NCERT Solutions for Class 12 Chemistry

NCERT Solutions for Class 12 Chemistry

Class 12 holds much importance in every student’s life, after which they get to enter the real world in their chosen career. What you study in class 11 and class 12 forms the base of your further studies in the course of your choice. Hence, having a strong grip on all the subjects becomes extremely important. If you’re planning to pursue engineering or enter the field of science with any course, Chemistry is a critical subject. 

While several useful resources are available online that can help you  study effectively, nothing can match the value of NCERT Solutions for any student. At Extramarks, you can get easy access to study materials for any class, from NCERT solutions class 1 to NCERT class 12. Moreover, it also makes sense to refer to NCERT solutions class 11 as well, as you start preparing for class 12. Available in a simple downloadable format, these solutions will not only help you study better, but smarter!

Once you are thorough with the syllabus, you can access the NCERT solutions for Chemistry class 12 to revise and strengthen your understanding of the various concepts. Moreover, the easy-to-read language will allow you to understand the concepts clearly and help build your confidence. The solutions are based on the latest CBSE syllabus, so you can start your preparations without any worries!

Download Chapter-Wise NCERT Solutions for Class 12 Chemistry

The NCERT textbook for CBSE Class 12 Chemistry is divided into two parts with a total of 16 chapters. These chapters cover concepts of organic, inorganic, and physical Chemistry. Every chapter helps you gain in-depth knowledge of Chemistry basics, chemical compounds, biomolecules, etc., and their application in daily life. In addition, it also includes several topics that test your practical and numerical skills.

In our NCERT solutions for class 12 Chemistry, you can find systematic answers to every question from the NCERT textbook which will even help you with your last-minute revisions. Moreover, our solutions can also come in handy while solving any numerical questions, thus offering you maximum convenience. 

You can easily access the chapter-wise NCERT solutions for class 12 Chemistry by clicking the links given below. 

NCERT Solutions Class 12 Chemistry

For students in Class 12, Chemistry can be a tricky subject to study. Yet it is a critical subject as it consists of various theoretical, practical, and numerical concepts relating to chemical reactions and other problems. Many students find this subject difficult, yet it can be one of the top-scoring subjects if you approach it smartly.

Considering the vast syllabus of Chemistry, it is natural to be overwhelmed under examination pressure. However, to ease your worries, Extramarks brings you NCERT solutions for Chemistry class 12, designed by experts to support your exam preparations. With the right study resources at your disposal, you can ace the class 12 examinations with high scores. 

To begin with, you must go through the various concepts on a regular basis to refresh your memory. Secondly, you can access the numerical questions mentioned in the NCERT solutions for class 12 Chemistry solution and note down the important formulas and understand the steps involved in solving a particular question. The NCERT solutions for class 12 Chemistry solutions consist of a detailed explanation of each question. This will help you revise better and accurately recall the solutions while attempting complex questions in the examination. 

NCERT Solutions Class 12 Chemistry Chapter Details and Exercises

NCERT solutions for class 12 Chemistry can help build the conceptual foundation you need to ace the examinations. If you want to achieve higher scores, then the chapter-wise solutions and NCERT exercises will be of great assistance to you. It covers all the important topics, illustrations, formulas, and NCERT exercises for each chapter to help with your exam preparations. 

You can download the chapter-wise solutions for all the NCERT exercises by clicking on the links given.

Chapter 1: The Solid State – Term I

This chapter deals with the solid state of matter along with the characteristics of the crystal lattice, imperfections in solids and unit cells, and the classification of solids. You will come across various theoretical questions relating to the classification of solids into various categories – metallic, covalent, ionic, and amorphous solids. In addition, the numerical questions deal with the atomic mass of silver as well as the efficiency of packaging in a face-centred cubic metal crystal, simple cubic, and more. 

<Link for exercises 1.1-1.26>

Chapter 2: Solutions – Term I

The second chapter of the NCERT textbook consists of 12 questions dealing with the formation and properties of solutions.  Here, you will learn about types of solutions, the concentration of solutions, solubility of gases and solids in a liquid, the vapour pressure of liquid solutions, mole fraction, and more. You will also have numerical questions relating to the evaluation of the mass of Benzene, the molarity of solutions, and carbon tetrachloride. This chapter is not only critical from the exam point of view but is also an important part of the upcoming competitive exams. 

<Link for exercises 2.1-2.41>

Chapter 3: ElectroChemistry – Term II

Electrochemistry is the branch of Chemistry that deals with the relationship between chemical and electrical energy generated in a redox reaction and its conversion. The solutions NCERT Chemistry class 12 consists of 15 questions that evaluates the conceptual understanding of the students. It includes topics like Molar conductivity, Kohlrausch’s law, and their numerical questions. Moreover, there are 18 additional exercises where you have to deal with numerical and theoretical problems.

<Link for exercises 3.1-3.18>

Chapter 4: Chemical Kinetics – Term II

Chemical Kinetics provides an understanding of the rate of chemical reactions, the collision theory of chemical reaction, dependence on the rate of reaction, and the Arrhenius equation. Typically, there are three types of chemical reactions – instantaneous reactions, slow reactions, and moderately slow reactions. With the NCERT solution for class 12 Chemistry, you can get enough clarity on these concepts and practice them properly to score higher in the exams.

<Link for exercises 4.1-4.30>

Chapter 5: Surface Chemistry – Term II

The fifth chapter of the NCERT Chemistry textbook, Surface Chemistry deals with several features like absorption, catalysis, and colloids which include gel and emulsion. After completing the chapter, you will learn about absorption and its classification, Tyndall effects, reversible solutions, interfacial phenomenon and its significance, factors controlling absorption, and more. 

<Link for exercises 5.1-5.27>

Chapter 6: General Principles and Processes of Isolation of Elements

In this chapter, you will be able to learn about the process of extracting and the occurrence of certain metals, along with some basics of the field. This chapter covers a wide range of topics, from metallurgy, ores, causes of reaction, refining crude metals, extracting copper from copper pyrites, and more. Solving the various exercises in this chapter along with the previous years’ question papers will help boost your preparation for the class 12 examinations.

<Link for exercises 6.1-6.27>

Chapter 7: The p-Block Elements – Term I

This chapter covers the history of p-block elements which dates back to the 19th century. Along with this, it talks about the Group 13, 14, 15, 16, 17, and 18 elements, which are the p-block elements. Here, you will learn about the various properties of these elements and answer theoretical questions related to them. To get a clear idea of these complex concepts, you can refer to the NCERT solutions for class 12 Chemistry available at Extramarks. The detailed explanation will help you score better marks in final exams.

<Link for exercises 7.1-7.40>

Chapter 8: The d & f-Block Elements – Term II

With a better grip on the concepts of p-block elements, you can also easily study this chapter. The d-block or transition elements are the ones that lie between the s and p-block elements. The f-block elements are the inner transition series. In this chapter, you will be introduced to  various concepts such as the variation in ionic and atomic size of transition metals, their physical properties, general properties of transition elements, magnetic properties, and more. Here, you will come across various theoretical questions based on these topics.

<Link for exercises 8.1-8.38>

Chapter 9: Coordination Compounds – Term II

Coordination compounds, a part of modern inorganic chemistry,can be a difficult subject for many students. This is because it is an important chapter for both the Class 12 exams and competitive exams, so you must put more focus while studying for this chapter. Here, you will learn various concepts of Werner’s Theory of Coordination Compounds, nomenclature, stability, bonding, importance, and application of coordination compounds. 

<Link for exercises 9.1-9.32>

Chapter 10: Haloalkanes and Haloarenes – Term I

It is one of the most important chapters of class 12 Chemistry. Here you will learn about the physical and chemical properties of haloalkanes along with their preparation. The halogen derivatives of the hydrocarbons are called haloalkanes, while the aromatic compounds where halogens are directly attached to the carbon atom are the Haloarenes. Some of the important topics covered in this chapter include the nomenclature of Haloalkanes and Haloarenes, nature of the C – X bond, physical and chemical properties, and substitution reactions. 

To form a better understanding of these complex topics, you can refer to the NCERT solutions for class 2 Chemistry solutions where you can get a detailed explanation of all the solutions created by experts in a simple language.

<Link for exercise 10.1-10.22>

Chapter 11: Alcohols, Phenols, and Ethers – Term I

The questions in this chapter cover a wide range of topics, including reactions, properties, nomenclature, hydroboration oxidation, IUPAC names of 12 different compounds, etc. You will study the reactions involved in the process of making alcohols from phenol, alcohols, and ethers. This chapter primarily contains numerous theoretical questions, so it is important to polish your concepts with regular revisions and previous years’ question papers.

<Link for exercises 11.1-11.33>

Chapter 12: Aldehydes, Ketones, and Carboxylic Acids – Term II

From the exam point of view, this chapter is of the utmost importance. The twelfth chapter includes several in-text questions, exercises, as well as two questionnaires. Additionally, it also has more weightage in the term II exams. In this chapter, you will come across questions relating to topics including, chemical tests to differentiate terms of a compound, physical and chemical properties of Aldehydes, Ketones, and Carboxylic Acid, structural formula, uses, names of four prospective condensations, etc. You can download the NCERT solutions for class 12 Chemistry for this chapter and make revision notes to add another layer to your preparation.

<Link for exercises 12.1-12.20>

Chapter 13: Amines – Term II

In chapter 13, you will study the structure, nomenclature, preparation, physical & chemical properties, and uses of amines. Amines are the derivatives of ammonia which are obtained by the replacement of hydrogen. To offer you greater conceptual knowledge of amines, several examples relating to the determination of basicity of amines, reactions, and synthesis of amines are explained in this chapter. Answering the back exercises will boost your confidence in appearing for the exam and you can answer any type of question asked.

<Link for exercises 13.1-13.14>

Chapter 14: Biomolecules – Term I

The organic compounds that are present in different cells of living organisms are called biomolecules. The interaction of biomolecules makes up the molecular logic of life processes. The Biomolecules chapter covers various topics like the hydrolysis product of sucrose and lactose, monosaccharides, differentiating starch from cellulose and secondary proteins, amino acids, peptide bonds, etc. As per the CBSE guidelines, this chapter covers the structure and function of biomolecules.

<Link for exercises 14.1-14.25>

Chapter 15: Polymers

When studying for chapter 15 of the Chemistry NCERT textbook, you will learn concepts like polymer, monomer, and polymerisation, along with their classification, which is based on their source and structure. Additionally, you will study the classification of polymers based on their mode of polymerisation. You can easily score higher marks in the exam if you gain enough clarity on the various topics in this chapter. 

<Link for exercise 15.1-15.20>

Chapter 16: Chemistry in Everyday Life

With a little practice and revision, you can easily gain full marks in this chapter. This chapter is entirely theoretical, where you learn about the various chemicals and their applications in real life. Some of the topics covered in this chapter are drugs and their classifications, the therapeutic action of different classes of drugs, drug-target interaction, chemicals in food and cleansing agents, pesticides, and insecticides. 

<Link for exercises 16.1-16.27>

Overall, Chemistry can be a difficult subject to study, as it requires a lot of practice of various chemical formulas and numericals. However, with the right guidance and study material, you can score really well. All you need to do is download the NCERT solutions for Chemistry class 12 and start preparing!

Merits of Extramarks NCERT Solutions for Class 12 Chemistry

  • With Extramarks, you can get the NCERT solutions for Chemistry class 12, which you can easily access anytime and anywhere
  • The solutions are created by experienced teachers who are masters of their subject matter. This ensures that you learn only from the best
  • Every chapter and topic is explained in a comprehensive manner using practical examples for your ease of understanding
  • Most of the syllabus for any engineering competitive exam such as JEE Main or JEE Advanced is nearly similar to the class 11 and class 12 syllabus. Thus, having a strong grip on the various subjects and topics will also help you crack these competitive exams with ease
  • The NCERT solutions will also prove to be useful in assisting your classroom education
  • The language is simple and easy to read which helps in a quick understanding of the subject

CBSE Marking Scheme 2022-23

Because class 12 is such an important part of academics, the CBSE has divided the entire syllabus into two terms of equal marks to make it easier for students to learn and understand the various concepts in 2022, owing to the coronavirus pandemic, CBSE has also reduced the syllabus by 30% for the academic year 2022-23.

Of a total of 16 chapters in the Chemistry textbook, 6 chapters are part of term 1. These are- Solutions, p-block elements, the solid state, Haloalkanes and Haloarenes, Biomolecules, and Alcohols, Phenols, and Ethers. The marking scheme for the theory exam is as follows-

Unit No. Name of Unit  Marks
1 Solid State 10
2 Solutions
3 p-Block Elements 10
4 Haloalkanes and Haloarenes
5 Alcohols, Phenols, and Ether 15
6 Biomolecules
Total 35

In Chemistry term 2 of class 12 examinations, CBSE prescribes 7 chapters out of a total of 16 chapters. These include Chemical Kinetics, Coordination Compound, d & f-Block Elements, Surface Chemistry, ElectroChemistry, Amines, Aldehydes, Ketones, and Carboxylic Acids. You can refer to the marking scheme for the theory exam as given below-

Unit No. Name of Unit Marks
1 ElectroChemistry
2 Chemical Kinetics 13
3 Surface Chemistry
4 d and f-Block Elements 9
5 Coordination Compounds
6 Aldehydes, Ketones, and Carboxylic Acids 13
7 Amines
Total 35

As NCERT covers about 50% of the questions asked in the board exam for Class 12, it is highly recommended to finish the book thoroughly, with extra focus on the exercises given at the end of each chapter. Moreover, the difficult topics are also explained with the help of relevant diagrams, images, graphs, and illustrations. Therefore, understanding the fundamentals through the NCERT textbook will only give you better results as you appear for your examination. 

Think NCERT Solutions, Think Extramarks

The NCERT solutions provided by Extramarks are undoubtedly the best study resources to assist your class 12 board exam preparations. All the solutions are accurately designed and edited, keeping in mind the latest CBSE syllabus for Class 12. Once you are done with the Class 12 syllabus, you can download the NCERT solutions for class 12 Chemistry and start revising the various important concepts to score well in the final exams. Some key features of the NCERT solutions offered by Extramarks are-

  • Best reference material

More often than not, students find it difficult to concentrate in  class and retain each and every topic discussed. This can only affect their studies if it is not supported with additional study material. The NCERT solutions for class 12 Chemistry from Extramarks will act as reference material for your classroom education and NCERT textbook questions. It will help strengthen the different concepts discussed in the class and help you answer them easily. These solutions will also give you a general idea of how to approach the complex questions in the exam.

  • Based on the CBSE syllabus

At Extramarks, we make sure that your exam preparations only get better and stronger with our help. So we design our solutions based on the latest CBSE syllabus and marking scheme to ensure that none of the important topics or questions are left behind. The main purpose of the class 12 Chemistry solutions is to help you self-analyse your preparedness so that you can improve your weak areas. Moreover, all the questions, answers, and examples you will find are based on the CBSE curriculum and have a high probability of being asked in the exam.

  • Created by subject matter experts

Want to ensure excellent performance and higher scores in my class 12 final exams. If Chemistry is the subject you struggle with, Extramarks’ easy-to-download solutions NCERT Chemistry class 12 is drafted by top-notch faculty members who are experts in their subject matter. They draft the solutions, keeping in mind the convenience of the students, to help them score well. The simple language and use of practical examples to explain complex topics help you grasp and retain the concepts well enough. Moreover, all the questions, whether numerical, MCQs, or theoretical,  are 100% accurate. So you can rest assured of the viability of these solutions and start preparing now!

  • Fit for revision

Reading the entire NCERT book more than once or twice can be a difficult task, especially when you want to revise. However, with the student-friendly study material from Extramarks, you can understand the difficult topics, clear your doubts, and even make a quick last-minute revision because it will be easy to retain these solutions. Moreover, questions are explained using everyday, real-life examples, which will make it easier for you to apply and grasp various concepts. And not only that, since many engineering competitive exams share the syllabus for classes 11 and 12, these solutions will act as revision notes while you prepare for exams like JEE Main and JEE Advanced in the future.

Benefits of Referring to Chemistry Class 12 NCERT Solutions 

  • By using the NCERT solutions for class 12, you will gain more confidence in attempting the questions as they are based on the CBSE marking scheme and prescribed weightage
  • You can easily assess your key weak areas and work harder with regular revision and consistent practice to improve and perform well in the exam. It will strengthen your efforts and allow you to achieve higher scores
  • Chemistry is a difficult but rewarding subject, the NCERT solutions will point you in the right direction in terms of how to approach a specific question or which formulas to use in various chemical reactions and numerical questions.This will ensure that you retain the answers easily
  • Along with this study material, you must also practise as many previous years’ question papers as possible as it will give you an idea of the types of questions asked, the exam pattern, the marking scheme, important topics to focus on, and more

The NCERT solutions class 12 will help you gain in-depth knowledge of the several models of Chemistry and their applications in daily life. Not only will this help you understand the subject better, it will also allow you to ace the exam!

Q.1 Determine the amount of CaCl2 (i = 2.47) dissolved in 2.5 litre of water such that its osmotic pressure is 0.75 atm at 27°C.

Ans.

Osmotic pressure,

π=iCRT=i(n/V)RTOr,n=π×Vi×R×T=0.75atm×2.5L2.47×0.0821LatmK1mol1×300K

= 0.0308 mol

Molar mass of CaCl2= 40 + (2 x 35.5) = 111 g/mol

∴ Amount of CaCl2 dissolved = 0.0308 x 111= 3.42 g.

Q.2 Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K2SO4 in 2 litre of water at 25° C, assuming that it is completely dissociated.

Ans.

Given,

w = 25 mg = 0.025 g

V = 2 L; T = 25°C = (25 + 273) K = 298 K

R = 0.0821 L atm K–1mol–1

M = Molar mass of K2SO4 = (2 × 39) + (1 × 32) + (4 × 16) = 174 g/mol

When K2SO4 is dissolved in water,

K2SO42K++SO42Total number of ions produced = 3  i = 3π=i(w/M)(1/V)RT=3×0.025×1×0.0821×298174×2=5.27×103atm

Q.3 The air is a mixture of a number of gases. The major components are oxygen and nitrogen with approximate proportion of 20% is to 79% by volume at 298 K. The water is in equilibrium with air at a pressure of 10 atm. At 298 K if the Henry’s law constants for oxygen and nitrogen are 3.30 × 107 mm and 6.51 × 107 mm respectively, calculate the composition of these gases in water.

Ans.

Percentage of oxygen (O2) in air = 20 %

Percentage of nitrogen (N2) in air = 79%

Also, it is given that water is in equilibrium with air at

a total pressure of 10 atm, i.e.,

Partial pressure of oxygen,

p o 2 = 20 100 × 10atm MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVC0dg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeaacaGaaiaabaqaamaabaabaaGcbaGaaeiCamaaBa aaleaacaqGVbWaaSbaaWqaaiaabkdaaeqaaaWcbeaakiaabccacaqG 9aGaaeiiamaalaaabaGaaeOmaiaabcdaaeaacaqGXaGaaeimaiaabc daaaGaaeiiaiaabEnacaqGGaGaaGjbVlaabgdacaqGWaGaaGjbVlaa bggacaqG0bGaaeyBaaaa@4831@

= 2 atm = 2 × 760 mm = 1520 mm of Hg

Partial pressure of nitrogen,

p N 2 = 79 100 × 10atm MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVC0dg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeaacaGaaiaabaqaamaabaabaaGcbaGaaeiCamaaBa aaleaacaqGobWaaSbaaWqaaiaabkdaaeqaaaWcbeaakiaabccacaqG 9aGaaeiiamaalaaabaGaae4naiaabMdaaeaacaqGXaGaaeimaiaabc daaaGaaeiiaiaabEnacaqGGaGaaeymaiaabcdacaaMe8Uaaeyyaiaa bshacaqGTbaaaa@4691@

= 7.9 atm = 7.9 × 760 mm = 6004 mm of Hg

Now, according to Henry’s law:

p = KH x

In case of oxygen,

     pO2=KH.xO2  xO2=pO2/KH            =1520mmHg3.30×107mmHg

= 4.61 x 10–5

In case of nitrogen,

     pN2=KH.xN2  xN2=pN2/KH            =6004mmHg6.51×107mmHg

= 9.22 × 10–5

Hence, the mole fractions of oxygen and nitrogen in water are 4.61 ×10−5 and 9.22 ×10−5 respectively.

Q.4 Benzene and toluene form ideal solution over the entire range of composition. The vapour pressure of pure benzene and toluene at 300 K are 50.71 mm Hg and 32.06 mm Hg respectively. Calculate the mole fraction of benzene in vapour phase if 80 g of benzene is mixed with 100 g of toluene.

Ans.

Molar mass of benzene (C6H6) = 78 g/mol

Molar mass of toluene (C6H5CH3) = 92 g/mol

No. of moles present in 80 g of benzene = 80/78

= 1.026 mol

No. of moles present in 100g of toluene= 100/92

= 1.087 mol

Mole fraction of benzene, xb

= 10.26 1.026 + 1.087 = 0.486 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVC0dg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeaacaGaaiaabaqaamaabaabaaGcbaGaaeypaiaabc cadaWcaaqaaiaabgdacaqGWaGaaeOlaiaabkdacaqG2aaabaGaaeym aiaab6cacaqGWaGaaeOmaiaabAdacaqGGaGaae4kaiaabccacaqGXa GaaeOlaiaabcdacaqG4aGaae4naaaacaqGGaGaaeypaiaabccacaqG WaGaaeOlaiaabsdacaqG4aGaaeOnaaaa@493A@

And, mole fraction of toluene, xt = (1 – xb)

= (1 – 0.486)

= 0.514

Given that, p°benzene= 50.71 mm Hg and

toluene = 32.06 mm Hg

Applying Raoult’s law,

pbenzene= xbenzene x p°benzene

= 0.486 x 50.71 mm = 24.65 mm Hg

ptoluene = xtoluene x p°toluene

= 0.514 x 32.06 mm = 16.48 mm Hg

Hence, mole fraction of benzene in vapour phase is given by:

= p b p b + p t = 24.65 24.65+16.48 =0.60 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVC0dg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeaacaGaaiaabaqaamaabaabaaGceaqabeaacqGH9a qpdaWcaaqaaiaadchadaWgaaWcbaGaamOyaaqabaaakeaacaWGWbWa aSbaaSqaaiaadkgaaeqaaOGaey4kaSIaamiCamaaBaaaleaacaWG0b aabeaaaaaakeaacqGH9aqpdaWcaaqaaiaaikdacaaI0aGaaiOlaiaa iAdacaaI1aaabaGaaGOmaiaaisdacaGGUaGaaGOnaiaaiwdacaaMe8 Uaey4kaSIaaGjbVlaaigdacaaI2aGaaiOlaiaaisdacaaI4aaaaiab g2da9iaaicdacaGGUaGaaGOnaiaaicdaaaaa@51ED@

Q.5 Vapour pressure of pure acetone and chloroform at 328 K are 741.8 mm Hg and 632.8 mm Hg respectively. Assuming that they form ideal solution over the entire range of composition, plot ptotal, pchloroform, and pacetone as a function of xacetone. The experimental data observed for different compositions of mixture is.

xacetone 0 11.8 23.4 36 50.8 58.2 64.5 72.1
pacetone/ mmHg 0 54.9 110.1 202.4 322.7 405.9 454.1 521.1
pchloroform/mmHg 632.8 548.1 469.4 359.7 257.7 193.6 161.2 120.7

Plot this data also on the same graph paper. Indicate whether it has positive deviation or negative deviation from the ideal solution.

Ans.

100 x xacetone 0 0.118 0.234 0.36 0.508 0.582 0.645 0.721
pacetone/ mmHg 0 54.9 110.1 202.4 322.7 405.9 454.1 521.1
pchloroform/mmHg 632.8 548.1 469.4 359.7 257.7 193.6 161.2 120.7
p total 632.8 603.1 579.5 562.1 580.4 599.5 615.3 641.8

As the plot for ptotal dips downwards, hence the solution shows negative deviation from the ideal behaviour.

Q.6 100 g of liquid A (molar mass 140 g mol−1) was dissolved in 1000 g of liquid B (molar mass 180 g mol−1). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 Torr.

Ans.

Number of moles of liquid A,

nA= 100/140 = 0.714 mol

Number of moles of liquid B,

nB= 1000/180 = 5.556 mol

Then, mole fraction of A

x A = n A n A + n B = 0.714 0.714 + 5.556 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVC0dg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeaacaGaaiaabaqaamaabaabaaGcbaGaaeiEamaaBa aaleaacaqGbbaabeaakiaabccacaqG9aGaaeiiamaalaaabaGaaeOB amaaBaaaleaacaqGbbaabeaaaOqaaiaab6gadaWgaaWcbaGaaeyqaa qabaGccaaMe8Uaae4kaiaaysW7caqGUbWaaSbaaSqaaiaabkeaaeqa aaaakiaabccacaqG9aGaaeiiamaalaaabaGaaeimaiaab6cacaqG3a GaaeymaiaabsdaaeaacaqGWaGaaeOlaiaabEdacaqGXaGaaeinaiaa bccacaqGRaGaaeiiaiaabwdacaqGUaGaaeynaiaabwdacaqG2aaaaa aa@51E8@

= 0.114

Then, mole fraction of B, xB= 1 – 0.114= 0.886

Given that, vapour pressure of pure liquid B, pB°

= 500 torr

Therefore, vapour pressure of liquid B in the solution,

pB= pB° xB

⇒ pB= 500 × 0.886 = 443 torr

Total vapour pressure of the solution, ptotal = 475 torr

Therefore, vapour pressure of liquid A in the solution,

pA= ptotal− pB = 475 – 443 = 32 torr

Now from Raoult’s law,

pA = pA° xA

⇒ pA° = pA / xA

= 32/0.114

= 280.7 torr

Hence, the vapour pressure of pure liquid A is 280.7 torr.

Q.7 Henry’s law constant for the molality of methane in benzene at 298 K is 4.27 × 105 mm Hg. Calculate the solubility of methane in benzene at 298 K under 760 mm Hg.

Ans.

Given values are

p = 760 mm Hg

KH = 4.27 × 105mm Hg

According to Henry’s law,

p = KHx

x=p/KH=760mmHg4.27×105

= 178 × 10–5

Hence, the mole fraction of methane in benzene is 1.78×10−3.

Q.8 Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water.

Ans.

Given data are,

P1° = 17.535 mm, w2= 25 g, w1= 450 g

Molar mass of glucose (C6H12O6), M2= 180 g/mol

Molar mass of water, M1= 18 g/mol

Then, number of moles of glucose, n2 = 25 /180

= 0.139 mol

And, number of moles of water n1= 450/18 = 25 mol

According to Raoult’s law,

p 1 o p 1 p 1 o = n 2 n 1 + n 2 17.535 p 1 17.535 = 0.139 0.139+25 17.535 p 1 = 0.139×17.535 0.139+25 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVC0dg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeaacaGaaiaabaqaamaabaabaaGceaqabeaadaWcaa qaaiaadchadaqhaaWcbaGaaGymaaqaaiaad+gaaaGccaGGtaIaamiC amaaBaaaleaacaaIXaaabeaaaOqaaiaadchadaqhaaWcbaGaaGymaa qaaiaad+gaaaaaaOGaeyypa0ZaaSaaaeaacaWGUbWaaSbaaSqaaiaa ikdaaeqaaaGcbaGaamOBamaaBaaaleaacaaIXaaabeaakiabgUcaRi aad6gadaWgaaWcbaGaaGOmaaqabaaaaaGcbaGaeyO0H49aaSaaaeaa caaIXaGaaG4naiaac6cacaaI1aGaaG4maiaaiwdacaaMe8Uaai4eGi aaysW7caWGWbWaaSbaaSqaaiaaigdaaeqaaaGcbaGaaGymaiaaiEda caGGUaGaaGynaiaaiodacaaI1aaaaiabg2da9maalaaabaGaaGimai aac6cacaaIXaGaaG4maiaaiMdaaeaacaaIWaGaaiOlaiaaigdacaaI ZaGaaGyoaiabgUcaRiaaikdacaaI1aaaaaqaaiabgkDiElaaigdaca aI3aGaaiOlaiaaiwdacaaIZaGaaGynaiaaysW7caGGtaIaaGjbVlaa dchadaWgaaWcbaGaaGymaaqabaGccqGH9aqpdaWcaaqaaiaaicdaca GGUaGaaGymaiaaiodacaaI5aGaaGjbVlabgEna0kaaysW7caaIXaGa aG4naiaac6cacaaI1aGaaG4maiaaiwdaaeaacaaIWaGaaiOlaiaaig dacaaIZaGaaGyoaiaaysW7cqGHRaWkcaaMe8UaaGOmaiaaiwdaaaaa aaa@855F@

⇒ 17.535 − p1= 0.097

⇒ p1= 17.44 mm of Hg

Hence, the vapour pressure of water is 17.44 mm of Hg.

Q.9 19.5 g of CH2FCOOH is dissolved in 500 g of water. The depression in the freezing point of water observed is 1.0° C. Calculate the van’t Hoff factor and dissociation constant of fluoroacetic acid.

Ans.

