Bhagya

Balbharti Maharashtra State Board 12th Chemistry Textbook Solutions Chapter 13 Amines Textbook Exercise Questions and Answers.

Maharashtra State Board Class 12 Chemistry Solutions Chapter 13 Amines

1. Choose the most correct option.

Question i.
The hybridization of nitrogen in primary amine is ………………………..
a. sp
b. sp²
c. sp³
d. sp³d
Answer:
c. sp³

Question ii.
Isobutylamine is an example of ………………………..
a. 2° amine
b. 3° amine
c. 1° amine
d. quaternary ammonium salt.
Answer:
a. 2° amine

Question iii.
Which one of the following compounds has the highest boiling point?
a. n-Butylamine
b. sec-Butylamine
c. isobutylamine
d. tert-Butylamine
Answer:
a. n-Butylamine

Question iv.
Which of the following has the highest basic strength?
a. Trimethylamine
b. Methylamine
c. Ammonia
d. Dimethylamine
Answer:
d. Dimethylamine

Question v.
Which type of amine does produce N₂ when treated with HNO₂?
a. Primary amine
b. Secondary amine
c. Tertiary amine
d. Both primary and secondary amines
Answer:
a. Primary amine

Question vi.
Carbylamine test is given by
a. Primary amine
b. Secondary amine
c. Tertiary amine
d. Both secondary and tertiary amines
Answer:
a. Primary amine

Question vii.
Which one of the following compounds does not react with acetyl chloride?
a. CH₃-CH₂-NH₂
b. (CH₃-CH₂)₂NH
c. (CH₃-CH₂)₃N
d. C₆H₅-NH₂
Answer:
Answer:
c. (CH₃ – CH₂)₃N

Question viii.
Which of the following compounds will dissolve in aqueous NaOH after undergoing reaction with Hinsberg reagent?
a. Ethylamine
b. Triethylamine
c. Trimethylamine
d. Diethylamine
Answer:
a. Ethyl amine

Question ix.
Identify ‘B’ in the following reactions

Answer:
d. CH₃-CH₂-OH

Question x.
Which one of the following compounds contains azo linkage?
a. Hydrazine
b. p-Hydroxyazobenzene
c. N-Nitrosodiethylamine
d. Ethylenediamine
Answer:
b. p-Hydroxyazobenzene

2. Answer in one sentence.

Question i.
Write reaction of p-toluenesulfonyl chloride with diethylamine.
Answer:

Question ii.
How many moles of methylbromide are required to convert ethanamine to N, N-dimethyl ethanamine?
Answer:
2 moles of methylbromide are required to convert ethanamine to N, N-dimethyl ethanamine.

Question iii.
Which amide does produce ethanamine by Hofmann bromamide degradation reaction?
Answer:
Propanamide (CH₃ – CH₂ – CONH₂) produces ethanamine by Hofmann bromamide degradation reaction.

Question iv.
Write the order of basicity of aliphatic alkylamine in gaseous phase.
Answer:
The order of basicity of aliphatic alkyl amines in the gaseous follows the order : tertiary amine > secondary amine > primary amine > NH₃.

Question v.
Why are primary aliphatic amines stronger bases than ammonia?
Answer:
The alkyl group tends to increase the electron density on the nitrogen atom. As a result, amines can donate the lone f pair of electrons on nitrogen more easily than ammonia. Hence, aliphatic amines are stronger bases than ammonia.

Question vi.
Predict the product of the following reaction. Nitrobenzene Sn/Conc. HCl?
Answer:
The product is aniline/

Question vii.
Write the IUPAC name of benzylamine.
Answer:
The IUPAC name is Phenylmethanamine.

Question viii.
Arrange the following amines in an increasing order of boiling points. n-propylamine, ethylmethyl amine, trimethylamine.
Answer:
Amines in an increasing order of boiling points : trimethyl amine, ethyl methyl amine, n-propyl amine

Question ix.
Write the balanced chemical equations for the action of dil H₂SO₄ on diethylamine.
Answer:

Question x.
Arrange the following amines in the increasing order of their _pK_b values. Aniline, Cyclohexylamine, 4-Nitroaniline
Answer:
Cyclohexyl amine (_pK_A 3.34), aniline (_pK_A 9.13) 4-nitroaniline (_pK_A 12.99)

3. Answer the following

Question i.
Identify A and B in the following reactions.

Answer:

Question ii.
Explain the basic nature of amines with suitable example.
Answer:
The basic strength of amines is expressed in terms of K_b or _pK_b value. According to Lowry-Bron-sted theory the basic nature of amines is explained by the following equilibrium equation.

In this equilibrium amine accepts H⁺, hence an amine is a Lowry-Bronsted base.

According to Lewis theory, the species which donates a pair of electrons is called a base.

The nitrogen atom in amiqes has a lone pair of electrons, which can be donated to suitable acceptor like proton H⁺.

The aqueous solutions of amines are basic in nature due to release of free OH^– ions in solutions. Hence amines are Lewis bases. There exists an equilibrium in their aqueous solutions as follows :

R – NH₂ + H₂O ⇌ RNH₃ + OH^–

Since OH^– is a stronger base, equilibrium shifts towards left-hand side giving less concentration of OH^–.

Here, K_b value is smaller and _pK_b value is larger.

Hence amines are weak bases.

Question iii.
What is diazotisation? Write diazotisation reaction of aniline.
Answer:
Aryl amines react with nitrous acid in cold condition (273 – 278 K) forms arene diazonium salts. The conversion of primary aromatic amine into diazonium salts is called diazotisation.

Diazotisation of aniline :

Question iv.
Write reaction to convert acetic acid into methylamine.
Answer:

Question v.
Write a short note on coupling reactions.
Answer:

Reactions involving retention of diazo group : (Coupling reactions) :

Question vi.
Explain Gabriel phthalimide synthesis.
Answer:
Phthalimide is reacted with alcoholic KOH to form potassium phthalimide. Further potassium phthalimide is treated with an ethyl iodide. The product N-ethylphthalimide is hydrolysed with aq NaOH to form ethyl amine. This reaction is known Gabriel phthalimide synthesis.

Question vii.
Explain carbylamine reaction with suitable examples.
Answer:
Aliphatic or aromatic primary amines on heating with chloroform and alcoholic potassium hydroxide solution form carbyl amines or isocyanides with extremely unpleasant smell. This reaction is a test for primary amines.

Secondary and tertiary amines do not give this test.

Question viii.
Write reaction to convert
(i) methanamine into ethanamine
(ii) Aniline into p-bromoaniline.
Answer:
(1) Methanamine into ethanamine

(2) Aniline into p-bromo aniline

Question ix.
Complete the following reactions :
a. C₆H₅N₂ Cl + C₂H₅OH →
b. C₆H₅NH₂ + Br₂(aq) → ?
Answer:
(a)

(b)

Question x.
Explain Ammonolysis of alkyl halides.
Answer:
When an alkyl halide is heated with alcoholic ammonia in a sealed tube under pressure at 373 K, a mixture of primary, secondary, tertiary amines and a quaternary ammonium salt is obtained. In this reaction, breaking of C – X bond by ammonia is called ammonolysis of alkyl halides. The reaction is also known as alkylation. For example, when methyl bromide is heated with alcoholic ammonia at 373 K, it gives a mixture of methylamine (a primary amine), dimethylamine (a secondary amine), trimethyl amine (a tertiary amine) and tetramethylam- monium bromide (a quaternary ammonium salt).

The order of reactivity of alkyl halides with ammonia is R – I > R – Br > R – Cl.

Question xi.
Write reaction to convert ethylamine into methylamine.
Answer:

4. Answer the following.

Question i.
Write the IUPAC names of the following amines :

Answer:

Question ii.
What are amines? How are they classified?
Answer:
Amines are classified on the basis of the number of hydrogen atoms of ammonia that are replaced by alkyl group. Amines are classified as primary (1°), secondary (2°) and tertiary (3°).

(1) Primary amines (1° amines) : The amines in which only one hydrogen atom of ammonia is replaced by an alkyl group or aryl group are called primary (1°) amines.

Examples :
(i) CH₃ – NH₂ methylamine
(ii) CH₃ – CH₂ – NH₂ ethylamine

(2) Secondary amines (2° amines) : The amines in which two hydrogen atoms of ammonia are replaced by two, same or different alkyl or aryl groups are called secondary (2°) amines.

Examples :
(i) C₂H₅ – NH – CH₃ ethylmethylamine
(ii) CH₃ – NH – CH₃ dimethylamine

(3) Tertiary amines (3° amines) : The amines in which all the three hydrogen atoms of ammonia are replaced by three same or different alkyl or aryl groups are called tertiary (3°) amines.

Examples :

Secondary and tertiary amines are further classified as (1) Simple or symmetrical amines (2) Mixed or unsymmetrical amines.

(i) Simple or symmetrical amines : In simple amines same alkyl groups are attached to the nitrogen e.g.

(ii) Mixed or unsymmetrical amines : In mixed amines different alkyl groups are attached to the nitrogen.

Question iii.
Write IUPAC names of the following amines.

Answer:

Question iv.
Write reactions to prepare ethanamine from
a. Acetonitrile
b. Nitroethane
c. Propionamide
Answer:
a. Ethanamine from acetonitrile :

b. Ethanamine from nitroethane :

c. Ethanamine from Propionamide :

Question v.
What is the action of acetic anhydride on ethylamine, diethylamine and triethylamine?
Answer:
Acetylation of amines : The reaction in which the H atom attached to nitrogen in amine is replaced by acetyl group is called acetylation of amines.

(1) Ethylamine on reaction with acetic anhydride forms monoacetyl derivative, N-acetylethylamine.

(2) Diethylamine (a secondary amine) on reaction with acetic anhydride forms a monoacetyl derivative, N-acetyldiethyl amine (or N,N-diethyl acetamide).

(3) Triethylamine does not react with acetic anhydride as it does not have any H atom attached nitrogen atom of amin e

Question vii.
Distinguish between ethylamine, diethylamine and triethylamine by using Hinsberg’s reagent?
Answer:
This reaction is useful for the distinction of primary, secondary and tertiary amines.

(i) Primary amine (like ethyl amine) is treated with Hinsberg’s reagent (benzene sulphonyl chloride) forms N-alkyl benzene sulphonamide which dissolve in aqueous KOH solution to form a clear solution of potassium salt and upon acidification gives insoluble N-alkyl benzene sulphonamide.

(ii) Secondary amine like diethyl amine is treated with benzene sulphonyl chloride forms N,N-diethyl benzene which sulphonyl amide remains insoluble in aqueous KOH and does not dissolve in acid.

(iii) Tertiary amine like triethyl amine does not react with benzene sulphonyl chloride and remains insoluble in KOH, however it dissolves in dil. HCl to give a clear solution due to formation of ammonium salt.

Question viii.
Write reactions to bring about the following conversions :
a. Aniline into p-nitroaniline
b. Aniline into sulphanilic acid?
Answer:
(1) Aniline into p-nitroaniline

(2) Aniline into sulphanilic acid.

Activity :

• Prepare a chart of azodyes, colours and its application.
• Prepare a list of names and structures of N-containing ingredients of diet.

12th Chemistry Digest Chapter 13 Amines Intext Questions and Answers

Use your brain power! (Textbook Page No 282)

Question 1.
Classify the following amines as simple/mixed; 1°, 2°, 3° and aliphatic or aromatic. (C₂H₅)₂NH, (CH₃)₃N, C₂H₅ – NH – CH₃,

Answer:

(A) Common Names : Rules

1. According to common naming system, the amines are named as alkylamines.
2. The common name of a primary amine is obtained by writing the name of the alkyl group followed by the word ‘amine’.
   Example : CH₃ – NH₂ : methyl-amine
3. The simple {symmetrical) secondary and tertiary amines are written by adding prefix ‘di- (forpresence of two alkyl groups) and ‘tri’- (for presence of three alkyl groups) respectively to the name of alkyl groups.
   Examples: (i) CH₃ – NH – CH₃ dimethylamine, (ii) (C₂H₅)₃ N triethylamine
4. The mixed (or unsymmetrical) secondary and tertiary amines are given names by writing the names of alkyl groups in alphabetical order, followed by the word ‘amine’.
   Example : CH₃ – CH₂ – NH – CH₃ ethyhnethylamine

(B) IUPAC names : Rules

1. According to IUPAC system of nomenclature of amines, aliphatic amines are named as alkanarnines.
2. The name of the amine is obtained by replacing the suffix ‘e’ from parent alkane’s name by ‘amine’.
3. The position of the amino group is indicated by the lowest possible locant.
   Example :
   
4. In case of secondary and tertiary amines, the largest alkyl group is considered to be the parent alkane and other alkyl groups are written as N-substituents.
   Example : ClH₅NH – CH₃ N – Methylethanamine
5. A complete name of amine is written as one word.

Try this….. (Textbook Page No 283)

Question 1.
Draw possible structures of all the isomers of C₄H₁₁N. Write their common as well as IUPAC names.
Answer:

Use your brain power! (Textbook Page No 283)

Question 1.
Write chemical equations for

(i) reaction of alc. NH with C₂H₅I.
Answer:

(ii) Amonolysis of benzyl chloride followed by the reaction with 2 moles of CH3I.
Answer:

(2) Ammonolysis of alkyl halides is not suitable method to prepare primary amines.
Answer:
In the laboratory, ammonolysis of alkyl halides is not a suitable method to prepare primary amines as it gives a mixture of primary, secondary, tertiary amines and quaternary ammonium salts. (Refer to the reaction in answer to Question 16). The separation of primary amine becomes difficult.

Problem 13.1 : (Textbook Page No 285)

Question 1.
Write reaction to convert methyl bromide into ethyl amine? Also, comment on the number of carbon atoms in the starting compound and the product.
Solution :
Methyl bromide can be converted into ethyl amine in two stage reaction sequence as shown below.

The starting compound methyl bromide contains one carbon atom while the product ethylamine contains two carbon atoms. A reaction in which number of carbons increases involves a step up reaction. The overall conversion of methyl bromide into ethyl amine is a step up conversion.

Use your brain power! (Textbook Page No 285)

Identify ‘A’ and ‘B’ in the following conversions.

Answer:

Use your brain power! (Textbook Page No 286)

Question 1.
Write the chemical equations for the following conversions :
(1) Methyl chloride to ethylamine.
(2) Benzamide to aniline.
(3) 1, 4-Dichlorobutane to hexane-1, 6-diamine.
(4) Benzamide to benzylamine.
Answer:
(1) Methyl chloride to ethylamine

(2) Benzamide to aniline

(3) 1, 4-Dichlorobutane to hexane-1, 6-diamine

(4) Benzamide to benzylamine

Use your brain power! (Textbook Page No 287)

Question 1.
Arrange the following :
(1) In decreasing order of the boiling point C₂H₅ – OH, C₂H₅ – NH₂, (CH₃)₂ NH
(2) In increasing order of solubility in water: C₂H₅ – NH₂, C₃H₇ – NH₂, C₆H₅ – NH₂
Answer:
(1) Decreasing order of the boiling point : C₂H₅ — OH, C₂H₅ — NH₂, (CH₃)₂ NH
(2) Increasing order of solubility in water : C₆H₅NH₂, C₃H₇ — NH₂, C₂H₅ — NH₂

Use your brain power! (Textbook Page No 288)

Question 1.
Refer to pK_b values and answer which compound from the following pairs is the stronger base?
(1) CH₃ – NH₂ and (CH₃)₂ NH
(2) (C₂H₅)₂ NH and (C₂H₅)₃ N
(3) NH₃ and (CH₃)₂ CH – NH₂
Answer:
(1) CH₃ -NH₂ and (CH₃)₂ NH
(CH₃)₂ NH is a stronger base

(2) (C₂H₅)₂ NH and (C₂H₅)₃ N
(C₂H₅)₂ NH is a stronger base

(3) NH₃ and (CH₃)₂ CHNH₂
(CH₃)₂ CHNH₂ is a stronger base

Use your brain power! (Textbook Page No 290)

Question 1.
Arrange the following amines in decreasing order of their basic strength :
NH₃, CH₃ – NH₂, (CH₃)₂ NH, C₆H₅NH₂
Answer:
Decreasing order of basic strength :
(CH₃)₂NH, CH₃ -NH₂, NH₃, C₆H₅NH₂

Use your brain power! (Textbook Page No 291)

Question 1.

Answer:

Use your brain power! (Textbook Page No 291)

Question 1.
Complete the following reaction :

Answer:

Use your brain power! (Textbook Page No 292)

Answer:

Use your brain power! (Textbook Page No 292)

Question 1.
Write the carbylamine reaction by using aniline as starting material.
Answer:

Can you tell? (Textbook Page No 292)

(1) What is the formula of nitrous acid ?
(2) Can nitrous acid be stored in bottle ?
Answer:
(1) Formula of nitrous acid : H – O – N = O
(2) Nitrous acid cannot be stored in bottle.

Use your brain power! (Textbook Page No 294)

Question 1.
How will you distinguish between methyl amine, dimethylamine and trimethylamine by Hinsberg’s test?
Answer:
(1) Methyl amine (primary amine) reacts with benzene sulphonyl chloride to form N-methylbenzene sulphona- mide

(2) Dimethyl amine reacts with benzene sulphonyl chloride to give N, N – dimethylbenzene sulphonamide.

(3) Trimethyl amine does not react with benzene sulphonyl chloride and remains insoluble in KOH

Problem 13.1 : (Textbook Page No 295)

Question 1.
Write the scheme for preparation of p-bromoaniline from aniline. Justify your answer.
Solution :
NH₂ – group in aniline is highly ring activating and o – /p – directing due to involvement of the lone pair of electrons on ‘N’ in resonance with the ring. As a result, on reaction with Br₂ it gives 2,4,6-tribromoaniline. To get a monobromo product, it is necessary to decrease the ring activating effect of – NH₂ group. This is done by acetylation of aniline. The lone pair of ‘N’ in acetanilide is also involved in resonance in the acetyl group. To that extent, ring activation decreases.

Hence, acetanilide on bromination gives a monobromo product p-bromoacetanilide. After monobromination the original – NH₂ group is regenerated. The protection of – NH₂ group in the form of acetyl group is removed by acid catalyzed hydrolysis to get p-bromoaniline, as shown in the following scheme.

Use your brain power! (Textbook Page No 296)

Question 1.
(1) Can aniline react with a Lewis acid?
(2) Why aniline does not undergo Frledel – Craft’s reaction using aluminium chloride?
Answer:
(1) Aniline reacts with a Lewis acid, forms salt.
(2) Aniline does not undergo Friedcl-Crafr’s reaction (alkylation and acetylation) due to salt formation with aluminium chloride (Lewis acid), which is used as catalyst. Due to this, nitrogen of anime acquires + ve charge and hence acts as strong deactivating effect on the ring and makes it difficult for electrophilic attack.

Can you tell? (Textbook Page No 294)

(1) Do tertiary amines have ‘H’ bonded to ‘N?
(2) Why do tertiary amines not react with benzene sulfonyl chloride?
Answer:
(1) Tertiary amines do not have ‘H’ bonded to ‘N’.
(2) Tertiary amine does not undergo reaction with benzene sulphonyl chloride as it does not have any H atom attached to nitrogen atom of amine.

Balbharti Maharashtra State Board 12th Chemistry Textbook Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids Textbook Exercise Questions and Answers.

Maharashtra State Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids

1. Choose the most correct option.

Question i.
In the following resonating structures A and B, the number of unshared electrons in valence shell present on oxygen respectively are

Answer:
c. 4, 6

Question ii.
In the Wolf -Kishner reduction, alkyl aryl ketones are reduced to alkylbenzenes. During this change, ketones are first converted into
a. acids
b. alcohols
c. hydrazones
d. alkenes
Answer:
c. hydrazones

Question iii.
Aldol condensation is
a. electrophilic substitution reaction
b. nucleophilic substitution reaction
c. elimination reaction
d. addition – elimination reaction
Answer:
d. addition-elimination reaction

Question iv.
Which one of the following has the lowest acidity?


Answer:
(c)

Question v.
Diborane reduces
a. ester group
b. nitro group
c. halo group
d. acid group
Answer:
d. acid group

Question vi.
Benzaldehyde does NOT show positive test with
a. Schiff reagent
b. Tollens’ ragent
c. Sodium bisulphite solution
d. Fehling solution
Answer:
d. Fehling solution

2. Answer the following in one sentence

Question i.
What are aromatic ketones?
Answer:
The compounds in which group is attached to either two aryl groups or one aryl and one alkyl group are called aromatic ketones.

For example :

Question ii.
Is phenylacetic acid an aromatic carboxylic acid?
Answer:
Phenylacetic acid is not an aromatic carboxylic acid.

Question iii.
Write reaction showing conversion of ethanenitrile into ethanol.
Answer:

Question iv.
Predict the product of the following reaction:
\(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{COOCH}_{3} \frac{\mathrm{i} \cdot \mathrm{AlH}(\mathrm{i}-\mathrm{Bu})_{2}}{\text { ii. } \mathrm{H}_{3} \mathrm{O}^{\oplus}}{\longrightarrow} ?\)
Answer:

Question v.
Name the product obtained by reacting toluene with carbon monoxide and hydrogen chloride in presence of anhydrous aluminium chloride.
Answer:

Question vi.
Write reaction showing conversion of Benzonitrile into benzoic acid.
Answer:

Question vii.
Name the product obtained by the oxidation of 1,2,3,4-tetrahydronaphthalene with acidified potassium permanganate.
Answer:

Question viii.
What is formalin?
Answer:
The aqueous solution of formaldehyde (40%) is known as formalin.

Question ix.
Arrange the following compounds in the increasing order of their boiling points : Formaldehyde, ethane, methyl alcohol.
Answer:
Ethane, formaldehyde, methyl alcohol.

Question x.
Acetic acid is prepared from methyl magnesium bromide and dry ice in presence of dry ether. Name the compound which serves not only reagent but also as cooling agent in the reaction.
Answer:
The cooling agent used in the above reaction is dry ice (O = C = O).

3. Answer in brief.

Question i.
Observe the following equation of reaction of Tollens’ reagent with aldehyde. How do we know that a redox reaction has taken place. Explain.
\(\begin{array}{r}
\mathrm{R}-\mathrm{CHO}+2 \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}+\mathrm{OH}^{-} \stackrel{\Delta}{\longrightarrow} \\
\mathrm{R}-\mathrm{COO}^{-}+2 \mathrm{Ag} \downarrow+4 \mathrm{NH}_{3}+2 \mathrm{H}_{2} \mathrm{O}
\end{array}\)
Answer:
Tollen’s reagent oxidises acetaldehyde to acetic acid (carboxylate ion) and Ag in Tollen’s reagent complex are reduced to silver. In this reaction, oxidation and reduction takes place simultaneously hence, it is a redox reaction.

Question ii.
Formic acid is stronger than acetic acid. Explain.
Answer:

In acetic acid, the methyl group is an electron-donating group. The acetate ion formed gets destabilized due to the electron releasing effect of methyl group ( +1 effect) which is higher than that of H-atom in the corresponding formic acid. As a result, acetic acid dissociates to a lesser extent. Thus decreasing the acidity of acetic acid.

Formic acid having lower pK_a value than acetic acid. Hence, formic acid is a stronger acid than acetic acid.

Question iii.
What is the action of hydrazine on cyclopentanone in presence of —. KOH in ethylene glycol?
Answer:

Question iv.
Write reaction showing conversion of Acetaldehyde into acetaldehyde dimethyl acetal.
Answer:

Question v.
Aldehydes are more reactive toward nucleophilic addition reactions than ketones. Explain.
Answer:
The reactivity of aldehydes and keones is due to the polarity of carbonyl group which results in electrophilicity of carbon. The reactivity is further explained on the basis of electronic effect and steric effects.

(1) Influence of electronic effects: A ketone has two electron-donating aJ.kyl groups ( + I effect) bonded to carbonyl carbon which are responsible for decreasing its positive polarity and electrophilicity. In contrast. aldehydes have only electron-donating group bonded to carbonyl carbon. This shows aldehydes are more electrophilic than ketones.

(2) Steric effects : Two bulky alkyl groups in ketone come in the way of the incoming nucleophile. This is called steric hindrance to nucleophilic attack.

On the other hand. nucleophile can easily attack the carbonyl carbon in aldehyde because has one alkyl group and is less crowded or sterically less hindered.

Hence aldehydes are more reactive and can easily be attacked by nucleophiles.

Question vi.
Write reaction showing the action of the following reagent on propane nitrile
a. Dilute NaOH
b. Dilute HCl ?
Answer:
(1) Action of dil NaOH:

(2) Action of dil HCl:

Question vi.
Arrange the following carboxylic acids with increasing order of their acidic strength and justify your answer.

Answer:
The increasing order of acidity will be 1 <3 <2. Acidity depends on mainly two factors : (1) ease of proton release (2) stability of conjugate base formed. In example (3) the ether O exerts a I effect and is closer to COOH group than in 2 (1 effect diminishes). Also the conjugate base formed will be stabilized by the same – I effect by delocalization of charge.

4. Answer the following

Question i.
Write a note on
a. Cannizaro reaction
b. Stephen reaction.
Answer:
(1) The carbon atom adjacent to carbonyl carbon atom is called a-carbon atom (α – C) and the hydrogen atom attached to a-carbon atom is called α-hydrogen atom (α – H).

(2) The α-hydrogen of aldehydes and ketones is acidic in nature due to (i) the strong-I effect of carbonyl group (ii) resonance stabilization of the carbanion.

(3) Aldol condensation reaction is characteristic reaction of aldehydes and ketones containing active α-hydrogen atom.

(4) When aldehydes or ketones containing α – H atoms are warmed with a dilute base or dilute acid, two molecules of them undergo self condensation to give β-hydroxy aldehyde (aldol) or β-hydroxy ketone (ketol) respectively. The reaction is known as Aldol addition Reaction.

