case study questions chemistry class 12

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CBSE Board Class 12 Chemistry Case Based Questions With Solutions

CBSE Board Class 12 Chemistry Case Based Questions With Solutions to prepare for the upcoming term 1 board exams are given here. It is extremely helpful since the expert team has crafted this by following the Syllabus of class 12th. Case study questions are also known as passage based problems because it includes the paragraph from which 5 or more problems are curated. 

To solve these types of problems students need to have a thorough understanding of all the basics and fundamental concepts. Also, a student who is good in problem solving skills can easily answer such questions.

Those who are preparing for their CBSE Class 12 board exam term 1 can use it to practice the questions on a daily basis. Also, the experts have given step by step solutions to all those problems so that students can cope up with the given problem very easily.

CBSE Class 12 Chemistry Case Based Questions, Assertion & Reason, MCQs

CBSE Class 12 Chemistry includes many chemical reactions, chemical bondings, theories, experiments, discoveries, etc. All of these types of things should be well versed by the students to answer the Case study, Multiple Choice Questions, Assertion and Reason. So, practicing the given CBSE Class 12 Chemistry Case Study PDF will be very handy in boosting the basic knowledge and preparing for the final papers. 

We have provided the complete set of PDFs so that candidates can prepare for the board examination very easily. Before attempting the given PDF candidates are suggested here to complete their syllabus for the term 1 exam. Doing so will aid in self-assessment and give a sense of board exam preparation.

Class 12 Chemistry Case Study Questions For Term 1 Exam  

Class 12 Chemistry Case Study Questions for Term 1 exam includes The Solid State, The P block elements, Haloalkanes and Haloarenes, Biomolecules, etc. Questions for all these chapters are given in the PDF file that are available here for free to download.

Term 1 exam is about to be held in November-December this year. So, students should start their board exam preparation early. For that purpose we have provided the complete chapter wise case based questions, assertion and reason and a lot of Objective type problems for the practice purposes.

CBSE 12 Case Based Question With Answers

CBSE Case Based Questions with Answers are designed and developed for the term 1 exam which is developed by the subject matter experts. The answers are provided to help the learners to solve the problem if they are stuck at some questions and don’t know how to tackle that.

Not only answers are given but step by step explanations are given too. By taking help of those explanations students can easily get a good grasp over the concepts.

To Download case study based questions class 12 chemistry PDF visit here or download from Selfstudys.com.

To solve Class 12 Chemistry Assertion and Reason Questions read the given statement and then reason. Now, you need to verify both the assertion and reason. If both are given correctly then you have to check whether the given reason supports the statement or not.

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CBSE 12th Standard Chemistry Subject Solution Case Study Questions With Solution 2021

By QB365 on 21 May, 2021

QB365 Provides the updated CASE Study Questions for Class 12 , and also provide the detail solution for each and every case study questions . Case study questions are latest updated question pattern from NCERT, QB365 will helps to get  more marks in Exams

QB365 - Question Bank Software

12th Standard CBSE

Final Semester - June 2015

Read the passage given below and answer the following questions: The concentration of a solute is very important in studying chemical reactions because it determines how often molecules collide in solution and thus indirectly determine the rate of reactions and the conditions at equilibrium. There are several ways to express the amount of solute present in a solution. The concentration of a solution is a measure of the amount of solute that has been dissolved in a given amount of solvent or solution. Concentration can be expressed in terms of molarity, molality, parts per million, mass percentage, volume percentage, etc. The following questions are multiple choice questions. Choose the most appropriate answer: (i) The molarity (in mol L -1 ) of the given solution will be

(ii) Which of the following is correct relationship between mole fraction and molality?

(iii) Which of the following is temperature dependent?

(iv) Which of the following is true for an aqueous solution of the solute in terms of concentration?

Read the passage given below and answer the following questions: At 298 K, the vapour pressure of pure benzene, C 6 H 6 is 0.256 bar and the vapour pressure of pure toluene C 6 H 5 CH 3 is 0.0925 bar. Two mixtures were prepared as follows: (i) 7.8 g of C 6 H 6 + 9.2 g of toluene (ii) 3.9 g of C 6 H 6 + 13.8 g of toluene The following questions are multiple choice questions. Choose the most appropriate answer: (i) The total vapour pressure (bar) of solution 1 is

(ii) Which of the given solutions have higher vapour pressure?

(iii) Mole fraction of benzene in vapour phase in solution 1 is

(iv) Solution I is an example of a/an

Read the passage given below and answer the following questions: An ideal solution may be defined as the solution which obeys Raoult's law exactly over the entire range of concentration. The solutions for which vapour pressure is either higher or lower than that predicted by Raoult's law are called non-ideal solutions. Non-ideal solutions can show either positive or negative deviations from Raoult's law depending on whether the A-B interactions in solution are stronger or weaker than A - A and B - B interactions. The following questions are multiple choice questions. Choose the most appropriate answer: (i) Which of the following solutions is/are ideal solution(s)? (i) Bromoethane and iodoethane (ii) Acetone and chloroform (iii) Benzene and acetone (iv)n-heptane and n-hexane

(ii) Which of the following is not true for positive deviations?

(iii) For water and nitric acid mixture which of the given graph is correct?

(iv) Water- HCl mixture I. shows positive deviations  II. forms minimum boiling azeotrope III. shows negative deviations  IV. forms maximum boiling azeotrope

Read the passage given below and answer the following questions: The properties of the solutions which depend only on the number of solute particles but not on the nature of the solute are called colligative properties. Relative lowering in vapour pressure is also an example of colligative properties. For an experiment, sugar solution is prepared for which lowering in vapour pressure was found to be 0.061 mm of Hg. (Vapour pressure of water at 20°C is 17.5 mm of Hg.) The following questions are multiple choice questions. Choose the most appropriate answer: (i) Relative lowering of vapour pressure for the given solution is

(ii) The vapour pressure (mm of Hg) of solution will be

(iii) Mole fraction of sugar in the solution is

(iv) The vapour pressure (mm of Hg) of water at 293 K when 25 g of glucose is dissolved in 450 g of water is

Read the passage given below and answer the following questions: Few colligative properties are: (a) relative lowering of vapour pressure: depends only on molar concentration of solute (mole fraction) and independent of its nature. (b) depression in freezing point: it is proportional to the molal concentration of solution. (c) elevation of boiling point: it is proportional to the molal concentration of solute. (d) osmotic pressure: it is proportional to the molar concentration of solute. A solution of glucose is prepared with 0.052 g at glucose in 80.2 g of water. (K f = 1.86 K kg mol -1  and K b = 5.2 K kg mol -1 ) The following questions are multiple choice questions. Choose the most appropriate answer: (i) Molality of the given solution is

(ii) Boiling point for the solution will be

(iii) The depression in freezing point of solution will be

(iv) Mole fraction of glucose in the given solution is

*****************************************

Cbse 12th standard chemistry subject solution case study questions with solution 2021 answer keys.