Given values are,

w2 = 19.5 g, w1 = 500 g, Kf= 1.86 K kg mol–1

ΔTf = 1.0° C

M 2 = 1000× K f × w 2 w 1 × ΔT f M 2 = 1000×1.86×19.5 500×1.0 = 72.54g/mol MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVC0dg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pse9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeaacaGaaiaabaqaamaabaabaaGceaqabeaacaqGnb WaaSbaaSqaaiaabkdaaeqaaOGaaeiiaiaab2dacaqGGaWaaSaaaeaa caqGXaGaaeimaiaabcdacaqGWaGaaGjbVlaabEnacaaMe8Uaae4sam aaBaaaleaacaqGMbaabeaakiaaysW7caqGxdGaaGjbVlaabEhadaWg aaWcbaGaaeOmaaqabaaakeaacaqG3bWaaSbaaSqaaiaabgdaaeqaaO GaaGjbVlaabEnacaaMe8UaaeiLdiaabsfadaWgaaWcbaGaaeOzaaqa baaaaaGcbaaabaGaaeytamaaBaaaleaacaqGYaaabeaakiaabccaca qG9aGaaeiiamaalaaabaGaaeymaiaabcdacaqGWaGaaeimaiaaysW7 caqGxdGaaGjbVlaabgdacaqGUaGaaeioaiaabAdacaaMe8Uaae41ai aaysW7caqGXaGaaeyoaiaab6cacaqG1aaabaGaaeynaiaabcdacaqG WaGaaGzaVlaaysW7caqGxdGaaGjbVlaabgdacaqGUaGaaeimaaaaca qGGaGaaeypaiaabccacaqG3aGaaeOmaiaab6cacaqG1aGaaeinaiaa ysW7caqGNbGaae4laiaab2gacaqGVbGaaeiBaaaaaa@7D67@

Thus M2 (observed) = 72.54 g/mol

M2 (calculated) for CH2FCOOH = 14 + 19 + 45

= 78 g/mol

van’t Hoff factor, i= M2 (cal)/ M2 (obs)

= 78/72.54 = 1.0753

Let  α be the degree of dissociation of CH2FCOOH

Initial conc.At eq.CH2FCOOH          C mol/L  C(1 – α) CH20FCOO + H+0Total = C (1 +α )Therefore, i=C(1+α)C=1+αα=i1=1.07531=0.0753Ka=[CH2FCOO][H+][CH2FCOOH]=.C(1α)=2C(1α)Taking the volume of the solution as 500 mL, we have the concentration:C = 19.578×1500×1000=0.5MKa=2(1α)=(0.5)(0.0753)210.0753=3.07×103

Q.10 Calculate the depression in the freezing point of water when 10 g of CH3CH2CHClCOOH is added to 250 g of water. Ka= 1.4 × 10−3, Kf = 1.86 K Kg mol-1

Ans.

Molar mass of CH 3 CH 2 CHClCOOH = ( 15 + 14 + 13 + 35.5 + 12 + 32 + 1 ) = 122.5 g/mol No . of moles present in 10 g of CH 3 CH 2 CHClCOOH = 10g 122.5g/mol =0.0816mol Molality of the solution ( m ) = 8.16× 10 –2 mol 250g ×1000g/kg=0.3264 Letαb e the degree of dissociation of CH 3 CH 2 CHClCOOH Initialconc. Ateq. CH 3 CH 2 CHClCOOH Cmol/L C( 1α ) CH 3 C H 2 0 CHClCOO + H + 0 K a = . C( 1α ) 2 α= K a /C α= 1.4× 10 –3 0.3264 = 0.0655 Initialconc. Ateqm. CH 3 CH 2 CHClCOOH Cmol/L C( 1α ) CH 3 CH 2 0 CHClCOO + H + 0

van’t Hoff factor, i= (1+α)/ 1 = 1 + α = 1 + 0.0655
= 1.065
Thus depression of freezing point,
ΔTf = i x Kf x m
= (1.065) (1.86) (0.3264) = 0.65 K

Q.11 The depression in freezing point of water observed for the same amount of acetic acid, tri-chloroacetic acid and tri-fluoroacetic acid increases in the order given above. Explain briefly.

Ans.

The structures as given below:

The electro-negativity goes in the following order

F > Cl > H

H is least electronegative while F is most electronegative.

Therefore, F can withdraw electrons towards itself more than Cl and H. Consequently, tri-fluoroacetic acid is the strongest acid while acetic acid the weakest acid. Thus, tri-fluoroacetic acid can easily lose H+ ions i.e., tri-fluoroacetic acid ionizes to the largest extent.In contrast; acetic acid ionizes to the minimum extent to give ions in the aqueous solution.

If more ions are produced, then depression of the freezing point will be greater. Hence, the depression in the freezing point increases in the order:

Acetic acid < tri-chloroacetic acid < tri-fluoroacetic acid

Q.12 Calculate the amount of benzoic acid (C6H5COOH) required for preparing 250 mL of 0.15 M solution in methanol.

Ans.

0.15 M solution of benzoic acid in methanol =

1000 mL of solution contains 0.15 mol of benzoic acid

∴ 250 mL of solution contain = (0.15 × 250)/1000

=0.0375 mol of benzoic acid

Molar mass of benzoic acid (C6H5COOH) = 7 × 12 + 6 × 1 + 2 × 16 = 122 g/mol

Hence, required benzoic acid =0.0375 mol ×122 gmol−1

= 4.575 g

Q.13 Nalorphene (C19H21NO3), similar to morphine, is used to combat withdrawal symptoms in narcotic users. Dose of nalorphene generally given is 1.5 mg. Calculate the mass of 1.5 × 10−3 m aqueous solution required for the above dose.

Ans.

The molar mass of nalorphene (C19H21NO3), is given as: (19 × 12 + 21 × 1 + 1 × 14 + 3 × 16)= 311 g/mol

1.5 × 10−3 m aqueous solution of nalorphene means:

1 kg (1000 g) of water contains 1.5 ×10−3mol

= 1.5 × 10–3 × 311 g = 0.4665 g of nalorphene

Therefore, total mass of the solution = (1000 + 0.4665) g

= 1000.4665 g nalorphene

Thus the mass of the solution containing 0.4665 g of nalorphene is 1000.4665 g.

The mass of the solution containing 1.5 mg of nalorphene is:

= 1000.4665×1.5× 10 –3 0.4665 g MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVC0dg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pse9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeaacaGaaiaabaqaamaabaabaaGcbaGaaeypaiaays W7daWcaaqaaiaabgdacaqGWaGaaeimaiaabcdacaqGUaGaaeinaiaa bAdacaqG2aGaaeynaiaaysW7caqGxdGaaGjbVlaabgdacaqGUaGaae ynaiaaysW7caqGxdGaaGjbVlaabgdacaqGWaWaaWbaaSqabeaacaqG taIaae4maaaakiaaysW7caaMe8oabaGaaeimaiaab6cacaqG0aGaae OnaiaabAdacaqG1aaaaiaabEgaaaa@54B8@

= 3.22 g
Hence, the mass of aqueous solution required is 3.22 g.

Q.14 Calculate the mass percentage of aspirin (C9H8O4) in acetonitrile (CH3CN) when 6.5 g of C9H8O4 is dissolved in 450 g of CH3CN.

Ans.

Mass%of aspirin = Mass of aspirin Mass of aspirin+Massofactonitrile ×100 = 6.5 6.5+450 ×100=1.424% MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVC0dg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pse9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeaacaGaaiaabaqaamaabaabaaGceaqabeaacaqGnb GaaeyyaiaabohacaqGZbGaaGjbVlaabwcacaaMe8Uaae4BaiaabAga caqGGaGaaeyyaiaabohacaqGWbGaaeyAaiaabkhacaqGPbGaaeOBai aabccacaqG9aGaaeiiamaalaaabaGaaeytaiaabggacaqGZbGaae4C aiaabccacaqGVbGaaeOzaiaabccacaqGHbGaae4CaiaabchacaqGPb GaaeOCaiaabMgacaqGUbaabaGaaeytaiaabggacaqGZbGaae4Caiaa bccacaqGVbGaaeOzaiaabccacaqGHbGaae4CaiaabchacaqGPbGaae OCaiaabMgacaqGUbGaaGjbVlaabUcacaaMe8UaaeytaiaabggacaqG ZbGaae4CaiaaysW7caqGVbGaaeOzaiaaykW7caqGHbGaae4yaiaabs hacaqGVbGaaeOBaiaabMgacaqG0bGaaeOCaiaabMgacaqGSbGaaeyz aaaacaaMe8Uaey41aqRaaGjbVlaaigdacaaIWaGaaGimaaqaaaqaai aaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8Ua aGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7ca aMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaa ysW7caaMe8UaaeypaiaabccadaWcaaqaaiaabAdacaqGUaGaaeynaa qaaiaabAdacaqGUaGaaeynaiaaysW7caqGRaGaaGjbVlaabsdacaqG 1aGaaeimaaaacaaMe8Uaae41aiaaysW7caqGXaGaaeimaiaabcdaca aMe8UaaeypaiaaysW7caqGXaGaaeOlaiaabsdacaqGYaGaaeinaiaa bwcaaaaa@C34A@

Q.15 If the solubility product of CuS is 6 × 10−16, calculate the maximum molarity of CuS in aqueous solution.

Ans.

Maximum molarity of CuS in aqueous solution

= Solubility of CuS in mol/L
Let s be the solubility of CuS in mol L–1

CuS Cu 2+ s + S s 2– MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVC0dg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pse9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeaacaGaaiaabaqaamaabaabaaGcbaGaae4qaiaabw hacaqGtbGaaeiiamaaoOaaleqabaaakiaawcCicaGL9gcacaaMe8+a aCbeaeaacaqGdbGaaeyDamaaCaaaleqabaGaaeOmaiaabUcaaaaaba qcLbwacaWGZbaaleqaaOGaaGjbVlaabUcacaaMe8+aaCbeaeaacaqG tbaaleaajugybiaabohaaSqabaGcdaahaaWcbeqaaiaabkdacaqGta caaaaa@4B2B@

Ksp = [Cu2+] [S2]

= s × s = s2 = 6 × 10–16

s = 6× 10 –16 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVC0dg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pse9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeaacaGaaiaabaqaamaabaabaaGcbaGaae4Caiaabc cacaqG9aGaaeiiamaakaaabaGaaeOnaiaaysW7caqGxdGaaGjbVlaa bgdacaqGWaWaaWbaaSqabeaacaqGtaIaaeymaiaabAdaaaaabeaaaa a@4194@

= 2.45 × 10–8 mol L–1

Thus the maximum molarity of CuS in an aqueous solution is 2.45 × 10–8 mol L–1

Q.16 If the density of some lake water is 1.25 g mL−1and contains 92 g of Na+ ions per kg of water, calculate the molality of Na+ ions in the lake.

Ans.

Number of moles present in 92 g of Na+ ions

= 92g 23g/mol = 4mol MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVC0dg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pse9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeaacaGaaiaabaqaamaabaabaaGcbaGaaeypaiaays W7daWcaaqaaiaabMdacaqGYaGaae4zaaqaaiaabkdacaqGZaGaaGjb VlaabEgacaqGVaGaaeyBaiaab+gacaqGSbaaaiaab2dacaqGGaGaae inaiaaysW7caqGTbGaae4BaiaabYgaaaa@4839@

Molality (m) = 4 mol/ 1 kg = 4 m

Thus the molality of Na+ ions in the lake is 4 m.

Q.17 Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water?

(i) Phenol (ii) Toluene (iii) Formic acid (iv) Ethylene glycol (v) Chloroform (vi) Pentanol.

Ans.

(i) Phenol (C6H5OH) has the polar group −OH and non-polar group −C6H5. So phenol is partially soluble in water.

(ii) Toluene (C6H5−CH3) has no polar groups. Thus, toluene is insoluble in water.

(iii) Formic acid (HCOOH) has the polar group −OH and can form H-bond with water.Thus, formic acid is highly soluble in water.

(iv)Ethylene glycol (HOCH2–CH2OH) has polar −OH group and can form H−bond. Thus, it is highly soluble in water.

(v)Chloroform is non-polar, so it is insoluble in water.

(vi) Pentanol (C5H11OH) has polar −OH group, but it also contains a very bulky non-polar −C5H11 group. Thus, pentanol is partially soluble in water.

Q.18 Based on solute-solvent interactions, arrange the following in order of increasing solubility in n-octane and explain. Cyclohexane,KCl,CH3OH,CH3CN.

Ans.

. n-octane is a non-polar solvent. Therefore, the solubility of a non-polar solute is more than that of a polar solute in the n-octane.

The order of increasing polarity is

Cyclohexane < CH3CN < CH3OH < KCl

Therefore, the order of increasing solubility is:

KCl < CH3OH < CH3CN < Cyclohexane

Q.19 Suggest the most important type of intermolecular attractive interaction in the following pairs.

(i) n-hexane and n-octane

(ii) I2and CCl4

(iii) NaClO4and water

(iv) methanol and acetone

(v) acetonitrile (CH3CN) and acetone (C3H6O)

Ans.

(i) Both are non- polar molecules. Hence, intermolecular interactions in them will be London dispersion forces.

(ii) Both are non- polar molecules. Hence, intermolecular interactions in them will be London dispersion forces.

(iii) NaClO4 gives Na+ and ClO4 ions in the solution and water is polar molecule. Hence, intermolecular interactions in them will be ion- dipole interactions.

(iv)Both are polar molecules, hence intermolecular interactions in them will be dipole-dipole interactions.

(v) Both are polar molecules, hence intermolecular interactions in them will be dipole-dipole interactions.

Q.20 At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be its concentration?

Ans.

The given values are,

T= 300 K, π1 = 4.98 bar,

T = 300 K, π2 = 1.52 bar,

Applying the relation,

π = CRT

In first case,

4.98 = (36/180) x R x 300 = 60 R—– (i)

In second case,

1.52 = C x R x 300——- (ii)

Dividing (ii) by (i), we get C= 0.061 M

Q.21 Two elements A and B form compounds having formula AB2 and AB4. When dissolved in 20 g of benzene (C6H6), 1 g of AB2 lowers the freezing point by 2.3 K whereas 1.0 g of AB4 lowers it by 1.3 K. The molar depression constant for benzene is 5.1 K kg mol−1. Calculate atomic masses of A and B.

Ans.

Applying the formula,

M 2 = 1000× K f × w 2 w 1 × ΔT f MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVC0dg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pse9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeaacaGaaiaabaqaamaabaabaaGcbaGaaeytamaaBa aaleaacaqGYaaabeaakiaabccacaqG9aGaaeiiamaalaaabaGaaeym aiaabcdacaqGWaGaaeimaiaaysW7caqGxdGaaGjbVlaabUeadaWgaa WcbaGaaeOzaaqabaGccaaMe8Uaae41aiaaysW7caqG3bWaaSbaaSqa aiaabkdaaeqaaaGcbaGaae4DamaaBaaaleaacaqGXaaabeaakiaays W7caqGxdGaaGjbVlaabs5acaqGubWaaSbaaSqaaiaabAgaaeqaaaaa aaa@525A@

For AB2 compound,

M AB 2 = 1000×5.1×1 20×2.3 = 110.87g/mol MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVC0dg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pse9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeaacaGaaiaabaqaamaabaabaaGcbaGaaeytamaaBa aaleaacaqGbbGaaeOqamaaBaaameaacaqGYaaabeaaaSqabaGccaqG GaGaaeypaiaabccadaWcaaqaaiaabgdacaqGWaGaaeimaiaabcdaca aMe8Uaae41aiaaysW7caqG1aGaaeOlaiaabgdacaaMe8Uaae41aiaa ysW7caqGXaaabaGaaeOmaiaabcdacaaMe8Uaae41aiaaysW7caqGYa GaaeOlaiaabodaaaGaaeiiaiaab2dacaqGGaGaaeymaiaabgdacaqG WaGaaeOlaiaabIdacaqG3aGaaGjbVlaabEgacaqGVaGaaeyBaiaab+ gacaqGSbaaaa@5DF5@

For AB4 compound,

M AB4 = 1000×5.1×1 20×1.3 = 196.15g/mol MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVC0dg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pse9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeaacaGaaiaabaqaamaabaabaaGcbaGaaeytamaaBa aaleaacaqGbbGaaeOqaiaabsdaaeqaaOGaaeiiaiaab2dacaqGGaWa aSaaaeaacaqGXaGaaeimaiaabcdacaqGWaGaaGjbVlaabEnacaaMe8 Uaaeynaiaab6cacaqGXaGaaGjbVlaabEnacaaMe8Uaaeymaaqaaiaa bkdacaqGWaGaaGjbVlaabEnacaaMe8Uaaeymaiaab6cacaqGZaaaai aabccacaqG9aGaaeiiaiaabgdacaqG5aGaaeOnaiaab6cacaqGXaGa aeynaiaaysW7caqGNbGaae4laiaab2gacaqGVbGaaeiBaaaa@5DC3@

Let the atomic masses of A and B be a and b respectively.

Then, we can write:

Molar mass of AB2 = a + 2b= 110.87 g/mol—- (i)

Molar mass of AB4 = a + 4b= 196.15 g/mol—- (ii)

Eq. (i) – Eq. (ii) gives

2b = 85.28

Or, b = 42.64

Substituting this in equation (i)

a + 2 x 42.64 = 110.87

⇒ a = 25.59

Hence, the atomic masses of A and B are 25.59 u and 42.64 u respectively.

Q.22 A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15 K.

Ans.

Here, ΔTf = (273.15 − 271) K = 2.15 K

Molar mass of sugar (C12H22O11)

= 12 × 12 + 22 × 1 + 11 × 16

= 342 g/mol

5% solution (by mass) of cane sugar in water means 5 g of cane sugar is present in (100 − 5)g = 95 g of water.

Thus number of moles of cane sugar = (5/342)

= 0.0146 mol

Therefore, molality of the solution,

(m) = (0.0146 mol/ 0.095 kg)

= 0.1537 mol kg–1

Applying the relation,

ΔTf = Kf × m

⇒ Kf = ΔTf / m = 2.15 K / (0.1537 mol/kg)

= 13.99 K kg/ mol

Molar mass of glucose (C6H12O6)

= 6 × 12 + 12 × 1 + 6 × 16

= 180 g/mol

5% glucose in water means 5 g of glucose is present in (100 − 5) g = 95 g of water.

Number of moles of glucose = 5/180 = 0.0278 mol

Thus the molality of the glucose solution

M = (0.0278 mol/ 0.095 kg) = 0.2926 mol kg–1

Applying the relation,

ΔTf = Kf × m

= 13.99 K kg mol−1×0.2926 mol kg–1

= 4.09 K

∴ The freezing point of 5% glucose solution is

(273.15 − 4.09) K= 269.06 K

Q.23 The resistance of a conductivity cell containing 0.001M KCl solution at 298 K is 1500. What is the cell constant if conductivity of 0.001M KCl solution at 298 K is 0.146 × 10–3 S cm–1.

Ans.

Given,

Conductivity, κ = 0.146 × 10–3 S cm–1

Resistance, R = 1500

Cell constant = κ × R

= 0.146 × 10–3 × 1500

= 0.219 cm–1

Q.24 How much electricity is required in coulomb for the oxidation of

(i) 1 mol of H2O to O2.
(ii) 1 mol of FeO to Fe2O3

Ans.

(i) 

,H2>OH2+12O2Now, we can write:O2 12O2+ 2 eElectricity required for the oxidation of1 mol of H2O to O2 = 2F                             2×96487C                              = 192974C(ii) <mtext

,Fe2+Fe3++ e1Electricity required for the oxidation of1mol of FeO to Fe2O3= 1F= 96487C

Q.25 A solution of Ni(NO3)2 is electrolysed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode?

Ans.

Given,Current=5ATime=20×60=1200sCharge=Current×Time                     =5×1200                     =6000CAccordingtothereaction,Ni2+(aq)+2eNi(s)                                     58.7gNickeldepositedby2×96487C=58.71gTherefore,nickeldepositedby6000C=58.71×60002×96487g                                                                               =1.825gHence,1.825gofnickelwillbedepositedatthe  cathode.

Q.26 Three electrolytic cells A, B, C containing solutions of ZnSO4, AgNO3 and CuSO4, respectively are connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited?

Ans.

According to the reaction:Ag+(aq)+eAg(s)                                 108g         i.e.,108g of Ag is deposited by 96487 C.=96487×1.45108=1295.43CGiven,Current= 1.5A=1295.431.5s Time= 863.6s= 864s= 14.40 minAgain,Cu2+(aq)+2eCu(s)                                     63.5gi.e.,2×96487 C of charge deposit=63.5g of CuTherefore,1295.43C of charge will deposit=63.5 x 1295.432 x 96487g= 0.426g of CuZn(aq)2++2eZn(s)                65.4 gi.e.,  2×96487C of charge deposit= 65.4g of ZnTherefore,1295.43C of charge will deposit=65.4 x 1295.432 x 96487g= 0.439g of Zn

Q.27 Using the standard electrode potentials given in Table 3.1, predict if the reaction between the following is feasible:

(i) Fe3+(aq) and I (aq)
(ii) Ag+ (aq) and Cu(s)
(iii) Fe3+ (aq) and Br (aq)
(iv) Ag(s) and Fe3+ (aq)
(v) Br2 (aq) and Fe2+ (aq)

Ans.

(i)Fe3+(aq)+eFe2+(aq)            ×2;      E°=+0.77V                  2I(aq)  I2(s)+2e                     E°=0.54V2Fe3+(aq)+2I(aq)    2Fe2+(aq)+I2(s);  E°=+0.23V     Since E°for the overall reaction is positive, the reaction between Fe3+(aq)and I(aq) is feasible.(ii)Ag+(aq)+eAg(s)            ×2;                   E°=+0.80V                     Cu(s)  Cu2+(aq)  +2e                  E°=0.34V2Ag+(aq)+Cu(s)    2Ag(s)+Cu2+(aq);       E°=+0.46V     Since E°for the overall reaction is positive, the reaction between Ag+(aq)and Cu(s) is feasible.(iii)Fe3+(aq)+eFe2+(aq)            ×2;            E°=+0.77V               2Br(aq)  Br2(l)+2e                         E°=1.09V2Fe3+(aq)+2Br(aq)    2Fe2+(aq)+Br2(l);  E°=0.32V     Since E°for the overall reaction is negative, the reaction between Fe3+(aq) and Br(aq) is not feasible.(iv)Ag(s)  Ag+(aq)+e                                    E°=0.80V        Fe3+(aq)+eFe2+(aq)                               E°=+0.77VAg(s)+Fe3+(aq)    Ag+(aq)+Fe2+(aq)     E°=0.03VSinceE°fortheoverallreactionisnegative,thereactionbetweenAg(s)andFe3+(aq)isnotfeasible. (iv)Ag(s)A g + (aq)+ e E°=0.80V F e 3+ (aq)+ e F e 2+ (aq)E°=+0.77V Ag(s)+F e 3+ (aq)A g + (aq)+F e 2+ (aq)E°=0.03V Since E° fortheoverallreactionispositive,thereactionbetween Br 2 ( aq )and Fe 2+ ( aq )is feasible. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sspeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@CB50@

Q.28 A solution containing 30 g of non volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 K. Calculate:

i. Molar mass of the solute
ii. Vapour pressure of water at 298 K.

Ans.

Let, the molar mass of the solute be M g mol1n1(Solvent,H2O)=90g18g/mol=5moln2(solute)=30Mmolp1=2.8KPa;p1°p1p1°=n2n1+n2p1°2.8p1°=30/M5+30/M12.8P1°=30/M(5M+30)/M2.8p1°=130(5M+30)2.8p1°=5M(5M+30)p1°2.8=(5M+30)5M(i)After the addition of 18 g of water:n1= (90+18)/18 = 6 molp1= 2.9 kPa;Again, applying the relation:p1°2.9p1°=n2n1+n2p1°2.9p1°=30/M6+30/M12.9p1°=30/M(6M+30)/M2.9p1°=130(6M+30)2.9p1°=6M(6M+30)p1°2.9=(6M+30)6M(ii)Dividing equation (i) by (ii), we have:2.92.8=5M+305M6M+306M2.92.8×6M+306M=5M+305M2.9×5×(6M+30)=2.8×6¯(5M+30)3M=69M=23uii) Putting the value of M in equation (i), we havep1°2.8=(5×23+30)5×23p1°2.8=145115p1°=3.53kPa

Q.29 Calculate the mass of a non-volatile solute (molar mass 40 g mol–1) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%.

Ans.

Molar mass of solute, M2= 40 g mol–1

Mass of octane, w1= 114 g

Molar mass of octane, (C8H18), M1= 8 × 12 + 18 × 1

= 114 g/mol

Let the vapour pressure of pure octane be p1° .

Then, the vapour pressure of the octane after dissolving the non-volatile solute is

80100P1°=0.8p1°Applying the relation:p1°p1p1°=w2×M1M2×w1p1°0.8p1°p1°=w2×11440×1140.2p1°p1°=w240Or, w2=40×0.2=8g

Q.30 Predict the products of electrolysis in each of the following:

(i) An aqueous solution of AgNO3 with silver electrodes.
(ii) An aqueous solution of AgNO3 with platinum electrodes.
(iii) A dilute solution of H2SO4 with platinum electrodes.
(iv) An aqueous solution of CuCl2 with platinum electrodes.

Ans.

( i ) Atcathode:Thefollowingreductionreactionscompetetotakeplaceatthecathode. Ag (aq) + + e Ag (s) ; E ° =0.80 V H (aq) + + e 1 2 H 2( g ) ; E ° =0.00 V Thereactionwithahighervalueof E ° takesplaceatthecathode. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@4391@ Therefore,deposition ofsilverwilltakeplaceatthecathode. At anode: TheAganodeisattackedby NO 3 ions.Therefore,thesilverelectrode attheanode dissolves in the solution to from Ag. ( ii ) Atcathode:Thefollowingreductionreactionscompetetotakeplaceatthecathode Ag (aq) + + e Ag (s) ; E ° =0.80 V H (aq) + + e 1 2 H 2(g) ; E ° =0.00 V Thereactionwithahighervalueof E ° takesplaceatthecathode. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@C39F@ At cathode:The following oxidation reactions are possible at the anode.Cl(aq) 12Cl2(g)+ e1 ; E°=1.36 V2H2O(l) O2(g)+4H(aq)++ 4e ; E°=1.23 VAttheanode,thereactionwithalowervalueofE°ispreferred. Butduetotheoverpotential of oxygen, Cl gets oxidised at the anode to produce Cl 2 gas. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@883D@

Q.31 The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it.

Ans.

1 molal solution means 1 mol of the solute is present in 1000 g of the solvent (water).

Molar mass of water = 18 g mol−1

∴ Number of moles present in 1000 g of water= 1000/18

= 55.56 mol

Therefore, mole fraction of the solute in the solution is

X2=11+55.56=0.0177Now, we knowp1°p1p1°=X212.3p112.3=0.0177

⇒12.3 − p1= 0.2177

⇒p1 = 12.0823

Hence, the vapour pressure of the solution is 12.08 kPa.

Q.32 Heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components are 105.2 kPa and 46.8 kPa respectively. What will be the vapour pressure of a mixture of 26.0 g of heptane and 35 g of octane?

Ans.

Molar mass of heptane (C7H16) = (7 × 12 + 16 × 1)

= 100 g mol−1

Number of moles of heptane= 26 / 100 = 0.26 mol

Molar mass of octane (C8H18) = (8 × 12 + 18 × 1)

= 114 g mol–1

Number of moles of octane = 35/114 = 0.31 mol

Mole fraction of heptane (x1)

X 1 = 0.26 0.26+0.31 =0.456 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbujxyKLgDP9MBHXgibjxyIL2yaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVy0xg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeaacaGaaiaabaWaaqaabaqbaaGcbaGaamiwamaaBaaaleaacaaIXaaabeaakiabg2da9maalaaabaGaaGimaiaac6cacaaIYaGaaGOnaaqaaiaaicdacaGGUaGaaGOmaiaaiAdacqGHRaWkcaaIWaGaaiOlaiaaiodacaaIXaaaaiabg2da9iaaicdacaGGUaGaaGinaiaaiwdacaaI2aaaaa@4E10@

Mole fraction of octane (x2) = (1 –0.456) = 0.544

Vapour pressure of heptanes (p1°) = 105.2 kpa

Vapour pressure of octane (p2°) = 46.8 kPa

Now, partial pressure of heptane, p1 = x1 p1°

p1= 0.456 x 105.2 = 47.97 kPa

Partial pressure of octane, p2 = x2 p2°

p2 = 0.544 × 46.8 = 25.46 kPa

P(total) = p1 + p2 = 47.97 + 25.46 = 73.43 kPa

Q.33 An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?

Ans.

Vapour pressure of pure water at normal boiling point (p1°) = 1 atm = 1.013 bar

Vapour pressure of the solution at normal boiling point (p1) = 1.004 bar

Mass of solute (w2) = 2 g

Mass of solvent (water), (w1) = 98 g

Molar mass of solvent (water), (M1) = 18 g mol−1

According to Raoult’s law,

p1°p1p1°=w2×M1M2×w11.0131.0041.013=2×18M2×98M2=1.013×2×180.0009×98=41.35g/mol

Q.34 What is meant by positive and negative deviations from Raoult’s law and how is the sign of ΔmixH related to positive and negative deviations from Raoult’s law?

Ans.

According to Raoult’s law, the partial vapour pressure of each volatile component in any solution is directly proportional to its mole fraction.

The solutions which obey Raoult’s law are known as ideal solutions. The solutions that do not obey Raoult’s law (non ideal solutions) have vapour pressures either higher or lower than that predicted by Raoult’s law.

i) If the vapour pressure is higher, then the solution is said to exhibit positive deviation from Raoult’s law.
ii) If it is lower, then the solution is said to exhibit negative deviation from Raoult’s law.

In the case of an ideal solution, the enthalpy of the mixing of the pure components for forming the solution is zero. Thus, ΔmixH = 0.

In the case of solutions showing positive deviations, absorption of heat takes place. Thus, ΔmixH = +ve

In the case of solutions showing negative deviations, evolution of heat takes place. Thus, ΔmixH= –ve.

Q.35 The partial pressure of ethane over a solution containing 6.56 × 10−2g of ethane is 1 bar. If the solution contains 5.00 × 10−2g of ethane, then what shall be the partial pressure of the gas?

Ans.

