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Page 3 : EBD_8336, , Corporate, Office, , DISHA PUBLICATION, , No part of this publication may be reproduced in, any form without prior permission of the publisher., The author and the publisher do not take any legal, responsibility for any errors or misrepresentations that, might have crept in. We have tried and made our best, efforts to provide accurate up-to-date information in, this book., , All Right Reserved, , © Copyright, Disha, , 45, 2nd Floor, Maharishi Dayanand Marg,, Corner Market, Malviya Nagar, New Delhi - 110017, Tel : 49842349 / 49842350, , By:, Kalpana Bhargav, Typeset by Disha DTP Team, , www.dishapublication.com, , www.mylearninggraph.com, , Books &, ebooks for, School &, Competitive, Exams, , Etests, for, Competitive, Exams, , Write to us at
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CONTENTS, 1., , Some Basic Concepts of Chemistry 1-10, Topic 1 : Significant Figures, Laws of Chemical, Combinations and Mole Concept, Topic 2: Percent Composition and Empirical, Formula, Topic 3: Stoichiometric Calculations, , 2., , Structure of Atom, 11-20, Topic 1 : Atomic Models and Dual Nature of, Electromagnetic Radiation, Topic 2 : Bohr's model for Hydrogen Atom, (Emission and Absorption Spectra), Topic 3: Dual Behaviour of Matter and, Heisenberg Uncertainty Principle, Topic 4 : Quantum Mechanical Model of Atom, , 3., , 4., , 5., , 6., , Classification of Elements and, Periodicity in Properties, 21-27, Topic 1: Modern Periodic Table, Topic 2: Periodic Trends in Properties of, Elements, Chemical Bonding and Molecular, Structure28-48, Topic 1: Electrovalent, Covalent and, Co‑ordinate Bonding, Topic 2: Octet rule, Resonance and Hydrogen, Bonding, Topic 3: Dipole Moment and Bond Polarity, Topic 4: VSEPR Theory and Hybridisation, Topic 5: Valence Bond and Molecular Orbital, Theory, States of Matter, 49-57, Topic 1: Gas laws and Ideal gas Equation, Topic 2: Kinetic Theory of Gases and Molecular, Speeds, Topic 3 : van der Waal's Equation and, liquefaction of Gases, Topic 4: Liquid State, Thermodynamics, 58-69, Topic 1: First Law and Basic Fundamentals of, Thermodynamics, Topic 2: Laws of Thermochemistry, Topic 3: Entropy and Second Law of, Thermodynamics, Topic 4: Spontaneity and Gibb's Free Energy, , 7., , Equilibrium, 70-89, Topic 1: Law of Mass Action, Equilibrium, Constant (Kc and Kp) and its Application, Topic 2: Relation between K, Q and G and, Factors Effecting Equilibrium, Topic 3: Theories of Acids and Bases, Ionic, Product of Water and pH Scale, Topic 4: Ionisation of Weak Acids and Bases, and Relation between Ka and Kb, Topic 5: Common Ion Effect, Salt Hydrolysis,, Buffer Solutions and Solubility Product, , 8., , Redox Reactions, 90-93, Topic 1: Oxidation and Reduction Reactions, Topic 2: Oxidation Number, Topic 3: Disproportionation and Balancing of, Redox Reactions, Topic 4: Electrode Potential and Oxidising,, Reducing Agents, , 9., , Hydrogen, 94-97, Topic 1: Preparation and Properties of, Hydrogen, Topic 2: Preparation and Properties of Water, Topic 3: Preparation and Properties of, Hydrogen Peroxide, , 10. The s-Block Elements, 98-104, Topic 1: Preparation and Properties of Alkali, Metals and their Compounds, Topic 2: Some Important Compounds of, Sodium, Topic 3: Preparation and Properties of Alkaline, Earth Metals and their Compounds, Topic 4: Some Important Compounds of, Calcium, 11. The p-Block Elements, (Group 13 & 14), Topic 1: Boron Family, Topic 2: Carbon Family, , 105-109, , 12. Organic Chemistry - Some Basic, Principles and Techniques, 110-122, Topic 1: Classification and Nomenclature of, Organic Compounds, Topic 2: Isomerism in Organic Compounds, Topic 3: Concept of Reaction Mechanism in, Organic Compounds
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EBD_8336, 13. Hydrocarbons, Topic 1: Alkanes, Topic 2: Alkenes, Topic 3: Alkynes, Topic 4: Aromatic Hydrocarbons, , 123-141, , 14. Environmental Chemistry, Topic 1: Air Pollution, Topic 2: Water and Soil Pollution, , 142-144, , 15. The Solid State, 145-151, Topic 1: Properties and Types of Solids, Topic 2: Crystal Structure of Solids, Topic 3: Cubic System and Bragg's Equation, Topic 4: Imperfection in Solids, 16. Solutions, 152-162, Topic 1: Solubility and Concentration of, Solutions, Topic 2: Vapour Pressure, Laws of Solutions, and Ideal, Non-ideal Solutions, Topic 3: Colligative Properties and Abnormal, Molecular Masses, 17. Electrochemistry, 163-174, Topic 1: Conductance and Conductivity, Topic 2: Electrolysis and Types of Electrolysis, Topic 3: Cells and Electrode Potential, Nernst, Equation, Topic 4: Commercial Cells and Corrosion, 18. Chemical Kinetics, 175-187, Topic 1: Rate of Reaction, Rate Laws and Rate, Constant, Topic 2: Order of Reaction and Half Life Period, Topic 3: Theories of Rate of Reaction, 19. Surface Chemistry, 188-192, Topic 1: Adsorption, Topic 2: Catalysis and Theories of Catalysis, Topic 3: Colloids and Emulsions, 20. General Principles and Processes, of Isolation of Elements, 193-195, Topic 1: Occurrence of Metals, Topic 2: Metallurgical Processes, Topic 3: Purification and Uses of Metals, 21. The p-Block Elements, (Group 15, 16, 17 and 18), Topic 1: Nitrogen Family, Topic 2: Oxygen Family, Topic 3: Halogen Family, Topic 4: Noble Gases, , 196-209, , 22. The d-and f-Block Elements, 210-222, Topic 1: Characteristics of d-Block Elements, Topic 2: Compounds of Transition Metals, Topic 3: Lanthanoids and Actinoids, 23. Coordination Compounds, 223-240, Topic 1: Coordination Number, Nomenclature, and Isomerism of Coordination Compounds, Topic 2: Magnetic Moment, Valence Bond, Theory and Crystal Field Theory, Topic 3: Organometallic Compounds, 24. Haloalkanes and Haloarenes 241-254, Topic 1: Preparation and Properties of, Haloalkanes, Topic 2: Preparation and Properties of, Haloarenes, Topic 3: Some Important Polyhalogen, Compounds, 25. Alcohols, Phenols and Ethers 255-268, Topic 1: Preparation and Properties of Alcohols, Topic 2: Preparation and Properties of Phenols, Topic 3: Preparation and Properties of Ethers, 26. Aldehydes, Ketones and, Carboxylic Acids, 269-290, Topic 1: Methods of Preparation of Carbonyl, Compounds, Topic 2: Properties of Carbonyl Compounds, Topic 3: Preparation and Properties of, Carboxylic Acids, 27. Amines, 291-304, Topic 1: Aliphatic and Aromatic+ Amines, Topic 2: Amides, Cyanides and Isocyanides, Topic 3: Nitrocompounds, Alkyl Nitrites and, Diazonium Salts, 28. Biomolecules, 305-315, Topic 1: Carbohydrates and Lipids, Topic 2: Amino Acids and Proteins, Topic 3: Nucleic Acid and Enzymes, Topic 4: Vitamins and Hormones, 29. Polymers, 316-322, Topic 1: Classification of Polymers, Topic 2: Preparation and Properties of, Polymers, Topic 3: Uses of Polymers, 30. Chemistry in Everyday Life, , 323-324, , 31. Nuclear Chemistry, , 325-328
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NEET, , What you should know about NEET, NEET ALL INDIA MERIT LIST, AIR will include the rank numbers and names of all the candidates who appear for, NEET. It will not take into accord the candidate’s category or their stance towards, the AIQ reservation., • With the help of all the merit list, candidates can check the overall positions of, all the candidates who have appeared for NEET., • In this merit list, the factors like qualifying status and the state from which a, candidate belongs will not be considered while making this merit list., • NTA will provide the All India Rank and AIQ Merit List along with the other, required information to DGHS., • State medical councils prepare their own merit lists for counselling based, on factors such as the applicant’s eligibility on the basis of their domicile, requirement., • A state merit list will be released by the state medical councils and it will depict, the state rank of candidates on a comparative scale with applicants from the, same state., , MERIT LIST FOR 15% OF ALL INDIA QUOTA SEATS, NTA will prepare the merit list for 15% of the All India Quota (AIQ) admissions, by numbering the candidates who qualify NEET 2020 by scoring equal to or more, marks equivalent to the cutoff., • The merit list will have equal number of names as seats under All India Quota., • For such candidates, counselling will be done on the basis of their All India, Quota Rank and not their state rank., • There will also be a waiting list equal to 4 times the merit list or so if required, by the DGHS (Director General of Health Services), • Andhra Pradesh and Telangana will now participate in 15% All India Quota, Counselling and do not require a self declaration form anymore., • Candidates from Jammu and Kashmir still need to produce the self declaration, from for participating in All India Quota Counselling., (i)
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EBD_8336, , NEET, , • Merit List 85% of Seats, • For the remaining 85% of the seats, NTA will forward the results to DGHS, to be, forwarded to the state medical councils and mandated authorities., • Admitting authorities will invite candidates for counselling and a separate merits, list will be released on the basis of All India rank., • State counselling will be conducted by the respective counselling authorities., Candidates must check their websites to stay updated., • Deemed private universities will conduct their own counselling procedure, • Candidates who wish to take admission in AFMC must apply separately on their, website since they will hold a second screening process., , NEET TIE-BREAKING CRITERIA, If two or more candidates score equal marks, than the following process of, comparison shall be followed to resolve the situation:, • Candidate with higher marks in Zoology and Botany will be prioritized., • Candidate with a better score in Chemistry shall be preferred., • Candidate with lesser incorrect answers shall be prioritized., • As the last resort to, if the tie persists, candidate older in age will be preferred., , ELIGIBILITY CRITERIA FOR NEET 2021, As per NTA, candidates willing to appear in NEET 2021 should qualify the eligibility, criteria mentioned below., • Age Limit – Candidates must attain 17 years of age by December 31, 2021 to, sit for NEET 2021. There is no upper age limit for NEET 2021 as per an interim, order of the Supreme Court. However, the candidature of aspirants more than, 25 years of age is subject to the Supreme Courts’ orders., • Educational Qualification – Candidates who have completed or appearing in class, 12th with minimum 5 subjects including English, Chemistry, Physics as compulsory, subjects can apply. Candidates of distance learning or private students and those, who have biology as an additional subject are also eligible for the exam., • Minimum Percentage Required – General candidates should score a minimum, 50%, general PwD – 45%, OBC/SC/ST – 45% and OBC/SC/ST PWD – 40%., • Number of Attempts – There is no limit on the number of attempts for NEET, 2021. Candidates can appear for the exam as many times as they want., (ii)
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NEET, , NEET 2021 – NEXT YEAR PLAN, • CBSE has decided to reduce the syllabus up to 30% from Class IX to XII. The, syllabus has been reduced from all the subjects including physics, Chemistry,, and Biology which are major subjects for NEET. This could affect the preparation, of the candidates as they will not be able to understand the basics of the, topics and will also not be able to solve the questions which are interrelated to, curtailed topics. Read More Syllabus reduction impact on NEET 2021, • With the reduction in the syllabus by CBSE, there are speculations that the, syllabus for NEET 2021 could also be revised but no official announcement has, been made yet by the respective authorities., • The official notification of NEET 2021 is expected to release in the month of, December 2020., , REASONS THAT COULD AFFECT NEET 2021 EXAM DATE, • Due to COVID-19 schools are closed and because of the high strength of students, in the schools, we can expect that the school could be the last place to open, after the lockdown is over., • However, all the students are attending the classes in the online mode but still, to complete the whole syllabus online is the biggest challenge., • This may lead to the delay in the final exams of class 12 which could further lead, to the delay of NEET 2021., , MAJOR CHANGES IN NEET, The NEET 2020 will be experiencing few major striking changes in its pattern and, conducting nature with the focus being on more accuracy of identification of, applicants to ensure 100% transparency in its examining procedure. Applicants now, have to upload Class X certificates, left-hand thumb impressions and also provide, their roll numbers for Class XII Board Exams along with the standard signature and, scanned photograph - both in passport and postcard sizes., • One Exam for all Medical Aspirants: The merger of AIIMS and JIPMER will only, see all of 1207 AIIMS MBBS and 200 JIPMER MBBS seats in the respective, institutes to be contested for, in a single analysis of NEET-UG performance of, candidates., (iii)
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EBD_8336, , NEET, , • UT applicants need to self-declare their contesting consideration: With the new, formation of Union Territories of India, the likes of Kashmir and Ladakh have, thought of pulling off from the 15% allocation of the All India Quota Scheme., Therefore, aspirants from these UTs need to fill in the form for self-declaration, for seat allotment under AIQ., • Application Fee Hike: The fee for application has been hiked to ` 1500 for general, candidates with respect to ` 1400 from last year and applicants can now even, upload live photographs while they fill out their form. The amount respects to, ` 1400 for General-EWS applicants and ` 800 for reserved category applicants, for 2020., • Flexible Edit Window: As the maximum of Class 12 medical students will get, their admit cards for board exam after the registration for NEET is over. Thus,, the edit window will be re-opened by the authority for NEET Application form, 2020 for the students of Class 12 to punch in their roll numbers., • More Exam Centres: This year, the exam of NEET 2020 will be conducted across, the nation in exactly 155 cities in India with the inclusion of the union territory, of Ladakh as the latest entry., • Upper Age Limit Relaxation: For NEET 2020, the upper age limit is capped at 25, years as on 31st December 2019, while the lower age limit is capped at 17 years, as on 31st December 2019. There is a relaxation of flat 5 years for reserved, category candidates in the upper age limit., However, general applicants above the 25-age mark can appear for the NEET 2020, test with their candidature being considered or not, is still in a turmoil for the, awaiting verdict of the Supreme Court regarding the same., An aspirant who applies for NEET has to meet the eligibility criteria of NEET which, specifies the academic qualifications, age limit, pass percentage, nationality etc, that are mandatory to appear for the exam. The NEET eligibility criteria are based, on regulations of Graduate Medical Education 1997 Act. Candidates must meet the, NEET eligibility as given failing which they will not be allowed to appear for the, exam. It should be noted that merely fulfilling the eligibility criteria of NEET is not, sufficient for admissions. NEET admission criteria and cutoff for MBBS, BDS and, other medical courses will be separate for All India and state counselling., , (iv)
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1, , Some Basic Concepts of Chemistry, , 1, , Some Basic Concepts, of Chemistry, , Trend Analysis with Important Topics & Sub-Topics, 2020, Topic Name, , Sub-Topic, , Significant figures,law, of chemical, combinations and, mole concept, , mole concept, , Stoichiometric, calculations, LOD - Level of Difficulty, , 1, , 2., , 3., , 4., , E, , 1, , stoichiometric, calculations, E - Easy, , Topic 1 : Significant Figures, Laws of Chemical, Combinations and Mole Concept, 1., , 2019, , 2018, , 2017, , 2016, , QNS. LOD QNS. LOD QNS. LOD QNS. LOD QNS. LOD, , Which one of the followings has maximum, number of atoms ?, [2020], (a) 1 g of Mg(s) [Atomic mass of Mg = 24], (b) 1 g of O2(g) [Atomic mass of O = 16], (c) 1 g of Li(s) [Atomic mass of Li = 7], (d) 1 g of Ag(s) [Atomic mass of Ag = 108], The number of moles of hydrogen molecule, required to produce 20 moles of ammonia, through Haber’ s process is:, [2019], (a) 10, (b) 20, (c) 30, (d) 40, In which case is number of molecules of water, maximum?, [2018], (a) 18 mL of water, (b) 0.18 g of water, (c) 10–3 mol of water, (d) 0.00224 L of water vapours at 1 atm and, 273 K, A mixture of gases contains H2 and O2 gases in, the ratio of 1 : 4 (w/w). What is the molar ratio of, the two gases in the mixture ?, [2015], , A - Average, , 5., , 6., , 7., , E, , 1, , E, , 1, , A, , D - Difficult, , Qns - No. of Questions, , (a) 4 : 1, (b) 16 : 1, (c) 2 : 1, (d) 1 : 4, The number of water molecules is maximum in :, [2015 RS], (a) 18 molecules of water, (b) 1.8 gram of water, (c) 18 gram of water, (d) 18 moles of water, If Avogadro number NA, is changed from, 6.022 × 1023 mol–1 to 6.022 × 1020 mol–1 this, would change :, [2015 RS], (a) the definition of mass in units of grams, (b) the mass of one mole of carbon, (c) the ratio of chemical species to each other, in a balanced equation., (d) the ratio of elements to each other in a, compound, When 22.4 litres of H2(g) is mixed with 11.2 litres, of Cl2(g), each at S.T.P., the moles of HCl(g), formed is equal to :, [2014], (a) 1 mole of HCl (g), (b) 2 moles of HCl (g), (c) 0.5 moles of HCl (g), (d) 1.5 moles of HCl (g)
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EBD_8336, 2, , 8., , 9., , 10., , 11., , 12., , 13., , 14., , 15., , 16., , 17., , CHEMISTRY, 1.0 g of magnesium is burnt with 0.56 g O2 in a, closed vessel. Which reactant is left in excess, and how much ?, [2014], (At. wt. Mg = 24 ; O = 16), (a) Mg, 0.16 g, (b) O2, 0.16 g, (c) Mg, 0.44 g, (d) O2, 0.28 g, Which has the maximum number of molecules, among the following ?, [2011 M], (a) 44 g CO2, (b) 48 g O3, (c) 8 g H2, (d) 64 g SO2, The number of atoms in 0.1 mol of a triatomic, gas is :, [2010], (NA = 6.02 ×1023 mol–1), (a) 6.026 × 1022, (b) 1.806 × 1023, (c) 3.600 × 1023, (d) 1.800 × 1022, What volume of oxygen gas (O2) measured at, 0°C and 1 atm, is needed to burn completely 1L, of propane gas (C3H8) measured under the same, conditions ?, [2008], (a) 7 L, (b) 6 L, (c) 5 L, (d) 10 L, Number of moles of MnO-4 required to oxidize, one mole of ferrous oxalate completely in acidic, medium will be :, [2008], (a) 0.6 moles, (b) 0.4 moles, (c) 7.5 moles, (d) 0.2 moles, Volume occupied by one molecule of water, (density = 1 g cm–3) is :, [2008], –23, 3, –, (a) 9.0 × 10 cm, (b) 6.023 × 10 23 cm3, (c) 3.0 × 10–23 cm3, (d) 5.5 × 10– 23 cm3, The number of moles of KMnO4 that will be, needed to react with one mole of sulphite ion in, acidic solution is, [2007], (a) 4/5, (b) 2/5, (c) 1, (d) 3/5, An element, X has the following isotopic, composition :, [2007], 200X : 90%, 199X : 8.0 %, 202X : 2.0%, The weighted average atomic mass of the, naturally occuring element X is closest to, (a) 201 amu, (b) 202 amu, (c) 199 amu, (d) 200 amu, The number of moles of KMnO4 reduced by one, mole of KI in alkaline medium is:, [2005], (a) one, (b) two, (c) five, (d) one fifth, The maximum number of molecules is present in, (a) 15 L of H2 gas at STP, [2004], (b) 5 L of N2 gas at STP, , 18., 19., , 20., , 21., , 22., , 23., , 24., , 25., , 26., , (c) 0.5 g of H2 gas, (d) 10 g of O2 gas, Which has maximum number of molecules?, (a) 7 g N2, (b) 2 g H2, [2002], (c) 16 g NO2, (d) 16 g O2, Specific volume of cylindrical virus particle is, 6.02 × 10–2 cc/g whose radius and length are 7 Å, & 10 Å respectively. If NA = 6.02 × 1023, find, molecular weight of virus, [2001], (a) 3.08 × 103 kg/mol (b) 3.08 × 104 kg/mol, (c) 1.54 × 104 kg/mol (d) 15.4 kg/mol, Assuming fully decomposed, the volume of CO2, released at STP on heating 9.85 g of BaCO3, (Atomic mass, Ba = 137) will be, [2000], (a) 2.24 L, (b) 4.96 L, (c) 1.12 L, (d) 0.84 L, Haemoglobin contains 0.334% of iron by weight., The molecular weight of haemoglobin is, approximately 67200. The number of iron atoms, (at. wt. of Fe is 56) present in one molecule of, haemoglobin are, [1998], (a) 1, (b) 6, (c) 4, (d) 2, The number of significant figures for the three, numbers 161 cm, 0.161 cm, 0.0161 cm are, (a) 3, 4 and 5 respectively, [1998], (b) 3, 4 and 4 respectively, (c) 3, 3 and 4 respectively, (d) 3, 3 and 3 respectively, The weight of one molecule of a compound, C60H122 is, [1995], (a) 1.2 × 10–20 gram (b) 1.4 × 10–21 gram, (c) 5.025 × 1023 gram (d) 6.023 × 1023 gram, In the final answer of the expression, ( 29.2 - 20.2) (1.79 ´ 10 5 ), 1.37, [1994], the number of significant figures is :, (a) 1, (b) 2, (c) 3, (d) 4, If NA is Avogadro’s number then number of, valence electrons in 4.2g of nitride ions (N3–) is, (a) 2.4 NA, (b) 4.2 NA, [1994], (c) 1.6 NA, (d) 3.2 NA, The molecular weight of O2 and SO2 are 32 and, 64 respectively. At 15°C and 150 mm Hg pressure,, one litre of O2 contains ‘N’ molecules. The, number of molecules in two litres of SO2 under, the same conditions of temperature and pressure, will be :, [1990]
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3, , Some Basic Concepts of Chemistry, , 27., , 28., , 29., , 30., , 31., , 32., , 33., , (b) N, (a) N/2, (c) 2N, (d) 4N, Boron has two stable isotopes, 10B (19%) and, 11B (81%). Average atomic weight for boron in, the periodic table is, [1990], (a) 10.8, (b) 10.2, (c) 11.2, (d) 10.0, The number of oxygen atoms in 4.4 g of CO2 is, [1990], (a) 1.2 × 1023, (b) 6 × 1022, (c) 6 × 1023, (d) 12 × 1023, The number of gram molecules of oxygen in, 6.02 × 1024 CO molecules is, [1990], (a) 10 g molecules, (b) 5 g molecules, (c) 1 g molecules, (d) 0.5 g molecules, What is the weight of oxygen required for the, complete combustion of 2.8 kg of ethylene ?, (a) 2.8 kg, (b) 6.4 kg, [1989], (c) 9.6 kg, (d) 96 kg, Ratio of Cp and Cv of a gas ‘X’ is 1.4. The number, of atoms of the gas ‘X’ present in 11.2 litres of it, at NTP will be, [1989], 23, 23, (a) 6.02 ×10, (b) 1.2 × 10, (c) 3.01 × 1023, (d) 2.01 × 1023, 1 c.c. N2O at NTP contains :, [1988], 1, ., 8, (a), ´ 10 22 atoms, 224, 6.02, (b), ´ 10 23 molecules, 22400, 1.32, (c), ´ 10 23 electrons, 224, (d) all the above, At S.T.P. the density of CCl4 vapours in g/L will, be nearest to :, [1988], (a) 6.87, (b) 3.42, (c) 10.26, (d) 4.57, Topic 2: Percent Composition and, Empirical Formula, , 34., , An organic compound contain s carbon,, hydrogen and oxygen. Its elemental analysis, gave C, 38.71% and H, 9.67%. The empirical, formula of the compound would be :, [2008], (a) CH3O, (b) CH2O, (c) CHO, (d) CH4O, , 35., , 36., , 37., , 38., , 39., , Percentage of Se in peroxidase anhydrase, enzyme is 0.5% by weight (at. wt. = 78.4) then, minimum molecular weight of peroxidase, anhydrase enzyme is, [2001], (a) 1.568 × 103, (b) 15.68, (c) 2.136 × 104, (d) 1.568 × 104, An organic compound containing C, H and O, gave on analysis C – 40% and H – 6.66%. Its, empirical formula would be, [1999, 94], (a) C3H6O, (b) CHO, (c) CH2O, (d) CH4O, An organic compound containing C, H and N, gave the following analysis :, C = 40% ; H = 13.33% ; N = 46.67%, Its empirical formula would be, [1998], (a) C2H7N2, (b) CH5N, (c) CH4N, (d) C2H7N, The percentage weight of Zn in white vitriol, [ZnSO 4.7H2O] is approximately equal to, ( Zn = 65, S = 32, O = 16 and H = 1), [1995], (a) 33.65 %, (b) 32.56 %, (c) 23.65 %, (d) 22.65 %, A metal oxide has the formula Z2O3. It can be, reduced by hydrogen to give free metal and, water. 0.1596 g of the metal oxide requires 6 mg, of hydrogen for complete reduction. The atomic, weight of the metal is, [1989], (a) 27.9, (b) 159.6, (c) 79.8, (d) 55.8, , Topic 3: Stoichiometric Calculations, 40. A mixture of 2.3 g formic acid and 4.5 g oxalic, acid is treated with conc. H2SO4. The evolved, gaseous mixture is passed through KOH pellets., Weight (in g) of the remaining product at STP, will be, [2018], (a) 1.4, (b) 3.0, (c) 4.4, (d) 2.8, 41. 20.0 g of a magnesium carbonate sample, decomposes on heating to give carbon dioxide, and 8.0 g magnesium oxide. What will be the, percentage purity of magnesium carbonate in, the sample ?, [2015 RS], (a) 75, (b) 96, (c) 60, (d) 84, 42. What is the mass of precipitate formed when 50, mL of 16.9% solution of AgNO3 is mixed with 50, mL of 5.8% NaCl solution ?, [2015 RS], (Ag = 107.8, N = 14, O = 16, Na = 23, Cl = 35.5), (a) 28 g, (b) 3.5 g, (c) 7 g, (d) 14 g
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EBD_8336, 4, , 43., , 44., , 45., , 46., , 47., , 48., , 49., , 1, 2, 3, 4, 5, 6, , CHEMISTRY, In an experiment it showed that 10 mL of 0.05 M, solution of chloride required 10 mL of 0.1 M, solution of AgNO3, which of the following will, be the formula of the chloride (X stands for the, symbol of the element other than chlorine):, [NEET Kar. 2013], (a) X2Cl, (b) X2Cl2, (c) XCl2, (d) XCl4, 6.02 × 1020 molecules of urea are present in, 100 mL of its solution. The concentration of, solution is :, [NEET 2013], (a) 0.01 M, (b) 0.001 M, (c) 0.1 M, (d) 0.02 M, What is the [OH] in the final solution prepared, by mixing 20.0 mL of 0.050 M HCl with 30.0 mL of, 0.10 M Ba(OH)2?, [2009], (a) 0.40 M, (b) 0.0050 M, (c) 0.12 M, (d) 0.10 M, 10 g of hydrogen and 64 g of oxygen were filled, in a steel vessel and exploded. Amount of water, produced in this reaction will be:, [2009], (a) 3 mol, (b) 4 mol, (c) 1 mol, (d) 2 mol, How many moles of lead (II) chloride will be, formed from a reaction between 6.5 g of PbO, and 3.2 g of HCl ?, [2008], (a) 0.044, (b) 0.333, (c) 0.011, (d) 0.029, 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 litre of 0.1M, H2SO4 solution is, [2007], (a) 16.65 mL, (b) 22.20 mL, (c) 5.55 mL, (d) 11.10 mL, The mass of carbon anode consumed (giving, only carbon dioxide) in the production of, 270 kg of aluminium metal from bauxite by the, Hall process is (Atomic mass: Al = 27) [2005], (a) 270 kg, (b) 540 kg, (c) 90 kg, (d) 180 kg, (c), (c), (a), (a), (d), (b), , 7, 8, 9, 10, 11, 12, , (a), (a), (c), (b), (c), (b), , 13, 14, 15, 16, 17, 18, , (c), (b), (d), (b), (a), (b), , 19, 20, 21, 22, 23, 24, , 50., , 51., , 52., , 53., , 54., , In Haber process 30 litres of dihydrogen and, 30 litres of dinitrogen were taken for reaction, which yielded only 50% of the expected product., What will be the composition of gaseous mixture, under the aforesaid condition in the end? [2003], (a) 20 litres ammonia, 25 litres nitrogen, 15 litres, hydrogen, (b) 20 litres ammonia, 20 litres nitrogen, 20 litres, hydrogen, (c) 10 litres ammonia, 25 litres nitrogen, 15 litres, hydrogen, (d) 20 litres ammonia, 10 litres nitrogen, 30 litres, hydrogen, In the reaction, 4 NH3 (g) + 5 O2 (g) ® 4 NO(g) + 6 H2O(l), When 1 mole of ammonia and 1 mole of O2 are, made to react to completion,, [1998], (a) 1.0 mole of H2O is produced, (b) 1.0 mole of NO will be produced, (c) all the oxygen will be consumed, (d) all the ammonia will be consumed, Liquid benzene (C6H6) burns in oxygen according, to the equation, 2C6H6(l) + 15O2(g) ¾® 12CO2(g) + 6H2O(g), How many litres of O2 at STP are needed to, complete the combustion of 39 g of liquid, benzene? (Mol. wt. of O2 = 32, C6H6 = 78), [1996], (a) 74 L, (b) 11.2 L, (c) 22.4 L, (d) 84 L, A 5 molar solution of H2SO4 is diluted from, 1 litre to a volume of 10 litres, the normality of, the solution will be :, [1991], (a) 1 N, (b) 0.1 N, (c) 5 N, (d) 0.5 N, One litre hard water contains 12.00 mg Mg2+., Mili-equivalents of washing soda required to, remove its hardness is :, [1988], (a) 1, (b) 12.16, (c) 1 × 10–3, (d) 12. 16 × 10–3, , (d), (c), (c), (d), , ANSWER KEY, 25 (a) 31 (a), 26 (c) 32 (d), 27 (a) 33 (a), 28 (a) 34 (a), , (b), (c), , 29, 30, , (b), (c), , 35, 36, , (d), (c), , 37, 38, 39, 40, 41, 42, , (c), (d), (d), (d), (d), (c), , 43, 44, 45, 46, 47, 48, , (c), (a), (d), (b), (d), (c), , 49, 50, 51, 52, 53, 54, , (c), (c), (c), (d), (a), (a)
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5, , Some Basic Concepts of Chemistry, , Hints & Solutions, (c) Number of atoms, , 3., , ol, ., , D, , 6., , 0.00224, = 10–4, 22.4, Molecules of water = mole × NA = 10–4 NA, (a) Ratio of weight of gases = wH : wO = 1 : 4, 2, 2, 1 4, Ratio of moles of gases = nH : nO = :, 2 32, 2, 2, 1 32, = 4 :1, \ Molar Ratio = ´, 2 4, (d) No. of moles of water, In 1.8 g of H2O = 0.1 moles, In 18 g of H2O = 1 moles, 1 mole contains 6.022 × 1023 molecules of water, therefore maximum number of molecules is in 18, moles of water., , 7., , 5., , (b) If 6.022 × 1023 changes to 6.022 × 1020/mol, then this would change mass of one mole of, carbon., , (a) H2(g) +, 22.4 L, = 1 mole, 1 – 0.5, , (4) Moles of water =, , 4., , Number of, molecules, , Mass of 12C carbon is used to define the atomic, mass unit. In this system, 12C is assigned a mass, of exactly 12 a.m.u., Earlier one mole was defined as the amount of, substance that contains as many particles as there, are atoms in exactly 12 g of the 12C isotope. From, November 2018, one mole is defined as exactly, 6.02214076 × 1023 constitutive particles, which, may be atoms, molecules, ions or electrons., Hence, if we change the value of NA from 6.022 ×, 1023 mol–1 to 6.022 × 1020 mol–1 then mass of one, mole of carbon will also change., The definition of mass is independent of Avogadro, number NA., , 18, NA = N, A, 18, (2) Molecules of water = mole × NA, 0.18, NA = 10–2 N, A, 18, (3) Molecules of water = mole × NA = 10–3 NA, , Divide by mol., mass, Multiply by mo l., mass, , =, , =, , iv, id, e, m by, as m, s o, l, , ul, tip, m ly b, as y, s m, , M, , Mass in, grams, , ol ., ym, yb s, i pl a s, u lt m, o l., M, ym, e b ss, v id m a, , 2., , Number of, moles, , Di, , W, ´ N A ´ atomicity, Molar mass, 1, ´ NA ´1, (a) Number of Mg atoms =, 24, 1, ´ NA ´ 2, (b) Number of O atoms =, 32, 1, (c) Number of Li atoms = ´ N A ´ 1, 7, 1, ´ N A ´1, (d) Number of Ag atoms =, 108, (c) N2 + 3H2 ¾® 2NH3, 3, 1 Mol NH3 = mol H2, 2, 3, 20 mol NH3 = × 20 mol H2 = 30 mol H2, 2, \ 30 moles of H2 are required., (a), (1) Mass of water = 18 × 1 = 18 g, Molecules of water = mole × NA, =, , ., , 1., , 8., , 11.2 L, = 0.5 mole, 0.5 – 0.5, , 0, , initial, , 2 × 0.5, , final, , Moles of HCl formed = 2 × 0.5 = 1, 1, (a) Initially Mg +, O 2 ¾® MgO, 2, 1g, or, , 9., , Cl2(g) ¾® 2HCl(g), , 0.56, 1, 0.56, mole, mole, 24, 32, 0.041 7 mole 0.0175 mole, (0.0417 – 2, 0 mole, × 0.0175), = 0.0067 mole, , \ Mass of Mg = 0.0067 × 24 = 0.16g, (c) No. of molecules, 44, = 1, N, Moles of CO2 =, A, 44
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EBD_8336, 6, , 10., 11., , CHEMISTRY, Moles of O3 =, , 48, =1, N, A, 48, , Moles of H2 =, , 8, = 4 , 4N, A, 2, , 14., , 5vol, 5L, , 2MnO -4 + 6H + + 5SO32 - ¾¾, ®, , 15., , From the above equation we find that we need, 5 L of oxygen at N.T.P. to completely burn 1 L of, propane at N.T.P., If we change the conditions for both the gases, from N.T.P. to same conditions of temperature, and pressure. The same results are obtained., i.e. 5 L is the correct answer., COO–, –, +, (b), 5, +2 MnO4 + 16 H, COO–, , 200 ´ 90 + 199 ´ 8 + 202 ´ 2, 90 + 8 + 2, 18000 + 1592 + 404, =, 100, , =, , In KMnO 4 the oxidation state of Mn is +7. In, acidic medium Mn takes up five electrons and, making it an oxidizing agent. In strongly alkaline, solution it takes up only 1 electrons and making it, much weaker oxidising agent. In neutral medium it, gives up 3 electrons to form MnO2., Acidic medium:, –, MnO4 + 8H+ + 5e– ® Mn2+ + 4H2O, –, 2–, Alkaline medium: MnO4 + e– ® MnO4, Neutral medium:, –, MnO4 + 4H+ + 3e– ® MnO2 + 2H2O, , (c) \ Volume occupied by 1 gram water = 1 cm3, or Volume occupied by, , 6.023 ´ 1023, molecules of water = 1 cm3, 18, 1, [\ 1g water =, moles of water], 18, , 19996, = 199.96 ; 200 amu, 100, , Average atomic mass, = (Mass of isotope A × % natural abundance) +, (Mass of isotope B × % natural abundance) +, ................., (% Natural abundance of A + % natural abundance, of B) + ........, , 2+, , 2 Mn +10CO2+8H2O, From above equation 2 moles of MnO4 required, to oxidise 5 moles of oxalate., Thus number of moles of MnO4– required to, oxidise one mole of oxalate = 2 × 2/5 = 0.4 moles, , 2Mn 2 + + 5SO4 2 - + 3H 2O, From the equation it is clear that, Moles of MnO4– require to oxidise 5 moles of, SO32 – are 2, Moles of MnO4– require to oxidise 1 mole of, SO32– are 2/5., (d) Average isotopic mass of, X =, , oxalate ion, , 13., , 1 ´ 18, cm 3 = 3.0×10–23 cm3., 6.023 ´ 10 23, i.e. the correct answer is option (c)., (b) The balance chemical equation is :, , =, , 64, =1, N, Moles of SO2 =, A, 64, (b) The number of atoms in 0.1 mol of a triatomic, gas = 0.1 × 3 × 6.023 × 1023., = 1.806 × 1023, (c) Writing the equation of combustion of, propane (C3H8), we get, C3 H 8 + 5O 2 ® 3CO 2 + 4H 2 O, 1vol, IL, , 12., , Thus, volume occupied by 1 molecule of water, , 16., , 17., , (b) In weak alkaline medium, the equation is:, –, –, 2Mn SO4– + H2O + I ¾® 2MnO2 + IO4– + 2O H, Hence, one mole of KI reduce, 2 moles of KMnO4., (a) No. of molecules in different cases, (a) Q 22.4 litre at STP contains, = 6.023×1023molecule of H2, \15 litre at STP contains =, , 15, ´ 6.023 ´ 10 23, 22.4, , (b) Q 22.4 litre at STP contains, = 6.023×1023 molecule of N2, \ 5 litre at STP contians =, , 5, ´ 6.023 ´ 10 23, 22.4, , (c) Q 2 g of H2= 6.023×1023 molecules of H2, 0.5, ´ 6.023 ´ 10 23, \ 0.5 g of H2=, 2, (d) Similarly 10 g of O2 gas, =, , 10, ´ 6.023 ´ 10 23 molecules, 32, , Thus (a) will have maximum number of molecules
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7, , Some Basic Concepts of Chemistry, 18., 19., , (b) 2g of H2 means one mole of H2, hence, contains 6.023 × 1023 molecules. Others have, less than one mole, so have less no. of molecules., (d) Specific volume (volume of 1 g) of, cylindrical virus particle = 6.02 × 10–2 cc/g, Radius of virus (r) = 7 Å = 7 × 10–8 cm, Length of virus = 10 × 10–8 cm, Volume of virus =, 22, pr 2 l = ´ (7 ´ 10 -8 ) 2 ´10 ´10 -8, 7, = 154 × 10–23 cc, Wt. of one virus particle =, , 154 ´ 10, , =, , 20., , -2, , 26., , 6.02 ´ 10, = 15400 g/mol = 15.4 kg/mol, (c) BaCO3 ® BaO + CO2, , As 197 g of BaCO3 will release, 22.4 litre of CO2 at STP, \ 1 g of BaCO3 will release, 22.4, =, litre of CO2, 197, And 9.85 g of BaCO3 will release carbon dioxide, 22 .4, ´ 9.85 = 1.12 litre of CO, =, 2, 197, (c) Given : Percentage of the iron = 0.334%;, Molecular weight of the haemoglobin, = 67200 and atomic weight of iron = 56. We know, that the number of iron atoms, , 27., 28., , 29., , 30., , Molecular wt. of haemoglobin ´ % of iron, 100 ´ Atomic weight of iron, , =, , 67200 ´ 0.334, =4, 100 ´ 56, (d) Number 161 cm, 0.161 cm and 0.0161cm have, 3, 3 and 3 significant figures respectively., , =, , 22., , All non-zero digits are significant and the zeros at, the beginning of a number are not significant., , 23., , (b) Molecular weight of C60H122, = (12 × 60) + 122 = 842., Therefore, weight of one molecule, =, =, , Molecular weight of C60 H122, Avogadro's number, 842, 23, , 6.023 ´ 10, , = 1.36 ´ 10-21 g ; 1.4 ´ 10-21 g, , (29.2 – 20.2)(1.79 ´ 105 ), 1.37, As the least precise number contains 3, significant figures therefore, answers should also, contains 3 significant figures., (a) No of moles of nitride ion, 4.2, = 0.3 mol = 0.3 ´ N A nitride ions., =, 14, Valence electrons = 8 × 0.3 NA = 2.4 NA, , (c), , Nitride ion has seven protons in the nucleus and, ten electrons surrounding the nucleus. Therefore, total no. of electrons is 10. Number of valence, electrons is (5 + 3) = 8., , ´ 6.02 ´ 10 23, , 197 g, , 21., , 25., , volume, specific volume, , Mol. wt. of virus = Wt. of NA particle, -23, , 24., , 31., , (c) According to Avogadro's law "equal, volumes of all gases contain equal numbers of, molecules under similar conditions of, temperature and pressure". Thus if 1 L of one, gas contains N molecules, 2 L of any gas under, the same conditions will contain 2N molecules., 19 ´ 10 + 81 ´ 11, (a) Average atomic mass =, 100, = 10.81, (a) 1 mol of CO2 = 44 g of CO2, 4.4, 4.4 g CO2 =, = 0.1 mol CO2, 44, = 6 × 1022 molecules, = 2 × 6 × 1022 atoms or 1.2 × 10 23 atoms of oxygen., (b) 6.02 × 1023 molecules of CO =1mole of CO, 6.02 × 1024 CO molecules = 10 moles of CO, = 10 g atoms of O = 5 g molecules of O2, ¾® 2CO + 2H O, (c) C2H4 + 3 O2 ¾, 2, 2, 28 kg 96 kg, As 28 kg of C2H4 undergo complete combustion, by 96 kg of O2, \ 2.8 kg of C2H4 undergo complete combustion, by 9.6 kg of O2., (a) Cp / Cv = 1.4 shows that the gas is diatomic., 22.4 litre at NTP º 6.02 × 1023 molecules, 11.2 L at NTP = 3.01 × 1023 molecules, No. of atoms in gas = 3.01 × 1023 × 2 atoms, = 6.02 × 1023 atoms, Cp, , 2, = 1 + , where F = degree of freedom of, Cv, F, the gas molecules, For mono atomic gas, F = 3, Cp, 2 5, = 1 + = = 1.67, \ g=, Cv, 3 3, , r=
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EBD_8336, 8, , CHEMISTRY, For diatomic gas, F = 5, Cp, 2 7, = 1 + = = 1.40, \ g=, Cv, 5 5, For triatomic gas, F = 6 or 7 (depending upon the, nature of the gas), Cp, 2, = 1 + = 1.33, \ g=, Cv, 6, Cp, 2, g=, = 1 + = 1.29, Cv, 7, , 32., , 36., , (d) At NTP, 22400 cc of N2O, = 6.02 × 1023 molecules, \ 1 cc of N O contain, 2, =, , 6.02 ´ 1023, molecules, 22400, 3 ´ 6.02 ´ 10, 22400, , 37., , 23, , 1. 8, ´ 1022 atoms, 224, (Q N2O molecule has 3 atoms), No. of electrons in a molecule of N 2O, = 7 + 7 + 8 = 22, Hence, no. of electrons, , =, , =, , 6.02 ´ 1023, 1.32 ´ 10 23, electrons, ´ 22 =, 224, 22400, (a) 1 mol CCl4 vapour = 12 + 4 × 35.5, = 154 g º 22.4 L at STP, \ Density = 154 gL-1 = 6.875 gL-1, 22.4, (a), , 34., , Element, , %, , Atomic, weight, , C, , 38.71, , 12, , H, , 9.67, , 1, , O, , 100 (38.71 + 9.67), = 51.62, , 16, , 35., , Atomic, ratio, , Thus, empirical formula is CH3O., (d) Suppose the mol. wt. of enzyme = x, Given 100 g of enzyme wt of Se = 0.5 g, 0.5, ´x, \ In x g of enzyme wt. of Se =, 100, 0.5 ´ x, Hence 78.4 =, 100, \ x = 15680 = 1.568 × 104, , 40, , 12, , 40, = 3.33, 12, , 3.33, =1, 3.33, , H, , 6.66, , 1, , 6.66, = 6.66, 1, , 6.66, =2, 3.33, , O, , 53.34 16, , 53.34, 3.33, = 3.33, =1, 16, 3.33, , (% of O in organic compound, = 100 – ( 40 + 6.66 ) = 53.34 % ), \ Empirical formula of organic compound = CH2O., (c) As the sum of the percentage of C, H & N, is 100. Thus, it does not contains O atom., Table for empirical formula, At. Relative, wt. Number, , Ratio, , 40, 12, , 3.33, , C, , 40.00 1 2, , H, , 13.33 1, , N, , 46.67 1 4, , = 3.33, , 13.33, 1, 46.67, 14, , = 13.33, = 3.33, , 3.33, , =1, , 13.33, 3.33, 3.33, 3.33, , =4, , =1, , 38., , Hence, empirical formula = CH4N, (d) Molecular weight of ZnSO 4 .7H 2O, = 65 + 32 + (4 × 16) + 7(2 × 1 + 16) = 287., \ percentage mass of zinc (Zn), , 39., , 65, ´ 100 = 22.65%, 287, (d) The reaction can be given as, ¾® 2Z + 3H2O, Z2 O3 + 3H2 ¾, , Simple, ratio, , 38.71, 3.23, = 3.23, =1, 12, 3.23, 9.67, 9.67, = 9.67, =3, 1, 3.23, 51.62, 3.23, = 3.23, =1, 16, 3.23, , C, , Element %, , =, , 33., , (c) Table for empirical formula :, At. Relative, Ratio, Element %, wt number, , =, , 0.1596 g of Z2O3 react with H2 = 6 mg = 0.006 g, \ 1 g of H2 react with, 0.1596, = 26.6 g of Z2O3, 0.006, \ Eq. wt. of Z O = 26.6, 2 3, (from the definition of eq. wt.), Eq. wt. of Z + Eq. wt. of O (8) = 26.6, Þ Eq. wt. of Z = 26.6 – 8 = 18.6, Valency of metal in Z2O3 = 3, =
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9, , Some Basic Concepts of Chemistry, Atomic wt., valency, \ At. wt. of Z = 18.6 × 3 = 55.8, , Eq. wt.of metal =, , 40., , (d), , At start, (moles), , =, , Final moles, , HCOOH, 2.3 1, =, 46 20, , nNaCl =, , AgNO3 + NaCl ® AgCl ¯ + Na+ + NO3–, , H2SO4, ¾¾¾¾®, CO + H2O, Dehydrating, agent, [H 2O absorbed by H 2SO 4], , 0, H2SO4, , 0, , 0, , 1, 20, , 1, 20, , 1 mole, \ 0.049 mole, , 43., , [H 2O absorbed by H 2SO 4], , (moles), , Final moles, , 0, , 0, , 0, , 1, 20, , 1, 20, , 0, , 1, 20, , CO2 is absorbed by KOH., So, the remaning product is only CO., Moles of CO formed from both reactions, 1, 1, 1, =, +, =, 20 20 10, Left mass of CO = moles × molar mass, 1, = ´ 28 = 2.8 g, 10, 41., , (d), , 44., 45., , ® MgO + CO 2, MgCO 3 ¾¾, , 84 g of MgCO3 form 40 g of MgO, \, , 42., , 20g of MgCO3 form, , 40 ´ 20, g of MgO, 84, , = 9.52 g of MgO, Since 8.0 g of MgO is formed, 8, ´ 100 = 84.0%, Purity of sample =, 9.52, (c) 50 mL of 16.9% solution of AgNO3, , æ 16.9, ö, ç 100 ´ 50 ÷ = 8.45 g of Ag NO3, è, ø, 8.45g, nmole =, (107.8 + 14 + 16 ´ 3) g / mol, æ 8.45 g ö, =ç, ÷ = 0.0497 moles, è 169.8g / mol ø, 50 mL of 5.8% solution of NaCl contain, , æ 5.8, ö, ´ 50 ÷ = 2.9 g, NaCl = ç, è 100, ø, , 1 mole, 0.049 mole, , 1 mole, 0.049 mole of AgCl, , w, ® w = (nAgCl) × Molecular Mass, M, = (0.049) × (107.8 + 35.5) = 7.02 g, (c) Millimoles of solution of chloride, = 0.05 × 10 = 0.5, Millimoles of AgNO3 solution = 10 × 0.1 = 1, So, the millimoles of AgNO3 are double than the, chloride solution., \ XCl2 + 2AgNO3 ¾® 2AgCl + X (NO3)2, , n=, , H2C2O4 ¾¾¾® CO + CO2 + H2O, 4.5 1, =, At start =, 90 20, , 2.9g, = 0.0495 moles, (23 + 35.5) g/ mol, , (a) M =, , 100 ´ 6.02 ´ 10, , 23, , =, , 6.02 ´ 1021, 6.02 ´ 1023, , = 0.01 M, (d) No. of milli equivalent of HCl, = 20 × 0.05 = 1.0, No. of milli equivalent of Ba (OH)2, = 30 × 0.1 × 2 = 6.0, After neutralization, no. of milli equivalents in, 50 mL. of solution = (6 – 1) = 5, Total volume of the solution = 20 + 30 = 50 mL, \ No. of equivalent of OH– in 50 mL is, [OH - ] =, , 46., , 6.02 ´ 10 20 ´ 1000, , (b), , 5 ´ 1000, ´ 10 -3 = 0.1 M, 50, , H2 +, 10g, , (5 mol), , 1, ® H2O, O2 ¾¾, 2, 64g, , (2 mol), , In this reaction oxygen is the limiting agent., Hence, amount of H2O produced depends on, the amount of O2 taken., Q 0.5 mole of O2 gives H2O = 1 mol, \ 2 mole of O2 gives H2O = 4 mol, When there is not enough of one reactant in a, chemical reaction, the reaction stops. To find the, amount of product produced, we must determine, reactant that will limit the chemical reaction (the, limiting reagent) we can find the limiting reagent, by calculating the amount of product that can be, formed by each reactant, the one that produces, less product is the limiting reagent.
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EBD_8336, 10, , 47., , CHEMISTRY, (d) Writing the equation for the reaction, we get, ® PbCl2 + H2O, PbO + 2HCl ¾¾, 207 + 16 2 × 36.5, 207 + 71, = 223 g = 73g, = 278g, From this equation we find 223 g of PbO reacts, with 73 g of HCl to form 278 g of PbCl2., If we carry out the reaction between 3.2 g HCl, and 6.5 g PbO., Amount of PbO that reacts with 3.2 g HCl, 223, =, ´ 3.2 =9.77 g., 73, Since amount of PbO present is only 6.5 g so, PbO is the limiting reagent., Amount of PbCl 2 formed by 6.5 g of PbO, 278, ´ 6.5g, 223, Number of moles of PbCl2 formed, , =, , 48., , 278 6.5, ´, moles = 0.029 moles., =, 223 278, (c) Molarity of H2SO4 solution, 98 ´ 1000, =, ´ 1.80 = 18.0, 98 ´ 100, Suppose V mL of this H2SO4 is used to prepare, 1 lit. of 0.1M H2SO4, , Again,, No. of gram equivalent of C, mass in gram, =, gram equivalent weight, mass, Þ 30 × 103 =, 3, Þ mass = 90 × 103 g = 90 kg, 50., , 51., , 52., , No. of gram equivalent of Al =, , 270 ´ 10, 9, , = 30 × 103, Hence,, No. of gram equivalent of C = 30 × 103, , 6 moles, 1.2 moles, , 15(32), , \, , 156 g of benzene, required oxygen, = 15 × 22.4 litre, 1g of benzene, required oxygen, , \, , 15´ 22.4, litre, 156, 39 g of benzene, required oxygen, , 15 ´ 22.4 ´ 39, = 84.0 litre, 156, (a) 5 M H2SO4 = 10 N H2SO4,, (Q Basicity of H2SO4 = 2), N1V1 = N2V2,, 10 × 1 = N2 × 10 or N2 = 1 N, , =, , 54., 3, , 4 moles, 0.8 moles, , =, , 53., , 0, +4, 12, = 3 ( C ® C O2 ), 4, , 5 moles, 1 mole, , ®12CO2 (g) + 6 H 2 O(g), (d) 2C6 H 6 + 15O2 (g) ¾¾, 2(78), , Equivalent weight of C, =, , 2 vol., 20 litre, , It is given that only 50% of the expected product, is formed hence, only 10 litre of NH3 is formed., N2 used = 5 litres,, left = 30 – 5 = 25 litres, H2 used = 15 litres, left = 30 – 15 = 15 litres, (c) According to Stoichiometry, they should, react as follow:, 4 NH 3 + 5O 2 ¾, ¾® 4 NO + 6H 2 O, , Q, , 1000 ´ 0.1, = 5.55 mL., 18.0, (c) 2Al 2O 3 + 3C ¾¾, ® Al + 3CO 2, Gram equivalent of Al2O3 º g equivalent of C, , 27, =9, 3, , ® 2NH 3, , Thus, for 1 mole of O2 only 0.8 moles of NH3 is, consumed. Hence O2 is consumed completely., , or V =, , Now equivalent weight of Al =, , N 2 + 3H 2, 1 vol., 3 vol., 10 litre 30 litre, , 4 moles, 0.8 moles, , \ V ´ 18.0 = 1000 ´ 0.1, , 49., , (c), , (a) Mg2+ + Na 2 CO3 ¾¾, ® MgCO3 + 2Na +, 1 g eq., 1g eq., 1 g eq. of Mg2+ = 12 g of Mg2+ = 12000 mg, = 1000 milli eq. of Na 2CO3, \ 12 mg Mg2+ = 1 milli eq. Na2CO3
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2, , Structure of Atom, , Trend Analysis with Important Topics & Sub-Topics, 2020, , 2019, , 2018, , 2017, , 2016, , Topic Name, Bohr's model for H, atom, , Sub-Topic, QNS. LOD QNS. LOD QNS. LOD QNS. LOD QNS. LOD, Absortion/emission, 1, E, spectra, uncertainity, Dual behaviour of, principle/orbital, matter and Heisenberg, 1, A, stability/de-Broglie, uncertainity principle, wavelength, 1, A, 1, E, Quantum mechanical Quantum numbers, model of atom, orbital energy, 1, A, LOD - Level of Difficulty, , E - Easy, , Topic 1 : Atomic Models and Dual Nature of, Electromagnetic Radiation, 1., , 2., , Calculate the energy in joule corresponding to, light of wavelength 45 nm :, (Planck’s constant h = 6.63 × 10–34 Js; speed of, light c = 3 × 108 ms–1), [2014], (a) 6.67 × 1015, (b) 6.67 × 1011, (c) 4.42 × 10–15, (d) 4.42 × 10–18, According to law of photochemical equivalence, the energy absorbed (in ergs/mole) is given as, (h = 6.62 × 10–27 ergs, c = 3 × 1010 cm s–1,, NA = 6.02 × 1023 mol–1), [NEET Kar. 2013], , 1.196 ´ 1016, 1.196 ´ 108, (b), l, l, 2.859 ´ 105, 2.859 ´ 1016, (c), (d), l, l, The value of Planck’s constant is 6.63 × 10–34 Js., The speed of light is 3 × 1017 nm s–1.. Which, value is closest to the wavelength in nanometer, of a quantum of light with frequency of, 6 × 1015 s–1?, [NEET 2013], (a) 25, (b) 50, (c) 75, (d) 10, , (a), , 3., , A - Average, , 4., , D - Difficult, , The energies E1 and E2 of two radiations are 25, eV and 50 eV, respectively. The relation between, their wavelengths i.e., l1 and l2 will be: [2011], (a) l1 = l2, (b) l1 = 2l2, 1, l2, 2, The energy absorbed by each molecule (A2) of, a substance is 4.4 × 10–19 J and bond energy, per molecule is 4.0 × 10–19 J. The kinetic energy, of the molecule per atom will be:, [2009], (a) 2.2 × 10–19 J, (b) 2.0 × 10–19 J, (c) 4.0 × 10–20 J, (d) 2.0 × 10–20 J, The value of Planck's constant is 6.63 × 10–34 Js., The velocity of light is 3.0 × 108 m s–1. Which value, is closest to the wavelength in nanometers of a, quantum of light with frequency of 8 × 1015 s–1 ?, [2003], (a) 3 × 107, (b) 2 × 10–25, (c) 5 × 10–18, (d) 4 × 101, If the energy of a photon is given as 3.03 × 10–19 J, then, the wavelength (l) of the photon is :, (a) 6.56 nm, (b) 65.6 nm [2000], (c) 656 nm, (d) 0.656 nm, , (c) l1 = 4l2, , 5., , 6., , 7., , Qns - No. of Questions, , (d) l1 =
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EBD_8336, 12, , 8., , 9., , 10., , CHEMISTRY, In the photo-electron emission, the energy of, the emitted electron is, [1994], (a) greater than the incident photon, (b) same as than of the incident photon, (c) smaller than the incident photon, (d) proportional to the intensity of incident, photon., The electron was shown experimentally to have, wave properties by, [1994], (a) de Broglie, (b) Davisson and Germer, (c) N. Bohr, (d) Schrodinger., Which of the following is never true for cathode, rays ?, [1994], (a) They possess kinetic energy, (b) They are electromagnetic waves, (c) They produce heat, (d) They produce mechanical pressure., , 14., , 15., , (a) 1.54 ´1015 s -1, 16., , 12., , 13., , Which of the following series of transitions in, the spectrum of hydrogen atom falls in visible, region ?, [2019], (a) Lyman series, (b) Balmer series, (c) Paschen series, (d) Brackett series, æ Z2 ö, Based on equation E = – 2.178 × 10-18 çç 2 ÷÷ J,, èn ø, certain conclusions are written. Which of them, is not correct ?, [NEET 2013], (a) Larger the value of n, the larger is the orbit, radius., (b) Equation can be used to calculate the change, in energy when the electron changes orbit., (c) For n = 1, the electron has a more negative, energy than it does for n = 6 which mean, that the electron is more loosely bound in, the smallest allowed orbit., (d) The negative sign in equation simply, means that the energy or electron bound, to the nucleus is lower than it would be if, the electrons were at the infinite distance, from the nucleus., According to the Bohr Theory, which of the, following transitions in the hydrogen atom will give, rise to the least energetic photon?, [2011 M], , 17., , 18., , 19., , 20., , 21., , (b) 1.03 ´1015 s -1, , (d) 2.00 ´1015 s -1, In hydrogen atom, energy of first excited state, is –3.4 eV. Find out KE of the same orbit of, Hydrogen atom, [2002], (a) + 3.4 eV, (b) + 6.8 eV, (c) – 13.6 eV, (d) + 13.6 eV, According to Bohr’s theory the energy required, for an electron in the Li2+ ion to be emitted from, n = 2 state is (given that the ground state ionization, energy of hydrogen atom is 13.6 eV), [1999], (a) 61.2 eV, (b) 13.6 eV, (c) 30.6 eV, (d) 10.2 eV, The Bohr orbit radius for the hydrogen atom (n = 1), is approximately 0.530 Å. The radius for the first, excited state (n = 2) orbit is (in Å), [1998], (a) 0.13, (b) 1.06, (c) 4.77, (d) 2.12, The radius of hydrogen atom in the ground state, is 0.53 Å. The radius of Li2+ ion (atomic number = 3), in a similar state is, [1995], (a) 0.17 Å, (b) 0.265 Å, (c) 0.53 Å, (d) 1.06 Å, If ionization potential for hydrogen atom is, 13.6 eV, then ionization potential for He+ will be, (a) 54.4 eV, (b) 6.8 eV, [1993], (c) 13.6 eV, (d) 24.5 eV, The energy of an electron in the nth Bohr orbit, of hydrogen atom is, [1992], 13.6, 13.6, (a) – 4 eV, (b) – 3 eV, n, n, 13.6, 13.6, (c) – 2 eV, (d) –, eV, n, n, (c), , Topic 2 : Bohr's model for Hydrogen Atom, (Emission and Absorption Spectra), 11., , (a) n = 6 to n = 1, (b) n = 5 to n = 4, (c) n = 6 to n = 5, (d) n = 5 to n = 3, The energy of second Bohr orbit of the, hydrogen atom is -328 kJ mol -1; hence the, energy of fourth Bohr orbit would be:, [2005], (a) -41 kJ mol-1, (b) -82 kJ mol-1, (c) -164 kJ mol-1, (d) -1312 kJ mol-1, The frequency of radiation emitted when the, electron falls from n = 4 to n = 1 in a hydrogen, atom will be (Given ionization energy of, H=2.18 ×10–18J atom–1and h = 6.625 × 10–34 J s ), [2004], 3.08 ´1015 s -1
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13, , Structure of Atom, 22., , 23., , If r is the radius of the first orbit, the radius of, nth orbit of H-atom is given by, [1988], (a) rn2, (b) rn, (c) r/n, (d) r2 n2, The spectrum of He is expected to be similar to, that, [1988], (a) H, (b) Li+, (c) Na, (d) He+, , 29., , 30., , Topic 3: Dual Behaviour of Matter and, Heisenberg Uncertainty Principle, 24., , 25., , In hydrogen atom, the de-Broglie wavelength, of an electron in the second Bohr orbit is, [Given that Bohr radius, a0 = 52.9 pm], [NEET Odisha 2019], (a) 105.8 pm, (b) 211.6 pm, (c) 211.6 p pm, (d) 52.9 p pm, Which one is the wrong statement ?, [2017], (a) The uncertainty principle is, , 31., , DE ´ Dt ³ h / 4p, , 26., , (b) Half filled and fully filled orbitals have, greater stability due to greater exchange, energy, greater symmetry and more, balanced arrangement., (c) The energy of 2s orbital is less than the energy, of 2p orbital in case of Hydrogen like atoms, h, (d) de-Broglie's wavelength is given by l =, ,, mn, where m = mass of the particle, n = group, velocity of the particle, A 0.66 kg ball is moving with a speed of 100 m/s., The associated wavelength will be, , ( h = 6.6 ´10, , -34, , 27., , 28., , ), , Js :, , 10–32m, , (c), , 1, 2m, 1, m, , h, p, h, p, , 33., , [2010], 10–32m, , (a) 1.0 ´, (b) 6.6 ´, (c) 6.6 ´ 10–34m, (d) 1.0 ´ 10–35m, The measurement of the electron position if, associated with an uncertainty in momentum,, which is equal to 1 × 10–18 g cm s– 1 . The, uncertainty in electron velocity is,, [2008], (mass of an electron is 9 × 10– 28 g), (a) 1 × 109 cm s–1, (b) 1 × 106 cm s–1, 5, –1, (c) 1 × 10 cm s, (d) 1 × 1011 cm s–1, If uncertainty in position and momentum are, equal, then uncertainty in velocity is : [2008], (a), , 32., , (b), (d), , h, 2p, h, p, , Given : The mass of electron is 9.11 × 10–31 kg, Planck constant is 6.626 × 10–34 Js,, the uncertainty involved in the measurement of, velocity within a distance of 0.1 Å is [2006], (a) 5.79 × 107 ms–1, (b) 5.79 × 108 ms–1, 5, –1, (c) 5.79 × 10 ms, (d) 5.79 × 106 ms–1, The position of both, an electron and a helium, atom is known within 1.0 nm. Further the, momentum of the electron is known within, 5.0 × 10–26 kg ms–1. The minimum uncertainty, in the measurement of the momentum of the, helium atom is, [1998], (a) 50 kg ms–1, (b) 80 kg ms–1, (c) 8.0 × 10–26 kg ms–1 (d) 5.0 × 10–26 kg ms–1, The momentum of a particle having a de Broglie, wavelength of 10–17 metres is, [1996], (Given h = 6.625 × 10–34 Js), (a) 3.3125 × 10–7 kg ms–1, (b) 26.5 × 10–7 kg ms–1, (c) 6.625 × 10–17 kg ms–1, (d) 13.25 × 10–17 kg ms–1, Uncertainty in position of an electron, (mass = 9.1 × 10–28 g) moving with a velocity of, 3 × 104 cm/s accurate upto 0.001% will be, (use h/4(p) in uncertainty expression where, h = 6.626 ×10–27 erg-second), [1995], (a) 1.93 cm, (b) 3.84 cm, (c) 5.76 cm, (d) 7.68 cm, When an electron of charge ‘e’ and mass ‘m’, moves with a velocity ‘v’ about the nuclear, charge ‘Ze’ in circular orbit of radius ‘r’, the, potential energy of the electrons is given by, [1994], (a), , 34., , Ze2 / r, , (b) - Ze2 / r, , 2, (d) mv / r, (c) Ze2 / r, Which of the following statements do not form, a part of Bohr’s model of hydrogen atom ?, [1989], (a) Energy of the electrons in the orbits are, quantized, (b) The electron in the orbit nearest the nucleus, has the lowest energy, (c) Electrons revolve in different orbits around, the nucleus, (d) The position and velocity of the electrons, in the orbit cannot be determined, simultaneously.