(i) (d) : Density of solution = 1.202 g/mL Volume of solution =  \(\frac{100 \mathrm{~g}}{1.202 \mathrm{~g} / \mathrm{mL}}=83.2 \mathrm{~mL}\) Molarity =  \(\frac{n_{\mathrm{KI}}}{\text { Volume of solution in } \mathrm{L}}\) \(=\frac{0.120 \mathrm{~mol}}{0.0832 \mathrm{~L}}=1.4423 \mathrm{~mol} \mathrm{~L}^{-1}\) (ii) (a):  \(x_{2}=\frac{n_{2}}{n_{1}+n_{2}} ; x_{1}=\frac{n_{1}}{n_{1}+n_{2}} ; \frac{x_{2}}{x_{1}}=\frac{n_{2}}{n_{1}}\) \(\frac{x_{2}}{x_{1}}=\frac{m_{2} / M_{2}}{m_{1} / M_{1}}=\frac{m_{2}}{m_{1}} \times \frac{M_{1}}{M_{2}}\)           ...(i) Molality =  \(\frac{n_{2}}{m_{1}}=\frac{m_{2}}{M_{2} \times m_{1}}\)               ...(ii) From(i) and (ii), m =  \(\frac{x_{2}}{x_{1}} \times \frac{1}{M_{1}} ; x_{1}=1-x_{2}\) Hence. x 2  =  \(\frac{m M_{1}}{1+m M_{1}}\) (iii) (a) : Mass does not depend on temperature while volume does. Hence, molarity depends on temperature. (iv) (b): 1M solution contains 1 mole of solute in less than 1000 g of the solvent whereas 1 m solution has 1 mole of the solute in 1000 g of the solvent.

(i) (b) : Moles of C 6 H 6  =  \(\frac{7.8}{78}=0.1\) Mole C 6 H 5 CH 3  =  \(\frac{9.2}{92}=0.1\) Mole fraction of C 6 H 6  =  \(\frac{0.1}{0.1+0.1}=0.5\) => Mole fraction of C 6 H 5 CH 3 = 0.5 Vapour pressure of toluene = Vapour pressure of pure toluene x mole fraction of toluene = 0.0925 x 0.5 = 0.04625 Vapour pressure of benzene = 0.256 x 0.5 = 0.128 Total vapour pressure of solution = 0.17425 (ii) (a) :  Moles of benzene in solution-II =  \(\frac{3.9}{78}=0.05\) Moles of toluene in solution-II =  \(\frac{13.8}{92}=0.15\) Vapour pressure of solution = 0.256 x 0.05 + 0.0925 x 0.15 = 0.0128 + 0.013875 = 0.026675 (iii) (c) : Mole fraction of benzene in vapour phase \(y_{\text {benzene }}=\frac{p_{\text {benzene }}}{P_{\text {total }}}=\frac{0.128}{0.17425}=0.734\) (iv) (a) : Benzene and toluene form an ideal solution.

(i) (d) : II represents negative deviations and III represents positive deviations. (ii) (b): For positive deviations  \(p_{A}>p_{A}^{\circ} x_{A} \text { and } p_{B}>p_{B}^{\circ} x_{B}\) (iii) (b) : Water and nitric acid mixture shows negative deviations from Raoult's law, hence  \(p_{A}<p_{A}^{\circ} x_{A} \text { and } p_{B}<p_{B}^{\circ} x_{B}\) (iv) (d): Water-HCl mixture shows negative deviations from Raoult's law and solutions showing negative deviations from ideal behaviour form maximum boiling azeotrope.

(i) (a) : Vapour pressure of water  \(\left(p_{A}^{\circ}\right)\)  = 17.5 mm of Hg Lowering of vapour pressure  \(\left(p_{A}^{\circ}-p_{A}\right)\) = 0.061 Relative lowering of vapour pressure   \(=\frac{p_{A}^{\circ}-p_{A}}{p_{A}^{\circ}}=\frac{0.061}{17.5}=0.00348\) (ii) (c): P = Vapour pressure of solvent - lowering in vapour pressure = 17.5 - 0.061 = 17.439 mm of Hg (iii) (a):  \(\frac{p_{A}^{\circ}-p_{A}}{p_{A}^{\circ}}=x_{B}=0.00348\) Hence, mole fraction of sugar = 0.00348 (iv) (b):  \(\frac{p_{A}^{\circ}-p_{A}}{p_{A}^{\circ}}=x_{B}=\frac{w_{B} \times M_{A}}{M_{B} \times w_{A}}\) \(\frac{17.5-p_{A}}{17.5}=\frac{25 \times 18}{450 \times 180}=5.56 \times 10^{-3}\) \(17.5-p_{A}=17.5 \times 5.56 \times 10^{-3}\) \(17.5-p_{A}=0.0973\) P = 17.40 mm Hg

(i) (b) : m  \(=\frac{0.052}{180} \times \frac{1000}{80.2}=0.0036\) (ii) (c):   \(\Delta T_{b}=K_{b} \times m=5.2 \times 0.0036=0.0187 \mathrm{~K}\) \(T_{b}=373+0.0187=373.0187 \mathrm{~K} \approx 373.02 \mathrm{~K}\) (iii) (d):  \(\Delta T_{f}=K_{f} \times m=1.86 \times 0.0036=0.067 \mathrm{~K}\) (iv) (a): Moles of glucose  \(=\frac{0.052}{180}=0.00028\) Moles 0f water =  \(\frac{80.2}{18}=4.455\) Mole fraction of glucose =  \(\frac{0.00028}{4.45+0.00028}=6.28 \times 10^{-5}\)

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CBSE Class 12 Chemistry Case Study Questions PDF

Case studies play a pivotal role in CBSE Class 12 Chemistry, as they enable students to apply theoretical knowledge to real-life scenarios. CBSE Class 12 Chemistry Case Study Questions PDF section introduces the significance of case studies in enhancing analytical skills and understanding complex chemical reactions.

Case studies challenge students to think critically, analyze experimental data, and devise problem-solving strategies. They provide a deeper understanding of chemical principles and their practical applications, fostering a holistic learning experience. Familiarize yourself with the structure of case study questions to streamline your preparation. Each case study presents a unique chemical problem, encouraging students to identify relevant concepts and devise accurate solutions.

Table of Contents

Class 12 Chemistry Case Study Questions

CBSE Class 12 Chemistry question paper will have case study questions too. These case-based questions will be objective type in nature. So, Class 12 Chemistry students must prepare themselves for such questions. First of all, you should study NCERT Textbooks line by line, and then you should practice as many questions as possible.

Chapter-wise Solved Case Study Questions for Class 12 Chemistry

Class 12 students should go through important Case Study problems for Chemistry before the exams. This will help them to understand the type of Case Study questions that can be asked in Grade 12 Chemistry examinations. Our expert faculty for standard 12 Chemistry have designed these questions based on the trend of questions that have been asked in last year’s exams. The solutions have been designed in a manner to help the grade 12 students understand the concepts and also easy-to-learn solutions.

Tips to Excel in CBSE Class 12 Chemistry Examinations

Excel in your Chemistry exams with these practical tips.

A. Regular Practice with Case Studies

Consistent practice with case study questions enhances your ability to tackle complex problems. Dedicate time to solving various case studies to build confidence.

B. Understanding Analytical Skills

Develop strong analytical skills to approach case studies logically. Break down complex problems into simpler components and analyze them step-by-step.

C. Time Management Strategies

Allocate sufficient time for each case study during the exam. Practice time management in mock tests to complete the paper within the stipulated time.