Applying the Henry’s law

m = KH x p

⇒ (6.56 x 10–2 g) = KH x 1 bar [In the first case]

⇒ KH = 6.56 x 10–2 g/bar

Again,

5.00 x 10–2g = (6.56 x 10–2 g/bar) p [In the second case]

p= 5.000× 10 2 g 6.5× 10 2 g/bar =0.762bar MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbujxyKLgDP9MBHXgibjxyIL2yaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVy0xg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeaacaGaaiaabaWaaqaabaqbaaGceaqabeaacqGHshI3caWGWbGaeyypa0ZaaSaaaeaacaaI1aGaaiOlaiaaicdacaaIWaGaaGimaiabgEna0kaaigdacaaIWaWaaWbaaSqabeaacqGHsislcaaIYaaaaOGaam4zaaqaaiaaiAdacaGGUaGaaGynaiabgEna0kaaigdacaaIWaWaaWbaaSqabeaacqGHsislcaaIYaaaaOGaam4zaiaac+cacaWGIbGaamyyaiaadkhaaaaabaGaeyypa0JaaGimaiaac6cacaaI3aGaaGOnaiaaikdacaaMc8UaamOyaiaadggacaWGYbaaaaa@604B@

Q.36 State Henry’s law and mention some important applications?

Ans.

Henry’s law states that,

The mass of gas dissolved in a given volume of the liquid at constant temperature is directly proportional to the pressure of the gas present in the equilibrium with the liquid.

Mathematically, m µ p or, m = K x p

Where, m = mass of the gas dissolved in a unit volume of the solvent

p = pressure of the gas in equilibrium with the solvent

K= constant of proportionality

Some important applications of Henry’s law are given as follows:
(a) In packing of soda cans and bottles: – To increase the solubility of CO2 gas in soda water, bottles of soda water are always sealed under higher pressure.
(b) In deep see diving: – Nitrogen is a more soluble gas in our blood and in deep sea where pressure increases, its solubility also increases. When scuba diver tries to come rapidly towards the surface of water, pressure decreases and dissolved N2 gas comes back from the blood and makes bubbles in his veins. It causes bends .To avoid bends divers use oxygen diluted with helium as helium is less soluble in blood.
(c) In the function of lungs: When air enters the lungs, the partial pressure of O2 is high. It combines with haemoglobin in blood to form oxyhaemoglobin. In tissues, the partial pressure of O2 is low, hence O2 is released from oxyhaemoglobin which is utilized for the functions of the cells.

Q.37 Why do gases always tend to be less soluble in liquids as the temperature is raised?

Ans.

Considering the equilibrium,

Gas + Solvent Solution + Heat MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbujxyKLgDP9MBHXgibjxyIL2yaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVy0xg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeaacaGaaiaabaWaaqaabaqbaaGcbaGaae4raiaabggacaqGZbGaaeiiaiaabUcacaqGGaGaae4uaiaab+gacaqGSbGaaeODaiaabwgacaqGUbGaaeiDaiaabccacqWImhYGcaqGGaGaae4uaiaab+gacaqGSbGaaeyDaiaabshacaqGPbGaae4Baiaab6gacaqGGaGaae4kaiaabccacaqGibGaaeyzaiaabggacaqG0baaaa@583D@

Dissolution of gas in liquid is an exothermic process. As the temperature is increased, equilibrium will shift towards the backward direction and the solubility will decrease

Q.38 What role does the molecular interaction play in a solution of alcohol and water?

Ans.

In water as well as in alcohol, strong hydrogen bonding is present. When both of these are mixed together, the molecular interactions get weakened. Hence they show positive deviation from the ideal behaviour. Thus, the solution will have higher vapour pressure and lower boiling point as compared to water and alcohol.

Q.39 A sample of drinking water was found to be severely contaminated with chloroform (CHCl3) supposed to be a carcinogen. The level of contamination was 15 ppm (by mass):

(i) Express this in percent by mass

(ii) Determine the molality of chloroform in the water sample.

Ans.

(i) The term ppm means parts per million.

15 ppm = (15/106) [1 million= 106]

Thus % by mass= (15/106) x 100 = 15 x 10–4

(ii) Since 15 g of chloroform is in 106 g of solution

Mass of the solvent = 106 g

Molar mass of CHCl3 = 12 + 1 + (3 x 35.5)= 119.5 g/mol

Number of moles of CHCl3 = 15/119.5= 0.1255 mole

Molality = (0.1255/106) x1000 = 1.25 x 10–4 M

Q.40 An antifreeze solution is prepared from 222.6 g of ethylene glycol (C2H6O2) and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g mL–1, then what shall be the molarity of the solution?

Ans.

Mass of the solute (ethylene glycol) = 222.6g

Molecular mass of C2H6O2 = 62 g/mol

Moles of the solute= (222.6/62) = 3.59 mol

Mass of the solvent = 200g = 0.200 kg

Molality of the solution

= Moles of the solute/ Mass of solvent

= 3.59/ 0.200

= 17.95 mol/kg

Total mass of the solution = 422.6 g

Volume of the solution = Mass/Density

=(422.6g/ 1.072 g/ mL)

= 394.2 mL

= 0.3942 L

Molarity of the solution

= Moles of the solute/ Mass of solvent

=3.59 moles/ 0.3942 L

= 9.11 mol/L

Q.41 A solution is obtained by mixing 300 g of 25% solution and 400 g of 40% solution by mass. Calculate the mass percentage of the resulting solution.

Ans.

300g of 25% solution contains solute

= 300 x (25/100) = 75 g of solute

400 g of 40% solution contains solute

= 400 x (40/100) = 160 g of solute

Total amount of solute = (75 + 160) = 235 g and

Total amount of solution = (300+400) = 700g

Therefore, mass percentage (w/w) of the solute in the resulting solution,

= 235 700 ×100 =33.57% MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbujxyKLgDP9MBHXgibjxyIL2yaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVy0xg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeaacaGaaiaabaWaaqaabaqbaaGceaqabeaacqGH9aqpdaWcaaqaaiaaikdacaaIZaGaaGynaaqaaiaaiEdacaaIWaGaaGimaaaacqGHxdaTcaaIXaGaaGimaiaaicdaaeaacqGH9aqpcaaIZaGaaG4maiaac6cacaaI1aGaaG4naiaacwcaaaaa@4C12@

And, mass percentage (w/w) of the solvent in the resulting solution = (100 − 33.57) % = 66.43%

Q.42 How many mL of 0.1 M HCl are required to react completely with 1 g mixture of Na2CO3 and NaHCO3 containing equimolar amounts of both?

Ans.

Let x g of Na2CO3 be present in the mixture.

So, (1 – x) g of NaHCO3 is present in the mixture.

Molar mass of Na2CO3 = (2 x 23 + 12 + 3 x 16)

= 106 g/mol

Molar mass of NaHCO3 = (23 + 1 + 12 + 3 x 16)

84 g mol–1

Moles of Na2CO3 in x g= x/106

And Moles of NaHCO3 in (1 – x) g = (1 – x)/ 84

As the mixture contains equimolar amounts of two components, therefore,

x106=1x84Or, 106-106x=84xOr, x=106190g=0.558gThus, moles of Na2CO3=0.558/106=00.00526And moles of NaHCO3=(10.558)/84=0.442/84=0.00526Reaction of Na2CO3 and NaHCO3 with HCl are as follows:Na2CO3+2HCl2NaCl+H2O+CO2NaHCO3+HClNaCl+H2O+CO21 mole of Na2CO3 requires 2 moles of HCl 0.00526 moles of Na2CO3 requires =(0.00526×2)moles of HCl=0.01052 mole1 mole of NaHCO3 requires 1 mole of HCl 0.00526 moles of NaHCO3 requires 0.00526 moles of HCl=0.01052 mole1 mole of NaHCO3 requires 1 mole of HCl 0.00526 moles of NaHCO3 requires 0.00526 moles of HClTotal HCl required =(0.01052+0.00526) moles=0.01578moles 0.1 mole of 0.1 M HCl are present in 1000 mL of HCl 0.01578 moles of HCl will be present in =(10000.1)×0.01578=157.8mL

Thus, 157.8 mL of 0.1 M HCl are required to react completely with 1 g mixture of Na2CO3 and NaHCO3 containing equimolar amounts of both.

Q.43 A solution of glucose in water is labelled as 10% w/w, what would be the molality and mole fraction of each component in the solution? If the density of solution is 1.2 g mL–1, then what shall be the molarity of the solution?

Ans.

10g glucose is present in 100 g solution, i.e. 90 g of water =0.090 kg of water.Number of moles of gulcose=(10180)=0.0555molNumber of moles of water =(9018)=5molesMolality =(0.05555mol0.090kg)=0.617mMole fraction of glucose(x);X(glucose)=0.05550.0555+5=0.01Thus, X(H2O)=(10.01)=0.99Density of the solution =1.2g/ml100 g solution = 100/1.2ml=83.33ml=0.0833LMolarity=0.0555mol0.08333L=0.67M

Q.44 Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1.504 g mL–1?

Ans.

Let the mass of the solution = 100g

The mass of nitric acid = 68 g

Molar mass of HNO3 = 63 g mol–1

68 g HNO3 = 68/63 mol = 1.079 mol.

Given, density of the solution =1.504 g/mL

Volume of the solution = 100/1.504 mL= 66.5 mL

Molarity of the solution = (1.079/0.0665)M

= 16.23 M

Q.45 Define the following terms:

  1. Mole fraction
  2. Molality
  3. Molarity
  4. Mass percentage.

Ans.

i) Mole Fraction: Mole fraction of a constituent is defined as the fraction obtained by dividing number of moles of that constituent by the total number of moles of all the constituents present in the solution and is expressed as:

Mole fraction of a component= Number of moles of the component Total number of moles of all the components ( ii )Molality: Molality ( m ) is defined as the number of moles of the solute per kilogram ( kg ) of the solvent and is expressed as: Molality= Moles of solute Mass of solvent in kg ( iii )Molarity: Molarity ( M ) is defined as number of moles of solute dissolved in one litre ( or one cubic decimetre ) of solution and is expressed as: Molarity= Moles of solute volume of solution in litre ( iv )Mass percentage: The mass percentage of a component of a solution is defined as the mass of the component in grams present in 100 grams of the solution and is expressed as: Mass% of a component= Mass of the component in the solution Total mass of the solution ×100 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Q.46 Give an example of a solid solution in which the solute is a gas.

Ans.

The example of a solid solution where the solute is gas is the solution of hydrogen in palladium.

Q.47 Define the term solution. How many types of solutions are formed? Write briefly about each type with an example.

Ans.

Solution is defined as a homogeneous mixture of two or more chemically non-reacting substances. There are mainly three types of solutions. Some details about them are as follows:

Type of solution Solute Solvent Common examples
Gaseous Solutions Gas Gas Mixture of oxygen and nitrogen gases
Liquid Gas Chloroform mixed with nitrogen gas
Solid Gas Camphor in nitrogen gas
Liquid Solutions Gas Liquid Oxygen dissolved in water
Liquid Liquid Ethanol dissolved in water
Solid Liquid Glucose dissolved in water
Solid Solutions Solid Solid Copper dissolved in gold
Liquid Solid Amalgam of mercury with sodium
Gas Solid Solution of hydrogen in palladium

Q.48 Explain the following with suitable examples:

(i) Ferromagnetism
(ii) Paramagnetism
(iii) Ferrimagnetism
(iv) Antiferromagnetism
(v) 12-16 and 13-15 group compounds.

Ans.

(i) Ferromagnetism: The substances that are strongly attracted by a magnetic field are called ferromagnetic substances. Ferromagnetic substances can be permanently magnetized even in the absence of a magnetic field. Some examples of ferromagnetic substances are iron, cobalt, nickel, gadolinium, and CrO2.

In solid state, the metal ions of ferromagnetic substances are grouped together into small regions called domains and each domain acts as a tiny magnet. In an un-magnetised piece of a ferromagnetic substance, the domains are randomly-oriented and so, their magnetic moments get cancelled. However, when the substance is placed in a magnetic field, all the domains get oriented in the direction of the magnetic field. As a result, a strong magnetic effect is produced. This ordering of domains persists even after the removal of the magnetic field. Thus, the ferromagnetic substance becomes a permanent magnet.

Schematic alignment of magnetic moments in ferromagnetic substances

(ii) Paramagnetism: The substances that are attracted by a magnetic field are called paramagnetic substances. Some examples of paramagnetic substances are O2, Cu2t, Fe3t, and Cr3t.

Paramagnetic substances get magnetized in a magnetic field in the same direction, but lose magnetism when the magnetic field is removed. To undergo paramagnetism, a substance must have one or more unpaired electrons. This is because the unpaired electrons are attracted by a magnetic field, thereby causing paramagnetism.

(iii) Ferrimagnetism: The substances in which the magnetic moments of the domains are aligned in parallel and anti-parallel directions, in unequal numbers, are said to have ferrimagnetism. Examples include Fe3O4 (magnetite), ferrites such as MgFe2O4 and ZnFe2O4.

Ferrimagnetic substances are weakly attracted by a magnetic field as compared to ferromagnetic substances. On heating, these substances become paramagnetic.

Schematic alignment of magnetic moments in ferromagnetic substances

(iv) Antiferromagnetism: Antiferromagnetic substanceshave domain structures similar to ferromagnetic substances, but are oppositely-oriented. The oppositely-oriented domains cancel out each other’s magnetic moments.

Schematic alignment of magnetic moments in antiferromagnetic substances

(v) 12-16 and 13-15 group compounds: The 12-16 group compounds are prepared by combining group 12 and group 16 elements and the 13-15 group compounds are prepared by combining group 13 and group 15 elements. These compounds are prepared to stimulate average valence of four as in Ge or Si. Indium (III) antimonide (InSb), aluminium phosphide (AlP), and gallium arsenide (GaAs) are typical compounds of groups 13-15. GaAs semiconductors have a very fast response time and have revolutionized the designing of semiconductor devices. Examples of group 12-16 compounds include zinc sulphide (ZnS), cadmium sulphide (CdS), cadmium selenide (CdSe), and mercury (II) telluride (HgTe). The bonds in these compounds are not perfectly covalent. The ionic character of the bonds depends on the electronegativities of the two elements.

Q.49 If NaCl is doped with 10−3 mol % of SrCl2, what is the concentration of cation vacancies?

Ans.

It is given that NaCl is doped with 10−3 mol % of SrCl2.

This means that 100 mol of NaCl is doped with 10−3 mol of SrCl2.

Therefore,1 mol of NaCl is doped with 103100 mol of SrCl2=105 mol of SrCl2Cation vacancies produced by one Sr2+ion=1concentration of the cation vacancies produced by 105 mol of Sr2+ ions =105×6.022×1023=6.022×1018mol1Hence, the concentration of cation vacancies created by SrCl2 is 6.022×1018per mol of NaCl.

Q.50 Aluminium crystallizes in a cubic close-packed structure. Its metallic radius is 125 pm.

(i) What is the length of the side of the unit cell?
(ii) How many unit cells are there in 1.00 cm3 of aluminium?

Ans.

(i)For cubic close-packed structure:a=22r=22×125pm=353.55pm(approximately)(ii)Volume of one unit cell=(354pm)3=4.4×107pm3=4.4×107×1030cm3=4.4×1023cm3Therefore, number of unit cells in 1.00 cm3=1.00cm34.4×1023cm3=2.27×1022

Q.51 Explain the following terms with suitable examples:

(i) Schottky defect
(ii) Frenkel defect
(iii) Interstitials and
(iv) F-centres

Ans.

(i) Schottky defect: Schottky defect is basically a vacancy defect shown by ionic solids. In this defect, an equal number of cations and anions are missing to maintain electrical neutrality. It decreases the density of a substance. Significant number of Schottky defects is present in ionic solids. For example, in NaCl, there are approximately 106 Schottky pairs per cm3 at room temperature. Ionic substances containing similar-sized cations and anions show this type of defect. For example: NaCl, KCl, CsCl, AgBr, etc.

(ii) Frenkel defect: Ionic solids containing large differences in the sizes of ions show this type of defect. When the smaller ion (usually cation) is dislocated from its normal site to an interstitial site, Frenkel defect is created. It creates a vacancy defect as well as an interstitial defect. Frenkel defect is also known as dislocation defect. Ionic solids such as AgCl, AgBr, AgI, and ZnS show this type of defect.

(iii) Interstitials: Interstitial defect is shown by non-ionic solids. This type of defect is created when some constituent particles (atoms or molecules) occupy an interstitial site of the crystal. The density of a substance increases because of this defect.

(iv) F-centres: When the anionic sites of a crystal are occupied by unpaired electrons, the ionic sites are called F-centres. These unpaired electrons impart colour to the crystals. For example, when crystals of NaCl are heated in an atmosphere of sodium vapour, the sodium atoms are deposited on the surface of the crystal. The Cl ions diffuse from the crystal to its surface and combine with Na atoms, forming NaCl. During this process, the Na atoms on the surface of the crystal lose electrons. These released electrons diffuse into the crystal and occupy the vacant anionic sites, creating F-centres.

Q.52 In terms of band theory, what is the difference

(i) Between a conductor and an insulator
(ii) Between a conductor and a semiconductor

Ans.

(i) The valence band of a conductor is partially-filled or it overlaps with a higher energy, unoccupied conduction band.

On the other hand, in the case of an insulator, the valence band is fully-filled and there is a large gap between the valence band and the conduction band.

(ii) In the case of a conductor, the valence band is partially-filled or it overlaps with a higher energy, unoccupied conduction band. So, the electrons can flow easily under an applied electric field.

On the other hand, the valence band of a semiconductor is filled and there is a small gap between the valence band and the next higher conduction band. Therefore, some electrons can jump from the valence band to the conduction band and conduct electricity.

Q.53 Gold (atomic radius = 0.144 nm) crystallizes in a face-centred unit cell. What is the length of a side of the cell?

Ans.

For a face-centred unit cell:
a= √2 r
It is given that the atomic radius, r = 0.144 nm
So, a=√2 × 0.144
=0.407nm
Hence, length of a side of the cell =0.407 nm

Q.54 Ferric oxide crystallises in a hexagonal close-packed array of oxide ions with two out of every three octahedral holes occupied by ferric ions. Derive the formula of the ferric oxide.

Ans.

Let the number of oxide (O2−) ions be x.

So, number of octahedral voids = x
It is given that two out of every three octahedral holes are occupied by ferric ions.
So, number of ferric (Fe3+) ions=2/3 x
Therefore, ratio of the number of Fe3+ ions to the number of O2− ions,
Fe3+ : O2- = 2/3 x : x
= 2/3 : 1
= 2 : 3
Hence, the formula of the ferric oxide is Fe2O3.

Q.55 How much electricity in terms of Faraday is required to produce

(i) 20.0 g of Ca from molten CaCl2.
(ii) 40.0 g of Al from molten Al2O3.

Ans.

(i)

,Ca2++ 2e –Ca                            40gElectricity required to produce 40g of calcium = 2F Electricity required to produce 20g of calcium = 2 × 2040FTherefore, electricity required to produce 20g of calcium = 1F(ii)

,Al3++ 3e Al                           27gElectricity required to produce 27g of Al = 3F= 3×4027 FTherefore, electricity required to produce 40g of Al = 4.44 F.

Q.56 Non-stoichiometric cuprous oxide, Cu2O can be prepared in laboratory. In this oxide, copper to oxygen ratio is slightly less than 2:1. Can you account for the fact that this substance is a p-type semiconductor?

Ans.

In the cuprous oxide (Cu2O) prepared in the laboratory, copper to oxygen ratio is slightly less than 2:1. This means that the number of Cu+ ions is slightly less than twice the number of O2− ions. This is because some Cu+ ions have been replaced by Cu2+ ions.

Every Cu2+ ion replaces two Cu+ ions, thereby creating holes. As a result, the substance conducts electricity with the help of these positive holes. Hence, the substance is a p-type semiconductor.

Q.57

How much charge is required for the following reductions: ( i ) 1 mol of Al 3+ to Al. ( ii ) 1 mol of Cu 2+ to Cu. ( iii ) 1 mol of MnO 4 to Mn 2+ . MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVC0dg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pse9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeaacaGaaiaabaqaamaabaabaaGceaqabeaaqaaaaa aaaaWdbiaabIeacaqGVbGaae4DaiaabccacaqGTbGaaeyDaiaaboga caqGObGaaeiiaiaabogacaqGObGaaeyyaiaabkhacaqGNbGaaeyzai aabccacaqGPbGaae4CaiaabccacaqGYbGaaeyzaiaabghacaqG1bGa aeyAaiaabkhacaqGLbGaaeizaiaabccacaqGMbGaae4Baiaabkhaca qGGaGaaeiDaiaabIgacaqGLbGaaeiiaiaabAgacaqGVbGaaeiBaiaa bYgacaqGVbGaae4DaiaabMgacaqGUbGaae4zaiaabccacaqGYbGaae yzaiaabsgacaqG1bGaae4yaiaabshacaqGPbGaae4Baiaab6gacaqG ZbGaaeOoaaWdaeaadaqadaqaa8qacaqGPbaapaGaayjkaiaawMcaa8 qacaqGGcGaaeymaiaabccacaqGTbGaae4BaiaabYgacaqGGaGaae4B aiaabAgacaqGGcGaaeiiaiaabgeacaqGSbWdamaaCaaaleqabaWdbi aabodacaqGRaaaaOGaaeiOaiaabshacaqGVbGaaeiiaiaabgeacaqG SbGaaeOlaaWdaeaadaqadaqaa8qacaqGPbGaaeyAaaWdaiaawIcaca GLPaaapeGaaeiOaiaabgdacaqGGaGaaeyBaiaab+gacaqGSbGaaeii aiaab+gacaqGMbGaaeiiaiaaboeacaqG1bWdamaaCaaaleqabaWdbi aabkdacaqGRaaaaOGaaeiOaiaabshacaqGVbGaaeiiaiaaboeacaqG 1bGaaeOlaaWdaeaadaqadaqaa8qacaqGPbGaaeyAaiaabMgaa8aaca GLOaGaayzkaaWdbiaabckacaqGXaGaaeiiaiaab2gacaqGVbGaaeiB aiaabccacaqGVbGaaeOzaiaabckacaqGGaGaaeytaiaab6gacaqGpb Waa0baaSqaaiaabsdaaeaacaqGtacaaOGaaeiiaiaabshacaqGVbGa aeiiaiaab2eacaqGUbWdamaaCaaaleqabaWdbiaabkdacaqGRaaaaO GaaeOlaaaaaa@AD5A@

Ans.

(i) Al 3+ +3 e Al Required charge = 3F = 3 × 96487 C = 289461C (ii) Cu 2+ +2 e Cu Required charge = 2 F = 2 × 96487 C = 192974C (iii) MnO 4 Mn 2+ i.e ., Mn 7+ + 5 e Mn 2+ Required charge = 5 F = 5 × 96487 C = 482435C MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVC0dg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pse9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeaacaGaaiaabaqaamaabaabaaGceaqabeaaqaaaaa aaaaWdbiaabIcacaqGPbGaaeykaiaabccacaqGbbGaaeiBamaaCaaa leqabaGaaG4maiabgUcaRaaakiaabUcacaqGZaWdamaaDaaaleaape GaaeyzaaWdaeaapeGaae4eGaaak8aacaaMe8+dbiabgkziUkaaysW7 caqGbbGaaeiBaaqaaiabgsJiCjaaysW7caaMe8UaaeOuaiaabwgaca qGXbGaaeyDaiaabMgacaqGYbGaaeyzaiaabsgacaqGGaGaae4yaiaa bIgacaqGHbGaaeOCaiaabEgacaqGLbGaaeiiaiaab2dacaqGGaGaae 4maiaabAeaaeaacaqG9aGaaeiiaiaabodapaGaaeiiaiaabEnacaqG GaWdbiaabMdacaqG2aGaaeinaiaabIdacaqG3aGaaeiiaiaaboeaca qGGaaabaGaaeypaiaabccacaqGYaGaaeioaiaabMdacaqG0aGaaeOn aiaabgdacaqGdbaabaGaaeikaiaabMgacaqGPbGaaeykaiaaysW7ca qGdbGaaeyDamaaCaaaleqabaGaaGOmaiabgUcaRaaakiaabUcacaqG YaWdamaaDaaaleaapeGaaeyzaaWdaeaapeGaae4eGaaakiabgkziUk aaysW7caqGdbGaaeyDaaqaaiabgsJiCjaaysW7caaMe8UaaeOuaiaa bwgacaqGXbGaaeyDaiaabMgacaqGYbGaaeyzaiaabsgacaqGGaGaae 4yaiaabIgacaqGHbGaaeOCaiaabEgacaqGLbGaaeiiaiaab2dacaqG GaGaaeOmaiaabccacaqGgbGaaeiiaiaab2dacaqGGaGaaeOma8aaca qGGaGaae41aiaabccapeGaaeyoaiaabAdacaqG0aGaaeioaiaabEda caqGGaGaae4qaaqaaiaabccacaqG9aGaaeiiaiaabgdacaqG5aGaae OmaiaabMdacaqG3aGaaeinaiaaboeaaeaacaqGOaGaaeyAaiaabMga caqGPbGaaeykaiaabccacaqGnbGaaeOBaiaab+eadaqhaaWcbaGaaG inaaqaaiaacobiaaGccqGHsgIRcaqGGaGaaeytaiaab6gadaahaaWc beqaaiaaikdacqGHRaWkaaaakeaacaqGPbGaaeOlaiaabwgacaqGUa GaaeilaiaabccacaqGnbGaaeOBamaaCaaaleqabaGaaG4naiabgUca RaaakiaabccacaqGRaGaaeiiaiaabwdadaqhaaWcbaGaamyzaaqaai aacobiaaGccqGHsgIRcaqGnbGaaeOBamaaCaaaleqabaGaaGOmaiab gUcaRaaaaOqaaiaabkfacaqGLbGaaeyCaiaabwhacaqGPbGaaeOCai aabwgacaqGKbGaaeiiaiaabogacaqGObGaaeyyaiaabkhacaqGNbGa aeyzaiaabccacaqG9aGaaeiiaiaabwdacaqGGaGaaeOraiaabccaae aacaqG9aGaaeiiaiaabwdapaGaaeiiaiaabEnacaqGGaWdbiaabMda caqG2aGaaeinaiaabIdacaqG3aGaaeiiaiaaboeacaqGGaaabaGaae ypaiaabccacaqG0aGaaeioaiaabkdacaqG0aGaae4maiaabwdacaqG dbaaaaa@EA0B@

Q.58 Analysis shows that nickel oxide has the formula Ni0.98O1.00. What fractions of nickel exist as Ni2+ and Ni3+ ions?

Ans.

The formula of nickel oxide is Ni0.98O1.00.

Therefore, the ratio of the number of Ni atoms to the number of O atoms,

Ni : O = 0.98 : 1.00 = 98 : 100

Now, total charge on 100 O2−ions = 100 × (−2)

= −200

Let the number of Ni2+ ions be x.

So, the number of Ni3+ ions is 98 − x.

Now, total charge on Ni2+ ions = x(+2)

= +2x

And, total charge on Ni3+ ions = (98 − x)(+3)

= 294 − 3x

Since, the compound is neutral, we can write:

2x + (294 − 3x) + (−200) = 0

⇒ −x + 94 = 0

⇒ x = 94

Therefore, number of Ni2+ ions = 94

And, number of Ni3+ ions = 98 – 94 = 4

Hence, fraction of nickel that exists as Ni2+ = 94/98
= 0.959
And, fraction of nickel that exists as Ni3+ = 4/98
= 0.041
Alternatively, fraction of nickel that exists as Ni3+ = 1 − 0.959
= 0.041

Q.59 Conductivity of 0.00241 M acetic acid is 7.896 × 10-5 S cm-1. Calculate its molar conductivity and if ∧m0 for acetic acid is 390.5 S cm2 mol-1, what is its dissociation constant?

Ans.

Given, κ = 7.896 × 10–5 Scm–1c = 0.00241 mol L–1Then, molar conductivity, Λm = KC= 7.896 × 10–5 Scm–10.00241 mol L–1 × 1000 cm3L= 32.76Scm2mol1Again,Λm0= 390.5Scm2mol1α = ΛmΛm0 = 32.76 Scm2 mol–1390.5 Scm2 mol–1Now,=0.084Dissociation constant,Ka=2(1α)= (0.00241 molL–1)(0.084)2(1 – 0.084)=1.86×105molL1

Q.60 Copper crystallises into a fcc lattice with edge length 3.61×10−8 cm. Show that the calculated density is in agreement with its measured value of 8.92 g cm−3.

Ans.

Edge length, a = 3.61 × 10−8 cm

As the lattice is fcc type, the number of atoms per unit cell, z=4

Atomic mass, M = 63.5 g mol−1

We also know that, NA = 6.022 × 1023 mol−1

Applying the relation:

d=zMa3NA=4×63.5g mol1(3.61×108)3×6.022×1023mol1=8.97g cm3

The measured value of density is given as 8.92g cm−3. Hence, the calculated density 8.97g cm−3 is in agreement with its measured value.

Q.61 If the radius of the octachedral void is r and radius of the atoms in close packing is R, derive relation between r and R.

Ans.

A sphere with centre O, is fitted into the octahedral void as shown in the above figure. It can be observed from the figure that ∆POQ is right-angled

∠POQ = 900

Now, applying Pythagoras theorem, we can write:

PQ 2 = PO 2 + OQ 2 ( 2R ) 2 = ( R+r ) 2 + ( R+r ) 2 ( 2R ) 2 =2 ( R+r ) 2 2R 2 = ( R+r ) 2 2 R=R+r r= 2 R−R r=0.414 R MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbujxyKLgDP9MBHXgibjxyIL2yaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVy0xg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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jfaa8aabaGaeyO0H4TaaeOCaiabg2da9iaacIcacaqGsbaabaGaeyO0H4TaaiiOaiaabkhacqGH9aqpcaaIWaGaaiOlaiaabsdacaqGXaGaaeinaiaabkfaaaaa@932B@

Q.62 Niobium crystallizes in body-centred cubic structure. If density is 8.55g cm−3, calculate atomic radius of niobium using its atomic mass 93 u.

Ans.