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EBD_8336, 14, , CHEMISTRY, , Topic 4 : Quantum Mechanical Model of Atom, 35., , 36., , 37., , Orbital having 3 angular nodes and 3 total nodes is, [NEET Odisha 2019], (a) 6 d, (b) 5 p, (c) 3 d, (d) 4 f, 4d, 5p, 5f and 6p orbitals are arranged in the, order of decreasing energy. The correct option, is:, [2019], (a) 5f > 6p > 5p > 4d (b) 6p > 5f > 5p > 4d, (c) 6p > 5f > 4d > 5p (d) 5f > 6p > 4d > 5p, Which one is a wrong statement?, [2018], (a) Total orbital angular momentum of electron, in 's' orbital is equal to zero, (b) An orbital is designated by three quantum, numbers while an electron in an atom is, designated by four quantum numbers, (c) The value of m for dz2 is zero, (d) The electronic configuration of N atom is, 1s2, , 38., , 39., , 40., , 41., , 2s2, , 2p1x 2p1y 2p1z, , Two electrons occupying the same orbital are, distinguished by, [2016], (a) Principal quantum number, (b) Magnetic quantum number, (c) Azimuthal quantum number, (d) Spin quantum number, What is the maximum number of orbitals that, can be identified with the following quantum, numbers?, [2014], n = 3, l = 1, ml = 0, (a) 1, (b) 2, (c) 3, (d) 4, What is the maximum numbers of electrons that, can be associated with the following set of, quantum numbers?, [NEET 2013], n = 3, l = 1 and m = –1, (a) 6, (b) 4, (c) 2, (d) 10, The orbital angular momentum of a p-electron, is given as :, [2012 M], (a), , h, 2p, , (b), , 3, , (c), , 3h, 2p, , (d), , 6., , h, 2p, h, 2p, , 42., , 43., , 44., , 45., , 46., , 47., , 48., , 49., , The correct set of four quantum numbers for the, valence electron of rubidium atom (Z = 37) is, [2012], (a) 5, 1, 1 + 1/2, (b) 6, 0, 0, + 1/2, (c) 5, 0, 0, + 1/2, (d) 5, 1, 0, + 1/2, Maximum number of electrons in a subshell with :, l = 3 and n = 4 is :, [2012], (a) 14, (b) 16, (c) 10, (d) 12, If n = 6, the correct sequence for filling of, electrons will be :, [2011], (a) ns ® (n – 2) f ® (n – 1) d ® np, (b) ns ® (n – 1) d ® (n – 2) f ® np, (c) ns ® (n – 2) f ® np ® (n – 1) d, (d) ns ® np (n – 1) d ® (n – 2) f, The total number of atomic orbitals in fourth, energy level of an atom is :, [2011], (a) 8, (b) 16, (c) 32, (d) 4, Which of the following is not permissible, arrangement of electrons in an atom? [2009], (a) n = 5, l = 3, m = 0, s = + 1/2, (b) n = 3, l = 2, m = – 3, s = – 1/2, (c) n = 3, l = 2, m = – 2, s = – 1/2, (d) n = 4, l = 0, m = 0, s = – 1/2, Maximum number of electrons in a subshell of, an atom is determined by the following: [2009], (a) 2 l + 1, (b) 4 l – 2, (c) 2 n2, (d) 4 l + 2, Consider the following sets of quantum numbers:, n, l, m, s, (i) 3, 0, 0, + 1/2, (ii) 2, 2, 1, + 1/2, (iii) 4, 3, –2, – 1/2, (iv) 1, 0, –1, – 1/2, (v) 3, 2, 3, + 1/2, Which of the following sets of quantum number, is not possible?, [2007], (a) (i), (ii), (iii) and (iv) (b) (ii), (iv) and (v), (c) (i) and (iii), (d) (ii), (iii) and (iv), The orientation of an atomic orbital is governed, by, [2006], (a) Spin quantum number, (b) Magnetic quantum number, (c) Principal quantum number, (d) Azimuthal quantum number
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15, , Structure of Atom, 50., , 51., , 52., , 53., , 54., , 55., , 56., , 57., , The ions O2– , F– , Na +, Mg2+ and Al 3+ are, isoelectronic. Their ionic radii show, [2003], 2–, –, (a) A decrease from O to F and then increase, from Na+ to Al3+, (b) A significant increase from O2– to Al3+, (c) A significant decrease from O2– to Al3+, (d) An increase from O 2– to F– and then, decrease from Na+ to Al3+, Which of the following is isoelectronic?[2002], (a) CO2, NO2, (b) NO2–, CO2, –, (c) CN , CO, (d) SO2, CO2, The following quantum numbers are possible, for how many orbital (s) n = 3, l = 2, m = +2 ?, [2001], (a) 1, (b) 3, (c) 2, (d) 4, Set of isoelectronic species is, [2000], (a), , N 2 , CO2 , CN- , O2 (b) N, H 2S, CO, , (c), , N 2 , CO, CN - , O 2+2 (d) Ca , Mg, Cl, , The ion that is isoelectronic with CO is [1997], (a) CN–, (b) O2+, –, (d) N2+, (c) O2, The orbitals are called degenerate when, [1996], (a) they have the same wave functions, (b) they have the same wave functions but, different energies, (c) they have different wave functions but, same energy, (d) they have the same energy, If electron has spin quantum number + 1/2 and, a magnetic quantum number – 1, it cannot be, present in, [1994], (a) d-orbital, (b) f-orbital, (c) p-orbital, (d) s-orbital., For which one of the following sets of four, quantum numbers, an electron will have the, highest energy?, [1994], n, l, m, s, (a) 3, 2, 1, 1/2, (b) 4, 2, –1, 1/2, (c) 4, 1, 0, –1/2, (d) 5, 0, 0, –1/2, , 58., , 59., , Which of the following species has four lone, pairs of electrons ?, [1993], (a) I, , (b) O -, , (c) Cl–, , (d) He, , The order of filling of electrons in the orbitals of, an atom will be, [1991], (a) 3d, 4s, 4p, 4d, 5s, (c) 5s, 4p, 3d, 4d, 5s, , 60., , For azimuthal quantum number l = 3, the, maximum number of electrons will be [1991], (a) 2, , 61., , (b) 4s, 3d, 4p, 5s, 4d, (d) 3d, 4p, 4s, 4d, 5s, , (b) 6, , (c) 0, (d) 14., In a given atom no two electrons can have the, same values for all the four quantum numbers., This is called, [1991], (a) Hund’s Rule, (b) Aufbau principle, (c) Uncertainty principle, , 62., , 63., , (d) Pauli’s Exclusion principle., An ion has 18 electrons in the outermost shell, it, is, [1990], (a) Cu+, , (b) Th4+, , (c) Cs+, , (d) K+, , The total number of electrons that can be, accommodated in all the orbitals having principal, quantum number 2 and azimuthal quantum, number 1 is, [1990], (a) 2, (c) 6, , 64., , 65., , The maximum number of electrons in a subshell, is given by the expression, [1989], (a) 4l – 2, , (b) 4l + 2, , (c) 2l + 2, , (d) 2n2, , Number of unpaired electrons in N2+ is [1989], (a) 2, (b) 0, (c) 1, , 66., , (b) 4, (d) 8, , (d) 34, , The number of spherical nodes in 3p orbitals are, (a) one, (c) none, , (b) three, (d) two, , [1988]
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EBD_8336, 16, , CHEMISTRY, ANSWER KEY, , 1, 2, 3, 4, 5, 6, 7, , (d), (b), (b), (b), (d), (d), (c), , 8, 9, 10, 11, 12, 13, 14, , 15, 16, 17, 18, 19, 20, 21, , (c), (b), (b), (b), (c), (c), (b), , (c), (a), (c), (d), (a), (a), (c), , 22, 23, 24, 25, 26, 27, 28, , (a), (b), (c), (c), (d), (a), (a), , 29, 30, 31, 32, 33, 34, 35, , (d), (d), (c), (a), (b), (d), (d), , 36, 37, 38, 39, 40, 41, 42, , (a), (d), (d), (a), (c), (a), (c), , 43, 44, 45, 46, 47, 48, 49, , (a), (a), (b), (b), (d), (b), (b), , 50, 51, 52, 53, 54, 55, 56, , (c), (c), (a), (c), (a), (d), (d), , 57, 58, 59, 60, 61, 62, 63, , (b), (c), (b), (d), (d), (a), (c), , 64, 65, 66, , (b), (c), (a), , Hints & Solutions, 1., 2., , - 34, ´ 3 ´ 108, hc 6.63 ´ 10, =, l, 45 ´ 10 - 9, –, 18, = 4.42 × 10 J, hc, (b) E =, ´ NA, l, 6.62 ´ 10-27 ´ 3 ´ 1010 ´ 6.02 ´ 10 23, =, l, , (d) E =, , =, , 7., , 8., , (b) c = u l, c, 3 ´ 10, =, = 50 nm, u 6 ´ 1015, (b) Given E1 = 25eV E2 = 50 eV, , hc, hc, E =, l1 2 l 2, , l, 25 1, \ 2 =, =, l1 50 2, , 5., , \, , E1 l 2, =, E2 l1, , \ l1 = 2l 2, , 9., , (d) K.E per atom, , ( 4.4 ×10 ) – ( 4.0 ×10 ), =, –19, , –19, , 2, , 0.4×10, =, 2, , –19, , = 2.0 ´ 10, , –20, , J, , 3 ´ 10 8, 15, , = 3.75 ´ 10 - 8 m, , 6.626 ´ 10 -34 ´ 3 ´ 108, , 3.03 ´ 10 -19, 19.878, or l =, ´ 10 -7 = 6.56 × 10–7m= 656 nm, 3.03, (c) K.E. of emitted electron is smaller than the, incident photon., K.E of emitted electron = hu – hu0, where hu0 = work function., [work function is the minimum energy required to, remove an electron from the surface of the, material.], Energy of the incident photon = hu, maximum kinetic energy of the photoelectrons, depend upon the frequency of the light not on the, intensity of light., , l=, , E1 =, , hc, or l = c, l, n, , 8 ´ 10, In nanometer l = 3.75× 10 which is closest to 4 × 101, (c) The energy of photon,, hc, E =, = 3.03 ´ 10-19, l, , or l =, , 17, , 4., , (d) E = hn =, Þl=, , 1.19 ´ 108, ergs mol–1, l, Law of photochemical equivalence is the, fundamental principle relating to chemical reactions, induced by light which states that for every mole, of a substance that reacts 6.022 × 1023 quanta of, light are absorbed. The photochemical equivalence, law is also sometimes called the stark-Einstein law., , 3., , 6., , 10., 11., , (b) Wave nature of electron was shown by, Davisson and Germer. Davisson and Germer, demonstrated the physical reality of the wave, nature of electrons by showing that a beam of, electrons could also be diffracted by crystals, just like light of x-rays., (b) Cathode rays are not electromagnetic waves., (b) Balmer series
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17, , Structure of Atom, 12., , (c) The electrons has more negative energy in, lower orbits that in higher orbit and also the, electrons are more tightly bound in the smallest, allowed orbit., Energy of an electron at infinite distance from the, nucleus is zero. As an electron approaches the, nucleus, the electron attraction increases and hence, the energy of electron decreases and thus becomes, negative. Thus as the value of n decreases, i.e., lower the orbit is, more negative is the energy of, the electron in it., , 13. (c) Energy of photon obtained from th e, transition n = 6 to n = 5 will have least energy., , DE = 13.6Z 2, , 1 ö, æ 1, çè n 2 n 2 ÷ø, 1, 2, , Therefore the centripetal force is supplied by, electrostatic force of attraction between the, electron and nucleus i.e., , mn2 Ze2, = 2, r, r, Ze2, r, 1, 1 Ze 2, or mn2 =, = K.E, 2, 2 r, now total energy (En) = K.E + P.E, in first excited state, , or mn2 =, , é Ze 2 ù, 1, E = mv 2 + ê ú, 2, êë r úû, =+, , 2, , æ Zö, 14. (b) E n = E1 ´ ç ÷, è nø, E1, E 2 = 2 = -328kJ/mol, 2, E, E 4 = 21, 4, Þ, , Þ, , E 4 22, 1, = 2 = 2, E2 4, 2, E, -328, E4 = 2 =, kJ /mol = -82 kj /mol, 22, 4, , 15. (c) I.E. = E1 – E ¥, 2.18 × 10–18 = E1 – 0, E1 = 2.18 × 10–18 J-atom–1, æ 1, 1 ö, DE = E1 ç 2 - 2 ÷, è n1 n 2 ø, , 1ö, æ1, Þ hv = 2.18 × 10–18 ç 2 - 2 ÷, è1, 4 ø, , 15, Þ 6.625 × 10–34 × v = 2.18 × 10–18 ×, 16, , Þ v=, , 2.18 ´ 10, , –18, , ´ 15, , -34, , ´ 16, , 6.625 ´ 10, , Þ, 16. (a) Suppose the nucleus of hydrogen atom, have charge of one proton +e. The electron, revolves in an orbit of radius r around it., v = 3.08 × 1015 s–1, , 1 Ze2 Ze2, 2 r, r, , 1 Ze 2, 2 r, 1 Ze 2, \ K .E =, = +3.4 eV, 2 r, -3.4 eV = -, , For Bohr orbit, kinetic energy = –Total energy and, Potential energy = 2 × Total energy., , 17. (c) Energy of electron in 2nd orbit of Li +2, Z2, = -13.6 2, n, , =, , - 13.6 ´ (3) 2, , = –30.6 eV, (2) 2, Energy required = 0 – (–30.6) = 30.6 eV, 18. (d) Radius of hydrogen atom = 0.530 Å, Number, of excited state (n) = 2 and atomic number of, hydrogen atom (Z) = 1. We know that the Bohr, radius., (r ) =, , (2)2, n2, ´ Bohr Radius of atom =, ´ 0.530, 1, Z, , = 4 ´ 0.530 = 2.12 Å, 19. (a) State of hydrogen atom (n) = 1, (due to ground state), Radius of hydrogen atom (r) = 0.53 Å. Atomic, number of Li (Z) = 3., Radius of Li2+ ion, =r´, , (1) 2, n2, = 0.53 ´, = 0.17Å., 3, Z
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EBD_8336, 18, , CHEMISTRY, , 20. (a) The ionization energy of any hydrogen like, species (having one electron only) is given by, the equation, , I.E =, , 2 2, , 2p Z m e, , 4, , 28. (a) We know Dp.Dx ³, h, 4p, since Dp = Dx (given), , or m.Dv.Dx =, , or I.E µ Z 2, , h2, Since the atomic number of H is 1 and that of He, is 2, therefore, the I.E. of He+ is four times (22) the, I.E. of H i.e., 13.6 × 4 = 54.4 eV, , \ Dp.Dp =, or ( Dv ) =, 2, , For H-like particles,, , 21.8 ´ 10-19 2, Z J/atom, h2, 1312 2, = - 2 Z kJ/mol, n, I.E = Ea o – E1 = 0 – (– I.EHZ2), En = -, , or Dv =, , = Z2 × I.EH, , Now for, , Z=2, , \, , I.E = 4 × I.EH, , for, , Li2+,, , Z=3, , \, , I.E = 9 × I.EH, , Dv >, , 21. (c) Energy of an electron in Bohr's orbit is given, by the relationship. En = -, , 13.6, , Dv =, , eV., , n2, th, 2, 22. (a) Radius of n orbit = r1 n . (for H-atom), 23. (b) Both He and Li+ contain 2 electrons each,, therefore their spectrum will be similar., 24. (c) nl = 2pr, , 30., , 2p, r, n, rn = a0n2, r2 = 52.9 × (2)2 pm, , 2p, × 52.9 × 4, 2, l = 211.6 p pm, 25. (c) For hydrogen like atoms energy of, 2s-orbital and 2p-orbital is equal., , 31., , h, 6.6 ´ 10 -34, =, = 1 ´ 10 -35 m, mv 0.66 ´ 100, 27. (a) Dp = mDv, Substituting the given values of Dx and m, we get, 1×10–18 gcms–1 = 9×10–28 g × Dv, , 32., , l=, , 1 ´ 10-18, 9 ´ 10-28, = 1.1 × 109 cm s–1 ; 1×109 cm s–1, , or Dv =, , 4 pm, , 2, , =, , 1 h, 2m p, h, 4p, , h, 4p, , h, 4 pDxm, , 6.626 ´ 10 -34, 4 p ´ 0.1 ´ 10 -10 ´ 9.11 ´ 10 -31, , 66, ´ 10 7 = 5.79 ´ 10 6 m / sec, 4p ´ 9, (d) By Heisenberg uncertainty Principle, h, Dx ´ Dp =, (which is constant), 4p, As Dx for electron and helium atom is same thus, momentum of electron and helium will also be, same and the uncertainty in momentum of helium, atom is equal to 5 × 10–26 kg ms–1., (c) Acc. to de-Broglie,, =, , l=, , 26. (d) l =, , 4pm2, h, , Dx . m Dv >, , [\ Dp = mDv], , h, h, or mDv mDv =, 4p, 4p, h, , 29. (d) We know that Dx . Dp ³, , He+,, , \, , h, 4p, , h 6.625 ´ 10-34, h, Þ mv = =, mv, l, 10-17, Þ p = 6.625 × 10–17 kg m/s, (a) Given mass of an electron(m) = 9.1 × 10–28 g;, Velocity of electron (v ) = 3 ´ 10 4 cm /s;, l=, , Accuracy in velocity = 0.001% = 0.001 ;, 100, Actual velocity of the electron, 0.001, = 0.3cm/s ., 100, Planck’s constant (h) = 6.626×10–27erg-sec., ( Dv ) = 3 ´ 104 ´
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19, , Structure of Atom, \ Uncertainty in the position of the electron, ( Dx ) =, , h, 6.626´10, ´7, <, 28, ,, 4 pmDv 4´ 22´(9.1´10 )´ 0.3, =1.93cm, r, , 33., , 35., , 36., , 37., , (d) It is uncertainty principle which was given, by Heisenberg and not Bohr’s postulate., (d) Total number of nodes = (n – 1), 3= n–1, Þ n=4, Number of angular nodes = l = 3 Þ f-subshell, (a) 5 f, 5+3=8, 6p, 6+1=7, 5p, 5+1=6, 4d, 4+2=6, 5f > 6p > 5p > 4d, , 38., 39., , 2s, , 42., , 2p1x, , 2p1y, , 0, , 43., , 44., , 2p1z, , +1, , hence the number of orbital identified by m = 0, can be one only., (c) n = 3 ® 3rd shell, l = 1 ® p sub shell., m = – 1 is possible for two electrons present in, an orbital., m = 0 is fixed for Pz orbital but Px and Py orbital, can take any value among ± 1., , h, 2p, , 2 =, , h, 2p, , (c) Electronic configuration of Rb = [Kr] 5s1, Set of quantum numbers, n = 5, l = 0, \ s-orbital, m = 0, s = + 1/2, The other possible answer is 5, 0, 0, –1/2 In and, orbital any electron can take the value of ± 1/2, and the other elctron of the same orbital, takes the, value with opposite sign., , (d) Two electrons occupying the same orbital, should have opposite spins i.e. they differ in, spin quantum number., (a) Given: n = 3, l = 1, m = 0, Hence orbital is 3p, –1, , 40., , So, =, , (d) The correct configuration of 'N' is, 2, , h, l(l + 1), 2p, , For p orbital l = 1, , Ze2, ., (b) P.E. = work done = ò - 2 dr = r, r, ¥, , 1s2, , (a) Orbital angular momentum, =, , Ze 2, , Kinetic and potential energy of atoms results from, the motion of electrons. When electrons are excited, they move to a higher energy orbital farther away, from the atom. The further the orbital is from the, nucleus, the higher the potential energy of an, electron at that energy level. When the electron, returns to a low energy state, it releases potential, energy in the form of kinetic energy., , 34., , 41., , ,27, , 45., , 46., , 47., , (a) (n = 4, l = 3) Þ 4f subshell, Since, maximum no. of electrons in a subshell, = 2(2l + 1), So, total no. of electron in 4f subshell, = 2 (2 × 3 + 1) = 14 electrons., (a), 1s, , 2s, , 2p, , 3s, , 3p, , 3d, , 4s, , 4p, , 4d, , 4f, , 5s, , 5p, , 5d, , 5f, , 6s, , 6p, , 6d, , 6f, , When n = 6, the sequence of filling of electrons:, 6s ¾¾, ® 4f ¾¾, ® 5d ¾¾, ® 6p, or ns ¾¾, ® (n – 2)f ¾¾, ® (n – 1)d ¾¾, ® np, (b) Total no. of atomic orbital in a shell = n2., Given n = 4; hence number of atomic orbitals in, 4th shell will be 16., (b) m = 2 l +1, thus for l = 2, m = 5, hence values, of m will be – 2, –1, 0, + 1, + 2., Therefore for l = 2, m cannot have the value – 3., (d) The number of subshell is (2 l + 1). The, maximum number of electrons in the sub shell is, 2 (2 l + 1) = (4 l + 2)
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EBD_8336, 20, , 48., , 49., , 50., , 51., 52., , CHEMISTRY, (b) (ii) is not possible for any value of n because, l varies from 0 to (n – 1) thus for n = 2, l can be, only 0 and 1., (iv) is not possible because for l = 0, m = 0., (v) is not possible because for l = 2, m varies, from –2 to +2., (b) Magnetic quantum no. represents the, orientation of atomic orbitals in an atom. For, example px, py & pz have orientation along, X-axis, Y-axis & Z-axis., (c) Amongst isoelectronic species, ionic radii, of anion is more than that of cations. Further, size of anion increases with increase in –ve, charge and size of cation decreases with increase, in + ve charge. Hence, ionic radii decreases from, O– to Al3+., (c) Both CN– and CO have 14 electrons., (a) Quantum number n = 3, l = 2, m = +2 represent, one of the 3d-orbitals with, s=±, , 53., , 1, 2, , which is possible only for one orbital., (c) The molecule which contains same number, of electrons are called isoelectronic. eg., N2 = CO = CN - = O +2 2 = 14e–, , 54., , 55., 56., , 57., , (a) We know that ions which have the same, number of electrons are called isoelectronic. We, also know that both CO and CN– have 14, electrons, therefore these are isoelectronic., (d) The orbitals which have same energy are, called degenerate orbitals eg. px , p y , pz ., (d) For s-orbital; l = 0, Thus value of m must be zero, which is given as, –1 in the question. Hence, the electron can not, be present in s-orbital., (b) The sub-shell are 3d, 4d, 4p and 5s, 4d has, highest energy as n + l value is maximum for, this., , 58., , (c) Outer electronic configuration of Cl, 2, 2, 2, 1, = 3s 3 p x 3 p y 3 p z, Outer electronic configuration of, Cl– =3s 2 3px2 3p y 2 3pz 2, , 59., , 60., 61., 62., , hence, Cl– contains four lone pairs of electrons., (b) The sub-shell with lowest value of (n + l) is, filled up first. When two or more sub-shells have, same (n + l) value the subshell with lowest value, of 'n' is filled up first therefore the correct order is, orbital, 4s, 3d, 4p, 5s, 4d, n+l, 4 + 0 3 + 2 4 +1 5 + 0 4 + 2, value, =4 5, 5, 5, 6, (d) l = 3 means f-subshell. Maximum no. of, electrons = 4l + 2 = 4 × 3 + 2 = 14, (d) This is as per the definition of Pauli’s, Exclusion principle., (a) Cu+ = 29 – 1 = 28 e–, thus the electronic configuration of Cu+ is, = 1s 2 2s 2 2p 6 3s 2 3p6 3d10, 144244, 3, 18e -, , (c) n = 2, l = 1 means 2 p – orbital. Electrons that, can be accommodated = 6 as p sub-shell has 3, orbitals and each orbital contains 2 electrons., 64. (b) No. of orbitals in a sub-shell = 2l + 1, Þ No. of electrons = 2(2l + 1) = 4l + 2, 65. (c) N(7) = 1s2 2s2 2p3, 63., , N 2+ = 1s 2 , 2s 2 2 p1x, , Unpaired electrons = 1., 66. (a) No. of radial nodes in 3p-orbital, = (n – l – 1), [for p ortbital l = 1], =3–1–1=1, There are two types of nodes, angular nodes and, radial notes. Angular nodes are typically flat plane., The quantum number l determines the number of, angular nodes in an orbital. Radial nodes are spheres, that occurs as the principal quantum number, increases. Total nodes of an orbital is the sum of, angular and radial nodes which is given by, N=n–l–1
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3, , Classification of Elements, and Periodicity in Properties, , Trend Analysis with Important Topics & Sub-Topics, 2020, Topic Name, Modern periodic, table, , Periodic trends in, properties of, elements, LOD - Level of Difficulty, , 2., , 2018, , 2017, , 2016, , Sub-Topic, QNS. LOD QNS. LOD QNS. LOD QNS. LOD QNS. LOD, modern periodic, 2, E, 1, E, table, Electronic, 1, A, configuration, I.E/ionic,atomic, 1, A, 1, A, size/electron gain, enthalpy trend, Ionisation enthalpy, 1, A, trend, E - Easy, , Topic 1: Modern Periodic Table, 1., , 2019, , Identify the incorrect match., [2020], Name, IUPAC Official, Name, (A) Unnilunium, (i) Mendelevium, (B) Unniltrium, (ii) Lawrencium, (C) Unnilhexium, (iii) Seaborgium, (D) Unununnium, (iv) Darmstadtium, (a) (B), (ii), (b) (C), (iii), (c) (D), (iv), (d) (A), (i), Match the following :, [2020], Oxide, Nature, (A) CO, (i) Basic, (B) BaO, (ii) Neutral, (C) Al2O3, (iii) Acidic, (D) Cl2O7, (iv) Amphoteric, Which of the following is correct option?, (A), (B), (C), (D), (a) (ii), (i), (iv) (iii), (b) (iii), (iv) (i), (ii), (c) (iv) (iii), (ii), (i), (d) (i), (ii), (iii), (iv), , A - Average, , 3., , 4., , 5., 6., , 7., , D - Difficult, , Qns - No. of Questions, , The element Z = 114 has been discovered recently., It will belong to which of the following family/, group and electronic configuration ? [2017], (a) Carbon family, [Rn] 5f 14 6d10 7s2 7p2, (b) Oxygen family, [Rn] 5f 14 6d10 7s2 7p4, (c) Nitrogen family, [Rn] 5f 14 6d10 7s2 7p6, (d) Halogen family, [Rn] 5f 14 6d10 7s2 7p5, An atom has electronic configuration 1s2 2s2 2p6, 3s2 3p6 3d3 4s2, you will place it in which group?, (a) Fifth, (b) Fifteenth [2002], (c) Second, (d) Third, The element, with atomic number 118, will be, (a) alkali, (b) noble gas [1996], (c) lanthanide, (d) transition element, The electronic configuration of an element is, 1s 2 2s 2 2 p6 3s 2 3 p3. What is the atomic number, of the element, which is just below the above, element in the periodic table?, [1995], (a) 33, (b) 34, (c) 36, (d) 49, If the atomic number of an element is 33, it will, be placed in the periodic table in the, [1993], (a) First group, (b) Third group, (c) Fifth group, (d) Seventh group.
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EBD_8336, 22, , CHEMISTRY, , Topic 2: Periodic Trends in Properties of, Elements, 8., , 9., , 10., , 11., , 12., , 13., , Match the oxide given in column A with its, property given in column B [NEET Odisha, 2019], Column-A, Column-B, (i) Na2O, (A) Neutral, (ii) Al2O3, (B) Basic, (iii) N2O, (C) Acidic, (iv) Cl2O7, (D) Amphoteric, Which of the following options has all correct, pairs?, (a) (i)-(B), (ii)-(D), (iii)-(A), (iv)-(C), (b) (i)-(B), (ii)-(A), (iii)-(D), (iv)-(C), (c) (i)-(C), (ii)-(B), (iii)-(A), (iv)-(D), (d) (i)-(A), (ii)-(D), (iii)-(B), (iv)-(C), For the second period elements the correct, increasing order of first ionisation enthalpy is :, (a) Li < Be < B < C < N < O < F < Ne [2019], (b) Li < B < Be < C < O < N < F < Ne, (c) Li < B < Be < C < N < O < F < Ne, (d) Li < Be < B < C < O < N < F < Ne, The correct order of atomic radii in group 13, elements is, [2018], (a) B < Al < In < Ga < Tl, (b) B < Al < Ga < In < Tl, (c) B < Ga < Al < In < Tl, (d) B < Ga < Al < Tl < In, In which of the following options the order of, arrangement does not agree with the variation, of property indicated against it ?, [2016], (a) Al3+ < Mg2+ < Na+ < F– (increasing ionic, size), (b) B < C < N < O (increasing first ionisation, enthalpy), (c) I < Br < F < Cl (increasing electron gain, enthalpy), (d) Li < Na < K < Rb (increasing metallic radius), The species Ar, K+ and Ca2+ contain the same, number of electrons. In which order do their radii, increase ?, [2015], (a) Ca 2+ < Ar < K + (b) Ca 2+ < K + < Ar, (c) K + < Ar < Ca 2+ (d) Ar < K + < Ca 2+, The formation of the oxide ion O2–(g), from, oxygen atom requires first an exothermic and, then an endothermic step as shown below :, O(g) + e– ® O–(g); Df H = –141 kJ mol–1, O– (g) + e– ® O2– (g); Df H = +780 kJ mol–1, , 14., , 15., , 16., , 17., , 18., , 19., , Thus process of formation of O2– in gas phase, is unfavourable even though O2– is isoelectronic, with neon. It is due to the fact that [2015 RS], (a) Electron repulsion outweighs the stability, gained by achieving noble gas configuration, (b) O– ion has comparatively smaller size than, oxygen atom, (c) Oxygen is more electronegative, (d) Addition of electron in oxygen results in, larger size of the ion., Which of the following orders of ionic radii is, correctly represented ?, [2014], (a) H– > H+ > H, (b) Na+ > F– > O2–, (c) F– > O2– > Na+ (d) Al3+> Mg2+> N3–, Which one of the following arrangements, represents the correct order of least negative to, most negative electron gain enthalpy for C, Ca,, Al, F and O?, [NEET Kar. 2013], (a) Ca < Al < C < O < F(b) Al < Ca < O < C < F, (c) Al < O < C < Ca < F (d) C < F < O < Al < Ca, Identify the wrong statement in the following:, [2012], (a) Amongst isoelectronic species, smaller the, positive charge on the cation, smaller is, the ionic radius., (b) Amongst isoelectronic species, greater the, negative charge on the anion, larger is the, ionic radius., (c) Atomic radius of the elements increases as, one moves down the first group of the, periodic table., (d) Atomic radius of the elements decreases, as one moves across from left to right in, the 2nd period of the periodic table., What is the value of electron gain enthalpy of, Na+ if IE1 of Na = 5.1 eV ?, [2011M], (a) –5.1 eV, (b) –10.2 eV, (c) +2.55 eV, (d) +10.2 eV, Among the elements Ca, Mg, P and Cl, the order, of increasing atomic radii is :, [2010], (a) Ca < Mg < P < Cl (b) Mg < Ca < Cl < P, (c) Cl < P < Mg < Ca (d) P < Cl < Ca < Mg, Which of the following represents the correct, order of increasing electron gain enthalpy with, negative sign for the elements O, S, F and Cl ?, [2005, 2010], (a) Cl < F < O < S, (b) O < S < F < Cl, (c) F < S < O < Cl, (d) S < O < Cl < F
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Classification of Elements and Periodicity in Properties, 20., , 21., , 22., , 23., , 24., , 25., , 26., , 27., , 28., , The correct order of the decreasing ionic radii, among the following isoelectronic species are :, (a) Ca 2+ > K + > S2– > Cl –, [2010], –, 22+, +, (b) Cl > S > Ca > K, (c) S2– > Cl – > K + > Ca 2+, (d) K + > Ca 2+ > Cl – > S 2–, Amongst the elements with following electronic, configurations, which one of them may have the, highest ionization energy?, [2009], (a) Ne [3s23p2], (b) Ar [3d104s24p3 ], (c) Ne [3s23p1], (d) Ne [3s23p3], The stability of + 1 oxidation state increases in, the sequence:, [2009], (a) Tl < In < Ga < Al (b) In < Tl < Ga < Al, (c) Ga < In < Al < Tl (d) Al < Ga < In < Tl, Which one of the following ionic species has the, greatest proton affinity to form stable compound?, (a) NH -2, (b) F–, [2007], (c) I–, (d) HS–, Which of the following electronic configuration, of an atom has the lowest ionisation enthalpy?, [2007], (a) 1s2 2s2 2p3, (b) 1s2 2s2 2p5 3s1, (c) 1s2 2s2 2p6, (d) 1s2 2s2 2p5, Identify the correct order of the size of the following:, (a) Ca2+ < K+ < Ar < Cl– < S2–, [2007], (b) Ar < Ca2+ < K+ < Cl– < S2–, (c) Ca2+ < Ar < K+ < Cl– < S2–, (d) Ca2+ < K+ < Ar < S2– < Cl–, Which one of the following oxides is expected, to exhibit paramagnetic behaviour? [2005], (a) CO2, (b) SiO2, (c) SO2, (d) ClO2, Ionic radii are, [2004], (a) inversely proportional to effective nuclear, charge, (b) inversely proportional to squar e of, effective nuclear charge, (c) directly proportional to effective nuclear, charge, (d) directly proportional to square of effective, nuclear charge, Among K, Ca, Fe and Zn, the element which can, form more than one binary compound with, chlorine is, [2004], (a) Fe, (b) Zn, (c) K, (d) Ca, , 29., , 30., , 31., , 32., , 33., , 34., , 35., , 36., , 37., , 23, , Which of the following statements is true? [2002], (a) Silicon exhibits 4 coordination number in, its compound, (b) Bond energy of F2 is less than Cl2, (c) Mn(III) oxidation state is more stable than, Mn(II) in aqueous state, (d) Elements of 15th group shows only +3 and, +5 oxidation states, Which of the following order is wrong? [2002], (a) NH3 < PH3 < AsH3 – Acidic, (b) Li < Be < B < C – First IP, (c) Al2O3 < MgO < Na2O < K2O – Basic, (d) Li+ < Na+ < K+ < Cs+ – Ionic radius, Correct order of first IP among following, elements Be, B, C, N, O is, [2001], (a) B < Be < C < O < N (b) B < Be < C < N < O, (c) Be < B < C < N < O (d) Be < B < C < O < N, Of the given electronic configurations for the, elements, which electronic configuration, indicates that there will be abnormally high, difference in the second and third ionization, energy for the element?, [1999], (a) 1s2 2s2 2p6 3s2, (b) 1s2 2s2 2p6 3s1, (c) 1s2 2s2 2p6 3s2 3p1(d) 1s2 2s2 2p6 3s2 3p2, The first ionization potentials (eV) of Be and B, respectively are, [1998], (a) 8.29, 9.32, (b) 9.32, 9.32, (c) 8.29, 8.29, (d) 9.32, 8.29, Which of the following does not represent the, correct order of the properties indicated [1997], (a) Ni2+ > Cr2+ > Fe2+ > Mn2+ (size), (b) Sc > Ti > Cr > Mn (size), (c) Mn2+ > Ni2+ < Co2+ <Fe2+ (unpaired electron), (d) Fe2+ > Co2+ > Ni2+ > Cu2+ (unpaired electron), Which one of the following ions will be the, smallest in size?, [1996], (a) Na+, (b) Mg2+, (c) F–, (d) O2 –, Among the following oxides, the one which is, most basic is, [1994], (a) ZnO, (b) MgO, (c) Al2O3, (d) N2O5, One of the characteristic properties of nonmetals is that they, [1993], (a) Are reducing agents, (b) Form basic oxides, (c) Form cations by electron gain, (d) Are electronegative
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EBD_8336, 24, , 38., , 39., , 40., , 41., , 42., , CHEMISTRY, Which electronic configuration of an element, has abnormally high difference between second, and third ionization energy?, [1993], (a) 1 s2, 2 s2, 2 p6, 3 s1, (b) 1 s2, 2 s2, 2 p6, 3 s1 3 p1, (c) 1 s2, 2 s2, 2 p6, 3 s2 3 p2, (d) 1 s2, 2 s2, 2 p6, 3 s2, In the periodic table from left to right in a period,, the atomic volume, [1993], (a) Decreases, (b) Increases, (c) Remains same, (d) First decrease then increases, Na+, Mg++, Al3+ and Si4+ are isoelectronic. The, order of their ionic size is, [1993], (a) Na+ > Mg++ < Al3 + < Si4+, (b) Na+ < Mg++ > Al3+ > Si4+, (c) Na+ > Mg++ > Al3+ > Si4+, (d) Na+ < Mg++ > Al3+ < Si4+, One would expect proton to have very large, [1993], (a) Charge, (b) Ionization potential, (c) Hydration energy (d) Radius., Which of the following sets has strongest, tendency to form anions ?, [1993], , (a) Ga, In, Tl, (b) Na, Mg, Al, (c) N, O, F, (d) V, Cr, Mn, 43. Elements of which of the following groups will, form anions most readily ?, [1992], (a) Oxygen family, (b) Nitrogen family, (c) Halogens, (d) Alkali metals, 44. In the periodic table, with the increase in atomic, number, the metallic character of an element, [1989], (a) Decreases in a period and increases in a group, (b) Increases in a period and decreases in a group, (c) Increases both in a period and the group, (d) Decreases in a period and the group., 45. The electronic configuration of four elements are, given below. Which element does not belong to, the same family as others ?, [1989], (a) [Xe]4f 145d101s2 (b) [Kr]4d 10 5s 2, (d) [Ar] 3d10 4s2, (c) [Ne]3s23p5, 46. Pauling’s electronegativity values for elements, are useful in predicting, [1989], (a) Polarity of the molecules, (b) Position in the E.M.F. series, (c) Coordination numbers, (d) Dipole moments., , ANSWER KEY, 1, , (c), , 6, , (a) 11, , (b) 16, , (a) 21, , (d) 26, , (d) 31, , (a) 36, , (b) 41, , (c) 46, , 2, , (a), , 7, , (c) 12, , (b) 17, , (a) 22, , (d) 27, , (a) 32, , (a) 37, , (a) 42, , (c), , 3, , (a), , 8, , (a) 13, , (a) 18, , (c) 23, , (a) 28, , (a) 33, , (d) 38, , (d) 43, , (c), , 4, , (a), , 9, , (b) 14 (N) 19, , (b) 24, , (b) 29, , (b) 34, , (a) 39, , (d) 44, , (a), , 5, , (b) 10, , (c) 15, , (c) 25, , (a) 30, , (b) 35, , (b) 40, , (c) 45, , (c), , (a) 20, , (a)
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25, , Classification of Elements and Periodicity in Properties, , Hints & Solutions, 1., 2., , 3., , 4., , 5., , 6., , 7., , 8., , (c) Unununium (Z = 111), it is Roentgenium, (Rg) not Darmstadtium., (a) CO : Neutral oxide, BaO : Basic oxide, Al2O3 : Amphoteric oxide, Cl2O7 : Acidic oxide, (a) After 86Rn (Group 18), elements from atomic, number 89 to 103 (actinides) are in group 3, 114 = 103 + 11. Thus, element with atomic, number 114 will be in group (3 + 11) or group 14, (carbon family)., The outer shell configuration of group 14 is, ns2 = np2., Hence, only option (a) is correct., (a) The electronic configuration clearly suggest, that it is a d-block element (having configuration, (n – 1) d 1– 10 ns 0 – 2 ) which starts from group 3, and goes till group 12. Hence with d3 configuration, it would be classified in the fifth group., (b) Electronic configuration of element with, atomic number 118 will be [Rn]5f 146d10 7s27p6., Since its elctronic configuration in the outer most, orbit (ns2np6) resemble with that of inert or noble, gases, therefore it will be noble gas element., (a) Atomic number of the given element is 15, and it belongs to 5th group. Therefore atomic, number of the element below the above element, = 15 + 18 = 33., (c) Element with Z = 33, , (1s 2 2s2 p6 3s2 p6d10 4s2 p3 ) lies in fifth (or 15th), group., (a) Na2O basic oxide, Al2O3 amphoteric oxide, N2O neutral, Cl2O7 acidic oxide, , 10., , 11., , Halogens have high electron affinities which, decreases on moving down the group. However,, fluorine has lower value than chlorine which is due, to its small size and repulsion between the electron, added and electrons already present., , 12., , 13., , 14., , 15., , Elements of gp.1 and 2 form besicoxides while, , gp. 13 and 14 elements form amphoteric, oxides. Elements of gp. 16 and 17 form acidic, oxides. Acidic character of oxides increases, from left to right in the periodic table., , 9., , (b) Consider the stability of electronic, configuration after loss of one electron., , (c) Due to poor shiel ding effect of 3d, electrons in Ga, the atomic radii of Ga < Al., Thus, the correct order of atomic radii is B <, Ga < Al < In < Tl., (b) The correct order is B < C < O < N, Generally ionisation energy increases across a, period. But here first I.E. of O is less than the first, I.E. of N. This is due to the half-filled 2p orbital in, N(1s2, 2s2, 2p3) which is more stable than the 2p, orbital in O (1s2, 2s2, 2p4)., (c) The correct order of electron affinity is, I < Br < F < Cl, , 16., , (b) In isoelectronic species the radius decrease, with increase in nuclear charge hence increasing, order of radius is Ca+2 < K+ < Ar, (a) Incoming electrons occupies the smaller n, = 2 shell, also negative charge on oxygen (O–) is, another factor due to which incoming electron, feel repulsion., Hence, electron repulsion outweigh the stability, gained by achieving noble gas configuration., (N) All answers are incorrect., H– > H > H+; O2– > F– < Na+, O2– > F– > Na+; N3– > Mg2+ > Al3+, For isoelectronic species the size is, determined by Zeffe. Higher the Zeffe. lower is, the size of the ions/species., (a) As the nuclear charge increases, the force, of attraction between the nucleus and the, incoming electron increases and hence the, elecron gain enthalpy becomes more negative,, hence the correct order is, Ca < Al < C < O < F, (a) As the positive charge increases on metal, cation, radius decreases. This is due to the fact, that nuclear charge in the case of a cation is, acting on lesser number of electrons and pulls, them closer.
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EBD_8336, 26, , 17., , CHEMISTRY, (a) IE1 of Na = – Electron gain enthalpy of Na +, = – 5.1 eV., , 25., , In this question, temperature is to be defined as, absolute zero. This is due to the fact that ionization, energy and electron affinity are defined at absolute, zero temperature., , 5, RT, 2, 5, Electron gain enthalpy = electron affinity - RT, 2, (c) 12 Mg, P, Cl, Ca, 15, 17, 20, 110, 99, 197 (pm), 160, So, the order will be: Cl < P < Mg < Ca, (b) O < S < F < Cl, Electron gain enthalpy of given elements are, – 141, – 200, – 333 and – 349 kJ mol–1 respectively., , Ionization enthalpy = ionization energy +, , 18., , 19., , Due to small size of atom, addition of an electron, is not easy. This is the reason why the magnitude, of electron gain enthalpy of oxygen and fluorine is, less than that of sulphur and chlorine respectively., , 20., , 21., , 22., , (c) Among the isoelectronic species, size, increases with the increase in negative charge., Thus S2– has the highest negative charge and, hence largest in size followed by Cl–, K+ and Ca2+., (d) The smaller the atomic size, larger is the, value of ionisation potential. Further the atoms, having half filled or fully filled orbitals are, comparatively more stable, hence more energy, is required to remove the electron from such, atoms., (d) The stability of +1 oxidation state increases, from aluminium to thallium i.e., Al < Ga < In < Tl, Inert pair effect is generally exhibited by some, heavier nucleus p-block elements with common, molecular formula ns2np1–6. These elements have, less tendency to leave their outer most s-electrons, at the time of chemical reaction i.e. electrons, present in s-orbital does not participate in the bond, formation. For example Tl, Po, Sn, Pb, Bi., , 23., , 24., , (a) Proton affinity decreases in moving across, the period from left to right due to increase in, charge, within a group the proton affinities, decreases from top to bottom., Nitrogen family > Oxygen family > Halogens, (b) 1s 2 , 2s 2 , 2p 5 , 3s 1 is an unstable, configuration with the outermost electron in 3rd, orbit. Hence, it has lowest ionisation enthalpy., , 26., , (a) For isoelectronic species, size of anion, increases as negative charge increases whereas, size of cation decreases with increase in positive, charge. Further ionic radii of anions is more than, that of cations. Thus the correct order is, Ca2+ < K+ < Ar < Cl– < S2–, (d) Due to odd number of electrons in ClO2, it, is expected to exhibit paramagnetic behaviour., Cl, O, , 27., , O, , Paramagnetic, , (a) Ionic radii are inversely proportional to, effective nuclear charge., Ionic radii in the nth orbit is given as, 1, n2 a0, or rn µ, Z, Z, when n = principal quantum number, Z=effective nuclear charge., (a) Among the given options, only Fe shows, variable oxidation states so it can form two, chlorides, viz. FeCl2 and FeCl3., (b) This is because of inter-electronic, replusions between lone pairs., B.E. :, F–F, Cl – Cl, (kJ mol–1) :, 158.8, 242.6, .., .., : .F. - F, ..:, , rn =, , 28., 29., , «, , «, , 30., , 31., , 32., , 33., , (b) Along the period, I.P. generally increases, but not regularly. Be and B are exceptions. First, I.P. increases in moving from left to right in a period,, but I.P. of B is lower than Be., (a) Be – 1s22s2; B – 1s22s22p1; C – 1s22s22p2;, N – 1s22s22p3; O – 1s22s22p4. IP increases along, the period. But IP of Be > B. Further IP of O < N, because atoms with fully or partly filled orbitals are, most stable and hence have high ionisation energy., (a) Mg = 1s 2 2 s 2 2 p 6 3s 2, After the removal of 2 electrons, the magnesium, ion will acquire noble gas configuration hence, removal of 3rd electron will require large amount, of energy., (d) First ionisation potential of Be is greater, than boron due to following configuration, 2 2, 2 2 1, 4Be=1s ,2s, 5B=1s ,2s 2p, Order of attraction of electrons towards nucleus, 2s > 2p, so more amount of energy is required to
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Classification of Elements and Periodicity in Properties, , 34., , 35., 36., 37., , 38., , 39., , 40., , 41., 42., , remove the electron in 2s-orbital in comparison, to 2p orbital., (a) In a period on moving from left to right ionic, radii decreases., So order of cationic radii is Cr2+ > Mn2+ > Fe2+ > Ni2+, (b) Greater is the positive charge on atom, larger, will be the effective nuclear charge. Hence smaller, is the size., (b) N2O5 is strongly acidic, ZnO and Al2 O3 are, amphoteric, therefore, MgO is most basic., (a) Non metals form oxides with oxygen and, thus reduce oxides of metals behaving as, reducing agents., (d) Abnormally high difference between 2nd, and 3rd ionization energy means that the element, has two valence electrons., (d) Atomic volume is the volume occupied by, one mole of an element. Within a period from left, to right, atomic volume first decreases and then, increases due to increase in nuclear charge and, increase in molar mass., (c) Amongst isoelectronic ions, the size of the, cation decreases as the magnitude of the charge, increases., (c) Proton (H+) being very small in size would, have very large hydration energy., (c) N, O and F (p-block elements) are highly, , 43., , 27, , electronegative non metals and will have the, strongest tendency to form anions by gaining, electrons from metal atoms., (c) Elements of halogen group form anions most, readily., Electron affinity values are high in case of halogen, because halogens have seven electrons ( ns 2 np 5 ), in the valence shell, they have a strong tendency, to acquire the nearest inert gas configuration by, gaining an electron from the metallic atom and form, halide ions easily., , 44., 45., , 46., , (a) Metallic character decreases in a period and, increases in a group., (c) Elements (a), (b) and (d) belong to the same, group since each one of them has two electrons, in the s-sub shell. In contrast, element (c) has, seven electrons in the valence shell and hence, does not lie in the same group., (a) Paulings electronegativity values for elements, are useful in predicting polarity of the molecule., Pauling scale of electronegativity was helpful in, predicting :, (i) Nature of bond between two atoms, (ii) Stability of bond, By calculating the difference in electro-negativities,, polarity of bond can be calculated.
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EBD_8336, 28, , 4, , CHEMISTRY, , Chemical Bonding and, Molecular Structure, , Trend Analysis with Important Topics & Sub-Topics, 2020, Topic Name, , Sub-Topic, , Dipole moment and, bond polarity, , dipole moment, , VSEPR theory and, hybridisation, , structure and, hybridisation, VSEPR theory, molecular orbital, theory/bond order, , Valence bond and, molecular orbital, theory, , valence bond theory, , LOD - Level of Difficulty, , E - Easy, , 1., , 2., , 3., , 4., , 2019, , 2018, , 2017, , 2016, , QNS. LOD QNS. LOD QNS. LOD QNS. LOD QNS. LOD, 1, , A, 2, , A, , 2, , A, 2, , 1, , Topic 1: Electrovalent, Covalent and, Co-ordinate Bonding, Which of the following is electron - deficient?, [NEET 2013], (a) (SiH3)2, (b) (BH3)2, (c) PH3, (d) (CH3)2, The correct sequence of increasing covalent, character is represented by, [2005], (a) LiCl < NaCl < BeCl2, (b) BeCl2 < LiCl < NaCl, (c) NaCl < LiCl < BeCl2, (d) BeCl2 < NaCl < LiCl, Which of the following is the electron, deficient molecule?, [2005], (a) C2H6, (b) B2H6, (c) SiH4, (d) PH3, Cation and anion combines in a crystal to form, following type of compound., [2000], (a) ionic, (b) metallic, (c) covalent, (d) dipole-dipole, , A, , 1, , A, , 1, , A, , A - Average, , 5., , 6., , 7., , 8., , 9., , 10., , 1, , A, , D - Difficult, , 1, , E, , A, , Qns - No. of Questions, , Among the following electron deficient, compound is :, [2000], (a) BCl3, (b) CCl4, (c) PCl5, (d) PCl3, Which of the following compounds has a, 3-centre bond?, [1996], (a) Diborane, (b) Carbon dioxide, (c) Boron trifluroide (d) Ammonia, Linus Pauling received the Nobel Prize for his, work on, [1994], (a) atomic structure (b) photosynthesis, (c) chemical bonds, (d) thermodynamics, Which of the following pairs will form the most, stable ionic bond ?, [1994], (a) Na and Cl, (b) Mg and F, (c) Li and F, (d) Na and F, The weakest among the following types of bonds, is, [1994], (a) ionic, (b) covalent, (c) metallic, (d) H–bond., Strongest bond is in between, [1993], (a) CsF, (b) NaCl, (c) Both (a) and (b) (d) None of above
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29, , Chemical Bonding and Molecular Structure, 11., , 12., , 13., , Among the following which compound will, show the highest lattice energy ?, [1993], (a) KF, (b) NaF, (c) CsF, (d) RbF, Among LiCl, BeCl2 BCl3 and CCl4, the covalent, bond character follows the order, [1990], (a) LiCl < BeCl2 > BCl3 > CCl4, (b) BeCl2 < BCl3 < CCl4 < LiCl, (c) LiCl < BeCl2 < BCl3 < CCl4, (d) LiCl > BeCl2 > BCl3 > CCl4, Which of the following does not apply to metallic, bond ?, [1989], (a) Overlapping valence orbitals, (b) Mobile valency electrons, (c) Delocalized electrons, (d) Highly directed bonds., Topic 2: Octet rule, Resonance and, Hydrogen Bonding, , 14., , 17., , 18., , 19., , 20., , Which of the following structures is the most, preferred and hence of lowest energy for SO3?, [2011 M], 21., , 15., , 16., , (a), , (b), , (c), , (d), , What is the dominant intermolecular force or, bond that must be overcome in converting liquid, CH3OH to a gas?, [2009], (a) Dipole-dipole interaction, (b) Covalent bonds, (c) London dispersion force, (d) Hydrogen bonding, Which of the following is not a correct, statement?, [2006], (a) The cannonical structures have no real, existence, (b) Every AB 5 molecule does in fact have, square pyramidal structure, (c) Multiple bonds are always shorter than, corresponding single bonds, (d) The electron-deficient molecules can act, as Lewis acids, , 22., , 23., , In X — H — Y, X and Y both are electronegative, elements, (a) Electron density on X will increase and on, H will decrease, [2001], (b) In both electron density will decrease, (c) In both electron density will increase, (d) Electron density will decrease on X and will, increase on H, Which one of the following molecules will form, a linear polymeric structure due to hydrogen, bonding?, [2000], (a) NH3, (b) H2O, (c) HCl, (d) HF, In PO43– ion, the formal charge on each oxygen, atom and P—O bond order respectively are, [1998], (a) –0.75, 0.6, (b) – 0.75, 1.0, (c) – 0.75, 1.25, (d) –3, 1.25, The low density of ice compared to water is due, to, [1997], (a) hydrogen-bonding interactions, (b) dipole-dipole interactions, (c) dipole-induced dipole interactions, (d) induced dipole-induced dipole interactions, The boiling point of p-nitrophenol is higher than, that of o-nitrophenol because, [1994], (a) NO2 group at p-position behave in a, different way from that at o-position., (b) intramolecular hydrogen bonding exists in, p-nitrophenol, (c) there is intermolecular hydrogen bonding, in p-nitrophenol, (d) p-nitrophenol has a higher molecular, weight than o-nitrophenol., Which one of the following is the correct order, of interactions ?, [1993], (a) Covalent < hydrogen bonding < van der, Waals < dipole-dipole, (b) van der Waals < hydrogen bonding <, dipole-dipole < covalent, (c) van der Waals < dipole-dipole < hydrogen, bonding < covalent, (d) Dipole-dipole < van der Waals < hydrogen, bonding < covalent., Strongest hydrogen bond is shown by [1992], (a) Water, (b) Ammonia, (c) Hydrogen fluoride, (d) Hydrogen sulphide.
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EBD_8336, 30, , 24., , CHEMISTRY, Which one of the following formulae does not, correctly represent the bonding capacities of the, two atoms involved ?, [1990], , (a), , +, H, é, ù, |, ê, ú, êH — P — H ú, |, ê, ú, H, ë, û, , (b) F, , F, O, , (c), , O, , O¬ N, O–H, , O, , (d) H – C = C, , 25., , 30., , O–H, Which one shows maximum hydrogen bonding?, (a) H2O, (b) H2Se, [1990], (c) H2S, (d) HF., , 31., , 32., , Topic 3: Dipole Moment and Bond Polarity, 26., , 27., , 28., , 29., , Which of the following set of molecules will, have zero dipole moment?, [2020], (a) Boron trifluoride, hydrogen fluoride,, carbon dioxide, 1,3-dichlorobenzene, (b) Nitrogen trifluoride, beryllium difluoride,, water, 1,3-dichlorobenzene, (c) Boron trifluoride, beryllium difluoride,, carbon dioxide, 1,4-dichlorobenzene, (d) Ammonia, beryllium difluoride, water,, 1,4-dichlorobenzene, Which of the following is the correct order of, dipole moment?, [NEET Odisha 2019], (a) H2O < NF3 < NH3 < BF3, (b) NH3 < BF3 < NF3 < H2O, (c) BF3 < NF3 < NH3 < H2O, (d) BF3 < NH3 < NF3 < H2O, Which of the following molecules has the, maximum dipole moment ?, [2014], (a) CO2, (b) CH4, (c) NH3, (d) NF3, Which of the following is a polar molecule ?, [NEET 2013], (a) SF4, (b) SiF4, (c) XeF4, (d) BF3, , 33., , 34., , 35., , The electronegativity difference between N and, F is greater than that between N and H yet the, dipole moment of NH3 (1.5 D) is larger than that, of NF3 (0.2D). This is because, [2006], (a) in NH3 the atomic dipole and bond dipole, are in the same direction whereas in NF3, these are in opposite directions, (b) in NH3 as well as NF3 the atomic dipole, and bond dipole are in opposite directions, (c) in NH3 the atomic dipole and bond dipole, are in the opposite directions whereas in, NF3 these are in the same direction, (d) in NH3 as well as in NF3 the atomic dipole, and bond dipole are in the same direction, Wh ich of th e followi ng woul d ha ve a, permanent dipole moment?, [2005], (a) SiF4, (b) SF 4, (c) XeF4, (d) BF3, The correct order of the O–O bond length in O2,, H2O2 and O3 is, [1995, 2005], (a), , O 2 > O 3 > H 2O 2, , (b), , O 3 > H 2O 2 > O 2, , (c), , O 2 > H 2O 2 > O 3, , (d) H 2 O 2 > O 3 > O 2, H2O is dipolar, whereas BeF2 is not. It is because, [2004], (a) the electronegativity of F is greater than, that of O, (b) H2O involves hydrogen bonding whereas, BeF2 is a discrete molecule, (c) H2O is linear and BeF2 is angular, (d) H2O is angular and BeF2 is linear, The dipole moments of diatomic molecules AB, and CD are 10.41D and 10.27 D, respectively, while their bond distances are 2.82 and 2.67 Å,, respectively. This indicates that, [1999], (a) bonding is 100% ionic in both the molecules, (b) AB has more ionic bond character than CD, (c) AB has lesser ionic bond character than, CD, (d) bonding is nearly covalent in both the, molecules, Which of the following bonds will be most polar?, [1992], (a) N – Cl, (b) O – F, (c) N – F, (d) N – N
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31, , Chemical Bonding and Molecular Structure, 36., , H2O has a non zero dipole moment while BeF2, has zero dipole moment because, [1989], (a) H2O molecule is linear while BeF2 is bent, (b) BeF2 molecule is linear while H2O is bent, (c) Fluorine has more electronegativity than, oxygen, (d) Beryllium has more electronegativity than, oxygen., Topic 4: VSEPR Theory and Hybridisation, , 37., , 38., , 39., , In the structure of ClF3, the number of lone pair, of electrons on central atom ‘Cl’ is, [2018], (a) One, (b) Two, (c) Three, (d) Four, Which of the following molecules represents, the order of hybridisation sp2, sp2, sp, sp from, left to right atoms?, [2018], (a) HC º C – C º CH, (b) CH2 = CH – C º CH, (c) CH3 – CH = CH – CH3, (d) CH2 = CH – CH = CH2, Which of the following pairs of compounds is, isoelectronic and isostructural ?, [2017], (a) TeI2,XeF2, , 40., , 41., , 42., , (b) IBr2- , XeF2, , (d) BeCl2,XeF2, (c) IF3, XeF2, The species, having bond angles of 120° is :[2017], (a) CIF3, (b) NCl3, (c) BCl3, (d) PH3, Consider the molecules CH4, NH3 and H2O., Which of the given statements is false? [2016], (a) The H–C–H bond angle in CH4, the H–N–H, bond angle in NH3, and the H–O–H bond, angle in H2O are all greater than 90°, (b) The H–O–H bond angle in H2 O is larger, than the H–C–H bond angle in CH4., (c) The H–O–H bond angle in H2O is smaller, than the H–N–H bond angle in NH3., (d) The H–C–H bond angle in CH4 is larger, than the H–N–H bond angle in NH3., Predict the correct order of electron repulsion, among the following :, [2016], (a) lone pair- lone pair > lone pair - bond pair >, bond pair - bond pair, , 43., , 44., , 45., , (b) lone pair - lone pair > bond pair - bond pair, > lone pair - bond pair, (c) bond pair - bond pair > lone pair - bond, pair > lone pair - lone pair, (d) lone pair - bond pair > bond pair - bond, pair > lone pair - lone pair, Which of the following pairs of ions are, isoelectronic and isostructural ?, [2015], (a), , ClO3– , CO32–, , (b) SO32– , NO3–, , (c), , ClO3– , SO32–, , (d) CO32– , SO32–, , Maximum bond angle at nitrogen is present in, which of the following ?, [2015], (a), , NO 2–, , (b) NO +2, , (c), , NO3–, , (d) NO 2, , In which of the following pairs, both the species, are not isostructural ?, [2015 RS], (a), , 46., , 47., , 48., , 49., , 50., , SiCl4 , PCl+, 4, , (b) diamond, silicon carbide, (c) NH3, PH3, (d) XeF4, XeO4, Be2+ is isoelectronic with which of the following, ions?, [2014], (a) H+, (b) Li+, (c) Na+, (d) Mg2+, Which one of the following species has plane, triangular shape ?, [2014], –, –, (b) NO3, (a) N3, (d) CO2, (c) NO2–, In which of the following pair both the species, have sp3 hybridization?, [NEET Kar. 2013], (a) H2S, BF3, (b) SiF4, BeH2, (c) NF3, H2O, (d) NF3, BF3, XeF2 is isostructural with, [NEET 2013], (b) SbCl3, (a) ICl2–, (c) BaCl2, (d) TeF2, Which of the following species contains three, bond pairs and one lone pair around the central, atom ?, [2012], (a) H2O, (b) BF3, –, (c) NH2, (d) PCl3
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EBD_8336, 32, , 51., , CHEMISTRY, Which one of the following pairs is isostructural, (i.e., having the same shape and hybridization)?, , é BCl3 and BrCl3- ù, ë, û, -ù, é, (b) ë NH3 and NO3 û, (c) [ NF3 and BF3 ], , (a), , [2012], , é BF4- and NH +4 ù, ë, û, Which of the two ions from the list given below, that have the geometry that is explained by the, same hybridization of orbitals, NO2–, NO3– ,, [2011], NH2–, NH4+, SCN– ?, (a) NO2– and NO3– (b) NH4+ and NO3–, (c) SCN– and NH2– (d) NO2– and NH2–, Considering the state of hybridization of carbon, atoms, find out the molecule among the following, which is linear ?, [2011], (a) CH3– CH = CH–CH3, (b) CH3 – C º C – CH3, (c) CH2 = CH – CH2 – C º CH, (d) CH3 – CH2 – CH2 – CH3, In which of the following molecules the central, atom does not have sp3 hybridization? [2010], (a) NH +4, (b) CH4, (d), , 52., , 53., , 54., , 55., , 56., , 57., , (d) BF4–, (c) SF4, Some of the properties of the two species, NO3and H3O+ are described below. Which one of, them is correct?, [2010], (a) Similar in hybridization for the central atom, with different structures., (b) Dissimilar in hybridization for the central, atom with different structures., (c) isostructural with same hybridization for, the central atom., (d) Isostructural with different hybridization, for the central atom., In which one of the following species the central, atom has the type of hybridization which is not, the same as that present in the other three?, (a) SF4, (b) I3–, [2010], (c) SbCl52–, (d) PCl5, In which of the following pairs of molecules/, ions, the central atoms have sp2 hybridization?, [2010], , NO -2 and NH 3 (b) BF3 and NO -2, (c) NH -2 and H 2 O (d) BF3 and NH -2, In which of the following molecules / ions, BF3, NO2- , NH 2- and H2O ,, [2009], (a), , 58., , the central atom is sp2 hybridized ?, (a) NH 2- and H2O, (b) NO2- and H2O, , BF3 and NO 2(d) NO 2- and NH 2The correct order of increasing bond angles in, the following triatomic species is :, [2008], (a) NO-2 < NO+2 < NO 2, (b) NO-2 < NO2 < NO2+, (c), , 59., , 60., , 61., , 62., 63., , 64., , 65., , 66., , (c) NO+2 < NO 2 < NO2(d) NO +2 < NO-2 < NO, In which of the following pairs, the two species, are isostructural?, [2007], (a) SO32– and NO3– (b) BF3 an NF3, (c) BrO3– and XeO3 (d) SF4 and XeF4, Which of the following is not isostructural with, SiCl4?, [2006], (b) PO43–, (a) SO42–, (d) SCl4, (c) NH4+, Which of the following species has a linear shape ?, [2006], (a) SO2, (b) NO2+, (c) O3, (d) NO2–, In which of the following molecules all the bonds, are not equal?, [2006], (a) BF3, (b) AlF3, (c) NF3, (d) ClF3, Which of the following molecules has trigonal, planar geometry?, [2005], (a) BF3, (b) NH3, (c) PCl3, (d) IF3, In BrF 3 molecule, the lone pairs occupy, equatorial positions to minimize, [2004], (a) lone pair - bond pair repulsion only, (b) bond pair - bond pair repulsion only, (c) lone pair - lone pair repulsion and lone pair, - bond pair repulsion, (d) lone pair - lone pair repulsion only, In an octahedral structure, the pair of d orbitals, involved in d 2 sp3 hybridization is, [2004], (a), , d, , (c), , d, , d, , x2 - y 2 , z 2, , (b) d xz, d 2 2, x -y, , d, z 2 , xz, , (d) d xy , d yz
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33, , Chemical Bonding and Molecular Structure, 67., , 68., 69., , 70., , 71., , 72., , In a regular octahedral molecule, MX6 the number, of X - M - X bonds at 180° is, [2004], (a) three, (b) two, (c) six, (d) four, Which of the following has pp – dp bonding?, [2002], (b) SO32–, (a) NO3–, (d) CO32–, (c) BO33–, Main axis of a diatomic molecule is z, molecular, orbital px and py overlap to form which of the, following orbital?, [2001], (a) p - molecular orbital, (b) s - molecular orbital, (c) d - molecular orbital, (d) No bond will be formed, Which of the following two are isostructural?, [2001], (a) NH3, BF3, (b) PCl5, ICl5, (d) CO3–2, SO3–2, (c) XeF2, IF2–, In which of the following, the bond angle is, maximum?, [2001], (a) NH3, (b) SCl2, (d) PCl3, (c) NH4+, Among the following ions the pp–dp overlap, could be present in, [2000], (a), , 73., , 74., 75., , NO -2, , 78., , 79., , 80., , 81., , (b) NO 3-, , (c) PO34(d) CO 23Which one of the following has the pyramidal, shape?, [1999], (b) SO3, (a) CO32–, (c) BF3, (d) PF3, Which of the following molecules is planar?, (a) SF4, (b) XeF4, [1998], (c) NF3, (d) SiF4, The AsF5 molecule is trigonal bipyramidal. The, hybrid orbitals used by the As atom for bonding are, , d, , 77., , 2 , d 2 , s, p x , p y, , (a), x2 - y, z, (b) dxy, s, px, py, pz, , 82., , 83., , 84., , [1997], , (c) s, px, py, pz, d, , z2, , (d), 76., , d, , x2 - y 2, , , s, px, py, pz, , The cylindrical shape of an alkyne is due to the, fact that it has, [1997], (a) three sigma C – C bonds, (b) two sigma C – C and one 'p' C – C bond, (c) three 'p' C – C bonds, (d) one sigma C– C and two 'p' C – C bonds, , 85., , The BCl3 is a planar molecule whereas NCl 3 is, pyramidal because, [1995], (a) B-Cl bond is more polar than N-Cl bond, (b) N-Cl bond is more covalent than B-Cl bond, (c) nitrogen atom is smaller than boron atom, (d) BCl3 has no lone pair but NCl3 has a lone, pair of electrons, The distance between the two adjacent carbon, atoms is largest in, [1994], (a) benzene, (b) ethene, (c) butane, (d) ethyne, Among the following orbital bonds, the angle is, minimum between, [1994], (a) sp3 bonds, (b) px and py orbitals, (c) H – O – H in water (d) sp bonds., Which of the following does not have a, tetrahedral structure?, [1994], (a) BH –4, (b) BH3, (c) NH +4, (d) H 2 O., Which of the following statements is not correct ?, [1993], (a) Double bond is shorter than a single bond, (b) Sigma bond is weaker than a p (pi) bond, (c) Double bond is stronger than a single bond, (d) Covalent bond is stronger than hydrogen, bond., Which structure is linear ?, [1992], (a) SO2, (b) CO2, (c) CO 32(d) SO 24 Which one of the following has the shortest, carbon carbon bond length ?, [1992], (a) Benzene, (b) Ethene, (c) Ethyne, (d) Ethane, In compound X, all the bond angles are exactly, 109°28; X is, [1991], (a) Chloromethane, (b) Carbon tetrachloride, (c) Iodoform, (d) Chloroform., Which statement is NOT correct ?, [1990], (a) A sigma bond is weaker than a p -bond., (b) A sigma bond is stronger than a p -bond., (c) A double bond is stronger than a single, bond., (d) A double bond is shorter than a single, bond.
