressions from the extract.
Answer:
1. ‘torrent of grief’. The river sweeps along with great speed as if it is pouring out sorrow.
2. ‘Wind howling down the gorge’. The wind blowing through the narrow gorges creates a sound exactly like howling.
3. ‘The river has a soul’. The river is spoken of as a living being, a human.
4. ‘It holds its breath’. The river may be choking with debris and filth.

Question (iii)
‘The river has a soul.’
‘Life and death.’
These are the two expressions that are repeated in the poem; but both of them indicate different figures of speech. Find out and discuss.
Answer:
(a) ‘Life and death’
In the first usage it is used as antithesis, to emphasise the beginning and end.
In the second instance it is irony to indicate that neither life nor death is permanent. Ironically the rituals are permanent.
(b) ‘The river has a soul’
In the first instance it is used to personify the river. Several human-like qualities are attributed to the river,
= it cuts through the land
= it is cascading in grief
= it holds its breath
= it seeks a land

Question (iv)
Find out the beauty of the free verse reflected in this poem.
Answer:
“Small Towns and Rivers’ is written in free- verse. Since there are irregular lengths of lines and no rhyme, the reading of the poem is almost like a story-telling. Each stanza has a different number of lines and there is no order for mixing up the short and long lines.

The poem is not confined by an obvious rhythm so we feel there is a kind of freedom.We are free to imagine the widespread setting of the North-eastern terrain of mountains and rivers, mists, golden sunlight and the town by the river.

(A5)

Question (i)
Prepare the arguments for group discussion on the topic
‘A balanced progress never harms the Nature’.
For:

• Growth should be planned to take place in stages
• Planning vital for the growth
• Sustainability must be ensured
• Pros and cons of the damage to environment must be weighed
• Short-term gains in progress must not harm long-term eco-factors

Against:

• Growth must not be halted for issues of environment
• Costs will go up due to delays
• Delays in progress will slow down economy and employment
• Slowdown in economy will cause public to protest
• Sacrifices have to be made – one can’t have the cake and eat it too.

Question (ii)
Compose 4 to 6 lines on ‘Gift of the Seasons’.
Answer:
Gift of the Seasons
Each season brings a sweetly wrapt gift;
We can gift her back : no water pollution in
the season of the Sun.
No air pollution when
The rains come down.
And no degradation the rest of the year!

Question (iii)
Write an appreciation of the poem ‘Small Towns and Rivers’. Use the points given below:

• About the poem/poet/title
• Theme
• Poetic devices, language, style
• Special features – tone and type
• Values, message
• Your opinion about the poem

Answer:
The poem ‘Small Towns and Rivers’ written by Mamang Dai is a beautiful word-picture. It is also a lament of the poet about her beautiful native land of Arunachal Pradesh.

This theme shows in the way she begins the poem that small towns remind her of death. It is shocking. She implies the town is unchanging in all weathers, but development comes along and changes everything. There is irony in that the cycle of life and death shows that life is not permanent, but the rituals are permanent.

She uses metaphor that the rivers are not only alive like us humans, but actually immortal. She personifies the river by way the river ‘holds its breath’ because it is choking. It is flowing in search of a place where it will flow clean and clear. The poet uses metaphor of the water-cycle to illustrate the river has a soul and its waters are immortal.

The poet builds a climax with ‘shrine of happy’ childhood memories. This becomes growing up -‘grow with anxiety’. Then she speaks of how the dead are placed pointing west so the soul can ascend directly into the sun’s golden home in the east. This tells us about the traditions of her region.

The poem is in free verse and seems to be in easy language, but we can understand the full depth of meaning only after reading it more than once. The poem is a lament about the destruction of nature for development. We all will feel the sorrow of the poet when we read about how nature’s beauty is damaged for man’s greed called ‘progress’.

Question (iv)
Write a dialogue between two friends on ‘Importance of the rivers’
Answer:
Priya: Jai, it goes without saying that fresh, clean water is essential for humans and nature to survive. Rivers are precious sources of fresh drinking water for people across the world. And when rivers are so badly polluted by industry or by poor water management practices, it can be a case of life-or-death. This unfortunately happens across the world.

Jai: Yes Priya. Freshwater habitats account for some of the richest biodiversity in the world, and rivers are a vital, vibrant ecosystem for many species. Only those who live by the river know about this wealth of nature. Those who live far away and damaging the system with the poisons are not aware.

Priya: People depend on rivers for their way of life and their livelihoods. From fishing to agriculture, the way our waterways are managed has a direct impact on people’s lives. There are millions of people who follow their ancestors’ way of living and earning a livelihood. But modern technology has wrecked the very source of these.

Jai: Rivers are absolutely vital: for fresh drinking water, for people’s livelihoods and for nature. Unfortunately, they’re still threatened. We must commit to recovering freshwater biodiversity, restoring natural river flows and cleaning up polluted water for people and nature to thrive.
Priya: Yes Jai. I agree. It is the crying need of the day.

(A6)

Question (i)
Collect information about rivers in Maharashtra.

Question (ii)
Further reading:
‘The River Poems’ – Mamang Dai
‘The World Is Too Much With Us’ – William Wordsworth

Yuvakbharati English 12th Digest Chapter 2.8 Small Towns and Rivers Additional Important Questions and Answers

Read the extract and complete the activities given below:

Global Understanding:

Question 1.
Describe the river in the 3rd stanza.
Answer:
The river flows with great force -‘like a torrent’-. The river has life and soul. It breathes. But it seems as if the river is holding its breath. It seems to be in search of fishes which will live in it and stars which will be reflected in its waters.

Question 2.
What is meant by immortality?
Answer:
Immortality means the ability to live forever, without death.

Question 3.
Give reasons-The dead are placed pointing to the west.
Answer:
The people of Arunachal Pradesh believe that it will be possible for the soul of the departed person enter ‘the house of the sun’. They hope the soul will be able to ‘walk into the golden east’. So they place the dead (with feet) pointing to the west.

Question 4.
The poet has described her small town in Arunachal Pradesh. Pick out from the extract the lines that describe the poet’s town.
Answer:

1. The town has ‘A shrine of happy pictures’ to mark the days of childhood.
2. Small towns grow with anxiety for the future.
3. Like her town, ‘small towns’ are ‘by the river’.
4. Make a list of natural elements mentioned in the extract.

Question 5.
Make a list of natural elements mentioned in the extract.
Answer:
River, Earth, Mountaintops, Sun, Sunlight, Bamboo

Question 6.
The poet is convinced with the thought of immortality of water. Pick out the relevant lines from the extract.
Answer:
1. The river has a soul.
2. from the first drop of rain to dry earth
3. mist on the mountaintops
4. the immortality of water.

Interpretation/Inference/Analysis:

Question 1.
The land of fish and stars.
Answer:
The poet says she thinks the river is holding its breath. One has to hold one’s breath when he/she is unable to breathe or does not want to breathe. The river may be choking with garbage and is not able to breathe. The river may be stinking and may not be able to breathe.

As the river is so filthy there are no fish. It is dirty; the water is not sparkling in the day and cannot reflect the stars at night. So the river is in search of a land where there it can flow clean, it will have fish and where its clear water will sparkle in the sun and glitter with stars at night.

Question 2.
The river has a soul.’ Elaborate the concept in your words as the poet has explained in the extract.
Answer:
The poet states the river has a soul. The soul is deathless. The water that flows in the river came from the drops of rain. The water evaporated, rose as mist to the mountaintops. Then it formed clouds and poured down as rain to the dry earth and flowed again. Thus the river goes on, immortal, deathless.

Question 3.
The poet has used some unconventional expressions. Illustrate them in your words. Illustrate them in your words. Shrine of happy pictures
Answer:
There is a shrine probably in the town which has pictures inside. The pictures may be those of the happy moments experienced by the people in the town. Those memories are so sacred that it is a shrine to them. They protect and guard it because they have only sad and grim things happening at present.

Personal Response:

Question 1.
Rivers are our lifeline. They are an extremely important part of the eco-system and even considered sacred. Many major rivers and smaller ones have been misused and almost destroyed. Write down how we can restore our rivers to their original state.
Answer:
The condition of rivers worldwide is horrific. Everything from industrial chemical-waste to garbage is being thrown into rivers. Melted snow or the rainwater from springs come down from hills and mountains as sweet water for our survival. Polluting this is a crime against humanity.

Strict laws should be made and enforced to stop industrial activity near rivers. Wastes from industry, city, town or village must not reach the river. Only channels of rainwater must be allowed to reach the river. Awareness should be created for maintaining cleanliness along the banks.

In the olden days lakes, water bodies were considered precious and were guarded, Houses had wells. But people have lost the respect for water sources. Rivers are treated as sewage channels. Fines and punishments must be imposed and security put in for safeguarding our beautiful rivers.

Poetic Creativity:

Question 1.
Compose 4 to 6 lines on ‘River’.
Answer:
The River is our Mother

Like a mother the river soothes us.
When dying of thirst she revives us.
When tired and dirty she cleans us.
When growing our grain she waters the green.
Why! Oh why can’t we keep her clean!

Balbharti Maharashtra State Board 12th Chemistry Important Questions Chapter 9 Coordination Compounds Important Questions and Answers.

Maharashtra State Board 12th Chemistry Important Questions Chapter 9 Coordination Compounds

Question 1.
What are double salts?
Answer:
Double salts are crystalline molecular or addition compounds containing more than one salt in simple molecular proportions soluble in water and in solution they ionise and exhibit all the properties of the constituent ions.

For example, K₂SO_4⁺ A1₂(SO₄)_3^– 24H₂O
\(\mathrm{K}_{2} \mathrm{SO}_{4} \cdot \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} \cdot 24 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{aq})} \longrightarrow 2 \mathrm{~K}_{(\mathrm{aq})}^{+}+2 \mathrm{Al}_{(\mathrm{aq})}^{3+}+4 \mathrm{SO}_{4(\mathrm{aq})}^{2-}+24 \mathrm{H}_{2} \mathrm{O}_{0}\)

Question 2.
Define coordination compound.
Answer:
Coordination compound : It consists of a central metal ion or atom surrounded by atoms, molecules or anions called ligands by coordinate bonds, e.g. cisplatin Pt(NH₃)₂Cl₂, [Cu(NH₃)₄]SO₄.

Question 3.
Define Lewis bases and Lewis acids with respect to a coordination compound.
Answer:

• Lewis bases : In a coordination compound the ligands being electron pair donors they are Lewis bases.
• Lewis acids : The central metal atom or ion being electron acceptor behaves as a Lewis acid.
• For example, in the coordination compound, [Cu(NH₃)₄]²⁺, NH₃ is a Lewis base and Cu²⁺ is a Lewis acid.

Question 4.
Define coordination sphere. Give example.
Answer:
Coordination sphere : A coordination entity consisting of a central metal atom or ion and the coordinating groups like neutral molecules or anions (ligands) written inside a square bracket is together called coordination sphere. This is a discrete structural unit. The ionisable groups (generally ions) called counter ions are written outside the bracket.

For example, in the coordination compound K₄[Fe(CN)₆], the coordination sphere is [Fe(CN)₆]^4- while K⁺ represents counter ion.

Question 5.
Define and explain charge number of a complexion.
Answer:
Charge number of a complexion : The net charge carried by a complexion or a coordination entity is called its charge number.

Explanation :
(i) Charge number is equal to the algebraic sum of the charges carried by central metal atom or ion and all the ligands attached to it.
(ii) E.g. consider anionic complex, [Fe(CN)₆]^4-.
Charge number of [Fe(CN)₆]^4- = Charge on Fe²⁺ ions + 6 x charge on CN^– = ( + 2) + 6( -1) = – 4 Hence charge number of [Fe(CN)₆]^4- is – 4.

Question 6.
Explain the oxidation state of a metal in a complex.
Answer:

• The oxidation state of a metal atom or ion in the complex is the apparent charge carried by it in the complex.
• It depends upon the atomic number and electronic configuration of the metal atom or ion.
• The coordination number, the formula and geometry of a complex depend upon the oxidation state of the metal
  atom or ion.

Question 7.
What is the charge on a monodentate ligand X in the complex, [NiX₄]^2-?
Answer:
The charge number of the complex ion is – 2. Nickel being divalent, its oxidation state is + 2. If the charge on monodentate ligand X is y, then Charge number = charge on Ni²⁺ charge on 6X – 2 = + 2 + 4 xy
∴ y = – 1
Hence the charge on ligand X is – 1.

Question 8.
Calculate the oxidation state of a metal in the following complexes :
(a) [Fe(NH₃)₆](NO₃)₃
(b) Ni(CO)₅.
Answer:
(a) [Fe(NH₃)₆](NO₃)₃ ⇌ [Fe(NH₃)₆]³⁺ + 3NO₃^–

NH₃ is a neutral ligand, and the charge number of complex ion is + 3.
If the oxidation state of Fe is x then,
+ 3 = x + 6(0)
∴ x = + 3
∴ The oxidation state of Fe is +3.

(b) Ni(CO)₅ is a neutral complex and CO is a neutral ligand. If the oxidation state of Ni is x, then zero = x + 5 x (zero)
∴ x = zero.
The oxidation state of Ni is zero.

Question 9.
Define and explain the term coordination number (C.N.) of a metal in the complex.
Answer:
Coordination number or legancy (C.N.) : The number of (monodentate) ligands which are directly bonded by coordinate bonds to central metal atom or ion in a coordination compound is called coordination number (C.N.) of the metal atom or ion.

Explanation :

• The coordination number (C.N.) is a characteristic property of the metal and its electronic configuration.
• C.N. takes the values from 2 to 10, of which 4 and 6 are very common.
• The light transition metals show C.N. 4 and 6 while the heavier transition metals show C.N. 8.
• The geometry and shape of a complex compound depends upon C.N. of the metal.

Question 10.
Mention primary valence, secondary valence and coordination number in the following complexes :
(a) [Cu(NH₃)JCI₂
(b) [Co(NH₃)₃CI₃]
(C) K4[Fe(CN)₆]
(d) [CoF₆]₃
(e) [Pt(NH₃)₂Cl₂]
(f) [Pt(NH₃)₂(P_y)₃CI₂]
(g) Cr(CO)₆
(h) [Ni(CN)₄]^2-
Answer:

Question 11.
Classify the following complexes as homoleptic and heteroleptic complex :
(a) [Cu(NH₃)₄]SO₄;
(b) [Cu(en)₂(H₂O)CI]²⁺
(c) [Fe(H₂O)₅(NCS)]²⁺
(d) Tetraaminezinc(II) nitrate.

Question 12.
Summarise the rules of IUPAC nomenclature of coordination compounds.
Answer:
Following rules are followed for naming coordination compounds recommended by IUPAC :

1. In case of a complexion or a neutral molecule, name the ligand first and then the metal.
2. The names of anionic ligands are obtained by changing the ending -ide to -o and -ate to -ato.
3. The name of a complex is one single word. There must not be any space between different ligand names as well as between ligand name and the name of the metal.
4. After the name of the metal, write its oxidation state in Roman number which appears in parentheses without any space between metal name and parentheses.
5. If complex has more than one ligand of the same type, the number is indicated with prefixes, di-, tri-, tetra-, penta-, hexa- and so on.
6. For the complex having more than one type of ligands, they are written in an alphabetical order. Suppose two ligands with prefixes are tetraaqua and dichloro. While naming in alphabetical order, tetraaqua is written first and then dichloro.
7. If the ligand itself contains numerical prefix in its name, then display number by prefixes bis for 2, tris for 3, tetrakis for 4 and so forth. Put the ligand name in parentheses. For example, (ethylenediamine)3 or (en)3 would appear as tris (ethylenediamine) or tris(ethane-l, 2-diamine).
8. The metal in cationic or neutral complex is specified by its usual name while in the anionic complex the name of metal ends with ‘ate’.

Question 13.
State effective atomic number (EAN).
OR
State and explain effective atomic number (EAN). How is it calculated?
Answer:
Effective atomic number (EAN) : It is the total number of electrons present around the central metal atom or ion and calculated as the sum of electrons of metal atom or ion and the number of electrons donated by ligands.

It is calculated by the formula : EAN = Z – X + Y where.
Z = Atomic number of metal atom
X = Number of electrons lost by a metal atom forming a metal ion
Y = Total number of electrons donated by all ligands in the complex.

Generally the value of EAN is equal to the atomic number of the nearest inert element.

Explanation : Consider a complex ion [Co(NH₃)₆]³⁺
Oxidation state of cobalt is + 3 hence X = 3.
There are six ligands, hence Y = 2 x 6 = 12
Atomic number of cobalt. Z = 27
∴ EAN = Z – X + Y = 27 – 3 + 12 = 36.

Question 14.
Find effective atomic number (EAN) in the following complexes :
(1) [Ni(CO)₄]
(2) [Fe(CN)₆]^4-
(3) [Co(NH₃)₆]³⁺
(4) [Zn(NH₃)J²⁺
(5) [Pt(NH₃)₆]⁴⁺
Answer:

Question 15.
What is effective atomic number (EAN) in the following complexes ?
(1) [Fe(CN)₆]^3-
(2) [CU(NH₃)₄]²⁺
(3) [Pt(NH₃)₄]²⁺
Answer:

Question 16.
Calculate EAN in the following complexes :
(1) [Cr(H₂O)₂(NH₃)₂(en)]CI₃;
(2) [Ni(en)₂]SO₄;
(3) Na3[Cr(C₂O₄)₃].
Answer:

Question 17.
Define in coordination compounds :
(1) Isomerism
(2) Isomers.
Answer:

1. Isomerism : It is the phenomenon in coordination compounds having same molecular formula but different physical and chemical properties due to different arrangements of the ligands around the central metal atom or ion in the space.
2. Isomers : The isomers are the coordination compounds having same molecular formula but different physical and chemical properties due to the difference in arrangements of the ligands in the space.

Question 18.
Mention the types of isomerisms in coordination compounds.
Answer:
There are two principal types of isomerisms in coordination compounds as follows :
(A) Stereoisomerism
(B) Structural isomerism (OR Constitutional isomerism)

(A) Stereoisomerism is further classified as :

• Geometrical isomerism
• Optical isomerism

(B) Structural isomerism is further classified as :

• Ionisation isomerism
• Linkage isomerism
• Coordination isomerism
• Solvate (or hydrate) isomerism

Question 19.
Why does stereoisomerism arise in the coordination compounds?
Answer:
In the coordination compounds (complexes) the ligands are linked to the central metal atom or ion by coordinate bonds which are directional in nature and hence give rise to the phenomenon of stereoisomerism.

In this isomerism, the different stereoisomers have different arrangements of ligands (atoms, molecules or ions) in space around the central metal atom or ion. Hence they have different physical and chemical properties and give rise to the phenomenon of stereoisomerism.

Question 20.
Define, in coordination compounds : (1) Stereoisomerism (2) Stereoisomers.
Answer:
(1) Stereoisomerism The phenomenon of isomerism in the coordination compounds arising due to different spatial positions of the ligands in the space around the central metal atom or ion is called stereoisomerism.