It is given that the density of niobium, d = 8.55g cm−3

Atomic mass, M = 93g mol−1

As the lattice is bcc type, the number of atoms per unit cell, z = 2

We also know that, NA = 6.022 × 1023 mol−1

Applying the relation:

d=zMa3NAa3=zMdNA=2×93gmol18.55gcm3×6.022×1023mol1=3.612×1023cm3so, a=3.306×108cmFor body-centred cubic unit cell:r=34a=34×3.306×108cm= 1.432 × 109 cm= .1432 nm

Q.63 Silver crystallises in fcc lattice. If edge length of the cell is 4.07×10−8 cm and density is 10.5g cm−3, calculate the atomic mass of silver.

Ans.

It is given that the edge length, a = 4.077 × 10−8 cm

Density, d = 10.5g cm−3

As the lattice is fcc type, the number of atoms per unit cell, z=4

We also know that, NA = 6.022 × 1023 mol−1

Using the relation:

d=zMa3NAM=da3NAz=10.5gcm3×(4.077×108cm)3×6.022×1023mol14=107.13gmol1Therefore, atomic mass of silver =107.13u

Q.64 Calculate the efficiency of packing in case of a metal crystal for

(i) simple cubic
(ii) body-centred cubic
(iii) face-centred cubic (with the assumptions that atoms are touching each other).

Ans.

(i) Simple cubic

In a simple cubic lattice, the particles are located only at the corners of the cube and touch each other along the edge.

Let the edge length of the cube be ‘a’ and the radius of each particle be r.

So, we can write:

a = 2r

Now, volume of the cubic unit cell =a3

=(2r)3

=8r3

We know that the number of particles per unit cell is 1.

=43πr3Therefore, packing efficiency=Valume of one particleVolume of cubic unit cell×100%Hence, packing efficiency=16π×100%=16×227×100%=52.4%

(ii) Body-centred cubic

It can be observed from the above figure that the atom at the centre is in contact with the other two atoms diagonally arranged.

From ∆FED, we have:

b2=a2+a2b2=2a2b=2aAgain, from ΔAFD, we have:c2=a2+b2c2=a2+2a2(Since b2=2a2)c2=3a2c=3 Let the radiusoftheatomber.Lengthofthebodydiagonal, c = 4r3 a=4ra=4r3orr=3a4Volume of cube,a3=(4r3)3A bodycentred cubic lattice contains 2 atoms.So, volume of the occupied cubic lattice=2π43r3=83πr3Packingefficiency=volumeoccupiedbytwospheresintheunitcellTotalvolumeoftheunitcell×100%=83πr3(43r)3×100%=83πr36433r3×100%=68%

(iii) Face-centred cubic

Let the edge length of the unit cell be ‘a’ and the length of the face diagonal AC be b.

From ΔABC, we have:AC2=BC2+AB2b2=a2+a2b2=2a2b=2 aLet r be the radius of the atom.Now, from the figure, it can be observed that:b=4r2a=4ra=2r2Now, volume of the cube,a3=(2r2)3We know that the number of atoms per unit cell is 4.So, volume of the occupied unit cell=4π43r3Packingefficiency=VolumeoccupiedbysheresintheunitcellTotalvolumeoftheunitcell×100%=4π43r3(2r2)3×100%=163πr32r316×100%=74%

Q.65 The conductivity of sodium chloride at 298 K has been determined at different concentrations and the results are given below:

Concentration/M 0.001 0.010 0.020 0.050 0.100

102 × κ/S m-1 1.237 11.85 23.15 55.53 106.74

Calculate ∧m for all concentration and draw a plot between ∧m and c½. Find the value of ∧m0.

Ans.

Given,
κ = 1.237 × 10-2 S m-1, c = 0.001 M
Then, κ = 1.237 × 10-4 S cm-1, c½ = 0.0316 M1/2
∴∧m = κ/C
=(1.237 x 10-4 Scm-1 x 1000cm3/ 0.001 mol L-1 x L
=123.7 Scm2 L-1
Given,
κ = 11.85 × 10-2 S m-1, c = 0.010M
Then, κ = 11.85 × 10-4 S cm-1, c½ = 0.1 M1/2
∴∧m = κ/C
=118.5 Scm2 L-1
Given,
κ = 23.15 × 10-2 S m-1, c = 0.020 M
Then, κ = 23.15 × 10-4 S cm-1, c1/2 = 0.1414 M1/2
∴∧m = κ/CL
=115.8 Scm2 L-1
Given,
K = 55.53 × 10-2 S m-1, c = 0.050 M
Then, κ = 55.53 × 10-4 S cm-1, c1/2 = 0.2236 M1/2
∴∧m = κ/C
=111.1 1 Scm2 L-1
Given,
κ = 106.74 × 10-2 S m-1, c = 0.100 M
Then, κ = 106.74 × 10-4 S cm-1, c1/2 = 0.3162 M1/2
∴∧m = κ/C
=106.74 Scm2 L-1
Now, we have the following data:

C1/2 ⁄M1/2 0.0316 0.1 0.1414 0.2236 0.3162

m (Scm2 mol-1) 123.7 118.5 115.8 111.1 106.74

Q.66 The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S cm-1. Calculate its molar conductivity.

Ans.

Given,κ = 0.0248 S cm1c = 0.20M   Molar conductivity,m=k ×1000c                              =0.0248×10000.2                              = 124S cm2mol1

Q.67 ‘Stability of a crystal is reflected in the magnitude of its melting point’. Comment Collect melting points of solid water, ethyl alcohol, diethyl ether and methane from a data book. What can you say about the intermolecular forces between these molecules?

Ans.

Higher the melting point, greater is the intermolecular force of attraction and greater is the stability. A substance with higher melting point is more stable than a substance with lower melting point.

The melting points of the given substances are:

Solid water → 273 K

Ethyl alcohol → 158.8 K

Diethyl ether → 156.85 K

Methane → 89.34 K

Now, on observing the values of the melting points, it can be said that among the given substances, the intermolecular force in solid water is the strongest and that in methane is the weakest.

Q.68 Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.

Ans.

Conductivity of a solution is defined as the conductance of an electrolyte in which electrodes are unit distance apart and have area of cross section between them as unity. If ρ is resistivity, then we can write:

κ=1ρ

The conductivity of a solution at any given concentration is the conductance (G) of one unit volume of solution kept between two platinum electrodes with the unit area of cross-section and at a distance of unit length.

i.e., G=kal=k.1=k

(Since a = 1, l = 1)

Conductivity always decreases with a decrease in concentration, both for weak and strong electrolytes. This is because the number of ions per unit volume that carry the current in a solution decreases with a decrease in concentration.

Molar conductivity:

Molar conductivity of a solution at a given concentration is the conductance of volume V of a solution containing 1 mole of the electrolyte kept between two electrodes with the area of cross-section A and distance of unit length.

m=kAl

Now, l = 1 and A = V (volume containing 1 mole of the electrolyte).

    m=kV

Molar conductivity increases with a decrease in concentration. This is because the total volume V of the solution containing one mole of the electrolyte increases on dilution.

The variation of ∧_m with √C for strong and weak electrolytes is shown in the following plot:

Q.69 How can you determine the atomic mass of an unknown metal if you know its density and the dimension of its unit cell? Explain.

Ans.

By knowing the density of an unknown metal and the dimension of its unit cell, the atomic mass of the metal can be determined.

Let ‘a’ be the edge length of a unit cell of a crystal, ‘d’ be the density of the metal, ‘m’ be the atomic mass of the metal and ‘z’ be the number of atoms in the unit cell.

Now, density of the unit cell = mass of the unit cell / volume of the unit cell

d= Zm a 3 ( i ) [ Since mass of the unit cell = Number of atoms in the unit cell ×Atomic mass ]

Q.70 The rate constant for the first order decomposition of H2O2 is given by the following equation:
log k = 14.34 − 1.25 × 104 K/T
Calculate Ea for this reaction and at what temperature will its half-period be 256 minutes?

Ans.

Q.71 The following data were obtained during the first order thermal decomposition of SO2Cl2 at a constant volume.

SO2Cl2(g)SO2(g)+ Cl2(g)

Experiment Time/s-1 Total pressure/atm
1 0 0.5
2 100 0.6

 Calculate the rate of the reaction when total pressure is 0.65 atm.

Ans.

Q.72 A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is
(i) doubled
(ii) reduced to half?

Ans.

Let the concentration of the reactant be [A] = a

Rate of reaction, R = k [A]2

= ka2

(i) If the concentration of the reactant is doubled, i.e. [A] = 2a, then the rate of the reaction would be

R’= k(2a)2

= 4ka2

=4R

Therefore, the rate of the reaction would increase by 4 times.

iiIf the concentration of the reactant is reduced to half, i.e. A = 12a then the rate of the reaction would be R*=k12a2=14=14RTherefore, the rate of the reaction would be reduced to =1th4

Q.73 Mention the factors that affect the rate of a chemical reaction.

Ans.

The factors that affect the rate of a reaction are as follows.

(i) Concentration of reactants (pressure in case of gases)

(ii) Temperature

(iii) Nature of reactants

(iv) Presence of a catalyst

(v) Surface area

Q.74 The decomposition of dimethyl ether leads to the formation of CH4, H2 and CO and the reaction rate is given by Rate = k [CH3OCH3]3/2
The rate of reaction is followed by increase in pressure in a closed vessel, so the rate can also be expressed in terms of the partial pressure of dimethyl ether, i.e.,

Rate=k ( P CH 3 OCH 3 ) 3/2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbujxyKLgDP9MBHXgibjxyIL2yaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVy0xg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeaacaGaaiaabeWaaqaabaqbaaGcbaaeaaaaaaaaa8qacaqGsbGaaeyyaiaabshacaqGLbGaeyypa0Jaae4saiaabckacaGGOaGaaeiua8aadaWgaaWcbaWdbiaaboeacaqGibWdamaaBaaameaapeGaaG4maaWdaeqaaSWdbiaab+eacaqGdbGaaeisa8aadaWgaaadbaWdbiaaiodaa8aabeaaaSqabaGcpeGaaiyka8aadaahaaWcbeqaa8qacaaIZaGaai4laiaaikdaaaaaaa@4EEA@

If the pressure is measured in bar and time in minutes, then what are the units of rate and rate constants?

Ans.

If pressure in measured in bar and time in minutes, then

Unit of rate = bar min-1

Q.75 The decomposition of NH3 on platinum surface is zero order reaction. What are the rates of production of N2 and H2 if k = 2.5 × 10-4 mol-1 L s-1?

Ans.

Q.76 For the reaction:
2A + B → A2B
The rate = k[A][B]2 with k = 2.0 × 10-6 mol-2 L2 s-1. Calculate the initial rate of the reaction when [A] = 0.1 mol L-1, [B] = 0.2 mol L-1. Calculate the rate of reaction after [A] is reduced to 0.06 mol L-1.

Ans.

The initial rate of the reaction is

Rate = k [A][B]2

= (2.0 × 10-6 mol-2 L2 s-1) (0.1 mol L-1) (0.2 mol L-1) 2

= 8.0 × 10-9 mol L-1 s-1

When [A] is reduced from 0.1 mol L-1 to 0.06 mol-1, the concentration of A reacted = (0.1 − 0.06) mol L-1 = 0.04 mol L-1

Therefore, concentration of B reacted

= 1 2 ×0.04 mo l 1 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbujxyKLgDP9MBHXgibjxyIL2yaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVy0xg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeaacaGaaiaabeWaaqaabaqbaaGcbaaeaaaaaaaaa8qacqGH9aqpdaWcaaWdaeaapeGaaGymaaWdaeaapeGaaGOmaaaacqGHxdaTcaaIWaGaaiOlaiaaicdacaaI0aacbmGaa8hOaiaa=1gacaWFVbGaa8hBa8aadaahaaWcbeqaa8qacqGHsislcaaIXaaaaaaa@4ABA@

= 0.02 mol L-1

Then, concentration of B available, [B] = (0.2 − 0.02) mol L-1 = 0.18 mol L-1

After [A] is reduced to 0.06 mol L-1, the rate of the reaction is given by, Rate = k [A][B]2

= (2.0 × 10-6 mol-2 L2 s-1) (0.06 mol L-1) (0.18 mol L-1) 2

= 3.89×10-9mol L-1 s-1

Q.77 From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants.

(i) 3 NO(g) → N2O (g); Rate = k[NO]2
(ii) H2O2 (aq) + 3 I (aq) + 2 H+ → 2 H2O (l) + I3 ; Rate = k[H2O2][I]
(iii) CH3CHO(g) → CH4 (g) + CO(g); Rate = k [CH3CHO]3/2
(iv) C2H5Cl(g) → C2H4(g) + HCl(g); Rate = k [C2H5Cl]

Ans.

(i)Given Rate=k[NO]2Therefore, order of the reaction=2k=Rate[NO]2Dimension of k=mol L-1s-1(mol L-1)2=mol L-1s-1mol2 L-2=L mol-1 s-1(ii)Given rate=k[H2O2][I]Therefore, order of the reaction=2k=Rate[H2O2][I]Dimension of k=mol L-1s-1(mol L-1)(mol L-1)=L mol-1s-1(iii)Given rate = k[CH3CHO]32=32Therefore,order of reaction= k=Rate[CH3CHO]32Dimension of k=mol L-1 s-1(mol L-1)32=mol L-1 s-1mol32 L32=L12 mol-12 s-1(iv)Given rate=k[C2H5Cl]Therefore, order of the reaction=1k=Rate[C2H5Cl]Dimension of k=mol L-1 s-1mol L-1=s-1

Q.78 Explain the terms with suitable examples:
(i) Alcosol (ii) Aerosol (iii) Hydrosol

Ans.

(i) Alcosol:

A colloidal solution having alcohol as the dispersion medium and a solid substance as the dispersed phase is called an alcosol.

For example: colloidal sol of cellulose nitrate in ethyl alcohol is an alcosol.

(ii) Aerosol:

A colloidal solution having a gas as the dispersion medium and a solid as the dispersed phase is called an aerosol.

For example: fog

(iii) Hydrosol

A colloidal solution having water as the dispersion medium and a solid as the dispersed phase is called a hydrosol.

For example: starch sol or gold sol

Q.79 Comment on the statement that “colloid is not a substance but a state of substance”.

Ans.

Common salt (a typical crystalloid in an aqueous medium) behaves as a colloid in a benzene medium. Hence, we can say that a colloidal substance does not represent a separate class of substances. When the size of the solute particle lies between 1 nm and 1000 nm, it behaves as a colloid.

Hence, we can say that colloid is not a substance but a state of the substance which is dependent on the size of the particle. A colloidal state is intermediate between a true solution and a suspension.

Q.80 What are micelles? Give an example of a micellers system.

Ans.

Micelle formation is done by substances such as soaps and detergents when dissolved in water. The molecules of such substances contain a hydrophobic and a hydrophilic part. When present in water, these substances arrange themselves in spherical structures in such a manner that their hydrophobic parts are present towards the centre, while the hydrophilic parts are pointing towards the outside (as shown in the given figure). This is known as micelle formation.

Q.81 Explain the following terms:
(i) Electrophoresis (ii) Coagulation
(iii) Dialysis (iv) Tyndall effect.

Ans.

(i) Electrophoresis:

The movement of colloidal particles under the influence of an applied electric field is known as electrophoresis. Positively charged particles move to the cathode, while negatively charged particles move towards the anode. As the particles reach oppositely charged electrodes, they become neutral and get coagulated.

(ii) Coagulation:

The process of settling down of colloidal particles i.e., conversion of a colloid into a precipitate is called coagulation.

(iii) Dialysis

The process of removing a dissolved substance from a colloidal solution by the means of diffusion through a membrane is known as dialysis. This process is based on the principle that ions and small molecules can pass through animal membranes unlike colloidal particles.

(iv) Tyndall effect:

When a beam of light is allowed to pass through a colloidal solution, it becomes visible like a column of light. This is known as the Tyndall effect. This phenomenon takes place as particles of colloidal dimensions scatter light in all directions.

Q.82 What is shape selective catalysis?

Ans.

A catalytic reaction which depends upon the pore structure of the catalyst and on the size of the reactant and the product molecules is called shape-selective catalysis. For example, catalysis by zeolites is a shape-selective catalysis. The pore size present in the zeolites ranges from 260-740 pm. Thus, molecules having a pore size more than this cannot enter the zeolite and undergo the reaction.

Q.83 Describe some features of catalysis by zeolites.

Ans.

Zeolites are alumino-silicates that are micro-porous in nature. Zeolites have a honeycomb-like structure, which makes them shape-selective catalysts. They have an extended 3D-network of silicates in which some silicon atoms are replaced by aluminium atoms, giving them an Al−O−Si framework. The reactions taking place in zeolites are very sensitive to the pores and cavity size of the zeolites. Zeolites are commonly used in the petrochemical industry.

Q.84 What do you mean by activity and selectivity of catalysts?

Ans.

(a) Activity of a catalyst:

The activity of a catalyst is its ability to increase the rate of a particular reaction. Chemisorption is the main factor in deciding the activity of a catalyst. The adsorption of reactants on the catalyst surface should be neither too strong nor too weak. It should just be strong enough to make the catalyst active.

(b) Selectivity of the catalyst:

The ability of the catalyst to direct a reaction to yield a particular product is referred to as the selectivity of the catalyst. For example, by using different catalysts, we can get different products for the reaction between H2 and CO.

(i)CO(g)+3H2(g)NiCH4(g)+H2O(g)(ii)CO(g)+2H2(g)CuZnOCrOCH3OH(g)(iii)CO(g)+H2(g)CuHCHO(g)

Q.85 Give four examples of heterogeneous catalysis.

Ans.

(i) Oxidation of sulphur dioxide to form sulphur trioxide. In this reaction, Pt acts as a catalyst.

2SO2(g)Pt(s)2SO3(g)(ii)Formation of ammonia by the combination of dinitrogen and dihydrogen in the presence of finely divided iron.N2(g)+3H2(g)Fe(s)2NH3(g)This process is called the Habers process.(iii)Oswalds process:Oxidation of ammonia to nitric oxide in the presence of platinum.4NH3(g)+5O2(g)Pt(s)4NO(g)+6H2O(g)(iv)Hydrogenation of vegetable oils in the presence of Ni.Vegetable oil(l)+H2(g)Ni(s)Vegetable ghee(s)

Q.86 Action of soap is due to emulsification and micelle formation. Comment.

Ans.

The cleansing action of soap is due to emulsification and micelle formation. Soaps are basically sodium and potassium salts of long chain fatty acids, R-COO-Na+. The end of the molecule to which the sodium is attached is polar in nature, while the alkyl-end is non-polar. Thus, a soap molecule contains a hydrophilic (polar) and a hydrophobic (non-polar) part.

When soap is added to water containing dirt, the soap molecules surround the dirt particles in such a manner that their hydrophobic parts get attached to the dirt molecule and the hydrophilic parts point away from the dirt molecule. This is known as micelle formation. Thus, we can say that the polar group dissolves in water while the non-polar group dissolves in the dirt particle. Now, as these micelles are negatively charged, they do not coalesce and a stable emulsion is formed.

Q.87 Give four uses of emulsions.

Ans.

Four uses of emulsions:

(i) Cleansing action of soaps is based on the formation of emulsions.

(ii) Digestion of fats in intestines takes place by the process of emulsification.

(iii) Antiseptics and disinfectants when added to water form emulsions.

(iv) The process of emulsification is used to make medicines.

Q.88 What are emulsions? What are their different types? Give example of each type.

Ans.

The colloidal solution in which both the dispersed phase and dispersion medium are liquids is called an emulsion. There are two types of emulsions:

(a) Oil in water type:

Here, oil is the dispersed phase while water is the dispersion medium. For example: milk, vanishing cream, etc.

(b) Water in oil type:
Here, water is the dispersed phase while oil is the dispersion medium. For example: cold cream, butter, etc.

Q.89 Explain what is observed

(i) When a beam of light is passed through a colloidal sol.

(ii) An electrolyte, NaCl is added to hydrated ferric oxide sol.

(iii) Electric current is passed through a colloidal sol?

Ans.

(i) When a beam of light is passed through a colloidal solution, then scattering of light is observed. This is known as the Tyndall effect. This scattering of light illuminates the path of the beam in the colloidal solution.

(ii) When NaCl is added to ferric oxide sol, it dissociates to give Na+ and Cl ions. Particles of ferric oxide sol are positively charged. Thus, they get coagulated in the presence of negatively charged Cl ions.

(iii) The colloidal particles are charged and carry either a positive or negative charge. The dispersion medium carries an equal and opposite charge. This makes the whole system neutral. Under the influence of an electric current, the colloidal particles move towards the oppositely charged electrode. When they come in contact with the electrode, they lose their charge and coagulate.

Q.90 How are colloids classified on the basis of
(i) Physical states of components
(ii) Nature of dispersion medium and
(iii) Interaction between dispersed phase and dispersion medium?

Ans.

Colloids can be classified on various bases:

(i) On the basis of the physical state of the components (by components we mean the dispersed phase and dispersion medium). Depending on whether the components are solids, liquids, or gases, we can have eight types of colloids.

(ii) On the basis of the dispersion medium, sols can be divided as:

Dispersion medium Name of sol
Water Aquasol or hydrosol
Alcohol Alcosol
Benzene Benzosol
Gases Aerosol

(iii) On the basis of the nature of the interaction between the dispersed phase and dispersion medium, the colloids can be classified as lyophilic (solvent attracting) and lyophobic (solvent repelling).

Q.91 What are enzymes? Write in brief the mechanism of enzyme catalysis.

Ans.

Enzymes are basically protein molecules of high molecular masses. These form colloidal solutions when dissolved in water. These are complex, nitrogenous organic compounds produced by living plants and animals. Enzymes are also called ‘biochemical catalysts’.

Mechanism of enzyme catalysis:

On the surface of the enzymes, various cavities are present with characteristic shapes. These cavities possess active groups such as −NH2, −COOH, etc. The reactant molecules having a complementary shape fit into the cavities just like a key fits into a lock. This leads to the formation of an activated complex. This complex then decomposes to give the product.

Hence,

Step 1: E + S → ES

(Activated complex)

Step 2: ES → E + P

Q.92 What is the difference between multimolecular and macromolecular colloids? Give one example of each. How are associated colloids different from these two types of colloids?

Ans.

(i) In multi-molecular colloids, the colloidal particles are an aggregate of atoms or small molecules with a diameter of less than 1 nm. The molecules in the aggregate are held together by van der Waal’s forces of attraction. Examples of such colloids include gold sol and sulphur sol.

(ii) In macro-molecular colloids, the colloidal particles are large molecules having colloidal dimensions. These particles have a high molecular mass. When these particles are dissolved in a liquid, sol is obtained. For example: starch, nylon, cellulose, etc.

(iii) Certain substances tend to behave like normal electrolytes at lower concentrations. However, at higher concentrations, these substances behave as colloidal solutions due to the formation of aggregated particles. Such colloids are called aggregated colloids.

Q.93 What are lyophilic and lyophobic sols? Give one example of each type. Why are hydrophobic sols easily coagulated?

Ans.

(i) Lyophilic sols:

Colloidal sols that are formed by mixing substances such as gum, gelatin, starch, etc. with a suitable liquid (dispersion medium) are called lyophilic sols. These sols are reversible in nature i.e., if two constituents of the sol are separated by any means (such as evaporation), then the sol can be prepared again by simply mixing the dispersion medium with the dispersion phase and shaking the mixture.

(ii) Lyophobic sols:

When substances such as metals and their sulphides etc. are mixed with the dispersion medium, they do not form colloidal sols. Their colloidal sols can be prepared only by special methods. Such sols are called lyophobic sols. These sols are irreversible in nature. For example: sols of metals.

Now, the stability of hydrophilic sols depends on two things- the presence of a charge and the solvation of colloidal particles. On the other hand, the stability of hydrophobic sols is only because of the presence of a charge. Therefore, the latter are much less stable than the former. If the charge of hydrophobic sols is removed (by addition of electrolytes), then the particles present in them come closer and form aggregates, leading to precipitation.

Q.94 Discuss the effect of pressure and temperature on the adsorption of gases on solids.

Ans.

Effect of pressure

Adsorption is a reversible process and is accompanied by a decrease in pressure. Therefore, adsorption increases with an increase in pressure.

Effect of temperature

Adsorption is an exothermic process. Thus, in accordance with Le-Chatelier’s principle, the magnitude of adsorption decreases with an increase in temperature.

Q.95 How are the colloidal solutions classified on the basis of physical states of the dispersed phase and dispersion medium?

Ans.

One criterion for classifying colloids is the physical state of the dispersed phase and dispersion medium. Depending upon the type of the dispersed phase and dispersion medium (solid, liquid, or gas), there can be eight types of colloidal systems.

Dispersed phase Dispersion medium Type of colloid Example
1. Solid Solid Solid Sol Gemstone
2. Solid Liquid Sol Paint
3. Solid Gas Aerosol Smoke
4. Liquid Solid Gel Cheese
5. Liquid Liquid Emulsion Milk
6. Liquid Gas Aerosol Fog
7. Gas Solid Solid foam Pumice stone
8. Gas Liquid Foam Froth

Q.96 Why is adsorption always exothermic?

Ans.

Adsorption is always exothermic. This statement can be explained in two ways.

(i) Adsorption leads to a decrease in the residual forces on the surface of the adsorbent. This causes a decrease in the surface energy of the adsorbent. Therefore, adsorption is always exothermic.

(ii)H of adsorption is always negative. When a gas is adsorbed on a solid surface, its movement is restricted leading to a decrease in the entropy of the gas i.e., ∆S is negative. Now for a process to be spontaneous, ∆G should be negative.

ΔG=ΔHTΔS

Since ∆S is negative, ∆H has to be negative to make ∆G negative. Hence, adsorption is always exothermic.

Q.97 What role does adsorption play in heterogeneous catalysis?

A.er

Heterogeneous catalysis:

A catalytic process in which the catalyst and the reactants are present in different phases is known as a heterogeneous catalysis. This heterogeneous catalytic action can be explained in terms of the adsorption theory. The mechanism of catalysis involves the following steps:

(i) Adsorption of reactant molecules on the catalyst surface.

(ii) Occurrence of a chemical reaction through the formation of an intermediate.

(iii) De-sorption of products from the catalyst surface

(iv) Diffusion of products away from the catalyst surface. In this process, the reactants are usually present in the gaseous state and the catalyst is present in the solid state. Gaseous molecules are then adsorbed on the surface of the catalyst. As the concentration of reactants on the surface of the catalyst increases, the rate of reaction also increases. In such reactions, the products have very less affinity for the catalyst and are quickly desorbed, thereby making the surface free for other reactants.

Q.98 What do you understand by activation of adsorbent? How is it achieved?

Ans.

By activating an adsorbent, we tend to increase the adsorbing power of the adsorbent. Some ways to activate an adsorbent are:

(i) By increasing the surface area of the adsorbent. This can be done by breaking it into smaller pieces or powdering it.

(ii) Some specific treatments can also lead to the activation of the adsorbent. For example, wood charcoal is activated by heating it between 650 K and 1330 K in vacuum or air. It expels all the gases absorbed or adsorbed and thus, creates a space for adsorption of gases.

Q.99 What is an adsorption isotherm? Describe Freundlich adsorption isotherm.

Ans.

The plot between the extent of adsorption x/m against the pressure of gas (P) at constant temperature (T) is called the adsorption isotherm.

Freundlich adsorption isotherm: Freundlich adsorption isotherm gives an empirical relationship between the quantity of gas adsorbed by the unit mass of solid adsorbent and pressure at a specific temperature. From the given plot it is clear that at pressure Ps, x/m reaches the maximum value. Ps is called the saturation pressure. Three cases arise from the graph now.

Case I- At low pressure: The plot is straight and sloping, indicating that the pressure is directly proportional to


Q.100 What are the factors which influence the adsorption of a gas on a solid?

Ans.

There are various factors that affect the rate of adsorption of a gas on a solid surface.

(1) Nature of the gas:

Easily liquefiable gases such as NH3, HCl etc. are adsorbed to a great extent in comparison to gases such as H2, O2 etc. This is because Van der Waal’s forces are stronger in easily liquefiable gases.

(2) Surface area of the solid

The greater the surface area of the adsorbent, the greater is the adsorption of a gas on the solid surface.

(3) Effect of pressure

Adsorption is a reversible process and is accompanied by a decrease in pressure. Therefore, adsorption increases with an increase in pressure.

(4) Effect of temperature

Adsorption is an exothermic process. Thus, in accordance with Le-Chatelier’s principle, the magnitude of adsorption decreases with an increase in temperature.

Q.101 Give reason why a finely divided substance is more effective as an adsorbent.

Ans.

Adsorption is a surface phenomenon. Therefore, adsorption is directly proportional to the surface area. A finely divided substance has a large surface area. Both physisorption and chemisorption increase with an increase in the surface area. Hence, a finely divided substance behaves as a good adsorbent.

Q.102 What is the difference between physisorption and chemisorption?

Ans.

Physisorption Chemisorption
1. In this type of adsorption, the adsorbate is attached to the surface of the adsorbent with weak van der Waal’s forces of attraction. In this type of adsorption, strong chemical bonds are formed between the adsorbate and the surface of the adsorbent.
2. No new compound is formed in the process. New compounds are formed at the surface of the adsorbent.
3. It is generally found to be reversible in nature. It is usually irreversible in nature.
4. Enthalpy of adsorption is low as weak van der Waal’s forces of attraction are involved. The values lie in the range of 20-40 kJmol -1. Enthalpy of adsorption is high as chemical bonds are formed. The values lie in the range of 40-400 kJmol-1.
5. It is favoured by low temperature conditions. It is favoured by high temperature conditions.
6. It is an example of multi-layer adsorption. It is an example of mono-layer adsorption.

Q.103 Distinguish between the meaning of the terms adsorption and absorption.

Give one example of each.

Ans.

Adsorption is a surface phenomenon of accumulation of molecules of a substance at the surface rather than in the bulk of a solid or liquid. The substance that gets adsorbed is called the ‘adsorbate’ and the substance on whose surface the adsorption takes place is called the ‘adsorbent’. Here, the concentration of the adsorbate on the surface of the adsorbent increases. In adsorption, the substance gets concentrated at the surface only. It does not penetrate through the surface to the bulk of the solid or liquid. For example, when we dip a chalk stick into an ink solution, only its surface becomes coloured. If we break the chalk stick, it will be found to be white from inside.