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EBD_8336, 34, , 86., , 87., , 88., , 89., , CHEMISTRY, In which one of the following molecules the, central atom said to adopt sp2 hybridization?, (a) BeF2, (b) BF3, [1989], (c) C2H2, (d) NH3, Which of the following molecule does not have, a linear arrangement of atoms ?, [1989], (a) H2S, (b) C2H2, (c) BeH2, (d) CO2, Equilateral shape has, [1988], (a) sp hybridisation (b) sp2 hybridisation, (c) sp3 hybridisaiton (d) sp3 hybridisation, The angle between the overlapping of one, s-orbital and one p-orbital is, [1988], (a) 180°, (b) 120°, (c) 109°28', (d) 120° 60', , 95., , 96., , 97., , Topic 5: Valence Bond and Molecular, Orbital Theory, 90., , Identify a molecule which does not exist. [2020], (a) Li2, (b) C2, (c) O2, (d) He2, , 91., , Which of the following is paramagnetic?, [NEET Odisha 2019], (a) O2, (b) N2, (c) H2, (d) Li2, The manganate and permanganate ions are, tetrahedral, due to:, [2019], (a) The p-bonding involves overlap of porbitals of oxygen with d-orbitals of, manganese, (b) There is no p-bonding, (c) The p-bonding involves overlap of p-orbital, of oxygen with p-orbitals of manganese, (d) The p- bonding involves overlap of dorbital of oxygen with d-orbitals of, manganese, Which of the following diatomic molecular, species has only p–bonds accordin g to, Molecular Orbital Theory ?, [2019], (a) O2, (b) N2, (c) C2, (d) Be2, Consider the following species :, CN+, CN–, NO and CN, Which one of these will have the highest bond, order?, [2018], , 92., , 93., , 94., , 98., , (a) NO, (b) CN–, (c) CN, (d) CN+, Which of the following pairs of species have, the same bond order ?, [2017], +, –, (a) O2, NO, (b) CN , CO, (c) N 2 , O 2(d) CO, NO, The correct bond order in the following species, is:, [2015], (a), , O 22+ < O 2– < O2+, , (b) O +2 < O 2– < O 22+, , (c), , O 2– < O +2 < O 22+, , (d) O 22+ < O 2+ < O 2–, , Which of the following options represents the, correct bond order ?, [2015], (a), , O 2– < O 2 < O 2+, , (b) O –2 > O2 < O 2+, , (c), , O 2– < O2 > O 2+, , (d) O –2 > O2 > O2+, , Which of the following species contains equal, number of s- and p-bonds :, [2015], (a) XeO4, (b) (CN)2, (c) CH2(CN)2, , 99., , (d) HCO3–, , +, Decreasing order of stability of O2, O2 ,O 2 and, , O 22- is :, , (a), , 2O+, 2 > O2 > O2 > O2, , (b), , +, O 22- > O2 > O2 > O2, , (c), , 2O2 > O+, 2 > O2 > O2, , (d), , +, 2O2 > O2 > O2 > O2, , [2015 RS], , 100. The hybridization in volved in complex, [Ni(CN)4]2–. is (At. No. Ni = 28), [2015 RS], (a) dsp 2, (b) sp3, (c) d 2 sp2, (d) d 2 sp3, 101. The outer orbitals of C in ethene molecule can, be considered to be hybridized to give three, equivalent sp2 orbitals. The total number of sigma, (s) and pi (p) bonds in ethene molecule is, [NEET Kar. 2013], (a) 1 sigma (s) and 2 pi (p) bonds, (b) 3 sigma (s) and 2 pi (p) bonds, (c) 4 sigma (s) and 1 pi (p) bonds, (d) 5 sigma (s) and 1 pi (p) bonds
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35, , Chemical Bonding and Molecular Structure, 102. In which of the following ionisation processes, the bond energy increases and the magnetic, behaviour changes from paramagnetic to, diamagnetic?, [NEET Kar. 2013], (b) O2 ® O2+, (a) N2 ® N2+, (d) NO ® NO+, (c) C2 ® C2+, 103. The pair of species that has the same bond order, in the following is:, [NEET Kar. 2013], (a) O2, B2, (b) CO, NO+, (c) NO–, CN–, (d) O2, N2, 104. Which of the following is paramagnetic ?, [NEET 2013], (a) O -2, (b) CN–, , (c) NO+, (d) CO, 105. Which one of the following molecules contains, no p bond?, [NEET 2013], (a) H2O, (b) SO2, (c) NO2, (d) CO2, 106. Four diatomic species are listed below. Identify, the correct order in which the bond order is, increasing in them:, [2008, 2012 M], (a) NO <, < C 22 - < He +2, (b) O -2 < NO < C22 - < He 2+, (c) C22 - < He 2+ < O2- < NO, (d) He+2 < O2- < NO < C 22 During change of O2 to O 2O2-, , 107., , ion, the electron, adds on which one of the following orbitals ?, [2012 M], (a) p* orbital, (b) p orbital, (c) s* orbital, (d) s orbital, 108. The pair of species with the same bond order is :, , (b) O+2 , NO+ [2012], O2–, 2 , B2, (c) NO, CO, (d) N2, O2, 109. Bond order of 1.5 is shown by :, [2012], +, (a) O 2, (b) O 2, (a), , (c) O 22 (d) O2, 110. The pairs of species of oxygen and their, magnetic behaviours are noted below. Which of, the following presents the correct description ?, (a) O -2 , O 22 - – Both diamagnetic [2011 M], (b), , O + ,O22 - –, , Both paramagnetic, , (c), , O+2 ,O2, , – Both paramagnetic, , (d), , O,O 22 -, , – Both paramagnetic, , 111. Which of the following has the minimum bond, length ?, [2011], (b) O2 –, (a) O2+, (d) O2, (c) O2 2–, 112. Which one of the following species does not, exist under normal conditions?, [2010], (a) Be+2, (b) Be2, (c) B 2, (d) Li 2, 113. According to MO theory which of the following, lists ranks the nitrogen species in terms of, increasing bond order?, [2009], –, –, 2–, <, <, N, N, (a) N 2–, (b), <, N, N, 2, 2, 2, 2 < N2, 2, 2–, –, (c) N 2– < N 2–, 2 < N 2 (d) N 2 < N 2 < N 2, 114. The angular shape of ozone molecule (O 3 ), consists of :, [2008], (a) 1 sigma and 2 pi bonds, (b) 2 sigma and 2 pi bonds, (c) 1 sigma and 1 pi bonds, (d) 2 sigma and 1 pi bonds, 115. The correct order of C–O bond length among, CO, CO32 - , CO2 is, [2007], 2–, 2–, (a) CO < CO3 < CO2 (b) CO3 < CO2 < CO, (c) CO < CO2 < CO32– (d) CO2 < CO32– < CO, 116. The number of unpaired electrons in a, paramagnetic diatomic molecule of an element, with atomic number 16 is, [2006], (a) 3, (b) 4, (c) 1, (d) 2, 117. Among the following the pair in which the two, species are not isostructural is, [2004], (a) SiF4 and SF4, , (b) IO3- and XeO 3, , (c) BH -4 and NH +4, (d) PF6- and SF6, 118. Which of the following statements is not correct, for sigma and pi-bonds formed between two, carbon atoms?, [2003], (a) Sigma-bond determines the direction, between carbon atoms but a pi-bond has, no primary effect in this regard, (b) Sigma-bond is stronger than a pi-bond, (c) Bond energies of sigma- and pi-bonds are, of the order of 264 kJ/mol and 347 kJ/mol,, respectively, (d) Free rotation of atoms about a sigma-bond, is allowed but not in case of a pi-bond, 119. In NO3– ion, number of bond pair and lone pair, of electrons on nitrogen atom respectively are, (a) 2, 2, (b) 3, 1, [2002], (c) 1, 3, (d) 4, 0
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EBD_8336, 36, , CHEMISTRY, , 120. The relationship between the dissociation, [2000], energy of N2 and N2+ is :, (a) Dissociation energy of N2+ > dissociation, energy of N2, (b) Dissociation energy of N2 = dissociation, energy of N2+, (c) Dissociation energy of N2 > dissociation, energy of N2+, (d) Dissociation energy of N2 can either be, lower or higher than the dissociation energy, of N2+, 121. Which one of the following arrangements represents the increasing bond orders of the given, species?, [1999], (a) NO+ < NO < NO– < O2–, (b) O2– < NO– < NO <NO+, (c) NO– < O2– < NO < NO+, (d) NO < NO+ < O2– < NO–, 122. Amon g the followin g which one is not, paramagnetic? [Atomic numbers : Be = 4,, Ne = 10, As = 33, Cl = 17], [1998], (a) Cl– (b) Be+, (c) Ne+2 (d) As +, 123. The number of anti-bonding electron pairs in, O 2- 2 molecular ion on the basis of molecular, orbital theory is, (Atomic number of O is 8) [1998], (a) 5, (b) 2, (c) 3, (d) 4, 124. N2 and O2 are converted into monoanions, N2–, and O2– respectively. Which of the following, statements is wrong ?, [1997], (a) In N2, the N—N bond weakens, (b) In O2, the O—O bond order increases, (c) In O2, bond length decreases, (d) N2– becomes diamagnetic, , 125. The correct order of N–O bond lengths in NO,, [1996], NO2–, NO3– and N2O4 is, (a) N 2 O 4 > NO -2 > NO 3- > NO, (b) NO > NO 3- > N 2 O 4 > NO -2, (c) NO 3- > NO -2 > N 2 O 4 > NO, (d) NO > N 2 O 4 > NO-2 > NO3126. The ground state electronic configuration of, valence shell electrons in nitrogen molecule (N2), is written as KK s 2 s 2 , s * 2 s 2 , p 2 p x2 , p 2 p 2y s 2 pz2, Bond order in nitrogen molecule is, [1995], (a) 0, (b) 1, (c) 2, (d) 3, 127. Which of the following species is paramagnetic?, (a) O 22(b) NO, [1995], (c) CO, (d) CN–, 128. Mark the incorrect statement in the following, [1994], (a) The bond order in the species O2, O2+ and, O2– decreases as O +2 > O 2 > O -2, (b) The bond energy in a diatomic molecule, always increases when an electron is lost, (c) Electrons in antibonding M.O. contribute, to repulsion between two atoms., (d) With increase in bond order, bond length, decreases and bond strength increases., 129. Linear combination of two hybridized orbitals, belonging to two atoms and each having one, electron leads to a, [1990], (a) Sigma bond, (b) Double bond, (c) Co-ordinate covalent bond, (d) Pi bond., , ANSWER KEY, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, , (b), (c), (b), (a), (a), (a), (c), (b), (d), (a), (b), (c), (d), , 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, , (d), (d), (b), (a), (d), (c), (a), (c), (b), (c), (d), (d), (c), , 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, , (c), (c), (a), (a), (b), (d), (d), (c), (c), (b), (b), (b), (b), , 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, , (c), (b), (a), (c), (b), (d), (b), (b), (c), (a), (d), (d), (a), , 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, , (b), (c), (b), (c), (b), (c), (b), (c), (d), (b), (d), (a), (c), , 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, , (a), (a), (b), (a), (c), (c), (c), (d), (b), (c), (d), (d), (c), , 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, , (b), (b), (b), (b), (c), (b), (a), (b), (a), (b), (a), (d), (a), , 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 102, 103, 104, , (a), (c), (b), (b), (c), (a), (a), (a), (a), (d), (d), (b), (a), , 105, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117, , (a), (d), (a), (a), (b), (c), (a), (b), (a), (d), (c), (d), (a), , 118, 119, 120, 121, 122, 123, 124, 125, 126, 127, 128, 129, , (c), (d), (c), (b), (a), (d), (b), (c), (d), (b), (b), (a)
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37, , Chemical Bonding and Molecular Structure, , Hints & Solutions, H, , B, H, , 3., 4., , B, H, , H, , It contains two, 3 centre - 2 electron bonds that, are present above and below the plane of, molecules. Compounds which do not have, sufficient number of electrons to form normal, covalent bonds are called electron deficient, molecules., (c) As difference of electronegativity increases, % ionic character increases and covalent, character decreases i.e., as electronegativity, difference decreases covalent character increases., Further greater the charge on the cation more, will be its covalent character. Be has maximum, (+2) charge., (b) The compound, of which central atom is, octetless known as electron deficient compound., Hence B2H6 is electron deficient compound., (a) Cation and anion form ionic bond in, crystal., The electrostatic force that binds the oppositely, charged ions which are formed by transfer of, electron from one atom to another is called ionic, bond., , 5., , B, , B, , H, H, H, The bond represented by dots form the 3-centred, electron pair bond., The idea of three centred electron pair bond, B–H–B bridges is necessary because diborane does, not have sufficient electrons to form normal, covalent bonds. It has only 12 electrons instead of, 14 that are required to give simple ethane like, structure., , 7., 8., , (c) Chemical bonds., (b) The stability of the ionic bond depends upon, the lattice energy which is expected to be more, between Mg and F due to +2 charge on Mg atom., , 12., , 13., , 14., , Option (a) :, (0), , :O:, ××, S, O (+2) O:(–1), Formal charge = total no. of valence e– in the, free atom – total no. of non-bonding (lone pair), 1, e - - (total no. of bonding e–)., 2, 1, Formal charge on S stom = 6 – 0 – (8) = +2, 2, Similarly, formal charges on coordinated bonded, oxygen atom and covalently bonded oxygen atom, are –1 and 0 respectively., :O:, , (–1), , Option (d) :, , :, , 6., , (a) Boron in BCl3 has 6 electrons in outer most, shell. Hence BCl3 is an electron pair deficient, compound., H, H, (a) H, , 11., , (d) H-bond is the weakest., (a) According to Fajan's rule, ionic character, increases with increase in size of the cation and, decrease in size of the anion. Thus, CsF has, higher ionic character than NaCl and hence bond, strength of CsF is stronger than NaCl., (b) For compounds containing ions of same, charge, lattice energy increases as the size the, ions decrease. Thus, NaF has highest lattice, energy. The size of cation is in the order Na+ <, K+ < Rb+ < Cs+., (c) As we move in period from Li ® Be ®, B ® C, the electronegativity (EN) increases and, hence the EN difference between the element, and Cl decreases and accordingly the covalent, character increases.Thus LiCl < BeCl2 < BCl3<, CCl4 is correct., (d) In metallic bonds each ion is surrounded, by equal no. of oppositely charged ions hence, have electrostatic attraction on all sides and, hence do not have directional characteristics., (d) The lowest energy structure is the one with, the smallest formal charges on the atoms., Option (b) and (c) are not possible structures of, SO3., , S, :, O, O, :, Formal charge on S atom, 1, = 6 – 0 – (12) = 0, 2, Formal charge on O atom, 1, = 6 – 4 – (4) = 0, 2, :, , H, , :, , H, , 2., , 9., 10., , (b) (BH3)2 or (B2H6), , :, , 1.
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EBD_8336, 38, , 15., 16., , CHEMISTRY, (d) Due to intermolecular hydrogen bonding, in methanol, it exist as assosiated molecule., (b) Statement (a), (c), (d) are correct. Statement, (b) is incorrect statement., AB5 may have two structures as follows :, , Intramolecular Hydrogen bonding in o-nitrophenol, is a interaction with in the molecule whereas the, intermolecular H-bonding in p-nitrophenol is a, interaction between the molecules which results, into higher boiling point. Intermolecular hydrogen, bonding also results in a stronger driving force for, cystal formation in other molecules generating, higher melting temperature e.g. p-hydroxy benzoic, acid., , B, , B, B, , B, , B, , A, , B, B, , B, , Square Pyramidal, , A, , 22., B, , Individual cannonical structures do not exist. The, molecule as such has a single structure, which is, the resonance hybrid of the cannonical forms., , 17., , 18., 19., , (a) In X — H - - - Y, X and Y both are, electronegative elements (i.e attracts the, electron pair) then electron density on X will, increase and on H will decrease., (d) F—H-----F—H-----F—H-----F, HF form linear polymer structure due to, hydrogen bonding., (c) Bond order between P – O, =, , 25., , With the increase of electronegativity and decrease, in size of the atom to which hydrogen is covalently, linked, the strength of hydrogen bond increases., , 26., , O–, , O P O–, , O P O–, , O–, , O–, , O–, , O P O, O, , –, , –, , (c) BF3, BeF2, CO2 and 1, 4-dichlorobenzene, all are symmetrical molecules., F, , µ=0, , B, , O, , –, , F, , O–, , O P O, O–, , 3, = -0.75, 4, (a) We know that due to polar nature, water, molecules are held together by intermolecular, hydrogen bonds. The structure of ice is open, with large number of vacant spaces, therefore, the density of ice is less than water., (c) The b.p. of p-nitrophenol is higher than, that of o-nitrophenol because in p-nitrophenol, there is intermolecular H-bonding but in, o-nitrophenol it is intramolecular H-bonding., , or Formal charge on oxygen = -, , 21., , 24., , no. of bonds in all possible direction, 5, = = 1.25, total no. of resonating structures, 4, , –, , 20., , 23., , B, Trigonal Bipyramidal, , (b) The strength of the interactions follows the, order: van der Waal’s < hydrogen – bonding <, dipole-dipole < covalent., (c) Hydrogen fluoride shows str ongest, hydrogen bonding due to high electronegativity, of fluorine., O, ||, (d) H - C = C* - O - H, The star marked carbon has a valency of 5 and, hence this formula is not correct., (d) As F is most electronegative thus HF shows, maximum strength of hydrogen bond., , F, , F, , Be, , F, , µ=0, , O, , C, , O, , µ=0, , Cl, 27., , Cl µ = 0, , (c) Dipole moment of a molecule is the vector, sum of dipoles of bonds. So based on molecular, geometry of following molecules,, , Three equal vectors, at 120° has resultant, dipole moment = 0, so, non-polar molecule, , Vectors not alligned in, the same direction of, lone pair, so less dipole, m o m en t
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39, , Chemical Bonding and Molecular Structure, , 33., , 28., , Vectors alligned, , Maximum dipole, , in same direction, , moment of lone pair, , (c) Dipole moment of NH3 > NF3, d –. ., , N, , N, , d+ H, , 29., , 30., , H, H, m = 1.4D, , –F, (d ), , F, , –, (d ), , 34., , d+, , F, , –, (d ), , m = 0.23D, , (F is more electronegative than N), (a) SF4 has 4 bond pairs and 1 lone pair, of electrons, sp3d hybridisation leads to irregular, F, F, |, shape S, and resultant m ¹ 0., |, F, F, (a) In NH3 the atomic dipole and bond dipole, are in the same direction whereas in NF3 these, are in opposite direction so in the former case, they are added up whereas in the latter case net, result is reduction of dipole moment. It has been, shown in the following figure :, .., N, , 35., , .., N, , 31., , 32., , H, , H, , F, , F, , (b) SF4 has permanent dipole moment., SF4 has sp3d hybridization and see saw shape, (irregular geometry)., m¹ 0, Whereas XeF4 shows square planar geometry,, SiF4 has tetrahedral shape and BF3 has Trigonal, planar shape. All these are symmetric molecules., Hence m = 0., (d) In H — O — O — H, the O – O bond is, purely single. O3 molecule is a resonance hybrid, of two structures : O, O ® O and O ¬ O, O. Thus, its bond length will be longer than, , = 4.8 ´ 10 -10 ´ 2.67 ´ 10 -10 = 12.81D, now % ionic character, Actual dipole moment of the bond, =, Dipole moment of pure ionic compound, % ionic character in, 10.41, ´ 100 = 76.94%, AB =, 13.53, % ionic character in, 10.27, ´ 100 = 80.23%, CD =, 12.81, (c) As the electronegativity difference, between N and F is maximum hence, this bond is, most polar., Polarity of the bond depends upon the, electronegativity difference of the two atoms, forming the bond. Greater the electronegativity, difference, more is the polarity of the bond., N – Cl, O–F, N–F, N–N, 3.0–3.0, 3.5–4.0 3.0–4.0, 3.0–3.0, , F, H, , O2 molecule which has purely double bond, (O, O). Hence, the correct order is:, H2O2 > O3 > O2 (bond length of O — O), (d) In a linear symmetrical molecule like BeF2,, the bond angle between three atoms is 180°,, hence the polarity due to one bond is cancelled, by the equal polarity due to other bond. H2O is, angular., (c) As dipole moment = electric charge × bond, length, D. M. of AB molecules, = 4.8 ´ 10 -10 ´ 2.82 ´ 10 - 8 = 13.53D, D.M. of CD molecules, , 36., 37., , (b) BeF2 is linear and hence has zero dipole, moment while H2O, being a bent molecule, has a, finite or non-zero dipole moment., (b) The structure of ClF3 is, , F, , F, , Cl, F, , The number of lone pair of electrons on, central Cl is 2., 38., , sp2, , sp2, , sp, , (b) CH2 = CH – C, , sp, , CH
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EBD_8336, 40, , 39., , CHEMISTRY, (b) IBr2- , XeF2, , 47., , Total number of valence electrons are equal in, both the species and both the species exhibit, linear shape., 40., , 6, = 3 = sp 2 hence geometry is trigonal, 2, planner., NO2– (nitrite ion) also has sp2 hybridization and, gives a trigonal planner geometry but because there, are only two outer atoms, the molecular geometry, is bent with Ð120º bond angles., , Cl, 120°, Cl, , Cl, , 41., , (b), , CH4, , NH3, , H2O, , 48., , (c) Applying VSEPR theory, both NF3 and H2O, are sp3 hybridized., , 49., , (a), , 50., , CI ¾ I ¾ CI, (d) PCl3, , Tetrahedral;, , H, , 10, 7°, , H, , N, , H, , H, , Trigonal, pyramidal, , 104.3°, , 43., , 44., 45., , P, , H, , Cl, , Bent, , The geometry of H2O should have been tetrahedral, if there are all bond pairs. But due to presence of, two lone pairs the shape is distorted tetrahedral., Hence bond angle reduced to 104.5° from 109.5°., , 42., , F, , F, F, (Square planar), 46., , BF4- hybridisation sp3, tetrahedral structure., , 52., , NO2– ,, , H=, , 1, [5 + 0 + 1 - 0] = 3 = sp 2, 2, , NO3– ,, , H=, , 1, [5 + 0 + 1 - 0] = 3 = sp 2, 2, , NH2– ,, , H=, , 1, [5 + 2 + 1 + 0] = 4 = sp3, 2, , NH4+,, , H=, , 1, [5 + 4 + 0 - 1] = 4 = sp3, 2, , 1, H = [4 + 0 + 0 + 0] = 2 = sp, 2, \ NO2– and NO3– have same hybridisation., SCN–,, , 1, [No. of valence electrons of, 2, central atom + no. of monovalent atoms attached, to it + Negative charge if any – positive charge if, any], , Hybridisation =, , O, , O, [Tetrahedral], , (b) Be2+ = (4 – 2) = 2, is isoelectronic with Li+ (3 – 1 = 2), Since both have same number of electrons in, their outermost shell., , Cl, , NH +4 hybridisation sp3, tetrahedral structure., (a), , Xe, O, , sp3d and Linear, , (d), , O, , Xe, , Cl, , sp3d and Linear, , 51., , (a) According to VSEPR theory order of, repulsion in between lp – lp, lp – bp and bp –, bp is as under, lp – lp > lp – bp > bp – bp, (c) ClO3– and SO3–2 both have same number, of electrons (42) and central atom in each being, sp3 hybridised. Both are having one lone pair, on central atom hence they are pyramidal., (b) NO2+ has sp hybridisation so it is linear, with bond angle = 180°., (d) XeF4, XeO4, F, , F ¾ Xe ¾ F, –, , H, C 109°28', H, H, H, , 1, (5 + 0 + 1 - 0), 2, , =, , (c) BCl3 is trigonal planar and hence the bond, angle is 120°., , B, , (b) Hybridization of NO3– =, , 53., , sp 3 sp sp, sp 3, (b) H3C — C º C — CH3, linear
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41, , Chemical Bonding and Molecular Structure, It is important to note that in but-2-yne, the, hydrogen atoms are not linear., , 60., , H sp3 sp3 sp3 sp3 H, H C–C C–C H, H, H, Linear part of molecule, , 54., , (c) (a) NH +4 : sp3 hybridisation, (b) CH4: sp3 hybridisation, (c) SF4: sp3 d hybridisation, (d), , 55., , 56., , 61., , BF4– : sp3 hybridisation, , (b) In NO3- , nitrogen is in sp2 hybridisation,, thus planar in shape. In H3 O + , oxygen is in sp3, hybridisation, thus trigonal pyramidal in shape., 5+5+2, (c) SbCl52– :, = 6 means sp 3 d 2, 2, hybridisation, , 62., 63., , I3– , SF4 , and PCl5 ; all have sp3d hybridization., , 57., , (b), , BF3 :, , 3+3, = 3 means sp2 hybridisation, 2, , 5 +1, = 3 means sp2 hybridisation, 2, (c) We find that in the given molecules, hybridisation is, BF3 ® sp2, NO -2 ® sp2, , Cl, , NO-2 :, , 58., , 59., , NH 2- ® sp3, H2O ® sp3, (b) From the structure of three species we can, determine the number of lone pair electrons on, central atom (i.e. N atom) and thus the bond, angle., , N, , ...O... .., O.., NO 2, , N, O, , +, , O, , O¬ N=O, NO+2, , NO 2, , We know that higher the number of lone pair of, electron on central atom, greater is the lp – lp, repulsion between Nitrogen and oxygen atoms., Thus smaller is bond angle., The correct order of bond angle is, +, , NO2- < NO2 < NO2, , (c) Hybridization of Br in BrO3–, 1, H = ( 7 + 0 - 0 +1) = 4, 2, hybridization = sp3, Hybridization of Xe in XeO3., 1, H = (8 + 0 - 0 + 0 ) = 4, 2, Hybridization = sp3, In both cases, the structure is trigonal pyramidal., (d) In SiCl4 there is sp3 hybridisation, so the, structure is tetrahedral. In SO42–, PO43–, NH4+, the structure is tetrahedral with sp 3, hybridisation. But in SCl4, sp3d hybridisation is, present, so its shape is different i.e., see saw., (b) NO2+ will have linear shape as it will have, sp hybridisation., (d) In BF3, AlF3 & NF3 all fluoride atoms are, symmetrically oriented with respect to central, metal atom but in ClF3 three fluorine atoms are, arranged as follows :, F, , 64., , 65., , F, , F, Here two bonds are in equatorial plane & one, bond is in axial plane., (a) BF3 is sp2 hybridised. So, it is trigonal, planar. NH3, PCl3 has sp3 hybridisation hence, has trigonal pyramidal shape, IF3, has sp3d, hydridization and is T-shaped., (c) The possible structures are:, F, F, F, Br F, Br F, Br, F, F, F, F, , I, II, III, 90° lp – lp, repulsion 0, 0, 1, 90° lp – bp, repulsion 4, 6, 3, 90° bp – bp, repulsion 2, 0, 2, The first structure is minimizing the lp – lp, repulsion if we compare it with III. The same is, minimizing the lp – bp repulsions if we compare, it with II. Also it can be noted that, I structure is, not minimizing the bp – bp repulsions.
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EBD_8336, 42, , 66., , CHEMISTRY, (a), , d, , x2 – y 2, , , d, , z2, , orbitals are involved in d2sp3, , 69., , (a) For p-overlap the lobes of the atomic, orbitals are perpendicular to the line joining the, nuclei., , 70., , (c) In XeF2 and IF2- . Both XeF2 and IF2– are, sp3d hybridized and have linear shape., (c) In all the species, the center atom is sp3, hybridized. But only in NH4+, no lone pair of, electrons are present. Thus, it has maximum, bond angle., (c) In P–O bond, p bond is formed by the, sidewise overlapping of d-orbital of P and porbital of oxygen. Hence it is formed by pp and, dp overlapping., , hybridisation., Only those d orbitals whose lobes are directed, along X, Y and Z directions hybridise with s and p, orbitals. In other three d orbitals namely, dxy , d yz and d xz , the lobes are at an angle of 45°, , 71., , from both axis, hence the extent of their overlap, with s and p orbitals is much lesser than d 2 2, x -y, , and d 2 orbitals., z, , 67., , 72., X5, , (a), , X1, , X4, , O, s p, P, , M, –, , O, , X3, , X6, , X2, , 73., , (d) PF3 has pyramidal shape, Phosphorus exist in sp3 hybridiation state hence, it exist in tetrahedral geometry. But due to, presence of lone pair its shape is pyramidal., , 74., , (b) XeF4 hybridisation is =, , Thus here bond angles between, X 4 - M - X 2 = 180°, X1 - M - X 3 = 180°, X 5 - M - X 6 = 180°, 68., , (b) In SO3-2, –p, , O, , s, , O, , O s=S :, S is, , –, , sp3 hybridised,, , so, , 2 2, 6, 2 1, 1, 1, 16 S = 1s 2 s 2 p 3s 3 p x 3 p y 3 p z, 1442443, 3, , sp hybridisation, , F, , 3d1xy, {, , The shape is, , unhybride, , In 'S' unhybride d-orbital is present, which will, involved in p bond formation with oxygen atom., = 1s 2 2 s 2 2 p 2x 2 p1y 2 p1z, In oxygen two unpaired p-orbital is present in, these one is involved in s bond formation while, other is used in p bond formation, 8O, , Thus in SO 32 - , pp and dp orbitals are involved, for pp - d p bonding., , 1, (V + X - C + A), 2, hence V = 8 (no. of valence e– ), X = 4 (no of monovalent atom), , 1 (8 + 4 + 0 - 0) = 6, sp3d2, 2, C = 0 charge on cation, A = 0 (charge on anion)., , –, , s, , –, , O, , –, , O, , Square planar shape., , Xe, F, , 75., , F, , F, , (c) The electronic configuration of As is, 1, 1, 1 1, 2 2, 6 1, As = 1s 2 s 2 p 3s 3 p x 3 p y 3 p z 3d, 144424443, , ¯ sp3d hybridisation, , So, the hybrid orbitals used by As atom in AsF5, molecule are s, px, py, pz, dz2., In the formation of a stable trigonal pyramid, the, two axial bonds above and below the equatorial, xy-plane have to be equally strong. Thus, the dz2, orbital is used.
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43, , Chemical Bonding and Molecular Structure, 76., , (d) In alkynes the hybridisation is sp i.e each, carbon atom undergoes sp hybridisation to form, two sp-hybrid orbitals. The two 2p-orbitals, remain unhybridised. Hybrid orbitals form one, sigma and two unhybridised orbitals form pbonds., ., , ., , . ., , C, , ., , 77., , 78., , 79., 80., 81., , 82., , 83., , 84., , ., , ., , 86., 87., ., , Hence two p bond and one sigma bond between, C — C lead to cylindrical shape., (d) As there is no lone pair on boron in BCl 3, therefore no repulsion takes place. But there is, a lone pair on nitrogen in NCl 3. Therefore, repulsion takes place. Thus, BCl3 is planar, molecule but NCl3 is a pyramidal molecule., (c) The C–C bond distance decreases as the, multiplicity of the bond increases. Thus, bond, distance decreases in the order: butane (1.54 Å), > benzene (1.39 Å) > ethene (1.34) Å > ethyne, (1.20 Å). Thus in butane, C – C bond distance is, the largest., (b) The angle between the bonds formed by, px and py orbitals is the minimum i.e. 90°., (b) BH3 has sp2 hybridization and hence does, not have tetrahedral structure while all others, have tetrahedral structures., (b) Sigma bond is stronger than p-bond. The, electrons in the p bond are loosely held. The, bond is easily broken and is more reactive than, s -bond. Energy released during sigma bond, formation is always more than p bond because, of greater extent of overlapping., (b), , CO 2 has sp-hybridization and is linear. SO2, , and CO 32- are planar (sp2 ) while SO 24 - is, tetrahedral (sp3)., (c) The bond length decreases in the order, sp3 > sp2 > sp., Because of the triple bond, the carbon-carbon, bond distance in ethyne is shortest., (b) All compounds have sp3 hybridisation, but, in CCl4, all bonded atoms are same. Hence, the, bond angle will be exactly 109°28°, , (a) A s - bond is stronger than a p -bond, hence option (a) is not correct., Sigma (s) bonds are formed by head on overlap of, unhybridised s–s, p–p or s–p orbitals and, hybridised orbitals (sp, sp2, sp3, sp3d and sp3d 2), hence s bonds are strong bonds where as Pi (p)bonds are formed by side ways overlap of, unhybridised p- and d-orbitals hence p bonds are, weak bonds., , ., , C, , ., , 85., , 88., , (b) BF3 involves sp2-hybridization., (a) For linear arrangement of atoms the, hybridisation should be sp(linear shape, 180°, angle). Only H2 S has sp3 -hybridization and, hence has angular shape while C2H2, BeH2 and, CO2 all involve sp-hybridization and hence have, linear arrangement of atoms., (b) Equilateral or triangular planar shape, involves sp2 hybridization., , 89., , (a), , 90., , p-orbital, s-orbital, The overlap between s- and p-orbitals occurs, along internuclear axis and hence the angle is 180°., (d) For He2 molecule, Electronic configuration, is s1s 2 , s *1s 2, , 1, 1, ( N b - N a ) = (2 - 2) = 0, 2, 2, Since, bond order of He2 is zero, so it does not, exist., 91. (a) Molecular orbital configuration of O2 is, given as :, O2 (16 e–) : s1s2 s*1s2 s2s2 s*2s2 s2p2z, , Bond order =, , p 2 p 2x = p 2 p 2y p * 2 p1x = p * 2 p1y, , 92., , So, in O2 molecule, there are two (2) unpaired, electrons, so, it is a “paramagnetic”, substance in nature., (a), O, O–, , Mn, , Mn, O, , –, O– O, , Manganate ion, , 93., , O, , O, , O, , Permanganate ion, , (c) Only p bond is present in C2 molecule., s1s2 s*1s2 s2s2 s*2s2 p2p2x = p2p2y
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EBD_8336, 44, , 94., , CHEMISTRY, O 2– (Super oxide ion): Total number of electrons, (16 + 1) = 17 ., Electronic configuration, , (b) NO : (s1s)2, (s*1s)2, (s2s)2,(s*2s)2,(s2pz)2,, (p2px)2 = (p2py)2,(p*2px)1 = (p*2py)0, 10 - 5, B.O. =, = 2.5, 2, –, 2, CN : (s1s) , (s*1s)2, (s2s)2,(s*2s)2,, (p2px)2 = (p2py)2, (s2pz)2, , s1s 2 < s*1s 2 < s 2s 2 < s* 2s 2 < s 2 p x2, , < p 2 p 2y = p 2 pz2 < p* 2 p y2 = p* 2 p1z, , 10 - 4, =3, 2, CN : (s1s)2, (s*1s)2, (s2s)2,(s*2s)2,, (p2px)2 = (p2py)2,(s2pz)1, , Bond order =, , B.O. =, , O+22 ion: Total number of electrons, = (16 – 2) = 14 Electronic configuration, s1s2 < s*1s2 < s2s2 < s*2s2 < s2px2 < p2py2, = p2pz2, , 9-4, B.O. =, = 2.5, 2, +, 2, CN : (s1s) , (s*1s)2, (s2s)2,(s*2s)2,, (p2px)2 = (p2py)2, , 8-4, =2, 2, Hence, option (2) should be the right, answer., (b) CN– and CO have same no. of electrons and, have same bond order equal to 3., , Bond order =, , B.O. =, , 95., , Short cut trick to calculate the bond order:, N2 has 14 electrons and its bond order is 3. We, have to remember this concept that every electron, added or subtracted to 14, reduces the bond order, by 0.5. For example, CN– Þ no. of electrons = 6 + 7 + 1 = 14, \ bond order = 3, CO Þ no. of electrons = 6 + 8 = 14, \ bond order = 3, NO Þ no. of electrons = 7 + 8 = 15, \ bond order = 3 – 0.5 = 2.5, NO+ Þ no. of electrons = 7 + 8 – 1 = 14, \ bond order = 3, O2– Þ no. of electrons = 8 + 8 + 1 = 17, \ bond order = 3 – 1.5, Please note that this method will work for any, species that have electrons between 10 and 18., , 96., , (c), , O +2 ion - Total number of electrons, , (16 – 1) = 15., Electronic configuration, s1s 2 < s*1s 2 < s 2 s 2 < s * 2 s 2 < s 2 p x2, , < p 2 p 2y = p2 pz2 < p * 2 p1y, Bond order =, , N b - N a 10 - 5 5, 1, =, = =2, 2, 2, 2, 2, , (Nb - Na ) 10 - 7 3 1, =, = =1, 2, 2, 2 2, , 97., , (N b – Na ) 10 – 4 6, =, = =3, 2, 2, 2, , So bond order: O2– < O2+ < O22+, (a) Oxygen molecule (O2) – Total number of, electrons = 16 and electronic configuration is, s1s 2 < s*1s 2 < s 2s 2 < s* 2s 2 < s 2 p x2, , < p 2 p 2y = p 2 pz2 < p* 2 p1y = p* 2 p1z, Bond order =, , N b - N a 10 - 6 4, =, = =2, 2, 2, 2, , O +2 ion - Total number of electrons (16 – 1) = 15., , Electronic configuration, s1s 2 < s*1s 2 < s 2s 2 < s* 2s 2 < s 2 p x2, , < p 2 p 2y = p2 pz2 < p* 2 p1y, Bond order =, , N b - Na 10 - 5 5, 1, =, = =2, 2, 2, 2, 2, , O2– (Super oxide ion) Total number of electrons, , (16 + 1) = 17 . Electronic configuration, σ1s2 < σ*1s2 < σ2s2 < σ* 2s 2 < σ2p2x, , < π2p2y = π2p2z < π* 2p 2y = π* 2p1z, Bond order =, , (Nb - Na ) 10 - 7 3 1, =, = =1, 2, 2, 2 2
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45, , Chemical Bonding and Molecular Structure, (c), , Xe, , B.O. 2, 2.5, Bond energy decreases, Magnetic behaviour changes from diamagnetic, to paramagnetic, , 98., , (a), , 99., , O O O, Number of s bonds = 4, Number of p bonds = 4, (a) According to molecular orbital theory as, bond order decreases stability of the molecule, decreases, 1, Bond order = (N b – Na ), 2, , 1, 2, , Bond order for O2+ = (10- 5) = 2.5, 1, (10 - 6) = 2, 2, 1, Bond order for O -2 = (10 - 7) = 1.5, 2, 1, 2Bond order for O2 = (10 - 8) = 1.0, 2, hence the correct order is, , Bond order for O 2 =, , O2+, , O-2 (17) = s1s2, s*1s2, s2s2, s*2s2,, s2pz2, p2px2 = p2py2, p*2px2 = p*2py1, , 105. (a), , > O 2–, > O2 >, 2, Ni2+ = [Ar] 3d 8, 4s0, O2–, , 2+, , Ni =, 4s, , 4p, , E.S., dsp 2, s, , s C = C —H, s, H—, p, |, |, s, s, H H, , 2.5, B.O. 3, Bond energy decreases, Magnetic behaviour changes from diamagnetic, to paramagnetic, , (b), , ® O 2+, O 2 ¾¾, B.O. 2, , 2.5, , Bond energy increases, Magnetic behaviour does not change., , s, , O, , s, , H, , s, , s, , s, , s, , p, , p, , p, , p, , O ¬ S=O O ¬ N =O O = C = O, , 106. (d) Calculating the bond order of various, species., , p 2 px2 p 2 p 2y p* 2 px2 p* 2 p1y, Number of electrons in bonding Number of electrons in non-bonding, B.O. =, 2, , =, , ® N +2, N 2 ¾¾, , 102. (d) (a), , H, , O2- : KK s 2s 2 s* 2 s 2 s2 p 2z, , G.S., 3d, , ® NO+, NO ¾¾, 2.5, B.O. 2, bond energy increases, Magnetic behaviour changes from paramagnetic, to diamagnetic, 103. (b) No. of electrons in CO = 6 + 8 = 14, No. of electrons in NO+ = 7 + 8 – 1 = 14, \ CO and NO+ are isoelectronic species., Isoelectronic species have identical bond order., , (d), , 104. (a) Molecular orbital configuration of O-2 is, , 100. (a), In the presence of strong ligand CN–, pairing of, electrons will occur:, , 101. (d), , ® C2+, C2 ¾¾, , O, , 8-5, or 1.5, 2, , 2 * 2, 2, NO : KK s 2s s 2s p2 px » p2 p2y s 2 pz2 p* 2 p1x, , B.O. =, , Nb - Na 8 - 2, or 2.5, =, 2, 2, , C 22- : kk s 2 s 2 s* 2 s 2 p 2 p x2 p 2 p 2y s 2 p 2z, , B.O. =, , Nb - Na 8 - 2, or 3, =, 2, 2, , He +2 = s1s 2 s *1s1
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47, , Chemical Bonding and Molecular Structure, 114. (d) The shape of ozone molecule is, +, O, , .. .., , .. .., , .., , .., , O, , O, , . .O..–, , .., , .., , – O, , .., , .., , +, O, , In it we find 2 s and 1p bond, i.e., option (d) is, correct., 115. (c) All these structures exhibits resonance, and can be represented by the following, resonating structures., (i), , +, , O, , C, , N, O, , –, , O, –, , O, , .., , : C º O : ¬¾, ® : C = O:, , (ii), , To form NO 3- , nitrogen uses one p-electron, for p-bond formation and two p-electrons for, s-bond formation. 2s electrons are used for, coordinate bond formation. Thus there is no, lone pair on nitrogen and four bond pairs are, present., , –, , O, , O, , C, , C, , –, , O, , –, , O, , –, , O, , O, , O, , .., -, , +, , ®:O -C º O, =C=O, : ¬¾, :, (iii) : O, .., .., .., -, , .., , O, , 120. (c) Dissociation energy of any molecules, depends upon bond order. Bond order in N2, , +, , ¬¾, ® :O, º C-O:, .., , More is the single bond character. More will be, the bond length. Hence, the corret order is :, CO < CO2 < CO32–, 116. (d) Electronic configuration of the molecule, according to molecular orbital theory, is, s1s2s*1s2s2s2s*2s2s2pz2 (p2px2 = s2py2), (p*2px2 = p2py2) s*2pz2s3s2s*3s2s3pz2, (p3px2 = p3py2) (p*3px1 = 3py1), Last two electrons are unpaired. So no. of, unpaired electron is 2., 117. (a) SiF4 has symmetrical tetrahedral shape which, is due to sp3 hybridisation of silicon atom in its, excited state while SF4 has distorted tetrahedral, or sea-saw geometry which arises due to sp3d, hybridisation of sulphur atom and one lone pair, of e–s in one of the equatorial hybrid orbital., 118. (c) As sigma bond is stronger than the p (pi), bond, so it must be having higher bond energy, than p (pi) bond., , molecule is 3 while bond order in N +2 is 2.5., Further we know that more the Bond order, more, is the stability and more is the BDE., 121. (b), , NO+ = s1s 2 s *1s 2 s 2 s 2s * 2 s 2 s2 p x 2, p 2 p y 2 = p 2 pz 2, , Bond order of NO+ =, , 1, (N b - N a ), 2, , 1, 1, = (10 - 4) = ´ 6 = 3, 2, 2, Similarly, Bond order of NO =, , 1, (10 - 5), 2, 1, = (5) = 2.5, 2, , 1, 1, Bond order of NO– = (10 - 6) = (4) = 2, 2, 2, 1, 1, Bond order of O -2 = (10 - 7) = (3) = 1.5, 2, 2, By above calculation, we get, Decreasing bond order, , NO+ > NO > NO– > O -2, 122. (a) Paramagnetic character is based upon, presence of unpaired electron, Cl – = 1s 2 2s 2 2 p6 3s 2 3 px2 3 p 2y 3 p 2z, , 119. (d) N :, 1s 2, , 2s 2, , 2p3, , Be + = 1s 2 2 s1
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EBD_8336, 48, , CHEMISTRY, , As+ = 1s 2 2 s 2 2 p 6 3s 2 3 p 6 3d 10, 4 s 2 4 p1x 4 p1y 4 p1z, , Hence only Cl– do not have unpaired electrons., 123. (d) Total no. of electrons in O 2 2 - = 16 + 2 = 18, Distribution of electrons in molecular orbital, 2, , *, , 2, , 2, , *, , s1s , s 1s , s2s , s 2s, , 2, , , s2 px2 , p2 p 2y, , p2 pz2 , p* 2 px2 p* 2 p 2y, , —, —, , Anti bonding electron = 8 (4 pairs), 124. (b) We know that in O2 bond, the order is 2, and in O2– bond, the order is 1.5. Therefore, the, wrong statements is (b)., 125. (c) The N–O bond length decreases in the order, O, , O, , —N, > O—, , —, , —, , N, , O, , O, , O, — O > N—, —O, — N—N —, > O—, O, , 126. (d) In this configuration, there are four, completely filled bonding molecular orbitals and, one completely filled antibonding molecular, orbital. So that Nb = 8 and N a = 2 ., 1, 2, , 1, 2, , \ Bond order = ( N b - N a ) = (8 - 2) = 3., 127. (b) The structure of NO:, ×, ×, ×, , N, , : :, , Ne 2+ = 1s 2 2s 2 2 px2 2 p1y 2 p1z, , O, , In NO, the presence of unpaired electron is clear., Therefore, it is paramagnetic., 128. (b) The removal of an electron from a diatomic, molecule may increase or decrease the bond, order., Removal of an electron from bonding orbital results, in decrease of bond order, hence reduces bond, strength while removal of an electron from, antibonding orbital results into increase in bond, order hence increases the bond strength., , 129. (a) Linear combination of two hybridized, orbitals leads to the formation of sigma bond.
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49, , States of Matter, , 5, , States of Matter, , Trend Analysis with Important Topics & Sub-Topics, 2020, Topic Name, Gas law and ideal gas, equation, van der Waal's, equation and, liquefaction of gases, LOD - Level of Difficulty, , Sub-Topic, Graham's law of, diffusion, Dalton's law of, partial pressure, van der Waal's, equation, liquefaction of, gases, , 2019, , 2., , 3., , 4., , 2017, , 2016, , QNS. LOD QNS. LOD QNS. LOD QNS. LOD QNS. LOD, 1, 1, , E - Easy, , A mixture of N2 and Ar gases in a cylinder, contains 7 g of N2 and 8 g of Ar. If the total, pressure of the mixture of the gases in the, cylinder is 27 bar, the partial pressure of N2 is:, [Use atomic masses (in g mol–1) : N = 14, Ar = 40], (a) 12 bar, (b) 15 bar, [2020], (c) 18 bar, (d) 9 bar, The volume occupied by 1.8 g of water vapour, at 374°C and 1 bar pressure will be, [Use R = 0.083 bar LK–1 mol–1], [NEET Odisha 2019], (a) 5.37 L, (b) 96.66 L, (c) 55.87 L, (d) 3.10 L, Equal moles of hydrogen and oxygen gases are, placed in a container with a pin-hole through, which both can escape. What fraction of the, oxygen escapes in the time required for one-half, of the hydrogen to escape ?, [2016], (a) 1/8, (b) 1/4, (c) 3/8, (d) 1/2, Equal masses of H2,O2 and methane have been, taken in a container of volume V at temperature, , A, , A, 1, , A - Average, , Topic 1: Gas laws and Ideal gas Equation, 1., , 2018, , 5., , 6., , 7., , 8., , A, , 1, , A, , 1, , A, , D - Difficult, , Qns - No. of Questions, , 27°C in identical conditions. The ratio of the, volumes of gases H2 : O2 : methane would be :, (a) 8 : 16 : 1, (b) 16 : 8 : 1 [2014], (c) 16 : 1 : 2, (d) 8 : 1 : 2, Dipole-induced dipole interactions are present, in which of the following pairs : [NEET 2013], (a) Cl2 and CCl4, (b) HCl and He atoms, (c) SiF4 and He atoms (d) H2O and alcohol, 50 mL of each gas A and of gas B takes 150 and, 200 seconds respectively for effusing through a, pin hole under the similar condition. If molecular, mass of gas B is 36, the molecular mass of gas A, will be :, [2012], (a) 96, (b) 128, (c) 32, (d) 64, A certain gas takes three times as long to effuse, out as helium. Its molecular mass will be :, (a) 27 u, (b) 36 u, [2012 M], (c) 64 u, (d) 9 u, Two gases A and B having the same volume, diffuse through a porous partition in 20 and 10, seconds respectively. The molecular mass of A, is 49 u. Molecular mass of B will be : [2011]
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EBD_8336, 50, , 9., , 10., , 11., , 12., , 13., , 14., , 15., , 16., , CHEMISTRY, (b) 12.25 u, (a) 50.00 u, (c) 6.50 u, (d) 25.00 u, A bubble of air is underwater at temperature 15°C, and the pressure 1.5 bar. If the bubble rises to, the surface where the temperature is 25°C and, the pressure is 1.0 bar, what will happen to the, volume of the bubble ?, [2011M], (a) Volume will become greater by a factor of 1.6., (b) Volume will become greater by a factor of 1.1., (c ) Volume will become smaller by a factor of 0.70., (d) Volume will become greater by a factor of 2.5., From a heated mixture of nitrogen, oxygen and, carbon, two compounds (out of the many, obtained) are isolated. The rates of diffusion of, the two isolated compounds are almost identical., The two compounds are, [1999], (a) N2O and CO2, (b) CO and NO, (c) CO2 and NO2, (d) N2O and CO, If 500 mL of gas A at 400 torr and 666.6 mL of B, at 600 torr are placed in a 3 litre flask, the pressure, of the system will be, [1999], (a) 200 torr, (b) 100 torr, (c) 550 torr, (d) 366 torr, A gaseous mixture contains H2 and O2 in the, molar ratio 8 : 1. The ratio of H2 : O2 by weight in, this mixture would be, [1999], (a) 4 : 1, (b) 1 : 8, (c) 8 : 1, (d) 1 : 2, 500 mL of nitrogen at 27°C is cooled to –5°C at, the same pressure. The new volume becomes, [1995], (a) 326.32 mL, (b) 446.66 mL, (c) 546.32 mL, (d) 771.56 mL, 600 c.c. of a gas at a pressure of 750 mm is, compressed to 500 c.c. Taking the temperature, to remain constant, the increase in pressure, is, (a) 150 mm, (b) 250 mm [1995], (c) 350 mm, (d) 450 mm, The correct value of the gas constant ‘R’ is close, to :, [1992], (a) 0.082 litre-atmosphere K, (b) 0.082 litre-atmosphere K–1 mol–1, (c) 0.082 litre – atmosphere–1 K mol–1, (d) 0.082 litre –1 atmosphere – 1 K mol, At constant temperature, for a given mass of an, ideal gas, [1991], (a) The ratio of pressure and volume always, remains constant., , 17., , (b) Volume always remains constant., (c) Pressure always remains constant., (d) The product of pressure and volume, always remains constant., Pressure remaining the same, the volume of a, given mass of an ideal gas increases for every, degree centigrade rise in temperature by definite, fraction of its volume at, [1989], (a) 0°C, (b) its critical temperature, (c) absolute zero, (d) its Boyle temperature, Topic 2: Kinetic Theory of Gases and, Molecular Speeds, , 18., , 19., , 20., , By what factor does the average velocity of a, gaseous molecule increase when the temperature, (in Kelvin) is doubled ?, [2011], (a) 2.0, (b) 2.8, (c) 4.0, (d) 1.4, If a gas expands at constant temperature, it, indicates that :, [2008], (a) kinetic energy of molecules decreases, (b) pressure of the gas increases, (c) kinetic energy of molecules remains the same, (d) number of the molecules of gas increases, Which of the following expressions correctly, represents the relationship between the average, molar kinetic energy, KE, of CO and N2 molecules, at the same temperature ?, [2000], (a) KECO < KE N 2, (b), , 21., , 22., , KECO > KE N 2, , (c) KECO = KE N2, (d) cannot be predicted unless volumes of the, gases are given, The temperature of the gas is raised from 27°C to, 927°C, the root mean square velocity is [1994], (a), 927 / 27 time the earlier value, (b) same as before, (c) halved, (d) doubled, At STP, 0.50 mol H2 gas and 1.0 mol He gas, (a) have equal average kinetic energies [1993], (b) have equal molecular speeds, (c) occupy equal volumes, (d) have equal effusion rates
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51, , States of Matter, 23., , 24., , Internal energy and pressure of a gas per unit, volume are related as :, [1993], 3, (a) P = 2 E, (b) P = E, 2, 3, 1, (c) P = E, (d) P = 2 E, 2, The ratio among most probable velocity, mean, velocity and root mean square velocity is given, by, [1993], (a) 1 : 2 : 3, , 25., , 26., , 27., , 28., , 29., , 30., , (b) 1 :, , 31., , 32., , 2. 3, , (d), (c), 2 : 3 : 8/ p, 2 : 8/p : 3, A closed flask contains water in all its three, states solid, liquid and vapour at 0°C. In this, situation, the average kinetic energy of water, molecules will be, [1992], (a) the greatest in all the three states, (b) the greatest in vapour state, (c) the greatest in the liquid state, (d) the greatest in the solid state, In a closed flask of 5 litres, 1.0 g of H2 is heated, from 300 to 600 K. Which statement is not correct?, (a) Pressure of the gas increases, [1991], (b) The rate of collision increases, (c) The number of moles of gas increases, (d) The energy of gaseous molecules increases, The root mean square speeds at STP for the, gases H2, N2, O2 and HBr are in the order : [1991], (a) H2< N2< O2 < HBr (b) HBr < O2 < N2 < H2, (c) H2 < N2 = O2< HBr (d) HBr < O2 < H2 < N2., Root mean square velocity of a gas molecule is, proportional to, [1990], (a) m1/2, (b) m0, (c) m–1/2, (d) m, Absolute zero is defined as the temperature, (a) at which all molecular motion ceases, (b) at which liquid helium boils, [1990], (c) at which ether boils, (d) all of the above, Topic 3 : van der Waal's Equation and, liquefaction of Gases, A gas at 350 K and 15 bar has molar volume 20, percent smaller than that for an ideal gas under the, same conditions. The correct option about the gas, and its compressibility factor (Z) is:, [2019], (a) Z > 1 and attractive forces are dominant, (b) Z > 1 and repulsive forces are dominant, , 33., , 34., , 35., , 36., , 37., , (c) Z < 1 and attractive forces are dominant, (d) Z < 1 and repulsive forces are dominant, The correction factor ‘a’ to the ideal gas equation, corresponds to, [2018], (a) Density of the gas molecules, (b) Volume of the gas molecules, (c) Forces of attraction between the gas molecules, (d) Electric field present between the gas molecules, Given van der Waals constants for NH3, H2, O2, and CO2 are respectively 4.17, 0.244, 1.36 and, 3.59, which one of the following gases is most, easily liquefied?, [2018], (a) NH3, (b) H2, (c) CO2, (d) O2, A gas such as carbon monoxide would be most, likely to obey the ideal gas law at : [2015 RS], (a) high temperatures and low pressures., (b) low temperatures and high pressures., (c) high temperatures and low pressures., (d) low temperatures and low pressures., Maximum deviation from ideal gas is expected, from :, [NEET 2013], (a) N2(g), (b) CH4(g), (c) NH3 (g), (d) H2(g), What is the density of N2 gas at 227°C and, 5.00 atm pressure? (R = 0.0821 atm K–1 mol–1), [NEET Kar. 2013], (a) 0.29 g/ml, (b) 1.40 g/ml, (c) 2.81 g/ml, (d) 3.41 g/ml, For real gases, van der Waals equation is written as, æ, an2 ö, çè p + 2 ÷ø (V - nb) = nRT, V, where ‘a’ and ‘b’ are van der Waals constants., Two sets of gases are :, (I) O2, CO2, H2 and He (II) CH4, O2 and H2, The gases given in set-I in increasing order of, ‘b’ and gases given in set-II in decreasing order, of ‘a’, are arranged below. Select the correct, order from the following :, [2012 M], (a) (I) He < H2 < CO2 < O2 (II) CH4 > H2 > O2, (b) (I) O2 < He < H2 < CO2 (II) H2 > O2 > CH4, (c) (I) H2 < He < O2 < CO2 (II) CH4 > O2 > H2, (d) (I) H2 < O2 < He < CO2 (II) O2 > CH4 > H2, A gaseous mixture was prepared by taking equal, mole of CO and N2. If the total pressure of the, mixture was found 1 atmosphere, the partial, pressure of the nitrogen (N2) in the mixture is :
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EBD_8336, 52, , 38., , 39., , 40., , 41., , 42., , 43., , 44., , CHEMISTRY, (a) 0.5 atm, (b) 0.8 atm, [2011], (c) 0.9 atm, (d) 1 atm, The pressure exerted by 6.0 g of methane gas in, a 0.03 m3 vessel at 129°C is (Atomic masses :, C = 12.01, H = 1.01 and R = 8.314 JK–1 mol –1), (a) 31684 Pa, (b) 215216 Pa [2010], (c) 13409 Pa, (d) 41648 Pa, van der Waal's real gas, act as an ideal gas, at, which conditions?, [2002], (a) High temperature, low pressure, (b) Low temperature, high pressure, (c) High temperature, high pressure, (d) Low temperature, low pressure, Cyclopropane and oxygen at partial pressures, 170 torr and 570 torr respectively are mixed in a, gas cylinder. What is the ratio of the number of, moles of cyclopropane to the number of moles, of oxygen (nC3H6/nO2)?, [1996], 170 ´ 42, = 0.39, (a), 570 ´ 32, 170 æ 170 570 ö, (b), +, ç, ÷ » 0.19, 42 è 42, 32 ø, 170, = 0.23, (c), 740, 170, = 0.30, (d), 570, At which one of the following temperature pressure conditions the deviation of a gas from, ideal behaviour is expected to be minimum? [1996], (a) 350 K and 3 atm. (b) 550 K and 1 atm., (c) 250 K and 4 atm. (d) 450 K and 2 atm., Under what conditions will a pure sample of an, ideal gas not only exhibit a pressure of 1 atm but, also a concentration of 1 mole litre–1 ?, (R = 0.082 litre atm mol–1deg–1), [1993], (a) At STP, (b) When V = 22.4 litres, (c) When T = 12 K, (d) Impossible under any conditions, When is deviation more in the behaviour of a, gas from the ideal gas equation PV = nRT ? [1993], (a) At high temperature and low pressure, (b) At low temperature and high pressure, (c) At high temperature and high pressure, (d) At low temperature and low high pressure, Which is not true in case of an ideal gas ? [1992], (a) It cannot be converted into a liquid, , 45., , 46., , 47., , 48., , (b) There is no interaction between the molecules, (c) All molecules of the gas move with same speed, (d) At a given temperature, PV is proportional, to the amount of the gas, An ideal gas can’t be liquefied because [1992], (a) its critical temperature is always above 0°C, (b) its molecules are relatively smaller in size, (c) it solidifies before becoming a liquid, (d) forces operated between its molecules are, negligible, Select one correct statement. In the gas equation,, PV = nRT, [1992], (a) n is the number of molecules of a gas, (b) V denotes volume of one mole of the gas, (c) n moles of the gas have a volume V, (d) P is the pressure of the gas when only one, mole of gas is present., A gas is said to behave like an ideal gas when, the relation PV/T = constant. When do you, expect a real gas to behave like an ideal gas ?, (a) When the temperature is low, [1991], (b) When both the temperature and pressure, are low, (c) When both the temperature and pressure, are high, (d) When the temperature is high and pressure, is low, In van der Waal's equation of state for a nonideal gas, the term that accounts for, intermolecular forces is :, [1990], (a) (V – b), (b) (RT)–1, a ö, æ, (d) RT, çè P + 2 ÷ø, V, If P, V, M, T and R are pressure, Volume, molar, mass, temperature and gas constant respectively,, then for an ideal gas, the density is given by, [1989], P, RT, (a), (b), PM, RT, PM, M, (c), (d), RT, V, Correct gas equation is :, [1989], , (c), , 49., , 50., , (a), (c), , V1T2 V2T1, =, P1, P2, PT, 1 2 = P2V2, V1, T2, , (b), , PV, 1 1 = T1, P2V2 T2, , (d), , V1V2, = P1P2, T1T2
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53, , States of Matter, Topic 4: Liquid State, 51., , 52., , The surface tension of which of the following, liquid is maximum?, [2005], (a) C2H5OH, (b) CH3OH, (c) H2O, (d) C6H6, In a pair of immiscible liquids, a common solute, dissolves in both and the equilibrium is reached., Then the concentration of the solute in upper, layer is, [1994], , 53., , (a) In fixed ratio with that in the lower layer, (b) Same as the lower layer, (c) Lower than the lower layer, (d) Higher than the lower layer., A liquid can exist only :, [1994], (a) between triple point and critical temperature, (b) at any temperature above the melting point, (c) between melting point and critical, temperature, (d) between boiling and melting temperature., , ANSWER KEY, (b) 31 (c), , 1, , (b), , 7, , (b), , 13, , (b), , 19, , (c), , 25, , 2, , (a), , 8, , (b), , 14, , (a), , 20, , (c), , 26, , (c), , 32, , (a), , 37, , (a), , 43, , (b), , 49, , (d), , 38, , (d), , 44, , (c), , 50, , (b), , 3, , (a), , 9, , (a), , 15, , (b), , 21, , (d), , 27, , (b), , 33, , (a), , 39, , (a), , 45, , (d), , 51, , (c), , 4, , (c), , 10, , (a), , 16, , (d), , 22, , (a), , 28, , (c), , 34, , (c), , 40, , (d), , 46, , (c), , 52, , (a), , 5, , (b), , 11, , (a), , 17, , (a), , 23, , (a), , 29, , (a), , 35, , (d), , 41, , (b), , 47, , (d), , 53, , (d), , 6, , (N), , 12, , (d), , 18, , (d), , 24, , (d), , 30, , (c), , 36, , (c), , 42, , (c), , 48, , (c), , Hints & Solutions, 1., , (b), , nN 2 =, , 7 1, = = 0.25, 28 4, , rH 2, , MO 2, v /t, = 1 1 Þ, =, MH 2, rO2 v2 / t2, , 8 1, = = 0.20, 40 5, Now, applying Dalton's law of partial pressure,, nAr =, , 1/ 2, = 16 = 4, 1/ x, , pN 2 = mole fraction of N 2 × PTotal, , x, =4, 2, \x=8, \ Fraction of O2 = 1/8, (c) According to Avogadro’s law "At same, temperature and pressure. Volume µ no. of moles", w, w, w, nH 2 = ; nO = ; nCH =, 2, 4, 2, 32, 16, , 0.25, 5, ´ 27 = ´ 27 = 15 bar, 0.45, 9, , 2., , (a) According to ideal gas equation., PV = nRT, V=, V=, , 3., , 32, 2, , 4., , W æ RT ö, ç, ÷, Mè P ø, , Q VH : VO : VCH = nH2 : nO2 : nCH4, 2, 2, 4, , 1.8 0.083 ´ 647, ´, = 5.37 L, 18, 1, , (a) Given, nH2 = nO2 and tH2 = tO2, According to Graham's law of diffusion for two, different gases., , w w, w, :, :, = 16 : 1 : 2, 2 32 16, (b) HCl is polar (m ¹ 0) and He is non-polar (m = 0),, thus gives dipole-induced dipole interaction., , =, , 5.
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EBD_8336, 54, , CHEMISTRY, Induced dipole forces result when an ion or a dipole, induces a dipole in an atom or a molecule with no, dipole. These are weak forces. These are of two, types namely. Ion-induced dipole force and dipoleinduced force. Ion-induced dipole force: when the, approach of an ion induces a dipole in an atom by, disturbing the arrangement of electrons., Dipole induced dipole force: when a polar molecule, induces a dipole in an atom or in a non-polar, molecule., , VB, =, tB, , VA, tA, , (N), , 7., , 200, 36, tB, 4, 36, =, =, Þ, Þ, 150, MA, 3, tA, MA, 16, 36, 81, =, Þ, Þ MA =, = 20.25, 9, MA, 4, (b) According to Graham’s law of diffusion, , 3r1, =, r1, , 1, d, , 1, M, , M2, 4, , 11., , MB, MA, , 6., , rµ, , r1, =, r2, , r1, M2, Þ r = M, 2, 1, , Þ 9=, , 8., , (b), V, 20 =, V, 10, , 12., , 10., , (d), , Moles of H 2 8, = (Given), Moles of O 2 1, , æ M.W. of O 2 ö æ weight of H 2 ö 8, ç, ÷ç, ÷, ç M.W. of H ÷ . ç weight of O ÷ = 1, 2ø è, 2ø, è, weight of H 2 8 ´ 2 1, =, =, weight of O 2 32 ´ 1 2, , 13., MB, 49, , 1, ´ 49 = 12.25u, 4, (a) Given, P1 = 1.5 bar, T1 = 273 + 15 = 288 K, V1 = V, P2 = 1.0 bar, T2 = 273 + 25 = 298K, V2 = ?, PV, 1 1 = P2V2, T1, T2, , 1.5 ´ V 1 ´ V2, =, 288, 298, V2 = 1.55 V i.e., volume of bubble will be almost, 1.6 time to initial volume of bubble., (a) Rate of diffusion depend upon molecular, weight, , (b) Given initial volume (V1) = 500 mL ; Initial, temperature (T1) = 27ºC = 300 K and final, temperature (T2) = –5ºC = 268 K., From Charle’s law :, V1 V2, 500 V2, =, or, =, 300 268, T1 T2, Where V2 = New volume of gas, , MB =, , 9., , 200 400 600, +, =, = 200 torr, 3, 3, 3, , Real life application of Boyle's law:, when we fill bike tier with air with pump, the gas, molecules inside the tire get compressed and, packed closer together. This increase the pressure, of the gas and it starts to push against the wall of, the tire and the tires becomes tighter., , M2, 4, , MB, MA, , 1, MB, Þ =, 2, 49, , Hence compounds are N2O and CO2 as both, have same molar mass i.e. 22, (a) Applying Boyle's law P1 V1 = P2 V2 for both, gases, 200, 500, ´ 400 = P ´ 3 Þ P =, 1000, 3, 666.6, 400, 600 ´, = P¢ ´ 3 Þ P¢ =, 1000, 3, Þ PT = P + P ¢ =, , M2 = 36 u, , rA, =, rB, , M2, Þ r1 = r2 if M1 = M2, M1, , 500, ´ 268 = 446.66 ml., 300, (a) Given initial volume (V1) = 600 c.c.; Initial, pressure (P 1 ) = 750 mm and final volume, (V2) = 500 c.c. According to Boyle’s law,, P1V1 = P2V2, Þ 750 × 600 = P2 × 500, V2 =, , 14., , 750 ´ 600, = 900 mm ., 500, Therefore increase in pressure = (900 – 750), = 150 mm., , or P2 =
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55, , States of Matter, 15., 16., , (b) R = 0.082 litre atm K–1 mol–1 ., (d) According to Boyle’s law at constant, , 17., , 1, or PV = constant, V, (a) Charle’s Law - The volume of the given, , Mean velocity (v ) =, , temperature, P µ, , 1, mass of a gas increases or decreases by, of, 273, its volume at 0°C for each degree rise or fall of, temperature at constant pressure., , t ö, æ, V t = V 0 ç1 +, at constant Pressure, è 273 ÷ø, 18., , (d) Average velocity =, , Root mean square velocity (u) =, \ a :v :u =, , 25., 26., , \, , 27., , 3, kT, 2, As temperature is same hence average kinetic, energy of CO and N2 will be same., , 28., , 22., , 27 + 273, 300, 1, =, =, 927 + 273, 1200 2, \ u2 = 2u1, (a) Average kinetic energy depends only on, =, , 23., , 24., , 3, temperature æç K.E. = kT ö÷, è, 2 ø, 1, 1, (a) PV = mnu 2 = Mu 2, 3, 3, 2 1, 2, 2, 2, = . M u = E or P = E per unit vol., 3 2, 3, 3, 2 RT, (d) Most probable velocity (a ) =, M, , 1, M, i.e. higher will be the molar mass, lower will be, the value of urms., Molecular masses of H2, N2, O2 and HBr are 2,, 28, 32 and 81. Hence the correct order of urms, will be HBr < O2 < N2 < H2, (c) According to kinetic gas equation, , 1, mNu 2 , u = root mean square velocity, 3, 1, 1, 2 3PV, or u µ, i.e u µ m 2, Þ u =, mN, m, (a) Absolute zero is the temperature at which, kinetic energy of gas molecules becomes zero, i.e. all molecular motion ceases., , PV =, , u µ T or u1 / u 2 = T1 / T2, , (d), , 1, 2 1, 2, (b) PV = m N u = M u, 3, 3, , At STP, u µ, , (c) Average molar kinetic energy =, , 21., , 8, : 3, p, (b) In vapour state, the molecules are free to, move with highest average kinetic energy., (c) The number of moles of gas do not changes., , or u = 3PV / M ., , The molar kinetic energy of a gas is proportional, to its temperature and the proportionality constant, 3, is, times the gas constant R., 2, 3, E m = RT, 2, , 20., , 2 RT, 8RT, 3RT, :, :, pM, M, M, , Volume is constant and the mass of H2 is fixed so, the number of moles of the gas do not change. As, temperature increases the pressure also increases, and the rate of collision among the gas molecules, and their energy also increases., , i.e., v µ T, , 19., , 3RT, M, , = 2:, , 8RT, pM, , V2, 2T, =, = 1.41, V1, T, (c) At any constant temperature the K.E. of, gaseous molecules remains same., , 8RT, pM, , 29., , 30., , (c) Compressibility factor, Z =, , PV, nRT, , Given: At 350 K and 15 bar,, molar volume < volume of ideal gas, \ Z< 1, Therefore, attractive forces are dominant and, the gas can be compressed easily.
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EBD_8336, 56, , 31., 32., , 33., 34., , 35., 36., , 37., , CHEMISTRY, (c) In real gas equation, van der Wall constant, (a) µ forces of attraction., (a) van der Waal constant ‘a’, signifies, intermolecular forces of attraction., Higher is the value of ‘a’, easier will be the, liquefaction of gas., (a) At high temperature and low pressure., (c) Higher the critical temperature more easily, will be the gas liquify. Now since most easily, liquefiable gas show larger deviation, NH3 will, show maximum deviation from ideal behaviour., 5 ´ 28, PM, =, = 3.41g/mL, RT 0.0821 ´ 500, (c) Set-I : O2, CO2, H2, He for 'b' values, Set-II : CH4, O2, H2,, for 'a' values, Value of van der Waals constant 'b' increases, with increase in molecular volume. Clearly, the, increasing order of molecular volume (size of, molecule) is:, He < H2 < O2 < CO2, Value of 'a' increases with increase in, intermolecular attraction. CH4 is a polar molecule,, thus, it will possess highest value of 'a'. O2 and, H2, both are non-polar molecules. The van der, Waals force µ molecular mass. Hence, the correct, order of value of 'b' is : CH4 > O2 > H2., From the given options, (c) is most suitable., , PV, 1 = n1RT, n1 µ P1 and n2 µ P2, n1 P1, n, 170, =, Þ 1 =, = 0.30, n2 P2, n2 570, , 41., , AlAt high temperature and low pressure, the, correction terms (an2/V2) and (nb) become negligible, and thus, the deviation from ideal behaviour will, be minimum., , (d) Density =, , 42., , Partial pressure of a gas = mole fraction of gas ×, total pressure, nCO + nN 2, , 43., 44., 45., 46., , ´1 =, , nN 2, 2nN 2, , 47., , ´1, , 38., , nRT w RT, =, (d) P =, V, m V, , 39., , 6 ´8.314 ´ 402, ; 41648Pa, 16.05, 0.03, (a) At higher temperature and low pressure, real gas acts as an ideal gas., (d) By Ideal gas equation, , 40., , (d) At low pressure and high temperature, correction for 1 mole of a gas is negligible, i.e, the effect of a/v2 and b becomes negligible., Thus, the gas equation becomes, PV = nRT, PV = RT or, , 1, = = 0.5 atm., 2, , =, , 1, = 12 K, 0.082, (b) At low temperature and high pressure., (c) Molecules of an ideal gas move with, different speeds., (d) In an ideal gas, the intermolecular forces of, attraction are negligible and hence, it cannot be, liquefied., (c) In the equation PV = nRT, n moles of the, gas have volume V., T=, , In van der Waals equation, n moles of gas occupies, volume V and exert pressure P at a temperature T., , PCO + PN2 = 1 atm, , nN 2, , n, RT = CRT., V, Hence 1 = 1 × 0.082 × T, , (c) PV = nRT or P =, , \, , (a) Given nCO = nN 2, , \ PN 2 =, , (b) At low pressure and high temperature, real, gas nearly behave like ideal gas. Hence, deviation, is minimum from ideal behaviour., , 48., , PV, = 1 [Ideal gas equation], RT, , \ Z = 1, and gas shows ideal behaviour., (c) van der Waal's equation for 1 mole :, a ö, æ, çè P + 2 ÷ø (V - b) = RT, V, Here, æç P + a ö÷ represent the intermolecular, è, V2ø, forces and (V – b) is the corrected volume.