(2) Stereoisomers : The coordination compounds having same molecular formula but different stereoisomerism due to different spatial arrangements of the ligand groups in the space around the central metal atom or ion are called stereoisomers.

Question 21.
Define :
(1) Geometrical isomerism and
(2) Geometrical isomers.
Answer:
(1) Geometrical isomerism : The phenomenon of isomerism in the heteroleptic coordination compounds with the same molecular formula but different spatial arrangement of the ligands in the space around the central metal atom or ion is called geometrical isomerism.

(2) Geometrical isomers : The heteroleptic coordination compounds having same molecular formula but different geometrical isomerism due to different spatial arrangements of the ligands in the space around the central metal atom or ion are called geometrical isomers.

Question 22.
Define cis and trails isomers in the coordination compounds.
Answer:
(1) Cis-isomer : A heteroleptic coordination compound in which two similar ligands are arranged adjacent to each other is called cis-isomer. For example,

Gis-Diamminedichloroplatinum(II)

(2) Trans-isomer : A heteroleptic coordination compound in which two similar ligands are arranged diagonally opposite to each other is called trans-isomer. For example,

Trans-Diamminedichloroplatinum(II)

Question 23.
Write structures for geometrical isomers of Diamminebromochloroplatinum(II).
Answer:

Question 24.
Explain the geometrical isomerism of the octahedral complex of the type [MA₄B₂] with a suitable example.
Answer:

• Consider an octahedral complex of a metal M with coordination number six and monodentate ligands a and b having formula [MA₄B₂],
• CA-isomer is obtained when both the B ligands occupy adjacent (1,2) positions.
• Trans-isomer is obtained when the ligands B occupy the opposite (1,6) positions.
• For example, consider a complex [CO(NH₃)₄CI₂]⁺. The structures of cis and trans isomers are
  

Question 25.
Explain the geometrical isomerism of the octahedral complex of the type [M(AA)₂B₂]^n± with a suitable example.
Answer:

• Consider an octahedral complex of metal M with coordination number six and a bidentate ligand AA and monodentate ligand B having molecular formula [M(AA)₂B₂]^n± .
• Bidentate ligand AA has two identical coordinating atoms.
• Cis- isomer is obtained when two bidentate AA ligands as well as two ‘B’ ligands are at adjacent positions.
• Trans-isomer is obtained when two AA ligands and two B ligands are at opposite positions.
• For example, consider a complex [Co(en)₂CI₂]⁺.
  

Question 26.
Explain the geometrical isomerism of the octahedral complex of the type [MA4BC] with suitable example.
Answer:

• Consider an octahedral complex of metal M with coordination number six and monodentate ligands A, B andC.
• Cis-isomer is obtained when both the ligands B and C occupy adjacent (1,2) positions.
• Trans-isomer is obtained when the ligands B and C occupy opposite positions.
• For example, consider a complex [Pt(NH₃)₄BrCI] of the type [MA₄BC],
  

Question 27.
Define : (1) Optical isomerism (2) Optical isomers.
Answer:
(1) Optical isomerism : The phenomenon of isomerism in which different coordination compounds having same molecular formula have different optical activity is called optical isomerism.

(2) Optical isomers : Different coordination compounds having same molecular formula but different optical activity
are called optical isomers.

Question 28.
Explain : (1) Plane polarised light (2) Optical activity.
Answer:
(1) Plane polarised light : A monochromatic light having vibrations only in one plane is called a plane polarised light. This light is obtained by passing monochromatic light through NICOL prism.

(2) Optical activity : A phenomenon of rotating a plane of a plane polarised light by an optically active substance is
called optical activity. This substance is said to be optically active.

Question 29.
Explain : (1) Dextrorotatory substance (2) Laevorotatory substance. (1 mark each)
Answer:

1. Dextrorotatory substance : An optically active substance which rotates the plane of a plane polarised light to right hand side is called dextrorotatory or d isomer denoted by d.
2. Laevorotatory substance : An optically, active substance which rotates the plane of a plane polarised light to the left hand side is called laevorotatory or l isomer and denoted by l.

Question 30.
What are the conditions for the optical isomerism in coordination compounds?
Answer:

• Optical isomerism is exhibited by those coordination compounds which possess chirality.
• There should not be the presence of element of symmetry which makes the complex optically inactive.
• The mirror images of the complex molecule or ion must be non-superimposable with the molecule or ion. B

Question 31.
What are enantiomers?
Answer:
Enantiomers : The two forms of the optical active complex molecule which are mirror images of each other are called enantiomers.

There are two forms of enantiomers, d form and l form.

Question 32.
Draw diagrams for the optical isomers of a complex, [Co(en)₃]³⁺.
Answer:
The complex [Co(en)₃]³⁺ has two optical isomers.

Question 33.
Explain the optical isomerism in the octahedral complex with two symmetrical bidentate chelating ligands.
Answer:
The octahedral complexes of the type [M(AA)2Q2]”±, in which two symmetrical bidentate chelating ligands like AA and two monodentate ligands like a are coordinated to the central metal atom or ion exhibit optical isomerism and two optical isomers d and l can be resolved. For example, [Pt Cl2(en)2]2.

The cis-form is unsymmetrical and optically active while the trans-form is symmetrical and hence optically inactive. The optical isomers of cis-form (d and ) of this complex along with trans-form are shown below,

Question 34.
When are optical isomers called chiral?
Answer:
When the mirror images of optical isomers of the complex are nonsuperimposable they are said to be chiral. For example, [Co(en)₂(NH₃)₂]³⁺.

Question 35.
Define and explain ionisation isomerism.
Answer:
Ionisation isomerism : The phenomenon of isomerism in the metal complexes in which there is an exchange of ions between coordination (or inner) sphere and outer sphere is known as ionisation isomerism.

Explanation :

• Ionisation isomers have same molecular formula but different arrangement of ions in the inner sphere and outer sphere in the complex,
• Hence on ionisation, these ionisation isomers produce different ions in the solution. This ionisation isomerism is also called ion-ion exchange isomerism.

Examples:
(A) [CO(NH₃)₄CI₂]Br and (B) [Co(NH₃)₄CIBr] Cl

Ionisation :
(A) [CO(NH₃)₄CI₂] Br ⇌ [CO(NH₃)₄CI₂]⁺ + Br-
(B) [Co(NH₃)₄CIBr]Cl ⇌ [CO(NH₃)₄ClBr]⁺ + CP
In the isomers (A) and (B), there is an exchange of ions namely Br^– and CI^–.

Question 36.
Define : (1) Linkage isomerism (2) Linkage isomers.
OR
What is linkage isomerism ? Explain with an example.
Answer:
(1) Linkage isomerism : The phenomenon of isomerism in which the coordination compounds have same metal atom or ion and same ligand but bonded through different donor atoms or linkages is known as linkage isomerism.

(2) Linkage isomers : The coordination compounds having same metal atom or ion and ligand but bonded through different donor atoms or linkages are called linkage isomers.
For example : Nitro complex [CO(NH₃)₅NO₂]CI₂ = (Yellow) and nitrito complex [CO(NH₃)₅ONO]CI₂ (Red)

Question 37.
Explain linkage isomers with NO₂^– group as a ligand.
Answer:
(1) Nitro group (_–NO₂^–) is an ambidentate ligand. NO₂^– group may link to central metal atom, through N or O.
(2) The two linkage isomers are, [CI: → Ag ← : NO₂]^– and [CI: → Ag ← O-NO]^–
Choloronitroargentate(I) ion and Chloronitritoargentate(I) ion

Question 38.
Write linkage isomers of a complex having constituents Co³⁺, 5NH3 and NO^–₂.
Answer:
(i) NO^–₂ is an ambidentate ligand which can be linked through N or O.
(ii) The linkage isomers are as follows :
(a) [CO(NH₃)₅(NO₂)]²⁺ Pentaamminenitrocobalt(III) ion
(b) [CO(NH₃)₅(ONO)]²⁺ Pentaamminenitritocobalt(III) ion

Question 39.
Define: (1) Coordination isomerism (2) Coordination isomers.
Answer:
(1) Coordination isomerism : The phenomenon of isomerism in the ionic coordination compounds having the same molecular formula but different complex ions involving the interchange of ligands between cationic and anionic spheres of different metal. ions is called coordination isomerism.

(2) Coordination isomers : The ionic coordination compounds having same molecular formula but different complexions duc to interchange of ligands between cationic and anionic spheres of different metal ions are called coordination isomers.

For example,

Question 40.
Give three examples of coordination isomers. (I mark each)
Answer:

• [Cu(NH₃)₄] [PICI₄] and I Pt(NH₃)₄] [ICuCl4₄]
• [Cr(NH₃)₆] [Cr(CN)₆] and [Cr(NH₃)₄(CN)₂] [Cr(NH₃)₂(CN)₄]
• [Cr(NH₃)₆] [Cr(SCN)₆] and [Cr(NH₃)₄(SCN)₂] [Cr(SCN)₄(NH₃)₂]

Question 41.
Define Solvate or Hydrate isomerism.
Answer:
Solvate ate or Hydrate isomerism : The phenomenon of isomerism in the coordination compounds arising due to the exchange of solvent or H₂O molecules inside the coordination sphere and outer sphere of the complex is known as solvate or hydrate isomerism.

Question 42.
Define solvate or hydrate isomers.
OR
What are hydrate isomers? Explain with examples.
Answer:
Solvate or Hydrate isomers : The coordination compounds having the same molecular formula but differ in the number of solvent or H₂O molecules inside the coordination sphere and outer sphere of the complexes are called solvate or hydrate isomers.

For example : [Cr(H₂O)₆] CI₃; [Cr(H₂O)₅CI]CI₂ H₂O; and [Cr(H₂O)₄CI₂] CI 2H₂O.

Question 43.
A coordination compound has the formula COCI₃ 6H₂O. Write the hydrate isomers of the complex.
Answer:
The possible hydrate isomers of the coordination compounds having molecular formula COCI₃ 6H₂O are as follows :
(1) [CO(H₂O)₆]CI₃;
(2) [CO(H₂O)₅CI]CI₂ H₂O
(3) [CO(H₂O)₄CI₂] Cl – 2H₂O
(4) [Co(H₂O)₃ CI₃] 3H₂O.

Question 44.
Explain the steps involved in describing the bonding in coordination compounds using valence bond theory.
Answer:

• Vacant d-orbitais of metal ion form coordination bonds with ligands.
• s, p orbitais along with vacant d-orbitais of metal ion take part in hybridisation.
• The number of vacant hybrid orbitais formed is equal to number of hybridising orbitais which is equal to the number of ligand donor atoms or coordination number of the metal.
• The metal-ligand coordination bonds are formed by the overlap between the vacant hybrid orbitais of metal and the filled orbitais of the ligands.
• The hybrid orbitais used by the metal ion point in the direction of the ligands.
• When inner (n – 1)d orbitais of metal ion are used in the hybridisation then the complex is called (a) inner orbital complex while when outer nd orbitais are used, complexes are called (b) outer orbital complexes. .

Question 45.
Explain the steps involved ¡n the metal-ligand bonding.
Answer:

• Find the oxidation state of central metal ion in the complex.
• Write the valence shell electronic configuration of metal ion.
• From the formula of the complex determine the number of ligands and find the number of metal ion orbitais required for bonding.
• Find the orbitais of metal ion available for hybridisation and the type of hybridisation involved.
• Represent the electronic configuration of metal ion after hybridisation.
• Exhibit filling of hybrid orbitais after complex formation.
• Determine the nunther of unpaired electrons and predict magnetic property of the complex.
• Find whether the complex is low spin or high spin (applicable for octahedral complexes with d⁴ or d⁸ electronic configuration.)

Question 46.
What are the salient features of valence bond theory (VBT)?
Answer:
The salient features of valence bond theory (VBT) are as follows :

1. According to this theory, a central metal atom or ion present in a complex provides a definite number of vacant orbitals (s, p, d and) to accommodate the electrons from the ligands for the formation coordinate bonds with the metal ion atom.
2. The number of vacant orbitals provided by the central metal atom or ion is the same as the coordination number of the metal. For example : Cu²⁺ provides 4 vacant orbitals in the complex. [Cu(NH₃)₄]²⁺.
3. The vacant orbitals of metal atom or ion undergo hybridisation forming the same number of hybridised orbitals, since the bonding with the hybrid orbitals is stronger.
4. Each ligand has one or more orbitals containing one or more lone pairs of electrons.
5. The shape or geometry of the complex depends upon the type of hybridisation of the metal atom.
6. When inner orbitals namely (n – 1) d orbitals in transition metal atom or ion hybridise, the complex is called inner complex and when outer orbitals i.e., nd orbitals hybridise then the complex is called outer complex.
7. When the central metal atom or ion in the complex contains one or more unpaired electrons the complex is
   paramagnetic while if all the electrons are paired, the complex is diamagnetic.

Question 47.
What is the spin pairing process in the coordination compound?
Answer:
When the ligands approach the metal atom or ion for the formation of a complex, they influence the valence electrons of metal atom or ion. Accordingly the ligands are classified as (A) strong ligands and (B) weak ligands.

(A) Strong ligands :

• They cause the pairing of unpaired electrons present in the metal atom or ion.
• Spin pairing process :
  • The process of pairing of unpaired electrons in metal atom or ion due to the presence of ligands in the complex is called spin pairing process.
  • This spin pairing process decreases the number of unpaired electrons and hence decreases the paramagnetic character of the complex.
  • The strong ligands also promote the outer ns electrons to the vacant inner (n – 1)d orbitals.

(B) Weak ligands : The weak ligands have no effect on the electrons in the valence shell of a metal atom or ion.
Strong ligands : CO, CN, ethylenediammine (en), NH₃, EDTA, etc.
Weak ligands : CI^–, I^–, OH^–, etc.

[Note : If a complex has n number of unpaired electrons then the magnetic moment, μ is given by ‘spin only’ formula μ = n(n + 2) B.M. where B.M. (Bohr Magneton) is the unit of magnetic moment. Hence from the magnitude of p, the number of unpaired electrons in the complex and its structure can be evaluated.]

Question 48.
Explain the structure of octahedral complex, [CO(NH₃)₆]³⁺ on the basis of valence bond theory.
Answer:
(1) Hexaamminecobalt(III) ion, [CO(NH₃)₆]³⁺ is a cationic complex, the oxidation state of cobalt is + 3 and the coordination number is 6.

(2) Electronic configuration : ₂₇CO [Ar]¹⁸ 3d⁷ 4s²
Electronic configuration : Co³⁺ [Ar]¹⁸ 3d⁶ 4s° 4p°

(3) Since NH₃ is a strong ligand, due to spin pairing effect, all the four unpaired electrons in 3d orbital are paired giving two vacant 3d orbitals.

(4) Since the coordination number is Co³⁺ ion gets six vacant orbitais by hybridisation of two 3d vacant orbitais, One 4s and three 4p orbitais forming six d²sp³ hybrid orbitais giving octahedral geometly. It is an inner complex.

(5) 6 lone pairs of electrons from 6NH₃ ligands are accommodated in the six vacant d²sp³ hybrid orbitals. Thus six hybrid orbitals of Co³⁺ overlap with filled orbitals of NH³ forming 6 coordinate bonds giving octahedral geometry to the complex.

Since the complex has all electrons paired, it is diamagnetic.

Question 49.
Explain the geometry of [CoF₆]^3- on the basis of valence bond theory.
Answer:
(1) Hexafluorocobaltate(III) ion, [CoF]^3- is an anionic complex, the oxidation state of cobalt is +3 and the coordination number is 6.
(2) Electronic configuration : ₂₇Co [Ar]¹⁸ 3d⁷ 4s² 4p° 4d°
Electronic configuration : Co³⁺ [Ar]18 3d⁶ 4s° 4p° 4d°

(3) Since F is a weak ligand, there is no spin pairing effect and Co³⁺ possesses 4 unpaired electrons.
(4) Since the coordination number is 6, the Co³⁺ ion gets six vacant orbitals by hybridisation of one 45 orbital, three 4p orbitals and two 4d orbitals forming six sp³d² hybrid orbitals giving octahedral geometry.

(5) 6 lone pairs of electrons from 6F^– ligands are accommodated in the six vacant sp³d² hybrid orbitals. Thus six hybrid orbitals of Co³⁺ overlap with filled orbitals of F^– forming 6 coordinate bonds giving octahedral geometry to the complex. It is an outer complex.

As the complex has 4 unpaired electrons it is paramagnetic.
Magnetic movement μ is,

Question 50.
Explain the structure of tetrachloronickelate(II) [NiCI]^2- on the basis of valence bond theory.
Answer:
(1) Tetrachloronickelate(II) ion is an anionic complex, oxidation state of Ni is +2 and the coordination number is 4.
(2) Electronic configuration : ₂₈Ni [Ar]¹⁸ 3d⁸ 4s² 4p°
Electronic configuration : Ni²⁺ [Ar]¹⁸ 3d⁸ 4s° 4p°

(3) Since the coordination number is 4, it gets 4 vacant hybrid orbitals by sp³ -hybridisation of one 4s and three 4p orbitals giving tetrahedral geometry to the complex.

(4) As Cl is a weak ligand, 2 unpaired electrons in Ni²⁺ remain undisturbed.
(5) 4 lone pairs of electrons from 40 ligands are accommodated in the vacant four sp³ hybrid orbitals. Thus four sp³ hybrid orbitals of Ni²⁺ overlap with filled orbitals of Cl^– forming 4 coordination bonds, giving tetrahedral geometry to the complex.
Since the complex has 2 unpaired electrons, it is paramagnetic.

Magnetic moment \(\mu \text { is, } \mu=\sqrt{n(n+2)}=\sqrt{2(2+2)}=2.83 \text { B.M. }\)

Question 51.
Explain the structure of [Ni(CN)₄]^2- on the basis of valence bond theory.
Answer:
(1) Tetracyanonickelate (II) ion, INi(CN)₂]^2- is an anionic complex, oxidation state of Ni is + 2 and the coordination number is 4.
(2) Electronic configuration : ₂₂Ni [Ar]¹⁸ 3d⁸ 4s² 4p°
Electronic configuration: Ni²⁺ [Ai]¹⁸ 3d⁸ 4s° 4p°

(3) Since CN^– is a sa-ong ligand, one of the unpaired electrons in 3d orbital is promoted giving two paired electrons and one vacant 3d orbital.

(4) Since the coordination number is 4, Ni²⁺ gets 4 vacant hybrid orbitais by hybridisation of one 3d, one 4s and two 4p orbitais forming four dsp² hybrid orbitaIs. This has square planar geometry.

(5) 4 lone pairs from 4CN^– ligands are accommodated in the vacant four dsp² hybrid orbitais. Thus four dps² hybrid orbitais of Ni²⁺ overlap with filled orbitais of CN forming 4 coordinate bonds giving square planar geometry to the complex. It is an inner complex.

