NTA CUET (UG) 10 Mock Test Sample Question Papers Chemistry (2024) 9789357288118, 9357288112

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For 2024 Exam

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CHEMISTRY Section II (Domain Specific Subject) Strictly ar per the Latest Examination Pattern issued by NTA

The ONLY book you need to Ace CUET (UG)

SECTION II (Domain Specific)

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5

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Valuable Exam Insights

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Extensive Practice

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With On-Tips Notes & Updated Mind Maps

With Latest Solved Paper 2023

With 10 Solved Sample Question Papers

With 650+ NCERT - based MCQs

With 450+ Explanations & Smart Answer Keys

(i)

3rd EDITION

ISBN SYLLABUS COVERED

YEAR 2023-24 “9789357288118”

CUET (UG) CERTIFICATE OF COMMON UNIVERSITY ENTRANCE TEST

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This book is published by Oswaal Books and Learning Pvt Ltd (“Publisher”) and is intended solely for educational use, to enable students to practice for examinations/tests and reference. The contents of this book primarily comprise a collection of questions that have been sourced from previous examination papers. Any practice questions and/or notes included by the Publisher are formulated by placing reliance on previous question papers and are in keeping with the format/pattern/ guidelines applicable to such papers. The Publisher expressly disclaims any liability for the use of, or references to, any terms or terminology in the book, which may not be considered appropriate or may be considered offensive, in light of societal changes. Further, the contents of this book, including references to any persons, corporations, brands, political parties, incidents, historical events and/or terminology within the book, if any, are not intended to be offensive, and/or to hurt, insult or defame any person (whether living or dead), entity, gender, caste, religion, race, etc. and any interpretation to this effect is unintended and purely incidental. While we try to keep our publications as updated and accurate as possible, human error may creep in. We expressly disclaim liability for errors and/or omissions in the content, if any, and further disclaim any liability for any loss or damages in connection with the use of the book and reference to its contents”.

( iiEdition ) Kindle

Preface National Testing Agency (NTA) has been established in November 2017 under the Societies Registration Act (1860) by the Ministry of Education as a premier, specialist, autonomous, and self-sustained testing organization to conduct entrance examinations for admission/fellowship in higher educational institutions. The Common University Entrance Test (CUET (UG) - 2022) is being introduced for admission into all UG Programmes in all Central Universities for the academic session 2023 under the Ministry of Education, (MoE). The Common University Entrance Test (CUET) will provide a common platform and equal opportunities to candidates across the country, especially those from rural and other remote areas, and help establish a better connection with the Universities. A single examination will enable the Candidates to cover a wide outreach and be part of the admissions process to various Central Universities. CUET – UG Computer Based Test (CBT) for the Central Universities is to be conducted by the National Testing Agency (NTA). The curriculum for CUET is based on the National Council of Educational Research and Training (NCERT) syllabus for class 12 only. CUET scores are mandatorily required while admitting students to undergraduate courses in 44 central universities. A merit list will be prepared by participating Universities/organizations. Universities may conduct their individual counselling on the basis of the scorecard of CUET (UG)-2023 provided by NTA.

A few benefits of studying from Oswaal Sample Question Papers • • • • •

Crisp Revision With On-Tips Notes & Updated Mind Maps Valuable Exam Insights With Latest Solved Paper 2023 100% Exam Readiness With 10 Solved Sample Question Papers Extensive Practice With 650+ NCERT - based MCQs Concept Clarity With 450+ Explanations & Smart Answer Keys

Our Heartfelt Gratitude! Finally, we would like to thank our authors, editors, and reviewers. Special thanks to our students who send us suggestions and constantly help improve our books. We promise to always strive towards ‘Making Learning Simple’ for all of you. Wish you all Happy Learning!

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-Team Oswaal Books

Oswaal Expert Tips toinCrack Oswaal Books ExpertBooks Tips to Crack CUET (UG) the First A empt CUET (UG) in the First Attempt Excited about your UG but unsure if you will get admission to your preferred university? In a major announcement by the chairman of the University Grants Commission, the Naonal Tesng Agency will be conducng the Common Universies Entrance Test (CUET (UG) 2022) for undergraduate programs in Central Universies for the upcoming academic session. However, the UGC Chairperson also stated that CUET (UG) will not just be limited to admissions to Central Universies. Many prominent private universies have indicated that they would also like to adopt a common entrance exam for undergraduate admissions and take admissions on the basis of CUET (UG) scores. This makes CUET (UG) a very important examinaon in itself and hence it becomes mandatory to be aware of the ps & tricks that could help you ace the exam on the first a† empt.

The first step is to understand The pa† ern of the examinaon. CUET includes three secons, secon 1 includes queson based on languages, secon 2 includes 27 domain-based subjects and secon 3 includes General Test. The syllabus of the upcoming Common University Entrance Test, CUET 2022, will be completely based on the syllabus of class 12 th . No queson will be asked from class 11th syllabus.

While preparing for the exam, it is i m p o r t a n t t o i d e n f y t h e important topics and pracce important quesons from those t o p i c s . P r a c c e i m p o r t a n t q u e s o n s t h r o u g h O s w a a l Q u e so n B a n k a n d S a m p l e Q u e s o n P a p e r s , L i s n g topics also helps in idenfying the weak areas that need special effort and me. The aspirants can start preparing to focus on the areas that they consider to be tough, followed by the ones that are their strengths.

Make a habit of preparing notes f ro m t h e b e g i n n i n g o f t h e preparaon. It will not only help in making the study systemac but also make the revision of the syllabus easy even when you might have limited me to revise.

Collecng and preparing from the appropriate study material cannot be ignored as irrelevant. The books chosen by the aspirants to study from should be on the lines of the current syllabus and the ones that could help you with swi• revision before the examinaon.

Make sure to revise as much as possible. The revision will help the aspirants in keeping the concepts fresh in their minds unl the day of the final examinaons. They may refer to a few good pracce quesons and concise revision notes to achieve their desired results.

Devote a sufficient amount of me to all the secons of the examinaons. This requires a wellmade plan and an honest adherence to the said plan. Priorize the most important topics or the topics that the aspirants are not familiar with to be able to master them in me.

With this said, an important queson that is gaining ground amongst students who will be appearing for this exam is if they should take coaching to get themselves ready for the exams. The answer is a simple no, the exam will simply not require any coaching as it is completely based on the Class 12th syllabus which will be quite fresh in students' minds as they will be just out of school. All they need is a good revision and pracce of quesons from Oswaal Queson Bank and Sample Queson Papers for CUET (UG) preparaons.

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Contents

l Oswaal Books Expert Tips to Crack CUET (UG) in the first Attempt l Latest Syllabus l CUET Solved Paper 2023 (26th May 2023) l CUET Solved Paper 2022 (10th August 2022—Slot-1, 17th August 2022—Slot-1, 18th August 2022—Slot-1, 20th August 2022—Slot-1, 21st August 2022—Slot-2, 23rd August 2022—Slot-1, 30th August 2022—Slot-2) l CUET Solved Paper 2021 (23rd SEP. 2021–Slot-2 UIQP02)

iv - iv vi - viii 6 - 13 14 - 82 83 - 87

Sample Question Papers

1 - 4 5 - 8 9 - 13 14 - 17 18 - 21 22 - 25 26 - 30 31 - 34 35 - 39 40 - 44

l Sample Question Paper - 1 l Sample Question Paper - 2 l Sample Question Paper - 3 l Sample Question Paper - 4 l Sample Question Paper - 5 l Sample Question Paper - 6 l Sample Question Paper - 7 l Sample Question Paper - 8 l Sample Question Paper - 9 l Sample Question Paper - 10

Solutions l l l l l l l l l l

45 - 48 49 - 52 53 - 56 57 - 60 61 - 64 65 - 68 69 - 73 74 - 78 79 - 83 84 - 88

Sample Question Paper - 1 Sample Question Paper - 2 Sample Question Paper - 3 Sample Question Paper - 4 Sample Question Paper - 5 Sample Question Paper - 6 Sample Question Paper - 7 Sample Question Paper - 8 Sample Question Paper - 9 Sample Question Paper - 10



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Latest Syllabus CHEMISTRY - 306 Note: There will be one Question Paper which will have 50 questions out of which 40 questions need to be attempted.

Unit I: Solid State

Unit V: Surface Chemistry

Classification of solids based on different binding forces:

Adsorption – physisorption and chemisorption; factors

molecular, ionic covalent, and metallic solids, amorphous

affecting adsorption of gases on solids; catalysis:

and crystalline solids(elementary idea), unit cell in two

homogenous and heterogeneous, activity and selectivity:

dimensional and three-dimensional lattices, calculation of

enzyme catalysis; colloidal state: the distinction between

density of unit cell, packing in solids, packing efficiency,

true solutions, colloids, and suspensions; lyophilic,

voids, number of atoms per unit cell in a cubic unit

lyophobic multimolecular and macromolecular colloids;

cell, point defects, electrical and magnetic properties,

properties of colloids; Tyndall effect, Brownian movement,

Band theory of metals, conductors, semiconductors and

electrophoresis, coagulation; emulsions – types of

insulators and n and p-type semiconductors.

emulsions.

Unit II: Solutions

Unit VI: General Principles and Processes of Isolation of

Types of solutions, expression of concentration of solutions

Elements

of solids in liquids, the solubility of gases in liquids, solid

Principles and methods of extraction – concentration,

solutions, colligative properties – the relative lowering of

oxidation, reduction electrolytic method, and refining;

vapour pressure, Raoult’s law, elevation

occurrence and principles of extraction of aluminum,

of B.P., depression of freezing point, osmotic pressure,

copper, zinc, and iron.

determination of molecular masses using colligative

Unit VII: p-Block Elements

properties, abnormal molecular mass, Vant Hoff factor.

Group 15 elements: General introduction, electronic

Unit III: Electrochemistry

configuration, occurrence, oxidation states, trends in

Redox reactions; conductance in electrolytic solutions,

physical and chemical properties; nitrogen – preparation,

specific and molar conductivity variations of conductivity

properties, and uses; compounds of nitrogen: preparation

with concentration, Kohlrausch’s Law, electrolysis and

and properties of ammonia and nitric acid, oxides of

laws of electrolysis (elementary idea), dry cell – electrolytic

nitrogen

cells and Galvanic cells; lead accumulator, EMF of a cell,

forms; compounds of phosphorous: preparation and

standard electrode potential, Nernst equation and its

properties of phosphine, halides (PCl3, PCl5) and oxoacids

application to chemical cells. Relation between Gibbs

(elementary idea only).

energy change and EMF of a cell, fuel cells; corrosion.

Group 16 elements: General introduction, electronic

Unit IV: Chemical Kinetics

configuration, oxidation states, occurrence, trends in

Rate of a reaction (average and instantaneous), factors

physical and chemical properties; dioxygen: preparation,

affecting rates of reaction: concentration, temperature, catalyst; order and molecularity of a reaction; rate law and specific rate constant, integrated rate equations, and half-life (only for zero and first-order reactions); concept

(

structure

only);

Phosphorous-allotropic

properties, and uses; classification of oxides; ozone. Sulphur – allotropic forms; compounds of sulphur: preparation, properties, and uses of sulphur dioxide; sulphuric acid: industrial process of manufacture,

of collision theory (elementary idea, no mathematical

properties and uses, oxoacids of sulphur (structures only).

treatment).Activation energy, Arrhenius equation.

Group 17 elements: General introduction, electronic configuration,

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oxidation

states,

occurrence,

trends

Contd... in physical and chemical properties; compounds of

Unit XI: Alcohols, Phenols, and Ethers

halogens: preparation, properties and uses of chlorine

Alcohols: Nomenclature, methods of preparation, physical

and hydrochloric acid, interhalogen compounds, oxoacids of halogens (structures only).

and chemical properties (of primary alcohols only); identification of primary, secondary, and tertiary alcohols;

Group 18 elements: General introduction, electronic

mechanism of dehydration, uses, with special reference to

configuration, occurrence, trends in physical and chemical

methanol and ethanol.

properties, uses.

Phenols: Nomenclature, methods of preparation, physical

Unit VIII: d and f Block Elements

and chemical properties, acidic nature of phenol,

General introduction, electronic configuration, occurrence

electrophilic substitution reactions, uses of phenols.

and characteristics of transition metals, general trends in

Ethers: Nomenclature, methods of preparation, physical

properties of the first-row transition metals – metallic

and chemical properties, uses.

character, ionization enthalpy, oxidation states, ionic radii, colour, catalytic property, magnetic properties, interstitial compounds, alloy formation. Preparation and properties of K2Cr2O7 and KMnO4. Lanthanoids – electronic configuration, oxidation states, chemical reactivity, and lanthanoid contraction and its consequences. Actinoids – Electronic configuration, oxidation states, and comparison with lanthanoids.

Unit XII: Aldehydes, Ketones, and Carboxylic Acids Aldehydes and Ketones: Nomenclature, nature of carbonyl group, methods of preparation, physical and chemical properties, mechanism of nucleophilic addition, the reactivity of alpha hydrogen in aldehydes; uses. Carboxylic Acids: Nomenclature, acidic nature, methods of preparation, physical and chemical properties; uses. Unit XIII: Organic Compounds Containing Nitrogen Amines: Nomenclature, classification, structure, methods

Unit IX Coordination Compounds Coordination compounds: Introduction, ligands, coordination number, colour, magnetic properties and shapes, IUPAC nomenclature of mononuclear coordination compounds, bonding, Werner’s theory VBT, CFT; isomerism (structural

of preparation, physical and chemical properties, uses, identification of primary secondary, and tertiary amines. Cyanides and Isocyanides – will be mentioned at relevant places in context.

and stereo)importance of coordination compounds (in

Diazonium salts: Preparation, chemical reactions, and importance in synthetic organic chemistry.

qualitative analysis, extraction of metals and biological

Unit XIV: Biomolecules

systems). Unit X: Haloalkanes and Haloarenes Haloalkanes: Nomenclature, nature of C-X bond, physical and chemical properties, mechanism of substitution reactions. Optical rotation. Haloarenes: Nature of C-X bond, substitution reactions (directive influence of halogen for monosubstituted

Carbohydrates – Classification (aldoses and ketoses), monosaccharide (glucose and fructose), D-L configuration, oligosaccharides (sucrose, lactose, maltose), polysaccharides (starch, cellulose, glycogen): importance. Proteins - Elementary idea of a-amino acids, peptide bond, polypeptides, proteins, primary structure, secondary structure, tertiary structure and quaternary structure (qualitative idea only), denaturation of proteins; enzymes.

compounds only).

Hormones –Elementary idea (excluding structure).

Uses and environmental effects of – dichloromethane,

Vitamins – Classification and functions.

trichloromethane, tetrachloromethane, iodoform, freons,

Nucleic Acids: DNA and RNA

DDT.

( vii )

Contd... Unit XV: Polymers Classification of

–

1. Chemicals in medicines – analgesics, tranquilizers, Natural

polymerization

and

(addition

synthetic, and

antiseptics, disinfectants, antimicrobials, antifertility

methods

drugs, antibiotics, antacids, antihistamines.

condensation),

copolymerization. Some important polymers: natural

2. Chemicals in food – preservatives, artificial sweetening

and synthetic like polythene, nylon, polyesters, bakelite, rubber. Biodegradable and non-biodegradable polymers.

agents, elementary idea of antioxidants. 3. Cleansing agents – soaps and detergents, cleansing

Unit XVI: Chemistry in Everyday Life

action. 

Don't Stop Reading !

You never know what might be asked in the exam. To download Chapter-wise Mind Maps scan the code below

To download On Tips Notes scan the code below

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Be mindful. Be grateful. Be positive. Be true. Be kind Three things that make you special

Three people you are grateful for and why

Three simple things you are grateful for

A challenging experience that made you stronger

Three ways to inject gratitude into a current challenge

Describe the last time you did something nice for someone

A fear you have overcome

Three activities you enjoy most and why

What made you smile today?

Three things you love about your family

What is your favorite place, and why?

Three things you love most about yourself

The last time you were overcome with joy

A risk you are grateful you took and why

Three everyday items you are grateful for

Three songs that bring you joy

What skill do you have that you are grateful for and why?

One luxury you are thankful for

Describe a rejection you are grateful for

Three things about your body you are grateful for

What are you most grateful for in your daily life?

Three things you are grateful for about where you live

Three items in your home you are grateful for

Say thank you to someone

Something in nature you are grateful for

A person in your past you are grateful for

Something at school you’re grateful for

Describe the last time you laughed so hard you cried

What is your proudest accomplishment?

Three things you want to manifest

(1)

Positive Affirmations

(2)

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(4)

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0808

NAGPUR

Writing Your Notes

(5)

CUET (UG) Exam Paper 2023 National Testing Agency Held on 26th May 2023

CHEMISTRY Solved

(This includes Questions pertaining to Domain Specific Subject only)

Max. Marks: 200

Time allowed: 45 Minutes

General Instructions: (i) (ii) (iii) (iv) (v)

This paper consists of 50 MCQs, attempt any 40 out of 50 . Correct answer or the most appropriate answer: Five marks (+5) . Any incorrect option marked will be given minus One mark (–1) . Unanswered/Marked for Review will be given No mark (0) . If more than one option is found to be correct then Five marks (+5) will be awarded to only those who have marked any of the correct options . (vi) If all options are found to be correct then Five marks (+5) will be awarded to all those who have attempted the question . (vii) If none of the options is found correct or a Question is found to be wrong or a Question is dropped then all candidates who have appeared will be given five marks (+5). (viii) Calculator / any electronic gadgets are not permitted . OCOCH3 1. The compound



COOH is used as:

(1) Antiseptic (3) Analgesic Ans. Option (3) is correct

(2) Antibiotic (4) Pesticide

Choose the correct answer from the options given below (1) (A)-(I), (B)-(II), (C)-(III), (D)-(IV) (2) (A)-(II), (B)-(III), (C)-(lV), (D)-(I) (3) (A)-(III), (B)-(IV), (C)-(I), (D)-(II) (4) (A)-(IV), (B)-(III), (C)-(I), (D)-(II) Ans. Option (4) is correct

Explanation: The given compound is "Aspirin (Salicylic acid), which is used for the treatment of fever and mild pain. 2. A small amount of CaF2 is added in electrolytic reduction of Al2O3 dissolved in fused cryolite (A) To decrease the rate of oxidation of carbon at anode (B) To act as catalyst (C) To make fused mixture conducting (D) To lower the fusion temperature of melting. Choose the correct answer from the options given below: (1) (A), (B), (D) only (2)  (B), (C) only (3) (A), (B) only (4)  (C), (D) only Ans. Option (4) is correct

Explanation: A small amount of CaF2 is added in electrolytic reduction of Al2O3 dissolved in fused cryolite because it enhance the electrical conductivity of fused mixture of Al2O3 and it lowers the fusion temperature of melting. 3. Match List -I with List - II. List -1 Complex (A) (B) (C) (D)

[CoF6]3– [Co(NH3)6]3+ [NiCl4]2– [Ni(CN)4]2–

List - II hybridisation (I) (II) (III) (IV)

sp3 dsp2 d2sp3 sp3d2

Explanation: Hybridisation is a concept of intermixing of two atomic orbitals which give rise to a new atomic orbital and define the geometry of the complex compounds. 4. Match List -I with List - II. List -I (A) (B) (C) (D)

Chloroxylenol + terpineol Penicillin Iproniazid Aspirin

List - II (I) (II) (III) (IV)

Tranquillizer Analgesic Antiseptic Antibiotic



Choose the correct answer from die options given below (1) (A)-(III), (B)-(II), (C)-(I), (D)-(IV) (2) (A)-(III), (B)-(IV), (C)-(I), (D)-(II) (3) (A)-(II), (B)-(IV), (C)-(III), (D)-(I) (4) (A)-(l), (B)-(III), (C)-(II), (D)-(IV) Ans. Option (2) is correct Explanation: Mixture of Chloroxylenol and terpineol is called Dettol which is used as an antiseptic. Penicillin act as antibiotic which kill/ inhibit the growth of microorganism.Iproniazid act as Tranquillizer which was initially used for treatment of Tuberculosis and later on, used as antidepressant. Aspirin used for relieving mild pain and fever. 5. Zone refining is mainly used for: (1) Ni (2) Ga (3) Zr (4) Ta Ans. Option (2) is correct

CUET Solved Paper 2023 Explanation: Zone refining method is used for purification of metals like Ga, Si and Ge, which acts as semiconductors. 6. 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 molecular mass of the solute? (1) 41.35 g/mol (2) 10 g/mol (3) 23.4 g/mol (4) 20.8 g/mol Ans: Option (1) is correct Explanation: Mass of Solute (w2) = 2g Mass of solution = 100g Mass of solvent = 100-2 = 98g vapour pressure of pure water = 1 atm. or 1.013 bar According to Raoult's LawP - P W ´ M1 = X2 = 2 P M 2 ´ W1 1.013 - 1.004 2 ´ 18 = M 2 ´ 98 1.013 M 2 = 41.35g/ mol 7. Solution of KMnO4 is reduced to various products P , Q , R depending upon the pH of the solution. At pH 7 it forms a green coloured solution R . P , Q , R , will be: P (1) (2) (3) (4)

2+

Mn MnO2 MnO42– Mn2+

Q

R

MnO2 MnO42– Mn2+ Mn4+

MnO42– Mn2+ MnO2 MnO2

Ans. Option (1) is correct Explanation: At pH 7, it forms a green coloured compound name as MnO42–. 8. Molecules are the constituent particles of molecular solids. Identify the molecular solid amongst the following; (1) SO2 (2) C (3) Fe (4) NaCl Ans. Option (1) is correct Explanation: Molecular solids are the solids in which smaller number of atoms strongly bonded together i.e., CO2, SO2, H2O, Naphthalene, gasoline etc. 9. Which of the following reaction will give t-Butyl ethyl ether by Williamson synthesis? CH3 |

(1) CH3–C–ONa + CH3 CH2Cl |

CH3

7

CH3 |

(2) CH3–CH2ONa + CH3–C–Cl |

CH3 CH3 |

CH3 |

(3) CH3–C–Br + CH3–C–ONa |

|

H

H CH3 |

(4) CH3–CH2CH2ONa + CH3–CH–Br Ans. Option (1) is correct Explanation: Williamson synthesis involve the reaction of alkoxide with alkyl halide to form ether. Here, the correct pair of reactant used for the preparation of t-butyl ethyl ether are sodium t-butoxide and ethyl chloride. 10. Electrolysis of molten NaCl gives: (1) H2 at cathode, CI2 at anode and NaOH solution (2) Na at cathode, Cl2 at anode and NaOH solution (3) H2 at anode, Cl2 at cathode (4) Na at cathode and Cl2 at anode Ans. Option (4) is correct Explanation: NaCl → Na+(l) + Cl–(l) At cathode: reduction of 2Na+(l) + 2e– → 2Na(l) At anode: oxidation of 2CI–(l) → Cl2(g) + 2eNet reaction is written as: 2Na+(l) + 2CI–(l) → 2Na(l) + Cl2(g) 11. Which two transition metal ions has same 3d electronic configuration? (A) Mn3+ (B) Fe2+ 2+ (C) Mn (D) Fe3+ (E) Cr 3+ Choose the correct answer from the options given below: (1) (A) and (B) only (2) (B) and (C) only (3) (C) and (D) only (4) (D) and (E) only Ans. Option (3) is correct Explanation: (A) The electronic configuration for a Mn3+ ion is [Ar] 3d4.

(B) The electronic configuration for a Fe2+ ion is [Ar] 3d6. (C) The electronic configuration for a Mn2+ ion is [Ar] 3d5. (D) The electronic configuration for a Fe3+ ion is [Ar] 3d5. (E) The electronic configuration for a Cr3+ ion is [Ar] 3d54s1

12. Limiting molar conductivity of H2O is equal to: (1) Λ°m NaCl + Λ°m HCl – Λ°m NaOH (2) Λ°m HC1 + Λ°m NaOH – Λ°m NaCl (3) Λ°m HCl + Λ°m NH4OH – Λ°m NaCl (4) Λ°m NaCl + Λ°m HCl – Λ°m NH4OH Ans. Option (2) is correct Explanation: Λ m HCl  → Λ m H + + Λ m Cl − + Λ m NaOH  → Λ m Na + + Λ m OH − − Λ m NaCl  → Λ m Na + + Λ m Cl Λ m H 2 O  → Λ m H + + Λ m OH −

8

OSWAAL CUET (UG) Sample Question Papers, CHEMISTRY

13. A cell is prepared by dipping a Cu rod in 1 M CuSO4 solution and Sn rod in 1 M SnCl2 solution. The standard electrode potential of Cu is +0.34 V and Sn is -0.14 V. Emf of cell will be: (1) 0.48 V (2) 0.20 V (3) 0.34 V (4) 0.14 V Ans. Option (1) is correct Explanation: Eo=Ec-Ea =0.34–(-0.14) =0.48V 14. On passing ammonia gas through a solution of copper sulphate, a deep blue solution is obtained. The deep blue colour of the solution is due to the formation of: (1) [Cu(NH3)2]2+ (2) [Cu(NH3)4]2+ + (3) [Cu(NH3)6] (4) [Cu(NH3)6]2+ Ans. Option (2) is correct Explanation: Deep blue colour developed by addition of excess of ammonia gas in copper Sulphate solution and the formed complex is [Cu(NH3)4]2+. 15. The correct increasing basicity in aqueous medium for (I) NH3 (II) CH3CH2CH2  (III) (CH3CH2)2NH (IV) (CH3CH2)3N is: (A) (II) < (I) < (IV) < (III) (B) (II) < (I) < (III) < (IV) (C) (I) < (II) < (IV) < (III) (D) (I) > (II) > (IV) > (III) Choose the correct answer from the options given below: (1) (C) (2) (B) (3) (D) (4) (A) Ans. Option (3) is correct Explanation: The basicity of amine and alkyl amine depends upon the +I effect and Steric effect. 16. OH (1) NaOH (2) CO2 (3) HCl

A

Product 'A' is: (1) Salicylaldehyde (2) Benzoic acid (3) p-Chloro benzoic acid (4) Salicylic acid Ans. Option (4) is correct Explanation:

17. Match List-I with List-II. List – I (A) (B) (C) (D)

List – II

[Ni(CN)42– bidentate chelate ligand [NiCl4]2-

(I) (II) (III) (IV)

EDTA paramagnetic diamagnetic oxalate



Choose the correct answer from the options given below (1) (A)-(III), (B)-(IV), (C)-(l), (D)-(II) (2) (A)-(II), (B)-(IV), (C)-(I), (D)-(III) (3) (A)-(III), (B)-(I), (C)-(IV), (D)-(II) (4) (A)-(II), (B)-(I), (C)-(IV), (D)-(III) Ans. Option (1) is correct Explanation: (A) In [Ni(CN)4]2–, CN– is a strong field ligand which allows the pairing of the electrons and diamagnetic in nature. (B) Oxalate is an example of bidentate ligand. (C) EDTA is a chelating agent. (D) In [Ni(Cl)4]2-, Cl– is a weak field ligand which do not allow the pairing of the electrons in an orbital. Hence, it is paramagnetic in nature. 18. Identify the biodegradable polymer in the following: (1) Nylon -6, 6 (2) PVC (3) Polythene (4) PHBV Ans. Option (4) is correct Explanation: PHBV is poly -3-hydroxy butyrate Co-3-hydroxy valerate. It is biodegradable, nontoxic and bio-compatible plastic polymer. 19. Which of the following statements are correct about alkyl halides: (A) Alkylhalides are polar so they are soluble in water (B) RX on reaction with alc.KOH give alkenes (C) RX on reaction with AgNO2 give RONO (D) SNl proceeds with inversion in configuration (E) R—X have higher boiling point than parent alkanes Choose the correct answer from the options given below: (1) (B), (E) only (2) (A), (C) only (3) (D), (E) only (4) (B), (C) only Ans. Option (1) is correct Explanation: Their boiling point is higher than the parent alkanes because of high molecular weight of halogens, high intermolecular forces of attractions and polarizability of the halogens. H H H H | β |α

H – C – C – H  |

|

H X (X = Cl, Br, I)

alc.KOH ∆

   C = C H H

20. When 1.5g of a non-volatile solute is dissolved in 30g of a solvent, the boiling point of the solvent is raised by 2 K. Calculate the molar mass of the solute, given that Kb for the solvent is 1.85 K kg mol-1. (1) 46.25 g mol–1 (2) 103 g mol–1 (3) 94 g mol–1 (4) 23 g mol–1 Ans. Option (1) is correct

CUET Solved Paper 2023 Explanation:

9

Explanation: Methyl propanoate on reduction gives Propanol and Methanol.

K × WB × 1000 M B= b ∆Tb × WA 1.85 × 1.5 × 1000 = 2 × 30 = 46.25 g mol -1

CH3CH3COOCH3

LiAlH4

Methyl propanoate

CH3CH2CH2OH+CH3OH Propanol (1o alcohol)

Methanol (1o alcohol)

21. Which of the following statements are true? (A) Noble gases have low M.P. or B.P. due to weak dispersion forces. (B) All noble gases have completely filled ns2np6 electronic configuration. (C) XeF2 and XeF6 are colourless crystalline solid. (D) XeO3 has pyramidal geometry. Choose the correct answer from the options given below: (1) (A), (C) only (2) (B), (C), (D) only (3) (A), (B), (C), (D) only (4) (A), (C), (D) only

24. Arrange the following in the increasing order of their van't Hoff factors (A) Very dilute MgCl2 solution (B) Very dilute AlCl3 solution (C) Very dilute NaCl solution (D) Very dilute Al2O3 solution (E) Very dilute urea solution Choose the correct answer from the options given below: (1) (E) < (C) < (A) < (B) < (D) (2) (E) < (C) < (A) < (D) < (B) (3) (E) < (C) < (B) < (A) < (D) (4) (E) < (B) < (A) < (D) < (C)

Ans. Option (4) is correct

Ans. Option (1) is correct Explanation: Urea, i=1 NaCl Na++Cl–, i=2 MgCl2 Mg 2++2Cl–, i=3 AlCl3 Al3++3Cl–, i=4 Al2O3 2Al3++3O–, i=5

Explanation: All noble gases have low melting and boiling point due to weak dispersion forces of attraction.XeF2 and XeF4 are colourless crystalline solids. XeO3 has pyramidal symmetry. F Xe

Xe O

O O

F

F F

F Xe

F

25. Match List - I with List - II.

F

List - I

F

(A) Denaturation

XeO3= Pyramidal   Symmetry    XeF2     XeF6

(B) Primary structure of protein (C) Quaternary structure of protein (D) Secondary structure of protein

22. The entities which do not dissociate into simple ions are: (A) [Fe(CN)6]4- (B) KCl∙MgCl2∙6H2O (D) KAl(SO4)2∙12H2O



(C) K4[Fe(CN)6]



(E) FeSO4∙(NH4)2SO4∙6H2O



Choose the correct answer from the options given below: (1) (A) and (C) only (2) (B) and (D) only (3) (D) and (E) only (4) (A) and (B) only Ans. Option (1) is correct Explanation: Both are the complex coordination compounds and remains unchanged. They do not have water of crystallization in their structures. 23. Ethanol cannot be prepared from: COOC2H5 (1) (2) CH3CH2CH2COOC2H5 (3) C2H5COOCH3 (4) CH3COOC2H5 Ans. Option (3) is correct

List - II (I)

Sequence of amino acids in a protein (II) a-Helix and b-pleated sheets (III) 2o and 3o structures gets destroyed (IV) The spatial arrangement of subunits (containing 2 or more polypeptide chain) with respect to each other.



Choose the correct answer from the options given below: (1) (A)-(I), (B)-(II), (C)-(III), (D)-(IV) (2) (A)-(III), (B)-(I), (C)-(IV), (D)-(II) (3) (A)-(III), (B)-(II), (C)-(IV), (D)-(I) (4) (A)-(I), (B)-(IV), (C)-(II), (D)-(III) Ans. Option (2) is correct Explanation: Denaturation is a process in which proteins denature at high temperature, loose its identity and remains in primary structure Primary structure is a sequence of amino acid. Secondary structure consist of a-helix and β-pleated sheets. Quaternary struture is the spatial arrangement of sub-units with respect to each other (consist of two or more polypeptide chain).

10

OSWAAL CUET (UG) Sample Question Papers, CHEMISTRY

26. Match List -1 with List - II.

Explanation:

List -I

List -II

(A)

NH2

(B)

N2Cl

+

–

Br2 /H2O

+ (I) NH3

– SO3

t 90%

Br

(1) (CH3CO)2O, Pyridine (2) HNO3, H2SO4 (II) Br (3) OH–

– (1) HBF 4

(C)

N2Cl

NH2

(D)

NH2

H2SO4 ∆

2 n 10 K From equation (1) and (2) t 99% =2 × t 90%

NH2

(II) NO2

(2) NaNO2, ∆

=

NO2

(IV)

Choose the correct answer from the options given below: (1) (A)-(II), (B)-(IV), (C)-(III), (D)-(I) (2) (A)-(III), (B)-(II), (Q-(IV), (D)-(I) (3) (A)-(II), (B)-(III), (C)-(IV), (D)-(I) (4) (A), (I), (B)-(II), (C)-(III), (D)-(IV)

30. Match List -1 with List - II. List -I

28. Glylosidic linkage is formed by: (1) Elimination of water molecule (2) Addition of water molecule (3) Elimination of ammonia molecule (4) Addition of ammonia molecule Ans. Option (1) is correct Explanation: Glycosidic linkage formed by two monosaccharide units connected by a oxygen atom with the loss of one water molecule. CH2OH H OH

O

H OH

H

H

OH

H

HOH2C O Glycosidic linkage

H OH

– D – glucose

O

H OH CH2OH H

– D – fructose Surcose

29. t99%with respect to t90% for a first order reaction is (1) four times of t50% (2) one and half time (3) It is same (4) Double Ans. Option (4) is correct

List - II

(A) Molarity

Number of moles of solute/kilogram of solvent (B) Molality (II) Osmotic pressure (C) Coiligative property (III) Number of moles of solute/litre of solution (D) Non-ideal solution (IV) Deviation from Raoult's Law

Ans. Option (3) is correct

Explanation: 2Mn 2++5S 2O8 2-+8H 2O → 2MnO4 -+10SO4 2-+16H+

(2)

x=2



27. Manganese ions on reaction with peroxydisulphate ions form: (1) MnO42– (2) MnO2 (3) Mn2O7 (4) MnO4– Ans. Option (4) is correct

(1)

1 For t 99% , a t = ao 100 ao 1 t 99% =  n K a o / 100

Br +

10a o a o = 100 10 ao 1 n10 = n = K a o / 100 K

for t 90% , at =

(I)



Choose the correct answer from the options given below: (1) (A)-(III), (BHD, (C)'(IV), (D)-(II) (2) (A)-(III), (B)-(I), (C)-(II), (D)-(IV) (3) (A)-(I), (B)-(III), (C)-(II), (D)-(IV) (4) (A)-(I), (B)-(III), (C)-(IV), (D)-(II) Ans. Option (2) is correct Explanation: Molarity is number of moles of solute dissolve per litre of solution. Molality is number of moles of solutes dissolve per kilogram of solvent. Osmosis is one of the colligative property. Non-ideal solutions do not follows Raoult’s Law. 31. The decomposition of ammonia on platinum surface is a zero order reaction. How much time it will take for 1 × 10–4 mol L–1 of ammonia to reduce into half of its concentration? (K=0.5×l0–4 mol L–1 s–1) (1) 1 s (2)  10 s (3) 100 s (4)  5 s Ans. Option (1) is correct Explanation:

( A0 ) =

1 × 10 –4

1 × 10 –4 = 0.5 × 10 –4 2 A −A t= 0 K 1 × 10-4 – 0.5 × 10 –4 t= 0.5 × 10-4 t=1s

( A=)

CUET Solved Paper 2023 32. Identify the correct statement: (1) CrO is acidic, Cr2C3 is basic, CrO3 is amphoteric. (2) CrO is basic, Cr2C3 is acidic, CrO3 is amphoteric. (3) CrO is basic, Cr2C3 is amphoteric, CrO3 is acidic. (4) CrO is amphoteric, Cr2C3 is basic, CrO3 is acidic. Ans. Option (3) is correct Explanation: Oxidation state of CrO is +2, Oxidation state of Cr2O3 is +3 and oxidation state of CrO3 is +6. Lower oxidation states forms basic oxide, higher oxidation state forms amphoteric and acidic oxides. 33. Which of the following is not a property of α-sulphur? (1) It is readily soluble in CS2 (2) It has monoclinic structure. (3) This allotrope is yellow in colour. (4) It's melting point is 385.8 K. Ans. Option (2) is correct

Ans. Option (3) is correct Explanation: Many properties of metal like lustre, electrical and thermal conductivity depends upon free electrons. 36. The integrated rate equation for a first order reaction kt is log Ro— log Rt. The straight line graph is 2.303 obtained by plotting: (1) time vs Rt (2) time vs log Rt 1 (3) time vs Rt (4) time vs Ro Ans. Option (2) is correct Explanation:

Explanation: a-sulphur is known as rhombic sulphur while b-sulphur is known as monoclinic sulphur.

O H

(1)

H OH O H

H H

OH

H O

CH2OH OH

OH OH H

H H

H

(4)

H O

CH2OH

37. The correct order of reactivity of following halides towards SN1 reaction is: Cl CH2Cl CH2Br CH2I

OH

H

(3)

Time

H

OH

(2)

Slope = –k

OH

H CH2OH

First order

ln[A]t

ln[A]0

34. The following pentose sugar is obtained by complete hydrolysis of DNA, along with phosphoric acid and nitrogenous base: CH2OH

11

OH H

H

H

H H Ans. Option (2) is correct Explanation: Complete hydrolysis of DNA give inorganic phosphate, 2-deoxyribose and one of four different heterocyclic bases. 35. Lustre, electrical and thermal conductivity of metals can be explained by the theory of: (1) dipole-dipole interactions (2) Coulombic (electrostatic) forces (3) Sea of free electrons (4) London forces

(I) (II) (III) (1) (IV) > (III) > (II) > (I) (2) (I) > (II) > (III) > (IV) (3) (IV) > (II) > (III) > (I) (4) (I) > (III) > (II) > (IV) Ans. Option (1) is correct

(IV)

Explanation: In SN1 mechanism, reactivity decreases by intermediate benzyl carbocation formed in the slowest steps which is stabilized by resonance. 38. The reaction of 3-methylbutan-2-ol with HBr, will give: (1) 2-Bromo-2-methylbutane (2) 3-Bromo-2-methylbutane (3) 3-Bromo-3-methylbutane (4) 2-Bromo-3-methylbutane Ans. Option (1) is correct Explanation: CH3—CH—CH—CH3

HBr

CH3 OH 39. The shape of XeF6 is: (1) Regular octachedral (3) Distorted octahedral

Br CH3—C—CH2—CH3 CH3 (2) Square planar (4) Square pyramidal

12

OSWAAL CUET (UG) Sample Question Papers, CHEMISTRY

Ans. Option (3) is correct

(1) (A) (3) (C)

Explanation:

F F

Ans. Option (4) is correct F

Explanation: cyclic hexanone undergo aldol condensation.