(5) In aldol condensation, the product is formed by the nucleophilic addition of α-carbon atom of a second molecule which gets attached to carbonyl carbon atom of the first molecule and α-hydrogen atom of the second molecule gets attached to carbonyl oxygen atom of the first molecule forming (- OH) group to give β-hydroxy aldehyde or ketone.

(6) This is a reversible reaction, establishing an equilibrium favouring aldol formation to a greater extent than ketol formation.

(7) For aldehyde :

Acetaldol on heating undergoes subsequent elimination of water giving rise to α, β unsaturated aldehyde.

The overall reaction is called aldol condensation. It is a nucleophilic addition-elimination reaction. For ketone :

Diacetone alcohol on dehydration by heating forms α, β unsaturated ketone.

Question ii.
What is the action of the following reagents on toluene ?
a. Alkaline KMnO₄, dil. HCl and heat
b. CrO₂Cl₂ in CS₂
c. Acetyl chloride in presence of anhydrous AlCl₃.
Answer:
(1) Action of alkaline KMnO₄ : When toluene is heated with alkaline KMnO₄. (methyl group gets oxidised to earboxy lic group) benzoic acid is obtained

(2) Action of CrO₂Cl₂ in C₂:
Answer:
When Loluenc is ircated with soluion of chromyl chloride (CrO₂Cl₂) in Cs₂, brown chromium complex is obtained, which on acid hydrolysis gives benzaldehyde.

(3) Action of acetyl chloride in presence of anhyd. AlCl₃.
Answer:
When toluene is treated with acetyl chloride in presence of anhydrous AlCl₃ 4-methyl acetophenone is obtained.

Question iii.
Write the IUPAC names of the following structures :

Question iv.
Write reaction showing conversion of p- bromoisopropyl benzene into p-Isopropyl benzoic acid (3 steps).
Answer:

Question v.
Write reaction showing aldol condensation of cyclohexanone.
Answer:

Activity :
Draw and complete the following reaction scheme which starts with acetaldehyde. In each empty box, write the structural formula of the organic compound that would be formed.

12th Chemistry Digest Chapter 12 Aldehydes, Ketones and Carboxylic Acids Intext Questions and Answers

Use your brain power! (Textbook Page No 254)

Question 1.
Classify the following as aliphatic and aromatic aldehydes.

Answer:

Use your brain power! (Textbook Page No 255)

Question 1.
Classify the following as simple and mixed ketones. Benzoplienone, acetoneq hutanoneq acetophenone.

Answer:

Use your brain power! (Textbook Page No 264)

Write IUPAC names for the following compounds.

Try this ….. (Textbook Page No 260)

Question 1.
Draw structures for the following :
(1) 2-Methylpentanal
(2) Hexan-2-one
Answer:

Can you tell? (Textbook Page No 260)

Question 1.
Which is the reagent that oxidizes primary alcohols to only aldehydes and does not oxidize aldehydes further into carboxylic acid ?
Answer:
The reagent that is used to make only aldehydes is-heated Cu at 573 K.

Use your brain power! (Textbook Page No 261)

Question 1.
Write the structure of the product formed on Rosenmund reduction of ethanoyl chloride and benzoyl chloride.
Answer:

Can you think? (Textbook Page No 261)

Question 1.
What is the alcohol formed when benzoyl chloride is reduced with pure palladium as the catalyst ?
Answer:

Use your brain power! (Textbook Page No 262)

Question 1.
Name the compounds which are used for the preparation of benzophenone by Friedel-Crafts acylation reaction. Draw their structures.
Answer:
The compounds which are used in preparation of benzophenone by Friedel – Crafts reaction are :

Use your brain power! (Textbook Page No 263)

Identify the reagents necessary to achieve each of the following transformations:

Answer:

Use your brain power! (TextBook Page No 264)

Predict the products (name and structure) in the following reactions.

Answer:

Problem 12.1 : (Textbook Page No 276)

Question 1.
Alcohols (R – OH), phenols (Ar – OH) and carboxylic acids (R – COOH) can undergo ionization of O – H bond to give away proton H⁺; yet they have different pK_a values, which are 16, 10 and 4.5 respectively. Explain.
Solution :
pK_a value is indicative of acid strength. Lower of pK_a value stronger the acid. Alcohols, phenols and carboxylic acids, all involve ionization of an O – H bond. But their different pK_a values indicate that their acid strengths are different. This is because the resulting conjugate bases are stabilized to different extents.

As the conjugate base of carboxylic acid is best stabilized, among the three, carboxylic acids are strongest and have the lowest pKa value. As conjugate base of alcohols is destabilized, alcohols are weakest acids and have highest pIQ value. As conjugate base of phenols is moderately stabilized, phenols are moderately acidic and have intermediate pBQ value.

Try this….. (Textbook Page No 277)

Question 1.
Compare the following two conjugate bases and answer.

(1) Indicate the inductive effects of CH₃ – group in (a) and Cl-group in (b) by putting arrowheads in the middle of appropriate covalent bonds.
(2) Which species is stabilized by inductive effect, (a) or (b) ?
(3) Which species is destabilized by inductive effect, (a) or (b) ?
Answer:

(2) The monochloroacetate ion formed gets stabilised due to electron-withdrawing of Cl atom (- I effect).
(3) The acetate ion formed gets destabilised due to electron releasing effect of methyl group

Use your brain power! (Textbook Page No 277)

Question 1.
(1) Compare the pK_a values and arrange the following in an increasing order of acid strength. CI₃CCOOH, ClCH₂COOH, CH₃COOH, Cl₂CHCOOH
(2) Draw structures of conjugate bases of monochloroacetic acid and dichloroacetic acid. Which one is more stabilized by – 1 effect?
Answer:

The dichloroacetate ion formed gets stabilised due electron-withdrawing effect of two chlorine atoms. (- 1 effect)

Try this….. (Textbook Page No 277)

Question 1.
Arrange the following acids in order of their decreasing acidity.

Answer:
Acidity in the decreasing order

Try this ….. (Textbook Page No 267)

Question 1.
Draw the structure of propanone and indicate its polarity.
Answer:

Can you tell? (Textbook Page No 268)

Question 1.
Simple hydrocarbons, ethers, ketones and alcohols do not get oxidized by Tollen’s reagent. Explain, Why?
Answer:
(1) Due to the presence of hydrogen atom in aldehyde group , an aldehyde is oxidised to carboxylic acid which is not possible in case of ethers, ketones, alcohols and hydrocarbons.
(2) In ketones, carbonyl atom is attached to C-atom, hence C – C bond in can’t be broken easily.
(3) H atom attached to carbonyl carbon can be oxidised to – OH group giving carboxylic group Therefore, aldehyde reduces Tollen’s reagent, whereas simple hydrocarbons, ethers, ketones and alcohols do not reduce Tollen’s reagent.

Use your brain power! (Textbook Page No 269)

Question 1.
Sodium bisulfite is sodium salt of sulfurous acid, write down its detailed bond structure.
Answer:

Bond structure of sodium bisulfite

Use your brain power! (Textbook Page No 270)

Predict the product of the following reaction:

Answer:

Use your brain power! (Textbook Page No 271)

Question 1.
Draw the structures of
(1) The semicarbazone of cyclohexanone
(2) The imine formed in the reaction between 2-methylhexanal and ethyl amine
(3) 2, 4-dinitrophenyl hydrazone of acetaldehyde.
Answer:
(1) The semi carbazone of cyclohexanone.

(2) The imine formed between 2-methyl hexanal and ethyl amine.

(3) 2, 4-dinitrophenylhydrazone of acetaldehyde.

Try this….. (Textbook Page No 272)

Question 1.
Write chemical reactions taking place when propan-2-ol is treated with iodine and sodium hydroxide.
Answer:

Question 2.
When acetaldehyde Is treated with dilute NaOH, the following reaction is observed.

(1) What are the functional groups in the product?
(2) Can another product be formed during the same reaction? (Deduce the answer by doing atomic audit of reactant and product)
(3) Is this an addition reaction or condensation reaction?
Answer:
(1) There arc two functiona’ groups in the product: -CRO and -OH
(2) No other product can be formed in the same reaction.
(3) This is an addition reaction.

Use your brain power! (Textbook Page No 273)

Question 1.
Observe the following reaction.
Answer:

Question 2.
Will this reaction give a mixture of products like a cross aldol reaction ?
Answer:
No, since benzaldehyde, does not have a-hydrogen atom, it will not undergo self aldol condensation.

Use your brain power! (Textbook Page No 274)

Question 1.
Can isobutyraldehyde undergo a Cannizzaro reaction? Explain.
Answer:
Since isobutyraldehydecontains a-carbon atom, it cannot undergo Cannizzaro reaction.

Can you tell? (Textbook Page No 279)

What is the term used for elimination of water molecule ?
Answer:
Dehydration.

Use your brain power! (Textbook Page No 278)

Question 1.
Fill in the blanks and rewrite the balanced equations.

Answer:

Balbharti Maharashtra State Board 12th Chemistry Textbook Solutions Chapter 11 Alcohols, Phenols and Ethers Textbook Exercise Questions and Answers.

Maharashtra State Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers

1. Choose the correct option.

Question i.
Which of the following represents the increasing order of boiling points of (1), (2), and (3)?
(1) CH₃ – CH₂ – CH₂ – CH₂ – OH
(2) (CH₃)₂ CH – O – CH₃
(3) (CH₃)₃COH
A. (1) < (2) < (3)
B. (2) < (1) < (3)
C. (3) < (2) < (1)
D. (2) < (3) < (1)
Answer:
(a) (1) < (2) < (3)

Question ii.
Which is the best reagent for carrying out following conversion ?

A. LiAlH₄
B. Conc. H₂SO₄, H₂O
C. H₂/Pd
D. B₂H₆, H₂O₂ – NaOH
Answer:
B. Conc. H₂SO₄, H₂O

Question iii.
Which of the following reaction will give ionic organic product on reaction ?
A. CH₃ – CH₂ – OH + Na
B. CH₃ – CH₂ – OH + SOCl₂
C. CH₃ – CH₂ – OH + PCl₅
D. CH₃ – CH₂ – OH + H₂SO₄
Answer:
C. CH₃ – CH₂ – OH + PCl₅

Question iv.
Which is the most resistant alcohol towards oxidation reaction among the follwoing ?

Answer:
(c)

Question v.
Resorcinol on distillation with zinc dust gives
A. Cyclohexane
B. Benzene
C. Toluene
D. Benzene-1, 3-diol
Answer:
(b) Benzene

Question vi.
Anisole on heating with concerntrated HI gives
A. Iodobenzene
B. Phenol + Methanol
C. Phenol + Iodomethane
D. Iodobenzene + methanol
Answer:
B. Phenol + Methanol

Question vii.
Which of the following is the least acidic compound ?

Answer:
(b)

Question viii.
The compound incapable of hydrogen bonding with water is ……

Answer:
(b)

Question ix.
Ethers are kept in air tight brown bottles because
A. Ethers absorb moisture
B. Ethers evaporate readily
C. Ethers oxidise to explosive peroxide
D. Ethers are inert
Answer:
C. Ethers oxidise to explosive peroxide

Question x.
Ethers reacts with cold and concentrated H₂SO₄ to form
A. oxonium salt
B. alkene
C. alkoxides
D. alcohols
Answer:
A. oxonium salt

2. Answer in one sentence/ word.

Question i.
Hydroboration-oxidation of propene gives…..
Answer:
n-propyl alcohol (CH₃ – CH₂ – CH₂ – OH)

Question ii.
Write the IUPAC name of alcohol having molecular formula C₄H₁₀O which is resistant towards oxidation.
Answer:

Question iii.
Write the structure of optically active alcohol having molecular formula C₄H₁₀O
Answer:

Question iv.
Write name of the electrophile used in Kolbe’s Reaction.
Answer:
Electrophile : Carbon dioxide (O = C = O)

3. Answer in brief.

Question i.
Why phenol is more acidic than ethyl alcohol ?
Answer:
(1) In ethyl alcohol, the -OH group is attached to sp³ – hybridised carbon while in phenols, it is attached to sp² – hybridised carbon.

(2) Due to higher electronegativity of sp² – hybridised carbon, electron density on oxygen decreases. This increases the polarity of O-H bond and results in more ionization of phenol than that of alcohols.

(3) Electron donating inductive effect (+1 effect) of the alkyl group destabilizes alkoxide ion. As a result alcohol does not ionize much in water, therefore alcohol is neutral compound in aqueous medium.

(4) In alkoxide ion, the negative charge is localized on oxygen, while in phenoxide ion the negative charge is delocalized. The delocalization of the negative charge (structure I to V) makes phenoxide ion more stable than that of phenol.

The delocalization of charge in phenol (structures VI to X), the resonating structures have charge separation (where oxygen atom of OH group to be positive and delocalization of negative charge over the ortho and para positions of aromatic ring) due to which phenol molecule is less stable than phenoxide ion. This favours ionization of phenol. Thus phenols are more acidic than ethyl alcohol.

Question ii.
Why p-nitrophenol is a stronger acid than phenol ?
Answer:
(1) In p-nitrophenol, nitro group (NO₂) is an electron withdrawing group present at para position which enhances the acidic strength (-1 effect). The O-H bond is under strain and release of proton (H⁺) becomes easy. Further p-nitrophenoxide ion is more stabilised due to resonance.

(2) Since the absence of electron withdrawing group (like – NO₂) in phenol at ortho and para position, the acidic strength of phenol is less than that of p-nitrophenol.

Question iii.
Write two points of difference between properties of phenol and ethyl alcohol.
Answer:

Question iv.
Give the reagents and conditions necessary to prepare phenol from
a. Chlorobenzene
b. Benzene sulfonic acid.
Answer:
(1) From chlorobenzene : Reagents required : NaOH and dil. HC1 Temperature : 623 K, Pressure : 150 atm
(2) From Benzene sulphonic acid : Reagents required : aq NaOH, caustic soda, dil. HC1 Temperature : 573 K

Question v.
Give the equations of the reactions for the preparation of phenol from isopropyl benzene.
Answer:
Preparation of phenol from cumene (isopropylbenzene) : This is the commercial method of preparation of phenol. When a stream of air is passed through cumene (isopropylbenzene) suspended in aqueous Na₂CO₃ solution in the presence of cobalt naphthenate catalyst, isopropyl benzene hydroperoxide or cumene hydroperoxide is formed. Isopropylbenzene hydroperoxide on warming with dil. H₂SO₄ gives phenol and acetone. Acetone is an important by-product of the reaction and is separated by distillation. The reaction is called auto oxidation.

Question vi.
Give a simple chemical test to distinguish between ethanol and ethyl bromide.
Answer:
When ethyl bromide is heated with aq NaOH; ethyl alcohol is formed whereas ethanol does not react with aq NaOH

4. An ether (A), C₅H₁₂O, when heated with excess of hot HI produce two alkyl halides which on hydrolysis form compound (B)and (C), oxidation of (B) gave and acid (D), whereas oxidation of (C) gave a ketone (E). Deduce the structural formula of (A), (B), (C), (D) and (E).
Answer:

5. Write structural formulae for

a. 3-Methoxyhexane
b. Methyl vinyl ether
c. 1-Ethylcyclohexanol
d. Pentane-1,4-diol
e. Cyclohex-2-en-1-ol
Answer:

6. Write IUPAC names of the following

Answer:

Activity :
• Collect information about production of ethanol as byproduct in sugar industry and its importance in fuel economy.
• Collect information about phenols used as antiseptics and polyphenols having antioxidant activity.

12th Chemistry Digest Chapter 11 Alcohols, Phenols and Ethers Intext Questions and Answers

Use your brain power! (Textbook Page No 235)

Question 1.
Classify the following alcohols as l0/2°/3° and allylic/benzylic

Answer:
(1) Ally lie alcohol (primary)
(2) Allylic alcohol (secondary)
(3) Allylic alcohol (tertiary)
(4) Benzylic alcohol (primary)
(5) Benzylic alcohol (secondary)

Use your brain power ….. (Textbook Page No 236)

Question 1.
Name t-butyl alcohol using carbinol system of nomenclature.
Answer:
Trimethyl carbinol.

Problem 11.1 (Textbook Page No 238)

Question 1.
Draw structures of following compounds:
(i) 2,5-DiethIphenoI
(ii) Prop-2-en-I-oI
(iii) 2-methoxypropane
(iv) Phenylmethanol
Solution :

Try this ….. (Textbook Page No 238)

Write IUPAC names ol (he following compounds.

Do you know (Textbook Page No 238)

Question 1.
The mechanism of hydration of ethylcnc to ethyl alcohol.
Answer:
The mechanism of hydration of ethylene involves three steps:

Step 1: Ethylene gets protonated to form carbocation by electrophilic attack of H₃O (Formation of carbocation intermediate).

Step 2 : Nucleophilic attack of water on carbocation

Step 3 : Deprotonation to form an alcohol

Problem 11.2 : (Textbook Page No 239)

Question 1.
Predict the products for the following reaction.

Solution:
The substrate (A) contains an isolated and an aldehyde group. H₂/Ni can reduce both these functional groups while LiAlH₄ can reduce only – CHO of the two, Hence

Try this ….. (Textbook page 240)

Question 1.
Arrange O – H, C – H and N – H bonds in increasing order of their bond polarity.
Answer:
Increasing order of polarity :C – H, N – H, O – H

Problem 11.3 : (Textbook Page No 241)

Question 1.
The boiling point of n-butyl alcohol, isobutyl alcohol, sec-butyl alcohol and tert-butyl alcohol are 118 °C, 108 °C. 99 °C and 82 °C respectively. Explain.
Solution:
As branching increases, intermolecular van der Waal’s force become weaker and the boiling point decreases. Therefore, n-butyl alcohol has highest boiling point 118 °C and tert-butyl alcohol has lowest boiling point 83 °C. Isobutyl alcohol is a primary alcohol and hence its boiling point is higher than that of sec-butyl alcohol.

Problem 11.4 : (Textbook Page No 242)

The solubility of o-nitrophenol and p-nitrophenol is 0.2 g and 1.7 g/100 g of H₂O respectively. Explain the difference.
Solution :

p-Nitrophenol has strong intermolecular hydrogen bonding with solvent water. On the other hand, o-nitrophenol has strong intramolecular hydrogen bonding and therefore the intermolecular attraction towards solvent water is weak. The stronger the intermolecular attraction between solute and solvent higher is the solubility. Hence p-nitrophenol has higher solubility in water than that of o-nitrophenol.

Problem 11.5 : (Textbook Page No 243 & 244)

Question 1.
Arrange the following compounds in decreasing order of acid strength and justify.
(1) CH₃ – CH₂ – OH
(2) (CH₃)₃ C – OH
(3) C₆H₅ – OH
(4) p-NO₂ – C₆H₄ – OH
Solution :
Compounds (3) and (4) are phenols and therefore are more acidic than the alcohols (1) and (2). The acidic strengths of compounds depend upon stabilization of the corresponding conjugate bases. Hence let us compare electronic effects in the conjugate bases of these compounds :

The conjugate base of the alcohol (1) is destabilized by + 1 effect of one alkyl group, whereas conjugate base of the alcohol (2) is destabilized by +1 effect of three alkyl groups. Hence (2) is weaker acid than (1)

Phenols : The conjugate base of p-nitrophenol (4) is better resonance stabilized due to six resonance structures compared to the five resonance structure of conjugate base of phenol (3). The resonance structure VI has – ve charge on only electronegative oxygens. Hence the phenol (4) is stronger acid than (3). Thus the decreasing order of acid strength is (4), (3), (1), (2).

Use your brain power (Textbook Page No 244)

Question 1.
What are the electronic effects exerted by – OCH₃ and – Cl? Predict the acid strength of

Answer:
The electronic effects exerted by – Cl and – O CH₃ are as follows :
(1) Cl being more electronegative atom it pulls the bonding electrons towards itself. This is known as negative inductive effect (- I).

(2) – OCH₃ is less electronegative group which repels the bonding electrons away from it. This is known as positive inductive effect ( + I).

(3) The relative to parent phenol, is more acidic than .

Problem 11.6 : (Textbook Page No 245)

Question 1.
Mechanism of acid catalyzed dehydration of ethanol to give ethene.
Answer:
The mechanism of dehydration of ethanol involves the following order :
Step 1 : Formation of protonated alcohols : Initially ethyl alcohol gets protonated to form ethyl oxonium ion.

Step 2 : Formation of carbocation : It is the slowest step and hence, the rate determining step of the reaction.

Steps 3: Formation of ethene: Removal of a proton (H⁺) from carbocation.

The acidused in step I is released in step 3, the equilibrium is shifted to the right, ethene is removed as it is formed.

Problem 11.6 : (Textbook Page No 245)

Question 1.
Write the reaction showing major and minor products formed on heating butan-2-ol with concentrrated sulphuric acid.
Solution :
In the reaction described butan-2-ol undergoes dehydration to give but-2-ene (major) and but-l-ene (minor) in accordance with Saytzeff rule.

Problem 11.7 : (Textbook Page No 246)

Question 1.
Write and explain reactions to convert propan-l-ol into propan-2-ol.
Solution :
The dehydration of propane-l-ol to propene is the first step. Markownikoff hydration of propene is the second step to get the product propan-2-ol. This is brought about by reaction with concemtrated H₂SO₄ followed by hydrolysis.

Problem 11.8 : (Textbook Page No 246)

Question 1.
An organic compound gives hydrogen on reaction with sodium metal. It forms an aldehyde having molecular formula C₂H₄O on oxidation with pyridinium chlorochromate. Name the compounds and give equations of these reactions.
Solution :
The given molecular formula C₂H₄O of aldehyde is written as CH₃ – CHO. Hence the formula of alcohol from which this is obtained by oxidation must be CH₃ – CH₂ – OH. The two reactions can, therefore, be represented as follows :

(Do you know? Textbook Page No 248)

Question 90.
Write the mechanism of dehydration of alcohol to give ether.
Answer:
Dehydration of alcohols to form ether is SN₂ reaction. The mechanism of dehydration of ethanol involves the following steps.

Step 1 (Protonation) : Initially ethyl alcohol gets protonated in the presence of acid to form ethyl oxonium ion.

Step 2 (SN² mechanism) : Protonated alcohol species undergoes a backside attack by second molecule of alcohol is a slow step.

Step 3 (Deprotonation) : Formation of diethyl ether by elimination of proton

Problem 11.9 : (Textbook Page No 249)

Question 1.
Ethyl isopropyl ether does not form on reaction of sodium ethoxide and isopropyl chloride.

(i) What would be the main product of this reaction?
(ii) Write another reaction suitable for the preparation of ethyl isopropyl ether.
Solution :
(i) Isopropyl chloride is a secondary chloride. On treating with sodium ethoxide it gives elimination reaction to form propene as the main product.

(ii) Ethyl isopropyl ether can be prepared as follows using ethyl chloride (1⁰ chloride) as substrate.

Do you know? (Textbook Page No 250)

Question 1.
The mechanism of the reaction of HI with methoxy ethane.
Answer:
The reaction mechanism takes place as follows :
Step 1 : Protonation of ether Initially the ether molecule (methoxy ethane) protonated by cone. HI to form oxonium ion.

Step 2 : Iodide is a good nucleophile. It attacks the least substituted carbon of the oxonium ion formed in step 1 and displaces an alcohol molecule by SN² mechanism.

For example :
• Use of excess HI converts the alcohol into alkyl iodide.
• In case of ether having one tertiary alkyl group the reaction with hot HI follows SN1 mechanism, and tertiary iodide is formed rather than tertiary alcohol.

Step 1 :

Step 2 :

Balbharti Maharashtra State Board 12th Chemistry Textbook Solutions Chapter 8 Transition and Inner Transition Elements Textbook Exercise Questions and Answers.

Maharashtra State Board Class 12 Chemistry Solutions Chapter 8 Transition and Inner Transition Elements

1. Choose the most correct option.

Question i.
Which one of the following is diamagnetic
a. Cr^3⊕
b. Fe^3⊕
c. Cu^2⊕
d. Sc^3⊕
Answer:
d. Sc^3⊕

Question ii.
Most stable oxidation state of Titanium is
a. +2
b. +3
c. +4
d. +5
Answer:
c. +4

Question iii.
Components of Nichrome alloy are
a. Ni, Cr, Fe
b. Ni, Cr, Fe, C
c. Ni, Cr
d. Cu, Fe
Answer:
(c) Ni, Cr

Question iv.
Most stable oxidation state of Ruthenium is
a. +2
b. +4
c. +8
d. +6
Answer:
(b) +4

Question v.
Stable oxidation states for chromium are
a. +2, +3
b. +3, +4
c. +4, +5
d. +3, +6
Answer:
d. +3, +6

Question vi.
Electronic configuration of Cu and Cu⁺¹
a. 3d¹⁰, 4s⁰; 3d⁹, 4s⁰
b. 3d⁹, 4s¹; 3d⁹4s⁰
c. 3d¹⁰, 4s¹; 3d¹⁰, 4s⁰
d. 3d⁸, 4s¹; 3d¹⁰, 4s⁰
Answer:
c. 3d¹⁰, 4s¹; 3d¹⁰, 4s°

Question vii.
Which of the following have d0s0 configuration
a. Sc^3⊕
b. Ti^4⊕
c. V^5⊕
d. all of the above
Answer:
d. All of the above

Question viii.
Magnetic moment of a metal complex is 5.9 B.M. Number of unpaired electrons in the complex is
a. 2
b. 3
c. 4
d. 5
Answer:
d. 5

Question ix.
In which of the following series all the elements are radioactive in nature
a. Lanthanoids
b. Actinoids
c. d-block elements
d. s-block elements
Answer:
b. Actinides

Question x.
Which of the following sets of ions contain only paramagnetic ions
a. Sm^3⊕, Ho^3⊕, Lu^3⊕
b. La^3⊕, Ce^3⊕, Sm^3⊕
c. La^3⊕, Eu^3⊕, Gd^3⊕
d. Ce^3⊕, Eu^3⊕, Yb^3⊕
Answer:
d. Ce^3⊕, Eu^3⊕, Yb^3⊕

Question xi.
Which actinoid, other than uranium, occur in a significant amount naturally?
a. Thorium
b. Actinium
c. Protactinium
d. Plutonium
Answer:
a. Thorium

Question xii.
The flux added during extraction of Iron from hematite are its?
a. Silica
b. Calcium carbonate
c. Sodium carbonate
d. Alumina
Answer:
b. Calcium carbonate

2. Answer the following

Question i.
What is the oxidation state of Manganese in
\(\text { (i) } \mathrm{MnO}_{4}^{2-}(\mathrm{ii}) \mathrm{MnO}_{4}^{-} \text {? }\)
Answer:
Oxidation state of Manganese in
\((i) \mathrm{MnO}_{4}^{2-} is +6
(ii) \mathrm{MnO}_{4}^{-}is +7\)

*Question ii.
Give uses of KMnO₄

Question iii.
Why salts of Sc^3⊕, Ti^4⊕, V^5⊕ are colorless?
Answer:
(i) Sc³⁺ salts are colourless :

• The electronic configuration of ₂₁Sc [Ar| 3d¹ 4s² and Sc³⁺ [Ar] d°.
• Since there are no unpaired electrons in 3d subshell, d → d transition is not possible.
• Therefore, Sc³⁺ ions do not absorb the radiations in the visible region. Hence salts of Sc³⁺ are colourless (or white).