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57, , States of Matter, 49., , (d) PV = nRT =, , m, RT, M, , m, PM, or PM = RT = dRT Þ d =, V, RT, , 50., , (b), , 52., , PV, PV, PV, = constant or 1 1 = 2 2, T, T1, T2, , PV, 1 1 = T1, P2V2 T2, (c) Due to intermolecular H-bonding the, surface tension of H2O is more than other given, liquids. One H2O molecule is joined with 4, another H2O molecule through H–bond., Þ, , 51., , 53., , Hydrogen bonding is in order H2O > C2H5OH >, CH3OH., (a) By definition of Nernst distribution law., When a solute is shaken with two immiscible, liquids, having solubility in both, the solute, distributes itself between the two liquids in such, a way that the ratio of its concentrations in two, liquids is constant at a given temperature,, provided the molecular state of the solute, remains the same in both the liquids., (d) A substance exists as a liquid above its, m. pt. and below its b. pt.
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EBD_8336, 58, , CHEMISTRY, , 6, , Thermodynamics, , Trend Analysis with Important Topics & Sub-Topics, 2020, Topic Name, First law and basic, fundamentals of, thermodynamics, Laws of, thermochemistry, Entropy and second, law of thermodynamics, , Sub-Topic, first law of, thermodynamics, , 2019, , 2018, , 1, , A, , 1, , A, , bond di ssociation, energy, second law of, thermodynamics, entropy, , Spontaneity and Gibb's, free energy, , spontaniety, , LOD - Level of Difficulty, , E - Easy, , 2017, , 2016, , QNS. LOD QNS. LOD QNS. LOD QNS. LOD QNS. LOD, , 1, , 1, , 1, , A, , 1, , A, , D, , A, 1, , A - Average, , E, , D - Difficult, , 1, , E, , Qns - No. of Questions, , Topic 1: First Law and Basic Fundamentals, of Thermodynamics, 1., , The correct option for free expansion of an ideal, gas under adiabatic condition is, [2020], (a) q = 0, DT < 0 and w > 0, (b) q < 0, DT = 0 and w = 0, (c) q > 0, DT > 0 and w > 0, (d) q = 0, DT = 0 and w = 0, , 2., , 3., , An ideal gas expands isothermally from 10–3 m3, to 10–2 m3 at 300 K against a constant pressure, of 105 Nm–2. The work done on the gas is, [NEET Odisha 2019], (a) –900 kJ, (b) +270 kJ, (c) –900 J, (d) +900 kJ, Reversible expansion of an ideal gas under, isothermal and adiabatic conditions are as, shown in the figure., [NEET Odisha 2019], , 4., , AB ® Isothermal expansion, AC ® Adiabatic expansion, Which of the following options is not correct?, (a) TC > TA, (b) DSisothermal > DSadiabatic, (c) TA = TB, (d) Wisothermal > Wadiabatic, Under isothermal condition, a gas at 300 K, expands front 0.1 L to 0.25 L against a constant, external pressure of 2 bar. The work done by, the gas is [Given that 1 L bar = 100 J] [2019], (a) –30 J, (b) 5 kJ, (c) 25 J, (d) 30 J
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59, , Thermodynamics, 5., , 6., , 7., , 8., , 9., , 10., , 11., , 12., , A gas is allowed to expand in a well insulated, container against a constant external pressure, of 2.5 atm from an initial volume of 2.50 L to a, final volume of 4.50 L. The change in internal, energy DU of the gas in joules will be:- [2017], (a) – 500 J, (b) – 505 J, (c) + 505 J, (d) 1136.25 J, The heat of combustion of carbon to CO2 is, –393.5 kJ/mol. The heat released upon formation, of 35.2 g of CO2 from carbon and oxygen gas is, [2015 RS], (a) –315 kJ, (b) +315 kJ, (c) –630 kJ, (d) –3.15 kJ, When 5 litres of a gas mixture of methane and, propane is perfectly combusted at 0°C and 1, atmosphere, 16 litre of oxygen at the same, temperature and pressure is consumed. The, amount of heat released from this combustion in, kJ (DHcomb (CH4) = 890 kJ mol–1, DHcomb (C3H8) =, 2220 kJ mol–1) is, [NEET Kar. 2013], (a) 32, (b) 38, (c) 317, (d) 477, Which of the following is correct option for free, expansion of an ideal gas under adiabatic, condition ?, [2011], (a) q = 0, DT ¹ 0, w = 0, (b) q ¹ 0, DT = 0, w = 0, (c) q = 0, DT = 0, w = 0, (d) q = 0, DT < 0, w ¹ 0, Three moles of an ideal gas expanded, spontaneously into vacuum. The work done will, be :, [2010], (a) Zero, (b) Infinite, (c) 3 Joules, (d) 9 Joules, Which of the following are not state functions ?, (I) q + w, (II) q, [2008], (III) w, (IV) H - TS, (a) (I) and (IV), (b) (II), (III) and (IV), (c) (I), (II) and (III), (d) (II) and (III), In a closed insulated container, a liquid is stirred, with a paddle to increase the temperature, which, of the following is true?, [2002], (a) DE = W ¹ 0, q = 0 (b) DE = W = q ¹ 0, (c) DE = 0, W = q ¹ 0 (d) W = 0, DE = q ¹ 0, When 1 mol of a gas is heated at constant, volume, temperature is raised from 298 to 308 K., If heat supplied to the gas is 500 J, then which, statement is correct ?, [2001], (a) q = w = 500 J, DU = 0, (b) q = DU = 500 J, w = 0, (c) q = –w = 500 J, DU = 0, (d) DU = 0, q = w = –500 J, , 13., , Adiabatic expansions of an ideal gas is, accompanied by, [1999], (a) decrease in DE, (b) increase in temperature, (c) decrease in DS, (d) no change in any one of the above properties, Topic 2: Laws of Thermochemistry, , 14., , 15., , 16., , 17., , 18., , The bond dissociation energies of X2, Y2 and, XY are in the ratio of 1 : 0.5 : 1. DH for the formation, of XY is –200 kJ mol–1. The bond dissociation, energy of X2 will be, [2018], (a) 200 kJ mol–1, (b) 100 kJ mol–1, (c) 400 kJ mol–1, (d) 800 kJ mol–1, Three thermochemical equations are given below:, (i) C(graphite) + O2(g) ® CO2(g);, DrH° = x kJ mol–1, 1, (ii) C(graphite) + O2(g) ® CO(g);, 2, DrH° = y kJ mol–1, 1, (iii) CO(g) + O2(g) ® CO2(g);, 2, DrH° = z kJ mol–1, Based on the above equations, find out which, of the relationship given below is correct?, [NEET Kar. 2013], (a) x = y – z, (b) z = x + y, (c) x = y + z, (d) y = 2z – x, Standard enthalpy of vapourisation Dvap H° for, water at 100°C is 40.66 kJ mol–1. The internal, energy of vaporisation of water at 100°C, (in kJ mol–1) is :, [2012], (a) + 37.56, (b) – 43.76, (c) + 43.76, (d) + 40.66, (Assume water vapour to behave like an ideal gas)., Equal volumes of two monoatomic gases, A and, B, at same temperature and pressure are mixed., The ratio of specific heats (Cp/Cv) of the mixture, will be :, [2012 M], (a) 0.83, (b) 1.50, (c) 3.3, (d) 1.67, Enthalpy change for the reaction,, [2011], 4H(g) ¾¾, ® 2H 2 (g) is – 869.6 kJ., The dissociation energy of H–H bond is :, (a) – 434.8 kJ, (b) – 869.6 kJ, (c) + 434.8 kJ, (d) + 217.4 kJ
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61, , Thermodynamics, 31., , 32., , 33., , 34., , 35., , 36., , 37., , 38., , 39., , The densities of graphite and diamond at 298 K, are 2.25 and 3.31 g cm–3, respectively. If the, standard free energy difference (DGº) is equal, to 1895 J mol–1, the pressure at which graphite, will be transformed into diamond at 298 K is, [2003], (a) 9.92 × 105 Pa, (b) 9.92 × 108 Pa, (c) 9.92 × 107 Pa, (d) 9.92 × 106 Pa, For which one of the following equations is, DHºreact equal to DHf º for the product? [2003], (a) 2 CO(g) + O 2 (g) ® 2 CO 2 (g), (b) N 2 (g) + O3 (g) ® N 2 O 3 (g), (c) CH 4 ( g ) + 2Cl ( g ) ® CH 2 Cl 2 l + 2HCl ( g ), (), (d) Xe(g) + 2F2 (g) ® XeF4 (g), For the reaction, C3 H8 ( g ) + 5O2 ( g ) ® 3CO2 ( g ) + 4 H 2O(l ), at constant temperature, DH – DE is [2003], (a) – RT, (b) + RT, (c) – 3 RT, (d) + 3 RT, Heat of combustion DHº for C (s), H2 (g) and, CH4 (g) are –94, –68 and –213 kcal/mol, then, DHº for C(s) + 2H 2 (g) ® CH 4 (g) is [2002], (a) –17 kcal, (b) – 111 kcal, (c) –170 kcal, (d) –85 kcal, 1, Enthalpy of CH 4 + O 2 ® CH 3 OH is, 2, negative. If enthalpy of combustion of CH4 and, CH3OH are x and y respectively, then which, relation is correct, [2001], (a) x > y, (b) x < y, (c) x = y, (d) x ³ y, What is the enthalpy change for,, 2H 2O 2 (l) ® 2 H 2 O(l) + O 2 (g) if heat of, formation of H2O2 (l) and H2O (l) are –188 and, –286 kJ/mol respectively?, [2001], (a) –196 kJ/mol, (b) + 948 kJ/mol, (c) + 196 kJ/mol, (d) –948 kJ/mol, The values of heat of formation of SO2 and SO3, are –298.2 kJ and –98.2 kJ. The heat of formation, of the reaction, [2000], SO 2 + (1 / 2) O 2 ® SO 3 will be, (a) –200 kJ, (b) –356.2 kJ, (c) + 200 kJ, (d) – 396.2 kJ, For a cyclic process, which of the following is, not true?, [1999], (a) DH = 0, (b) DE = 0, (c) DG = 0, (d) Total W = 0, For a reaction in which all reactants and, products are liquids, which one of the following, equations is most applicable ?, [1999], , 40., , 41., , (a) DH < DE, (b) DH = DS, (c) DH = DE, (d) DH = DG, One mole of an ideal gas at 300 K is expanded, isothermally from an initial volume of 1 litre to, 10 litres. The DE for this process is, (R = 2 cal. mol–1 K–1), [1998], (a) 163.7 cal, (b) zero, (c) 1381.1 cal, (d) 9 lit. atm, Given that C + O2 ® CO 2 : DHº = - x kJ, 2CO + O2 ® 2CO 2 : DHº = - y kJ, the enthalpy of formation of carbon monoxide, will be, [1997], 2x - y, y - 2x, (b), 2, 2, (c) 2x – y, (d) y – 2x, Hydrogen has an ionisation energy of 1311 kJ, mol –1 and for chlorine it is 1256 kJ mol –1., Hydrogen forms H+ (aq) ions but chlorine does, not form Cl+ (aq) ions because, [1996], (a) H+ has lower hydration enthalpy, (b) Cl+ has lower hydration enthalpy, (c) Cl has high electron affinity, (d) Cl has high electronegativity, If enthalpies of formation of C2 H 4( g ) , CO2(g), , (a), , 42., , 43., , 44., , 45., , 46., , and H 2 O(l) at 25°C and 1atm pressure are 52,, – 394 and – 286 kJ/mol respectively, the, enthalpy of combustion of C2H4 is equal to, [1995], (a) – 141.2 kJ/mol, (b) – 1412 kJ/mol, (c) + 14.2 kJ/mol, (d) + 1412 kJ/mol, Equal volumes of molar hydrochloric acid and, sulphuric acid are neutralized by dil. NaOH, solution and x kcal and y kcal of heat are liberated, respectively. Which of the following is true ?, [1994], 1, (a) x = y, (b) x = y, 2, (c) x = 2y, (d) None of these, For the reaction, [1994], N 2 + 3H 2, 2NH3, D H = ?, (a) D E + 2RT, (b) D E –2RT, (c) D H = RT, (d) D E – RT., During isothermal expansion of an ideal gas, its, [1991, 94], (a) internal energy increases, (b) enthalpy decreases, (c) enthalpy remains unaffected, (d) enthalpy reduces to zero.
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EBD_8336, 62, , 47., , CHEMISTRY, If DH is the change in enthalpy and DE , the, change in internal energy accompanying a, gaseous reaction, then, [1990], (a) DH is always greater than DE ,, , DH < D E only if the number of moles of, the products is greater than the number of, moles of the reactants, (c) D H is always less than D E, (d) D H < D E only if the number of moles of, products is less than the number of moles, of the reactants., Topic 3: Entropy and Second Law of, Thermodynamics, , 54., , (b), , 48., , 49., , 50., , 51., , 52., , 53., , For the reaction, 2Cl(g) ¾® Cl2(g), the correct, option is :, [2020], (a) DrH > 0 and DrS < 0, (b) DrH < 0 and DrS > 0, (c) DrH < 0 and DrS < 0, (d) DrH > 0 and DrS > 0, In which case change in entropy is negative ?, [2019], (a) Evaporation of water, (b) Expansion of a gas at constant temperature, (c) Sublimation of solid to gas, (d) 2H(g) ® H2(g), The enthalpy of fusion of water is 1.435 kcal/mol., The molar entropy change for the melting of ice, at 0°C is :, [2012], (a) 10.52 cal / (mol K) (b) 21.04 cal / (mol K), (c) 5.260 cal / (mol K) (d) 0.526 cal / (mol K), If the enthalpy change for the transition of liquid, water to steam is 30 kJ mol–1 at 27ºC, the entropy, change for the process would be :, [2011], (a) 10 J mol –1 K–1, (b) 1.0 J mol–1 K–1, (c) 0.1 J mol–1 K–1, (d) 100 J mol–1 K–1, Standard entropies of X2, Y2 and XY3 are 60, 40, and 50 JK–1mol–1 respectively. For the reaction, 1, 3, X 2 + Y2 XY3 , DH = – 30 kJ, 2, 2, to be at equilibrium, the temperature should be:, [2010], (a) 750 K, (b) 1000 K, (c) 1250 K, (d) 500 K, For the gas phase reaction,, [2008], , PCl5(g) PCl3(g) + Cl2(g), which of the following conditions are correct ?, (a) DH = 0 and DS < 0 (b) DH > 0 and DS > 0, (c) DH < 0 and DS < 0 (d) DH > 0 and DS < 0, , 55., , 56., 57., , 58., , 59., , 60., , 61., , What is the entropy change (in JK–1 mol–1) when, one mole of ice is converted into water at 0º C?, (The enthalpy change for the conversion of ice, to liquid water is 6.0 kJ mol–1 at 0ºC) [2003], (a) 21.98, (b) 20.13, (c) 2.013, (d) 2.198, 2 mole of an ideal gas at 27ºC temperature is, expanded reversibly from 2 lit to 20 lit. Find the, entropy change (R = 2 cal/mol K), [2002], (a) 92.1, (b) 0, (c) 4, (d) 9.2, Unit of entropy is, [2002], (a) JK–1 mol–1, (b) J mol–1, (c) J–1 K–1 mol–1, (d) JK mol–1, The entropy change in the fusion of one mole, of a solid melting at 27ºC (Latent heat of fusion,, 2930 J mol–1) is :, [2000], –1, –1, –1, (a) 9.77 J K mol, (b) 10.73 J K mol–1, –1, –1, (c) 2930 J K mol, (d) 108.5 J K–1 mol–1, Identify the correct statement regarding, entropy:, [1998], (a) At absolute zero of temperature, entropy, of a perfectly crystalline substance is taken, to be zero, (b) At absolute zero of temperature, the, entropy of a perfectly crystalline substance, is +ve, (c) At absolute zero of temperature, the, entropy of all crystalline substances is, taken to be zero, (d) At 0ºC, the entropy of a perfectly, crystalline substance is taken to be zero, According to the third law of thermodynamics, which one of the following quantities for a, perfectly crystalline solid is zero at absolute, zero?, [1996], (a) Free energy, (b) Entropy, (c) Enthalpy, (d) Internal energy, Given the following entropy values (in J K–1 mol–1), at 298 K and 1 atm :H2 (g) : 130.6, Cl2 (g) : 223.0,, HCl (g) : 186.7. The entropy change (in J K–1, mol–1) for the reaction, H 2 (g) + Cl2 (g) ¾¾, ® 2HCl(g) is, [1996], (a) +540.3, (b) +727.0, (c) –166.9, (d) +19.8, A chemical reaction will be spontaneous if it is, accompanied by a decrease of, [1994], (a) entropy of the system, (b) enthalpy of the system, (c) internal energy of the system, (d) free energy of the system
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63, , Thermodynamics, Topic 4: Spontaneity and Gibb's Free Energy, 62., , 63., , 64., , 65., , For a given reaction, DH = 35.5 kJ mol-1 and, DS = 83.6 JK-1 mol-1. The reaction is spontaneous, at: (Assume that DH and DS do not vary with, temperature), [2017], (a) T > 425 K, (b) All temperatures, (c) T > 298 K, (d) T < 425 K, The correct thermodynamic conditions for the, spontaneous reaction at all tempertures is, [2016], (a) DH > 0 and DS > 0 (b) DH > 0 and DS < 0, (c) DH < 0 and DS > 0 (d) DH < 0 and DS < 0, For the reaction :, [2014], X2O4(l) ® 2XO2(g), DU = 2.1 k cal, DS = 20 cal K–1 at 300 K, Hence DG is:(a) 2.7 k cal, (b) – 2.7 k cal, (c) 9.3 k cal, (d) – 9.3 k cal, In which of the following reactions, standard, entropy change (DS°) is positive and standard, Gibb’s energy change (DG°) decreases sharply, with increasing temperature ?, [2012], 1, (a) C (graphite) + O2(g) ® CO(g), 2, , (b) CO(g) +, , 1, O (g) ® CO2(g), 2 2, , (c) Mg(s) +, , 1, O (g) ® MgO(s), 2 2, , 1, 1, 1, C (graphite) + O2(g) ® CO2(g), 2, 2, 2, For vaporization of water at 1 atmospheric, pressure, the values of DH and DS are 40.63, kJmol–1 and 108.8 JK–1 mol–1, respectively. The, temperature when Gibbs energy change (DG), for this transformation will be zero, is: [2010], (a) 293.4 K, (b) 273.4 K, (c) 393.4 K, (d) 373.4 K., Match List -I (Equations) with List-II (Type of, processes) and select the correct option. [2010], List I, List II, Equations, Type of processes, (1) Kp > Q, (i) Non-spontaneous, (2) DG° < RT In Q (ii) Equilibrium, (3) Kp = Q, (iii) Spontaneous and, endothermic, DH, (4) T >, (iv) Spontaneous, DS, , (d), , 66., , 67., , 68., , 69., , 70., , 71., , 72., , 73., , Options:, (1), (2), (3), (4), (a) (ii), (i), (iv), (iii), (b) (i), (ii), (iii), (iv), (c) (iii), (iv), (ii), (i), (d) (iv), (i), (ii), (iii), The values of DH and DS for the reaction,, C(graphite) + CO2(g) ® 2CO(g) are 170 kJ and, 170 JK–1, respectively. This reaction will be, spontaneous at, [2009], (a) 910 K, (b) 1110 K, (c) 510 K, (d) 710 K, The enthalpy and entropy change for the, reaction, Br2(l) + Cl2 (g) ® 2 BrCl (g), are 30kJ mol–1 and 105 JK–1 mol–1 respectively., The temperature at which the reaction will be in, equilibrium is, [2006], (a) 273 K, (b) 450 K, (c) 300 K, (d) 285.7 K, Identify the correct statement for change of Gibbs, energy for a system (DG system) at constant, temperature and pressure :, [2006], (a) If DGsystem = 0, the system has attained, equilibrium, (b) If DGsystem = 0, the system is still moving in, a particular direction, (c) If DGsystem < 0, the process is not spontaneous, (d) If DGsystem > 0, the process is spontaneous, A reaction occurs spontaneously if, [2005], (a) TDS < DH and both DH and DS are + ve, (b) TDS > DH and DH is + ve and DS is - ve, (c) TDS > DH and both DH and DS are + ve, (d) TDS = DH and both DH and DS are + ve, Which of the following pairs of a chemical, reaction is certain to result in a spontaneous, reaction?, [2005], (a) Exothermic and increasing disorder, (b) Exothermic and decreasing disorder, (c) Endothermic and increasing disorder, (d) Endothermic and decreasing disorder, Stan dard enthalpy and standard entropy, changes for the oxidation of ammonia at 298 K, are – 382.64 kJ mol–1 and -145.6 JK–1 mol–1,, respectively. Standard Gibb's energy change for, the same reaction at 298 K is, [2004], (a) –22.1 kJ mol–1, (b) –339.3 kJ mol–1, (c) –439.3 kJ mol–1, (d) –523.2 kJ mol–1
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EBD_8336, 64, , 74., , 75., , CHEMISTRY, Considering entropy (S) as a thermodynamic, parameter, the criterion for the spontaneity of, any process is, [2004], (a) DSsystem + DSsurroundings > 0, (b) DSsystem - DSsurroundings > 0, (c) DSsystem > 0 only, (d) DSsurroundings > 0 only, The factor of DG values is important in, metallurgy. The DG values for the following, reactions at 800ºC are given as :, S2 (s) + 2O 2 (g) ¾¾, ® 2SO2 (g) ; DG = –544 kJ, 2Zn(s) + S2 (s) ¾¾, ® 2ZnS(s) ; DG = –293 kJ, 1, 2, 3, 4, 5, 6, 7, 8, , (d), (c), (a), (a), (b), (a), (c), (c), , 9, 10, 11, 12, 13, 14, 15, 16, , (a), (d), (a), (b), (a), (d), (c), (a), , 17, 18, 19, 20, 21, 22, 23, 24, , (d), (c), (b), (a), (b), (c), (d), (b), , 2Zn(s) + O2 (g) ¾¾, ® 2ZnO(s) ; DG = –480 kJ, , Then DG for the reaction :, , 2ZnS(s) + 3O2 (g) ¾¾, ® 2ZnO(s) + 2SO2 (s), , 76., , will be :, [2000], (a) –357 kJ, (b) –731 kJ, (c) –773 kJ, (d) –229 kJ, Consider the following reaction occurring in an, automobile, [1994], 2C8H18 (g) + 25 O 2 (g) ¾¾, ® 16CO 2 +18H 2 O(g), the sign of D H, D S and D G would be, (a) +, –, +, (b) –, +, –, (c) –, +, +, (d) +, +, –., , ANSWER KEY, 25 (a) 33 (c) 41 (b), 26 (d) 34 (a) 42 (b), 27 (c) 35 (a) 43 (b), 28 (d) 36 (a) 44 (b), 29 (b) 37 (c) 45 (b), 30 (c) 38 (d) 46 (c), 31 (N) 39 (c) 47 (d), 32 (d) 40 (b) 48 (c), , 49, 50, 51, 52, 53, 54, 55, 56, , (d), (c), (d), (a), (b), (a), (d), (a), , 57, 58, 59, 60, 61, 62, 63, 64, , (a), (a), (b), (d), (d), (a), (c), (b), , 65, 66, 67, 68, 69, 70, 71, 72, , (a), (d), (d), (b), (d), (a), (c), (a), , 73 (b), 74 (a), 75 (b), 76 (b), , Hints & Solutions, 1., , (d) Free expansion of ideal gas, Pex = 0, \ w = - Pex DV = 0, , 5., , Q Adiabatic process Þ q = 0, DE = q + w (first law of thermodynamics), , \ DE = 0, DE = nCv dT Þ DE = 0, So, q = 0, DT = 0, w = 0., , 2., 3., , 4., , 6., , (c) W = –Pext (Vf – Vi) = –10 (10 – 10 ) = –900 J, (a) Since graph A to C represents adiabatic, reversible expansion, so work is done on, the expense of internal energy, therefore,, there is decrease in internal energy. So the, temperature decreases., i.e., TC < TA, (a) W = – Pext (V2 – V1) (Irreversible isothermal, expansion), = –2 (0.25 – 0.1), = – 2 (0.15) = – 0.3 L Bar, = – 0.3 × 100 J = – 30 J, 5, , –2, , –3, , (b) The system is in isolated state., Q For an adiabatic process, q = 0, DU = q + w, \ DU = w, = –pDV, = –2.5 atm × (4.5 – 2.5) L, = –2.5 × 2 L-atm, = –5 × 101.3 J, = –506.5 J » –505J, (a) C + O2 ® CO2 + 393.5 kJ/mol, 12g, 44g, 44g CO2 is formed from 12g of carbon, 12 ´ 35.2, g of C, 44, = 9.6 g of C = 9.6/12 = 0.8 mole, 1 mole release heat 393.5 kJ, 0.8 mole release heat = 393.5 × 0.8, = 314.8 kJ » 315 kJ, (c) CH 4 + 2O 2 ¾® CO2 + 2H2O, , 35.2g is formed from, , 7., , x, , 2x, , C3 H8 + 5O 2 ¾® 3CO2 + 4H2O, (5- x ), , 5(5- x )
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65, , Thermodynamics, 2x + 5(5– x) = 16, Þ x = 3L, \ Heat released, 3, 2, ´ 890 +, ´ 2220 = 317, 22.4, 22.4, (c) For adiabatic process DQ = 0, In free expansion, external pressure is zero., Thus, work done W = 0 and DE = W = 0. If, DE = 0, then DT = 0., (a) Ideal gas during spontaneous expansion, into vacuum does not do any external work., (d) We know that q (heat) and work (w) are not, state functions but (q + w) is a state functions., H – TS (i.e. G) is also a state function., (a) Internal energy is dependent upon, temperature and according to first law of, thermodynamics total energy of an isolated, system remains same, i.e., in a system of, constant mass, energy can neither be created, nor destroyed by any physical or chemical, change but can be transformed from one form, to another, DE = q + w, , =, , 8., , 9., 10., 11., , 17., , For monoatomic gases, diatomic gases, , For closed insulated container, q = 0, so,, DE = +W, as work is done on the system, , 12., 13., , 14., , (b) As volume is constant hence, work done, in this process is zero hence, heat supplied is, equal to change in internal energy., (a) DE =DQ – W, For adiabatic expansion, DQ = 0, Þ DE = –W, The negative sign shows decrease in Internal, energy, which is equal to the work done by the, system on the surroundings., (d) Let B.E of X2, Y2 and XY are x kJ mol–1,, 0.5 x kJ mol–1 and x kJ mol–1 respectively, , triatomic gases, , Cp, Cv, , 18., , 19., , 1, 1, X 2 + Y2 ® XY; DH = –200 kJ mol–1, 2, 2, DH = –200 = S (B.E)Reactants – S(B.E)Product, , 15., 16., , 1, é1, ù, = ê ´ ( x) + ´ (0.5 x ) ú – [1 ´ ( x )], 2, 2, ë, û, On solving, x = 800 kJ mol–1, (c) Applying Hess’s law, equation (i) can be, obtained by adding equations (ii) and (iii)., \x=y+z, (a) Dvap H° = 40.66 kJ mol–1, T = 100 + 273 = 373 K, DE = ?, , DH = DE + DngRT Þ DE = DH – DngRT, Dng = number of gaseous moles of products, – number of gaseous moles of reactants, H2Ol H2O(g), Dng = 1 – 0 = 1, DE = DH – RT, DE = (40.66 × 103) – (8.314 × 373), = 37559 J/mol or 37.56 kJ / mol, (d) For a monoatomic gas, 3, Cv = R, 2, 3, 5, \ Cp = R + R = R, 2, 2, 5, Cp 2 R 5, =, = = 1.67, Cv 3 R 3, 2, , 20., , Cp, Cv, Cp, Cv, , Cp, Cv, , = 1.67, , = 1.40, = 1.29 if degree of freedom = 7, , = 1.33 if degree of freedom = 6, , (c) Given, ® 2H 2 (g); DH = -869.6 kJ, 4H(g) ¾¾, or 2H 2 (g) ¾¾, ® 4H(g); DH = 869.6 kJ, 869.6, ® 2H(g); DH =, H 2 (g) ¾¾, = 434.8 kJ, 2, (b) Given, DH, 1, A ¾¾, ®B, + 150, ...(a), 2, ...(b), 3B ¾¾, ® 2C + D –125, , ...(c), ® 2 D +350, E + A ¾¾, To calculate DH operate, 2 × eq. (a) + eq. (b) – eq. (c), DH = 300 – 125 – 350 = – 175, (a) Fe2O3(s)+ CO(g) ® 2FeO(s)+ CO2(g), Given, Fe2O3(s)+ 3CO(g) ® 2Fe(s)+ 3CO2(g) ;, DH = – 26.8 kJ … (i), FeO(s) + CO(g) ® Fe(s) + CO2(g) ;, DH = –16.5 kJ … (ii)
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EBD_8336, 66, , 21., , CHEMISTRY, \ DH for the reaction, Fe2O3(s) + CO2(g) ® 2FeO(s) + CO2(g), can be calculate as, eqn (i) — 2 × eqn (ii), \ D H = –26.8 + 33.0 = + 6.2 kJ, (b) Enthalpy of reaction, = SB.E(Reactant)– SB.E(Product), = éë B.E (C= C) + 4 B.E.(C–H) + B.E.(H - H) ùû, - éë B.E.(C - C) + 6 B.E.(C - H) ùû, , 22., , = [606.1 + (4 × 410.5) + 431.37)] – [336.49, + (6 × 410.5)], = –120.0 kJ mol–1, (c) The reaction for formation of HCl can be, written as, H2 + Cl2 ® 2HCI, H – H + Cl – Cl ® 2 (H – Cl), Substituting the given values, we get enthalpy, of formation of, DHf (HCl) = [(B.E)H–H + (B.E)Cl–Cl] –, [2 × B.E. (H–Cl)], = 434 + 242 – 2 × (431) = –186 kJ, \ Enthalpy of formation for, -186, kJ = –93 kJ., 2, (d) This reaction shows the formation of H2O,, and the X2 represents the enthalpy of formation, of H2O., , The resonance energy provides extra stability, to the benzene molecule so it has to overcome, for hydrogenation to take place. So DH = – 358.5, – (–150.4) = –208.1 kJ, Resonance reduces the energy of molecule and thus,, increases the stability. Hence, the magnitude of, enthalpy of hydrogenation will be less than the, expected value for benzene., , 26., , If the container is closed then pressure is not, constant and DH = DE + V DP, , 27., , 28., , 1 mol HCl =, , 23., , The enthalpy of formation is the heat evolved or, absorbed when one mole of substance is formed, from its constituent atoms., , 24., , (b), , 29., , 30., , 1, 1, H 2 + Cl2 ¾¾, ® HCl, 2, 2, , D H HCl =, , å B.E. of reactant, - å B.E. of products, , 31., , (a), , + H2 ¾ ¾, ®, ®, + 3 H2 ¾¾, , = – 358.5 kJ, , = -6 ´101.32 = -608 J, (c) Given Cp = 75 JK–1 mol–1, 100, n=, mole , Q = 1000 J, DT = ?, 18, Q = nCpDT, 1000 ´ 18, Þ DT =, = 2.4 K, 100 ´ 75, (N) Change in volume from 1 mole of graphite, to 1 mole of diamond, 12, 12, = -1.71 cm3, 3.31 2.25, = –1.71 × 10–6 m3, Gibb's free energy is the measure of useful work, done at constant temperature and pressure., DG = –p.DV, , ; DH = – 119.5 kJ, , ; DH = 3(– 119.5), , (c) As MgO is a oxide of weak base hence, some energy is lost to break MgO (s). Hence, enthalpy is less than –57.33 kJ mol–1., (d) H 2 (g) + Br2 (g) ® 2HBr (g), DH ° = (B E)reactant – (B E) product, = (433 + 192) - (2 ´ 364), = 625 – 728 = – 103 kJ, (b) W = – PDV, = -3(6 - 4) = -6 litre atmosphere, , DV =, , 1, 1, -90 = ´ 430 + ´ 240 - B.E. of HCl, 2, 2, \ B.E. of HCl = 215 + 120 + 90, = 425 kJ mol–1, , 25., , (d) We know that, DH = DE + PDV, In the reaction, H2 + Br2 ® 2HBr this is no, change in volume or DV = 0, So, DH = DE for this reaction, , -DG, 1895J, =, DV 1.71 ´ 10-6 m3, = 1.1 × 109 Pa, (d) N2(g) + O3(g) ¾¾, ® N2O3(g), , Þ p=, , 32.
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67, , Thermodynamics, , 33., , The state of N2O3 is liquid in NTP conditions., Thus, this reaction is not equal to DHf° for the, product., (d) Xe(g) + 2F2(g) ¾¾, ® XeF4(g), For this reaction DHr° is equal to DHf° of XeF4(g)., (c) DH = DE + DnRT, Dn = 3 – (1 + 5), = 3 – 6 = –3, , 38., , (a) C(s) + O2 (g) ® CO2 (g), , 40., 41., , DH - DE = (-3RT ), , 34., , DH = -94 kcal/mole ..(i), 1, H 2 (g) + O2 ® H 2 O(g), 2, DH = -68 kcal/mole ..(ii), CH 4 + 2O2 ® CO2 + 2H 2O,, DH = -213 k cal/mole ...(iii), , 35., , 36., , C(s) + 2H 2 ® CH 4 (g), DH = ?, ...(iv), Eqn. (iv) can be obtained by, eq. (i) + eq. (ii) × 2 – eq.(iii), C(s) + O2 (g) ® CO 2 (g), 2H 2 (g) + O2 (g) ® 2H 2O(g), CO2 (g) + 2H 2 O(g) ® CH 4 (g) + 2O 2 (g), C(s) + 2H 2 (g) ® CH 4 (g), So, DH CH 4 = -94 + 2( -68) - ( -213), = –94 – 136 + 213 = –17 k cal/mole, (a) The enthalpy of combustion is always, negative., CH4 + 2O2 ¾¾, ® CO2 + 2H2O; DH1 = –x kJ, 3, CH3OH + O2 ¾¾, ® CO2 + 2H2O; D H2 = - ykJ, 2, 1, CH4 + O2 ¾¾, ®CH2OH; + DH = DH1- DH2, 2, It is given that DH = is negative., Thus, –x –(–y) = – ve, y – x = –ve, Hence, x > y., (a) 2H 2 O 2 (l) ¾¾, ® 2H 2 O(l) + O 2 (g) DH = ?, , 39., , DH = DE - DnRT, , = [(2 ´ ( -286)) + (0) - (2 ´ (-188))], = [-572 + 376] = -196 kJ / mole, 37., , (c) SO 2 +, DH, , 1, O2 ¾, ¾® SO 3, 2, , = DH fo (SO ), 3, , - DH fo (SO ), 2, , = –98.2 + 298.2 = 200 kJ/Mole, , (Q Dn = 0), DH = DE, (b) For an isothermal process DE = 0, (b) Given C + O 2 = CO 2 , DH º = - x kJ ....(a), 2CO 2 = 2CO + O 2 DH º = + y kJ, , … (b), , or CO2 = CO + 1/ 2 O2 , DH º = + y / 2 kJ ...(c), From eq. no. (a) and (c), , 42., 43., , 1, y - 2x, kJ, C + O 2 = CO, DH º = y / 2 - x =, 2, 2, (b) Hydration energy of Cl+ is very less than, H+, hence it doesn’t form Cl+ (aq) ion., (b) Enthalpy of formation of C 2 H 4 , CO2 and, , H 2 O ar e 52, – 394 an d – 286 kJ/ mol, respectively. (Given), The reaction is, C 2 H 4 + 3O 2 ® 2CO 2 + 2H 2O., change in enthalpy,, ( DH ) = DH products - DH reactants, , 44., , = 2 ´ (-394) + 2 ´ (-286) - (52 + 0), = – 1412 kJ/ mol., (b) 1 M H2SO4 = 2g eq. of HCl, , 45., , 1, y., 2, (b) D ng = 2 – 4 = – 2, D H = D E – 2RT., , Hence y = 2x or x =, , 46., , DH = [2 ´ DH f of H 2 O ( l ) + ( DH f of O 2 ), , -(2 ´ DH f of H 2O2(l) )], , (d) For a cyclic process, DE = 0, DH = 0 & DG = 0. As all depend upon, final state and initial state,W doesn’t depend, on path followed., (c) As all reactant and product are liquid, Dn ( g ) = 0, , 47., , 48., , (c) During isothermal expansion of ideal gas,, D T = 0. Now H = E + PV, Q DH = DE + D( PV ), \ D H = D E + D (nRT);, Thus, if D T = 0., DH = DE, i.e., remain unaffected., (d) As DH = DE + DngRT, if np < nr; D ng = np – nr = – ve., Hence D H < D E., ® 2Cl (g) is, (c) We know that, Cl 2 (g) ¾¾, endothermic reaction because it required energy, to break bond.
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EBD_8336, 68, , 49., , 50., 51., , CHEMISTRY, So reverse reaction, 2Cl (g) ¾¾, ® Cl2 (g) will, be, exothermic, D r H < 0 ., Also, two gaseous atom combine together to, form 1 gaseous molecule., (d) In 2H(g) ¾ ® H 2 (g), no. of moles, decreases, therefore entropy decreases., DH 1.435 ´ 103, =, T, 273, = 5.260 cal / mol - K, (d) Given D H = 30 kJ mol–1 T = 273 + 27 = 300 K, , (c) DS =, , DST =, , 52., , DHT 3 ´ 104, =, J mol–1, 300, T, = 100 J mol–1 K–1, , 1, 3, X 2 + Y2 XY3, 2, 2, ΔS = 50 – (30 + 60) = – 40 J, For equilibrium D G = 0 = D H – T D S, , (a) ΔS for the reaction, , H - 30000, =, = 750 K, S, - 40, (b) For the reaction, PCl5 (g) PCl3 (g) + Cl2 (g), The reaction given is an example of, decomposition reaction and we know that, decomposition reactions are endothermic in, nature, i.e, D H > 0., Further, D n = (1 + 1) – 1= + 1, Hence more number of molecules are present in, products which shows more randomness i.e., D S > 0 (D S is positive), T, , =, , = 2 × 2 × T × 2.303 × 1 = 9.2 T, Entropy change, DS =, , q, T, q ¾¾, ® required heat per mole, T ¾¾, ® constant absolute temperature, Unit of entropy is JK–1 mol–1, , 56., , (a) DS =, , 57., , (a) DS =, , 54., , 2930, J K–1 mol–1 = 9.77 J K–1 mol–1, 300, (a) We know from the third law of, thermodynamics, the entropy of a perfectly, crystalline substance at absolute zero, temperature is taken to be zero., (b) Entropy is zero for perfectly crystaline solid, at absolute zero., , 58., , 59., , Entropy states the randomness or disorderness of, the system. At absolute zero, the movement of, molecules of the system or randomenss of the, system is zero, hence entropy is also zero., , D, , (a) DS =, , 55., , 61., 62., , -, , -, , w = q = nRT ´ 2.303 log, , v2, v1, , = 2 RT ´ 2.303 log, , 20, 2, , (d) Entropy change, DS = DSproduct - DSreactant, = 2 (186.7) – (223 + 130.6), = 373.4 – 353.6, = 19.8 JK–1 mol–1, (d) D G is negative for a spontaneous process., (a) Given D H 35.5 kJ mol–1, –1 mol–1, D S = 83.6 JK, Q D G = D H – TD S, For a reaction to be spontaneous, D G = –ve, i.e., D H < TD S, DH 35.5 ´ 103 Jmol -1, =, DS, 83.6 JK -1, So, the given reaction will be spontaneous at, T > 425 K, (c) D G = D H – T · D S, For a spontaneous reaction D G = –ve (always), which is possible only if, D H < 0 and D S > 0, \ spontaneous at all temperatures., (b) Given D U = 2.1 k cal, D S = 20 cal. K– 1, T = 300 K, Q D H = D U + D ngRT, Putting the values given in the equation, \, , DH per mole 6000, =, T, 273, , 21.98 JK 1mol 1, (d) For isothermal reversible expansion, =, , 60., , DH, T, , DS (per mole) =, , DH y, Latent heat of fusion, =, T, Melting point, , =, , D, , 53., , q 9.2T, =, = 9.2 cal., T, T, , 63., , 64., , T>
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EBD_8336, 70, , CHEMISTRY, , 7, , Equilibrium, , Trend Analysis with Important Topics & Sub-Topics, 2020, Topic Name, , Sub-Topic, , Law of mass, action,equilibrium, constant and its, application, , equilibrium, constant/Kp and Kc, , Relation between K,Q, and G and factors, affecting equilibrium, , Theories of acid and, bases,ionic product of, water and pH scale, , relation between K, and G, Le-chatelier, principle, Clausius and, Clapeyron's, equation, Bronsted acid and, bases, , solubility, , LOD - Level of Difficulty, , E - Easy, , 1, , 1, , Ksp, , The equilibrium constant of the following are :, N2 + 3H2 2NH3, K1, [2017], N2 + O2 2NO, K2, K3, , 2017, , 2, , Topic 1: Law of Mass Action, Equilibrium, Constant (Kc and Kp) and its Application, , 1, H 2 + O 2 ® H 2O, 2, , 2018, , 2016, , QNS. LOD QNS. LOD QNS. LOD QNS. LOD QNS. LOD, , A, , pH scale, , Common ion effect,salt, hydrolysis, buffer, solution and solubility, product, , 1., , 2019, , 1, , E, , 1, , E, , 1, , D, , 1, , A, , 1, , A, , A - Average, , A, , 1, , E, , 1, , A, , D, , D, 1, , 1, , A, , D - Difficult, , 1, , A, , Qns - No. of Questions, , The equilibrium constant (K) of the reaction :, K, 5, , , 2NO + 3H2O, will be;, 2NH3 + O 2 , 2, (a) K 2 K33 / K1, (b) K2K 3/K 1, , (c), , K 23 K3 / K1, , (d) K1K33 / K 2
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71, , Equilibrium, 2., , 3., , 4., , A 20 litre container at 400 K contains CO2(g) at, pressure 0.4 atm and an excess of SrO (neglect, the volume of solid SrO). The volume of the, container is now decreased by moving the, movable piston fitted in the container. The, maximum volume of the container, when pressure, of CO2 attains its maximum value, will be :, (Given that : SrCO3 (s) SrO (s) + CO2(g), Kp, = 1.6 atm), [2017], (a) 10 litre, (b) 4 litre, (c) 2 litre, (d) 5 litre, If the value of an equilibrium constant for a, particular reaction is 1.6 × 10 12 , then at, equilibrium the system will contain :- [2015], (a) mostly reactants, (b) mostly products, (c) similar amounts of reactants and products, (d) all reactants, If the equilibrium constant for, 2NO(g) is K,, N2(g) + O2(g) , the equilibrium constant for, [2015 RS], 1, 1, NO(g) will be:, N (g) + O2(g) , 2 2, 2, , 7., , (a) 1 / (2K1K2), (c) [1 / K1K2, 8., , 1, K, 2, (c) K, (d) K 2, Given that the equilibrium constant for the, 2SO (g) has a, reaction 2SO2(g) + O2(g) , 3, value of 278 at a particular temperature. What is, the value of the equilibrium constant for the, following reaction at the same temperature ?, , 5., , 6., , K2, , 9., , (b), , 1, SO2 (g) + O2 (g), SO3 (g) , [2012 M], 2, (a) 1.8 × 10–3, (b) 3.6 × 10–3, –2, (c) 6.0 × 10, (d) 1.3 × 10–5, Given the reaction between 2 gases represented, by A2 and B2 to give the compound AB(g)., 2 AB(g)., A2(g) + B2(g) , At equilibrium, the concentration, of A2 = 3.0 × 10–3 M, of B2= 4.2 × 10–3 M, of AB = 2.8 × 10–3 M, lf the reaction takes place in a sealed vessel at, 527°C, then the value of Kc will be : [2012 M], (a) 2.0, (b) 1.9, (c) 0.62, (d) 4.5, , ]½, , (b) 1 / (4K1K2), (d) 1 / (K1K2), , In which of the following equilibrium Kc and Kp, are not equal?, [2010], , (a) 2 NO(g), N 2 (g) + O2 (g), (b) SO2 (g) + NO 2 (g) SO3 (g) + NO(g), (c) H 2 (g) + I 2 (g) 2HI(g), , 1, , (a), , 2NO(g),, For the reaction N2(g) + O2(g) , the equilibrium constant is K1. The equilibrium, constant is K2 for the reaction, [2011], 2NO (g)., 2NO(g) + O2(g) , 2, What is K for the reaction, 1 N (g) + O (g) ?, NO2(g) , 2, 2 2, , 10., , (d) 2C(s) + O2(g) 2CO2(g), The reaction 2A(g) + B(g) 3C(g) + D(g) is, begun with the concentrations of A and B both, at an initial value of 1.00 M. When equilibrium is, reached, the concentration of D is measured and, found to be 0.25 M. The value for the equilibrium, constant for this reaction is given by the, expression, [2010], (a) [(0.75)3 (0.25)] ¸ [(0.75)2 (0.25)], (b) [(0.75)3 (0.25)] ¸ [(1.00)2 (1.00)], (c) [(0.75)3 (0.25)] ¸ [(0.50)2 (0.75)], (d) [(0.75)3 (0.25)] ¸ [(0.50)2 (0.25)], If the concentration of OH– ions in the reaction, Fe3+ (aq) + 3OH– (aq) is, Fe(OH) (s) , 3, , 1, times, then equilibrium, 4, concentration of Fe3+ will increase by : [2008], (a) 8 times, (b) 16 times, (c) 64 times, (d) 4 times, The dissociation equilibrium of a gas AB2 can, be represented as :, [2008], 2 AB ( g ) + B2 ( g ), 2 AB2 ( g ) , , decreased by, , 11., , The degree of dissociation is ‘x’ and is small, compared to 1. The expression relating the, degree of dissociation (x) with equilibrium, constant Kp and total pressure P is :, (a) (2Kp/P), (b) (2Kp/P)1/3, (c) (2Kp /P)1/2, (d) (Kp /P)
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EBD_8336, 72, , 12., , CHEMISTRY, The values of Kp1 and Kp2 for the reactions, Y + Z, X , ...(a), [2008], 2 B, and A , , 13., , ...(b), , are in the ratio of 9 : 1. If degree of dissociation, of X and A be equal, then total pressure at, equilibrium for (a) and (b) are in the ratio :, (a) 3 : 1, (b) 1 : 9, (c) 36 : 1, (d) 1 : 1, The value of equilibrium constant of the reaction, 1, 1, [2008], HI (g ) H 2 (g) + I 2 is 8.0, 2, 2, The equilibrium constant of the reaction, 2HI(g) will be:, H 2 (g) + I 2 (g) , 1, 16, , Then,, , (c) K1 =, 17., , 2NO; K 2, N 2 + O 2 , , 15., , (a), , K 2 K32, K1, , (b), , K 22 K3, K1, , (c), , K1 K 2, K3, , (d), , K 2 K33, K1, , For the reaction, [2006], CO (g) + 2H O(l),, CH 4 (g) + 2O 2 (g) , 2, 2, DrH = –170.8 k J mol–1, Which of the following statements is not true ?, (a) The equilibrium constant for the reaction, [CO 2 ], is given by K p =, [CH 4 ][O2 ]2, (b) Addition of CH4(g) or O2(g) at equilibrium, will cause a shift to the right, (c) The reaction is exothermic, (d) At equilibrium, the concentrations of, CO2(g) and H2O(l) are not equal, , 1, 1, N 2 (g) + O 2 (g) ......(ii), 2, 2, [1989, 94, 2005], , æ 1 ö, (a) K1 = ç, è K 2 ÷ø, , (b), , 1, H 2O; K3, H 2 + O2 , 2, The equilibrium constant for the oxidation of 2, moles NH3 by oxygen to give NO is, , K1 and K2 are equilibrium constant for reactions, (i) and (ii), 2 NO (g) .............(i), N2(g) + O2 (g) , , NO(g) , , 1, 64, 1, (c) 16, (d), 8, The following equilibrium constants are given:, [2003, 2007], 2NH3 ; K1, N 2 + 3H 2 , , (a), , 14., , 16., , 18., , 19., , 20., , 2, , 1, K2, , (b) K1 = K22, (d) K1 = (K2)0, , Value of KP in the reaction, MgCO3 (s ) ® MgO( s) + CO2 ( g ) is, (a) KP = PCO, 2, P, ´ PMgO, (b) K P = PCO ´ CO2, 2, PMgCO3, (c), , KP =, , (d), , KP =, , [2000], , PCO2 ´ PMgO, PMgCO3, PMgCO3, PCO2 ´ PMgO, , For dibasic acid correct order is, [2000], (a) Ka1 < Ka 2, (b) K a1 > K a2, (c) Ka1 = Ka2, (d) not certain, If K1 and K2 are the respective equilibrium, constants for the two reactions, XeOF (g) + 2HF(g), XeF6(g) + H2O(g) , 4, XeOF (g) + XeO F (g), XeO4(g) + XeF6(g) , 4, 3 2, the equilibrium constant of the reaction, XeO F (g) + H O(g), XeO4(g) + 2HF(g) , 3 2, 2, will be, [1998], (a) K1/(K2)2, (b) K1 . K2, (c) K1/K2, (d) K2/K1, If a is the fraction of HI dissociated at, H, equilibrium in the reaction, 2 HI (g) , 2, (g) + I2 (g), starting with 2 moles of HI, the total, number of moles of reactants and products at, equilibrium are, [1996], (a) 2 + 2a, (b) 2, (c) 1 + a, (d) 2 – a
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73, , Equilibrium, 21., , The rate constant for forward and backward, reaction of hydrolysis of ester are 1.1 ´ 10 -2 and, 1 .5 ´ 10 -3 per minute respectively. Equilibrium, constant for the reaction, [1995], + , CH COOC H + H , 3, , (a) 4.33, (c) 6.33, , 2 5, , CH3COOH + C 2 H5OH is, (b) 5.33, (d) 7.33, , Topic 2: Relation between K, Q and G and, Factors Effecting Equilibrium, 22., , 23., , 24., , 25., , 26., , 27., , Hydrolysis of sucrose is given by the following, reaction., [2020], Glucose + Fructose, Sucrose + H2O , If the equilibrium constant (KC) is 2 × 1013 at, –, 300 K, th e value of DrG at the same, temperature will be :, (a) 8.314 J mol–1K–1 × 300 K × ln(2 × 1013), (b) 8.314 J mol–1K–1 × 300 K × ln(3 × 1013), (c) –8.314 J mol–1K–1 × 300 K × ln(4 × 1013), (d) –8.314 J mol–1K–1 × 300 K × ln(2 × 1013), Which one of the following conditions will favour, maximum formation of the product in the, reaction,, [2018], X 2 (g) D r H = – X kJ :, A2 (g) + B2 (g) , (a) Low temperature and high pressure, (b) Low temperature and low pressure, (c) High temperature and low pressure, (d) High temperature and high pressure, Consider the following liquid - vapour, equilibrium., [2016], Vapour, Liquid , Which of the following relations is correct ?, dlnP DH v, dlnG DH v, =, =, (a), (b), 2, 2, dT, RT, dT, RT, dlnP -DH v, dlnP -DH v, =, =, (c), (d), 2, 2, dT, RT 2, dT, T, Which of the following statements is correct for, a reversible process in a state of equilibrium ?, [2015], (a) DG = 2.30 RT log K, (b) DGº = –2.30 RT log K, (c) DGº = 2.30 RT log K, (d) DG = –2.30 RT log K, , For the reversible reaction,, [2014], , N2(g) + 3H2(g) 2NH3(g) + heat, The equilibrium shifts in forward direction, (a) By increasing the concentration of NH3(g), (b) By decreasing the pressure, (c) By decreasing concentration of N2(g) and, H2(g), (d) By increasing pressure and decreasing, temperature., For a given exothermic reaction, Kp and KP¢ are, the equilibrium constants at temperatures T1 and, T2, respectively. Assuming that heat of reaction, is constant in temperature range between T1 and, T2, it is readily observed that:, [2014], (a) Kp > K P¢, , (b) Kp < K P¢, , 1, K ¢p, The dissociation constants for acetic acid and, HCN at 25°C are 1.5 × 10–5 and 4.5 × 10–10, respectively. The equilibrium constant for the, equilibrium, [2009], HCN + CH COO–, CN– + CH3COOH , 3, would be:, (a) 3.0 × 10– 5, (b) 3.0 × 10– 4, 4, (c) 3.0 × 10, (d) 3.0 × 105, The reaction quotient (Q) for the reaction, 2NH (g), N2(g) + 3H2(g) , 3, (c) Kp = K P¢, , 28., , 29., , is given by Q =, , 30., , 31., , (d) Kp =, , [NH 3 ]2, , [N 2 ][H 2 ]3, , . The reaction will, , proceed from right to left if, [2003], (a) Q = 0, (b) Q = Kc, (c) Q < Kc, (d) Q > Kc, where Kc is the equilibrium constant, For the reaction, 2BaO(s) + O (g);, 2BaO2(s) , 2, DH = +ve. In equilibrium condition, pressure of, O2 is dependent on, [2002], (a) mass of BaO2, (b) mass of BaO, (c) temperature of equilibrium, (d) mass of BaO2 and BaO both, In a two-step exothermic reaction, , , , , A2(g) + B2(g) , D(g), 3C(g) , Step 1, , Step 2, , Steps 1 and 2 are favoured respectively by, [1997]
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EBD_8336, 74, , 32., , 33., , 34., , 35., , CHEMISTRY, (a) high pressure, high temperature and low, pressure, low temperature, (b) high pressure, low temperature and low, pressure, high temperature, (c) low pressure, high temperature and high, pressure, high temperature, (d) low pressure, low temperature and high, pressure, low temperature, The equilibrium constant for the reaction, 2A at 500 K and 700 K are 1 × 10–10, A2 , and 1 × 10–5 respectively. The given reaction is, [1996], (a) exothermic, (b) slow, (c) endothermic, (d) fast, Stan dar d Gibb’s free energy change for, isomerization reaction, [1995], trans-2-pentene, cis-2- pentene , is – 3.67 kJ/mol at 400 K. If more trans-2-pentene, is added to the reaction vessel, then, (a) more cis-2-pentene is formed, (b) equilibrium remains unaffected, (c) additional trans-2-pentene is formed, (d) equilibrium is shifted in forward direction, According to Le-chatelier’s principle, adding heat, liquid equilibrium will cause the, to a solid , (a) temperature to increase, [1993], (b) temperature to decrease, (c) amount of liquid to decrease, (d) amount of solid to decrease., Which one of the following information can be, obtained on the basis of Le Chatelier principle?, [1992], (a) Dissociation constant of a weak acid, (b) Entropy change in a reaction, (c) Equilibrium constant of a chemical reaction, (d) Shift in equilibrium position on changing, value of a constraint., , 38., , 39., , 40., , 41., , 42., , 43., , Topic 3: Theories of Acids and Bases, Ionic, Product of Water and pH Scale, 36., , 37., , The pH of 0.01 M NaOH (aq) solution will be, [NEET Odisha 2019], (a) 9, (b) 7.01, (c) 2, (d) 12, Which of the following cannot act both as, Bronsted acid and as Bronsted base?, [NEET Odisha 2019], (a) HSO4–, (b) HCO3–, (c) NH3, (d) HCl, , 44., , 45., , Conjugate base for Brönsted acids H2O and HF, are :, [2019], (a) OH– and H2F+, respectively, (b) H3O+ and F–, respectively, (c) OH– and F–, respectively, (d) H3O+ and H2F+, respectively, Following solutions were prepared by mixing, different volumes of NaOH and HCl of different, concentrations :, [2018], M, M, a. 60 mL, HCl + 40 mL, NaOH, 10, 10, M, M, HCl + 45 mL, NaOH, b. 55 mL, 10, 10, M, M, c. 75 mL, HCl + 25 mL, NaOH, 5, 5, M, M, HCl + 100 mL, NaOH, d. 100 mL, 10, 10, pH of which one of them will be equal to 1?, (a) b, (b) a, (c) c, (d) d, What is the pH of the resulting solution when, equal volumes of 0.1 M NaOH and 0.01 M HCl, are mixed ?, [2015 RS], (a) 12.65, (b) 2.0, (c) 7.0, (d) 1.04, Which of the following salts will give highest, pH in water ?, [2014], (a) KCl, (b) NaCl, (c) Na2CO3, (d) CuSO4, Which of these is least likely to act as Lewis, base?, [NEET 2013], (a) F–, (b) BF3, (c) PF3, (d) CO, Which of the following is least likely to behave, as Lewis base ?, [2011], (a) H2O, (b) NH3, (c) BF3, (d) OH–, Which one of the following molecular hydrides, acts as a Lewis acid?, [2010], (a) NH3, (b) H2O, (c) B2H6, (d) CH4, Which of the following molecules acts as a Lewis, acid ?, [2009], (a) (CH3)2 O, (b) (CH3)3 P, (c) (CH3)3 N, , (d) (CH3)3 B
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75, , Equilibrium, 46., , 47., , 48., , 49., , 50., , 51., , The ionization constant of ammonium hydroxide, is 1.77 × 10–5 at 298 K. Hydrolysis constant of, ammonium chloride is:, [2009], (a) 6.50 × 10– 12, (b) 5.65 × 10–13, (c) 5.65 × 10–12, (d) 5.65 × 10–10, Equal volumes of three acid solutions of pH 3, 4, and 5 are mixed in a vessel. What will be the H+, ion concentration in the mixture ?, [2008], (a) 1.11 × 10–4 M, (b) 3.7 × 10–4 M, (c) 3.7 × 10– 3 M, (d) 1.11× 10–3 M, Calculate the pOH of a solution at 25°C that, contains 1× 10– 10 M of hydronium ions, i.e., H3O+., [2007], (a) 4.000, (b) 9.0000, (c) 1.000, (d) 7.000, The hydrogen ion concentration of a 10–8 M, HCl aqueous solution at 298 K (Kw = 10–14) is, [2006], –8, –8, (a) 11 × 10 M, (b) 9.525 × 10 M, (c) 1.0 × 10–8 M, (d) 1.0 × 10–6 M, What is the correct relationship between the, pHs of isomolar solutions of sodium oxide, (pH 1 ), sodium sulph ide (pH 2 ), sodium, selenide (pH3) and sodium telluride (pH4)?, [2005], (a) pH1 > pH2 > pH3 > pH4, (b) pH1 > pH2 » pH3 > pH4, (c) pH1 < pH2 < pH3 < pH4, (d) pH1 < pH2 < pH3 » pH4, The rapid change of pH near the stoichiometric, point of an acid-base titration is the basis of, indicator detection. pH of the solution is related, to ratio of the concentrations of the conjugate, acid (HIn) and base (In–) forms of the indicator, by the expression, [2004], [ In - ], = pK In - pH, [ HIn ], [ HIn ], = pK In - pH, (b) log, [ In - ], [ HIn ], = pH - pK In, (c) log, [ In - ], [ In - ], = pH - pK In, (d) log, [ HIn ], Which one of the following orders of acid, strength is correct?, [2003], , (a), , 52., , 53., , 54., , 55., , 56., , 57., , 58., , log, , 59., , 60., , (a) RCOOH > HC º CH > HOH > ROH, (b) RCOOH > ROH > HOH > HC º CH, (c) RCOOH > HOH > ROH > HC º CH, (d) RCOOH > HOH > HC º CH > ROH, Which one of the following compounds is not a, protonic acid?, [2003], (a) SO2 (OH)2, (b) B (OH)3, (c) PO (OH)3, (d) SO (OH)2, In HS–, I–, RNH2 and NH3, order of proton, accepting tendency will be, [2001], (a) I– > NH3 > RNH2 > HS–, (b) HS– > RNH2 > NH3 > I–, (c) RNH2 > NH3 > HS– > I–, (d) NH3 > RNH2 > HS– > I–, A base when dissolved in water yields a solution, with a hydroxyl ion concentration of 0.05 mol litre–1., The solution is, [2000], (a) basic, (b) acidic, (c) neutral, (d) either 'b' or 'c', Conjugate acid of NH -2 is :, [2000], (b) NH3, (a) NH4+, (c) NH2, (d) NH, Which of the following statements about pH, and H+ ion concentration is incorrect? [2000], (a) Addition of one drop of concentrated HCl in, NH4OH solution decreases pH of the solution., (b) A solution of the mixture of one equivalent of, each of CH3COOH and NaOH has a pH of 7, (c) pH of pure neutral water is not zero, (d) A cold and concentrated H2SO4 has lower, H+ ion concentration than a dilute solution, of H2SO4, Among boron trifluoride, stannic chloride and, stannous chloride, Lewis acid is represented by, [1999], (a) only stannic chloride, (b) boron trifluoride and stannic chloride, (c) boron trifluoride and stannous chloride, (d) only boron trifluoride, What is the H+ ion concentration of a solution, prepared by dissolving 4 g of NaOH (Atomic, weight of Na = 23 amu) in 1000 mL?, [1999], (a) 10–10 M, (b) 10–4 M, (c) 10–1 M, (d) 10–13 M, The pH value of a 10 M solution of HCl is [1995], (a) less than 0, (b) equal to 0, (c) equal to 1, (d) equal to 2
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EBD_8336, 76, , CHEMISTRY, Topic 4: Ionisation of Weak Acids and, Bases and Relation between K a and K b, , 61., , 62., , 63., , 64., , 65., , 66., , At 100°C the Kw of water is 55 times its value at, 25°C. What will be the pH of neutral solution?, (log 55 = 1.74), [NEET Kar. 2013], (a) 6.13, (b) 7.00, (c) 7.87, (d) 5.13, Accumulation of lactic acid (HC3H5 O3 ), a, monobasic acid in tissues leads to pain and a, feeling of fatigue. In a 0.10 M aqueous solution,, lactic acid is 3.7% dissociated. The value of, dissociation constant, Ka, for this acid will be:, [NEET Kar. 2013], (a) 2.8 × 10–4, (b) 1.4 × 10–5, (c) 1.4 × 10–4, (d) 3.7 × 10–4, A weak acid, HA, has a Ka of 1.00 × 10–5. If, 0.100 mol of this acid is dissolved in one litre of, water, the percentage of acid dissociated at, equilbrium is closest to, [2007], (a) 1.00%, (b) 99.9%, (c) 0.100%, (d) 99.0%, At 25°C, the dissociation constant of a base,, BOH, is 1.0 ´ 10-12. The concentration of, hydroxyl ions in 0.01 M aqueous solution of, the base would be, [2005], (a) 1.0 ´ 10- 5 mol L-1 (b) 1.0 ´ 10-6 mol L-1, (c) 2.0 ´ 10-6 mol L-1 (d) 1.0 ´ 10-7 mol L-1, Ionisation constant of CH3COOH is 1.7 × 10–5., If concentration of H+ ions is 3.4 × 10–4M, then, find out initial concentration of CH3COOH, molecules, [2001], (a) 3.4 × 10–4M, (b) 3.4 × 10–3M, (c) 6.8 × 10–3M, (d) 6.8 × 10–4M, Aqueous solution of acetic acid contains, [1991], (a) CH3COO– and H+, (b) CH3COO–, H3O+ and CH3COOH, (c) CH3COO–, H3O+and H+, (d) CH3COOH, CH3COO– and H+, , Topic 5: Common Ion Effect, Salt Hydrolysis,, Buffer Solutions and Solubility Product, 67., , Find out the solubility of Ni(OH)2 in 0.1 M, NaOH. Given that the ionic product of Ni(OH)2, is 2 × 10–15, [2020], (a) 2 × 10–8 M, (b) 1 × 10–13 M, (c) 1 × 108 M, (d) 2 × 10–13 M, , 68., , 69., , 70., , 71., , 72., , 73., , The molar solubility of CaF2(Ksp = 5.3 × 10–11), in 0.1 M solution of NaF will be, [NEET Odisha 2019], –10, –1, (a) 5.3 × 10 mol L, (b) 5.3 × 10–11 mol L–1, (c) 5.3 × 10–8 mol L–1, (d) 5.3 × 10–9 mol L–1, pH of a saturated solution of Ca(OH)2 is 9. The, solubility product (Ksp) of Ca(OH)2 is: [2019], (a) 0.5 10–15, (b) 0.25 10–10, –15, (c) 0.125 10, (d) 0.5 10–10, The solubility of BaSO4 in water is, 2.42 × 10–3 gL–1 at 298 K. The value of its, solubility product (Ksp) will be, (Given molar mass of BaSO4 = 233 g mol–1), (a) 1.08 × 10–10 mol2L–2, [2018], –12, 2, –2, (b) 1.08 × 10 mol L, (c) 1.08 × 10–8 mol2L–2, (d) 1.08 × 10–14 mol2L–2, Concentration of the Ag+ ions in a saturated, solution of Ag2C2O4 is 2.2 × 10–4 mol L–1., Solubility product of Ag2C2O4 is :[2017], (a) 2.66 × 10–12, (b) 4.5 × 10–11, (c) 5.3 × 10–12, (d) 2.42 × 10–8, MY and NY3, two nearly insoluble salts, have, the same Ksp values of 6.2 × 10–13 at room, temperature. Which statement would be true in, regard to MY and NY3 ?, [2016], (a) The molar solubilities of MY and NY3 in, water are identical., (b) The molar solubility of MY in water is less, than that of NY3, (c) The salts MY and NY3 are more soluble in, 0.5 M KY than in pure water., (d) The addition of the salt of KY to solution, of MY and NY3 will have no effect on their, solubilities., Consider the nitration of benzene using mixed, conc of H2SO4 and HNO3. If a large amount of, KHSO4 is added to the mixture, the rate of, nitration will be, [2016], (a) faster, (b) slower, (c) unchanged, (d) doubled
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77, , Equilibrium, 74., , 75., , 76., , 77., , 78., , 79., , 80., , 81., , The Ksp of Ag2 CrO4, AgCl, AgBr and AgI, are respectively, 1.1 × 10 –12 , 1.8 × 10–10 ,, 5.0 × 10–13, 8.3 × 10–17. Which one of the, following salts will precipitate last if AgNO3, solution is added to the solution containing equal, moles of NaCl, NaBr, NaI and Na2CrO4?[2015], (a) AgCl, (b) AgBr, (c) Ag2CrO4, (d) AgI, Which one of the following pairs of solution is, not an acidic buffer ?, [2015 RS], (a) HClO4 and NaClO4, (b) CH3COOH and CH3 COONa, (c) H2CO3 and Na2CO3, (d) H3PO4 and Na3PO4, Using the Gibbs energy change, DG° = + 63.3kJ,, for the following reaction,, [2014], 2Ag+ (aq) + CO 2–(aq), Ag2CO3 , 3, the Ksp of Ag2CO3(s) in water at 25°C is:(R = 8.314 J K–1 mol–1), (a) 3.2 × 10–26, (b) 8.0 × 10–12, –3, (c) 2.9 × 10, (d) 7.9 × 10–2, Identify the correct order of solubility in aqueous, medium:, [NEET 2013], (a) ZnS > Na2S > CuS (b) Na2S > CuS > ZnS, (c) Na2S > ZnS > CuS (d) CuS > ZnS > Na2S, The values of Ksp of CaCO3 and CaC2O4 are, 4.7 × 10–9 and 1.3 × 10–9 respectively at 25°C. If, the mixture of these two is washed with water,, what is the concentration of Ca 2+ ions in water?, [NEET Kar. 2013], (a) 7.746 × 10–5 M, (b) 5.831 × 10–5 M, (c) 6.856 × 10–5 M, (d) 3.606 × 10–5 M, The dissociation constant of a weak acid is, 1 × 10– 4. In order to prepare a buffer solution, with a pH = 5 the [Salt]/[Acid] ratio should be, [NEET Kar. 2013], (a) 1 : 10, (b) 4 : 5, (c) 10 : 1, (d) 5 : 4, pH of a saturated solution of Ba(OH)2 is 12., The value of solubility product (Ksp) of Ba (OH)2, is :, [2012], (a) 3.3 × 10– 7, (b) 5.0 × 10–7, (c) 4.0 × 10–6, (d) 5.0 × 10–6, Equimolar solutions of the following substances, were prepared separately. Which one of these, will record the highest pH value ?, [2012], , 82., , 83., , 84., , 85., , (a) BaCl2, (b) AlCl3, (c) LiCl, (d) BeCl2, Buffer solutions have constant acidity and, alkalinity because, [2012], (a) these give unionised acid or base on, reaction with added acid or alkali., (b) acids and alkalies in these solutions are, shielded from attack by other ions., (c) they have large excess of H+ or OH– ions, (d) they have fixed value of pH, A buffer solution is prepared in which the, concentration of NH3 is 0.30 M and the, concentration of NH4+ is 0.20 M. If the equilibrium, constant, Kb for NH3 equals 1.8 × 10–5, what is, the pH of this solution ? (log 2.7 = 0.433)., [2011], (a) 9.08, (b) 9.43, (c) 11.72, (d) 8.73, In qualitative analysis, the metals of Group I can, be separated from other ions by precipitating, them as chloride salts. A solution initially, contains Ag+ and Pb2+ at a concentration of, 0.10 M. Aqueous HCl is added to this solution, until the Cl– concentration is 0.10 M. What will, the concentrations of Ag+ and Pb2+ be at, equilibrium?, (Ksp for AgCl = 1.8 × 10–10,, Ksp for PbCl2 = 1.7 × 10–5), [2011M], (a) [Ag+] = 1.8 × 10–7 M ; [Pb2+] = 1.7 × 10–6 M, (b) [Ag+] = 1.8 × 10–11 M ; [Pb2+] = 8.5 × 10–5 M, (c) [Ag+] = 1.8 × 10–9 M ; [Pb2+] = 1.7 × 10–3 M, (d) [Ag+] = 1.8 × 10–11 M ; [Pb2+] = 8.5 × 10–4 M, If pH of a saturated solution of Ba(OH)2 is 12,, the value of its K(sp) is :, [2010], (a) 4.00 × 10–6 M3, (c) 5.00 × 10–6 M3, , 86., , 87., , (b) 4.00 × 10–7 M3, (d) 5.00 × 10–7 M3, , What is [H+] in mol/L of a solution that is 0.20 M, in CH3COONa and 0.10 M in CH3COOH? Ka for, CH3COOH = 1.8 × 10-5 ., [2010], (a) 3.5 × 10–4, (b) 1.1 × 10–5, (c) 1.8 × 10–5, (d) 9.0 × 10–6, In a buffer solution containing equal, concentration of B– and HB, the Kb for B–, is 10–10. The pH of buffer solution is : [2010], (a) 10, (b) 7, (c) 6, (d) 4
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EBD_8336, 78, , 88., , 89., , 90., , 91., , 92., , 93., , 94., , 95., , CHEMISTRY, Equimolar solutions of the following were, prepared in water separately. Which one of the, solutions will record the highest pH ? [2008], (a) SrCl2, (b) BaCl2, (c) MgCl2, (d) CaCl2, Which of the following pairs constitutes a, buffer?, [2006], (a) NaOH and NaCl, (b) HNO3 and NH4NO3, (c) HCl and KCl, (d) HNO2 and NaNO2, H2S gas when passed through a solution of, cations containing HCl, precipitates the, cati ons of secon d group of qualitative, analysis but not those belonging to the fourth, group. It is because, [2005], (a) presence of HCl decreases the sulphide, ion concentration., (b) solubility product of group II sulphides, is more than that of group IV sulphides., (c) presence of HCl increases the sulphide, ion concentration., (d) sulphides of group IV cations are, unstable in HCl., The solubility product of a sparingly soluble, salt AX2 is 3.2 × 10–11. Its solubility ( in moles/, litre) is, [2004], –6, –4, (a) 5.6 × 10, (b) 3.1 × 10, (c) 2 × 10–4, (d) 4 × 10–4, The solubility product of AgI at 25ºC is, 1.0 × 10–16 mol 2 L–2 . The solubiliy of AgI in, 10–4 N solution of KI at 25ºC is approximately, (in mol L–1 ), [2003], (a) 1.0 × 10–8, (b) 1.0 × 10–16, (c) 1.0 × 10–12, (d) 1.0 × 10–10, Solubility of MX2-type eletrolytes is 0.5 × 10–4, mol/lit, then find out Ksp of electrolytes [2002], (a) 5 × 10–12, (b) 25 × 10–10, –13, (c) 1 × 10, (d) 5 × 10–13, Which has the highest value of pH? [2002], (a) CH3COOK, (b) Na2CO3, (c) NH4Cl, (d) NaNO3, Solution of 0.1 N NH4OH and 0.1 N NH4Cl has, pH 9.25. Then find out pKb of NH4OH [2002], (a) 9.25, (b) 4.75, (c) 3.75, (d) 8.25, , Solubility of a M2S salt is 3.5 × 10–6, then its, solubility product will be, [2001], (a) 1.7 × 10–16, (b) 1.7 × 10–6, (c) 1.7 × 10–18, (d) 1.7 × 10–12, 97. The solubility product of a sparingly soluble, salt BA2 is 4 × 10–12. The solubility of BA2 is, [1999], (a) 4 × 10–4, (b) 4 × 10–12, (c) 4 × 10–3, (d) 1 × 10–4, 98. The solubility products of CuS, Ag2S and HgS, are 10 –31 , 10 –44 , 10 –54 respectively. The, solubilities of these sulphides are in the order, [1997], (a) Ag2S > HgS > CuS (b) Ag2S > CuS > HgS, (c) HgS > Ag2S > CuS (d) CuS > Ag2S > HgS, 99. A physician wishes to prepare a buffer solution, of pH = 3.58 that efficiently resists a change in, pH yet contains only small concentrations of, the buffering agents. Which one of the following, weak acids together with its sodium salt would, be the best to use ?, [1997], (a) m-chlorobenzoic acid (pKa = 3.98), (b) p-chlorocinnamic acid (pKa = 4.41), (c) 2, 5-dihydroxy benzoic acid (pKa = 2.97), (d) Acetoacetic acid (pKa = 3.58), 100. The pH value of blood does not appreciably, change by a small addition of an acid or a base,, because the blood, [1995], (a) is a body fluid, (b) can be easily coagulated, (c) contains iron as a part of the molecule, (d) contains serum protein which acts as buffer, 101. Which of the following is most soluble ? [1994], (a) Bi2 S3 (Ksp = 1 × 10–17), (b) MnS (Ksp = 7 × 10–16), (c) CuS (Ksp = 8 × 10–37), (d) Ag2 S (Ksp = 6 × 10–51)., 102. 0.1 M solution of which one of these substances, will be basic ?, [1992], (a) Sodium borate, (b) Ammonium chloride, (c) Calcium nitrate, (d) Sodium sulphate., 96.