Since the complex ion has all electrons paired, it is diamagnetic.

Question 52.
What are the limitations of valence bond theory?
Answer:
In case of the coordination compounds, the valence bond theory has the following limitations :
(1) It cannot explain the spectral properties (colours) of the complex compounds.
(2) Even if the magnetic moments can be calculated from the number of unpaired electrons, it cannot explain the magnetic moment arising due to orbital motion of electrons.
(3) It cannot explain why the metal ions with the same oxidation state give inner complexes and outer complexes with different ligands.
(4) In every complex, it cannot explain magnetic properties based on geometry of the complex.
(5) Quantitative interpretations of thermodynamic and kinetic stabilities of the coordination compounds cannot be accounted.
(6) The complexes with weak field ligands and strong field ligands cannot be distinguished.
(7) It cannot predict the tetrahedral and square planar geometry of complexes with coordination number 4.
(8) The order of reactivity of inner complexes of d³, d⁴, d⁵ and d⁶ metal ions cannot be explained.
(9) It cannot explain the rates and mechanisms of reactions of the coordination compounds.

Question 53.
What are the assumptions of Crystal Field Theory (CFT)?
Answer:
Bethe and van Vleck developed Crystal Field Theory (CFT) to explain various properties of coordination compounds. The salient features of CFT are as follows :

1. In a complex, the central metal atom or ion is surrounded by various ligands which are either negatively charged ions (F^–, CI^–, CN^–, etc.) or neutral molecules (H₂O, NH₃, en, etc.) and the most electronegative atom in them points towards central metal ion.
2. The ligands are treated as point charges involving purely electrostatic attraction between them and metal ion.
   • The central metal ion has five, (n – 1)d degenerate orbitals namely d_xy, d_yz, d_zx, d_{(x² – y²)} and d_z².
   • When the ligands approach the metal ion, due to repulsive forces, the degeneracy of <i-orbitals is destroyed and they split into two groups of different energy, t_2g and e_g orbitals. This effect is called crystal field splitting which depends upon the geometry of the complex.
   • The T-orbitals lying in the direction of ligands are affected to a greater extent while those lying in between the ligands are affected to a less extent.
   • Due to repulsion, the orbitals along the axes of ligands acquire higher energy while those lying in between the ligands acquire less energy.
   • Hence repulsion by ligands give two sets of split up orbitals of metal ion with different energies.
   • The energy difference between two sets of d-orbitais after splitting by ligands is called crystal field splitting energy (CFSE) and represented by Δ₀ or by arbitrary term 10D_q. The value of Δ or 10D_q depends upon the geometry of the complex.
     
3. The electrons of metal ion occupy the split d-orbitais according to Hund’s rule. aufbau principle and those orbitais with minimum repulsion and the farthest away from the ligands.
4. CFI’ does not account for overlapping of orbitais of central metal ion and ligands, hence does not consider covalent nature of the complex.
5. From the crystal field stability energy, the stability of the complexes can he known.
   

Question 54.
What is crystal field splitting?
Answer:
The splitting of five degenerate d-orbitals of the transition metal ion into different sets of orbitals (to_2g and e_g) having different energies in the presence of ligands in the complex is called crystal field splitting.

Question 55.
What is crystal field stabilisation energy?
Answer:
Crystal field stabilisation energy (CFSE) It is the change in energy achieved by preferential filling up of the orbitals by electrons in the complex of metal atom or ion.

CFSE is expressed as a negative .quantity i.e., CFSE < 0. Higher the negative value more is the stability of the complex. m

Question 56.
Explain the factors affecting Crystal Field Splitting parameter (Δ₀).
Answer:
Crystal Field Splitting parameter (Δ₀) depends on. (a) Strength of the ligands and (b) Oxidation state of the metal.

(a) Strength of the ligands : Since strong field ligands like CN^–, en, etc. approach closer to the central metal ion, it results in a large crystal field splitting and hence Δ₀ has higher values.
(b) Oxidation state of the metal A metal ion with the higher positive charge draws the ligands closer to it which results in large separation of t_2g and e_g set of orbitals. The complexes involving metal ions with low oxidation state have low values of Δ₀. For example [Fe(NH₃)₆]³⁺ has higher Δ₀ than [Fe(NH₃)₆]²⁺. H

Question 57.
Explain the octahedral geometry of complexes using crystal field theory.
Answer:
(1) In an octahedral complex [MX₆]ⁿ⁺, the metal atom or ion is placed at the centre of regular octahedron while six ligands occupy the positions at six vertices of the octahedron.

(2) Among five degenerate d-orbitals. two orbitals namely d_{x² – y²} and d_z² are axial and have maximum electron density along the axes, while remaining three c-orbitals namely d_xy, d_yz and d_yz are planar and have maximum electron density in the planes and in-between the axes.

(3) Hence, when the ligands approach a metal ion, the orbitals d_{x² – y²} and d_z² experience greater repulsion and the orbitals d_zy, d_yz and d_zx experience less repulsion.

(4) Therefore the energy of d_{x² – y²} and d_z² increases while the energy of d_xy, d_yz and d_zx decreases and five d-orbital lose degeneracy and split into two point groups. The orbitals d_xy, d_yz and d_yz form t_2g group of lower energy while d_{x² – y²} and d_z² form e group of higher energy.

Thus t_2g has three degenerate orbitals while e_g has two degenerate orbitals.

(5) Experimental calculations show that the energy of t_2g orbitals is lowered by 0.4Δ₀ or 4D_q and energy of e_g. orbital is increased by 0.6 Δ₀ or 6D_q Thus energy difference between t_2g and e_g orbitals is Δ₀ or 10D_g which is crystal field splitting energy.

(6) CFSF increases with the increasing strength of ligands and oxidation stale of central metal ion.

Question 58.
Explain the tetrahedral geometry of complexes using crystal field theory.
Answer:
(1) In the tetrahedral complex, [MX₄]^n±, the metal atom or ion is placed at the centre of the regular tetrahedron and the four ligands, are placed at four corners of the tetrahedron.

(2) The ligands approach the central metal atom or ion in-between the three coordinates x, y and The orbitals d_xy, d_yz and d are pointed towards ligands and experience greater repulsion while the axial orbitals d_x² – _y² and d_z² lie in-between metal-ligand bond axes and experience comparatively less experience.

Fig. 9.12 (a) and (b) : Tetrahedral geometry hasing central metal atoll) (M) at the centre and four ligands (L) al the four corners

(3) Therefore energy of d_xy, d_yz and dd_zx orbitals is increased while that of d_x²y² and d_z² is lowered. Hence 5d-orbitals lose their degeneracy and split into two point groups, namely f2? of higher energy (d_xy, d_yz and d_zx) and e_g of lower energy (d_{x² – y²} and d_z²).

(4) The experimental calculations show that the energy of t_2g orbitals is increased by 0.4 Δ₀ or 4D_q and energy of e_g is lowered by 0.6 Δ₀ or 6D_g. Thus energy difference between t_2g and e_g orbitals is Δ₀ or 10D_g which is crystal field splitting energy (CFSE).

(5) This explains that the entry of each electron in e_g orbitals, stabilises the complex by 0.6 Δ₀ or 6D_q While the entry of each electron in t_2g orbitals destabilises the tetrahedral complex by 0.4 Δ₀ or 4D_g.

(6) In case of strong field ligands, the electrons prefer to pair up in e_g orbitals giving low spin (LS) complexes while in case of weak held ligands, the electrons prefer to enter higher energy t_2g orbitals giving more unpaired electrons and hence form high spin (HS) complexes.

Table 9.4 : Properties of complexes

Question 59.
Give valence bond description for the bonding in the complex [VCI₄]^–. Draw box diagrams for free metal ion. Which hybrid prbitals are used by the metal? State the number of unpaired electrons.
Answer:

Since CI^– is a weak ligand, there is no pairing of electrons.
Number of unpaired electrons = 2
Type of hybridisation = sp³

Geometry of complex ion = Tetrahedral
The complex ion is paramagnetic.

Question 60.
Write a note on colour in coordination compounds.
Answer:

• A large number of coordination compounds show wide range of colours due to d – d transition of electron and this can be explained by crystal field theory (CFT).
• The complex absorbs the light in one visible region (400 nm to 700 nm) and transmits the light in different visible region giving complementary colour.
• Consider an octahedral purple coloured complex of [Ti(H₂O)₆]³⁺ which absorbs green light and transmits purple colour. Similarly [Cu(H₂O)₆]²⁺ absorbs the light in the red region of radiation spectrum and transmits in the blue region, hence the complex appears blue.
• The absorption of light arises due to d-d transition of electron from lower energy level (t_2g) to higher energy level (e_g) in octahedral complex.
• The energy required for transition depends upon crystal field splitting energy Δ₀. If Δ₀ = ΔE, then the energy of an absorbed photon (hv) is \(\Delta E=h v=\frac{h c}{\lambda}\) where λ, v and c are wavelength, frequency and velocity of the absorbed light.
• Higher the magnitude of Δ₀ or ΔE, higher is the frequency or lower is the wavelength of the absorbed radiation.
• Since Δ₀ depends upon nature of metal atom or ion, its oxidation state, nature of ligands and the geometry of the complex, different coordination compounds have different colours.

Question 61.
Explain the purple colour of the complex, [Ti(H₂O)₆]³⁺ with the help of crystal field theory.
Answer:
(1) [Ti(H₂O)₆]³⁺ is an octahedral complex, oxidation state of titanium is +3 (Ti³⁺) and the coordination number is 6.

(2) Electronic configuration = ₂₂Ti [Ar]¹⁸ 3d² 4s²
Electronic configuration = Ti³⁺ + [Ar]¹⁸ 3d¹
OR

(3) According to crystal field theory, 3d orbitals undergo crystal field splitting giving higher energy e_g, two orbitals and lower energy t_2g, three orbitals.

(4) The crystal field splitting energy (CFSE), Δ₀ is found to be 3.99 x 10^-19 J/ion from the spectrochemical studies.

(5) The absorption of radiation of wavelength λ or frequency v results in the transition of one unpaired electron from Photon energy t_2g orbital to e_g orbital.

(6) The wavelength of the absorbed radiation will be.

(7) Hence the complex [Ti(H₂O)₆]³⁺ absorbs the green radiation of wavelength 498 nm in the visible region and transmits the complementary purple light. Therefore the complex is purple coloured.

Question 62.
An octahedral complex absorbs the radiation of wavelength 620 nm. Find the crystal field splitting energy.
Answer:
Crystal field splitting energy Δ₀ is given by,

Question 63.
What are the applications of coordination compounds ?
Answer:
(1) In biology : Several biologically important natural compounds are metal complexes which play an important role in number of processes occurring in plants and animals.

For example, chlorophyll in plants is a complex of Mg²⁺ ions, haemoglobin in blood is a complex of iron, vitamin B₁₂ is a complex of cobalt.

(2) In medicine : The complexes are used on a large scale in medicine. Many medicines in the complex form are more stable, more effective and can be assimilated easily.

For example, platinum complex [Pt(NH₃)₂CI₂] known as cisplatin is effectively used in cancer treatment. EDTA is used to treat poisoning by heavy metals like lead.

(3) To estimate hardness of water :

• The hardness of water is due to the presence Mg²⁺ and Ca²⁺ ion in water.
• The strong field ligand EDTA forms stable complexes with Mg²⁺ and Ca²⁺. Hence these ions can be removed by adding EDTA to hard water.

Similarly these ions can be selectively estimated due to the difference in their stability constants.

(4) Electroplating : This involves deposition of a metal on the other metal. For smooth plating, it is necessary to supply continuously the metal ions in small amounts.

For this purpose, a solution of a coordination compound is used which dissociates to a very less extent. For example, for uniform and thin plating of silver and gold, the complexes K[Ag(CN)₂] and K[Au(CN)₂] are used.

Multiple Choice Questions

Select and write the most appropriate answer from the given alternatives for each subquestion :

Question 1.
The coordination number of cobalt in the complex [Co(en)₂Br₂]CI₂ is
(a) 4
(b) 5
(c) 6
(d) 7
Answer:
(c) 6

Question 2.
EDTA combines with cations to form
(a) chelates
(b) polymers
(c) clathrates
(d) non-stoichiometric compounds
Answer:
(a) chelates

Question 3.
Which one of the following compounds can exhibit coordination isomerism?
(a) [Co(en)₂CI₂]Br
(b) [CO(NH₃)₆] [Cr(CN)₆]
(c) [Co(en)₃]CI₃
(d) [CO(NH₃)₅NO₂]CI₂
Answer:
(b) [CO(NH₃)₆] [Cr(CN)₆]

Question 4.
Which of the following compounds can exhibit linkage isomerism?
(a) [Co(en)₃]CI₃
(b) [Co(en)₂CI₂]CI
(c) [Co(en)₂NO₂Br]CI
(d) [Co(NH₃)₅CI]Br₂
Answer:
(c) [Co(en)₂NO₂Br]CI

Question 5.
Oxidation number of cobalt in K[COCI₄] is
(a) +1
(b) -1
(c) +3
(d) -3
Answer:
(b) -1

Question 6.
The correct structure of [Cr(H₂O)₆]³⁺ is ……………………
(a) octahedral
(b) tetrahedral
(c) square pyramidal
(d) trigonal bipyramidal
Answer:
(b) tetrahedral

Question 7.
Amongst the following ions which one has the highest paramagnetism?
(a) [Cr(H₂O)₆]³⁺
(b) [Fe(H₂O)₆]²⁺
(c) [CU(H₂O)₆]²⁺
(d) [Zn(H₂O)₆]²⁺
Answer:
(b) [Fe(H₂O)₆]²⁺

Question 8.
The geometry of [Ni(CN)₄]^3- and [NiCI₄]^-2 are
(a) both tetrahedral
(b) both square planar
(c) tetrahedral and square planar respectively
(d) square planar and tetrahedral respectively
Answer:
(d) square planar and tetrahedral respectively

Question 9.
The complex cis-[Pt(NH₃)₂CI₂] is used in treatment of cancer under the name.
(a) Aspirin
(b) Eqanil
(c) cisplatin
(d) transplatin
Answer:
(c) cisplatin

Question 10.
[CO(NH₃)₆]³⁺ is an orbital complex and is in nature.
(a) inner, paramagnetic
(b) inner, dimagnetic
(c) outer, paramagnetic
(d) outer, dimagnetic
Answer:
(b) inner, dimagnetic

Question 11.
The IUPAC name of [Ni(Co)₄] is
(a) tetra carbonyl nickel (O)
(b) tetra carbonyl nickel (II)
(c) tetra carbonyl nickelate (O)
(d) tetra carbonyl nickelate (II)
Answer:
(a) tetra carbonyl nickel (O)

Question 12.
The number of ions produced by the complex [CO(NH₃)₄CI₂] Cl is
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(b) 2

Question 13.
The dimagnetic species is
(a) [Ni(CN)₄]^2-
(b) [NiCl₄]^2-
(c) [CoCI₄]^2-
(d) [CoF₆]^2-
Answer:
(a) [Ni(CN)₄]^2-

Question 14.
Which one of the following is an inner orbital complex as well as diamagnetic in behaviour (Atomic no. Zn = 30, Cr = 24, Co = 27, Ni = 28)
(a) [Zn(NH₃)₆]²⁺
(b) [Cr(NH₃)₆]³⁺
(c) [CO(NH₃)₆]³⁺
(d) [Ni(NH₃)₆]²⁺
Answer:
(c) [CO(NH₃)₆]³⁺

Question 15.
Among [Ni(Co)J, [Ni(CN)4]2A [NiClJ2- Species, the hybridisation states at the Nickel atom are respectively
(a) sp³, dsp², sp³
(b) sp³, dsp², dsp²
(c) dsp², sp³, sp³
(d) sp³, sp³, dsp²
Answer:
(a) sp³, dsp², sp³

Question 16.
The strongest ligand in the following is
(a) CN^–
(b) Br^–
(c) HO^–
(d) F^–
Answer:
(a) CN^–

Question 17.
Magnetic moment of (NH4)2 (MnBr4) is BM
(a) 5.91
(b) 4.91
(c) 3.91
(d) 2.91
Answer:
(a) 5.91

Question 18.
The complex which violates EAN rule is
(a) Fe(CO)₅
(b) [Fe(CN)₆]^3-
(c) Ni(CO)₄
(d) [Zn(NH₃)₄]CI₂
Answer:
(b) [Fe(CN)₆]^3-

Question 19.
EDTA is a ligand of the type
(a) bidentate
(b) tridentate
(c) tetradentate
(d) hexadentate
Answer:
(d) hexadentate

Question 20.
The cationic complex among the following is
(a) K₃[Fe(CN)₆]
(b) Ni(CO)₄
(c) K₂HgI₄
(d) [CO(NH₃)₆]CI₂
Answer:
(d) [CO(NH₃)₆]CI₂

Question 21.
If Z is the atomic number of a metal, X is number of electrons lost forming metal ion and Y is the number of electrons from the ligands then EAN is
(a) Z + X + Y
(b) X – Z + Y
(c) Z – X + Y
(d) X + Z – Y
Answer:
(c) Z – X + Y

Question 22.
Octahedral complex has hybridisation,
(a) dsp²
(b) d³sp³
(c) dsp³
(d) d²sp³
Answer:
(d) d²sp³

Question 23.
Inner complex has hybridisation,
(a) d²sp³
(b) sp³d²
(c) sp³d
(d) sp³d³
Answer:
(a) d²sp³

Question 24.
The number of unpaired electrons in [CO(NH3)6]3+ is
(a) 0
(b) 1
(c) 2
(d) 4
Answer:
(a) 0

Question 25.
The number of unpaired electrons in [NiClJ2- and [Ni(CN)4]2_ are respectively,
(a) 2, 2
(b) 2, 0
(c) 0, 0
(d) 1, 2
Answer:
(b) 2, 0

Question 26.
Among the following complexes, the highest magnitude of crystal field stabilisation energy will be for [Co(H₂O)₆]³⁺, [CO(CN)₆]^3-, [Co(NH₃)₆]³⁺, [CoF₆]^3-
(a) [Co(H₂O)₆]³⁺
(b) [CO(CN)₆]^3-
(c) [Co(NH₃)₆]³⁺
(d) [CoF₆]^3-
Answer:
(b) [CO(CN)₆]^3-

Question 27.
The number of unpaired electrons in a low spin octahedral complex ion of d1 is
(a) 0
(b) 1
(c) 2
(d) 3
Answer:
(b) 1

Question 28.
The number of unpaired electrons in a high spin octahedral complex ion of d⁷ is
(a) 0
(b) 1
(c) 2
(d) 3
Answer:
(d) 3

Question 29.
Ligand used in the estimation of hardness of water is
(a) EDTA
(b) DBG
(c) chloride
(d) bromo
Answer:
(a) EDTA

Question 30.
Which of the following complexes will give a white precipitate on treatment with a solution of barium nitrate?
(a) [Cr(NH₃)₄SO₄] CI
(b) [CO(NH₃)₄CI₂] NO₂
(c) [Cr(NH₃)₄CI₂] SO₄
(d) [CrCI₃(H₂O)₄]CI
Answer:
(c) [Cr(NH₃)₄CI₂] SO₄

Question 31.
What is effective atomic number of Fe (z = 26) in [Fe(CN)₆]^4-?
(a) 12
(b) 30
(c) 26
(d) 36
Answer:
(d) 36

Question 32.
Cisplatin compound is used in the treatment of
(a) malaria
(b) cancer
(c) AIDS
(d) yellow fever
Answer:
(b) cancer

Balbharti Maharashtra State Board 12th Chemistry Textbook Solutions Chapter 9 Coordination Compounds Textbook Exercise Questions and Answers.