(ii) Ti⁴⁺ salts are colourless :

• The electronic configuration of ₂₂Ti [Ar] 3d²4s² and Ti⁴⁺ : [Ar] d°
• Since there are no unpaired electrons in 3d subshell, d-*d transition is not possible.
• Therefore, Ti³⁺ ions do not absorb the radiation in visible region. Hence salts of Ti³⁺ are colourless.

(iii) Vs⁵⁺ salts are eolourless :

• The electronic configuration of ₂₃V : [Ar] 3d³4s² and V⁵⁺ : [Ar] 3d°
• Since there are no unpaired electrons in 3d-subshell, d – d transition is not possible.
• Therefore, V⁵⁺ ions do not absorb the radiations in the visible region. Hence, V⁵⁺ salts are colourless, a

Question iv.
Which steps are involved in the manufacture of potassium dichromate from chromite ore?
Answer:
Steps in the manufacture of potassium dichromate from chromite ore are :

• Concentration of chromite ore.
• Conversion of chromite ore into sodium chromate (Na₂CrO₄).
• Conversion of Na₂CrO₄ into sodium dichromate (Na₂Cr₂O₇).
• Conversion of Na₂Cr₂O₇ into K₂Cr₂O₇.

Question v.
Balance the following equation
(i) KMnO₄ + H₂C₂O4 + H₂SO₄ → MnSO₄ + K₂SO₄ + H₂O + O₂
(ii) K₂Cr₂O₇ + KI + H₂SO₄ → K₂SO₄ + Cr₂(SO₄)₃ + 7H₂O + 3I₂
Answer:
(i) 2KMnO₄ + 3H₂SO₄ + 5H₂C₂O₄ → K₂SO₄ + 2MnSO₄ + 8H₂O + 10CO₂
(ii) Acidified potassium dichromate oxidises potassium iodide (KI) to iodine (I2). Potassium dichromate is reduced to chromic sulphate. Liberated iodine turns the solution brown K₂Cr₂O₇ + 6KI + 7H₂SO₄ → 4K₂SO₄ + Cr₂(SO₄)₃ + 7H₂O + 3I₂ [Oxidation state of iodine increases from – 1 to zero]

Question vi.
What are the stable oxidation states of plutonium, cerium, manganese, Europium?
Answer:
Stable oxidation states :
Plutonium + 3 to + 7
Cerium + 3, + 4
Manganese + 2, + 4, + 6, + 7
Europium +2, +3

Question vii.
Write the electronic configuration of chromium and copper.
Answer:
Chromium (₂₄Cr) has electronic configuration,
₂₄Cr (Expected) : Is² 2s² 2p⁶ 3s² 3p⁶ 3d⁴ 4s²
(Observed) : Is² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹

Explanation :

• The energy difference between 3d- and 45-orbitals is very low.
• d-orbitals being degenerate, they acquire more stability when they are half-filled (3d⁵).
• Therefore, there arises a transfer of one electron from 45-orbital to 3d-orbital in Cr giving more stable half-filled orbital. Hence, the configuration of Cr is [Ar] 3d⁵ 4s¹ and not [Ar] 3d⁴ 4s².

Copper (₂₉CU) has electronic configuration,
₂₉Cu (Expected) : Is² 2s³ 2p⁶ 3s³ 3p⁶ 3d⁹ 4s²
(Observed) : Is² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹

Explanation :

• The energy difference between 3d- and 45-orbitals is very low.
• d-orbitals being degenerate, they acquire more stability when they are completely filled.
• Therefore, there arises a transfer of one electron from 45-orbital to 3d-orbital in Cu giving completely filled more stable d-orbital.

Hence, the configuration of Cu is [Ar] 3d¹⁰ 4s¹ and not [Ar] 3d⁹ 4s².

Question viii.
Why nobelium is the only actinoid with +2 oxidation state?
Answer:

• Nobelium has the electronic configuration ₁₀₂NO : [Rn] 5f¹⁴6d°7s²
• No²⁺ : [Rn] 5f¹⁴6d°
• Since the 4f subshell is completely filled and 6d° empty, + 2 oxidation state is the stable oxidation state.
• Other actinoids in + 2 oxidation state are not as stable due to incomplete 4f subshell.

Question ix.
Explain with the help of balanced chemical equation, why the solution of Ce(IV) is prepared in acidic medium.
Answer:
Ce⁴⁺ undergoes hydrolysis as, Ce⁴⁺+ 2H₂O → Ce(OH)₄ + 4H⁺.
Due to the presence of H⁺ in the solution, the solution is acidic.

Question x.
What is meant by ‘shielding of electrons’ in an atom?
Answer:
The inner shell electrons in an atom screen or shield the outermost electron from the nuclear attraction. This effect is called the shielding effect.

The magnitude of the shielding effect depends upon the number of inner electrons.

Question xi.
The atomic number of an element is 90. Is this element diamagnetic or paramagnetic?
Answer:
The 90th element thorium has an electronic configuration, [Rn] 6d²7s². Since it has 2 unpaired electrons it is paramagnetic.

3. Answer the following

Question i.
Explain the trends in atomic radii of d-block elements
Answer:

1. The atomic or ionic radii of 3-d series transition elements are smaller than those of representative elements, with the same oxidation states.
2. For the same oxidation state, there is an increase in nuclear charge and a gradual decrease in ionic radii of 3d-series elements is observed. Thus ionic radii of ions with oxidation state + 2 decreases with increase in atomic number.
3. There is slight increase is observed in Zn²⁺ ions. With the higher oxidation states, effective nuclear charge increases. Therefore ionic radii decrease with increase in oxidation state of the same element. For example, Fe²⁺ ion has ionic radius 77 pm whereas Fe³⁺ has 65 pm.

Question ii.
Name different zones in the Blast furnace. Write the reactions taking place in them.
Answer:
(i) Zone of combustion : The hot air oxidises coke to CO which is an exothermic reaction, due to which the temperature of furnace rises.
C + ^1/2 O₂ → CO ΔH= – 220kJ
Some part of CO dissociates to give finely divided carbon and O₂.
2CO → 2C + O₂
The hot gases with CO rise up in the furnace and heats the charge coming down. CO acts as a fuel as well as a reducing agent.

(ii) Zone of reduction : At about 900 °C, CO reduces Fe₂O₃ to spongy (or porous) iron.
Fe₂O₃ + 3CO → 2Fe + 3CO₂
Carbon also reduces partially Fe₂0₃ to Fe.
Fe₂O₃ + 3C → 2Fe + 3CO

(iii) Zone of slag formation : At 1200 K limestone, CaCO₃ in the charge, decomposes and forms a basic flux CaO which further reacts at 1500 K with gangue (SiO₂, Al₂O₃) and forms a slag of CaSiO₃ and Ca₃AlO₃.
CaCO₃ + CaO + CO₂.
CaO + SiO₂ → CaSiO₃
12CaO + 2Al₂O₃ → 4Ca₃AlO₃ + 3O₂

The slag is removed from the bottom of the furnace through an outlet.

(iv) Zone of fusion : The impurities in ore like MnO₂ and Ca₃(PO₄)₂ are reduced to Mn and P while SiO₂ is reduced in Si. The spongy iron moving down in the furnace melts in the fusion zone and dissolves the impurities like C, Si, Mn, phosphorus and sulphure. The molten iron collects at the bottom of furnace. The lighter slag floats on the molten iron and prevents its oxidation.

The molten iron is removed and cooled in moulds. It is called pig iron or cast iron. It contains about 4% carbon.

Question iii.
What are the differences between cast iron, wrought iron and steel?
Answer:

Question iv.
Iron exhibits +2 and +3 oxidation states. Write their electronic configuration. Which will be more stable? Why?
Answer:
The electronic configuration of Fe^{2 +} and Fe³⁺ :
Fe²⁺ : Is² 2s² 2p⁶ 3s² 3p⁶ 3d⁶
Fe³⁺ : Is² 2s² 2p⁶ 3s² 3p⁶ 3d⁵

Due to loss of two electrons from the 4.v-orbital and one electron from the 3d-orbital, iron attains 3⁺ oxidation state. Since in Fe³⁺, the 3d-orbital is half-filled, it gets extra stability, hence Fe³⁺ is more stable than Fe²⁺.

Question v.
Give the similarities and differences in elements of 3d, 4d and 5d series.
Answer:
Similarity :

• They are placed between .s-block and p-block of the periodic table.
• All elements are metals showing metallic characters.
• Some are paramagnetic.
• Most of them give coloured compounds.
• They have catalytic properties.
• They form complexes.
• They have variable oxidation states.

Differences :

• In 4d and 5d series lanthanide and actinoid contraction is observed. In 3d series atomic size changes are less marked.
• 4d and 5d elements have high coordination numbers compared to 3d elements.
• 4d and 5d series have similar properties whereas 3d series have different properties.

Question vi.
Explain trends in ionisation enthalpies of d-block elements.
Answer:

1. The ionisation enthalpies of transition elements are quite high and lie between those of 5-block and p-block elements. This is because the nuclear charge and atomic radii of transition elements lie between those of 5-block and p-block elements.
   
2. As atomic number of transition elements increases along the period and along the group, first ionisation enthalpy increases even though the increase is not regular.
3. If IE_1; IE₂ and IE₃ are the first, second and third ionisation enthalpies of the transition elements, then IE₁ < IE₂ < IE₃.
4. In the transition elements, the added last differentiating electron enters into (n – 1) d-orbital and shields the valence electrons from the nuclear attraction. This gives rise to the screening effect of (n – 1) d-electrons.
5. Due to this screening effect of (n – 1) d electrons, the ionisation enthalpy increases slowly and the increase is not very regular.

Question vii.
What is meant by diamagnetic and paramagnetic metal? Give one example of diamagnetic and paramagnetic transition metal and lanthanoid metal.
Answer:

1. Paramagnetic substances : When a magnetic field is applied, substances which are attracted towards the applied magnetic field are called paramagnetic substances. Example : Ni²⁺, Pr⁴⁺
2. Diamagnetic substances : When a magnetic field is applied, substances which are repelled by the magnetic fields are called diamagnetic substances. Example : Zn²⁺, La³⁺
3. Ferromagnetic substances : When a magnetic field is applied, substances which are attracted very strongly are called ferromagnetic substances. These substances can be magnetised. For example, Fe, Co, Ni are ferromagnetic.

Question viii.
Why the ground-state electronic configurations of gadolinium and lawrencium are different than expected?

Question ix.
Write steps involved in the metallurgical process
Answer:
The various steps and principles involved in the extraction of pure metals from their ores are as follows.:

• Concentration of ores in which impurities (gangue) are removed.
• Conversion of ores into oxides or other reducible compounds of metals.
• Reduction of ores to obtain crude metals.
• Refining of metals giving pure metals.

Question x.
Cerium and Terbium behaves as good oxidising agents in +4 oxidation state. Explain.
Answer:

• The most stable oxidation state of lanthanoids is +3.
• Hence, Ce⁴⁺ (cerium) and Tb⁴⁺ (terbium) tend to get + 3 oxidation state which is more stable.
• Since they get reduced by accepting electron, they are good oxidising agents in + 4 oxidation state.

Question xi.
Europium and Ytterbium behave as good reducing agents in +2 oxidation state explain.
Answer:

• The most stable oxidation state of lanthanoids is + 3.
• Hence, Eu²⁺ and Yb²⁺ tend to get + 3 oxidation states by losing one electron.
• Since they get oxidised, they are good reducing agents in + 2 oxidation state.

Activity :
Make groups and each group prepare a PowerPoint presentation on the properties and applications of one element. You can use your imagination to create some innovative ways of presenting data.

You can use pictures, images, flow charts, etc. to make the presentation easier to understand. Don’t forget to cite the reference(s) from where data for the presentation is collected (including figures and charts). Have fun!

12th Chemistry Digest Chapter 8 Transition and Inner Transition Elements Intext Questions and Answers

Do you know? (Textbook Page No 165)

Question 1.
In which block of the modern periodic table are the transition and inner transition elements placed?
Answer:
The transition elements are placed in d-block and inner transition elements are placed in f-block of the modern periodic table.

Use your brain power! (Textbook Page No 167)

Question 1.
Fill in the blanks with correct outer electronic configurations.
Answer:
Answers are given in bold.

Try this….. (Textbook Page No 168)

Question 1.
Write the electronic configuration of Cr and Cu.
Answer:
₂₄Cr : [Ar] 3d⁵4s¹ ₃₀Cu : [Ar] 3d¹⁰4s¹

Can you tell? (Textbook Page No 168)

Question 1.
Which of the first transition series element shows the maximum number of oxidation states and why?
Answer:

• ₂₅Mn shows the maximum number of oxidation states, + 2 to +7.
• ₂₅Mn : [Ar] 3d⁵4s³
• Mn has incompletely filled J-subshell.
• Due to small difference in energy between 3d and 4s -orbitals, Mn can lose (or share) electrons from both the orbitals.
• Hence Mn shows oxidation states from + 2 to +7.

Question 2.
Which elements in the 4d and 5d-series will show maximum number of oxidation states?
Answer:
In 4d-series maximum number of oxidation states are for Ruthenium Ru ( + 2, +3, + 4„ +6, +7, + 8). In 5d-series, maximum number of oxidation states are for Osmium, Os ( + 2 to + 8).

Try this ….. (Textbook Page No 168)

Question 1.
Write the electronic configuration of Mn⁶⁺, Mn⁴⁺, Fe⁴⁺, Co⁵⁺, Ni²⁺.
Answer:

Try this ….. (Textbook Page No 171)

Question 1.
Pick up the paramagnetic species from the following : Cu¹⁺, Fe³⁺, Ni²⁺, Zn²⁺, Cd²⁺, Pd²⁺.
Answer:
The following ions are paramagnetic : Fe³⁺, Ni²⁺, Pd²⁺

Try this ….. (Textbook Page No 171)

Question 1.
What will be the magnetic moment of transition metal having 3 unpaired electrons?
(a) equal to 1.73 B.M.?
(b) less than 1.73 BM.
(c) more than 1.73 B.M.?
Answer:
By spin-only formula, \(\mu=\sqrt{n(n+2)}\) where n is number of unpaired electrons.
\(\mu=\sqrt{3(3+2)}=\sqrt{3(5)}=3.87 \mathrm{~B} . \mathrm{M}\)
Thus the value is more than 1.73 B.M.

Use your brain power! (Textbook Page No 171)

Question 1.
A metal ion from the first transition series has two unpaired electrons. Calculate the magnetic moment.
Answer:
\(\)\begin{aligned}
\mu &=\sqrt{n(n+2)} \\
&=\sqrt{2(2+2)} \\
&=\sqrt{8} \\
&=2.84 \text { B.M. }
\end{aligned}\(\)

Problem (Textbook Page No 172)

Question 1.
Calculate the spin-only magnetic moment of divalent cation of a transition metal with atomic number 25.
Answer:
For element with atomic number 25. electronic configuration of its divalent cation will be : [Ar] 3d⁵.

Try this….. (Textbook Page No 172)

Question 1.
Calculate the spin-only magnetic moment of a divalent cation of element Slaving atomic number 27.
Answer:
Electronic configuration of divalent ion of an element with atomic number 27 : [Ar] 3d⁷;

Can you tell? (Textbook Page No 172)

Question 1.
Compounds of s and p-block elements are almost white. What could be the absorbed radiation? (uv or visible)?
Answer:
The white colour of a compound indicates the absorption of uv radiation.

Can you tell? (Textbook Page No 181)

Question 1.
Why f-block elements are called inner transition metals?
Answer:
f-block elements are called inner transition elements since f-orbital lies much inside the f-orbital in relation to the transition metals, These elements have 1 to 14 electrons in their f-orbital.

Question 2.
Are there an similarities between transition and inner transition metals?
Answer:
There are some properties similarity between transition and inner transition metals.

• They are placed between s and p-block elements.
• They are metals with filling of inner suhshells in their electronic configuration.
• They show variable oxidation slates.
•  They show magnetism.
• They form coloured compounds.
• They have catalytic property.

Problem (Textbook Page No 184)

Question 1.
Which of the following will have highest fourth ionisation enthalpy, La⁴⁺, Gd⁴⁺, Lu⁴⁺.
Answer:
La : 4f°5d¹6s²
Gd : 4f¹5d¹6s²
Lu : 4f¹⁴5d¹6s²
Lu will have the highest fourth ionisation enthalpy since Lu³⁺ has the most stable configuration of 4f¹⁴.

Use your brain power! (Textbook Page No 185)

Question 1.
Do you think that lanthanoid complex would show magnetism?
Answer:
Lanthanoid complexes may show magnetism.

Question 2.
Can you calculate the spin only magnetic moment of lanthanoid complexes using the same formula that you used for transition metal complexes?
Answer:
You cannot calculate magnetic moment of lanthanoid complexes using spin only formula as you have to consider orbital momentum also.

Question 3.
Calculate the spin only magnetic moment of La³⁺. Compare the value with that given in the table.
Answer:
La³⁺ ion has no unpaired electron.

La³⁺ ion has zero value of magnetic moment same as given in the table.

Balbharti Maharashtra State Board 12th Chemistry Textbook Solutions Chapter 7 Elements of Groups 16, 17 and 18 Textbook Exercise Questions and Answers.

Maharashtra State Board Class 12 Chemistry Solutions Chapter 7 Elements of Groups 16, 17 and 18

1. Select appropriate answers for the following.

Question i.
Which of the following has the highest electron gain enthalpy?
A. Fluorine
B. Chlorine
C. Bromine
D. Iodine
Answer:
B. Chlorine

Question ii.
Hydrides of group 16 are weakly acidic. The correct order of acidity is
A. H₂O > H₂S > H₂Se > H₂Te
B. H₂Te > H₂O > H₂S > H₂Se
C. H₂Te > H₂Se > H₂S > H₂O
D. H₂Te > H₂Se > H₂O > H₂S
Answer:
C. H₂Te > H₂Se > H₂S > H₂O

Question iii.
Which of the following element does not show oxidation state of +4 ?
A. O
B. S
C. Se
D. Te
Answer:
A. O

Question iv.
HI acid when heated with conc. H₂SO₄ forms
A. HIO₃
B. KIO₃
C. I₂
D. KI
Answer:
C. I₂

Question v.
Ozone layer is depleted by
A. NO
B. NO₂
C. NO₃
D. N₂O₅
Answer:
A. NO

Question vi.
Which of the following occurs in liquid state at room temperature?
A. HIO₃
B. HBr
C. HCl
D. HF
Answer:
D. HF

Question vii.
In pyrosulfurous acid oxidation state of sulfur is
A. Only +2
B. Only +4
C. +2 and +6
D. Only +6
Answer:
B. Only + 4

Question viii.
Stability of interhalogen compounds follows the order
A. BrF > IBr > ICl > ClF > BrCl
B. IBr > BeF > ICl > ClF > BrCl
C. ClF > ICl > IBr > BrCl > BrF
D. ICl > ClF > BrCl > IBr > BrF
Answer:
C. ClF > ICl > IBr > BrCl > BrF

Question ix.
BrCl reacts with water to form
A. HBr
B. Br₂ + Cl₂
C. HOBr
D. HOBr + HCl
Answer:
D. HOBr + HCl

Question x.
Chlorine reacts with excess of fluorine to form.
A. ClF
B. ClF₃
C. ClF₂
D. Cl₂F₃
Answer:
B. ClF₃

Question xi.
In interhalogen compounds, which of the following halogens is never the central atom.
A. I
B. Cl
C. Br
D. F
Answer:
D. F

Question xii.
Which of the following has one lone pair of electrons?
A. IF₃
B. ICl
C. IF₅
D. ClF₃
Answer:
C. IF₅

Question xiii.
In which of the following pairs, molecules are paired with their correct shapes?
A. [I₃] : bent
B. BrF₅ : trigonal bipyramid
C. ClF₃ : trigonal planar
D. [BrF₄] : square planar
Answer:
A. [I₃] : bent

Question xiv.
Among the known interhalogen compounds, the maximum number of atoms is
A. 3
B. 6
C. 7
D. 8
Answer:
D. 8

2. Answer the following.

Question i.
Write the order of the thermal stability of the hydrides of group 16 elements.
Answer:
The thermal stability of the hydrides of group 16 elements decreases in the order of H₂O > H₂S > H₂Se > H₂Te.

Question ii.
What is the oxidation state of Te in TeO₂?
Answer:
The oxidation state of Te in TeO₂ is + 4.

Question iii.
Name two gases which deplete ozone layer.
Answer:
Nitrogen oxide (NO) released from exhaust systems of car or supersonic jet aeroplanes and chlorofluorocarbons (Freons) used in aerosol sprays and refrigerators deplete ozone layer.

Question iv.
Give two uses of ClO₂
Answer:
(i) ClO₂ is used as a bleaching agent for paper pulp and textiles.
(ii) It is also used in water treatment.

Question v.
What is the action of bromine on magnesium metal?
Answer:
Bromine reacts instantly with magnesium metal to give magnesium bromide.

Question vi.
Write the names of allotropic forms of selenium.
Answer:
Selenium has two allotropic forms as follows :
(i) Red (non-metallic) form
(ii) Grey (metallic) form

Question vii.
What is the oxidation state of S in H₂SO₄.
Answer:
The oxidation state of S in H₂SO₄ is + 6.

Question viii.
The pKa values of HCl is -7.0 and that of HI is -10.0. Which is the stronger acid?
Answer:
For HCl, pK_a = -7.0, hence its dissoClation constant is, K_a = 1 x 10^-7.
For HI pK_a = – 10.0, hence its dissoClation constant is K_a = 1 x 10^-7. Hence HCl dissoClates more than HI.
Therefore HCl is a stronger acid than HI.

Question ix.
Give one example showing reducing property of ozone.
Answer:
Ozone decomposes to liberate nascent oxygen, hence it is a powerful oxidising agent. O_3(g) → O_2(g) + O

For example :
(i) It oxidises lead sulphide (PbS) to lead sulphate (PbSO₄).
pbS_(s) + 4O_3(g) → PbSO_(s) + 4O_2(g)
(ii) Potassium iodide, KI is oxidised to iodine, I₂ in the solution.
2KI_(aq) + H₂O₍₁₎ + O_3(g) → 2KOH_(aq) + I_2(s) + O_2(g)

Question x.
Write the reaction of conc. H₂SO₄ with sugar.
Answer:
Concentrated sulphuric acid when added to sugar, it is dehydrated giving carbon.

The carbon that is left behind is called sugar charcoal and the process is called char.

Question xi.
Give two uses of chlorine.
Answer:
Chlorine is used for :

• for sterilization of drinking water.
• bleaching wood pulp required for the manufacture of paper and rayon, cotton and textiles are also bleached using chlorine.
• in the manufacture of organic compounds like CHCl₃, CCl₄, DDT, dyes and drugs.
• in the extraction of metals like gold and platinum.
• in the manufacture of refrigerant like Freon (i.e., CCl₂F₂).
• in the manufacture of several poisonous gases like mustard gas (Cl-C₂H₄-S-C₂H₄-Cl), phosgene (COCl₂) used in warfare.
• in the manufacture of tear gas (CCl₃NO₂).