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79, , Equilibrium, 103. In which of the following solvents, AgBr will, have the highest solubility ?, [1992], , 10-3 M NaBr, (c) Pure water, , (b) 10 -3 M NH 4 OH, (d) 10 -3 M HBr, , (a), , 104. The compound whose aqueous solution has the, highest pH is, [1988], (a) NaCl, (b) NaHCO3, (c) Na2CO3, (d) NH4Cl., , ANSWER KEY, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, , (a), (d), (b), (a), (c), (c), (c), (d), (c), (c), (b), , 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, , (c), (b), (d), (a), (a), (a), (b), (d), (b), (d), (d), , 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, , (a), (d), (b), (d), (a), (c), (d), (c), (d), (b), (a), , 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, , (d), (d), (d), (d), (c), (c), (a), (c), (b), (c), (c), , 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, , (d), (d), (b), (a), (a), (a), (d), (c), (b), (c), (a), , 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, , (b), (b), (c), (d), (a), (a), (c), (a), (d), (c), (b), , 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, , (d), (d), (a), (a), (c), (b), (b), (c), (a), (b), (c), , 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, , (a), (c), (b), (a), (a), (b), (c), (d), (d), (d), (b), , 89 (d) 100 (d), 90 (a) 101 (a), 91 (c) 102 (a), 92 (c) 103 (b), 93 (d) 104 (c), 94 (b), 95 (b), 96 (a), 97 (d), 98 (b), 99 (d), , Hints & Solutions, 1., , (a), , Kp = PCO = 1.6 atm = maximum pressure of CO2, 2, volume of container at this stage., , [ NH3 ], [ N 2 ][ H 2 ]3, 2, , 2NH3 ; K1 =, (i) N 2 + 3H 2 , 2NO; K 2 =, (ii) N 2 + O 2 , , [ NO ]2, , [ N 2 ][ O 2 ], 1, [ H 2O ], ® H 2 O; K 3 =, (iii) H 2 + O 2 ¾¾, 2, [ H 2 ][ O 2 ]1/2, Applying (II + 3 × III – I) we will get, 5, K, , , 2NH3 + O 2 , 2NO + 3H 2O;, 2, , 2, 3, NH3 ], H 2O ], [, [, K=, [ N2 ][ O 2 ] [ H 2 ]3 ´ [ O2 ]3 / 2 [ N 2 ][ H 2 ]3, , [ NO]2, , 2., , 3., , ´, , \ K = K2 × K33 / K1, (d) Max. pressure of CO2 = Pressure of CO2 at, equilibrium, For reaction,, SrO(s) + CO2(g), SrCO3(s) , , nRT, …(i), P, Since container is sealed and reaction was not, earlier at equilibrium., \ n = constant., PV 0.4 ´ 20, =, n=, …(ii), RT, RT, Put equation (ii) in equation (i), é 0.4 ´ 20 ù RT, = 5L, V= ê, ë RT úû 1.6, (b) Equilibrium constant for reaction:, [Product], K = 1.6 × 1012 =, [Reactant], The value of K is very high so the system will, contain mostly products at equilibrium., 2NO(g), (a) N2(g) + O2(g) , , V=, , 4., , K=, , [NO]2, [N 2 ][O2 ]
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EBD_8336, 80, , CHEMISTRY, 9., , 1, 1, NO, N 2 + O 2 , 2, 2, [NO], K¢ =, = K1 2, 12, [N 2 ] [O 2 ]1 2, , 5., , 6., , 2SO3 K = 278 (given), 2SO2 + O2 , æ 1ö, SO2 + 1 O2 K' =, SO3 , ç K÷, 2, è, ø, 1, =, = 6 × 10–2, 278, , (c), , 7., , Kc =, , 10., , 2, , (2.8 ´ 10-3 )2, -3, , -3, , =, , (2.8)2, = 0.62, 3 ´ 4.2, , [NO 2 ], , 11., , 1, 1, ´, K1 K 2, 1, , 1, , [N 2 ] 2 [O 2 ] 2 [NO][O2 ] 2 [N 2 ] 2 [O2 ], K=, =, ´, [NO], [NO2 ], [NO2 ], 2C(s)+O2 (g) 2 CO 2 (g), , If Dn = 0, then (RT)Dn = 1, Þ Kc = Kp, , (b) For the reaction, 2 AB2 (g) 2 AB(g) + B2 (g), initially 2, 0, 0, 2x, x, at equi 2(1- x), , [N 2, 2], K=, [NO2 ], , Dn = 2 – 1 = + 1, \ Kc and Kp are not equal., , [\ [solid]=1]., , is increased by the factor of (n)x, where x is the, stoichiometric coeficient., , 1, ] 2 [O, , (d), , 3, , 1, times then the concentration of another product, n, , [NO]2 [O 2 ], For the reaction, 1 N 2 (g) + O2 (g), NO2 (g) , 2, , 8., , (c) For this reaction Keq. is given by, , To maintain the equilibrium constant if, concentration of one of the product decreases by, , 2, , 1, , (0.50) 2 (0.75), , 1, times then for given, 4, reaction, equilibrium constant to remain, constant, we have to increase the concentration, of [Fe3+] by a factor of 43 i.e. 4× 4 × 4 = 64. Thus,, option (c) is correct answer., , [NO]2, [N 2 ][O 2 ], For the reaction, 2NO2 (g), 2NO(g) + O2 (g) , , 1, , (0.75)3 (0.25), , If (OH–) is decreased by, , K1 =, , Hence, K =, , [ A]2 [ B ], , =, , = (Fe3+) (OH–)3, , 3 ´ 10 ´ 4.2 ´ 10, (c) For the reaction, 2NO(g), N 2 (g) + O 2 (g) , , K2 =, , [C ]3 [ D ], , éFe3+ ù éOH - ù, ûë, û, K=ë, [ Fe(OH)3 ], , 2AB ; K c = [ AB], (c) A2 + B2 , [ A2 ][ B2 ], , Kc =, , (c), 2A(g) + B(g) 3C(g) + D(g), Mole ratio, 2, 1, 3, 1, Molar concentration 1, 1, 0, 0, at t = 0, Molar, 0.50 0.75, 0.75 0.25, concentration at equilibrium, , The partial pressure at equilibrium are calculated, on the basis of total number of moles at, equilibrium., Total number of moles, = 2 (1–x) + 2x + x = (2 + x), 2(1 - x), \ PAB =, ´ P where P is the total, 2, (2 + x ), pressure., 2x, x, ´ P , PB =, PAB =, ´P, 2, (2 + x ), (2 + x ), Since x is very small so can be neglected in, denominator, Thus, we get, PAB = x × P, PAB = (1 – x) × P, 2
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EBD_8336, 82, , CHEMISTRY, Now operate,, =, =, , [N ] [H 2 ]3, [NO]2, . 2, ´, [N 2 ] [O2 ] [H 2 ]3 [O2 ]3/ 2, [NH3 ]2, , [NO]2 [H 2 O]3, [NH3 ]2 [O2 ]5 / 2, , \ K=, , 15., , 16., , K 2 . K33, K1, [H 2O]3, , K 2 . K 33, K1, , (a) First option is incorrect as the value of KP, given is wrong. It should have been, PCO2, KP =, PCH4 ´ [PO2 ]2, (a) For reaction (i), , [NO]2, [N 2 ][O2 ], and for reaction (ii), , 21., , 22., , 23., , (a) MgCO3 (s) ® MgO(s) + CO 2 (g), MgO & MgCO3 are solid and they do not exert, any pressure and hence only pressure exerted, is by CO2. Therefore KP = PCO2, (b) In polyprotic acids the loss of second, proton occurs much less readily than the first., Usually the Ka values for successive loss of, protons from these acids differ by at least a factor, , 24., , of 10–3 i.e., K a1 > K a2, , ( 1), H+ + X 2- ( K a ), HX- , 2, H+ + HX- K a, H 2X , , 19., , (d) For the reaction, XeOF (g) + 2HF(g), XeF6(g) + H2O(g) , 4, , [XeOF4 ][HF]2, ....(a), [XeF6 ][H 2 O], and for the reaction, XeOF (g) + HeO F (g), XeO4(g) + XeF6(g) , 4, 3 2, [XeOF4 ][XeO 3 F2 ], K2 =, ....(b), [XeO 4 ][XeF6 ], For reaction :, , 25., 26., , K1 =, , XeO 4 (g) + 2HF(g) ® XeO3F2(g) + H 2 O(g), , [XeO3F2 ][H 2O], , [XeO4 ][HF]2, \ From eq. no. (a) and (b), K = K 2 / K1, (b) According to equation, 2HI, At t = 0 (2 moles), At equilibrium (2 – 2a) moles, , 1, [N 2 ]½ [O2 ]½, therefore K1 = 2, [NO], K2, , K2 =, , 18., , 20., , =K, , K1 =, , 17., , K=, , 27., , +, , 0, a mole, , I2, 0, a mole, , Total moles at equilibrium = 2 – 2a + a + a = 2, mole, (d) Rate constant of forward reaction (Kf ), = 1.1 × 10–2 and rate constant of backward, reaction (Kb) = 1.5 × 10–3 per minute. Equilibrium, constant (Kc), K f 1.1 ´ 10 -2, =, =, = 7.33, K b 1.5 ´ 10 -3, (d) DG = DG° + RT ln Q, At equilibrium DG = 0, Q = Keq, So, DrG° = –RT ln Keq, DrG° = –8.314 J mol–1 K–1 × 300 K × ln(2 × 1013), X (g); DH = –X kJ, (a) A2(g) + B2(g) , 2, On increasing pressure equilibrium shifts in a, direction where number of moles decreases i.e., forward direction., On decreasing temperature, equilibrium shifts, in exothermic direction i.e., forward direction., So, high pressure and low temperature favours, maximum formation of product., (d) Clausius – Clapeyron's equation, d ln P DH v, =, dT, RT 2, (b) DG° = –2.30RT log K, because at equilibrium DG = 0, (d) Given reaction is exothermic reaction., Hence according to Le-Chatelier's principle low, temperature favours the forward reaction and, on increasing pressure equilibrium will shift,, towards lesser number of moles i.e. forward, direction., (a) In exothermic reactions on increasing, temperature value of Kp decreases, \ Kp > Kp¢, , 28., , H2, , (Assuming T1 < T2), , (c) Given, CH3COOH CH3COO– + H+ ;, K a1 , = 1.5 × 10– 5, ....(i)
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83, , Equilibrium, H++ CN–; K = 4.5 × 10–10, HCN , a2, +, –, , or H + CN HCN;, 1, 1, K a' 2 =, =, ...(ii), Ka2 4.5 ´ 10 –10, \ From (i) and (ii), we find that the equilibrium, constant (Ka) for the reaction,, CH COO– + HCN, is, CN– + CH3COOH , 3, K a = K a1 ´ K a' 2, , 29., , 1.5×10 –5, , èrate, , –10, , ø, , èrate, , 35., , 36., , 1, ´ 105 = 3.33 ´ 10 4, 3, 4.5×10, (d) For reaction to proceed from right to left, Q, > Kc, æbackward ö æforward ö i.e the reaction will be fast, , =, , 34., , =, , 37., 38., , ø, , in backward direction i.e rb > rf ., When Q = Kc, the system is at equilibrium, Q < Kc, reaction shift to the right from left., Q > Kc, reaction proceed from right to left., , 30., , 31., , (c) For the reaction, BaO(s) + O2 (g); DH = +ve., BaO 2 (s) , At equilibrium K p = PO2, [For solid and liquids concentration term is taken, as unity], Hence, the value of equilibrium constant, depends only upon partial pressure of O2 ., Further on increasing temperature, formation of, O2 increases as this is an endothermic reaction., , , , , (d) A2(g)+ B2(g) , D(g), 3C(g) , , 32., , since the steps 1 and 2 are exothermic hence, low temprature will favour both the reactions. In, step - 1, moles are increasing hence low pressure, will favour it. In step 2, moles are decreasing,, hence high pressure will favour it., 2A Equilibrium constant is given by, (b) A2 , , Step 1, , Kc =, , 33., , 39., , Since the value given is very small, hence conc., of products is less. It means the reaction is slow., (a) If more trans-2-pentene is added, then its, concentration in right hand side will increase., But in order to maintain the K constant,, concentration of cis-2-pentene will also increase., Therefore more cis-2-pentene will be formed., , (c) Meq. of HCl = 75 ´, , 1, ´ 1 = 15, 5, , 1, ´ 1= 5, 5, Meq. of HCl in resulting solution = 10, 10, 1, Molarity of [H+] in resulting mixture =, =, 100 10, é1ù, +, pH = –log[H ] = –log ê ú = 1.0, ë 10 û, (a) g eq of NaOH = 0.1 × V = 0.1V, g eq of HCl = 0.01 × V = 0.01V, g eq of NaOH > g eq. HCl, hence, resultant solution should be basic, hence, from the eqn, M1V1 – M2V2 = MV, 0.1V – 0.01V = MV, , Meq. of NaOH = 25 ´, , 40., , Step 2, , [ A]2, [ A2 ], , Liquid, (d) Solid , It is an endothermic process. So when, temperature is raised, more liquid is formed., Hence adding heat will shift the equilbrium in, the forward direction., (d) According to Le-chatelier's principle, whenever a constraint is applied to a system in, equilibrium, the system tends to readjust so as, to nullify the effect of the constraint., (d) [OH–] = 0.01 M = 10–2 M, pOH = –log[OH–] = –log(10–2) = 2, pH = 14 – pOH = 12, (d) HCl cannot accept H+ therefore cannot act, as Bronsted base., (c) When a proton is removed from an acid,, we obtain its conjugate base., H + + OH H 2 O , H + + F HF , , 0.09, = 0.045 = 4.5 × 10–2, 2, Now, pOH = – log [OH–], = – log 4.5 × 10–2 = 1.34, Q pH + pOH = 14, \ pH = 14 – 1.34 = 12.65, (c) Na2CO3 is a salt of strong base (NaOH), and weak acid (H2CO3). On hydrolysis this salt, will produce strongly basic solution. i.e. pH will, be highest (pH > 7) for this sotluion. Others are, combination of, , M=, , 41.
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EBD_8336, 84, , 42., 43., 44., 45., 46., , CHEMISTRY, KCl = Strong acid + Strong base, ® neutral solution (pH » 7), NaCl = Strong acid + Strong base, ® neutral solution (pH » 7), CuSO4 = Strong acid + wake base, ® Acidic solution (pH < 7), (b) BF3 acts as Lewis acid., (c) BF3 behaves as lewis acid as it is an, electron deficient species., (c) Boron in B2H6 is electron deficient., (d) (CH3)3 B – is an electron deficient, thus, behave as a lewis acid., (d) Ammonium chloride is a salt of weak base, and strong acid. In this case, hydrolysis, constant, Kh can be calculated as, , 50., , 51., , \ K In =, , (b) [H3O]+ for a solution having pH = 3 is given, by, [H3O]+ = 1×10–3 moles/litre [\ [H3O]+ = 10–pH], Similarly for solution having pH = 4,, [H3O]+ = 1 × 10–4 moles/ litre and for pH=5, [H3O+] = 1×10–5 moles/ litre, Let the volume of each solution in mixture be 1L,, then total volume of mixture solution, = (1 + 1 + 1) L =3L, Total [H3O]+ ion present in mixture solution, = (10–3 + 10–4 + 10–5) moles, Then [H3 O]+ ion concentration of mixture, solution, -4, , -5, , 10 + 10 + 10, 0.00111, M=, M, 3, 3, = 0.00037 M = 3.7 ×10–4 M., (a) Given [H3O+] = 1 × 10–10 M, at 25ºC [H3O+] [OH–] = 10–14, =, , 48., , -3, , \ [OH - ] =, , 10 -14, 10 -10, , 49., , \ pOH = 4, (a) For a solution of 10–8 M HCl, [H+] = 10–8, [H+] of water = 10–7, Total [H+] = 10–7 + 10–8 = 10 × 10–8 + 10–8, 10–8 (10 + 1) = 11 × 10–8, , [HIn], [In - ], , Taking negative on both sides, [HIn], - log[H + ] = - log K In - log [In ], [In - ], [HIn ], , or we can write pH = pK In + log, or log, 52., , [In - ], = pH - pK In, [HIn], , (c) The higher is the tendency to donate, proton, stronger is the acid. Thus, the correct, order is R – COOH > HOH > R – OH > CH º CH, depending upon the rate of donation of proton., The stability order of conjugate base is, , RCOOQ > H - OQ > R – OQ > HC º CQ, , 53., , (b) B(OH)3 does not provide H+ ions in water, instead it accepts OH– ion and hence it is Lewis, acid, , 54., , [B(OH) ]- + H +, B(OH )3 + H 2O , 4, (c) Strong base has higher tendency to accept, the proton. Increasing order of base and hence, the order of accepting tendency of proton is, , 55., , I- < HS- < NH3 < RNH 2, (a) Given : Hydroxyl ion concentration, [OH–] = 0.05 mol L–1. We know that the, , = 10-4, , Now, [OH - ] = 10 - pOH = 10–4, , [HIn], [H + ][In - ], or [H + ] = K In ´, [HIn], [In - ], , or log H + = log K In + log, , K, 1 ´ 10 -14, Kh = w =, = 5.65 ´ 10 -10, K b 1.77 ´ 10 -5, , 47., , (a) The solution formed from isomolar, solutions of sodium oxide, sodium sulphide,, sodium selenide and sodium telluride are H2O,, H2S, H2Se & H2Te respectively. As the acidic, strengh increases from H2O to H2Te thus pH, decreases and hence the correct order of pHs is, pH1 > pH2 > pH3 > pH4., (d) For an acid-base indicator, H+ + In–, HIn , , [H + ][OH - ] = 1´10 -14, , or [H + ] = 1´ 10, , -14, , 0.05, , –1, , = 2 ´ 10 -13 mol L
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85, , Equilibrium, , 56., , Since [OH–] > [H+], Therefore the solution is basic., (b) Because NH3 after losing a proton (H+), gives NH2–, NH 2– + H 3O +, NH + H O , 3, , 57., , 58., , On taking log on both side, – log [H+] = –log (55 × 10–14)1/2, 1, ( - log 55 + 14log10), 2, pH = 6.13, , pH =, , 2, , (Conjugate acid-base pair differ only by a proton), (b) CH3COOH is weak acid while NaOH is, strong base, so one equivalent of NaOH can, not be neutralized with one equivalent of, CH3COOH. Hence the solution of one equivalent, of each does not have pH value as 7. Its pH will be, towards basic side as NaOH is a strong base, hence conc. of OH– will be more than the conc., of H+., (c) Because of electron deficiency, boron, trifluoride and stannous chloride are lewis acid., , Calculation of pOH in this question: value of pH, and pOH must be same for a neutral solution., Thus, pOH = 6.13, also pH + pOH = –log (55 × 10–14), , 62., , 4, = 0.1, 40, [Molecular weight of NaOH = 40], No. of moles of OH– = 0.1, , 63., , 64., , 60., , 61., , -13, , –, , 65., , 1 ´ 10 –5, = 10-2 = 1%, 0.100, , (d) Given Kb = 1.0 × 10–12, [BOH] = 0.01 M, , =, , –1, , \ [H ] = 10, (Q[OH ] = 10 ), (a) Molarity (M) = 10M. HCl is a strong acid, and it is completely dissociated in aqueous, solutions as : HCl (10), H+(10) + Cl–., So, for every moles of HCl, there is one H+., Therefore [H+] = [HCl] or [H+] = 10., pH = – log[H+] = – log [10] = – 1., (a) Kw at 25°C = 1 × 10–14, At 25ºC, Kw = [H+] [OH–] = 10–14, At 100°C (given), Kw = [H+] [OH–] = 55 × 10–14, Q for a neutral solution, [H+] = [OH–], \ [H+]2 = 55 × 10–14, or [H+] = (55 × 10–14)1/2, Q pH = – log [H+], , Ka, C, , a=, , 0.1, = 0.1 mol / L, 1 litre, , As we know that, [H + ] [OH - ] = 10 -14, +, , ; 1.4 × 10–4, (a) Given Ka = 1.00×10–5, C= 0.100 mol, for a weak electrolyte,, degree of dissociation, a=, , (d) No. of moles of NaOH =, , Concentration of OH– =, , Ka, ;, 0.10, , K a = (0.037) 2 ´ 0.10 = 1.37 ×10–4, , Vacant d-orbitals are available with stannic, chloride, hence it can also accept pair of electrons, and act as a lewis acid., , 59., , Ka, = 0.037 =, c, , (c) a =, , [OH] = ?, , Kb / c, , 1 ´ 10-12, = 1.0 × 10–5, 0.01, , Now [OH–] = c.a = 0.01 × 10–5, = 1 × 10–7mol L–1, CH COO– + H+, (c) CH3COOH , 3, [CH3COO- ][H+ ], [CH 3COOH], Given that,, Ka =, , [CH 3 COO - ] = [H + ] = 3.4 ´ 10 -4 M, Ka for CH3COOH = 1.7 × 10–5, CH3COOH is weak acid, so in it [CH3COOH] is, equal to initial concentration. Hence, , 1.7 ´10 -5 =, , (3.4 ´ 10-4 )(3.4 ´10 -4 ), [CH 3COOH], , [ CH 3 COOH ] =, , 3.4 ´ 10 -4 ´ 3.4 ´ 10 -4, , 1.7 ´ 10 -5, = 6.8 × 10–3M
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EBD_8336, 86, , 66., , CHEMISTRY, , 67., , (b) An aqueous solution of acetic acid, dissociates as, CH COO– + H O+, CH3COOH + H2O , 3, 3, Ni2+ + 2OH–, (d) Ni(OH)2 , , 68., , NaOH ¾¾, ® Na + OH–, Total [OH–] = 2s + 0.1 » 0.1, Ionic product = [Ni]2+[OH]2, 2 × 10–15 = s(0.1)2, s = 2 × 10–13, Solubility of Ni(OH)2 = 2 × 10–13 M, Ca2+ + 2F–, (d) CaF2 , , s, , s, , t=0, , 0, , 0, , s, , 2s, , s = 6.2´ 10- 13, s = 7.87 × 10–7 mol L–1, | N3+ + 3Y –, NY3 , Ksp = s × (3s)3 = 27s4 = 6.2 × 10–13, 1/4, , 73., , NaF ¾¾, ® Na+ + F–, 0.1, , 0.1, , 0.1, , Due to common ion effect of NaF, solubility of, CaF2 is further supressed. Therefore, the, concentration of F– will be mainly due to NaF., Ksp = [Ca2+] [F–]2, Ksp = (s) (0.1 + 2s) 2, 0.1 + 2s » 0.1, –11, Ksp, 5.3 ´ 10, =, s=, = 5.3 × 10–9 mol L–1, 2, (0.1), (0.1)2, 69., , 70., , 74., , [Ag+] =, , [Ag + ] =, , [Cl – ], , 5.3 ´ 10 –13, [Br – ], , AgI, Ksp = [Ag+] [I–] = 8.3 × 10–17, , [Ag + ] =, , s, , -4, éë C2 O24 - ùû = 2.2 ´ 10 M = 1.1 ´ 10 -4 M, 2, , 1.8 ´ 10 –10, , AgBr, Ksp = [Ag+] [Br–] = 5.0 × 10–13, , 2.42 ´ 10-3, = 1.038 ´ 10-5 mol L-1, 233, Ksp = s2 = (1.038 × 10–5)2 = 1.08 × 10–10 mol2 L–2, (c), 2Ag + ( aq ) + C 2O 42- ( aq ), Ag 2C 2 O 4 (s) , , Ksp = [Ag+]2 [C2O42–], [Ag+] = 2.2 × 10–4 M, Given that:, \ Concentration of C2O42– ions,, , 1.1 ´ 10 –12, éCrO4–2 ù, ë, û, , AgCl, Ksp = [Ag+] [Cl–] = 1.8 × 10–10, , s=, , 2s, , æ 6.2 ´ 10-13 ö, s= ç, ÷, 27, è, ø, –4, s = 3.89 × 10 mol L–1, \ molar solubility of NY3 is more than MY in, water., (b) The presence of large amount of KHSO4, will decrease ionisation of H2SO4 that result in, lesser ionisation of nitric acid and lesser, formation of nitronium ion [NO2+]. Hence the, rate of nitration will be slower., (c) Ag2CrO4, Ksp = [Ag+]2 é CrO 4–2 ù = 1.1 × 10–12, ë, û, , [Ag + ] =, , Ca 2 + + 2OH (a) Ca(OH) 2 , pH = 9, pOH = 14 – 9 = 5, [OH–] = 10–5, -5, é Ca 2 + ù = 10, ë, û, 2, æ 10-5 ö, -5 2, Ksp = [Ca2+] [OH–]2 = ç 2 ÷ ´ (10 ), è, ø, = 0.5 × 10–15, (a) Solubility of BaSO4 = 2.42 × 10–3 gL–1, , \, , 71., , 72., , 2s, , At eqm., , Ksp = (2.2 × 10–4)2 (1.1 × 10–4), = 5.324 × 10–12, | M+ + Y–, (b) MY , 2, Ksp = s = 6.2 × 10–13, \, , 75., , 8.3 ´ 10–17, [I – ], , –2, –, –, –, If we take éë CrO 4 ùû = [Cl ] = [Br ] = [I ] = 1, then maximum [Ag+] will be required in case of, Ag2CrO4., (a) Among the given acids, HClO4 is a very, strong acid, completely dissociates, \ [HA] ® 0 in this case, hence cannot be used, for acidic buffer.
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87, , Equilibrium, 76., , (b) DG = – 2.303 log K, , Ba(OH) 2, , here K = [Ag+]2 [CO2–, 3 ] = Ksp, +, 3, \ 63.3 × 10 = – 2.303 × 8.314 × 298 log Ksp, +3, , 63.3 ´ 10, = - 11.09, 5705.8, \ Ksp = Antilog (– 11.09) = 8 × 10– 12, (c) Solubility of alkali metal is maximum among, the following. Among ZnS (1.7 × 10–5) & CuS, (8 × 10–37) ZnS has higher value of Ksp., , [OH–] = 10–2, 2s = 10–2, , \ log Ksp = 77., , 78., , 10-2, = 5 ´ 10-3 m , Ksp = s . (2s)2, 2, Þ Ksp = 5 × 10–7, s=, , 81., , (a) CaCO3 ¾® Ca 2+ + CO32x, , x, , CaC2O4 ¾® Ca 2+ + C 2 O 24y, , 82., , y, , Lets take an example of an acidic buffer, CH3COOH and CH3COONa., , Now, Ksp (CaCO3) = [Ca2+] [CO32-], , CH3COOH, , or, , 1.3 × 10 –9 = (x + y) y, , On solving, we get, [Ca2+] = 7.746 × 10–5 M, (c) Given, Ka = 1 × 10– 4, \ pKa = – log (1× 10– 4) = 4, Now from Handerson equation, , [Salt], [Acid], Putting the values, , pH = pKa + log, , 5 = 4 + log, , 83., , [Salt], [Acid], , [Salt], = 5–4=1, [Acid], Taking antilog, [Salt]/[Acid] = 10 = 10 : 1, (b) Given pH = 12, or [H+] = 10–12, Since, [H+] [OH–] = 10–14, [OH–] =, , 10-14, 10, , -12, , = 10–2, , (b) Given : [NH3] = 0.3 M, [NH4+] = 0.2 M,, Kb = 1.8 × 10–5 ., [salt], pOH = pKb + log, [base], [pKb = –log Kb; pKb = –log 1.8 × 10–5], \ pKb = 4.74, 0.2, 0.3, = 4.74 + 0.3010 – 0.4771 = 4.56, pH = 14 – 4.56 = 9.436, (c) Ksp = [Ag+] [Cl–], , pOH = 4.74 + log, , log, , \, , CH3COO– + H+ ;, , CH3COONa, CH3COO– + Na+, when few drops of HCl are added to this buffer,, the H + of HC l immediatly combine with, CH3COO– ions to form undissociated acetic acid, molecules. Thus there will be no H + ions to, combine with CH 3 COO – ions to form, undissociated acetic acid molecules. Thus there, will be no appreciable change in its pH value., Likewise if few drops of NaOH are added, the, OH – ions will combine with H + ions to form, unionised water molecule. Thus pH of solution, will remain constant., , 4.7 × 10 –9 = (x + y) x, , similarly, Ksp (CaC2O4) = [Ca2+] [C2O42–], , 80., , (a) (AlCl3, LiCl & BeCl2) ) all these solutions, are acidic due to cationic hydrolysis, whereas, BaCl2, is salt of strong base (Ba(OH)2) and strong, acid (HCl), hence it will have maximum pH., (a) Constant acidity and alkalinity of buffer, solution is due to the unionised acid or base., , \ [Ca2+] = x + y, , or, , 79., , Ba 2+ + 2OH, s, 2s, , 84., , 1.8 × 10–10 = [Ag+] [0.1], [Ag+] = 1.8 × 10–9 M, Ksp = [Pb+2] [Cl–]2, 1.7 × 10–5 = [Pb+2] [0.1]2, [Pb+2] = 1.7 × 10–3 M
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EBD_8336, 88, , 85., , CHEMISTRY, (d) Ba (OH)2 (s) ¾ ¾, ® Ba 2+ (aq) + 2 OH - (aq), pH = 12 or pOH = 2, , 91., , [OH - ] = 10-2 M, , Ba(OH) 2 ¾¾, + 2 OH–, ® Ba 2 +, 10-2, 0.5× 10-2, [\ Concentration of Ba 2 + is half of OH - ], , 92., , = [0.5 × 10-2 ] [1 × 10-2 ]2, = 0.5 × 10 -6 = 5 × 10-7 M3, (d), , é Salt ù, pH = p Ka+ log ê, ë Acid úû, , log é H = log Ka – log é Salt ù, ë û, êë Acid úû, +, é Acid ù, log éë H ùû = log Ka + log ê, ë Salt úû, , or, , +ù, , 93., , é H + ù = K é Acid ù, aê, ë û, ë Salt úû, 0.1, = 9 × 10-6 M, 0.2, (d) Kb = 10-10 ; Ka = 10-4 or pKa = 4, For the buffer solution containing equal, concentration of B– and HB, pH = pKa + log 1, , = 1.8 × 10 -5 ´, , 87., , 88., , 89., 90., , pH = pKa = 4, (b) The highest pH will be recorded by the, most basic solution. The basic nature of, hydroxides of alkaline earth metals increase as, we move from Mg to Ba and thus the solution of, BaCl2 in water will be most basic and so it will, have highest pH., (d) HNO2 is a weak acid and NaNO2 is salt of, that weak acid and strong base (NaOH)., (a) IV th g r oup n eeds h i gh er S 2– i on, concentration., In presence of HCl, the dissociation of H 2 S, decreases hence produces less amount of sulphide, ions due to common ion effect, thus HCl decreases, the solubility of H 2 S which is sufficient to, precipitate IInd group radicals., , 3.2 ´ 10-11, = 2 ´ 10-4, 4, (c) Ksp for AgI = 1 × 10–16, In solution of KI, I– would be due to the both, AgI and KI, 10–4 solution KI would provide, = 10–4 I–, AgI would provide, say = x I– (x is solubility of, AgI), Total I– = (10–4 + x) , Ksp of AgI = (10–4 + x) x, Þ Ksp = 10–4x + x2, as x is very small, \ x2 can be ignored, \ 10–4 x = 10–16, , or s = 3, , K sp = [Ba 2+ ] [OH - ]2, , 86., , (c) For AX 2 ; K sp = 4s3 \ 3.2 ´ 10-11 = 4 s3, , 94., , x, , (solubility), , =, , 10 -16, 10 -4, , = 10 -12 (mol L-1 ), , (d) Given s = 0.5 × 10–4 moles/L, M2+ + 2X], [MX2 , Q For MX2, Ksp = s × (2s)2 = 4s3, Ksp = 4 × (0.5 × 10–4)3 = 4 × 0.125 ×10–12, = 0.5 × 10–12 = 5 × 10–13, (b) Na2CO3 is a salt of weak acid H2CO3 and, strong base NaOH, therefore, its aqueous, solution will be basic hence has pH more than 7., Na 2 CO 3 + 2H 2 O ¾¾, ® 2NaOH + 2H 2 CO 3, strong base, , 95., , (b), , pOH = pKb + log, , weak acid, , [Salt], [Base], , [salt], [Base], but pOH+ pH = 14 or pOH = 14 – pH, , or pKb = pOH–log, , [Salt], = pK b, [Base], 0.1, 14 - 9.25 - log, = pK b, 0.1, 14 – 9.25 – 0 = pKb, pKb = 4.75, , \ 14– pH – log, , 96., , (a), , 2 M + + S––, M 2S , 2s, , s, , Solubility product = (2s)2 (s) = 4s3, = 4 (3.5 × 10–6)3 = 1.7 × 10–16
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89, , Equilibrium, 97., , (d), , 101. (a) For Bi2S3 . Ksp = (2s)2.(3s)3, = 4s2. 27s3 = 108s5, , BA2 ® B+ 2 A, s, , 2s, , Solubility product = [s] [2s]2 = 4s3, 4´ 10, 4, , =, , 5, , 1 ´ 10 –17, 108, , \ s = 10–4, (b) For Binary salts like CuS & HgS, solubilty,, , or s = K sp =, , 7 ´ 10-, , s = K sp, , for CuS, , Þ sCuS = 10-31 = 3.16 × 10–16,, , For Ag2S Ksp = 2s2. s = 4s3, , 3, , sHgS = 10 -54 = 10–27, +, , or s = 3, , 2s, , s, , K sp = 4s3 or sAg 2S = 3, , =, , \, , 3, , K sp, 4, , -45, , =3, , s = K sp = 8 ´ 10-37, K sp, , =, , 3, , 6 ´ 10 -51, 4, , The solubility of Bi2S3 will be in the order of 5th, root of 10–17, thus, it is most soluble., , 10-44, 4, , 0.25 ´ 10 ´ 10, = 13.54 ´ 10, The order is Ag2S > CuS > HgS, , -16, , [Salt], (d) pH = pKa + log, [Acid], For acetoacetic acid, [Salt], 3.58 = 3.58 + log, [Acid], [Salt], =0, Þ log, [Acid], , [Salt], =1, [Acid], This is the required condition for maximum buffer, capacity., 100. (d) The buffer system present in serum is, H2CO3 + NaHCO3 and as we known that a buffer, solution resist the change in pH therefore pH, value of blood does not change by a small, addition of an acid or a base., Þ, , 16, , 4, thus Bi2S3 has maximum solubility., , 2, , For Ag 2S ® 2 Ag + S, , 99., , K sp, , 108, For MnS. Ksp = s2, , 4 × 10–12 = 4s3 or s =, 98., , or s = 5, , - 12, , 102. (a) Sodium borate is a salt of strong base, (NaOH) and weak acid (H3BO3). Hence, its, aqueous solution will be basic., 103. (b) AgBr has the highest solubility in 10–3 M, NH4OH. AgBr is less soluble in NaBr and HBr, due to presence of common ion (Br). With, NH4OH, AgBr will react as:, AgBr(s) + 2NH4OH(aq.), ® [Ag(NH3)2]Br(aq.) + 2H2O(l), Hence, the highest solubility with NH4 OH, solution., 104. (c) NaCl is a salt of strong acid and strong, base hence its aqueous solution will be neutral, i.e. pH = 7. NaHCO3 is an acidic salt hence pH <, 7. Na2CO3 is a salt of weak acid and strong base., Hence its aqueous solution will be strongly, basic i.e. pH > 7., NH4Cl is salt of weak base and strong acid, hence, its aqueous solution will be strongly acidic i.e., pH < 7.
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EBD_8336, 90, , CHEMISTRY, , 8, , Redox Reactions, , Trend Analysis with Important Topics & Sub-Topics, 2020, Topic Name, , Sub-Topic, , Oxidation number, , oxidation number, disproportionation, Disproportionation, reaction, and balancing of redox, balancing of redox, reaction, reaction, LOD - Level of Difficulty, , 2019, , 2., , 3., , 1, , E - Easy, , Zn gives H2 gas with H2SO4 and HCl but not, with HNO3 because, [2002], (a) Zn acts as an oxidising agent when it reacts, with HNO3, (b) HNO3 is weaker acid than H2SO4 and HCl, (c) In electrochemical series, Zn is above, hydrogen, (d) NO3- is reduced in preference to hydronium, ion, Which of the following involves a redox reaction?, (a) Reaction of H2SO4 with NaOH, [1997], (b) Production of ozone from oxygen in the, atmosphere by lightning, (c) Production of nitrogen oxides from nitrogen, and oxygen in the atmosphere by lightning, (d) Evaporation of water, The loss of electron is termed as, [1995], (a) oxidation, (b) reduction, (c) combustion, (d) neutralization, Topic 2: Oxidation Number, , 4., , 2017, , 2016, , QNS. LOD QNS. LOD QNS. LOD QNS. LOD QNS. LOD, E, 1, , A - Average, , Topic 1: Oxidation and Reduction Reactions, 1., , 2018, , What is the change in oxidation number of, carbon in the following reaction?, [2020], CH4(g) + 4Cl2(g) ¾® CCl4(l) + 4HCl(g), , 5., , 6., , 7., , A, , 1, , D, , 1, , A, , D - Difficult, , Qns - No. of Questions, , (a) 0 to + 4, (b) – 4 to + 4, (c) 0 to – 4, (d) + 4 to + 4, The oxidation state of Cr in CrO6 is, [NEET Odisha 2019], (a) +4, (b) –6, (c) +12, (d) +6, 2–, Oxidation numbers of P in PO3–, 4 , of S in SO 4, and that of Cr in Cr2 O 72– are respectively, [2009], (a) + 3, + 6 and + 5, (b) + 5, + 3 and + 6, (c) – 3, + 6 and + 6, (d) + 5, + 6 and + 6, The oxidation states of sulphur in the anions, SO32–, S2O42– and S2O62– follow the order[2003], , (a), , S2 O 6 2 - < S2 O 4 2- < SO 3 2-, , (b) S2 O 4 2 - < SO 3 2- < S2 O 6 2(c), 8., , SO32- < S2 O4 2- < S2O 6 2-, , (d) S2 O 4 2 - < S 2 O 6 2- < SO 3 2A compound contains atoms of three elements, A, B and C. If the oxidation number of A is +2, B, is +5, and that of C is –2, the possible formula of, the compound is :, [2000], (a) A2(BC3)2, (b) A3(BC4)2, (c) A3(B4C)2, (d) ABC2
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91, , Redox Reactions, 9., , 10., , 11., , The oxidation number of phosphorus in pyrophosphoric acid is, [1999], (a) +3, (b) +1, (c) +4, (d) +5, The oxidation number of chromium in potassium, dichromate is, [1988, 1995], (a) + 6, (b) – 5, (c) – 2, (d) + 2, Phosphorus has the oxidation state of + 3 in, (a) Phosphorous acid, [1994], (b) Orthophosphoric acid, (c) Hypophosphorous acid, (d) Metaphosphoric acid., , Topic 3: Disproportionation and Balancing, of Redox Reactions, 12., , 15., , 16., , 17., , Which of the following reactions are disproportionation reaction?, [2019], (a) 2Cu+ ® Cu2+ + Cu, (b) 3MnO42– + 4H+ ® 2MnO4– + MnO2 + 2H2O, , aZn + bNO 3- + cH + ® dNH +4 + eH 2 O + fZn 2+, , D, , (c) 2KMnO4 ¾¾® K2MnO4 + MnO2 + O2, D, , 14., , +, 2+, MnO4– + C2 O2–, + CO2 + H 2 O, 4 + H ® Mn, The correct coefficients of the reactants for the, balanced equation are, H+, MnO4–, C2 O2–, 4, 5, 2, (a) 16, (b) 2, 5, 16, (c) 5, 16, 2, (d) 2, 16, 5, Consider the change in oxidation state of, bromine corresponding to different emf values, as shown in the diagram below :, [2018], , BrO4–, , 1.82 V, , ¾¾¾® BrO3–, , 1.5 V, , ¾¾¾® HBrO, ¾¾, , 13., , (d) 2MnO4–+ 3Mn2+ + 2H2O ¾¾® 5MnO2 + 4H+, Select the correct option from the following:, (a) (a) and (b) only, (b) (a), (b) and (c), (c) (a), (c) and (d), (d) (a) and (d) only, For the redox reaction, [2018], , Br –¬¾¾¾ Br2 ¬¾¾¾, 1.0652 V, , 1.595 V, , Then the species undergoing disproportionation is, (a) BrO3–, (b) BrO 4–, (c) HBrO, (d) Br2, , A mixture of potassium chlorate, oxalic acid and, sulphuric acid is heated. During the reaction, which element undergoes maximum change in, the oxidation number ?, [2012], (a) S, (b) H, (c) Cl, (d) C, When Cl2 gas reacts with hot and concentrated, sodium hydroxide solution, the oxidation number, of chlorine changes from :, [2012], (a) zero to +1 and zero to –5, (b) zero to –1 and zero to +5, (c) zero to –1 and zero to +3, (d) zero to +1 and zero to –3, The following redox reaction is balanced by, which set of coefficients ?, [1999], , 18., , 19., , a, b, c, d, e, f, (a) 1, 1, 10, 1, 3, 1, (b) 2, 2, 10, 2, 3, 2, (c) 4, 2, 10, 1, 3, 4, (d) 4, 1, 10, 1, 3, 4, In which of the following reactions, there is no, change in valency ?, [1994], ¾® 3KClO4 + KCl, (a) 4 KClO3 ¾, ¾® 2H2O + 3S, (b) SO2 + 2H2S ¾, (c) BaO2 + H2SO4 ¾, ¾® BaSO4 + H2O2, ¾® 2 BaO2., (d) 3 BaO + O2 ¾, Which substance serves as a reducing agent in, the following reaction ?, [1994], 14H+ + Cr2 O 27 - + 3Ni ® 2Cr3+ + 7H2O + 3Ni2+, (a) H2O, (b) Ni, (c) H+, , (d) Cr2 O 27-, , Topic 4: Electrode Potential and Oxidising,, Reducing Agents, 20., , The standard electrode potential (E°) values of, Al3+/Al, Ag+/Ag, K+/K and Cr3+/Cr are –1.66 V,, 0.80 V, –2.93V and –0.74 V, respectively. The, correct decreasing order of reducing power of, the metal is, [NEET Odisha 2019], (a) Al > K > Ag > Cr, (b) Ag > Cr > Al > K, (c) K > Al > Cr > Ag, (d) K > Al > Ag > Cr
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EBD_8336, 92, , 21., , CHEMISTRY, Standard reduction potentials of the half, reactions are given below :, F2(g) + 2e– ® 2F– (aq); E° = + 2.85 V, Cl2(g) + 2e– ® 2Cl–(aq); E° = + 1.36 V, Br2(l) + 2e– ® 2Br–(aq); E° = + 1.06 V, I2(s) + 2e– ® 2I–(aq); E° = + 0.53 V, , The strongest oxidising and reducing agents, respectively are :, [2012 M], (a) F2 and I–, (b) Br2 and Cl–, (c) Cl2 and Br–, (d) Cl2 and I2, 22. The oxide, which cannot act as a reducing agent,, is, [1995], (a) NO2, (b) SO2, (c) CO2, (d) ClO2, , ANSW E R K E Y, 1, , (d ), , 4, , (b ), , 7, , (b ), , 10, , (a ), , 13, , (b ), , 16, , (b ), , 19, , (b ), , 2, , (c ), , 5, , (d ), , 8, , (b ), , 11, , (a ), , 14, , (c ), , 17, , (d ), , 20, , (c ), , 3, , (a ), , 6, , (d ), , 9, , (d ), , 12, , (a ), , 15, , (c ), , 18, , (c ), , 21, , (a ), , 22, , (c ), , Hints & Solutions, 1., , (d) Zinc gives H2 gas with dil H2SO4/HCl but, not with HNO3 because in HNO3, NO3– ion is, reduced and give NH4NO3, N2O, NO and NO2, , 5., , (d) +6 is the most appropriate oxidation state, of Cr in CrO6., –1, –1, O, O, , [Zn + 2HNO3 ¾¾, ® Zn(NO3 )3 + 2H] ´ 4, HNO3 + 8H ¾® NH3 + 3H2O, NH3 + HNO3 ¾® NH4NO 3, 4Zn+10HNO3 ¾® 4Zn(NO3)2+NH4NO3+3H2O, Zn is above of hydrogen in electrochemical series., So, Zn displaces H2 from dilute H2SO4 and HCl, with liberation of H2., Zn + H2SO4 ® ZnSO4 + H2, , 2., , O, –1, 6., , (b), (c), , 0, , 0, , 7., , +2 -2, , The structure of S 2O 42– and S 2 O 6 2– are, symmetrical. Thus, both sulphur atoms are in same, oxidation state. This is not the case with S2O32–, or S4O62– ions., , here oxidation of N2 & reduction of O2 is taking, place, D, , 3., 4., , (b), , -4, , +4, , CH 4 (g) + 4Cl2 (g) ® CCl 4 (l ) + 4HCl (g), , Change in oxidation state of carbon is –4 to +4., , 2x =12; x = + 6, (b) SO 32 - ® S is in + 4 oxidation state, S2 O 62- ® S is in + 5 oxidation state, , N 2 + O2 ¾¾¾® 2 N O (redox reaction), , (d) H 2O(l) ¾¾, ® H 2 O(g) (not redox reaction), (a) Losing of electron is called oxidation., , PO3–, 4 = x + 4 (– 2) = – 3; x – 8 = – 3; x = + 5, , S2 O 24- ® S is in + 3 oxidation state, , Light, , 0, , O–1, , Cr2 O72– = 2x + 7 (– 2) = – 2; 2x – 14 = – 2;, , Light, 3 O 2 ¾¾¾® 2 O 3 (not redox reaction), , 0, , (d), , –1, O, , SO 2–, 4 = x + 4 (– 2) = – 2; x – 8 = – 2; x = + 6, , ® Na 2SO 4 +2H 2 O, (c) (a) 2NaOH+H 2SO 4 ¾¾, (neutralization), , +6, Cr, , –1, O, , (nearly 6%), , 8., , (b) Oxidation number of a compound must be, 0. Using the values for A, B and C in the four, options we find that A3(BC4)2 is the answer., Check : (+2)3 + [(+5) + 4(–2)]2 = 6 + (5 – 8)2 = 0
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93, , Redox Reactions, 9., , 10., , 11., , 12., , (d) Pyrophosphoric acid H4P2O7, Let oxidation state of phosphorus is x, (4 × 1 + (– 2) × 7 + 2 x) = 0, \ 2x = 10 or x = +5, (a) Let x = oxidation no. of Cr in K2Cr2O7., \ (2 × 1) + (2 × x) + 7 (– 2) = 0, or 2 + 2x – 14 = 0 or x = + 6., (a) O.N. of P in H3PO3 (phosphorous acid), 3 × 1 + x + 3 × (– 2) = 0 or x = + 3, In orthophosphoric acid (H3PO4) O.N. of P is, + 5, in hypophosphorous acid (H3PO2) it is + 1, while in metaphosphoric acid (HPO3), it is + 5., (a) In a disproportionation reaction, one species, undergoes both oxidation and reduction., Reduction, , Reaction in (d) involves comproportionation or, synproportionation. When two reactants, each, containing the same element but with a different, oxidation number, form a product in which the, element involved reach the same oxidation number., Oxidation, , Reduction, , It is opposite to disproportionation., , 15., , 16., , +, , 2+, , 2 Cu ¾¾¾¾® Cu + Cu°, , +5, +6, +6, KClO 3 + H 2C 2 O 4 + H 2SO 4 ® K 2SO 4, –1, + KCl + CO 2 + H 2 O, i.e. maximum change in oxidation number is, observed in Cl (+5 to –1)., (b) On reaction with hot and concentrated alkali, a mixture of chloride and chlorate is formed, , (c), , Hot, 3Cl2 + 3 NaOH(excess) ¾¾¾, ®, , Oxidation, , -1, , Reduction, +6, , 2–, , +, , +7, , –, , +4, , 3MnO4 + 4H ¾® 2MnO4 + MnO2 + 2H2O, , 17., , (b), , +7, , +2, , Mn O 4– ¾¾, ® Mn2+; 5e– gain, , +3, , +4, C2O 2–, ® CO2, 4 ¾¾, , ...(ii), ; 2e loss, Multiplying (i) by 2 and (ii) by 5 to balance e–, 2+, ® 2 Mn + 10 CO2, 2 MnO 4– + 5 C 2 O 2–, 4 ¾¾, , 18., , 19., , On balancing charge;, +, ®, 2 MnO 4– + 5 C2 O 2–, 4 + 16 H ¾¾, , 14., , 2 Mn 2 + + 10 CO 2 + 8 H 2O, (c) Calculate E°cell corresponding to each, compound undergoing disproportionation, reaction. The reaction for which E°cell comes out, + ve is spontaneous., HBrO ¾® Br2 E° = 1.595 V, SRP (cathode), HBrO ¾® BrO 3– E° = –1.5 V, SOP (anode), , 2HBrO ¾® Br2 + BrO 3–, E°cell = SRP (cathode) – SRP (anode), = 1.595 – 1.5, = 0.095 V, E°cell > 0 Þ DG° < 0 [spontaneous], , Zn ® Zn +2, , 4Zn + 10H + + NO3- ® 4Zn 2 + + NH +4 + 3H 2O, , ...(i), , -, , (d), , +5, , 5NaCl + NaClO3 + 3H 2O, + 2e ....(1), , 8e - + 10H + + NO 3- ® NH +4 + 3H 2 O .....(2), operate eq. (1) × 4 + eq. (2) × 1, , Oxidation, , 13., , +4, +, 5MnO2 + 4H, , +7, 2+, 2MnO–4 + 3Mn + 2H2O, , 20., 21., , +2 -1, , +1 +6 -2, , +2 +6 -2, , +1 - 1, , ® Ba S O4 + H 2 O2, (c) Ba O 2 + H 2 S O4 ¾¾, In this reaction, none of the elements undergoes, a change in oxidation number or valency., (b) The element undergo oxidation itself and, reduces others is known as reducing agent. In, this reaction O. N. of Ni Changes from 0 to + 2, and hence Ni acts as a reducing agent., (c) Lesser is the reduction potential greater is, the reducing power., Reducing power : K > Al > Cr > Ag, (a) F2 is the strongest oxidising agent as it, has highest reduction potential while I– is the, strongest reducing agent since it has lowest, reduction potential., Higher the value of reduction potential, higher will, be the oxidising power whereas the lower the value, of reduction potential higher will be the reducing, power., , 22., , (c) Carbon has the maximum oxidation state, of + 4, therefore carbon dioxide (CO2) cannot, act as a reducing agent.