Maharashtra State Board Class 12 Chemistry Solutions Chapter 9 Coordination Compounds

1. Choose the most correct option.

Question i.
The oxidation state of cobalt ion in the complex [Co(NH₃)₅Br]SO₄ is ……………………….
a. + 2
b. + 3
c. + 1
d. + 4
Answer:
(b) + 3

Question ii.
IUPAC name of the complex [Pt(en)₂(SCN)₂]²⁺ is ………………………
a. bis (ethylenediamine dithiocyanatoplatinum (IV) ion
b. bis (ethylenediamine) dithiocyantoplatinate (IV) ion
c. dicyanatobis (ethylenediamine) platinate IV ion
d. bis (ethylenediammine)dithiocynato platinate (IV) ion
Answer:
(a) bis(ethylenediamine dithiocyanatoplatinum (IV) ion

Question iii.
Formula for the compound sodium hexacynoferrate (III) is
a. [NaFe(CN)₆]
b. Na₂[Fe(CN)₆]
c. Na[Fe(CN)₆]
d. Na₃[Fe(CN)₆]
Answer:
(d) Na₃[Fe(CN)₆]

Question iv.
Which of the following complexes exist as cis and trans isomers?
1. [Cr(NH₂)₂Cl₄]^⊕
2. [Co(NH₃)₅Br]^2⊕
3. [PtCl₂Br₂]^2⊕ (square planar)
4. [FeCl₂(NCS)₂]^2⊕ (tetrahedral)
a. 1 and 3
b. 2 and 3
c. 1 and 3
d. 4 only
Answer:
(a) 1 and 3

Question v.
Which of the following complexes are chiral?
1. [Co(en)₂Cl₂]^⊕
2. [Pt(en)Cl₂]
3. [Cr(C₂O₄)₃]^3⊕
4. [Co(NH₃)4CI₂]^⊕
a. 1 and 3
b. 2 and 3
c. 1 and 4
d. 2 and 4
Answer:
(a) 1 and 3

Question vi.
On the basis of CFT predict the number of unpaired electrons in [CrF₆]^3⊕.
a. 1
b. 2
c. 3
d. 4
Answer:
(c) 3

Question vii.
When an excess of AgNO₃ is added to the complex one mole of AgCl is precipitated. The formula of the complex is ……………..
a. [CoCl2(NH₃)₄]Cl
b. [CoCl(NH₃)₄] Cl₂
c. [CoCl₃(NH₃)₃]
d. [Co(NH₃)₄]Cl₃
Answer:
(a) [COCI₃(NH₃)₄]CI

Question viii.
The sum of coordination number and oxidation number of M in [M(en)₂C₂O₄]Cl is
a. 6
b. 7
c. 9
d. 8
Answer:
(c) 9

2. Answer the following in one or two sentences.

Question i.
Write the formula for tetraammineplatinum (II) chloride.
Answer:
Formula of tetraamineplatinum(II) chloride : [Pt(NH₃)₄]CI₂

Table 9.1 : IUPAC names of anionic and neutral ligands

Table 9.2: IUPAC names of anionic complexes

Metal	Name
A1 Cr Cu Co Au(Gold) Fe Pb Mn Mo Ni Zn Ag Sn	Aluminate Chromate Cuprate Cobaltate Aurate Ferrate Plumbate Manganate Molybdate Nickelate Zincate Argentate Stannate

Table 9.3 : IUPAC names of some complexes

Complex	IUPAC name
(i) Anionic complexes :
(a) [Ni(CN)J2- (b) [Co(C204)3]3- (c) [Fe(CN)6]4-	Tetracyanonickelate(II) ion Trioxalatocobaltate(III) ion Hexacyanoferrate(II) ion
(ii) Compounds containing complex anions and metal cations :
(a) Na3[Co(N02)6] (b) K3[A1(C204)3] (c) Na3[AIF6]	Sodium hexanitrocobaltate(III) Potassium trioxalatoaluminate(III) Sodium hexafluoroaluminate(III)
(iii) Cationic complexes :
(a) [Cu(NH3)4]2+ (b) [Fe(H20)5(NCS)]2+ (c) [Pt(en)2(SCN)2]2+	Tetraamminecopper(II) ion Pentaaquai sothiocyanatoiron(III) ionBis(ethylenediamine)dithiocyanatoplatinum(IV)
(iv) Compounds containing complex cation and anion :
(a) [PtBr2(NH3)4]Br2 (b) [Co(NH3)5C03]CI (c) [Co(H20)(NH3)5]I3	Tetraamminedibromoplatinum(IV) bromide, Pentaamminecarbonatocobalt(III) chloride, Pentaammineaquacobalt(III) iodide
(v) Neutral complexes :
(a) Co(N02)3(NH3)3 (b) Fe(CO)5 (c) Rh(NH3) 3(SCN) 3	Triamminetrinitrocobalt(III) Pentacarbonyliron(0) Triamminetrithiocyanatorhodium(III)

Question ii.
Predict whether the [Cr(en)₂(H₂O)₂]³⁺ complex is chiral. Write structure of its enantiomer.
Answer:
(i) Complex is chiral.
(ii) The following are its enantiomers

Question iv.
Name the Lewis acids and bases in the complex [PtCl₂(NH₃)₂].
Answer:
Lewis acid : Pt²⁺
Lewis bases : Cl^– and NF₃

Question v.
What is the shape of a complex in which the coordination number of central metal ion is 4?
Answer:
A complex with the coordination number of central metal ion equal to 4 may be tetrahedral or square planar.

Question vi.
Is the complex [CoF₆] cationic or anionic if the oxidation state of cobalt ion is +3?
Answer:
In the complex, Co carries + 3 charge while 6F^– carry – 6 charge. Hence the net charge on the complex is – 3.
Therefore it is an anionic complex.

Question vii.
Consider the complexes [Cu(NH₃)₄][PtCl₄] and [Pt(NH₃)₄] [CuCl₄]. What type of isomerism these two complexes exhibit?
Answer:
Since in these two given complexes, there is an exchange of ligands between cationic and anionic constituents, they exhibit coordination isomerism.

Question viii.
Mention two applications of coordination compounds.
Answer:
(1) In biology : Several biologically important natural compounds are metal complexes which play an important role in number of processes occurring in plants and animals.

For example, chlorophyll in plants is a complex of Mg²⁺ ions, haemoglobin in blood is a complex of iron, vitamin B₁₂ is a complex of cobalt.

(2) In medicine : The complexes are used on a large scale in medicine. Many medicines in the complex form are more stable, more effective and can be assimilated easily.

For example, platinum complex [Pt(NH₃)₂CI₂] known as cisplatin is effectively used in cancer treatment. EDTA is used to treat poisoning by heavy metals like lead.

(3) To estimate hardness of water :

• The hardness of water is due to the presence Mg²⁺ and Ca²⁺ ion in water.
• The strong field ligand EDTA forms stable complexes with Mg²⁺ and Ca²⁺. Hence these ions can be removed by adding EDTA to hard water.

Similarly these ions can be selectively estimated due to the difference in their stability constants.

(4) Electroplating : This involves deposition of a metal on the other metal. For smooth plating, it is necessary to supply continuously the metal ions in small amounts.
For this purpose, a solution of a coordination compound is used which dissociates to a very less extent. For example, for uniform and thin plating of silver and gold, the complexes K[Ag(CN)₂] and K[Au(CN)₂] are used.

3. Answer in brief.

Question i.
What are bidentate ligands? Give one example.
Answer:
Bidentate ligand : This ligand has two donor atoms in the molecule or ion. For example, ethylenediamine, H₂N – (CH₂)₂ – NH₂.

Question ii.
What is the coordination number and oxidation state of metal ion in the complex [Pt(NH₃)Cl₅]²?
Answer:
Coordination number = 6
Oxidation state of Pt = +4.

Question iii.
What is the difference between a double salt and a complex? Give an example.
Answer:

Double salt	Coordination compound (complex)
(1) Double salts exist only in the solid state and dissociate into their constituent ions in the aqueous solutions.	(1) Coordination compounds exist in the solid-state as well as in the aqueous or non-aqueous solutions.
(2) Double salts lose their identity in the solution.	(2) They do not lose their identity completely.
(3) The properties of double salts are same as those of their constituents.	(3) The properties of coordination compounds are different from their constituents.
(4) Metal ions in the double salts show their normal valence.	(4) Metal ions in the coordination compounds show two valences namely primary valence and second­ary valence satisfied by anions or neutral molecules called ligands.
(5) For example in K2SO4. K2SO4. A12(SO4)3. 24H2O. The ions K+, Al3 + and SO4 show their properties.	(5) In K4[Fe(CN)6], ions K+ and [Fe(CN)6]4‘~ ions show their properties.

Question iv.
Classify the following complexes as homoleptic and heteroleptic
[Cu(NH₃)₄]SO₄, [Cu(en)₂(H₂O)Cl]^3⊕, [Fe(H₂O)₅(NCS)]^2⊕, tetraammine zinc (II) nitrate.
Answer:
Homoleptic complex :
(a) [Cu(NH₃)₄]SO₄
(d) Tetraaminezinc (II) nitrate : [Zn(NH₃)₄](NO₃)₂

Heteroleptic Complex :
(b) [Cu(en)₂(H₂O)CI]²⁺
(c) [Fe(H₂O)₅(NCS)]²⁺

Question v.
Write formulae of the following complexes
a. Potassium ammine-tri chloroplatinate (II)
b. Dicyanoaurate (I) ion
Answer:
(a) Potassium amminetrichloroplatinate(II) K[Pt(NH₃)CI₃]
(b) Dicyanoaurate (I) ion [AU(CN)₂]^–

Question vi.
What are ionization isomers? Give an example.
Answer:
Ionisation isomers : The coordination compounds having same molecular composition but differ in the compositions of coordination (or inner) sphere and outer sphere and produce different ions on ionisation in the solution are called ionisation isomers. For example, Pentaamminesulphatocobalt (III) bromide [Co(NH₃)₅SO₄] Br, Pentaamminebromocobalt(III) sulphate [Co(NH₃)₅Br] SO₄.

Question vii.
What are the high-spin and low-spin complexes?
Answer:
(1) High spin complex (HS) :

• The complex which has greater iwmher of unpaired electrons and hence a higher value of resultant spin and magnetic moment is called high spin (or spin free) or IlS complex.
• It is formed with weak field ligands and the complexes have lower values for crystal field splitting energy (CFSE). Δ₀
• The paramagnetism of HS complex is larger.

(2) Low spin complex (LS) :

• The complex which has the Icasi number of unpaired electrons or all electrons paired and hence the lowest
  (or no) resultant spin or magnetic moment is called low spin (or spin paired) or LS complex.
• It is formed with strong tickl ligands and the complexes have higher values of crystal field splitting energy (Δ₀).
• Low spin complex is diamagnetic or has low paramagnetism.

Table 9.5 : d-orbitai diagrams fir high spin and low spin complexes

(Only the electronic configurations c4 to d1 render the high spin and low spin complexes)

Question viii.
[CoCl₄]^2⊕ is a tetrahedral complex. Draw its box orbital diagram. State which orbitals participate in hybridization.
Answer:
₂₇Co [Ar] 3d⁷4s²
Oxidation state of Co = +2 Co²⁺ [Ar] 3d⁷ 4s°
Since CI^– is a weak ligand, there is no pairing of electrons. Since C.N. is 4, there is sp³ hybridisation.

Question ix.
What are strong field and weak field ligands? Give one example of each.
Answer:
The ligands are then classified as (a) strong field and (b) weak field ligands. Strong field ligands are those in which donor atoms are C,N or P. Thus CN^–, NC^–, CO, HN₃, EDTA, en (ethylenediammine) are considered to be strong ligands. They cause larger splitting of d orbitals and pairing of electrons is favoured. These ligands tend to form low spin complexes. Weak field ligands are those in which donor atoms are halogens, oxygen or sulphur.

For example, F^–, CI^–, Br^–, I^–, SCN^–, C₂O₄^2-. In case of these ligands the A0 parameter is smaller compared to the energy required for the pairing of electrons, which is called as electron pairing energy. The ligands then can be arranged in order of their increasing field strength as
I^– < Br^– < CI^– < S^2- < F^– < OH^– < C₂O₄^2- < H₂O < NCS^– < EDTA < NH₃ < en < CN^– < CO.

Question x.
With the help of a crystal field energy-level diagram explain why the complex [Cr(en)₃]^3⊕ is coloured?
Answer:

Since (en) is a strong field ligand there is pairing of electrons. The electrons occupy the t_2g orbitals of lower energy. It has one unpaired electron. Due to d-d transition, it is coloured.

4. Answer the following questions.

Question i.
Give valence bond description for the hybrid orbitals are used by the metal? State the number of unpaired electrons.
Answer:
Since CI^– is a weak ligand, there is no pairing of electrons.
Number of unpaired electrons = 2
Type of hybridisation = sp³

Geometry of complex ion = Tetrahedral
The complex ion is paramagnetic.

Question ii.
Draw a qualitatively energy-level diagram showing d-orbital splitting in the octahedral environment. Predict the number of unpaired electrons in the complex [Fe(CN)₆]^4⊕. Is the complex diamagnetic or paramagnetic? Is it coloured? Explain.
Answer:
(A) r-orbital splitting in the octahedral environment :

(B) [Fe (CN)₆]^4- is an octahedral complex.
(C) Since CN^– is a strong ligand, there is pairing of electrons and the complex is diamagnetic.
(D) The complex exists as lemon yellow crystals.
(In the complex all electrons in t_2g are paired and requires high radiation energy for excitation.)

Question iii.
Draw isomers in each of the following
a. [Pt(NH₃)₂ClNO₂]
b. [Ru(NH₃)4Cl₂]
c. [Cr(en₂)Br₂]^⊕
Answer:
(a) [Pt(NH₃)₂CINO₂]

(b) [RU(NH₃)₄CI₂]

(c) [Cr(en₂)Br₂]⁺

Question iv.
Draw geometric isomers and enantiomers of the following complexes.
a. [Pt(en)₃]^4⊕
b. [Pt(en)₂ClBr]^2⊕
Answer:
The complex [Pt(en)₃]⁴⁺ has two optical isomers.

Question v.
What are ligands? What are their types? Give one example of each type.
Answer:
Ligands : The neutral molecules or negatively charged anions (or rarely positive ions) which are bonded by coordinate bonds to the central metal atom or metal ion in a coordination compound are called ligands or donor groups. For example in [Cu(CN)₄]^2-, four CN^– ions are ligands coordinated to central metal ion Cu²⁺. Ligands can be classified on the basis of number of electron donor atoms in the ligand i.e. denticity.

(1) Monodentate or unidentate ligand : A ligand molecule or an ion which has only one donor atom with a lone pair of electrons forming only one coordinate bond with metal atom or ion in the complex is called monodentate or unidentate ligand. For example NH₃, Cl^–, OH^–, H₂O, etc.

(2) Polydentate or multidentate ligand : A ligand molecule or an ion which has two or more donor atoms with the lone pairs of electrons forming two or more coordinate bonds with the central metal atom or ion in the complex is called polydentate or multidentate ligand. For example, ethylene diamine, H₂N – (CH₂)₂ – NH₂.
According to the number of donor atoms they are classified as follows :

• Bidentate ligand : This ligand has two donor atoms in the molecule or ion. For example, ethylenediamine, H₂N – (CH₂)₂ – NH₂.
• Tridentate ligand : This ligand molecule has three donor atoms or three sites of attachment.
  E.g. Diethelene triamine, H₂N – CH₂ – CH₂ – NH – CH₂ – CH₂ – NH₂. This has three N donor atoms.
• Tetradentate (or quadridentate) ligand : This ligand molecule has four donor atoms.
  Eg. Triethylene tetraamine which has four N donor atoms.
• Hexdentate ligand : This ligand molecule has six donor atoms. E.g. Ethylenediamine tetracetato.

(3) Ambidentate ligand : A ligand molecule or an ion which has two or more donor atoms, however in the formation of a complex, only one donor atom is attached to the metal atom or an ion is called ambidentate ligand. For example, NO^–₂ which has two donor atoms N and O forming a coordinate bond, M ← ONO (nitrito) or M ← NO₂ (nitro).

(4) Bridging ligand : A monodentate ligand having more than one lone pairs of electrons, hence can attach to two or more metal atoms or ions and hence acts as a bridge between different metal atoms is called bridging ligand. For example : OH^–, F^–, SO₄^-2, etc.

Question vi.
What are cationic, anionic and neutral complexes? Give one example of each.
Answer:
(1) Cationic sphere complexes : A positively charged coordination sphere or a coordination compound having a positively charged coordination sphere is called cationic sphere complex.

For example : [Zn(NH₃)₄]²⁺ and [Co(NH₃)₅CI] SO₄ are cationic complexes. The latter has coordination sphere [Co(NH₃)₅CI]²⁺, the anion SO₄²⁺ makes it electrically neutral.

(2) Anionic sphere complexes : A negatively charged coordination sphere or a coordination compound having negatively charged coordination sphere is called anionic sphere complex. For example, [Ni(CN)₄]²⁺ and K₃[Fe(CN)₆] have anionic coordination sphere; [Fe(CN)₆]^3- and three K+ ions make the latter electrically neutral.

(3) Neutral sphere complexes : A neutral coordination complex does not possess cationic or anionic sphere.

[Pt(NH₃)₂CI₂] or [Ni(CO)₄] are neither cation nor anion but are neutral sphere complexes.

Question vii.
How stability of the coordination compounds can be explained in terms of equilibrium constants?
Answer:
Stability of the coordination compounds : The stability of coordination compounds can be explained on the basis of their stability constants. The stability of coordination compounds depends on metal-ligand interactions. In the complex, metal serves as electron-pair acceptor (Lewis acid) while the ligand as Lewis base (since it is electron
donor). The metal-ligand interaction can be realized as the Lewis acid-Lewis base interaction. Stronger the interaction greater is stability of the complex.