Question xii.
Complete the following.
1. ICl₃ + H₂O …….. + …….. + ICl
2. I₂ + KClO₃ ……. + KIO₂
3. BrCl + H₂O ……. + HCl
4. Cl₂ + ClF₃ ……..
5. H₂C = CH₂ + ICl …….
6. XeF₄ + SiO₂ ……. + SiF₄
7. XeF₆ + 6H₂O …….. + HF
8. XeOF₄ + H₂O ……. + HF
Answer:
1. 2ICI₃ + 3H₂O → 5HCl + HlO₃ + ICl
2. I₂ + KCIO₃ → ICl + KIO₃
3. BrCl + H₂O → HOBr + HCl
4. Cl₂ + C1F₃ → 3ClF
5. CH₂ = CH₂ + ICl → CH₂I – CH₂Cl
6. 2XeF₆ + SiO₂ → 2XeOF₄ + SiF₄
7. XeF₆ + 3H₂O → XeO₃ + 6HF
8. XeOF₄ + H₂O→  XeO₂F₂ + 2HF

Question xiii.
Match the following
A – B
XeOF₂ – Xenon trioxydifluoride
XeO₂F₂ – Xenon monooxydifluoride
XeO₃F₂ – Xenon dioxytetrafluoride
XeO₂F₄ – Xenon dioxydifluoride
Answer:
XeOF₂ – Xenon monooxydifluoride
XeO₂F₂ – Xenon dioxydifluoride
XeO₃F₂ – Xenon trioxydifluoride
XeO₂F₄ – Xenon dioxytetrafluoride

Question xiv.
What is the oxidation state of xenon in the following compounds?
XeOF₄, XeO₃, XeF₅, XeF₄, XeF₂.
Answer:

3. Answer the following.

Question i.
The first ionisation enthalpies of S, Cl and Ar are 1000, 1256 and 1520 kJ/mol^-1, respectively. Explain the observed trend.
Answer:
(i) The atomic number increases as, ₁₆S < ₁₇Cl < ₁₈Ar₁.
(ii) Due to decrease in atomic size and increase in effective nuclear charge, Cl binds valence electrons strongly.
(iii) Hence ionisation enthalpy of Cl (1256 kJ mol^-1) is higher than that of S(1000 kJ mol^-1)
(iv) Ar has electronic configuration 3s²3p⁶. Since all electrons are paired and the octet is complete, it has the highest ionisation enthalpy, (1520 kJ mol^-1)

Question ii.
“Acidic character of hydrides of group 16 elements increases from H₂O to H₂Te” Explain.
Answer:
(i) The thermal stability of the hydrides of group 16 elements decreases from H₂O to H₂Te. This is because the bond dissociation enthalpy of the H-E bond decreases down the group.
(ii) Thus, the acidic character increases from H₂O to H₂Te.

Question iii.
How is dioxygen prepared in laboratory from KClO₃?
Answer:
By heating chlorates, nitrates and permanganates.
Potassium chlorate in the presence of manganese dioxide on heating decomposes to form potassium chloride and oxygen.

Question iv.
What happens when
a. Lead sulfide reacts with ozone (O₃).
b. Nitric oxide reacts with ozone.
Answer:
(i) It oxidises lead sulphide (PbS) to lead sulphate (PbSO₄) changing the oxidation state of S from – 2 to +6.
PbS_(s) + 4O_3(g) → PbSO_(s) + 4O_2(g)

(ii) Ozone oxidises nitrogen oxide to nitrogen dioxide.
NO_(g) + O_3(g) → NO_2(g) + O_2(g)

Question v.
Give two chemical reactions to explain oxidizing property of concentrated H₂SO₄.
Answer:
Hot and concentrated H₂SO₄ acts as an oxidising agent, since it gives nascent oxygen on heating.

Question vi.
Discuss the structure of sulfur dioxide.
Answer:
(i) SO₂ molecule has a bent V shaped structure with S-O-S bond angle 119.5° and bond dissoClation enthalpy is 297 kJ mol^-1.
(ii) Sulphur in SO₂ is sp² hybridised forming three hybrid orbitals. Due to lone pair electrons, bond angle is reduced from 120° to 119.5°.
(iii) In SO₂, each oxygen atom is bonded to sulphur by σ and a π bond.
(iv) a bond between S and O are formed by sp²-p overlapping.
(v) One of π bonds is formed by pπ – pπ overlapping while other n bond is formed by pπ – dπ overlap.
(vi) Due to resonance both the bonds are identical having observed bond length 143 pm due to resonance,

Question vii.
Fluorine shows only -1 oxidation state while other halogens show -1, +1, +3, +5 and +7 oxidation states. Explain.
Answer:

• Halogens have outer electronic configuration ns² np⁵.
• Halogens have tendency to gain or share one electron to attain the stable configuration of nearest inert element with configuration ns²np⁶.
• Hence they are monovalent and show oxidation state – 1.
• Since fluorine does not have vacant d-orbital, it shows only one oxidation state of – 1 while all other halogens show variable oxidation states from – 1 to +7.
• These oxidation states are, – 1, +1, + 3, +5 and + 7. Cl and Br also show oxidation states + 4 and + 6 in their oxides and oxyaClds.

Question viii.
What is the action of chlorine on the following
a. Fe
b. Excess of NH₃
Answer:
(a) Chlorine reacts with Fe to give ferric chloride.
2Fe + 3Cl₂ → 2FeCl₃

(b) Chlorine reacts with the excess of ammonia to form ammonium chloride, NH₄Cl and nitrogen.

Question ix.
How is hydrogen chloride prepared from sodium chloride?
Answer:

1. In the laboratory, hydrogen chloride, HCl is prepared by heating a mixture of NaCl and concentrated H₂SO₄.
   
2. Hydrogen chloride gas, is dried by passing it through a dehydrating agent like concentrated H₂SO₄ and then collected by upward displacement of air.

Question x.
Draw structures of XeF₆, XeO₃, XeOF₄, XeF₂.
Answer:

Question xi.
What are interhalogen compounds? Give two examples.
Answer:
Interhalogen compounds : Compounds formed by the combination of atoms of two different halogens are called interhalogen compounds. In an interhalogen compound, of the two halogen atoms, one atom is more electropositive than the other. The interhalogen compound is regarded as the halide of the more electropositive halogen.
For example ClF, BrF₃, ICl

Question xii.
What is the action of hydrochloric acid on the following?
a. NH₃
b. Na₂CO₃
Answer:
a. Hydrochloric acid reacts with ammonia to give white fumes of ammonium chloride.
NH₃ + HCl → NH₄Cl

b. Hydrochloric acid reacts with sodium carbonate to give sodium chloride, water with the liberation of carbon dioxide gas.
Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂

Question xiii.
Give two uses of HCl.
Answer:
Hydrogen chloride (OR hydrochloric acid) is used :

• in the manufacture of chlorine and ammonium chloride,
• to manufacture glucose from com, starch
• to manufacture dye
• in mediClne and galvanising
• as an important reagent in the laboratory
• to extract glue from bones and for the purification of bone black.
• for dissolving metals, Fe + 2HCl_(aq) → FeCl₂ + H_2(g)

Question xiv.
Write the names and structural formulae of oxoacids of chlorine.
Answer:

Question xv.
What happens when
a. Cl₂ reacts with F₂ in equal volume at 437 K.
b. Br₂ reacts with excess of F₂.
Answer:
(a) Cl₂ reacts with F₂ in equal volumes at 437 K to give chlorine monofluoride ClF.

(b) Br₂ reacts with excess of F₂ to give bromine trifluoride BF₃.

Question xvi.
How are xenon fluorides XeF₂, XeF₄ and XeF₆ obtained ? Give suitable reactions.
Answer:
Xenon fluorides are generally prepared by the direct reaction of xenon and fluorine in different ratios and under appropriate experimental conditions, such as temperature, in the presence of an electric discharge and by a photochemical reaction.
(i) Preparation of XeF₂ :

(ii) Preparation of XeF₄ :

(iii) Preparation of XeF₆ :

Question xvii.
How are XeO₃ and XeOF₄ prepared?
Answer:
Preparation of XeO₃ : Xenon trioxide (XeO₃) is prepared by the hydrolysis of XeF₄ or XeF₆.

• By hydrolysis of XeF₄ :
  3XeF₄ + 6H₂0 → 2Xe + XeO₃ + 12 HF + \(1 \frac{1}{2} \mathrm{O}_{2}\)
• By hydrolysis of XeF₆ :
  XeF₆ + 3H₂O → XeO₃ + 6HF
• Preparation of XeOF₄ :
  Xenon oxytetrafluoride (XeOF₄) is prepared by the partial hydrolysis of XeF₆.
  XeF₆ + H₂O → XeOF₄ + 2HF

Question xviii.
Give two uses of neon and argon.
Answer:
Uses of neon (Ne) :

• Neon is used in the production of neon discharge lamps and signs by filling Ne in glass discharge tubes.
• Neon signs are visible from a long distance and also have high penetrating power in mist or fog.
• A mixture of neon and helium is used in voltage stabilizers and current rectifiers.
• Neon is also used in the production of lasers and fluorescent tubes.

Uses of argon (Ar) :

• Argon is used to fill fluorescent tubes and radio valves.
• It is used to provide inert atmosphere for welding and production of steel.
• It is used along with neon in neon sign lamps to obtain different colours.
• A mixture of 85% Ar and 15% N2 is used in electric bulbs to enhance the life of the filament.

Question xix.
Describe the structure of Ozone. Give two uses of ozone.
Answer:
(A)

• Ozone has molecular formula O₃.
• The lewis dot and dash structures for O₃ are :
  
• Infrared and electron diffraction spectra show that O₃ molecule is angular with 0-0-0 bond angle 117°.
• Both 0-0 bonds are identical having bond length 128 pm which is intermediate between single and double bonds.
• This is explained by considering resonating structures and resonance hybrid.
  

(B) Uses of Ozone :

• Ozone sterilises drinking water by oxidising germs and bacteria present in it.
• It is used as a bleaching agent for ivory, oils, starch, wax and delicate fabrics like silk.
• Ozone is used to purify the air in crowded places like Clnema halls, railways, tunnels, etc.
• In industry, ozone is used in the manufacture of synthetic camphor, potassium permanganate, etc.

Question xx.
Explain the trend in following atomic properties of group 16 elements.
i. Atomic radii
ii. Ionisation enthalpy
iii. Electronegativity.
Answer:
(1) Atomic and ionic radii :

1. As compared to group 15 elements, the atomic and ionic radii of group 16 elements are smaller due to higher nuclear charge.
2. The atomic and ionic radii increase down the group from oxygen to polonium. This is due to the addition of a new shell at each successive elements on moving down the group. The atomic radii increases in the order O < S < Se < Te < Po

(2) Ionisation enthalpy :

• The ionisation enthalpy of group 16 elements has quite high values.
• Ionisation enthalpy decreases down the group from oxygen to polonium. This is due to the increase in atomic volume down the group.
• The first ionisation enthalpy of the lighter elements of group 16 (O, S, Se) have lower values than those of group 15 elements in the corresponding periods. This is due to difference in their electronic configurations.

Group 15 : (valence shell) ns² np_x¹ np_y¹ np_z¹
Group 16 : (valence shell) ns² np_x² np_y¹ np_z¹

Group 15 elements have extra stability of half-filled and more symmetrical orbitals, while group 16 elements acquire extra stability by losing one of paired electrons from npx- orbital forming half-filled p-orbitals.

Hence group 16 elements have lower first ionisation enthalpy than group 15 elements.

(3) Electronegativity :

• The electronegativity values of group 16 elements have higher values than corresponding group 15 elements in the same periods.
• Oxygen is the second most electronegative elements after fluorine. (O = 3.5, F = 4)
• On moving down the group electronegativity decreases from oxygen to polonium.
• On moving down the group atomic size increases, hence nuclear attraction decreases, therefore electro-negativity decreases.

4. Answer the following.

Question i.
Distinguish between rhombic sulfur and monoclinic sulfur.
Answer:

Question ii.
Give two reactions showing oxidizing property of concentrated H₂SO₄.
Answer:
Hot and concentrated H₂SO₄ acts as an oxidising agent, since it gives nascent oxygen on heating.

Question iii.
How is SO₂ prepared in the laboratory from sodium sulfite? Give two physical properties of SO₂.
Answer:
(A) Laboratory method (From sulphite) :

• Sodium sulphite on treating with dilute H₂SO₄ forms SO₂.
  Na₂SO₃ + H₂SO_4(aq) → Na₂SO₄ + H₂O₍₁₎ + SO_2(g)
• Sodium sulphite, Na₂SO₃ on reaction with dilute hydrochloric acid solution forms SO₂.
  Na₂SO_3(aq) + 2HCl_(aq) → 2NaCl_9aq0 + H₂O₍₁₎ + SO_2(g)

(B) Physical properties of SO₂

• It is a colourless gas with a pungent smell.
• It is highly soluble in water and forms sulphurous acid, H₂SO₃ SO_2(g) + H₂O₍₁₎ → H₂SO_3(aq)
• It is poisonous in nature.
• At room temperature, it liquefies at 2 atmospheres. It has boiling point 263K.

Question iv.
Describe the manufacturing of H₂SO₄ by contact process.
Answer:
Contact process of the manufacture of sulphuric acid involves following steps :

(1) Preparation of SO₂ : Sulphur or pyrite ores like iron pyrites, FeS₂ on burning in excess of air, form SO₂.

(2) Oxidation of SO₂ to SO₃ : SO₂ is oxidised to SO₃ in the presence of a heterogeneous catalyst V₂O₅ and atmospheric oxygen. This oxidation reaction is reversible.

To avoid the poisoning of a costly catalyst, it is necessary to make SO₂ free from the impurities like dust, moisture, As₂O₃ poison, etc.

The forward reaction is exothermic and favoured by increase in pressure. The reaction is carried out at high pressure (2 bar) and 720 K temperature. The reacting gases, SO₂ and O₂ are taken in the ratio 2:3.

(3) Dissolution of SO₃ : SO₃ obtained from catalytic converter is absorbed in 98%. H₂SO₄ to obtain H₂S₂O₇, oleum or fuming sulphuric acid.


Flow diagram for the manufacture of sulphuric acid

Question 7.1 (Textbook Page No 141)

12th Chemistry Digest Chapter 7 Elements of Groups 16, 17 and 18 Intext Questions and Answers

Question 1.
Elements of group 16 generally show lower values of first ionisation enthalpy compared to the elements of corresponding period of group 15. Why?
Answer:
Group 15 elements have extra stable, half filled p-orbitals with electronic configuration (ns²np³). Therefore more amount of energy is required to remove an electron compared to that of the partially filled orbitals (ns²np⁴) of group 16 elements of the corresponding period.

Question 7.2 (Textbook Page No 141)

Question 1.
The values of first ionisation enthalpy of S and Cl are 1000 and 1256 kJ mol^-1, respectively. Explain the observed trend.
Answer :
The elements S and Cl belong to second period of the periodic table.
Across a period effective nuclear charge increases and atomic size decreases with increase in atomic number. Therefore the energy required for the removal of electron from the valence shell (I.E.) increases in the order S < Cl.

Question 7.4 (Textbook Page No 141)

Question 1.
Fluorine has less negative electron gain affinity than chlorine. Why?
Answer :
The size of fluorine atom is smaller than chlorine atom. As a result, there are strong inter electronic repulsions in the relatively small 2p orbitals of fluorine and therefore, the incoming electron does not experience much attraction. Thus fluorine has less negative electron gain affinity than chlorine.

Try this… (Textbook Page No 140)

Question 1.
Explain the trend in the following properties of group 17 elements.

(1) Atomic size,
(2) Ionisation enthalpy,
(3) Electronegativity,
(4) Electron gain enthalpy.
Answer:
(1) Atomic size :

• Atomic and ionic radii increase down the group as atomic number increases due to the addition of new electronic valence shell to each succeeding element.
• The atomic radii increase in the order F < Cl < Br < 1
• Halogens possess the smallest atomic and ionic radii in their respective periods since the effective nuclear charge experienced by valence electrons in halogen atoms is the highest.

(2) Ionisation enthalpy :

• The ionisation enthalpies of halogens are very high due to their small size and large nuclear attraction.
• The ionisation ethalpies decrease down the group since the atomic size increases.
• The ionisation enthalpy decreases in the order F > Cl > Br > I.
• Among halogens fluorine has the highest ionisation enthalpy due to its smallest size.

(3) Electronegativity :

• Halogens have the highest values for electronegativity due to their small atomic radii and high effective nuclear charge.
• Each halogen is the most electronegative element of its period.
• Fluorine has the highest electronegativity as compared to any element in the periodic table.
• The electronegativity decreases as,
  F > Cl > Br > I
  4.0 3.2 3.0 2.7 (electronegativity)

(4) Electron gain enthalpy (Δ_egH) :

• The halogens have the highest negative values for electron gain enthalpy.
• Electron gain enthalpies of halogens are negative indicating release of energy.
• Halogens liberate maximum heat by gain of electron as compared to other elements.
• Since halogens have outer valence electronic configuration, ns² np⁵, they have strong tendency to accept an electron to complete an octet and acquire electronic configuration of the nearest inert elements.
• In case of fluorine due to small size of 2 p-orbitals and high electron density, F has less negative electron gain enthalpy than Cl.
  F_(g) + e^– → F^–_(g) Δ_egH = – 333 klmol^-1
  Cl_(g) + e^– → Cl^–_(g) Δ_egH = – 349 kJ mol^-1
• The variation in electron gain enthalpy is in the order of, Cl > F > Br > I.

Question 2.
Oxygen has less negative electron gain enthalpy than sulphur. Why?
Answer:

• Oxygen has a smaller atomic size than sulphur.
• It is more electronegative than sulphur.
• It has a larger electron density.
• Due to high electron density, oxygen does not accept the incoming electron easily and therefore has less electron gain enthalpy than sulphur.

Question 7.3 (Textbook Page No 141)

Question 1.
Why is there a large difference between the melting and boiling points of oxygen and sulphur?
Answer :
Oxygen exists as diatomic molecule (O₂) whereas sulphur exists as polyatomic molecule (S₈). The van der Waals forces of attraction between O₂ molecules are relatively weak owing to their much smaller size. The large van der Waals attractive forces in the S₈ molecules are due to large molecular size. Therefore oxygen has low m.p. and b.p. as compared to sulphur.

Question 7.5 (Textbook Page No 141)

Question 1.
Bond dissoClation enthalpy of F₂ (158.8 kj mol^-1) is lower than that of Cl₂ (242.6 kj mol^-1) Why?
Answer :
Fluorine has small atomic size than chlorine. The lone pairs on each F atom in F₂ molecule are so close together that they strongly repel each other, and make the F – F bond weak. Thus, it requires less amount of energy to break the F – F bond. In Cl₂ molecule the lone pairs on each Cl atom are at a larger distance and the repulsion is less.

Thus Cl – Cl bond is comparatively stronger. Therefore bond dissoClation enthalpy of F₂ is lower than that of Cl₂.

Question 7.6 (Textbook Page No 142)

Question 1.
Noble gases have very low melting and boiling points. Why?
Answer :
Noble gases are monoatomic, the only type of interatomic interactions which exist between them are weak van der Waals forces. Therefore, they can be liquefied at very low temperatures and have very low melting or boiling points.

Can you tell? (Textbook Page No 142)

Question 1.
The first member of the a group usually differs in properties from the rest of the members of the group. Why?
Answer:
The first member of a group usually differs in properties from the rest of the members of the group for the following reasons :

• Its small size
• High electronegativity
• Absence of vacant d-orbitals in its valence shell.

Use your brain power! (Textbook Page No 142)

Question 1.
Oxygen forms only OF₂ with fluorine while sulphur forms SF₆. Explain. Why?
Answer:

• Oxygen combines with the most electronegative element fluorine to form OF₂ and exhibits positive oxidation state (+ 2). Since, oxygen does not have vacant J-orbitals it cannot exhibit higher oxidation states.
• Sulphur has vacant d-orbitals and hence can exhibit + 6 oxidation state to form SF₆.

Question 2.
Which of the following possesses hydrogen bonding? H₂S, H₂O, H₂Se, H₂Te
Answer:

• Oxygen being more electronegative, is capable of forming hydrogen bonding in the compound H₂O.
• The other elements S, Se and Te of Group 16, being less electronegative do not form hydrogen bonds.
• Thus, hydrogen bonding is not present in the other hydrides H₂S, H₂Se and H₂Te.

Question 3.
Show hydrogen bonding in the above molecule with the help of a diagram.
Answer:

Try this….. (Textbook Page No 143)

Question 1.
Complete the following tables :

Answer:

Can you tell? (Textbook Page No 146)

Question 1.
What is allotropy?
Answer:
The property of some elements to exist in two or more different forms in the same physical state is called allotropy.

Question 2.
What is the difference between allotropy and polymorphism?
Answer:

• Allotropy is the existence of an element in more than one physical form. It means that under different conditions of temperature and pressure an element can exist in more than one physical forms.
• Coal, graphite and diamond etc., are different allotropic forms of carbon.
• Polymorphism is the existence of a substance in more than one crystalline form.
• It means that under different conditions of temperature and pressure, a substance can form more than one type of crystal. For example, mercuric iodide exists in the orthorhombic and trigonal form.

Question 7.7 (Textbook Page No 146)

Which form of sulphur shows paramagnetic behaviour?
Answer :
In the vapour state, sulphur partly exists as S₂ molecule, which has two unpaired electrons in the antibonding π^* orbitals like O₂. Hence it exhibits paramagnetism.

Try this….. (Textbook Page No 149)

Question 1.
Why water in a fish pot needs to be changed from time to time?
Answer:
A fish pot is an artificial ecosystem and the fish in it are selective and maintained in a restricted environment.

In a fish pot, the unwanted food and waste generated by the fish mix with the water and remain untreated due to lack of decomposers.

Accumulation of waste material will decrease the levels of dissolved oxygen in the water pot.

Hence, it is necessary to change the water from time to time.

Question 7.8 (Textbook Page No 149)

Dioxygen is paramagnetic in spite of having an even number of electrons. Explain.
Answer :
Dioxygen is a covalently bonded molecule.
The paramagnetic behaviour of O₂ can be explained with the help of molecular orbital theory.
Electronic configuration O₂
KK σ(2s)² σ(2s)² σ^*(2p_z)² π(2p_x)² π(2p_x)² π(2p_y)² π^*(2p_x)¹ π^*(2p_y)¹. Presence of two unpaired electrons in antibonding orbitals explains paramagnetic nature of dioxygen.

Question 7.9 (Textbook Page No 150)

High concentration of ozone can be dangerously explosive. Explain.
Answer :
Thermal stability : Ozone is thermodynamically unstable than oxygen and decomposes into O₂. The decomposition is exothermic and results in the liberation of heat (ΔH is – ve) and an increase in entropy (ΔS is positive). This results in large negative Gibbs energy change (ΔG). Therefore high concentration of ozone can be dangerously explosive. Eq O₃ → O₂ + O

Try this…… (Textbook Page No 151)

(a) Ozone is used as a bleaching agent. Explain.
Answer:

• Ozone due to its oxidising property can act as a bleaching agent. O_3(g) → O_2(g) + O
• It bleaches coloured matter. coloured matter + O → colourless matter
• Ozone bleaches in the absence of moisture, so it is also known as dry bleach.
• Ozone can bleach ivory and delicate fabrics like silk.

(b) Why does ozone act as a powerful oxidising agent?
Answer:
Ozone decomposes to liberate nascent oxygen, hence it is a powerful oxidising agent. O_3(g) → O_2(g) + O
For example :

• It oxidises lead sulphide (PbS) to lead sulphate (PbSO₄).
  pbS_(s) + 4O_3(g) → PbSO_(s) + 4O_2(g)
• Potassium iodide, KI is oxidised to iodine, I₂ in the solution.
  2KI_(aq) + H₂O₍₁₎ + O_3(g) → 2KOH_(aq) + I_2(s) + O_2(g)

Question 7.10 : (Textbook Page No 154)

What is the action of concentrated H₂SO₄ on (a) HBr (b) HI
Answer :
Concentrated sulphuric acid oxidises hydrobromic acid to bromine.

2HBr + H₂SO₄ → Br₂ + SO₂ + 2H₂O
It oxidises hydroiodic acid to iodine.
2HI + H₂SO₄ → I₂ + SO₂ + 2H₂O

Try this….. (Textbook Page No 156)

Question 1.
Give the reasons for the bleaching action of chlorine.
Answer:

• Chlorine acts as a powerful bleaching agent due to its oxidising nature.
• In moist conditions or in the presence of water it forms unstable hypochlorous acid, HOCl which decomposes giving nascent oxygen which oxidises the vegetable colouring matter of green leaves, flowers, litmus, indigo, etc.
  Cl₂ + H₂O → HCl + HOCl
  HOCl → HCl + [O]
  Vegetable coloured matter + [O] → colourless matter.

Question 2.
Name two gases used in war.
Answer:
Phosgene : COCl₂
Mustard gas: Cl – CH₂ – CH₂ – S – CH₂ – CH₂ – Cl

Use your brain power! (Textbook Page No 157)

Question 1.
Chlorine and fluorine combine to form interhalogen compounds. The halide ion will be of chlorine or fluorine?
Answer:
Among the- two halogens, chlorine is more electropositive than fluorine (Electronegativity values: F = 4.0, Cl = 3.2)

The interhalogen compound is regarded as the halide of the more electropositive halogen. Hence, the interhalogen compound is the fluoride of chlorine, i.e. chlorine monofluoride, CiF.

Question 2.
Why does fluorine combine with other halogens to form maximum number of fluorides?
Answer:
Since fluorine is the most electronegative element and has the smallest atomic radius compared to other halogen fluorine forms maximum number of fluorides.

Use your brain power! (Textbook Page No 158)

Question 1.
What will be the names of the following compounds: ICl, BrF?
Answer:
ICl : Iodine monochloride
BrF : Bromine monofluoride

Question 2.
Which halogen (X) will have maximum number of other halogen (X ) attached?
Answer:
The halogen Iodine (I) will have the maximum number of other halogens attached.

Question 3.
Which halogen has tendency to form more interhalogen compounds?
Answer:
The halogen fluorine (F) has the maximum tendency to form more interhalogen compounds as it has a small size and more electronegativity.

Question 4.
Which will be more reactive?
(a) ClF₃ or ClF,
(b) BrF₅ or BrF
Answer:
ClF₃ is more reactive than ClF, while BrF₅ is more reactive than BrF. Both ClF₃ and BrF₅ are unstable compared to ClF and BrF respectively due to steric hindrance hence are more reactive.

Question 5.
Complete the table :

Answer:

Use your brain power! (Textbook Page No 159)

Question 1.
In the special reaction for ICl, identify the oxidant and the reductant? Denote oxidation states of the species.
Answer:

Potassium chlorate, KClO₃ is the oxidising agent or oxidant and iodine is the reducing agent or reductant.

Use your brain power! (Textbook Page No 162)

Question 1.
What are missing entries?