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EBD_8336, 94, , CHEMISTRY, , 9, , Hydrogen, , Trend Analysis with Important Topics & Sub-Topics, 2020, Topic Name, Preparation and, properties of hydrogen, Preparation and, properties of water, LOD - Level of Difficulty, , 2., , 3., , 2018, , 2017, , 2016, , Sub-Topic, QNS. LOD QNS. LOD QNS. LOD QNS. LOD QNS. LOD, properties of, 1, A, hydrogen, 1, A, hardness of water, E - Easy, , Topic 1: Preparation and Properties of, Hydrogen, 1., , 2019, , Which of the following statements about, hydrogen is incorrect ?, [2016], (a) hydrogen has three isotopes of which, tritium is the most common., (b) Hydrogen never acts as cation in ionic salts, (c) Hydronium ion, H3 O+ exists freely in, solution, (d) Dihydrogen does not act as a reducing agent, When a substance A reacts with water, it, produces a combustible gas B and a solution of, substance C in water. When another substance, D reacts with this solution of C, it also produces, the same gas B on warming but D can produce, gas B on reaction with dilute sulphuric acid at, room temperature. A imparts a deep golden yellow, colour to a smokeless flame of Bunsen burner., A, B, C and D respectively are, [1998], (a) Na , H2, NaOH, Zn, (b) K, H2, KOH, Al, (c) Ca, H2, Ca(OH)2, Sn, (d) CaC2, C2H2, Ca(OH)2, Fe, Which one of the following pairs of substances, on reaction will not evolve H2 gas?, [1998], (a) Iron and H2SO4 (aqueous), (b) Iron and steam, (c) Copper and HCl (aqueous), (d) Sodium and ethyl alcohol, , A - Average, , 4., , D - Difficult, , Qns - No. of Questions, , The hydride ion, H–, is a stronger base than the, hydroxide ion, OH–. Which one of the following, reactions will occur if sodium hydride (NaH) is, dissolved in water?, [1997], (a) H - (aq) + H 2 O(l) ® H 3O - (aq), (b), , H - (aq) + H 2 O(l) ® OH - (aq) + H 2 (g), , H – (aq) + H 2 O(l) ®, OH– (aq) + 2H+ (aq) + 2e–, –, (d) H (aq) + H2O(l) ® reaction, The ionization of hydrogen atom would give rise, to, [1990], (a) Hydride ion, (b) hydronium ion, (c) Proton, (d) hydroxyl ion., , (c), , 5., , Topic 2: Preparation and Properties of Water, 6., , 7., , The number of hydrogen bonded water, molecules(s) associated with CuSO4·5H2O is, [NEET 2019 Odisha], (a) 5, (b) 3, (c) 1, (d) 2, The method used to remove temporary hardness, of water is:, [2019], (a) Calgon’s method, (b) Clark’s method, (c) Ion-exchange method, (d) Synthetic resins method
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95, , Hydrogen, 8., , 9., , 10., , 11., , 12., , Some statements about heavy water are given, below:, (a) Heavy water is used as a moderator in, nuclear reactors., (b) Heavy water is more associated than, ordinary water., (c) Heavy water is more effective solvent than, ordinary water., Which of the above statements are correct?, (a) (a) and (c), (b) (a) and (b)[2010], (c) (a), (b) and (c), (d) (b) and (c), Which of the following groups of ions makes, the water hard ?, [1994], (a) Sodium and bicarbonate, (b) Magnesium and chloride, (c) Potassium and sulphate, (d) Ammonium and chloride., The dielectric constant of H2 O is 80. The, electrostatic force of attraction between Na+ and, Cl– will be, [1994], (a) reduced to 1/40 in water than in air, (b) reduced to 1/80 in water than in air, (c) will be increased to 80 in water than in air, (d) will remain unchanged., At its melting point, ice is lighter than water, because, [1992], (a) H2O molecules are more closely packed in, solid state, (b) Ice crystals h ave hollow hexagonal, arrangement of H2O molecules., (c) On melting of ice the H2O molecule shrinks, in size, (d) Ice froms mostly heavy water on first, melting., Calgon used as a water softener is, [1989], , (c) Na 4 [Na 4 (PO4 )5 ], (d) Na 4 [Na 2 (PO 4 )6 ], Topic 3: Preparation and Properties of, Hydrogen Peroxide, 13., , 14., , 15., , 16., , 17., , (i) H2O2 + O3 ® H2O + 2O2, (ii) H2O2 + Ag2O ® 2Ag + H2O + O2, Role of hydrogen peroxide in the above reactions, is respectively [2014], (a) Oxidizing in (i) and reducing in (ii), (b) Reducing in (i) and oxidizing in (ii), (c) Reducing in (i) and (ii), (d) Oxidizing in (i) and (ii), When H2O2 is oxidised, the product is [1999], (a) OH–, (b) O2, (c) O2–, (d) HO2–, The volume strength of 1.5 N H2O2 solution is, (a) 4.8 L, (b) 5.2 L, [1996], (c) 8.4 L, (d) 8.8 L, The O – O – H bond angle in H2O2 is [1994], (a) 106°, (b) 109° 28', (c) 120°, (d) 97°, Which of the following is the true structure of, H2O2 ?, [1989], H, (b) O, , (a) H– O – O – H, , 18., , (a) Na 2 [Na 4 (PO3 )6 ], (b) Na 4 [Na 2 (PO3 )6 ], , O, , H, H, H, (c), (d), O, O=O, O, H, H, The reaction of H2O2 with sulphur is an example, of ........reaction, [1988], (a) Addition, (b) Oxidation, (c) Reduction, (d) Redox, , ANSWER KEY, 1, 2, , (a, d), (a), , 3, 4, , (c), (b), , 5, 6, , (c), (c), , 7, 8, , (b), (b), , 9, 10, , (b), (b), , 11, 12, , (b), (a), , 13, 14, , (c), (b), , 15, 16, , (c), (d), , 17, 18, , (b), (d)
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EBD_8336, 96, , CHEMISTRY, , Hints & Solutions, 1., , (a, d) Among the three isotopes of hydrogen,, , 8., , ( ), 1, , Protium 1 H is most common. It is an energetic, reducing agent. It reduces oxides, chlorides and, sulphides of certain metals and produce free, metals at ordinary temperature., CuO + 2H ® Cu + H2O, 2., , (a), , 2Na + 2H 2 O ® 2NaOH + H 2 , 'A', , 'B', , 'C', , Zn + 2 NaOH ® Na 2 ZnO 2 + H 2 , 'D ', , 'C', , 9., , 10., , 'B', , Zn + dil. H 2SO 4 ® ZnSO 4 + H 2 , 'D ', , 'B', , Dielectric constant or relative permittivity is the, factor by which the electric field between the, charges is decreased relative to vacuum (or air as, mentioned in this question)., , Na produces golden yellow colour with, smokeless flame of Bunsen burner., 3., , (c), , Fe + dil. H 2SO 4 ® FeSO 4 + H 2 , , 11., , 3Fe + 4H 2 O ® Fe 3O 4 + 4H 2 , Steam, , Cu + dil. HCl ® No reaction, Copper does not evolve H2 from acid as it is, below hydrogen in electrochemical series., , ® OH - (aq) + H 2 (g), (b) H - (aq)+ H 2 O(l) ¾¾, base 1, , acid 1, , base 2, , acid2, , H–, , In this reaction, acts as bronsted base as it, accepts one proton (H+) from H2O to form H2., H (g ) ¾, ¾® H + (g ) + e - ., , 5., , (c), , 6., , (c) The actual structure of CuSO4· 5H2O is, [Cu(H2O)4]SO4 · H2O, so only one water molecule, is associated with the molecule via hydrogen, bond., , 7., , (b) In this method Ca(OH)2 is used., , (b) In liquid state, the H2O molecules are held, together by intermolecular H-bonds., When water freezes at atmospheric pressure it, crystalises in normal hexagonal form. In it, each, oxygen atom is surrounded by 4 hydrogen atoms,, two by strong covalent bonds and two by weak, hydrogen bonds. Since hydrogen bonds are longer, than covalent bonds. The molecules of water are, not closely packed in crystal lattice. There exists a, number of hollow spaces in the crystal lattice,, which decreases the density of ice., , 2 Na + C 2 H 5 OH ® 2C 2 H 5 ONa + H 2 , , 4., , (b) Heavy water is used for slowing down the, speed of neutrons and used as moderators., Boiling point of heavy water is more than that, of ordinary water, so it is more associated. The, dielectric constant of heavy water is slightly les, than that of ordinary water. Hence, ordinary water, is more effective solvent than heavy water., (b) Temporary hardness is due to presence of, bicarbonates of calcium and magnesium and, permanent hardness is due to the sulphates or, chlorides of both of calcium and magnesium., (b) Electrostatic forces of attraction are, reduced to 1/80th in water., , 12., , (a) The complex salt of metaphosphoric acid, sodium hexametaphosphate (NaPO3)6, is known, as calgon. It is represented as Na2[Na4(PO3)6], Reduction, , 13., , (c) (i) H2–1O2 +, , –2, o, H2O + 2O2, , o, O3, Oxidation
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EBD_8336, 98, , CHEMISTRY, , 10, , The s -Block, Elements, , Trend Analysis with Important Topics & Sub-Topics, 2020, Topic Name, Preparation and, properties of alkali, metals and their, compounds, Some important, compounds of sodium, Preparation and, properties of alkaline, earth metal and their, compounds, LOD - Level of Difficulty, , Sub-Topic, biological, importance of, alkali metals, properties of, sodium chloride, properties of, alkaline earth, metals compounds, biololgical, importance of, alkaline earth, metals, , 2., , 3., , 2018, , 2017, , 2016, , QNS. LOD QNS. LOD QNS. LOD QNS. LOD QNS. LOD, 1, , E, , 1, , E, , E - Easy, , Topic 1: Preparation and Properties of Alkali, Metals and their Compounds, 1., , 2019, , The following metal ion activates many, enzymes, participates in the oxidation of glucose, to produce ATP and with Na, is responsible for, the transmission of nerve signals., [2020], (a) Copper, (b) Calcium, (c) Potassium, (d) Iron, Which of the alkali metal chloride (MCl) forms, its dihydrate salt (MCl × 2H2O) easily?, [NEET Odisha 2019 ], (a) KCl, (b) LiCl, (c) CsCl, (d) RbCl, The function of "Sodium pump" is a biological, process operating in each and every cell of all, animals. Which of the following biologically, important ions is also a consituent of this, pump :, [2015], (a) Mg2+, (b) K+, (c) Fe2+, (d) Ca2+, , 1, , A, , 1, , E, , A - Average, , 4., , 5., , 6., , 7., , 2, , A, , 1, D - Difficult, , E, , Qns - No. of Questions, , Which one of the alkali metals, forms only, the, normal oxide, M2O on heating in air ?, [2012], (a) Rb, (b) K, (c) Li, (d) Na, The ease of adsorption of the hydrated alkali, metal ions on an ion-exchange resins follows the, order :, [2012], (a) Li+ < K+ < Na+ < Rb+, (b) Rb+ < K+ <Na+ < Li+, (c) K+ < Na+ < Rb+ < Li+, (d) Na+ < Li+ < K+ < Rb+, In the replacement reaction, , CI + MF, CF + MI, The reaction will be most favourable if M happens, to be :, [2012 M], (a) Na, (b) K, (c) Rb, (d) Li, The sequence of ionic mobility in aqueous, solution is :, [2008], (a) K+ > Na+ > Rb+ > Cs+, (b) Cs+ > Rb+ > K+ > Na+
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99, , The s-Block Elements, , 8., , 9., , 10., , 11., , 12., , 13., 14., , 15., , (c) Rb+ > K+ > Cs+ > Na+, (d) Na+ > K+ > Rb+ > Cs+, The alkali metals form salt-like hydrides by the, direct synthesis at elevated temperature. The, thermal stability of these hydrides decreases in, which of the following orders ?, [2008], (a) CsH > RbH > KH > NaH > LiH, (b) KH > NaH > LiH > CsH > RbH, (c) NaH > LiH > KH > RbH > CsH, (d) LiH > NaH > KH > RbH > CsH, The correct order of the mobility of the alkali metal, ions in aqueous solutions is, [2006], (a) Na+ > K+ > Rb+ > Li+, (b) K+ > Rb+ > Na+ > Li+, (c) Rb+ >K+ > Na+ > Li+, (d) Li+ > Na+ > K+ > Rb+, In crystals of which one of the following ionic, compounds would you expect maximum distance, between centres of cations and anions? [1998], (a) LiF, (b) CsF, (c) CsI, (d) LiI, Which of the following metal ions plays an, important role in muscle contraction ? [1994], (a) K+, (b) Na+, (c) Mg2+, (d) Ca2+, Which of the following statement is false ? [1994], (a) Strontium decomposes water readily than, beryllium, (b) Barium carbonate melts at a higher, temperature than calcium carbonate, (c) Barium hydroxide is more soluble in water, than magnesium hydroxide, (d) Beryllium hydroxide is more basic than, barium hydroxide., Which of the following has largest size ?[1993], (a) Na, (b) Na+, –, (c) Na, (d) Can’t be predicted, Compared with the alkaline earth metals, the alkali, metals exhibit, [1990], (a) Smaller ionic radii, (b) Highest boiling points, (c) Greater hardness, (d) Lower ionization energies., Which one of the following properties of alkali, metals increases in magnitude as the atomic, number rises ?, [1989], (a) Ionic radius, (b) Melting point, (c) Electronegativity, (d) First ionization energy., , Topic 2: Some Important Compounds of Sodium, 16. HCl was passed through a solution of CaCl2,, MgCl 2 and NaCl. Which of the following, compound(s) crystallise(s)?, [2020], (a) Only NaCl, (b) Only MgCl2, (c) NaCl, MgCl2 and CaCl2, (d) Both MgCl2 and CaCl2, 17. Crude sodium chloride obtain ed by, crystallisation of brine solution does not contain, [NEET Odisha 2019], (a) CaSO4, 18., , 19., , 20., , 21., , 22., , (b) MgSO4, , (c) Na2SO4, (d) MgCl2, In Castner-Kellner cell for production of sodium, hydroxide:, [NEET Kar. 2013], (a) Brine is electrolyzed with Pt electrodes, (b) Brine is electrolyzed using graphite, electrodes, (c) Molten sodium chloride is electrolysed, (d) Sodium amalgam is formed at mercury, cathode, Which of the following statements is incorrect?, [2011M], (a) Pure sodium metal dissolves in liquid, ammonia to give blue solution., (b) NaOH reacts with glass to give sodium silicate, (c) Aluminium reacts with excess NaOH to give, Al(OH)3, (d) NaHCO3 on heating gives Na2CO3, Which of the following oxides is not expected to, react with sodium hydroxide?, [2009], (a) CaO, (b) SiO2, (c) BeO, (d) B2O3, In which of the following processes, fused, sodium hydroxide is electrolysed at a 330ºC, temperature for extraction of sodium? [2000], (a) Castner's process (b) Down's process, (c) Cyanide process (d) Both 'b' and 'c', Aqueous solution of sodium carbonate absorbs, NO and NO2 to give, [1996], (a) CO2 + NaNO3, (b) CO2 + NaNO2, (c) NaNO2 + CO, (d) NaNO3 + CO
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EBD_8336, 100, , CHEMISTRY, , 23. Which of the following is known as fusion, mixture?, [1994], (a) Mixture of Na2CO3 + NaHCO3, (b) Na2CO3.10H2O, (c) Mixture of K2CO3 + Na2CO3, (d) NaHCO3, 24. Washing soda has formula, [1990], (a) Na2CO3.7H2O, (b) Na2CO3.10H2O, (c) Na2CO3.3H2O, (d) Na2CO3, Topic 3: Preparation and Properties of Alkaline, Earth Metals and their Compounds, 25., , 26., , 27., , 28., , 29., , 30., , 31., , Which of the following is an amphoteric, hydroxide ?, [2019], (a) Sr(OH)2, (b) Ca(OH)2, (c) Mg(OH)2, (d) Be(OH)2, Enzymes that utilize ATP in phosphate transfer, require an alkaline earth metal (M) as the cofactor., M is:, [2019], (a) Be, (b) Mg, (c) Ca, (d) Sr, Which of the following oxides is most acidic in, nature?, [2018], (a) MgO, (b) BeO, (c) CaO, (d) BaO, Among CaH2, BeH2, BaH2, the order of ionic, character is, [2018], (a) BeH2 < CaH2 < BaH2, (b) CaH2 < BeH2 < BaH2, (d) BaH2 < BeH2 < CaH2, (d) BeH2 < BaH2 < CaH2, Which of the following statements is false ?, [2016], (a) Mg2+ ions form a complex with ATP, (b) Ca2+ ions are important in blood clotting, (c) Ca2+ ions are not important in maintaining, the regular beating of the heart., (d) Mg2+ ions are important in the green parts, of plants., Solubility of the alkaline earth's metal sulphates, in water decreases in the sequence :- [2015], (a) Ca > Sr > Ba > Mg (b) Sr > Ca > Mg > Ba, (c) Ba > Mg > Sr > Ca (d) Mg > Ca > Sr > Ba, Which of the following compounds has the, lowest melting point ?, [2011], (a) CaCl2, (b) CaBr2, (c) CaI2, (d) CaF2, , 32. Which of the following alkaline earth metal, sulphates has hydration enthalpy higher than, the lattice enthalpy?, [2010], (a) CaSO 4, (b) BeSO 4, (c) BaSO 4, (d) SrSO 4, 33. Property of the alkaline earth metals that increases, with their atomic number, [2010], (a) Solubility of their hydroxides in water, (b) Solubility of their sulphates in water, (c) Ionization energy, (d) Electronegativity, 34. The correct order of increasing thermal stability, of K2CO3, MgCO3, CaCO3 and BeCO3 is [2007], (a) BeCO3< MgCO3 < CaCO3 < K2CO3, (b) MgCO3 < BeCO3 < CaCO3 < K2CO3, (c) K2CO3 < MgCO3 < CaCO3 < BeCO3, (d) BeCO3 < MgCO3 < K2CO3 < CaCO3, 35. In which of the following the hydration energy is, higher than the lattice energy?, [2007], (a) MgSO4, (b) RaSO4, (c) SrSO4, (d) BaSO4, 36. Calcium is obtained by the, [1997], (a) electrolysis of solution of calcium chloride, in water, (b) electrolysis of molten anhydrous calcium, chloride or fused calcium chloride, (c) roasting of limestone, (d) reduction of calcium chloride with carbon, 37. For two ionic solids CaO and KI, identify the, wrong statement amongst the following : [1997], (a) The lattice energy of CaO is much large, than that of KI, (b) KI is more soluble in water, (c) KI has higher melting point, (d) CaO has higher melting point, 38. Which one is the correct statement with reference, to solubility of MgSO4 in water?, [1996], (a) SO 4 2– ion mainly contributes towards, hydration energy, (b) Sizes of Mg2+ and SO42– are similar, (c) Hydration energy of MgSO4 is higher in, comparison to its lattice energy, (d) Ionic potential (charge/radius ratio) of Mg2+, is very low
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101, , The s-Block Elements, 39. Sodium is made by the electrolysis of a molten, mixture of about 40% NaCl and 60% CaCl2, because, [1995], (a) Ca2+ can reduce NaCl to Na, (b) Ca2+ can displace Na from NaCl, (c) CaCl2 helps in conduction of electricity, (d) this mixture has a lower melting point than NaCl, 40. All the following substances react with water., The pair that gives the same gaseous product is, [1994], (a) K and KO2, (b) Na and Na2O2, (c) Ca and CaH2, (d) Ba and BaO2, 41. Which one of the following has minimum value, of cation/anion ratio., [1993], (a) NaCl, (b) KCl, (c) MgCl2, (d) CaF2, 42. Electronic configuration of calcium atom can be, written as, [1992], 2, 2, (a) [Ne], 4p, (b) [Ar], 4s, (c) [Ne], 4s2, (d) [Kr], 4p2, 43. Which of the following atoms will have the, smallest size ?, [1989], (a) Mg, (b) Na, (c) Be, (d) Li, Topic 4: Some Important Compounds of Calcium, 44., , On heating which of the following releases CO2, most easily ?, [2015 RS], (a) K2CO3, (b) Na2CO3, (c) MgCO3, (d) CaCO3, 45. Which one of the following is present as an active, ingredient in bleaching powder for bleaching, action ?, [2011], (a) CaOCl2, (b) Ca(OCl)2, (c) CaO2Cl, (d) CaCl2, 46. Match List – I with List –II for the compositions, of substances and select the correct answer using, the code given below the lists :, [2011M], , List - I, Substances, (1) Plaster of paris, (2) Epsomite, (3) Kieserite, (4) Gypsum, , List - II, Composition, (i) CaSO4.2H2O, (ii) CaSO4.½ H2O, (iii) MaSO4.7 H2O, (iv) MgSO4. H2O, (v) CaSO4, , Code :, (1), (2), (3), (4), (a) (iii), (iv), (i), (ii), (b) (ii), (iii), (iv), (i), (c) (i), (ii), (iii), (v), (d) (iv), (iii), (ii), (i), 47. The compound A on heating gives a colourless, gas and a residue that is dissolved in water to, obtain B. Excess of CO2 is bubbled through, aqueous solution of B, C is formed which is, recovered in the solid form. Solid C on gentle, heating gives back A. The compound is [2010], (a) CaSO4.2H2O, (b) CaCO3, (c) Na2CO3, (d) K2CO3, 48. 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’, a compound ‘X’ is reformed. The compound, ‘X’ is, [2004], (a) Ca(HCO3)2, (b) CaCO3, (c) Na2CO3, (d) K2CO3, 49. Identify the correct statement, [1995], (a) gypsum is obtained by heating plaster of, Paris, (b) plaster of Paris can be obtained by hydration, of gypsum, (c) plaster of paris is obtained by partial, oxidation of gypsum, (d) gypsum contains a lower percentage of, calcium than plaster of Paris, , ANSWER KEY, 1, 2, 3, 4, 5, , (c), (b), (b), (c), (b), , 6, 7, 8, 9, 10, , (c), (b), (d), (c), (c), , 11, 12, 13, 14, 15, , (d), (d), (c), (d), (a), , 16, 17, 18, 19, 20, , (a), (b), (d), (c), (a), , 21, 22, 23, 24, 25, , (a), (b), (c), (b), (d), , 26, 27, 28, 29, 30, , (b), (b), (a), (c), (d), , 31, 32, 33, 34, 35, , (c), (b), (a), (a), (a), , 36, 37, 38, 39, 40, , (b), (c), (c), (d), (c), , 41, 42, 43, 44, 45, , (c), (b), (c), (c), (b), , 46, 47, 48, 49, , (b), (b), (b), (d)
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EBD_8336, 102, , CHEMISTRY, , Hints & Solutions, 1., , (c) Potassium (K) activates many enzymes to, participate in oxidation of glucose to produce, ATP and helps in the transmission of nerve, signal along with Na., , 2., , (b) Only LiCl forms a dihydrate, other metal, chlorides do not form hydrates., (b) In sodium pump, high concentration of, potassium ions and a low concentration of, sodium ions are maintained within a cell by a, plasma membrane protein., (c) All the alkali metals when heated with, oxygen form different types of oxides for example, lithium forms lithium oxide (Li2O), sodium forms, sodium peroxide (Na2O2), while K, Rb and Cs, form their respective superoxides., , 3., , 4., , Li+ being smallest, combines with small anion, O2– to form stable Li2O., The Na+ being relatively larger than Li+ and weaker, positive field than Li+ , combines with O 22 (peroxide ion) to form Na2, O2. K+, Rb+ and Cs+, being relative larger than Na+ and with weaker, positive field than Na + combines with O2(superoxide ion) to for KO 2, RbO 2 and CsO 2, respectively., , 5., , 6., 7., , (b) All alkali metal salts are ionic (except, Lithium) and soluble in water due to the fact, that cations get hydrated by water molecules., The degree of hydration depends upon the size, of the cation. Smaller the size of a cation, greater is its, hydration energy., Relative ionic radii :, Cs + > Rb + > K + > Na + > Li +, Relative ionic radii in water or relative degree of, hydration:, Li + > Na + > K + > Rb + > Cs +, (c) Tertiary halide can show ionic reaction with, MF so, MF should be most ionic for reaction to, proceed forward. Hence ‘M’ should be ‘Rb’., (b) Smaller the ion more is its ionic mobility in, aqueous solution. Ionic radii of the given alkali, metals is in the order Na+ < K+ < Rb+ < Cs+ and, thus expected ionic mobility will be in the same, , 8., , 9., , 10., , 11., 12., 13., , 14., 15., , order Cs+ < Rb+ < K+ < Na+. However due to, high degree of solvation (or hydration) and, because of lower size or high charge density,, the hydrated ion size follows the same order, Cs+ < Rb+ < K+ < Na+ and thus conductivity, order is Cs+ > Rb+ > K+ > Na+., (d) The stability of alkali metal hydrides, decreases from Li to Cs. It is due to the fact that, M–H bonds becomes weaker with increase in, size of alkali metals as we move down the group, from Li to Cs. Thus the order of stability of, hydrides is, LiH > NaH > KH > RbH > CsH, (c) Hydrated Ionic radii of alkali metals in water, follows the order Li+ > Na+ > K+ > Rb+ > Cs+, Thus in aqueous solution due to larger ionic, radius Li+ has lowest mobility and hence the, correct order of ionic mobility is, Li + < Na + < K + < Rb +, (c) As Cs+ ion has larger size than Li+ and I–, has larger size than F–, therefore maximum, distance between centres of cations and anions, is in CsI., (d) Ca2+ ions is an essential element for the, contraction of muscles., (d) Be(OH)2 is amphoteric, but the hydroxides, of other alkaline earth metals are basic. The basic, strength increases gradually., (c) A cation is always much smaller than the, corresponding atom, whereas an anion is always, larger than the corresponding atom, hence the, size decreases in the order, Na - > Na > Na +, (d) Because of larger size and smaller nuclear, charge, alkali metals have low ionization, potential relative to alkaline earth metals., (a) Within a group, ionic radius increases with, increase in atomic number. The melting point, decrease down the group due to weakening of, metallic bond. The electronegativity and the 1st, ionization energy also decreases down the, group.
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103, , The s-Block Elements, 16., , 17., , 18., , (a) When HCl is passed through the solution, Cl– ion concentration increases. Hence ionic, product becomes more than solubility product., Only NaCl is crystallised due to less solubility, than MgCl2 and CaCl2., (b) Crude sodium chloride which is obtained, by crystallisation of brine solution contains, Na2SO4, CaSO4, CaCl2 and MgCl2 are present, as impurities in crude., (d) In castner kellner cell, sodium amalgam is, formed at mercury cathode., Sodium is discharged from brine at a mercury, cathode in preference to hydrogen because of the, high hydrogen overpotential at mercury surface., This is the reason of using mercury cathode instead, of graphite or platinum in castner-kellner cell., , 19., , (c) 2Al(s) + 2NaOH (aq) + 2H2O (l) ¾¾, ®, 2NaAlO2 + 3H2, sod. meta aluminate, , 20., , 21., , (a) NaOH is a strong alkali. It combines with, acidic and amphoteric oxides to form salts. Since, CaO is a basic oxide hence does not react with, NaOH., (a) In Castner process, for production of (Na), sodium metal, sodium hydroxide (NaOH) is, electrolysed at 330ºC., , 22., , (b) Na 2 CO3 + NO + NO 2 ® 2 NaNO2 + CO2, , 23., , (c) Mixture of K2CO3 and Na2CO3 is called as, fusion mixture., (b) Washing soda is Na2CO3. 10H2O., (d) Amphoteric hydroxide means it can react, with both acid and base., , 24., 25., , 28., , 29., , 30., , 26., 27., , (b) Enzyme that utilise ATP in phosphate, transfer require an alkaline earth metal (M), Mg as the cofactor., (b) In metals, moving down the group, metallic, character increases, so basic nature, increases hence most acidic will be BeO., BeO < MgO < CaO < BaO, increasingbasic character, , ¾¾¾¾¾¾¾¾¾¾, ®, , ionic., (c) Calcium regulates muscle contraction,, including beating of heart muscle, so that it can, contract and pump out blood to all our body., (d) Solubility of alkaline earth metal sulphates, decreases down the group due to decrease in, hydration energy., MgSO 4 > CaSO4 > SrSO 4 > BaSO 4, Hydration, energy, , Solubility, , 31., , (c) Melting point of metal halides decreases, as the size of the halogen increases. The correct, order is, CaF2 > CaCl2 > CaBr2 > CaI2, , 32., , (b) Be 2+ is very small, hence its hydration, enthalpy is greater than its lattice enthalpy. Also,, hydration energy decreases down the group., (a) The magnitude of hydration energy for the, hydroxides of alkaline earth metals remains, almost same whereas lattice energy decreases, appreciably down the group. Hence, solubility, increases down the group., (a) As the cation size increases down the, group, the metal carbonates become more ionic, in nature. Hence, the thermal stability increases, as: BeCO3 < MgCO3 < CaCO3., The ionic character of group 1 carbonates is, more than that of group 2, thus they possess, more thermal stability. Hence, the correct order, of thermal stability : BeCO3 < MgCO3 < CuCO3, < K2CO3., (a) The solubility and the hydration energy of, sulphates of alkaline earth metals decreases as, we move down the group from Be to Ba due to, the reason that ionic size increases down the, group. The lattice energy remains constant, because sulphate ion is so large, so that small, change in cationic sizes do not make any, difference. Thus the order will be:, , 33., , 34., , Be ( OH ) 2 + 2HCl ¾¾, ® BeCl2 + 2H 2O, Be ( OH ) 2 + 2NaOH ¾¾, ® Na 2 éë Be ( OH ) 4 ùû, , (a) BeH2 < CaH2 < BaH2, Smaller the size of cation, more will be its, polarising power. Hence, BeH2 will be least, , 35., , BeSO4 > MgSO4 > CaSO4 > SrSO4 > BaSO4
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EBD_8336, 104, , 36., , 37., , 38., , 39., , 40., , CHEMISTRY, (b) Calcium is obtained by electrolysis of a, fused mass consisting six parts calcium chloride, and one part calcium fluoride at about 700°C in, an electrolytic cell., (c) CaO has higher lattice energy because, of higher charge on Ca 2+ and O2– , which, results in higher attraction. KI is more soluble, in water because of low lattice energy and, higher hydration energy. Clearly (c) is wrong, because CaO has higher melting point as, compared to KI., (c) MgSO4 is the only alkaline earth metal, sulphate which is soluble in water and for, solubility hydration energy should be greater, than lattice energy., (d) Sodium is obtained by electrolytic, reduction of its chloride. Melting point of, chloride of sodium is high (803°C) so in order, to lower its melting point(600°C), calcium, chloride is added to it., (c) Ca and CaH2 both react with water to form, H2 gas,, , 44., , 45., , Bleaching powder is a mixture of calcium, hypochlorite Ca (OCl) 2 , dibasic calcium, hypochlorite Ca (OCl)2. 2 Ca (OH)2 and dibasic, calcium chloride CaCl2 . 2Ca(OH)2., , 46., , 47., , Ñ, , Ca (HCO3 )2(s) ¾¾, ® CaCO3(s) + CO2(g) + H2 O, (A), (C), , 48., , 42., 43., , (b), , X, , heat, , Solid, boil, , Z, , CO2 + Residue, H2 O, , excess CO, ¬¾ ¾ ¾¾2¾ Y, , Clear solution, , The given properties coincide with CaCO3, , Na 2 O 2 + 2H 2 O ¾, ¾® 2 NaOH + H 2 O 2, Also, Ba gives H2 while BaO2 gives H2O2, ¾® Ba (OH)2 + H2, Ba + 2 H2O ¾, , ¾® Ba (OH)2 + H2 O2, BaO2 + 2H2O ¾, (c) Atomic size of K+ > Ca2+ > Mg2+ and that, of Cl– > F–. Therefore, Mg2+/Cl– ratio has the, minimum value., (b) 20Ca = 1s22s22p63s23p64s2 = [Ar] 4s2, (c) Within a period, the atomic size decreases, from left to right.Further atomic size increases, down the group. Hence the correct order is, i.e. Na > Mg > Li > Be., , D, (b) CaCO3 (s) ¾¾® CO2 (g) + CaO(s), (A), colourless residue, , CaO(s) + H 2 O ¾ ¾, ® Ca (OH )2, (B), Ca (OH )2 + 2CO2 + H2O ¾ ¾, ® Ca (HCO3 )2, (C), , 2KO 2 + 2H 2 O ¾¾, ® 2KOH + O 2 + H 2 O 2, Similarly, Na gives H2 while Na2O2 gives H2O2, 2Na + 2H 2O ¾¾, ® 2NaOH + H 2, , 41., , 1, H2O, 2, , Plaster of paris can be more accurately written as, (CaSO4)2 . H2O, , CaH 2 + 2H 2 O ¾, ¾® Ca (OH ) 2 + 2 H 2, whereas, K gives H2 while KO2 gives O2 and H2O2, , 2KO2 + 2H 2 O ¾¾, ® 2KOH + O2 + H 2 O2, , (b) (A) Plaster of paris = CaSO4., (B) Epsomite = MgSO4.7H2O, (C) Kieserite = MgSO4.H2O, (D) Gypsum = CaSO4.2H2O, , Ca + 2 H 2 O ¾, ¾® Ca ( OH ) 2 + H 2, , ¾® 2KOH + H2, 2K + 2 H2O ¾, , (c) Carbonates becomes more thermally stable, down the group, therefore MgCO3 will leave, CO2 easily., (b) Active ingredient in bleaching powder for, bleaching action is Ca (OCl)2., , heat, CaCO3 ¾ ¾ ¾® CO 2 + CaO, Residue, heat ' X ', H 2O, excess, Ca(HCO3 )2 ¬¾, ¾¾ Ca(OH)2, CO2, 'Z ', 'Y ', , 49., , (d) Gypsum is CaSO4. 2H2O and plaster of, Paris is (CaSO 4 ) 2 .H 2O. Therefore, gypsum, contains a lower percentage of calcium than, plaster of Paris.
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105, , The p-Block Elements (Group 13 & 14), , 11, , The p-Block Elements, (Group 13 & 14), , Trend Analysis with Important Topics & Sub-Topics, 2020, Topic Name, , Boron family, , Carbon family, , 2019, , oxides of carbon, multiple concept, (zeolites, allotropes,, oxides of carbon), inert pair effect, , 1, , A, , 1, , A, , E - Easy, , Topic 1: Boron Family, 1., , 2017, , 1, , properties of halides, of gp 14, LOD - Level of Difficulty, , 2018, , 2016, , Sub-Topic, QNS. LOD QNS. LOD QNS. LOD QNS. LOD QNS. LOD, multiple concept, 1, A, (structure,formulae), physical/chemical, properties of, 1, E, elements of gp 13, , Match the following and identify the correct, option., [2020], (A) CO(g) + H2(g), (i) Mg(HCO3)2 +, Ca(HCO3)2, (B) Temporary, (ii) An electron, hardness of, deficient, water, hydride, (C) B2H6, (iii) Synthesis gas, (D) H2O2, (iv) Non-planar, structure, (A), (B), (C), (D), (a) (iii), (ii), (i), (iv), (b) (iii), (iv) (ii), (i), (c) (i), (iii), (ii), (iv), (d) (iii), (i), (ii), (iv), , 2, , A, , A - Average, , 2., , A, , D - Difficult, , Qns - No. of Questions, , Aluminium chloride in acidified aqueous, solution forms a complex ‘A’, in which, hybridisation state of Al is ‘B’. What are ‘A’ and, ‘B’, respectively?, [NEET Odisha 2019], (a) [Al(H2O)6]3+, d2sp3, , (b) [Al(H2O)6]3+, sp3d 2, , (c) [Al(H2O)4]3+, sp3, , (d) [Al(H2O)4]3+, dsp2, 3., , Which one of the following elements is unable, to form ion?, [2018], (a) Ga, , 4., , (b) Al, , (c) In, (d) B, The stability of +1 oxidation state among Al, Ga,, In and Tl increases in the sequence :[2015 RS], (a) Ga < In < Al < Tl, , (b) Al < Ga < In < Tl, , (c) Tl < In < Ga < Al, , (d) In < Tl < Ga < Al
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EBD_8336, 106, , 5., , 6., , 7., , 8., , 9., , 10., , 11., , 12., , CHEMISTRY, Which of the following structure is similar to, graphite?, [NEET 2013], (a) B, (b) B4C, (c) B2H6, (d) BN, Aluminium is extracted from alumina (Al2O3 ), by electrolysis of a molten mixture of : [2012], (a) Al2O3 + HF + NaAlF4, (b) Al2O3 + CaF2 + NaAlF4, (c) Al2O3 + Na3AlF6 + CaF2, (d) Al2O3 + KF + Na3AlF6, The tendency of BF3, BCl3 and BBr3 to behave, as Lewis acid decreases in the sequence: [2010], (a) BCl3 > BF3 > BBr3, (b) BBr3 > BCl3 > BF3, (c) BBr3 > BF3 > BCl3, (d) BF3 > BCl3 > BBr3, Al2O3 can be converted to anhydrous AlCl3 by, heating, [2006], (a) Al2O3 with NaCl in solid state, (b) a mixture of Al2O3 and carbon in dry Cl2 gas, (c) Al2O3 with Cl2 gas, (d) Al2O3 with HCl gas, In borax bead test which compound is formed?, [2002], (a) Ortho-borate, (b) Meta-borate, (c) Double oxide, (d) Tetra-borate, Which of the following statements about H3BO3, is not correct ?, [1994], (a) It is a strong tribasic acid, (b) It is prepared by acidifying an aqueous, solution of borax, (c) It has a layer structure in which planar BO3, units are joined by hydrogen bonds, (d) It does not act as proton donor but acts as, a Lewis acid by accepting a lone pair of, electrons, Which of the following elements is extracted, commercially by the electrolysis of an aqueous, solution of its compound ?, [1993], (a) Cl, (b) Br, (c) Al, (d) Na, An example of a double salt is, [1989], (a) Bleaching powder (b) K4[Fe(CN)6], (c) Hypo, (d) Potash Alum, , 13., , 14., , 15., , 16., , 17., , 18., , Topic 2: Carbon Family, Which of the following is not correct about, carbon monoxide ?, [2020], (a) It reduces oxygen carrying ability of blood., (b) The carboxyhaemoglobin (haemoglobin, bound to CO) is less stable than, oxyhaemoglobin., (c) It is produced due to incomplete, combustion., (d) It forms carboxyhaemoglobin, Identify the correct statements from the, following :, [2020], (A) CO2(g) is used as refrigerant for ice-cream, and frozen food., (B) The structure of C60 contains twelve six, carbon rings and twenty five carbon rings., (C) ZSM-5, a type of zeolite, is used to convert, alcohols into gasoline., (D) CO is colorless and odourless gas., (a) (A) and (C) only, (b) (B) and (C) only, (c) (C) and (D) only, (d) (A), (B) and (C) only, Which of the following compounds is used in, cosmetic surgery?, [NEET Odisha 2019], (a) Zeolites, (b) Silica, (c) Silicates, (d) Silicones, Which of the following is incorrect statement?, (a) PbF4 is covalent in nature, [2019], (b) SiCl4 is easily hydrolysed, (c) GeX4 (X = F, Cl, Br, I) is more stable than GeX2, (d) SnF4 is ionic in nature, Which of the following species is not stable ?, [2019], (a) [SiF6]2–, (b) [GeCl6]2–, (c) [Sn(OH)6]2–, (d) [SiCl6]2–, It is because of inability of ns2 electrons of the, valence shell to participate in bonding that:[2017], 2+, 4+, (a) Sn is oxidising while Pb is reducing, (b) Sn 2+ and Pb2+ are both oxidising and, reducing, (c) Sn4+ is reducing while Pb4+ is oxidising, (d) Sn2+ is reducing while Pb4+ is oxidising
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107, , The p-Block Elements (Group 13 & 14), 19., , 20., , 21., , 22., , 23., , 24., , 1, 2, 3, , Which of these is not a monomer for a high, molecular mass silicone polymer?[NEET 2013], (a) Me2SiCl2, (b) Me3SiCl, (c) PhSiCl3, (d) MeSiCl3, The basic structural unit of silicates is :, [NEET 2013], 4–, (b) SiO32–, (a) SiO4, 2–, (d) SiO, (c) SiO4, Which statement is wrong? [NEET Kar. 2013], (a) Feldspars are not aluminosilicates, (b) Beryl is an example of cyclic silicate, (c) Mg2SiO4 is orthosilicate, (d) Basic structural unit in silicates is the SiO4, tetrahedron, Name the type of the structure of silicate in which, one oxygen atom of [SiO4]4– is shared ?[2011], (a) Linear chain silicate, (b) Sheet silicate, (c) Pyrosilicate, (d) Three dimensional, The straight chain polymer is formed by [2009], (a) hydrolysis of CH3 SiCl 3 followed by, condensation polymerisation, (b) hydrolysis of (CH 3 ) 4 Si by addition, polymerisation, (c) hydrolysis of (CH3)2SiCl 2 followed by, condensation polymerisation, (d) hydrolysis of (CH3 )3 SiCl followed by, condensation polymerisation, Which one of the following statements about, the zeolites is false ?, [2004], (a) They are used as cation exchangers, (b) They have open structure which enables, them to take up small molecules, , (d), (b), (d), , 4, 5, 6, , (b), (d), (c), , 7, 8, 9, , (b), (b), (b), , 10, 11, 12, , (a), (a), (d), , 25., , 26., , 27., , 28., , 29., , 30., , (c) Zeolites are aluminosilicates having three, dimensional network, (d) Some of the SiO44- units are replaced by, AlO 54- and AlO 96- ions in zeolites, Glass reacts with HF to produce, [2000], (a) SiF4, (b) H2SiF6, (c) H2SiO3, (d) Na3AlF6, In graphite, electrons are, [1993, 1994], (a) Localised on every third C-atom, (b) Present in anti-bonding orbital, (c) Localised on each C-atom, (d) Spread out between the structure, Which of the following types of forces bind, together the carbon atoms in diamond ?[1992], (a) Ionic, (b) Covalent, (c) Dipolar, (d) van der Waals., Water gas is produced by, [1992], (a) Passing steam through a red hot coke bed, (b) Saturating hydrogen with moisture, (c) Mixing oxygen and hydrogen in the ratio, of 1 : 2, (d) Heating a mixture of CO2 and CH4 in, petroleum refineries., Glass is a, [1991], (a) Liquid, (b) Solid, (c) Supercooled liquid, (d) Transparent organic polymer, The substance used as a smoke screen in, warfare is, [1989], (a) SiCl4, (b) PH3, (c) PCl5, (d) Acetylene, , ANSWER KEY, 13 (b) 16 (a) 19, 14 (c) 17 (d) 20, 15 (d) 18 (d) 21, , (b) 22, (a) 23, (a) 24, , (c), (c), (c), , 25, 26, 27, , (b), (d), (b), , 28, 29, 30, , (a), (c), (a)
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EBD_8336, 108, , CHEMISTRY, , Hints & Solutions, 1., , (d), A. Mixture of CO and H2 gases is known as, water gas or synthesis gas., B. Temporary hardness of water is due to, bicarbonates of calcium and magnesium., C. Diborane (B2H6) is an electron deficient, hydride., D. H2O2 is non-planar molecule having open, book like structure., , 2., , Acidified aq. sol., (b) AlCl3 ¾¾¾¾¾¾, ® [Al (H2O)6]3+, , 3., , (d) MF63–, Boron belongs to 2nd period and it does not, have vacant d-orbital., (b) Lower oxidation state become more stable, on moving down the group, Al < Ga < ln < Tl, , 9., , 5., , 6., , 7., , 8., , (d) Boron nitride (BN) is known as inorganic, graphite. The most stable form is hexagonal one., It has layered structure similar to graphite., (c) Fused alumina (Al2O3) is a bad conductor, of electricity. Therefore, cryolite (Na 3AlF6) and, fluorspar (CaF2) are added to purified alumina, which not only make alumina a good conductor, of electricity but also reduce the melting point, of the mixture to around 1140 K., (b) In BF3, p-p overlap between B and F is, maximum due to identical size and energy of, p-orbitals, so electron deficiency in boron of BF3, is neutralized partially to the maximum extent by, back donation. Also, the tendency to back, donate decreases from F to I. So the order will be:, BF3 < BCl3 < BBr3., (b) Al2O3 can be converted to anhydrous AlCl3, by heating a mixture of Al2O3 and carbon in dry, Cl2 gas., , Al2O3 + 3C + 3Cl2, , 1000°C, , 2AlCl3 + 3CO, , vapours, , cooled, , Solid anhydrous, aluminium, chloride, , 2, , D, , Na 2 B4O7 ¾¾, ® 2 NaBO2 +, anhydrous, , B2 O3, sod.metaborate Boric anhydride, , ®, CuO + B2 O3 ¾¾, 10., 11., , sp 3 d 2, , 4., , D, , ®, (b) Na2B4O7. 10 H2O ¾¾¾¾, -10H O, , Cu(BO 2 )2, cupric meta borate(Blue beed), , (a) H3BO3 is a weak monobasic acid., (a) On a commercial scale, chlorine is prepared, by electrolysis of an aqueous solution of sodium, chloride (brine solution). Cl2 is evolved at the, anode., (b) Br2 is prepared by passing Cl2 gas through, sea water containing NaBr, kBr etc., (c) Al is extracted by electrolysis of fused, mixture of alumina, cryolite and fluorspar., Products of electrolysis depend on the, concentration of electrolyte, overvoltage factor and, the electrodes used. Also, the commercial viability, is an important factor., , 12., 13., , 14., , (d) Potash Alum, K2SO4. Al2(SO4). 24H2O is a, double salt., (b) The carboxyhaemoglobin (haemoglobin, bound to CO), is about 300 times more stable, than oxygaemoglobin., (c), A. Dry ice, CO2(s), is used as refrigerant, B. C60 contains 20 six membered rings, 12 five, membered rings, Zeolites are crystalline, 3D aluminosilicates. ZSM5 contain two sets of perpendicular, intersecting, channel, one is 10 membered another is 8, membered., , 15., 16., 17., , (d) Silicones are used in cosmetic surgery., (a) PbF 4 is ionic in nature due to high, electronegativity difference., (d) [SiCl6]2– does not exist because six large, chloride ions cannot be accommodated around, Si4+, due to its small size.
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109, , The p-Block Elements (Group 13 & 14), 18., , (d) Due to inert pair effect, Pb(II) is more stable, than Pb(IV), Sn(IV) is more stable than Sn(II), \ Pb(IV) is easily reduced to Pb(II) and can, acts as an oxidising agent whereas Sn(II) is easily, oxidised to Sn(IV) and can acts as a reducing, agent., Inertness of ns2 electrons of the valence shell to, participate in bonding on moving down the group, in heavier p-block elements is called inert pair effect., It occurs due to poor shielding of the ns2 electrons, of the valence shell by the intervening d and f, electrons., , 19., , 20., 21., 22., , (b) Since Me3SiCl contains only one Cl,, therefore it can’t form high molecular mass, silicon polymer. It can form only dimer., (a) SiO44– is basic structural unit of silicates., (a) Feldspars are 3-dimensional aluminosilicates., O–, (c), O–, , Si, , –, O, , –, O, , Si, O, , Pyrosilicate (Si 2 O 7, , 23., , O–, , CH3, , Si, , CH3, , CH3, , 24., , CH3, , 5–, Zeolites have SiO4–, 4 and AlO4 tetrahedrons linked, together in a 3-D open structure in which 4 or 6, membered rings predominate. Due to open, structure they have cavities and can take up water, and other small molecules., , 25., 26., , 28., , (b) 6 HF + SiO 2 ® H 2SiF6 + 2 H 2 O, (d) In graphite, each carbon is sp2 -hybridized, and the single occupied unh ybridized, p-orbital of C-atoms overlap side ways to give, p -electron cloud which is delocalized and thus, the electrons are spread out between the, structure., (b) In diamond, each carbon atom is sp 3, hybridized and thus, forms covalent bonds with, four other carbon atoms lying at the corners of a, regular tetrahedron., (a) Water gas is made by blowing steam, through the layer of incandescent coal., Steam, , 29., OH, , Polymerisation of dimethylsilanol yields linear, thermoplastic polymer., , C, Red hot, , ¾¾, ® H 2 + CO, water gas, , (c) Glass is a super cooled liquid., Supercooling is the process of lowering the, temperature of a liquid or gas below its freezing, point without becoming a solid., , OH, , Dimethylsilanol, , CH3, , (c) Zeolites are crystaline solid structures, made up of silicon, aluminium and oxygen making, a framework with cavities., , H 2O +, , Si, , CH3, , HO — Si — O — Si — OH, , 27., , CH3, , CH3, CH3, , ) 6–, , Cl H OH – 2HCl, ¾¾®, +, Cl H OH, , CH3, , HO — Si — OH + H O —Si — OH ¾ ®, , O–, , (c) Hydrolysis of substituted chlorosilanes, yield corresponding silanols which undergo, polymerisation., CH3, , CH3, , 30., , (a) SiCl4 gets hydrolysed in moist air and gives, white fumes which are used as a smoke screen, in warfare.
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113, , Organic Chemistry - Some Basic Principles and Techniques, (c), , (d), , 25., , 26., 27., , 28., , 29., , 30., , 31., , (a) 120° and 120°, (c) 109.5° and 90°, , CH 3 - CH 2 - C = CH 2, |, CH 3, , CH3 - CH - CH = CH 2, |, CH3, , A straight chain hydrocarbon has the molecular, formula C8H10. The hybridization of the carbon, atoms from one end of the chain to the other are, respectively sp3, sp2, sp2, sp3, sp2, sp2, sp and, sp. The structural formula of the hydrocarbon, would be :, [1991], (a) CH3C º CCH2 – CH = CHCH = CH2, (b) CH3CH2 – CH = CHCH = CHC º CH, (c) CH3CH = CHCH2 – C º CCH = CH2, (d) CH3CH = CHCH2 – CH = CH – C º CH., An sp3 hybrid orbital contains, [1991], (a) 1/4 s-character, (b) 1/2 s-character, (c) 1/3 s-character, (d) 2/3 s-character., The shortest C – C bond distance is found in, [1991], (a) Diamond, (b) Ethane, (c) Benzene, (d) Acetylene, An organic compound X (molecular formula, C6H7O2N) has six carbon atoms in a ring system,, two double bonds an d a nitr o group as, substituent, X is, [1990], (a) Homocyclic but not aromatic, (b) Aromatic but not homocyclic, (c) Homocyclic and aromatic, (d) Heterocyclic and aromatic, Which of the following possesses a sp-carbon, in its structure ?, [1989], (a) CH2 = CCl – CH = CH2, (b) CCl2 = CCl2, (c) CH2 = C = CH2, (d) CH2 = CH – CH = CH2, Cyclic hydrocarbon ‘A’ has all the carbon and, hydrogen atoms in a single plane. All the carbon, carbon bonds have the same length, less than, 1.54 Å, but more than 1.34 Å. The C – C – C, bond angle will be, [1989], (a) 109°28', (b) 100°, (c) 180°, (d) 120°, The Cl – C – Cl angle in 1,1,2,2- tetrachloroethene, and tetrachloromethane respectively will be, about, [1988], , (b) 90° and 109.5°, (d) 120° and 109.5°, , Topic 2: Isomerism in Organic Compounds, 32., , Given, , CH3, , 33., , 34., , 35., , 36., , 37., , 38., , CH3, , CH3, CH3, , O, , O, , O, , (I), , (II), , (III), , CH3, CH3, , Which of the given compounds can exhibit, tautomerism?, [2015], (a) I and III, (b) II and III, (c) I, II and III, (d) I and II, Which one of the following pairs represents, stereoisomerism?, [2005], (a) Structural isomerism and Geometrical, isomerism, (b) Optical isomerism and Geometrical, isomerism, (c) Chain isomerism and Rotational isomerism., (d) Linkage isomerism and Geometrical, isomerism, .., , C H 2 - C - CH 3 and CH 2 = C - CH 3 are, ||, |, O, :O, . .:, [2002], (a) Resonating structures, (b) Tautomers, (c) Geometrical isomers, (d) Optical isomers, Tautomerism will be exhibited by, [1997], (a) (CH3)2NH, (b) (CH3)3CNO, (c) R3CNO2, (d) RCH2NO2, The number of possible isomers of the compound, with molecular formula C7H8O is, [1995], (a) 3, (b) 5, (c) 7, (d) 9, Isomers of a substance must have the same, [1991], (a) Structural formula, (b) Physical properties, (c) Chemical properties, (d) Molecular formula, How many chain isomers could be obtained from, the alkane C6H14?, [1988], (a) Four, (b) Five, (c) Six, (d) Seven
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EBD_8336, 114, , CHEMISTRY, , Topic 3: Concept of Reaction Mechanism in, Organic Compounds, 39., , 40., , A tertiary butyl carbocation is more stable than, a secondary butyl carbocation because of which, of the following ?, [2020], (a) + R effect of – CH3 groups, (b) – R effect of – CH3 groups, (c) Hyperconjugation, (d) – I effect of – CH3 groups, Paper chromatography is an example of, [2020], (a) Partition chromatography, (b) Thin layer chromatography, (c) Column chromatography, (d) Adsorption chromatography, , 41. The most stable car bocation, among the, following, is, [NEET Odisha, 2019], , 43., , Å, , (a), , (c), 44., , 45., , Å, , 46., , (c) CH3 – CH2 – CH – CH2 – CH3, Å, , (d) CH3 – CH – CH2 – CH2 – CH3, 42. Among the following, the reaction that proceeds, through an electrophilic substitution, is: [2019], (a), , (b), , +, , Cu Cl, , 2 2, N 2Cl– ¾¾®, , Cl + N 2, , AlCl3, , + Cl2 ¾¾®, , Cl + HCl, , Cl, (c), , Cl, , UV light, , + Cl2 ¾¾® Cl, , Cl, Cl, , 47., , Cl, 48., , (d), , Heat, , CH2OH + HCl ¾®, CH2Cl + H2O, , H, , (b), Y, , NO2, , Å, , Å, , Å, , Y, , (a) CH3 – CH2 – CH, (b) (CH3)3 C – CH – CH3, , Which of the following carbocations is expected, to be most stable?, [2018], NO2, NO2, , H, Y, , Å, , (d) H, Y, , H, NO2, , Å, , Which of the following is correct with respect, to – I effect of the substituents? (R = alkyl), [2018], (a) – NH2 < – OR < – F, (b) – NR2 < – OR < – F, (c) – NR2 > – OR > – F, (d) – NH2 > – OR > – F, The most suitable method of separation of 1 : 1, mixture of ortho and para-nitrophenols is :, [1994, 2017], (a) Chromatography (b) Crystallisation, (c) Steam distillation (d) Sublimation, The correct statement regarding electrophile is:[2017], (a) Electrophile is a negatively charged species, and can form a bond by accepting a pair of, electrons from another electrophile, (b) Electrophiles are generally neutral species, and can form a bond by accepting a pair of, electrons from a nucleophile, (c) Electrophile can be either neutral or, positively charged species and can form a, bond by accepting a pair of electrons from, a nucleophile, (d) Electrophile is a negatively charged species, and can form a bond by accepting a pair of, electrons from a nucleophile, The pair of electron in the given carbanion,, CH3C º C- , is present in which of the following, orbitals ?, [2016], (a) 2p, (b) sp3, (c) sp2, (d) sp, Which of the following statements is not correct, for a nucleophile?, [2015 RS], (a) Nucleophile is a Lewis acid, (b) Ammonia is a nucleophile, (c) Nucleophiles attack on less e– density sites, (d) Nucleophiles are not electron seeking.
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115, , Organic Chemistry - Some Basic Principles and Techniques, 49., , Consider the following compounds, CH3, , Ph, , CH3—C—CH—, , Ph—C—PH, , [2015], , 54., , CH3, , CH3, (I), , 50., , (II), , (III), , Hyperconjugation occurs in :, (a) II only, (b) III only, (c) I and III, (d) I only, Which of the following is the most correct, electron displacement for a nucleophilic reaction, to take place?, [2015], H, H2, H3C—C = C – C – Cl, H, H, H2, (b) H 3C—C = C – C – Cl, H, H, H2, (c) H 3C—C = C – C – Cl, H, H, H2, (d) H 3C—C = C – C – Cl, H, In Duma's method for estimation of nitrogen,, 0.25 g of an organic compound gave 40 mL of, nitrogen collected at 300 K temperature and 725, mm pressure. If the aqueous tension at 300 K is, 25 mm, the percentage of nitrogen in the, compound is :, [2015], (a) 18.20, (b) 16.76, (c) 15.76, (d) 17.36, In the Kjeldahl’s method for estimation of, nitrogen present in a soil sample, ammonia, evolved from 0.75 g of sample neutralized 10 mL, of 1 M H2SO4. The percentage of nitrogen in, the soil is :, [2014], (a) 37.33, (b) 45.33, (c) 35.33, (d) 43.33, Arrange the following in increasing order of, stability, [NEET Kar. 2013], , 52., , 53., , Å, , g, , Å, , (C) (CH3 )2 - CH, (E), , 55., , 56., , 57., , 58., , Å, , (D) CH3 - CH 2, , Å, , CH 3, , (a) E < D < C < B < A (b) E < D < C < A < B, (c) D < E < C < A < B (d) A < E < D < C < B, , 59., , g, , (c) CH3 – C H2 < CH3 – C H – CH3 <, g, g, (CH3)2 C – CH2 – CH3 < (CH3)3 C, g, g, (d) CH3 – C H2 < CH3 – C H – CH3 < (CH3)3, g, g, C < (CH3)2 C – CH2CH3, What is the hybridisation state of benzyl, carbonium ion, , Å, , (A) (CH3 )2 C - CH2CH3 (B) (CH3 )3 - C, , g, , g, , (b) (CH3)2 C – CH2CH3 < CH3 – C H – CH3 <, g, g, CH3 – C H2 < (CH3)3 C, , (a), , 51., , Homolytic fission of the following alkanes forms, free radicals CH3 – CH3, CH3 – CH2 – CH3,, (CH3)2 CH – CH3, CH3 – CH2 – CH (CH3)2., Increasing order of stability of the radicals is, [NEET Kar. 2013], g, g, (a) (CH3)3 C < (CH3)2 C – CH2CH3 <, g, g, CH3 – C H – CH3 < CH3 – C H2, , +, , CH2?, , [NEET Kar. 2013], (a) sp3, (b) sp2, (c) spd2, (d) sp2 d, Nitrogen detection in an organic compound is, carried out by Lassaigne’s test. The blue colour, formed corresponds to which of the following, formulae?, [NEET Kar. 2013], (a) Fe3[Fe(CN)6]3, (b) Fe3[Fe(CN)6]2, (c) Fe4[Fe(CN)6]3, (d) Fe4[Fe(CN)6]2, The correct order of increasing bond length of, C – H, C – O, C – C and C = C is :, [2011], (a) C – H < C = C < C – O < C – C, (b) C – C < C = C < C – O < C – H, (c) C – O < C – H < C – C < C = C, (d) C – H < C – O < C – C < C = C, In Duma's method of estimation of nitrogen, 0.35 g of an organic compound gave 55 mL of, nitrogen collected at 300 K temperature and, 715 mm pressure. The percentage composition, of nitrogen in the compound would be :, (Aqueous tension at 300 K = 15 mm) [2011], (a) 15.45, (b) 16.45, (c) 17.45, (d) 14.45, The Lassaigne’s extract is boiled with conc., HNO3 while testing for halogens. By doing so it, [2011], (a) decomposes Na2S and NaCN, if formed., (b) helps in the precipitation of AgCl., (c) increases the solubility product of AgCl., (d) increases the concentration of NO3– ions.
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EBD_8336, 116, , 60., , CHEMISTRY, The stability of carbanions in the following :, (a), , 61., , RC º C, , (b), , (c) R 2C = CH, (d) R 3C - CH 2, is in the order of :, [2008], (a) (a) > (b) > (c) > (d) (b) (b) > (c) > (d) > (a), (c) (d) > (b) > (c) > (a) (d) (a) > (c) > (b) > (d), Which amongst the following is the most stable, carbocation?, [2005], (a), , Å, , (b), , CH3, , 62., , 63., , 64., , 66., , Å, , CH3 CH 2, , 67., , CH3, |, , Å, , (c), , 65., , CH3- CH, |, CH3, , (d), , CH3 C Å, |, , CH3, , The best method for the separati on of, naphthalene and benzoic acid from their, mixture is, [2005], (a) distillation, (b) sublimation, (c) chromatography (d) crystallisation, In steam distillation of toluene, the pressure for, toluene in vapour is, [2001], (a) Equal to pressure of barometer, (b) Less than pressure of barometer, (c) Equal to vapour pressure of toluene in, simple distillation, (d) More than vapour pressure of toluene in, simple distillation, Which one of the following order is correct, regarding the – I effect of the substituents ?, [1998], (a) –NR2 < – OR > –F (b) –NR2 > –OR > –F, (c) –NR2 < –OR < –F (d) –NR2 > –OR < –F, , 68., , Which one of the following is a technique most, suitable for purification of cyclohexanone from, a mixture containing benzoic acid, isoamyl, alcohol, cyclohexane and cyclohexanone? [1997], (a) Crystallization, (b) Sublimation, (c) IR spectroscopy, (d) Gas chromatography, Lassaigne’s test for the detection of nitrogen, fails in, [1994], (a) NH2CONHNH2.HCl (b) NH2NH2.HCl, (c) NH2CONH2, (d) C6H5NHNH2.HCl, A is a lighter phenol and B is an aromatic, carboxylic acid. Separation of a mixture of A and, B can be carried out easily by using a solution, of, [1992], (a) Sodium hydroxide (b) Sodium sulphate, (c) calcium chloride (d) Sodium bicarbonate, Which of the following is the most stable, carbocation (carbonium ion) ?, [1991], (a), , +, , (b), , CH3 C H2, , +, , ( CH 3 ) 2 C H, , +, , 69., , 70., , 71., , +, , (c) ( CH 3 ) 3 C, (d) C 6 H 5 C H 2, In sodium fusion test of organic compounds,, the nitrogen of the organic compound is, converted into, [1991], (a) Sodamide, (b) Sodium cyanide, (c) Sodium nitrite, (d) Sodium nitrate, Kjeldahl’s method is used in the estimation of, [1990], (a) Nitrogen, (b) Halogens, (c) Sulphur, (d) Oxygen, Lassaigne’s test is used in qualitative analysis, to detect, [1989], (a) Nitrogen, (b) Sulphur, (c) Chlorine, (d) All of these, , ANSWER KEY, 1, , (d), , 9, , (c), , 17, , (b ), , 25, , (d), , 33, , (b), , 41, , (d), , 49, , (b), , 57, , (a), , 65, , (c), , 2, , (a), , 10, , (c), , 18, , (d), , 26, , (a), , 34, , (a), , 42, , (b), , 50, , (b), , 58, , (b ), , 66, , (b), , 3, , (d), , 11, , (a), , 19, , (c), , 27, , (d), , 35, , (d), , 43, , (d), , 51, , (b), , 59, , (a), , 67, , (d), , 4, , (a), , 12, , (a), , 20, , (b), , 28, , (a), , 36, , (b), , 44, , (a), , 52, , (a), , 60, , (a), , 68, , (c), , 5, , (a), , 13, , (b), , 21, , (d), , 29, , (c), , 37, , (d), , 45, , (c), , 53, , (b), , 61, , (d ), , 69, , (b), , 6, , (a), , 14, , (b), , 22, , (d), , 30, , (d), , 38, , (b), , 46, , (c), , 54, , (c), , 62, , (b ), , 70, , (a), , 7, , (d), , 15, , (a), , 23, , (b), , 31, , (d), , 39, , (c), , 47, , (d), , 55, , (b), , 63, , (b ), , 71, , (d), , 8, , (b), , 16, , (c), , 24, , (b), , 32, , (c), , 40, , (a), , 48, , (a), , 56, , (c), , 64, , (c)
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117, , Organic Chemistry - Some Basic Principles and Techniques, , Hints & Solutions, O, 1., , O, 2, , C, , (d) H, , 9., , 4, , 3, , 1, , 5, 6, 3-keto-2-methylhex-4-enal, , Aldehydes get higher priority over ketone and, alkene in numbering of principal carbon chain., OH, , 2., , (a), , 3, , H3C, , 4, , 6, , CH 3, , 5, , 3., 4., , IUPAC name of the structure is, 3-ethyl-2-hydroxy-4-methylhex-3-en-5-ynoic acid, CH 3, (d), CH - CH 2 - (isobutyl group), CH 3, (a) Br CH2 CH, CH 2, 3, , 5., , 1, COOH, , 2, , 4 ethyl-3-propylhex-1-ene, , (a), , Cl, CH3, 1, , 4, , C=C, 2, , 3, , 4, , 10. (c), , 5, , CH3, |, , CH 3CH 2 CH 2 - C H - C H - CH 2 CH 3, |, , CH 2CH3, , is 4-Ethyl-3-methylheptane, 13. (b) CH3, , 3, , 1, , CH3, 4, , (b) C – 1 is sp hybridized (C º C), C – 3 is sp3 hybridized (C– C), , 5, , 2, , 7, 8 CH3, , 6, , I, , CH3, CH3, 3 |, sp, sp, sp 2, sp3 |, (d) CH3, C, CH = CH, CH — C º C H, 6|, 7, 5, 4, 3, 2, 1, CH3, , Cl, , 12. (a) Correct IUPAC name of, , CH3, 4–ethyl-3¢–methyloctane, , 14. (b) The IUPAC name of, 1, , 8., , 1, , C, ||, O, , Contribution of s in sp hybrid orbital is maximum, (50%) so this orbital is closer to nucleus. It will, have greater tendency to pull electron towards it., Hence it becomes more electronegative and sp 3, becomes least electronegative as it has only 25%, s-character., , CH2CH3, , Correct IUPAC name of above compound is, trans-2-chloro-3-iodo-2-pentene, , 7., , 2, , 3, , 5, , It is 2, 3-dimethylpentanoyl chloride., 11. (a) Among the three given hybrid orbitals, sp, hybrid orbital is most electronegative., , 1, , 2, , The correct name is 3-Bromoprop-1-ene., (a) The given compound is, 2, 1, CH = CH2, CH3 — CH2 — CH2 — C — C — CH2 — CH3, 3, 4, CH2 – CH3, 5, 6, , 6., , C – 5 is sp2 hybridized (C = C), Thus the correct sequence is sp, sp3, sp2., (c) General molecular formula of alkanols is, CnH2n + 2O , (CnH2n + 1OH), , 2, , 3, , 4, , 5, , 6, , C H2 = C H — C H2 — C H2 — C º C H, is Hex-1-en-5-yne or 1-hexene-5-yne, The lowest number is given to the C = C double, bond.
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EBD_8336, 118, , CHEMISTRY, , 1, 2 3, 4, 15. (a) CH3 C CH CH3, , The vital force theory suffered first death blow in, 1828 when Wohler synthesized the Ist organic, compound urea in the laboratory from inorganic, compounds reported below :, , O CH3, 3 Methyl-2- butanone, , to isomeric charge, , NH 4 CNO ¾¾ ¾ ¾ ¾ ¾ ¾¾® NH 2 CONH2, , 1, 2 3, CH3 CH CH CH3, 5, 4, CH3 CH2 CH3, , rearrangement leading, , 2, 3 dimethyl pentane, , 4, 3 2, 1, CH3 CH CH CH3, Cl, , 20. (b), , 1, , Br, , 2 Bromo-3-chlorobutane, , 2, 1, CH3 C, , 21. (d) Resonance structures are separated by a, double headed arrow ( ¬¾® ), 22. (d) Angle increases progressively, sp3 ( 109°28' ), sp2 (120°), sp (180°), 23. (b) 5 CH - 4 CH - 3CH = 2C - 1CHO, 3, |, |, OH, CH3, , 4-Methyl-2-pentyne, sp 2, , sp, , sp 2, , 2, , 3, , 4, , 16. (c) CH 3 - CH = C = CH 2, 1, , CH3 - 2CH - 3CH - 4 CH 2 - 5CH3, |, |, CH3 CH 2 CH3, 3-Ethyl-2-methylpentane, , 5CH, 3, 3 4, C CH, CH3, , sp3, , 5, , 6, , 4-Hydroxy-2-methylpent-2-en-1-al, , 7, , CH3, |, , 17. (b) C H 3 – C H – C H – C H 2 – C H 2 – C H – C H 3, |, |, CH3 CH3, , |, CH 3, , 24. (b), , When many substituents are present, the, numbering is done from the end where the first, point of difference is the lowest number., CH 3 – 6 CH – 5 CH – 4 CH 2 – 3 CH 2 – 2 CH – 1CH3, |, |, |, CH3 CH 3, CH 3, , Set of locants = 2, 5, 6, 1, , CH3 – 2 CH – 3 CH – 4 CH 2 – 5 CH 2 – 6 CH – 7 CH 3, |, |, |, CH3 CH 3, CH 3, , Set of locants = 2, 3, 6, The first point of difference 3 is lower than the, number 5 of first set. Thus, the correct name is 2,, 3, 6-trimethylheptane., , 18. (d) Huckel’s rule states that for aromaticity, there must be (4n + 2) p electrons present in a, compound, where n is an integer., 19. (c) Urea is the first organic compound, synthesized in the laboratory by wohler., , 1, , C H 3 - 2 C = 3 C H - 4 CH 3, 2-Methyl-2-butene, , 2, 3, 6 – trimethylheptane, , 7, , Urea, , Later on a further blow to vital force theory was, given by Kolbe (1845) who prepared acetic acid,, the first organic compound, in laboratory from its, elements., , sp3, , sp2, , sp 2, , sp3, , sp2, , sp 2, , sp, , sp, , 25. (d) CH3 – CH = CH – CH2 - CH = CH– C º CH, 26. (a) sp3 orbital has 1/4(25%) s-character & 75%, p-character., 27. (d) Shortest C – C distance (1.20 Å) is in, acetylene, as acetylene has sp hybridisation., The bond length increases in the order, Cº C < C= C < C-C, , sp (1.20Å), , sp 2 (1.34Å), , sp (1.54Å), , NO 2, NO 2, , 28. (a), , or, , Hence it is homocyclic (as the ring system is made, of one type of atoms, i.e. carbon) but not aromatic., As it does not follow (4n +2)p electron rule of, aromaticity.
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119, , Organic Chemistry - Some Basic Principles and Techniques, sp 2, , sp 2, , sp, , 29. (c) CH 2 = C = CH 2, 30. (d) All the properties mentioned in the question, suggest that it is a benzene molecule. Since in, benzene all carbons are sp2–hybridized, therefore,, C – C – C angle is 120°, H, 1.09Å, , H, , C, , C, , Tautomerism is exhibited by the oscillation of, hydrogen atom between two polyvalent atoms, present in the molecule., , H, 1.40Å, , 120°, , C, , C, , 120°, , C, , 36. (b) The possible isomers of the compound with, molecular formula C 7 H8 O is 5. These are, C6H5OCH3, C6H5CH2OH and, , H, , C, , (Anisole), , H, , H, , 31. (d) Tetrachloroethene being an alkene has sp2 hybridized C– atoms and hence the angle, Cl – C – Cl is 120° while in tetrachloromethane,, carbon is sp3 hybridized, therefore the angle Cl –, C – Cl is 109.5°., 32. (c) All of these compounds show tautomerism, H3C, , CH3, , H3C, , CH3, , H, O, , OH, CH3, , CH3, CH3, , H, O, O, H, H, CH3, , CH3, , CH3, , CH3, OH, , 33. (b) Optical and geometrical isomerism pair up, to exhibit stereoisomerism. This is because the, isomers differ only in their orientation in space., –, , CH2 = C — CH3, , O, , O, , I, , –, II, , both are resonating structures., , o- cresol, , OH, , CH3, , CH3, , OH, , OH, , p-cresol, , m-cresol, , 37. (d) Organic compounds having same molecular, formula but differ from each other in physical or, chemical properties or structural formula are, known as isomers., 38. (b) Five chain isomers are possible which are –, (i) CH 3 - CH 2 - CH 2 - CH 2 - CH 2 - CH 3, (ii) CH3 - CH - CH 2 - CH 2 - CH3, |, CH3, 2-methyl pentane, , OH, OH, , O, , CH3, , benzyl alcohol, , n-hexane, , CH3, , 34. (a) CH2 — C — CH3, , 35. (d) As option (d) has a-hydrogen atom., Therefore it shows tautomerism whereas other, structures do not., O, OH, R CH2 N, R CH N, O, O, Nitroform, Aciform, , CH3, |, (iii) CH3 - C - CH 2 - CH3, |, CH3, 2, 2-dimethyl butane, , (iv) CH3 - CH - CH - CH 3, |, |, CH3 CH3, 2, 3-dimethyl butane, , (v) CH3 - CH 2 - CH - CH 2 - CH3, |, CH3, 3-methyl pentane
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EBD_8336, 120, , 39., , CHEMISTRY, 48. (a) nucleophiles are usually lewis bases., , (c), , CH3, , Å, , H3C – C – CH3, , Å, Tertiary butyl carbocation, (9 a-H atoms), , H3C – CH – CH2 – CH3, Secondary butyl carbocation, (5 a-H atom), , t-Butyl carbocation is more stable due to hyperconjugation., 40. (a) Paper chromatography is a type of partition, chromatography., 41. (d) Stability of carbocation µ No. of a-H, Among the given carbocations,, a -H, , 42., , Å, , a -H, , CH3 – CH– CH 2 – CH 2 – CH3 is the most, stable carbocation (5 a-H), (b) (i) AlCl3 + Cl – Cl ® [AlCl4]– + Cl+, Cl, + Cl ¾®, , (d) –NO2 group is meta-directing group, NO2, NO2, , Y, , +, H, , (Less stable due to more, , 44., , +, , ¬¾®, H, Y, , Å, CH 3—CH—CH=CH 2, , Å, CH3 – CH=CH—CH2, , Hyperinflation results from the interaction of s, bond electrons with an adjacent empty or partially, filled p-orbital or a orbital to give an extended, molecular othital that increase the stability., , V2 (Volume of nitrogen at STP), , H, , withdrawing effect of –NO2), , NO2, , NO2, H, Y, , Y, e–, , CH 3 – CH=CH.CH2 —Cl, , PV, 1 1 = P2V2, T1, T2, , +, , ¬¾®, , 49. (b) Only structure (III) has H in conjugation with, free radical., So, hyperconjugation is possible in III only., –, 50. (b) p bond is transferred after leaving Cl, , 51. (b) Wt. of organic substance = 0.25 g, V1 = 40 mL, T1 = 300 K, P1 = 725 – 25 = 700 mm of Hg, P2 = 760 mm of Hg (at STP), T2 = 273 K, , +, , (ii), 43., , Electrophile, , Nucleophile is a species that provide electron while, species which are deficient of electrons are termed, as lewis acid., , NO2, , +, +, ¬¾®, H, Y, –, , (More stable due to less e withdrawing effect of –, NO 2) greater no. of resonating structures., , (a) –I effect increases on increasing, electronegativity of atom. So, correct order of, –I effect is –NH2 < – OR < – F., 45. (c) The o-isomer is steam volatile due to intramolecular H-bonding. The p-isomer is not steam, volatile due to intermolecular H-bonding or, association of molecules. Thus, both can be, separated by steam distillation., 46. (c) Electrophite is a electron deficient species, and can accpet pair of electrons from nucleophite, 47. (d) CH3—C º C, No.of s bp - 1 ù, lp –1 û 2 & hybridisation is sp, , =, , 273 ´ 700 ´ 40, = 33.52 mL, 300 ´ 760, , Percentage of nitrogen, , 28× volume of N 2 at STP ×100, = 22400 × wt. of organic substance, =, , 28 ´ 33.52 ´ 100, = 16.76%, 22400 ´ 0.25, , 52. (a) 10 mL, 1 M H2SO4 = 20 mL, 1 M NH3, Q wt of N in one mole NH3 = 14, \ 20 × 10– 3 mol NH3 ¾®, 20 × 10– 3 × 14 nitrogen, \ 0.75 g of sample contains, 14 ´ 20 ´ 10 - 3, ´ 100 = 37.33%, 0.75, 53. (b) Greater the number of e– donating alkyl, groups (+I effect), greater will be the stability of, carbocations., , =
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121, , Organic Chemistry - Some Basic Principles and Techniques, 54. (c) Stability depends on, hyperconjugative structure., , number, , of, , 55. (b) In the carbonium ion the carbon atom, carrying the positive charge is sp2 hybridized., 56. (c) The prussian blue colour is of Fe4 [Fe(CN)6]3, ferric ferrocyanide, 57. (a) Bond length order is, C-H < C =C < C-O < C-C, , 1.10A °, , 1.34A °, , 1.40A °, , 1.54A °, , 58. (b) Given wt of compound taken (w) = 0.35 g, Volume of nitrogen collected (V) = 55 mL, Room temperature (t K) = 300 K, Atmospheric pressure (P) = 715 mm, Aq. tension (r) = 15 mm, Calculation Volume of N2 at NTP =, =, , ( P - ρ) ´ V 273, ´, mL, t, 760, , (715 - 15) ´ 55 273, ´, = 46.098 mL, 300, 760, , % of nitrogen, 28 ´ vol.of N 2 at NTP ´ 100, =, 22400 ´ wt of organic compound, =, , 28 ´ 46.098 ´ 100, = 16.46%, 22400 ´ 0.35, , 59. (a) Conc. HNO3 decomposes NaCN and Na2S, to avoid their interference., NaCN + HNO 3 ¾¾, ® NaNO 3 + HCN , , Na 2S + 2HNO3 ¾¾, ® 2NaNO 3 + H 2S , , 60. (a) As 's' character increases, stability of, carbanion increases. Also, electron donating, groups decreases the stability of carbanion., Hence, the correct stability, order:, sp 2, , > R2, , C >, sp, , R3, , +1, , C CH2, , C=CH, 2, sp, , 61. (d) More the number of alkyl groups, greater, will be the dispersal of positive charge and, therefore more the stability of carbocation., CH3, |, , Thus CH3 –C +, |, , is most stable, , CH3, , 62. (b) Among the given compounds naphthalene, is volatile but benzoic acid is non-volatile (it, forms a dimer). So, the best method for their, separation is sublimation, Sublimation method is applicable to compounds, which can be converted directly into the vapour, phase from its solid state on heating and back to, the solid state on cooling., , 63. (b) The principle of steam distillation is based, on Dalton's law of partial pressures. Suppose p1, and p2 be the vapour pressures of water vapour, and the toluene at the distillation temperature., Toluene boils when total pressure is equal to, atmospheric pressure p i.e p = p1 + p2, or p2 = p – p1, As a result, when toluene boil in the presence of, steam its partial pressure p 2 is less than, atmospheric pressure., 64. (c) As electronegativity of N, O & F follow the, order N < O < F hence based upon, electronegative character order of-I effect is, – NR2 < – OR < – F., The atom or group which has more power to attract, electrons in comparision to hydrogen is said to, have -I effect. Thus higher the electronegativity of, atom, stronger will be the -I effect., , 65. (c) IR spectroscopy is used for the purification of, cyclohexanone from a mixture of benzoic, acid, isoamyl alcohol, cyclohexane and, cyclohexanone. Because in this method, each, functional group appear at a certain peak. So,, cyclohexanone can be identified by carbonyl, peak., An amyl alcohol is any one of 8 alcohols with the, formula C5H12O (alkanol with 5 carbons)., , 66. (b) Hydrazine (NH2 NH2 ) does not contain, carbon and hence on fusion with Na metal, it, cannot form NaCN; so it does not show, Lassaigne’s test for nitrogen.