On the other hand, the process of absorption is a bulk phenomenon. In absorption, the substance gets uniformly distributed throughout the bulk of the solid or liquid.

Q.104 How do emulsifiers stabilise emulsion? Name two emulsifiers.

Ans.

An emulsion is a colloidal solution of two immiscible liquids in which one of the liquids acts as the dispersed phase and the other acts as the dispersion medium. An emulsifiers or emulsifying agent is a substance that is added to an emulsion to stabilise it. For example, proteins, gums, natural and synthetic soaps are the emulsifier for the oil in water emulsions. Heavy metal salts of fatty acids, long chain alcohols, lamp-black are the emulsifiers for the water in oil emulsions.

Soaps and detergents are the commonly used emulsifiers.

Q.105 Predict conditions under which Al might be expected to reduce MgO.

Ans.

Above 1350°C, the standard Gibbs free energy of formation of Al2O3 from Al is less than that of MgO from Mg. Therefore, above 1350°C, Al can reduce MgO.

Q.106 Outline the principles of refining of metals by the following methods:
(i) Zone refining
(ii) Electrolytic refining
(iii) Vapour phase refining

Ans.

(i) Zone refining:

This method is based on the principle that impurities are more soluble in the molten state of metal (the melt) than in the solid state. In the process of zone refining, a circular mobile heater is fixed at one end of a rod of impure metal. As the heater moves, the molten zone of the rod also moves along with it. As a result, pure metal crystallizes out of the melt and the impurities pass to the adjacent molten zone. This process is repeated several times, which leads to the segregation of impurities at one end of the rod. Then, the end with the impurities is cut off. Silicon, boron, gallium, indium etc. can be purified by this process.

(ii) Electrolytic refining:

Electrolytic refining is the process of refining impure metals by using electricity. In this process, impure metal is made the anode and a strip of pure metal is made the cathode. A solution of a soluble salt of the same metal is taken as the electrolyte. When an electric current is passed, metal ions from the electrolyte are deposited at the cathode as pure metal and the impure metal from the anode dissolves into the electrolyte in the form of ions. The impurities present in the impure metal gets collected below the anode. This is known as anode mud.

Anode :MMn++neCathode : Mn++neM

(iii) Vapour phase refining :

Vapour phase refining is the process of refining metal by converting it into its volatile compound and then, decomposing it to obtain a pure metal. To carry out this process,

(i) the metal should form a volatile compound with an available reagent, and

(ii) the volatile compound should be easily decomposable so that the metal can be easily recovered.

Nickel, zirconium, and titanium are refined using this method.

Q.107 What is the role of graphite rod in the electrometallurgy of aluminium?

Ans.

In the electrometallurgy of aluminium, a fused mixture of purified alumina (Al2O3), cryolite (Na3AlF6) and fluorspar (CaF2) is electrolysed. In this electrolysis, graphite is used as the anode and graphite-lined iron is used as the cathode. During the electrolysis, Al is liberated at the cathode, while CO and CO2 are liberated at the anode, according to the following equation.

Cathode: Al3+(melt) + 3e→ Al(l)

Anode: C(s) + O2- (melt) → CO(g) + 2e

C(s) + 2O2- (melt) → CO2(g) + 4e

Q.108 Name the processes from which chlorine is obtained as a by-product. What will happen if an aqueous solution of NaCl is subjected to electrolysis?

Ans.

In the electrolysis of molten NaCl, Cl2 is obtained at the anode as a by product.

NaClmelt Namelt++ ClmeltAt cathode: Namelt++e NasAt anode: Clmelt+ 1/2Cl2g+ e2Clg Cl2gThe overall reaction is as follows:NaClmelt ElectrolysisNas+ 12Cl2gIf an aqueous solution of NaCl is electrolyzed, Cl2 will be obtained at the anode but at the cathode, H2​will be obtained instead of Na. This is because the standard reduction potential of Na E°=2.71V is more negative than of H2O E°=0.83V. Hence, H2O will get preference to get reduced at the cathode and as a result, H2 is evolved.NaClaq Naaq++ ClaqAt cathode:2H2Ol+2e H2g+ 2OHaqAt anode:Clmelt Clg+ e2Clg Cl2g

Q.109 The choice of a reducing agent in a particular case depends on thermodynamic factor. How far do you agree with this statement? Support your opinion with two examples.

Ans.

The above figure is a plot of Gibbs energy (∆GΦ) vs. T for formation of some oxides. It can be observed from the above graph that a metal can reduce the oxide of other metals, if the standard free energy of formation (∆fGΦ) of the oxide of the former is more negative than the latter. For example, since ∆fGΦ(Al, Al2,O3) is more negative than, ∆fGΦ(Cu, Cu2,O), Al can reduce Cu2O to Cu, but Cu cannot reduce Al2O3. Similarly, Mg can reduce ZnO to Zn, but Zn cannot reduce MgO because ∆fGΦ(Mg, MgO) is more negative than ∆fGΦ(Zn, ZnO)

Q.110 Out of C and CO, which is a better reducing agent for ZnO?

Ans.

Reduction of ZnO to Zn is usually carried out at 1673 K. From the above figure, it can be observed that above 1073 K, the Gibbs free energy of formation of CO from C and above 1273 K, the Gibbs free energy of formation of CO2 from C is lesser than the Gibbs free energy of formation of ZnO. Therefore, C can easily reduce ZnO to Zn. On the other hand, the Gibbs free energy of formation of CO2 from CO is always higher than the Gibbs free energy of formation of ZnO. Therefore, CO cannot reduce ZnO. Hence, C is a better reducing agent than CO for reducing ZnO.

Q.112 The value of ∆fGΦ for formation of Cr2O3 is − 540 kJmol-1 and that of Al2 O3 is – 827 kJmol-1. Is the reduction of Cr2O3 possible with Al?

Ans.

The value of ∆fGΦ for the formation of Cr2O3 from Cr (−540 kJmol-1) is higher than that of Al2O3 from Al (−827 kJmol-1). Therefore, Al can reduce Cr2O3 to Cr. Hence, the reduction of Cr2O3 with Al is possible.

Alternatively,

Al+32O2Al2O3 ΔfG=827 kJmol12Cr+32O2Cr2O3 ΔfG=540 kJmol1Subtracting equation(ii)from(i),we have2Al+Cr2O3Al2O3+2CrΔfG=827(540)=287 kJmol1

As ∆fGΦ for the reduction reaction of Cr2O3 by Al is negative, this reaction is possible.

Q.113 Why is zinc not extracted from zinc oxide through reduction using CO?

Ans.

The standard Gibbs free energy of formation of zinc oxide from zinc is lower than that of CO2 from CO. Therefore, CO cannot reduce zinc oxide to zinc. Hence, zinc is not extracted from zinc oxide through reduction using CO.

Q.114 How is leaching carried out in case of low grade copper ores?

Ans.

In case of low grade copper ores, leaching is carried out using acid or bacteria in the presence of air. In this process, copper goes into the solution as Cu2+ ions.

Cu(s)+2H(aq)++12O2(g)Cu(aq)2++2H2O(l)The resulting solution Is treated with scrap iron or H2to get metallic copper..Cu(aq)2++H2(g)Cu(s)+2H(aq)+

Q.115 What is the role of cryolite in the metallurgy of aluminium?

Ans.

Cryolite (Na3AlF6) has two roles in the metallurgy of aluminium:

1. To decrease the melting point of the mixture from 2323 K to 1140 K.

2. To increase the electrical conductivity of Al2O3.

Q.116 Why copper matte is put in silica lined converter?

Ans.

Copper matte contains Cu2S and FeS. Copper matte is put in a silica-lined converter to remove the remaining FeO and FeS present in the matte as slag (FeSiO3). Also, some silica is added to the silica-lined converter. Then, a hot air blast is blown. As a result, the remaining FeS and FeO are converted to iron silicate (FeSiO3) and Cu2S is converted into metallic copper.

2FeS+3O22FeO+2SO2FeO+SiO2FeSiO32Cu2S+3O22Cu2O+2SO22Cu2O+Cu2S 6Cu+SO2

Q.117 Differentiate between “minerals” and “ores”.

Ans.

Minerals are naturally occurring chemical substances containing metals. They are found in the Earth’s crust and are obtained by mining.

Ores are rocks and minerals viable to be used as a source of metal.

For example, there are many minerals containing zinc, but zinc cannot be extracted profitably (conveniently and economically) from all these minerals. Zinc can be obtained from zinc blende (ZnS), calamine (ZnCO3), Zincite (ZnO) etc. Thus, these minerals are called ores of zinc.

Q.118 How is ‘cast iron’ different from ‘pig iron”?

Ans.

The iron obtained from blast furnaces is known as pig iron. It contains around 4% carbon and many impurities such as S, P, Si, Mn in smaller amounts.

Cast iron is obtained by melting pig iron and coke using a hot air blast. It contains a lower amount of carbon (3%) than pig iron. Unlike pig iron, cast iron is extremely hard and brittle.

Q.119 Giving examples, differentiate between ‘roasting’ and ‘calcination’.

Ans.

Roasting is the process of converting sulphide ores to oxides by heating the ores in a regular supply of air at a temperature below the melting point of the metal. For example, sulphide ores of Zn, Pb, and Cu are converted to their respective oxides by this process.

2ZnsZinc blende+3O2Δ2ZnO+2SO22PbSGalena+3O2Δ2PbO+2SO22Cu2SGalena+3O2Δ2Cu2O+2SO2On the other hand, calcination is the process of converting hydroxide and carbonate ones to oxides by heating the ones either in the absence or in a limited supply of air at a temperature below the melting point of the metal.This process causes the escaping of volatile matter leaving behind the metal oxide. For example, hydroxide of Fe, carbonates of Zn, Ca, Mg are converted to their respective oxides by this process.Fe2O33H2OLimoniteΔFe2O3+3H2OZnCO3sCalamineΔZnOs+CO2gCaMgCO32DolomiteΔCaOs+MgOs2CO

Q.120 How can you separate alumina from silica in bauxite are associated with silica? Give equations, if any?

Ans.

To separate alumina from silica in bauxite are associated with silica, first the powdered one is digested with a concentrated NaOH solution at 473 – 523 K and 35 – 36 bar pressure. This results in the leaching out of alumina (Al2O3) as sodium aluminate and silica (SiO2) as sodium silicate leaving the impurities behind.

Al2O38Alumina+2NaOHaq+3H2Ol2NaAlOH4aqSodium aluminateSiO2Silica+2NaOHaqNa2SiO3aqSodium silicate+H2OlThen,CO2gas is passed through the resulting solution to neutralize the aluminate in the solution, which resultsi n the precipitation of hydrated alumina.To induce precipitation,the solution is seeded with freshly prepared samples of hydrated alumina.2NaAlOH4aq+ CO2gSodium aluminate Hydrated alumina Al2O3.xH2Os+ 2NaHCO3aq Sodiumhydrogen carbonateDuring this process, sodium silicate remains in the solution. The obtained hydrated alumina is filtered, dried, and heated to get back pure aluminaAl2O3.xH2OsHydrated alumina1470KAl2O3sAlumina+xH2Og

Q.121 Describe a method for refining nickel.

Ans.

Nickel is refined by Mond’s process. In this process, nickel is heated in the presence of carbon monoxide to form nickel tetracarbonyl, which is a volatile complex.

Ni+4CO330350KNi(CI)4NickelteracarbonylThen,the obtained nickel tetracarbonyl is decomposed by subjecting it to a higher temperature (450470K) to obtain pure nickel metal.Ni(CI)4Nickelteracarbonyl450470KNi+4CO

Q.122 Which method of refining will you suggest for an element in which impurities present have chemical properties close to the properties of that element?

Ans.

Zone refining

Q.123 State the role of silica in the metallurgy of copper.

Ans.

During the roasting of pyrite one, a mixture of FeO and Cu2O is obtained.

2CuFeS2+O2ΔCu2S+2FeS+SO22Cu2S+3O2Δ2Cu2O+2SO22FeS+3O2Δ 2FeO+2SO2The role of silica in the metallurgy of copper is to remove the iron oxide obtained during the process of roasting as slag.If the sulphide ore of copper contains iron,then silica SiO2 is added as flux before roasting.Then,FeO combines with silica to form iron silicate,FeSiO3slag.FeO+SiO2FeSiO3

Q.124 Write chemical reactions taking place in the extraction of zinc from zinc blende.

Ans.

The different steps involved in the extraction of zinc from zinc blende (ZnS) are given below:

(i) Concentration of ore

First, the gangue from zinc blende is removed by the froth floatation method.

(ii) Conversion to oxide (Roasting)

Sulphide one is converted into oxide by the process of roasting. In this process, ZnS is heated in a regular supply of air in a furnace at a temperature, which is below the melting point of Zn.

2ZnS + 3O2 → 2ZnO + 2SO2

(iii) Extraction of zinc from zinc oxide (Reduction)

Zinc is extracted from zinc oxide by the process of reduction. The reduction of zinc oxide is carried out by mixing it with powdered coke and then, heating it at 673 K.

ZnO +Ccoke,673kZn+CO

(iv) Electrolytic Refining

Zinc can be refined by the process of electrolytic refining. In this process, impure zinc is made the anode while a pure copper strip is made the cathode. The electrolyte used is an acidified solution of zinc sulphate (ZnSO4). Electrolysis results in the transfer of zinc In pure from the anode to the cathode.

Anode: Zn → Zn2+ + 2e

Cathode: Zn2+ + 2e → Zn

Q.125 Which method of refining may be more suitable if element is obtained in minute quantity?

Ans.

Column chromatograhy

Q.126 Write down the reactions taking place in different zones in the blast furnace during the extraction of iron.

Ans.

During the extraction of iron, the reduction of iron oxides takes place in the blast furnace. In this process, hot air is blown from the bottom of the furnace and coke is burnt to raise the temperature up to 2200 K in the lower portion itself. The temperature is lower in the upper part. Thus, it is the lower part where the reduction of iron oxides (Fe2O3 and Fe3O4) takes place. The reactions taking place in the lower temperature range (500 – 800 K) in the blast furnace are:

3Fe2O3 + CO → 2Fe3O4 + CO2

Fe3O4 + 4CO → 3FeO + 4CO2

Fe2O3 + CO → 2Fe3 + CO2

The reactions taking place in the higher temperature range (900 – 1500 K) in the blast furnace are:

C + CO2 → 2CO

FeO + CO → Fe + CO2

The silicate impurity of the one is removed as slag by calcium oxide (CaO), which is formed by the decomposition of limestone (CaCO3).

CaCO3 → CaO + CO2

CaO + SiO2 → CaSiO3

Calcium silicate

(Slag)

Q.127 Name the common elements present in the anode mud in electrolytic refining of copper. Why are they so present?

Ans.

In electrolytic refining of copper, the common elements present in anode mud are selenium, tellurium, silver, gold, platinum, and antimony.

These elements are very less reactive and are not affected during the purification process. Hence, they settle down below the anode as anode mud.

Q.128 Out of C and CO, which is a better reducing agent at 673 K?

Ans.

At 673 K, the value of

ΔG(CO,CO2)

is less than that of ∆G(C,CO) . Therefore, CO can be reduced more easily to CO2 than C to CO. Hence, CO is a better reducing agent than C at 673 K.

Q.129 Explain: (i) Zone refining (ii) Column chromatography.

Ans.

(i) Zone refining:

This method is based on the principle that impurities are more soluble in the molten state of metal (the melt) than in the solid state. In the process of zone refining, a circular mobile heater is fixed at one end of a rod of impure metal. As the heater moves, the molten zone of the rod also moves with it. As a result, pure metal crystallizes out of the melt and the impurities pass onto the adjacent molten zone. This process is repeated several times, which leads to the segregation of impurities at one end of the rod. Then, the end with the impurities is cut off. Silicon, boron, gallium, indium etc. can be purified by this process.

(ii) Column chromatography:

Column chromatography is a technique used to separate different components of a mixture. It is a very useful technique used for the purification of elements available in minute quantities. It is also used to remove the impurities that are not very different in chemical properties from the element to be purified. Chromatography is based on the principle that different components of a mixture are differently adsorbed on an adsorbent. In chromatography, there are two phases: mobile phase and stationary phase. The stationary phase is immobile and immiscible. Al2O3 column is usually used as the stationary phase in column chromatography. The mobile phase may be a gas, liquid, or supercritical fluid in which the sample extract is dissolved. Then, the mobile phase is forced to move through the stationary phase. The component that is more strongly adsorbed on the column takes a longer time to travel through it than the component that is weakly adsorbed. The adsorbed components are then removed (eluted) using a suitable solvent (eluant).

Q.130 Why is the extraction of copper from pyrites more difficult than that from its oxide ore through reduction?

Ans.

The Gibbs free energy of formation (ΔfG) of CS2 and H2S is higher than that of Cu2S. Therefore, C and H2 cannot reduce Cu2S to Cu. On the other hand, the Gibbs free energy of formation (ΔfG) of Cu2O is higher than that of CO and thus C can reduce Cu2O to Cu.

Q.131 What is the role of depressant in froth floatation process?

Ans.

In the froth floatation process, the role of the depressants is to separate two sulphide ores by selectively preventing one ore from forming froth. For example, to separate two sulphide ores (ZnS and Pbs), NaCN is used as depressant which selectively allows PbS to come with froth, but prevents ZnS from coming to froth. This happens because NaCN reacts with ZnS to form Na2[Zn(CN)4].

4NaCN + Zns → Na2 [Zn(CN)4] + Na2S

Q.132 Copper can be extracted by hydrometallurgy but not zinc. Explain.

Ans.

The reduction potentials of zinc and iron are lower than that of copper. In hydrometallurgy, zinc and iron can be used to displace copper from their solution.

Fe(s) + Cu2+ (aq) → Fe2+(aq) + Cu(s)

But to displace zinc, more reactive metals i.e., metals having lower reduction potentials than zinc such as Mg, Ca, K, etc. are required. But all these metals react with water with the evolution of H2 gas.

2K(s) + 2H2O(l) → 2KOH(aq) + H2(g)

As a result, these metals cannot be used in hydrometallurgy to extract zinc.

Hence, copper can be extracted by hydrometallurgy but not zinc.

Q.133 List the uses of Neon and argon gases.

Ans.

Followings are the uses of neon gas:

(i) It is mainly used in discharge tubes and fluorescent lamps which are used in advertising purposes.
(ii) It is used in beacon lights.
(iii) It is filled in discharge tubes with characteristic colours.

Followings are the uses of argon gas:

(i) It is used to provide an inert temperature in a high metallurgical process. (ii) It is also used in laboratories to handle air-sensitive substances.
(iii)Argon along with nitrogen is used in gas-filled electric lamps. This is because Ar is more inert than N.

Q.134 Why do noble gases have comparatively large atomic sizes?

Ans.

Noble gases are inert in nature; they don’t form molecules so the atomic radii of the noble gasses correspond to van der Waal’s radii. By definition, van der Waal’s radii are larger than covalent radii. That is why noble gases have comparatively large atomic sizes.

Q.135 Give the formula and describe the structure of a noble gas species which is iso-structural with:
(i)ICl4 (ii) IBr2 (iii) BrO3

Ans.

(i) ICl4 is iso-electronic with XeF4 and a square planar geometry

(ii) XeF2 is iso-electronic to IBr2has a linear structure.

(iii)XeO3 is iso-structural to BrO3 and has a pyramidal molecular structure.

Q.136 Which one of the following does not exist?
(i)XeOF4 (ii) NeF2 (iii) XeF2 (iv) XeF6

Ans.

The sum of 1st and 2nd ionization enthalpies of Ne is much higher than those of Xe. Thus F2 can oxidize Xe to Xe2+ but cannot oxidize Ne to Ne2+. So,NeF2 does not exist but all the xenon fluorides (XeF4 and XeF6) and xenon oxyfluoride (XeOF4) do exist.

Q.137 Arrange the following in the order of property indicated for each set:
(i) F2, Cl2, Br2, I2 : increasing bond dissociation enthalpy.
(ii) HF, HCl, HBr, HI : increasing acid strength.
(iii) NH3, PH3, AsH3, SbH3, BiH3 − increasing base strength.

Ans.

(i) Bond dissociation energy usually decreases on moving down a group as the atomic size increases. The bond dissociation energy is decreased as it requires less energy to break the bonds as we moving down the group (bond distance increases from F2 to I2). The bond dissociation energy of F2 molecule is very less than Br2 and Cl2, it is due to the fact that F-atom is very small in size and hence the three lone pairs of electrons on each F atom repel the bond pairs holding the F-atom in the F2 molecule.

Thus, the increasing order for bond dissociation energy among halogens is as follows:

I2< F2< Br2< Cl2

(ii) The relative acid strength of HF, HCl, HBr, and HI depends on the bond dissociation enthalpies. Since the bond dissociation enthalpies of H–X bond decreases from H–F to H–I as the size of the atom increases from F to I. Thus the acid strength increases in the following order.

HF < HCl < HBr < HI

(iii)Due to the presence of lone pair of electrons on the central atom in AH3, all behave as Lewis bases. On moving from nitrogen to bismuth, the size of the atom increases while the electron density on the atom and as well electronegativity also decreases and hence the basic strength decreases as we move from NH3 to BiH3.

BiH3< SbH3< AsH3< PH3< NH3

Q.138 How are XeO3 and XeOF4 prepared?

Ans.

(i) XeO3 can be prepared in two ways as shown

6XeF4 + 12H2O → 4Xe + 2XeO3 +24HF + 3O2

XeF6 + 3H2O → XeO3 + 6HF

(ii) XeOF4 can be prepared using XeF6.

XeF6 + H­2O → XeOF4 +2HF

Q.139 With what neutral molecule is ClO isoelectronic? Is that molecule a Lewis base?

Ans.

ClO has (17 + 8 + 1) = 26 electrons and OF2 has (8 + 2 x 9) = 26 electron system. ClO is isoelectronic to OF2.

If we replace O(9 elecrtons) in ClO by F (9 electrons) atom, the resulting neutral molecule is ClF.

Since Cl in ClF can further donate electrons to two more F atoms to form ClF3. Therefore, ClF acts as a Lewis base.

Q.140 How are xenon fluorides XeF2, XeF4 and XeF6 obtained?

Ans.

XeF2, XeF4, and XeF6 are obtained by a direct reaction between Xe and F2. The products obtained are dependent on the reaction conditions.

Xe(g)xenoninexcess+F2(g)673K,1barXeF2(s)Xe(g)(1:5ratio)+F2(g)873K,7bsrXeF4(s)Xe(g)(1:20ratio)+F2(g)573K,6070barXeF6(s)

Q.141 Write balanced equations for the following:
(i) NaCl is heated with sulphuric acid in the presence of MnO2.
(ii) Chlorine gas is passed into a solution of NaI in water.

Ans.

The required reactions are as follows:

(i) 4NaCl +MnO2+4H2SO4MnCl2+4NaHSO4+Cl2+2H2O(ii) Cl2(g)+2NaI(aq)2NaCl(aq)+I2(s)

Q.142 What are the oxidation states of phosphorus in the following:
(i)H3PO3, (ii) PCl3, (iii) Ca3P2, (iv)Na3PO4 , (v)POF3?

Ans.

Let the oxidation state of P be x.

(i) H3PO3 :
3(+1) + x +3(–2) = 0
Or, x = +3

(ii) PCl3 :
x + 3(–1) = 0
Or, x = +3

(iii) Ca3P2:
3(+2) + 2 x = 0Or, x = –3

(iv) Na3PO4 :
3(+1) + x + 4 (–2) = 0
Or, x = +5

(v) POF3 :
x + 1(–2) + 3(–1) = 0
Or, x = +5

Q.143 What inspired N. Bartlett for carrying out reaction between Xe and PtF6?

Ans.

Neil Bartlett initially carried out a reaction in which oxygen reacts with PtF6 to yield an ionic solid, O2+PtF6.

In this reaction, O2 gets oxidized to O2+ by PtF6.

O2(g) + PtF6(g) → O2+[PtF6]

Later, he realized that the first ionization energy of oxygen (1175 kJ/mol) and Xe (1170 kJ/mol) is almost the same. He thought that PtF6 should oxidize Xe to Xe+. Thus, he tried to prepare a compound with Xe and PtF6. When Xe and PtF6 were mixed, a rapid reaction took place and a red solid with formula, Xe+PtF6 was formed.

Xe+Pt F 6 X e + [ Pt F 6 ] MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbujxyKLgDP9MBHXgibjxyIL2yaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVy0xg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeaacaGaaiaabeWaaqaabaqbaaGcbaGaamiwaiaadwgacqGHRaWkcaWGqbGaamiDaiaadAeadaWgaaWcbaGaaGOnaaqabaGccqGHsgIRcaWGybGaamyzamaaCaaaleqabaGaey4kaScaaOWaamWaaeaacaWGqbGaamiDaiaadAeadaWgaaWcbaGaaGOnaaqabaaakiaawUfacaGLDbaadaahaaWcbeqaaiabgkHiTaaaaaa@4E88@

Q.144 How can you prepare Cl2 from HCl and HCl from Cl2? Write reactions only.

Ans.

(a) HCl can be oxidized to Cl2 by a number of oxidizing agents such as MnO2, KMnO4, K2Cr2O7 etc.

MnO2 + 4 HCl → MnCl2 + Cl2 2H2O

(b) HCl can be prepared from Cl2 by dissolving it in water.

Cl 2 ( g ) H 2 O( I )HCl( aq )+ HOCl( aq ) Hypochlorousacid MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbujxyKLgDP9MBHXgibjxyIL2yaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVy0xg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeaacaGaaiaabeWaaqaabaqbaaGcbaGaae4qaiaabYgadaWgaaWcbaGaaeOmaaqabaGcdaqadaqaaiaabEgaaiaawIcacaGLPaaacaqGibWaaSbaaSqaaiaabkdaaeqaaOGaae4tamaabmaabaGaaeysaaGaayjkaiaawMcaaiabgkziUkaabIeacaqGdbGaaeiBamaabmaabaGaaeyyaiaabghaaiaawIcacaGLPaaacaqGRaWaaCbeaeaacaqGibGaae4taiaaboeacaqGSbWaaeWaaeaacaqGHbGaaeyCaaGaayjkaiaawMcaaaWcbaGaaeisaiaabMhacaqGWbGaae4BaiaabogacaqGObGaaeiBaiaab+gacaqGYbGaae4BaiaabwhacaqGZbGaaGPaVlaabggacaqGJbGaaeyAaiaabsgaaeqaaaaa@669B@

Q.145 Write the reactions of F2 and Cl2 with water.

Ans.

Reactions are as follows:

2F 2 ( g ) +2H 2 O( I ) 4H + ( aq ) +4F ( aq ) +O 2 ( g ) 3F 2 ( g ) +3H 2 O( I ) 6H + ( aq ) +6F ( aq ) +O 3 ( g ) Cl 2 ( g ) +H 2 O( I )HCl( qa )+ HOCl( aq ) Hypochlorousacid MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbujxyKLgDP9MBHXgibjxyIL2yaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVy0xg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeaacaGaaiaabeWaaqaabaqbaaGceaqabeaacaqGYaGaaeOramaaBaaaleaacaqGYaaabeaakmaabmaabaGaae4zaaGaayjkaiaawMcaaiaabUcacaqGYaGaaeisamaaBaaaleaacaqGYaaabeaakiaab+eadaqadaqaaiaabMeaaiaawIcacaGLPaaacqGHsgIRcaqG0aGaaeisamaaCaaaleqabaGaae4kaaaakmaabmaabaGaaeyyaiaabghaaiaawIcacaGLPaaacaqGRaGaaeinaiaabAeadaahaaWcbeqaaiaab2caaaGcdaqadaqaaiaabggacaqGXbaacaGLOaGaayzkaaGaae4kaiaab+eadaWgaaWcbaGaaeOmaaqabaGcdaqadaqaaiaabEgaaiaawIcacaGLPaaaaeaacaqGZaGaaeOramaaBaaaleaacaqGYaaabeaakmaabmaabaGaae4zaaGaayjkaiaawMcaaiaabUcacaqGZaGaaeisamaaBaaaleaacaqGYaaabeaakiaab+eadaqadaqaaiaabMeaaiaawIcacaGLPaaacqGHsgIRcaqG2aGaaeisamaaCaaaleqabaGaae4kaaaakmaabmaabaGaaeyyaiaabghaaiaawIcacaGLPaaacaqGRaGaaeOnaiaabAeadaahaaWcbeqaaiaab2caaaGcdaqadaqaaiaabggacaqGXbaacaGLOaGaayzkaaGaae4kaiaab+eadaWgaaWcbaGaae4maaqabaGcdaqadaqaaiaabEgaaiaawIcacaGLPaaaaeaacaqGdbGaaeiBamaaBaaaleaacaqGYaaabeaakmaabmaabaGaae4zaaGaayjkaiaawMcaaiaabUcacaqGibWaaSbaaSqaaiaabkdaaeqaaOGaae4tamaabmaabaGaaeysaaGaayjkaiaawMcaaiabgkziUkaabIeacaqGdbGaaeiBamaabmaabaGaaeyCaiaabggaaiaawIcacaGLPaaacaqGRaWaaCbeaeaacaqGibGaae4taiaaboeacaqGSbWaaeWaaeaacaqGHbGaaeyCaaGaayjkaiaawMcaaaWcbaGaaeisaiaabMhacaqGWbGaae4BaiaabogacaqGObGaaeiBaiaab+gacaqGYbGaae4BaiaabwhacaqGZbGaaGPaVlaabggacaqGJbGaaeyAaiaabsgaaeqaaaaaaa@A3A2@

Q.146 Why are halogens coloured?

Ans.

The halogens are coloured because halogens absorb radiations in the visible region. This results in the excitation of valence electrons to a higher energy region while the remaining light is transmitted. Since the amount of energy required for excitation differs for each halogen, each halogen displays a different colour.