Xe F

F F

40. The relative order of reactivity of 1°, 2°, 3° alcohol towards dehydration is: (1) 1° > 2° > 3° (2) 2° > 1° > 3° (3) 2° > 3° > 1° (4) 3° > 2° > 1° Ans. Option (4) is correct Explanation: Reactivity is 3°> 2° >1°, because tertiary carbocation is more stable than 2° and 2° is more stable than 1° due to inductive effect of alkyl group. Read the passage below and answer the question: Aldehydes and Ketones, having atleast one methyl group linked to the carbonyl carbon atom (methyl Ketones), are oxidised by sodium hypohalite to sodium salts of corresponding carboxylic acids having one carbon atom less them that of carbonyl compound is converted to haloform. This oxidation does not affect carbon – carbon double bond, if present in a molecule. 41.

List - I (A)

CH3 H

(B)

CH3 H

(C)

CH3 H

(D)

CH3 H

List - II (I)

Hydrazones

(II)

oximes

C=O+H2NOH

C=O+H2NNH2 (III)

Ethane

(IV)

phenyl hydrazones

C=O+H2NNHC6H5

C=O+Zn – Hg/HCl

Choose the correct answer from the options given belowr: (1) (A)-(II), (B)-(IV), (C)-(III), (D)-(I) (2) (A)-(II), (B)-(I), (C)-(IV), (D)-(III) (3) (A)-(III), (B)-(IV), (C)-(I), (D)-(II) (4) (A)-(III), (B)-(I), (C)-(II), (D)-(IV) Ans. Option (2) is correct 42. Which of the following aldehyde will give Aldol condensation: O || CH3 CH=C–C–H | O | || CH3–C–CHO CH 3 | —CHO CH3 (A)

(B)

(2) (B) (4) (D)

(C)

(D)

43. Arrange in decreasing order of nuelophilic addition: (A) CH3CHO (B) C2H5COCH3 (C) CH3COCH3 (D) HCHO Choose the correct answer (1) (D) > (A) > (C) > (B) (2) (D) > (A) > (B) > (C) (3) (C) > (B) > (A) > (D) (4) (B) > (C) > (A) > (D) Ans. Option (1) is correct Explanation: The reactivity of nucleophilic addidtion in carbonyl compound decreases with increse in electron density on carbonyl carbon. The inductive effect increase the electron density on carbonyl carbon atoms. Hence, +I effect is HCHO H2S > H2Se> H2Te > H2Po This is because the H-E bond length increases down the group, hence the bond dissociation enthalpy decreases down the group. 20. Match list I with list II List I

List II

A. Ammonia

I. Ostwald’s process

B. Chlorine

II. Contact process

C. Sulphuric Acid

III. Deacon process

D. Nitric Acid

IV. Haber’s process

Choose the correct answer from the options given below: (A) A-IV, B-III, C-II, D-I (B) A-IV, B-I, C-II, D-III (C) A-IV, B-III, C-I, D-II (D) A-IV, B-I, C-III, D-II Ans. Option (A) is correct Explanation: Ammonia: Haber’s process Nitric acid: Ostwald’s process Sulphuric acid: Contact process Chlorine: Deacon’s process

OSWAAL CUET (UG) Sample Question Papers, Chemistry 21. The formula of a noble gas species which is isostructural with BrO-3 is: (A) XeOF4 (B) XeF2 (C) XeO3 (D) XeF4 Ans. Option (C) is correct – Explanation: In BrO3 , the hybridisation of Br is 3 sp and has a pyramidal structure while in XeO3, the hybridisation of Xe is also sp3 with pyramidal – structure. BrO3 is isostructural with XeO3. Br O

Xe –

O O BrO3

O

O O XeO3

22. Match list I with list II List I (Transition Metals)

List II (Maximum Oxidation State)

A. Ti

I. 7

B. V

II. 4

C. Mn

III. 5

D. Cu IV. 2 Choose the correct answer from the options given below: (A) A-II, B-III, C-I, D-IV (B) A-I, B-II, C-III, D-IV (C) A-III, B-I, C-II, D-IV (D) A-II, B-I, C-III, D-IV Ans. Option (A) is correct Explanation: Electronic configuration of Ti is [Ar] 3d24s2 So, maximum oxidation state of Ti is +4. Electronic configuration of V is [Ar] 3d34s2 So, maximum oxidation state of V is +5. Electronic configuration of Mn is [Ar] 3d54s2 So, maximum oxidation state of Mn is +7. Electronic configuration of Cu is [Ar] 3d104s1 So, maximum oxidation state of Cu is +2.

Electronic configuration of X2+ : [Ar]3d54s0 Number of unpaired electrons, n=5

 = n(n + 2) = 5(5 + 2) = 35 = 5.92 BM 25. Which one of the following transition metal ion is colourless? (A) Sc3+ (C) Mn2+ Ans. Option (A) is correct

(B) V2+ (D) Co3+

Explanation: The ion will show colour if it has unpaired electrons in its electronic configuration. In Sc3+, electronic configuration [Ar]3d04s0, there are no unpaired electrons due to completely vacant d orbital that cannot show d-d transition for colour spectra. Hence, it will be colourless. 26. Among the following statements, choose the correct statements. (A) SN2 reaction proceeds with stereochemical inversion. (B) The process of conversion of a racemic mixture into enantiomer is known as racemisation. (C) A mixture containing 2 enantiomers in equal proportions is known as a racemic mixture. (D) The stereoisomers related to each other as superimposable mirror image are called enantiomers. (E) The objects which are non-superimposable on their mirror image are said to be chiral and this property is known as chirality. Choose the correct answer from the options given below: (A) A, B and C only (B) A, C and E only (C) B, C and E only (D) C, D and E only Ans. Option (B) is correct

23. The metal from first transition series having positive

Explanation:

value: (A) Cr (B) V (C) Cu (D) Ni Ans. Option (C) is correct Explanation: The reduction potential of Cu2+ is positive in value in 3d series and the rest Ni2+, V2+, Cr2+ have negative values. 24. Magnetic moment of a divalent ion in aqueous solution of an element with atomic number 25 is:

Inversion: Stereochemical inversion is shown in SN2 reaction mechanism.

E°M

2+

M

(A) 2.84 BM (B) 3.87 BM (C) 4.90 BM (D) 5.92 BM Ans. Option (D) is correct Explanation: Atomic number of element X=25 Electronic configuration of X: [Ar]3d54s2

Racemic mixture: A mixture of equal quantities of two enantiomers, or substances that have dissymmetric molecular structures that are mirror images of one another. Chirality: No-superimposable mirror images are said to be chiral

Thus, options A, C and E are correct 27. IUPAC name of neopentyl chloride is: (A) 1-Chloro – 2, 2-dimethylpropane (B) 2-Chloro – 1, 2-dimethylpropane (C) 2-Chloro – 2 – Methylbutane (D) 2-Chloro – 2 – Methylpentane

Solved Paper-2022 19 Ans. Option (A) is correct Explanation: Common name: Neopentylchloride IUPAC name:

CH3 | H 3C – C – CH 2 – Cl | CH 3 1-chloro-2, 2-dimethyl propane 28. The structure of major mono halo product in the following reaction is _________ . CH 2OH + HCl heat

HO

Cl

CH 2Cl

(A)

(B) HO

Cl CH 2Cl

(C)

CH 2OH

CH 2OH

(D)

OH

Cl

Ans. Option (A) is correct Explanation: Nucleophilic substitution of chloride on sp3 carbon is faster than sp2 carbon.

CH2OH  + HCl HO

CH2—O

H H

HO +

CH2

–

Cl

HO

CH2—Cl

(E) The high boiling points of alcohols are mainly due to the presence of intramolecular hydrogen bonding. Choose the correct answer from the options given below: (A) A, D and E only (B) A, B and C only (C) B, C and D only (D) C, D and E only Ans. Option (A) is correct Explanation: Boiling points of alcohols increase with increase in the number of carbon atoms due to increase in van Der Waals forces. Also, it decreases with increase in branching due to decreased surface area. Alcohols have higher boiling point than hydrocarbons of their comparable mass due to intermolecular hydrogen bonding. (Note: Option E is correct if intramolecular is replaced by intermolecular) 30. Arrange the following compounds in increasing order of their acid strength: (A) Propan-1-ol (B) 3-nitrophenol (C) 3, 5-dinitrophenol (D) Phenol (E) 4-Methylphenol Choose the correct answer from the options given below: (A) A < D < C < B < E (B) C < B < D < E < A (C) A < B < C < D < E (D) A < E < D < B < C Ans. Option (D) is correct Explanation: Alcohols and phenols are electronrich compounds and when any electronwithdrawing group is attached, their acidity increases. Alkyl groups show +I effect and –NO2 group show –M effect. Acidic nature of electron-withdrawing groups is as follows: Dinitro- > nitro > Phenyl > methyl > hydride 31. The structure of the product of the following reaction is: O

CH 2—C—O—CH3

HO

O O

29. Among the following statements, choose the correct statements. (A) Boiling point of alcohols increases with increase in the number of carbon atoms. (B) In alcohols, boiling points increases with increase of branching in carbon chain. (C) Boiling points comparison to molecular mass. (D) Boiling points comparison to molecular mass.

of alcohols are lesser in haloalkanes of comparable

(A)

(B)

CH 2—C—O—CH3 O

OH

(C)

(D)

CH 2—CH—OCH3 OH

OH

OH

of alcohols are higher in hydrocarbons of comparable

NaBH4

O CH 2—C—H O CH 2—C—H

Ans. Option (B) is correct

OSWAAL CUET (UG) Sample Question Papers, Chemistry Explanation: Sodium borohydride is a good reducing agent. Although not as powerful as lithium aluminium hydride (LiAlH4), it is very effective for the reduction of aldehydes and ketones to alcohols. By itself, it will generally not reduce esters, carboxylic acids, or amides (although it will reduce acyl chlorides to alcohols).

O

NaBH4

CH2—C—O—CH3

acids behave both as acids and bases so they are amphoteric in nature.

O

O +

NH2

NH3 (zwitter ion)

34. Match List-I with List-II List I (Nomenclature)

List II (Structure) CHO

i.

OH

O

2. Benzaldehyde

CH2—C—O—CH2 O

ii.

32. The product of the following reaction is:

iii.

+ HBr (B)

Br

iv.

+ C2H5OH

+ C2 H5Br

OC2H5

(D)

OC2H5 Br

Br Explanation: Nucleophilic substitution of HBr gives phenol and bromoethane as a product.

Acetophenone:

Ketone

group:

O

Benzaldehyde: Aldehyde group on benzene:

H +

O–CH2–CH 3

+ HBr

Choose the correct answer from the options given below: (A) 1 – III, 2 – I, 3 – II, 4 – IV (B) 1 – II, 2 – III, 3 – IV, 4 – I (C) 1 – I, 2 – II, 3 – III, 4 – IV (D) 1 – IV, 2 – III, 3 – II, 4 – I Ans. Option (B) is correct Explanation:

Ans. Option (A) is correct

O–CH2–CH3

C

COOH

4. Benzophenone

OH

C— CH3

O

3. Benzoic acid

OC2H5

(C)

R—CH—C—O

1. Acetophenone

O

(A)

–

R—CH—C—O—H

OH

CHO

Br– CH3–CH2–Br

33 . Amino acid in zwitter ionic form show:

Benzoic acid: Carboxylic acid on benzene: COOH

(A) Acid Behaviour (B) Basic Behaviour (C) Amphoteric Behaviour (D) Neutral Behaviour Ans. Option (C) is correct Explanation: Amino acids have amino (−NH2) group, basic in nature and accept a proton and COOH group loses a proton forming a dipolar ion, called the Zwitter ion. In this form, amino

Benzophenone: ketone group:

Two benzene attached

by

O C

35. Which simple chemical test is used to distinguish between ethanal and propanal? (A) Iodoform test (B) Tollen’s test (C) Fehling’s test (D) Lucas test

Solved Paper-2022 21 Ans. Option (A) is correct Explanation: Iodoform test is used to detect aldehydes and ketones having CH3CO− (methyl ketone) group. I2 CH 3CHO ⎯⎯⎯ → CHI 3 + HCOONa  NaOH

Yellow Pr ecipitate I2 CH 3CH 2CHO ⎯⎯⎯ → No characteristic change NaOH

Tollens' test, also known as silver-mirror test is used to distinguish between an aldehyde and a ketone. Fehling’s test is used for the estimation or detection of reducing sugars and non-reducing sugars. Lucas test is used to differentiate and categorise primary, secondary and tertiary alcohols. 36. Which of the following compound would undergo Aldol condensation? (A) Methanal (B) Benzaldehyde (C) 2, 2-Dimethylbutanal (D) Phenylacetaldehyde Ans. Option (D) is correct Explanation: Aldehydes which contain at least 2 hydrogen atoms, undergo Aldol condensation reaction Phenylacetaldehyde contains 2 hydrogen atoms and undergoes Aldol condensation reaction in presence of dilute alkali.

Tranquilizers: A neurologically active drug that is used to reduce anxiety, fear, tension, agitation, and related states of mental disturbance. Disinfectants: Any substance used primarily on non-living objects to kill germs, such as viruses, bacteria, and other microorganisms that can cause infection and disease. Antiseptics: The substances that help to stop the growth of microorganisms on the skin used daily in medical settings to reduce the risk of infection and stop the spread of germs. 38. Out of the following artificial sweetening agents, which one has the highest sweetness value in comparison to cane sugar? (A) Saccharin (B) Alitame (C) Sucralose (D) Aspartame Ans. Option (B) is correct Explanation: Sweetness comparison with cane sugar of the following artificial sweetening agents is as follows Alitame > Sucralose > Saccharin > Aspartame 2000 > 600 > 550 > 100 39. Among the following polymers, which one is the copolymer? (A) Polypropene (B) Polystyrene (C) Polyvinyl chloride (D) Glyptal Ans. Option (D) is correct Explanation: Homopolymers consist of single species of repeating units whereas copolymers consist of two or more types of repeating units. Polypropylene, Polystyrene, and Polyvinyl chloride: Homopolymer Glyptal: Copolymer COOH

37. Among the following statements choose the correct statements. (A) Analgesics reduce or abolish pain without causing impairment of consciousness, mental confusion. (B) Tranquilizers are neurological inactive drugs. (C) Morphine is the example of non-narcotic analgesics. (D) Disinfectants are applied to inanimate objects whereas antiseptics are applied to the living tissues. (E) Same substance can act as an antiseptic as well as disinfectant by varying the concentration. Choose the correct answer from the options given below: (A) A, D and E only (B) B, C and D only (C) A, C and E only (D) B, C and E only Ans. Option (A) is correct Explanation: Analgesics: These are medications that relieve pain which do not turn off nerves, change the ability to sense your surroundings or alter consciousness. Morphine is narcotic analgesics.

CH 2 – CH 2 + n n HO HO Ehylene glycol

COOH Phthalic acid

O

O

O ||

O || n

Glyptal

40. Among the following, which one is a disaccharide? (A) Glucose (B) Glycogen (C) Maltose (D) Starch Ans. Option (C) is correct Explanation: Disaccharides are formed by the glycosidic linkage of two monosaccharide units. Glucose: Monosaccharide Glycogen: Polysaccharide Maltose: Disaccharide Starch: Polysaccharide 41. Structure of ammonium salt when ethylamine reacts with one mole of HCl: (A) C2H5 – NH+3Cl– (B) (C2H5)2 – NH+2Cl– + – (C) (C2H5)3 – NH Cl (D) (C2H5)4 – N+Cl– Ans. Option (A) is correct Explanation: When ethyl amine reacts with HCl, it forms ethyl ammonium chloride as a product.

OSWAAL CUET (UG) Sample Question Papers, Chemistry 

–

CH3 CH2 — NH2  HCl   CH3 CH2 NH3 Cl 42. Among the following amines, which one is most basic (in aqueous solution)? (A) NH3 (B) C2H5NH2 (C) (C2H5)2NH (D) (C2H5)3N Ans. Option (C) is correct Explanation: In aqueous solution, secondary amine is most basic among aliphatic amines due to inductive effect, solvation effect, and steric hindrance of the alkyl group in aqueous medium. Hence, the basic strength order of the amines is: (C2H5)2NH > (C2H5)3N > C2H5NH2 > NH3 43. The correct order of basicity of amines in gas phase (A) 1° < 3° < 2° (B) 3° < 1° < 2° (C) 2° < 3° < 1° (D) 1° < 2° < 3° Ans. Option (D) is correct Explanation: The order of basicity of amines in the gaseous phase increase with the + I effect of the attached alkyl groups to the amine group follows the order: 3o amine >2o amine >1o amine >NH3. 44. Among the following, which one has the highest pKb value? (A) C2H5NH2 (B) C6H5NHCH3 (C) (C2H5)2NH (D) C6H5NH2 Ans. Option (D) is correct Explanation: Higher is the value of pKb weaker is the basic strength. In aniline (C6H5NH2), due to delocalization of lone pairs, its availability for protonation is minimal. Hence, it is least basic amongst all other amines. 45. Among the following, which one has the highest Kb value? (A) C2H5NH2

(B) C6H5N(CH3)2

(C) (C2H5)2NH

(D) CH3NH2

Ans. Option (C) is correct Explanation: The basic nature of amines shows, it must have more value of Kb. (C2H5)2NH>C2H5NH2>CH3NH2>C6H5N(CH3)2 Kb : 10–3

10–3.29

10–3.38

10–8.92

Passage:

According to the valence bond theory, the metal atom or ion under the influence of ligands can use its (n-1)d, ns, np, nd orbitals for hybridisation to yield a set of equivalent orbitals of definite geometry. These hybridised orbitals are allowed to overlap with ligand orbitals that can donate electron pairs for bonding. It is usually possible to

predict the geometry of a complex from the knowledge of its magnetic behaviour on the basis of the valence bond theory. Consider the formation of [Co(NH3)5Cl]Cl2 . Answer question no. 46 to 50 as per directed. 46. The IUPAC name of the above coordination entity is: (A) Chloridopentaamminecobaltate (II) chloride (B) Chloridopentaamminecobaltate (II) dichloride (C) Pentaamminechloridocobalt (III) chloride (D) Pentaamminechloridocobalt (III) dichloride Ans. Option (C) is correct Explanation: There are 5 amine group, 1 chloride group attached to cobalt(III) in the coordination sphere. [Co(NH3)5Cl]Cl2 : Pentaamminechloridocobalt (III) chloride. 47. The spin only magnetic moment of the complex [Co(NH3)5Cl]Cl2 in BM is: (A) 1.7 (B) 0.0 (C) 3.8 (D) 4.9 Ans. Option (B) is correct Explanation: Due to strong field ligand, NH3 electrons of d-orbital pair up so there is zero unpaired electrons to show spin only magnetic moment. Hence, spin only magnetic moment is 0 BM. 48. The hybridisation of cobalt in the above coordination entity is: (A) sp3d2 (B) d2sp3 (C) sp3d Ans. Option (B) is correct

(D) dsp3

Explanation: In [Co(NH3)5Cl]Cl2, Co3+ Electronic configuration of Co3+: [Ar]3d54s04p0

•• ••

NH3 NH3

••

NH3

•• •• – NH3 NH3 Cl

(d2 sp3 hybridised)

49. The coordination number of cobalt in the above coordination entity is (A) 2 (B) 4 (C) 5 (D) 6 Ans. Option (D) is correct Explanation: Coordination number is equal to the total number of donating sites within the coordination sphere = 5 + 1 = 6. 50. The primary valence of Co in above coordination entity is (A) 1 (B) 2 (C) 3 (D) 4 Ans. Option (C) is correct Explanation: X+5(0) +3(-1)=0 X=+3 Primary valency = +3



CUET Question Paper 2022 NATIONAL TESTING AGENCY 17th August 2022—Slot-1

Chemistry [This includes Questions pertaining to Domain Specific Subject Only]

SOLVED . Time Allowed: 45 Mins.

Maximum Marks: 200

General Instructions: 1. The test is of 45 Minutes duration. 2.

The test contains 50 Questions out of which 40 questions need to be attempted.

3.

Marking Scheme of the test: a.

Correct answer or the most appropriate answer: Five marks (+5)

b.

Any incorrect option marked will be given minus one mark (–1).

c.

Unanswered/Marked for Review will be given no mark (0).

1. In a face-centred cubic unit cell of close packed atoms, the radius of atom (r) is related to the edge length ‘a’ of the unit cell by the expression: a (A) r = (B) r = a 2 2 (C) r =

a

(D) r =

2 2

3a 4

(A) (B) (C) (D)

Ans. Option (C) is correct Explanation: For FCC,

Ans. Option (A) is correct Explanation: For ferromagnetic substances, below Tc(Curie

4* radius

temperature),

spins

are

aligned

antiparallel but do not cancel.

edge

Ferrimagnetic

edge

Figure: Atoms in FCC arrangement

From the figure, Edge length (a) in terms of radius = 4r and relation between edge length (a) and radius of atom is given by:

2a = 4r So, from this, R = r =

2a 4 2a 2 2 2

Simplifying, we get r =

a 2 2

2. Which of the following arrangements represents alignment of magnetic moments of Ferrimagnetic substance?

Fig: Alignment of magnetic moment in Ferrimagnetic Substances Out of the given options, correct answer is option (A) as here, all spins do not cancel each other. 3. An alloy of Cu, Ag and Au is found to have Cu constituting the ccp lattice. If Ag atoms occupy edge centres, and Au is present at body centre, then the alloy has formula: (A) Cu4Ag2Au (B) Cu4Ag4Au (C) Cu4Ag3Au (D) CuAgAu Ans. Option (C) is correct Explanation: Cu is having CCP lattice, so total number of atoms in CCP lattice = 4 Hence, total no. of Cu atoms = 4

OSWAAL CUET (UG) Sample Question Papers, Chemistry Ag atom is occupying edge centre, As no. of atoms at the edges of CCP = 12 and each edge atom occupies 1/4th space, So total no. of Ag atoms = 12* ¼ = 3 Also, Gold (Au) is present at body centre, And no. of body centre atom in CCP = 1 Hence, total no. of Gold (Au) atoms = 1 So the formula of alloy will be Cu4 Ag3Au. 4. What is ∆G° for the given reaction? Zn (S) + Cu2+(aq) ⎯→ Zn2+ (aq) + Cu(S) 0 (Given : E

Zn 2+ /Zn

=–0.76 V & E0

Cu2 + / Cu

= +0.34 V)

F = 96500 C mol–1 (A) ∆G° = –212.3 kJ/mol (B) ∆G° = +212.3 kJ/mol (C) ∆G° = –312.5 kJ/mol (D) ∆G° = 0 Ans. Option (A) is correct Explanation: Relation between Gibb’s Free energy and cell potential is given by: Δ G = -nFE°cell Where, Δ G = change in Gibb’s free energy, kJ/mole E°cell = cell potential, V F= Faraday= 96500 C For given cell reaction, Zn (S) + Cu2+(aq) ⎯→ Zn2+ (aq) + Cu(S) n=2 (Given :

E0

Zn 2+ /Zn

=–0.76 V & E0

Cu2 + / Cu

= +0.34 V)

As reduction potential of Cu is higher, it will act at cathode and Zn will act at anode. E° cell = E°cathode- E°anode E° cell = 0.34- (-0.76) = 1.1 V Now, Δ G = -nFE°cell Hence, Δ G = - 2 * 96500* 1.1 Δ G = -212.3 kJ/mole 5. How many electrons flow when a current of 5 amperes is passed through a metal for 193 s? (Given: F = 96500 C mol–1, NA = 6.022 × 1023 mol–1) (A) 6.022 × 1023 electrons (B) 6.022 × 1021 electrons (C) 3.011 × 1021 electrons (D) 3.011 × 1023 electrons Ans. Option (B) is correct Explanation: When current flows through a conductor, the relation between charge and the current flowing through conductor per unit time is given by: Q = I* t Where, I = current flowing through the conductor, A Q= amount of charge flowing, C= charge t= time Q = 5 * 193 = 965 C

Now, 1 F= charge of 1 mole electrons Hence, 0.01 F = charge of 0.01 mole electrons = charge of 6.02 * 1023 * 0.01 electrons ( as 1 mole of electrons = (6.02 * 1023 atoms) = 6.02 * 1021 electrons 6. Kohlrausch law of independent migration of ions is applicable (A) Only to weak electrolytes at a certain concentration. to strong electrolytes at all (B) Only concentrations. (C) To both - strong and weak electrolytes. (D) To non-electrolytes Ans. Option (C) is correct Explanation: Kohlrausch's law of independent migration of ions states that the limiting molar conductivity of an electrolyte can be represented as the sum of individual contributions of its cations and anions. This law is applicable to both strong and weak electrolytes. So correct answer is option (C). 7. In the first-order reaction the concentration of the reactant is reduced 1 th in 60 minutes, what 4

will be its half-life? (A) 120 minutes (B) 40 minutes (C) 30 minutes (D) 25 minutes Ans. Option (C) is correct Explanation: Half of half life will be t1/4 t1/4 = 2t1/2 t1/4 = time for the concentration of reactant to be reduced to 1/4th of its original concentration = 60 min (given) So, 2 t1/2 = 60 Hence, t1/2 = 30 min 8. Which of the following is the correct relationship between the time required for completion of 99.9% of a first order reaction and its half-life? (A) t 1 = 5 × t 99.9% 2

(B) t 99.9%= 10 × t 1 2

(C) t 99.9%=2t 1 2

(D) t 99.9%= t 1 2 Ans. Option (B) is correct

Solved Paper-2022 25

(1)

Assuming, [A]0 = 100 M Then, at t= t99.9%, 99.95 reaction is completed, X = 99.9 M From (1),  100  t99.9 = 2.303/ k log  (2)   100 – 99.9  = 6.93 /k (3) Since, t1/2 = 0.693/k; Writing equation (3) in terms of half life, 0.693 t99.9 =  10 k t99.9 = 10  t1/ 2 9. A catalyst increases the rate of reaction by: (A) Decreasing enthalpy of reactants (B) Increasing internal energy of reactants (C) Decreasing activation energy of reaction (D) Increasing activation energy of reaction Ans. Option (C) is correct Explanation: A catalyst is a chemical substance that either increases or decreases the rate of a chemical reaction. In the case of activation energy, a catalyst lowers it. 10. Match List-I with List-II. List-I A. B. C. D. C.

List-II

A. B. C C.

Gemstone Milk Cloud

I. II. III.

Emulsion Solid sol Foam

D.

Froth

IV.

Aerosol

Choose the correct answer from the options given below: (A) A-IV, B-I, C-II, D-III (B) A-II, B-I, C-IV, D-III (C) A-II, B-IV, C-I, D-III (D) A-III, B-I, C-IV, D-II Ans. Option ( B ) is correct Explanation: The examples given belong to colloidal solutions. Gemstone is an example of solid sol where the dispersed phase as well as dispersion medium is solid. Milk is an example of emulsion where the dispersed phase as well as dispersion medium is liquid. Cloud is an example of aerosol where the dispersed phase is liquid and dispersion medium is gas. Froth is an example of foam where the

dispersed phase is gas and dispersion medium is liquid. 11. Consider the case when a highly diluted solution of KI is added to AgNO3 solution. Arrange the following in the increasing order of ease of coagulation of the resulting sol. (A) BaSO4 (B) NaCl (C) Na3PO4 (D) K4[Fe(CN)6] Choose the correct answers from the options given below: (A) A < C < B < D (B) D < C < A < B (C) A < B < C < D (D) B < A < C < D Ans. Option (D) is correct Explanation: The colloidal solution will be positively charged due to an excess of Ag+ ions. These positively charged metal ions will attract the anion of the electrolyte. The higher the negative charge on the electrolyte, the higher will be the coagulation. BaSO4 contains -2 charge on SO4 ion. NaCl contains -1 charge on Cl ion. Na3PO4 contains -3 charge on PO4 ion and K4 [Fe (CN)6] contains -4 charge on [Fe(CN)6] So, the order is: NaCl (B) < BaSO4 (A) < Na3PO4 (C) < K4 [Fe (CN)6] (D) 12. Which of the following conditions can be used to change the physical adsorption of gas to chemical adsorption? (A) Decrease in temperature (B) Increase in temperature (C) Using a catalyst (D) Increasing surface area of the adsorbent Ans. Option (B) is correct Explanation: Physisorption may transform to chemisorption at high temperatures because Physisorption is exothermic in nature. Therefore, in accordance with Le-Chatelier's principle, it decreases with an increase in temperature. With an increase in temperature the weak van Der Waals forces breaks and chemical bond formation starts and leads to chemisorption. 13. Which of the following graph(s) are for first order reactions? In [R] 0

(A)

In [R]

 [ A]o  2.303 k= log10   t  [ A]o − x 

t

(B)

Rate

Explanation:

[R]

OSWAAL CUET (UG) Sample Question Papers, Chemistry

[R]

[R]

log [R]0 / [R]

In [R] 0

(D)

Copper

Copper pyrites Malachite Cuprite Copper glance

CuFeS2 CuCO3.Cu(OH)2 Cu2O Cu2S

Zinc

Zinc ZnS blende or Sphalerite ZnCO3 Calamine ZnO Zincite

t

t

(A) A and D only (C) A, D and E only Ans. (NA)

(B) B and C only (D) C and E only

Explanation: For first order reaction:

log

FeS2

[ R ]o k = t [ R ] 2.303

Graph relating change in concentration w.r.t time t [A] 0

slope = - k

Malachite is an ore of copper, Bauxite is an ore of Aluminium, Calamine is an ore of Zinc, Haematite and Siderite are ores of Iron (Fe). Statement E stating siderite is an ore of Zinc is incorrect. 15. Arrange the following molecules in the increasing order of number of P – OH bonds present in it. (A) H4P2O6 (B) H3PO2 ( C ) H3PO4 (D) H3PO3 Choose the correct answer from the options given below: (A) B < C < D < A (B) B < D < C < A (C) D < B < A < C (D) D < C < B < A Ans. Option (B) is correct OH O | ||

Graph relating log [R0 /R] vs t

(A) H4P2O6 – no. of P-OH bonds = 4

Time

Option A , C and D all are correct. 14. Consider the statements for the metallurgical processes and select the correct statements: (A) Malachite is an ore of copper. (B) Bauxite is an ore of aluminium. (C) Calamine is an ore of Zinc. (D) Haematite is an ore of iron. (E) Siderite is an ore of Zinc. Choose the correct statement from the options given below (A) A, B, E and D only (B) A and B only (C) A, B, C and D only (D) A only Ans. Option (C) is correct Explanation: Metal

Ores Bauxite Kaolinite (a form of clay)

Iron

Haematite Magnetite Siderite

Composition AlOx(OH)3.2x [Where 0 B > A (C) A > B > C > D (D) A > B > D > C

Solved Paper-2022 27 Ans. Option (B) is correct Explanation: In a group, acidic character increases down the group as the M−H bond strength decreases. Acidic character increases from H2O to H2Te. H2Te has the highest acidic character, then comes H2Se , H2S and H2O with least acidic character.

3d

17. Which of the following molecules have linear shape? (A) XeF2 (B) XeF4 (C) XeF6 (D) XeO3 Ans. Option (A) is correct. F F

F F — Xe — F

Xe F

F

F Xe

F

F

F F

XeF2

XeF4

3d

4s

[Ar]

Table: Dissociation enthalpy for hydrides of Group 16 elements.

Explanation:

Explanation: configuration of Cr is [Ar] 3d⁵ 4s¹ [Ar]

configuration of Cr+3 is 3d3

dissH(H-E)/kJ mol-1 463 347 276 238

Hydride H2 O H2 S H2Se H2Te

19. The spin-only magnetic moment of Cr3+ion in BM is: (A) 1.73 (B) 3.87 (C) 4.89 (D) 3.57 Ans. Option (B) is correct

XeF6

||

Xe | O || O O (Pyramidal) (XeO3 )

Structure having linear geometry is XeF2. 18. Which of the following statements about dblock elements are NOT correct? (A) They show variable oxidation states. (B) They exhibit paramagnetic and diamagnetic properties. (C) All of their ions are coloured. (D) They exhibit catalytic property. Ans. Option (C) is correct Explanation: Transition elements have vacant d orbitals. So, due to the presence of vacant dorbital, they produce coloured compounds. All ions of d block elements are coloured except d block elements with d0 or d10 configuration which are colourless. Transition elements show variable oxidation states. Transition elements show catalytic properties.

Magnetic moment Where, electrons Here, So,

4s

µ = √𝑛(𝑛 + 2) n = no. of

unpaired

n= 3 µ = √3(3 + 2) = 3.87 BM

20. Select the correct statements for the d-block and f-block elements: A. The maximum oxidation state shown by manganese is +6. B. Sc3+ (Scandium) is colourless. C. Brass is an alloy of Copper and Zinc. D. Lanthanide series included a total of 15 elements. E. V2O5 (Vanadium Pentaoxide) is used in the manufacturing of sulphuric acid (by contact process). Choose the correct answer from the options given below: (A) A and B only (B) A, B and E only (C) B, C and E only (D) B, C and D only Ans. Option (C) is correct Explanation: (A) Maximum oxidation state shown by Mn is 6, it is incorrect. Maximum oxidation state shown by Mn is +7. (B) Sc+3 is colourless. This is a correct statement as it has d0 configuration and transition elements with d0 or d10 are colourless. (C) Brass is an alloy of Cu and Zn. This is a correct statement as brass contains 60% Cu and 40% Zn. (D) Lanthanide series consists of 15 elements. This is incorrect as it contains 14 elements from La to Lu. (E) Vanadium pentaoxide is used as catalyst in manufacture of sulphuric acid. This is a correct statement. This catalyst is used to convert sulphur dioxide to sulphur trioxide, which is further used to manufacture sulphuric acid. 21. The IUPAC name of [Pt(NH3)2Cl2] is: (A) Diammine dichlorido platinum (II) (B) Diammine dichlorido platinum (IV)

OSWAAL CUET (UG) Sample Question Papers, Chemistry (C) Diammine dichlorido platinum (0) (D) Diimmine dichlorido platinum (IV) Ans. Option (A) is correct Explanation: [Pt(NH3)2Cl2] IUPAC - Diamminedichlorido platinum (II) x + 0 + 2(-1) = 0 Oxidation no. of Pt is 2. 22. Match List I with List II List I: (Property) A.

B.

Transition metal can act as a catalyst.

Zr and Hf have similar atomic radii.

C. Transition metals form complex compounds.

D. Transition metal ions are coloured.

List II: (Reason) I.

II.

Due to their high ionic charges, small size and availability of dorbitals. Unpaired electrons in d-orbitals of metal ions.

III. Ability to adopt multiple oxidation states and to form complexes.

IV. As a consequence contraction of Lanthanoid contraction. Choose the correct answer from the option given below: (A) A-I, B-IV, C-II, D-III (B) A-II, B-III, C-IV, D-I (C) A-III, B-IV-, C-I, D-II (D) A-IV, B-I, C-III, D-III Ans. Option (C) is correct Explanation: (A) Transition element act as catalysts because they have a tendency to adopt multiple oxidation states and can form complexes. (B) Zr and Hf have the same atomic radius because of Lanthanoid contraction, the steady decrease in the size of the atoms and ions of the rare-earth elements with increasing atomic number from lanthanum (atomic number 57) through lutetium (atomic number 71). As a result of lanthanoid contraction, the size of Hf decreases and is similar to Zr (an element of the previous period). (C) Transition metal form complex compounds because they have empty vacant d orbitals that can accept pairs of electrons and forms coloured complexes due to the presence of vacant d -orbitals.

(D) Transition metal ions are coloured due to the presence of unpaired electrons in the d- orbital. These electrons absorb radiation from the visible region of light to promote an electron from a lower energy d-orbital to a higher energy dorbital. 23. The donor atoms in ethylenediamine tetraacetate ions are: (A) Two N and two O atoms (B) Two N and four O atoms (C) Four N and two O atoms (D) Three N and Three O atoms Ans. Option (B) is correct Explanation: –

OOCH2 C

–

OOCH2 C

–

N–CH2–CH2 –N

CH2COO

–

CH2COO

Ethylene diamienetetraacetate This structure shows nitrogen has 2 lone pairs of electrons and there are four lone pairs of electrons with each pair on one O atom. Hence, the donor atoms are two N and four O atoms. 24. Indicate the complex ion which does not show geometrical isomerism: (A) [Cr(H2O)4Cl2]+ (B) [Pt(NH3)2Cl2] (C) [Pt(NH3)6]3+ (D) [Co(CN)4(NC)2]3– Ans. Option (C) is correct

Explanation: [Pt(NH3)6] 3+ this type of complex does not show geometric isomerism because no different ligands are attached to show cis and trans form. Geometric isomerism in remaining is shown below : [Cr(H2O)4Cl2]+, [Pt(NH3)2Cl2], [Co(CN)4(NC)2]3– 25. Which of the following is the electronic configuration of the central metal atom/ion of [Co(H2O)6]2+ complex ion? (A) t2g6 eg1 (B) t2g4 eg3 5 2 (C) t2g eg (D) t2g6 eg0 Ans. Option (C) is correct Explanation: Co

3d 4s

In Co [(H2O)6]+2, oxidation state of Co is +2, so it loses two electrons from the 4s orbital and the electron in d-orbital is split into t2g and eg orbitals. eg [Co(H 2O) 6]

2+

O

t2g

Solved Paper-2022 29 26. Identify the compound Y in the following reaction:

N2Cl

Cu2Cl2

NaNO 2 + HCl 273 – 278 Cl

(A)

–

+

NH2

Y + N2

(B) Cl

Cl

(C)

List I : Names of drug Cl

Cl

Ans. Option (A) is correct Explanation: Given reaction is Sandmayer reaction for the preparation of Haloarenes. + –

N2X HX 1NaNO 2

X CuX 60 – 100°C

0°C

X = CN, Br, Cl, SO3H So, the Y is Chlorobenzene. 27. Consider the following bromides: Me

Me Me

Br

Me Br

Explanation: Jones reagent is a solution of chromium trioxide in aqueous sulphuric acid. This Jones reagent is used to oxidise primary alcohols to carboxylic acids. 30. Match List I with List II

(D)

NH2

29. Jones Reagent is: (A) KMnO4 | H2SO4 (B) K2Cr2O7 | H2SO4 (C) CrO3 | H2SO4 (D) KMnO4 | KOH Ans. Option (C) is correct

Br

The correct order of reactivity towards SN1 reaction is: (A) I > II > III (B) II > III > I (C) II > I > III (D) III > II > I Ans. Option (B ) is correct Explanation: More stable the carbocation, higher will be the reactivity of SN1 reaction. On removal of Br atom, carbocation will be: CH3-CH2-CH2-CH2+ - primary carbocation CH2= CH-CH-CH3 - conjugated carbocation CH3-CH2-CH+-CH3- 2° carbocation Conjugate carbocation is the most stable, then comes 3°, then 2° and then 1° in the decreasing order. 28. Aryl halides cannot be prepared by the reaction of aryl alcohols with PCl3, PCl5 or SOCl2 because: (A) Phenols are highly stable compounds. (B) Carbon-oxygen bond in phenols has a partial double bond character. (C) Carbon-oxygen bond is highly polar. (D) Phenyl cation is stabilised by resonance. Ans. Option (B) is correct Explanation: Aryl halides cannot be prepared by replacing the hydroxyl group of phenols because the carbon-oxygen bond in phenols has a partial double bond character and is stronger than a single bond. Hence, difficult to break.