Best Books for Class 12 Chemistry

Strictly as per the new term-wise syllabus for Board Examinations to be held in the academic session 2024 for class 12 Multiple Choice Questions based on new typologies introduced by the board- Stand-Alone MCQs, MCQs based on Assertion-Reason Case-based MCQs. Include Questions from CBSE official Question Bank released in April 2024 Answer key with Explanations What are the updates in the book: Strictly as per the Term wise syllabus for Board Examinations to be held in the academic session 2024. Chapter-wise -Topic-wise Multiple choice questions based on the special scheme of assessment for Board Examination for Class 12th Chemistry.

Mastering CBSE Class 12 Chemistry case study questions is crucial for excelling in the exams. Embrace case studies as a valuable learning tool, and with practice, you’ll ace your Chemistry exams with confidence.

Benefits of Utilizing the CBSE Class 12 Chemistry Case Study PDF

• Enhanced Learning Experience : The case study PDF offers practical examples and scenarios, making the learning process engaging and relatable for students.
• Application of Theoretical Concepts : It enables students to apply theoretical knowledge to practical situations, honing their problem-solving and analytical skills.
• Real-World Relevance : By connecting classroom learning to real-life applications, students can grasp the practical significance of chemistry in various industries.
• Critical Thinking Development : Analyzing case studies encourages students to think critically and make informed decisions based on chemical principles.
• Exam Preparation : Exposure to case studies aids in better preparation for chemistry examinations by providing a comprehensive understanding of the subject.

The CBSE Class 12 Chemistry case study PDF brings a refreshing perspective to the world of education. By intertwining theoretical knowledge with practical applications, it equips students to face real-world challenges with confidence. The diverse case studies provide invaluable insights, encouraging students to explore chemistry beyond the classroom and make a positive impact on society.

What is the CBSE Class 12 Chemistry case study PDF?

The CBSE Class 12 Chemistry case study PDF is a curated document by CBSE, presenting real-life applications of chemistry concepts for students to understand the subject’s practical relevance.

How does the case study PDF benefit students?

The case study PDF enhances the learning experience, fosters critical thinking, promotes application-based learning, and prepares students for examinations.

Are the case studies diverse in content?

Yes, the case studies cover various branches of chemistry, including organic, inorganic, physical, environmental, and analytical chemistry.

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Case Study Class 12 Chemistry With Questions Answers

In Coming Exams, CBSE will ask two Case Study Questions in the CBSE class 12 Chemistry questions paper. Each theme will have five questions and students will have a choice to attempt any four of them. Here are some example questions Based On Case Study Problems:

Question-1 Read the passage given below and answer any four out of the following questions: Ammonia is present in small quantities in air and soil where it is formed by the decay of nitrogenous organic matter e.g., urea. On a large scale, ammonia is manufactured by Haber’s process. In accordance with Le Chatelier’s principle, high pressure would favour the formation of ammonia. Ammonia is a colourless gas with a pungent odour. Its freezing and boiling points are 198.4 and 239.7 K respectively. In the solid and liquid states, it is associated with hydrogen bonds as in the case of water and that accounts for its higher melting and boiling points than expected on the basis of its molecular mass. Ammonia gas is highly soluble in water. Its aqueous solution is weakly basic due to the formation of OH– ions. The presence of a lone pair of electrons on the nitrogen atom of the ammonia molecule makes it a Lewis base.

The following questions are multiple-choice questions. Choose the most appropriate choice 1. On a small scale, ammonia is obtained from ammonium salts which decompose when treated with 1.caustic soda 2.calcium chloride 3.sodium hydroxide 4.sodium chloride

2.The optimum conditions for the production of ammonia are a pressure of 1. 200*105 Pa 2. 400*105 Pa 3. 100*105 Pa 4. 300*105 Pa

3. The catalyst which is used in the preparation of NH3 by Haber’s process 1. Mg2O3 + K2O 2. Al2O3 + K2O 3. NaO3 + K2O 4. None of these 4. The ammonium molecule has: 1. five bond pair and two lone pair 2. four lone pair and one bond pair 3. three bond pair and one lone pair 4. three bond pair and two lone pair 5. A compound reacts with ammonia to form deep colour solution, identify the compound 1. Au2+ 2. Cu2+ 3. Al3+

Questions-2 Read the passage and answer any four out of the following questions: Colloidal particles always carry an electric charge. The nature of this charge is the same on all the particles in a given colloidal solution and may be either positive or negative. The charge on the sol particles is due to one or more reasons, viz., due to electron capture by sol particles during electrodispersion of metals. When two or more ions are present in the dispersion medium, preferential adsorption of the ion common to the colloidal particle usually takes place. When silver nitrate solution is added to the potassium iodide solution, the precipitated silver iodide adsorbs iodide ions from the dispersion medium, and negatively charged colloidal solution results. acquired a positive or a negative charge by selective adsorption on the surface of a colloidal particle The combination of the two layers of opposite charges around the colloidal particle is called Helmholtz electrical double layer. The presence of equal and similar charges on colloidal particles is largely responsible for providing stability to the colloidal solution. In these questions, a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.

1. Assertion and reason both are correct statements and reason is correct explanation for assertion 2. Assertion and reason both are correct statements but reason is not correct explanation for assertion 3. Assertion is correct statement and reason is wrong statement 4. Assertion is wrong statement but reason is correct statement

1. Assertion: The presence of equal and similar charges on colloidal particles is largely responsible in providing stability to the colloidal solution. Reason: The repulsive forces between charged particles having the same charge prevent them from aggregating and provide stability.

2. Assertion: The first layer is mobile in Helmholtz electrical double layer. Reason: The potential difference between the fixed layer and the diffused layer of opposite charges is called zeta potential.

3. Assertion: The sol particle in the colloid has a charge. Reason: The charge in sol is due to electron capture by sol particles during the electrodispersion of metals.

4. Assertion: Methylene blue sol is a negatively charged sol. Reason: When KI solution is added to AgNO3 solution, positively charged sol formed.

5. Assertion: If FeCl3 is added to an excess of hot water, a positively charged sol of hydrated ferric oxide is formed. Reason: When ferric chloride is added to NaOH a negatively charged sol is obtained with adsorption of OH- ions.

Question-3 .Read the passage and answer the following questions: The crystal field theory (CFT) is an electrostatic model which considers the metal-ligand bond to be ionically arising purely from electrostatic interactions between the metal ion and the ligand. Ligands are treated as point charges in case of anions or point dipoles in case of neutral molecules. The five d orbitals in an isolated gaseous metal atom/ion have the same energy, i.e., they are degenerate. In an octahedral coordination entity with six ligands surrounding the metal atom/ion, there will be repulsion between the electrons in metal d orbitals and the electrons (or negative charges) of the ligands. This splitting of the degenerate levels due to the presence of ligands in a definite geometry is termed crystal field splitting and the energy separation is denoted by Δ0. The colour in the coordination compounds can be readily explained in terms of the crystal field theory. In these questions, a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. 1.Assertion and reason both are correct statements and reason is correct explanation for assertion. 2. Assertion and reason both are correct statements but reason is not correct explanation for assertion. 3. Assertion is correct statement but reason is wrong statement. 4. Assertion is wrong statement but reason is correct statement.

1. Assertion: The dx2-y2 and dz2 orbitals which point towards the axes along the direction of the ligand will experience more repulsion. Reason: The dxy, dyz and dxz orbitals which are directed between the axes will be lowered in energy.