Consider the equilibrium for the metal-ligand interaction :
M^a+ + nL^x- ⇌ [ML_n]^a+(-nx)
where a, x, [a + ( – nx)] denote the charge on the metal, ligand and the complex, respectively. Now, the equilibrium constant K is given by

Stability of the complex can be explained in terms of K. Higher the value of K larger is the thermodynamic stability of the complex hence K is called stability constant, and denoted by Kstah. The equilibria for the complex formation with the corresponding K values are given below.

Ag⁺ + 2CN^– ⇌ [Ag(CN)2]^– K = 5.5 x 10¹⁸
Cu²⁺ + 4CN^– ⇌ [CU(CN)4]^2- K = 2.0 x 10²⁷
Co³⁺ + 6NH₃ ⇌ [CO(NH3)6]³⁺ K = 5.0 x 10³³

From the above data, the stability of the complexes is [Co(NH₃)₆]³⁺ > [Cu(CN)₄]^2- > [Ag(CN)₂]^–.

Question viii.
Name the factors governing the equilibrium constants of the coordination compounds.
Answer:
The equilibrium constant of the complex depends on the following factors :

(a) Charge to size ratio of the metal ion : Higher the ratio greater is the stability. For the divalent metal ion complexes their stability shows the trend : Cu²⁺ > Ni²⁺ > Co²⁺ > Fe²⁺ > Mn²⁺ > Cd²⁺. The above stability order is called the Irving-William order. In the above list both Cu and Cd have the charge + 2, however, the ionic radius of Cu2 + is 69 pm and that of Cd2 + is 97 pm. The charge to size ratio of Cu²⁺ is greater than that of Cd²⁺. Therefore the Cu²⁺ forms stable complexes than Cd²⁺.

(b) Nature of the ligand : A second factor that governs stability of the complexes is related to how easily the ligand can donate its lone pair of electrons to the central metal ion that is, the basicity of the ligand. The ligands those are stronger bases tend to form more stable complexes.

Activity :
1. The reaction of chromium metal with H 2SO4 in the absence of air gives blue solution of chromium ion.
Cr(s) + 2H^⊕(aq) → Cr^2⊕(aq) + H₂(s)
Cr^2⊕ forms octahedral complex with H₂O ligands.
a. Write formula of the complex
b. Describe bonding in the complex using CFT and VBT.
Draw crystal field splitting and valence bond orbital diagrams.

2. Reaction of complex [Co(NH₃)₃(NO₂)₃ with HCl gives a complex [Co(NH₃)₃H₂OCl₂]^⊕ in which two chloride ligands are trans to one another.
a. Draw possible stereoisomers of starting material
b. Assuming that NH₃ groups remain in place, which of two starting isomers would give the observed product?

12th Chemistry Digest Chapter 9 Coordination Compounds Intext Questions and Answers

Use your brain power ……. (Textbook page 192)

Question 1.
Draw Lewis structures of the following ligands and identify the donor atom in them :
NH₃, H₂O.
Answer:

Try this ………. (Textbook page 193)

Question 1.
Can you write ionisation of [Ni (NH₃)₆] CI₂?
Answer:
[Ni (NH₃)₆] CI₂ → [Ni(NH₃)₆]²⁺ + 2CI^–

Question 2.
Identify coordination sphere and counter ions.
Answer:
Coordination sphere : [Ni(NH₃)₆]²⁺
Counter ions : CI^–

Can you tell ? (Textbook page 193)

Question 1.
A complex is made of Co (III) and consists of four NH₃ molecules and two CI^– ions as ligands. What is the charge number and formula of complexion?
Answer:
The complex ion has formula, [Co(NH₃)₄CI₂]⁺.
The charge number is + 1.

Use vour brain power ……………… (Textbook page 193)

Question 1.
Coordination number used in coordination of compounds is somewhat different than that used in solid state. Explain.
Answer:

• In a coordination compound the coordination number is the number of donor atoms of ligands directly attached to metal atom or ion.
• In a solid state, the number of closest constituent atoms or ions in contact with a particular atom in the crystal lattice is called coordination number.
• In a coordination compound, coordination number depends upon nature of metal atom or ion, and its electronic configuration.
• In a solid state, the coordination number depends upon the crystalline structure of the unit cell.

Can you tell? ………………. (Textbook page 194)

Question 1.
What is the coordination number of
(a) Co in [CoCl₂(en)₂]⁺ = 6
(b) Ir in [Ir(C₂O₄)₂Cl₂]³⁺ and
(c) Pt in [Pt(NO₂)₂(NH₃)₂] ?
Answer:
(a) Coordination number of Co in [CoCl₂(en)₂]⁺ = 6
(b) Coordination number of Ir in [Ir(C₂O₄)₂Cl₂]³⁺ = 6
(c) Coordination number of Pt in [Pt(NO₂)₂(NH₃)₂] = 4

Use your brain power ……… (Textbook page 195)

Question 1.
Classify the complexes as homoleptic and heteroleptic:
(a) [Co (NH₃)₅CI]SO₄,
(b) [CO(ONO)(NH₃)₅]CI₂,
(c) [CoCl(NH₃)(en)₂]²⁺ and
(d) [Cu(C2O₄)₃]^3-
Answer:
Homoleptic Complexes : (d) [Cu(C₂O₄)₃]^3-
Heteroleptic Complexes : (a) [CO(NH₃)₅CI]SO₄
(b) [CO(ONO)(NH₃)₅]CI₂,
(C) [CoCl(NH₃)(en)₂]²⁺

Use your brain power ……… (Textbook page 195)

Question 1.
Classify the complexes as cationic, anionic or Cr(H₂O)₂(C₂O₄)_2^3-, PtCI₂(en)₂ and Cr(CO)₆.
Answer:
Cationic complexes : [CO(NH₃)₆]CI₂
Anionic complexes : Na4[Fe(CN)₆], [Cr(H₂O)₂ (C₂O₄)2]^3-
Neutral complexes : Cr(CO)₆, Pt CI₂(en)₂

Try this ……. (Textbook page 197)

Question 1.
Write the representation of the following :
(i) Tricarbonatocobaltate(III) ion.
(ii) Sodium hexacyanoferrate(III).
(iii) Potassium hexacyafioferrate(II).
(iv) Aquachlorobis(ethylenediamine)cobalt(III).
(v) Tetraaquadichlorochromium(III) chloride.
(vi) Diamminedichloroplatinum(II).
Answer:
(i) [Co(C0₃)₃]^3-
(ii) Na₃[Fe(CN)₆]
(iii) K₄[Fe(CN)₆]
(iv) Co(en)₂(H₂O)(Cl)
(v) [Cr(H₂O)₄CI₂]CI
(vi) Pt(NH₃)₂CI₂

Try this …… (Textbook page 196)

Question 1.
Find out the EAN of
(a) [Zn(NH₃)₄]²⁺
(b) [Fe(CN)₆]⁴⁺
Answer:
(a) For the complex ion, [Zn(NH₃)₄]²⁺ :
Atomic number of Zn = Z = 30
Charge on metal ion = + 2
∴ Number of electrons lost by Zn atom = X = 2 Total number of electrons donated by 4NH^₂3
ligands = Y = 2 x 4 = 8
EAN = Z – X + Y
= 30 – 2 + 8
= 36

(Note : This is atomic number of the nearest inert element 3₆Kr.)

(b) For the complex ion, [Fe(CN)J^4- :
For Fe, Z = 26 (Atomic number)
X = 2 (Due to + 2 charge on Fe)
Y = 12 (Due to 6 CN^– ligands)
∴ EAN = Z – X + Y
= 26 – 2 + 12
= 36

Use your brain power …… (Textbook page 197)

Question 1.
Do the following complexes follow the EAN rule
(a) Cr(CO)₄,
(b) Ni(CO)₄,
(c) Mn(CO)₅,
(d) Fe(CO)₅?
Answer:
(a) Cr(CO)₄ : EAN = Z – X + Y
(b) Ni(C0)4 : EAN = Z – X + Y
= 24 – 0 + 8
= 28 – 0 + 8
= 32
= 36
(c) Mn(CO)₅ : EAN = Z – X + Y
= 25 – 0 + 10
= 35

(d) Fe(CO)5 : EAN = Z – X + Y
= 26 – 0 + 10
= 36

Conclusion :
(a) Cr(CO)₄ and (c) Mn(CO)₅ do not follow EAN Rule.

Try this ….. (Textbook page 199)

Question 1.
Draw structures of ci,c and trans isomers of [Fe(NH₃)₂(CN)₄]
Answer:

Remember ….. (Textbook page 199)

Our hands are non-superimposable mirror images. When you hold your left hand up to a mirror the image looks like right hand.

Try this ….. (Textbook page 199)

Question 1.
Draw enantiomers of [Cr(OX)₂]³ where OX = C₂O4 :
Answer:

Question 2.
Draw (A) enantiomers and (B) cis and trans isomers of [Cr(H₂O)₂(OX)₂] :
Answer:
(A) Enantiomers :

(B) as and trans isomers :

Can you tell ? ….. (Textbook page 200)

Question 1.
Can you write IUPAC names of isomers (I) [Co(NH₃)₅SO₄]Br and (II) [Co(NH₃)₅Br]SO₄?
Answer:

Question 2.
Write linkage isomers of [Fe(H₂O)₅SCN]⁺. Write their IUPAC names.
Answer:

Use your brain power …..(Textbook page 201)

Question 1.
The stability constant K of the [Ag(CN)₂]^– is 5.5 x 10 while that for the corresponding [Ag(NH₃)₂]⁺ is 1.6 x 10⁷. Explain why [Ag(CN)₂]^2- is more stable.
Answer:
Stability constant of [Ag(CN)₂]^2- is larger than that of [Ag(NH₃)₂]⁺ and hence [Ag(CN)₂]^2- is more stable. Also, CN is a stronger ligand than NH₃.

Remember …… (Textbook Page 202)

Question 1.
Complete the missing entries.

(Note : The missing entries are underlined.)

Table 9.3: Type of hybridisation and geometry of a complex

Try this ….. (Textbook page 204)

Question 1.
Based on the VBT predict structure and magnetic behaviour of the [Ni(NH₃)₆]
Answer:
₂₈Ni [Ar] 3d⁸ 4s²
Ni³⁺ [Ar] 3d⁷ 4s°
Hybridisation : sp³d²
Geometry : Octahedral
Magnetic property : Paramagnetic

Try this …… (Textbook page 202)

Question 1.
Give VBT description of bonding in each of following complexes. Predict their magnetic behaviour.
(a) [ZnCI₄]²⁺
(b) [CO(H₂O)₆]^2- (high spin)
(c) [Pt(CN)₄]^2- (square planar)
(d) [CoCI₄]^2- (tetrahedral)
(e) [Cr(NH₃)₆]³⁺

Try this ……. (Textbook page 206)

Question 1.
Sketch qualitatively crystal field d orbital energy level diagrams for each of the following complexes :
(a) [Ni(en)₃]²⁺ (b) [Mn(CN)₆]^3- (c) [Fe(H₂O)₆]²⁺
Predict whether each of the complexes is diamagnetic or paramagnetic.
Answer:
(a) The complex ion, [Ni(en)₃]²⁺ is octahedral.
₂₈Ni [Ar] 3d⁸ 4s²
Ni²⁺ [Ar] 3d⁸ 4s°.

Since en is a strong ligand there is pairing of electrons.
Number of unpaired electrons = n = 2 in t_2g, orbitals
Magnetic moment = \(\mu=\sqrt{n(n+2)}\)
\(=\sqrt{2(2+2)}=2.83 \mathrm{~B} . \mathrm{M} .\)

The complex ion is paramagnetic.

(b) The complex ion [Mn(CN)₆]^3- is octahedral.
₂₅Mn [Ar] 3d⁵ 4s²
Mn³⁺ [Ar] 3d⁴ 4s°

Since CN^– is a strong ligand there is pairing of electrons.
Number of unpaired electrons = n = 2 in t_2g, orbitals
Magnetic moment = \(\mu=\sqrt{n(n+2)}\)
\(=\sqrt{2(2+2)}=2.83 \mathrm{~B} . \mathrm{M}\).

The complex ion is paramagnetic.

(c) The complex ion [Fe(H₂O)₆]²⁺ is octahedral.
₂₆Fe [Ar] 3d⁶ 4s²
Fe²⁺ [Ar] 3d⁶ 45°

Since H₂O is a weak ligand, there is no pairing of electrons.
Number of unpaired electrons = n = 4 in t_2g and e_g orbitals.
Magnetic moment
\(\begin{aligned}
=\mu &=\sqrt{n(n+2)} \\
&=\sqrt{4(4+2)} \\
&=4.90 \mathrm{~B} . \mathrm{M} .
\end{aligned}\)
The complex ion is paramagnetic.

Balbharti Maharashtra State Board Class 12 Secretarial Practice Solutions Chapter 10 Dividend and Interest Textbook Exercise Questions and Answers.

Maharashtra State Board Class 12 Secretarial Practice Solutions Chapter 10 Dividend and Interest

1A. Select the correct answer from the options given below and rewrite the statements.

Question 1.
Dividend is paid to ____________
(a) Shareholder
(b) Debenture holder
(c) Depositor
Answer:
(a) Shareholder

Question 2.
____________ is profit shared by company with a shareholder.
(a) Interest
(b) Rent
(c) Dividend
Answer:
(c) Dividend

Question 3.
Dividend is recommended by ____________
(a) Managing Director
(b) Secretary
(c) Board of Directors
Answer:
(c) Board of Directors

Question 4.
Interim Dividend is declared by ____________
(a) Board of Directors
(b) Debenture holders
(c) Depositors
Answer:
(a) Board of Directors

Question 5.
Final Dividend is declared by ____________
(a) Board of Directors
(b) Shareholders
(c) Depositors
Answer:
(b) Shareholders

Question 6.
Dividend cannot be declared out of ____________
(a) Capital
(b) Profit
(c) Reserves
Answer:
(a) Capital

Question 7.
Dividend amount should be transferred in a separate Bank Account within ____________ days of its declaration.
(a) 5
(b) 15
(c) 50
Answer:
(a) 5

Question 8.
Dividend should be paid within ____________ days of its declaration.
(a) 3
(b) 13
(c) 30
Answer:
(c) 30

Question 9.
____________ holders get dividend from residual profits.
(a) Equity share
(b) Preference share
(c) Debenture
Answer:
(a) Equity Share

Question 10.
Dividend is paid first to ____________ shareholders.
(a) Equity
(b) Preference
(c) Deferred
Answer:
(b) Preference

Question 11.
____________ warrant is a cheque containing dividend amount sent by company to the shareholders.
(a) Dividend
(b) Share
(c) Interest
Answer:
(a) Dividend

Question 12.
IEPF is created by ____________ where unpaid dividend is transferred by company.
(a) Central Government
(b) Company
(c) Shareholders
Answer:
(a) Central Government

Question 13.
Payment of ____________ Dividend must be authorised by the Articles of Association.
(a) Interim
(b) Final
(c) Bonus
Answer:
(a) Interim

Question 14.
____________ is a return paid to creditors by the company.
(a) Dividend
(b) Interest
(c) Rent
Answer:
(b) Interest

Question 15.
____________ is not linked to profits of the company.
(a) Dividend
(b) Interest
(c) Bonus
Answer:
(b) Interest

1B. Match the pairs.

Question (I).

Group ‘A’	Group ‘B’
(a) Dividend	(1) Equity Shareholders
(b) Interest	(2) Appropriation of Profit
(c) Interim Dividend	(3) Recommendation of Secretary
(d) Final Dividend	(4) Registrar of Company
(e) Fluctuating Rate of Dividend	(5) Obligatory to pay
	(6) Decided and Declared by the Board of Directors
	(7) Decided by Board and Declared by Members
	(8) Decided by President of India
	(9) Company not allowed to pay
	(10) Declared by Government of India

Answer:

Group ‘A’	Group ‘B’
(a) Dividend	(2) Appropriation of Profit
(b) Interest	(5) Obligatory to pay
(c) Interim Dividend	(6) Decided and Declared by the Board of Directors
(d) Final Dividend	(7) Decided by Board and Declared by Members
(e) Fluctuating Rate of Dividend	(1) Equity Shareholders

Question (II).

Group ‘A’	Group ‘B’
(a) Dividend	(1) Inform stock exchange about dividend declaration
(b) Interest	(2) Creditors
(c) IEPF	(3) Registered Shareholders
(d) Unpaid Dividend Account	(4) Balance of Unpaid Dividend Transferred
(e) Listed Company	(5) Unregistered Company
	(6) Non-listed company
	(7) Unpaid/Unclaimed Dividend
	(8) Balance of unpaid bonus transferred here
	(9) Must inform the government about dividend declaration
	(10) General Public

Answer:

Group ‘A’	Group ‘B’
(a) Dividend	(3) Registered Shareholders
(b) Interest	(2) Creditors
(c) IEPF	(4) Balance of Unpaid Dividend Transferred
(d) Unpaid Dividend Account	(7) Unpaid/Unclaimed Dividend
(e) Listed Company	(1) Inform stock exchange about dividend declaration

1C. Write a word or term or a phrase that can substitute each of the following statements.

Question 1.
The return on investment is paid to the shareholders of the company.
Answer:
Dividend

Question 2.
The meeting where the final dividend is declared.
Answer:
AGM (Annual General Meeting)

Question 3.
The company has to intimate the stock exchange about the declaration of dividends.
Answer:
Listed company

Question 4.
The shareholders get dividends at a fixed rate.
Answer:
Preference

Question 5.
The shareholders get dividends at a fluctuating rate.
Answer:
Equity

Question 6.
Request by the shareholder in the prescribed form for payment of dividend into shareholders bank amount.
Answer:
Dividend Mandate

Question 7.
The number of days within which payment of dividend be completed by the company after its declaration.
Answer:
30 days

Question 8.
Dividend declared between two AGMs.
Answer:
Interim Dividend

Question 9.
Dividend decided and declared by the Board.
Answer:
Interim Dividend

Question 10.
The return is paid to the creditors by the company.
Answer:
Interest

1D. State whether the following statements are True or False.

Question 1.
The dividend is paid to registered shareholders of the company.
Answer:
True

Question 2.
The dividend is decided by the Board.
Answer:
True

Question 3.
The dividend is decided by the shareholders.
Answer:
False

Question 4.
Dividend once declared cannot be revoked.
Answer:
True

Question 5.
Dividend cannot be paid out of capital.
Answer:
True

Question 6.
Shareholders decide about the rate and amount of profit to be given as dividends.
Answer:
False

Question 7.
All categories of shareholders get a fixed-rate dividend.
Answer:
False

Question 8.
IEPF is the fund created by the company.
Answer:
False

Question 9.
Interest is a liability for the company.
Answer:
True

Question 10.
Interest is paid to shareholders of the company.
Answer:
False

1E. Find the odd one.