List II : Class of drugs

A. A. Ranitidine

I. I. Tranquilizer

B. B.Valium

II. II. Narcotic Analgesic

C. C.Codeine

III. III. Antibiotic

D. D.Chloramphenicol

IV IV. Antacid

Choose the correct answer from the options given below: (A) A-IV, B-I, C-II, D-III (B) A-IV, B-II, C-II, D-I (C) A-III, B-IV, C-II, D-I (D) A-I, B-II, C-III, D-IV Ans. Option (A) is correct Explanation: (A) Ranitidine - Antacid (B) Valium – Tranquilizer (C) Codeine - Narcotic Analgesic (D) Chloramphenicol – Antibiotic 31. Match List I with List II List-I Given pair of organic

List-II Tests

compounds can be distinguished by (1)f 1. Ethanal/ Propanal

I.

Sodium Hydrogen carbonate test

(2) 2. Ethanol/Ethanoic acidII. II. Fehling’s Test (3) 3. Butanal/Butan-2-one III III. Tollen’s Test (4) 4. Benzaldehyde/Ethanal IV. Iodoform Test Choose the correct answer from the options given below: (A) 1 – IV, 2 – I, 3 – II, 4 – III (B) 1 – III, 2 – II, 3 – IV, 4 – I (C) 1 – IV, 2 – I, 3 – III, 4 – II (D) 1 – I, 2 – II, 3 – III, 4 – IV Ans. Option (C) is correct

OSWAAL CUET (UG) Sample Question Papers, Chemistry Explanation: 1. Ethanal/Propanal – Iodoform test –Used to identify the presence of aldehyde or ketone in an unknown compound. 2. Ethanol/Ethanoic–acid - Sodium Hydrogen carbonate test – Used for the test of the carboxylic acid group. 3. Butanal/Butan-2-one Tollen’s Test- also known as the silver mirror test used to distinguish between Aldehyde and Ketone. 4. Benzaldehyde/Ethanal Fehling’s Test – For the detection of reducing and nonreducing sugar. 32. Identify A and B in the following reaction: NH3 2CH 3Cl C6H5CH2Cl ⎯⎯⎯→ A ⎯⎯⎯⎯ ⎯ → B (A) A = C6H5CH2NH2, B = C6H5CH2NH–CH3 (B) A = C6H5CH3, B = C6H5CH©3 (C) A = C6H5CH2NH2, B = C6H5CH2 —N —CH3 | CH3

(D) A = C6H5CH3, B = C6H5CH2CH2CH3 Ans. Option (C) is correct Explanation: When Benzyl chloride react with ammonia to give Benzylamine (A) which further reacts with 2 moles of chloromethane to give N, N-dimethyl Benzylamine. (B) NH

3 C6H5—CH2—Cl ⎯⎯⎯→ C6H5—CH2NH2

Benzyl chloride

Benzylamine

2CH 3Cl

⎯⎯⎯⎯⎯ → C6H5—CH2—N—CH3

| CH3 N, N-Dimethyl benzylamine

33. Out of the following compounds, which is the most basic in aqueous solution? (A) CH3—NH2 (B) (CH3)2NH (C) (CH3)3N (D) C6H5NH2 Ans. Option (B ) is correct Explanation: The correct order of basicity in aqueous solution is: NH3 < tertiary amine < primary amine < secondary amine. So, (CH3)2NH is the most basic. 34. Arrange the following in increasing order of their boiling points: A. (C2H5)2NH

B. n-C4H9NH2

C. C2H5N(CH3)2 D. n-C4H9OH Choose the correct answer from the options given below: (A) B < C < A < D (B) C < B < A < D (C) C < A < B < D (D) C < D < B
(ii) > (iii) (3) (iii) < (i) < (ii)

(1)

(4)

(2)

14. What type of isomerism is shown by the pair [Cr(H2O)6]Cl3 and [Cr(H2O)5Cl] Cl2 . H2O ? (1) Ionisation isomerism (2) Coordination isomerism (3) Solvate isomerism (4) Linkage isomerism 15. In the presence of a catalyst, heat evolved or absorbed during reaction (1) increases. (2) decreases. (3) remains unchanged. (4) may increase or decrease.

16. Identify the compound Y in the following reaction. + – N2Cl

NH2 NaNO2+HCI

Cu2Cl2

273–278 K



Cl

(a)



(2) Cl

Cl

(3)

Cl

(2) (ii) < (i) < (iii) (4) (iii) > (i) > (ii)

CN

OH

(iii)

with the molecule (a) given below?

OH —OH CN —

(1) No product formed (2)

(3)

(ii)

21. Which of the following structures is enantiomeric

+ HCN



(i)

(4)

Y+N2



(4)

(3)





72. Amongst the following, the most stable complex is (1) [Fe(H2O)6]3+ (2) [Fe(NH3)6]3+ (3) [Fe(C2O4)3]3− (4) [FeCl6]3−

23. KH value for Ar(g), CO2(g), HCHO(g) and CH4(g)

are 4.039, 1.67, 1.83 × 10–5, and 0.143, respectively. Arrange these gases in the order of their increasing solubility. (1) HCHO < CH4 < CO2 < Ar (2) HCHO < CO2 < CH4 < Ar (3) Ar < CO2 < CH4 < HCHO (4) Ar < CH4 < CO2 < HCHO

24. CH3CONH2 on reaction with NaOH and Br2 in

Cl

17. The oxidation state of Fe in K3 [Fe(C2O4)3] is (1) 0 (2) 2 (3) 3 (4) 4 18. Bond angle in ethers is slightly more than (1) Square planar angle (2) Trigonal bipyramidal angle (3) Tetrahedral angle (4) None of the above

alcoholic medium gives: (1) CH3CH2NH2 (2) CH3CH2Br (3) CH3NH2 (4) CH3COONa

25. Which of the following pairs represents anomers? CHO

(1) H HO H H

CHO HO H OH OH

CH2OH

HO HO H H

H H OH OH CH2OH

Sample Question Papers CHO

(2)



32. Which of the following statement is correct?

CHO OH H OH OH

H HO H H

CH2OH

HO H HO HO

3

H OH H H CH2OH

(3)

(1) The rate of a reaction decreases with passage of time as the concentration of reactants decreases. (2) The rate of a reaction is same at any time during the reaction. (3) The rate of a reaction is independent of temperature change. (4) The rate of a reaction decreases with increase in concentration of reactant(s).

33. Which of the following should be most volatile ? (i) CH3CH2NH2

(ii) (CH3)3N

CH3CH2

(iii)

NH (iv) CH3CH2CH3 H 3C

(4)

(1) (ii) (3) (i)

(2) (iv) (4) (iii)

34. A colloidal system having a solid substance as

26. Which of the following is not a favourable condition for physical adsorption?

a dispersed phase and a liquid as a dispersion . medium is classified as (1) solid sol (2) gel (3) emulsion (3) sol

35. Which of the following statements is not correct ?

(1) High pressure (2) Negative dH (3) Higher critical temperature of adsorbate (4) High temperature

27. Which of the following will give a white precipitate

(1) Copper liberates hydrogen from acids. (2) In its higher oxidation states, manganese forms stable compounds with oxygen and fluorine. (3) Mn3+ and Co3+ are oxidising agents in aqueous solution.

upon reacting with AgNO3? (1) K2 [Pt(en)2Cl2] (2) [Co(NH3)3Cl3] (3) [Cr(H2O)6]Cl3 (4) [Fe(H2O)3Cl3]

(4) Ti2+ and Cr2+ are reducing agents in aqueous solution.

28. Which one of the following is not a condensation

butanone on oxidation with alkaline KMnO4 solution ? (1) Butan-1-ol (2) Butan-2-ol (3) Both of these (4) None of these

polymer? (1) Melamine (3) Dacron

(2) Glyptal (4) Neoprene

29. Assertion (A): Rate of reaction doubles when

concentration of reactant is doubled if it is a first order reaction. Reason (R): Rate constant also doubles. (1) Both A and R are true and R is the correct explanation of A (2) Both A and R are true but R is NOT the correct explanation of A (3) A is true but R is false (4) A is false and R is true

30. The oxidation of Ni in [Ni(CO)4] is (1) 0  (3) 3  

(2) 2   (4) 4

31. Which of the following reagents would not be a

good choice for reducing an aryl nitro compound to an amine ?

36. Which of the following compounds will give

37. Value of Henry’s constant KH ________________. (1) increases with increase in temperature. (2) decreases with increase in temperature (3) remains constant (4) first increases then decreases.

38. Which one of the following elements constitutes a major impurity in pig iron? (1) Silicon (2) Oxygen (3) Sulphur (4) Graphite

39. Which of the following help in food preservation by retarding the action of oxygen on food. (1) Food colours (2) Antioxidants (3) Preservatives (4) Fat emulsifiers

40. Debye-Huckel Onsager equation for strong electrolytes:

∧ = ∧o − A C

(1) H2(excess)/Pt (2) LiAlH4 in ether



(3) Fe and HCl (4) Sn and HCl

Which of the following equality holds?

OSWAAL CUET (UG) Sample Question Papers, CHEMISTRY

4

→∞ (1) ∧ = ∧ o as C  → A (2) ∧ = ∧ o as C 



(3) ∧ = ∧ o as C  → 0 (4) ∧ = ∧ o as C  →1



41. For Lysine, H 2 N − (CH 2 )4 − CH− COOH , which of the following is incorrect?

 NH 2

(1) a-Amino acid. (2) Basic amino acid. (3) Amino acid synthesised in body. (4) b-Amino acid.

42. Which one of the following sets forms the biodegradable polymer? (1) CH2==CH—CN and CH2==CN—CH==CH2 (2) H2N—CH2—COOH and H2N—(CH2)5—COOH (3) HO—CH2—CH2—OH and (4)

HOOC—

—COOH

—CH=CH2 and

CH2==CH—CH==CH2

43. Which of the following can possibly be used as an

analgesic without causing addiction and mood modification? (1) Diazepam (2) Tetrahydrocatenol (3) Morphine (4) N-Acetyl-para-aminophenol

44. Rate law cannot be determined from balanced chemical equation if (1) reverse reaction is involved. (2) it is an elementary reaction. (3) it is a sequence of elementary reactions. (4) any of the reactants is in excess.

45. Glucose and fructose are

(1)  isomers of each other (2) homologous of each other

(3) anomers of each other (4) enantiomers of each other

46. Cassiterite is an ore of: (1) Mn (3) Sb

(2) Ni (4) Sn

Case Based Read the passage given below and answer the

following questions: In spite of the predictions of stable noble gas compounds since at least 1902, unsuccessful attempts at their synthesis gave rise to the widely held opinion that noble gases are not only noble but also inert. It was not until 1962 that this dogma was shattered when Bartlett in Canada published the first stable noble gas compound XePtF6. This discovery triggered a worldwide frenzy in this area, and within a short time span many new xenon, radon, and krypton compounds were prepared and characterized. The recent discoveries show the ability of xenon to act as a ligand . The discovery by Seppelt’s group that more than one xenon atom can attach itself to a metal center which in the case of gold leads to surprisingly stable Au- Xe bonds. The bonding in [AuXe4]2+ involves 4 Xe ligands attached by relatively strong bonds to a single Au(II) center in a square planar arrangement with a Xe-Au bond length of about 274 pm This discovery provides not only the first example of multiple xenon ligands but also represents the first strong metal - xenon bond.

47. In the complex ion [AuXe4]2+, Xe acts as: (1) central atom (3) chelating agent

(2) ligand (4) electrophile

48. Hybridisation shown by Au in [AuXe4]2+ is: (1) sp3 (3) sp3d2

(2) sp3d (4) sp2

49. Compounds of noble gases except _______ are known (1) Krypton (3) Helium

(2) Radon (4) Xenon

50. Xe is a ___________ ligand. (1) ambidentate (3) unidentate

(2) bidentate (4) hexadentate

qqq

SAMPLE

Question Paper Maximum Marks : 200

2

Time : 45 Minutes

General Instructions: (i) This paper consists of 50 MCQs, attempt any 40 out of 50 . (ii) Correct answer or the most appropriate answer: Five marks (+5) . (iii) Any incorrect option marked will be given minus one mark (– 1) . (iv) Unanswered/Marked for Review will be given no mark (0) . (v) If more than one option is found to be correct then Five marks (+5) will be awarded to only those who have marked any of the correct options . (vi) If all options are found to be correct then Five marks (+5) will be awarded to all those who have attempted the question . (vii) If none of the options is found correct or a Question is found to be wrong or a Question is dropped then all candidates who have appeared will be given five marks (+5). (viii) Calculator / any electronic gadgets are not permitted . [B]

solute in (1) 1000 g of the solvent

(4)

(2) one litre of the solvent (3) one litre of the solution (4) 22.4 litre of the solution

2. Consider the reaction A  B. The concentration

(1)

Concentration →

of both the reactants and the products varies exponentially with time. Which of the following figures correctly describes the change in concentration of reactants and products with time? [B]

[A] Time →

(2)

Concentration →

[B]

[A] Time →

(3)

Concentration →

[B]

[A] Time →



[A]

Concentration →

1. A molar solution is one that contains one mole of a

[A] [B] Time →



3. In the extraction of copper from its sulphide ore, the metal is finally obtained by the reduction of cuprous oxide with (1) copper (I) sulphide (Cu2S) (2) sulphur dioxide (SO2) (3) iron sulphide (FeS) (4) carbon monoxide (CO)

4. Which of the following is not true about the ionic

solids? (1) Bigger ions form the close packed structure. (2) Smaller ions occupy either the tetrahedral or the octahedral voids depending upon their size. (3) Occupation of all the voids is not necessary. (4) The fraction of octahedral or tetrahedral voids occupied depends upon the radii of the ions occupying the voids.

5. Which of the following statements are true?

(1) Only types of interactions between particles of noble gases are due to weak dispersion forces. (2) Hydrolysis of XeF6 is a redox reaction. (3) Xenon fluorides are not reactive. (4) None of the above.

6. How many ions are produced from the complex [Co(NH3)5Cl]Cl2 in solution? (1) 4   (2) 2   (3) 3   (4) 5

OSWAAL CUET (UG) Sample Question Papers, CHEMISTRY

6

7. Which of the following is an example of absorption? (1) Water on silica gel (2) Water on calcium chloride (3) Hydrogen on highly divided nickel (4) Oxygen on metal surface

8. Three cyclic structures of monosaccharides are given below which of these are anomers ? (i)

(ii)

(iii)

(1) (i) and (ii) (2) (ii) and (iii) (3) (i) and (iii) (4) (iii) is anomer of (i) and (ii)

9. When KMnO4 solution is added to oxalic acid solution, the decolourisation is slow in the beginning but becomes instantaneous after some time because (1) CO2 is formed as the product. (2) Reaction is exothermic. (3) MnO4– catalyses the reaction. (4) Mn2+ acts as auto-catalyst.

10. The correct order of the packing efficiency in different types of unit cells is ________. (1) fcc < bcc < simple cubic (2) fcc > bcc > simple cubic (3) fcc < bcc > simple cubic (4) bcc < fcc = simple cubic

11. The conversion of an alkyl halide into an alcohol by aqueous NaOH is classified as (1) a dehydrohalogenation reaction (2) a substitution reaction (3) an addition reaction (4) a dehydration reaction

12. The method of zone refining of metals is based on the principle of: (1) greater noble character of the solid metal than that of the impurity. (2) greater solubility of the impurity in the molten state than in the solid. (3) greater mobility of the pure metal than that of impurity. (4) higher melting point of the impurity than that of the pure metal.

13. T  he formula of the complex triamminetri(nitrito-O) Cobalt (III) is (1) [Co(ONO)3(NH3)3] (2) [Co(NO2)3(NH3)3] (3) [Co(ONO2)3(NH3)3] (4) [Co(NO2)(NH3)3]

14. In which of the following structures coordination number for cations and anions in the packed structure will be same? (1) Cl– ion form fcc lattice and Na+ ions occupy all octahedral voids of the unit cell. (2) Ca2+ ions form fcc lattice and F– ions occupy all the eight tetrahedral voids of the unit cell. (3) O2– ions form fcc lattice and Na+ ions occupy all the eight tetrahedral voids of the unit cell. (4) S2– ions form fcc lattice and Zn2+ ions go into alternate tetrahedral voids of the unit cell. 15. Which of the following oxidation state is common for all lanthanoids? (1) +2

(2) +3

(3) +4

(4) +5

16. How many alcohols with molecular formula C4H10O are chiral in nature?

(1) 1

(2) 2

(3) 3

(4) 4

17. Solid A is very hard electrical insulator in solid as well as in molten state and melts at an extremely high temperature. What type of solid is it? (1) Ionic solid

(2) Molecular solid

(3) Covalent solid

(4) Metallic solid

18. The carboxylic acid that does not undergo HVZ reaction is (1) CH3COOH (2) (CH3)2COOH (3) CH3CH2CH2CH2COOH (4) (CH3)3CCOOH

19. A brown ring is formed in the ring test for NO3– ion. It is due to the formation of:

(1) [Fe(H2O)5(NO)]2+

(2) FeSO4.NO2.

2+

(3) [Fe(H2O)4(NO)2] (4) FeSO4.HNO3.

20. Write IUPAC name of the following compound: CH3 CH3 — CH2 — CH2 — CH2 — N CH3 (1) N,N-Dimethylpropanamine (2) 1,1-Dimethylbutanamine (3) N-Methylpentan-1-amine (4) N,N-Dimethylbutan-1-amine

21. Which of the following alkyl halides will undergo SN1 reaction most readily? (1) (CH3)3C—F (2) (CH3)3C—Cl (3) (CH3)3C—Br (4) (CH3)3C—I

22. The unit of rate constant depends upon the (1) molecularity of the reaction (2) activation energy of the reaction

Sample Question Papers (3) order of the reaction (4) temperature of the reaction

23. Which of the following statements is false? (1) Artificial silk is derived from cellulose. (2) Nylon-6, 6 is an example of elastomer. (3) The repeat unit in natural rubber is isoprene. (4) Both starch and cellulose are polymer of glucose.

24. In physisorption, adsorbent does not show specificity for any particular gas because (1) involved van der Waal’s forces are universal. (2) gases involved behave like ideal gases. (3) enthalpy of adsorption is low. (4) it is a reversible process.

25. Which one of the following is employed as a tranquilizer drug? (1) Promthazine (3) Naproxen

(2) Valium (4) Mifepristone

26. Which one of the following statements is not true? (1) Buna-S is a copolymer of butadiene and styrene. (2) Natural rubber is a 1, 4-polymer of isoprene. (3) In vulcanization, the formation of sulphur bridges between different chains make rubber harder and stronger. (4) Natural rubber has the trans-configuration at every double bond.

27. Assertion (A): HNO3 makes iron passive. Reason (R): HNO3 forms a protective layer of ferric nitrate on the surface of iron. (1) Both A and R are true and R is the correct explanation of A (2) Both A and R are true but R is NOT the correct explanation of A (3) A is true but R is false (4) A is false and R is True

7

31. Hinsberg’s reagent which is used to test amines is (1) (2) (3) (4)

Benzene sulphonamide Benzene diazonium chloride Benzene sulphonyl chloride Acetanilide

32. Acrilan is a hard and a high melting material. Which one of the following represents its structure? —CH2—CH— (1) CN n CH3      CH  C    (2) 2      COOCH3  n  

    CH  CH    (3)  2     COOC2 H5  n  (4) —CH—CH— Cl n 33. At equilibrium, the rate of dissolution of a solid solute in a volatile liquid solvent is __________. (1) less than the rate of crystallisation (2) greater than the rate of crystallisation (3) equal to the rate of crystallisation (4) zero

34. In which of the following molecules, carbon atom marked with asterisk (*) is asymmetric? (i) (ii)

(iii)



(iv)

28. Gammexane is: (1) bromobenzene (2) benzyl chloride (3) chlorobenzene (4) benzene hexachloride 29. IUPAC name of the compound. CH3  CH  OCH3 | CH3 is ___________. (1) 1-methoxy-1-methylethane (2) 2-methoxy-2-methylethane (3) 2-methoxypropane (4) isopropylmethyl ether

30. Purification of aluminium by electrolytic refining is known as: (1) Hall’s process (3) Hoope’s process

(2) Baeyer’s process (4) Serpeck’s process

(1) (i), (ii), (iii), (iv) (3) (ii), (iii), (iv)

(2) (i), (ii), (iii) (4) (i), (iii), (iv)

35. Williamson synthesis is used to obtain (1) Primary alcohol (2) Ether (3) Aldehyde (4) Ketone

36. Compounds ‘A’ and ‘B’ react according to the following chemical equation:

A(g) + 2B(g) → 2C(g) Concentration of either ‘A’ or ‘B’ were changed keeping the concentration of one of the reactants constant and rates were measured as function of initial concentration. Following results were obtained. Choose the correct option for this reaction.

OSWAAL CUET (UG) Sample Question Papers, CHEMISTRY

8

1.

0.30

0.30

0.10

2.

0.30

0.60

0.40

(3) Iron powder along with Al2O3 and K2O is used as a catalyst in the preparation of NH3 by Haber ’s process. (4) Change in enthalpy is positive for the preparation of SO3 by catalytic oxidation of SO2.

3.

0.60

0.30

0.20

45. For a dilute solution, Raoult’s law states that

Initial Initial Initial Expericoncentration concentration concentration of ment of [A]/mol L−1 of [B]/mol L−1 [C]/mol L−1

(1) Rate = k [A]2[B] (3) Rate = k [A][B]

(2) Rate = k [A][B]2 (4) Rate = k [A]2[B]0

37. Which of the following compounds is most reactive towards nucleophilic addition reactions ?

O  (1) CH 3 C  H O  (3) C  H

O  (2) CH 3 C CH 3 O  (4) C CH 3

38. Which set of ions exhibit specific colours? (Atomic

number of Sc = 21, Ti = 22, V=23, Mn = 25, Fe = 26, Ni = 28 Cu = 29 and Zn =30) (1) Sc3+, Ti4+, Mn3+ (2) Sc3+, Zn2+, Ni2+ (3) V 3+, V2+, Fe3+ (4) Ti3+, Ti4+, Ni2+ 39. The best reagent for converting 2–phenylpropanamide into 2-phenylpropanamine is _____. (1) excess H2 (2) Br2 in aqueous NaOH (3) Iodine in the presence of red phosphorus (4) LiAlH4 in ether 40. Method by which lyophobic sol can be protected: (1) by addition of oppositely charged sol. (2) by addition of an electrolyte. (3) by addition of lyophilic sol. (4) by boiling. 41. In Clemmensen reduction carbonyl compound is treated with _________. (1) zinc amalgam + HCl (2) sodium amalgam + HCl (3) zinc amalgam + nitric acid (4) sodium amalgam + HNO3 42. Curdling of milk is an example of: (1) breaking of peptide linkage (2) hydrolysis of lactose (3) breaking of protein into amino acids (4) denaturation of protein 43. The CFSE for octahedral [CoCl6]4– is 18,000 cm–1. The CFSE for tetrahedral [CoCl4]2– will be (1) 18,000 cm–1. (2) 16,000 cm–1. (3) 8,000 cm–1. (4) 20,000 cm–1. 44. Which of the following statements is correct? (1) S–S bond is present in H2S2O6. (2) In peroxosulphuric acid (H2SO5) sulphur is in +8 oxidation state.

(1) The lowering of vapour pressure is equal to the mole fraction of solute. (2) The relative lowering of vapour pressure is equal to the mole fraction of solute. (3) The relative lowering of vapour pressure is proportional to the amount of solute in solution. (4) The vapour pressure of the solution is equal to the mole fraction of the solute.

46. Lanthanoid contraction is caused due to (1) (2) (3) (4)

Atomic number Size of 4f orbitals Effective nuclear charge Poor shielding effect of 4f electrons

Case Based Read the passage given below and answer the following questions:

The cell constant is usually determined by measuring

the resistance of the cell containing a solution whose conductivity is already known. For this purpose, we generally use KCl solutions whose conductivity is known accurately at various concentrations and at different temperatures. Consider the resistance of a conductivity cell filled with 0.1 M KCl solution is 200 Ohm. If the resistance of the same cell when filled with 0.02 M KCl solution is 420 Ohm.

(Conductivity of 0.1 M KCl solution is 1.29 S m–1.) 47. What is the conductivity of 0.02 M KCl solution ? (1) 0.452 S m–1 (3) 0.614 S m–1

(2) 0.215 S m–1 (4) 0.433 S m–1

48. What will happen to the conductivity of the cell with the dilution ? (1) First decreases then increases (2) Increases (3) First increases then decreases (4) Decreases

49. The cell constant of a conductivity cell ________. (1) changes with change of electrolyte. (2) changes with change of concentration of electrolyte. (3) changes with temperature of electrolyte. (4) remains constant for a cell.

50. SI unit for conductivity of a solution is ________. (1) S m–1 (3) mol cm–3

(2) S m2 mol–1 (4) S cm2 mol–1

qqq

SAMPLE

Question Paper Maximum Marks : 200

3

Time : 45 Minutes

General Instructions: (i) This paper consists of 50 MCQs, attempt any 40 out of 50 . (ii) Correct answer or the most appropriate answer: Five marks (+5) . (iii) Any incorrect option marked will be given minus one mark (– 1) . (iv) Unanswered/Marked for Review will be given no mark (0) . (v) If more than one option is found to be correct then Five marks (+5) will be awarded to only those who have marked any of the correct options . (vi) If all options are found to be correct then Five marks (+5) will be awarded to all those who have attempted the question . (vii) If none of the options is found correct or a Question is found to be wrong or a Question is dropped then all candidates who have appeared will be given five marks (+5). (viii) Calculator / any electronic gadgets are not permitted .

1. Proteins are found to have two different types of secondary structures, viz. a-helix and β-pleated sheet structure. a-helix structure of protein is stabilised by: (1) Peptide bonds (2) van der Waals forces (2) Hydrogen bonds (4) Dipole-dipole interactions

2. Which of the following complexes formed by Cu2+

(2) Terylene: —OC—

—COOCH2—CH2—O—

(3) Nylon 6, 6: [ NH(CH 2 )6 NHCO(CH 2 )4 —CO] n

n

(4) Teflon: [CF2 —CF2 ] n

6. Arrange the following in increasing order of basic strength:

ions is most stable? (1) Cu2+ + 4NH3 ® [Cu(NH3)4]2+, log K = 11.6 (2) Cu2+ + 4CN– ® [Cu(CN)4]2–, log K = 27.3 (3) Cu2+ + 2en ® [Cu(en)2]2+, log K = 15.4 (4) Cu2+ + 4H2O ® [Cu(H2O)4]2+, log K = 8.9

Aniline, p-nitroaniline and p-toluidine.

3. Which of the following acids does not form

7. The colour of the coordination compounds depends

anhydride ? (1) Formic acid (3) Propionic acid

(2) Acetic acid (4) n-butyric acid

4. In which of the following elements, 5f orbitals are

(1) Aniline < p-Nitroaniline < p-Toluidine (2) Aniline < p-Toluidine < p-Nitroaniline (3) p-Toluidine < p-Nitroaniline < Aniline (4) p-Nitroaniline < Aniline < p-Toluidine on the crystal field splitting. What will be the correct order of absorption of wavelength of light in the visible region, for the complexes, [Co(NH3)6]3+, [Co(CN)6]3–, [Co(H2O)6]3+.

progressively filled?

(1) [Co(CN)6]3– > [Co(NH3)6]3+ > [Co(H2O)6]3+

(1) Alkaline earth metals (2) Actinoids

(2) [Co(NH3)6]3+ > [Co(H2O)6]3+ > [Co(CN)6]3–

(3) Lanthanoids (4) Transition elements

(4) [Co(CN)6]3– > [Co(NH3)6]3+ > [Co(H2O)6]3+

5. Structures of some common polymers are given. Which one is not correctly presented? (1) Neoprene : — CH  C == CH  CH  CH — 2 2 2  Cl n

(3) [Co(H2O)6]3+ > [Co(NH3)6]3+ > [Co(CN)6]3–

8. Elemental silicon to be used as a semiconductor is purified by ________ . (1) heating under vacuum (2) floatation (3) zone refining (4) electrolysis

OSWAAL CUET (UG) Sample Question Papers, CHEMISTRY

10

9. Arrange the following compounds in the increasing order of their densities.

Cl

(i)

(ii) Cl

CO2

SO2

CH4

H2

Critical temp./K

304

630

190

33



(2) SO2 (4) H2

11. A first-order reaction is 50% completed in 1.26 × 1014 s. How much time would it take for 100% completion? (1) 1.26 × 1015 s (2) 2.52 × 1014 s 28 (3) 2.52 × 10 s (4) Infinite

12. Which of the following orders are correct as per the properties mentioned against each?

(3) (4)

(4) 340.4 Scm2mol–1

6

∆t



(4)

∆[Br − ] ∆[H + ] =6 ∆t ∆t

16. Bithionol is generally added to the soaps as an

Gas

(2)

(3) 390.5 Scm2mol–1

∆t

the following gases shows least adsorption on a definite amount of charcoal?

As2O3 > SiO2 > P2O3 > SO2 AsH3 < PH3 < NH3 S < O < Cl H2S > H2Se > H2Te

199.1 (2) 375.3 Scm2mol–1

∆[Br − ] 5 ∆[H + ] (3) =

Cl

10. On the basis of data given below, predict which of

(1)

OH

(1) 350 Scm2mol–1

rate of reaction given below? 5Br− (aq) + BrO3– (aq) + 6H+ (aq) ® 3Br2 (aq) + 3H2O (l) ∆[Br − ] 6 ∆[H + ] ∆[Br − ] ∆[H + ] (1) (2) = =5 ∆t 5 ∆t ∆t ∆t

(iv)

(1) CO2 (3) CH4

4

73.5



Br

Cl (1) (i)  < (ii) < (iii) < (iv) (2) (i)  < (iii) < (iv) < (ii) (3) (iv) < (iii) < (ii) < (i) (4) (ii) < (iv) < (iii) < (i)



K+

15. Which of the following expressions is correct for the



(iii)

3

Acid strength Enthalpy of vaporisation More negative electron gain enthalpy Thermal stability

additive to function as a/an: (1) buffering agent (2) antiseptic (3) softener (4) dryer.

17. In an electrochemical process, a salt bridge is used (1) as a reducing agent (2) as an oxidizing agent (3) to complete the circuit so that current can flow  (4) None of these

18. Propanamide on reaction with bromine in aqueous NaOH gives: (1) Propanamine (2) Ethanamine (3) N-Methyl ethanamine (4) Propane nitrile

19. The total number of tetrahedral voids in the face

centred unit cell is _____. (1) 6 (2) 8 (3) 10 (4) 12 20. Which reagent will you use for the following reaction? CH3CH2CH2CH3¾® CH3CH2CH2CH2Cl + CH3CH2CHClCH3 (1) Cl2/UV light (2) NaCl + H2SO4 (3) Cl2 gas in dark (4) Cl2 gas in the presence of iron in dark

13. How many ions are produced from the complex

21. Considering Ellingham diagram, which of the

14. Which of the following option will be the limiting

22. Identify the name of the given reaction:

[Co(NH3)6] Cl2 in solution? (1) 6 (2) 4 (3) 3 (4) 2

following metals can be used to reduce alumina? (1) Mg (2) Zn (3) Fe (4) Cu

molar conductivity of CH3COOH if the limiting molar conductivity of CH3COONa is 91 Scm2mol–1? Limiting molar conductivity for individual ions are given in the following table. S.No 1 2

Ions H

+ +

Na

Limiting molar conductivity / Scm2mol–1

(1) Etard reaction

349.6

(2) Hell-Volhard-Zelinsky reaction

50.1

(3) Stephen reaction (4) None of the above

Sample Question Papers 23. Which of the following solutions in water has

highest boiling point? (1) 1 M NaCl (2) 1 M MgCl2 (3) 1 M urea (4) 1 M glucose 24. Which of the following are peroxoacids of sulphur? (1) H2SO5 and H2S2O8 (2) H2SO5 and H2S2O7 (3) H2S2O7 and H2S2O8 (4) H2S2O6 and H2S2O7 25. Consider a first order gas phase decomposition reaction given below: A(g) B(g) + C(g) The initial pressure of the system before decomposition of A was ‘pi’. After lapse of time ‘t’, total pressure of the system increased by x units and became ‘pt’. The rate constant k for the reaction is given as: (1) k =

pi 2.303 log t pi − x

(2) k =

pi 2.303 log 2 pi − pt t

(3) k =

pi 2.303 log t 2 pi + pt

(4) k = 2.303 log pi t pi + x 26. What is ‘A’ in the following reaction?  CH2–CH=CH2

11

27. Examine the given defective crystal

A+ B– A+ B– A+

B–

0

B–

A+ B–

A+ B– A+ 0

A+

B– A+ B– A+ B– How is the density of the crystal affected by this defect? (1) Density increases (2) Density decreases (3) No effect on density (4) Density first increases then decreases

28. The cell constant of a conductivity cell __________. (1) Changes with change of electrolyte. (2) Changes with change of concentration of electrolyte. (3) Changes with temperature of electrolyte. (4) Remains constant for a cell.

29. The electronic configuration of Cu(II) is 3d9 whereas

that of Cu(I) is 3d10. Which of the following is correct?  (1) Cu(II) is more stable. (2) Cu(II) is less stable. (3) Cu(I) and Cu(II) are equally stable. (4) Stability of Cu(I) and Cu(II) depends on nature of copper salts. 

30. Amino acids are: + HCl

(1) Acidic (3) Amphoteric

A

CH2–CH=CH2 Cl

(1)

(2) Basic (4) Neutral

31. ‘‘Metals are usually not found as nitrates in their ores.’’ Out of the following two (I and II) reasons which is/are true for the above observation? I. Metal nitrates are highly unstable. II. Metal nitrates are highly soluble in water. (1) I and II are true (2) I and II are false (3) I is false but II is true (4) I is true but II is false

32. Chloropicrin is obtained by the reaction of: (1) steam on carbon tetrachloride (2) nitric acid on chlorobenzene (3) chlorine on picric acid (4) nitric acid on chloroform.

(2)



CH2– CH – CH3

33. If two liquids A and B form minimum boiling

Cl

(3)

(4)



azeotrope at some specific composition, then (1) A–B interactions are stronger than those between A–A or B–B. (2) vapour pressure of solution increases because more number of molecules of liquids A and B can escape from the solution. (3) vapour pressure of solution decreases because less number of molecules of only one of the liquids escape from the solution. (4) A–B interactions are weaker than those between A–A or B–B.

OSWAAL CUET (UG) Sample Question Papers, CHEMISTRY

12

34. The crystal showing frenkel defect is:

39. When 1 mole of benzene is mixed with 1 mole of

(1)

toluene the vapour will contain: (Given: vapour of benzene = 12.8kPa and vapour pressure of toluene = 3.85 kPa). (1)  equal amount of benzene and toluene as it forms an ideal solution (2) unequal amount of benzene and toluene as it forms a non ideal solution

(2)

(3) higher percentage of benzene (4) higher percentage of toluene

40. (3)

 CH3     Monomer of   C  CH2   is:     CH3 

(1) 2-methylpropene

(2) styrene

(3) propylene

(4) ethene

41. Λ 0m [NH 4OH] is equal to __________ 0 0 0 (1) Λ m[NH4 OH] + Λ m[NH4 Cl] − Λ [HCl]

(4)

0 0 0 (2) Λ m[NH4 Cl] + Λ m[NaOH] − Λ [NaCl] 0 0 0 (3) Λ m[NH4 Cl] + Λ m[NaCl] − Λ [NaOH]

(4) Λ0 m[NaOH] + Λ0 m[NaCl] − Λ0[NH4 Cl]

42. Interstitial compounds are formed when small 35. Assertion (A): Aromatic 1° amines can be prepared

by Gabriel Phthalimide synthesis. Reason (R): Aryl halides do not undergo nucleophilic substitution with anion formed by phthalimide. (1) Both A and R are true and R is the correct explanation of A (2) Both A and R are true but R is NOT the correct explanation of A (3) A is true but R is false (4) A is false and R is True

36. Which of the following is a diamagnetic ion ? (Atomic numbers of Sc, V, Mn and Cu are 21, 23, 25 and 29 respectively) (1) V2+ (3) Cu2+

(2) Sc3+ (4) Mn3+

37. The slope in the plot of ln[R] vs. time gives (1) +k

(2)

+k 2.303

(3) –k

(4)

−k 2.303

(where [R] is the final concentration of reactant.)

38. Extent of adsorption of adsorbate from solution phase increases with (1) increase in amount of adsorbate in solution. (2) decrease in surface area of adsorbent. (3) increase in temperature of solution.

(4) decrease in amount of adsorbate in solution.

atoms are trapped inside the crystal lattice of metals. Which of the following is not the characteristic property of interstitial compounds? (1) They have high melting points in comparison to pure metals. (2) They are very hard. (3) They retain metallic conductivity. (4) They are chemically very reactive.