2. Assertion: The complex [Ti(H2O)6]3+, which is red in colour. Reason: The crystal field theory attributes the colour of the coordination compounds to d-d transition of the electron.

3. Assertion: Ligands for which Δ0Δ0 < P are known as weak field ligands and form high spin complexes. Reason: If Δ0 > P, then the fourth electron enters one of the eg orbitals giving the configuration t2g3 eg1.

4. Assertion: In tetrahedral coordination entity formation, the d orbital splitting is inverted and is smaller as compared to the octahedral field splitting. Reason: Spectrochemical series is based on the absorption of light by complexes with different ligands.

5. Assertion: The crystal field model is successful in explaining the formation, structures, colour and magnetic properties of coordination compounds. Reason: The anionic ligands are found at the low end of the spectrochemical series.

Answer Key:

1. (b) Assertion and reason both are correct statements and reason is not correct explanation for assertion. 2. (c) Assertion is correct statement but reason is wrong statement. 3. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion. 4. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion. 5. (d) Assertion is wrong statement but reason is correct statement.

Question-4 Read the passage and answer any four out of the following question Alfred Werner (1866-1919), a Swiss chemist was the first to formulate his ideas about the structures of coordination compounds. Werner proposed the concept of a primary valence and a secondary valence for a metal. The coordination entity constitutes a central metal atom or ion bonded to a fixed number of ions or molecules. In a coordination entity, the atom/ion to which a fixed number of ions/groups are bound in a definite geometrical arrangement around it is called the central atom or ion. The ions or molecules bound to the central atom/ion in the coordination entity are called ligands. Ligands may be simple ions such as Cl-, small molecules such as H2O or NH3, larger molecules such as H2NCH2CH2NH2 or N(CH2CH2NH2)3 or even macromolecules, such as protein. Ligands are unidentate, bidentate and polydentate. The coordination number (CN) of a metal ion in a complex is the number of ligand donor atoms to which the metal is directly bonded.

In these questions, a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.

1. Assertion and reason both are correct statements and reason is correct explanation for assertion 2. Assertion and reason both are correct statements but reason is not correct explanation for assertion 3. Assertion is correct statement but reason is wrong statement 4. Assertion is wrong statement but reason is correct statement

1. Assertion: Binary compounds such as CrCl3, have a primary valence of 3. Reason: Coordinate compound metals show only one type of linkage that is primary linkage.

2. Assertion: CoCl3(NH3)3 is a coordination entity in which the cobalt ion is surrounded by three ammonia molecules and three chloride ion. Reason: The central atom/ion in the coordination entities: [NiCl2(H2O)4] is Ni2+.

3. Assertion: H2NCH2CH2NH2 (ethane-1,2-diamine) ligand is said to be didentate. Reason: Didentate ligands are bind through two donor atoms.

4. Assertion: The complex ions, [PtCl6]2- the coordination number of Pt is 4. Reason: Ligand which can ligate through two different atoms is called ambidentate ligand.

5. Assertion: EDTA can bind through two nitrogen and four oxygen atoms to a central metal ion. Reason: The number of ligating groups attach to an atom is called the denticity of the ligand.

1. (c) Assertion is correct statement but reason is wrong statement 2. (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion 3. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion 4. (d) Assertion is wrong statement but reason is correct statement 5. (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion

Question-5 Read the passage given below and answer any four out of the following questions: Nitrogen differs from the rest of the members of group 15 due to its smaller size, high electronegativity, high ionisation enthalpy, and non-availability of d orbitals. Nitrogen has a unique ability to form pπ-pπ multiple bonds with itself. Nitrogen exists as a diatomic molecule with a triple bond one s and two p between the two atoms. Phosphorus, arsenic and antimony from single bonds as P–P, As–As and Sb–Sb while bismuth forms metallic bonds in an elemental state. Dinitrogen is produced commercially by the liquefaction and fractional distillation of air. Liquid dinitrogen (b.p. 77.2 K) distils out first leaving behind liquid oxygen (b.p. 90 K). In the laboratory, dinitrogen is prepared by treating an aqueous solution of ammonium chloride with sodium nitrite. Dinitrogen is a colourless, odourless, tasteless and non-toxic gas. It has two stable isotopes 14N and 15N. It has very low solubility in water. The main use of dinitrogen is in the manufacture of ammonia and other industrial chemicals containing nitrogen.

The following questions are multiple-choice questions. choose the most appropriate answer

1. N–N bond is weaker than the single P–P bond because 1. high interelectronic repulsion of the bonding electrons 2. high interelectronic repulsion of the non-bonding electrons 3. no repulsion between bonding electrons 4. no repulsion between non-bonding electrons 2. Very pure nitrogen can be obtained by the 1. thermal decomposition of sodium 2. thermal decomposition of barium azide 3. thermal decomposition of ammonium dichromate 4. both (a) and (b) 3. Dinitrogen is rather inert at room temperature because of 1. low bond enthalpy of N≡≡N bond 2. high bond enthalpy of N≡≡N bond 3. low freezing point 4. low boiling point 4. Dinitrogen combines with dioxygen only at very high temperature (at about 2000 K) to form 1. nitric oxide 2. nitrate 3. nitrites 4. nitric acid 5. Liquid dinitrogen is used as a refrigerant to 1. preserve biological materials 2. preserve food items 3. in cryosurgery 4. all of these

1. (b) high interelectronic repulsion of the non-bonding electrons 2. (d) both (a) and (b) 3. (b) high bond enthalpy of N≡≡N bond 4. (a) nitric oxide 5. (d) all of these

Question-6 Read the following passage and answer any four out of the following questions: Transition metal oxides are generally formed by the reaction of metals with oxygen at high temperatures. The highest oxidation number in the oxides coincides with the group number. In vanadium, there is a gradual change from the basic V2O3 to less basic V2O4 and to amphoteric V2O5. V2O4 dissolves in acids to give VO2+ salts. Potassium dichromate is a very important chemical used in the leather industry and as an oxidant for the preparation of many azo compounds. Dichromates are generally prepared from chromate. Sodium dichromate is more soluble than potassium dichromate. The latter is, therefore, prepared by treating the solution of sodium dichromate with potassium chloride. Sodium and potassium dichromates are strong oxidising agents; sodium salt has a greater solubility in water and is extensively used as an oxidising agent in organic chemistry. Potassium dichromate is used as a primary standard in volumetric analysis.

The following questions are multiple-choice questions. Choose the most appropriate answer.

1. All transition metal reacts with oxygen to form MO oxide except 1. scandium 2. vanadium 3. cupper 4. zinc 2. As the oxidation number of a metal increases, ionic character 1. increases 2. decreases 3. remain the same 4. none of these 3. The shape of chromate ion is 1. tetrahedral 2. pyramidal 3. square planer 4. triangular 4. Dichromates are generally prepared from chromate, which in turn are obtained by the fusion of 1. FeCr2O 2. FeCr2O 4 3. Na2CrO 4 4. Na2Cr2O 7 5. The oxo cations stabilise VIV 1. VO 2. VO 4+ 3. VO 2+ 4. all of these

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CBSE Class 12 Chemistry Case Study Based Important Questions with Solutions, Download PDF Here

Cbse class 12 chemistry case study based questions: for the cbse class 12 chemistry exam that is scheduled for february 27, 2024, important case study-based questions in chemistry are provided in this article. also, get information related to the chemistry exam pattern and marking scheme in the article..