Question 1.
Final Dividend, Interim Dividend, Interest
Answer:
Interest

Question 2.
Out of Capital, Out of free reserve, Out of money given by the government
Answer:
Out of Capital

Question 3.
Dividend Account, Dividend Mandate, Unpaid/ Unclaimed Dividend Account
Answer:
Dividend Mandate

Question 4.
Dividend warrant, Dividend Mandate, Cheque
Answer:
Dividend Mandate

1F. Complete the sentences.

Question 1.
Word dividend is derived from Latin term ____________
Answer:
Dividendum

Question 2.
Dividend is paid to ____________
Answer:
registered shareholders

Question 3.
Dividend can be declared only on recommendation of ____________
Answer:
Board of Directors

Question 4.
Dividend must be paid in ____________
Answer:
cash

Question 5.
The meeting at which final dividend is approved is ____________
Answer:
Annual General Meeting

Question 6.
Dividend cannot be paid out of ____________
Answer:
capital

Question 7.
Interim dividend is decided and declared by ____________
Answer:
Board of Directors

Question 8.
Predecided and a fixed rate of dividend is paid to ____________
Answer:
preference shareholder

Question 9.
Payment of dividend must be completed within ____________
Answer:
30 days

Question 10.
Payment of Interim Dividend needs to be authorized by ____________
Answer:
Articles of Association

Question 11.
The obligatory payment made by company to its creditors is called as ____________
Answer:
Interest

1G. Select the correct option from the bracket.

Question 1.

Group ‘A’	Group ‘B’
(a) Dividendum	(1) ………………..
(b) Interest	(2) ………………..
(c) …………………	(3) Final Dividend
(d) ………………..	(4) Interim Dividend
(e) Government Fund	(5) …………………

(Latin term, Creditors, At AGM, At Board Meeting, IEPF)
Answer:

Group ‘A’	Group ‘B’
(a) Dividendum	(1) At Board Meeting
(b) Interest	(2) Latin term
(c) IEPF	(3) Final Dividend
(d) At AGM	(4) Interim Dividend
(e) Government Fund	(5) Creditors

Question 2.

Group ‘A’	Group ‘B’
(a) Preference Shares	(1)  ……………………
(b) Equity Shares	(2) ……………………..
(c) Deposit holders	(3) ……………………..
(d) …………………….	(4) Payment of Dividend
(e) …………………….	(5) Dividend Declared but not paid/claimed

(Fixed rate Dividend, Dividend at a Fluctuating Rate, Interest, within 30 days, Unclaimed/Unpaid Dividend)
Answer:

Group ‘A’	Group ‘B’
(a) Preference Shares	(1) Unclaimed/Unpaid Dividend
(b) Equity Shares	(2) Within 30 days
(c) Deposit holders	(3) Interest
(d) Fixed rate of Dividend	(4) Payment of Dividend
(e) Dividend at a Fluctuating Rate	(5) Dividend Declared but not paid/claimed

1H. Answer in one sentence.

Question 1.
What is Dividend?
Answer:
The dividend is a share in distributable profits of the company to which the shareholder in entitled when formally declared by the company.

Question 2.
Who has the right to recommend dividends?
Answer:
The Board of directors has the right to recommend.

Question 3.
What is a Final Dividend?
Answer:
The final dividend is that dividend that is declared and paid after the close of the financial year.

Question 4.
What is an Interim Dividend?
Answer:
The interim dividend is that dividend that is declared and paid between two AGMs of an accounting year.

Question 5.
Who declares Interim Dividend?
Answer:
The Board of directors declares Interim dividends.

Question 6.
Which shares get dividends at a fixed rate?
Answer:
Preference shares get dividends at a fixed rate.

Question 7.
Which shares get dividends at a fluctuating rate?
Answer:
Equity shares get dividends at a fluctuating rate.

Question 8.
At which meeting Interim Dividend is decided and declared?
Answer:
Interim Dividend is decided and declared in Board Meeting.

Question 9.
What is Interest?
Answer:
Interest is the price paid for the productive services rendered by capital

Question 10.
State the time within which unpaid dividends be transferred to the unpaid dividend account.
Answer:
The unpaid dividend should be transferred within 7 days of the end of 30 days within which payment has to be made.

1I. Correct the underlined word and rewrite the following sentences.

Question 1.
The dividend is paid to creditors.
Answer:
The dividend is paid to shareholders.

Question 2.
Interest is paid to shareholders.
Answer:
Interest is paid to creditors.

Question 3.
The final Dividend is paid between two AGM.
Answer:
Interim Dividend is paid between two AGM.

Question 4.
Special Resolution must be passed to declare the Final Dividend.
Answer:
An ordinary resolution must be passed to declare the Final Dividend.

Question 5.
The dividend must be paid within 60 days of its declaration.
Answer:
The dividend must be paid within 30 days of its declaration.

Question 6.
The Dividend to be paid should be transferred to Dividend A/c within 30 days of its declaration.
Answer:
The Dividend to be paid should be transferred to Dividend A/c within 5 days of its declaration

Question 7.
The dividend is an obligation to be paid by a company every year.
Answer:
Interest is an obligation to be paid by a company every year.

Question 8.
Preference shareholders are given the last priority in the payment of dividends.
Answer:
Equity shareholders are given the last priority in the payment of dividends.

Question 9.
Preference shareholders get dividends from residual profits.
Answer:
Equity shareholders get dividends from residual profits.

Question 10.
Dividend is payable every year irrespective of profits made by the company.
Answer:
Interest is payable every year irrespective of profits made by the company.

1J. Arrange in proper order.

Question 1.
(a) Recommendation of Dividend.
(b) Checking sufficiency of profits
(c) Board Meeting
Answer:
(a) Checking sufficiency of profits
(b) Board Meeting
(c) Recommendation of Dividend

Question 2.
(a) Transfer to Dividend Account
(b) Transfer to IEPF
(c) Transfer to Unpaid Dividend Account
Answer:
(a) Transfer to Dividend Account
(b) Transfer to Unpaid Dividend Account
(c) Transfer to IEPF

Question 3.
(a) Closure of Register of Members.
(b) Intimate Stock Exchange of Board Meeting.
(c) Intimate Stock Exchange of declaration of dividend.
Answer:
(a) Intimate Stock Exchange of Board Meeting
(b) Intimate Stock Exchange of declaration of dividend
(c) Closure of Register of Members

Question 4.
(a) Decision on Rate of Dividend
(b) Transfer of IEPF
(c) Payment of Dividend
Answer:
(a) Decision on Rate of Dividend
(b) Payment of Dividend
(c) Transfer to IEPF

Question 5.
(a) Payment of Interim Dividend
(b) Board meeting deciding and declaring Interim Dividend
(c) Authorization of Articles of Association
Answer:
(a) Authorization of Articles of Association
(b) Board meeting deciding and declaring Interim Dividend
(c) Payment of Interim Dividend

2. Explain the following terms/concepts.

Question 1.
Profit
Answer:
Profit is the financial gain from business activity minus expenses. Profit is the income remaining after deducting total costs from total revenue. It is also called financial gain. Profit is the difference between revenues and expenses for a given period. It is the tool for measuring the success of the business. Without profit, the company cannot survive in the market.

Question 2.
Dividend
Answer:
The term dividend is derived from the Latin word ‘Dividendum’ which means that which is to be divided. A dividend is the portion of the company’s earnings distributed to the shareholders decided and managed by the company’s board of directors. The dividend is a share in distributable profits of the company Shareholder is entitled to receive the dividend when it is formally declared by the company

Question 3.
Interest
Answer:
Interest is the cost of borrowing money typically expressed as an annual percentage of a loan. The money people earn on their savings is called Interest. Interest is paid to the lender by the borrower, in case of a loan or from the financial institution to the depositor, in the case of a savings account. In financial terms, it is a payment made for using the money of another i.e. borrower takes money from the lender. Interest is the cost of renting the money for the borrower and it is the income from lending money for the lender.

Question 4.
Final Dividend
Answer:
The final dividend is declared and paid after the financial year is closed. The final dividend is decided and recommended by the Board of Directors. The final dividend is approved by the shareholder in the AGM. The declaration of the final dividend does not require authorization of Articles of Association. The rate of the final dividend is always higher than the Interim dividend. It is declared from sources like the current year’s profits, money provided by Government for dividends, etc.

Question 5.
Interim Dividend
Answer:
The interim dividend is the dividend that is declared and paid in the middle of an accounting year i.e. before the finalization of accounts for the year. Dividend declared by the Board of Directors between two Annual General Meetings is called Interim Dividend. The interim dividend is paid in the middle of the accounting year.

Question 6.
Unpaid Dividend
Answer:
Dividend declared by the company but neither paid to nor claimed by a shareholder within 30 days of its declaration is termed as Unpaid and Unclaimed Dividend.

Question 7.
Unpaid Dividend Account
Answer:
A dividend declared by the company but neither paid to nor claimed by a shareholder is called an Unpaid/Unclaimed Dividend. This unpaid/unclaimed dividend should be transferred to the Unpaid/Unclaimed Account within 30 days of its declaration. This ‘Unpaid Dividend Account’ is opened in a scheduled Bank by the company.

Question 8.
Dividend Mandate
Answer:
The dividend is paid by different modes of payment like cash, cheque, or warrant or by electronic mode. It can also be paid by using the Dividend Mandate. If the shareholder wishes to get dividend credited directly in the Bank Account he is required to send a request in a prescribed form which is called ‘Dividend Mandate’. The dividend Mandate authorizes the company to pay dividends directly to shareholders’ bankers.

Question 9.
IEPF
Answer:
IEPF means Investors Education and Protection Fund. Any amount in the Unpaid Dividend Account of a company that is unpaid/ unclaimed for a period of 7 (seven years) from the date of such a transfer shall be, transferred to ‘Investors Education and Protection Fund’. The claimant can claim his dividend by filling the prescribed form and submitting the necessary documents. The claimant needs to follow the procedure.

Question 10.
Rate of Dividend
Answer:
The return that a shareholder receives on his investment from the company is called a dividend. The dividend is always declared by the company on the face value of a share irrespective of its market value. The rate of dividend is expressed as a percentage of the face value of a share per annum.

3. Study the following case/situation and express your opinion.

1. LMN Co Limited decides to declare a dividend for the financial year 2018-19 in which it has earned profits less than their expectations.

Question (a).
Is Board right in recommending a dividend of Rs. 5/- per share out of free reserves?
Answer:
Yes, Board is right in recommending a dividend of Rs. 5/- per share out of free reserves, as dividends can be paid out of free reserves.

Question (b).
Can Board declare the dividend though it is not approved by AGM?
Answer:
No Board cannot declare the divided if it is not approved by AGM as dividends should be approved by shareholders by passing an ordinary resolution in AGM.

Question (c).
Can the Board give dividends in the form of gifts?
Answer:
No Board cannot give dividends in the form of gifts. It must be paid in cash, not in kind.

2. ABC Co. Ltd. decides to pay Interim Dividend.

Question (a).
Can it be paid out of free reserves?
Answer:
No. the Interim Dividend cannot be paid out of free reserves.

Question (b).
Is the Board right in declaring the same at the Board Meeting?
Answer:
Yes. Board is right in declaring the same at the Board meeting as it has the power to declare an Interim Dividend.

Question (c).
Can the company distribute the same within 30 days of its declaration?
Answer:
Yes, after the declaration, the Interim dividend should be paid within 30 days of its declaration.

3. RAJ Company limited decides to pay Interim Dividend.

Question (a).
Is the Board justified to decide Interim Dividend of Rs. 5/per share even though profits to date are insufficient?
Answer:
The interim dividend is paid out of profits between two annual general meetings. It cannot be paid out of any reserves. So it is not justified.

Question (b).
Can the Board declare it out of Free Reserves?
Answer:
No Board cannot declare an Interim dividend out of free reserves.

Question (c).
Can the Board declare it out of Capital?
Answer:
No Board cannot declare out of capital.

4. DIAMOND Co. Ltd. is considering declaring an Interim Dividend.

Question (a).
In how many days of the declaration it should transfer the funds to Dividend Account?
Answer:
The interim dividend must be transferred to the Dividend Account within 5 days of its declaration.

Question (b).
In how many days it must pay it to shareholders?
Answer:
The interim dividend should be paid within 30 days of its declaration to shareholders.

Question (c).
In how many days of the declaration it must transfer the funds to the Unpaid Dividend A/c?
Answer:
Unpaid/unclaimed Interim dividend should be transferred to ‘Unpaid Dividend Account’ within 7 days of the expiry of 30 days of declaration i.e. 37 days of its declaration.

5. The Board of Directors of STAR Co. Ltd. which is a listed company recommends a dividend of Rs. 15/- per share to be paid in cash.

Question (a).
Is it justified to pay the dividend firstly to its Preference Shareholders and then after to Equity Shareholders?
Answer:
Yes, because preference shares are entitled to the dividend before it is paid to the equity shareholder. Equity shareholders get dividends from residual profits i.e. after paying to preference shareholders.

Question (b).
Is the AGM required to approve the same?
Answer:
Yes for declaration of final divided Approval of AGM is a must.

Question (c).
Can the company pay dividends in cash?
Answer:
Yes, the company pay a dividend in cash and not in kind.

6. GOLD Co. Ltd. declares a dividend of Rs. 10/- per share for F.Y. 2018-19.

Question (a).
Is the company under default, if the dividend was not paid within 30 days of its declaration?
Answer:
Yes, the company is to default as the time limit within which the company must pay dividends after the declaration is 30 days.

Question (b).
Is the company right in transferring the unpaid dividend to its Debenture Reserve Account?
Answer:
No, the company has to transfer the total amount of dividend which remains unpaid/unclaimed to the ‘Unpaid Dividend Account.

Question (c).
Does the company have to transfer the amount of unpaid dividend to IEPF after 30 days?
Answer:
No, any amount in the unpaid dividend account of a company that remains unpaid/unclaimed for a period of 7 years from the date of such a transfer, should be transferred to (IEPF), ‘Investors Education and Protection Fund’.

4. Distinguish between the following:

Question 1.
Final Dividend and Interim Dividend
Answer:

Points	Interim Dividend	Final Dividend
1. Meaning	Interim Dividend is the dividend that is declared between two Annual General Meetings of a company.	The final Dividend is the dividend that is declared at the Annual General Meeting of a company.
2. When declared?	It is declared between two Annual General Meetings.	It is declared after the completion of the financial year of the company.
3. Who declares?	The interim dividend is declared by the Board of Directors by passing a resolution.	The final dividend is decided and recommended by the Board of Directors. It is declared by the shareholders.
4. Authorization	Authorization of Articles is necessary for the declaration of interim dividends.	Authorization of Articles is not necessary for the declaration of the final dividend.
5. Rate of Dividend	The rate of the Interim dividend is lower than the final dividend.	The rate of the final dividend is always higher than the Interim dividend.
6. Source	It is declared out of profits of the current accounting year.	It is declared from different sources like the current year’s profits, free reserves, capital profits, money provided by Government for dividends, etc.
7. Accounting Aspect	It is declared before the preparation of the final accounts of the company.	It is declared only after the accounts of the year are prepared and finalized.

Question 2.
Dividend and Interest
Answer:

Points	Dividend	Interest
1. Meaning	The dividend is the return payable to the shareholders of the company for their investment in the share capital.	It is the return payable to the creditors of the company. For e.g. Debenture holder, Deposit holders.
2. Intervals	Dividends need not be paid on regular basis and they can vary according to the company’s profits.	Interest has to be paid at regular intervals at a fixed rate.
3. Given to whom	It is paid to the member i.e. the owners of the company.	It is paid to the creditor of the company.
4. Expense	Dividends are not the expense as they are based on the profit made. If no profit, they are not paid for that period.	Interest is the expense to the company.
5. Rate of Dividend	The rate of the Interim dividend is lower than the final dividend.	The rate of the final dividend is always higher than the Interim dividend.
6. Obligation	It has to be paid only when the company made profits.	It is not linked to the Profits of the company. It is an obligation for the company.
7. When payable	It is payable when a company earns sufficient profit.	It is payable every year irrespective of the profits of the company.
8. Rate	It is paid at a fluctuating rate to the equity shareholders.	The rate of Interest is Fixed and pre-determined at the time of issue of the security.

5. Answer in brief.

Question 1.
State any four points to be kept in mind by a listed company with respect to dividends.
Answer:
When a company lists its shares on Stock Exchange, additional listing agreements are to be followed which are as follows:

• Stock exchange should be informed if the securities are listed 2 days prior to the Board meeting in which recommendation of final dividend is to be considered.
• Stock Exchange should be informed immediately regarding the declaration of dividend as soon as the Board meeting gets over.
• Notice of closing book should be informed at least 7 (seven) working days before the closure to the stock exchange.
• Transfer Register and Register of Members should be closed.

Question 2.
Discuss any four features of dividend.
Answer:

• It is the portion of profits of the company paid to its shareholders.
• It is payable out of profits of the company.
• It is an unconditional payment made by the company.
• If the company has issued equity shares with differential rights as to dividend, the terms of issue of such shares will govern the rights of shareholders about receiving the dividend.

Question 3.
Explain the features of interest.
Answer:
Interest is the cost of borrowing money typically expressed as an annual percentage of a loan. The money people earn on their savings is called Interest. Interest is paid to the lender by the borrower in case of a loan or from the financial institution to the depositor in the case of a savings account. In financial terms, it is a payment made for using the money of another i.e. borrower takes money from the lender. Interest is the cost of renting the money for the borrower and it is the income from lending money for the lender.
Features:

• Interest is the price paid for the productive services rendered by capital.
• Interest has a direct relation with risk. The higher the risk, the higher is the interest.
• The rate of interest is expressed as the annual percentage of the principal.
• The rate of interest is determined by various factors like money supply, fiscal policy, the volume of borrowings, rate of inflation, etc.
• Interest is a charge against the profit of the Company. Even if, the company makes no profit, interest should be paid.
• The rate of interest is fixed and pre-determined.

6. Justify the following statements.

Question 1.
The dividend is paid out of the profits of the company.
Answer:

• The dividend is the portion of profits of the company paid to its shareholders.
• It is payable out of profits of the company.
• Dividend can be paid out of capital profits on fulfilling these conditions.
• Capital Profits are realized in cash.
• Articles of Association of the company permit such a distribution.
• It remains as profits after revaluation of all assets and liabilities.
• Thus, it is rightly justified that dividend is paid out of profits of the company.

Question 2.
Interim dividends cannot be paid out of free reserves.
Answer:

• Dividend declared by the Board of Directors between two Annual General Meetings is called Interim Dividend.
• The interim dividend shall not be declared out of free reserves.
• In the event of a loss or inadequacy of profits during a financial year, no interim dividend shall be declared.
• The declaration of an interim dividend does not create a debt against a company.
• The board of directors can cancel an interim dividend after declaring it.
• Thus, it is rightly justified that Interim dividends cannot be paid out of reserves.