For example, F2 absorbs violet light (higher excitation energy) and hence appears pale yellow while I2 absorbs yellow and green light (lower excitation energy), so it appears deep violet. For the same reason Br2 shows orange red colour and Cl2 shows greenish yellow colour.

Q.147 Write two uses of ClO2.

Ans.

The uses of ClO2:

(i) It is an excellent bleaching agent.
(ii) It is a powerful oxidizing agent and chlorinating agent.

Q.148 Explain why inspite of nearly the same electronegativity, oxygen forms hydrogen bonding while chlorine does not.

Ans.

Both chlorine and oxygen have almost the same electronegativity values, but the size of chlorine is larger than oxygen atom. Thus electron density per unit volume on oxygen is higher than that of chlorine. So oxygen is able to form hydrogen bond but chlorine cannot.

Q.149 Explain why fluorine forms only one oxoacid, HOF.

Ans.

Fluorine forms only one oxoacid (HOF), because of its high electronegativity and small size and in this compound oxidation state of F is +1.

Again, due to the absence of d-orbitals, F can’t expand its valency beyond +1 (i.e. +3, +5, and +7); hence F does not form higher oxoacids such as HOFO, HOFO2, HOFO3.

F 2 +H 2 O( ice ) HOF +1 + HF -1 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbujxyKLgDP9MBHXgibjxyIL2yaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVy0xg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeaacaGaaiaabeWaaqaabaqbaaGcbaGaaeOramaaBaaaleaacaqGYaaabeaakiaabUcacaqGibWaaSbaaSqaaiaabkdaaeqaaOGaae4tamaabmaabaGaaeyAaiaabogacaqGLbaacaGLOaGaayzkaaGaeSiZHm4aaCbiaeaacaqGibGaae4taiaabAeaaSqabeaacaqGRaGaaeymaaaakiaabUcadaWfGaqaaiaabIeacaqGgbaaleqabaGaaeylaiaabgdaaaaaaa@5020@

Q.150 Why are halogens strong oxidising agents?

Ans.

The general electronic configuration of halogens is np5, where n =2–6. Thus, halogens need only one more electron to complete their octet and to attain the stable noble gas configuration.

X 2 +2 e 2 X MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbujxyKLgDP9MBHXgibjxyIL2yaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVy0xg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeaacaGaaiaabeWaaqaabaqbaaGcbaGaamiwamaaBaaaleaacaaIYaaabeaakiabgUcaRiaaikdacaWGLbWaaWbaaSqabeaacqGHsislaaGccqGHsgIRcaaIYaGaamiwamaaCaaaleqabaGaeyOeI0caaaaa@4703@

Due to low bond dissociation enthalpy, high electronegetivity and large electron gain enthalpy halogens have a strong tendency to accept electrons and get reduced. They only can act as strong oxidising agents.

According to the electrode potentials of halogen atoms it is evident that F2 is the strongest oxidising agent and I2 is the weakest oxidising agent.

Q.151 How is SO2 an air pollutant?

Ans.

SO2 can act as an air pollutant because of the following reasons:

(i) Even in very low concentrations (5 ppm), SO2 causes irritation in the respiratory tract. It causes throat and eye irritation and can also affect the larynx to cause breathlessness.
(ii) It combines with rain water present to form sulphurous acid. This causes acid rain. Acid rain damages soil, plants, and buildings, especially those made of marble (CaCO3).

CaCO3+H2SO3CaSO3+H2O+CO2

(iii) Even in low concentration (0.03 ppm) of SO2 is extremely harmful to plants. Plants exposed to sulphur dioxide for a long time lose colour from their leaves (loss of green colour). This condition is known as chlorosis. This happens because the formation of chlorophyll is affected by the presence of sulphur dioxide.

Q.152 Describe the manufacture of H2SO4 by contact process?

Ans.

Sulphuric acid is manufactured by the Contact Process which involves three steps:

(i) Production of SO2 by burning sulphur or roasting of sulphur pyrites.

S8+8O28SO24FeS2IronPyrites+11022Fe2O3+8SO2(ii) Catalytic conversion of SO2to SO3by the reaction with oxygen in the presence of a catalyst (V2O5),2SO2(g)+O2(g)2So3(g)ΔfHο=196.6Kj/molSO3 produced is absorbed on 98%H2SO4to give H2S2O7(oleum).SO3+H2SO4H2S2O7

This oleum is then diluted to obtain H2SO4 of the desired concentration. In practice, the plant is operated at 2 bar (pressure) and 720 K (temperature). The sulphuric acid thus obtained is 96-98% pure.

Q.153 Which aerosols deplete ozone?

Ans.

Aerosols such as chlorofluorocarbons (CFCs) and Freon (CCl2F2) accelerate the depletion of ozone. In the presence of ultraviolet radiations, molecules of CFCs break down to form chlorine free radicals that combine with ozone to form oxygen. The ozone layer depletes By the free radical mechanism, initiated by chlorine free radicals. This causes hole in the layer.

Q.154 Knowing the electron gain enthalpy values for

O →O and O →O2− as −141 and 702 kJmol−1 respectively, how can you account for the formation of a large number of oxides having O2− species and not O?

Ans.

Stability of an ionic compound depends on its lattice energy. More the lattice energy of a compound, more stable will be the ionic compound.

Consider the following reaction of a divalent metal (M) with oxygen. The formation of MO2 and MO involves the following steps:

M(g)ΔiH1M+(g)ΔiH2M2+(g)O(g)ΔegH1O(g)M2+(g)+2O(g)LatticeenergyM2+(O)2(S)M2+(g)+2O2(g)LatticeenergyM2+O2(S)

If electron gain-enthalpy were the only factor involved in the formation of O2– ions, we would expect that O will prefer to form O rather than O2– ions. But a large number of oxides have O2– species and not O is present. The reasons behind it are

  1. O2– is more stable than O because it has stable noble gas configuration.
  2. Due to the higher charge on O2− than on Oion, the lattice energy released during the formation of oxides containing O2− species in the solid state is much higher than lattice energy produced during the formation of oxides containing O species. The lattice energy of formation of MO type of species is quite higher that can compensates the higher electron gain enthalpy (ΔegH2) needed during the formation of O2– to O ions. Thus the formation of MO variety is energetically more feasible than the compound of Oions. For that reason, oxygen forms a large number of oxides having O2– species and not O.

Q.155 Why is di-oxygen a gas but sulphur a solid?

Ans.

Due to small size and high electronegativity, oxygen forms pπ-pπ multiple bonds. As a result oxygen can exist as diatomic (O2) molecules. These molecules are held together by weak van der Waals forces of attraction which can easily overcome by the collision of the molecule at room temperature. So O2 exist as gas at room temperature.

On the other hand, due to the bigger size and lower electronegativity, S does not form multiple bonds. Instead, it prefers S–S single bond. Again, S–S single bond is stronger than O–O single bond; more energy is required to break this bond. S has a much greater tendency to undergo catenation and lower tendency for pπ–pπ multiple bonds. It exists as a puckered structure held together by strong covalent bonds. Hence, it is a solid at room temperature.

Q.156 Justify the placement of O, S, Se, Te and Po in the same group of the periodic table in terms of electronic configuration, oxidation state and hydride formation.

Ans.

The elements of group 16 are collectively called chalcogens.

(i) Electronic configuration: All these elements have the same ns2np4 (n=2 to 6) valence shell electronic configuration and hence are justified to place in group 16 of the periodic table.

8O=[He]2s22p4         16S=[Ne]3s23p434Se=[Ar]3d104s24p4    52Te=[Kr]4d105s25p484Po=[Xe]4f145d106s26p4

(ii)Oxidation states: As these elements have six valence electrons (ns2np4), they accept 2 electrons to complete their octets and they can display an oxidation state of −2. However, only oxygen predominantly shows the oxidation state of −2 owing to its high electronegativity. It also exhibits the oxidation state of −1 (H2O2), zero (O2), and +2 (OF2). However, the stability of the −2 oxidation state decreases on moving down a group due to a decrease in the electronegativity of the elements. The heavier elements of the group show an oxidation state of +2, +4, and +6 due to the availability of d-orbitals.

(iii) Formation of hydrides: These elements form hydrides of formula H2E, where E = O, S, Se, Te, Po (Like H2O, H2S, H2Se, H2Te and H2Po).

Q.157 Can PCl5 act as an oxidising as well as a reducing agent? Justify.

Ans.

The oxidation state of P in PCl5 is +5. Since P has five electrons in its valence shell thus it can donate its 5 valence electrons and cannot increase its oxidation state beyond +5. Therefore, PCl5 can only act as an oxidising agent (which can donate electrons). It can decrease its oxidation state from +5 to +3 and act as an oxidising agent. For example,

2Ag0+PCl5+52AgCl+1+PCl3+3PCl5+5+H20PCl3+3+2HCl+1

In the above reactions, oxidation number of P decreases from +5 in PCl5 to +3 in PCl3 and that of Ag and H increases from 0 to +1. Thus PCl5 acts as an oxidising agent.

Q.158 Give the disproportionation reaction of H3PO3.

Ans.

Orthophosphorus acid (H3PO3) on heating undergoes self-oxidation reduction to give orthophosphoric acid (H3PO4) and phosphine (PH3). This kind of reaction is called disproportionation reaction. The oxidation states of P in various species involved in the reaction are mentioned below.

4H3PO3+3Δ3H3PO4+5+PH33

Q.159 Why does nitrogen show catenation properties less than phosphorus?

Ans.

Catenation property depends upon the strength of the element-element bond (same elements). Since the N–N bond strength is much weaker than P–P bond strength, N does not show catenation property. Again, due to the smaller size of nitrogen atom, there is greater repulsion of electron density of two nitrogen atoms, thereby weakening the N−N single bond.

Q.160 Write main differences between the properties of white phosphorus and red phosphorus.

Ans.

White Phosphorus Red Phosphorus
  1. It so soft, waxy white solid with graphic odour.
  1. It is a hard and crystalline solid, without any smell.
  1. It is insoluble in water but soluble in carbon disulphide.
  1. It is insoluble in both water and carbon disulphlde.
  1. It is polsonous.
  1. It is non-polsonous
  1. It undergoes spontaneous combustion in air.
  1. It is relatively less reactive.
  1. In both solid and vapour states, it exists as a p4­­­­­­ molecule.
  1. It exists as a chain of tetrahedral p4 units.

Q.161 Nitrogen exists as diatomic molecule and phosphorus as P4. Why?

Ans.

Nitrogen owing to its small size has a tendency to form pπ−pπ multiple bonds with itself. Nitrogen thus forms a very stable diatomic molecule (NºN).

On the other hand, due to the larger size of P and low electronegativity, its pπ−pπ bonding is not feasible. Instead it prefers to form P–P single bonds; hence it exists as P4 molecules.

Q.162 Explain why NH3 is basic while BiH3 is only feebly basic.

Ans.

Among the group 15 elements, N has small size due to which the lone pair of electrons is concentrated in a small region. This means that the charge density per unit volume is high. On moving down a group, the size of the central atom increases and the charge gets distributed over a large area decreasing the electron density. Hence, the electron donating capacity of group 15 element hydrides decreases on moving down the group. So, NH3 is highly basic but BiH3 is feebly basic.

Q.163 Why does R3P=O exist but R3N=O does not (R = alkyl group)?

Ans.

Due to the absence of d-orbital N cannot participate in pπ-dπ bonding. For this reason nitrogen cannot expand its coordination number beyond four. In R3N=O, N has a covalency of 5.

The compound R3N=O does not exist.

In contrast, P due to presence of d-orbitals form pπ-dπ bonds and hence can expand its covalency beyond 4.

P form R3P=O in which the covalency of P is 5.

Q.164 The H-N-H angle value is higher than H-P-H, H-As-H and H-Sb-H angles. Why?

Ans.

Hydride H-E-H bond angle
NH3 107.8°
PH3 93.6°
AsH3 91.8°
SbH3 91.3°
BiH3 90°

Since nitrogen is highly electronegative, there is high electron density around nitrogen. This causes greater repulsion between the electron pairs around nitrogen, resulting in maximum bond angle. As we move down the group size of the central atom goes on increasing and its electronegativity goes on decreasing. Consequently, the repulsive interactions between the electron pairs decrease, thereby decreasing the H−E−H bond angle.

Q.165 Give the resonating structures of NO2 and N2O5.

Ans.

The resonating structure of NO2 is:

The resonating structure of N2O5 are as follows:

Q.166 Illustrate how copper metal can give different products on reaction with HNO3.

Ans.

Concentrated nitric acid is a strong oxidising agent and attacks most metals except noble metals such as gold and platinum.

Cu gives different products on reaction with HNO3 depending upon the conc. of nitric acid.

3Cu+8HNO 3 ( dilute )3Cu ( NO 3 ) 2 +2NO+4H 2 O Cu+4HNO 3 ( conc. )Cu ( NO 3 ) 2 +2NO 2 +2H 2 O MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbujxyKLgDP9MBHXgibjxyIL2yaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVy0xg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@788F@

Q.167 How is ammonia manufactured industrially?

Ans.

On a large scale, ammonia is manufactured by Haber’s process.

N2(g)+3H2(g)2NH3(g);

The optimum conditions for manufacturing ammonia are:

  1. Pressure: 200 × 105 Pa.
  2. Temperature: ~700 K.
  3. Catalyst is used such as iron oxide with small amounts of A2O3 and K2O.

Q.168 How is nitrogen prepared in the laboratory? Write the chemical equations of the reactions involved.

Ans.

In the laboratory, dinitrogen is prepared by treating an aqueous solution of ammonium chloride with sodium nitrite.

NH4Cl(aq)+NaNO2(aq)ΔN2(g)+2H2O(l)+NaCl(aq)

Small amounts of NO and HNO3 are also formed in this reaction; these impurities can be removed by passing the gas through aqueous sulphuric acid containing potassium dichromate.

Q.169 Why does NH3 form hydrogen bond but PH3 does not?

Ans.

The electronegativity of N (3.0) is much higher than that of H (2.1). Thus N–H bond is polar and NH3 undergoes intermolecular H-bonding.

In contrast, both P and H both have an electronegativity of 2.1. Thus P–H bond is not-polar and hence PH3 does not undergo H-bonding because one main condition of H-bonding is H atom should be attached to an electronegative atom.

Q.170 Discuss the trends in chemical reactivity of group 15 elements.

Ans.

General trends in chemical reactivity of group-15 are as follows:

(i) Reactivity towards metals: The group 15 elements react with metals to form binary compounds in which metals exhibit −3 oxidation states.

(ii) Reactivity towards hydrogen: The elements of group 15 react with hydrogen to form hydrides of type EH3, where E = N, P, As, Sb, or Bi. The stability of hydrides decreases on moving down from NH3 to BiH3.

(iii) Reactivity towards halogens: The group 15 elements react with halogens (Cl, Br, I) to form two series of salts: EX3 and EX5. However, nitrogen does not form NX5 as it lacks the d-orbital. The trihalides of N are unstable, however NF3 is stable. Other trihalides are known only as their unstable ammoniates, i.e., NBr3.NH3 and NI3.NH3. The instability of NCl3, NBr3 and NI3 is because of weak N–X bond due to the large difference in size.

(iv) Reactivity towards oxygen: The elements of group 15 form two types of oxides: E2O3 and E2O5, where E = N, P, As, Sb, or Bi. The acidic character decreases on moving down a group. Oxides of N are acidic in nature; Oxides of As and Sb are amphoteric, while oxide of Bi is basic in nature.

Q.171 Why does the reactivity of nitrogen differ from phosphorus?

Ans.

Nitrogen is less reactive than phosphorus. Nitrogen exists as a diatomic molecule. In N2, the two nitrogen atoms are attached via triple bond (N≡N). This triple bond has very high bond strength, which is very difficult to break. It is because of small size of nitrogen, it is able to form pπ−pπ bonds with itself. As a result, nitrogen is inert and unreactive in its elemental state.

On the other hand, phosphorus exists as a tetra-atomic molecule (P4). Since the P—P single bond is much weaker than N≡N triple bond; therefore, phosphorus is much more reactive than nitrogen in its elemental state.

Q.172 Discuss the general characteristics of Group 15 elements with reference to their electronic configuration, oxidation state, atomic size, ionization enthalpy and electronegativity.

Ans.

The general characteristics of Group 15 elements:

(i)Electronic configuration:

All the elements in group 15(N, P, As, Sb, Bi) have 5 valence electrons.

The general electronic configuration: ns2 np3

(ii) Oxidation states: All these elements have 5 valence electrons and require three more electrons to complete their octets. The gain of electron is very difficult as the nucleus have to attract three more electrons. Only nitrogen accepts electrons in its valance shell due to its small size and the distance between the nucleus and the valence shell is relatively small. The remaining elements of this group show a formal oxidation state of −3 in their covalent compounds. In addition to the −3 state, N and P also have −2 oxidation states (in NH2NH2 and P2H4 respectively). N only shows oxidation state of –1 in NH2OH.

All the elements present in this group show +3 and +5 oxidation states. The stability of +5 oxidation state decreases down a group, whereas the stability of +3 oxidation state increases due to the inert pair effect.

(iii) Ionization energy and electronegativity:

First ionization energy decreases on moving down a group due to the increase in atomic size. As we move down a group, electronegativity decreases, owing to an increase in atomic radii.

(iv)Atomic size: On moving down a group, the atomic size increases. As the number of shells increases down the group, atomic size also increases.

Q.173 What can be inferred from the magnetic moment values of the following complex species?

Example Magnetic moment(BM)
K4[Mn(CN)6] 2.2
[Fe(H2O)6]2+ 5.3
K2[MnCl4] 5.9

Ans.

Magnetic moment (m): μ= n( n+2 ) For value n= 1, μ= 1( 1+2 ) = 3 =1.723 For value n= 2, μ= 2( 2+2 ) = 8 =2.83 For value n= 3, μ= 3( 3+2 ) = 15 =3.87 For value n= 4, μ= 4( 4+2 ) = 32 =5.66 For value n= 5, μ= 5( 5+2 ) = 35 =5.95 ( i ) K 4 [ Mn ( CN ) 6 ]:In this complex, Mn is in +2 oxidation state, i.e. Mn 2+ . For in transition metals, the magnetic moment is calculated from the spinonly formula. n( n+2 ) =2.2 We can see from the above calculation that the given value is closest to n=1. This means that Mn has only one unpaired electron. CN is strong field ligand which paired up the electrons in 3d. ( ii ) [ Fe ( H 2 O ) 6 ] 2+ : n( n+2 ) =5.3 We can see from the above calculation that the given value is closest to n=4. Also, in this complex, Fe is in the +2 oxidation state. This means that Fe has 4 unpaired electrons in the dorbital. H 2 O is a weak ligand which cannot pair up the delectrons. ( iii ) K 2 [ MnC l 4 ]: n( n+2 ) =5.9 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We can see from the above calculation that the given value is closest to n=5. Also, in this complex, Mn is in the +2 oxidation state. This means that Mn has 5 unpaired electrons in the d-orbital. Thus Cl is weak ligand.

Q.174 Comment on the statement that elements of the first transition series possess many properties different from those of heavier transition elements.

Ans.

The properties of the elements of the first transition series differ from those of the heavier transition elements are discussed in the following way:

(i)The atomic sizes of the elements of the first transition series are smaller than those of the heavier elements (elements of 2nd and 3rd transition series).

The atomic sizes of the elements in the 3rd transition series are almost same as those of the corresponding members in the 2nd transition series because of the lanthanoid contraction.

(ii) +2 and +3 oxidation states are more common for elements in the first transition series (+2 oxidation state is stable), while higher oxidation states are more common for the heavier elements.

(iii)Ionization enthalpies of 5d series are higher than the corresponding elements of 3d and 4d series.

(iv)The enthalpies of atomisation of the elements in the first transition series are lower than those of the corresponding 4d and 5d elements in the second and third transition series).

(v) The melting and boiling points of the first transition series are lower than those of the heavier transition elements due to the stronger metallic bonding (M−M bonding).

(vi)The elements of the first transition series form low-spin or high-spin complexes depending upon the strength of the ligand field. But the heavier transition elements form low spin complexes and that is independent of the strength of ligand field.

Q.175 Write down the number of 3d electrons in each of the following ions:

Ti2+, V2+, Cr3+, Mn2+, Fe2+, Fe3+, Co2+, Ni2+ and Cu2+.

Indicate how would you expect the five 3d orbitals to be occupied for these hydrated ions (octahedral).

Ans.

Metal Ion No. of 3d electrons Occupancy of 3d Orbitals
Ti2+ 2
V2+ 3
Cr3+ 3
Mn2+ 5
Fe2+ 6
Fe3+ 5
Co2+ 7
Ni2+ 8
Cu2+ 9

Q.176 Compare the general characteristics of the first series of the transition metals with those of the second and third series metals in the respective vertical columns. Give special emphasis on the following points:
(i) Electronic configurations,
(ii) Oxidation states,
(iii) Ionisation enthalpies, and
(iv) Atomic sizes.

Ans.

(i) In the 1st, 2nd and 3rd transition series, the 3d, 4d and 5d orbital are respectively filled.

We know that elements in the same vertical column generally have similar electronic configurations.

The elements with unusual electronic configurations:

1st transition series 2nd transition series 3rd transition series
Cr(24)=3d5 4s1 Mo(42)=4d5 5s1 W(74)=5d4 6s2
Cr(29)=3d10 4s1 Tc(43)=4d6 5s1 Pt(78)=5d9 6s1
Ru(44)=4d7 5s1 Au(79)=5d10 6s1
Rh(45)=4d8 5s1
Pd(46)=4d10 5s0
Pd(47)=4d10 5s1

As a result of these exceptions, it happens many times that the electronic configurations of the elements present in the same group don’t match with each other.

(ii) In each of the three transition series the number of oxidation states shown by the elements is the maximum in the middle and the minimum at the extreme ends.

+2 and +3 oxidation states are quite stable for all elements present in the first transition series. For example,

II                                       III                                   III[Fe(CN)6]4,           [CrCl6]3,                 [Co(NH3)6]3+

are stable complexes formed by the 1st transition series.

But no such complexes are known for the second and third transition series such as Mo, W, Rh, In. They form complexes in which higher oxidation states are more prominent. For example: WCl6, ReF7, RuO4, etc.

(iii) In each of the three transition series, the first ionization enthalpy increases from left to right. The first ionization enthalpies of the third transition series are higher than those of the first (3d) and second (4d) transition series. This occurs due to the weak shielding effect of 4f electrons in the third transition (5d) series.

There are also elements in the 2nd transition series whose first ionization enthalpies are lower than those of the elements corresponding to the same vertical column in the 1st transition series.

(iv) Atomic size generally decreases from left to right across a period. Now, among the three transition series, atomic sizes of the elements in the 2nd transition series are greater than those of the elements corresponding to the same vertical column in the 1st transition series. However, the atomic sizes of the elements in the 3rd transition series are virtually the same as those of the corresponding members in the 2nd transition series due to lanthanoid contraction.

Q.177 Write the electronic configurations of the elements with the atomic numbers 61, 91, 101, and 109.

Ans.

Atomic Number Electronic configuration
61 (Pm) =[Xe]54 4f5 5d0 6s2
91 (Pa) =[Rn]86 5f2 6d0 7s2
101 (Md) =[Rn]86 5f13 6d0 7s2
109 (Mt) =[Rn]86 5f14 6d7 7s2

Q.178 Compare the chemistry of the actinoids with that of lanthanoids with reference to:

(i) Electronic configuration

(ii) Oxidation states and

(iii) Chemical reactivity.

Ans.

(i) Electronic configuration:

The general electronic configuration for lanthanoids

= [ Xe ] 54 4 f 114 5 d 01 6 s 2 The general electronic configuration foractinoids = [ Rn ] 86 5 f 114 6 d 01 7 s 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbujxyKLgDP9MBHXgibjxyIL2yaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVy0xg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@8DE8@

(ii) Oxidation states:

The principal oxidation state of lanthanoids is (+3), but +2 and +4 oxidations states also observed. Actinoids exhibit a greater range of oxidation states. This is because the 5f, 6d and 7s levels are of comparable energies. For actinoids also (+3) is the common oxidation state.

(iii) Chemical reactivity:

In the lanthanide series, the earlier members of the series are more reactive. They have reactivity that is comparable to Calcium. With an increase in the atomic number, the lanthanides start behaving similar to Al. Actinoids, on the other hand, are highly reactive metals, especially in finely divided form. When they are added to boiling water, they give a mixture of oxide and hydride. Actinoids combine with most of the non-metals at moderate temperatures.

Q.179 Name the members of the lanthanoid series which exhibit +4 oxidation state and those which exhibit +2 oxidation state. Try to correlate this type of behavior with the electronic configurations of these elements.

Ans.

+2 oxidation state +4 oxidation state
Nd (60) Ce(58)
Sm(62) Pr(59)
Eu(63) Nd(60)
Tm(69) Tb(65)
Yb(70) Dy(66)

Ce after forming Ce4+attains a stable electronic configuration of [Xe].

Tb after forming Tb4+attains a stable electronic configuration of [Xe] 4f7.

Eu after forming Eu2+attains a stable electronic configuration of [Xe] 4f7.

Yb after forming Yb2+attains a stable electronic configuration of [Xe] 4f14.

Q.180 Use Hund’s rule to derive the electronic configuration of Ce3+ion and calculate its magnetic moment on the basis of ‘spin-only’ formula.

Ans.

Electronic configuration of58Ce:58Ce=[Xe]544f26s2And for Ce3+it is as follows: =[Xe]544f1Thereis only one unpaired electron, i.e. n=1μ=n(n+2)μ=1(1+2)μ=3BM=1.73BM

Q.181 Which is the last element in the series of the actinoids? Write the electronic configuration of this element. Comment on the possible oxidation state of this element.

Ans.

The last actinoids: Lawrencium (Z = 103)

Electronic configuration 103Lr:

[ Rn ] 86 5 f 14 6 d 1 7 s 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbujxyKLgDP9MBHXgibjxyIL2yaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVy0xg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeaacaGaaiaabeWaaqaabaqbaaGcbaWaamWaaeaacaWGsbGaamOBaaGaay5waiaaw2faamaaCaaaleqabaGaaGioaiaaiAdaaaGccaaI1aGaamOzamaaCaaaleqabaGaaGymaiaaisdaaaGccaaI2aGaamizamaaCaaaleqabaGaaGymaaaakiaaiEdacaWGZbWaaWbaaSqabeaacaaIYaaaaaaa@4AF4@

The most common oxidation state displayed by it is +3; because after losing 3 electrons it attains stable filled f-orbital (f14) configuration.

Q.182 The chemistry of the actinoid elements is not so smooth as that of the Lanthanoids. Justify this statement by giving some examples from the oxidation state of these elements.

Ans.

Lanthanoids primarily show three oxidation states (+2, +3, +4). Among these oxidation states, +3 state is the most common. Due to the energy difference between 4f, 5d, and 6s orbitals is quite large, lanthanoids display a limited number of oxidation states. On the other hand, the energy difference between 5f, 6d, and 7sorbitals is small. Hence, actinoids can display a large number of oxidation states.

For example, uranium and plutonium display +3, +4, +5, and +6 oxidation states while neptunium displays +3, +4, +5, and +7.

Like lanthanides, most common oxidation state in case of actinoids is also +3.

Q.183 What are inner transition elements? Decide which of the following atomic numbers are numbers of the inner transition elements: 29, 59, 74, 95, 102, 104.

Ans.

Inner transition metals are those elements in which the last electron enters the f-orbital.

These include lanthanoids(58-71) and actinoids (90-103).

Thus elements with atomic numbers 59, 95, and 102 are inner-transition elements.

Q.184 What are alloys? Name an important alloy which contains some of the lanthanoid metals. Mention its uses.

Ans.

An alloy is a homogeneous mixture of two or more metals or metals and non-metals.

An important alloy which contains lanthanoid metals is Mischmetal. It contains lanthanoids (94−95%), iron (5%), and traces of S, C, Ca, and Al.

Application of the alloy:

(a) Misch metal is used in cigarettes and gas lighters.

(b) It is used in tracer bullets and shells.

(c) It is used in flame throwing tanks.

Q.185 Indicate the steps in the preparation of:

  1. K2Cr2O7 from chromite ore.
  2. KMnO4 from pyrolusite ore.

Ans.

(i) Potassium dichromate is prepared from chromite ore (FeCr2O4) in the following steps.

(i)Preparation of sodium chromate:

The reaction with sodium carbonate occurs as follows:

4FeCr2O4+8Na2CO3+7O28Na2CrO4+3Fe2O3+8CO2iiConversion of sodium chromate into sodium dichromate:Thesolution of sodium chromate is treated with conc. sulphuric acid to form sodium dichromate.2Na2CrO4+H2SO4Na2Cr2O7+Na2SO4+H2OSod.ChromateConc.            Sod.dichromateiiiConversion of sodium dichromate to potassium dichromate:The solution of sodium dichromate is treated with potassium chloride to convert it into potassium dichromate.Na2Cr2O7+2KClK2Cr2O7+NaClThe dichromate ion Cr2O72 exists in equilibrium with chromate ion CrO42 at pH 4. However, by changing the pH, they can be interconverted.Cr2O72+H2OpH=42CrO42+2HOrangered                                       Yellowdichromate                                   chromateiiPotassium permanganate can be prepared from pyrolusite MnO2.The ore is fused with KOH in the presence of either atmospheric oxygen or an oxidizing agent, such as KNO3to give KMnO4.2MnO2+4KOH+O2Δ2K2MnO4+2H2O                                                                    GreenThe green mass can be extracted with water and then oxidized either electrolytically or by passing chlorine/ozone into the solution.Electrolytic oxidation occurs as follows:K2MnO42K++MnO42H2OH++OHAtanode:MnO42MnO42MnO4+emanganate                              Permanganateiona  Oxidationbychlorine:2K2MnO4+Cl22KMnO4+2KClbOxidationbyozone:2K2MnO4+ H2O+O32KMnO4+2KOH+O2

Q.186 Give examples and suggest reasons for the following features of the transition metal chemistry:

(i)The lowest oxide of transition metal is basic, the

highest is amphoteric/acidic.