43. Compounds A and C in the following reaction are CH CHO 3

(i) CH3MgBr (ii) H2O

(A)

 (1) identical (3) functional isomers

(i) H2SO4,D

(B)

Hydroboration oxidation

(C)

(2) positional isomers (4) optical isomers

44. Bond dissociation enthalpy of E-H (E = element) bond is given below. Which of the compounds will act as strongest reducing agent? Compound

NH3

PH3

∆ diss (E − H)/kJ mol.−1

389

322

(1) NH3 (3) AsH3

AsH3 SbH3 297

255

(2) PH3 (4) SbH3

45. Freshly prepared precipitate sometimes gets converted to colloidal solution by _________ . (1) coagulation (2) electrolysis (3) diffusion (4) peptisation

Sample Question Papers 46. Which of the following is the most stable complex? (1) [Fe(CO)5]   (3) [Fe(C2O4)3]3–

(2) [Fe(H2O)6]3+ (4) [Fe(CN)6]3–

Case Based

13

47. Write down the decreasing order of reactivity of

sodium metal towards primary, secondary and tertiary alcohols. (1) 1oalc3oalc (3) 3oalc RR’CHOH >> RR’R”COH Protonated alcohols as electrophiles (2) RCH2OH > RR’R”COH > RR’CHOH R-CH2-OH+H→R-CH2+OH2 (3) RCH2OH < RR’CHOH (ii) > (iii) (2) (iii) > (ii) > (i) (3) (ii) > (i) > (iii) (4) (i) > (iii) > (ii) 7. Which of the following interface cannot be obtained?

15

Sample Question Papers (1) Liquid-liquid (2) Solid-liquid (3) Liquid-gas (4) Gas-gas

8. Dinucleotide is obtained by joining two nucleotides together by phosphodiester linkage. Between which carbon atoms of pentose sugars of nucleotides are these linkages present ? (1) 5’ and 3’ (2) 1’ and 5’ (3) 5’ and 5’ (4) 3’ and 3’

9. Write the IUPAC name of CH3—CH2—CHO  |  NH2 (1) 1-Aminopropanaldehyde (2) 2-Aminopropanal (3) 1-Aminoethan-1-al (4) None of the above

10. In the preparation of compounds of Xe, Bartlett had

taken O2+PtF6– as a base compound. This is because (1) both O2 and Xe have same size. (2) both O2 and Xe have same electron gain enthalpy. (3) both O2 and Xe have same ionisation enthalpy. (4) both Xe and O2 are gases.

11. The sharp melting point of crystalline solids is due

to ___________. (1) a regular arrangement of constituent particles observed over a short distance in the crystal lattice. (2) a regular arrangement of constituent particles observed over a long distance in the crystal lattice. (3) same arrangement of constituent particles in different directions. (4) different arrangements of constituent particles in different directions.

12. Which of the following alcohols will yield the

corresponding alkyl chloride on reaction with concentrated HCl at room temperature? (1) CH3CH2 – CH2 – OH (2) CH 3CH 2  CH  OH | CH 3 (3) CH 3CH 2  CH  CH 2OH | CH 3 CH 3 | (4) CH 3CH 2  C  OH | CH 3

(1) Carbon (3) Phosphorus

(2) Nitrogen (4) Boron

15. Relative lowering of vapour pressure is a colligative

property because _________ . (1) It depends on number of particles of electrolyte solute in solution and does not depend on the nature of the solute particles. (2) It depends on the concentration of a nonelectrolyte solute in solution as well as on the nature of the solute molecules. (3) Is depends on the concentration of an electrolyte or non-electrolyte solute is solution as well on the nature of solute molecules. (4) None of the above

16. Sulphide ores of metals are usually concentrated

by froth floatation process. Which one of the following sulphide ores offers an exception and is concentrated by chemical leaching? (1) Argentite (2) Galena (3) Copper pyrite (4) Sphalerite

17. Which of the following amines can be prepared by Gabriel synthesis ? (1) Isobutylamine (2) Toluene (3) N-methylbenzylamine (4) Aniline

18. Which of the following statements is not correct? (1) La is actually transition element. (2) In Lanthanide series, Ionic radii decrease from La3+ to Lu3+. (3) La(OH)3 is less basic than Lu(OH)3 (4) Ionic radii of Zr and Hf are almost similar due to Lanthanoid contraction.

19. Which one of the following statements is not true? (1) Ampicillin is a natural antibiotic (2) Aspirin is both analgesic and antipyretic (3) Sulphadiazine is a synthetic antibacterial drug (4) Some disinfectants can be used as antiseptics.

20. Which of the following will show Tyndall effect? (1) Aqueous solution of soap below critical micelle concentration. (2) Aqueous solution of soap above critical micelle concentration. (3) Aqueous solution of sodium chloride. (4) Aqueous solution of sugar.

21. Which of the following compounds is aromatic alcohol? OH

13. In a chemical reaction X → Y, it is found that the rate

CH2OH

CH2OH

OH

of reaction doubles when the concentration of X is increased four times. The order of the reaction with respect to X is (1) 1 (2) 0 (3) 2 (4) 1/2

(A) (B) (1) A, B, C, D (3) B, C

14. Which of the following elements can be involved in

22. An electrochemical cell behaves like an electrolytic

pp–dp bonding?

cell when

CH3 (C) (2) A, D (4) A

(D)

CH3

OSWAAL CUET (UG) Sample Question Papers, CHEMISTRY

16

(1) Ecell = Eexternal (3) Eexternal > Ecell

(2) Ecell = 0 (4) Eexternal < Ecell

23. Which one of the following is employed as a tranquilizer? (1) Naproxen (2) Tetracycline (3) Chlorpheniramine (4) Equanil

30. W  hich one is the complementary base of cytosine in one strand to that in other strand of DNA? (1) Adenine (2) Guanine (3) Thymine (4) Uracil

31. General electronic configuration of actinoids is

24. Gadolinium belongs to 4f series. Its atomic number

(n – 2)f1–14 (n – 1)d0–2 ns2. Which of the following actinoids have one electron in 6d orbital? (1) U (Atomic no. 92) (2) Np (Atomic no.93) (3) Pu (Atomic no. 94) (4) Bk (Atomic no. 97)

(1) [Xe] 4f 75d16s2

32. Arrange the following compounds in increasing

is 64. Which of the following is the correct electronic configuration of Gadolinium? 8

(2) [Xe] 4f 65d26s2

2

9

(3) [Xe] 4f 6d

(4) [Xe] 4f 5s

1

25. Assertion (A): Amorphous solids possess a long-

range order in the arrangement of their particles. Reason (R): The formation of amorphous solids involves very rapid cooling. (1) Both A and R are true and R is the correct explanation of A (2) Both A and R are true but R is NOT the correct explanation of A (3) A is true but R is false (4) A is false and R is True

26. Which of the following is halogen exchange

(3) R – OH + HX CH3 (4)

+ X2

——

RI + NaX — C==C—— — ZnCl2 H X R – X + H 2O CH3 Ch CH33 ——

reaction? (1) RX + NaI — (2) C==C ——+ HX —

Fe dark

X

+ X

27. The value of rate constant of a pseudo-first-order reaction: (1) depends on the concentration of reactants present in small amount. (2) depends on the concentration of reactants present in excess. (3) is independent of the concentration of reactants. (4) depends only on temperature.

28. A solid compound X on heating gives CO2 gas and a

residue. The residue mixed with water forms Y. On passing an excess of CO2 through Y in water, a clear solution Z is obtained. On boiling Z, compound X is reformed. The compound X is: (1) Ca(HCO3)2 (2) CaCO3 (3) Na2CO3 (4) K2CO3

29. Consider the following reaction:

Cu(s) + 2Ag+(aq) → 2Ag(s) + Cu2+(aq) Depict the galvanic cell in which the given reaction takes place. (1) Cu2+ (aq)|Cu(s) ||Ag+(aq)|Ag(s) (2) Cu(s) | Cu2+(aq) || Ag+ (aq) | Ag(s) (3) Ag+(aq)|Ag(s)||Cu2+ (aq)|Cu(s) (4) Ag(s)|Ag+(aq)||Cu2+ (aq)|Cu(s)

order of boiling point.:

Propan-1-ol, butan-1-ol, butan-2-ol, pentan-1-ol (1) Propan-1-ol, butan-2-ol, butan-1-ol, pentan-1ol (2) Propan-1-ol, butan-1-ol, butan-2-ol, pentan-1ol (3) Pentan-1-ol, butan-2-ol, butan-1-ol, propan-1ol (4) Pentan-1-ol, butan-1-ol, butan-2-ol, propan-1ol

33. Which of the following is most effective in

coagulating negatively charged hydrated ferric oxide sol? (1) NaNO3 (2) MgSO4 (3) AlCl3 (4) KCl

34. Which of the following structures represents neoprene polymer? = CH  CH 2 ] (1) [CH 2  C = n  Cl Cl  (2) [CH 2  CH ] n

CN  (3) [CH 2  CH] n (4) [ CH  CH 2 ] n  C6 H 5

35. What will happen during the electrolysis of aqueous solution of CuSO4 by using platinum electrodes? (1) Copper will deposit at cathode. (2) Copper will deposit at anode. (3) Oxygen will be released at anode. (4) Copper will dissolve at anode.

36. The correct IUPAC name for CH2 = CHCH2NHCH3 is: (1) Allylmethylamine (2) 2-amino-4-pentene (3) 4-aminopent-1-ene (4) N-methylprop-2-en-1-amine

37. The half-life period for a zero order reaction is equal to

(1)

0.693 k

(2)

2k [ R]0

Sample Question Papers (3)

2.303 k

(4)

46. Which of the following statements is not true about

[ R]0 2k

(where [R]0 is initial concentration of reactant and k is rate constant).

38. Which of the following species can act as the strongest base? (1) –OH (2) –OR

NO2

39. Which one of the following polymers is prepared by condensation polymerisation? (1) Teflon (2) Natural rubber (3) Styrene (4) Nylon-6, 6

40. Gadolinium belongs to 4f series. Its atomic number is 64. Which of the following is the correct electronic configuration of Gadolinium? (1) [Xe] 4f 75d16s2 (2) [Xe] 4f 65d26s2 8 2 (3) [Xe] 4f 6d (4) [Xe] 4f 95s1

41. The increase in the temperature of the aqueous solution will result in its (1) Molarity to increase. (2) Molarity to decrease. (3) Mole fraction to increase. (4) Mass % to increase.

42. Which of the following compounds will react with sodium hydroxide solution in water? (1) C6H5OH (2) C6H5CH2OH (3) (CH3)3COH (4) C2H5OH

43. In mRNA, the complementary bases of AAT is: (1) CCG (2) UUA (3) AUU (4) UUU

44. In the electrolysis of aqueous sodium chloride solution which of the half cell reaction will occur at anode?

(1) Na ( aq )+e → Na ( s ) ; −

Θ



E Cell = 2.71 V

(2) 2H 2O(l) → O2 ( g )+4H+ ( aq ) + 4e− ; EΘCell = 1.23 V 1 (3) H ( aq )+e → H 2 ( g ) ; 2 1 (4) Cl − ( aq ) → Cl 2 (g)+e − ; 2 +

−

the hexagonal close packing? (1) The coordination number is 12. (2) It has 74% packing efficiency. (3) Tetrahedral voids of the second layer are covered by the spheres of the third layer. (4) In this arrangement spheres of the fourth layer are exactly aligned with those of the first layer.

Case Based

(3) –OC6H5 (4) – O

+

17

Θ

E Cell = 0.00 V EΘCell = 1.36 V

45. The best method for preparing primary amines

from alkyl halides without changing the number of carbon atoms in the chain is: (1) Hoffmann Bromamide reaction (2) Gabriel phthalimide synthesis (3) Sandmeyer reaction (4) Reaction with NH3

Read the passage given below and answer the following questions:

The crystal field theory (CFT) is an electrostatic model which considers the metal-ligand bond to be ionic arising purely from electrostatic interactions between the metal ion and the ligand. Ligands are treated as point charges in case of anions or dipoles in case of neutral molecules. The five d orbitals in an isolated gaseous metal atom/ion have same energy, i.e., they are degenerate. This degeneracy is maintained if a spherically symmetrical field of negative charges surrounds the metal atom/ ion. However, when this negative field is due to ligands (either anions or the negative ends of dipolar molecules like NH3 and H2O) in a complex, it becomes asymmetrical and the degeneracy of the d orbitals is lifted. It results in splitting of the d orbitals.

47. The colour of the coordination compounds depends on the crystal field splitting. What will be the correct order of absorption of wavelength of light in the visible region, for the complexes, [Co(NH3)6]3+, [Co(CN)6]3−, [Co(H2O)6]3+ (1) [Co(CN)6]3− > [Co(NH3)6]3+ > [Co(H2O)6]3+ (2) [Co(NH3)6]3+ > [Co(H2O)6]3+ > [Co(CN)6]3− (3) [Co(H2O)6]3+ > [Co(NH3)6]3+> [Co(CN)6]3− (4) [Co(CN)6]3−> [Co(NH3)6]3+> [Co(H2O)6]3+

48. The CFSE for octahedral [CoCl6]4− is 18,000 cm−1. The CFSE for tetrahedral [CoCl4]2− will be (1) 18,000 cm−1 (2) 16,000 cm−1 (3) 8,000 cm−1 (4) 20,000 cm−1

49. An aqueous pink solution of cobalt(II) chloride changes to deep blue on addition of excess of HCl. This is because _____________. (1) [Co(H2O)6]2+ is transformed into [CoCl6]4− (2) [Co(H2O)6]2+ is transformed into [CoCl4]2− (3) tetrahedral complexes have larger crystal field splitting than octahedral complex. (4) None of the above

50. A chelating agent has two or more than two donor atoms to bind to a single metal ion. Which of the following is not a chelating agent? (1) Thiosulphato (2) Oxalato (3) Glycinato (4) Ethane-1,2-diamine

qqq

SAMPLE

Question Paper Maximum Marks : 200

5

Time : 45 Minutes

General Instructions: (i) This paper consists of 50 MCQs, attempt any 40 out of 50 . (ii) Correct answer or the most appropriate answer: Five marks (+5) . (iii) Any incorrect option marked will be given minus one mark (– 1) . (iv) Unanswered/Marked for Review will be given no mark (0) . (v) If more than one option is found to be correct then Five marks (+5) will be awarded to only those who have marked any of the correct options . (vi) If all options are found to be correct then Five marks (+5) will be awarded to all those who have attempted the question . (vii) If none of the options is found correct or a Question is found to be wrong or a Question is dropped then all candidates who have appeared will be given five marks (+5). (viii) Calculator / any electronic gadgets are not permitted .

1. Which of the following is a 3° amine ? (1) 1-methylcyclohexylamine (2) Triethylamine (3) tert-butylamine (4) N-methylaniline

2. Which of the following forms cationic micelles

above certain concentration? (1) Sodium dodecyl sulphate (2) Sodium acetate (3) Urea (4) Cetyltrimethylammonium bromide

3. The formula of the coordination compound Tetraammineaquachloridocobalt(III) chloride is (1) [Co(NH3)4(H2O)Cl]Cl2 (2) [Co(NH3)4(H2O)Cl]Cl3 (3) [Co(NH3)2(H2O)Cl]Cl2 (4) [Co(NH3)4(H2O)Cl]Cl

4. Which of the following statements is wrong? (1) Single N–N bond is stronger than the single P–P bond. (2) PH3 can act as a ligand in the formation of coordination compound with transition elements. (3) NO2 is paramagnetic in nature. (4) Covalency of nitrogen in N2O5 is four.

5. Phenol is less acidic than ______________. (1) ethanol (3) o-methylphenol

(2) o-nitrophenol (4) o-methoxyphenol

6. Calcium is obtained by the: (1) roasting of limestone. (2) electrolysis of solution of calcium chloride in H2O.

(3) electrolysis of molten anhydrous calcium chloride. (4) reduction of calcium chloride with carbon.

7. Which of the following is most effective in coagulating positively charged hydrated ferric oxide sol? (1) NaNO3 (2) Na2SO4 (3) (NH4)3PO4 (4) LiAlH4

8. Give IUPAC name of the compound given below. (1) 2-chloro-5-hydroxyhexane (2) 2-hydroxy-5-chlorohexane (3) 5-chlorohexan-3-ol (4) 2-chlorohexan-5-ol 9. Affinity for hydrogen decreases in the group from fluorine to iodine. Which of the halogen acids should have highest bond dissociation enthalpy? (1) H-F (2) HCl (3) HBr (4) HI 10. Arrange the following compounds in increasing order of their boiling points: CH3 — CH—CH2Br (i) CH3 — (ii) CH3CH2CH2CH2Br CH3 | (iii) CH3  C  CH3 | Br (1) (ii) < (i) < (iii) (3) (iii) < (i) < (ii)

(2) (i) < (ii) < (iii) (4) (iii) < (ii) < (i)

Sample Question Papers 11. For a zero order reaction, the slope in the plot of [R]

+k 2.303

(1) K kg mol-1 or K (molality)-1 (2) –k

(3) kg mol-1 K-1 or K-1 (molality)-1 (4) +k

(where [R] is the final concentration of reactant) 12. Which of the statements about solutions of

electrolytes is not correct ? (1) Conductivity of solution depends upon size of ions. (2) Conductivity depends upon viscosity of solution. (3) Conductivity does not depend upon solvation of ions present in solution. (4) Conductivity of solution increases with temperature.

13. Natural rubber has: (1) alternate cis- and trans-configuration (2) random cis- and trans-configuration (3) all cis-configuration (4) all trans-configuration

14. Which of the following aqueous solutions should have the highest boiling point? (1) 1.0 M NaOH (2) 1.0 M Na2SO4 (3) 1.0 M NH4NO3 (4) 1.0 M KNO3

15. The IUPAC name of anisole is (1) 2-methyltoluene (2) Methyl phenyl ether (3) Methoxybenzene (4) Ethoxybenzene

22. Assertion

(A): Aromatic aldehydes formaldehyde undergo Cannizzaro reaction.



and

Reason (R): Aromatic aldehydes are almost as reactive as formaldehyde.

(1) Both A and R are true and R is the correct explanation of A. (2) Both A and R are true but R is NOT the correct explanation of A. (3) A is true but R is false. (4) A is false and R is true.

23. Atomic number of Mn, Fe and Co are 25, 26, 27

respectively. Which of the following inner orbital octahedral complex ions are diamagnetic?

(1) [Co(NH3)6]3+ 

(2) [Mn(CN)6]3−

(3) [Fe(CN)6]3−   (4) None of the above

24. Toluene reacts with a halogen in the presence of iron (III) chloride giving ortho and para halo compounds. The reaction is.

(1) Electrophilic elimination reaction (2) Electrophilic substitution reaction (3) Free radical addition reaction

25. Electrode potential for Mg electrode varies

(2) 6 (4) 5

17. Graphite cannot be classified as __________. (1) Conducting solid (3) Covalent solid

(4) K mol kg-1 or K (molality)

(4) Nucleophilic substitution reaction

16. T  he coordination number of ‘Co’ in the complex [Co(en)3]3+ is (1) 3  (3) 4  

(2) mol kg-1 K-1 or K-1 (molality)

(2) Network solid (4) Ionic solid

18. Methylamine reacts with HNO2 to form _________. (1) CH3 — O — N = O (2) CH3 — O — CH3 (3) CH3OH (4) CH3CHO

according to the equation: E Mg 2 + /Mg =E° Mg 2+ /Mg −

0.059 1 log . 2 [Mg 2+ ]

The graph of E Mg 2+ /Mg vs. log [Mg 2+ ] is



(1)

EMg2+/ Mg

(3)

21. The unit of ebullioscopic constant is:

EMg2+/ Mg

vs. time is (1) − k 2.303

19

(2) log[Mg2+]



log[Mg2+]

20. Roasting of sulphides gives the gas X as a by-product. This is a colourless gas with chocking smell of burnt sulphur and causes great damage to respiratory organs as a result of acid rain. Its aqueous solution is acidic, acts as a reducing agent and its acid has never been insolated. The gas X is  : (1) H2S (2) SO2 (3) CO2 (4) SO3

(3)

(4)

EMg2+/ Mg

1.26 × 1014 s. How much time would it take for 100% completion ? (1) 1.26 × 1015 s (2) 2.52 × 1014 s 28 (3) 2.52 × 10 s (4) Infinite

EMg2+/ Mg

19. A first order reaction is 50% completed in

log[Mg2+]

26. Aspirin is an acetylation product of: (1) m-Hydroxybenzoic acid (2) o-Dihydroxybenzoic acid (3) o-Hydroxybenzoic acid (4) p-Dihydroxybenzene

log[Mg2+]

20

OSWAAL CUET (UG) Sample Question Papers, CHEMISTRY

27. Which of the following units is useful in relating

concentration of solution with its vapour pressure? (1) Mole fraction (2) Parts per million (3) Mass percentage (4) Molality 28. In which of the following reactions conc. H2SO4 is used as an oxidising reagent? (1) CaF2 + H2SO4 ® CaSO4 + 2HF (2) 2HI + H2SO4 ® I2 +SO2 + 2H2O (3) Cu + 2H2SO4 ® CuSO4 + SO2+ 2H2O (4) NaCl + H2SO4 ® NaHSO4 + HCl 29. A compound is formed by two elements M and N. The element N forms ccp lattice and atoms of M occupy two atoms an Mercury 1/3rd of tetrahedral voids. What is the formula of the compound (1) MN2 (2) M2N3 (3) M3N2 (4) M2N2 30. Which of the following is most effective in coagulating positively charged methylene blue sol? (1) Na3PO4 (2) K4 [Fe(CN)6]

(3) Na2SO4

(4) Al2(SO4)3

31. Reaction of C6H5CH2Br with aqueous sodium

hydroxide follows: (1) SN1 mechanism (2) SN2 mechanism (3) Any of the above two depending upon the temperature of reaction (4) Saytzeff rule 32. Considering the formation, breaking and strength of hydrogen bond, predict which of the following mixtures will show a positive deviation from Raoult’s law? (1) Methanol and acetone. (2) Chloroform and acetone. (3) Nitric acid and water. (4) Phenol and aniline.

33. The crystal field splitting energy for octahedral (Do) and tetrahedral (Dt) complexes is related as 2 5 (1) ∆ t = ∆ o (2) ∆ t = ∆ o 9 9 (3) ∆ t =

4 ∆o 9

(4) ∆ t = 2 ∆ o

34. For the reaction A → B, the rate of reaction becomes three times when the concentration of A is increased by nine times. What is the order of reaction ? (1) 1 (2) 2 (3) 1/2 (4) 0

35. There are 14 elements in actinoid series. Which of the following element does not belong to this series? (1) U (2) Np (3) Tm (4) Fm

36. Reduction potentials of some ions are given below. Arrange them in decreasing order of oxidising power. Ion

ClO4–

IO4–

BrO4–

Reduction EQ=1.19V EQ=1.65V EQ=1.74V potential EQ/V

(1) ClO–4 > IO–4 > BrO– 4 (2) IO–4 > BrO–4 > ClO–4 (3) BrO–4> IO–4 > ClO– 4 (4) BrO–4 > ClO–4 > IO–4 37. Biodegradable polymer which can be produced from glycine and aminocaproic acid is: (1) buna-N

(2) nylon 6, 6

(3) nylon 2-nylon 6

(4) PHBV

38. When the rate constant has same units as the rate of reaction, the order of the reaction is: (1) Zero order (2) First order (3) Second order (4) Fractional order

39. Write the IUPAC name of the following compounds CH3 | CH3 — O — C — CH3 | CH3 (1) 2-Methoxy-isopropane (2) 2-Methyl-2-methoxypropane (3) 2-Methoxy-2-methylpropane (4) 2-Methoxy-2,2 -dimethyl ethane

40. Which is the example of emulsions? (1) Smoke (3) Gold Sol

(2) Starch (4) Hair Cream

41. The electronic configuration of Cu(II) is 3d9 whereas

that of Cu(I) is 3d10. Which of the following is correct? (1) Cu(II) is more stable. (2) Cu(II) is less stable. (3) Cu(I) and Cu(II) are equally stable. (4) Stability of Cu(I) and Cu(II) depends on nature of copper salts.

42. In the extraction of copper from its sulphide ore, the

metal finally obtained by the reduction of cuprous oxide with: (1) iron (II) sulphide (2) carbon monoxide (3) copper (I) sulphide (4) sulphur dioxide

43. The correct increasing order of basic strength for the following compounds is __________. (I)  (III) NH2   (II)   NH2

  (1) II d > b > a > c (2) b > d > a > c > e (3) d > e > c > b > a (4) e > d > c > b > a

22. What is the coordination number in a square close packed structure in two dimensions? (1) 2 (2) 3 (3) 4 (4) 6

23. 4L of 0.02 M aqueous solution of NaCl was diluted by adding one litre of water. The molality of the resultant solution is __________ (1) 0.004 (2) 0.008 (3) 0.012 (4) 0.016

24. Indicate the complex ion which shows geometrical isomerism. (1) [Cr(H2O)4Cl2]+ (3) [Co(NH3)6]3+

(2) [Pt(NH3)3Cl] (4) [Co(CN)5(NC)]3+

25. Name the major monohalo product of the following reaction:



OSWAAL CUET (UG) Sample Question Papers, CHEMISTRY

24

(1) 1-Iodo-1-methyl cyclohexane (2) 1-Iodomethyl cyclohexane (3) 1-Chloro cyclohexane (4) None of the above 26. Interstitial compounds are formed when small atoms are dropped under the curved lattice of metals. Whether the following is not the characteristics property of interstitial compounds? (1) They have high melting points in to pure metals (2) They are very hard (3) They retain metallic conductivity (4) They are chemically very reactive

27. IUPAC name for the given compound is CH 3 − CH − O − CH 2 − CH 2 − CH 3 | CH 3



(1) 2-ethoxy-2-methylethane. (2) 2-propoxypropane. (3) 2-methyl-2-ethoxypropane (4) None of the above

28. Which of the following are characteristics of thermosetting polymers? (1) Heavily branched cross-linked polymers. (2) Linear slightly branched long-chain molecules. (3) Become infusible on moulding so cannot be reused. (4) Both (1) and (3).

29. Cellular oxidation of glucose is a: (1) spontaneous and endothermic process (2) non spontaneous and exothermic process (3) non spontaneous and endothermic process (4) spontaneous and exothermic process

30. Assertion (A): At the equilibrium position in the

process of adsorption ∆H = T∆S. Reason (R): Adsorption is accompanied by decrease in surface energy. (1) Both A and R are true and R is the correct explanation of A (2) Both A and R are true but R is NOT the correct explanation of A (3) A is true but R is false (4) A is false and R is True

31. Predict the product of the following reaction: (1) CH3CH2CH3 (3) CH3CH2CHO

(2) CH3CHOHCH3 (4) CH3CONHCH3 

32. Which of the following is not a target molecule for drug function in body? (1) Carbohydrates (3) Vitamins

(2) Lipids (4) Proteins

33. When 0.1 mol CoCl3(NH3)5 is treated with excess

of AgNO3, 0.2 mol of AgCl are obtained. The conductivity of solution will correspond to

(1) 1 : 3 electrolyte. (3) 1 : 1 electrolyte.

(2) 1 : 2 electrolyte. (4) 3 : 1 electrolyte.

34. How long will 5g of a reactant take to reduce to 3g? (Given: Rate constant, k = 1.15 × 10–3 s–1) (1) 222.189 s (2) 444.379 s (3) 111.095 s (4) 888.789 s

35. When 0.1 mol CoCl3(NH3)5 is treated with excess

of AgNO3, 0.2 mol of AgCl are obtained. The conductivity of solution will correspond to (1) 1: 3 electrolyte (2) 1: 2 electrolyte (3) 1: 1 electrolyte (4) 3: 1 electrolyte

36. Which of the following statements are true? (1) Only types of interactions between particles of noble gases are due to weak dispersion forces. (2) Ionisation enthalpy of molecular oxygen is very close to that of xenon. (3) Hydrolysis of XeF6 is a redox reaction. (4) Xenon fluorides are not reactive.

37. Under which condition a bimolecular reaction is kinetically first order reaction: (1) When two reactants are involved. (2) When one of the reactants is in excess. (3) When one of the reactants does not involve in reaction. (4) None of these.

38. What kind of compounds undergo Cannizaro reactions ? (1) Ketones with no α-hydrogen (2) Aldehydes with α-hydrogen (3) Carboxylic acids with α-hydrogen (4) Aldehydes with no α-hydrogen

39. Which of the following statement is correct? (1) Ecell and DrG of cell reaction both extensive properties. (2) Ecell and DrG of cell reaction both intensive properties. (3) Ecell is an intensive property while DrG of cell reaction is an extensive property. (4) Ecell is an extensive property while DrG of cell reaction is an intensive property.

40. The values of colligative properties of colloidal solution are of small order in comparison to those shown by true solutions of same concentration because of colloidal particles . (1) exhibit enormous surface area. (2) remain suspended in the dispersion medium. (3) form lyophilic colloids. (4) are comparatively less in number.

41. A narrow spectrum antibiotic is active against. (1) Gram-positive or Gram-negative bacteria. (2) Gram-negative bacteria only. (3) single organism or one disease. (4) both Gram-positive and Gram-negative bacteria.

Sample Question Papers 42. The reagent which does not react with both, acetaldehyde and benzaldehyde. (1) Sodium hydrogen sulphite (2) Phenyl hydrazine (3) Fehling's solution (4) Grignard reagent

43. For a reaction, A + H 2O → B Rate ∝ [A] The order of the reaction is: (1) Zero order (2) Fractional order (3) Pseudo first order (4) Second order 44. Using the data given below find strongest reduction agent.

E - Cr O2- /Cr3+ =1.33 V, E - Cl 2

7

2

/C1-

=1.36 V

E - MnO- /Mn 2+ =1.51V, E - Cr3+ /Cr = - 0.74 V



4

(1) Cl- (3) Cr3+

(2) Cr (4) Mn2+

45. Which of the following is most acidic? (1) Benzyl alcohol (3) Phenol

(2) Cyclohexanol (4) m-Chlorophenol

46. Brine electrolysed by using inert electrodes. The reaction at anode is: 1 − − Θ (1) Cl (aq.) → Cl2 (g)+e ; ECell = 1.36 V. 2 Θ = 1.23 V . (2) 2H 2O (l ) → O 2 (g) + 4H+ + 4e − ; ECell + − Θ (3) Na (aq.) + e → Na (s); ECell = 2.71 V . 1 + − Θ (4) H (aq.) + e → H 2 (g); ECell = 0.00 V. 2

25

Case Based Read the passage given below and answer the following questions:

Transition elements are those that have partially

filled d-orbitals. The configuration of these elements corresponds to (n – 1)d1 – 10 ns1 – 2. It is important to note that elements like mercury, cadmium and zinc are not considered transition elements because of their electronic configurations, which corresponds to (n – 1)d1 – 10 ns2. These elements can form coloured compounds and ions due to d – d transition. A large variety of ligands can bind to these elements, due to which, a wide variety of stable complexes are formed. The boiling and melting point of these elements are high. These elements have a large ratio of charge to the radius.

47. The element of second row of transition element beneath Titanium in periodic table is: (1) Scandium (2) Copper (3) Zinc (4) Zirconium

48. Ions formed from ligands that surrounds the central atom is: (1) cation (3) interstitial ion

(2) anion (4) complex ion

49. Which of the following is NOT a characteristic of transition element? (1) Lustrous (3) Malleability

(2) Low boiling point (4) Ductility

50. Transition metals are generally coloured because (1) they absorb electro-magnetic radiations. (2) they undergo d – d – transition. (3) their penultimate d– subshells are fully filled. (4) completely filled d– orbitals.

qqq

SAMPLE

Question Paper Maximum Marks : 200

7

Time : 45 Minutes

General Instructions: (i) This paper consists of 50 MCQs, attempt any 40 out of 50 . (ii) Correct answer or the most appropriate answer: Five marks (+5) . (iii) Any incorrect option marked will be given minus one mark (– 1) . (iv) Unanswered/Marked for Review will be given no mark (0) . (v) If more than one option is found to be correct then Five marks (+5) will be awarded to only those who have marked any of the correct options . (vi) If all options are found to be correct then Five marks (+5) will be awarded to all those who have attempted the question . (vii) If none of the options is found correct or a Question is found to be wrong or a Question is dropped then all candidates who have appeared will be given five marks (+5). (viii) Calculator / any electronic gadgets are not permitted .

1. 2-Bromopentane, 2-Bromo-2-methylbutane, 1-Bromopentane Write the compound which is most reactive towards β-elimination reaction. (1) 2-Bromopentane (2) 1- Bromopentane (3) 2-Bromo-2-methylbutane (4) None of the above

2. In calcium fluoride, having the fluorite structure, 2+

the coordination number for calcium ion (Ca ) and fluoride ion (F–) are: (1) 4 and 2 (2) 6 and 6 (3) 8 and 4 (4) 4 and 8

3. Which of the following reactions is an example of auto-reduction? (1) Fe3O 4 + 4CO → 3Fe+ 4CO 2 (2) Cu 2O+C → 2Cu + CO (3) Cu 2+(aq.)+Fe (s) → Cu (s)+Fe2+(aq.) 1 1 (4) Cu 2O + Cu 2S → 3Cu + SO 2 2 2

4. In the following sequence of reactions KCN

H 3 O+

LiAlH 4

CH 3 − Br → A → B → C



ether

the end product (C) is: (1) acetone (3) acetaldehyde

(2) methane (4) ethyl alcohol

5. A brown ring is formed in the ring test for NO3– ion. It is due to the formation of:

(1) [Fe(H2O)5(NO)]2+. (3) [Fe(H2O)4(NO)2]

2+.

(2) FeSO4.NO2. (4) FeSO4.HNO3.

6. The d-electron configurations of Cr2+, Mn2+ Fe2+

and Co2+ are d4, d5, d6 and d7 respectively. Which one of the following will exhibit minimum paramagnetic behaviour?

(At. no. Cr = 24, Mn = 25, Fe = 26, Co = 27) (1) [Fe(H2O)6]2+ (3) [Cr(H2O)6]2+

(2) [Co(H2O)6]2+ (4) [Mn(H2O)6]2+

7. The reaction in which the aqueous solution of

sodium salt of carboxylic acids on electrolysis give alkanes: (1) Soda lime decarboxylation (2) Kolbe’s electrolysis decarboxylation (3) Dry distillation of calcium formate (4) Reduction of carboxylic acid.

8. Structure of a mixed oxide is cubic close packed

(ccp). The cubic unit cell of mixed oxide is composed of oxide ions. One fourth of the tetrahedral voids are occupied by divalent metal A and the octahedral voids are occupied by a monovalent metal B. The formula of the oxide is : (1) ABO2 (2) A2BO2 (3) A2B3O4 (4) AB2O2

9. Arrange the following compounds in increasing order of their boiling points.

(i) (ii) (iii) (1) (ii) < (iii) < (i) (3) (iii) < (i) < (ii)

(2) (i) < (ii) < (iii) (4) (iii) < (ii) < (i)

Sample Question Papers 10. Bell-metal is an alloy of: (1) Cu + Pb (3) Cu + Zn

(2) Cu + Sn (4) Cu + Ni

11. Which of the following statements is not true about

enzyme inhibitors? (1) Inhibit the catalytic activity of the enzyme. (2) Prevent the binding of substrate. (3) Generally a strong covalent bond is formed between an inhibitor and an enzyme. (4) Inhibitors can be competitive or noncompetitive.

12. Which of the following are peroxoacids of sulphur? (1) H2SO5 and H2S2O8 (2) H2SO5 and H2S2O7 (3) H2S2O7 and H2S2O8 (4) H2S2O6 and H2S2O7

13. In the metallurgy of aluminium,

(1) Al3+ is oxidised to Al (s). (2) graphite anode is oxidised to carbon monoxide and carbon dioxide. (3) oxidation state of oxygen changes in the reaction at anode. (4) oxidation state of oxygen changes in the overall reaction involved in the process.

14. Formaldehyde reacts with methyl magnesium bromide followed by hydrolysis to form. (1) Methanol (4) Ethanol (3) Propanol (4) Butanol

15. In solid state, PCl5 is a:

(1) covalent solid. (2) octahedral structure. (3) ionic solid with [PCl6]+ octahedral and [PCl4]− tetrahedral. (4) ionic solid with [PCl4]+ tetrahedral and [PCl6]– octahedral.

16. The fraction of total volume occupied by the atoms present in a simple cube is: π π (1) (2) 3 2 4 2 (3)

π 4

(4)

π 6

17. Stainless steel contains iron and: (1) Cr + Ni (3) Zn + Pb

(2) Cr + Zn (4) Fe + Cr + Ni

18. In the following reactions: CH3 | (i) CH3  CH  CH  CH3 | OH

H+ / heat



A + B Major  Minor  product  product     

 HBr, dark

(ii) A → in absence of peroxide

C + D Major  Minor  product  product     

27

The major products (A) and (C) are respectively: CH 3  (1) CH 2 =C  CH 2  CH 3 and CH 3 | CH 2  CH  CH 2  CH 3 | Br CH3 | (2) CH3—C=CH—CH3 and CH3  CH 3  C  CH 2  CH 3  Br CH3  (3) CH 2 = C CH 2  CH 3 and CH3 | CH 3  CH  CH  CH 3  Br CH3  (4) CH 3  CH = CH  CH 3 and CH3  CH 3  C  CH 2  CH 3  Br

19. Zone refining is based on the principle that: (1) impurities of low boiling metals can be separated by distillation. (2) impurities are more soluble in molten metal than in solid metal. (3) different components of a mixture are differently adsorbed on an adsorbent. (4) vapours of volatile compound can be decomposed in pure metal.

20. In a cyclo-trimetaphosphoric acid molecule, how many single and double bonds are present? (1) 3 double bonds; 9 single bonds (2) 6 double bonds; 6 single bonds (3) 3 double bonds; 12 single bonds (4) Zero double bonds; 12 single bonds

21. Which of the following reaction? (1) (2)

is halogen exchange

(3) (4)

+ x

OSWAAL CUET (UG) Sample Question Papers, CHEMISTRY

28

22. A compound formed by elements X and crystallizes

in a cubic structure in which the X atoms are at the corners of a cube and the Y atoms are at the facecentres. The formula of the compound is: (1) XY3 (2) X3Y (3) XY (4) XY2

23. Which of the following polymers of glucose is stored by animals? (1) Cellulose (3) Amylopectin

(2) Amylose (4) Glycogen

24. A solution has 1 : 4 mole ratio of pentane to hexane. The vapour pressure of the pure hydrocarbons at 20°C are 440 mm of Hg for pentane and 120 mm of Hg for hexane. The mole fraction of pentane in the vapour phase would be : (1) 0.549 (2) 0.200 (3) 0.786 (4) 0.478

25. Conductivity k, is equal to __________________. (1) Ùm

(2)

l (3) A



G* R

(4) All of the above NaOH/I2 26. C6H5 – CO – CH3 → ? + ? (1) C6H5COOH + CH4 (2) C6H5COONa + CHI3 (3) C6H6 + CH3COONa + HI (4) C6H5CH2COOH 

27. The

experimental data 2A + B2 → 2AB is:

for

the

Exp.