CBSE Class 12 Chemistry Paper Pattern

Important case study based questions for class 12th chemistry with solutions.

The following questions are case-based questions. Each question has an internal choice and carries 4 marks each. Read the passage carefully and answer the questions that follow.

Q.1. Many people believe that James Watson and Francis Crick discovered DNA in the 1950s. In reality, this is not the case. Rather, DNA was first identified in the late 1860s by Swiss chemist Friedrich Miescher. Then, in the decades following Miescher's discovery, other scientists--notably, Phoebus Levene and Erwin Chargaff--carried out a series of research efforts that revealed additional details about the DNA molecule, including its primary chemical components and the ways in which they joined with one another. Without the scientific foundation provided by these pioneers, Watson and Crick may never have reached their groundbreaking conclusion of 1953: that the DNA molecule exists in the form of a three-dimensional double helix. Chargaff, an Austrian biochemist, as his first step in this DNA research, set out to see whether there were any differences in DNA among different species. After developing a new paper chromatography method for separating and identifying small amounts of organic material, Chargaff reached two major conclusions: (i) the nucleotide composition of DNA varies among species. (ii) Almost all DNA, no matter what organism or tissue type it comes from maintains certain properties, even as its composition varies. In particular, the amount of adenine (A) is similar to the amount of thymine (T), and the amount of guanine (G) approximates the amount of cytosine (C). In other words, the total amount of purines (A + G) and the total amount of pyrimidines (C + T) are usually nearly equal. This conclusion is now known as "Chargaff's rule." Chargaff’s rule is not obeyed in some viruses. These either have single- stranded DNA or RNA as their genetic material. Answer the following questions: a. A segment of DNA has 100 adenine and 150 cytosine bases. 

a. What is the total number of nucleotides present in this segment of DNA? 

Ans. A = 100 so T = 100

C=150 so  G = 150

Total nucleotides = 100+100+150+150 =500 

b. A sample of hair and blood was found at two sites. Scientists claim that the samples belong to the same species. How did the scientists arrive at this conclusion? 

Ans. They studied the nucleotide composition of DNA. It was the same so they concluded that the samples belong to the same species. 

c. The sample of a virus was tested and it was found to contain 20% adenine, 20% thymine, 20 % guanine and the rest cytosine. Is the genetic material of this virus (a) DNA- double helix (b) DNA-single helix (c) RNA? What do you infer from this data? 

Ans.  A = T = 20%

But G is not equal to C so double helix is ruled out. 

 The bases pairs are ATGC and not AUGC so it is not RNA.

 The virus is a single helix DNA virus.

Importance of Solving Case Study Questions for Class 12 Chemistry

• Case study-based questions in Class 12 are pivotal for good scores. Practicing these questions equips students to tackle these types of questions effectively.
• Solving case study-based questions helps students develop an efficient time management strategy.
• Excelling in Case study-based questions cultivates a holistic subject comprehension essential for future pursuits.

Important Study Resources for CBSE Class 12 Chemistry Exam 2024

• CBSE Class 12 Chemistry Syllabus 2023-24
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• CBSE Chemistry Previous Year Question Paper Class 12
• Top 30+ CBSE Class 12 Chemistry MCQs for Board Exam 2024 with Answers to Secure Good Marks
• CBSE Board Exam Mistakes To Avoid During Exam Preparations For 100% Success
• CBSE Class 12 Preparation Tips: TOP 10 Ways to Score High in CBSE Class 12th Board Exam

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Case Study Questions Class 12 Chemistry

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Download the app to get CBSE Sample Papers 2023-24, NCERT Solutions (Revised), Most Important Questions, Previous Year Question Bank, Mock Tests, and Detailed Notes.

CBSE will ask Case Study Questions class 12 Chemistry in session 2020-21. These will be the first two questions in the board exam question paper. The first question will have 5 MCQs out of which students will attempt any 4 questions. The second question will carry 5 Assertion & Reason type questions with the choice to attempt any four.

Case Study Questions

As you know, CBSE will hold exams in May-June this year. There is already a reduction of 30% in the syllabus. Now, the case study questions have been added. So, this year the question paper is going to be a bit easier. Although it is easy yet these case study questions need special attention and regular practice.

We have added around 10 sample questions based on the latest pattern in myCBSEguide App. These all questions include two case study questions.

Class 12 Chemistry Question Bank

If you go through the previous year question papers, you will analyze that many questions are repeated word by word and many others are almost similar. So, it is always recommended to check all questions asked in previous years. This will not only help you to get an idea about the question pattern but also help you to understand the difficulty level of the questions.

myCBSEguide App has the previous year’s question bank. These questions are arranged chapter-wise. If you are preparing a particular chapter, you will get all questions asked from that chapter in the last 10 years.

Case Study Questions Examples

Here are two examples of case study questions. To get more such questions download the myCBSEguide App and browse Sample Papers there.

Read the passage given below and answer any four out of the following questions: Ammonia is present in small quantities in air and soil where it is formed by the decay of nitrogenous organic matter e.g., urea. On a large scale, ammonia is manufactured by Haber’s process. In accordance with Le Chatelier’s principle, high pressure would favour the formation of ammonia. Ammonia is a colourless gas with a pungent odour. Its freezing and boiling points are 198.4 and 239.7 K respectively. In the solid and liquid states, it is associated through hydrogen bonds as in the case of water and that accounts for its higher melting and boiling points than expected on the basis of its molecular mass. Ammonia gas is highly soluble in water. Its aqueous solution is weakly basic due to the formation of OH– ions. The presence of a lone pair of electrons on the nitrogen atom of the ammonia molecule makes it a Lewis base.

• caustic soda
• calcium chloride
• sodium hydroxide
• sodium chloride
• ​200 10 5 Pa
• 400 10 5 Pa
• 100 10 5 Pa
• 300 10 5 Pa
• Mg 2 O 3  + K 2 O
• Al 2 O 3  + K 2 O
• NaO 3  + K 2 O
• None of these
• five bond pair and two lone pair
• four lone pair and one bond pair
• three bond pair and one lone pair
• three bond pair and two lone pair

Read the passage and answer any four out of the following questions: Colloidal particles always carry an electric charge. The nature of this charge is the same on all the particles in a given colloidal solution and may be either positive or negative. The charge on the sol particles is due to one or more reasons, viz., due to electron capture by sol particles during electrodispersion of metals. When two or more ions are present in the dispersion medium, preferential adsorption of the ion common to the colloidal particle usually takes place. When silver nitrate solution is added to the potassium iodide solution, the precipitated silver iodide adsorbs iodide ions from the dispersion medium, and negatively charged colloidal solution results. acquired a positive or a negative charge by selective adsorption on the surface of a colloidal particle The combination of the two layers of opposite charges around the colloidal particle is called Helmholtz electrical double layer. The presence of equal and similar charges on colloidal particles is largely responsible for providing stability to the colloidal solution.

In these questions, a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. 