Question 3.
Annual General Meeting (AGM) is crucial for Final Dividend.
OR
The final Dividend is declared only after the accounts are prepared and finalized.
Answer:

• The final dividend is that dividend that is declared and paid after the closing of the financial year.
• It is decided and recommended by the Board of Directors.
• The rate of final dividend is declared by the shareholders in the AGM.
• It is declared only after the account of the year is prepared and finalized.
• Thus, it is rightly said that AGM is crucial for the Final Dividend. OR Final Dividend is declared only after the accounts are prepared and finalized.
• The final dividend is declared from different sources, and its declaration does not need the authorization of articles.
• Thus, it is rightly said that AGM is crucial for the Final Dividend. OR Final Dividend is declared only after the accounts are prepared and finalized.

Question 4.
Listed Company has to follow additional guidelines on dividend matters.
Answer:

• Notify stock exchange where company’s securities are listed at least 2 (two) days in advance of the date of the meeting of the Board at which recommendation of final dividend is to be considered.
• Intimate Stock Exchange immediately about the declaration of the dividend after the Board Meeting.
• Give notice of Book closure to the stock exchange at least 7(seven) working days before the closure.
• Close the Register of members and the Transfer Register.
• It must use an electronic mode of payment such as Electronic Clearing Services (ECS) or National Electronic Fund Transfer (NEFT); as approved by the Reserve Bank of India (RBI)
• The listed company has to express the dividend on a per-share basis only.

Question 5.
Equity shareholders get the last priority in receiving dividends.
Answer:

• The dividend is the portion of profits of the company paid to its shareholders.
• The dividend is payable only to the registered shareholders of the company.
• Preference shareholders are entitled to the dividend before it is paid to the equity shareholders.
• The equity shares do not enjoy a preference for dividends.
• They do not have priority for the payment of capital at the time of liquidation.
• Equity shareholders will get dividends from residual profit i.e. after paying to preference shareholders and arrears of dividend on cumulative preference shares.
• The equity shares get the last priority in dividends and thus are the residual claimants.
• Thus, it is rightly said that the equity shareholders get the last priority in receiving dividends.

Question 6.
Unpaid dividends cannot be used by the company.
Answer:

• The dividend declared by the company but has not been paid to or claimed by a shareholder within 30 days of its declaration is termed as an unpaid dividend.
• The total amount of dividend which remains unpaid should be transferred to ‘Unpaid Dividend Account’.
• Any amount in the Unpaid Dividend Account of a Company that remains unpaid/unclaimed for a period of 7 years will be transferred to ‘Investors Education and Protection Fund’.
• The company cannot use unpaid dividends. The only claimant of money can claim for it by following certain procedures.
• Thus, it is rightly said that unpaid dividends cannot be used by the company.

Question 7.
Interest is a liability/obligation of the company.
OR
Interest is paid to the creditor of the company.
Answer:

• Interest is a payment made for using another money So it is the cost of renting the money for the borrower and it is the income from lending money for the lender.
• The company has to pay interest, if it has borrowed money from creditors like Debenture holders, Depositors, Bondholders, etc.
• Interest is the liability of the company as it is a payment made for using money from the lender.
• Interest is a charge against the profit of the company.
• Even if, the company makes no profit, it has to pay interest to borrowers.
• Thus, it is rightly said that interest is a liability/ obligation for the company.

Question 8.
Approval of members is not needed for Interim Dividends.
Answer:

• Dividend declared by the Board of Directors between two Annual General Meetings is called Interim Dividend.
• It is paid in the middle of the accounting year.
• It is declared out of profits of the current account year.
• It is declared before the preparation of final accounts of the company.
• The Board of Directors has the power to declare Interim Dividend.
• Articles of Association’ of the Company must authorize the Board of Directors to declare an interim dividend.
• The Board Meeting has to pass a resolution for declaring the Interim dividend.
• Thus, it is rightly said that approval of members is not needed for Interim dividends.

7. Answer the following questions.

Question 1.
Define Dividend and explain its features.
Answer:
The term dividend is derived from the Latin word ‘Dividendum’ which means that which is to be divided. A dividend is the portion of the company’s earnings distributed to the shareholders decided and managed by the company’s board of directors.
The dividend is a share of distributable profits of the company. A shareholder is entitled to receive the dividend when it is formally declared by the company.
Definitions:

• The Institute of Chartered Accountants of India has defined Dividend “as a distribution to shareholders out of profits or reserves available for this purpose”.
• The Supreme Court has defined it as “In case of going – concern, it means the portion of profits of a company, which is allotted to the holders of shares in a company”.

Features of Dividend:

• It is the portion of profits of the company paid to its shareholders.
• It is payable out of profits of the company.
• It is an unconditional payment made by the company.
• The company pays dividends to the equity shareholders and preference shareholders only.
• If the company has issued equity shares with differential rights as to dividend, the terms of issue of such shares will govern the rights of shareholders about receiving the dividend.
• A dividend cannot be declared out of capital.
• Recommendation of the Board of Directors is necessary for the declaration of dividends.
• The dividend is recommended and approved by the Board of Directors by passing a resolution at the Annual General Meeting.
• The previous year’s dividend cannot be declared if that particular year’s Annual Account has been approved in the AGM.
• Dividend once approved and declared by shareholders, creates a debt. It cannot be revoked.
• The dividend includes the interim dividend.
• The dividend must be paid in cash, cheque or transferred through ECS or NEFT and not in kind.
• The dividend is to be paid on the paid-up value of shares.
• Dividend cannot be paid on calls paid in advance.

Question 2.
What is Interest? Explain its features.
Answer:
Interest is the cost of borrowing money typically expressed as an annual percentage of a loan. The money people earn on their savings is called Interest. Interest is paid to the lender by the borrower in case of a loan or from the financial institution to the depositor in the case of a savings account. In financial terms, it is a payment made for using the money of another i.e. borrower takes money from the lender. Interest is the cost of renting the money for the borrower and it is the income from lending money for the lender.
Features:

• Interest is the price paid for the productive services rendered by capital.
• Interest has a direct relation with risk. The higher the risk, the higher is the interest.
• The rate of interest is expressed as an annual percentage of the principal.
• The rate of Interest is determined by various factors like money supply, fiscal policy, the volume of borrowings, rate of inflation, etc.
• Interest is a charge against the profit of the Company. Even if the company makes no profit, interest should be paid.
• The rate of interest is fixed and pre-determined.

Question 3.
Discuss legal provisions for declaration of dividend.
Answer:
The term dividend is derived from the Latin word ‘Dividendum’ which means that which is to be divided.
A dividend means the profit of a company that is not retained in Legal Provisions for declaration of Dividend.
(i) Board Meeting:

• The Board of Directors has the power and authority to declare the dividend.
• The board meeting is called to pass a resolution to discuss the following points.
• Rate of Dividend and amount of Dividend to be paid.
• Book closure date for dividend.
• Date of Annual General Meeting.
• Bank with which a separate account should be opened to remit the dividend amount.

(ii) Shareholders’ Approval:

• The dividend is approved by shareholders by passing an Ordinary Resolution at the Annual General Meeting.
• Shareholders can declare a lower rate of dividend than what is recommended by the Board but not higher than that.
• Once the dividend is declared at the General Meeting, it cannot be canceled. Hence, the company cannot declare dividends for the second time in that year.

(iii) Separate Bank Account:
The company must deposit the dividend amount in a separate bank account i.e. “Dividend Account” opened in a scheduled bank. The dividend must be transferred to this account within 5 days of its declaration.

(iv) Prohibition to pay Dividend:

• A company cannot declare any dividend on equity shares if the company has failed to repay the deposit or any interest on the deposit.
• If the company is found guilty at the time of Payment of Interest to debenture holders, Redemption of Debentures and Preference Shares, Payment of Interest to a financial institution, etc. in that case no dividend can be declared.

Question 4.
Explain Interim Dividend.
Answer:

• The interim dividend is the dividend that is declared and paid in the middle of an accounting year i.e. before the finalization of accounts for the year. Dividend declared by the Board of Directors between two Annual General Meetings is called Interim Dividend.
• The interim dividend is paid in the middle of the accounting year.
• The interim dividend is declared by the Board of directors during any financial year out of surplus in the profit and loss account and out of profits of the financial year.

Features of Interim Dividend:

• The Board of Directors has the power to declare an interim dividend.
• Interim Dividend is only payment on account of the whole dividend for the year.
• The company should provide depreciation for the entire year and not for a part of the year before declaring an interim dividend.
• Interim dividends cannot be paid out of any reserves.
• The Board of directors can declare interim dividend only when it is mentioned in the Articles of Association of the Company.
• A resolution has to be passed in the Board Meeting for declaring the Interim Dividend.
• A separate Bank account should be maintained in a scheduled bank to credit the interim dividend within 5 (five) days of its declaration.
• Interim Dividend should be paid within 30 days of its declaration.
• Unpaid/Unclaimed dividend should be transferred to ‘Unpaid Dividend Account’ within 7 days of the expiry of 30 days of declaration i.e. 37 days of its declaration.
• Any amount remaining Unpaid/Unclaimed in the ‘Unpaid Dividend Account’ for 7 (seven) years should be transferred to IEPF.

Balbharti Maharashtra State Board Class 12 Secretarial Practice Solutions Chapter 9 Depository System Textbook Exercise Questions and Answers.

Maharashtra State Board Class 12 Secretarial Practice Solutions Chapter 9 Depository System

1A. Select the correct answer from the options given below and rewrite the statements.

Question 1.
In physical mode, securities are held in ____________ form.
(a) Paper
(b) Dematerialization
(c) Electronic
Answer:
(a) Paper

Question 2.
Risk of losing certificates exist in ____________ mode.
(a) Physical
(b) Dematerialization
(c) Digital
Answer:
(a) Physical

Question 3.
In Depository system, securities are held in ____________ form.
(a) Scrip based
(b) Electronic
(c) Physical
Answer:
(b) Electronic

Question 4.
____________ is the institute which facilitates electronic holding of securities.
(a) Depository Participant
(b) Issuer
(c) Depository
Answer:
(c) Depository

Question 5.
There is no payment of ____________ when securities are demated.
(a) Octroi
(b) Wealth Tax
(c) Stamp Duty
Answer:
(c) Stamp Duty

Question 6.
Depository Act was passed in ____________
(a) 1919
(b) 1996
(c) 1999
Answer:
(b) 1996

Question 7.
India has a ____________ depository system.
(a) Sole
(b) Multi
(c) Single
Answer:
(b) Multi

Question 8.
____________ is a constituent of depository system.
(a) Government
(b) Issuer
(c) Trust
Answer:
(b) Issuer

Question 9.
____________ is the oldest depository in India.
(a) Dow Jones
(b) NSDL
(c) CDSL
Answer:
(b) NSDL

Question 10.
Demat account is opened by ____________
(a) Beneficial owner
(b) CDSL
(c) SEBI
Answer:
(a) Beneficial owner

Question 11.
Demated shares are ____________
(a) Non-transferable
(b) Fungible
(c) Bearer
Answer:
(b) Fungible

Question 12.
____________ is a unique code given to a security.
(a) IBM
(b) BBM
(c) ISIN
Answer:
(c) ISIN

Question 13.
In India ISIN for corporate securities is allotted by ____________
(a) NSDL
(b) Central Government
(c) State Government
Answer:
(a) NSDL

Question 14.
____________ has to apply for ISIN.
(a) Company
(b) Depository Participant
(c) Depositors
Answer:
(a) Company

Question 15.
____________ has to pay charges to maintain Demat Account.
(a) Investor
(b) Issuer
(c) Depositor
Answer:
(a) Investor

Question 16.
NSDL is promoted by ____________
(a) NSE
(b) BSE
(c) FTSE
Answer:
(a) NSE

Question 17.
CDSL is promoted by ____________
(a) NSE
(b) BSE
(c) FTSE
Answer:
(b) BSE

1B. Match the pairs.

Question 1.

Group ‘A’	Group ‘B’
(a) Bad delivery	(1) 1956
(b) Depository Act	(2) A 12 digits number code
(c) ISIN	(3) Connects Government and Bank
(d) Depository Participant	(4) Second Depository in India
(e) CDSL	(5) The issuer company
(f) Depository	(6) Problem faced in physical mode
(g) Beneficial owner	(7) A 10 digits number/code
	(8) Connect depository and investor
	(9) First depository in the world
	(10) Custodian of securities in electronic form
	(11) Problem faced in electronic mode
	(12) 1996
	(13) Government organization
	(14) The investor

Answer:

Group ‘A’	Group ‘B’
(a) Bad delivery	(6) Problem faced in physical mode
(b) Depository Act	(12) 1996
(c) ISIN	(2) A 12 digits number code
(d) Depository Participant	(8) Connect depository and investor
(e) CDSL	(4) Second Depository in India
(f) Depository	(10) Custodian of securities in electronic form
(g) Beneficial owner	(14) The investor

1C. Write a word or a term or a phrase that can substitute each of the following statements.

Question 1.
This mode of holding securities may result in loss and theft of certificates.
Answer:
Physical mode of securities

Question 2.
The organization holds the securities in electronic mode.
Answer:
Depository

Question 3.
This system eliminates storing of certificates.
Answer:
Depository system

Question 4.
This system allows faster and easier transfer of securities.
Answer:
Depository system

Question 5.
The oldest Depository of India.
Answer:
NSDL (National Security Depository limited)

Question 6.
The country where the Depository system started for the first time.
Answer:
Germany

Question 7.
The registered owner of securities.
Answer:
Beneficial Owner

Question 8.
The agent of the Depository.
Answer:
Depository Participant

Question 9.
This process converts securities into an electronic form from a physical form.
Answer:
Dematerialization

Question 10.
This process converts securities into physical form from electronic form.
Answer:
Rematerialization

Question 11.
This means securities are without distinctive identity numbers.
Answer:
Fungibility

Question 12.
This is the unique code for security given in the Depository system.
Answer:
International Securities Identification Number

1D. State whether the following statements are true or false.

Question 1.
The physical mode of holding Securities is risky.
Answer:
True

Question 2.
Allotment of Securities takes a longer time when in physical mode.
Answer:
True

Question 3.
Transfer of Securities is easier in electronic mode.
Answer:
True

Question 4.
Bad delivery is likely in the Depository system.
Answer:
False

Question 5.
The depository system began in the USA for the first time in the world.
Answer:
False

Question 6.
India has a multi-depository system.
Answer:
True

Question 7.
The depository system is very similar to the banking system.
Answer:
True

Question 8.
DP is a constituent of the Depository system.
Answer:
True

Question 9.
DP is an Agent of the Depository.
Answer:
True

Question 10.
A bank can work as a DP.
Answer:
True

Question 11.
DRF is required for conversion from physical to electronic.
Answer:
True

Question 12.
ISIN is a unique code given to the specific Securities.
Answer:
True

1E. Find the odd one.

Question 1.
Elimination of storage of certificates, theft of certificates, torn certificates
Answer:
Elimination of storage of certificates

Question 2.
NSDL, CDSL, NBFC
Answer:
NBFC

Question 3.
Depository, DP, RBI
Answer:
RBI

Question 4.
DP, BO, State Government
Answer:
State government

Question 5.
Issuer, BO, Central Government
Answer:
Central government

Question 6.
DRF, RRF, PPF
Answer:
PPF

1F. Complete the sentences.

Question 1.
Central location for keeping Securities in demated form is ____________
Answer:
Depository

Question 2.
Freezing of debit/credit of Securities is possible in ____________
Answer:
Dematerialized Securities

Question 3.
First Depository of the world started in the year ____________
Answer:
1947

Question 4.
The Indian Depository Act was passed in the year ____________
Answer:
1996

Question 5.
Link between Depository and investor is ____________
Answer:
Depository participant

Question 6.
Account of Securities of the investor is maintained by ____________
Answer:
Depository Participant

Question 7.
The process which converts physical Securities in electronic form is ____________
Answer:
Dematerialization

Question 8.
The process which converts digital Securities in physical form is ____________
Answer:
Rematerialization

Question 9.
The issuer company must register with ____________
Answer:
Depository

Question 10.
The unique code identifying a security is ____________
Answer:
(ISIN) International Securities Identification Number

Question 11.
The first Depository of India is ____________
Answer:
NSDL (National Security Depository Limited)

1G. Select the correct option from the bracket.

Question (I).

Group ‘A’	Group ‘B’
(a) Dematerialization	(1) ………………..
(b) ……………………..	(2) DP
(c) First Depository of world	(3) …………………
(d) CDSL	(4) ………………..

(1999, Agent of Depository, Germany, Physical to electronic)
Answer:

Group ‘A’	Group ‘B’
(a) Dematerialization	(1) Physical to Electronic
(b) Agent of Depository	(2) DP
(c) First Depository of world	(3) Germany
(d) CDSL	(4) 1999

Question (II).

Group ‘A’	Group ‘B’
(a) First Depository in India	(1) …………………………
(b) ……………………..	(2) Rematerialisation
(c) Fungibility	(3) …………………………
(d) ……………………	(4) ISIN

(12 digital code, NSDL, Electronic to physical, No distinctive number)
Answer:

Group ‘A’	Group ‘B’
(a) First Depository in India	(1) NSDL
(b) Electronic to Physical	(2) Rematerialisation
(c) Fungibility	(3) No distinctive number
(d) 12 digital code	(4) ISIN

1H. Answer in one sentence.

Question 1.
What is the Depository system?
Answer:
A depository system is a system where Securities are held in electronic form.

Question 2.
Give examples of actions termed as corporate action.
Answer:
Payment of dividend, issue of Bonus Shares, offering of right Shares, Early Redemption of Debentures, Mergers and Acquisitions, etc.

Question 3.
When was Depository Act passed in India?
Answer:
Depository Act was passed in India in the year 1996.

Question 4.
What is a DP?
Answer:
DP means Depository participant, who is an agent of Depository.

Question 5.
What is Dematerialisation?
Answer:
Dematerialization is the process of converting physical Securities into electronic.

Question 6.
What is Rematerialization?
Answer:
Rematerialization is the process of converting electronic Securities into physical.

Question 7.
What is ISIN?
Answer:
ISIN is the unique code given to the specific Securities of the company. ISIN refers to ‘International Securities Identification Number’.

Question 8.
Name the Depositories in India?
Answer:

• National Security Depository Limited (NSDL)
• Central Depository Services Limited (CDSL)

1I. Correct the underlined words and rewrite the following sentences.

Question 1.
The electronic mode of holding Securities is risky.
Answer:
The physical mode of holding Securities is risky.

Question 2.
Allotment and Transfer of Securities are time-consuming in electronic mode.
Answer:
Allotment and Transfer of Securities are time-consuming in physical mode.

Question 3.
Banking system leads to a script less capital market.
Answer:
Depository system leads to a scriptless capital market.

Question 4.
Storage of certificates is not required in the physical mode of holding.
Answer:
Storage of certificates is not required in the electronic mode of holding.

Question 5.
India has a single Depository system.
Answer:
India has a multi Depository system.

Question 6.
Depository participant in India has to register under the Partnership Act.
Answer:
Depository participant in India has to register under the Depository Act.