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EBD_8336, 122, , CHEMISTRY, This test fails in case of diazo compounds, dinitro, compounds and compounds containing nitrogen, in the ring., , 67. (d) Carboxylic acids dissolve in NaHCO 3 and, evolve CO2 gas but phenols do not., NaHCO, , 3, RCOOH ¾¾¾¾, ® RCOONa + H 2O + CO2, 68. (c) Higher the +I effect greater is stability of the, species. Thus,, , +, , +, , +, , (CH3 )3 C + > (CH3 )2 C H > C6 H 5 C H 2 > CH 3 C H 2, , Also, primary benzyl carbocation have almost, the same stability as 2°-alkyl carbocations., 69. (b) Sodium cyanide (Na + C + N ® NaCN)., (Lassaigne's test), 70. (a) Kjeldahl's method is suitable for estimating, nitrogen in those compounds in which nitrogen, is linked to carbon and hydrogen. This method, , is basically used for estimating nitrogen in food, fertilizers and agricultural products., Kjeldahl's method is not used in case of nitro, azo, and azoxy compound., , 71. (d) Nitrogen, sulphur and halogens are tested, in an organic compound by Lassaigne's test., The organic compound is fused with sodium metal, as to convert these elements into ionisable inorganic, substances,, , Na + C + N ¾¾, ® NaCN, 2Na + S ¾¾, ® Na 2S, 2Na + X 2 ¾¾, ® 2NaX, The cyanide, sulphide or halide ions can be, confirmed in aqueous solution by usual test.
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13, , Hydrocarbons, , Trend Analysis with Important Topics & Sub-Topics, 2020, Topic Name, , Alkanes, , Alkenes, , 2019, , LOD - Level of Difficulty, , E - Easy, , 1, , A - Average, , Topic 1: Alkanes, 1., , 2., , 3., , 2017, , 2016, , Sub-Topic, QNS. LOD QNS. LOD QNS. LOD QNS. LOD QNS. LOD, conversion, 1, E, (substitution/ wurtz, reaction), conformations of, 1, A, 1, E, ethane, ozonolysis, 1, A, 1, A, reduction of alkyne, order of, acidity/acidic, character, hydration reaction, , Alkyne, , 2018, , The alkane that gives only one mono-chloro, product on chlorination with Cl2 in presence of, diffused sunlight is, [NEET Odisha 2019], (a) Isopentane, (b) 2, 2-dimethylbutane, (c) neopentane, (d) n-pentane, Hydrocarbon (A) reacts with bromine by, substitution to form an alkyl bromide which by, Wurtz r eaction is converted to gaseous, hydrocarbon containing less than four carbon, atoms. (A) is, [2018], (a) CH º CH, (b) CH2 = CH2, (c) CH4, (d) CH3 – CH3, With respect to the conformers of ethane, which, of the following statements is true ? [2017], (a) Bond angle changes but bond length, remains same, (b) Both bond angle and bond length change, , 4., , E, , D - Difficult, , 1, , E, , 1, , A, , 1, , A, , Qns - No. of Questions, , (c) Both bond angles and bond length remains, same, (d) Bond angle remains same but bond length, changes, The correct statement regarding the comparison, of staggered and eclipsed conformation of, ethane, is, [2016], (a) The staggered conformation of ethane is, less stable than eclipsed conformation,, because staggered conformation has, torsional strain, (b) The eclipsed conformation of ethane is, more stable than staggered conformation,, because eclipsed conformation has no, torsional strain, (c) The eclipsed conformation of ethane is, more stable than staggered conformation, even though the eclipsed conformation has, torsional strain, (d) The staggered conformation of ethane is, more stable than eclipsed conformation,, because staggered conformation has no, torsional strain.
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EBD_8336, 124, , 5., , CHEMISTRY, Given, , CH3, , H, H, , Br, , Br, , CH3, , and, , H, H, , CH3, , H, , CH3, , (a), , CH3, , 8., , H, CH3, CH3, H, , H, , CH3, CH3, , (c), , 10., , (d), H, H, , HO, H, , 9., H, , H, , (d), , H, , (b), H, , H, H, , HH, , H3C, , H, , Which of the following conformers for ethylene, glycol is most stable?, [2010], , 11., , OH, , H, , OH, , 12., , (a), H, , H, H, , 13., , OH, , H, , H, , (b), H, , H, OH, , H H, OH, H, , II, , [NEET Kar. 2013], I and II are, (a) A pair of optical isomers, (b) Identical, (c) A pair of conformers, (d) A pair of geometrical isomers, In the following the most stable conformation, of n-butane is:, [2010], H, , (c), , Br, CH3, , CH3, , I, , 7., , H, , H, , Br, , 6., , OH, OH, , H H, , Liquid hydrocarbons can be converted to a, mixture of gaseous hydrocarbons by : [2010], (a) Oxidation, (b) Cracking, (c) Distillation under reduced pressure, (d) Hydrolysis, A compound of molecular formula of C7H16, shows optical isomerism, compound will be, (a) 2, 3-Dimethylpentane, [2001], (b) 2,2-Dimethylbutane, (c) 2-Methylhexane, (d) None of the above, The reaction of ethyl magnesium bromide with, water would give, [1999], (a) Ethane, (b) Ethyl alcohol, (c) Ethyl bromide, (d) Ethyl ether, In commercial gasolines the type of, hydrocarbons which are more desirable, is, (a) branched hydrocarbons, [1997], (b) straight-chain hydrocarbons, (c) aromatic hydrocarbons such as toluene, (d) linear unsaturated hydrocarbons, The most stable conformation of n-butane is, [1997], (a) skew boat, (b) gauche, (c) staggered-anti, (d) eclipsed, Which one of the following reactions is, expected to readily give a hydrocarbon product, in good yields ?, [1997], (a), , Electrolyt ic, , RCOOK ¾¾ ¾ ¾¾®, -, , oxidation, Br2, +, , (b), , ¾®, RCOO Ag ¾¾, , (c), , 2®, CH 3 CH 3 ¾¾, ¾, , (d), , (CH 3 ) 3 CCl ¾¾ ¾ ¾®, , Cl, , hu, C 2 H 5OH
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EBD_8336, 128, , 36., , 37., , CHEMISTRY, Correct order of stability is :, [2000], (a) cis -2- butene > 1-butene > trans -2-butene, (b) trans-2-butene > cis-2-butene > 1-butene, (c) 1-butene > cis-2-butene > trans-2- butene, (d) cis-2-butene > trans-2-butene > 1-butene, The correct structure of trans-2 hexenal is [1999], CHO, (a), (c), , 38., , 39., , 40., , 41., , (b), , CHO, , (d), , CHO, , O, In the presence of platinum catalyst,, hydrocarbon A adds hydrogen to form n-hexane., When hydrogen bromide is added to A instead, of hydrogen, only a single bromo compound is, formed. Which of the following is A? [1996], (a) CH 3 — CH 2 — CH = CH — CH 2 — CH 3, (b) CH 3 — CH 2 — CH 2 — CH = CH — CH 3, (c) CH 3 — CH = CH — CH 2 — CH 2 — CH 3, (d) CH 2 = CH — CH 2 — CH 2 — CH 2 — CH 3, Which of the following will not show cis-trans, isomerism?, [1996], , CH 3 — CH = CH — CH 3, , (b) CH 3 — CH 2 — CH = CH — CH 2 — CH 3, (c) CH — CH = CH — CH — CH, 3, 2, 3, |, CH3, (d), , CH 3— C H — CH = CH — CH 2 — CH 3, , |, CH 3, , 43., , CHO, , A hydrocarbon ‘A’ on chlorination gives ‘B’, which on heating with alcoholic potassium, hydroxide changes into another hydrocarbon, ‘C’. The latter decolourises Baeyer's reagent and, on ozonolysis forms formaldehyde only. ‘A’ is, (a) Ethane, (b) Butane [1998], (c) Methane, (d) Ethene, In reaction sequence, [1997], CH2OH, Hypochlorous, R, CH 2 = CH 2 ¾¾ ¾¾¾, ¾® M ¾¾® |, acid, CH2OH, molecule 'M' and reagent 'R' respectively are, (a) CH3CH2Cl and NaOH, (b) CH3CH2OH and H2SO4, (c) CH2Cl – CH2OH and aqueous NaHCO3, (d) CH2 — CH2 and heat, , (a), , 42., , When 3, 3-dimethyl –2-butanol is heated with, H2SO4, the major product obtained is [1995], (a) 2, 3-dimethyl –2-butene, (b) 3, 3-dimethyl –1- butene, (c) 2, 3-dimethyl –1- butene, (d) cis & trans isomers of 2, 3-dimethyl – 2-butene, The alkene R – CH = CH2 reacts readily with, B2H6 and formed the product B which on, oxidation with alkaline hydrogen peroxides, produces, [1995], (a) R – CH2 – CHO, (c), , 44., , 45., , 46., , 47., , R-C =O, |, CH3, , (b) R– CH2 – CH2 – OH, (d) R - CH - C H 2, |, |, OH OH, , Which of the following compounds has the, lowest boiling point ?, [1994], (a) CH3CH2CH2CH2CH3, (b) CH3CH = CHCH2CH3, (c) CH3CH = CH – CH = CH2, (d) CH3CH2CH2CH3, When hydrochloric acid gas is treated with, propene in presence of benzoyl peroxide, it gives, [1993], (a) 2-Chloropropane (b) Allyl chloride, (c) No reaction, (d) n-Propyl chloride., The restricted rotation about carbon carbon, double bond in 2-butene is due to, [1993], 2, (a) Overlap of one s- and sp - hybridized, orbitals, (b) Overlap of two sp2 - hybridized orbitals, (c) Overlap of one p- and one sp2 - hybridized, orbitals, (d) Sideways overlap of two p- orbitals., Which one of the following can exhibit cis-trans, isomerism ?, [1989], (a), , CH 3 - CHCl - COOH, , (b), , H - C º C - Cl, , (c), , ClCH = CHCl, , (d), , ClCH 2 - CH 2Cl
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EBD_8336, 130, , 58., , CHEMISTRY, Predict the product C obtained in the following, reaction of butyne-1., [2007], HI, , CH 3CH 2 - C º CH + HCl ¾¾, ® B ¾¾® C, , 59., , I, |, (a) CH3 - CH 2 - CH 2 - C - H, |, Cl, I, |, (b) CH 3 - CH 2 - CH - CH 2 Cl, I, |, (c) CH3CH 2 - C - CH3, |, Cl, (d) CH3 - CH - CH 2CH 2 I, |, Cl, Products of the following reaction: [2005], (1) O, , 3, ® ...... are:, CH3CºC·CH2 CH3 ¾¾¾¾¾¾, (2) Hydrolysis, , 60., , 61., , 62., , 63., , (a) CH3COOH + CO2, (b) CH3COOH + HOOC ×CH2CH3, (c) CH3CHO + CH3CH2CHO, (d) CH3COOH + CH3COCH3, A compound is treated with NaNH2 to give, sodium salt. Identify the compound, [1993], (a) C2H2, (b) C6H6, (c) C2H6, (d) C2H4., Reduction of 2-butyne with sodium in liquid, ammonia gives predominantly, [1993], (a) cis-2-butene, (b) No reaction, (c) trans-2-butene, (d) n-butane., Re agent, , R - CH2 - CCl2 - R ¾¾¾¾® R - C º C - R, , The reagent is, [1993], (a) Na, (b) HCl in H2O, (c) KOH in C2H5OH (d) Zn in alcohol., Which is the most suitable reagent among the, following to distinguish compound (3) from rest, of the compounds ?, [1989], 1., , CH 3 - C º C - CH 3, , 2., , CH 3 - CH 2 - CH 2 - CH 3, , 3., , CH 3 - CH 2C º CH, , 4., , CH 3 - CH = CH 2., , 64., , (a) Bromine in carbon tetrachloride, (b) Bromine in acetic acid, (c) Alk. KMnO4, (d) Ammoniacal silver nitrate., Acetylenic hydrogens are acidic because [1989], (a) Sigma electron density of C – H bond in, acetylene is nearer to carbon, which has, 50% s-character, (b) Acetylene has only open hydrogen in each, carbon, (c) Acetylene contains least number of, hydrogens among the possible, hydrocarbons having two carbons, (d) Acetylene belongs to the class of alkynes, with molecular formula, CnH2n – 2., Topic 4: Aromatic Hydrocarbons, , 65., , Given:, CH3 H3C, , H3C, , CH2 H2C, , CH2, , CH3, , CH3, , CH2, , (I), , (II), , (III), , 66., , The enthalpy of the hydrogenation of these, compounds will be in the order as :, [2015], (a) III > II > I, (b) II > III > I, (c) II > I > III, (d) I > II > III, The oxidation of benzene by V2O5 in presence, of air produces :, [2015 RS], (a) benzoic anhydride (b) maleic anhydride, (c) benzoic acid, (d) benzaldehyde, , 67., , The radical,, , 68., , ., CH2 is aromatic because, , it has :, [NEET 2013], (a) 7 p-orbitals and 6 unpaired electrons, (b) 7 p-orbitals and 7 unpaired electrons, (c) 6 p-orbitals and 7 unpaired electrons, (d) 6 p-orbitals and 6 unpaired electrons, Which of the following chemical system is non, aromatic?, [NEET Kar. 2013], (a), , (b), , (c), , (d), , S
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131, , Hydrocarbons, 69., , 70., , 71., , 72., , Some meta-directing substituents in aromatic, substitution are given. Which one is most, deactivating?, [NEET 2013], (a) –SO3H, (b) –COOH, (c) –NO2, (d) –C º N, Which of the following compounds will not, undergo Friedal-Craft’s reaction easily :, [NEET 2013], (a) Xylene, (b) Nitrobenzene, (c) Toluene, (d) Cumene, Among the following compounds the one that, is most reactive towards electrophilic nitration is:, [2012], (a) Benzoic acid, (b) Nitrobenzene, (c) Toluene, (d) Benzene, Which one of the following is most reactive, towards electrophilic reagent ?, [2011], CH3, (a), OCH3, CH3, (b), OH, CH3, (c), NHCOCH3, CH3, , 75., , 76., , 1.Mg, Ether, 2.H 3O, , 74., , OH, (a), , (b), , CH3, (c), , CH2OH, , Cl, , (c), , BH3, , (d), , H3O, , OCH3, , Å, , Which one of the following is most reactive, towards electrophilic attack ?, [2008], NO2, , (b), , OH, (c), 77., , 78., , 79., , (d), , The order of decreasing reactivity towards an, electrophilic reagent, for the following would be, [2007], (i) benzene, (ii) toluene, (iii) chlorobenzene, (iv) phenol, (a) (ii) > (iv) > (i) > (iii) (b) (iv) > (iii) > (ii) > (i), (c) (iv) > (ii) > (i) > (iii) (d) (i) > (ii) > (iii) > (iv), Using anhydrous AlCl3 as catalyst, which one, of the following reactions produces, ethylbenzene (PhEt)?, [2004], (a), , H 3 C - CH 2 OH + C 6 H 6, , (b), , CH 3 - CH = CH 2 + C6 H 6, , (c), , H 2 C = CH 2 + C 6 H 6, , (d), , H 3C - CH 3 + C 6 H 6, , Which one of the following is a free-radical, substitution reaction?, [2003], ¾® CH 3CH ( OH ) CN, (a) CH 3CHO + HCN ¾, , (b), (c), (d), , CH3, NHCOCH3, (d), , (b), , Cl, , [2010], , (a) C6H5CH2CH2C6H5, (b) C6H5CH2OCH2C6H5, (c) C6H5CH2OH, (d) C6H5CH3, Which one is most reactive towards electrophilic, reagent?, [2010], CH3, CH3, , Å, , NO 2, , (a), , CH2OH, In the following reaction, C 6 H 5 CH 2 Br, ¾¾¾¾¾, ® X, the product ‘X’ is, +, , Å, , (a), , CH2OH, , (d), , 73., , Which of the following species is not, electrophilic in nature?, [2010], , CH3, Boiling, +Cl2 ¾¾®, Anh. AlCl, , 3, +CH 3Cl ¾¾¾®, , CH2Cl, , ¾¾®, , + AgNO2, CH2NO2, , CH2Cl, , CH 3
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EBD_8336, 132, , 80., , 81., , 82., , 83., , CHEMISTRY, The correct order of reactivity towards the, electrophilic substitution of the compounds, aniline (I), benzene (II) and nitrobenzene (III) is, [2003], (a) I > II > III, (b) III > II > I, (c) II > III > I, (d) I < II > III, For the formation of toluene by Friedal Craft, reaction, reactants used in presence of, anhydrous AlCl3 are, [2000], (a) C2H2 and CCl4, (b) CH4 and CaCN2, (c) C6H6 and CH3Cl (d) C2H5 OH and Zn, Which of the following compounds reacts, slower in electrophilic substitution?, [2000], (a) C6H5NO2, (b) C6H5OH, (c) C6H5CH3, (d) C6H5NH2, Which one of the following compounds will be, most easily attacked by an electrophile?[1998], Cl, (a), , CH3, , 84., , 85., , 86., , 87., , OH, (d), , Which one of these is not compatible with, arenes?, [1997], (a) Greater stability, (b) Delocalisation of p-electrons, (c) Electrophilic additions, (d) Resonance, Among the following compounds (I-III), the, ease of their reaction with electrophiles is,, [1997], , NO2, , I, , 88., , (b), , (c), , OCH3, , 89., , II, , III, , (a) II > III > I, (b) III > II > I, (c) II > I > III, (d) I > II > III, When CH3Cl and AlCl3 are used in Friedel-Crafts, reaction, the electrophile is, [1994], –, +, (a) Cl, (b) AlCl4, (d) AlCl2+, (c) CH3+, An example of electrophilic substitution reaction, is, [1994], (a) Chlorination of methane, (b) Conversion of methyl chloride to methyl alcohol, (c) Nitration of benzene, (d) Formation of ethylene from ethyl alcohol., Select the true statement about benzene amongst, the following, [1992], (a) Because of unsaturation benzene easily, undergoes addition, (b) There are two types of C – C bonds in, benzene molecule, (c) There is cyclic delocalisation of pi-electrons, in benzene, (d) Monosubstitution of benzene gives three, isomeric products., The most reactive compound for electrophilic, nitration is, [1992], (a) Benzene, (b) Nitrobenzene, (c) Benzoic acid, (d) Toluene., , ANSWER KEY, 1, 2, 3, 4, 5, 6, 7, 8, 9, , (c), (c), (c), (d), (c), (b), (a), (b), (a), , 10 (a), 11 (a), 12 (c), 13 (a), 14 (a), 15 (d), 16 (b), 17 (c), 18 (b), , 19, 20, 21, 22, 23, 24, 25, 26, 27, , (c), (d), (a), (d), (a), (b), (c), (d), (b), , 28, 29, 30, 31, 32, 33, 34, 35, 36, , (d), (a), (d), (c), (c), (a), (a), (d), (b), , 37, 38, 39, 40, 41, 42, 43, 44, 45, , (a), (a), (c), (a), (c), (a), (b), (d), (a), , 46, 47, 48, 49, 50, 51, 52, 53, 54, , (d), (c), (b), (b), (a), (c), (a), (b), (d), , 55, 56, 57, 58, 59, 60, 61, 62, 63, , (a), (c), (d), (c), (b), (a), (c), (c), (d), , 64, 65, 66, 67, 68, 69, 70, 71, 72, , (a), (a), (b), (d), (a), (c), (b), (c), (b), , 73, 74, 75, 76, 77, 78, 79, 80, 81, , (d), (d), (d), (c), (c), (c), (b), (a), (c), , 82, 83, 84, 85, 86, 87, 88, 89, , (a), (d), (c), (d), (c), (c), (c), (d)
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133, , Hydrocarbons, , Hints & Solutions, 1., , (c) Chlorination of neopentane with Cl2 in the, presence of diffused sunlight gives only one, mono-chloro compound, CH3, |, UV light, ®, CH3 — C — CH 3 + Cl 2 ¾¾¾¾, |, CH3, , CH3, |, CH3 — C — CH 2Cl + HCl, |, CH3, , 6., , (b) The bulky methyl groups are maximum away, from each other., Eclipsed conformation is the highest energy, conformation because of unfavorable interactions, between the electrons in the front and back C – H, bonds. The energy of eclipsed conformation is, approx. 3 kcal/mol higher than stagged, conformation., , 7., , 1-chloro-2, 2-dimethyl propane, , (a) Due to hydrogen bonding between the two, OH groups, gauche conformation of ethylene, glycol (a) is the most stable conformation., H, O, O, H, H, , All the hydrogen atoms are similar hence only, one mono-chloro compound is formed., , 2., , (c), , Br, hv, , H, , Na, ether, , 2 ® CH – Br ¾¾¾, CH 4 ¾¾¾, ®, 3, , 8., , CH 3 – CH 3, , (less than four 'C'), , 3., 4., , (c) There is no change in bond angles and bond, lengths in the conformations of ethane., (d) In staggered conformation any two, hydrogen atoms on adjacent carbon atoms are, as far apart as possible, thereby minimising, repulsion between the electron clouds of sbonds of two non-bonded H-atoms (torsional, strain), H, , H, H, , H, , Staggered form, No torsional strain, , 5., , HH, , H, , H, , H, , H, , 9., , H, , H, (b) During cracking, higher hydrocarbons, (liquid) are converted to lower gaseous, hydrocarbons., (a) A compound is said to exhibit optical, isomerism if it atleast contains one chiral carbon, atom, which is an atom bonded to 4 different, atoms or groups., CH3, |, (CH 3 ) 2 CH - C* - CH 2 CH 3, |, H, Br, , 10., H, H, , Eclipsed form, , (c) Conformers are form of stereoisomers in, which isomers can be interconverted by rotation, about single bonds. I and II are staggered and, eclipsed conformers respectively., , (a) C2 H 5 MgBr + H 2 O ¾, ¾® C 2 H 6 + Mg, , OH, (a) Gasoline (petrol) is a mixture of alkanes,, alkenes and aromatic hydrocarbons. The quality, of a gasoline is determined by the amount of, branched chain hydrocarbons (2,2,4trimethylpentane, commonly known as isooctane) present in it., 12. (c) Order of stability : staggered (anti) > gauche, > skew boat > eclipsed., , 11.
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135, , Hydrocarbons, 20., , (d), , CH3, , H3C, , H3C, , O3, ¾¾¾®, CH3 Zn, + H 2O, , CH3— C = CH — CH 3 + CH3— CH — CH = CH2, , O=C, OHC, , 22., , (d) C6H5—CH=CH—CH3 + HBr, , (80%), (A), , In this case, dehydration is governed by, Saytzeff’s rule, according to which hydrogen is, preferentially eliminated from the carbon atom, with fewer number of hydrogen atoms i.e., poor, becomes poorer. Thus, 2-methyl but-2-ene is the, major product., CH3, , ¾®, , (a) No. of double bonds = 4, No. of p bond electrons, = 2 × no. of double bond, =2× 4=8, , (ii) CH3— C = CH — CH, , Å, C6H5—CH—CH2—CH3, , (A), , (Benzyl carbocation), , ¾®, CH, , H, , CH2, , Br, , +, , (Major), , This reaction is governed by Markownikoff’s, rule according to which when an unsymmetrical, reagent e.g. HBr adds to an unsymmetrical, alkene, then the negative part of the reagent is, added to that carbon atom of the double bond, which bears the least number of hydrogen atom., Thus, in above case 2-methyl-2-bromobutane, will be the major product., , CH3, –, , Å, , OH, , CH3 ¾¾®, , ¾¾®, , CH3, 1,2-methyl shift, , C, , CH, , H3C, , OH, , CH3, , (Minor), , (A), , Peroxide effect is shown by HBr only because, H–Cl bond is stronger than H-Br and is not cleaved, by the free radicals, whereas H–I bond is weaker, and Iodine free radical combine to form iodine, molecule., , OH CH3, , C, , CH, , CH3, , OH–, , C, , CH3 CH3, , CH3, , 3° cabocation, (more stable), , (Major), , CH, , CH3, , 25., , B, , Tertiary carbocation is more stable hence gives, major product., , CH3, , 24., , +, , CH3 — C — CH2 – CH3, , CH3, , CH, , 3, , CH 3, , CH3, , C, , in absence, of peroxide, , Br, , CH3, , C, , ¾¾¾®, , (Minor), , Br, , (a), , HBr, dark, , 3, , (CH3) 2— CH — CH — CH, , –, , C6H5—CH—CH2—CH3, , 23., , 3-methyl, butene-1 (20%), (B), , 2-methyl-but-2-ene, , CH3, , 21., , Br, , CH3, , +, , H/ Heat, (b) (i) CH3— CH — CH — CH3 ¾¾®, , OH, , (c) So the IUPAC name of the compound is 1butene-3-yne., If both the double and triple bonds are present in, the compound, it is regarded as derivative of, alkyne. Further if double and triple bonds are at, equidistance from either side, the preference is given, to double bond., , 26. (d) Alkenes with double bonds cannot undergo, free rotation and can have different geometrical, shapes with two different groups on each end of, the double bond.
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EBD_8336, 138, , CHEMISTRY, Peroxide effect is observed only in case of, HBr., , 52., , 47. (c) Such isomers, which possess the same, molecular and structural formula but differ in, the arrangement of atoms around the double, bonded carbon atoms are known as geometrical, isomers., H - C - Cl, ||, H - C - Cl, , H - C - Cl, ||, Cl - C - H, , (cis), , Å, , NaNH, liq. NH 3, , 2 ® HC º CNa, HC º CH ¾¾¾¾, , H C–CH · Br, , 46. (d) Restricted rotation about carbon carbon, double bond in 2-butene is due to sideways, overlap of two p-orbital., Rotation around p bond is not possible. If any, attempt is made to rotate one of the carbon atoms,, the lobes of p-orbital will no longer remain, coplanar i.e no parallel overlap will be possible, and thus p-bond will break . This is known as, concept of restricted rotation. In other words, the presence of p-bonds fix the position of two, carbon atoms., , (a), , 3, 2, ¾¾¾¾¾¾, ® HC º C – CH 2 – CH3, , (X), 1-Butyne, , NaNH 2, , liq. NH3, , Å, , H C -CH – Br, , 3, 2, ¾ H3C – CH 2 – C º C N a, ¬¾¾¾¾¾, , H3C – CH 2 – C º C – CH 2 – CH3, 3-Hexyne (Y), , 53., , 54., 55., , (trans), , (b) The combustion reaction of ethylene is, 5, C 2 H 2 + O2 ® 2CO2 + H 2 O, 2, Both HC CH and CO2 have sp hybridization., O, P, ® CH3 - C - H, (d) CH º CH ¾¾¾¾¾¾, Hg +2 / H 2SO 4, , CH3CHO does not give Victor Meyer test., (a) 1-Butyne and 2-butyne are distinguished, by NaNH2 because 1-Butyne react with NaNH2, due to presence of terminal hydrogen., , 48. (b), , CH3, , CH2, , C, , CH + NaNH2, , 1– Butyne, , 3, , 3, , h, , i, ¾¾¾®, , 56., , 3, , 50., , (b) Alkynes can be reduced to cis-alkenes, with the use of Lindlar’s catalyst, (a) Correct order is, H - C º C - H > H 3C - C º C - H > H 2 C, , (, , 51., , Two acidic, hydrogens, , ), , (, , One acidic, hydrogen, , ), , = CH 2 > CH3 - CH3, (c) Hydration of alkynes give ketones., , OH, , H3 C - C º CH ¾¾, ® H3C - C = CH 2, (A), O, H3 C - C - CH3, (B), , Tautomerism, , ¾¾¾¾®, , 49., , Number of s, bonds = 21, , CH3CH 2 C CNa + NH 3, (c) When double and triple, both bonds are, present, then ‘lowest set of locants’ rule will, decide the IUPAC name., 1, , 2, , 3, , 4, , 5, , 5, , 4, , 3, , 2, , 1, , C H3 - C H = C H - C º C H, , (incorrect), (correct), , Hence, the correct name will be pent-3-en-1-yne., 57. (d) The amount of s-character in various hybrid, orbitals is as follows., sp = 50%, sp2 = 33% and sp3 = 25%, Therefore s-character of the C – H bond in, acetylene (sp) is greater than that of the, C – H bond in alkene (sp2 hybridized) which in, turn has greater s-character of the C – H bond in, alkanes. Thus, owing to a high s-character of the, C – H bond in alkynes, the electrons constituting, this bond are more strongly held by the carbon, nucleus with the result the hydrogen present on, such a carbon atom can be easily removed as, proton. The acidic nature of three types of C – H, bonds follows the following order, º C - H > = C - H > - C - H.
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139, , Hydrocarbons, Further, as we know that conjugate base of a, strong acid is a weak base, hence the correct, order of basicity is, (- ), , (- ), , 62., , (c) On heating ethylene chloride (1, 1 dichloro, ethane) with alcoholic KOH followed by, sodamide, alkyne is obtained, , (-), , R - CH 2 - CCl 2 - R, , H - C º C < CH 2 = CH < CH 2 - CH3, 58., , NaNH, , ® CH3 - CH 2 - C = CH 2, ¾¾, |, Cl, I, |, HI, ¾¾® CH3 - CH 2 - C - CH3, |, Cl, According to Markownikoff’s rule which states, that when an unsymmetrical alkene undergo, hydrohalogenation, the negative part goes to, the C-atom which contain lesser no. of H-atom., 59., , alc.KOH, , ¾¾¾¾® R - CH = CCl - R, , (c) CH3 - CH 2 - C º CH + HCl, 63., , 64., , 2®R - C º C - R, ¾¾¾¾, (d) Br2 in CCl4 (a), Br2 in CH3 COOH (b) and, alk. KMnO4 (c) will react with all unsaturated, compounds, i.e., 1, 3 and 4 while ammoniacal, AgNO3 (d) reacts only with terminal alkynes,, i.e., 3 and h ence compound 3 can be, distinguished from 1, 2 and 4 by. ammoniacal, AgNO3 (d)., (a) The acidity of acetylene or 1–alkynes can, be explained on the basis of molecular orbital, concept according to which formation of, C–H bond in acetylene involves sp-hybridised, carbon atom., , O, , (b) CH 3 - C º C - CH 2 - CH 3 ¾¾3 ®, , s electrons are closer to the nucleus than p electrons,, the electrons present in a bond having more scharacter will be correspondingly more closer to, the nucleus., Thus, owing to high s character of the C—H bond, in alkynes (s = 50%), the electrons constituting this, bond are more strongly held by the carbon nucleus, (i.e., the acetylenic carbon atom) or the sp orbital, acts as more electronegative species than the sp2, and sp 3 with the result the hydrogen present on, such a carbon atom (º C—H) can be easily removed, as a proton., , O, 2H O, , 2, CH3– C – C – CH2 – CH3 ¾®, , O, , O, , CH 3 - C O, |, , + H 2O2, , (O ), , ¾¾, ¾®, , CH3CH 2CO, Methylethy lglyoxal, CH3COOH + CH3CH 2COOH, Acetic acid, , 60., , 61., , Propionic acid, , The glyoxal formed as an intermediate is, oxidised by H2O2 to give the acids.`, (a) Only C2H2 (acetylene) has acidic H-atoms, and hence reacts with NaNH2 to form sodium, salt, i.e.,, ¾® HC º CNa + NH3., HC º CH + NaNH2 ¾, (c) Reduction of alkynes with Na/liq. NH3, gives trans-alkenes. This reaction is called Birch, reduction, CH 3 - C º C - CH 3 + H 2, Na / liq. NH, , 3®, ¾¾¾¾¾¾, , 65., , H3C, H, , C=C, , H, CH3, , trans-Butene or trans-2-Butene, , (a) Enthalpy of hydrogenation, 1, stability of alkene, \ III > II > I, , µ, , 66., , (b), , VO, 410°C, , 2 5, 2C6 H 6 (g) + 9O 2 (g) ¾¾¾®, , O, CH—C, 2, CH—C, , O + 4CO2(g) + 4H2O(g), , O, , Maleic anhydride, , 67., , (d) Presence of 6 p orbitals, each containing, one unpaired electron, in a six membered cyclic, structure is in accordance with Huckel rule of, aromaticity.
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EBD_8336, 140, , 68., , CHEMISTRY, (a) Huckel rule is not obeyed. It has only four, electrons. Further it does not have continous, conjugation., , 76. (c) Out of the given compounds the most, reactive towards nucleophilic attack is, , OH, , In thiophene, one lone pair of electrons are involved, in delocalisation., S, , Thus, the effective number of p-electrons become, 6 which is of the form (4n + 2). Also the ring is, planar, hence this is an aromatic compound., , 69. (c) Decreasing order of deactivating effect of, the given m-directing group is, > NO2 > – CN > – SO3H > – COOH, —NO2 group is most deactivating group due, to strong – E, – I and – M effects., 70. (b) –NO2 is a powerful electron withdrawing, group. Its presence on ring makes the ring less, active., 71. (c) Electrophilic rate order, , >, Toluene, , NO2, , COOH, , CH3, , AlCl, , 3®C H CH CH, (c) C6H6 + CH2 = CH2 ¾¾¾, 6 5, 2, 3, , 79. (b) In the presence of UV rays or energy O2 by, boiling, free radical is generated which attack the, methyl carbon atom of toluene., , Benzoic acid, , CH2, +H, , CH3, D, ¾®, Nitrobenzene, , Since nitration is an electrophilic substitution, hence presence of electron releasing group like, CH3 in the nucleus facilitates nitration., 72. (b) Due to + M effect of – OH group and, hyperconjugation of – CH3 group., 73., , 78., , >, , >, Benzene, , Phenoxide ion is stable due to resonance., 77. (c) Electrophiles have high affinity for electrons., They attack at the site where electron-density is, highest. Electron donating groups increases the, electron density. The electron donating tendency, decreases in the order :, –OH > –CH3 > –H > –Cl, Therefore, the correct order of reactivity towards, electrophile is, C6H5OH > C6H5CH3 > C6H6 > C6H5Cl, , benzyl free, radical, , hn, Cl2 ¾¾® 2Cl•, •CH, , •, + Cl ¾®, , ether, (d) C6H5CH2Br + Mg ¾¾¾®, +, , H2O/H, C6H5CH2MgBr ¾¾¾¾, ® C6H5CH3 + Mg Br(OH), , CH3, OCH3, 74. (d), Among –OH, –CH2OH, –NHCOCH3 and –OCH3,, methoxy group has the highest +M effect., H, 75. (d) H3O + H – O+– H has a lone pair of, gg, electrons on oxygen atom, thus it is not an, electrophile. Also the octet is complete., , CH2Cl, , 2, , 80. (a) Amino group is ring activating while nitro, group is deactivating. Hence, correct order is, aniline > benzene > nitrobenzene., I > II > III, NO2, , NH2, , I, , II, , III, , –NO2 is an electron attracting group hence, decrease the electron density on ring, whereas –, NH2 group is electron releasing group hence, increase electron density on ring. Benzene is also, e– rich due to delocalization of electrons.
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141, , Hydrocarbons, 81., , (c), , CH3, anhydrous AlCl, , 3, + CH3Cl ¾¾¾¾¾®, , + HCl, , 82. (a) CH3 and NH2 and OH are electron donating, groups, whereas NO2 is an electron withdrawing, group and leaves the benzene ring deactivated., Due to stronger electron attracting (–I effect), effect of NO2 group, C6 H5NO2 shows least, reactivity towards electrophilic substitution., 83. (d) –Cl atom shows o/p-directive influence, but deactivate the benzene ring, while, [–OH/–CH3] groups show o/p influence and, activate the benzene ring but –OH group is more, activating than –CH3., Hence order of electrophilic substitution, OH, , CH3, , Cl, , 84. (c) In arenes, p electrons are delocalised, hence, arenes do not undergo addition reactions easily., Aromatic compounds (Arenes) are highly stable, and show resonance eg. Benzene is the simplest, example., , 85. (d) –OCH3 activates the benzene ring. –NO2, deactivates the ring. Hence, the reaction of the, given compounds with electrophiles is in the, order, I > II > III., 86., , Å, , ¾® CH ·3 + AlCl4–, (c) CH3Cl + AlCl3 ¾, Electrophile, , 87. (c) Chlorination of methane proceeds via free, radical mechanism. Conversion of methyl, chloride to methyl alcohol proceeds via, nucleophilic substitution. Formation of ethylene, from ethyl alcohol proceeds via dehydration, reaction. Nitration of benzene is electrophilic, substitution reaction., 88. (c) Benzene do not show addition reaction like, other un saturated hydrocarbons. Due to, resonance all the C – C bonds have the same, nature, which is possible because of the cyclic, delocalisation of p-electrons in benzene., Monosubstitution will give only a single product., 89. (d) Due to + I-effect of the CH3-group, toluene, has much higher electron density in the ring than, benzene, Nitrobenzene and benzoic acid., Nitro and carboxylic group show- I-effect and, hence deactivate the ring towards electrophillic, substutution .
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EBD_8336, 142, , CHEMISTRY, , 14, , Environmental, Chemistry, , Trend Analysis with Important Topics & Sub-Topics, 2020, Topic Name, Air pollution, , Sub-Topic, , LOD - Level of Difficulty, , E - Easy, , 2., , 3., , 4., , 1, , air pollution, soil pollution, , 2018, , 2017, , 2016, , QNS. LOD QNS. LOD QNS. LOD QNS. LOD QNS. LOD, , green house effect, , Water and soil, pollution, , 1., , 2019, , Topic 1: Air Pollution, Among the following, the one that is not a green, house gas is:, [2019], (a) nitrous oxide, (b) methane, (c) ozone, (d) sulphur dioxide, Which oxide of nitrogen is not a common, pollutant introduced into the atmosphere both, due to natural and human activity?, [2018], (a) N2O5, (b) NO2, (c) NO, (d) N2O, Which of the following is not a common, component of Photochemical Smog? [2014], (a) Ozone, (b) Acrolein, (c) Peroxyacetyl nitrate, (d) Chlorofluorocarbons, Which one of the following statements regarding, photochemical smog is not correct?, [2012], (a) Carbon monoxide does not play any role in, photochemical smog formation., (b) Photochemical smog is an oxidising agent in, character., (c) Photochemical smog is formed through, photochemical reaction involving solar energy., (d) Photochemical smog does not cause, irritation in eyes and throat., , E, 1, , E, 1, , A - Average, , 5., , 6., , D - Difficult, , E, , Qns - No. of Questions, , The greenhouse effect is because of the, (a) presence of gases, which in general are, strong infrared absorbers, in the atmosphere, (b) presence of CO2 only in the atmosphere, (c) pressure of O3 and CH4 in the atmosphere, (d) N2O and chlorofluorohydrocarbons in the, atmosphere, [1996], Which of the following is/are the hazardous, pollutant(s) present in automobile exhaust gases?, (i) N2, (ii) CO, (iii) CH4, (iv) Oxides of nitrogen, (a) (ii) and (iii), (b) (i) and (ii) [1996], (c) (ii) and (iv), (d) (i) and (iii), Topic 2: Water and Soil Pollution, , 7., , 8., , The liquified gas that is used in dry cleaning, along with a suitable detergent is, [NEET Odisha 2019], (a) CO2, (b) Water gas, (c) Petroleum gas, (d) NO2, Which of the following is a sink for CO? [2017], (a) Microorganism present in the soil, (b) Oceans, (c) Plants, (d) Haemoglobin
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143, , Environmental Chemistry, (a) CO2, (c) Petroleum gas, 8., , 9., , (b) Water gas, (d) NO2, , (c) Fluoride deficiency in drinking water is, harmful. Soluble fluoride is often used to, bring its concentration upto 1 ppm., , Which of the following is a sink for CO? [2017], (a), , Microorganism present in the soil, , (b), , Oceans, , (c), , Plants, , (d), , Haemoglobin, , (d) When the pH of rain water is higher than 6.5,, it is called acid rain., 11. Which one of the following statement is not, true ?, [2011], (a) pH of drinking water should be between, 5.5 – 9.5., , Roasting of sulphides give the gas X as a by, product. This is colorless gas with choking smell, of burnt sulphur and caused 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 isolated. The, gas X is :, [NEET 2013], (a) SO2, (c) SO3, , (b) CO2, (d) H2S, , (b) Concentration of DO below 6 ppm is good, for the growth of fish., (c) Clean water would have a BOD value of less, than 5 ppm., (d) Oxides of sulphur, nitrogen and carbon are, the most widespread air pollutant., 12. Green chemistry means such reactions which, , 10. Which one of the following statements is not, true?, [NEET Kar. 2013], (a) Dissolved oxygen (DO) in cold water can, reach a concentration upto 10 ppm., , (a) produce colour during reactions, , (b) reduce the use and production of hazardous, chemicals, (c) are related to the depletion of ozone layer, , (b) Clean water would have a BOD value of 5, ppm., , 1, 11, , (d) 2, (b) 12, , (a), (b), , 3, , (d), , 4, , (d), , [2008], , (d) study the reactions in plants, , ANSWER KEY, 5 (a) 6 (c), , 7, , (a), , 8, , (a), , 9, , (a), , 10, , (d), , Hints & Solutions, 1., 2., , 3., , (d) Sulphur dioxide is not a green house gas., (a) Nitrous oxide (N2O) occurs naturally in, environment., In automobile engine, when fuel is burnt, dinitrogen and dioxygen combine to yield NO, and NO2., Thus, N2O5 is the answer., (d) The oxidised hydrocarbons and ozone in, presence of humidity cause photochemical, smong., , 4., , Hydrocarbons + O 2 , NO 2 , NO, O, O 3 ®, Peroxides, formaldehyde, peroxyacetyl-nitrate, (PAN), acrolein etc. Hence, chlorofluoro carbons, are not common component of photochemical, smog., (d) The oxidised hydrocarbons and ozone in, presence of humidity cause photochemical, smog., Hydrocarbons + O 2 , NO 2 , NO, O, O 3, ® Peroxides, formaldehyde, peroxyacetylnitrate, (PAN), acrolein etc.
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EBD_8336, 144, , 5., , CHEMISTRY, It is oxidising in nature and causes irritation to, eyes, lungs, nose, asthamatic attack and damage, plants., (a) Green house gases such as CO2, ozone,, methane, the chlorofluoro carbon compounds, and water vapour form a thick cover around the, earth which prevents the IR rays emitted by the, earth to escape. It gradually leads to increase in, temperature of atmosphere., Water vapour is the largest contributor to the earth's, greenhouse effect. However water vapour does not, control the Earth's temperature., , 6., , (c) CO and oxides of Nitrogen are poisonous, gases present in automobile exhaust gases., , 7., , (a) Liquified CO2 (carbon dioxide) with a, suitable detergent is used in dry cleaning., (a) Microorganisms present in the soil is a sink, for CO., , 8., , Microorganism present in soil convert CO to CO2,, thus it's sink., , 9., 10., , 11., 12., , (a) Based on the features given, gas must be SO2., (d) Acid rain is the rain water containing, sulphuric acid and nitric acid which are formed, from the oxides of sulphur and nitrogen present, in the air as pollutants. Rain water has a pH, below 5.6., (b) The growth of fishes get hindered if the, concentration of D.O. is below 6 ppm., (b) Green chemistry are the reaction, which, reduces the use of hazardous chemicals., Green chemistry is the programme of developing, new chemical products and chemical processes or, making improvements in the already existing, compounds and processes so as to make them less, harmful to human health and environment. This, means the same as to reduce the use and production, of hazardous chemicals.
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145, , The Solid State, , 15, , The Solid State, , Trend Analysis with Important Topics & Sub-Topics, 2020, Topic Name, , Cubic system and, Bragg's equation, , LOD - Level of Difficulty, , 2019, , 2., , 3., , 4., , 2017, , 2016, , Sub-Topic, QNS. LOD QNS. LOD QNS. LOD QNS. LOD QNS. LOD, relation between r, 1, A, and a, formula of, compound and, 1, A, voids, density ratio and, 1, D, structure, E - Easy, , A - Average, , Topic 1: Properties and Types of Solids, 1., , 2018, , A solid with high electrical and thermal, conductivity from the following is, [1994], (a) Si, (b) Li, (c) NaCl, (d) Ice, The pure crystalline substance on being heated, gradually first forms a turbid liquid at constant, temperature and still at higher temperature, turbidity completely disappears. The behaviour, is a characteristic of substance forming [1993], (a) Allotropic crystals, (b) Liquid crystals, (c) Isomeric crystals, (d) Isomorphous crystals., Most crystals show good cleavage because their, atoms, ions or molecules are, [1991], (a) weakly bonded together, (b) strongly bonded together, (c) spherically symmetrical, (d) arranged in planes., The ability of a substances to assume two or, more crystalline structures is called, [1990], (a) Isomerism, (b) Polymorphism, (c) Isomorphism, (d) Amorphism, , D - Difficult, , Qns - No. of Questions, , Topic 2: Crystal Structure of Solids, 5., , 6., , 7., , 8., , A solid compound XY has NaCl structure. If the, radius of the cation is 100 pm, the radius of the, anion (Y –) will be :, [2011 M], (a) 275.1 pm, (b) 322.5 pm, (c) 241.5 pm, (d) 165.7 pm, The pyknometric density of sodium chloride, crystal is 2.165 × 103 kg m–3 while its X-ray, density is 2.178 × 103 kg m–3. The fraction of, unoccupied sites in sodium chloride crystal is, [2003], –3, (a) 5.96 × 10, (b) 5.96, (c) 5.96 × 10–2, (d) 5.96 × 10–1, When molten zinc is converted into solid state,, it acquires hcp structure. The number of nearest, neighbours will be, [2001], (a) 6, (b) 12, (c) 8, (d) 4, In the solid state, MgO has the same structure, as that of sodium chloride. The number of, oxygens surrounding each magnesium in MgO, is, [1999], (a) 6, (b) 1, (c) 2, (d) 4
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EBD_8336, 146, , 9., , CHEMISTRY, For orthorhombic system, axial ratios are, a ¹ b ¹ c and the axial angles are, [1991], (a) a = b = g ¹ 90°, (b) a = b = g = 90°, (c) a = g = 90°, b ¹ 90°, (d) a ¹ b ¹ g = 90°, , is 2.72 g cm–3. The molar, , 16., , Topic 3: Cubic System and Bragg's Equation, 10., , An element has a body centered cubic (bcc), structure with a cell edge of 288 pm. The atomic, radius is:, [2020], , 2, 4, ´ 288pm, (b), ´ 288pm, 4, 3, 4, 3, ´ 288pm, ´ 288pm, (c), (d), 2, 4, A compound is formed by cation C and anion A., The anions form hexagonal close packed (hcp), lattice and the cations occupy 75% of octahedral, voids. The formula of the compound is:, [2019], (a) C2A 3, (b) C3A 2, (c) C3A 4, (d) C4A 3, Iron exhibits bcc structure at room temperature., Above 900°C, it transforms to fcc structure. The, ratio of density of iron at room temperature to, that at 900°C (assuming molar mass and atomic, radii of iron remains constant with temperature), is, [2018], (a), , 11., , 12., , (a), , 14., , 15., , 2, , (b), , 4 3, , 18., , 19., , 20., , 3 2, , 3 3, 1, (d), 2, 4 2, Which of the following statements about the, interstitial compounds is incorrect ?, [NEET 2013], (a) They are chemically reactive., (b) They are much harder then the pure metal., (c) They have higher melting points than the, pure metal., (d) They retain metallic conductivity., The number of carbon atoms per unit cell of, diamond unit cell is :, [NEET 2013], (a) 8, (b) 6, (c) 1, (d) 4, A metal has an fcc lattice. The edge length of, the unit cell is 404 pm. The density of the metal, , (c), , 13., , 3, , 17., , 21., , 22., , 23., , mass of the metal is :, (NA, Avogadro’s constant = 6.02 × 1023 mol-1), [NEET 2013], (a) 30 g mol–1, (b) 27 g mol–1, (c) 20 g mol–1, (d) 40 g mol–1, Structure of a mixed oxide is cubic close-packed, (c.c.p). 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 :, [2012 M], (a) ABO2, (b) A2BO2, (c) A2B3O4, (d) AB2O2, The number of octahedral void(s) per atom, present in a cubic close-packed structure is :, [2012], (a) 1, (b) 3, (c) 2, (d) 4, A metal crystallizes with a face-centered cubic, lattice. The edge length of the unit cell is 408, pm. The diameter of the metal atom is : [2012], (a) 288 pm, (b) 408 pm, (c) 144 pm, (d) 204 pm, AB crystallizes in a body centred cubic lattice, with edge length ‘a’ equal to 387 pm. The, distance between two oppositely charged ions, in the lattice is :, [2010], (a) 335 pm, (b) 250 pm, (c) 200 pm, (d) 300 pm, Copper crystallises in a face-centred cubic lattice, with a unit cell length of 361 pm. What is the, radius of copper atom in pm?[2009], (a) 157, (b) 181, (c) 108, (d) 128, Lithium metal crystallises in a body centred cubic, crystal. If the length of the side of the unit cell of, lithium is 351 pm, the atomic radius of the lithium, will be:, [2009], (a) 151.8 pm, (b) 75.5 pm, (c) 300.5 pm, (d) 240.8 pm, Percentage of free space in a body centred cubic, unit cell is :, [2008], (a) 30%, (b) 32%, (c) 34%, (d) 28%, If ‘a’ stands for the edge length of the cubic, systems : simple cubic, body centred cubic and, face centred cubic, then the ratio of radii of the, spheres in these systems will be respectively,, [2008]
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147, , The Solid State, (a), , 1, 1, 1, 3, 1, a : 3a :, a, a:, a:, a (b), 2, 2, 4, 2, 2 2, , 1, 3, 3, a:, a:, a (d) a : 3a : 2a, 2, 2, 2, The fraction of total volume occupied by the, atoms present in a simple cube is, [2007], (c), , 24., , p, p, (b), 3 2, 4 2, p, p, (c), (d), 6, 4, CsBr crystallises in a body centered cubic lattice., The unit cell length is 436.6 pm. Given that the, atomic mass of Cs = 133 and that of Br = 80 amu, and Avogadro number being 6.02 × 1023 mol–1,, the density of CsBr is, [2006], (a) 0.425 g/cm3, (b) 8.25 g/cm3, (c) 4.25 g/cm 3, (d) 42.5 g/cm3, In face-centred cubic lattice, a unit cell is, shared equally by how many unit cells? [2005], (a) 2, (b) 4, (c) 6, (d) 8, A compound formed by elements X and Y, crystallizes in a cubic structure in which the X, atoms are at the corners of a cube and the Y, atoms are at the face-centres. The formula of the, compound is, [2004], (a) XY3, (b) X3Y, (c) XY, (d) XY2, A compound is formed by elements A and B., The crystalline cubic structure has the A atoms, at the corners of the cube and B atoms at the, body centre. The simplest formula of the, compound is, [2000], (a) AB, (b) A6 B, (c) AB6, (d) A8 B4, The second order Bragg diffraction of X-rays, with l = 1.00 Å from a set of parallel planes in a, metal occurs at an angle 60º. The distance, between the scattering planes in the crystal is, [1998], (a) 0.575 Å, (b) 1.00 Å, (c) 2.00 Å, (d) 1.15 Å, The intermetallic compound LiAg crystallizes in, a cubic lattice in which both lithium and silver, atoms have coordination number of eight. To, what crystal class does the unit cell belong ?, (a), , 25., , 26., , 27., , 28., , 29., , 30., , 31., , 32., , 33., , (a) Simple cubic, [1997], (b) Face-centred cubic, (c) Body-centred cubic, (d) None, The edge length of a face centred unit cubic cell, is 508 pm. If the radius of the cation is 100 pm,, the radius of the anion is, [1996], (a) 288 pm, (b) 398 pm, (c) 154 pm, (d) 618 pm, In the fluorite structure, the coordination number, of Ca2+ ion is :, [1993], (a) 4, (b) 6, (c) 8, (d) 3, The number of atoms contained in an fcc unit, cell of a monoatomic substance is, [1993], (a) 1, (b) 2, (c) 4, (d) 6, Topic 4: Imperfection in Solids, , 34., , 35., , 36., , 37., , 38., , Formula of nickel oxide with metal deficiency, defect in its crystal is Ni 0.98 O. The crystal, contains Ni2+ and Ni3+ ions. The fraction of, nickel existing as Ni2+ ions in the crystal is, [NEET Odisha 2019], (a) 0.31, (b) 0.96, (c) 0.04, (d) 0.50, With which one of the following elements silicon, should be doped so as to give p-type, semiconductor ?, [2008], (a) Germanium, (b) Arsenic, (c) Selenium, (d) Boron, If NaCl is doped with 10– 4 mol % of SrCl2, the, concentration of cation vacancies will be, (NA = 6.02 × 1023 mol–1), [2007], (a) 6.02 × 1016 mol–1 (b) 6.02 × 1017 mol–1, (c) 6.02 × 1014 mol–1 (d) 6.02 × 1015 mol–1, The appearance of colour in solid alkali metal, halides is generally due to, [2006], (a) Schottky defect (b) Frenkel defect, (c) Interstitial positions (d) F-centres, Schottky defect in crystals is observed when, [1998], (a) an ion leaves its normal site and occupies, an interstitial site, (b) unequal number of cations and anions are, missing from the lattice, (c) density of the crystal is increased, (d) equal number of cations and anions are, missing from the lattice
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EBD_8336, 148, , 39., , CHEMISTRY, When electrons are trapped into the crystal in, anion vacancy, the defect is known as [1994], (a) Schottky defect, (b) Frenkel defect, (c) Stoichiometric defect, (d) F-centres, , 40., , On doping Ge metal with a little of In or Ga, one, gets, [1993], (a) p-type semi conductor, (b) n-type semi conductor, (c) insulator, (d) rectifier, , ANS WER KEY, 1, 2, 3, 4, , (b ), (b ), (d ), (b ), , 5, 6, 7, 8, , (c) 9, (a) 10, (b) 11, (a) 12, , (b), (d), (c), (d), , 13, 14, 15, 16, , (a), (a), (b ), (d ), , 17, 18, 19, 20, , (a), (a), (a), (d), , 21, 22, 23, 24, , (a), (b), (a), (d), , 25, 26, 27, 28, , (c), (c), (a), (a), , 29, 30, 31, 32, , (d), (c), (c), (c), , 33, 34, 35, 36, , (c), (b), (d), (b), , 37, 38, 39, 40, , (d ), (d ), (d ), (a), , Hints & Solutions, 1., 2., 3., 4., 5., , 6., , (b) Out of the given substances, only Li has, high electrical and thermal conductivity., (b) Liquid crystals on heating first become, turbid and on further heating they become clear., (d) Crystals show good cleavage because their, constituent particles are arranged in planes., (b) The phenomenon of occurrence of a, substance in two or more crystalline structures, is called polymorphism., (c) Radius ratio of NaCl like crystal, r+, = - = 0.414, r, 100, r- =, = 241.5 pm (r + = 100), 0.414, (a) Fraction of unoccupied sites in NaCl crystal, x-ray density–pyknometric density, =, x-ray density, =, , =, , 2.178 ´ 103 - 2.165 ´ 10 3, 2.178 ´ 103, , =, , 0.013 ´ 103, , 8., , (b) For orthorhombic system a = b = g = 90°, , 10., , (d) For bcc,, , 3a = 4r Þ r =, , (b) hcp is a closed packed arrangement in, which the unit cell is hexagonal and coordination, number is 12., (a) Since MgO has a rock salt structure. In this, structure each cation is surrounded by six anions, and vice versa., , 3a, 4, , Given, a = 288 pm, , r=, 11., , (c), , 3, ´ 288 pm, 4, Oh void (C) : HCP (A), No. of ions Þ 6 ×, , 75, :, 100, 3, :, 4, 3 :, , 12., , (d) For bcc lattice : Z = 2, a =, , 6, 1, 4, , 4r, 3, , For fcc lattice : Z = 4, a = 2 2r, , 2.178 ´ 103, , 5.96 × 10–3, Pyknometric density is nearest to true density, calculated from the molecular mass and crystalline, lattice of the product., , 7., , 9., , \, , d25 °C, d900 °C, , =, , æ ZM ö, çè N a3 ÷ø, A, bcc, æ ZM ö, çè N a3 ÷ø, A, fcc, 3, , 2 æ 2 2r ö, 3 3, = ç, =, ÷, 4, r, 4, 4 2, ç, ÷, è 3 ø, , C3 A4
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149, , The Solid State, 13., , (a) Interstitial compounds are chemically inert., In interstitial compounds small atoms like H, B &, C enter into the void sites between the packed, atoms of crystalline metal. They retain metallic, conductivity and are chemically inert., , 14., , (a) Diamond is like ZnS. In diamond cubic unit, cell, there are eight corner atoms, six face, centered atoms and four more atoms inside the, structure (tetrahedral voids)., Total no. of atoms present per unit cell, 1, 1, = 8´ + 6´ + 4 = 8, 8, 2, (corners), , 15., , (face, centered), , 20., , 21., , 22., , Z ´M, , 23., , (Q 1pm = 10-10 cm), , M=, 16., , 17., , 18., , 2.72 ´ 6.02 ´ (404)3, , = 26.99 = 27 g mol–1, , 4 ´ 107, , 1, (d) No. of A2 + ions = ´ 8 = 2, 4, No. of B+ ions = 4 × 1 = 4, No. of atoms in ccp,, 1, 1, O2 - = 8 ´ + 6 ´ = 4, 8, 2, Hence the formula of the oxide will be, A2B4O4 or AB2O2., (a) Number of octahedral voids in ccp is equal, to effective number of atoms., In ccp, effective number of atoms are 4 so, 4, octahedral voids., So, 1 octahedral voids per atom., , (a) For fcc structure,, , a 3 351 ´ 1.732, =, =151.98 pm, 4, 4, (b) Percentage of occupied space in a bcc is 68%., \ Percentage of free space in a bcc is, (100 – 68) = 32%, (a) Following generalization can be easily, derived for various types of lattice arrangements, in cubic cells between the edge length (a) of the, cell and r the radius of the sphere., a, For simple cubic : a = 2r or r =, 2, For body centred cubic :, , 24., , (2r = Diameter), , (a) For bcc lattice, diagonal = a 3 ., The distance between the two oppositely, , 25., , r=, , 4, 3, r or r =, a, 4, 3, For face centred cubic :, 1, a = 2 2 r or r =, a, 2 2, Thus the ratio of radii of spheres will be simple :, bcc : fcc, 1, a 3, = :, a:, a, 2, 2, 2 4, (d) Number of atoms per unit cell = 1, Atoms touch each other along edges. Hence, a, r=, 2, (r = radius of atom and a = edge length), 4 3, pr, p, Therefore % fraction = 3, = = 0.52, 3, 6, (2r ), (c) In body centred cubic lattice one molecule, of CsBr is within one unit cell., Atomic mass of unit cell = 133 + 80 = 213 a.m.u, Volume of cell = (436.6 × 10–10)3 cm3, a=, , 2 a = 4r, , 2 ´ 408, = 2r, 2, Diameter = 288.5 ; 288 pm, 19., , 387 ´ 1.732, ; 335pm, 2, (d) Since Cu metal crystallises in a face centred, cubic lattice, a, \ r=, 2 2, 361, r=, = 127.6 ; 128pm, 2 2, (a) Since lithium metal crystallises in a body, centred cubic crystal, , \, , ; where Z = number of formula units, N Aa 3, present in unit cell, which is 4 for fcc, a = edge length of unit cell. M = Molecular mass, 4´ M, 2.72 =, 23, 6.02 ´ 10 ´ (404 ´ 10 -10 )3, , a, 3, 2, , =, , (inside, body), , (b) Density is given by, d=, , charged ions =
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EBD_8336, 150, , CHEMISTRY, Density =, , Density =, =, , Z ´ at.wt., Av.no. ´ vol.of unit cell, , 29., , 1´ 213, , 23, , 3, , 6.02 ´10 ´ (436.6) ´ 10, , 213 ´ 107, 3, , 6.02 ´ (436.6), , -30, , = 4.25 g / cm3, , In the density formula, Z is number of particles, present in a unit cell and M is molar mass of that, particle. For any diatomic molecule like CsBr, Z =, 2 but for any mono atomic molecule like Li, Z = 1, for bcc lattice., , 26., , 30., , (c), , 32., , 33., , 28., , An isolated fcc cell is shown here. Each face, of the cell is common to two adjacent cells., Therefore, each face centre atom contributes, only half of its volume and mass to one cell., Arranging six cells each sharing the remaining, half of the face centred atoms, constitutes, fcc cubic lattice. e.g., Cu and Al., (a) For the given cubic structure,, No. of X atoms at the corners = 8 ´ 1 = 1, 8, 1, No. of Y atoms at the face-centres = 6 ´ = 3, 2, \ Formula of the compound = XY3, (a) Given: Atoms are present at the corners of, the cube = A and atoms present at body, centre = B. We know that a cubic unit cell has 8, corners. Therefore contribution of each atom at, 1, the corner = . Since number of atoms per unit, 8, 1, cell is 8, therefore total contribution = 8 ´ = 1 ., 8, We also know the one atom is in the body centre,, , 2, 3, Þ d=, = 1.15 Å, 2, 3, (where d = difference between the scattering, planes), (c) A body-centred cubic system consists of, eight atoms at the corners plus one atom at the, centre of cube., (c) For fcc, the edge length of the unit cell,, a = r + 2R + r, where, R = Radius of anion & r = radius of cation, \ 508 = 2 × 100 + 2R Þ R = 154 pm, (c) In fluorite structure, each F – ion is, surrounded by four Ca2+ ions whereas, each, Ca2+ is surrounded by eight F– ions, giving a, body centred cubic arrangement. Thus the, co-ordination number of Ca2+ = 8., (c) The no. of atoms is a unit cell may be, calculated by the the formula, n f ne, n, n, Z= c+ b+, +, 8, 1, 2, 4, where nc = no. of atom at the corner nb = no. of, atoms at body centre nf = no. of atoms at face, centre, ne = no. of atoms at edge centre., An fcc crystal contains, Þ 2 ´ 1 = 2.d ., , 31., , 27., , therefore number of atoms per unit cell = 1. Thus, formula of the compound is AB., (d) Given : Order of Bragg diffraction (n) = 2;, Wavelength (l) = 1 Å and angle (q) = 60º. We, know from the Bragg’s equation nl = 2d sin q, or 2 × 1 = 2d sin 60º, , 34., , 35., , 8 6, = + = 4 atoms in a unit cell., 8 2, (b) Ni0.98O = (Ni2+)x (Ni3+)0.98–x (O2–)1, Net charge = 0, [x × 2] + [(0.98 – x) × 3] + [–2 × 1] = 0, x = 0.94, Fraction of nickel existing as, 0.94, Ni2+ =, = 0.959 » 0.96, 0.98, (d) The semiconductors formed by the, introduction of impurity atoms containing one, elecron less than the parent atoms of insulators, are termed as p-type semiconductors. Therefore,, silicon containing 14 electrons has to be doped, with boron containing 13 electrons to give a, p-type semi-conductor.
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151, , The Solid State, 36., , Sr 2+, , Na+, , (b) For each, ion added, one, ion is, removed to maintain the electrical neutrality., Hence concentration of cation vacancies, = mole % of SrCl2 added = 10–4 mole %, , 39., , -4, , 10, ´ 6.023 ´ 1023 = 6.023 ´ 1017, 100, (d) When the crystal is irradiated with white, light, the trapped electrons in a hole absorbs, photon for excitation from ground state to excited, state. This gives colour to the compund., , =, , 37., , Alkali metal halide may have excess metal ion if a, negative ion is absent from its lattice site, leaving a, hole, which is occupied by electron to maintain, electrical neutrality. The 'holes' occupied by, electrons are called F-centres or colour centres., , 38., , (d) If in an ionic crystal of the type A+. B–,, equal number of cations and anions are missing, , from their lattice sites so that the electrical, neutrality is maintained. The defect is called, Schottky defect., (d) When electrons are trapped in anion, vacancies, these are called F-centres., , e–, , +ve –ve, ion ion, , F- centre in crystal, , 40., , (a) p-type of semiconductors are produced, by adding impurity containing less electrons, (i.e. atoms of group 13)., Ge belongs to Group 14 and In to Group 13., Hence, on doping, p-type semiconductor is, obtained.