• Assertion and reason both are correct statements and reason is correct explanation for assertion
• Assertion and reason both are correct statements but reason is not correct explanation for assertion
• Assertion is correct statement and reason is wrong statement
• Assertion is wrong statement but reason is correct statement
• Assertion:  The presence of equal and similar charges on colloidal particles is largely responsible in providing stability to the colloidal solution. Reason:  The repulsive forces between charged particles having the same charge prevent them from aggregating and provide stability.
• Assertion: The first layer is mobile in Helmholtz electrical double layer. Reason:  The potential difference between the fixed layer and the diffused layer of opposite charges is called zeta potential.
• Assertion: The sol particle in colloid has a charge. Reason:  The charge in sol is due to electron capture by sol particles during the electrodispersion of metals.
• Assertion:  Methylene blue sol is a negatively charged sol. Reason: When KI solution is added to AgNO 3 solution, positively charged sol formed.
• Assertion:  If FeCl3 is added to an excess of hot water, a positively charged sol of hydrated ferric oxide is formed. Reason: When ferric chloride is added to NaOH a negatively charged sol is obtained with adsorption of OH- ions.

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CBSE Class 12 Chemistry Case study Questions & Answers For Chapter 1, 2, 4, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16

Understudies can discover the chapter astute vital questions for course 12th Chemistry within the table underneath. These imperative questions incorporate questions that are regularly inquired in a long time. Moreover, arrangements are to give for these questions, with extraordinary accentuation on ease-of-study. Tap on the joins underneath to begin investigating.

Below we posted all the Case Study Questions & Answers for Class 12 Chemistry all Chapters –

CBSE Class 12 Case Study Question for Chemistry

Case study 1.

(1) Read the passage given below and answer the questions that follow.

 Are there nuclear reactions going on in our bodies?

There are nuclear reactions constantly occurring in our bodies, but there are very few of them compared to the chemical reactions, and they do not affect our bodies much. All of the physical processes that take place to keep a human body running are chemical processes. Nuclear reactions can lead to chemical damage, which the body may notice and try to fix. The nuclear reaction occurring in our bodies is radioactive decay. This is the change of a less stable nucleus to a more stable nucleus. Every atom has either a stable nucleus or an unstable nucleus, depending on how big it is and on the ratio of protons to neutrons. The ratio of neutrons to protons in a stable nucleus is thus around 1:1 for small nuclei (Z < 20). Nuclei with too many neutrons, too few neutrons, or that are simply too big are unstable. They eventually transform to a stable form through radioactive decay. Wherever there are atoms with unstable nuclei (radioactive atoms), there are nuclear reactions occurring naturally. The interesting thing is that there are small amounts of radioactive atoms everywhere: in your chair, in the ground, in the food you eat, and yes, in your body.

The most common natural radioactive isotopes in humans are carbon-14 and potassium-40. Chemically, these isotopes behave exactly like stable carbon and potassium. For this reason, the body uses carbon-14 and potassium-40 just like it does normal carbon and potassium; building them into the different parts of the cells, without knowing that they are radioactive. In time, carbon-14 atoms decay to stable nitrogen atoms and potassium-40 atoms decay to stable calcium atoms. Chemicals in the body that relied on having a carbon-14 atom or potassium-40 atom in a certain spot will suddenly have a nitrogen or calcium atom. Such a change damages the chemical. Normally, such changes are so rare, that the body can repair the damage or filter away the damaged chemicals. The natural occurrence of carbon-14 decay in the body is the core principle behind carbon dating. As long as a person is alive and still eating, every carbon-14 atom that decays into a nitrogen atom is replaced on average with a new carbon-14 atom. But once a person dies, he stops replacing the decaying carbon-14 atoms. Slowly the carbon-14 atoms decay to nitrogen without being replaced, so that there is less and less carbon-14 in a dead body. The rate at which carbon-14 decays is constant and follows first order kinetics. It has a half – life of nearly 6000 years, so by measuring the relative amount of carbon-14 in a bone, archeologists can calculate when the person died. All living organisms consume carbon, so carbon dating can be used to date any living organism, and any object made from a living organism. Bones, wood, leather, and even paper can be accurately dated, as long as they first existed within the last 60,000 years. This is all because of the fact that nuclear reactions naturally occur in living organisms. (source: The textbook Chemistry: The Practical Science by Paul B. Kelter, Michael D. Mosher and Andrew Scott states)

(a)  Why is Carbon -14 radioactive while Carbon -12 not? (Atomic number of Carbon: 6)

Ans: Ratio of neutrons to protons is 2.3: 1 which is not the stable ratio of 1:1

(b) Researchers have uncovered the youngest known dinosaur bone, dating around 65 million years ago. How was the age of this fossil estimated?

Ans: Age of fossils can be estimated by C-14 decay. All living organisms have C-14 which decays without being replaced back once the organism dies.

(c)  Which are the two most common radioactive decays happening in human body?

Ans: carbon-14 atoms decay to stable nitrogen atoms and potassium-40 atoms decay to stable calcium.

(d) Suppose an organism has 20 g of Carbon -14 at its time of death. Approximately how much Carbon -14 remains after 10,320 years? (Given antilog 0.517 = 3.289)

Ans: t = 2.303/ k log (Co/Ct)

Co = 20 g Ct = ?

t = 10320 years k = 0.693/6000 (half-life given in passage)

substituting in equation:

10320 = 2.303 / (0.693/6000) log 20/ Ct

0.517 = log 20 / Ct anlilog (0.517) = 20/Ct

3.289 = 20/Ct

Ct = 6.17 g

(d) Approximately how old is a fossil with 12 g of Carbon -14 if it initially possessed 32 g of Carbon -14? (Given log 2.667 = 0.4260)

Co = 32 g Ct = 12

t = ? k = 0.693/6000 (half life given in passage) substituting in equation:

t = 2.303 / (0.693/6000) log 32/ 12

t = 2.303 x 60000 /0.693 log 2.667

t = 2.303x6000x0.4260 /0.693

= 8494 years

Key questions for 12th review Biology are outlined agreeing to the CBSE NCERT program. All address sorts are accessible within the PDF, from one-word to one-line answers, brief reply sorts to five point long reply sorts. Hence, understudies can plan for exams and indeed clarify their concepts through them. On the off chance that they refer to these questions, it’ll get ready their minds to pick up a competitive advantage. Understudies will gotten to be commonplace with question patterns and the sorts of questions that will show up on exams.

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MTG ScoreMore Case Study Based Sample Questions Chemistry Class 12 is specially designed to help students get familiar with solving these new patterns of questions. The book covers 800 + sample questions for practice with detailed explanations to each question. Practising these questions will definitely help students to get an edge in their CBSE preparations.

For the academic year 2020-21 CBSE has incorporated more Objective type/MCQ based questions which will focus on measuring critical thinking ability of students.

The new pattern of questions includes Case study based questions, Passage based questions, Assertion and Reason type questions. In Case study based/ Passage based questions, students will be expected to answer questions after reading a given paragraph or a passage. Assertion and Reason type question is just another way of checking the clarity of one’s concept.