(ii) A transition metal exhibits highest oxidation state in oxides and fluorides.

(iii)The highest oxidation state is exhibited in oxoanions of a metal.

Ans.

(i) In the case of a lower oxide of a transition metal, the metal atom is in lower oxidation state. This means that some of the valence electrons of the metal atom are not used in bonding. As a result, it can donate electrons and behave as a base. On the other hand, in the case of a higher oxide of a transition metal, the metal atom is in higher oxidation state. This means that all the valence electrons are involved in bonding and so, they are unavailable. Due to high effective nuclear charge it can accept electrons and behave as an acid.

For example, Mn(II)O is basic, whereas Mn2(VII)O7 is acidic.

(ii) Oxygen and fluorine bring out the highest oxidation states from the transition metals because of their high electronegativities and small sizes. In other words, a transition metal exhibits higher oxidation states in oxides and fluorides. For example, in OsF6 and V2O5, the oxidation states of Os and V are +6 and +5 respectively.

(iii) Oxygen is a strong oxidizing agent due to its high electronegativity and small size. So, oxo-anions of a metal have the highest oxidation state.

For example, in MnO4 the oxidation state of Mn is +7 and in Cr2O72– oxidation state of Cr is +6.

Q.187 Calculate the number of unpaired electrons in the following gaseous ions: Mn3+, Cr3+, V3+and Ti3+. Which one of these is the most stable in aqueous solution?

Ans.

Gaseous ions Number of unpaired

electrons

(i) Mn3+=[Ar]3d4 4
(ii) Cr3+= [Ar]3d3 3
(iii) V3+ = [Ar]3d2 2
(iv) Ti3+= [Ar]3d1 1

Cr3+ is the most stable in aqueous solutions owing to a half-filled t2g3 configuration.

Q.188 Which metal in the first series of transition metals exhibits +1 oxidation state most frequently and why?

Ans.

In the first transition series, Cu exhibits +1 oxidation state very frequently. It is because Cu (+1) has an electronic configuration of [Ar] 3d10

The completely filled d-orbital makes it highly stable, so it can easily lose one electron to get this stable d-orbital configuration.

Q.189 What is meant by ‘disproportionation’? Give two examples of disproportionation reaction in aqueous solution.

Ans.

When in a reaction, the oxidation state of an element in a compound increases in one of the products and decreases in the other product, it is said to undergo disproportionation of oxidation state.

For example 1:

3MnO42+4H+2MnO4+MnO2+2H2OMn(VI)                          Mn(VII)      Mn(IV)Cr(V) is oxidized to Cr(VI) and reduced to Cr(III)For example 2:2CrO42+8H+CrO42+Cr3++4H2OCr(V)                            Cr(VI)      Cr(III)

Mn (VI) is oxidized to Mn (VII) and reduced to Mn (IV).

Q.190 How would you account for the following?

(i) Of the d4species, Cr2+is strongly reducing while manganese(III) is strongly oxidising.

(ii) Cobalt(II) is stable in aqueous solution but in the presence of complexing reagents it is easily oxidised.

(iii) The d1configuration is very unstable in ions.

Ans.

(i) Cr2+is strongly reducing in nature. It has a d4 configuration. While acting as a reducing agent, it gets oxidized to Cr3+ (3d3 configurations) by losing one electron. As it has half-filled t2g level.

In the case of Mn3+ (d4), it acts as an oxidizing agent and gets reduced to Mn2+ (d5). This has an exactly half-filled d-orbital and is highly stable. It behaves like a strong oxidizing agent.

(ii) Co (II) is stable in aqueous solutions, but in the presence of strong field ligands, it is oxidized to Co (III). In the presence of strong field ligands the amount of crystal field stabilization energy (CFSE) released is higher which overcomes the 3rd ionization energy for Co.

(iii) The ions in d1configuration tend to lose one more electron to acquire stable d0 configuration. Also, the hydration or lattice energy is more than sufficient to remove the only electron present in the d-orbital of these ions. Therefore, they act as reducing agents.

Q.191 Compare the chemistry of actinoids with that of the lanthanoids with special reference to:

(i) Electronic configuration

(iii)Oxidation state

(ii) Atomic and ionic sizes and

(iv)Chemical reactivity.

Ans.

(i) Electronic configuration

The general electronic configuration of lanthanoids

=[Xe]544f1145d016s2The general electronic configuration ofactinoids=[Rn]865f1146d017s2

Lanthanoids belongs to 4f-series and actinoids belongs to 5f-series.

(ii)Oxidation state:

The principal oxidation state of lanthanoids is (+3), but +2 and +4 oxidation states are also observed. This is because of extra stability of fully-filled and half-filled orbitals. Actinoids exhibit a greater range of oxidation states. This is because the 5f, 6d, and 7s levels are of comparable energies.

(iii)Atomic and lonic sizes:

Similar to lanthanoids, actinoids also exhibit actinoid contraction, overall sizes of atomic and ionic radii decreases. The contraction is greater in actinoid elements due to the poor shielding effect of 5f orbitals.

(iv)Chemical reactivity:

In the lanthanide series, the earlier members of the series are more reactive. They have reactivity that is comparable to that of Calcium. With an increase in the atomic number, the lanthanides start behaving similar to Al.

For example, they react with dilute nitric acid to liberate H2 gas.

Actinoids, on the other hand, are highly reactive metals, especially in finely divided state.

For example, when they are added to boiling water, they give a mixture of oxide and hydride.

Q.192 Compare the stability of +2 oxidation state for the elements of the first transition series.

Ans.

Sc +3
Ti +1 +2 +3 +4
V +1 +2 +3 +4 +5
Cr +1 +2 +3 +4 +5 +6
Mn +1 +2 +3 +4 +5 +6 +7
Fe +1 +2 +3 +4 +5 +6
Co +1 +2 +3 +4 +5
Ni +1 +2 +3 +4
Cu +1 +2 +3
Zn +2

From the above table it is evident that the maximum number of oxidation states is shown by Mn (varying from +2 to +7). On moving from Sc to Mn, the number of oxidation states increases and on moving from Mn to Zn, the number of oxidation states decreases due to a decrease in the number of available unpaired electrons. On moving down the above chart the relative stability of the +2 oxidation state increases.

Q.193 Predict which of the following will be coloured in aqueous solution?

Ti3+, V3+, Cu+, Sc3+, Mn2+, Fe3+ and Co2+. Give reasons for each.

Ans.

Only the ions that have electrons in d-orbital will be coloured. The ions in which d-orbital is empty will be colourless.

The electronic configuration of the ions is given below:

Ti3+:[Ar]183d1                                  Sc3+:[Ar]18V3+:[Ar]183d2                                                       Mn2+:[Ar]183d5Cu+:[Ar]183d10                                Fe3+:[Ar]183d5Co2+:[Ar]183d7

In the above elements only Sc3+has an empty d-orbital.

All other ions, except Sc3+, will be coloured in aqueous solution because of d−d transitions.

Q.194 For M2+/M and M3+/M2+ systems, the E°values for some metals are as follows:

Cr2+/Cr −0.9V

Cr3+/Cr2+ −0.4 V

Mn2+/Mn −1.2V

Mn3+/ Mn2+ +1.5V

Fe2+/Fe −0.4V

Fe3+/Fe2+ +0.8 V

Q.195 Use this data to comment upon:

(i) The stability of Fe3+in acid solution as compared to that of Cr3+ or Mn3+ and
(ii) The ease with which iron can be oxidized as compared to a similar process for either chromium or manganese metal.

Ans.

(i)

Cr3+/Cr2+ has a –ve reduction potential. Cr3+ cannot be reduced to Cr2+, i.e, Cr3+ is most stable. Mn3+/ Mn2+ has large +ve E° value, hence Mn3+ can easily reduce to Mn2+. Thus Mn3+ is least stable. E° value for Fe3+/Fe2+ is +ve but small.

These metal ions can be arranged in the increasing order of their stability as: <fe<cr</cr</fe Mn3+ < Fe3+ < Cr3+ <fe<cr</cr</fe

(ii)

The oxidation potentials for the given pairs increase in the following order.

Mn2+ / Mn > Cr2+ / Cr >Fe2+ /Fe

So, the oxidation of Fe to Fe2+ is not as easy as the oxidation of Cr to Cr2+ and the oxidation of Mn to Mn2+. Thus, these metals can be arranged in the increasing order of their ability to get oxidized as: Fe < Cr < Mn

Q.196 Describe the preparation of potassium permanganate. How does the acidified permanganate solution react with (i) iron (II) ions (ii) SO2 and (iii) oxalic acid? Write the ionic equations for the reactions.

Ans.

Potassium permanganate can be prepared from pyrolusite (MnO2). The ore is fused with KOH in the presence of either atmospheric oxygen or an oxidizing agent, such as KNO3 to give KMnO4.

2MnO2+44KOH+O2Δ2K2MnO4+2H2O                                                                GreenThe green mass can be extracted with water and then oxidized either electrolytically or by passing chlorine/ozone into the solution.Electrolytic oxidation occurs as follows:K2MnO42K++MnO42-H2OH++OHAtanode:MnO42MnO4+e                     maganate          Permanganateion(a)     Oxidation by Cl2:2K2Mno4+Cl22KMnO4+2KCl(b)     Oxidation by O3:2K2MnO4+H2O+O32KMnO4+2KOH+O2(i) Acidified KMnO4solution oxidizes Fe (II) ions to Fe (III) ions i.e., ferrous ions to ferric ions.MnO48H++5eMn2++4H2O                              Fe2+Fe3++e]×5¯5Fe2++MnO4+8H+Mn2++4H2O+5Fe3+(ii)   Acidified potassium permanganate oxidizes SO2to sulphuric acid (H2SO4).MnO4+8H++5eMn2++4H2O]×2     SO2+2H2O2e4H++SO42]×5        ¯2MnO4+5SO2+2H2O2Mn2++5SO42+4H+(iii)Acidified potassium permanganate oxidizes oxalic acid to carbon dioxide.MnO4+8H++5eMn2++4H2O]×2                                 C2O422CO2+2e]×5¯5C2O42+2MnO4+16H+2Mn2++8H2O+10CO2

Q.197 Describe the oxidizing action of potassium dichromate and write the ionic equations for its reaction with:

(i)Iodide, (ii) Iron (II) solution, (iii) H2S

Ans.

K2Cr2O7 acts as a very strong oxidizing agent in the acidic medium.

Cr2O72+14H++6e2Cr3+7H2OK2Cr2O7takes up electrons to get reduced and acts as an oxidizing agent. The reactions of K2Cr2O7with other iodide, iron II solution, and H2S are given below.iOxidizes iodideKI to iodineI2:Cr2O72+14H++6e2Cr3+7H2O                                       2II2+2e]×3             ¯  Cr2O72+14H++6I2Cr3+3I2+7H2OiiK2Cr2O7 oxidizes iron II solution to iron III solution i.e., Fe2+ions to Fe3+ions.Cr2O72+14H++6e2Cr3++7H2O                                       Fe2+Fe3++e]×6             ¯Cr2O72+14H++6Fe2+2Cr3++6Fe3++7H2OiiiK2Cr2O7oxidizes H2S to sulphur.Cr2O72+14H++6e2Cr3++7H2O                                      H2S2H++S+2e]×3            ¯Cr2O72+8H++3H2S2Cr3++3S+7H2O

Q.198 Describe the preparation of potassium dichromate from iron chromite ore. What is the effect of increasing pH on a solution of potassium dichromate?

Ans.

Potassium dichromate is prepared from chromite ore (FeCr2O4) in the following steps.

(i)Preparation of sodium chromate:

The reaction with sodium carbonate occurs as follows:

4FeCr2O4+8Na2CO3+7O28Na2CrO4+2Fe2O3+8CO2(ii)Conversion of sodium chromate into sodium dichromate: The solution of sodium chromate is treated with conc. sulphuric acid to form sodium dichromate.2Na2CrO4+conc. H2SO4Na2Cr2O7+Na2SO4+H2OSodiumchromateSodiumdichromate(iii)Conversion of sodium dichromate to potassium dichromate:The hot concentrated solution of sodium dichromate is treated with potassium chloride to convert it into potassium dichromate.Na2Cr2O7+2KClK2Cr2OL7+2NaclThe chromates and dichromates are interconvertible in aqueous solution depending upon pH of the solution.2CrO42+2H+Cr2O72+H2OCr2O72+2OH2CrO42+H2OThe dichromate ion exists in equilibrium with chromate ion at pH 4.pH=4Cr2O72+H2O2CrO42+2H+OrangeredYellow(dichromate)(chromate)

Q.199 How is the variability in oxidation states of transition metals different from that of the non-transition metals? Illustrate with examples.

Ans.

In transition elements, the oxidation state can vary from +1 to the highest oxidation state by removing all its valence electrons (from s- and d-electrons).

In transition elements, the oxidation states differ by 1, for example, Cu+ and Cu2+; Fe2+ and Fe3+.

In non-transition elements, the oxidation states differ by 2, for example, +2 and +4 or +3 and +5, etc.

In transition elements, the higher oxidation states are more stable for heavier elements in group. For example, in group 6 Mo(VI) is more stable than Cr(VI).

Q.200 What are interstitial compounds? Why are such compounds well known for transition metals?

Ans.

Transition metals are large in size and contain many interstitial sites. Transition elements can trap atoms of other elements of small size. Such as H, C, N, in the interstitial sites of their crystal lattices. The resulting compounds are called interstitial compounds.

Most of the transition metals form interstitial compounds with small non-metal atoms (C, H, N). These small atoms enter the voids sites of the crystals of the transition metals and form chemical bonds with transition metals. In transition metal series, the formation of such compounds can be possible.

Q.201 Explain giving reasons:

(i)Transition metals and many of their compounds show paramagnetic behaviour.

(ii) The enthalpies of atomisation of the transition metals are high.

(iii) The transition metals generally form coloured compounds.

(iv)Transition metals and their many compounds act as good catalyst.

Ans.

(i) Paramagnetism arises due to the presence of unpaired electrons with each electron having a magnetic moment associated with its spin angular momentum and orbital angular momentum. In case transition metals, as they contain unpaired electrons in (n–1) d-electrons, most of the transition metal ions and their compounds show the paramagnetic property.

(ii) Transition elements have high effective nuclear charge, so the valence electrons are tightly bonded. So, they form very strong metallic bonds. As a result, the enthalpy of atomization of transition metals is high.

(iii)In transition elements, due to crystal field splitting d-orbitals are splitted into two sets, one consisting of lower energy orbitals and the other consisting of higher energy orbitals. Thus, the electron can jump from lower energy d-orbital to higher energy d-orbital by absorption of light in the visible region. The energy required for the electron transitions is quite small and falls in the visible region of radiation. The ions of transition metals absorb the radiation of a particular wavelength and the rest is reflected, imparting colour to the solution.

(iv)The catalytic activity of the transition elements is due to the following reasons:

  • Owing to their ability to show variable oxidation states (due to the presence of incomplete d-orbitals), transition metals form unstable intermediate compounds. Thus, they provide a new path with lower activation energy, Ea, for the reaction.
  • Transition metals also provide a suitable surface for the reactions to occur.

For example, V2O5 acts as catalyst.

Q.202 What are the different oxidation states exhibited by the lanthanoids?

Ans.

In the lanthanide series, +3 oxidation state is most common i.e., Ln(III) compounds are predominant.

Other than that +2 and +4 oxidation states can also be found in the solution or in solid compounds. For example, Eu shows an oxidation state of +2 and Ce shows an oxidation state of +4.

Q.203 Give plausible explanation for each of the following:

(i) Why are amines less acidic than alcohols of comparable molecular masses?
(ii) Why do primary amines have higher boiling point than tertiary amines?
(iii) Why are aliphatic amines stronger bases than aromatic amines?

Ans.

(i) Amines undergo protonation to give amide ion.

R NH 2 R N H Amideion + H + MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVy0xg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeaacaGaaiaabeqaamaabaabaaGceiqabeaaG6qaaiaabkfacaaMc8Uaae4eGiaaykW7caqGobGaaeisamaaBaaaleaacaqGYaaabeaakiaaykW7daGdKaWcbaaabeGccaGLsgcacaaMc8+aaCbeaeaacaqGsbGaaGPaVlaabobicaaMc8+aaCbiaeaacaqGobaaleqabaGaaeylaaaakiaabIeaaSqaaiaabgeacaqGTbGaaeyAaiaabsgacaqGLbGaaGPaVlaabMgacaqGVbGaaeOBaaqabaGccaaMc8Uaae4kaiaaykW7caqGibWaaWbaaSqabeaacaqGRaaaaaGcbaaaaaa@583F@

Similarly, alcohol loses a proton to give alkoxide ion.

R-OH → R-O + H+

Alkoxide ion

In an amide ion, the negative charge is on the N-atom whereas in alkoxide ion, the negative charge is on the O-atom. Since O is more electronegative than N, O can accommodate the negative charge more easily than N. As a result, the amide ion is less stable than the alkoxide ion. Hence, amines are less acidic than alcohols of comparable molecular masses.

(ii) In a molecule of tertiary amine, there are no H-atoms whereas in primary amines, two hydrogen atoms are present. Due to the presence of H-atoms, primary amines undergo extensive intermolecular H-bonding.

As a result, extra energy is required to separate the molecules of primary amines. Hence, primary amines have higher boiling points than tertiary amines.

(iii) Due to the -R effect of the benzene ring, the electrons on the N- atom are less available in case of aromatic amines. Therefore, the electrons on the N-atom in aromatic amines cannot be donated easily. This explains why aliphatic amines are stronger bases than aromatic amines.

Q.204 Write the reactions of (i) aromatic and (ii) aliphatic primary amines with nitrous acid.

Ans.

(i) Aromatic amines react with nitrous acid (prepared in situ from NaNO2 and a mineral acid such as HCl) at 273 – 278K to form stable aromatic diazonium salts, NaCl and H2O.

(ii) Aliphatic primary amines react with nitrous acid (prepared in situ from NaNO2 and a mineral acid such as HCl) to form unstable aliphatic diazonium salts, which further produce alcohol and HCl with the evolution of N2 gas.

Q.205 Why cannot aromatic primary amines be prepared by Gabriel phthalimide synthesis?

Ans.

Gabriel phthalimide synthesis is used for the preparation of aliphatic primary amines. It involves nucleophilic substitution (SN2) of alkyl halides by the anion formed by the phthalimide.

But aryl halides do not undergo nucleophilic substitution with the anion formed by the phthalimide.

Hence, aromatic primary amines cannot be prepared by this process.

Q.206 Complete the following reactions:

(i) C6H5NH2 + CHCl3 + alc.KOH →
(ii) C6H5N2Cl + H3PO2+ H2O →
(iii) C6H5NH2 + H2SO4(conc.) →
(iv) C6H5N2Cl + C2H5OH →
(v) C6H5NH2 + Br2(aq) →
(vi) C6H5NH2 + (CH3CO)2O →

( vii ) C 6 H 5 N 2 Cl (ii) NaNO 2 /CuΔ (i) HBF 4 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVy0xg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeaacaGaaiaabeqaamaabaabaaGcbiqaaaQddaqadaqaaiaabAhacaqGPbGaaeyAaaGaayjkaiaawMcaaiaaykW7caqGdbWaaSbaaSqaaiaabAdaaeqaaOGaaeisamaaBaaaleaacaqG1aaabeaakiaab6eadaWgaaWcbaGaaeOmaaqabaGccaqGdbGaaeiBaiaaykW7daGdSaWcbaGaaeikaiaabMgacaqGPaGaaGPaVlaabIeacaqGcbGaaeOramaaBaaameaacaqG0aaabeaaaSqaaiaabIcacaqGPbGaaeyAaiaabMcacaaMc8UaaeOtaiaabggacaqGobGaae4tamaaBaaameaacaqGYaaabeaaliaab+cacaqGdbGaaeyDaiaabs5aaOGaayPKHaaaaa@5A2F@

Ans.

( i ) C 6 H 5 NH 2 Aniline + CHCl 3 +3alc.KOH Carbylamine reaction 3H 2 O+3KCl+ C 6 H 5 NC Phenyl isocyanide ( ii ) C 6 H 5 N 2 Cl Benzenediazonium chloride + H 3 PO 2 + H 2 O C 6 H 6 Benzene + N 2 + H 3 PO 3 +HCl ( iii ) C 6 H 5 NH 2 Aniline +conc .H 2 SO 4 C 6 H 5 N + H 3 HS O 4 Aniliniumhydrogensulphate ( iv ) C 6 H 5 N 2 Cl Benzenediazonium chloride + C 2 H 5 OH Ethanol C 6 H 6 Benzene + CH 3 CHO Ethanal + N 2 +HCl MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVy0xg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@8B9B@

(vii)

Q.207 An aromatic compound ‘A’ on treatment with aqueous ammonia and heating forms compound ‘B’ which on heating with Br2 and KOH forms a compound ‘C’ of molecular formula C6H7N. Write the structures and IUPAC names of compounds A, B and C.

Ans.

It is given that compound ‘C’ having the molecular formula, C6H7N is formed by heating compound ‘B’ with Br2 and KOH. This is a Hoffmann bromamide degradation reaction. Therefore, compound ‘B’ is an amide and compound ‘C’ is an amine. The only amine having the molecular formula, C6H7N is aniline, (C6H5NH2).

Therefore, compound ‘B’ (from which ‘C’ is formed) must be benzamide, (C6H5CONH2).

Further, benzamide is formed by heating compound ‘A’ with aqueous ammonia. Therefore, compound ‘A’ must be benzoic acid.

The given reactions can be explained with the help of the following equations:

Q.208 Give the structures of A, B and C in the following reactions:

Ans.

(i)

Q.209

Accomplish the following conversions:

(i) Nitrobenzene to benzoic acid
(ii) Benzene to m-bromophenol
(iii) Benzoic acid to aniline
(iv) Aniline to 2,4,6-tribromofluorobenzene
(v) Benzyl chloride to 2-phenylethanamine
(vi) Chlorobenzene to p-chloroaniline
(vii) Aniline to p-bromoaniline
(viii) Benzamide to toluene
(ix) Aniline to benzyl alcohol.

Ans.

Q.210 Write short notes on the following:
(i) Carbylamine reaction
(ii) Diazotisation
(iii) Hofmann’s bromamide reaction
(iv) Coupling reaction
(v) Ammonolysis
(vi) Acetylation
(vii) Gabriel phthalimide synthesis.

Ans.

(i) Carbylamine reaction

Carbylamine reaction is used as a test for the identification of primary amines. When aliphatic and aromatic primary amines are heated with chloroform and ethanolic potassium hydroxide, carbylamines (or isocyanides) are formed. These carbylamines have very unpleasant odours. Secondary and tertiary amines do not respond to this test.

R NH 2 Primary amine Forexample, + CHCl 3 Chloroform + 3KOH( alc. ) Potassium hydroxide Δ RNC Carbylamine +3KCl+ 3H 2 O CH 3 NH 2 Methanamine + CHCl 3 +3KOH( alc. ) Δ CH 3 NC Methylcarblylamine ormethylisocyamide +3KCl+ 3H 2 O MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqk0xg9LrFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@117A@

(ii) Diazotisation
Aromatic primary amines react with nitrous acid (prepared in situ from NaNO2 and a mineral acid such as HCl) at low temperatures (273-278 K) to form diazonium salts. This conversion of aromatic primary amines into diazonium salts is known as diazotization. For example, on treatment with NaNO2 and HCl at 273-278 K, aniline produces benzenediazonium chloride, with NaCl and H2O as by-products.


(iii) Hoffmann bromamide reaction

When an amide is treated with bromine in an aqueous or ethanolic Solution of sodium hydroxide, a primary amine with one carbon atom less than the original amide is produced. This degradation reaction is known as Hoffmann bromamide reaction. This reaction involves the migration of an alkyl or aryl group from the carbonyl carbon atom of the amide to the nitrogen atom.

(iv) Coupling reaction

The reaction of joining two aromatic rings through the -N=N-bond is known as coupling reaction. Arenediazonium salts such as benzene diazonium salts react with phenol or aromatic amines to form coloured azo compounds.

It can be observed that, the para-positions of phenol and aniline are coupled with the diazonium salt. This reaction proceeds through electrophilic substitution.

(v) Ammonolysis

When an alkyl or benzyl halide is allowed to react with an ethanolic Solution of ammonia, it undergoes nucleophilic substitution reaction in which the halogen atom is replaced by an amino (-NH2) group. This process of cleavage of the carbon-halogen bond is known as ammonolysis.

When this substituted ammonium salt is treated with a strong base such as sodium hydroxide, amine is obtained.

R N + H 3 X +NaOH R NH 2 Amine + H 2 O+NaX MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqk0xg9LrFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeaacaGaaiaabeqaamaabaabaaGcbaGaaeOuaiaaykW7caqGtaIaaGPaVpaaxacabaGaaeOtaaWcbeqaaiabgUcaRaaakiaabIeadaWgaaWcbaGaae4maaqabaGccaaMc8+aaCbiaeaacaqGybaaleqabaGaeyOeI0caaOGaaGPaVlaabUcacaaMc8UaaeOtaiaabggacaqGpbGaaeisaiaaykW7daGdKaWcbaaabeGccaGLsgcacaaMc8+aaCbeaeaacaqGsbGaaGPaVlaabobicaaMc8UaaeOtaiaabIeadaWgaaWcbaGaaeOmaaqabaaabaGaaeyqaiaab2gacaqGPbGaaeOBaiaabwgaaeqaaOGaaGPaVlaabUcacaaMc8UaaeisamaaBaaaleaacaqGYaaabeaakiaab+eacaaMc8Uaae4kaiaaykW7caqGobGaaeyyaiaabIfaaaa@6868@

Though primary amine is produced as the major product, this process produces a mixture of primary, secondary and tertiary amines, and also a quaternary ammonium salt as shown.

RNH 2 ( ) RX R 2 NH ( ) RX R 3 N ( ) RX R 4 N + X Quaternary ammoniumsalt MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVy0xg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@79C0@

(vi) Acetylation

Acetylation (or ethanoylation) is the process of introducing an acetyl group into a molecule.

Aliphatic and aromatic primary and secondary amines undergo acetylation reaction by nucleophilic substitution when treated with acid chlorides, anhydrides or esters. This reaction involves the replacement of the hydrogen atom of -NH2 or > NH group by the acetyl group, which in turn leads to the production of amides. To shift the equilibrium to the right hand side, the HCl formed during the reaction is removed as soon as it is formed. This reaction is carried out in the presence of a base (such as pyridine) which is stronger than the amine.

When amines react with benzoyl chloride., the reaction is also known as benzoylation. For example,

(vii) Gabriel phthalimide synthesis

Gabriel phthalimide synthesis is a very useful method for the preparation of aliphatic primary amines. It involves the treatment of phthalimide with ethanolic potassium hydroxide to form potassium salt of phthalimide. This salt is further heated with alkyl halide, followed by alkaline hydrolysis to yield the corresponding primary amine.

Q.211 Describe a method for the identification of primary, secondary and tertiary amines. Also write chemical equations of the reactions involved.

Ans.

Primary, secondary and tertiary amines can be identified and distinguished by Hinsberg’s test. In this test, the amines are allowed to react with Hinsberg’s reagent, benzenesulphonyl chloride (C6H5SO2Cl). The three types of amines react differently with Hinsberg’s reagent. Therefore, they can be easily identified using Hinsberg’s reagent.

Primary amines: They react with benzenesulphonyl chloride to form N-alkylbenzenesulphonyl amide which is soluble in alkali.

Due to the presence of a strong electron-withdrawing sulphonyl group in the sulphonamide, the H-atom attached to nitrogen can be easily released as proton. So, it is acidic and dissolves in alkali.

Secondary amines: They react with Hinsberg’s reagent to give a sulphonamide which is insoluble in alkali.

There is no H-atom attached to the N-atom in the sulphonamide. Therefore, it is not acidic and insoluble in alkali.

Tertiary amines: They do not react with Hinsberg’s reagent at all.

Q.212 How will you convert:

(i) Ethanoic acid into methanamine
(ii) Hexanenitrile into 1-aminopentane
(iii) Methanol to ethanoic acid
(iv) Ethanamine into methanamine
(v) Ethanoic acid into propanoic acid
(vi) Methanamine into ethanamine
(vii) Nitromethane into dimethylamine
(viii) Propanoic acid into ethanoic acid

Ans.

(i)

Q.213 Arrange the following:
(i) In decreasing order of the pKbvalues:
C2H5NH2, C6H5NHCH3, (C2H5)2NH and C6H5NH2

(ii) In increasing order of basic strength:
C6H5NH2, C6H5N(CH3)2, (C2H5)2NH and CH3NH2

(iii) In increasing order of basic strength:
Aniline, p-nitroaniline and p-toluidine
C6H5NH2, C6H5NHCH3, C6H5CH2NH2.

(iv) In decreasing order of basic strength in gas phase:
C2H5NH2, (C2H5)2NH, (C2H5)3N and NH3

(v) In increasing order of boiling point:
C2H5OH, (CH3)2NH, C2H5NH2

(vi) In increasing order of solubility in water:
C6H5NH2, (C2H5)2NH, C2H5NH2.

Ans.

(i) In C2H5NH2, only one -C2H5 group is present while in (C2H5)2NH, two -C2H5 groups are present. Thus, the +I effect is more in (C2H5)2NH than in C2H5NH2. Therefore, the electron density over the N-atom is more in (C2H5)2NH than in C2H5NH2. Hence, (C2H5)2NH is more basic than C2H5NH2.