Answer:

Balbharti Maharashtra State Board 12th Chemistry Textbook Solutions Chapter 6 Chemical Kinetics Textbook Exercise Questions and Answers.

Maharashtra State Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics

1. Choose the most correct option.

Question i.
The rate law for the reaction aA + bB → P is rate = k[A] [B]. The rate of reaction doubles if
a. concentrations of A and B are both doubled.
b. [A] is doubled and [B] is kept constant
c. [B] is doubled and [A] is halved
d. [A] is kept constant and [B] is halved.
Answer:
b. [A] is doubled and [B] is kept constant

Question ii.
The order of the reaction for which the units of rate constant are mol dm^-3 s^-1 is
a. 1
b. 3
c. 0
d. 2
Answer:
c. 0

Question iii.
The rate constant for the reaction 2N₂O₅(g) → 2N₂O₄(g) + O₂(g) is 4.98 × 10^-4 s^-1. The order of reaction is
a. 2
b. 1
c. 0
d. 3
Answer:
b. 1

Question iv.
Time required for 90 % completion of a certain first order reaction is t. The time required for 99.9 % completion will be
a. t
b. 2t
c. t/2
d. 3t
Answer:
d. 3t

Question v.
Slope of the graph ln[A]_t versus t for first order reaction is
a. -k
b. k
c. k/2. 303
d. -k/2. 303
Answer:
a. -k

Question vi.
What is the half life of a first order reaction if time required to decrease concentration of reactant from 0.8 M to 0.2 M is 12 h?
a. 12 h
b. 3 h
c. 1.5 h
d. 6 h
Answer:
d. 6 h

Question vii.
The reaction, 3ClO^3Θ ClO₃^Θ + 2 Cl^Θ occurs in two steps,
(i) 2 ClO^– → ClO₂^Θ
(ii) ClO₂^Θ + ClO^Θ → ClO₃^Θ + Cl^Θ

The reaction intermediate is
a. Cl^Θ
b. ClO₂^Θ
c. ClO₃^Θ
d. ClO^Θ
Answer:
b. ClO₂^Θ

Question viii.
The elementary reaction O₂(g) + O(g) → 2O₂(g) is
a. unimolecular and second order
b. bimolecular and first order
c. bimolecular and second order
d. unimolecular and first order
Answer:
c. bimolecular and second order

Question ix.
Rate law for the reaction, 2NO + Cl₂ → 2 NOCl is rate = k[NO₂]²[Cl₂]. Thus k would increase with
a. increase of temperature
b. increase of concentration of NO
c. increase of concentration of Cl₂
d. increase of concentrations of both Cl₂ and NO
Answer:
a. increase of temperature

Question x.
For an endothermic reaction, X ⇌ Y. If E f is activation energy of the forward reaction and Er that for reverse reaction, which of the following is correct?
a. E_f = E_r
b. E_f < E_r
c. E_f > E_r
d. ∆H = E_f – E_r is negative
Answer:
(c) Ef → Er

2. Answer the following in one or two sentences.

Question i.
For the reaction,
N₂(g) + 3 H₂(g) → 2NH₃(g), what is the relationship among \(\frac{\mathrm{d}\left[\mathrm{N}_{2}\right]}{\mathrm{dt}}\)\(\frac{\mathrm{d}\left[\mathrm{H}_{2}\right]}{\mathrm{dt}} \text { and } \frac{\mathrm{d}\left[\mathrm{NH}_{3}\right]}{\mathrm{dt}} ?\)
Answer:
N_2(g) + 3H_2(g) → 2NH_3(g)
From the above reaction, when 1 mole of N₂ reacts, 3 moles of H₂ are consumed and 2 moles of NH₃ are formed.

If the instantaneous rate R of the reaction is represented in terms of rate of the consumption of N₂ then, \(R=-\frac{d\left[\mathrm{~N}_{2}\right]}{d t}\)

Hence the rate of reaction in terms of concentration changes in N₂, H₂ and NH₃ may be represented as,

Question ii.
For the reaction,
CH₃Br(aq) + OH-(aq) → CH₃OH^Θ (aq) +Br^Θ (aq), rate law is rate = k[CH₃Br][OH^Θ]
a. How does reaction rate changes if [OH^Θ] is decreased by a factor of 5?
b. What is change in rate if concentrations of both reactants are doubled?
Solution :
Given :
(a) Rate = R = k [CH₃Br] x [OH^–]
If R₁ and R₂ are initial and final rates of reaction then,

Rate will be increased 4 time.

Question iii.
What is the relationship between coeffients of reactants in a balanced equation for an overall reaction and exponents in rate law. In what case the coeffients are the exponents?
Answer:
Explanation : Consider the following reaction, aA + bB → products

If the rate of the reaction depends on the concentrations of the reactants A and B, then, by rate law,
R α [A]^a [B]^b
∴ R = k [A]^a [B^b
where [A] = concentration of A and
[B] = concentration of B

The proportionality constant k is called the velocity constant, rate constant or specific rate of the reaction.

a and b are the exponents or the powers of the concentrations of the reactants A and B respectively when observed experimentally.

The exponents or powers may not be necessarily a and b but may be different x and y depending on experimental observations. Then the rate R will be,
R = k [A]^x [B]^y
For example, if x = 1 and y = 2, then,
R = k [A] x [B]²

Question iv.
Why all collisions between reactant molecules do not lead to a chemical reaction?
Answer:
(i) Collisions of reactant molecules : The basic re-quirements of a reaction is that the reacting species (atoms, ions or molecules) must come together and collide for a reaction to occur. Therefore the rate of the reaction shall depend on the rate and frequency of collisions which increase with the amount of reacting species and temperature.

However it is observed that the rate of reaction is very low as compared to the rate of collisions between reacting species in gaseous phase or liquid phase. This suggests that all the collisions are not fruitful leading to a reaction. Hence it is necessary to consider another factor like energy of colliding species along with collision frequency.

(ii) Energy requirement (Activation energy) : The colliding molecules must possess a certain mini-mum energy called activation energy required far breaking and making bonds resulting in the reaction. This implies that the colliding molecules must have energy equal to or greater than the activation energy. The colliding molecules with less energy do not react.

(iii) Orientation of reactant molecules : The concept of activation energy is satisfactory in case of simple molecules or ions but not in case of complex or higher polyatomic molecules. It is observed that the rates of reaction are less as compared to the rates of collisions between activated molecules with activation energy.

This suggests that in addition, the colliding molecules must have proper orientations relative to each other during collisions. For example, consider the reaction, A – B-l-C → A + B – C. For the reaction to occur, C must collide with B while collisions with A will not be fruitful. Since B has to bond with C.

Question v.
What is the activation energy of a reaction?
Answer:
Activation energy : The energy required to form activated complex or transition state from the reactant molecules is called activation energy.
OR
The height of energy barrier in the energy profile diagram is called activation energy.

Question vi.
What are the units for rate constants for zero order and second order reactions if time is expressed in seconds and concentration of reactants in mol/L?
Answer:
(a) For a zero order reaction, the rate constant has units, molL^-1s^-1.
(b) For second order reaction,
Rate = k x [Reactant]²

Question vii.
Write Arrhenius equation and explain the terms involved in it.
Answer:
Arrhenius equation is represented as k = A x e^-E_a/RT
where
k = Rate constant at absolute temperature T
E_a = Energy of activation R = Gas constant
A = Frequency factor or pre-exponential factor.

Question viii.
What is the rate determining step?
Answer:
Many chemical reactions take place in a series of elementary steps. Among many steps of the reaction, one of the steps is the slowest step compared to other steps.

Rate determining step : The slowest step in the reaction mechanism which involves many steps is called the rate-determining step.

Question ix.
Write the relationships between rate constant and half life of fist order and zeroth order reactions.
Answer:
(a) For first order reaction, half-life period t_1/2 is, \(t_{1 / 2}=\frac{0.693}{k}\) where k is the rate constant.
(b) For zeroth-order reaction, half half period (t_1/2) is, \(t_{1 / 2}=\frac{[\mathrm{A}]_{0}}{2 k}\) where k is the rate constant and [A]₀ is initial concentration of the reactant.

Question x.
How do half lives of the fist order and zero order reactions change with initial concentration of reactants?
Answer:
(A) For the first order reaction, half life, t_1/2 is given by, \(t_{1 / 2}=\frac{0.693}{k}\) where k is rate constant. Hence it is independent of initial concentration of the reactant.

(B) Zero order reaction,
\(t_{1 / 2}=\frac{[\mathrm{A}]_{0}}{2 k}\) where [A]₀ is initial concentration of the reactant.

Hence, half life period increases with the increase in concentration of the reactant.

3. Answer the following in brief.

Question i.
How instantaneous rate of reaction is determined?
Answer:
(1) The instantaneous rate is expressed as an infinite¬simal change in concentration (- dc) of the reactant with the infinitesimal change in time (dt).
For a reaction, A → B, let an infinitesimal change in A be – dc in time dt, then Rate \(=\frac{d[\mathrm{~A}]}{d t}\).

Hence, it is represented as,
∴ Instantaneous rate \(=-\frac{d[\mathrm{~A}]}{d t}\)

The negative sign indicates a decrease in the concentration of A.

It is obtained by drawing a tangent to the curve obtained by plotting the concentration against the time. Hence, the slope at a given point represents the instantaneous rate of the reaction.

(2) The instantaneous rate can also be expressed as an infinitesimal change (or increase) in the concentration of the product with the infinitesimal change in time (dt).

Let dB be an infinitesimal change in the concentration of product B in time dt, then Rate \(=\frac{d[\mathrm{~B}]}{d t}=\frac{d x}{d t}\).

Hence,
Instantaneous rate \(=\frac{d x}{d t}\)

It is obtained from the slope of the curve obtained by plotting the concentration of the product against time.

The instantaneous rate is more useful in obtaining the rate law integrated equations.

Question ii.
Distinguish between order and molecularity of a reaction.
Answer:

Question iii.
A reaction takes place in two steps,
1. NO(g) + Cl₂(g) NOCl₂(g)
2. NOCl₂(g) + NO(g) → 2NOCl(g)
a. Write the overall reaction.
b. Identify reaction intermediate.
c. What is the molecularity of each step?
Solution :
Given :
(1) NO_(g) + Cl_2(g) → NOCl_2(g)
(2) NOCl_2(g) + NO_(g) → 2NOCl_(g)

(a) Overall reaction is obtained by adding both the reactions
2NO_(g) + Cl_2(g) → 2NOCl_2(g)
(b) The reaction intermediate is NOCl₂, since it is formed in first step and consumed in the second step.
(c) Since the first step is a slow and rate determin­ing step, the molecularity is two.

Since the second step is a fast step its molecularity is not considered.

Question iv.
Obtain the relationship between the rate constant and half-life of a fist order reaction.
Answer:
Consider the following reaction,

If [A]₀ and [A]t are the concentrations of A at start and after time t, then [A]₀ = a and [A]_t = a – x.

The velocity constant or the specific rate constant k for the first order reaction can be represented as,

where, a is the initial concentration of the reactant A, x is the concentration of the product B after time t, so that (a – x) is the concentration of the reactant A after time t.

Half-life of a reaction : The time required to reduce the concentration of the reactant to half of its initial value is called the half-life period or the half-life of the reaction.

If t_1/2 is the half-life of a reaction, then at t = t_1/2, x = a/2, hence a – x = a – a/2 = a/2

Hence, for a first order reaction, the half-life of the reaction is independent of the initial concentration of the reactant.

Question v.
How will you represent zeroth-order reaction graphically?
Answer:
(1) A graph of concentration against time : In case of a zero-order reaction, the rate of reaction is independent of the concentration of the reactant. The concentration [A]_t of the reactant at a time t is given by
[A]_t = – kt + [A]₀ (y = – mx + c)
where [A]₀ is the initial concentration of the reactant and k is a rate constant.

Hence in case of zero order reaction, when the concentration of the reactant is plotted against time, a straight line with the slope equal to – k is obtained. The concentration of the reactants de-crease with time. The intercept on the concentration axis gives the initial concentration, [A]₀.

(2) A graph of rate of a reaction against the concen-tration of the reactant: Rate of a zero order reaction is independent of the concentration of the reactant.

Rate, R = k [A]⁰ = k

Hence even if the concentration of the reactant decreases, the rate of the reaction remains constant.

Therefore if the rate of a zero order reaction is plotted against concentration, then a straight line with zero slope is obtained indicating, no change in the rate of the reaction with a change in the concentration of the reactants.

(3) A graph of half-life period against concentration : The half-life period of a zero-order reaction is given by, \(t_{1 / 2}=\frac{[\mathrm{A}]_{0}}{2 k}\) where [A]₀ is initial con-centration of the reactant and k is the rate constant. Hence the half-life period is directly proportional to the concentration.

When a graph of t_1/2 is plotted against concentration, a straight line passing through origin is obtained, and the slope gives \(\frac{1}{2 k}\), where k is the rate constant.

Question vi.
What are pseudo-fist order reactions? Give one example and explain why it is pseudo-fist order.
Answer:
Pseudo-first-order reaction : A reaction which has higher-order true rate law but is experimentally found to behave as first order is called pseudo first order reaction.
Explanation : Consider an acid hydrolysis reaction of an ester like methyl acetate.
CH₃COOCH_3(aq) + H₂O₍₁₎ \(\stackrel{\mathrm{H}_{(\mathrm{aq})}^{+}}{\longrightarrow}\) CH₃COOH_(aq) + CH₃OH_(aq)
Since the reaction involves two substances, ester and water, it is a bimolecular reaction and the true rate law should be, Rate = k’ [CH₃COOCH₃] x [H₂O]

Hence the reaction is expected to follow second order kinetics. However experimentally it is found that the reaction follows first order kinetics.

This is because solvent water being in a large excess, its concentration remains constant. Hence, [H₂O] = constant = k”
Rate = k [CH₃COOCH₃] x [H₂O]
= k [CH₃COOCH₃] x k”
= k’ x k” x [CH₃COOCH₃]
If k’ x k” = k, then Rate = k [CH₃COOCH₃],

This indicates that second-order true rate law is forced into first order rate law. Therefore this bimolecular reaction which appears of second order is called pseudo first order reaction.

Question vii.
What are the requirements for the colliding reactant molecules to lead to products?
Answer:
Collisions of reactant molecules : The basic re­quirements of a reaction is that the reacting species (atoms, ions or molecules) must come together and collide for a reaction to occur. Therefore the rate of the reaction shall depend on the rate and frequency of collisions which increase with the amount of reacting species and temperature.

However it is observed that the rate of reaction is very low as compared to the rate of collisions between reacting species in gaseous phase or liquid phase. This suggests that all the collisions are not fruitful leading to a reaction. Hence it is necessary to consider another factor like energy of colliding species along with collision frequency.

Energy requirement (Activation energy) : The colliding molecules must possess a certain mini­mum energy called activation energy required far breaking and making bonds resulting in the reac­tion. This implies that the colliding molecules must have energy equal to or greater than the activation energy. The colliding molecules with less energy do not react.

Orientation of reactant molecules : The concept of activation energy is satisfactory in case of simple molecules or ions but not in case of complex or higher polyatomic molecules. It is observed that the rates of reaction are less as compared to the rates of collisions between activated molecules with activa­tion energy.

This suggests that in addition, the colliding mole­cules must have proper orientations relative to each other during collisions. For example, consider the reaction, A – B + C → A + B – C. For the reaction to occur, C must collide with B while collisions with A will not be fruitful. Since B has to bond with C.

Question viii.
How catalyst increases the rate of reaction? Explain with the help of a potential energy diagram for catalyzed and uncatalyzed reactions.
Answer:
(i) A catalyst is a substance, when added to the reactants, increases the rate of the reaction without being consumed. For example, the decomposition of KClO₃ in the presence of small amount of MnO₂ is very fast but very slow in the absence of MnO₂.

2KClO_3(s) \(\frac{\mathrm{MnO}_{2}}{\Delta}\) 2KCl_(s) + 3O_2(g)

(ii) The phenomenon of catalysed reaction is called catalysis and depends on nature of the catalyst. In heterogeneous catalysis, the reactant molecules are adsorbed on the solid catalyst surface while in case of homogeneous catalysis, the catalyst reacts with one of the reactants, forms intermediate and decomposes reforming original catalyst and the products.

(iii) The catalyst provides alternative and lower energy path or mechanism for the reaction.

(iv) In the presence of the catalyst, the activation energy of the reaction is lowered. The height of activation energy barrier is less than that in the uncatalysed reaction.

(v) Due to lowering of energy of activation, (E_a) the number of molecules possessing E_a increases, hence the rate of the reaction increases.

(vi) The rate constant = k = A x e^-E_a/RT where A is a frequency factor and hence the rates of the catalysed reaction are higher than those of un-catalysed reactions.

(vii) The catalyst does not change the extent of the reaction but hastens the reaction.

(viii) The catalyst enters the reaction but does not appear in the balanced equation since it is consumed in one step and regenerated in the another.

Question ix.
Explain with the help of the Arrhenius equation, how does the rate of reaction changes with (a) temperature and (b) activation energy.
Answer:
(a) By Arrhenius equation, k = Ax e^-E_a/RT where k is rate constant, A is a frequency factor and Ed is energy of activation at temperature T. As E_a increases, the rate constant and rate of the reaction decreases.

(b) As temperature increases E_a/RT decreases but due to negative sign, k and rate increase with the increase in temperature.

Question x.
Derive the integrated rate law for first order reaction.
Answer:
Consider following gas phase reaction,

Let initial pressure of A be P₀ at t = 0. If after time t the pressure of a A decreases by jc then the partial pressures of the substances will be, P_A = P_Q – x; P_B = x and P_c = x

Total pressure will be,

P_T + P₀ – x + x + x = Po + x
∴ x = P_T – P_n

The partial pressures at time t will be,

Question xi.
How will you represent first-order reactions graphically.
Answer:
(1) A graph of rate of a reaction and concentra­tion : The differential rate law for first-order reac­tion, A → Products is represented as, Rate = [/latex]-\frac{d[\mathrm{~A}]}{d t}=k[\mathrm{~A}]\(

∴ Rate = k x [A]_t (y = mx). When the rate of a first order reaction is plotted against concentration, [A]_t, a straight line graph is obtained.

With the increase in the concentration [A]_t, rate R, increases. The slope of the line gives the value of rate constant k.

(2) A graph of concentration against time : When the concentration of the reactant is plotted against time t, a curve is obtained. The concentration [A], of the reactant decreases exponentially with time. The variation in the concentration can be represented as,

where [A]₀ and [A]_t are initial and final concentra­tions the reactant and k is the rate constant. The time required to complete the first order reaction is infinity.

(3) A graph of log₁₀ (a – x) against time t :

When log₁₀(a – x) is plotted against time t, a straight line with negative slope is obtained, from which the velocity constant k can be calculated.

(4) A graph of half-life period and concentration : The half-life period, t_1/2 of a first order reaction is given by, where k is the rate constant.

For the given reaction at a constant temperature, t_1/2 is constant and independent of the concentration of the reactant.

Hence when a graph of t_1/2 is plotted against concentration, a straight line parallel to the concen­tration axis (slope = zero) is obtained.

(5) A graph of log₁₀ [latex]\left(\frac{a}{a-x}\right)\) against time : The rate constant, for a first order reaction is represented as, where [A₀] and [A]_t are the respective initial and final concentrations of the reactant after time t.


When \(\log _{10}\left(\frac{a}{a-x}\right)\) is plotted against time t, a straight line graph passing through the origin is obtained and the slope gives the value of k/2.303. From this slope, the rate constant can be calculated.

Question xii.
Derive the integrated rate law for the first order reaction, A(g) → B(g) + C(g) in terms of pressure.
Answer:
Consider following gas phase reaction,

Let initial pressure of A be P₀ at t = 0. If after time t the pressure of a A decreases by jc then the partial pressures of the substances will be, P_A = P_Q – x; P_B = x and P_c = x

Total pressure will be,

P_T + P₀ – x + x + x = Po + x
∴ x = P_T – P_n

The partial pressures at time t will be,

Question xiii.
What is zeroth-order reaction? Derive its integrated rate law. What are the units of rate constant?
Answer:
Definition : Zero order reaction : A reaction in which the rate of the reaction does not depend on the concentration of any reactant taking part in the reaction is called zero order reaction.

Consider a zero order reaction, A → Products
The rate of the reaction is, Rate \(=\frac{-d[\mathrm{~A}]}{d t}\)

By rate law,
Rate = k x [A]⁰ = k
∴ – d[A] = k x dt

If [A]₀ is the initial concentration of the reactant A at t = 0 and [A]_t is the concentration of A present after time t, then by integrating above equation,

This is the integrated rate law expression for rate constant for zero order reaction.
∴ k x t = [A]₀ – [A]_t
∴ [A]_t = – kt + A₀

For a zero order reaction :
The rate of reaction is R = k [A]⁰ = k
Hence, the velocity constant k has the unit of the rate of the reaction, i.e., mol dm^-3 s^-1.

Question xiv.
How will you determine activation energy: (a) graphically using Arrhenius equation (b) from rate constants at two different temperatures?
Answer:
(a) By Arrhenius equation,
Rate constant = = A x e^-E_a/RT where A is a fre-quency factor.

When log₁₀k is plotted against 1/T a straight line with negative slope is obtained. From the slope of the graph, energy of activation E_a, is obtained as follows :
Slope = \(\frac{E_{\mathrm{a}}}{2.303 R}\)
∴ Ea = 2303R x sloPe

(b) For the given reaction, rate constants k₁ and k₂ are measured at two different temperatures T₁ and T₂ respectively. Then \(\log _{10} \frac{k_{2}}{k_{1}}=\frac{E_{\mathrm{a}}\left(T_{2}-T_{1}\right)}{2.303 R \times T_{1} \times T_{2}}\) where E_a is the energy of activation.

Hence by substituting appropriate values, energy of activation E_a for the reaction is determined.

Question xv.
Explain graphically the effect of temperature on the rate of reaction.
Answer:
(i) It has been observed that the rates of chemical reactions increase with the increase in temperature.
(ii) The kinetic energy of the molecules increases with the increase in temperature. The fraction of molecules possessing minimum energy barrier,
i. e. activation energy E_a increases with increase in temperature.
(iii) Hence the fraction of colliding molecules that possess kinetic energy (E_a) also increases, hence the rate of the reaction increases with increase in temperature.

(iv) The above figure shows that the area that represents the fraction of molecules with kinetic energy exceeding E_a is greater at higher temperature T₂ than at lower temperature T₁. This explains that the rate of the reaction increases at higher temperature.
(v) The shaded area to the right of activation energy E_a represents fraction of collisions of activated molecules having energy E_a or greater.

Question xvi.
Explain graphically the effect of catalyst on the rate of reaction.
Answer:
(i) The phenomenon of catalysed reaction is called catalysis and depends on nature of the catalyst. In heterogeneous catalysis, the reactant molecules are adsorbed on the solid catalyst surface while in case of homogeneous catalysis, the catalyst reacts with one of the reactants, forms intermediate and decomposes reforming original catalyst and the products.
(ii) The catalyst provides alternative and lower energy path or mechanism for the reaction.
(iii) In the presence of the catalyst, the activation energy of the reaction is lowered. The height of activation energy barrier is less than that in the uncatalysed reaction.

Question xvii.
For the reaction 2A + B → products, find the rate law from the following data.

Solution :
Given : 2A + B → Products
Rates : R₁ = 0.15 Ms^-1 R₂ = 0.3 Ms^-1
[A]₁ = 0.3 M [A]₂ = 0.6 M
[B]₁ = 0.05 M [B]₂ = 0.05 M
(i) If order of the reaction in A is x and in B is y then, by rate law,

∴ y = 1. Hence the reaction has order one in B.
The order of overall reaction = n = n_A + n_B = 1 + 1 = 2

Answer:
(i) Rate law : Rate = fc [A] x [B]
Rate constant = k = 10M^-1s^-1
Order of the reaction = 2

4. Solve

Question i.
In a first order reaction, the concentration of reactant decreases from 20 mmol dm^-3 to 8 mmol dm^-3 in 38 minutes. What is the half life of reaction? (28.7 min)
Solution :
Given: [A]₀ =20 mmol dm^-3;
[A]_t=8 mmol dm^-3; t=38 mm;

Answer:
Half life period = 28.74 min

Question ii.
The half life of a first order reaction is 1.7 hours. How long will it take for 20% of the reactant to react? (32.9 min)
Solution :
Given : t_1/2 = 1.7 hr; [A]₀ = 100;
[A]_t = 100 – 20 = 80; t =?
\(t_{1 / 2}=\frac{0.693}{k}\)

Answer:
Time required = t = 32.86 min

Question iii.
The energy of activation for a first order reaction is 104 kJ/mol. The rate constant at 25 0C is 3.7 × 10^-5 s^-1. What is the rate constant at 30⁰C? (R = 8.314 J/K mol) (7.4 × 10^-5)
Solution :

Answer:
k₂ = 7.382 x 10^-5 s^-1

Question iv.
What is the energy of activation of a reaction whose rate constant doubles when the temperature changes from 303 K to 313 K? (54.66 kJ/mol)
Solution :
Given : k₂ = 2k_t, T₁ = 303 K; T₂ = 313 K; E_a = ?

Answer:
Energy of activation = E_a = 54.66 kJ

Question v.
The rate constant of a reaction at 5000C is 1.6 × 10³ M^-1 s^-1. What is the frequency factor of the reaction if its activation energy is 56 kJ/mol. (9.72 × 106 M^-1 s^-1)
Solution :

Answer:
Frequency factor = A = 9.727 x 10⁶ M^-1s^-1

Question vi.
Show that time required for 99.9% completion of a first order reaction is three times the time required for 90% completion.
Solution :
Given : For 99.9 % completion, if [A]₀ = 100,

If t₁ and t₂ are the times required for 99.9 % and 90 % completion of reaction respectively, then

Answer:
Time required for 99.9 % completion of a first order reaction is three time the time required for 90 % completion of the reaction.