• 1. The Solid State
• 2. Solutions
• 3. Electrochemistry
• 4. Chemical Kinetics
• 5. Surface Chemistry
• 6. General Principles and Processes of Isolation of Elements*
• 7. The p-Block Elements
• 8. The d- and f-Block Elements
• 9. Coordination Compounds
• 10. Haloalkanes and Haloarenes
• 11. Alcohols, Phenols and Ethers
• 12. Aldehydes, Ketones and Carboxylic Acids
• 14. Biomolecules
• 15. Polymers*
• 16. Chemistry in Everyday Life*
• *This chapter is not a part of the Board Examination 2020-21 syllabus

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Class 12 Chemistry Case Study Questions Chapter 14 Biomolecules

• Post author: studyrate
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• Post category: Class 12 / 12 board
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In Class 12 Boards there will be Case studies and Passage Based Questions will be asked, So practice these types of questions. Study Rate is always there to help you. Free PDF Downloads of CBSE Class 12 Chemistry Chapter 14 Biomolecules Case Study and Passage-Based Questions with Answers were Prepared Based on the Latest Exam Pattern. Students can solve NCERT Class 12 Chemistry Case Study Questions Biomolecules  to know their preparation level.

Join our Telegram Channel, there you will get various e-books for CBSE 2024 Boards exams for Class 9th, 10th, 11th, and 12th.

In CBSE Class 12 Chemistry Paper, There will be a few questions based on case studies and passage-based as well. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

Biomolecules Case Study Questions With Answers

Here, we have provided case-based/passage-based questions for Class 12 Chemistry  Chapter 14 Biomolecules

Case Study/Passage-Based Questions

Case Study 1: Monosaccharides containing an aldehyde group are called aldoses while those containing a keto group are called ketoses. All monosaccharides containing five and six carbon atoms have cyclic structures, furanose (five-membered) and pyranose (six-membered). During ring formation, C 1  aldoses and C 2  in ketoses become chiral and hence all these monosaccharides exist in two forms called the α-anomer and β-anomer while C 1  and C2 are called glycosidic or anomeric carbon atoms. In contrast, stereoisomers, which differ in configuration at any other chiral carbon other than the glycosidic carbon are called epimers. Two molecules of the same or different monosaccharides combine together through glycosidic linkage to form disaccharides

Que 1. Which of the following compounds show furanose structures? 

a) Mannose b) Galactose c) Fructose

Answer:c) Fructose ​

Que 2. What is the relation between two molecules? (a) Enantiomers (b) Epimers (c) Functional groups (d) Anomers

Answer:(b) Epimers ​

Que 3. In disaccharides, the linkage connecting monosaccharide units is called (a) Glycogen linkage (b) Nucleoside linkage (c) Glycosidic linkage (d) Peptide linkage.

Answer:(c) Glycosidic linkage ​

Case Study 2: Proteins are the most abundant biomolecules in the living system. The chief sources of proteins are milk, cheese, pulses, fish, meat, peanuts, etc. They are found in every part of the body and form a fundamental basis of the structure and functions of life. These are also required for the growth and maintenance of the body. The word protein is derived from the Greek word, ‘proteins’ meaning ‘primary’ or of ‘prime importance’. Chemically, proteins are polymers in which the monomeric units are the α-amino acids. Amino acids contain amino (-NH2) and carboxylic (-COOH) functional groups. Depending upon the relative position of the amino group with respect to the carboxylic group, the amino acids can be classified as α, β, and γ-amino acids. Amino acids which are synthesized by the body are called non-essential amino acids. On the other hand, those amino acids which cannot be synthesized in the human body and are supplied in the form of diet (because they are required for proper health and growth) are called essential amino acids. Que 1. Amino acids show amphoteric behavior. Why? a) They have an amino group b) They have a carboxylic group c) Both (a) and (b) d) none of the above

Answer:c) Both (a) and (b) ​

Que 2. The name of linkage joining two amino acids a) Hydrogen bonding b) Peptide linkage c) Amino linkage d) Imino joints

Answer:b) Peptide linkage ​

Que 3. What are polypeptides? a) 10 < α-amino acids joined together b) amino acids joined together c) 20 < β-amino acids joined together d) None of the above

Answer: a) 10 < α-amino acids joined together ​

Que 4. What type of bonding helps in stabilizing the α-helix structure of proteins? a) Peptide linkage b) Hydrogen bonding c) Amino linkage d) Van der waals force

Answer: b) Hydrogen bonding ​

Case Study 3: When a protein in its native form, is subjected to physical changes like change in temperature or chemical changes like change in pH, the hydrogen bonds are disturbed. Due to this, globules unfold and helix gets uncoiled and protein loses its biological activity. This is called the denaturation of protein. The denaturation causes change in secondary and tertiary structures but primary structures remain intact. Examples of denaturation of protein are coagulation of egg white on boiling, curdling of milk, and formation of cheese when an acid is added to milk.

(i) Mark the wrong statement about denaturation of proteins

Answer:(c) Fibrous proteins are converted into globular proteins. ​

(ii) Which structure(s) of proteins remains(s) intact during denaturation process?

Answer: (b) Primary structure only ​

(iii) Cheese is a

Answer: (c) denatured protein ​

(iv) Secondary structure of protein refers to

Answer: (d) regular folding patterns of continuous portions of the polypeptide chain ​

Hope the information shed above regarding Case Study and Passage Based Questions for Class 12 Chemistry Chapter 14 Biomolecules with Answers Pdf free download has been useful to an extent. If you have any other queries about the CBSE Class 12 Chemistry Biomolecules Case Study and Passage-Based Questions with Answers, feel free to comment below so that we can revert back to us at the earliest possible. By Team Study Rate

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Case Study Questions Haloalkanes and Haloarenes Class 12 Chemistry

        Case Study Questions Haloalkanes and Haloarenes Class 12 Chemistry

1. Read the passage given below and answer the following questions: Nucleophilic substitution reactions are of two types; substitution nucleophilic bimolecular (S N 2) and substitution nucleophilic unimolecular (S N 1) depending on molecules taking part in determining the rate of reaction. The reactivity of alkyl halide towards S N 1 and S N 2 reactions depends on various factors such as steric hindrance, stability of intermediate or transition state, and polarity of the solvent. S N 2 reaction mechanism is favoured mostly by primary alkyl halide or transition state and polarity of the solvent, S N 2 reaction mechanism is favoured mostly by primary alkyl halide then secondary and then tertiary. This order is reversed in the case of S N 1 reactions. (i) Which of the following is most reactive towards nucleophilic substitution reaction? (a) C 6 H 5 Cl (b) CH 2 = CHCl (c) ClCH 2 CH = CH 2 (d) CH 3 CH = CHCl

(ii) Isopropyl chloride undergoes hydrolysis by (a) S N 1 mechanism (b) S N 2 mechanism (c) S N 1 and S N 2 mechanism (d) Neither S N 1 nor S N 2 mechanism

(iii) Tertiary alkyl halides are practically inert to substitution by S N 2 mechanism because of (a) Insolubility (b) Instability (c) Inductive effect (d) Steric Hindrance

(iv) Which of the following is the correct order of decreasing S N 2 reactivity? (a) RCH 2 X > R 2 CHX > R 3 CX (b) R 3 CX > R 2 CHX >RCH 2 X (c) R 2 CHX > R 3 CX > RCH 2 X (d) RCH 2 X > R 3 CX > R 2 CHX

(v) An organic molecule necessarily shows optical activity if it- a) Contains asymmetric carbon atoms b) Is non-polar c) Is non-superimposable on its mirror image d) Is superimposable on its mirror image.