[A]

[B]

Rate (Ms–1)

1.

0.50

0.50

1.6 × 10–4

2.

0.50

1.00

3.2 × 10–4

3.

1.00

1.00

3.2 × 10–4

reaction

The rate equation for the above data is: (1) rate = k[B2] (3) rate = k[A]2[B]2

(2) rate = k[B2]2 (4) rate = k [A]2[B]

28. The protecting power of lyophilic colloidal sol is expressed in terms of (1) coagulation value (2) gold number (3) critical micelle concentration (4) oxidation number

29. Name the linkage that holds the two units in the disaccharide ? (1) Nucleoside linkage (2) Glycosidic linkage (3) Peptide linkage (4) None of the above

30. Which statement about aspirin is not true? (1) Aspirin belongs to narcotic analgesics (2) It is effective in relieving pain (3) It has anti-blood clotting action (4) It is a neurologically active drug

31. For the reaction, 2N2O5 → 4NO2 + O2, rate and rate constant are 1.02 × 10–4 and 3.4 × 10–5 s–1 respectively, then concentration of N2O5 at that time will be : (1) 1.732 (2) 3 (3) 1.02 × 10–4 (4) 3.4 × 105

32. The correct order of the stoichiometrics of AgCl

formed when AgNO3 in excess is treated with the complexes, CoCl3.6NH3, CoCl3.5NH3, CoCl3.4NH3 respectively is: (1) 1 AgCl, 3 AgCl, 2 AgCl (2) 3 AgCl, 1 AgCl, 2 AgCl (3) 3 AgCl, 2 AgCl, 1 AgCl (4) 2 AgCl, 3 AgCl, 1 AgCl

33. Assertion (A): All naturally occurring α-amino acids

except glycine are optically active. Reason (R): Most naturally occurring α-amino acids have L-configuration. (1) Both A and R are true and R is the correct explanation of A (2) Both A and R are true but R is NOT the correct explanation of A (3) A is true but R is false (4) A is false and R is True

34. If molality of the dilute solution is doubled, the value of molal depression constant (Kf) will be: (1) doubled (2) halved (3) tripled (4) unchanged

35. Crystal field stabilization energy for high spin d4 octahedral complex is: (1) – 1.8D0 (3) – 1.2D0

(2) – 1.6D0 + P (4) – 0.6D0

36. The coagulation value in millimoles per liter of the

electrolytes used for the coagulation of As2S3 are given below: I. (NaCl) = 52   II. (BaCl2) = 0.69 III. (MgSO4) = 0.22

The correct order of their coagulating power is (1) I > II > III (3) III > II > I

(2) II > I > III (4) III > I > II

37. The freezing point depression constant for water

is – 1.86°C m–1. If 5.00 g Na2SO4 is dissolved in 45.0 g H2O, the freezing point is changed by –3.82°C. Calculate the van’t Hoff factor for Na2SO4. (1) 2.63 (2) 3.11 (3) 0.381 (4) 2.05

38. Which of the following nitrogen bases is not present in RNA? (1) Thymine (3) Guanine

(2) Adenine (4) Cytosine

39. The number of electrons delivered at the cathode

during electrolysis by a current of 1 ampere in 60 second is : (charge on electron = 1.60 × 10–19 C) (1) 6 × 1023 (2) 6 × 1020 20 (3) 3.75 × 10 (4) 7.48 × 1023

29

Sample Question Papers 40. Activation energy of a chemical reaction can be

determined by: (1) evaluating rate constant at standard temperature. (2) evaluating velocities of reaction at two different temperatures. (3) evaluating rate constants at two different temperatures. (4) changing concentration of reactants.

Case Based  Read the passage given below and answer the following questions:

The main problem encountered during electrophilic

41. In which of the following polymers, ethylene glycol is one of the monomer units? (1) —OCH ( CO)n 2CH2OOC

47.

(2) —CH ( )n 2—CH2— (3) | —CH )n ( 2—CH=CH—CH2—CH—CH2 — ( )n (4) —O—CH—CH 2—C—O—CH—CH2—C— | || | || O CH3 O CH2CH3

substitution reactions of aromatic amines is that of their very high reactivity. Substitution tends to occur at ortho and para-positions. If we have to prepare monosubstituted aniline derivative, how can the activating affect of –NH2 group be controlled? This can be done by protecting the –NH2 group by acetylation with acetic anhydride, then carrying out the desired substitution followed by hydrolysis of the substituted amide to the substituted amine. Give the major product of the following reaction: NH2 + 3Br2



Aniline NH2

(1) Br

a magnetic moment of 3.83 B.M., the correct distribution of 3d electrons in the chromium of the complex is:

(1)

1 3dxy ,

3d 1yz ,

3d 12 z

(2)



1 (3) 3dxy , 3d( x 2 − y 2 ) , 3d yz



3d( x 2 − y 2 ) , 3d z12

,

1 3dxz

(2)

Br NH2

(3)

NH2 Br



42. [Cr(H2O)6]Cl3 (Atomic number of Cr = 24) has

Br2/H2O

Br

(4) None of the above

Br

Br

1 1 1 (4) 3dxy , 3d yz , 3d zx

43. 5A current is passed through a solution of zinc sulphate for 40 min. The amount of zinc deposited at the cathode is (1) 40.65 g (2) 0.4065 g (3) 4.065 g (4) 65.04 g



Br

48. What is the major product A of the following reaction: O || H—N—C—CH3

44. Salvarsan is arsenic containing drug which was first used for the treatment of (1) syphilis (2) typhoid (3) meningitis (4) dysentery

NH2 (CH3CO)2O Pyridine

45. For adsorption of a gas on a solid, the plot of log x

m vs log p is linear with slope equal to (n being a whole number): (1) k (2) log k 1 (3) n (4) n

Aniline

N-Phenylethanamide (Acetanilide)

(1)

OH– or H+

Br2 CH3COOH

(2)

NH2

Br (Major)

NH2

Br

46. A solution contains non-volatile solute of molecular mass, M2. Which of the following can be used to calculate the molecular mass of solute in terms of osmotic pressure? m   m  RT (1) M2 =  2  VRT (2) M2 =  2   π  V  π

m  M2  2  πRT (3) = V 

m  π (4) M2 =  2   V  RT

2-bromoaniline

(3)

Br

Br

NH2



Br 4-bromoaniline (4) None of the above

30

OSWAAL CUET (UG) Sample Question Papers, CHEMISTRY

49. Why is the activating effect of -NHCOCH3 group in the above reaction less than the activating effect of amino group? (1) Due to mesomeric effect of benzene ring. (2) Due to inductive effect of alkyl group.

(3) Due to resonance effect of acetanilide. (4) All of the above.

50. Aniline is a resonance hybrid of (1) 3 structures (3) 2 structures

(2) 6 structures (4) 5 structures

qqq

SAMPLE

Question Paper Maximum Marks : 200

8

Time : 45 Minutes

General Instructions: (i) This paper consists of 50 MCQs, attempt any 40 out of 50 . (ii) Correct answer or the most appropriate answer: Five marks (+5) . (iii) Any incorrect option marked will be given minus one mark (– 1) . (iv) Unanswered/Marked for Review will be given no mark (0) . (v) If more than one option is found to be correct then Five marks (+5) will be awarded to only those who have marked any of the correct options . (vi) If all options are found to be correct then Five marks (+5) will be awarded to all those who have attempted the question . (vii) If none of the options is found correct or a Question is found to be wrong or a Question is dropped then all candidates who have appeared will be given five marks (+5). (viii) Calculator / any electronic gadgets are not permitted .

1. Concentrated aqueous sulphuric acid is 98% H2SO4

by mass and has a density of 1.80 g mL–1. Volume of acid required to make one liter of 0.1 M H2SO4 solution is:

(1) 11.10 mL

(2) 16.65 mL

(3) 22.20 mL

(4) 5.55 mL

2. Copper sulphate dissolves in excess of KCN to give: (2) [Cu(CN)4]3–

(1) CuCN (3) [Cu(CN)4]

2–

(4) Cu(CN)2

3. Extraction of gold and silver involves leaching the

metal with CN– ion. The metal is recovered by: (1) displacement of metal by some other metal from the complex ion. (2) roasting of metal complex. (3) calcination followed by roasting. (4) thermal decomposition of metal complex.

4. In which of the following coordination entities, the magnitude of D0 (CFSE in octahedral field) will be maximum?

(Atomic number of Co = 27) (1) [Co(H2O)6]3+ (3) [Co(CN6)]

3–

(2) [Co(NH3)6]3+ (4) [Co(C2O4)3]3–

5. For orthorhombic system axial ratios are a ≠ b ≠ c and the axial angles are: (1) α = β = γ ≠ 90°

(2) α = β = γ = 90°

(3) α = γ = 90°, β ≠ 90° (4) α ≠ β ≠ γ ≠ 90°

6. Structure of a disaccharide formed by glucose and fructose is given below. Identify anomeric carbon

atoms in monosaccharide units. f CH2OH a e O HOH2C O H H H H a b d e OH H HO O H HO c CH 2OH b f c d OH H OH H (1) ‘a’ carbon of glucose and ‘a’ carbon of fructose. (2) ‘a’ carbon of glucose and e’ carbon of fructose. (3) ‘a’ carbon of glucose and ‘b’ carbon of fructose. (4) ‘f ’ carbon of glucose and ‘f ’ carbon of fructose. 7. The commercial name of polyacrylonitrile is ______________. (1) dacron. (2) orlon (acrilan). (3) pvc. (4) bakelite.

8. Which of the following statement is correct for the spontaneous adsorption of a gas? (1) DS is negative and therefore, highly positive. (2) DS is negative and therefore, highly negative. (3) DS is positive and therefore, negative. (4) DS is positive and therefore, highly positive.

∆H should be ∆H should be ∆H should be ∆H should be

9. The basic character of the transition metal monoxides follow the order

(Atomic number Ti = 22, V = 23, Cr = 24, Fe = 26) (1) TiO > FeO > VO > CrO (2) TiO > VO > CrO > FeO (3) VO > CrO > TiO > FeO (4) CrO > VO > FeO > TiO

10. Which one of the following is an outer orbital complex and exhibits paramagnetic behaviour?

OSWAAL CUET (UG) Sample Question Papers, CHEMISTRY

32

(1) [Ni(NH3)6]2+ (3) [Cr(NH3)6]3+

(2) [Zn(NH3)6]2+ (4) [Co(NH3)6]3+

11. pA and pB are the vapour pressure of pure liquid

components A and B, respectively of an ideal binary solution. If χA represents the mole fraction of component A, the total pressure of the solution will be: (1) pA + χA (pB – pA) (2) pA + χA (pA – pB) (3) pB + χA (pB – pA) (4) pB + χA (pA – pB)

12. Aniline in a set of the following reactions yields a

(1) ΔG value for the overall reduction reaction with carbon monoxide is zero. (2) ΔG value for the overall reduction reaction with a mixture of 1 mol. carbon and 1 mol. oxygen is positive. (3) ΔG value for the overall reduction reaction with a mixture of 2 mol. carbon and 1 mol. oxygen will be positive. (4) ΔG value for the overall reduction reaction with carbon monoxide is negative.

coloured product Y.

–200 eO 2F

O2 D Fe + 2 C + O2 E

–300

2

O

–500

2C

+

The structure of Y would be:

CO2

B

2C 0

G°/kJ mol–1 of O

2

A

–400

+

O

2

–600

(1)

–700

2C

O

0

400

800 1200 1600 Temperature (°C)

2000

18. Atomic number of Cr and Fe are respectively 24 and

(2)

26, which of the following is paramagnetic with the spin of electron? (1) [Cr(CO)6] (2) [Fe(CO)5] (3) [Fe(CN)6]4– (4) [Cr(NH3)6]3+

(3)

19. Which of the following polymer is stored in the liver of animals? (1) Amylose (3) Amylopectin

(4)

13. Which of the following is an iso-electronic pair? BrO–2, –

BrF+ 2

(1) ICl2, ClO2

(2)

(3) ClO2, BrF

(4) CN , O3

14. How will you distinguish between aliphatic aldehydes and aromatic aldehydes ? (1) Fehling’s test (2) Benedict’s test (3) Iodoform test (4) Hinsberg reagent

15. In cube of any crystal A-atom placed at every corners and B-atom placed at every centre of face. The formula of compound is: (1) AB (2) AB3 (3) A2B2 (4) A2B3

16. Which of the following polymer can be formed by using the following monomer unit? H | N O C H2C H2C

CH2

(2) Cellulose (4) Glycogen

20. Name the main compounds A and B formed in the following reaction:

( ) ( i )CH3 MgBr CH 3 CN  → A  →B ( ii ) H O+ Zn Hg / conc. HCl

(1) CH3CH2COOH [A], CH3CH2CH3 [B] 3

(2) CH3CH2CHO [A], C2H4 [B] (3) CH3COCH3 [A], CH3CH2CH3 [B] (4) CH3COCH3 [A], C2H6 [B]

21. Assertion (A): Λm for weak electrolytes shows a

sharp increase when the electrolytic solution is diluted. Reason (R): For weak electrolytes degree of dissociation increases with dilution of solution. (1) Both A and R are true and R is the correct explanation of A (2) Both A and R are true but R is NOT the correct explanation of A (3) A is true but R is false (4) A is false and R is True

22. The ionic radii of A+ and B– ions are 0.98 × 10–10 m

(1) Nylon 6, 6 (3) Nylon 2–nylon 6 (2) Melamine polymer (4) Nylon-6

and 1.81 × 10–10 m. The coordination number of each ion in AB is : (1) 8 (2) 2 (3) 6 (4) 4

17. On basis of given graph, For the reduction of FeO at

23. The compound that causes general anti-depressant

H 2C

CH2

the temperature corresponding to point D, which of the following statements is correct?

action on the central nervous system belongs to the class of:

Sample Question Papers (1) analgesics. (2) tranquilizers. (3) narcotic analgesics. (4) antihistamines.

24. Phenol can be distinguished from ethanol by the reaction with _____ (1) Br2/water (3) Glycerol

(2) Na (4) All of the above

25. Which one of the following statements is incorrect

about enzyme catalysis? (1) Enzymes are mostly proteinous in nature. (2) Enzyme action is specific. (3) Enzymes are denaturated by UV-rays and at high temperature. (4) Enzymes are least reactive at optimum temperature.

26. Phosgene is a common name for: (1) phosphoryl chloride (2) thionyl chloride (3) carbon dioxide and phoshine (4) carbonyl chloride

27. A hypothetical electrochemical cell is shown below A | A+ (xM) || B+ (yM) | B The EMF measured is + 0.20 V. The cell reaction is  : (1) A + B+ → A+ + B (2) A+ – B → A + B+ (3) A+ + B– → A, B+ + e– → B (4) The cell reaction cannot be predicted

28. Which of the following statements is not true about low density polythene? (1) Tough (2) Hard (3) Poor conductor of electricity (4) Highly branched structure

29. The final product C, obtained in this reaction:

33

(2) Approximately at the temperature corresponding to point A. (3) Above temperature at point A but below temperature at point D. (4) Above temperature at point A.

31. Which of the following will exhibit maximum ionic conductivity? (1) K4[Fe(CN)6] (3) [Cu(NH3)4]Cl2

(2) [Co(NH3)6]Cl3 (4) [Ni(CO)4]

32. Which of the following statements is correct? (1) Some tranquilizers function by inhibiting the enzymes which catalyse the degradation of noradrenaline (2) Tranquilizers are narcotic drugs (3) Tranquilizers are chemical compounds that do not affect the message transfer from nerve to receptor (4) Tranquilizers are chemical compounds that can relieve pain and fever

33. Molarity of liquid HCl, if density of solution is 1.17 g/cc is: (1) 36.5 (3) 32.05

(2) 18.25 (4) 42.10

34. Phenol is less acidic than_________. (1) ethanol (3) o-methylphenol

(2) o-nitrophenol (4) o-methoxy phenol

35. Reduction potentials of some ions are given below.

Arrange them in decreasing order of oxidising power. ClO4-

Ion

IO4-

BrO4-

Reduction EQ=1.19V EQ=1.65V EQ=1.74V potential EQ/V (1) ClO–4 > IO–4 > BrO–4 (2) IO–4 > BrO–4 > ClO–4 (3) BrO–4> IO–4 > ClO–4 (4) BrO–4 > ClO–4 > IO–4

36. Grignard reagent is prepared by the reaction between: (1) magnesium and alkane (2) magnesium and aromatic hydrocarbon (3) zinc and alkyl halide (4) magnesium and alkyl halide



(1)

(2)

37. Which one of the following statements is not

(3)









(4)



30. Choose the correct option of temperature at which

correct? (1) Catalyst does not initiate any reaction. (2) The value of equilibrium constant is changed in the presence of a catalyst in the reaction equilibrium. (3) Enzymes catalyze mainly biochemical reaction. (4) Coenzymes increase the catalytic activity of enzyme.

38. The reaction

carbon reduces FeO to iron and produces CO.

(1) Below temperature at point A.



—OH

NaH

–

+

—O Na

34

OSWAAL CUET (UG) Sample Question Papers, CHEMISTRY Me-I

 can be classified as:

Me

(D) Zone refining

—O

(iv) Extraction of Au (v) Purification of Ni

Code: A B C D (1) (i) (ii) (iii) (iv) (2) (iii) (iv) (v) (i) (3) (iv) (ii) (iii) (i) (4) (ii) (iii) (i) (v)

(1) dehydration reaction (2) Williamson alcohol synthesis reaction (3) Williamson ether synthesis reaction (4) alcohol formation reaction.

39. Which of the following statements are correct? (a) Among halogens, radius ratio between iodine and fluorine is maximum. (b) Leaving F – F bond, all halogens have weaker X – X bond than X – X’ bond in inter-halogens. (c) Among inter-halogen compounds maximum number of atoms ate present in iodine fluoride. (d) Inter-halogen compounds are more reactive than halogen compounds. (1) (a) and (b) (2) (a), (c) and (d) (3) (b) and (c) (4) (a), (b) and (c)

44. In acidic medium, H2O2 changes Cr2O72– to CrO5

40. Through which of the following reactions number

46. A hydrogen gas electrode is made by dipping

41. On the basis of the information available from the

platinum wire in a solution of HCl of pH = 10 and by passing hydrogen gas around the platinum wire at 1 atm pressure. The oxidation potential of electrode would be: (1) 0.059 V (2) 0.59 V (3) 0.118 V (4) 1.18 V

of carbon atoms can be increased in the chain? (1) Grignard reaction (2) Cannizzaro reaction (3) Clemmenson reduction (4) HVZ reaction reaction: 4 2 Al + O2 → Al2O3 , ∆G = − 827 kJ mol −1 3 3

42. Which is the correct statement about birth control

and

43. Match items of Column I with the items of Column Column I (A) Cyanide process (B) Froth floatation process (C) Electrolytic reduction

Column II (i) Ultrapure Ge (ii) Dressing of ZnS (iii) Extraction of Al

named after: (1) Perkin (3) Hofmann



electrolysis of Al2O3 is: (F = 96500 C mol–1) (1) 2.14 V (2) 4.28 V (3) 6.42 V (4) 8.56 V

II and assign the correct code:

45. Amides can be converted into amines by a reaction (2) Claisen (4) Kekule

Case Based

of O2, the minimum EMF required to carry out the

pills? (1) Contain oestrogen only (2) Contain progesterone only (3) Contain a mixture of oestrogen progesterone derivatives (4) Progesterone enhances ovulation

which has two (—O—O—) bonds. Oxidation state of Cr in CrO5 is: (1) +5 (2) +3 (3) +6 (4) –10

Read the passage given below and answer the following questions:

The rate of the reaction is proportional to the concentration of the reactant. Hydrogenation of ethene results in the formation of ethane. The rate constant, k for the reaction was found to be 2.5 × 10–15s–1. The concentration of the reactant reduces to one-third of the initial concentration in 5 minutes.

47. Find the order of reaction: (1) Zero order (3) Second order

(2) First order (4) Fractional order

48. The rate law equation is: (1) Rate = k [C2H6] (3) Rate = k [C2H4]

(2) Rate = k [C2H4]2 (4) Rate = k [C2H4]2

49. The half-life for the reaction is: (1) 2.772 × 10–24 s (3) 1.386 × 10–24 s

(B) 2.772 × 10–12 s (D) 1.386 × 10–12 s

50. The rate constant of the reaction after 5 minutes is: (1) 0.4290 min–1 (3) 0.2197 min–1

(2) 0.1297 min–1 (4) 0.6591 min–1

qqq

SAMPLE

Question Paper Maximum Marks : 200

9

Time : 45 Minutes

General Instructions: (i) This paper consists of 50 MCQs, attempt any 40 out of 50 . (ii) Correct answer or the most appropriate answer: Five marks (+5) . (iii) Any incorrect option marked will be given minus one mark (– 1) . (iv) Unanswered/Marked for Review will be given no mark (0) . (v) If more than one option is found to be correct then Five marks (+5) will be awarded to only those who have marked any of the correct options . (vi) If all options are found to be correct then Five marks (+5) will be awarded to all those who have attempted the question . (vii) If none of the options is found correct or a Question is found to be wrong or a Question is dropped then all candidates who have appeared will be given five marks (+5). (viii) Calculator / any electronic gadgets are not permitted .

1. Number of elements are available in the Earth’s crust but most abundant elements are: (1) Al and Fe. (2) Al and Cu. (3) Fe and Cu. (4) Cu and Ag.

2. A ferromagnetic substance becomes a permanent

magnet when it is placed in a magnetic field because __________. (1) All the domains get oriented in the direction of magnetic field (2) All the domains get oriented in the direction opposite to the direction of magnetic field (3) Domains get oriented randomly (4) Domains are not affected by magnetic field

3. Which one of the following does not exist? (1) XeOF4 (3) XeF2

(2) the sign of the charge alone (3) the magnitude of its charge (4) both magnitude and sign of its charge

7. Which of the following pairs of compounds are enantiomers ?

(1)



(2)

(2) NeF2 (4) XeF6

4. The rate of first order reaction is 1.5 × 10–2 mol L–1 min–1 at 0.5 M concentration of the reactant. The half-life of the reaction is : (1) 0.383 min (2) 23.1 min (3) 8.73 min (4) 7.53 min

5. Among the following series of transition metal ions,

(3)

(4)

2

the one in which all metal ions have 3d electronic configuration is:

(Atomic number Ti = 22, V = 23, Cr = 24, Mn = 25) (1) Ti3+, V2+, Cr3+, Mn4+ (2) Ti+, V4+, Cr6+, Mn7+ (3) Ti4+, V3+, Cr2+, Mn3+ (4) Ti2+, V3+, Cr4+, Mn5+

6. The ability of anion to bring about coagulation of a given colloid depends upon (1) its charge

8. In the extraction of copper from its sulphide ore, the metal is formed by the reduction of Cu2O with: (1) FeS (2) CO (3) Cu2S (4) SO2

9. Complete the following reaction: Xe + PtF6 ® (1) Xe + PtF6 ® XeF4 + PtF2 (2) Xe + PtF6 ® XeF6 + Pt (3) Xe + PtF6 ® Xe+[PtF6]–

36

OSWAAL CUET (UG) Sample Question Papers, CHEMISTRY (A)       (B)

(4) Xe + PtF6 ® XeO2F4 + Pt

10. Which of the following is an example of liquid

(1)

dishwashing detergent?

(1) CH3(CH2)10 –CH2OSO3– Na+ (2) C9H19

| OH

(–CH2 –CH2 – O – ) – CH2CH2OH –

| SO 4

= H3C  C  CH3 , B= H3C  C ≡ CH (3) A

+

SO3 Na CH3

|| O

+

—

CH 9 19 (3) CH3

(2) A= H 3C  C= CH 2 , B= H 2 C  C= CH 2

(–CH2 –CH2 – O – ) – CH2CH2OH

– (4) CH3—(CH2)15 — N — CH3 Br

—

(4)



CH3

11. Schottky defect is observed in crystals when

__________. (1) some cations move from their lattice site to interstitial sites (2) equal number of cations and anions are missing from the lattice (3) some lattice sites are occupied by electrons (4) some impurity is present in the lattice CH 2  C H 2 12. Reaction of with RMgX leads to the O formation of (1) RCH2CH2OH (2) RCHOHCH3 R (3) RCHOHR (4)  CHCH 2OH R

18. At the critical micelle concentration (CMC) the sur-

13. The molar conductivity of a 0.5 mol/dm solution of

(2) CH3CH2OSO3H

3

AgNO3 with electrolytic conductivity of 5.76 × 10 S cm–1 at 298 K is: (1) 2.88 S cm2/mol (2) 11.52 S cm2/mol 2 (3) 0.086 S cm /mol (4) 28.8 S cm2/mol

–3

14. Among [Ni(CO)4], [Ni(CN)4]2–, [NiCl4]2– species, the hybridization states of the Ni atom are, respectively (Atomic number of Ni = 28) (1) sp3, dsp2, dsp2 (2) sp3, dsp2, sp3 3 3 2 (3) sp , sp , dsp (4) dsp2, sp3, sp3

15. A substance A decomposes by a first order reaction starting initially with [A] = 2.00 m and after 200 min, [A] becomes 0.15 m. For this reaction t1/2 is: (1) 53.49 min (2) 50.49 min (3) 48.45 min (4) 46.45 min

16. Which one of the following statements is correct

when SO2 is passed through acidified K2Cr2O7 solution? (1) SO2 is reduced. (2) Green Cr2(SO4)3 is formed. (3) The solution turns blue. (4) The solution is decolourised.

17. Predict the correct intermediate and product in the following reaction.

factant molecules: (1) decompose (2) dissociate (3) associate (4) become completely soluble

19. Consider the following reaction: PBr3

(i) H SO room, temperature

2 4  →Z (ii) H O, heat

The product Z is:

2

(1) CH3CH2 OCH2CH3 (3) CH3CH2OH (4) CH2 = CH2 Ni/H 2 20. CH3CH 2Cl NaCN  → X  → acetic anhydride

Y → Z Z in the above reaction sequence is: (1) CH3CH2CH2NHCOCH3 (2) CH3CH2CH2NH2 (3) CH3CH2CH2CONHCH3 (4) CH3CH2CH2CONHCOCH3

21. If a is the length of the side of a cube, the distance between the body centered atom and one corner atom in the cube will be: (1) (3)

2

a

(2)

3 a 4

(4)

3

4 3

a

3 a 2

22. The correct structure of the product ‘A’ formed in the reaction:

H O,H SO

2 2 4→ H 3C −− C ≡ CH 

HgSO4 Intermediate → Product

alc.KOH

Ethanol  → X → Y

(g),

  

Sample Question Papers

(1)

(2)







(3)

(4)







23. Which one of the most acidic compound? OH

(1) (a)



CH3 OH

37

molecules, one nitro group and two chlorine atoms for one cobalt atom. One mole of this compound produces three mole ions in an aqueous solution. On reacting this solution with excess of AgNO3 solution, we get two moles of AgCl precipitate. The ionic formula for this complex would be: (1) [Co(NH3)5(NO2)]Cl2 (2) [Co(NH3)5Cl] [Cl(NO2)] (3) [Co(NH3)4(NO2)Cl] [(NH3)Cl] (4) [Co(NH3)5] [(NO2)2Cl2]

28. Lanthanides are:

(1) 14 elements in the sixth period (Atomic number = 90 to 103) that are filling 4f sub-level (2) 14 elements in the seventh period (Atomic number = 90 to 103) that are filling 5f sub-level (3) 14 elements in the sixth period (Atomic number = 58 to 71) that are filling 4f sub-level (4) 14 elements in the seventh period (Atomic number = 58 to 71) that are filling 4f sub-level

29. For an endothermic reaction, energy of activation is Ea and enthalpy of reaction is ∆H (both of these in kJ/mol). Minimum value of Ea will be: (1) less than ∆H (2) equal to ∆H (3) more than ∆H (4) equal to zero

(2) (b) OH

30. Predict the products in the given reaction: CHO

(3) (c)

50% KOH



NO2



OH

(4)

O2N

NO2

(d)

CH2OH

(1)

24. Which is the monomer of neoprene in the following? (1) CH 2 = = C  CH = = CH 2  Cl

= C  CH = = CH 2 (4) CH 2 =  CH3

Cl

(2)

(2) Ge and Si (4) Zn and Hg

26. The equivalent conductance of Ba2+ and Cl– are 127 –1

–1

–1

and 76 Ω cm eq respectively at infinite dilution. The equivalent conductance of BaCl2 at infinite dilution will be: (1) 139.52 (2) 203 (3) 279 (4) 101.5

27. A coordination complex compound of cobalt has the molecular formula containing five ammonia

OH

+



OH

OH CH2OH

(3)

COO–K+

+



Cl

Cl COO–K+

CH2OH

25. Electrolytic refining is used to purify which of the following metals? (1) Cu and Zn (3) Zr and Ti

Cl CH2OH

(2) CH 2 == CH  C ≡≡ CH

= CH CH = = CH 2 (3) CH 2 =

CH2COO–K+

+



NO2



Cl

(4)

+ OH

OH

31. The Langmuir adsorption isotherm is deduced by

using the assumption that (1) the adsorption takes place in multi-layers. (2) the adsorption sites are equivalent in their ability to adsorb the particles. (3) the heat of adsorption varies with coverage. (4) the adsorbed molecules interact with each other.

OSWAAL CUET (UG) Sample Question Papers, CHEMISTRY

38

39. Assertion: Among [CO(NH3)6]3+ and [CO(en)3]3+,

C ≡≡N

32.

+ CH3MgBr → P OCH3

Product P in the above reaction is: (1)

(2)

coordination compound [CO(en)3]3+ is a more stable complex. Reason: Because (en) is a chelating ligand/bidentate ligand. (1) Both A and R are true and R is the correct explanation of A

(2) Both A and R are true but R is NOT the correct explanation of A (3) A is true but R is false (4) A is false and R is true (3)







40. Which of the following organic compounds

(4)

polymerized to form the polyester Dacron? (1) Propylene and para HO—(C6H4)—OH (2) Benzoic acid and ethanol



(3) Terephthalic acid and ethylene glycol



33. During reduction of aldehydes with hydrazine and potassium hydroxide, the first is the formation of (1) RCH == N — NH2 (2) RC == N (3) R  C  NH 2 (4) RCH == NH  O

(4) Benzoic acid and para HO—(C6H4)—OH

41. The reaction is described as H CH3(CH2)5 C Br

34. For the reduction of silver ions with copper metal

the standard cell potential was found to be + 0.46 V at 25°C. The value of standard Gibbs energy, ∆G° will be: (F = 96500 C mol–1) (1) – 89.0 kJ (2) – 89.0 J (3) – 44.5 kJ (4) – 98.0 kJ

35. What is the efficiency of glucose metabolism if 1

mole of glucose gives 38ATP energy? (Given: The enthalpy of combustion of glucose is 686 kcal, 1ATP = 7.3kcal) (1) 100% (2) 38% (3) 62% (4) 80%

36. When XeF6 is partially hydrolysed, it yields (1) XeSO3 (3) XeOF4

(2) XeOF2 (4) XeF2

37. The correct order of increasing reactivity of

CX bond towards nucleophile in the following compounds is:

(CH3)3C—X

    I

II (1) I < II < IV < III (3) IV < III < I < II

(CH2)CH—X

III IV (2) II < III < I < IV (4) III < II < I < IV

38. Compound, which is added to soap to impart antiseptic properties, is: (1) sodium laurylsulphate. (2) sodium dodecylbenzenesulphonate. (3) rosin. (4) bithional.

CH3

H

–OH

(CH2)5CH3 HO

(1)

C CH3

SE2

(3) SN2

(2) SN1 (4) SN0

42. Which of the following statement is true? (1) ATP is a nucleoside made up of nitrogenous base adenine and ribose sugar. (2) ATP consists the nitrogenous base, adenine and the sugar, deoxyribose. (3) ATP is a nucleotide which contains a chain of three phosphate groups bound to ribose sugar. (4) The nitrogenous base of ATP is the actual power source.

43. What is the activation energy for a reaction if its rate doubles when the temperature is raised from 20°C to 35°C?

(H = 8.314 J mol–1 K–1) (1) 342 kJ mol–1

(2) 269 kJ mol–1

(3) 34.7 kJ mol–1

(4) 15.1 kJ mol–1

44. A black compound of manganese reacts with a

halogen acid to give greenish yellow gas. When excess of this gas reacts with NH3 an unstable trihalide is formed. In this process the oxidation state of nitrogen changes from: (1) – 3 to +3. (2) – 3 to 0. (3) – 3 to +5. (4) 0 to – 3.

Sample Question Papers

39

45. A given nitrogen-containing aromatic compound

‘A’ reacts with Sn/HCl, followed by HNO2 to give an unstable compound ‘B’. ‘B’, on treatment with phenol, forms a beautiful coloured compound ‘C’ with the molecular formula C12H10N2O. The structure of compound ‘A’ is:

(1)



(2)



(3)

(4)

46. Match the items of Column I with items of Column II and assign the correct code: Column I

Column II

(A) Pendulum

(1)  Chrome steel

(B) Malachite

(2)  Nickel steel

(C) Calamine

(3) Na3AIF6

(D) Cryolite

(4) CuCO3. Cu(OH)2 (5) ZnCO3

Code: (1) A (1)  B (2)  C (3)  D (4) (2) A (2)  B (4)  C (5)  D (3) (3) A (2)  B (3)  C (3)  D (5) (4) A (4)  B (5)  C (3)  D (2)

Case Based

Read the passage given below and answer the following questions: Raoult’s law states that for a solution of volatile liquids, the partial vapour pressure of each component of the solution is directly proportional to its mole fraction present in solution. Dalton’s law of partial pressure states that the total pressure (ptotal) over the solution phase in the container will be the sum of the partial pressures of the components of the solution and is given as: ptotal = p1 + p2

47. What type of deviation from Raoult’s law does the above graph represent ? (1) First positive then negative (2) Negative deviation (3) Positive deviation (4) First negative then positive

48. In comparison to a 0.01 M solution of glucose, the depression in freezing point of a 0.01 M MgCl2 solution is _____________. (1) the same (2) about twice (3) about three times (4) about six times

49. A solution of two liquids boils at a temperature more

than the boiling point of either of them. What type of deviation will be shown by the solution formed in terms of Raoult’s law ? (1) Negative deviation (2) Positive deviation (3) First positive then negative (4) First negative then positive

50. Which of the following aqueous solutions should have the highest boiling point ? (1) 1.0 M NaOH (2) 1.0 M Na2SO4 (3) 1.0 M NH4NO3 (4) 1.0 M KNO3



SAMPLE

Question Paper Maximum Marks : 200

10

Time : 45 Minutes

General Instructions: (i) This paper consists of 50 MCQs, attempt any 40 out of 50 . (ii) Correct answer or the most appropriate answer: Five marks (+5) . (iii) Any incorrect option marked will be given minus one mark (– 1) . (iv) Unanswered/Marked for Review will be given no mark (0) . (v) If more than one option is found to be correct then Five marks (+5) will be awarded to only those who have marked any of the correct options . (vi) If all options are found to be correct then Five marks (+5) will be awarded to all those who have attempted the question . (vii) If none of the options is found correct or a Question is found to be wrong or a Question is dropped then all candidates who have appeared will be given five marks (+5). (viii) Calculator / any electronic gadgets are not permitted .

1. Coordination number of Ni in [Ni(C2O4)3]4– is:

(1) 3 (2) 6 (3) 4 (4) 2 2. The central dogma of molecular genetics states that the genetic information flows from: (1) Amino acids → Proteins → DNA (2) DNA → Carbohydrates → Proteins (3) DNA → RNA → Proteins (4) DNA → RNA → Carbohydrates 3. Reaction by which benzaldehyde cannot be prepared? CH CH3

(a) (1) (a)





(a)



(b) (2) (b) (b) (c) (c) (c) (3)



3

CH3 + CrO Cl and CS + CrO22Cl22 and CS22 + CrO2Cl 2 and CS2

COCl COCl COCl

+ followed followed by by H H33O O+ followed by H O+ 3

+ CO HCl presence of AlCl 33 + presence of anhy. AlCl +CO CO++ +HCl HClinin in presence of anhy. anhy. AlCl 3 COOH COOH COOH

++ and conc. HClHCl +Zn/Hg Zn/Hg and conc. HCl Zn/Hg and conc.



4. When (NH4)2Cr2O7 is heated, the gas evolved is: (1) N2 (3) O2

6. For carbylamine reaction, we need hot alc. KOH and (1) any primary amine and chloroform (2) chloroform and silver powder (3) a primary amine and an alkyl halide (4) a mono alkyl amine and trichloromethane

7. Match the compounds given in column I with the hybridization and shape given in column II and mark the correct option. Column I

Column II

(A)

XeF6

(i) Distorted octahedral

(B)

XeO3

(C)

XeOF4

+ H in pressure of Pd-BaSO

4 22 in pressure Pd-BaSO 4 ++HH pressure of of Pd-BaSO 4 2 in

(d) (d) (4) (d)

(2) The Df G° of the sulphide is greater than those for CS2 and H2S. (3) The DfG° is negative for roasting of sulphide ore to oxide. (4) Roasting of the sulphide to the oxide is thermodynamically feasible.

(2) NO2 (4) N2O

5. Which of the following statements, about the

advantage of roasting of sulphide ore before reduction is not true? (1) Carbon and hydrogen are suitable reducing agents for metal sulphides.