Question vii.
A first order reaction takes 40 minutes for 30% decomposition. Calculate its half life. (77.66 min)
Solution :

Answer:
Half life period = 77.70 min.

Question viii.
The rate constant for the first order reaction is given by log₁₀ k = 14.34 – 1.25 × 10⁴ T. Calculate activation energy of the reaction. (239.3 kJ/mol)
Solution :
Given : log₁₀ k = 14.34 – \(\frac{1.25 \times 10^{4}}{T}\) ……………………. (1)
From Arrhenius equation we can write,
\(\log _{10} k=\log _{10} A-\frac{E_{\mathrm{a}}}{2.303 R \times T}\) ……………………. (2)
By comparing equations (1) and (2),
\(\frac{E_{\mathrm{a}}}{2.303 \times R}\) = 1.25 x 10⁴
∴ Ea = 1.25 x 10⁴ x 2.303 x R
= 1.25 x 10⁴ x 2.303 x 8.314
= 23.93 x 10⁴ = 239.3 kJ mol^-1

[Note : Frequency factor A may also be calculated as follows : log₁₀ A = 14.34
∴ A = Antilog 14.34 = 2.188 x 10⁴
Answer:
Energy of activation = E_a = 239.3 kJ mol^-1.

Question ix.
What fraction of molecules in a gas at 300 K collide with an energy equal to activation energy of 50 kJ/mol? (2 × 10^-9)
Solution :
Given : T = 300 K; E_a = 50 kJ mol^-1
= 50 x 10³ mol^-1
The fraction of molecules undergoing fruitful collisions is

Answer:
Fraction of molecules undergoing collision = 2 x 10^-9

Activity :
1. If you wish to determine the reaction order and rate constant for the reaction, 2AB₂ → A₂ + 2B₂.
a) What data would you collect?
b) How would you use these data to determine whether the reaction is zeroth or first order?

2. The activation energy for two reactions are E_a and E’_a with E_a > E’_a. If the temperature of reacting system increases from T₁ to T₂, predict which of the following is correct?

k values are rate constants at lower temperatures and k values at higher temperatures.

12th Chemistry Digest Chapter 6 Chemical Kinetics Intext Questions and Answers

(Textbook Page No 121)

Question 1.
Write the expressions for rates of reaction for :
2N₂O_5(g) → 4NO_2(g) + O_2(g)?
Answer:
For the given reaction, Rate of reaction =
\(=R=-\frac{1}{2} \frac{d\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right]}{d t}\)
\(\begin{aligned}
&=+\frac{1}{4} \frac{d\left[\mathrm{NO}_{2}\right]}{d t} \\
&=+\frac{d\left[\mathrm{O}_{2}\right]}{d t}
\end{aligned}\)

Problem 6.1: (Textbook Page No 121)

Question 1.
For the reaction,
\(\mathbf{3 I}_{(a q)}^{-}+\mathbf{S}_{2} \mathbf{O}_{8(a q)}^{2-} \longrightarrow \mathbf{I}_{3(\text { (aq) }}^{-}+2 \mathbf{S O}_{4(\mathrm{aq})}^{2-}\)
Calculate (a) the rate of formation of I₃^–,
(b) the rates of consumption of 1 and S₂O and (c) the overall rate of reaction if the rate of formation of \(\mathrm{SO}_{4}^{2-}\) is 0.O22 moles dm^-3 sec^-1.
Answer:


∴ (a) Rate of formation of \(\mathrm{I}_{3}^{-}\) = 0.011 mol dm^-3 s^-1
(b) Rate of consumption of I^– = 0.033 mol dm^-3 s^-1
(c) Rate of consumption of \(\mathrm{S}_{2} \mathrm{O}_{8}^{2-}\) = 0.011 mol dm^-3 s^-1
(d) Overall rate of reaction = Rate of consumption of reactant = Rate of formation of product

Try this….. (Textbook Page No 122)

Question 1.
For the reaction :
NO_2(g) + CO_(g) → NO_(g) + CO_2(g), the rate of reaction is experimentally found to be proportional to the square of the concentration of NO2 and independent that of CO. Write the rate law.
Answer:
Since the rate of the reaction is proportional to [NO₂]² and [CO]⁰, the rate law is R = k[NO₂]² x [CO]⁰
∴ R = k[NO₂]².

Try this….. (Textbook Page No 124)

Question 1.
The reaction,
CHCl_3(g) + Cl_2(g) → CCl_4(g) + HCl_(g) is first order in CHCl₃ and 1/2 order in Cl₂. Write the rate law and overall order of reaction.
Answer:
Since the reaction is first order in CHCl₃ and 1/2 order in Cl₂, the rate law for the reaction will be, Rate = k[CHCl₃] X [Cl₂]^1/2
The overall order (n) of the reaction will be, n = l + = \(\frac{1}{2}=\frac{3}{2}\)

Use your brain power! (Textbook Page No 124)

Question 1.
The rate of the reaction 2A + B → 2C + D is 6 x 10^-4 mol dm^-3 s^-1 when [A] =[B] = O.3 mol dm^-3 If the reaction is of first order in A and zeroth order in B, what is the rate constant?
Answer:
For the reaction,
2A + B → 2C + D,

(Problem 6.7) (Textbook Page No 126)

Question 1.
A reaction occurs in the following steps :
(i) NO_2(g) + F_2(g) → NO₂F_(g) + F_(g) (slow)
(ii) F_(g) + NO_2(g) → NO₂F_(g) (fast)
(a) Write the equation of overall reaction.
(b) Write down rate law.
(c) Identify the reaction intermediate.
Solution :
(a) The addition of two steps gives the overall reaction as
2NO_2(g) + F_2(g) → 2NO₂ F_(g)
(b) Step (i) is slow. The rate law of the reaction is predicted from its stoichiometry. Thus, rate = k [NO₂] [F₂]
(c) F is produced in step (i) and consumed in step (ii) hence F is the reaction intermediate.

Try this….. (Textbook Page No 126)

Question 1.
A complex reaction takes place in two steps :
(i) NO_(g) + O_3(g) → NO_3(g) + O_(g)
(ii) NO_3(g) + O_(g) → NO_2(g) + O_2(g)
The predicted rate law is rate = k [NO] [O₃]. Identify the rate-determining step. Write the overall reaction. Which is the reaction inter-mediate? Why?
Answer:
(i) NO_(g) + O_3(g) → NO_3(g) + O_(g)
(ii) NO_3(g) + O_(g) → NO_2(g) + O_2(g)
(a) The first step is slow and rate determining step since the rate depends on concentrations of NO_(g) and O_3(g). (Given : Rate = k [NO] x [O])
(b) The overall reaction is the combination of two steps.
NO_(g) ⁺ O_3(g) → NO_2(g) + O_2(g)
(c) NO_3(g) and O_(g) are reaction intermediates. They are formed in first step (i) and removed in the second step (ii).

Try this….. (Textbook Page No 129)

Question 1.
The half-life of a first-order reaction is 0.5 min. Calculate (a) time needed for the reactant to reduce to 20% and (b) the amount decomposed in 55 s.
Answer:

Try this….. (Textbook Page No 123)

Question 1.
For the reaction 2A + 2B → 2C + D, if concentration of A is doubled at constant [B] the rate increases by a factor of 4. If the concentration of B is doubled with [A] being constant the rate is doubled. Write the rate law of the reaction.
Answer:
Rate = R₁ = k[A]^x [B]^y
When concentration of A = [2A] and

Hence order with respect to A is 2 and with respect to B is 1. By rate law,
Rate = A: [A]² [B]

Question 2.
The rate law for the reaction A + B → C is found to be rate = k [A]² x [B]. The rate constant of the reaction at 25 °C is 6.25 M^-2 S^-1. What is the rate of reaction when [A] = 1.0 mol dm^-3 and [B] = 0.2 mol dm^-3?
Answer:
Rate = k x [A]² x [B]
= 6.25 x 1² x 0.2
Rate = 1.25 x 10² mol dm^-3 s^-1

Balbharti Maharashtra State Board Marathi Yuvakbharati 12th Digest Chapter 8 रेशीमबंध Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board 12th Marathi Yuvakbharati Solutions Chapter 8 रेशीमबंध

12th Marathi Guide Chapter 8 रेशीमबंध Textbook Questions and Answers

कृती

1. कृती करा.

प्रश्न अ.

उत्तर :

प्रश्न आ.

उत्तर :

प्रश्न इ.

उत्तर :

प्रश्न ई.

उत्तर :

2. कारणे शोधा व लिहा.

प्रश्न अ.
पाखरांचा चिवचिवाट सुरू झालेला नसतो, कारण…
उत्तर :
पाखरांचा चिवचिवाट सुरू झालेला नसतो; कारण नुकते कुठे तीन-साडेतीन वाजलेले असतात.

प्रश्न आ.
मानवाला निसर्गाची ओढ लागते, कारण…
उत्तर :
मानवाला निसर्गाची ओढ लागते; कारण माणसाच्या मनात आदिमत्वही भरून राहिलेले असते.

3. अ. पहाटेच्या वेळी बागेत प्रवेश केल्यानंतर लेखकाला खालील फुलांसंदर्भात आलेले अनुभव लिहा.

प्रश्न 1.
सायली –
उत्तर :
सायलीच्या इवल्या इवल्या पानांतून एक वेगळीच हिरवाई वाहू लागते.

प्रश्न 2.
गुलमोहोर –
उत्तर :
गुलमोहोराजवळ जावं तर त्यानं स्वागतासाठी, केशरी सडाच शिंपून ठेवलेला असतो

प्रश्न 3.
जॅक्रांडा –
उत्तर :
जॅक्रांडाची निळीजांभळी फुलं रक्तचंदनी चाफ्याशी बिलगून गप्पागोष्टी करत असतात.

प्रश्न 4.
चाफा –
उत्तर :
चाफ्यांजवळ जावं तर त्यांच्या फुलांचा एक वेगळाच मंद मंद गंध येत असतो; पण तो निशिगंधासारखा मात्र नसतो.

आ. वर्णन करा.

प्रश्न 1.
उत्तररात्रीचे आगमन
उत्तर :
मध्यरात्र उलटली की उत्तररात्र हलकेच आकाशात पाऊल टाकते. उत्तररात्रीची पावले मुळातच मुलायम, त्यात ती रात्र आपली मुलायम पावले हळुवारपणे, अलगद ठेवीत येते. कुणालाही चाहूल लागणे कठीण. मात्र लेखकांचे उत्तररात्रीशी अत्यंत जवळिकेचे नाते आहे. त्यामुळे तिच्या पावलांची मंद मंद नाजूक स्पंदने लेखकांच्या मनात उमटत राहतात. लेखकांना झोप लागत नाही. आपली झोप जणू पूर्ण झाली आहे, असेच त्यांना वाटत राहते.

प्रश्न 2.
पहाट व पाखरे यांच्यातील नात
उत्तर :
लेखकांना भोवतालचा निसर्ग माणसासारखाच भावभावनांनी भरलेला भासतो. पहाटेची घटना तशी साधीशीच. पहाट होत आहे. पाखरांचा बारीक बारीक आवाज सुरू झाला आहे. त्यांच्या हालचालींना सुरुवात होत आहे. पहाट हळूहळू पुढे सरकत आहे. या प्रसंगात लेखकांना मानवी भावभावनांचे दर्शन घडते. पाखरांचा बारीक बारीक आवाज म्हणजे त्यांची कुजबुज होय. ती जणू एकमेकांना विचारताहेत, ” पहाट आली का? ” पहाटेचे हळुवार येणे पाहून लेखकांना वाटते की, पाखरांना त्रास होऊ नये म्हणूनच जणू पहाट हळूच पाखरांना विचारते की, “मी येऊ का?” त्या दोघांमधले हळुवार कोमल नातेच लेखकांना या वाक्यातून व्यक्त करायचे आहे.

इ. खालील घटकांच्या संदर्भात पाठात आलेल्या मानवी क्रिया लिहा.

प्रश्न 1.

1. वृक्ष – …………..
2. वेली – …………
3. फुले – ………..
4. पाखरे – ………….

उत्तर :

1. वृक्ष – डोळा लागलेला असतो.
2. वेली – डोळा लागलेला असतो.
3. फुले – विसावलेली, सुखावलेली असतात.
4. पाखरे – गाढ झोपलेली असतात.

4. व्याकरण.

अ. खालील वाक्प्रचारांचा अर्थ सांगून वाक्यात उपयोग करा.

प्रश्न 1.
मन समेवर येणे-
उत्तर :
अर्थ – मन शांत व एखाद्या गोष्टीशी एकरूप होणे.
वाक्य – राहुल रात्रभर भरकटणारे मन प्रभातफेरीसाठी बाहेर पडल्यावर एकदम समेवर आले.

प्रश्न 2.
साखरझोपेत असणे-
उत्तर :
अर्थ – पहाटेच्या गाढ स्वप्निल निद्रेत असणे.
वाक्य – पहाटे मोहन साखरझोपेत असताना बाहेर पाऊस पडत होता.

आ. खालील वाक्यांत योग्य विरामचिन्हांचा उपयोग करा.

प्रश्न 1.
खरं तर पहाटच त्यांना विचारत असते की मी येऊ का तुम्हांला भेटायला
उत्तर :
खरं तर पहाटच त्यांना विचारत असते, की ‘मी येऊ का तुम्हांला भेटायला?’

प्रश्न 2.
निशिगंध म्हणजे निशिगंधच
उत्तर :
निशिगंध म्हणजे निशिगंधच!

इ. खालील वाक्यांचा प्रकार ओळखून सूचनेप्रमाणे तक्ता पूर्ण करा.

प्रश्न 1.

उत्तर :
वाक्यप्रकार → उद्गारार्थी वाक्य
विधानार्थी → वृक्षवेली आपल्याला तजेला आणि विरंगुळा देतात.

वाक्यप्रकार → विधानार्थी वाक्य
उद्गारार्थी → किती अनावर भरती येते आल्हादाला आणि हर्षोल्हासाला!

वाक्यप्रकार → होकारार्थी वाक्य
नकारार्थी → वाफे तर ओले नाहीतच.

ई. खालील तक्ता पूर्ण करा.

प्रश्न 1.

उत्तर :

उ. खालील वाक्यांतील प्रयोग ओळखा व लिहा.

प्रश्न 1.

1. खिडकी हलकेच उघडतो.
2. मानवाला निसर्गाची ओढ लागून राहिली.
3. तुम्हांलाही त्यातला आल्हाद जाणवेल.

उत्तर :

1. कर्तरी प्रयोग
2. कर्मणी प्रयोग
3. भावे प्रयोग

5. स्वमत.

प्रश्न अ.
‘मानवाला निसर्गाची जी ओढ युगानुयुगांपासून लागून राहिली आहे, ती या आदिम, ॠजु, स्नेहबंधांमुळे तर नाही?…’ या विधानासंबंधी तुमचे मत लिहा.
उत्तर :
उत्तररात्रीचे दृश्य खरोखरच विलक्षण असते. सर्वत्र, सर्व काही शांतनिवांत असते. दिवसा इकडेतिकडे सतत धावणारी, कोणती ना कोणती कामे करीत राहणारी, एकमेकांशी बोलणारी, एकमेकांशी भांडणारी, एकमेकांवर प्रेम करणारी ही माणसे निवांत झोपलेली असतात. काहीजण दिवसा चिंतांनी ग्रासलेली असतात. काहीजण मिळालेल्या यशामुळे आनंदाच्या, सुखाच्या शिखरावर असतात. या सर्व भावभावना, सर्व सुखदु:खे उत्तररात्रीच्या क्षणांमध्ये विरून गेलेल्या असतात.

दुष्ट विचार, दुष्ट भावना आणि चांगल्या माणसांच्या मनातले चांगले विचार, चांगल्या भावना हे सर्व काही त्या क्षणी दूर निघून गेलेले असते. माणसे भांडतात तेव्हाचे त्यांचे भाव आठवून पाहा. सर्व त्वेष, द्वेष, राग, संताप उफाळून आलेला असतो. तीच माणसे उत्तररात्री या सर्व भावभावनांचे गाठोडे बाजूला ठेवून निवांत झालेली असतात. सज्जन व दुर्जन दोघेही शेजारी शेजारी झोपलेले असतील, तर त्यांच्यातला चांगला कोण व वाईट कोण हे नुसते पाहून ठरवताच येणार नाही. त्या क्षणी सर्वांचे मन निर्मळ, शुद्ध झालेले असते.

सर्व प्राणिमात्रांमध्ये, वनस्पतींमध्ये हाच शुद्ध भाव वसत असतो. आणि हा शुद्ध भाव अनादी काळापासून सर्वांच्या मनात वस्ती करून आहे. माणसाचे मन या मूळ भावनेकडेच धाव घेत असते. आदिम खूप खूप पूर्वीचे. ऋजू म्हणजे साधे, सरळ, निर्मळ, पारदर्शी. त्यात कोणतेही किल्मिष, वाईट भावनेचा लवलेशही नसतो. सगळ्यांच्याच ठायी हा भाव असल्याने सर्वजण एकमेकांशी हसतखेळत बोलू शकतात. एकमेकांच्या मदतीला धावतात. एकमेकांवर प्रेम करतात. त्या शुद्ध, निर्मळ भावनेने एकमेकांशी बांधले जातात. लेखकांना या वाक्यातून हेच सांगायचे आहे.

प्रश्न आ.
‘रेशीमबंध’ या शीर्षकाची समर्पकता तुमच्या शब्दांत लिहा.
उत्तर :
लेखकांचे निसर्गाशी अत्यंत कोमल, हळुवार, नाजूक नाते आहे. त्यांच्या मते, सर्व माणसांचेच तसे नाते असते. या हळुवार, कोमल नात्याचे दर्शन लेखक या पाठात घडवतात. हे नाते रेशमासारखे तलम, मुलायम आहे. म्हणून ते रेशीमबंध.

हे रेशीमबंध लेखकांनी अत्यंत मुलायमपणे, हळुवारपणे उलगडून दाखवले आहेत. नीरव शांततेत उत्तररात्र हळुवारपणे कोमल पावले टाकत येते. कोणाला चाहूलही लागत नाही. पण लेखकांच्या मनात त्या मुलायम पावलांची मंद मंद स्पंदने उमटतात. त्यांचे मन तितक्याच हळुवारपणे ती स्पंदने टिपते. त्यांच्या निसर्गाशी असलेल्या नात्याचा रेशमी मुलायम पण इथे जाणवतो.

साखरझोपेत जग विसावलेले असते. साऱ्या काळज्या-चिंता मिटून गेलेल्या असतात. मन सुखदुःखांच्या पलीकडे गेलेले असते. एक निर्मळ, शुद्ध असे स्वरूप मनाला प्राप्त होते. निसर्गाचा आत्माच त्यात असतो. लेखकांचे नाते या निर्मळपणाशी, त्या आत्म्याशी जडले आहे. त्यांना त्यांच्या नातीच्या शैशवातला नितळपणा जाणवतो.

या नितळपणाचा संबंध निसर्गाच्या आत्म्याशी, निर्मळपणाशी आहे. कोणालाही चाहूल लागू न देता पहाट अलगद अवतरते, पण लेखक अत्यंत संवेदनशील असल्याने त्यांना चाहूल लागते. त्या अनोख्या, नाजूक, तरल क्षणाचा लेखकांना अनुभव येतो. बागेतल्या वृक्षवेलींच्या रूपांनी, त्यांचे विविध रंग व सुगंध यांच्या रूपांनी लेखकांना स्वत:चे आदिमतेशी असलेले नाते जाणवते.

या पाठात लेखक निसर्गाशी असलेल्या स्वत:च्या नात्याचा शोध घेत आहेत. स्वतः प्रमाणे सगळीच माणसे निसर्गाशी कोमल, नाजूक, हळुवार भावनांनी बांधली गेली आहेत. ते बंध सहजासहजी दिसत नाहीत; दाखवून देता येत नाहीत. ते सूक्ष्म, तरल, कोमल भावनांचे बंध असतात. ते रेशमाप्रमाणे तलम, मुलायम असतात. म्हणून लेखक या बंधांना रेशीमबंध म्हणतात. संपूर्ण पाठाच्या केंद्रस्थानी हे रेशीमबंधच आहेत. म्हणून या पाठाला रेशीमबंध’ हे शीर्षक खूप साजते.

6. अभिव्यक्ती.

प्रश्न अ.
निसर्ग आणि मानव यांच्यातील परस्परसंबंध तुमच्या शब्दांत स्पष्ट करा.
उत्तर :
मला कळू लागले तेव्हापासूनचे सर्व आठवते. कधीही फिरायला जाण्याची कल्पना आली, सहलीला जाण्याची वेळ आली की, मला प्रचंड आनंद होतो. खरे सांगायचे तर मला एकट्यालाच असे वाटते, असे नाही. आमच्या वर्गातल्या सर्व मित्रमैत्रिणींना सहलीचा विषय आला की, अमाप आनंद होतो. सहलीला गेलो, निसर्गात गेलो की, खूप आनंद मिळतो. नदीत डुंबायला मिळाले तर कितीही वेळ डुंबत राहावेसे वाटते. तेच रानावनात भटकतानाही वाटत असते. झाडांच्या सोबत वावरताना कंटाळा येतच नाही. हिरव्यागार वृक्षवेलींनी सजलेला डोंगर पाहताना मन सुखावते. हे असे का होत असावे?

वनस्पती या सजीव आहेत; त्यांना माणसांसारख्याच भावभावना असतात. वनस्पतींनाही आनंद होतो, दुःख होते. त्यांच्यावर प्रेमाने हात फिरवला, तर त्याही सुखावतात, हे सर्व आता लहानथोरांपासून सर्वांनाच ठाऊक झाले आहे. प्रत्येक ऋतूशी माणसाचे भावनिक नाते निर्माण झाले आहे. म्हणूनच तर झाडे सुकून जाऊ लागली की माणसाचे मन कळवळते. वृक्ष नष्ट होऊ लागले की माणसाला दुःख होते. वृक्षतोड होताना दिसली की, माणसे खवळतात. शहरात माणसाला स्वत:चे वृक्षप्रेम जपता येत नाही, म्हणून माणसे कुंड्यांमध्ये रोपटी लावतात आणि त्यांचा आनंद घेतात. अमाप वृक्षतोड होऊ लागली, तेव्हा सुंदरलाल बहुगुणा या वृक्षप्रेमीने ‘चिपको आंदोलन’ उभारले. त्याला देशभर प्रचंड प्रतिसाद मिळाला.

लोकांच्या मनात हे एवढे वृक्षप्रेम दाटून आले, याचे कारणच हे की निसर्ग आणि आपण यांच्यात एक खोलवरचे नाते आहे. ज्या निसर्गाने माणसांना निर्माण केले, त्यानेच प्राण्यांना आणि वनस्पतींना निर्माण केले आहे. आपण सर्व निसर्गाची लेकरे आहोत. आपण, अन्य प्राणी आणि वनस्पती ही सर्व भावंडेच आहेत. आपणा सर्वांमध्ये हे असे रक्ताचेच नाते आहे. निसर्गच आपले पालनपोषण करतो. तोच अन्नपाणी देतो. तोच हवाही देतो. आपल्याला तोच जिवंत ठेवतो. आपल्या जगण्याचा आधारच निसर्ग हा आहे. हेच निसर्ग व मानव यांच्यातले नाते आहे.

प्रश्न आ.
डॉ. यू. म. पठाण यांच्या लेखनाची भाषिक वैशिष्ट्ये पाठाधारे स्पष्ट करा.
उत्तर :
डॉ. यू. म. पठाण यांच्या भाषेचे सर्वांत पहिले जाणवणारे वैशिष्ट्य म्हणजे त्यांची साधी, सरळ, भाषा स्वतःचा अनुभव ते पारदर्शीपणे व्यक्त करतात. उत्तररात्रीचे आकाशात पडणारे पाऊल हळुवारपणे, अलगद, मुलायम, कुणालाही चाहूल लागू न देणारे’ असे असते. पावलाचा हळुवारपणा, नाजुकपणा, मुलायम पण त्या वाक्यातून सहज प्रत्ययाला येतो. उदा., उत्तररात्र प्रवेश करते, त्याचे वर्णन करणारे वाक्य पाहा – ‘उत्तररात्रीने हलकेच आकाशात पाऊल ठेवलेलं असतं येथे पाऊल ‘पडत’ नाही किंवा ‘टाकले जात नाही. उत्तररात्र ‘हलकेच पाऊल ठेवते’ अत्यंत योग्य अशा क्रियापद योजनेमुळे भाषेतून व्यक्त होणाऱ्या अनुभवाचा प्रत्यय येतो. वाचकाला ती भाषा भावते. हे एक उदाहरण झाले. पण संपूर्ण पाठभर अशी अनुभवाचा प्रत्यय देणारी भाषा आढळते.

लेखकांना सृष्टीतले मानवेतर घटक कोरडे, भावनाविरहित वाटतच नाहीत. ते त्यांना माणसांप्रमाणेच भावभावनांनी भिजलेले, सजीव, चैतन्यपर्ण वाटतात. ते घटक वर्णनामध्ये मानवी रूपच घेऊन येतात. म्हणून पाखरे ‘हळूहळू डोळे किलकिले’ करून पाहतात. पहाटेच्या प्रकाशकिरणांना ‘खुणावतात ‘. पाखरे ‘कुजबुजली’. बोगनवेल ‘सळसळू’ लागते. दोन्ही वाफे ‘एकमेकांशी हितगुज’ करतात. मोगरा ‘खुदखुद हसतो’. तो कधी रुसून बसतो’. झाडे-वेली ‘भावुक होतात’ अशा रितीने सर्व घटक मानवी रूप घेऊन येतात. लेखन हृद्य बनते. लेखनातला अनुभव भावभावनांशी रसरशीत बनतो.