Question 7.
Demat accounts are opened and maintained by the Depository.
Answer:
Demat accounts are opened and maintained by the Depository Participant.

Question 8.
Securities are fungible in physical mode.
Answer:
Securities are fungible in electronic mode.

Question 9.
ISIN is a code given to a company.
Answer:
ISIN is a code given to the security of a company.

Question 10.
ISIN of Indian government Securities is issued by NSDL.
Answer:
ISIN of Indian government Securities is issued by RBI.

1J. Arrange in proper order.

Question 1.
(a) Gets statement of Account
(b) Open Demat Account
(c) Submit DRF
Answer:
(a) Submit DRF
(b) Open Demat Account
(c) Gets statement of Account

Question 2.
(a) Investor (BO) submits an application for Securities to the issuer company
(b) Depository intimates the DP about crediting Bo’s Account
(c) Issuer company gives details of allotment to Depository.
Answer:
(a) Investor (BO) submits an application for Securities to the issuer company
(b) Issuer company gives details of allotment to Depository
(c) Issuer company gives details of allotment to Depository

2. Explain the following terms/concepts.

Question 1.
Depository System
Answer:

• In the Depository system, securities are held in electronic form.
• The transfer and settlement of securities are done electronically.
• The Depository System maintains accounts of the shareholder, enables transfer, collects dividends, bonus shares, etc. on behalf of the shareholder.
• This system is also called a scriptless trading system.
• It keeps the securities safe. It offers scope for paperless trading by using state-of-art technology.

Question 2.
Dematerialization
Answer:

• Dematerialization is a process whereby a client can get physical certificates converted into electronic mode.
• For this client has to surrender the certificates along with the Demat Request Form (DRF) to the DP.

Question 3.
Rematerialization

• Rematerialization is the process whereby a client can get his electronic holdings of securities converted into physical certificates.
• For this client has to give a written request in the form of Remat Request Form (RRF) to the DP.

Question 4.
Fungibility
Answer:

• Fungibility means the state of being interchangeable. Securities issued by the same company of the same class have the same value no matter who owns them.
• The securities held in electronic form are fungible which means they don’t have distinctive numbers.

Question 5.
ISIN
Answer:

• ISIN is a standard numbering system that is accepted globally.
• ISIN consists of a 12 (twelve) digit alpha-numeric code which is divided into 3 (Three) parts.
• The company has to apply for ISIN for its securities with documents like a prospectus.

3. Study the following case/situation and express your opinion.

1. Mr. Z holds 100 shares of peculiar Co. Ltd. In physical mode and wishes to convert the same in electronic mode:

Question (a).
Mr. Z holds a savings bank account with CFDH Bank Ltd. Can he deposit his shares in this account for Demat?
Answer:
No. He can’t, as it is a savings bank account.

Question (b).
What type of account is needed for the same?
Answer:
For holding electronic securities, he needs to open a Demat A/c with the Depository Participant (DP).

Question (c).
Is it the RBI that will be the custodian of shares of Mr. Z after demating?
Answer:
No. RBI won’t be the custodian. After demating concerned ‘Depository’ (NSDL/CDSL) will be the custodian of the shares of Mr. Z.

2. Mr. R holds 100 shares of Peculiar Co. Ltd. In Demat mode.

Question (a).
He wants to transfer one share each to his wife, daughter, and son. Can he do so?
Answer:
Yes, Mr. R can do so, as under dematerialized securities market lot is of one share.

Question (b).
Does he need to submit DRF or DIS if he wants to transfer his shares?
Answer:
He needs to submit DIS (Delivery Instruction Slip) if he wants to transfer his shares.

Question (c).
Can he nominate his wife in his Demat account?
Answer:
Yes. Mr. R can nominate his wife for the Demat account.

3. Mrs. Z wishes to open a Demat account in her name.

Question (a).
Can she open the account going to the Mumbai office of NSDL?
Answer:
No, Mrs. 2 cannot go directly to the Mumbai office of NSDL.

Question (b).
Is she required to pay for the opening of the account and its maintenance?
Answer:
Yes, she is required to pay for the opening of the account and its maintenance.

Question (c).
Does she have to send the shares to the respective company for demating?
Answer:
Yes, she has to send her original share certificate through Depository Participant to the company.

4. Mr. L wants to demat his 25 shares of Peculiar Co. Ltd bearing certificate no 100 and distinctive no 76-100.

Question (a).
Which form is he required to fill as a written request to the DP?
Answer:
Mr. L is required to fill Demat Request Form (DRF) as a written request to DP.

Question (b).
Does he have to fill instrument of transfer if he wishes to transfer the same after demat?
Answer:
No, he needs to fill instrument of transfer after opening Demat A/c.

Question (c).
Does he have to quote certificate no. and distinctive no. if he wishes to transfer his shares after it is in Demat form?
Answer:
No, Mr. L need not quote certificate no. and distinctive no. if he wishes to transfer his shares after it is in Demat form.

5. Mr. S. holds 50 shares of Peculiar Co. Ltd in Demat form. The company has declared a dividend of Rs 5. Per-share and Bonus of 1 : 1 to its shareholders.

Question (a).
How will Mr. S get his dividend?
Answer:
Mr. S will get his dividend credited directly to his Demat A/c.

Question (b).
Will he get a Bonus share in Physical or Demat?
Answer:
He will get Bonus Share in his Demat A/c.

Question (c).
Who is entitled to dividend and Bonus: Mr. S or the depository? (NSDL in this case).
Answer:
Mr. S is entitled to dividends and Bonuses through the depository services.

4. Distinguish between the following.

Question 1.
Dematerialization and Rematerialization.
Answer:

Points	Dematerialization	Rematerialization
1. Meaning	Conversion of physical securities into electronic form is known as ‘Dematerialization’	Conversion of electronic securities into physical form is known as ‘Rematerialization’
2. Conversion	Securities in physical/paper form are converted into electronic form	The electronic form of securities are converted into physical form
3. Form Used	‘Demat Request Form (DRF) is submitted by the investor to the DP	‘Remat Request Form’ (RRF) is submitted by the investor to the DP
4. Process	It is an initial process	It is a reverse process
5. Function/Sequence	It is a primary and principal function of the depository	It is a secondary and supporting function of the depository
6. Distinctive numbers	Demat securities have no distinctive numbers	Remat securities will have certificates and distinctive numbers issued by the company
7. Securities Maintenance Authority	The depository is the custodian of securities and records	The issuing company keeps the record. The investor also maintains the record
8. Time Consumed	It is an easy and time-saving process	It is a complex and time-consuming process

5. Answer in brief.

Question 1.
Explain the disadvantages of the physical mode of holding securities.
Answer:
Securities mean shares, debentures, bonds, units of mutual funds, securities are held in physical form or paper form. Following are the problems/disadvantages of holding the securities in electronic form:
(i) Risk factor:
Paper certificates can be lost, damaged, torn, stolen, or misplaced during transit, etc.

(ii) Efforts in Duplicating:
If the original certificate is lost it becomes difficult to obtain duplicate certificates. It consumes time, effort, and money.

(iii) Delay in the allotment:
Allotment of new securities in physical form requires a longer time.

(iv) Delay in Transfer and Transmission:
Handling and recording physical certificates take a lot of time in the transfer and transmission of securities.

(v) Risk of Bad Delivery:
Buying and selling securities can create problems if the certificates are torn, forged, etc.

Question 2.
Explain any four advantages of the Depository system to investors.
Answer:
Investor refers to the Beneficial Owner, the one who invests in the company’s securities.
Advantages of Depository System to the Investor are as follows:
(i) No Risk:
Risks related to physical certificates like delays, loss of certificates, theft, mutilation, bad deliveries, etc. are totally eliminated.

(ii) Safety of Investment:
The safest and secure way of holding securities is in the form of DEMAT. It is controlled under Depositories Act and monitored by SEBI. For e.g. If we don’t want to deal in securities, for the time being, we can freeze the account instructing DP to avoid unexpected debit or credit.

(iii) Easy Transfer of Shares:

• There is no need of filling transfer form and submitting documents.
• Stamp duty is not applicable in this process.
• Cash realization and security transfer take place very fast.

(iv) Updated information:
The investor receives updated information about his transactions and holdings from DP and also sometimes from Depository.

(v) Loan against securities:
An investor can raise a loan from the banks against the security of Dematerialized securities.

(vi) No ‘Lots’:
The lot system has been abolished. The market lot is one share for dematerialized securities.

(vii) Nomination Facility:
Nomination Facility is allowed under the Depository system. In the event of the death of the investor, the process becomes simple.

(viii) Automatic Credit:
The investor’s account is automatically credited or debited by the company. This is called ‘Corporate Action.’
For e.g. Payment of dividend, Bonus Issue, Right Issue, Redemption, etc.

Question 3.
Explain any four advantages of the Depository system to the company.
Answer:
Benefits/Advantages of Depository System:
(A) Investors:
(i) Safety:
A safe and secure way of holding securities is in electronic form/mode. The depository system is monitored by SEBI. The Investor can keep his account in a ‘Freeze/Lock’ mode to avoid/prevent unexpected debit or credit or both by giving instructions to the DP.

(ii) Easy Transfer of Shares:
These securities can be easily transferred in electronic mode. Filling out the transfer form and uploading the documents can be easily done online. It is a time and efforts saving process. No stamp duty is levied on the depository system.

(iii) Update and Intimation:
The investor can check his status of holding securities at any time. Depository Participant provides investors with his statement of accounts periodically.

(iv) Automatic Credit:
Under, the depository system the investor’s account is automatically credited/debited. This credit or debit of accounts is usually in case of Payment of Dividend, Issue of Bonus Shares, Offering of Rights Shares, Early Redemption of Debentures, Mergers, and Acquisitions, etc.

(B) Companies:
(i) Up-to-date Information:
The depository system enables the company to maintain the information of the investors holding. It also helps the company to keep updated information about its shareholding pattern. The company is able to know the particulars of beneficial owners and their holdings periodically.

(ii) Reduction in costs and efforts:
Due to the depository system, maintaining the documents physically, the printing of certificates has saved a lot of time and cost.

(iii) Better Relationships:
The transfer process under the depository system is prompt and free from defects. So, complaints against the company by an investor are avoided in this regard. This helps the company build a good corporate image.

(iv) International Investment:
Paperless trading is a boon for the company management as it provides better and quicker services to the investors staying within the country and abroad. This attracts investment from abroad.

Question 4.
Explain Depository as a constituent of Depository System.
Answer:

• The depository is a company registered under the Companies Act, to deal in securities.
• It holds the securities in electronic form.
• It is responsible for safeguarding the investor’s securities and preventing any manipulation of records and transactions.
• There is no direct access of investors with the Depository.
• It works as a link between the company and investors.
• Statement of accounts is provided to the investor periodically.

Question 5.
Explain DP as the constituent of the Depository System.
Answer:

• Depository Participant acts as a link between the Depository and the Investor. In other words, it is an agent of the depository.
• DP is registered under the SEBI Act. It enjoys rights and obligations as specified under SEBI Regulations of 1996.
• It is an intermediary appointed by Depository.
• DP directly deals with customers and sends a statement of accounts periodically.
• The DP has a unique number for identification.
• Banks, Financial Institutions, Stock Brokers, etc. can function as Depository Participants.
• Thus, it is rightly justified that DP is an important constituent of the Depository system.

6. Justify the following statements.

Question 1.
Electronic holding of securities is safer than physical holding.
Answer:

• Electronic holding of securities means holding the securities in dematerialized form.
• Conversion of physical certificates into electronic form is known as ‘Dematerialization’.
• Holding securities in electronic form eliminates a huge volume of paperwork.
• The use of technology facilitates paperless trading which eliminates storage and handling of certificates. It also helps in reducing costs and efforts.
• There is no risk of getting lost, damaged, torn, stolen, misplaced during transit, etc.
• Delay in transfer and allotment of securities is also avoided.
• Thus, it is rightly justified that, electronic holding of securities is safer than physical holding.

Question 2.
Depository provides easy and quicker transfer of shares.
Answer:

• Under the Depository system, securities are held in electronic form.
• Not only the transfer and settlement of securities are done electronically but it also maintains the accounts of the shareholder, collects dividends, bonus shares, etc. on behalf of the shareholder.
• Depository Participant acts as a link between the Depository and the Investor.
• No stamp duty is levied on the depository system.
• Processing time in the transfer of securities is reduced and neither the securities nor the cash is held up for an unnecessarily long time.
• Hence it is also called a “Scripless trading system”. Thus, it is rightly justified that the Depository provides easy and quicker transfer of shares.

Question 3.
Depository System results in reduced time, cost and efforts.
Answer:

• The depository System has a very important role to play in the successful functioning of the Capital market.
• Under Depository System, securities are held in electronic form.
• The transfer and settlement of securities are done electronically.
• The Depository system maintains accounts of the shareholder, enables transfer, collects dividends, bonus shares, etc. on behalf of the shareholder.
• It eliminates a huge volume of paperwork, storage, and handling of physical certificates.
• Thus, it is rightly justified that, Depository system results in reduced time, cost and efforts.

Question 4.
The depository system is very similar to the Banking system.
Answer:

• The depository system is similar to the Banking system in many aspects.
• Depository keeps the securities safe, just like the bank keeps the money of the account holder safe.
• In banks, funds are held in accounts having unique numbers. Similarly, securities are held in accounts having unique IDs.
• There is no physical handling of money in the bank. Similarly, there is no physical handling of securities by the depository.
• Bank facilitates the transfer of funds between accounts. Depository facilitates the transfer of securities between accounts.
• Thus, it is rightly justified that the Depository system is very similar to the Banking system.

Question 5.
DP is an important constituent of the Depository system.
Answer:

• Depository Participant acts as a link between the Depository and the Investor. In other words, it is an agent of the depository.
• DP is registered under the SEBI Act. It enjoys rights and obligations as specified under SEBI Regulations of 1996.
• It is an intermediary appointed by Depository.
• DP directly deals with customers and sends a statement of accounts periodically.
• The DP has a unique number for identification.
• Banks, Financial Institutions, Stock Brokers, etc. can function as Depository Participants.
• Thus, it is rightly justified that DP is an important constituent of the Depository system.

Question 6.
Depository system allows both: Physical to electronic and electronic to physical conversion.
Answer:

• Under the Depository system, physical certificates can be converted into electronic ones known as ‘Dematerialization’.
• Similarly, the conversion of electronic securities into physical certificates is known as ‘Rematerialization’.
• Due to Dematerialization transfer of securities takes place fast and transactions are also settled immediately. Whereas in Rematerialization, settlement of transactions in the physical system takes more time.
• In Rematerialization, storing and handling physical certificates is more time-consuming. Whereas in dematerialization, there is no handling of certificates.
• There are chances of certificates getting lost, damaged, torn, stolen, misplaced during transit, etc.
• In this technological world, Demat securities are more preferred over Rematerialized securities.
• Thus, it is rightly justified that the Depository system allows both: Physical to electronic and electronic to physical conversion.

Question 7.
ISIN is a necessary component of Demat.
Answer:

• International Securities Identification Number (ISIN) is a code that uniquely identifies a specific securities issue.
• ISINs are allotted by that country’s NNA (National Numbering Agency).
• ISIN is a standard numbering system that is accepted globally.
• The ISIN’s structure is currently defined by the International Organization of Standardization (ISO).
• ISIN consists of a 12 (twelve) digit alpha-numeric code which is divided into 3 (three) parts.
• The company has to apply for ISIN for its securities with documents like a prospectus.
• In India, issuing ISIN to securities is assigned by SEBI to NSDL (for Demat shares). SEBI works as NNA in India.
• Thus, from the above points, it is rightly justified that ISIN is a necessary component of Demat.

7. Answer the following.

Question 1.
What is Depository System? Explain its advantages?
Answer:
The depository system came into existence in the year 1996. It is a system where securities are held in electronic form. It also maintains accounts of the shareholder, enables transfer, collects dividends, bonus shares, etc. on behalf of the shareholder. In other words, it is also called a scriptless trading system.

Benefits/Advantages of Depository System:
(A) Investors:
(i) Safety:
A safe and secure way of holding securities is in electronic form/mode. The depository system is monitored by SEBI. The Investor can keep his account in a ‘Freeze/Lock’ mode to avoid/ prevent unexpected debit or credit or both by giving instructions to the DP.

(ii) Easy Transfer of Shares:
These securities can be easily transferred in electronic mode. Filling out transfer forms and uploading the documents can be easily done online. It is a time and efforts saving process. No stamp duty is levied on the depository system.

(iii) Update and Intimation:
The investor can check his status of holding securities at any time. Depository Participant provides investors with his statement of accounts periodically.

(iv) Automatic Credit:
Under, the depository system the investor’s account is automatically credited/debited. This credit or debit of accounts is usually in case of Payment of Dividend, Issue of Bonus Shares, Offering of Rights Shares, Early Redemption of Debentures, Mergers, and Acquisitions, etc.

(B) Companies:
(i) Up-to-date Information:
The depository system enables the company to maintain the information of the investors holding. It also helps the company to keep updated information about its shareholding pattern. The company is able to know the particulars of beneficial owners and their holdings periodically.

(ii) Reduction in costs and efforts:
Due to the depository system, maintaining the documents physically, the printing of certificates has saved a lot of time and cost.

(iii) Better Relationships:
The transfer process under the depository system is prompt and free from defects. So, complaints against the company by an investor is avoided in this regard. This helps the company build a good corporate image.

(iv) International Investment:
Paperless trading is a boon for the company management as it provides better and quicker services to the investors staying within the country and abroad. This attracts investment from abroad.

Question 2.
Explain the constituents of the depository system.
Answer:

(i) Depository:

• Depository is an organization or a system where securities/shares are held in electric form.
• Depository transfers securities/shares between accounts on the instruction of the account holders.
• Depository contacts the customer through a depository participant.
• Transfer of shares is made through mere computerized book-entry in the depository.
• This becomes possible because shares are dematerialized.
• Only those securities which are held in the form of the share certificate are one’s names can be dematerialized.
• At present, there are two depositories. They are:
  • National Securities Depository Limited (NSDL)
  • Central Depository Services Limited (CDSL).

(ii) Depository Participant:

• The depository participant is the representative of the depository.
• Depository participant acts as an intermediary between investors and depositories.
• The depository participants have an identity number for identification.
• It has to maintain accounts of securities of each investor.
• Depository participant gives intimation about holdings from time to time by sending a statement of holding.
• According to SEBI guidelines financial institutions, banks, stockbrokers can act as a depository participant.

(iii) Beneficial Owner:

• An investor is known as ‘Beneficial Owner’ on whose name Demat account is opened.
• He enjoys the rights and benefits of members such as getting dividends, getting bonus shares, to vote at meetings.
• He is allotted an account number where securities are held.

(iv) Issuer Company:

• It is a company that makes an issue of securities.
• It must get registered with the depository.