(ii) Square planar (iii) Pyramidal

(D) XeF4 (iv) Square pyramidal Code: A B C D (1) (iv) (iii) (i) (ii) (2) (iv) (i) (ii) (iii) (3) (i) (iii) (iv) (ii) (4) (i) (ii) (iv) (iii)

8. Among the following complexes, the one which shows zero crystal field stabilization energy (CFSE) is: (1) [Mn(H2O)6]3+ (2) [Fe(H2O)6]3+ 2+ (3) [Co(H2O)6] (4) [Co(H2O)6]3+

41

Sample Question Papers 9. Which of the following pairs of metals is purified by van-Arkel method? (1) Zr and Ti (3) Ni and Fe

(2) Ag and Au (4) Ga and In

10. Which of the following structures represents the peptide chain? H O   | (1)  N  C  N  C  NH  C  NH    | O H

(2)

(3)

H H | | | | | | |  (2)  N  C  C  C  C  N  C  C  C         O H O H H  | | | | |  (3)  N  C  C  N  C  C  N  C  C       O O (4)

H O H | | | | |  | | ||  N  C C  C  N C C  N  C C C    ||   ||   H O

11. Which of the following forms cationic micelles above certain concentration? (1) Sodium ethyl sulphate (2) Sodium acetate (3) Urea (4) Cetyl trimethyl ammonium bromide

12. The transition elements have a general electronic configuration. (1) ns2np6nd1–10 (2) (n – 1)d1–10, ns0–2, np0–6 (3) (n – 1)d1–10, ns1–2 (4) nd1–10, ns2

13. Al2O3 is reduced by electrolysis at low potential

and high currents. If 4.0 × 104 A of current is passed through molten Al2O3 for 6h, what mass of aluminium is produced? [Assume 100% current efficiency, atomic mass of Al = 27 g mol–1] (1) 9.0 × 103 g (2) 8.1 × 104 g 5 (3) 2.4 × 10 g (4) 1.3 × 104 g

14. The rate of reaction between two reactants A and B decreases by a factor of 4, if the concentration of reaction B is doubled. The order of this reaction with respect to reaction B is: (1) – 1 (2) – 2 (3) 1 (4) 2

15. Predict the product:

(1)





(4)

16. Kohlrausch’s law states that at

(1) infinite dilution, each ion makes definite electrolyte, whatever be the nature of the other ion of the electrolyte. (2) infinite dilution each, ion makes definite contribution to equivalent conductance of an electrolyte depending on the nature of the other ion of the electrolyte. (3) infinite dilution, each ion makes definite contribution to conductance of an electrolyte, whatever be the nature of the other ion of the electrolyte. (4) infinite dilution, each ion makes definite contribution to equivalent conductance of an electrolyte, whatever be the nature of the other ion of the electrolyte.

17. For a first-order reaction, the half-life period is independent of (1) initial concentration of the reactant (2) cube root of initial concentration (3) first power of final concentration (4) square root of final concentration

18. In the reaction

–

OH

+

O Na

CHO + CHCl3 + NaOH

the electrophile involved is: ⊕

(1) dichloromethyl cation ( CHCl 2 ) ⊕

(2) formyl cation ( CHO) Θ

(3) dichloromethyl anion ( CHCl 2 ) (4) dichlorocarbene (: CCl2) 19. For the four successive transition elements (Cr, Mn, Fe and Co), the stability of +2 oxidation state will be there in which of the following order ? (Atomic number Cr = 24, Mn = 25, Fe =26, Co = 27) (1) Fe > Mn > Co > Cr (2) Co > Mn > Fe > Cr (3) Cr > Mn > Co > Fe (4) Mn > Fe > Cr > Co

(b) OSWAAL CUET (UG) Sample Question Papers, CHEMISTRY

42

20. Equivalent conductance of NaCl, HCl and

C2H5COONa at infinite dilution are 126.45, 426.16 and 91 Ω–1 cm2, respectively. The equivalent conductance of C2H5COOH is: (1) 201.28 Ω–1 cm2 (2) 390.71 Ω–1 cm2 –1 2 (3) 698.28 Ω cm (4) 540.48 Ω–1 cm2 21. D-(+)-glucose reacts with hydroxylamine and yields an oxime. The structure of the oxime would be: CH = NOH CH = NOH | | H  C  OH HO  C  H | | HO  C  H  H HO C (1) (2) | | HO  C  H H  C  OH | | H  C  OH H  C  OH | | CH 2OH CH 2OH CH = NOH | HO  C  H | H  C  OH (3) (4) | HO  C  H | H  C  OH | CH 2OH 22. Which of the following complex ions is not expected to absorb visible light? (1) [Ni(CN)4]2– (2) [Cr(NH3)6]3+ 2+ (3) [Fe(H2O)6] (4) [Ni(H2O)6]2+ 23. For the reaction, 2A + B → 3C + D

      which of the following does not express the reaction rate?

(1) − (3)

d[C] 3 dt

d[D] dt

(4) 25. Which of the following is an analgesic? (1) Streptomycin (2) Chloromycetin (3) Novalgin (4) Penicillin 26. The volume strength of 1.5 N H2O2 solution is: (1) 4.8 (2) 5.2 (3) 8.4 (4) 8.8

27. Which one of the following is a chain growth polymer? (1) Starch (3) Polystyrene

28. Identify the incorrect statement, regarding the molecule XeO4: (1) XeO4 molecule is square planar (2) there are four pπ - dπ bonds. (3) There are four sp3-p, σ bonds. (4) XeO4 molecule is tetrahedral

29. The method usually employed for the precipitation of a colloidal solution is: (1) dialysis (2) addition of electrolytes (3) diffusion through animal membrane (4) condensation

30. The number of carbon atoms per unit cell of diamond unit cell is: (1) 6 (3) 4



obtained.

(1)

—N

HCl

B

(A) The structure of C would be: (1) (2) (b)

CH3

Cold

(3)



CH3

NaNO2

(2) 1 (4) 8

following reaction

24. In a reaction of aniline a coloured product C was NH2

(2) Nucleic acid (4) Protein

31. The major organic product formed from the

d[B] dt d[A] (4) − 2 dt (2) −



(3)

C

(2)

(4)



32. In which of the following arrangements the given

sequence is not strictly according to the property indicated against it? (1) HF < HCl < HBr < HI: increasing acidic strength (2) H2O < H2S < H2Se < H2Te: increasing pKa values (3) NH3 < PH3 < AsH3 < SbH3: increasing acidic character (4) CO2 < SiO2 < SnO2 < PbO2 increasing acidizing

43

Sample Question Papers 33. Among the following four compounds (i) Phenol       (ii) Methyl phenol (iii) Meta-nitrophenol  (iv) Para-nitrophenol The acidity order is: (1) (iv) > (iii) > (i) > (ii) (2) (iii) > (iv) > (i) > (ii) (2) (i) > (iv) > (iii) > (ii) (4) (ii) > (i) > (iii) > (iv)



temperature. Reason (R): Molarity is dependent on volume of solution. (1) Both A and R are true and R is the correct explanation of A (2) Both A and R are true but R is NOT the correct explanation of A (3) A is true but R is false (4) A is false and R is True

36. Aluminium is extracted from alumina (Al2O3) by electrolysis of a molten mixture of: (1) Al2O3 + HF + NaAlF4 (2) Al2O3 + CaF2 + NaAlF4 (3) Al2O3 + Na3AlF6 + CaF2 (4) Al2O3 + KF + Na3AlF6

O

(d) (4)





O

O

42. In the silver plating of copper, K[Ag(CN)2] is used

instead of AgNO3. The reason is: (1) a thin layer of Ag is formed on Cu. (2) more voltage is required. (3) Ag+ ions are completely removed from solution. (4) less availability of Ag+ ions, as Cu cannot displace Ag from [Ag(CN)2]– ion.

43. All form ideal solution except (1) C6H6 and C6H5CH3 (2) C2H5Cl and C2H5I (3) C6H5Cl and C6H5Br (4) C2H5I and C2H5OH

44. Chloramphenicol is an:

37. Which one of the following monomers gives the polymer neoprene on polymerisation? (1) CH2==CHCl (2) CCl2==CCl2 Cl  (3) CH 2 = = C  CH = = CH 2 (4) CF2==CF2

38. A chemical reaction is catalysed by a catalyst X.

Hence, X (1) reduces enthalpy of the reaction (2) decreases rate constant of the reaction (3) increases activation energy of the reaction (4) does not affect equilibrium constant of the reaction

39. Which of the following compounds is most acidic? OH

(1) ClCH2CH2 OH (2) (b) OH

OH

(c) (3) OH

35. Assertion (A): Molarity of a solution changes with

(d) (4) NO2

(b) (2) OH

(2) 8 (4) 5

(c) (3)

when cyclohexanone undergoes aldol condensation followed by heating? O

(a) (1)

34. The fcc crystal contains how many atoms in each unit cell? (1) 6 (3) 4

41. Out of the following which is the product formed

CH3

40. When Zn converts from melted state to its solid

state, it has hcp structure, then find the number of nearest atoms? (1) 6 (2) 8 (3) 12 (4) 4

(1) antifertility drug (2) antihistamine (3) antiseptic and disinfectant (4) antibiotic-broad spectrum 45. Which is not correct regarding the adsorption of a gas on surface of solid? (1) On increasing temperature adsorption increase continuously. (2) Enthalpy and entropy change is negative. (3) Adsorption is more for some specific substance. (4) Reversible.

46. The complex ion [Co(NH3)6]3+ is formed by sp3d2 hybridisation. Hence, the ion should possess: (1) octahedral geometry (2) tetrahedral geometry (3) square planar geometry (4) tetragonal geometry

Case Based Read the passage given below and answer the following questions: Aryl halides are extremely less reactive towards nucleophilic substitution reactions due to the following reasons: (i) In haloarenes, the electron pairs on halogen atom are in conjugation with π-electrons of the ring. (ii) In haloalkane, the carbon atom attached to halogen is sp3 hybridised while in case of haloarene, the carbon atom attached to halogen is sp2 -hybridised. (iii) In case of haloarenes, the phenyl cation formed as a result of self-ionisation will not be stabilised by resonance.

OSWAAL CUET (UG) Sample Question Papers, CHEMISTRY

44



47. A primary alkyl halide would prefer to undergo

(1)

________. (1) SN1 reaction

(2) SN2 reaction

(3) α-Elimination

(4) Racemisation

48. Which of the following alkyl halides will undergoes SN1 reaction most readily? (1) (CH3)3C—F

(2) (CH3)3C—Cl

(3) (CH3)3C—Br

(4) (CH3)3C—I

49. What is ‘A’ in the following reaction?



(2)

(3)



(4)

50. Reaction of C6H5CH2Br with aqueous sodium

hydroxide follows _______. (1) SN1 mechanism (2) SN2 mechanism (3) Any of the above two depending upon the temperature of reaction (4) Saytzeff rule.

qqq

1

SOLUTIONS OF Question Paper 1. Option (2) is correct.

8. Option (3) is correct.

Explanation: At low temperature, substance exists in solid state. It is due to the decrease in molecular movement which leads to strong cohesive force, that is, the force which tightly holds the constituent particles together.

2. Option (3) is correct. Explanation:



0.059 [ Mg ] log n [Cu 2 + ]

= 2.71V −

0.059 0.1 log 2 0.001

= 2.71V −

0.059 log 10 2 2

Ecell = 2.651 V

3. Option (3) is correct. Explanation: Each point in a lattice is known as lattice point which can be either atom, molecule or ion. It is joined together by a straight line to bring out geometry of lattice in pure crystal constituents. They are arranged in fixed stoichiometric ratio. Hence, existences of free electrons are not possible.

4. Option (3) is correct. Explanation: Tm (Thulium) is a lanthanoid.

5. Option (3) is correct.

1

5

2 3

4 IUPAC name: 3-bromo-1-chlorocyclohexene

6. Option (1) is correct. Explanation: In Frenkel defect, one of the ion is missing from its lattice site and occupies an interstitial site. So, density of the crystal does not change.

9. Option (1) is correct. Explanation: Ag+(aq) + e– ® Ag (s); E° = + 0.80 V. 1 H+(aq) + e– ® H 2 (g) ; E° = 0.00 V. 2 n the basis of their standard reduction potential O (E°) values, cathode reaction is given by the one with higher E° values. Thus, Ag+(aq) + e– ® Ag(s) reaction will be more feasible at cathode.

10. Option (2) is correct. Explanation: It is formed by the loss of 3 electrons, the configuration of element X is [Ar] 3d64s2. Therefore, Atomic number = 26.

11. Option (1) is correct. Explanation: CH3 3

Explanation:

6

[Co(NH3)4Br2]Cl2 → [Co(NH3)4Cl2]+ + 2Cl−

7. Option (2) is correct. Explanation: Weakest acid has the strongest conjugate base. Since R-OH is the weakest acid, therefore, QOR is the strongest base.

2 1 OH (i) As -OH is functional group and –CH3 is substituent. (ii) IUPAC name: 3-methylphenol.

12. Option (4) is correct. Explanation: The molality of a solution does not change with temperature.

13. Option (2) is correct. Explanation:

O + HCN

—



Ecell = E°cell −

2+

Explanation: Ionization isomers have identical central ion and the other ligands except for a ligand that has exchanged places with an anion or neutral molecule that was originally outside the coordination complex. [Co(NH3)4Cl2]Br2 → [Co(NH3)4Cl2]+ + 2Br−

OH CN

—

It is a nucleophilic addition reaction.

OSWAAL CUET (UG) Sample Question Papers, CHEMISTRY

46

14. Option (3) is correct.

20. Option (3) is correct.

Explanation: Solvate / Hydrate isomerism Solvate or hydrate isomers have the same composition but differ with respect to the number of solvent ligand molecules as well as the counter ion in the crystal lattice.

15. Option (3) is correct.

Explanation: The electron withdrawing group (–NO2) increases the acid strength of aromatic acids while electron releasing group (–CH3) decreases the acid strength of aromatic acids. Hence, the increasing order of acid strength is given as

Explanation: There is no effect on heat evolved or absorbed during the reaction in the presence of a catalyst. It is because catalyst influence the rate of reaction and does not participate in the reaction.

21. Option (1) is correct.

16. Option (1) is correct. Explanation: When a primary aromatic amine is dissolved or suspended in cold aqueous mineral acid and treated with sodium nitrite, a diazonium salt is formed. When this freshly prepared diazonium salt is mixed with cuprous chloride, diazonium group is replaced by Cl. Then, chlorobenzene is formed which is Y in this reaction. + – N2Cl

NH2 NaNO2+HCl 273–278 K

Cl Cu2Cl2 (Sandmeyer reaction)

Explanation: Compound (a) is enantiomer of compound (1) because the configuration of two groups, that is, CH3 and C2H5 in them is reversed at chiral carbon.

22. Option (3) is correct. Explanation: We know that the stability of a complex increases by chelation. Therefore, the most stable complex is [Fe(C2O4)3]3−.

+N2 (Y)

17. Option (3) is correct.

c

o

o

c

o o

Then,

Explanation: Let oxidation state of Fe = x x + 3(–2) = –3 x – 6 = –3 x = +3

o

c

18. Option (3) is correct.

o

c

o

o

c

o

o

c o

Fe

o

o

c

c

Explanation: Bond angle in ether is slightly more than the tetrahedral angle due to repulsion between the two bulky alkyl groups.

19. Option (1) is correct.

o

o

o

23. Option (3) is correct.

Explanation: Adsorption is an exothermic process, so the ΔH of adsorption is always negative.

Explanation: According to Henry’s law,



ΔH < 0





ΔG = ΔH – TΔS





ΔG = Change in Gibbs free energy



C = Concentration of gas



ΔH = Change in enthalpy





T = Temperature in Kelvin

KH = Henry’s constant



ΔG = Change in entropy

It implies that as the value of KH increases, mole fraction of gas solute in solvent decreases.

Since, adsorption is a spontaneous process, the thermodynamic requirement is at constant temperature and pressure, ΔG must be negative. So, the enthalpy ΔH as well as entropy ΔS of the system is negative.



P = KH C KH ∝

1 C

Where P = Partial pressure of gas

Hence, higher the KH value, lower is the solubility of gas. The order of increasing solubility of gases in :

Ar < CO2 < CH4 < HCHO

Solutions 24. Option (3) is correct.

32. Option (1) is correct.

Explanation: CH3CONH2 + Br2 + 4NaOH Acetamide

Explanation: The rate of a reaction depends upon the concentration of reactants.

∆ CH 3CONH 2 + Br 2 + 4NaOH  → CH 3 NH 2 + 2NaBr Acetamide Methylamine + Na 2CO3 + 2H 2O

25. Option (3) is correct.

33. Option (2) is correct. Explanation: Primary and secondary amines form hydrogen bonds and hence are less volatile than corresponding alkanes.

34. Option (4) is correct.

Explanation: The isomers, which differ only in the configuration of the hydroxyl group at C-1, are called anomers and are referred to as a- and b-forms.

26. Option (4) is correct. Explanation: Physical adsorption is favoured at low temperature.

27. Option (3) is correct. Explanation: AgNO3 + [Cr(H2O)6]Cl3 → AgCl

Explanation: It is called as sol.

35. Option (1) is correct. Explanation: Copper does not liberate hydrogen from acids because copper lies below hydrogen in electrochemical series. So, copper does not have sufficient electrode potential to liberate elemental hydrogen form compounds in which oxidation state of hydrogen is +1.

36. Option (2) is correct. −

+ NO3 Since Cl is outside the coordination sphere, it can react with AgNO3 forming the white AgCl precipitate.

28. Option (4) is correct.

Explanation: Butan-2-ol is secondary alcohol which on oxidation with alkaline KMnO4 solution gives butanone (ketone). CH3CH2 CHOH + [O] CH3 Butan-2-ol

Explanation: Neoprene is an addition polymer.

alk.KMnO4

29. Option (3) is correct.

CH3CH2 CH3

Explanation: For first order reaction Rate1 = k[ A 1 ] [ A 2 ] = [2 A1 ] Rate 2 = k[ 2 A 1 ] Rate 2 = k × 2 Rate1 For a given reaction, rate constant is constant and independent of the concentration of reactant.

30. Option (1) is correct. Explanation: CO is a neutral ligand and its oxidation state is zero. Since the overall charge on the complex is zero too, hence oxidation state of Ni is 0.

31. Option (2) is correct.

C=O + H2O Butanone

37. Option (1) is correct. Explanation: Value of Henry’s constant increases with increase in temperature.

38. Option (4) is correct. Explanation: Graphite produces impurity in pig iron.

39. Option (2) is correct. Explanation: Antioxidant helps in food preservation by retarding the action of oxygen on food.

40. Option (3) is correct.

Explanation:

NO2

47

H2(excess)/Pt

NH2

Fe/HCl

NH2

Sn/HCl LiAlH4/ether

NH2 No reaction

Explanation: When c ® 0

Then ∧ = ∧o

41. Option (4) is correct. Explanation: Lysine whose structure formula is written as: (i) It is an a-amino acid.

48

OSWAAL CUET (UG) Sample Question Papers, CHEMISTRY (ii) It is basic amino acid because number of

44. Option (2) is correct. Explanation: In case of elementary reaction, the rate law can be determined from balanced chemical equation.

NH2 groups (2) is greater than number of COOH group. (iii) It is non-essential amino acid because it is synthesized in our body.

45. Option (1) is correct. Explanation: Glucose and fructose differ structurally and stereochemically. They have same molecular formula i.e., C6H12O6. Hence, these are isomers of each other.

42. Option (2) is correct. Explanation: Biodegradable polymer can be formed by H2N—CH2—COOH and H2N— (CH2)5—COOH.

46. Option (4) is correct. Explanation: Cassiterite is an ore of Sn with chemical composition SnO2.

43. Option (4) is correct. Explanation:

N-acetyl-para-aminophenol

i.e.,

paracetamol is an antipyretic which can also be used as an analgesic to relieve pains.

47. 48. 49. 50.

Case Based Option (2) is correct. Option (2) is correct. Option (3) is correct. Option (3) is correct.

qqq

2

SOLUTIONS OF Question Paper 1. Option (3) is correct. Explanation: A molar solution is one that contains one mole of a solute in one litre of the solution.



H H H H HOHO H H H H

Number of moles of solute Molarity ( M ) =   Volume of solution in L

2. Option (2) is correct. Explanation: As the reactant A’s concentration decreases with time, so the product B’s concentration increases. Also since the reaction is reversible, the increase and decrease in concentration with respect to time is similar.

3. Option (1) is correct. Explanation: In the extraction of copper from its sulphide ore, it is subjected to roasting. Some of it is oxidized to Cu2O which reacts with the remaining Cu2S to give copper metal. Cu2S + 2Cu2O → 6Cu + SO2 ↑ In this process, Cu2S behaves as a reducing agent.

4. Option (2) is correct. Explanation: In ionic solids, smaller ions occupy the voids, and this depends on stoichiometry of the compounds not on the radius of ions.

5. Option (1) is correct. Explanation: Only types of interactions between particles of noble gases are due to weak dispersion forces.

[Co(NH3)5Cl]Cl2 → [CO(NH3)5Cl]+(aq) + 2Cl−

7. Option (2) is correct. Explanation: Water on calcium chloride is an example of absorption, rest of all are examples of adsorption.

8. Option (1) is correct. Explanation: Cyclic structures of monosaccharides which differ in structure at carbon-1 are known as anomers.

CH2OH CH2OH

H H OHOH H O H O OHOH

CH2OH CH2OH

   (I)    (II) Structure (I) and (II) differ in structure at C-1.

9. Option (4) is correct. Explanation: When KMnO4 solution is added to oxalic acid solution, the decolourisation is slow in the beginning but becomes instantaneous after sometime because Mn2+ acts as an auto-catalyst. Reduction half-reaction: [MnO4− + 8 H+ + 5e − → Mn 2 + + 4 H 2O] × 2 Oxidation half-reaction: −

2 − [C 2O4 → 2CO2 + 2e ] × 5 Overall equation:

2 MnO4− + 16 H + + 5C 2O24 − → 2 Mn 2 + + 10CO2 + 8 H 2O

End point of this reaction: Colourless to light pink.

10. Option (2) is correct. Explanation: The correct order of the packing efficiency in different types of unit cells is given below: Unit Cell

6. Option (3) is correct. Explanation:

OHOH HOHO OHOH H H O HO H O H HO OHOH H H H H



Packing efficiency

fcc

74%

bcc

68%

Simple cubic

52.4%

fcc > bcc > simple cubic

11. Option (2) is correct. Explanation: R − X + NaOH → R − OH + NaX

12. Option (2) is correct. Explanation: Zone refining of metals is based on the principle that the impurities are soluble to greater extent in molten state than in the solid.

13. Option (1) is correct. Explanation: [Co(ONO)3 (NH3)3]

50

OSWAAL CUET (UG) Sample Question Papers, CHEMISTRY xA + yB → Product

14. Option (1) is correct. Explanation: NaCl unit cell has a fcc structure of Cl– ions, and Na+ ions occupy octahedral voids. The radius ratio of 0.524 for NaCl suggests an octahedral void.

r = k[A]x [B]y dx = k[A]x [B]y dt mol L−1s−1 = k(mol L−1 )x (mol L−1 )y k=

15. Option (2) is correct. Explanation: Most common oxidation state for lanthanoids is +3.

16. Option (1) is correct. Explanation: CH3 - CH 2 - C H - CH3 | OH Butan-2-ol is chiral in nature as it possesses chiral center.

17. Option (3) is correct. Explanation: Covalent solids are generally hard, act as insulators and melting points of such solids are extremely high.

18. Option (4) is correct. Explanation: The carboxylic acids having a-hydrogen atom undergo HVZ reaction. Since (CH3)3CCOOH does not contain a-H-atom; so, it does not undergo HVZ reaction.

19. Option (1) is correct.

20. Option (4) is correct.

23. Option (2) is correct. Explanation: Nylon 6, 6 is an example of fibers.

24. Option (1) is correct. Explanation: Physisorption is not specific to any gas since it involves van der Waal’s forces and no specific bonds formation takes place.

25. Option (2) is correct. Explanation: Valium is a tranquilizer.

26. Option (4) is correct. Explanation: Natural rubber is cis-1, 3 polyisoprene and has only cis-configuration about the double bond. CH2— C==C H

H3C cis-polyisoprene (Natural rubber)

n

27. Option (3) is correct. Explanation: HNO3 makes iron passive and its passivity is attained by formation of a thin film of oxide on iron.

28. Option (4) is correct.

Explanation: 3

= (mol L−1 ) 1 − ( x + y ) s−1 here (x + y) = order of the reaction W

—H2C

Explanation: When freshly prepared solution of ferrous sulphate (FeSO4) is added in a solution containing NO3− ion, formation of a browncoloured complex will take place. This is called as brown ring test of nitrate. Hence, two moles of ammonia will produce two moles of NO. NO3− + 3Fe2+ + 4H+ ® NO + 3Fe3+ + 2H2O [Fe(H2O)6]2+ + NO ® [Fe(H2O)5(NO)]2+ + H2O     Brown ring

4

mol L−1s−1 (mol L−1 )x (mol L−1 )y

2

1

CH CH CH 2 −2−CH CH 2 −2−CH CH 2 2−−N N 3 −3−CH

CH3

Explanation: Gammexane is an isomeric form of benzene hexachloride (BHC). Cl

CH3

Cl

IUPAC name: N, N – Dimethyl-butan-1-amine

21. Option (4) is correct. Explanation: (CH3)3C—I will undergo SN1 reaction most readily as C—I bond is weakest, due to the large difference in the size of carbon and iodine.

22. Option (3) is correct. Explanation: For the reaction, xA + yB → Product

r = k[A]x [B]y dx = k[A]x [B]y dt mol L−1s−1 = k(mol L−1 )x (mol L−1 )y

Cl Cl

Cl Cl BHC

29. Option (3) is correct. Explanation:

0.4 ( 0.6 ) = 0.1 ( 0.3)y

y

30. Option (3) is correct. Explanation: Purification of aluminium by electrolytic refining is known as Hoope’s process. In this process, a cell is used which consists of three layers. In this cell, pure Al acts as cathode and anode is made up of impure Al.

31. Option (3) is correct. Explanation: Hinsberg’s reagent which is used to test amines is benzene sulphonyl chloride. SO2 Cl

32. Option (1) is correct. Explanation: Acrilan is an addition polymer of acrylonitrile.

nCH 2 == CH  CN → [CH 2  CH] n  CN

33. Option (3) is correct. Explanation: In equilibrium state, the rate of dissolution of a solid solute in a volatile liquid solvent is equal to the rate of crystallization.

34. Option (2) is correct. Explanation: Asymmetric/Chiral carbon atom is that in which all of its four valencies lie with four different groups or atoms. In molecules (i), (ii) and (iii), all have asymmetric carbon as each carbon has satisfied all four valencies with four different groups or atoms. In molecule (iv), carbon satisfies two of its valencies with two hydrogen atoms i.e., similar atom. So, it is not an asymmetric carbon atom.

85. Option (2) is correct. Explanation: Williamson synthesis is used to obtain ether. For example, R – X + Na – O – R Alkyl halide

Sodium alkoxide

R – O – R' + NaX Ether

Explanation: Suppose order with respect to A and B are x and y respectively. Rate = k[A]x[B]y For experiment 1, 0.1 = k(0.3)x(0.3)y…(i) For experiment 2, 0.4 = k(0.3)x(0.6)y…(ii) For experiment 3, 0.2 = k(0.6)x(0.3)y…(iii) Dividing equation (ii) by (i)

0.4 ( 0.6 ) = 0.1 ( 0.3)y

y

y=2

51

y=2 ∴ Dividing equation (iii) by (i)

0.2 ( 0.6 ) = 0.1 ( 0.3)y

y

∴ x=1 Rate law Rate = k[A][B]2

37. Option (1) is correct. Explanation: Methyl benzaldehyde < Benzaldehyde < Propanone < Ethanal – reactivity towards nucleophilic substitution. Aldehydes are more reactive than aliphatic ketones. Aliphatic ketones are more reactive than aromatic ketones. The +I effect is more in ketone than in aldehyde. Thus, ketone will be least reactive in nucleophilic addition reactions. The presence of electron withdrawing group increases the reactivity towards the addition while the presence of electron donating group decreases the reactivity of compound towards nucleophilic addition. Benzaldehyde does not favour nucleophilic addition reaction due to resonance stabilisation.

38. Option (3) is correct. Explanation: V3+, V2+, Fe3+ ions exhibit specific colours. Atomic number of V = 23, Electronic configuration of V - [Ar]3d3 4s2 Electronic configuration of V2+ - [Ar]3d3 Electronic configuration of V3+ - [Ar] 3d2 Atomic number of Fe = 26 Electronic configuration of Fe - [Ar]3d6 4s2 Electronic configuration of Fe3+ - [Ar]3d5 Since these ions have partially filled d- subshells, they exhibit colour. Most transition-metal ions have a partially filled d subshell. As for other ions, Atomic number of Sc = 21

36. Option (2) is correct.



Solutions

Electronic configuration of Sc - [Ar]3d1 4s2 Electronic configuration of Sc3+ - [Ar]3d0 Since d subshell is empty, it shows no colour. Atomic number of Ti = 22 Electronic configuration of Ti- [Ar]3d2 4s2 Electronic configuration of Ti4+ - [Ar]3d0 Since d subshell is empty, it shows no colour. Atomic number of Mn =25 Electronic configuration of Mn- [Ar]3d5 4s2 Electronic configuration of Mn2+ [Ar]3d5 Since d subshell is partially filled, it shows colour.

52

OSWAAL CUET (UG) Sample Question Papers, CHEMISTRY 44. Option (4) is correct.

Atomic number of Ni = 28 Electronic configuration of Ni- [Ar]3d8 4s2 2+

Electronic configuration of Ni

[Ar]3d

Explanation: In H2S2O6, S-S linkage is present.

8

Since d subshell is partially filled, it shows colour. Atomic number of Zn =30 Electronic configuration of Zn- [Ar]3d10 4s2 Electronic configuration of Zn2+- [Ar]3d10 Since d subshell is full, it shows no colour.

39. Option (4) is correct.

Explanation: According to Raoult’s law, for a dilute solution, the relative lowering of vapour pressure is equal to the mole fraction of solute.

PA 00 − PA PA −0 PA PA 0 P Where 0 A PA 0 − PA PA −0 PA PA 0 PA XB  XB

Explanation:

Explanation: Lyophobic sol can be protected by adding lyophilic sol which is known as protective colloid.

= = = =

Relative lowering of vapour pressure Relative lowering of vapour pressure mole fraction of solute mole fraction of solute

Explanation: The lanthanoid contraction is due to poor shielding effect of 4f electrons.

Case Based 47. Option (3) is correct.

41. Option (1) is correct. Explanation: Clemmensen reduction is used to convert carbonyl group to CH2 group as follows: C=O

= XB = XB

46. Option (4) is correct.

40. Option (3) is correct.

Zn(Hg) + HCl

CH2

42. Option (4) is correct. Explanation: Curdling of milk is an example of denaturation of milk proteins.

43. Option (3) is correct. Explanation: CFSE for tetrahedral complex is 4 ∆t = ∆0 9

∆t =

45. Option (2) is correct.

4 × 18,000 = 8,000 cm −1 9

Explanation: Conductivity of 0.02 mol L–1 KCl Solution = Cell constant resistance 258 = 420 = 0.614 Sm −1

48. Option (4) is correct. Explanation: The conductivity decreases with dilution.

49. Option (4) is correct. Explanation: The cell constant of a conductivity cell remains constant for a cell.

50. Option (1) is correct. Explanation: SI unit for conductivity of a solution is S m–1.

qqq

SOLUTIONS OF Question Paper 1. Option (3) is correct. Explanation: a-helix structure of protein is stabilised by hydrogen bonds. A polypeptide chain forms all possible hydrogen bonds by twisting into right-handed helix with the –NH group of each amino acid residue hydrogen bonded to >C=O of an adjacent turn of helix.

2. Option (2) is correct. Explanation: The greater the value of log K, the greater will be stability of complex compound formed. For reaction, log K has the highest value. Cu 2 + + 4CN − → Cu ( CN )4 

2+

2+

Cu ( CN )4  and log K = 27.3 K=  4 Cu 2 +  CN − 

3. Option (1) is correct. Explanation: Formic acid (HCOOH) does not form anhydride because it does not contain a-C-atom.

4. Option (2) is correct. Explanation: Actinoids are 5f block elements so in actinoids, 5f orbitals are progressively filled.

5. Option (1) is correct. Explanation:

Cl  is of The structure [CH 2  CH = = C  CH 2 ] n neoprene.

It does not contain −I or +I group.

+I effect of CH3 group increases the electron density on the nitrogen atom of −NH2 group, so, it undergo protonation easily and hence, it is most basic amine.

7. Option (3) is correct. Explanation: ∆o values follow the order: [Co(H 2O)6 ]3+ < [Co(NH 3 )6 ]3+ < [Co(CN)6 ]3− and therefore, absorption wavelength follows the order: [Co(H 2O)6 ]3+>[Co(NH 3 )6 ]3+>[Co(CN)6 ]3−

8. Option (3) is correct. Explanation: Zone refining is used for metals which are required in very high purity. Semiconductor grade silicon is purified by this method.

9. Option (1) is correct. Explanation: Density increases with increase in molecular mass.

10. Option (4) is correct. Explanation: Higher critical temperatures indicate easily liquefiable gases which are readily adsorbed as the van der Waal’s forces responsible for adsorption of gases on solid surfaces are stronger near critical temperatures.

11. Option (4) is correct. Explanation: The reaction will be 100% complete only after infinite time.

6. Option (4) is correct. Explanation: The increasing order of basic strength is given as below:


1° as the stability of carbocations is of the order 3°>2°>1°.

7. Option (4) is correct. Explanation: Gas-gas interface cannot be obtained because they are completely miscible in nature. For example: Air is a mixture of various gases such as, O2, N2, CO2, etc.

8. Option (1) is correct. Explanation: These linkages are present between 5’ and 3’ of pentose sugars of nucleotides.

10. Option (3) is correct. Explanation: In the preparation of compounds of Xe, Bartlett had taken O2+PtF6– as a base compound. This is because both O2 and Xe have almost same ionisation enthalpy.

11. Option (2) is correct. Explanation: The sharp melting point of crystalline solids is due to a regular arrangement of constituent particles observed over a long distance in the crystal lattice.

12. Option (4) is correct. Explanation: As tertiary carbocation is more stable, so tertiary alcohols will yield the corresponding alkyl chloride on reaction with concentrated HCl at room temperature. While primary and secondary alcohols require the presence of a catalyst ZnCl2.

13. Option (4) is correct. Explanation: X → Y Rate(r) ∝ [X]n [Where n = Order of reaction] If the concentration X is increased by 4 times X’ = 4X Then, Rate(r’) ∝ [X’]n r ' [ 4X ] = = 2 n  r [X] r’ is new rate, X’ is a new concentration [4]n = 2 1 ∴n = 2 1 Order of reaction = 2 n

14. Option (3) is correct. Explanation: pp-dp bonding is present in phosphorus due to the presence of vacant d-orbitals and in carbon (C), nitrogen (N) and boron (B) do not have d orbitals.

58

OSWAAL CUET (UG) Sample Question Papers, CHEMISTRY

15. Option (1) is correct. Explanation: Colligative property depends on the number of solute particles and not on the nature of the particles.

16. Option (4) is correct. Explanation: Galena (PbS), copper pyrites (CuFeS2) and argentite (Ag2S) are concentrated by froth floatation process but sphalerite (ZnS) is concentrated by chemical leaching.

17. Option (1) is correct. Explanation: Gabriel phthalamide synthesis cannot be used for preparation of aromatic amines, as aromatic halides do not undergo nucleophilic substitution by salt formed by phthalamide.

18. Option (3) is correct. Explanation: La(OH)3 is more basic than Lu(OH)3. It is because of the fact that due to lanthanoid contraction the size of lanthanoid ion decreases regularly with increase in atomic size. Thus covalent character between lanthanoid ion and OH− increases from La3+ to Lu3+. Thus the basic character of hydroxides decreases from La(OH)3 to Lu(OH)3.

19. Option (1) is correct. Explanation: Ampicillin is a modification of penicillin and not a natural antibiotic. These semisynthetic penicillin (SSP) like ampicillin, cloxacillin, etc., are produced by chemically combining specific side chains (in place of benzyl side chain of penicillin group) or by incorporating specific precursors in the mould cultures.

20. Option (2) is correct. Explanation: Tyndall effect is a characteristic of colloidal solution in which colloidal particles show a coloured appearance when sunlight is passes through it and seen perpendicularly.

flow through the cell and if there is any further increase in the external potential (Eexternal), then reaction starts functioning in opposite direction i.e., an electrochemical cell behaves like an electrolytic cell. Eexternal > Ecell

23. Option (4) is correct. Explanation: Equanil is used for the treatment of stress mild and severe mental disease i.e., as a tranquilizer.

24. Option (1) is correct. Explanation: 64Gd : [Xe] 4f7 5d1 6s2

25. Option (4) is correct. Explanation: Amorphous solids do not possess a long-range order in the arrangement of their particles because their formation involves rapid cooling.

26. Option (1) is correct. Explanation: Halogen exchange reactions are those in which one halide replaces another. This reaction is known as Finkelstein reaction.

27. Option (2) is correct. Explanation: Rate constant of a pseudo-firstorder reaction depends on the concentration of reactants present in excess.

28. Option (2) is correct. Explanation: Here compound X is CaCO3 ∆

22. Option (3) is correct. Explanation: If an external opposite potential is applied on the electrochemical cell, the reaction continues to take place till the opposite voltage reaches the value 1.1V. At this stage, no current

Residue

X

  

CaO + H 2O → Ca(OH)2 Y Ca(OH)2 + CO2 + H 2O → Ca(HCO3 )2 Y

Excess

Z ∆

Ca(HCO3 )2 → CaCO3 + H 2O + CO2 ↑

21. Option (3) is correct. Explanation: Compound (A) i.e., phenol and compound (D), that is, a derivative of phenol cannot be considered as aromatic alcohol. As phenol is also known as carbolic acid and cannot be considered as aromatic alcohol. In compound (B) and (C), –OH group is bonded to sp3 hybridized carbon which in turn is bonded to benzene ring.

CaCO3  → CaO + CO2 ↑

Z

X

29. Option (2) is correct. Explanation: Oxidation half reaction:

Cu(s) → Cu 2 + (aq) + 2e − Reduction half reaction:

Ag + (aq) + e − → Ag(s)



Cu(s) | Cu2+(aq) || Ag+ (aq) | Ag(s)



Anode

(Oxidation)

salt bridge

Cathode (Reduction)

Solutions 30. Option (2) is correct.

37. Option (4) is correct.

Explanation: Guanine (G) is the complementary base of cytosine (C) in one stand to that in other stand of DNA. C º G

31. Option (1) is correct. Explanation: Uranium has an configuration of [Rn] 5f3 6d1 7s2.

electronic

32. Option (1) is correct. Explanation: Boiling point increases with increase in molecular mass of the alcohols. Among isomeric alcohols 1o alcohols have higher boiling point than 2o alcohols. Thus, correct order is: Propan-1-ol < Butan-2-ol < Butan-1-ol < pentan-1-ol

Explanation: The coagulating power of Al3+ is highest due to charge and small size, therefore, aluminium chloride will be required in minimum amount to coagulate negatively charged sol of As2S3.

34. Option (1) is correct.

Where [R]0 = initial concentration of reactant k = Rate constant

38. Option (2) is correct. Explanation: Weakest acid has the strongest conjugate base. Since R-OH is the weakest acid, therefore, -OR is the strongest base.