उत्तररात्री आगमन, सुख दुःखाच्या पलीकडे गेलेले मन, त्यांच्या नातीच्या शैशवातला नितळपणा, पहाटेचे अलवार आगमन असे अत्यंत तरल, मुलायम, नाजूक अनुभव प्रत्ययदर्शी होतात. आणखी एक वैशिष्ट्य बघा. असं कोणतं बरं नातं’, ‘का बरं म्हटलं असावं’ ही शब्दरूपे पाहा. ही छापील वळणाची शब्दरूपे नाहीत. ही दैनंदिन जीवनातली बोलण्यातली शब्दरूपे आहेत. एखादा माणूस आपल्या जिवलग मित्राशी जिवाभावाच्या गप्पागोष्टी करीत बसला असावा, तसे हा लेख वाचताना वाटत राहते. म्हणूनच संपूर्ण लेखात भाषेला एक वेगळाच गोडवा प्राप्त झाल्याचे दिसून येते.

प्रश्न इ.
संत तुकाराम महाराज यांनी वृक्षवल्लींना ‘सोयरी’ असे म्हटले आहे, यामागील तुम्हांला समजलेली कारणे लिहा.
उत्तर :
थोडा वेळ बागेत बसले, रानात फेरफटका मारला की, मनाला खूप आल्हाद मिळतो. शहरात आखीव रेखीव रस्ते आणि तशाच आखीव रेखीव इमारती. इमारतीतल्या प्रत्येक घराचा चेहरा सारखाच. याउलट, रानावनात सौंदर्याची मुक्त उधळण असते. तिथे एक झाड दुसऱ्या झाडासारखे नसते. एक पान दुसऱ्या पानासारखे नसते. एक हिरवा रंग पाहा.

त्या एका हिरव्या रंगांच्या शेकडो छटांचे दर्शन तिथे घडते. हजारो आकार, हजारो रंग, हजारो आवाज, हजारो गंध. तिथे पंचेद्रियांच्या सुखाची लयलूट असते. किती विविधता! पक्षी, प्राणी, किडेमुंग्या यांच्या हजारो जाती. त्यांच्यातही रंग, आकार, हालचाली यांचे हजारो प्रकार. निसर्गातली ही विविधता मनाला मोहवते. मन त्या सौंदर्यात बुडून जाते. कृत्रिमतेची चढलेली पुटे हळूहळू गळून पडतात. मन मोकळे होते. सौंदर्याचा, आनंदाचा अनुभव घेऊ लागते.

तुकाराम महाराजांनी विठ्ठलभक्तीसाठी वनाचाच आश्रय घेतला. माणसात असलेल्या सर्व कुभावनांपासून मुक्ती मिळावी; आपला आत्मा त्या कुभावनांपासून मुक्त व्हावा; ईश्वराचे निर्मळ, सोज्वळ रूप दिसावे; त्याच्याशी एकरूप होता यावे; म्हणून त्यांनी वन गाठले. वनात गेले की मन आपसूक मुक्त होते. मनाची ही अवस्था ईश्वराकडे जाण्यासाठी उत्तम अवस्था. वनाचे सौंदर्य म्हणजे ईश्वराचे एक रूपच, त्या रूपाच्या सान्निध्यात राहावे, षड्रिपू चा त्याग करावा म्हणजे आपण अलगद ईश्वराच्या जवळ जाऊन ठेपतो.

आपल्याला सर्व सुंदर, निर्मळ, चांगलेच दिसते. या चांगुलपणाचाच आस्वाद घेत राहावा, असे वाटू लागते. म्हणजे ईश्वराच्या दर्शनातच, त्याच्या स्मरणातच आकंठ बुडून जावे अशी अवस्था होऊन जाते. तुकाराम महाराजांना वृक्षवल्ली सोयरी वाटली ती या कारणाने. या वृक्षवल्लींच्या ठायी माणसाचे दुर्गुण नसतात. ती ईश्वराची रूपे होत. त्यांच्या सहवासातच ईश्वरभक्ती फुलते. वृक्षवल्ली, सर्व वनश्री आपल्याला ईश्वराच्या वाटेवर आणून सोडतात. तुकाराम महाराज म्हणूनच वृक्षवल्लींच्या सान्निध्यात गेले. वृक्षवल्लींना सोयरी’ मानले.

उपक्रम :

अ. तुमच्या परिसरातील देशी व विदेशी फुलांची माहिती इंटरनेटच्या माध्यमांतून मिळवून ती तुमच्या कनिष्ठ महाविदयालयातील काचफलकात प्रदर्शित करा.

आ. हिवाळ्यातील उत्तररात्री किंवा अगदी पहाटेच्या वेळी तुमच्या परिसराचे निरीक्षण करा. पाहिलेल्या निसर्गसौंदर्याचे वर्णन शब्दबद्ध करा. ते वर्गात वाचून दाखवा.

तोंडी परीक्षा.

शिक्षकांनी वाचून दाखवलेला उतारा ऐका. सारांश लिहा.

Marathi Yuvakbharati 12th Digest Chapter 8 रेशीमबंध Additional Important Questions and Answers

कृती करा.

प्रश्न 1.


उत्तर :

प्रश्न 2.

उत्तर :

कारणे शोधा व लिहा.

प्रश्न 1.
उत्तररात्र हळुवारपणे अलगद पाऊल टाकते; कारण ………
उत्तर :
उत्तररात्र हळुवारपणे अलगद पाऊल टाकते; कारण तिच्या आगमनाची चाहूल कोणलाही लागू नये, अशी तिची इच्छा असते.

प्रश्न 2.
लेखक डायनिंग टेबलजवळची खिडकी हलकेच उघडतात कारण ……….
उत्तर :
लेखक डायनिंग टेबलजवळची खिडकी हलकेच उघडतात कारण रात्रीच्या नीरव शांततेचा भंग होऊ नये, असे लेखकांना वाटत असते.

प्रश्न 3.
पाखरांनी पंख फडफडल्याशिवाय पहाटदेखील आकाशात येत नाही; कारण ……….
उत्तर :
पाखरांनी पंख फडफडवल्याशिवाय पहाटदेखील आकाशात येत नाही; कारण पाखरांची झोपमोड होऊ नये, असे पहाटेला मनोमन वाटत असते.

प्रश्न 4.
निसर्ग आणि मानव यांना एकमेकांची ओढ लागलेली असते; कारण …………..
उत्तर :
निसर्ग आणि मानव यांना एकमेकांची ओढ लागलेली असते; कारण त्या दोघांमध्ये युगानुयुगे आदिम, ऋजू स्नेहबंध निर्माण झाले आहेत.

वर्णन करा :

प्रश्न 1.
1. पहाटेचे आगमन.
2. मोगऱ्यात घडणारे मानवी भावभावनांचे दर्शन.
3. उजाडत जाणाऱ्या पहाटेसोबत आल्हादाला येणारी भरती.
4. पहाट व पाखरे यांच्यातील नाते.
उत्तर :
1. पहाटेचे आगमन : उत्तररात्रीचा काळोख हळूहळू विरत जातो. हळुवारपणे उजाडू लागते. गुलमोहोरावरील घरट्यांतून, जैक्रांडाच्या फांदयांवरून पाखरे डोळे किलकिले करून पाहू लागतात. त्यांचा आपापसातला आवाज कुजबुजीसारखा भासू लागतो. हळूहळू एखादया उत्सवासारखा हा चिवचिवाट वाढत जातो. सायली, बोगनवेल, क्रोटन्स, गुलमोहोर, जॅक्रांडा हे सर्वच सळसळू लागतात. हलू डोलू लागतात. अशा प्रकारे पहाटेचे आगमन होते.

2. मोगऱ्यात घडणारे मानवी भावभावनांचे दर्शन : लेखकांचे निसर्गाशी असलेले अत्यंत हृदय असे नाते या उताऱ्यातून व्यक्त झाले आहे. भोवतालच्या वृक्षवेली, पाखरे, पहाट, सकाळ हे सर्व मानवेतर घटक माणसासारख्याच भावभावनांनिशी वावरू लागतात. म्हणूनच पाणी न मिळाल्यामुळे कोमेजलेला मोगरा लेखकांना रुसल्यासारखा भासतो आणि पाणी मिळाल्यावर कळया आल्या की खुदखुद हसल्यासारखा भासतो. झाडेवेली भावुक होतात. कधीतरी थोडीफार रुसलीफुगली तरी त्यांच्यावरून प्रेमाने हात फिरवला, त्यांना पाणी दिले की गोंडस फुले देऊन आपल्याला केवडातरी विरंगुळा, तजेला देतात. अशी मानवी भावनांची देवाणघेवाण लेखकांना जाणवत राहते.

3. उजाडत जाणाऱ्या पहाटेसोबत आल्हादाला येणारी भरती : उत्तररात्रीचा काळोख विरत जातो आणि पहाटेचा उजेड सुरू होतो. हा काळोख नेमका कोणत्या क्षणी संपतो आणि उजेड कोणत्या क्षणी सुरू होतो हे सांगणे केवळ अशक्य असते. इतका तो क्षण तरल असतो. त्या क्षणी लेखकांना अनोख्या, लोभसवाण्या नाजूक अनुभवाचा प्रत्यय येतो. ही आल्हाददायकता निसर्गाच्या सर्व घटकांमध्ये लेखकांना दिसते. झाडांच्या, वेलींच्या सळसळण्यात, हलण्याडोलण्यात दिसते. सुरुवातीला मंद मंद असलेला चिवचिवाट हर्षोल्हासाचे रूप धारण करतो. जणू आनंदाला, आल्हादाला आलेली भरतीच वाटते.

4. पहाट व पाखरे यांच्यातील नाते : लेखकांना भोवतालचा निसर्ग माणसासारखाच भावभावनांनी भरलेला भासतो. पहाटेची घटना तशी साधीशीच. पहाट होत आहे. पाखरांचा बारीक बारीक आवाज सुरू झाला आहे. त्यांच्या हालचालींना सुरुवात होत आहे. पहाट हळूहळू पुढे सरकत आहे. या प्रसंगात लेखकांना मानवी भावभावनांचे दर्शन घडते. पाखरांचा बारीक बारीक आवाज म्हणजे त्यांची कुजबुज होय. ती जणू एकमेकांना विचारताहेत, ” पहाट आली का?” पहाटेचे हळुवार येणे पाहन लेखकांना वाटते की, पाखरांना त्रास होऊ नये म्हणूनच जणू पहाट हळूच पाखरांना विचारते की, “मी येऊ का?” त्या दोघांमधले हळुवार कोमल नातेच लेखकांना या वाक्यातून व्यक्त करायचे आहे.

केव्हा ते लिहा :

प्रश्न 1.
1. पहाट आकाशात हलकेच पाऊल टाकते, जेव्हा
2. पाखरांचा चिवचिवाट वाढत जातो, जेव्हा
उत्तर :
1. पहाट आकाशात हलकेच पाऊल टाकते, जेव्हा पाखरांच्या पंखांची फडफड त्याला ऐकू येते.
2. पाखरांचा चिवचिवाट वाढत जातो, जेव्हा पहाट हळूहळू उजाडत जाते.

म्हणजे काय ते लिहा :

प्रश्न 1.

1. गुलमोहोरावरची पाखरे आपापसात कुजबुजू लागतात, म्हणजे जणू काही ………
2. मोगऱ्याला चुकून एखादया दिवशी पाणी घालायचे राहून गेले, तर तो कोमेजू लागतो, म्हणजे जणू काही ……
3. पाण्याचा शिडकाव झाला की दुसऱ्या दिवशी मोगऱ्याला कळ्या येतात, म्हणजे जणू काही …………

उत्तर :

1. गुलमोहोरावरची पाखरे आपापसात कुजबुजू लागतात, म्हणजे जणू काही तो त्यांच्या आल्हादाचा उत्सव असतो.
2. मोगऱ्याला चुकून एखादया दिवशी पाणी घालायचे राहून गेले, तर तो कोमेजू लागतो, म्हणजे जणू काही तो लेखकांवर रुसून बसतो.
3. पाण्याचा शिडकाव झाला की दुसऱ्या दिवशी मोगऱ्याला कळ्या येतात, म्हणजे जणू काही तो कळ्यांच्या रूपाने लेखकांकडे पाहून खुदखुद हसू लागतो.

रेशीमबंध Summary in Marathi

शब्दार्थ :

1. उत्तररात्र – मध्यरात्रीनंतरचा काळ.
2. विरणे – विरविरीत होणे, विरळ होणे.
3. असोशी – ओढ.
4. आदिमत्व – आदिम म्हणजे पहिला, मूळचा, आदिमत्व म्हणजे मूळची अवस्था.
5. ऋजू – सरळ, साधा, निर्मळ, पारदर्शी.
6. आस – इच्छा.
7. नितळ – गाळ, गढूळपणा नसलेले, स्वच्छ, शुद्ध.
8. लोभस – उत्कटपणे आवडणारा.
9. नजाकत – सुबकपणा.

Balbharti Maharashtra State Board 12th Commerce Book Keeping & Accountancy Solutions Chapter 10 Computer in Accounting Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Book Keeping & Accountancy Solutions Chapter 10 Computer in Accounting

1. Objective questions:

A. Select the most appropriate alternatives from those given below and rewrite the statements.

Question 1.
The primary document for recording all financial transactions in Tally is the ______________
(a) Journal
(b) Trial sheet
(c) Voucher
(d) File
Answer:
(c) Voucher

Question 2.
______________ displays the balance day-wise for a selected voucher type.
(a) Record book
(b) Ledger book
(c) Journal book
(d) Daybook
Answer:
(d) Daybook

Question 3.
Fixed Deposit A/c comes under ______________ group.
(a) Investments
(b) Current liability
(c) Bank A/c
(d) Current asset
Answer:
(a) Investments

B. Give the word term or phrase which can substitute each of the following statements:

Question 1.
The details of Bills receivable are maintained in this record.
Answer:
Sundry Debtors

Question 2.
Tally software is classified into this category.
Answer:
Mercantile

Question 3.
The short key is used to save or accept the information.
Answer:
Ctrl + A

Question 4.
It is a damaged software, cracked, nearly fully functional.
Answer:
Pirated Software

Question 5.
The process by which all the calculations are automatically done by the accounting software.
Answer:
Automation

C. State whether the following statements are true or false with reason:

Question 1.
Alt + D is the short key for delete voucher entries.
Answer:
This statement is True.
To delete voucher entries, people use the Alt + D key.

Question 2.
In Tally, the F6 Function key is for the payment vouchers.
Answer:
This statement is False.
In Tally, the F6 Function key is useful for receipt vouchers.

Question 3.
Legal software is fully functional software without any restriction.
Answer:
This statement is True.
The base of the legal software is fully functional, safe, and legal, so one can use this kind of software without any hesitation and restriction.

Question 4.
Salary Account comes under Indirect expenses.
Answer:
This statement is True.
When the expenses are made for the purchase of goods, and for the manufacturing process, they are known as a direct expense. Salary does not fall in that category and so it comes under the indirect expense category.

Question 5.
Accounting software may not be customized to meet the special requirement of the user.
Answer:
This statement is False.
Customized Accounting software is prepared to meet the special requirement of the user which is not readily available in the market.

D. Answer in One Sentences:

Question 1.
What is CAS?
Answer:
CAS means Computerized Accounting System which helps business firms to implement accounting processes and makes it user friendly with automation.

Question 2.
Write the steps to create a ledger account in tally.
Answer:
Steps to create a ledger account in the tally are as follows:

• From Gateway of Tally Screen, click on accounts info.
• Path gateway to Tally – Accounts Info – Ledgers – Single ledger – Choses create.

Question 3.
How to view reports in Tally?
Answer:
For viewing accounting reports in accounting software to click on the report option and select the Display option.

Question 4.
State the various types of vouchers.
Answer:
Following are the various voucher types:

1. F4 Contra voucher – For cash deposited in the bank and cash withdrawn from the bank, Transfer from one cash A/c to another Cash A/c and Bank to Bank transfer.
2. F5 Payment voucher – For all types of payments are recorded through this voucher type (Cash and Bank) Cash or Bank.
3. F6 Receipt voucher – For Cash and Bank receipts
4. F7 Journal voucher – For non-cash transactions
5. F8 Sales voucher – For cash as well as credit sales
6. F9 Purchase voucher – For cash as well as a credit purchase

Question 5.
Write the steps to create a company.
Answer:
Following are the steps to create a company:

1. After entering into Accounting software Tally, double click on the option, create a company, under company information. Then follow the navigation path.
   Gateway of Tally > Company Info > Create Company
2. Fill in the detailed information in the company creation form, displayed on the screen – Company creation window.

Balbharti Maharashtra State Board 12th Commerce Book Keeping & Accountancy Solutions Chapter 9 Analysis of Financial Statements Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Book Keeping & Accountancy Solutions Chapter 9 Analysis of Financial Statements

Objective Questions

A. Select the most appropriate alternative from those given below and rewrite the sentences:

Question 1.
Gross Profit Ratio indicates the relationship of gross profit to the ____________
(a) Net cash
(b) Net sales
(c) Net purchases
(d) Gross sales
Answer:
(b) Net sales

Question 2.
Current ratio = \(\frac{……………….}{Current liabilities}\)
(a) Quick assets
(b) Quick liabilities
(c) Current assets
(d) None of these
Answer:
(c) Current assets

Question 3.
Liquid assets = ____________
(a) Current assets + Stock
(b) Current assets – Stock
(c) Current assets – Stock + Prepaid Expenses
(d) None of these
Answer:
(b) Current assets – Stock

Question 4.
Cost of goods sold = ____________
(a) Sales – Gross profit
(b) Sales – Net profit
(c) Sales proceeds
(d) None of these
Answer:
(a) Sales – Gross profit

Question 5.
Net profit ratio is equal to ____________
(a) Operating ratio
(b) Operating net profit ratio
(c) Gross profit ratio
(d) Current ratio
Answer:
(a) Operating ratio

Question 6.
The common size statement requires ____________
(a) common base
(b) journal entries
(c) cashflow
(d) current ratio
Answer:
(a) common base

Question 7.
Bill payable is ____________
(a) long-term loan
(b) current liabilities
(c) liquid assets
(d) net loss
Answer:
(b) current liabilities

Question 8.
Generally current ratio should be ____________
(a) 2 : 1
(b) 1 : 1
(c) 1 : 2
(d) 3 : 1
Answer:
(a) 2 : 1

Question 9.
From financial statement analysis the creditors are specially interested to know ____________
(a) Liquidity
(b) Profits
(c) Sale
(d) Share capital
Answer:
(a) Liquidity

B. Give one word/term/phrase for each of the following statements.

Question 1.
The statement showing the profitability of two different periods.
Answer:
Comparative Income Statement

Question 2.
The ratio measures the relationship between gross profit and net sales.
Answer:
Gross Profit Ratio

Question 3.
Critical evaluation of financial statement to measure profitability.
Answer:
Analysis of Financial Statement

Question 4.
A particular mathematical number showing the relationship between two accounting figures.
Answer:
Ratio

Question 5.
An asset that can be converted into cash immediately.
Answer:
Liquid Asset

Question 6.
The ratio measuring the relationship between net profit and ownership capital employed.
Answer:
ROCE

Question 7.
The statement showing financial position for different periods of the previous year and the current year.
Answer:
Comparative Balance Sheet

Question 8.
Statement showing changes in cash and cash equivalent during a particular period.
Answer:
Cash Flow Statement

Question 9.
Activity related to the acquisition of long-term assets and investment.
Answer:
Financing Activities

Question 10.
The ratio that establishes a relationship between Quick Assets and Current Liabilities.
Answer:
Liquid Ratio

C. State true or false with reasons.

Question 1.
Financial statements include only the Balance Sheet.
Answer:
This statement is False.
Financial statements include Balance Sheet and Profit and Loss A/c. This is because financial statements are prepared by business organisations to find out the efficiency, solvency, profitability, growth, strength, and status of the business. For this, they need information from the balance sheet as well as from Profit and Loss A/c.

Question 2.
Analysis of financial statements is a tool but not a remedy.
Answer:
This statement is True.
Based on analysis of the financial statement one can get an idea of the financial strength and weakness of the business. However, based on this one cannot take decisions about the business on various issues. Hence analysis of financial statements is a tool but not a remedy.

Question 3.
Purchase of fixed assets is operating cash flow.
Answer:
This statement is False.
Purchase of fixed assets is cash flow from investing activities. It is not a day-to-day operations activity like office/selling/distribution finance expenses/activities.

Question 4.
Dividend paid is not a source of funds.
Answer:
This statement is True.
The dividend is always paid on shares issued by a company as an expense. Shares itself is a source of funds. In payment of dividends, cash goes out from the company. It is an outflow of cash and not a source of funds.

Question 5.
Gross profit depends upon net sales.
Answer:
This statement is True.
The gross profit ratio discloses the relation between gross profit and total net sales. The gross profit ratio is an income-based ratio, where gross profit is an income. There is a direct relation between net sales and gross profit. Higher the net sales higher the gross profit.

Question 6.
Payment of cash against the purchase of stock is the use of funds.
Answer:
This statement is True.
Cash payment for the purchase of stock is made from cash balance or/and from bank balance which is a part of the business fund. When stock or materials we purchase we use cash for payment.

Question 7.
Ratio Analysis is useful for inter-firm comparison.
Answer:
This statement is True.
The comparison of the operating performance of a business entity with the other business entities is known as an inter-firm comparison. This ratio analysis assists to know-how and to what extent a business entity is strong or weak as compared to other business entities.

Question 8.
The short-term deposits are considered as cash equivalent.
Answer:
This statement is True.
The short-term deposits are liquid assets. It means deposits are kept for some period (usually less than one year) and they are kept with an intention to get money quickly as and when required.
They are as good as cash and considered as cash equivalent.

Question 9.
Activity ratios and Turnover ratios are the same.
Answer:
This statement is True.
Turnover ratio is an efficiency ratio to check how efficiently a company is using different assets to extract earnings from them.
Activity ratios are financial analysis tools used to measure a business’s ability to convert its assets into cash. From both these definitions, we can say that Activity ratios and Turnover ratios are the same.

Question 10.
The current ratio measures the liquidity of the business.
Answer:
This statement is True.
The current ratio shows the relationship between current assets and current liabilities. If the proportion of current assets is higher than current liabilities, the liquidity position of the business entity is considered good. More liquidity means more short-term solvency. From the above, it is proved that the current ratio measures the liquidity of the business.

Question 11.
Ratio analysis measures profitability efficiency and financial soundness of the business.
Answer:
This statement is True.
With the help of profitability ratios (Gross profit, Net profit, and Operating profit) one can get the idea of profitability efficiency of the firm, and with the help of liquidity ratios (Current ratio and liquid ratio) one can get the idea of solvency or financial soundness of the business.

Question 12.
Usually, the current ratio should be 3 : 1.
Answer:
This statement is False.
Usually, the current ratio should be 2 : 1. It means current assets are double of current liabilities. It shows the short-term solvency of business enterprises.

D. Answer in one sentence only.

Question 1.
Mention two objectives of the comparative statement.
Answer:
Objectives of comparative statements are:

• Compare financial data at two points of time and
• Helps in deriving the meaning and conclusions regarding the changes in financial positions and operating results.

Question 2.
State three examples of cash inflows.
Answer:
Examples of cash inflows are:

• Interest received
• Dividend received
• Sale of asset/investment
• Rent received.

Question 3.
State three examples of cash-out flows.
Answer:
Examples of cash outflows are:

• Interest paid
• Loss on sale of an asset
• Dividend paid
• Repayment of short-term borrowings.

Question 4.
Give the formula of Gross Profit Ratio.
Answer:
Gross profit ratio = \(\frac{\text { Gross profit }}{\text { Net sales }} \times 100\)
Where Gross profit = Net sales – Cost of goods sold.
Cost of goods sold = Opening stock + Purchase – Purchase return + Direct expense – Closing stock
Net sales = Sales – Sales return.

Question 5.
Give the formula of Gross profit.
Answer:
Gross profit = Net sales – Cost of goods sold.
Cost of goods sold = Opening stock + Purchase – Purchase return + Direct expense – Closing stock
Net sales = Sales – Sales return.

Question 6.
Give any three examples of current assets.
Answer:
Cash or cash equivalent short-term lending and advances, expenses paid in advance, taxes paid in advance, etc. are examples of current assets.

Question 7.
Give the formula of the current ratio.
Answer:
Current ratio = \(\frac{\text { Current assets }}{\text { Current liabilities }}\)

Question 8.
Give the formula of quick assets.
Answer:
Quick assets = Current assets – (Stock + Prepaid expense)

Question 9.
State the formula of Cost of goods sold.
Answer:
Cost of goods sold = Opening stock + Purchase – Purchase return + Direct expense – Closing stock

Question 10.
State the formula of Average stock.
Answer:
Average stock = \(\frac{\text { Opentng stock of goods }+\text { Closing stock of goods }}{2}\)

Practical Problems

Question 1.
From the Balance Sheet of Amar Traders as of 31st March 2018 and 31st March 2019 prepare a Comparative Balance Sheet.

Solution:
Comparative Balance Sheet of Amar Traders as of 31st March 2018 and 31st March 2019


Percentage change = \(\frac{\text { Amount of absolute change }}{\text { Amount of previous year }} \times 100\)

Question 2.
From the Balance Sheet of Alpha Limited prepare a Comparative Balance Sheet as of 31st March 2018 and 31st March 2019:
Balance Sheet as of 31st March 2018 and 31st March 2019

Solution:
Comparative Balance Sheet of Alpha Limited as of 31st March 2018 and 31st March 2019

Question 3.
Prepare Comparative Balance Sheet for the year ended 31-3-18 and 31-3-19. Assets & Liabilities as follows:

Solution:
Comparative Balance Sheet as of 31st March 2018 and 31st March 2019

Question 4.
Prepare Comparative Balance Sheet for the year ended 31-3-17 and 31-3-18.