Also, both C6H5NHCH3 and C6H5NH2 are less basic than (C2H5)2NH and C2H5NH2 due to the delocalization of the lone pair in the former two. Further, among C6H5NHCH3 and C6H5NH2, the former will be more basic due to the +I effect of -CH3 group. Hence, the order of increasing basicity of the given compounds is as follows:

C6H5NH2 < C6H5NHCH3 < C2H5NH2 < (C2H5)2NH

We know that the higher the basic strength, the lower is the pKb values.

C6H5NH2 > C6H5NHCH3 > C2H5NH2 > (C2H5)2NH

(ii) C6H5N(CH3)2 is more basic than C6H5NH2 due to the presence of the +I effect of two -CH3 groups in C6H5N(CH3)2. Further, CH3NH2 contains one -CH3 group while (C2H5)2NH contains two -C2H5 groups. Thus, (C2H5)2NH is more basic than C2H5NH2.

Now, C6H5N(CH3)2 is less basic than CH3NH2 because of the-R effect of -C6H5 group. Hence, the increasing order of the basic strengths of the given compounds is as follows:

C6H5NH2 < C6H5N(CH3)2 < CH3NH2 < (C2H5)2NH

(iii) (a)

In p-toluidine, the presence of electron-donating -CH3 group increases the electron density on the N-atom.

Thus, p-toluidine is more basic than aniline.

On the other hand, the presence of electron-withdrawing

-NO2 group decreases the electron density over the N-atom in p-nitroaniline. Thus, p- nitroaniline is less basic than aniline.

Hence, the increasing order of the basic strengths of the given compounds is as follows:

p-Nitroaniline < Aniline < p-Toluidine

(b) C6H5NHCH3 is more basic than C6H5NH2 due to the presence of electron-donating -CH3 group in C6H5NHCH3.

Again, in C6H5NHCH3, -C6H5 group is directly attached to the N-atom. However, it is not so in C6H5CH2NH2. Thus, in C6H5NHCH3, the -R effect of -C6H5 group decreases the electron density over the N-atom. Therefore, C6H5CH2NH2 is more basic than C6H5NHCH3. Hence, the increasing order of the basic strengths of the given compounds is as follows:

C6H5NH2 < C6H5NHCH3 < C6H5CH2NH2.

(iv) In the gas phase, there is no solvation effect. As a result, the basic strength mainly depends upon the +I effect. The higher the +I effect, the stronger is the base. Also, the greater the number of alkyl groups, the higher is the +I effect. Therefore, the given compounds can be arranged in the decreasing order of their basic strengths in the gas phase as follows:

(C2H5)3N > (C2H5)2NH > C2H5NH2 > NH3

(v) The boiling points of compounds depend on the extent of H-bonding present in that compound. The more extensive the H-bonding in the compound, the higher is the boiling point. (CH3)2NH contains only one H-atom whereas C2H5NH2 contains two H-atoms. Then, C2H5NH2 undergoes more extensive H-bonding than (CH3)2NH. Hence, the boiling point of C2H5NH2 is higher than that of (CH3)2NH.

Further, O is more electronegative than N. Thus, C2H5OH forms stronger H-bonds than C2H5NH2. As a result, the boiling point of C2H5OH is higher than that of C2H5NH2 and (CH3)2NH.

Now, the given compounds can be arranged in the increasing order of their boiling points as follows:

(CH3)2NH < C2H5NH2 < C2H5OH

(vi) The more extensive the H-bonding, the higher is the solubility. C2H5NH2 contains two H-atoms whereas (C2H5)2NH contains only one H-atom. Thus, C2H5NH2 undergoes more extensive H-bonding than (C2H5)2NH. Hence, the solubility in water of C2H5NH2 is more than that of (C2H5)2NH.

Further, the solubility of amines decreases with increase in the molecular mass. This is because the molecular mass of amines increases with an increase in the size of the hydrophobic part.

The molecular mass of C6H5NH2 is greater than that of C2H5NH2 and (C2H5)2NH.

Hence, the increasing order of their solubility in water is as follows:

C6H5NH2 < (C2H5)2NH < C2H5NH2

Q.214 What is a biodegradable polymer? Give an example of a biodegradable aliphatic polyester.

Ans.

A polymer that can be decomposed by bacteria is called a biodegradable polymer. Poly-β-hydroxybutyrate-co-β- hydroxyvalerate (PHBV) is a biodegradable aliphatic polyester.

Q.215 How is dacron obtained from ethylene glycol and terephthalic acid?

Ans.

The condensation polymerisation of ethylene glycol and terephthalic acid leads to the formation of dacron.

Q.216 Identify the monomer in the following polymeric structures.

Ans.

(i) The monomers of the given polymeric structure are decanoic acid [HOOC – (CH2)8 – COOH] and [H2N(CH2)6NH2] hexamethylene diamine.

(ii) The monomers of the given polymeric structure are

Q.317 Account for the following:

(i) pKb of aniline is more than that of methylamine.
(ii) Ethylamine is soluble in water whereas aniline is not.
(iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.
(iv) Although amino group is o, p- directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline.
(v) Aniline does not undergo Friedel-Crafts reaction.
(vi) Diazonium salts of aromatic amines are more stable than those of aliphatic amines.
(vii) Gabriel phthalimide synthesis is preferred for synthesising primary amines.

Ans.

(i) pKb of aniline is more than that of methylamine:

Aniline undergoes resonance and as a result, the electrons on the N-atom are delocalized over the benzene ring. Therefore, the electrons on the N-atom are less available to donate.

On the other hand, in case of methylamine (due to the +I effect of methyl group), the electron density on the N-atom is increased. As a result, aniline is less basic than methylamine.

Thus, pKb of aniline is more than that of methylamine.

(ii) Ethylamine is soluble in water whereas aniline is not:

Ethylamine when added to water forms intermolecular H-bonds with water. Hence, it is soluble in water.

But aniline does not undergo H-bonding with water to a very large extent due to the presence of a large hydrophobic -C6H 5 group. Hence, aniline is insoluble in water.

(iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide:

Due to the +I effect of -CH3 group, methylamine is more basic than water. Therefore, in water, methylamine produces OH ions by accepting H+ ions from water

CH 3 NH 2 +HOH CH 3 NH 3 + OH MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqk0xg9LrFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeaacaGaaiaabeqaamaabaabaaGcbaGaae4qaiaabIeadaWgaaWcbaGaae4maaqabaGccaaMc8Uaae4eGiaaykW7caqGobGaaeisamaaBaaaleaacaqGYaaabeaakiaaykW7caqGRaGaaGPaVlaabIeacaaMc8Uaae4eGiaaykW7caqGpbGaaeisaiaaykW7daGdKaWcbaaabeGccaGLsgcacaaMc8Uaae4qaiaabIeadaWgaaWcbaGaae4maaqabaGccaaMc8Uaae4eGiaaykW7caqGobGaaeisamaaBaaaleaacaqGZaaabeaakiaaykW7caqGRaGaaGPaVlaab+eacaqGibWaaWbaaSqabeaacaqGTaaaaaaa@5EB6@

Ferric chloride (FeCl3) dissociates in water to form Fe3+ and Cl ions.

FeCl 3 Fe 3+ + 3Cl MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqk0xg9LrFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeaacaGaaiaabeqaamaabaabaaGcbaGaaeOraiaabwgacaqGdbGaaeiBamaaBaaaleaacaqGZaaabeaakiaaykW7daGdKaWcbaaabeGccaGLsgcacaaMc8UaaeOraiaabwgadaahaaWcbeqaaiaabodacaqGRaaaaOGaaGPaVlaabUcacaaMc8Uaae4maiaaboeacaqGSbWaaWbaaSqabeaacaqGtacaaaaa@4B8F@

Then, OH ion reacts with Fe3+ ion to form a precipitate of hydrated ferric oxide.

2Fe 3+ + 6OH Fe 2 O 3 . 33H 2 O Hydrated ferricoxide MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqk0xg9LrFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@68CE@

(iv) Although amino group is o,p- directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m- nitroaniline: Nitration is carried out in an acidic medium. In an acidic medium, aniline is protonated to give anilinium ion (which is meta-directing).

For this reason, aniline on nitration gives a substantial amount of m-nitroaniline.

(v) Aniline does not undergo Friedel-Crafts reaction: A Friedel-Crafts reaction is carried out in the presence of AlCl3. But AlCl 3 is acidic in nature, while aniline is a strong base. Thus, aniline reacts with AlCl3 to form a salt (as shown in the following equation).

Due to the positive charge on the N-atom, electrophilic substitution in the benzene ring is deactivated. Hence, aniline does not undergo the Friedel-Crafts reaction.

(vi) Diazonium salts of aromatic amines are more stable than those of aliphatic amines: The diazonium ion undergoes resonance as shown below:

This resonance accounts for the stability of the diazonium ion. Hence, diazonium salts of aromatic amines are more stable than those of aliphatic amines.

(vii) Gabriel phthalimide synthesis is preferred for synthesising primary amines: Gabriel phthalimide synthesis results in the formation of 1° amine only. 2° or 3° amines are not formed in this synthesis. Thus, a pure 1° amine can be obtained. Therefore, Gabriel phthalimide synthesis is preferred for synthesizing primary amines.

Q.218 Write the names and structures of the monomers of the following polymers:

(i) Buna-S (ii) Buna-N
(iii) Dacron (iv) Neoprene

Ans.

Polymer Monomer Structure of monomer
(i) Buna-S 1, 3-butadiene CH2 = CH-CH=CH2
Styrene C6H5CH=CH2
(ii) Buna-N 1, 3-butadiene CH2 = CH-CH=CH2
Acrylonitrile CH2 =CH-CN
(iii) Dacron Ethylene glycol HOH2C-CH2OH
Terephthalic acid
(iv) Neoprene Chloroprene

Q.219 What are the monomeric repeating units of Nylon-6 and Nylon-6, 6?

Ans.

The monomeric repeating unit of nylon 6 is [NH – (CH2)5 – CO], which is derived from caprolactam. The monomeric repeating unit of nylon 6, 6 is [NH – (CH2)6 – NH – CO – (CH2)4 – CO], which is derived from hexamethylene diamine and adipic acid.

Q.220 Give one chemical test to distinguish between the following pairs of compounds.

(i) Methylamine and dimethylamine
(ii) Secondary and tertiary amines
(iii) Ethylamine and aniline
(iv) Aniline and benzylamine
(v) Aniline and N-methylaniline.

Ans.

(i) Methylamine and dimethylamine can be distinguished by the carbylamine test. Carbylamine test: Aliphatic and aromatic primary amines on heating with chloroform and ethanolic potassium hydroxide form foul-smelling isocyanides or carbylamines. Methylamine (being an aliphatic primary amine) gives a positive carbylamine test, but dimethylamine does not.

CH 3 NH 2 + CHCl 3 Methylamine( ) +3KOH Δ CH 3 NC+3KCl Methylisocyanide ( foulsmell ) + 3H 2 ( CH 3 ) 2 NH+ CHCl 3 +3KOH Δ Noreaction MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqk0xg9LrFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@B736@

(ii) Secondary and tertiary amines can be distinguished by allowing them to react with Hinsberg’s reagent (benzenesulphonyl chloride, C6H5SO2Cl).

Secondary amines react with Hinsberg’s reagent to form a product that is insoluble in an alkali. For example, N, N-diethylamine reacts with Hinsberg’s reagent to form N, N-diethylbenzenesulphonamide, which is insoluble in an alkali. Tertiary amines, however, do not react with Hinsberg’s reagent.

(iii) Ethylamine and aniline can be distinguished using the azo-dye test. A dye is obtained when aromatic amines react with HNO2 (NaNO2 + dil.HCl) at 0-5°C, followed by a reaction with the alkaline Solution of 2-naphthol. The dye is usually yellow, red, or orange in colour. Aliphatic amines give a brisk effervescence due (to the evolution of N2 gas) under similar conditions.

(iv) Aniline and benzylamine can be distinguished by their reactions with the help of nitrous acid, which is prepared in situ from a mineral acid and sodium nitrite. Benzylamine reacts with nitrous acid to form unstable diazonium salt, which in turn gives alcohol with the evolution of nitrogen gas.

On the other hand, aniline reacts with HNO2 at a low temperature to form stable diazonium salt. Thus, nitrogen gas is not evolved.

C 6 H 5 NH 2 273278K NaNO 2 +HCl C 6 H 5 N + 2 C l +NaCl+ 2H 2 O MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqk0xg9LrFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@7125@

(v) Aniline and N-methylaniline can be distinguished using the Carbylamine test. Primary amines, on heating with chloroform and ethanolic potassium hydroxide, form foul- smelling isocyanides or carbylamines. Aniline, being an aromatic primary amine, gives positive carbylamine test. However, N-methylaniline, being a secondary amine does not.

Q.221 Discuss the main purpose of vulcanisation of rubber.

Ans.

Natural rubber though useful has some problems associated with its use. These limitations are discussed below:

1. Natural rubber is quite soft and sticky at room temperature. At elevated temperatures (> 335 K), it becomes even softer. At low temperatures (< 283 K), it becomes brittle. Thus, to maintain its elasticity, natural rubber is generally used in the temperature range of 283 K-335 K.

2. It has the capacity to absorb large amounts of water.

3. It has low tensile strength and low resistance to abrasion.

4. It is soluble in non-polar solvents.

5. It is easily attacked by oxidising agents.

Vulcanisation of natural rubber is done to improve upon all these properties. In this process, a mixture of raw rubber with sulphur and appropriate additive is heated at a temperature range between 373 K and 415 K.

Q.222 How does the presence of double bonds in rubber molecules influence their structure and reactivity?

Ans.

Natural rubber is a linear cis-polyisoprene in which the double bonds are present between C2 and C3 of the isoprene units.

Because of this cis-configuration, intermolecular interactions between the various strands of isoprene are quite weak. As a result, various strands in natural rubber are arranged randomly. Hence, it shows elasticity.

Q.223 Label the hydrophilic and hydrophobic parts in the following compounds.

Ans.

(i)

(ii)

(iii)

Q.224 Write the name and structure of one of the common initiators used in free radical addition polymerisation.

Ans.

One common initiator used in free radical addition polymerisation is benzoyl peroxide. Its structure is given below:

Q.225 Write the monomers used for getting the following polymers.

(i) Polyvinyl chloride (ii) Teflon (iii) Bakelite

Ans.

(i) Vinyl chloride (CH2 = CHCl)

(ii) Tetrafluoroethylene (CF2 = CF2)

(iii) Formaldehyde (HCHO) and phenol (C6H5OH)

Q.226 Define thermoplastics and thermosetting polymers with two examples of each.

Ans.

Thermoplastic polymers are linear (slightly branched) long chain polymers, which can be repeatedly softened and hardened on heating. Hence, they can be modified again and again. Examples include polythene and polystyrene. Thermosetting polymers are cross-linked or heavily branched polymers which get hardened during the molding process. These plastics cannot be softened again on heating. Examples of thermosetting plastics include bakelite and urea-formaldehyde resins.

Q.227 Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.

(i) (CH3)2CHNH2 (ii) CH3(CH2)2NH2
(iii) CH3NHCH(CH3)2 (iv) (CH3)3CNH2
(v) C6H5NHCH3 (vi) (CH3CH2)2NCH3
(vii) m-BrC6H4NH2

Ans.

(i) 1-Methylethanamine (10 amine)

(ii) Propan-1-amine (10 amine)

(iii) N-Methyl-2-methylethanamine (20 amine)

(iv) 2-Methylpropan-2-amine (10 amine)

(v) N-Methylbenzamine or N-methylaniline (20 amine)

(vi) N-Ethyl-N-methylethanamine (30 amine)

(vii) 3-Bromobenzenamine or 3-bromoaniline (10 amine)

Q.228 Write the free radical mechanism for the polymerisation of ethene.

Ans.

Polymerisation of ethene to polythene consists of heating or exposing to light a mixture of ethene with a small amount of benzoyl peroxide as the initiator.

The reaction involved in this process is given below:

Q.229 Explain the term copolymerisation and give two examples.

Ans.

Monomers from two or more different monomeric units is called copolymerisation. Multiple units of each monomer are present in a copolymer. The process of forming polymer Buna−S from 1, 3-butadiene and styrene is an example of copolymerisation

Nylon 6, 6 is also a copolymer formed by hexamethylenediamine and adipic acid.

Q.230 How can you differentiate between addition and condensation polymerisation?

Ans.

S. No. Addition polymerisation Condensation polymerisation
(i) It is the process of repeated addition of monomers, possessing double or triple bonds to form polymers. It is the process of formation of polymers by repeated condensation reactions between two different bi-functional or tri-functional monomers. A small molecule such as water or hydrochloric acid is eliminated in each condensation.
(ii) For example, polythene is formed by addition polymerisation of ethene.

For example, nylon 6, 6 is formed by condensation polymerisation of hexamethylenediamine and adipic acid.

Q.231 Enumerate the reactions of D-glucose which cannot be explained by its open chain structure.

Ans.

(1) Aldehydes give Schiff’s test and react with NaHSO3 to form the hydrogen sulphite addition product. However, glucose does not undergo these reactions.

(2) The pentaacetate of glucose does not react with hydroxylamine. This indicates that a free −CHO group is absent from glucose.

(3) Glucose exists in two crystalline forms ∝ and β. The ∝-form (m.p. = 419 K) crystallises from a concentrated solution of glucose at 303 K and the β-form (m.p = 423K) crystallises from a hot and saturated aqueous solution at 371 K. This behavior cannot be explained by the open chain structure of glucose.

Q.232 If water contains dissolved calcium hydrogen carbonate, out of soaps and synthetic detergents which one will you use for cleaning clothes?

Ans.

Synthetic detergents are preferred for cleaning clothes. When soaps are dissolved in water containing calcium ions, these ions form insoluble salts that are of no further use. However, when synthetic detergents are dissolved in water containing calcium ions, these ions form soluble salts that act as cleansing agents.

Q.233 Explain the cleansing action of soaps.

Ans.

Soap molecules form micelles around an oil droplet (dirt) in such a way that the hydrophobic parts of the stearate ions attach themselves to the oil droplet and the hydrophilic parts project outside the oil droplet. Due to the polar nature of the hydrophilic parts, the stearate ions (along with the dirt) are pulled into water, thereby removing the dirt from the cloth.

Q.234 Can you use soaps and synthetic detergents to check the hardness of water?

Ans.

Soaps get precipitated in hard water, but not in soft water. Therefore, soaps can be used for checking the hardness of water. However, synthetic detergents do not get precipitated either in hard water or in soft water. Therefore, synthetic detergents cannot be used for checking the hardness of water.

Q.235 Why do elastomers posssess elastic properties?

Ans.

In elastomers, the polymer chains are held together by weak intermolecular forces which allow them to be stretched. Moreover, a few cross-links between the polymer chains allows them to retract when the forces are withdrawn. Thus, elastomers possess elastic properties.

Q.236 Is (–NH – CHR – CO)n, a homopolymer or copolymer?

Ans.

(–NH – CHR – CO)n is a homopolymer because it is obtained from a single monomer unit, NH2−CHR−COOH.

Q.237 How will you bring about the following conversions in not more than two steps?

(i) Propanone to Propene
(ii) Benzoic acid to Benzaldehyde
(iii) Ethanol to 3-Hydroxybutanal
(iv) Benzene to m-Nitroacetophenone
(v) Benzaldehyde to Benzophenone
(vi) Bromobenzene to 1-Phenylethanol
(vii) Benzaldehyde to 3-Phenylpropan-1-ol
(viii) Benazaldehyde to α-Hydroxyphenylacetic acid
(ix) Benzoic acid to m- Nitrobenzyl alcohol

Ans.

Q.238 Define the term polymerisation.

Ans.

Polymerisation is the process of forming high molecular mass (103 – 103u) macromolecules, which consist of repeating structural units derived from monomers. In a polymer, various monomer units are joined by strong covalent bonds.

Q.239 How do you explain the functionality of a monomer?

Ans.

The functionality of a monomer is the number of binding sites that is/are present in that monomer.

For example, the functionality of monomers such as ethene and propene is one and that of 1, 3-butadiene and adipic acid is two.

Q.240 Why do soaps not work in hard water?

Ans.

Soaps are sodium or potassium salts of long-chain fatty acids. Hard water contains calcium and magnesium ions. When soaps are dissolved in hard water, these ions displace sodium or potassium from their salts and form insoluble calcium or magnesium salts of fatty acids. These insoluble salts separate as scum.

This is the reason why soaps do not work in hard water.

Q.241 Distinguish between the terms homopolymer and copolymer and give an example of each.

Ans.

S. No. Homopolymer Copolymer
(i) The polymers that are formed by the polymerisation of a single monomer are known as homopolymers. In other words, the repeating units of homopolymers are derived only from one monomer. The polymers whose repeating units are derived from two types of monomers are known as copolymers.
(ii) For example, polythene is a homopolymer of ethene . For example, Buna−S is a copolymer of 1, 3-butadiene and styrene.

Q.242 What are natural and synthetic polymers? Give two examples of each type.

Ans.

Natural polymers are polymers that are found in nature. They are formed by plants and animals. Examples include protein, cellulose, starch, etc. Synthetic polymers are polymers made by human beings. Examples include plastics (polythene), synthetic fibres (nylon 6, 6), synthetic rubbers (Buna − S).

Q.243 Explain the terms polymer and monomer.

Ans.

Polymers are high molecular mass macromolecules composed of repeating structural units derived from monomers. Polymers have a high molecular mass (103 – 107 u). In a polymer, various monomer units are joined by strong covalent bonds. Polymers can be natural as well as synthetic. Polythene, rubber, and nylon 6, 6 are examples of polymers.

Monomers are simple, reactive molecules that combine with each other in large numbers through covalent bonds to give rise to polymers. For example, ethene, propene, styrene, vinyl chloride.

Q.244 What are biodegradable and non-biodegradable detergents? Give one example of each.

Ans.

Detergents that can be degraded by bacteria are called biodegradable detergents. Such detergents have straight hydrocarbon chains. For example: sodium lauryl sulphate. Detergents that cannot be degraded by bacteria are called non-biodegradable detergents. Such detergents have highly-branched hydrocarbon chains. For example: sodium -4- (1, 3, 5, 7- tetramethyloctyl) benzene sulphonate

Q.245 Explain the following terms with suitable examples

(i) cationic detergents
(ii) anionic detergents and
(iii) non-ionic detergents.

Ans.

(i) Cationic detergents

Cationic detergents are quaternary ammonium salts of acetates, chlorides or bromides. These are called cationic detergents because the cationic part of these detergents contains a long hydrocarbon chain and a positive charge on the N-atom. For example, cetyltrimethylammonium bromide.

(ii) Anionic detergents

Anionic detergents are of two types:

1. Sodium alkyl sulphates: These detergents are sodium salts of sulphonated long chain alcohols. They are prepared by first treating these alcohols with concentrated sulphuric acid and then with an alkali like sodium hydroxide. For example,sodium lauryl sulphate, C11H23CH2OSO3 Na+.

2. Sodium alkylbenzenesulphonates: These detergents are sodium salts of long chain alkylbenzenesulphonic acids. They are prepared by Friedel-Crafts alkylation of benzene with long chain alkyl halides or alkenes. The obtained product is first treated with concentrated sulphuric acid and then with sodium hydroxide. Sodium 4-(1-dodecyl) benzenesulphonate (SDS) is an example of anionic detergents.

(iii) Non-ionic detergents

Molecules of these detergents do not contain any ions. These detergents are esters of alcohols having high molecular mass. They are obtained by reacting polyethylene glycol and stearic acid.

Q.246 An organic compound contains 69.77% carbon, 11.63% hydrogen and rest oxygen. The molecular mass of the compound is 86. It does not reduce Tollens’ reagent but forms an addition compound with sodium hydrogensulphite and give positive iodoform test. On vigorous oxidation it gives ethanoic and propanoic acid. Write the possible structure of the compound.

Ans.

% of carbon = 69.77 %

% of hydrogen = 11.63 %

% of oxygen = {100 – (69.77 + 11.63)}%

= 18.6 %

Thus, the ratio of the number of carbon, hydrogen, and oxygen atoms in the organic compound can be given as:

C:H:O= 69.77 12 : 11.63 1 : 18.6 16 =5.81:11.63:1.16 =5:10:1 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqk0xg9LrFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeaacaGaaiaabeqaamaabaabaaGceaqabeaacaqGdbGaaeOoaiaabIeacaqG6aGaae4taiaaykW7caqG9aGaaGPaVpaalaaabaGaaeOnaiaabMdacaqGUaGaae4naiaabEdaaeaacaqGXaGaaeOmaaaacaqG6aWaaSaaaeaacaqGXaGaaeymaiaab6cacaqG2aGaae4maaqaaiaabgdaaaGaaeOoamaalaaabaGaaeymaiaabIdacaqGUaGaaeOnaaqaaiaabgdacaqG2aaaaaqaaiaab2dacaqG1aGaaeOlaiaabIdacaqGXaGaaeOoaiaabgdacaqGXaGaaeOlaiaabAdacaqGZaGaaeOoaiaabgdacaqGUaGaaeymaiaabAdaaeaacaqG9aGaaeynaiaabQdacaqGXaGaaeimaiaabQdacaqGXaaaaaa@5F94@

Therefore, the empirical formula of the compound is C5H10O. Now, the empirical formula mass of the compound can be given as:

5 x 12 + 10 x1 + 1 x 16 = 86

Molecular mass of the compound = 86

Therefore, the molecular formula of the compound is given by C5H10O.

Since the given compound does not reduce Tollen’s reagent, it is not an aldehyde. Again, the compound forms sodium hydrogen sulphate addition products and gives a positive iodoform test. Since the compound is not an aldehyde, it must be a methyl ketone. The given compound also gives a mixture of ethanoic acid and propanoic acid. Hence, the given compound is pentan-2-one.

The given reactions can be explained by the following equations:

Q.247 How are synthetic detergents better than soap?

Ans.

Soaps work in soft water. However, they are not effective in hard water. In contrast, synthetic detergents work both in soft water and hard water. Therefore, synthetic detergents are better than soaps.

Q.248 What problem arises in using alitame as artificial sweetener?

Ans.

Alitame is a high potency sweetener. It is difficult to control the sweetness of food while using alitame as an artificial sweetener.

Q.249 Name the sweetening agent used in the preparation of sweets for a diabetic patient.

Ans.

Artificial sweetening agents such as saccharin and aspartame can be used in preparing sweets for diabetic patients.

Q.250 What are artificial sweetening agents? Give two examples.

Ans.

Artificial sweetening agents are chemicals that sweeten food. However, unlike natural sweeteners, they do not add calories to our body. They do not harm the human body. Examples of artificial sweeteners are aspartame and saccharin.

Q.251 Why is use of aspartame limited to cold foods and drinks?

Ans.

Aspartame becomes unstable at cooking temperature. This is the reason why its use is limited to cold foods and drinks.

Q.252 What are food preservatives?

Ans.

Food preservatives are chemicals that prevent food from spoilage due to microbial growth. Table salt, sugar, vegetable oils and sodium benzoate (C6H5COONa) and salts of propanoic acid are some examples of food preservatives.

Q.253 Give plausible explanation for each of the following:

(i) Cyclohexanone forms cyanohydrin in good yield but 2, 2, 6 trimethylcyclohexanone does not.
(ii) There are two -NH2 groups in semicarbazide. However, only one is involved in the formation of semicarbazones.
(iii) During the preparation of esters from a carboxylic acid and an alcohol in the presence of an acid catalyst, the water or the ester should be removed as soon as it is formed.

Ans.

(i) Cyclohexanones form cyanohydrins according to the following equation.

In this case, the nucleophile CN can easily attack without any steric hindrance However, in the case of 2, 2, 6 trimethylcydohexanone, methyl groups at α-position offer steric hindrances and as a result, CN cannot attack effectively.

For this reason, it does not form a cyanohydrin.

(ii) Semicarbazide undergoes resonance involving only one of the two -NH2 groups, which is attached directly to the carbonyl-carbon atom.

Therefore, the electron density on -NH2 group involved in the resonance also decreases. As a result, it cannot act as a nucleophile. Since the other -NH2 group is not involved in resonance; it can act as a nucleophile and can attack carbonyl-carbon atoms of aldehydes and ketones to produce semicarbazones.

(iii) Ester along with water is formed reversibly from a carboxylic acid and an alcohol in presence of an acid.

RCOOH carboxylic acid + R’OH Alcohol H + RCOOR’ Ester + H 2 O Water MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqk0xg9LrFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@6F8C@

If either water or ester is not removed as soon as it is formed, then it reacts to give back the reactants as the reaction is reversible. Therefore, to shift the equilibrium in the forward direction i.e., to produce more ester, either of the two should be removed.

Q.254 What is tincture of iodine? What is its use?

Ans.

Tincture of iodine is a 2 − 3 percent solution of iodine in alcohol − water mixture. It is applied to wounds as an antiseptic.

Q.255 What are the main constituents of dettol?

Ans.

The main constituents of dettol are chloroxylenol and α-terpineol.

Q.256 Complete each synthesis by giving missing starting material, reagent or products

Ans.

Q.257 Describe the following:

  1. Acetylation
  2. Cannizzaro reaction
  3. Cross aldol condensation
  4. Decarboxylation

Ans.

(i) Acetylation

The introduction of an acetyl functional group into an organic compound is known as acetylation. It is usually carried out in the presence of a base such as pyridine, dimethylaniline, etc. This process involves the substitution of an acetyl group for an active hydrogen atom. Acetyl chloride and acetic anhydride are commonly used as acetylating agents.

CH 3 CH 2 OH Ethanal + CH 3 COCl AcetylChloride pyridine CH 3 COOC 2 H 5 Ethylacetate + HCl MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqk0xg9LrFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@820A@

(ii) Cannizzaro reaction:

The self oxidation-reduction (disproportionate) reaction of aldehydes having no a- hydrogens on treatment with concentrated alkalis is known as the Cannizzaro reaction. In this reaction, two molecules of aldehydes participate where one is reduced to alcohol and the other is oxidized to carboxylic acid.

For example, when ethanol

is treated with concentrated potassium hydroxide, ethanol and potassium ethanoate are produced.