39. Option (4) is correct. Explanation: Nylon 6, 6 is a condensation polymer. nHOOC—(CH2)4—COOH  

 

Adipic acid

+ nH2N—(CH2)6—NH2

  Hexamethylene diamine

H H     ∆ , − H2O         N (CH ) N C (CH ) C → 2 6 2 4     O O  n

40. Option (1) is correct.

Explanation: Neoprene is — CH 2  C == CH  CH 2 —  Cl n

Explanation: CuSO4  Cu 2 + + SO2-4

41. Option (2) is correct.

42. Option (1) is correct.

H 2O  H+ + OH− At cathode, Cu 2 + + 2e− → Cu ;

EΘCell = 0.34 V

1 H2 ; 2

EΘCell = 0.00 V

This reaction will take place due to higher reduction potential. At anode, 2SO24 − + 2e − → S 2O82 − ;

EΘCell = 1.96 V

2H 2O → O2 + 4 H + + 4 e −

EΘCell = 1.23 V

The reaction with lower value of E° will be preferred at anode, hence O2 is released at anode.

36. Option (4) is correct. 2

Explanation: Gadolinium has an electronic configuration of [Xe] 4f 75d16s2.

Explanation: An increase in temperature increase the volume of solution and therefore it will result in its molarity to decrease.

35. Option (3) is correct.

3

Explanation: Half life period of a zero order [R]0 reaction = 2k



33. Option (3) is correct.

H+ + e− →

59

1

Explanation: CH2 = CHCH2 NHCH3 IUPAC name: N-methylprop-2-en-1-amine

Explanation: Phenol being more acidic reacts with sodium hydroxide solution in water to give sodium phenoxide which is resonance stabilized. Alcohols are very weak acids. C 6 H 5OH + NaOH → C 6 H 5ONa + H 2O

43. Option (2) is correct. Explanation: In mRNA A::: U T::: A So, the complementary bases of AAT in mRNA is UUA.

44. Option (2) is correct. Explanation: During electrolysis NaCl → Na + + Cl − H 2O → H + + OH −

( (

) )

Na + + e − → Na EΘCell = −2.71V 1 H + + e − → H 2 EΘCell = 0.00 V 2

) = 0.00 V )

Na + + e − → Na EΘCell = −2.71V H+ + e−

2

Θ Cell

At cathode, 1 H 2 + OH − 2 At anode, two reactions are possib ble. 1 Θ = 1.36 V Cl − → Cl 2 + e − ; ECell 2 Θ 2 H 2O → O2 + 4 H + + 4 e − ; ECell = 1.23 V H 2O + e − →

45. Option (2) is correct. Explanation: Gabriel phthalimide synthesis is used to get primary amines from alkyl halides without changing the number of carbon atoms. O O O KOH (alc) NH NK +R – X O O

—

( 1 → H (E 2

(UG) Sample Question Papers, CHEMISTRY

—

H 2O → H + + OH −

— —

NaCl → Na + +CUET Cl − OSWAAL

—

60

Primary amine

46. Option (4) is correct.

A B A B A

Explanation: Hexagonal close packing (hcp) can be arranged by two layers A and B one over another which can be diagrammatically represented as: I n the above figure the first layer and fourth are not exactly aligned. Thus, statement (4) is not correct while other statements, i.e., (1), (2) and (3) are true.

Case Based 47. Option (3) is correct. Explanation: [Co(H2O)6]3+ > [Co(NH3)6]3+> [Co(CN)6]3−

48. Option (3) is correct. Explanation: 8,000 cm−1

49. Option (2) is correct. Explanation: [Co(H2O)6]2+ is transformed into [CoCl4]2−.

50. Option (1) is correct. Explanation: Thiosulphato is a monodentate ligand whereas oxalato, glycinato and ethylene diamine are bidentate ligands and can form rings with the central metal ion. So, they are also chelating ligands. Thiosulphato is a monodentate ligand and hence, cannot form chelate rings. Hence, it is not a chelating ligand.

qqq

SOLUTIONS OF Question Paper 1. Option (2) is correct. Explanation: Triethylamine [(C2H5)3N] is a 3° or tertiary amine as nitrogen atom contains three ethyl groups.

2. Option (2) is correct. Explanation: Sodium acetate forms cationic micelles. In molecules of soap and detergent the –ve ion aggregate to form a micelle of colloidal size. The negative ion has a long hydrocarbon chain and a polar group COO– at one end.

3. Option (1) is correct. Explanation: Tetraammineaquachloridocobalt (III) chloride is [Co(NH3)4(H2O)Cl]Cl2. Central atom – Cobalt(III) Coordination sphere ligandsTetraammine - 4 NH3 groups neutral ligand Aqua – 1 H2O groups neutral ligand Chlorido -1 Cl group, negatively charged ligand, one negative charge Counter ion -2 Chloride ions Since Cobalt is 3+, one valency is satisfied with Cl in coordination sphere and 2 by chlorine counter ions. Hence, the formula of the coordination compound is [Co(NH3)4(H2O)Cl]Cl2.

4. Option (1) is correct. Explanation: N−N single bond is weaker than P−P bond due to smaller size of N as compared to P. Smaller size of N leads to smaller N−N bond length. Because of larger size of P atom, P−P bond length is more and lone pair-lone pair repulsion between P atoms is less which makes the P−P bond stronger than N−N bond.

5

6. Option (3) is correct. Explanation: Calcium is obtained by electrolysis of molten anhydrous calcium chloride.

7. Option (3) is correct. Explanation: For the coagulating of positively charged hydrated ferric sol, the coagulating process of the anions are in the order: PO43– > SO42– > NO3–

8. Option (3) is correct. Explanation:



5-chlorohexan-3-ol

9. Option (1) is correct. Explanation: F being smallest has the shortest H-F bond and therefore HF has the highest bond dissociation energy.

10. Option (3) is correct. Explanation: Boiling points of isomeric haloalkane decrease with increase in branching as with increase in branching surface area decreases which leads to decrease in intermolecular forces.

11. Option (2) is correct. Explanation:

5. Option (2) is correct. Explanation: In o-nitrophenol, nitro group is present at ortho position. Presence of electron withdrawing group at ortho position increases the acidic strength. On the other hand, in o-methylphenol and in o-methoxyphenol electron releasing group (–CH3 or –OCH3), at ortho or para positions of phenol decreases the acidic strength of phenols. So, phenol is less acidic than o-nitrophenol.

12. Option (3) is correct. Explanation: Conductivity depends upon solvation of ions present in solution. Greater the solvation of ions of an electrolyte, lesser will be the electrical conductivity of the solution.

13. Option (3) is correct. Explanation: Natural rubber is a cis-polyisoprene.

62

OSWAAL CUET (UG) Sample Question Papers, CHEMISTRY

14. Option (2) is correct. Explanation: 1.0 M Na2SO4, Since it furnishes maximum number of ions (2Na+ + SO42-).

15. Option (3) is correct.

Number of unpaired electron = 0 Magnetic property = Diamagnetic

24. Option (2) is correct. Explanation:

Explanation: C6H5OCH3 IUPAC name: Methoxybenzene

16. Option (2) is correct. Explanation: Coordination number is the number of ligands joined to the central metal ion or atom. Since ethylenediamine is a bidentate ligand, Co has coordination number of 6.

17. Option (4) is correct. Explanation: Graphite is a conducting solid, network or covalent solid but it cannot be classified as ionic solid.

18. Option (3) is correct. Explanation: Methylamine reacts with HNO2 to form CH3OH.

25. Option (2) is correct. Explanation: E Mg 2+ / Mg = E0 Mg 2+ / Mg +

0.059 log  Mg 2 +  2

Compare this equation with the equation of straight line y = mx + c. The graph of E Mg 2+ /Mg vs. log [Mg2+] is a straight line with a positive slope and intercept

E Mg 2+ /Mg .

26. Option (3) is correct. Explanation: Aspirin is acetyl salicylic acid which is formed by acetylation of ortho-hydroxy benzoic acid. OH

19. Option (4) is correct. Explanation: The reaction will be 100% complete only after infinite time.

20. Option (2) is correct. Explanation: SO2 gas is liberated when any sulphide ore is roasted.

2M2S + 3O2 → 2M2O + 2SO2

SO2 is a colourless gas with chocking smell of burnt sulphur.

21. Option (1) is correct. Explanation: It is the unit of ebullioscopic constant (K). k = K kg mol-1 or K (molality)-1

22. Option (3) is correct. Explanation: Aromatic aldehydes and formalde­ hyde do not contain a-hydrogen and thus undergo Cannizzaro reaction. Formaldehyde is more reactive than aromatic aldehydes.

23. Option (1) is correct. Explanation: Molecular orbital electronic configuration of Co3+ in [Co(NH3)6]3+ is

OCOCH3 COOH

COOH

CH3COCl

Ortho-hydroxy benzoic acid

Aspirin

27. Option (1) is correct. Explanation: Mole fraction is used in relating vapour pressure with concentration of solution and according to the Raoult’s law, the partial vapour pressure of each component in the solution is directly proportional to its mole fraction.

28. Option (2, 3) is correct. Explanation: In given four reactions, option (2) and (3) represent oxidising behaviour of H2SO4 that oxidising agent reduces itself as oxidation state of central atom decreases. The reaction is given below: −1

−6

0

−4

2 H I + H 2SO4 → I 2 + SO2 + 2 H 2O 0

−6

+2

+4

Cu + 2 H 2 SO4 → C uSO4 + SO2 + 2 H 2O

29. Option (2) is correct. Explanation: Suppose the atoms N in the ccp = a \ No. of tetrahedral voids = 3 → 2a 2a No. of atoms M = : a = 2: 3 3 Hence compound as M2: N3

Solutions 30. Option (1) is correct.

63

36. Option (3) is correct.

Explanation: Greater the valence of the flocculating ion, greater its ability to bring coagulation as per hardy-Schulze rule.

31. Option (1) is correct. Explanation: C6H5CH2Br will follow SN1 mechanism on reaction with aqueous sodium hydroxide since the carbocation formed C6H5CH2 is a resonance stabilised cation. Benzylic halides show high reactivity towards the SN1 reaction. The carbocation thus formed gets stabilised through resonance as shown in the structure.

Explanation: Higher the reduction potential, higher is its tendency to get reduced. Hence, the order of oxidising power is: ClO–4 < IO–4 < BrO–4

37. Option (3) is correct. Explanation:

CH3 CH2

CH2

CH2

CH2

38. Option (1) is correct. 32. Option (1) is correct. Explanation: Mixture of methanol and acetone exhibits positive deviation because methanolmethanol and acetone-acetone interaction is more than methanol-acetone. The more number of hydrogen bonds are broken the less number of new hydrogen bonds are formed.

33. Option (3) is correct. Explanation: ∆t =

4 ∆o 9

34. Option (3) is correct. Explanation: A→B Rate of reaction, r → k [A]n…(i) If concentration of A is increased by nine times, then rate of reaction becomes three times, r ' = 3r A ' = 9A r ' ∝ [A ']n

3r ' ∝ [9A]n …(ii) From eq. (i) and (ii) [A] n r = 3r [9A] n



1  1 =  3  3 1 = 2n n=1 2

2n

∴ Order of reaction = [½]

Explanation: In case of zero order reaction, the rate constant has same units as the rate of reaction. r = k [A]0 r=k Unit of rate = mol L–1s–1 Unit of k = mol L–1s–1

39. Option (3) is correct. Explanation: 2-Methoxy-2-methylpropane. Longest carbon chain is taken as the parent alkane in the IUPAC nomenclature.

40. Option (4) is correct. Explanation: Smoke is colloidal solution of solid in gas. Starch is lyophobic sol of dispersion medium, water. Gold sol has dispersed phase of solid and dispersion medium as liquid. Hair cream is emulsion having disperse phase, liquid and dispersion medium of liquid.

41. Option (1) is correct. Explanation: Cu (I) should be stable as fully filled d-orbital (d10) than incompletely filled (d9) d-orbital of Cu(II). But Cu(II) has a greater charge density than Cu(I) ion and therefore forms much stronger bonds releasing more energy.

42. Option (3) is correct. Explanation: Cu2S + 2Cu2O → 6Cu + SO2 ↑

43. Option (4) is correct. Explanation: NH2

NH2

NH2

I

CH3 III

35. Option (3) is correct. Explanation: Tm is Thulium which belongs to Lanthanoids. Uranium (U), Neptunium (Np), Fermium (Fm) belong to Actinoid series.

NO2 II

64

OSWAAL CUET (UG) Sample Question Papers, CHEMISTRY Electron withdrawing group decreases the basic strength while electron releasing groups increases the basic strength of aniline.

44. Option (2) is correct. Explanation: It is formed by the loss of 3 electrons, the configuration of element X is [Ar] 3d6 4s2. Therefore, Atomic number = 26.

45. Option (1) is correct. Explanation: Structural formula of Ethane-1, 2 dioic acid is COOH | COOH  ∴ It is oxalic acid.

46. Option (3) is correct. Explanation: PH3 is less basic than NH3. P4 +3NaOH + 3H2O → PH3 + 3NaH2PO2 (Phosphine)         

Case Based 47. Option (4) is correct. Explanation: Adenine and guanine are associated to form the adenine–thymine and guaninecytosine base pairs.

48. Option (2) is correct. Explanation: DNA molecule has regular internucleotide linkage and irregular sequence of the different nucleotides.

49. Option (3) is correct. Explanation: the backbone of DNA is made up of Deoxyribose (Sugar) molecule connected to each other using Phosphates.

50. Option (4) is correct. Explanation: A chemical compound that is used to make one of the building blocks of DNA and RNA. It is a type of pyrimidine. i.e., thymine.

qqq

6

SOLUTIONS OF Question Paper 1. Option (3) is correct.

7. Option (1) is correct.

Explanation: The extraction of chlorine from brine is based on the oxidation reaction of Cl− ion. 2H 2O (l) + 2Cl− (aq.) → H 2 (g) + Cl 2 (g) + 2OH − (aq.)

2H 2O (l) + 2Cl− (aq.) → H 2 (g) + Cl 2 (g) + 2OH − (aq.) The DGQ for this reaction is + 422 kJ DG° = - nFE° −∆G° E° = nF = −2.2 V

Cl+ attacks the chlorobenzene. + Cl

benzene

SO2 is oxidized to sulphuric acid and releases nascent hydrogen which bleaches the material. But this is a temporary as atmospheric oxygen re-oxides the bleached matter after some time.

Explanation: Buna-S is prepared by copolymerisation of 1, 3-butadiene and styrene.

Explanation: Cl+ is an electrophile formed by the following reaction: AlCl 3 + Cl 2 → [AlCl 4 ] + Cl

SO2(g)+2H2O → H2SO4+ 2[H]

8. Option (4) is correct.

2. Option (2) is correct.

−

Explanation: SO2 acts as a bleaching agent under moist conditions.

+

ring

CH=CH2 | Na nCH2=CH–CH=CH2+

–[CH2–CH=CH–CH2–CH–CH2 ]–n Styrene butadiene Rubber (Buna S)

1, 3-butadiene Styrene

to

give

Cl +

Chlorobenzene

3. Option (1) is correct. Explanation: Frenkel defect is also known as dislocation defect because in this defect one of the ion is missing from its lattice site and occupies an interstitial site.

Neoprene is prepared by polymerisation of chloroprene: CI CI | | Polymerisation nCH2=C–CH=CH2 (CH2–C=CH–CH2 — )n O2/peroxides Neoprene Chloroprene

9. Option (2) is correct. 10. Option (1) is correct. Explanation: Due to delocalization of lone pair of electrons on the N-atom into the benzene ring, C6H5NH2 is the weakest base.

4. Option (3) is correct. Explanation: Copper ore when mixed with silica, iron oxide slags off as iron silicate and copper is produced in the form of copper matte which contains Cu2S (I) and FeS (II).

5. Option (1) is correct. Explanation: In the carbylamine test, a primary amine reacts with chloroform and KOH to form alkyl isocyanide (i.e., R–NC) having unpleasant smell.

6. Option (4) is correct. Explanation: Glucose is an example of aldohexose because it contains an aldehyde group.

..

NH2

NH2

NH2

NH2

Resonating Structure of Aniline

11. Option (3) is correct. Explanation: Z×M d= 3 a × NA M= =

d × N A × a3 Z

(

10.5 × 6.022 × 10 23 × 4.07 × 10 −8

1 = 106.6 g mol −

4

)

3

g cm −3

66

OSWAAL CUET (UG) Sample Question Papers, CHEMISTRY

12. Option (2) is correct.

21. Option (2) is correct.

Explanation: Water will move from side (B) to side (A) if a pressure greater than osmotic pressure is applied on piston (B). This is a process of reverse osmosis.

Explanation:

13. Option (3) is correct. Explanation: KCN is used to increase number of carbon atoms. RX + KCN → R − CN + KX

 

      (b)  >  (d)     >   (a)

H / Raney Ni

2 R − CN + 4 H  → R − CH 2 NH 2

14. Option (2) is correct. Explanation: The higher the value of KH, the lower is the solubility of the gas in the liquid. Therefore, the order of increasing solubility of gases will be : Ar < CO2 < CH4 < HCHO.

15. Option (2) is correct. Explanation: Linkage isomerism is the existence of coordination compounds that have the same composition differing with the connectivity of the metal to a ligand. Typical ligands that give rise to linkage isomers are: thiocyanate, SCN−; isothiocyanate, NCS−.

16. Option (1) is correct. Explanation: H3C-Br + AgF ® H3C-F + AgBr.

17. Option (3) is correct. Explanation: Biomolecules, such as carbohydrates, lipids, proteins and nucleic acids, are target molecules of drugs and usually interact with drugs. These drugs possess some common structural feature, may have the same mechanism of action on a specific drug target.

18. Option (2) is correct. Explanation: Each triplet of nucleotides, which have a specific sequence of bases, is called codon.

19. Option (4) is correct. Explanation: The strongest reducing agent is SbH3 due to the presence of minimum bond enthalpy.

20. Option (4) is correct. Explanation: High temperature will favour dissolution. Powdered sugar has large surface area and hence, it is favourable for dissolution.

  

   (c)      >    (e) The presence of electron withdrawing group viz. −NO2 group increases the acidity of phenol due to −I effect and electron releasing group viz. − OCH3 group decreases the acidity of phenol due to +I effect. Moreover, p-nitrophenol is more acidic than m-nitrophenol. While, p-methoxyphenol is less acidic than m-methoxyphenol.

22. Option (3) is correct. Explanation: In square close packing, in two dimensions each sphere is in contact with four of its neighbours. Thus, its coordination number is 4.

23. Option (4) is correct. Explanation: n n M = = 0.02 = V 4 n = 0.08 n 0.08 m= = = 0.016 5 Mass of water in kg

24. Option (1) is correct. Explanation: [Cr(H2O)4Cl2]+ Cl Cl + H 2O Cr OH2 H 2O OH2 cis-isomer

H 2O

Cl Cr

OH2

+

H 2O Cl trans-isomer

H 2O

25. Option (2) is correct. Explanation: According to Markovnikov’s rule, hydrogen will add to the carbon atom having less number of hydrogen atoms.

Solutions 26. Option (4) is correct. Explanation: Interstitial compounds are usually non-stoichiometric and are neither typically ionic nor covalent. Hence, interstitial compounds are chemically inert.



27. Option (2) is correct. Explanation: IUPAC name of the compound is 2-propoxypropane.

28. Option (4) is correct. Explanation: Thermosetting polymers are heavily branched cross-linked polymers. They cannot be reused and cannot melt on heating and cannot be remoulded.

29. Option (4) is correct. 30. Option (2) is correct. Explanation: ∆G = ∆H – T∆S Gibb’s Free energy equation Adsorption is a spontaneous process so ∆G is negative. At equilibrium it becomes zero and ∆H = T∆S is attained.

31. Option (1) is correct. Explanation:



67

. initial ] t = 2 303 log [ R initial k [ R final final ] 2.303 5 log = −33 − 1.15 × 10 3 .303 2 × 0.2219 = 1.15 × 10 −−33 = 444.379 s

35. Option (2) is correct. Explanation: When 0.1 mole of CoCl3(NH3)5 was reacted with excess of AgNO3, we get 0.2 moles of AgCl. So, there are two chloride ions that are free and not part of the complex. The formula for complex has to be [Co(NH3)5Cl]Cl2. [Co(NH3)Cl]Cl2 → [Co(NH3)5Cl]2+ + 2Cl− Therefore, the conductivity of the solution will be 1: 2 electrolyte.

36. Option (1) is correct. Explanation: Weak dispersion forces are present between particles of noble gases. Ionisation enthalpy of molecular oxygen is very close to that of xenon.

37. Option (2) is correct. Propane

Acetone

It is a Wolff-Kishner reduction which converts C=O group into –CH2– group.

32. Option (3) is correct. Explanation: Vitamins are not a target molecule for drug function in body.

33. Option (2) is correct. Explanation: One mole of AgNO3 precipitates one mole of chloride ion. In the above reaction, when 0.1 mole CoCl3(NH3)5 is treated with excess of AgNO3. 0.2 mole of AgCl are obtained and thus, there must be two free chloride ions in the solution of electrolyte. So, molecular formula of complex will be [Co(NH3)5Cl]Cl2 and electrolyte solution must contain [Co(NH3)5Cl2+] and two Cl– as constituent ions. Thus, it is 1 : 2 electrolyte. [Co(NH3Cl)]Cl 2 → [Co(NH 3 )5 Cl]2 + (aq) + 2Cl − (aq)

34. Option (2) is correct. Explanation: Initial amount= 5 g Final concentration = 3 g Rate constant= 1.15 × 10–3 s–1 We know that for a First order reaction

Explanation: When one of the reactants is in excess, a bimolecular reaction is kinetically first order reaction.

38. Option (4) is correct. Explanation: Aldehydes with no α-hydrogen undergo Canizzaro reaction.

39. Option (3) is correct. Explanation: Ecell is an intensive property and it does not depend upon number of particles but DrG of the cell reaction is an extensive property because this depends upon number of particles.

40. Option (4) is correct. Explanation: Colloidal particles are large in size and less in number.

41. Option (1) is correct. Explanation: A narrow spectrum antibiotic is active against Gram-positive or Gram-negative bacteria.

42. Option (3) is correct. Explanation: Aliphatic aldehydes(acetaldehyde) reduce the Fehling’s solution to red cuprous oxide.

68

OSWAAL CUET (UG) Sample Question Papers, CHEMISTRY CH 3CHO + 2CuO + 5OH − → CH 3COOH + Cu 2O ↓+ 3H 2O Red ppt.

Aromatic aldehydes (benzaldehyde) do not react with Fehling’s solution.

43. Option (3) is correct. Explanation: Slope : A + H2O → B r ∝ [A] (Q [H2O] = excess) It is called pseudo first order reaction.

44. Option (2) is correct. Explanation: The negative value of standard reduction potential for Cr3+ to Cr means that the redox couple is a stronger reducing agent.

45. Option (4) is correct. Explanation: m-chlorophenol is most acidic as electron withdrawing (−Cl) group increases the acidity of phenols.

46. Option (1) is correct. Explanation: When oxidation occurs at anode two possible reactions take place, i.e., Oxidation of chlorine and oxygen: Among these two elements the oxidation of chloride ion is preferred because oxidation of oxygen requires over-voltage. Through electrolysis process, chlorine is obtained by giving out hydrogen and aqueous sodium hydroxide as a by-product.

Case Based 47. Option (4) is correct. Explanation: Scandium, copper and zinc are from I series of transition elements whereas zirconium belong to second series beneath the titanium.

48. Option (4) is correct. Explanation: A complex ion has a central metal ion surrounded by a number of other molecules or ions. These are considered to be attached to the central ion by coordinate bonds. The molecules or ions surrounding the central metal ion are called ligands.

49. Option (2) is correct. Explanation: They have all characteristics of metals. Thus, transition metals have high boiling point because their atoms are closely packed by the strong metallic bonds.

50. Option (2) is correct. Explanation: Due to the presence of unpaired d electrons and d–d– transition, electrons get excited to higher energy d–orbitals which gives colour to compounds. When visible (white) light falls on a compound, it absorbs certain radiations of white light and transmit the rests as complementary colour to that of the absorbed light.

qqq

SOLUTIONS OF Question Paper 1. Option (3) is correct.

6. Option (2) is correct.

Explanation:

Explanation:

7

Cr+2 : d4  4 Unpaired electrons Mn+2 : d5 Fe+2 : d6

2. Option (3) is correct.

Co+2 : d7 2+

Explanation: In fluorite structure, Ca ions are in the face centered cubic arrangement. Each Ca2+ is connected to 4F– ions below it and to another set of 4F– ions above it i.e., Ca2+ has a coordination number of 8 and each F– ion has a coordination number 4.

3. Option (4) is correct. 1 1 Cu 2O + Cu 2S → 3Cu + SO2 2 2 Explanation: This reaction includes reduction of copper (I) oxide by copper (I) sulphide and in this process copper is reduced by itself. This process is called as auto-reduction. The solidified copper so obtained is known as blistered copper.

4. Option (4) is correct. H 3 O+

CH 3 Br → CH 3CN → CH 3COOH (A)

LiAlH 4

(B)

→ CH 3CH 2OH ether

(C)

5. Option (1) is correct. Explanation: When freshly prepared solution of ferrous sulphate (FeSO4) is added in a solution containing NO3− ion, formation of a browncoloured complex will take place. This is called as brown ring test of nitrate. Hence, two moles of ammonia will produce two moles of NO. NO3− + 3Fe2+ + 4H+ → NO + 3Fe3+ + 2H2O [Fe(H2O)6]

2+

 4 Unpaired electrons  3 Unpaired electrons

Less number of unpaired electrons means less paramagnetic behaviour.

7. Option (2) is correct. Explanation: It is Kolbe’s electrolytic decarboxylation. RCOONa(aq) → RCOO– + Na+ At anode, 2RCOO– → R–R + 2CO2 + 2e– Alkane At cathode, 2H2O + 2e– → H2 + 2OH–

8. Option (4) is correct. Explanation: Number of atoms in cubic close packing = 4 = O2– Number of tetrahedral voids = 2N = 2 × 4 = 8 1 Number of A2+ ions = 8 × = 2 4

Explanation: KCN

 5 Unpaired electrons

+ NO → [Fe(H2O)5(NO)] Brown ring

2+

+ H 2O

Number of octahedral voids = Number of B+ ions    = N = 4 Ratio of ions will be O2– : A2+ : B+ = 4 : 2 : 4 = 2 : 1 : 2 Formula of oxide = AB2O2

9. Option (3) is correct. Explanation: Boiling point of (i) is 364 K, boiling point of (ii) is 375 K. boiling point of (iii) is 346 K As the branching increases in the isomeric alkyl halides, the boiling point decreases.

10. Option (2) is correct. Explanation: Bell-metal contains Cu (80%) and Sn (20%).

OSWAAL CUET (UG) Sample Question Papers, CHEMISTRY

70

11. Option (3) is correct.

18. Option (2) is correct.

Explanation: Enzyme inhibitors tend to reduce the activity of a particular enzyme. Generally, a weak bond such as H-bonding, van der Waals interaction, etc., is formed between the enzyme and the inhibitor.

12. Option (1) is correct. Explanation: Peroxoacids of sulphur must contain one O−O bond as shown below:

Explanation:

CH 3 | H+ CH 3  CH  CH  CH 3 → heat  OH

CH 3 | CH 3  CH  CH  CH 3 ⊕

1, 2 hydride shift

→

CH 3 | − H+ CH 3  CH  CH 2  CH 3 → ⊕ CH 3 CH 3 | | CH 3  C = CH  CH 3 + CH 3  CH  CH = CH 2

13. Option (2) is correct. Explanation: In metallurgy of aluminium (Al), graphite anode is oxidised to carbon monoxide (CO) and carbon dioxide (CO2) C (s) + O − (melt) → CO (g) + 2e −





C (s) + 2O

−

(melt) → CO (g) + 4e

−

14. Option (2) is correct.

H−C−H + CH3MgBr Formaldehyde

(B) Minor

HBr, dark

(A) → in absence of peroxide

CH 3 CH 3 | | CH 3  CH  CH  CH 3 + CH 3  CH  CH  CH 3 |  Br Br (C) Major

Explanation: O

(A) Major

19. Option (2) is correct.

OMgBr

Explanation: Zone refining method is based on the principle that impurities are more soluble in molten metal as compared to the solid state of the metal.

H−C−H CH3

Adduct

H2O/H+

OH H−C−H + Mg CH3

(D) Minor

20. Option (1) is correct. Br OH

Explanation: In cyclo-trimetaphosphoric acid, 3 double bonds and 9 single bonds are present. The structure is given below:

Ethanol

15. Option (4) is correct. Explanation: Structure of PCl5 in solid state is:

Cyclo-trimetaphosphoric acid (HPO3)3

21. Option (1) is correct.

16. Option (4) is correct. Explanation: Fraction of total volume occupied by the atoms present in a simple cube is π/6 or 0.52

17. Option (1) is correct. Explanation: Stainless steel contains 73% iron, 18% chromium and 8% nickel.

Explanation: It is halogen exchange reaction as in this reaction both R and Na exchanges halogens.

22. Option (1) is correct. Explanation: X atoms at the corners = Y atoms at the face centres = Ratio of atoms, X : Y = 1 : 3 Hence, formula is XY3

1 ×6 =3 2

1 ×8 =1 8

Solutions 23. Option (4) is correct.

From equation (1) and (2),

Explanation: Glycogen is a polymer of glucose found in liver, brain and muscles of animals. Cellulose is a polymer found in plant while amylase and amylopectin are structural units of starch.

24. Option (4) is correct. Explanation: Total vapour pressure of mixture = Vapour pressure of pentane in mixture   + Vapour pressure of hexane in mixture As the ratio of pentane to hexane = 1 : 4 ∴ Mole fraction of pentane = 1/5 Mole fraction of hexane = 4/5 Total vapour pressure  Mole fraction of Vapour pressure =  × of pentane  pentane   Mole fraction of Vapour pressure + ×  of hexane hexane 

4 1  =  × 440 + × 120 5  5

Mole fraction of  Vapour pressure  =  × of mixture pentane in vapour phase  88 = 184 × mole fraction of pentane in vapour  phase ∴ Mole fraction of pentane in vapour phase 88 = = 0.478 184

25. Option (2) is correct.

NaOH / I 2 C 6 H 5COCH 3  → C 6 H 5COONa + CHI3

Iodoform

Acetophone

27. Option (1) is correct. Explanation: Rate of reaction = k [A]m [B2]n where m and n are the orders w.r.t. A and B2 respectively. In following experiments

1.6 × 10

−4

= k [0.50] [0.50] …(1)

3.2 × 10

−4

= k [0.50]m [1]n 

m

−4

m

n

n

3.2 × 10 = k [1.00] [1]  From equations (2) and (3),  

3.2 × 10

−4

=

[0.5]m [1]n [0.5]m [0.5]n

⇒ 2 = 2n ⇒ n = 1

28. Option (2) is correct. Explanation: Gold number is defined as the minimum amount of lyophilic colloid in milligrams which prevent the flocculation of 10 mL gold sol by the addition of 1 mL 10 NaCl solution. Lesser the gold number, higher is the protecting power.

29. Option (2) is correct. Explanation: The linkage which holds the two monosaccharide units through oxygen atom is called glycosidic linkage.

Explanation: Synthesis of prostaglandins compound inhibited by aspirin, which stimulates inflammation in tissue and causes pain. So, it is effective in relieving pain. It does not make a person addictive as it is non-narcotic drug.

31. Option (2) is correct. Explanation: For the reaction, 2N 2O5 → 4NO2 + O2

−

d [N 2O5 ] = k [N 2O5 ] dt 1.02 × 10 –4 3.4 × 10 − 5

=3

32. Option (3) is correct.

Explanation:

3.2 × 10

=

= [A]0 [B2 ]1 k[B 2 ] Hence, rate k=

∴   [N2O5] =

26. Option (2) is correct.

−4

1.6 × 10

 

−4

1.02 × 10–4 = 3.4 × 10–5 s–1 × [N2O5]

1 l G* Explanation: κ = . or R A R



3.2 × 10 − 4

30. Option (1) is correct.

= 184 mm of Hg ∴ Vapour pressure of pentane in mixture



71

m

[Co(NH 3 )6 ]3 + 3Cl + AgNO3 → 3AgCl (Excess)

 

n

(Excess)

+[Co(NH 3 )5 Cl]2 +

…(3)

k[0.5] [1]

⇒ 1 = 2m ⇒ m = 0

+[Co(NH 3 )6 ]3 +

[Co(NH 3 )5 Cl]2 + 2Cl − + AgNO3 → 2AgCl

[Co(NH 3 )4 Cl]3 + Cl − + AgNO3 → AgCl

n

k [1] [1] m

…(2)

Explanation: According to Werner’s theory, CoCl3 6NH3 → [Co(NH3)6]3+3Cl– CoCl3 5NH3 → [Co(NH3)6C]2+2Cl– CoCl34NH3 0 → [Co(NH3)4Cl2]+Cl– When AgNO3 in excess is treated with these complexes then following reactions take place.

(Excess)

 

+[Co(NH 3 )4 Cl 2 ]+

72

OSWAAL CUET (UG) Sample Question Papers, CHEMISTRY

33. Option (2) is correct.

40. Option (3) is correct.

Explanation: Most of the naturally occurring amino acids have L−configuration. All naturally occurring a−amino acids are optically active except glycine.

34. Option (4) is correct. Explanation: The value of Kf depends only on nature of the solvent and independent of composition of solute particles, i.e., does not depend on the molality of the solution.

35. Option (4) is correct. Explanation: For high spin d4 octahedral complex

Explanation: Activation energy can be calculated by using Arrhenius equation. log

 T1 − T2  k2 Ea = −   k1 2.303 R  T1T2 

where, k1, k2 are two different rate constants at temperature T1 and T2 respectively.

41. Option (1) is correct. Explanation: It can be obtained from condensation polymerisation of ethylene glycol and phthalic acid with the elimination of water molecule. nHOCH2—CH2OH+HOOC

0.6 0 0.4 0

Degenerate d-orbitals

COOH

Glycol

eg

n

420—400 K Zn(OCOCH3)2+SB2O3

—O—CH ( 2—CH2OOC

t2g

∴ CFSE = [(– 3 × 0.4) + (1 × 0.6)] D0 = – 0.6D0

36. Option (3) is correct.

O ||

C

Terylene or decron

42. Option (4) is correct.

Explanation: Coagulating power ∝

nH2O

Phthalic acid

1 Coagulating value

Lower the coagulating value, higher is the coagulating power so, the correct order is: MgSO4 > BaCl2 > NaCl   III     II I

37. Option (1) is correct.

Explanation: µ for Cr in [Cr(H2O)6]Cl3 = 3.83 B.M.   

= µ

n ( n + 2)

= 3.83

n ( n + 2) ⇒ n = 3

Thus, number of unpaired electrons in d-orbitals subshells of chromium (Cr = 24) = 3 ∴ Configuration of Cr = 1s2 2s2 2p6 3s2 3p6 3d3

Explanation: According to depression in freezing point,

In 3d3 the distribution of electrons

DTf = iKfm where, Kf = cryoscopic constant ∆Tf × Wsolvent 3.82 × 45 i= = k f × nsolute × 1000  5  1.86 ×  × 1000  142    i = 2.63



38. Option (1) is correct. Explanation: In RNA, thymine is not present. In place of thymine, uracil is present in RNA.

39. Option (3) is correct. Explanation: We know that, Q = It ⇒ Q = 1 × 60 = 60 C Now, 1.60 × 10–19 C ≡ 1 electron 60 60 C ≡ = 3.75 × 10 20 1.6 × 10 −19     = 3.75 × 1020 electrons

1 1 3dxy , 3d 1yz , 3d zx , 3dx02 − y 2 , 3d z02

43. Option (3) is correct. Explanation: Given that I = 5 ampere t = 40 min = 40 × 60 = 2400 sec Amount of electricity passed

Q = It = 5 × 2400 = 12000 C

Zn2+ + 2e– → Zn (n = 2e–) From Faraday first law, W = ZIt (Z = equivalent mass) Z=

Mass 65.39 = g of Zn 2 × 96500 nF

Therefore, 12000 C charge will deposit =

65.38 × 12000 2 × 96500

= 4.065 g of Zn

73

Solutions 44. Option (1) is correct.

Case Based

Explanation: Syphilis is an acute and chronic infectious disease caused by the bacterium Treponema pallidum. It is a sexually-transmitted infection. Primary route of transmission is through sexual contact but it may also be transmitted from mother to foetus during pregnancy or at birth.

47. Option (3) is correct. Explanation:

x vs. log p as m from Freundlich adsorption isotherm Taking log on both the sides x 1 log = log k + log p m n Explanation: For the plot of log

+ 3Br 2

Br 2 /H 2 O

1 n

+ 3HBr

Br 2.4.6-Tribromoaniline

48. Option (2) is correct. Explanation: O || H—N—C—CH3

NH2 (CH3CO)2O Pyridine

Aniline

Slope =

Br

Br

Aniline

45. Option (4) is correct.

log

NH2

NH2

Br2 CH3COOH

N-Phenylethanamide (Acetanilide)

–

NH2

OH or H

+

2 3

Br 4-Bromoaniline

Br (Major)

49. Option (3) is correct. Explanation: The lone pair of electrons on nitrogen of acetanilide interacts with oxygen atom due to resonance as shown below:

x m Intercept C= log k

log P

  y = c + mx 1 Thus, slope m = n

46. Option (2) is correct. Explanation: For ideal gases, PV = nRT For dilute solution, P = p pV = nRT m m RT pV = 2 RT ⇒ M 2 = 2 πV M2

Hence, the lone pair of electrons on nitrogen is less available for donation to benzene ring by resonance. Therefore, activating effect of –NHCOCH3 group is less than that of amino group.

50. Option (4) is correct. Explanation: Aniline is a resonance hybrid of 5 structures. NH2

NH2

NH2

ii

iii

NH2

NH2

where, m2 = mass of solute M2 = molecular mass of solute. i

iv

v

qqq

8

SOLUTIONS OF Question Paper 1. Option (4) is correct.

6. Option (3) is correct.

Explanation:

Weight percentage × density × 10 Normality = Equivalent weight (N 1)

98 × 1.8 × 10 = = 36 N 49 N 2 = 0.1 × 2 = 0.2 N    N V = N 2V2     1 1 36 × V = 0.2 × 1000   0.2 × 1000 = V = 5.55 mL 36     

7. Option (2) is correct. Explanation: The commercial polyacrylonitrile is orlon (acrilan).

Explanation: [CuSO4 + 2KCN → Cu(CN)2 + K2SO4] × 2 Cupric cyanide

2Cu(CN)2 → Cu2(CN)2 + NC-CN

  Cyanogens

Cu2(CN)2 + 6KCN → 2K3[Cu(CN)4]

Soluble complex salt

2CuSO4 + 10KCN → 2K3[Cu(CN)4]  + 2K2SO4 + (CN)2 Thus, copper sulphate dissolves in excess of KCN to give K3[Cu(CN)4] or [Cu(CN)4]3–.