Solution:
Comparative Balance Sheet as of 31st March 2017 and 31st March 2018

Question 5.
Prepare Comparative Income statement of Noha Limited for the year ended 31-3-17 and 31-3-18.

Solution:
Comparative Income Statement of Noha Limited
For the year ended 31st March 2017 and 31st March 2018

Question 6.
Prepare Comparative Income Statement of Sourabh Limited for years ended 31-3-17 and 31-3-18.

Solution:
Comparative Income Statement of Sourabh Limited
For the year ended on 31st March 2017 and 31st March 2018

Question 7.
Following is the Balance Sheet of Sakshi Traders for the year ended 31-3-17 and 31-3-18

Prepare Common Size Balance Sheet for the years 31-03-17 and 31-03-18.
Solution:
Common Size Statement of Balance Sheet of Sakshi Traders as of 31st March 2017 and 31st March 2018

Note: Taking Total borrowed funds and Tota Funds applied as a base (100), Calculation is done.

Question 8.
Prepare Common Size Income Statement for the year ended 31-3-17 and 31-3-18.

Solution:
Common Size Statement for the year ended on 31st March 2017 and 31st March 2018

Note: Taking the amount of sales as base (100) other percentage figures are calculated.

Question 9.
Following is the Balance Sheet of Sakshi Limited. Prepare Cash Flow Statement:

Solution:
Cash Flow Statement
For the year ended 31st March 2017 and 31st March 2018

Question 10.
From the following Balance Sheet of Konal Traders prepare a Cash Flow Statement.

Solution:
Cash Flow Statement
For the year ended on 31st March 2017 and 31st March 2018

Question 11.
A company had following Current Assets and Current Liabilities:
Debtors = ₹ 1,20,000
Creditors = ₹ 60,000
Bills Payable = ₹ 40,000
Stock = ₹ 60,000
Loose Tools = ₹ 20,000
Bank Overdraft = ₹ 20,000
Calculate Current ratio.
Solution:
1. Current Assets = Debtors + Stock + Loose Tools
= 1,20,000 + 60,000 + 20,000
= ₹ 2,00,000

2. Current liabilities = Creditors + Bills payable + Bank overdraft
= 60,000 + 40,000 + 20,000
= ₹ 1,20,000

3. Current ratio = \(\frac{\text { Current assets }}{\text { Current liabilities }}\)
= \(\frac{2,00,000}{1,20,000}\)
= \(\frac{5}{3}\)
= 5 : 3

Question 12.
Current assets of company ₹ 6,00,000 and its Current ratio is 2 : 1. Find Current liabilities.
Solution:
Current ratio = \(\frac{\text { Current assets }}{\text { Current liabilities }}\)
\(\frac{2}{1}=\frac{6,00,000}{\text { Current liabilities }}\)
2 × Current liabilities = 6,00,000 × 1
Current liabilities = \(\frac{6,00,000}{2}\) = ₹ 3,00,000

Question 13.
Current liabilities = ₹ 3,00,000
Working capital = ₹ 8,00,000
Inventory = ₹ 2,00,000
Calculate Quick ratio.
Solution:
Current assets = Current liabilities + Working capital
= 3,00,000 + 8,00,000
= ₹ 11,00,000
Quick assets = Current assets – Inventory
= 11,00,000 – 2,00,000
= ₹ 9,00,000
Quick liability = Current liabilities – Bank O/D = ₹ 3,00,000
Quick ratio = \(\frac{\text { Quick assets }}{\text { Quick liabilities }}\)
= \(\frac{9,00,000}{3,00,000}\)
= \(\frac{3}{1}\)
= 3 : 1

Question 14.
Calculate Gross Profit ratio
Sales = ₹ 2,70,000
Net purchases = ₹ 1,50,000
Sales Ratio = ₹ 20,000
Closing Stock = ₹ 25,000
Operating Stock = ₹ 45,000
Solution:
Net sales = Sales – Sales return
= 2,70,000 – 20,000
= ₹ 2,50,000
Cost of goods sold = Opening stock + Net purchase – Closing stock
= 45,000 + 1,50,000 – 25,000
= ₹ 1,70,000
Gross profit = Net sales – Cost of goods sold
= 2,50,000 – 1,70,000
= ₹ 80,000
Gross Profit ratio = \(\frac{\text { Gross profit }}{\text { Net sales }} \times 100\)
= \(\frac{80,000}{2,50,000} \times 100\)
Gross profit ratio = 32%

Question 15.
Calculate Net Profit ratio from the following:
Sales = ₹ 3,80,000
Cost of goods sold = ₹ 2,60,000
Indirect expense = ₹ 60,000
Solution:
Sales = ₹ 3,80,000
Less: Cost of goods sold = ₹ 2,60,000
Gross profit = ₹ 1,20,000
Less: Indirect expense = ₹ 60,000
Net profit = ₹ 60,000
Net profit ratio = \(\frac{\text { Net profit }}{\text { Sales }} \times 100\)
= \(\frac{60,000}{3,80,000} \times 100\)
= 15.79%

Question 16.
Calculate Operating ratio:
Cost of goods sold = ₹ 3,50,000
Operating expense = ₹ 30,000
Sales = ₹ 5,00,000
Sales return = ₹ 30,000
Solution:
Net sales = Sales – Sales return
= 5,00,000 – 30,000
= ₹ 4,70,000
Operating ratio = \(\frac{\text { Cost of goods sold }+\text { Operating expense }}{\text { Net sales }} \times 100\)
= \(\frac{3,50,000+30,000}{4,70,000} \times 100\)
= \(\frac{3,80,000}{4,70,000} \times 100\)
= 80.85%

Question 17.
Calculate Current ratio.
1. Current assets = ₹ 3,00,000
2. Current liabilities = ₹ 1,00,000
Solution:
Current ratio = \(\frac{\text { Current assets }}{\text { Current liabilities }}\)
= \(\frac{3,00,000}{1,00,000}\)
= \(\frac{3}{1}\)
= 3 : 1

Balbharti Maharashtra State Board 12th Commerce Book Keeping & Accountancy Solutions Chapter 8 Company Accounts – Issue of Shares Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Book Keeping & Accountancy Solutions Chapter 8 Company Accounts – Issue of Shares

1. Objective Questions:

A. Select the appropriate answer from the alternative given below and rewrite the sentence.

Question 1.
The balance of Share Forfeiture A/c is transferred to ______________ Account after re-issue of these share.
(a) Reserve Capital
(b) Capital Reserve
(c) Profit & Loss
(d) Share Capital
Answer:
(b) Capital Reserve

Question 2.
Premium received on issue of shares is shown to ______________
(a) Liability side of Balance Sheet
(b) Asset side of Balance Sheet
(c) Profit & Loss A/c debit side
(d) Profit & Loss A/c credit side
Answer:
(a) Liability side of Balance Sheet

Question 3.
Shareholders get ______________ on shares.
(a) interest
(b) commission
(c) rent
(d) dividends
Answer:
(d) dividends

Question 4.
The document inviting to subscribe the shares of a company is ______________
(a) Prospectus
(b) Memorandum of Association
(c) Articles of Association
(d) Share certificate
Answer:
(a) Prospectus

Question 5.
As per SEBI guidelines, minimum amount payable on share application should be ______________ Nominal Value of shares.
(a) 10%
(b) 15%
(c) 2%
(d) 5%
Answer:
(d) 5%

Question 6.
When shares are forfeited the Share Capital Account is ______________
(a) credited
(b) debited
(c) neither debited nor credited
(d) None of the given
Answer:
(b) debited

Question 7.
The liability of shareholder in Joint Stock Company is ______________
(a) joint and several
(b) limited
(c) unlimited
(d) huge
Answer:
(b) limited

Question 8.
The Share Capital which a company is authorized to issue by its Memorandum of Association is ______________
(a) Nominal Capital/Authorised Capital
(b) Issued Capital
(c) Paid-up Capital
(d) Reserve Capital
Answer:
(a) Nominal Capital/Authorised Capital

Question 9.
The unpaid amount on allotment and calls may be transferred to ______________ Account.
(a) Calls-in-Advance
(b) Calls
(c) Calls-in-Arrears
(d) Allotment
Answer:
(c) Calls-in-Arrears

Question 10.
There must be provision in ______________ for forfeiture of shares.
(a) Articles of Association
(b) Memorandum of Association
(c) Prospectus
(d) Balance Sheet
Answer:
(a) Articles of Association

B. Give one word/term/phrase for each of the following statements.

Question 1.
Amount called up on shares by the company but not received.
Answer:
Calls-in-Arrears

Question 2.
Issue of share at its face value.
Answer:
Issue at par

Question 3.
The person who purchases the shares of a company.
Answer:
Shareholder

Question 4.
The form of business organisation where a huge amount of capital can be raised.
Answer:
Joint-stock company

Question 5.
The capital is subscribed by the public.
Answer:
Subscribed capital

Question 6.
The shares having preferential rights at the time of winding up of the company.
Answer:
Preference shares

Question 7.
The shares on which dividend is not fixed.
Answer:
Equity shares

Question 8.
The part of subscribed capital is not called up by the company.
Answer:
Uncalled capital

C. State true or false with reasons.

Question 1.
Directors can forfeit the shares for any reason.
Answer:
This statement is False.
After paying money on share application, When share applicant fails to pay the call money or premium on shares in spite of repeated reminders and warnings directors/company can forfeit the shares.

Question 2.
Once the application money is received, directors can immediately proceed with the allotment of shares.
Answer:
This statement is False.
Directors can proceed for allotment of shares only after receiving the minimum subscription amount of the issued amount by cheque or other instrument complying with all legal requirements.

Question 3.
Joint-stock company forms of business organisations came into existence after the industrial revolution.
Answer:
This statement is True.
As the volume and scale of trade and industry expanded, especially after the industrial revolution, a very large unit of the commercial organisation requiring large capital and greater managerial skill, called Joint-stock company came into existence.

Question 4.
Equity shareholders get a guaranteed rate of dividend every year.
Answer:
This statement is False.
One of the features of equity shares is the rate of dividend payable on equity shares keeps on changing from one year to another. So, there is no question of guaranteed dividend every year for equity shareholders.

Question 5.
The face value of shares and market value of shares is always the same.
Answer:
This statement is False.
Face value of shares means the issue price of shares while the market value of shares means the trading price of shares at the stock exchange. The face value of shares remains the same and fixed. However, market price changes as per the performance of the company. Hence face value and market value of shares is not the same.

Question 6.
Sweat shares are issued to the public.
Answer:
This statement is False.
Sweat shares are issued by a company to its directors or employees at a discount or for consideration other than cash. Sweat shares are not issued to the public.

D. State whether you agree or disagree with the following statements.

Question 1.
In the case of Pro-rata allotment the excess application money received must be refunded.
Answer:
Disagree

Question 2.
Calls-in-Advance account is shown on the asset side of the Balance Sheet.
Answer:
Disagree

Question 3.
The Authorised Capital is also known as Nominal Capital.
Answer:
Agree

Question 4.
Paid-up capital can be more than Called-up Capital.
Answer:
Disagree

Question 5.
The joint-stock company can raise a huge amount of capital.
Answer:
Agree

Question 6.
When shares are Forfeited Shares Capital Account is credited.
Answer:
Disagree

Question 7.
Directors can re-issue forfeited shares.
Answer:
Agree

Question 8.
When the issued price of a share is ₹ 12 and face value is ₹ 10, the share is said to be issued at a premium.
Answer:
Agree

Question 9.
A public limited company can issue its share without issuing its prospectus.
Answer:
Disagree

Question 10.
Shares can be issued for consideration other than cash.
Answer:
Agree

E. Answer in one sentence only.

Question 1.
What are Preference Shares?
Answer:
Preference Shares are a type of share which enjoys priority or preference over equity share for the repayment of dividends at a predetermined fixed rate and for the repayment of capital.

Question 2.
What is Registered Capital?
Answer:
Registered Capital or Authorised Capital means the maximum limit up to which a company is authorized to raise share capital.

Question 3.
What is Reserve Capital?
Answer:
Reserve Capital is that part of the subscribed capital which is reserved to be called up only at the time of winding up or liquidation of the company.

Question 4.
What is Over Subscription of Shares?
Answer:
When a company received more applications of shares than those actually offered or issued to the public, known as Over Subscription of Shares.

Question 5.
Which account is debited when share first call money is received?
Answer:
The bank account will be debited when share first call money is received.

Question 6.
When are shares allotted on a pro-rata basis?
Answer:
Shares are said to be allotted on a pro-rata basis when the applications are received for more shares than the number of shares issued and shares are allotted in the proportion to the number of shares applied for.

Question 7.
What is Forfeiture of Shares?
Answer:
When a shareholder fails to pay the call money or premium on the shares in spite of repeated reminders and warnings, the company forfeits the shares of such defaulters known as forfeiture of shares.

Question 8.
What is Calls-in-Arrears?
Answer:
Non-payment of allotment or call money by the applicants in spite of repeated reminders are called Calls-in-Arrears.

Question 9.
What do you mean by Shares Issued at Premium?
Answer:
When shareholders are supposed to pay a price higher than the face value of the shares, their shares are said to be issued at a premium.

Question 10.
What is Paid-up Capital?
Answer:
Part of the called-up capital which is actually paid by the shareholders is called Paid-up Share Capital.

F. Complete the following sentences.

Question 1.
When the face value of the share is ₹ 100 and the issued price is ₹ 120, then it is said that the shares are issued at ______________
Answer:
premium

Question 2.
______________ Capital is the capital which a company is authorized to issue by its Memorandum of Association.
Answer:
Authorized

Question 3.
The difference between Called-up Capital and Paid-up Capital is known as ______________
Answer:
Calls-in-Arrears

Question 4.
______________ shareholders get fixed rate of dividend.
Answer:
Preference

Question 5.
______________ shareholders are the real owners of the company.
Answer:
Equity

Question 6.
______________ form of business organisation in which capital is raised through the issue of shares.
Answer:
Joint-stock company

Question 7.
______________ Capital is the part of Issued capital which is subscribed by the public.
Answer:
Subscribed

Question 8.
The part of Authorised Capital which is not issued to the public is known as ______________ Capital.
Answer:
Unissued

G. Calculate the following.

Question 1.
One shareholder holding 500 equity shares paid share application money @ ₹ 3, Allotment money @ ₹ 4 per share and failed to pay a final call of ₹ 3 per share his share was forfeited calculate the amount of forfeiture.
Solution:
Amount of forfeiture = Amount received by the company (In case of non-payment of ‘calls’)
Here, shareholders paid ₹ 3 per share on application and ₹ 4 per share on the allotment on 500 shares.
So, total amount received by company = 500 × ₹ 3 + 500 × ₹ 4
= 1,500 + 2,000
= ₹ 3,500
∴ Amount of share forfeiture = ₹ 3,500.

Question 2.
10,000 equity shares of ₹ 10 each issued at a 10% premium. Calculate the total amount of share premium.
Solution:
Equity shares = 10,000
Face value = ₹ 10 per share
Premium @ 10% = 10,000 × 10 × \(\frac{10}{100}\) = ₹ 10,000
So, premium 10,000 shares of ₹ 10 each at 10% = ₹ 10,000

Question 3.
The company received excess applications for 5000 shares @ ₹ 4 per share. The application of 1000 shares was rejected and a pro-rata allotment was made. Calculate the amount of application money adjusted with allotment.
Solution:
Excess application money received for 5000 shares @ ₹ 4 per share = ₹ 20,000
Less: Application of 1000 shares rejected and money refunded = ₹ 4,000
Excess money received to be adjusted with allotment = ₹ 16,000

Question 4.
80,000 equity shares of ₹ 10 each issued and fully subscribed and called up at 20% premium. Calculate the amount of Equity Share capital.
Answer:
Equity Share capital = No. of equity shares × face value of each share
= 80,000 × ₹ 10
= ₹ 8,00,000
Note: Equity Share capital has no concern with premium or discount amount.

Question 5.
Directors issued 20,000 equity shares of ₹ 100 each at par. These were fully subscribed and called up. All money was received except one shareholder holding 100 equity shares failed to pay a final call of ₹ 20 per share. Calculate the amount of Paid-up capital of the company.
Solution:
Fully subscribed and called-up amount = 20,000 equity shares × ₹ 100 each share
= ₹ 20,00,000
But one share holder failed to pay final call of ₹ 20 per share of 100 equity shares means
Non-payment of shares = 100 equity shares × ₹ 20 per share = ₹ 2,000
∴ Total Paid-up capital amount = ₹ 20,00,000 – ₹ 2,000 = ₹ 19,98,000

Question 6.
The company sends a regret letter for 100 shares and an Allotment letter to 25,000 shareholders. Application money per ₹ 20 per share. Calculate the amount of application money that the company is refunding.
Solution:
The company sends a Regret letter for 100 shares for ₹ 20 per share application money received i.e. only that much amount the company will refund.
Amount of refund = No. of shares × Value of per share
= 100 × ₹ 20
= ₹ 2,000

Practical Problems

Question 1.
Vijay Ltd. was registered with an authorized capital of ₹ 15,00,000 divided into 1,50,000 equity shares of ₹ 10 each.
The company issued 1,00,000 equity shares of ₹ 10 each at a premium of ₹ 2 per share. The company received applications for 80,000 equity shares and was allotted the shares.
The company received application money ₹ 3 per share, allotment money ₹ 4 per share
(Including premium) and first, call money ₹ 3 per share.
The Directors have not made the final call of ₹ 2 per share. All money was received except one shareholder holding 500 shares did not pay the first call.
Show Authorised Capital, Issued Capital, Subscribed Capital, Called-up Capital,
Paid-up Capital, Calls in Arrears, and Share Premium amount in the company balance sheet.
Solution:
In the books of Vijay Ltd.
Balance Sheet as on ______________

Working Notes:
1. Bank balance at the end = Amount received on application + Amount received on allotment + Amount received on 1st call + Premium amount received
= 80,000 × 3 + 80,000 × 2 × 79,500 × 3 + 80,000 × 2
= 2,40,000 + 1,60,000 + 2,38,500 + 1,60,000
= ₹ 7,98,500

2. Directors have not made the final call of ₹ 2 per share means total called-up amount = ₹ 10 – ₹ 2 = ₹ 8

3. Calls-in-Arrears on 500 shares at ₹ 3 = ₹ 1,500 of the first call

4. Share premium on 80,000 shares @ ₹ 2 received at allotment stage i.e. share premium amount = 80,000 x ₹ 2 = ₹ 1,60,000

Question 2.
Anand Company Limited issued 1,00,000 preference shares of ₹ 10 each payable as-
On Application ₹ 4
On Allotment ₹ 3
On First call ₹ 2
On Second & Final call?
The company received applications for all these shares and received all money.
Pass Journal Entries in the books of Anand Company Ltd.
Solution:
Journal Entries in the books of Anand Company Ltd.

Question 3.
Rohini Company Limited issued 25,000 equity shares of ₹ 100 each payable as follows:
On Application ₹ 20
On Allotment ₹ 30
On First call ₹ 20
On the Second & Final call ₹ 30
The application was received for 22,000 equity shares and allotment of shares was made to them. All money was received by the company.
Pass Journal Entries in the books of Rohini Co. Ltd.
Solution:
Journal Entries in the books of Rohini Company Limited

Question 4.
Deepak Manufacturing Co. Ltd. issued a prospectus inviting applications for 1,00,000 equity shares of ₹ 10 each payable as follows :
₹ 2 on Application
₹ 4 on Allotment
₹ 2 on the First call
₹ 2 on Final call
The application was received for 1,20,000 equity shares. The Directors decided to reject excess applications and refunded application money on that. The company received all money.
Pass Journal Entries in the books of a company.
Solution:
Journal Entries in the books of Deepak Manufacturing Co.Ltd

Question 5.
Sucheta Company Limited issued ₹ 20,00,000 new capital divided into ₹ 100 equity shares at a premium of ₹ 20 per share payable as ₹ 10 on Application, ₹ 40 on Allotment and ₹ 10 premium ₹ 50 on Final call and ₹ 10 premium.
The issue was oversubscribed to the extent of 26,000 equity shares. The applicants on 2,000 shares were sent a letter of regret and their application money was refunded.
The remaining applicants were allotted shares on a Pro-rata basis. All the money due on Allotment and Final call was only received.
Make necessary Journal Entries in the books of Sucheta Company Ltd.
Answer:
Solution:
Journal Entries in the books of Sucheta Company Limited

Working Note:
Calculation of Application money transferred to Share Allotment:
Application money received (26,000 × 10) = 2,60,000
Less: Application money refunded (2,000 × 10) = 20,000
Less: Application money transferred to Share Capital: (20,000 × 10) = 2,00,000
Excess money received on application transferred to Share Allotment = 40,000
Bifurcation of calls amount:

Question 6.
Suhas Limited issued 10,000 equity shares of ₹ 10 each at a premium of ₹ 2 per share payable ₹ 3 on application, ₹ 5 (including premium) on the allotment, and the balance in two calls of an equal amount. Applications were received for 11,000 equity shares and pro-rata allotment was made for all the applicants. The excess application money was adjusted towards allotment.
Mrs. Shobha who was allowed 200 equity shares failed to pay F/F/C and her shares were forfeited after the final call.
Show Journal Entries in the books of Suhas Ltd. and also show its presence in Balance Sheet.
Solution:
Journal Entries in the books of Suhas Limited

Balance Sheet of Suhas Limited

Working Notes:
1. Excess amount received at the time of application ₹ 3,000 adjusted at allotment stage, so allotment amount received in the bank is ₹ 47,000.

2. Amount called-up per share: ₹ 3 on application, ₹ 5 (including premium) on allotment i.e. ₹ 2 premium + ₹ 3 capital and balance amount ₹ 4 in two calls of the equal amount i.e. ₹ 2 on the first call and ₹ 2 on final call.

3. Mrs. Shobha was not able to pay F/F/C i.e. first and final call means 200 × ₹ 2 first call money = ₹ 400 and 200 × ₹ 2 final call money = ₹ 400.
Mrs. Shobha paid ₹ 6 per share towards capital which the company received and the company has the right to forfeit only paid amount means the company forfeited ₹ 1,200 of Mrs. Shobha.

Question 7.
Subhash Company Limited issues 2000 Equity shares of ₹ 100 each payable as ₹ 30 on application, ₹ 30 on the allotment, ₹ 40 on first and final call.
All the shares were subscribed and duly allotted. The company made all the calls. All cash was duly received except the first and final call on 100 equity shares. These shares were forfeited by the company and were re-issued as fully paid for ₹ 75 per share.
Show the Journal Entries in the books of Subhash Company Ltd.
Solution:
Journal Entries in the books of Subhash Company Limited

Working Notes:
1. Amount forfeited by the company on 100 shares forfeited = 100 × (30 + 30)
= 100 × 60
= ₹ 6,000

2. Calls-in-Arrears = 100 × 40 = ₹ 4,000.

3. Amount received on re-issue of 100 forfeited shares = 100 × 75 = ₹ 7,500.
Balance of ₹ 2,500 (i.e. loss 25 × 100) is transferred to Share Forfeiture A/c.

4. Amount transfer from Share Forfeiture A/c to Capital Reserve is ascertained by preparing Share Forfeiture A/c.

Question 8.
Pass Journal Entries for the forfeiture and re-issue of shares in the following cases:
(A) Asha Ltd. forfeited 100 equity shares of ₹ 20 each fully called-up for non-payment of the first call of ₹ 3 per share and final call of ₹ 5 per share. 80 shares of these were re-issued at ₹ 15 per share as fully paid.
(B) Bhakti Ltd. forfeited 100 equity shares of ₹ 10 each, ₹ 6 called-upon which the shareholder paid application and allotment of ₹ 5 per share. Of these 80 shares were re-issued as fully paid-up for ₹ 16 per share.
(C) Konark Ltd. forfeited 50 shares of ₹ 10 each, ₹ 8 called-up. The shareholder failed to pay the first call of ₹ 3 per share. Later on, 30 shares of these were re-issued at ₹ 7 per share.
Solution:
Journal Entries [For Asha Ltd.]

Working Notes for A:
1. Out of 100 forfeited shares, 80 shares were re-issued accordingly Equity Share Capital A/c is debited and credited.
2. To find the proportionate amount for Forfeiture A/c:
For 100 shares-share forfeiture amount = ₹ 1,200
∴ 80 shares – share forfeiture amount = ₹ 960
Now, out of this ₹ 960 we used ₹ 400 from Share Forfeiture A/c at the time of re-issue of shares.
So, balance of Share Forfeiture A/c = ₹ 960 – ₹ 400 = ₹ 560

Journal Entries [For Bhakti Ltd.]

Working Notes for B:
1. Out of 100 forfeited shares, 80 shares were re-issued accordingly Equity Share Capital A/c is debited for ₹ 600 and credited for ₹ 480.

2. The proportionate amount debited to Forfeiture A/c:
For 100 shares-share forfeiture amount debited = ₹ 500 1 Qn
∴ 80 shares – share forfeiture amount = ₹ \(\frac{80}{100} \times \frac{500}{1}\) = ₹ 400
Now, shares were re-issued at ₹ 6 per share which is a called-up amount.
∴ The proportionate amount for Forfeiture A/c ₹ 400 will be transferred to Capital Reserve A/c.

Journal Entries (For Konark Ltd.)

Working Note for C:
The proportionate amount debited to Forfeiture A/c:
For 50 shares – share forfeiture amount debited is ₹ 250
∴ 30 shares-share forfeiture amount = ₹ \(\frac{30}{50} \times 250\) = ₹ 150
Out of this ₹ 30 used for re-issue of forfeited shares.
∴ Balance of Share Forfeiture A/c = ₹ 150 – ₹ 30 = ₹ 120.