2. Read the passage given below and answer the following questions: The replacement of hydrogen atom in a hydrocarbon, aliphatic or aromatic results in the formation of haloalkanes and haloarenes respectively. Haloalkanes contain a halogen atom attached to sp 3 hybridized carbon atom of an alkyl group whereas haloarenes contain a halogen atom attached to sp 2 hybridized carbon atom of an aryl group. Haloalkanes and haloarenes may be classified on the basis of the number of halogen atoms in their structures as mono, di, or poly halogen compounds and also on the basis of the state of hybridization of the carbon atom to which the halogen atom is bonded. (i) Which of the following halide is 2°? (a) Isopropyl chloride (b) Isobutyl chloride (c) n-propyl chloride (d) n-butyl chloride

(ii) Which of the following is a Gem-dibromide is: (a) CH 3 CH(Br)CH 2 (Br) (b) CH 3 CBr 2 CH 3 (c) CH 2 (Br)CH 2 CH 2 (d) CH 2 BrCH 2 Br

(iii) IUPAC name of (CH 3 ) 3 CCl is: (a) 3-Chlorobutane (b) 2-Chloro-2-methylpropane (c) t-butyl chloride (d) n-butyl chloride

(iv) Which of the following is a primary halide? (a) Isopropyl iodide (b) Secondary butyl iodide (c) Tertiarybutyl bromide (d) Neohexyl chloride

(v) Which one of the following is not an allylic halide? (a) 4-Bromopent-2-ene (b) 3-Bromo-2-methylbut-1-ene (c) 1-Bromobut-2-ene (d) 4-Bromobut-1-ene

3. Read the passage given below and answer the following questions: Alkyl halides are prepared by the free radical halogenation of alkanes, addition of halogen acids to alkenes, replacement of -OH group of alcohols with halogens using phosphorus halides, thionyl chloride, or halogen acids. Aryl halides are prepared by electrophilic substitution to arene. Fluorine and iodides are best prepared by the halogen exchange method. These compounds find wide applications in industry as well as in day-to-day life. These compounds are generally used as solvents and as starting materials for the synthesis of a large number of organic compounds. (i) The best method for the conversion of an alcohol into an alkyl chloride is by treating the alcohol with (a) PCl 5 (b) Dry HCl in the presence of anhydrous ZnCl 2 (c) SOCl 2 in presence of pyridine (d) None of these

(ii) The catalyst used in the preparation of an alkyl chloride by the action of dry HCl on alcohol is (a) anhydrous AlCl 3 (b) FeCl 3 (c) anhydrous ZnCl 2 (d) Cu

(iii) An alkyl halide reacts with metallic sodium in dry ether. The reaction is known as: (a) Frankland’s reaction (b) Sandmeyer’s reaction (c) Wurtz reaction (d) Kolbe’s reaction

(iv) Fluorobenzene (C 6 H 5 F) can be synthesized in the laboratory (a) By direct fluorination of benzene with F 2 gas (b) By reacting bromobenzene with NaF solution (c) By heating phenol with HF and KF (d) From aniline by diazotization followed by heating the diazonium salt with HBF 4

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Case Study Question 1 on Solutions – Chapter 2 CBSE Class 12 Chemistry

• May 14, 2022
• Chemistry , CBSE , Class 12

Case-based Questions

Read the given passages and answer the questions that follow :

The spontaneous flow of the solvent through a semipermeable membrane from a pure solvent to a solution or from a dilute solution to a concentrated solution is called osmosis. The phenomenon of osmosis can be demonstrated by taking two eggs of the same size. In an egg, the membrane below the shell and around the egg material is semi- permeable. The outer hard shell can be removed by putting the egg in dilute hydrochloric acid. After removing the hard shell, one egg is placed in distilled water and the other in a saturated salt solution. After some time, the egg placed in distilled water swells-up while the egg placed in salt solution shrinks.

The external pressure applied to stop the osmosis is termed as osmotic pressure (a Colligative property). Reverse osmosis takes place when the applied external pressure becomes larger than the osmotic pressure.

1.      What do you expect to happen when red blood corpuscles (RBC’s) are placed in 0.5% NaCl solution?

Ans.   RBC’s are isotonic with 0.9% NaCl solution, so they will swell and may even burst when placed in 0.5% NaCl solution.

2.      Which one of the following will have higher osmotic pressure in 1 M KCl or 1 M urea solution?

Ans. 1 M KCl will have higher osmotic pressure because its dissociates to give K + and Cl – ions while urea does not dissociate into ions in the solution.

3. Name one SPM which can be used in the process of reverse osmosis.

Ans. Cellulose acetate placed on a suitable support.

4.      What are isotonic solutions?

Ans. Solutions having equal osmotic pressure are called isotonic solutions.

5.      Write van’t Hoff equation for dilute solution.

Ans. p V = nRT ,

Where, p = Osmotic pressure, n = Number of moles, V = Volume of solution in litre, R = Gas constant ,

T = Temperature

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Case Study Questions for Class 12 Chemistry Chapter 9 Coordination Compounds

• Last modified on: 3 years ago
• Reading Time: 4 Minutes

There is Case Study Questions in class 12 Chemistry in session 2020-21. For the first time, the board has introduced the case study questions in the board exam. The first two questions in the board exam question paper will be based on Case Study and Assertion & Reason. The first question will have 5 MCQs out of which students will have to attempt any 4 questions. The second question will carry 5 Assertion & Reason type questions with the choice to attempt any four. Here are the questions based on case study.

Case Study Question 1:

Read the passage given below and answer the following questions:

Coordination compounds are formulated and named according to the IUPAC system.

Few rules for naming coordination compounds are:

(I) In ionic complex, the cation is named first and then the anion.

(II) In the coordination entity, the ligands are named first and then the central metal ion.

(III) When more than one type of ligands are present, they are named in alphabetical order of preference with any consideration of charge

The following questions are multiple choice question. Choose the most appropriate answer:

(i) The IUPAC name of [Ni(CO) 4 ] is (a) tetracarbonylnickel(II) (b) tetracarbonylnickel(0) (c) tetracarbonylnickelate(II) (d) tetracarbonylnickelate(0)

(ii) The IUPAC name of the complex [Pt(NH 3 ) 3 Br(NO 2 )Cl]Cl is (a) triamminechlorobromonitroplatinum(IV) chloride (b) triamminebromonitrochloroplatinum(IV) chloride (c) triamminebromidochloronitroplatinum(IV) chloride (d) triamminenitrochlorobromoplatinum(IV) chloride

(iii) As per IUPAC nomenclature, the name of the complex [Co(H 2 O) 4 (NH 3 ) 2 ]Cl 3 is (a) tetraaquadiamminecobalt(II) chloride (b) tetraaquadiamminecobalt(III) chloride (c) diamminetetraaquacobalt(II) chloride (d) diamminetetraaquacobalt(III) chloride

(iv) Which of the following represents correct formula of dichloridobis(ethane-1,2-diamine)cobalt(III) ion? (a) [CoCl 2 (en)] 2+ (b) [CoCl 2 (en)2] 2+ (c) [CoCl 2 (en)] + (d) [CoCl 2 (en)2] +

Correct formula of pentaamminenitro-O-cobalt(III) sulphate is (a) [Co(NO 2 )(NH 3 ) 5 ]SO 4 (b) [Co(ONO)(NH 3 ) 5 ]SO 4 (c) [Co(NO2)(NH 3 ) 4 ](SO 4 ) 2 (d) [Co(ONO)(NH 3 ) 4 ](SO 4 ) 2

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