3. Option (1) is correct. Explanation: In cyanide process, the metal is recovered by displacement of metal by some more reactive metal from the complex. The reaction is given below: −

4 M + 8CN − + 2 H 2O+O2 → 4  M (CN )2  + 4OH − 2  M (CN )2  + Zn →  Zn (CN )4 

name

of

8. Option (2) is correct.

2. Option (2) is correct.

−

Explanation: C-adjacent to oxygen atom in the cyclic structure of glucose or fructose is known as anomericcarbon. In above structure ‘a’ and ‘b’ are present at adjacent to oxygen atom. Both carbons differ in configurations of the hydroxyl group.

2−

+ 2M

4. Option (3) is correct. Explanation: In all the given options, the central metal atom is same and contain same number of d electrons. Thus, CFSE is decided by ligands. In case of strong field ligand, CFSE is maximum, CN– is a strong field ligand, hence is [Co(CN)6]3– so CFSE is maximum.

5. Option (2) is correct. Explanation: For orthorhombic system, α = β = γ = 90°

Explanation: We know that ∆G = ∆H – TDS For adsorption of a gas, DS is negative because randomness decreases. Thus, for making ∆G negative, ∆H should be highly negative because reaction is exothermic. Thus, for adsorption of gas if DS is –ve than ∆H should be highly negative.

9. Option (2) is correct. Explanation: The order of basic character of the transition metal monoxide is TiO > VO > CrO > FeO because basic character of oxides decreases with increase in atomic number. Oxides of transitional metals in low oxidation state, i.e., +2 and +3 are generally basic except Cr2O3.

10. Option (1) is correct. Explanation: Those orbitals which utilizes 3d-orbitals for bonding and exhibit paramagnetic behaviour forms outer orbital complex. In [Ni(NH3)6]2+ : Ni+2 = [Ar] 3d8 4s0 3d

4s

In [Ni(NH3)6]2+ 3d

unpaired electrons 4s

××

4p

×× ×× ××

4d

×× ××

sp3d2 hybridisation

Solutions Thus, it forms outer orbital complex and is paramagnetic

75

14. Option (1) is correct.

(b) [Zn(NH3)6]2+ ⇒ outer orbital complex but diamagnetic

Explanation: On heating an aldehyde with Fehling’s reagent, a reddish brown precipitate is obtained. Aldehydes are oxidised to

(c) [Cr(NH3)6]3+ ⇒ inner orbital complex but paramagnetic

corresponding carboxylate anion. Aromatic aldehydes do not respond to this test.

(d) [Co(NH3)6]3+ ⇒ inner orbital complex but diamagnetic

R – CHO + 2Cu2+ + 5OH

– –

R –COO + Cu2O + 3H2O Reddish brown ppt

11. Option (4) is correct. Explanation: According to Raoult’s law, if volatile liquid is added to pure solvent then total pressure is equal to the sum of the partial pressure of volatile liquid and the solvent. p′A + pB′ …(i) Total pressure, p= T

p′A = p A x A

We have

1 = [9 − 4]= 2.5   2 Substituting these values in equation (i), we get

= pT p A x A + pB xB

Explanation: A atoms are at 8 corners of the cube. 1 No. of A atoms per unit cell = 8 × = 1 8 B atoms are at the face centre of six faces. 1 No. of B atoms per unit cell = 6 × = 3 2 Thus, formula of compound = AB3

16. Option (4) is correct. Explanation: Nylon-6 is the polymer formed by polymerisation of caprolactum.

Mole fraction x A + xB = 1

H | N

xB= 1 − x A

⇒

15. Option (2) is correct.

On substituting in above equation,

O 533–543 K H2O

pT = p A x A + pB (1 − x A )  

Caprolactam

= p A x A + pB − pB x A = pB + x A ( p A − p B )

– ( )n Polymerisation —NH–(CH 2)5 CO— Nylon-6 heat

12. Option (1) is correct.

17. Option (1) is correct.

Explanation: NH2

N==NCl NaNO2/HCl (273–278 K) (X) CH3

—N CH3

CH3 —N==N—

—N CH3

Coupling product (Y)

13. Option (2) is correct. Explanation: (A) ICl2 = 53 + 2 × 17 = 87 ClO2 = 17 + 16 = 33 (B) BrO2− = 35 + 2 × 8 + 1 = 52 BrF2+ = 35 +9 × 2 − 1 = 52 (C) ClO2 = 17 + 16 = 33

[H3N+–(CH2)5 –COO–]

BrF = 35 + 9 = 44

(D) CN− = 6 + 7 + 1 = 14 O3 = 8 × 3 = 24

Explanation: At point D both the curves of oxidation of iron (Fe) and oxidation of carbon monoxide (CO) meet each other. Thus, DGQ for the reaction FeO + CO → Fe + CO2 is zero. 2Fe + O2 → 2FeO; ∆G = −280 2CO + O2 → 2CO2 ; ∆G = − 280 2FeO + 2CO → 2Fe + 2CO2 ∆GΘ = ( − 280 + 280 ) = 0

18. Option (4) is correct. Explanation: Atoms, ions or molecules containing unpaired electrons are paramagnetic. In [Cr(NH3)6]3+, Cr+3 = [Ar] 3d3 4s0 In excited state ×× ××

××

×× ×× ××

d2sp3

Number of unpaired electrons = 3 Rest all others are diamagnetic.

76

OSWAAL CUET (UG) Sample Question Papers, CHEMISTRY

19. Option (4) is correct. Explanation: Glycogen is stored in the liver of animals.

20. Option (3) is correct. Explanation:

At negative pole: A → A+ + e–  (oxidation) At positive pole: B+ + e– → B  (reduction) Hence cell reaction will be A + B+ → A+ + B, E°cell = +0.20 V

O 28. Option (2) is correct.  (i) CH3 MgBr Zn(Hg ) / conc.HCl CH3CN  → CH − C − CH  → CH 3CH 2CH 3 + H 2O 3 3 (ii) H3O+ Explanation: Low density polythene is tough but                 [A] not hard (its flexible) and a poor conductor of O              Acetone electricity. It is highly branched structure.  (i) CH3 MgBr Zn(Hg ) / conc.HCl  → − −  → CH CH + H O CH C CH CH 3 3 3 2 3 2 (ii) H3O+ 29. Option (1) is correct.               [B]              Propane Explanation: O

21. Option (1) is correct. Explanation: Weak electrolytes dissociate partially in concentrated solution. On dilution, their degree of dissociation increases hence, their Λm increases sharply.

22. Option (3) is correct.

NH2

NH—C—CH3 Br2

Ac2O

CH2COOH

CH3

CH3

Explanation:

(A) +

− 10

0.98 × 10 r = 0.541 From radius= radio, − r 1.81 × 10 − 10

O NH—C—CH3

It lies in the range of 0.414 to 0.732 hence coordination number of each ion will be 6 as the compound will have NaCl type structure i.e., octahedral arrangement.

23. Option (2) is correct. Explanation: The compound that causes general anti-depressant action on the central nervous system belongs to the class of tranquilizers.

24. Option (1) is correct. Explanation: Phenol decolourises bromine water to form white precipitate of 2,4,6-tribromophenol whereas ethanol does not precipitate.

25. Option (4) is correct. Explanation: Enzymes are protein in nature. They are highly specific and get denaturated by high temperature or UV-rays. At optimum temperature (25–35°C) enzyme activity is maximum.

26. Option (4) is correct. Explanation: (phosgene).

Carbonyl

chloride

is

COCl2

27. Option (1) is correct. Explanation: Given electrochemical cell A|A+(xM)||B+(yM)|B The emf of cell is + 0.20 V. So, cell reaction is possible. (as ∆G = – nEF = –Ve spontaneous) The half-cell reaction are

NH2

Br

Br H2 O H+

CH3

CH3

(B)

(C)

30. Option (4) is correct. Explanation: FeO + C → Fe + CO It can be seen as a couple of two simpler reactions: FeO → Fe +

1 O2 (∆G°FeO, Fe ) 2

1 C + O2 → CO (∆G°c, co ) 2 Total Gibb’s energy change becomes ∆GΘ(c, co) < ∆(Fe, FeO) ∆ r G0 In DGQ versus T plotting Fe to FeO goes upwards and the plot for C to CO goes downwards. At temperature above point A, the C to CO lines comes below Fe to FeO and ∆GΘ (C, CO)< ∆GΘ (Fe, FeO) o, in this range, C will reduce FeO to Fe and S itself be oxidised to CO.

31. Option (1) is correct. Explanation: Ionic conductivity ∝ Number of ions In K4[Fe(CN)6], it produces maximum 5 ions thus show more conductivity. + 4–  K4[Fe(CN)6]     4K + [Fe(CN)6]

Solutions 32. Option (1) is correct. Explanation: Tranquilizers are neurologically active drugs. If the enzyme is inhibited, the neurotransmitter noradrenaline is slowly metabolised and can thus activate the receptor for longer periods thereby counter-acting the effect of depression. Tranquilizers form an essential component of sleeping pills.

fluoride. Inter-halogen compounds are more reactive than halogen compounds because A–B bond of dissimilar halogen is weaker than A–A or B–B bond of halogens.

40. Option (1) is correct. Explanation: The number of C-atoms can be increased in the chain by Grignard reaction. O OMgBr

33. Option (3) is correct.

H−C−H + CH3MgBr

Explanation: Density, ρ = 1.17 g cm–3 = 1170 g L–1

OH

Strength in g L−1 Molarity of solution = Molecular weight =

1170 = 32.05 M 36.5

34. Option (2) is correct. Explanation: Phenol is less acidic than o-nitrophenol as electron withdrawing (−NO2) group increases the acidity of phenols.

35. Option (3) is correct. Explanation: The higher the reduction potential, the higher is its tendency to get reduced. Hence, the order of their oxidising power is: BrO–4> IO–4 > ClO–4

36. Option (4) is correct.

+

H2 O / H  → H−C−H + Mg

CH3

Dry ether

R—X + Mg → R—Mg—X Grignard reagent

Here X = Cl, Br

37. Option (2) is correct. Explanation: A catalyst accelerates the forward and backward reaction to the same extent thus, the equilibrium constant does not change. Catalyst only speed up the reactions to attain equilibrium faster and does not alter equilibrium constant. Equilibrium constant varies only when temperature condition of reaction gets changed.

38. Option (3) is correct. Explanation: The treatment of sodium alkoxide with a suitable alkyl halide to form an ether is called as Williamson ether synthesis reaction.

39. Option (2) is correct. Explanation: In case of halogens, radius ratio between iodine and fluorine is maximum radius because iodine has maximum radius while fluorine has minimum radius. Also, due to highest ratio maximum numbers of atoms are present in iodine

H−C−H Br

CH3

OH

41. Option (1) is correct. Explanation: Given reaction, 4 2 –1 Al + O2 → Al 2O3 , ∆G = − 827 kJ mol 3 3 Here, n = 4 Because, for 1 mol of Al, n = 3 Given 4/3 moles, so 1 mol = 3 4/3 mol = 4/3 × 3 = 4   ∆G = – nFE° – 827 × 103 J = – 4 × E° × 96500 827 × 10 3 E= 4 × 96500

Explanation: Grignard reagent is prepared by heating an alkyl halide with dry magnesium powder in dry ether.

77

E° = 2.14 V

42. Option (3) is correct. Explanation: Both oestrogen and progesterone are sex hormones. Progesterone suppresses ovulation and oestrogen regulates the menstrual cycle.

43. Option (3) is correct. Explanation: Cyanide process is used for the extraction of gold. Froth floatation method is used for dressing of ZnS. Aluminium is extracted by electrolytic reduction. Zone refining is used for getting extra pure element like Ge.

44. Option (3) is correct. Explanation: H2O2 reacts with acidified solution of a dichromate Cr2O72– to give deep blue coloured complex, chromic peroxide CrO5. Cr2O72– + 2H+ + 4H2O → 2CrO(O2)2 + 5H2O Chromic Peroxide    (CrO5)

Oxidation state of Cr in CrO5 is +6.

45. Option (3) is correct. Explanation: Amides can be converted into amines by Hofmann’s bromamide reaction. O  R C  NH 2 + Br2 + 4KOH → RNH2 + 2KBr + K2CO3 + 2H2O

78

OSWAAL CUET (UG) Sample Question Papers, CHEMISTRY

46. Option (2) is correct. Explanation: d Oxidation half reaction is   H2 → 2H+ + 2e– (1 atm) (At pH 10) If pH = 10, [H+] = 10–10 From Nernst Equation, 0.0591 [H + ]2 log Ecell = E°cell − 2 PH2 For hydrogen electrode E°cell = 0 0.0591 (10 −10 )2 Ecell = − log 2 1 = 0.0591log 1010    Ecell = 0.591 V

47. Option (2) is correct. Explanation: Since the rate of the reaction is proportional to the concentration for the reactant i.e., ethene so, it is first order reaction.

48. Option (3) is correct. Catalyst Explanation: C 2 H 4 + H 2   → C2 H6 Ethene Ethane

49. Option (1) is correct. Explanation: For first order reaction,

0.693 k 0.693 = 2.5 × 10 −15 s −1 = 2.772 × 10 −24 s t1 2 =

50. Option (3) is correct. Explanation: t = 5 min

[R]0 =3 [R]

For first order reaction,

[R]0 2.303 log t [R] 2.303 = log 3 5 2.303 = × 0.4771 5 = 0.2197 min−1 K=

Rate law equation, Rate = K [C2H4]

qqq

SOLUTIONS OF Question Paper 1. Option (1) is correct. Explanation: Aluminium (Al) and iron (Fe) are most abundant elements in the Earth’s crust.

2. Option (1) is correct. Explanation: It can be magnetised permanently. So, the domains get oriented in the direction of applied magnetic field.

3. Option (2) is correct. Explanation: Xe has least ionisation energy among the noble gases and hence it forms chemical compounds with oxygen and fluorine, however, Ne cannot form compounds with oxygen and fluorine so NeF2 does not exist.

4. Option (2) is correct. Explanation: Rate for the first order reaction will be  dx  = = k [ A] Rate  dt  Now given that =

dx = 1.5 × 10 −2 mol L−1 min −1 dt

[A] = 0.5 M, k = ? ∴ 1. 5 × 10–2 = k × 0.5 ⇒ k = 3 × 10–2 min–1 0.693 0.693 t1 / 2 of reaction = = = 23.1 min k 3 × 10 –2

5. Option (4) is correct. Explanation: Ti2+ = [Ar]3d2 4s0 V3+ = [Ar]3d2 4s0 Cr4+ = [Ar]3d2 4s0 Mn5+ = [Ar]3d2 4s0

6. Option (4) is correct. Explanation: According to Hardy Schulze law Coagulating power ∝ charge of ion Thus, coagulation power depends upon both magnitude of charge and nature of charge. Greater the magnitude of charge quicker will be the coagulation.

7. Option (1) is correct. Explanation: Option (1) are the two nonsuperimposable mirror images of each other so they are enantiomers.

9

8. Option (3) is correct. Explanation: In the extraction of copper from its sulphide ore, the metal is formed by the reduction of Cu2O with Cu2S. The process of the reaction is finished by the process of autoreduction. In this reaction, copper appears as blistered copper. Chemical reaction occurring in this reaction is given below: 1 1 Cu 2O + Cu 2S → 3Cu + SO2 2 2

9. Option (3) is correct. Explanation: Xe + PtF6 ® Xe+[PtF6]–

10. Option (2) is correct. Explanation: Liquid dish washing detergents are non-ionic.

11. Option (2) is correct. Explanation: Schottky defect is observed in crystals when equal number of cations and anions are missing from the lattice.

12. Option (1) is correct. Explanation: R-MgX + CH 3  C H 2  

O → R—CH CH O—MgX 2 2 H2 O / H+  OH → RCH2CH2 OH + Mg X ∆

13. Option (2) is correct. Explanation: κ × 1000 Λm = Molarity

=

5.76 × 10 −3 × 1000 = 11.52 S cm2 mol–1 0.5

14. Option (2) is correct. Explanation: In Ni(CO)4, Nickel is sp3 hybridised because in it oxidation state of Ni is zero. 6 2 28Ni = [Ar] 3d 4s ××

×× ×× ×× sp3

OSWAAL CUET (UG) Sample Question Papers, CHEMISTRY

80

(CO is a strong field ligand hence does the pairing of electrons) In [Ni(CN)4]2– nickel is present as Ni+2, so its configuration is [Ar] 3d8 4s0 d

s ××

p

××

×× ××

OH  Therefore, A = CH 3  C == CH 2

    

O  B = CH 3  C  CH 3

18. Option (3) is correct.

dsp2

CN– is a strong field ligand so pairing occurs. In [NiCl4]2– Ni+2 = [Ar] 3d8 4s0

Explanation: At critical micelle concentration the surfactant molecules get associated with each other.

4s

4p

19. Option (3) is correct.

××

×× ×× ××

Explanation:

3d

PBr

alc

3 C 2 H 5OH  → C 2 H 5 Br  → CH 2 = CH 2 KOH

sp3

Cl– is a weak field ligand hence electrons are not paired.

(X)

Ethyne(Y)

H2 O

H 2 SO4

→ CH3CH2HSO4 → C2H5 OH   (Z) Hence Z is ethanol.

15. Option (1) is correct. Explanation: Given that, [A]0 = 2 m, t = 200 min,   [A] = 0.15 m

20. Option (1) is correct.

For first order reaction,

Explanation:

[A]0 2.303 log Rate constant, k = t [A]

CH 3CH 2Cl → CH 3CH 2CN

k= ⇒

t1 / 2

(X)

2.303 2 log t 0.15

k = 0.01295 min

Now

NaCN

Ni

      

–1

2

(Y)

Acetic anhydride

 → CH 3CH 2CH 2 NHCOCH 3

0.693 = K

= t1 / 2

→ CH 3CH 2CH 2 NH 2 H

(Z)

0.693 = 53.50 min 0.01295

21. Option (4) is correct. Explanation: The distance between the body centred atom and one corner atom is

16. Option (2) is correct. Explanation: K2Cr2O7 + H2SO4 + 3SO2 → K2SO4  + Cr2(SO4)3 + H2O

  (Green)

17. Option (4) is correct.

half of body diagonal.

22. Option (2) is correct. Explanation: O

O

Explanation: OH  H 2 O, H 2 SO4 CH 3  C ≡≡ CH  → CH 3  C == CH 2 HgSO4



Intermediate (Enol) (A)

Tautomerisation

O  CH 3  C  CH 3              

(Acetone) (B)

3a i.e., 2

H2(g), 1 atm Pd/C, C2H5OH

When H2 Pd/C catalyst is used then C=C is reduced at faster rate than C=O bond.

23. Option (4) is correct. Explanation: Electron withdrawing group at o and p-position w.r.t. OH group of phenol, increase the acidic strength.

81

Solutions Picric acid (2, 4, 6-trinitrophenol) is extremely more acidic among the given compounds due to the presence of three strong electron withdrawing groups (NO2 group) at ortho and para-positions,

24. Option (1) is correct. Explanation: Chloroprene is the monomer of neoprene. Cl  Polymerisation nH 2C ==C  CH ==CH 2  → Chloroprene

From the given diagram it is clear that Cl      CH 2  C = = CH  CH 2        n Neoprene





Ea = E’a + ∆H

Thus

Ea > ∆H

30. Option (3) is correct. Explanation: CH2OH

CHO

25. Option (1) is correct. Explanation: Copper (Cu) and zinc (Zn) are two metals which are purified by electrolytic refining. In this process, impure metal is used as anode and pure metal is used as cathode. Impurities from the blistered copper or impure zinc deposit as anode mud.

26. Option (1) is correct. Explanation: According to Kohlrausch’s law, the equivalent conductance of BaCl2 at infinite dilution, 1   λ ∞ of BaCl2 = λ ∞ of Ba2+ + λ ∞ of Cl– 2 127 + 76 =λ ∞ =139.5 2

27. Option (1) is correct. Explanation: As the complex gives 3 ions in aqueous solution thus, complex should be [Co(NH3)5NO2] Cl2 2+  [Co(NH3)5(NO2)]Cl2    [Co(NH3)5NO2] + 2Cl– –    2Cl + 2AgNO3 → 2AgCl + 2NO3

50% KOH

29. Option (3) is correct. Explanation: Here, Ea = activation energy of forward reaction E’a = activation energy of backward reaction ∆H = enthalpy of the reaction

Cl

Cl

Cl

This reaction is called cannizzaro’s reaction.

31. Option (2) is correct. Explanation: The main assumption of Langmuir adsorption isotherm are: (i) Adsorption takes place on the surface of the solid only till the whole of the surface is completely covered with a unimolecular layer of the adsorbed gas. (ii) Adsorption consists of two opposing processes condensation and evaporation. (iii) The rate of condensation depends upon the uncovered surface of the adsorbent available for condensation.

32. Option (2) is correct. Explanation: CH3 C ≡≡N

C ==N.MgBr

28. Option (3) is correct. Explanation: Lanthanides are the 14 elements of 3rd group and 6th period that are filling 4f subshell of antepenultimate shell from 1 to 14.

COO–K+

+

−δ

+δ

+ CH3MgBr → OCH3

OCH3

H3C—C ==O

OH 2H3O+

– NH3, – Mg Br

OCH3 (P)

OSWAAL CUET (UG) Sample Question Papers, CHEMISTRY

82

33. Option (1) is correct.

39. Option (1) is correct.

Explanation: H  C = O + NH2  NH2 →  R H R



 C= = N  NH 2 + H 2O 

Explanation: Since (ethylene diamine-en) is a chelating ligand/bidentate ligand, [CO(en3)]3+ is a more stable complex as compared to the other one. CH2 H2 C H2 N

NH2

34. Option (1) is correct. Explanation: The relation between standard Gibbs free energy and E°cell is ∆G° = – nFE°cell For the cell reaction 2Ag+ + Cu → Cu+2 + 2Ag

∆E°cell = 0.46 V



∆G° = –nF E°cell



∆G° = – 2 × 96500 × 0.46

M

40. Option (3) is correct. Explanation: nHO—CH2—CH2—OH + nHOOC—

Terephthalic acid

O 

= – 88780 J

∆G° = – 89.0 kJ

35. Option (2) is correct. Explanation: Glucose catabolism yields a total of 38 ATP. 38 ATP x 7.3 kcal/mol ATP = 262 kcal. Glucose has 686 kcal. Thus the efficiency of glucose metabolism is 262/686 x 100 = 38%.

—COOH

Ethylene glycol

O

—OCH2—CH2—O—C6H4—C— n



Terylene (Dacron)

41. Option (3) is correct. Explanation: Here the attack of nucleophile takes place on the opposite side of the leaving group in the substrate molecule. This results in the formation of inverted product.

36. Option (2) is correct. Explanation: Partial hydrolysis of XeF6 gives oxydifluorides, XeO2F2. XeF4 + H2O ® XeOF2 + 2HF Xenon oxydifluoride

37. Option (1) is correct. Explanation: Among the given halides aryl halides (C6H5X) is least reactive towards nucleophile as in the CX bond acquire some double bond character due to resonance. NO2 show –I effect hence facilitates the nucleophilic 3° halides are more reactive due to formation of more stable carbocation. Thus order of reactivity of CX bond towards nucelophile is X X | | NO2 < (CH3)2CH–X < (CH3)3C–X + | NO2 (I) (II) (III) (IV)

38. Option (4) is correct. Explanation: All soaps are made by boiling fats or oils with suitable hydroxide. Variations are made by adding different raw materials.

This occurs in SN2 reaction (bimolecular nucleophilic substitution reaction).

42. Option (3) is correct. 43. Option (3) is correct. Explanation: Initial temperature, T1 = 20 + 273 = 293 K Final temperature, T2 = 35 + 273 = 308 K R = 8.314 JK–1 mol–1 As rate becomes double on raising temperature r2 =2 r2 = 2r1 or ∴ r1 As rate constant, k ∝ r k2 =2 ∴ k1 According to Arrhenius equation we know that

log

 T1 − T2  k2 Ea = −   k1 2.303 R  T1T2 

83

Solutions log 2 =

⇒

 293 − 308  − Ea   2.303 × 8.314  293 × 308 

Ea = 34.7 kJ mol–1

(C)

(5)

Molecular formula of calamine is ZnCO3.

(D)

(3)

Molecular formula of cryolite is Na3AlF6.

44. Option (1) is correct. Explanation: MnO2 + 4HCl ® MnCl2 + 2H2O + Cl2 (Greenish yellow gas) NH3 + 3Cl2 ® NCl3 + 3HCl When excess of chlorine reacts with ammonia then NCl3 and HCl will form. In this reaction on left-hand side chlorine has (−3) oxidation state and on the right-hand chlorine has (+3) oxidation state.

Case Based 47. Option (2) is correct. Explanation: Negative deviation

48. Option (3) is correct. Explanation: ΔTf =iKf​m,  where i=1 for glucose. glucose​

Tf Δ = 1 × Kf​× 0.01 In case of MgCl2 → Mg2+ + 2Cl−, where i = 3, MgCl 2

∆Tf

45. Option (3) is correct. Explanation:

= 3 × 0.01 × Kf

MgCl 2

⇒ ∆Tf

glucose​

= 3 × ΔTf

ence, the depression in freezing point H of MgCl2​  is three times that of glucose.

49. Option (1) is correct. Explanation: Since the boiling point of the solution is more than the boiling point of the individual components in the solution, it indicates that the vapour pressure exerted by the solution is less than the expected, as boiling starts when vapour pressure equals the atmospheric pressure. Hence, the solution shows a negative deviation from the Raoult’s law.

46. Option (2) is correct. Explanation: Column Column I II

Explanation

(A)

(2)

Pendulum is made up of nickel steel.

(B)

(4)

Molecular formula of malachite is CuCO3.Cu(OH)2.

50. Option (2) is correct. Explanation: Na2SO4 will release 3 moles of ions/ moles of Na2SO4 in the aqueous solution, and Boiling point being a colligative property, the Boiling point of this solution will be the highest as other solutions release only 2 ions each.



10

SOLUTIONS OF Question Paper 1. Option (2) is correct.

O

Explanation: Coordination number of nickel in [Ni(C2O4)3]4– is 6 because C2O42– is a bidentate ligand.

F

F Xe

2. Option (3) is correct. Explanation: The central dogma of molecular genetics is Transcription

Translation

  

F Xe

F

DNA → RNA → Proteins

F

F

F

XeOF4 Square Pyramidal sp3d2

F XeF4 Square Planar sp3d2

8. Option (2) is correct.

3. Option (4) is correct.

Explanation: CFSE for octahedral complex is given by general formula as follows: CFSE = [– 0.4 (t2g electrons) + 0.6 (eg electrons)] D0

Explanation:

3 1 For Mn+3 ⇒ 3d4 → t2 g e g

In rest all options benzaldehyde is obtained.

4. Option (1) is correct. Explanation: When (NH4)2Cr2O7 is heated then N2 gas is evolved. (NH4)2Cr2O7 → Cr2O3 + 4H2O + N2 ↑

5. Option (1) is correct. Explanation: Carbon and hydrogen are not suitable reducing agents for metal sulphides.

6. Option (1) is correct. Explanation: CH3CH2NH2 + CHCl3 + 3KOH Warm

→ CH 3CH 2 NC + 3KCl + 3H 2O When any (aliphatic or aromatic) primary amines warmed with chloroform and an alcoholic solution of KOH form isocyanide or carbylamine.

7. Option (3) is correct. Explanation: F F

F Xe

F

Xe F

F XeF6 Distorted octahedral sp3d3

O

O

XeO3 Pyramidal sp3

O

CFSE = [(– 0.4 × 3) + (0.6 × 1)] D0 = – 0.6D0 3 2 For Fe3+, 3d5 → t2 g e g

CFSE = [(– 0.4 × 3) + (0.6 × 2)] = 0 5 2 For Co+2, 3d7 → t2 g e g

CFSE = [(– 0.4 × 5) + (0.6 × 2)] = – 0.8D0 4 2 For Cd3+, [3d6] → t2 g e g

CFSE = [(– 0.4 × 4) + (0.6 × 2)] = – 0.4D0

9. Option (1) is correct. Explanation: Zirconium and Titanium are purified by van-Arkel method. 1800°C

600°C

Zr + 2I 2 → ZrI 4  → Zr + 2I 2

Impure

Pure

10. Option (3) is correct.     Explanation: In peptide linkage   C  NH  ||    O the carboxyl group of one amino acids molecule forms an amide by combination with the amino group of next amino acid. H O O H R H 2

N C H

C

C

N

N

C

H

O

R'

C C

H

R3

85

Solutions 11. Option (4) is correct.

16. Option (4) is correct.

Explanation: Cetyl trimethyl ammonium bromide forms cationic micelles above certain concentration. In the molecules of detergents, the negative ions aggregate to form a micelle of colloidal size. In polar medium like water the negative ion has a long hydrocarbon chain and a polar group –Br– at one end and on the other hand it has N+ ion thus cationic micelle is formed.

12. Option (3) is correct.

17. Option (1) is correct.

Explanation: The transition elements have a general electronic configuration as (n – 1) d1–10 ns1–2.

13. Option (2) is correct. Explanation: Al2O3 ionises as 3– 3+  Al2O3     Al   +   AlO3 (Cathode) (Anode) So, the reaction at cathode will be Al+3 + 3e– → Al (3F) (27g) Mass of Al deposited by 3F of electricity = 27 g Mass of Al deposited by 4.0 × 104 × 6 × 3600 C of electricity 27 × 4 × 10 4 × 6 × 3600 g = 3F

= 8.1 × 10 4 g

Explanation: For the reaction A + B → Product Rate ∝ [A]x [B]y…(1) where, x and y are order w.r.t. A and B respectively. The rate of the reaction decreases by a factor of 4 if the concentration of reactant B is doubled. r ∝ [A]x [2B]y …(2) 4 From equation (1) and (2), we get y

⇒ y = –2 Hence order of reaction w.r.t. B is – 2

15. Option (1) is correct. Explanation: NH CH3 N-methyl aniline (2°-amine)

Explanation: For t1/2 of first order reaction 0.693 t1 / 2 = k i.e., independent of initial concentration of the reactant.

18. Option (4) is correct. Explanation: It is Reimer-Tiemann reaction. The electrophile formed is: CCl2 (Dichlorocarbene) according to the following reaction: Θ

 CHCl 3 + OH −    CCl 3 + H 2O Θ

CCl 3 →

• − • CCl 2 + Cl Electrophile

19. Option (4) is correct. Explanation: From E° values of M2+/M, we have

14. Option (2) is correct.

 1 4=   2

Explanation: Kohlrausch’s law states that the equivalent conductance of an electrolyte at infinite dilution is equal to the sum of the equivalent conductance of the component ions. λ∞ = λa + λc where, λ a = equivalent conductance of the anion lc = equivalent conductance of the cation.

E°/V   Cr   Mn   Fe   Co M2+/M

– 0.90 – 1.18 – 0.44 – 0.28

0

E value for Mn is more negative than expected from general trend due to extra stability of halffilled Mn2+ ion. Thus, correct order should be    Mn > Cr > Fe > Co But an examination of E° values for redox couple Mn3+/M2+ shows that Cr+2 is strong reducing agent (E°M3+/M2+ = 0.41 V) and liberates H2 from dilute acids. 2Cr+2(aq) + 2H+(aq) → 2Cr3+(aq) + H2 ↑ (g) Thus, correct order is Mn > Fe > Cr > Co.

20. Option (2) is correct. Explanation: From Kohlrausch’s Law, λ ∞ for NaCl = l + + λ – 

…(1)

for HCl = λH+ + λCl– 

…(2)

Na

+ NaNO2 + HCl or HNO2 CH3 N—N==O

N-nitroso-N-methylaniline (nitroso compound) (yellow only liquid)

∞

Cl

λ ∞ for C2H5COONa = lNa+ + λC Thus λ ∞ for C2H5COOH, Adding on (2) + (3) – (1), we get λ ∞ for CH3CH2COOH

λ ∞ (C

– 2H5COO

 …(3)

+ λ ∞ (HCl) – λ ∞ (NaCl)    = (91 + 426.16 – 126.45) S cm2 = 390.71 S cm2 2H5COONa)

86

OSWAAL CUET (UG) Sample Question Papers, CHEMISTRY

21. Option (4) is correct.

25. Option (3) is correct.

Explanation: = CH O = CH NOH | | H C  OH H  C  OH | | + NH 2 OH HO  C  H HO  C  H → | | − H2O H  C OH H  C  OH | | H  C  OH H  C  OH | | CH 2OH CH 2OH D-(+)-Glucose

Glucoxime

22. Option (1) is correct. Explanation: For the absorption of visible light presence of unpaired d-electrons is the necessity. In [Ni(CN)4]2– Ni+2 = [Ar] 3d8 4s0 ∴ [Ni(CN)4]2– orbitals 3d

××

4s

4p

××

×× ××

dsp2 –

Pairing occur because CN is a strong field ligand. Thus, [Ni(CN)4]2– does not contain any unpaired electron so it does not absorb visible light. All other in the option contains pair of electrons thus absorb visible light.

Explanation: Streptomycin, chloromycetin and penicillin are antibiotics while novalgin is an analgesic.

26. Option (3) is correct. Explanation: Normality = 1.5 N Equivalent weight of H2O2 = 17 g mol–1 So, strength of the solution, S = E × N = 17 × 1.5 = 25.5 2H2O2  → 2H2O + O2 2 × 34 = 68 g ∴ 68 g of H2O2 produce O2 at NTP = 22.4 L 22.4 25.5 g of H2O2 will produce = × 25.5 68    = 8.4 L of O2

27. Option (3) is correct. Explanation: Polystyrene is a chain growth polymer. CH2CH3 CH2=CH2

Fe2O3/Cr2O3

AlCl3

650°C

CH=CH2 (C6H5CO)2O2

23. Option (1) is correct. Explanation: Give reaction 2A + B → 3C + D 1 d[A] d[B] Rate of reaction = − = − 2 dt dt

1 d[C] d[D] = = 3 dt dt

n Polystyrene

28. Option (1) is correct. Explanation:

O

Hence, answer (1) is wrong.

24. Option (4) is correct. Explanation:

——CH—CH2—

Tetrahedral (sp3)

Xe O

O

O

29. Option (2) is correct. Explanation: On addition of electrolyte charge of colloidal particle will get neutralised and thus coagulation or precipitation of colloidal solution will occur.

30. Option (4) is correct. Explanation: Diamond is like ZnS (Zinc blende). Carbon forming ccp (fcc) and also occupying half of tetrahedral voids. Total number of carbon atoms per unit cell 1 1 = 8× + 6× + 4 =8 8 2 (tetrahedral (corners)



(face centered)

void)

87

Solutions 31. Option (3) is correct.

36. Option (3) is correct. Explanation: Alumina Al2O3 before subjecting to electrolytes, is mixed with fluorspar (CaF2) and cryolite (Na3AlF6) which lower its melting point and make it more conducting.

Explanation:

37. Option (3) is correct. Explanation: Chloroprene (2-chloro-1, 3-butadiene) on addition polymerisation gives neoprene.

32. Option (2) is correct. Explanation: Acidic strength of hydrides increases with increase in molar mass. Thus, order of acidic strength is HF < HCl < HBr < HI H2O < H2S < H2Se < H2Te NH3 < PH3 < AsH3 < SbH3 And as acidic strength increases, pKa decreases Thus, order of pKa is given as H2O > H2S > H2Se > H2Te

33. Option (1) is correct. Explanation: In phenols, the presence of electrons releasing group decreases the acidity whereas the presence of electron withdrawing groups increases the acidity as compared to phenol. Among meta and para nitrophenols the latter is more acidic due to presence of NO2 group at para position stabilizes the phenoxide ion to a greater extent than at meta position. Thus, order of acidic strength is: OH

OH

OH

OH

38. Option (4) is correct. Explanation: Catalyst speed up the reaction but it does not shift the position of equilibrium because catalyst reduces the height of barrier by providing an alternative path for the reaction and lowers the activation energy. However, the lowering in activation energy is to the some extent for the forward as well as the backward reaction.

39. Option (3) is correct. Explanation: Phenols are much more acidic than alcohols due to stabilization of phenoxide ion due to resonance. OH In case of due to +I effect of CH2 O–

CH3 it destabilise the

>

CH2

OH

CH3 >

>

NO2

ion but in

NO2

due to I effect of NO2

case of

NO2

34. Option (3) is correct. Explanation: In face centred cubic lattice The contribution of eight atoms of face centred 1 cubic cell = 8 × = 1 atom 8 The contribution of six face centred atoms 1 = 6× 2 =3 Therefore Z = 1 + 3 = 4 (no. of atoms per unit cell)

O–

NO2

is easy.

40. Option (3) is correct. Explanation: Hexagonal close packing has the coordination number equal to 12.

41. Option (2) is correct. Explanation: H

35. Option (1) is correct. Explanation: As molarity is dependent on volume of solution and volume rises with increase in temperature. Molarity is inversely proportional to temperature. So, as temperature increases, volume increases and molarity decreases.

thus release of H+

it stabilises the

2

O

OH

O

O

O–

OH H2O –OH

H O

O

H 2

88

O

OH

O

O

OSWAAL CUET (UG) Sample Question Papers, CHEMISTRY 46. Option (1) is correct.

O–

OH H2O –OH

H O

O

H2 O ∆

Explanation: As the hybridization of central metal in [Co(NH3)6]3+ complex ion is sp3d2 and coordination number of Co+3 is 6 so its geometry is octahedral.

47. Option (2) is correct. O

42. Option (4) is correct. Explanation: In the silver plating of copper K[Ag(CN)2] is used instead of AgNO3. Due to less availability of Ag+ ions Cu cannot displace Ag from [Ag(CN)2]– ion.

43. Option (4) is correct. Explanation: When C2H5I is mixed with C2H5OH, there is change in enthalpy and volume, so it is an example of non-ideal solution.

Explanation: A primary alkyl halide would prefer to undergo SN2 reaction.

48. Option (4) is correct. Explanation: (CH) 3C-I being a tertiary alkyl halide will most readily undergo S N1 reaction.

49. Option (3) is correct. Explanation:

44. Option (4) is correct. Explanation: Chloramphenicol is an antibiotic broad spectrum.

45. Option (1) is correct. Explanation: Adsorption is the property of a substance to hold or to concentrate liquid or dissolved substances upon its surface. Solids adsorb greater amount of substance at lower temperature and in general adsorption decreases with increasing temperature.

50. Option (1) is correct. Explanation: C6H5-C+H2 is stable carbocation so favours the progress of reaction by SN1 mechanism.

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