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EBD_8336, 152, , CHEMISTRY, , 16, , Solutions, , Trend Analysis with Important Topics & Sub-Topics, 2020, Topic Name, , Sub-Topic, , 2019, , 2018, , 2017, , 2016, , QNS. LOD QNS. LOD QNS. LOD QNS. LOD QNS. LOD, , molality and Kf, , 1, , E, , Solubiity and, concentration of, solutions, , effect of, temperature on, molality/molarity/, mole fraction, , 1, , E, , Vapour pressure,laws, of solutions and ideal, and non ideal, solutions, Colligative properties, and abnormal, molecular mass, , Raoult's l aw, , LOD - Level of Difficulty, , 1, , E, , azeoptropes, , 1, , 1, , E - Easy, , 2., , 3., , The density of 2 M aqueous solution of NaOH, is 1.28 g/cm3. The molality of the solution is, [Given that molecular mass of NaOH = 40 g mol–1], [NEET Odisha 2019 ], (a) 1.32 m, (b) 1.20 m, (c) 1.56 m, (d) 1.67 m, If molality of the dilute solutions is doubled, the, value of molal depression constant (Kf ) will be:[2017], (a) halved, (b) tripled, (c) unchanged, (d) doubled, Which of the following is dependent on, temperature?, [2017], , 1, , A, , A, A - Average, , Topic 1: Solubility and Concentration of, Solutions, 1., , A, , E, , vapour pressure, depression in, freezing point, , 1, , 4., , 5., , D - Difficult, , Qns - No. of Questions, , (a) Molarity, (b) Mole fraction, (c) Weight percentage, (d) Molality, What is the mole fraction of the solute in a 1.00, m aqueous solution ?, [2011, 2015 RS], (a) 0.177, (b) 1.770, (c) 0.0354, (d) 0.0177, How many grams of concentrated nitric acid, solution should be used to prepare 250 mL of, 2.0M HNO3 ? The concentrated acid is 70%, HNO3, [NEET 2013], (a) 90.0 g conc. HNO3, (b) 70.0 g conc. HNO3, (c) 54.0 g conc. HNO3, (d) 45.0 g conc. HNO3
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153, , Solutions, 6., , 7., , 8., , 9., , 10., , 11., , 12., , 13., , 25.3 g of sodium carbonate, Na 2CO3 is, dissolved in enough water to make 250 mL of, solution. If sodium carbonate dissociates, completely, molar concentration of sodium ions,, Na + and carbonate ions, CO32– are respectively, (Molar mass of Na2CO3 = 106 g mol–1) [2010], (a) 0.955 M and 1.910 M, (b) 1.910 M and 0.955 M, (c) 1.90 M and 1.910 M, (d) 0.477 M and 0.477 M, The mole fraction of the solute in one molal, aqueous solution is:, [2005], (a) 0.009, (b) 0.018, (c) 0.027, (d) 0.036, 1 M, 2.5 litre NaOH solution is mixed with another, 0.5 M, 3 litre NaOH solution. Then find out the, molarity of resultant solution, [2002], (a) 0.80 M, (b) 1.0 M, (c) 0.73 M, (d) 0.50 M, Molarity of liquid HCl will be, if density of, solution is 1.17 g/cc, [2001], (a) 36.5, (b) 32.05, (c) 18.25, (d) 42.10, Which of the following statements, regarding, the mole fraction (x) of a component in solution,, is incorrect?, [1999], (a) 0 < x < 1, (b) x < 1, (c) x is always non-negative, (d) –2 < x < 2, A 5% solution of cane sugar (mol. wt. =342) is, isotonic with 1% solution of a substance X. The, molecular weight of X is, [1998], (a) 34.2, (b) 171.2, (c) 68.4, (d) 136.8, The number of moles of oxygen in one litre of air, containing 21% oxygen by volume, in standard, conditions, is, [1995], (a) 0.186, (b) 0.21, (c) 0.0093, (d) 2.10, Which of the following modes of expressing, concentration is independent of temperature ?, [1992, 1995], , (a) Molarity, (c) Formality, , (b) Molality, (d) Normality, , Topic 2: Vapour Pressure, Laws of, Solutions and Ideal, Non-ideal Solutions, 14., , 15., , 16., , 17., , 18., , The mixture which shows positive deviation, from Raoult's law is, [2020], (a) Benzene + Toluene, (b) Acetone + Chloroform, (c) Chloroethane + Bromoethane, (d) Ethanol + Acetone, Which of the following statements is correct, regarding a solution of two components A and, B exhibiting positive deviation from ideal, behaviour?, [NEET Odisha 2019], (a) Intermolecular attractive forces between AA and B-B are equal to those between A-B., (b) Intermolecular attractive forces between AA and B-B are stonger than those between, A-B., (c) DmixH = 0 at constant T and P., (d) DmixV = 0 at constant T and P., In water saturated air, the mole fraction of water, vapour is 0.02. If the total pressure of the, saturated air is 1.2 atm, the partial pressure of, dry air is, [NEET Odisha 2019], (a) 0.98 atm, (b) 1.18 atm, (c) 1.76 atm, (d) 1.176 atm, The mixture that forms maximum boiling, azeotrope is:, [2019], (a) Water + Nitric acid, (b) Ethanol + Water, (c) Acetone + Carbon disulphide, (d) Heptane + Octane, Which of the following statement about the, composition of the vapour over an ideal 1 : 1, molar mixture of benzene and toluene is correct?, Assume that the temperature is constant at 25°C., (Given : Vapour pressure data at 25°C, benzene, = 12.8 kPa, toluene = 3.85 kPa), [2016], (a) The vapour will contain a higher, percentage of benzene, (b) The vapour will contain a higher, percentage of toluene, (c) The vapour will contain equal amounts of, benezene and toluene, (d) Not enough information is given to make a, predication
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EBD_8336, 154, , 19., , 20., , 21., , 22., , 23., , 24., , 25., , CHEMISTRY, At 100 °C the vapour pressure of a solution of, 6.5 g of a solute in 100 g water is 732 mm. If, Kb = 0.52, the boiling point of this solution will, be, [2016], (a) 101 °C, (b) 100 °C, (c) 102 °C, (d) 103 °C, Which one is not equal to zero for an ideal, solution:, [2015], (a) DSmix, (b) DVmix, (c) DP = Pobserved - PRaoult, (d) DHmix, PA and PB are the vapour pressure of pure liquid, components, A and B, respectively of an ideal, binary solution. If X A represents the mole, fraction of component A, the total pressure of, the solution will be, [2012], (a) pa + xa (pb – pa) (b) pa + xa (pa – pb), (c) pb + xa (pb – pa) (d) pb + xa (pa – pb), Vapour pressur e of chloroform (CHCl 3 ), and dichloromethane (CH2Cl2 ) at 25 ºC are, 200 mm Hg and 41.5 mm Hg respectively. Vapour, pressure of the solution obtained by mixing, 25.5 g of CHCl3 and 40 g of CH2Cl2 at the same, temperature will be : (Molecular mass of CHCl3, = 119.5 u and molecular mass of CH2Cl2 = 85 u)., [2012 M], (a) 173.9 mm Hg, (b) 615.0 mm Hg, (c) 347.9 mm Hg, (d) 285.5 mm Hg, An aqueous solution is 1.00 molal in KI. Which, change will cause the vapour pressure of the, solution to increase?, [2010], (a) Addition of NaCI, (b) Addition of Na 2SO4, (c) Addition of 1.00 molal KI, (d) Addition of water, A solution of acetone in ethanol, [2006], (a) shows a positive deviation from Raoult’s law, (b) behaves like a non ideal solution, (c) obeys Raoult’s law, (d) shows a negative deviation from Raoult’s law, The vapour pressure of two liquids ‘P’ and, ‘Q’ are 80 and 60 torr, respectively. The total, vapour pressure of solution obtained by, mixing 3 mole of ‘P’ and 2 mole of ‘Q’ would be, [2005], (a) 72 torr, (b) 140 torr, (c) 68 torr, (d) 20 torr, , 26., , 27., , 28., , 29., , 30., , Formation of a solution from two components, can be considered as, [2003], (i) Pure solvent, ® separated solvent, molecules, DH1, (ii) Pure solute, ® separated solute, molecules, DH2, (iii) Separted solvent ® Solution, DH3, & solute molecules, Solution so formed will be ideal if, (a) DH soln = DH 3 - DH1 - DH 2, (b) DH soln = DH1 + DH 2 + DH 3, (c) D H soln = D H 1 + D H 2 - D H 3, (d) DH soln = DH1 - DH 2 - DH 3, A solution containing components A and B, follows Raoult's law when, [2002], (a) A – B attraction force is greater than A – A, and B – B, (b) A – B attraction force is less than A – A and, B–B, (c) A – B attraction force remains same as, A–A and B –B, (d) Volume of solution is different from sum of, volume of solute and solvent, The beans are cooked earlier in pressure cooker,, because, [2001], (a) Boiling point increases with increasing, pressure, (b) Boiling point decreases with increasing, pressure, (c) Internal energy is not lost while cooking in, pressure cooker, (d) Extra pressure of pressure cooker, softens, the beans, The vapour pressure of a solvent decreased by, 10 mm of mercury when a non-volatile solute, was added to the solvent. The mole fraction of, the solute in the solution is 0.2. What should be, the mole fraction of the solvent if the decrease, in the vapour pressure is to be 20 mm of mercury?, [1998], (a) 0.8, (b) 0.6, (c) 0.4, (d) 0.2, The vapour pressure at a given temperature of, an ideal solution containing 0.2 mol of a nonvolatile solute and 0.8 mol of solvent is 60 mm of, Hg. The vapour pressure of the pure solvent at, the same temperature is, [1996]
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155, , Solutions, , 31., , 32., , 33., , 34., , 35., , (b) 60 mm of Hg, (a) 150 mm of Hg, (c) 75 mm of Hg, (d) 120 mm of Hg, Vapour pressure of benzene at 30 °C is 121.8 mm., When 15 g of a non volatile solute is dissolved, in 250 g of benzene, its vapour pressure, decreased to 120.2 mm. The molecular weight of, the solute (Mo. wt. of solvent = 78), [1995], (a) 356.2, (b) 456.8, (c) 530.1, (d) 656.7, According to Raoult's law, relative lowering of, vapour pressure for a solution is equal to, [1995], (a) moles of solute, (b) moles of solvent, (c) mole fraction of solute, (d) mole fraction of solvent, The relative lowering of the vapour pressure is, equal to the ratio between the number of, [1991], (a) solute molecules to the solvent molecules, (b) solute molecules to the total molecules in, the solution, (c) solvent molecules to the total molecules in, the solution, (d) solvent molecules to the total number of, ions of the solute., An ideal solution is formed when its components, [1988], (a) have no volume change on mixing, (b) have no enthalpy change on mixing, (c) have both the above characteristics, (d) have high solubility., All form ideal solution except, [1988], (a) C6H6 and C6H5 CH3, (b) C2H6 and C2H5I, (c) C6H5Cl and C6H5 Br, (d) C2H5 I and C2H5 OH., , 37., , 38., , 39., , 40., , 41., , Topic 3: Colligative Properties and, Abnormal Molecular Masses, 36., , The freezing point depression constant (Kf) of, benzene is 5.12 K kg mol–1. The freezing point, depression for the solution of molality 0.078 m, containing a non-electrolyte solute in benzene, is (rounded off upto two decimal places) :, , 42., , (a) 0.80K, (b) 0.40 K [2020], (c) 0.60 K, (d) 0.20 K, Which one of the following electrolytes has the, same value of van't Hoff's factor (i) as that of, the Al2(SO4)3 (if all are 100% ionised) ? [2015], (a) K3[Fe(CN)6], (b) Al(NO3)3, (c) K4[Fe(CN)6], (d) K2SO4, The boiling point of 0.2 mol kg–1 solution of X, in water is greater than equimolal solution of Y, in water. Which one of the following statements, is true in this case ?, [2015], (a) Molecular mass of X is greater than the, molecular mass of Y., (b) Molecular mass of X is less than the, molecular mass of Y., (c) Y is undergoing dissociation in water while, X undergoes no change., (d) X is undergoing dissociation in water., Of the following 0.10 m aqueous solutions,, which one will exhibit the largest freezing point, depression?, [2014], (a) KCl, (b) C6H12O6, (c) Al2(SO4)3, (d) K2SO4, The freezing point depression constant for water, is – 1.86 ºC m–1. If 5.00 g Na2SO4 is dissolved in, 45.0 g H2O, the freezing point is changed by, – 3.82 ºC. Calculate the van’t Hoff factor for, Na2SO4., [2011], (a) 2.05, (b) 2.63, (c) 3.11, (d) 0.381, The van’t Hoff factor i for a compound which, undergoes dissociation in one solvent and, association in other solvent is respectively :, [2011], (a) less than one and greater than one., (b) less than one and less than one., (c) greater than one and less than one., (d) greater than one and greater than one., A 0.1 molal aqueous solution of a weak acid is, 30% ionized. If K f for water is 1.86 °C/m, the, freezing point of the solution will be : [2011 M], (a) – 0.18 °C, (b) – 0.54 °C, (c) – 0.36 °C, (d) – 0.24 °C
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EBD_8336, 156, , 43., , 44., , 45., , 46., , 47., , 48., , CHEMISTRY, 200 mL of an aqueous solution of a protein, contains its 1.26 g. The osmotic pressure of this, solution at 300 K is found to be 2.57 × 10–3, bar. The molar mass of protein will be, (R = 0.083 L bar mol–1 K–1), [2011 M], (a) 51022 g mol–1, (b) 122044 g mol–1, (c) 31011 g mol–1, (d) 61038 g mol–1, A solution of sucrose (molar mass = 342 g mol–1), has been prepared by dissolving 68.5 g of, sucrose in 1000 g of water. The freezing point of, the solution obtained will be (Kf for water, = 1.86 K kg mol–1)., [2010], (a) – 0.372 °C, (b) – 0.520 °C, (d) + 0.372 °C, (d) – 0.570 °C, A 0.0020 m aqueous solution of an ionic, compound Co(NH 3 ) 5 (NO 2 )Cl freezes at, – 0.00732 °C. Number of moles of ions which, 1 mol of ionic compound produces on being, dissolved in water will be (Kf = – 1.86 °C/m), [2009], (a) 3, (b) 4, (c) 1, (d) 2, 0.5 molal aqueous solution of a weak acid (HX), is 20% ionised. If Kf for water is 1.86 K kg, mol–1 ,the lowering in freezing point of the, solution is, [2007], (a) 0.56 K, (b) 1.12 K, (c) – 0.56 K, (d) – 1.12 K, During osmosis, flow of water through a, semipermeable membrane is, [2006], (a) from both sides of semipermeable, membrane with equal flow rates, (b) from both sides of semipermeable, membrane with unequal flow rates, (c) from solution having lower concentration only, (d) from solution having higher concentration, only, A solution containing 10 g dm –3 of urea, (molecular mass = 60 g mol–1) is isotonic with a, 5% solution of a non-volatile solute. The, molecular mass of this non-volatile solute is, [2006], (a) 300 g mol–1, (b) 350 g mol–1, (c) 200 g mol–1, (d) 250 g mol–1, , 49., , 50., , 51., , 52., , 53., , 54., , 55., , 1.00 g of a non-electrolyte solute (molar mass, 250 g mol–1) was dissolved in 51.2 g of benzene., If the freezing point depression constant, Kf of, benzene is 5.12 K kg mol–1, the freezing point of, benzene will be lowered by, [2006], (a) 0.3 K, (b) 0.5 K, (c) 0.4 K, (d) 0.2, A solution of urea (mol. mass 56 g mol-1) boils, at 100.18 °C at the atmospheric pressure. If Kf, and Kb for water are 1.86 and 0.512 K kg mol-1, respectively, the above solution will freeze at, [2005], (a) 0.654 °C, (b) - 0.654 °C, (c) 6.54 °C, (d) - 6.54 °C, Camphor is often used in molecular mass, determination because, [2004], (a) it is readily available, (b) it has a very high cryoscopic constant, (c) it is volatile, (d) it is solvent for organic substances, 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 ?, [2002], (a), , æm ö, æ m ö RT, M 2 = ç 2 ÷ VRT (b) M 2 = ç 2 ÷, è p ø, èV ø p, , (c), , æm ö, M 2 = ç 2 ÷ pRT (d), èV ø, , æm ö p, M2 = ç 2 ÷, è V ø RT, , Which of the following colligative property can, provide molar mass of proteins (or polymers or, colloids) with greatest precision ?, [2000], (a) Osmotic pressure, (b) Elevation of boiling point, (c) Depression of freezing point, (d) Relative lowering of vapour pressure, Which of the following 0.10 m aqueous solutions, will have the lowest freezing point ?, [1997], (a) Al2(SO4)3, (b) C6H12O6, (c) KCl, (d) C12H22O11, At 25°C, the highest osmotic pressure is, exhibited by 0.1 M solution of, [1994], (a) CaCl2, (b) KCl, (c) Glucose, (d) Urea.
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157, , Solutions, 56., , 57., , 58., , Which one of the following salts will have the, same value of van’t Hoff factor (i) as that of, K4[Fe (CN)6]., [1994], (a) Al2(SO4)3, (b) NaCl, (c) Al (NO3)3, (d) Na2SO4., Which one is a colligative property ? [1992], (a) boiling point, (b) vapour pressure, (c) osmotic pressure (d) freezing point, If 0.1 M solution of glucose and 0.1 M solution, of urea are placed on two sides of the semipermeable membrane to equal heights, then it, will be correct to say that, [1992], (a) There will be no net movement across the, membrane, , 59., , 60., , (b) Glucose will flow towards urea solution, (c) urea will flow towards glucose solution, (d) water will flow from urea solution to glucose, Which of the following aqueous solution has, minimum freezing point ?, [1991], (a) 0.01 m NaCl, (b) 0.005 m C2H5OH, (c) 0.005 m MgI2, (d) 0.005 m MgSO4., Blood cells retain their normal shape in solution, which are, [1991], (a) hypotonic to blood, (b) isotonic to blood, (c) hypertonic to blood, (d) equinormal to blood., , ANS WER KEY, 1, , (d ), , 7, , (b), , 13, , (b), , 19, , (a), , 25, , (a), , 31, , (a), , 37, , (c), , 43, , (d), , 49, , (c), , 55, , (a), , 2, , (c), , 8, , (c), , 14, , (d), , 20, , (a), , 26, , (b), , 32, , (c), , 38, , (d ), , 44, , (a), , 50, , (b), , 56, , (a), , 3, , (a), , 9, , (b), , 15, , (b), , 21, , (d ), , 27, , (c), , 33, , (b), , 39, , (c), , 45, , (d), , 51, , (b), , 57, , (c), , 4, , (d ), , 10, , (d), , 16, , (d), , 22, , (N) 28, , (a), , 34, , (c), , 40, , (b ), , 46, , (b), , 52, , (b), , 58, , (a), , 5, , (d ), , 11, , (c), , 17, , (a), , 23, , (d ), , 29, , (b), , 35, , (d), , 41, , (c), , 47, , (c), , 53, , (a), , 59, , (a), , 6, , (b ), , 12, , (c), , 18, , (a), , 24, , (a), , 30, , (c), , 36, , (b), , 42, , (d ), , 48, , (a), , 54, , (a), , 60, , (b ), , Hints & Solutions, 1., , 2., , 3., , (d) Let, volume of solution = 1 L, Mole of NaOH = 2, Mass of NaOH solution = volume × density, = 1000 × 1.28 = 1280 g, Mass of H2O = (1280 – 2 × 40) = 1200 g = 1.2 kg, , Mole of NaOH, 2, Molality (m) =, =, =, Mass of solvent (kg), 1.2, 1.67 m, (c) K f (molal depression constant) only, depends on the nature of the solvent and is, independent of the concentration of the, solution., (a) Molarity depends on the volume of a, solution which can be changed with change in, temperature., , 4., , (d) Molality =, , =, , n, V (kg), , W, 1, ´, M V (kg), , (where nsolute = W/M), , n, i.e., 1 mole in 1 kg of water, V (kg), Moles of 1 kg H2O, , 1.00 m =, , 1000 g, = 55.55 mole, 18 g / mol, Moles of solute = 1, Mole fraction, =, , =, , nsolute, 1, =, nsolute + nwater (1 + 55.55), , = 0.01768 = 0.0177
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EBD_8336, 158, , 5., , CHEMISTRY, (d) Molarity (M) =, , \, , wt ´ 1000, mol. wt. ´ vol (mL), , wt. 1000, ×, 250, 63, 63, wt. =, g, 2, , 1000 mL of solution contain cane sugar, 5, =, ´ 1000 = 50 g/L, 100, 1 ´ 1000, Similarily 1% solution contains =, 100, = 10 g/L of X., , 2=, , wt. of 70% acid =, 6., , (b) Concentration of, Na 2CO3 =, , 100, ´ 31.5 = 45 g, 70, , 50 g / L, ´R´T, 342, Osmotic pressure of 1% solution of susbtance, , (p1) = C × R × T =, , 25.3 1000, ´, = 0.955 M, 106 250, , 10 g / L, ´R´T, M, As both are isotonic,, So p1 = p2, , (p2) =, , [Na + ] = 2 × 0.955 = 1.91 M, , 7., , é CO32– ù = 0.955 M, ë, û, (b) One molal solution means one mole solute, present in 1 kg (1000 g) solvent, i.e., mole of solute = 1, 1000g 1000, Mole of solvent (H2O) =, =, 18, 18g, 1, Mole fraction of solute =, æ 1000 ö, ç1 +, ÷, 18 ø, è, =, , 8., , 9., , 12., , 18, = 0.01768 ; 0.018., 1018, , (c) From molarity equation, M1V1 + M2V2 = MV, 1 × 2.5 + 0.5 × 3 = M × 5.5, 4, M=, = 0.727 ; 0.73M, 5.5, (b) Density = 1.17 g/cc, , Mass, As d =, Volume, volume = 1cc \ mass = 1.17g, , No. of moles of solute, Now molarity =, Volume of solution in litre, , 1.17 ´ 1000 1170, =, =32.05M, 36.5 ´ 1, 36.5, (d) The mole fraction can never be equal to –2, or 2. It is always between 0 and 1 i.e., 0 < x < 1., (c) 5% cane sugar solution means 100ml of, solution contain cane sugar = 5g, , 11., , 210, = 0.0093, 22400, (Q volume of 1 mole of gas at S.T.P. is 22400 mL), (b) The molality involves weights of the, solute and the solvent. Since weight does not, change with the temperature, therefore molality, does not depend upon the temperature., (d) Hydron bond of ethanol gets weakened, by addition of acetone. Thus the mixture of, ethanol and acetone show positive deviation, from Raoult's law., (b) In case of positive deviation from ideal, behaviour A-B interactions are weaker than AA and B-B interactions., (d) Mole fraction of water vapour = 0.02, \ Mole fraction of dry air = 1 – 0.02 = 0.98, Total pressure of saturated air = 1.2 atm, Acc. to Dalton’s law of partial pressure,, PDry air = PTotal xDry air = 1.2 × 0.98 = 1.176 atm, (a) The solutions which show a large negative, deviation from Raoult’s law form maximum, , Therefore, no. of moles =, , 13., , 14., , 15., , 16., , =, , 10., , 50, 10, ´R´T =, ´R´T, 342, M, 342, = 68.4, \ M (mol. wt. of X) =, 5, (c) Percentage volume of oxygen = 21%., So, 100 mL of air contains = 21mL of O2, \ Volume of oxygen in one litre of air, 21, ´ 1000 = 210mL., =, 100, , or, , 17.
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159, , Solutions, , 18., , boiling azeotrope. Nitric acid and water forms a, maximum boiling azeotrope with a boiling point, of 393.5 K., (a) Let us consider that A is benzene and B is, toluene, 1 : 1 molar mixture of A and B, 1, 1, \ xA = and xB =, 2, 2, 0, 0, Total pressure of solution (P) = PA xA + PB xB, 1, 1, P = 12.8 × + 3.85 × = 8.325 kPa, 2, 2, 1, 12.8 ´, 0, 2 = 0.768, yA = PA xA =, P, 8.325, \ yB = 1 – yA = 1 – 0.768 = 0.232, so, the vapour will contain higher percentage of, benzene., As the molar composition is same and benzene, has higher vapour pressure, the vapour will contain, higher percentage of benzene., , 19., , (a), , 21., , (d) P = PAXA + PBXB = PAXA + PB (1 – XA), (for binary sol. XA + XB = 1), Þ PAXA + PB – PBXA, Þ PB + XA (PA – PB), , 22., , (N) n CHCl3 =, , 760 - 732 6.5 ´ 18, =, M1 ´ 100, 760, , M1 = 31.75 g mol–1, DTb = m × Kb =, , 23., , 24., , 25., , 26., , W1 ´ 1000, × Kb, M 1 ´ W2, , 20., , Entropy will always increase upon mixing two, compounds as disorder increases., , 3, 2ù, é, = ê80 ´ + 60 ´ ú = 16 × 3 + 12 × 2, 5, 5û, ë, Ptotal = 48 + 24 = 72 torr, (b) For an ideal solution, DHmixing = 0, DH = DH1 + DH2 + DH3 (Accroding to Hess's, law), for ideal solutions there is no change in magnitude, of the attractive forces in the two components, present., , 0.52 ´ 6.5 ´1000, DTb =, = 1.06 °C, 31.75 ´100, , \ boiling point of solution, = 100 °C + 1.06 °C = 101 °C, (a) For an ideal solution DSmix > 0, , 40, = 0.47, 85, PT = PA° X A + PB° X B, 0.213, 0.47, + 41.5 ´, = 200 ´, 0.683, 0.683, = 62.37 + 28.55 = 90.92, (d) When the aqueous solution of one molal, KI is diluted with water, concentration decreases,, therefore the vapour pressure of the resulting, solution increases., (a) A solution of acetone in ethanol shows, positive deviation from Raoult's law. It is because, ethanol molecules are strongly hydrogen, bonded. When acetone is added, these, molecules break the hydrogen bonds and, ethanol becomes more volatile. Therefore, its, vapour pressure is increased., (a) Given V.PP = 80 torr, V.PQ = 60 torr, Ptotal = V·PP × XP + V·PQ × XQ, n CH 2Cl2 =, , æ P° - Ps ö n W1 M 2, ´, = =, èç P° ø÷ N M1 W2, , Where, W1 = wt of solute, W2 = wt of solvent, M1 = Mass of solute, M2 = Mass of solvent, at 100 °C, P° = 760 mm, , 25.5, = 0.213, 119.5, , 27., , 28., , (c) These two components A and B follows, the condition of Raoult’s law if the force of, attraction between A and B is equal to the force, of attraction between A and A or B and B., (a) The beans are cooked earlier in pressure, cooker because boiling point increases with, increasing pressure.
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EBD_8336, 160, , 29., , CHEMISTRY, (b) According to Raoult's law, , 33., , p° - p, n, =, (mole fraction of solute), p°, n+ N, 10, = 0.2 \ p° = 50 mm, p°, For other solution of same solvent, , 20, n, (mole fraction of solute), =, p° n + N, 20, = mole fraction of solute, Þ, 50, Þ Mole fraction of solute = 0.4, Hence, mole fraction of solvent = 1 – 0.4 = 0.6, Decrease in vapour pressure µ Mole fraction of, solute., Thus, depression of 2 × 0.2 mm Hg vapour, pressure requires = 2 × 0.2 mole fraction of solute., , 30., , (c) According to Raoult's law, Po - P, = xB, Po, , 34., , P, , o, , =, , 1, or 4 P o = ( P )´5, 5, , 60 ´ 5, = 75 mm of Hg, 4, (a) Given vapour pressure (p°) = 121.8 mm;, Weight of solute (w) = 15 g, Weight of solvent (W) = 250 g, From Raoult’s law,, , 35., , (d) Ethyl alcohol forms hydrogen bonding, with itself, hence it will not form ideal solution, with C2H5I., , 36., , (b) DTf = K f m = 5.12 × 0.078 = 0.399 K = 0.40 K, , 37., , ¾¾, ® 4K+ + [Fe(CN)6]–, (c) K4[Fe(CN)6] ¬¾, ¾, , and Al2(SO4)3 ® 2Al3+ + 3SO2–, 4, \ van’t Hoff factor is 5 for both Al2 (SO4)3 and, K4[Fe(CN)6], 38., , =, , Po - P, P, , o, , =, , 39., , 32., , 15 ´ 78 121.8, ´, = 356.2, 250, 1.6, (c) Relative lowering of vapour pressure, depends upon the mole fraction of solute., o, , i.e.,, , P -P, Po, , 40., , w M 121.8 - 120.2 15 78, ´, =, = ´, 121.8, m W, m 250, , or m =, , = mole fraction of solute, , (d) DTb = iKb m, Given, (DTb)x > (DTb)y, \ ix Kb m > iyKb m, (Kb is same for same solvent), , Þ Po =, , 31., , p° - p, n, = mole fraction of solute =, p°, n+ N, (c) For ideal solution,, , DVmixing = 0 and DH mixing = 0 ., , .2, 1ù, é, ê xB = Mole fraction of solute = .2 + .8 = 5 ú, ê, ú, ë P = 60 mm of Hg, û, , Po - P, , (b) According to Raoult's law, the relative, lowering in vapour pressure of a dilute solution, is equal to the mole fraction of the solute present, in the solution., , 41., 42., , ix > iy, So, x is undergoing dissociation in water., (c) Colligative properties µ no. of particles., Since Al2(SO4)3 contains maximum number of, particles, hence will have the largest value of, freezing point depression., (b) Given Kf = – 1.86º cm–1, mass of solute, = 5.00 g, mass of solvent = 45.0 g, DTf = i × Kf . m, 5 ´ 1000, 3.82 = i × 1.86 ×, 142 ´ 45, (Molecular mass of Na2SO4 = 142), \ i = 2.63, (c) If compound dissociates in solvent i > 1, and on association i < 1., (d) Given a = 30% i.e., 0.3, HA ¾¾, ® H+ + A–, 1–a, , a, , 1 – 0.3, , 0.3 0.3, , a, , i = 1 – 0.3 + 0.3 + 0.3, i = 1.3
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161, , Solutions, For isotonic solution, p1=p2, , DTf = 1.3 × 1.86 × 0.1 = 0.2418, Tf = 0 – 0.2418 = – 0.2418 °C, 43., , (d) p = CRT (Osmotic pressure), wt ´ 1000, RT = 2.57 ´ 10-3, p=, Molecular mass ´ V, 1.26 ´ 1000, ´ 0.083 ´ 300, =, Mol.mass ´ 200, Molecular mass =, , 50, 10, ´ .0821 ´ T, ´ .0821 ´ T =, 60, M, M = 300 gm mol–1, , (c), , 50., , (b) As DTf = Kf . m, DTb = Kb . m, , 1.26 ´ 0.083 ´ 300 ´1000, , (a), , DT f = K f, =, , 1000W2, M 2W1, , 1.86, = 0.654°C, 0.512, As the Freezing Point of pure water is 0°C,, , 1.86×1000×68.5, = 0.372°C, 342×1000, , = 0.18 ´, , T f = – 0.372°C Q (0 – DTf ), 45., , (d) DTf = 0 – (– 0.00732°) = 0.00732, DTf = i × Kf × m, ΔTf, 0.00732, =, =2, i=, K f × m 1.86× 0.002, , 46., , (b) As DTf = i Kf m, For, HX, 1, t=0, t eq, (1 - 0.20), , 47., 48., , DTf = 0 – Tf, , 51., , , , H+, 0, 0.20, , +, , X0, 0.20, , Total no. of moles = 1 – 0.20 + 0.20 + 0.20, = 1 + 0.20 = 1.2, \ DTf = 1.2 × 1.86 × 0.5 = 1.1160 » 1.12 K, (c) During osmosis, water flows through, semipermeable membrane from lower, concentration to higher concentration., (a) Osmotic pressure of urea from the formula, nRT, pV = nRT Þ p =, V, 10, Þ, ´.0821´T, [1 dm3 = 1 litre], 60, 5% solution means, 100 ml º 5 g, 1000 ml º 50 g/L, Osmotic pressure of solution having nonvolatile solute p2 =, , 50, × .0821 × T, M, , 1 1000, ´, = 0.4K, 250 51.2, , DT f, DT, Hence, we have m =, = b, Kf, Kb, K, or DT f = DTb f, Kb, Þ [ DTb = 100.18 - 100 = 0.18°C], , 200 ´ 2.57 ´10-3, = 61038 g mol–1, , 44., , DT = K f m = 5.12 ´, , 49., , 0.654 = 0 – Tf, \ Tf = – 0.654, Thus the freezing point of solution will be, – 0.654°C., (b) Solvent having high cryoscopic constant, can be used in determination of molecular mass, by cryoscopic method., , pV = nRT =, , m, æ m ö RT, RT or M 2 = ç 2 ÷, èV ø p, M, , 52., , (b), , 53., , Where M2 = molecular mass of solute, and m2 = mass of solute, (a) Molecular masses of polymers are best, determined by osmotic pressure method ., Because other colligative properties give so low, values that they cannot be measured accurately., Osmotic pressure measurements can be made at, room temperature and do not require heating which, may change the nature of the polymer., , 54., , (a) Depression in F.P. µ No. of particles., Al 2 (SO 4 )3 provides five ions on ionisation, Al 2 (SO 4 ) 3 ¾, ¾® 2 Al 3+ + 3SO 24 –, while KCl provides two ions, K + + ClKCl
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EBD_8336, 162, , CHEMISTRY, C6H12O6 and C12H22O11 are not ionised so they, have single particle., Hence, Al 2 (SO4 )3 have maximum value of, depression in F.P. or lowest F.P., , 55., , (a) Conc. of particles in CaCl2 sol. will be max., as i = 3 is max. So it exhibits highest osmotic, pressure., Glucose and urea do not dissociate into ions, as, they are non-electrolytes., , 56., , 57., , (a), , K 4 [Fe(CN) 6 ] and Al 2 (SO 4 )3 both, , dissociates to give 5 ions or i = 5, 4 K + + [ Fe(CN ) ]K 4 [Fe(CN )6 ] , 6, and, 2Al3+ + 3SO 42Al 2 (SO 4 )3 , , (c) Osmotic pressure, elevation in boiling point,, lowering of vapour pressure and depression in, freezing point are colligative properties., , 58., , 59., , 60., , (a) As both the solutions are isotonic hence, there will be no net movement of the solvent, through the semipermeable membrane between, two solutions., (a) DTf = i × Kf × m, Van't Hoff factor, i = 2 for NaCl,, hence DTf = 0.02 Kf which is maximum in the, present case., Hence DTf is maximum or freezing point is, minimum., (b) Blood cells neither swell nor shrink in, isotonic solution. As isotonic solution have, equal concentration therefore, there is no flow, of solvent., Blood cell shrink when placed in hypertonic, solution and swells when placed in hypotonic, solution.
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163, , Electrochemistry, , 17, , Electrochemistry, , Trend Analysis with Important Topics & Sub-Topics, 2020, Topic Name, , Sub-Topic, , Conductance and, conductivity, , ionic mobility, , Electrolysis and types, of electrolysis, , Cells and electrode, potential, Nernst, equation, , LOD - Level of Difficulty, , 2019, , 2., , 3., , 2017, , 2016, , QNS. LOD QNS. LOD QNS. LOD QNS. LOD QNS. LOD, , electrolysis, , 1, , E, , Faraday's l aw, relation between, equilibrium, , 1, , A, 1, , A, , 1, , E, , 1, , E, , 1, , D, , constant and Ecell, relation between, equilibrium, constant and Gibbs, energy, Nernst equation, E - Easy, , A - Average, , Topic 1: Conductance and Conductivity, 1., , 2018, , Following limiting molar conductivities are given, as, [NEET Odisha 2019], l°m(H2SO4) = x S cm2 mol–1, l°m(K2SO4) = y S cm2 mol–1, l°m(CH3COOK) = z S cm2 mol–1, l°m(in S cm2 mol–1) for CH3COOH will be, ( x – y), (a), +z, (b) x – y + 2z, 2, (c) x + y + z, (d) x – y + z, Ionic mobility of which of the following alkali, metal ions is lowest when aqueous solution of, their salts are put under an electric field ? [2017], (a) K, (b) Rb, (c) Li, (d) Na, Aqueous solution of which of the following, compounds is the best conductor of electric, current ?, [2015 RS], , 4., , 5., , D - Difficult, , Qns - No. of Questions, , (a) Acetic acid, C2H4O2, (b) Hydrochloric acid, HCl, (c) Ammonia, NH3, (d) Fructose, C6H12O6, At 25°C molar conductance of 0.1 molar aqueous, solution of ammonium hydroxide is 9.54, ohm-1 cm2mol-1 and at infinite dilution its molar, conductance is 238 ohm-1 cm2 mol-1. The degree, or ionisation of ammonium hydroxide at the same, concentration and temperature is : [NEET 2013], (a) 20.800%, (b) 4.008%, (c) 40.800%, (d) 2.080%, Limiting molar conductivity of NH4OH, (i.e., L°m(NH4OH)) is equal to :, [2012], (a), (b), (c), (d), , °, °, °, Lm, (NH 4Cl ) + Lm (NaCl) -L m(NaOH), , °, °, °, Lm, (NaOH) + L m (NaCl) -Lm (NH4 Cl ), , °, °, °, Lm, (NH 4OH) + L m(NH4Cl) -Lm( HCl), °, °, °, Lm, ( NH Cl) + Lm( NaOH ) -L m( NaCl), 4
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EBD_8336, 164, , 6., , 7., , 8., , CHEMISTRY, Molar conductivities (L°m) at infinite dilution of, NaCl, HCl and CH3COONa are 126.4, 425.9 and, 91.0 S cm2 mol–1 respectively. L°m for CH3COOH, will be :, [2012 M], (a) 425.5 S cm2 mol–1 (b) 180.5 S cm2 mol–1, (c) 290.8 S cm2 mol–1 (d) 390.5 S cm2 mol–1, An increase in equivalent conductance of a, strong electrolyte with dilution is mainly due to:, [2010], (a) increase in ionic mobility of ions, (b) 100% ionisation of electrolyte at normal, dilution, (c) increase in both i.e. number of ions and, ionic mobility of ions, (d) increase in number of ions, Which of the following expressions correctly, represents the equivalent conductance at infinite, dilution of Al2 (SO4)3,? Given that L °Al3+ and, , L°SO2 - are the equivalent conductances at, 4, infinite dilution of the respective ions. [2010], , 9., , 10., , (a), , 1 °, L, 3 Al3+, , +, , (b), , 2L °, , + 3L°, , (c), , L°, , (d), , æL °, + L ° 2- ö ´ 6, è Al3+, SO4 ø, , Al3+, , Al3+, , 1, 2, , 11., , 12., , 13., , L°, , SO 24-, , SO 24-, , + L°, , SO24 -, , M, solution of, 32, a weak monobasic acid is 8.0 mho cm2 and at, infinite dilution is 400 mho cm2. The dissociation, constant of this acid is:, [2009], (a) 1.25 × 10–6, (b) 6.25 × 10–4, (c) 1.25 × 10–4, (d) 1.25 × 10–5, Kohlrausch’s law states that at :, [2008], (a) finite dilution, each ion makes definite, contribution to equivalent conductance of, an electrolyte, whatever be the nature of, the other ion of the electrolyte., (b) infinite dilution each ion makes definite, contribution to equivalent conductance of, an electrolyte depending on the nature of, the other ion of the electrolyte., , The equivalent conductance of, , 14., , 15., , 16., , 17., , (c) infinite dilution, each ion makes definite, contribution to conductance of an, electrolyte whatever be the nature of the, other ion of the electrolyte., (d) infinite dilution, each ion makes definite, contriubtion to equivalent conductance of, an electrolyte, whatever be the nature of, the other ion of the electrolyte., The ionic conductance of Ba 2+ and Cl – are, respectively 127 and 76 ohm–1 cm2 at infinite, dilution. The equivalent conductance, (in ohm–1 cm2) of BaCl2 at infinite dilution will be :, (a) 139.5, (b) 203, [2000], (c) 279, (d) 101.5, Specific conductance of a 0.1 N KCl solution at, 23ºC is 0.012 ohm–1 cm–1. Resistance of cell, containing the solution at same temperature was, found to be 55 ohm. The cell constant is [2000], (a) 0.918 cm–1, (b) 0.66 cm–1, –1, (c) 1.142 cm, (d) 1.12 cm–1, Equivalent conductances of NaCl, HCl and, CH3COONa at infinite dilution are 126.45, 426.16, and 91 ohm–1 cm2 respectively. The equivalent, conductance of CH3COOH at infinite dilution, would be, [1997], (a) 101.38 ohm–1 cm2 (b) 253.62 ohm–1 cm2, (c) 390.71 ohm–1 cm2 (d) 678.90 ohm–1 cm2, If 0.01 M solution of an electrolyte has a, resistance of 40 ohms in a cell having a cell, constant of 0.4 cm–1, then its molar conductance, in ohm–1 cm2 mol–1 is, [1997], (a) 102, (b) 104, (c) 10, (d) 103, On heating one end of a piece of a metal, the, other end becomes hot because of, [1995], (a) resistance of the metal, (b) mobility of atoms in the metal, (c) energised electrons moving to the other end, (d) minor perturbation in the energy of atoms, If 0.5 A current is passed through acidified silver, nitrate solution for 100 minutes. The mass of, silver deposited on cathode, is (eq.wt.of silver, nitrate = 108), [1995], (a) 2.3523 g, (b) 3.3575 g, (c) 5.3578 g, (d) 6.3575 g, Which of the following is an insulator ?[1992], (a) Graphite, (b) Aluminium, (c) Diamond, (d) Silicon.
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165, , Electrochemistry, Topic 2: Electrolysis and Types of Electrolysis, 18., , 19., , 20., , 21., , 22., , 23., , 24., , 25., , On electrolysis of dil. sulphuric acid using, Platinum (Pt) electrode, the product obtained at, anode will be, [2020], (a) Oxygen gas, (b) H2S gas, (c) SO2 gas, (d) Hydrogen gas, The number of Faradays(F) required to produce, 20 g of calcium from molten CaCl2 (Atomic mass, of Ca = 40 g mol–1) is:, [2020], (a) 2, (b) 3, (c) 4, (d) 1, When 0.1 mol MnO42– is oxidised, the quantity, of electricity required to completely oxidise, [2014], MnO42– to MnO4– is:, (a) 96500 C, (b) 2 × 96500 C, (c) 9650 C, (d) 96.50 C, The weight of silver (at wt. = 108) displaced by a, quantity of electricity which displaces 5600 mL, of O2 at STP will be :[2014], (a) 5.4 g, (b) 10.8 g, (c) 54.9 g, (d) 108.0 g, How many grams of cobalt metal will be, deposited when a solution of cobalt (II) chloride, is electrolyzed with a current of 10 amperes for, 109 minutes (1 Faraday = 96,500 C; Atomic mass, of Co = 59 u), [NEET Kar. 2013], (a) 0.66, (b) 4.0, (c) 20.0, (d) 40.0, Al2O3 is reduced by electrolysis at low potentials, and high currents. If 4.0 × 104 amperes of current, is passed through molten Al 2 O 3 for, 6 hours, what mass of aluminium is produced?, (Assume 100% current efficiency and At. mass, of Al = 27 g mol–1), [2009], (a) 8.1 × 104 g, (b) 2.4 × 105 g, (c) 1.3 × 104 g, (d) 9.0 × 103 g, 4.5 g of aluminium (at. mass 27 amu) is, deposited at cathode from Al 3+ solution by a, certain quan tity of electric charge. The, volume of hydrogen produced at STP from, H+ ions in solution by the same quantity of, electric charge will be, [2005], (a) 44.8 L, (b) 22.4 L, (c) 11.2 L, (d) 5.6 L, In electrolysis of NaCl, when Pt electrode is, taken, then H2 is liberated at cathode while with, Hg cathode, it forms sodium amalgam. This is, because, [2002], , 26., , 27., , 28., , (a) Hg is more inert than Pt, (b) More voltage is required to reduce H+ at, Hg than at Pt, (c) Na is dissolved in Hg while it does not, dissolve in Pt, (d) Conc. of H+ ions is larger when Pt electrode, is taken, Standard electrode potentials are : Fe+2/Fe, [Eº = – 0.44]; Fe+3/Fe+2 [Eº = + 0.77] ; If Fe+2,, Fe+3 and Fe blocks are kept together, then [2001], (a) Fe+2 increases, (b) Fe+3 decreases, Fe +2, (c), remains unchanged, Fe +3, +2, (d) Fe decreases, On passing a current of 1.0 ampere for 16 min and, 5 sec through one litre solution of CuCl2, all, copper of the solution was deposited at cathode., The strength of CuCl2 solution was (Molar mass, of Cu= 63.5; Faraday constant = 96,500 Cmol–1), [1996], (a) 0.01 N, (b) 0.01 M, (c) 0.02 M, (d) 0.2 N, On electrolysis of dilute sulphuric acid using, platinum electrodes, the product obtained at the, anode will be, [1992], (a) hydrogen, (b) oxygen, (c) hydrogen sulphide (d) sulphur dioxide., Topic 3: Cells and Electrode Potential,, Nernst Equation, , 29., , !, For a cell involving one electron Ecell, = 0.59 V, , at 298 K, the equilibrium constant for the cell, reaction is :, [2019], , 30., , ù, é, 2.303RT, = 0.059 V at T = 298 Kú, êGiven that, F, û, ë, 2, 5, (a) 1.0 × 10, (b) 1.0 × l0, (c) 1.0 × l010, (d) 1.0 × l030, For the cell reaction, [2019], 2Fe3+(aq) + 2I– (aq) ® 2Fe2+ (aq) + I2(aq), 0.24 V at 298 K. The standard Gibbs energy (D,, G°) of the cell reaction is:, [Given that Faraday constant F = 96500 C mol–1], (a) –46.32 kJ mol–1, (b) –23.16 kJ mol–1, (c) 46.32 kJ mol–1, (d) 23.16 kJ mol–1
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EBD_8336, 166, , 31., , 32., , 33., , 34., , 35., , 36., , 37., , CHEMISTRY, In the electrochemical cell :[2003, 2017], Zn | ZnSO4 (0.01M) | | CuSO4 (1.0 M) | Cu, the emf, of this Daniel cell is E1. When the concentration, of ZnSO4 is changed to 1.0M and that of CuSO4, changed to 0.01M, the emf changes to E2. From, the followings, which one is the relationship, RT, between E1 and E2? (Given,, = 0.059), F, (a) E1 < E2, (b) E1 > E2, (c) E2 = 0 ¹ E1, (d) E1 = E2, The pressure of H2 required to make the potential, of H2-electrode zero in pure water at 298 K is :, [2016], –14, –12, (a) 10 atm, (b) 10 atm, (c) 10–10 atm, (d) 10–4 atm, A device that converts energy of combustion of, fuels like hydrogen and methane, directly into, electrical energy is known as :, [2015], (a) Electrolytic cell, (b) Dynamo, (c) Ni-Cd cell, (d) Fuel Cell, A hydrogen gas electrode is made by dipping, platinum wire in a solution of HCl of pH = 10 and, by passing hydrogen gas around the platinum, wire at one atm pressure. The oxidation potential, of electrode would be ?, [NEET 2013], (a) 0.59 V, (b) 0.118 V, (c) 1.18 V, (d) 0.059 V, Consider the half-cell reduction reaction :, Mn2+ + 2e– ® Mn, E° = –1.18 V, Mn2+ ® Mn3+ + e–, E° = –1.51 V, The E° for the reaction 3Mn 2+ ® Mn0 + 2Mn3+,, and possibility of the forward reaction are,, respectively, [NEET Kar. 2013], (a) – 2.69 V and no, (b) – 4.18 V and yes, (c) + 0.33 V and yes (d) + 2.69 V and no, Standard electrode potential of three metals X, Y, and Z are – 1.2 V, + 0.5 V and – 3.0 V, respectively., The reducing power of these metals will be :, [2011], (a) Y > Z > X, (b) X > Y > Z, (c) Z > X > Y, (d) X > Y > Z, The electrode potentials for, [2011], ® Cu+(aq), Cu2+(aq) + e– ¾¾, ® Cu(s), and Cu+(aq) + e– ¾¾, are + 0.15 V and + 0.50V, respectively. The value, of E°Cu 2+ /Cu will be :, , 38., , 39., , 40., , 41., , 42., , 43., , (a) 0.500 V, (b) 0.325 V, (c) 0.650 V, (d) 0.150 V, Standard electrode potential for Sn4+ / Sn2+ couple, is + 0.15 V and that for the Cr3+ / Cr couple is, – 0.74 V. These two couples in their standard state, are connected to make a cell. The cell potential, will be :, [2011], (a) + 1.19 V, (b) + 0.89 V, (c) + 0.18 V, (d) + 1.83 V, If the E°cell for a given reaction has a negative, value, then which of the following gives the, correct relationships for the values of DG° and, Keq ?, [2011], (a) DG° > 0 ; Keq > 1 (b) DG° < 0 ; Keq > 1, (c) DG° < 0 ; Keq < 1 (d) DG° > 0 ; Keq < 1, A solution contains Fe2+, Fe3+ and I– ions. This, solution was treated with iodine at 35°C. E° for, Fe3+ / Fe2+ is + 0.77 V and E° for I2/2I– = 0.536 V., The favourable redox reaction is : [2011 M], (a) I2 will be reduced to I–, (b) There will be no redox reaction, (c) I– will be oxidised to I2, (d) Fe2+ will be oxidised to Fe3+, 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, DG0 will be (F = 96500 C mol–1), (a) – 89.0 kJ, (b) – 89.0 J [2010], (c) – 44.5 kJ, (d) – 98.0 kJ, Consider the following relations for emf of a, electrochemical cell:, [2010], (i) emf of cell = (Oxidation potential of anode), – (Reduction potential of cathode), (ii) emf of cell = (Oxidation potential of anode), + (Reduction potential of cathode), (iii) emf of cell = (Reduction potential of anode), + (Reduction potential of cathode), (iv) emf of cell = (Oxidation potential of anode), – (Oxidation potential of cathode), Which of the above relations are correct?, (a) (ii) and (iv), (b) (iii) and (i), (c) (i) and (ii), (d) (iii) and (iv), Given:, [2009], (i) Cu2+ + 2e– ® Cu, Eo = 0.337 V, (ii) Cu2+ + e– ® Cu+, Eo = 0.153 V, Electrode potential, Eo for the reaction,, Cu + + e– ® Cu, will be :, (a) 0.90 V, (b) 0.30 V, (c) 0.38 V, (d) 0.52 V
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167, , Electrochemistry, 44., , 45., , On the basis of the following E° values, the, strongest oxidizing agent is :, [2008], [Fe(CN)6]4– ® [Fe(CN)6]3– + e–; E° = – 0.35 V, Fe2+ ® Fe3+ + e–, E° = – 0.77 V, (a) [Fe(CN)6]4–, (b) Fe2+, (c) Fe3+, (d) [Fe(CN)6]3–, The equilibrium constant of the reaction:, Cu(s) + 2Ag + (aq) ¾¾, ® Cu 2+ (aq) + 2Ag(s) ;, E° = 0.46 V at 298 K is, [2007], (a) 2.0 × 1010, (b) 4.0 × 1010, (c) 4.0 × 1015, (d) 2.4 × 1010, , E ° 2+, Fe /Fe, , = - 0.441V and, , If, , 47., , = + 0.771V, the standard EMF of the reaction, Fe + 2Fe3+ ® 3Fe2+ will be, [2006], (a) 1.653 V, (b) 1.212 V, (c) 0.111 V, (d) 0.330 V, A hypothetical electrochemical cell is shown below, A|A + (xM)||B+ (yM)|B, , 49., , 50., , 2KBr + I 2 ® 2KI + Br2, 2KBr + Cl 2 ® 2KCl + Br2, 2H 2O + 2F2 ® 4HF + O 2, , 53., , [2006], , The emf measured is +0.20 V. The cell reaction is, (a) A+ + e–® A; B++ e– ® B, (b) The cell reaction cannot be predicted, (c) A + B+ ® A+ + B, (d) A+ + B ® A + B+, The standard e.m.f. of a galvanic cell involving, cell reaction with n = 2 is found to be 0.295 V, at 25°C. The equilibrium constant of the, reaction would be, (Given F = 96500 C mol–1; R = 8.314JK–1mol–1), [2004], (a) 2.0 × 1011, (b) 4.0 × 1012, (c) 1.0 × 102, (d) 1.0 × 1010, On the basis of the information available from, the reaction, 4, 2, Al + O 2 ® Al 2 O 3 , DG = – 827 kJ mol–1 of, 3, 3, O2. The minimum e.m.f required to carry out an, electrolysis of Al2O3 is (F = 96500 C mol–1), (a) 8.56 V, (b) 2.14 V, [2003], (c) 4.28 V, (d) 6.42 V, Which reaction is not feasible?, [2002], (a) 2KI + Br2 ® 2KBr + I2, (b), (c), (d), , 52., , E ° 3+ 2+, Fe /Fe, , 46., , 48., , 51., , 54., , 55., , 56., , Cu(aq) is unstable in solution and undergoes, simultaneous oxidation and reduction according, to the reaction :, [2000], Cu 2+ (aq) + Cu( s ), 2Cu + (aq) , choose correct Eº for above reaction if Eº, Cu2+/Cu = 0.34 V and Eº Cu2+/Cu+ = 0.15 V, (a) –0.38 V, (b) +0.49 V, (c) +0.38 V, (d) –0.19 V, What is the Eºcell for the reaction, Cu(s) + Sn 4+ (aq), Cu 2+ (aq) + Sn 2+ (aq) + , at 25ºC if the equilibrium constant for the reaction, is 1 × 106?, [1999], (a) 0.5328 V, (b) 0.3552 V, (c) 0.1773 V, (d) 0.7104 V, For the cell reaction,, [1998], Cu2+ (C1, aq) + Zn(s) = Zn 2+ (C2, aq) + Cu(s) of, an electrochemical cell, the change in free energy,, DG, at a given temperature is a function of, (a) ln (C1), (b) ln (C2/C1), (c) ln (C2), (d) ln (C1 + C2), Without losing its concentration ZnCl2 solution, cannot be kept in contact with, [1998], (a) Au, (b) Al, (c) Pb, (d) Ag, Eº for the cell, Zn Zn 2+ (aq) Cu 2+ (aq) Cu is, 1.10 V at 25ºC. The equilibrium constant for the, cell reaction:, 2+, 2+, Zn + Cu (aq), Cu + Zn (aq) ,, is of the order of, [1997], (a) 10–18, (b) 10–37, (c) 1018, (d) 1037, Standard potentials (Eº) for some half-reactions, are given below :, [1997], 4, 2, +, +, (1) Sn + 2e ® Sn ; E º = +0.15 V, , (2), , 2 Hg2 + + 2e - ® Hg 2 2+ ; E º = +0.92 V, , (3), , PbO 2 + 4H + + 2e - ® Pb 2+ + 2H 2O ;, , E° + 1.45 V, Based on the above, which one of the following, statements is correct ?, (a) Sn4+ is a stronger oxidising agent than Pb4+, (b) Sn2+ is a stronger reducing agent than, Hg22+, (c) Hg2+ is a stronger oxidising agent than Pb4+, (d) Pb2+ is a stronger reducing agent than Sn2+
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EBD_8336, 168, , CHEMISTRY, , 57. The correct relationship between free energy and, equilibrium constant K of a reaction is [1996], (a) DG° = - RT ln K (b) DG = RT ln K, (c) DG° = RT ln K, (d) DG = - RT ln K, 58. An electrochemical cell is set up as: Pt; H2, (1atm) | HCl(0.1 M) || CH3COOH (0.1 M)| H2, (1atm); Pt. The e.m.f of this cell will not be zero,, because, [1995], (a) the temperature is constant, (b) e.m.f depends on molarities of acids used, (c) acids used in two compartments are different, (d) pH of 0.1 M HCl and 0.1 M CH3COOH is, not same, 59. The standard reduction potentials at 25°C of, Li+ / Li, Ba2+ / Ba, Na+ / Na and Mg2+ /Mg are, – 3.03, – 2.73, – 2.71 and – 2.37 respectively., Which one of the following is the strongest, oxidising agent?, [1994], (a) Na+, (b) Li+, (c) Ba2+, (d) Mg2+, , 61., , 62., , A button cell used in watches functions as, following, Zn(s) + Ag2O(s) + H2O(l), , 63., , 64., , 2Ag(s) + Zn2+(aq) + 2OH–(aq), If half cell potentials are :, Zn2+(aq) + 2e– ® Zn(s); Eo = – 0.76 V, Ag2O(s) + H2O (l) + 2e–, 2OH–(aq); Eo = 0.34 V, , (a) 1.968 V, (b) 2.0968 V [2008], (c) 1.0968 V, (d) 0.0968 V, The efficiency of a fuel cell is given by [2007], DG, DG, (b), DH, DS, DH, DS, (c), (d), DG, DG, In the silver plating of copper, K[Ag(CN)2] is, used instead of AgNO3. The reason is [2002], (a) A thin layer of Ag is formed on Cu, (b) More voltage is required, (c) Ag+ ions are completely removed from, solution, (d) Less availability of Ag+ ions, as Cu cannot, displace Ag from [Ag(CN)2]– ion, The most convenient method to protect the, bottom of ship made of iron is, [2001], (a) Coating it with red lead oxide, (b) White tin plating, (c) Connecting it with Mg block, (d) Connecting it with Pb block, The most durable metal plating on iron to protect, against corrosion is, [1994], (a) nickel plating, (b) copper plating, (c) tin plating, (d) zinc plating., , (a), , Topic 4: Commercial Cells and Corrosion, 60., , The cell potential will be :, [NEET 2013], (a) 0.42 V, (b) 0.84 V, (c) 1.34 V, (d) 1.10 V, Standard free energies of formation (in kJ/mol), at 298 K are – 237.2, – 394.4 and – 8.2 for H2O(l),, CO2(g) and pentane (g), respectively. The value, of E°cell for the pentane-oxygen fuel cell is :, , 65., , ® 2Ag(s) +, , 1, , (a), , 8, , (c), , 15, , (c), , 22, , (c), , ANSWER KEY, 29 (c) 36 (c), , 43, , (d), , 50, , (b), , 57, , (a), , 64, , (c), , 2, , (c), , 9, , (d), , 16, , (b), , 23, , (a), , 30, , (a), , 37, , (b), , 44, , (c), , 51, , (c), , 58, , (d), , 65, , (d), , 3, , (b), , 10, , (d), , 17, , (c), , 24, , (d), , 31, , (b), , 38, , (b), , 45, , (c), , 52, , (c), , 59, , (d), , 4, , (b), , 11, , (b), , 18, , (a), , 25, , (b), , 32, , (a), , 39, , (d), , 46, , (b), , 53, , (b), , 60, , (d), , 5, , (d), , 12, , (b), , 19, , (d), , 26, , (b), , 33, , (d), , 40, , (c), , 47, , (c), , 54, , (b), , 61, , (c), , 6, , (d), , 13, , (c), , 20, , (c), , 27, , (a), , 34, , (a), , 41, , (a), , 48, , (d), , 55, , (d), , 62, , (b), , 7, , (a), , 14, , (d), , 21, , (d), , 28, , (b), , 35, , (a), , 42, , (a), , 49, , (b), , 56, , (b), , 63, , (d)
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169, , Electrochemistry, , Hints & Solutions, 1., , (a) According to Kohlrausch’s law, l°m(AB) = l°m(A+) + l°m(B–), So, l°m(CH3COOH) = l°m(CH3COO–) + l°m(H+), So l°m(CH3COOH), = l°m(CH3COOK) +, , 1, 1, l° (H SO ) – l° (K SO ), 2 m 2 4 2 m 2 4, , x y, æ x – yö, – = z + çè, ÷, 2 ø, 2 2, (c) Li+ being smallest, has maximum charge, density., \ Li+ is most heavily hydrated among all, alkali metal ions. Effective size of Li+ in aqueous, solution is therefore, largest. So, moves slowest, under electric field., (b) HCl completely dissociates to give Hr and, Cl– ions, hence act as very good electrolyte., While others are non- electrolytes., , =z+, , 2., , 3., , 4., , (b) a =, , 5., , (d), , o, , Lm, , L ¥m, , =, , K Conductivity, =, Normality, N, On increasing dilution, both conductivity and, normality decreases. But the decreases in, conductivity is more than compensated by decrease, lceq =, , 9.54, = 0.04008 = 4.008 %., 238, o, , in its normality. Hence, effectively l ceq increases, on dilution., , 8., , 9., , Ka =, 10., , o, , L m( NH4 Cl ) = L mNH +4 + L mCl-, , o, , o, , o, , L m( NaOH ) = L mNa + + L mOH o, , o, , o, , L( NaCl ) = L mNa + + L mClo, , \ Lm, o, , = Lm, , (, , (, , NH 4+, , NH 4+, , ), , ), , o, , + Lm, o, , (, , OH -, , ), , o, , 11., , ( ) ( ), o, éo, ù, + L m( OH- ) - ê L m( Na + ) + L m( Cl- ) ú, ë, û, + Lm, , Cl-, , + Lm, , Na +, , o, , o, , o, , o, , L m( NH4 OH )= L m( NH 4Cl ) + L m ( NaOH ), o, , - L m( NaCl ), , 6., 7., , o, o, o, o, (d) L CH3COOH = L CH 3COONa + L HCl - L NaCl, , = 91 + 425.9 – 126.4 = 390.5, (a) Strong electrolytes are completely ionised, at all concentrations. On increasing dilution, the, number of ions remains the same but the ionic, mobility increases., , (c) Equivalent conductance of an electrolyte, at infinite dilution is given by the sum of, equivalent conductances of the respective ions, at infinite dilution., (d) Degree of dissociation,, L, 8.0, a=, =, = 2 ´ 10 -2, L ¥ 400, , (, , C a2, 1, » C a2 =, ´ 2 ´ 10-2, (1 - a), 32, , ), , 2, , = 1.25 ´ 10 -5, (d) Kohlrausch’s Law states that at infinite, dilution, each ion migrates independently of its, co-ion and contributes to the total equivalent, conductance of an electrolyte a definite share, which depends only on its own nature., From this definition we can see that option (d) is, the correct answer., (b) The ionic conductance or equivalent, conductance are given as:, , l¥, , Ba 2 +, , = 127 ohm–1 cm2 equivalent–1, , l ¥ – = 76 ohm–1 cm2 equivalent–1, Cl, , The equivalent conductance of BaCl2 at infinite, dilution will be given by:, ¥, Ù¥, BaCl 2 = l, , Ba 2, , + l ¥ – = 127 + 76, Cl, , = 203 ohm–1 cm2 equivalent–1, , If equivalent conductance (conductivity) of cation, and anion are l ¥+eq and l ¥–eq and equivalent, conductance of electrolyte is l¥eq in infinite dilution,, then,, ¥, ¥, Ù¥, eq =l + eq + l - eq
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EBD_8336, 170, , CHEMISTRY, At infinite dilution, if molar conductances of cation,, ¥, l ¥m, anion and electrolyte are l¥, + m , l - m and, respectively for the electrolyte Ax B y, th en, xAy+ + yBx–, AxBy , , 1 ¥, 1, l+ m + l¥, - m and, y, x, ¥, ¥, l¥, m = xl + m + yl - m, , ¥, l eq, =, , 12., , 13., , 17., 18., , 19., , (b) Given specific conductance of the solution, (k) = 0.012 ohm–1 cm–1 and resistance (R) = 55, ohm. We know that cell constant = Specific, conductance × Observed resistance = 0.012 ×, 55 = 0.66 cm–1., (c) By Kohlraush's law, Ù°eq NaCl = 126.45, l°, , Na +, , + l°, , Cl -, , l° + + l°, H, , Cl -, , l°, , CH3OO –, , = 126.45, , = 426.16, , + l°, , Na +, , = 91, , CH3COO, , Number of equivalents =, (Equivalent mass of Ca, =, , ...(2), ....(3), , 15., , 16., , W=, , Eit, 108 ´ 0.5 ´ 6000, =, = 3.3575 g., 96500, 96500, , ( +7), , ( +6), , Mn O 24 - ¾® MnO4- + e -, , 20., , (c), , 21., , Quantity of electricity required = 0.1F, = 0.1 × 96500 = 9650 C, (d) wO = nO × 32, , 22., , 5600, ´ 32 = 8g = 1 equivalent of O2, wO =, 2, 22400, = 1 equivalent of Ag = 108, (c) Applying,, , 0.1 mole, , H, , (d) Given molarity = 0.01 M, Resistance = 40 ohm;, l, -1, Cell constant = 0.4cm, A, Specific conductivity ( k), cell constant 0.4, =, =, = 0.01 ohm -1 cm -1, 40, resistance, 1000k, Molar conductance ( Ù m ) =, Molarity, 1000 ´ .01, =, = 103 ohm -1cm 2 mol -1, .01, (c) When one end of a metal is heated, the, free electrons are energised and move to the, other end. It heats up the other end of the metal., (b) Given current (i) = 0.5 A;, Time (t) = 100 minutes × 60 = 6000 sec, Equivalent weight of silver nitrate (E) = 108., According to Faraday's first law of electrolysis, , Atomic mass 40, =, = 20 ), Valency, 2, , =, , Ù°(CH3COOH) = 390.71ohm -1 cm 2, , 14., , Given mass, Equivalent mass, , 20, =1, 20, So, 1 Faraday of charge is required to deposit 1, equivalent of Ca., , ....(1), , on adding (2) and (3) then subtract (1) from it, °, l°, - + l + = 517.16 - 126.45, , (c) Diamond is an insulator, because of no free, electrons., (a) During the electrolysis of dil. sulphuric acid, using Pt electrodes following reaction occurs., At cathode : 4H+ (aq) + 4e– ¾® 2H2 (g), At anode : 2H2O (l) ¾® O2 (g) + 4H+ (aq) + 4e–, (d) 1 equivalent of any substance is deposited, by 1 F of charge., 20 g calcium contains,, , 2, , 2, , Eit, 96500, Equivalent weight of cobalt (II) = 59/2, I = 10 A, Time (t) = 109 min = 109 × 60 sec, Substituting these values we get,, , w = Zit =, , 59 ´ 10 ´ 109 ´ 60, = 20.0, 2 ´ 96500, (a) As Q = i × t, \ Q = 4.0 × 104 × 6 × 60 × 60 C, = 8.64 × 10 8 C, Now since 96500 C liberates 9 g of Al, 9, 8.64 × 108 C liberates, ´ 8.64 ´ 108 g Al, 96500, = 8.1 × 104 g of Al, w=, , 23.
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171, , Electrochemistry, 24., , (d) No. of gram, 4.5, = 0.5, Al3+ =, 9, , equivalent of H+ = No., , 2SO42–(aq), , of eq. of, , 27, = 9, 3, No. of gm eq. of H+ = no. of mole of H+, Hence, Mass of H+ = 0.5 × 1g = 0.5 g, We know that, 2g H2 at STP = 22.4 L, , For dilute sulphuric acid, first reaction is, preferred but for concentrated acid, second, reaction is preferred., , Eq. wt of Al3+ =, , 25., , 29., , 30., , 26., , 27., , (b) The reduction potential is higher in case of, E°(Fe3+/Fe2+) = +0.77V. Thus, Fe3+ will readily, reduce to Fe2+ and quantity of Fe3+ will decrease., , 31., , 0.059, log K, 1, , 0.59, = log K, 0.059, 10 = log k, K = 1010, (a) DG = –nFE°, = –2 × 96500 × 0.24 = – 46320 J/mol, = – 46.32 kJ/mol, (b) For cell,, Zn|ZnSO4(0.01 M) || CuSO4(1.0 M)|Cu, 2.303RT log [ Zn 2+ ], nF, [Cu 2+ ], , ( 0.01), 2.303RT, ´ log, 2´ F, 1, When concentrations are changed for ZnSO4, and CuSO4, we can write, \ E1 = E °cell -, , E2 = E °cell 32., , 33., 34., , 2.303RT, 1, ´ log, 2F, 0.01, , \ E1 > E2, (a) 2H+(aq) + 2e– ® H2(g), , P, E = E0 – 0.0591 log H 2, 2, [ H + ]2, PH 2, 0 = 0 – 0.0295 log, (10 -7 )2, PH 2, =1, (10-7 )2, PH 2 = 10–14 atm, (d) A device that converts energy of, combustion of fuels, directly into electrical, energy is known as fuel cell., (a) H2 ¾¾, ® 2H+ + 2e–, 1 atm, 10-10, (10-10 )2, 0.059, E, + =0–, log, H 2 /H, 2, 1, E, + = +0.59 V, , \, , =, , 28., , ° =, E cell, , Ecell = E °cell -, , W, q, =, (a) By Faraday's Ist Law,, E 96500, (where q = it = total electric charge), we know that no of equivalent, , W, it, 1 ´ 965, 1, =, =, =, E 96500 96500 100, (where i = 1 A, t = 16 × 60 + 5 = 965 sec.), Since, we know that, 1, no. of equivalent 100, Normality =, = 0.01 N, =, 1, Volume (in litre), (b) During electrolysis of sulphuric acid, the, following processes are possible at cathode:, ® O2(g) + 4H+(aq) + 4e–,, 2H2O(l) ¾¾, E° = +1.23 V, , 2.303RT, log K, nF, ° = 0.59 V , n = 1, Given : E cell, , (c), , 0.59 =, , 22.4, ´ 0.5 = 5.6 L, \ 0.5 g H2 at STP =, 2, (b) If mercury is used as cathode, H+ ions are, not discharged at mercury cathode because, mercury has a high over voltage for hydrogen., In electrolysis of NaCl, following ionisations take, place:, NaCl Na+ + Cl–, H2O H+ + OH–, Na+ and H+ ions move towards cathode., However, only H+ ions are discharged more readily, than Na+ ions because in electrochemical series,, hydrogen is lower than sodium., At cathode : 2H+ + 2e– ¾® H2(g), If mercury is used as cathode, then Na+ ions are, discharged at cathode in preference to H+ ions,, yielding sodium, which dissolves in mercury to, form sodium amalgam., At cathode : Na+ + e– ¾® Na(s), Na(s) + Hg(s) ¾® Na – Hg(s), , ¾¾, ®, , S2O82–(aq) + 2e–, E° = +1.96 V, , H 2 /H
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EBD_8336, 174, , 58., , 59., 60., 61., , CHEMISTRY, (d) For a concentration cell having different, concentrations of ions., 0.0591, C, log 1, E=, n, C2, If all the concentrations are identical then, obviously the cell voltage is zero. But as the pH, of 0.1 M HCl (strong acid) & pH of 0.1M, CH3COOH(weak acid) is not same, therefore, the cell voltage will not be zero., (d) Higher the reduction potential, stronger is, the oxidising agent., (d) E°Cell = E°OP + E°RP = 0.76 + 0.34 = 1.10 V, (c) Writing the equation for pentane-oxygen, fuel cell at respective electrodes and over all, reaction, we get, At Anode:, C 5 H12 + 10H 2 O ® 5CO 2 + 32H + + 32e -, , 62., , 63., , (pentane), , Calculation of DG° for the above reaction, DG° = [5×(–394.4) + 6× (–237.2)] – [–8.2], = – 1972.0 – 1423.2 + 8.2 = – 3387.0 kJ, = – 3387000 Joules., From the overall equation we find n = 32, 0, nFEcell, and, , Using the relation, DG° = –, substituting various values, we get, , 0, – 3387000 = –32×96500× Ecell, (F = 96500C), , 0, Ecell, =, , or, , 3387, V = 1.0968 V, 3088, , (b) Efficiency of a fuel cell ( h) =, , DG, DH, , DH is the required energy (input) and DG is the, useful energy obtained (output)., , At Cathode:, , 8O 2 + 32H + + 32e- ® 16H 2 O, Over all :C5 H12 + 8O 2 ® 5CO2 + 6H 2 O, , 3387000, 3387000, =, 32 ´ 96500 3088000, , or, , 64., , 65., , (d) In th e silver platin g of copper,, K[Ag(CN) 2 ] is used in stead of AgNO 3 ., Copper being more electropositive readily, precipitate silver from their salt solution, Cu + 2AgNO 3 ¾¾, ® Cu(NO3 ) 2 + Ag, whereas in K[Ag (CN)2] solution a complex anion, [Ag(CN)2]– is formed and hence Ag+ are less, available in the solution and therefore copper, cannot displace Ag from its complex ion., (c) For bottom of ship to be protected, it is, connected with more reactive metal than iron, like magnesium. This technique is called cathodic, protection., (d) This is because zinc has higher oxidation, potential than Ni, Cu and Sn., The process of coating of iron surface with zinc is, known as galvanization. Galvanized iron sheets, maintain their lustrue due to the formation of, protective layer of basic zinc carbonate.
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175, , Chemical Kinetics, , 18, , Chemical Kinetics, , Trend Analysis with Important Topics & Sub-Topics, 2020, , 2019, , 2018, , 2017, , 2016, , Topic Name, , Sub-Topic, QNS. LOD QNS. LOD QNS. LOD QNS. LOD QNS. LOD, first order reaction, 1, E, 1, A, rate law, rate of reaction, 1, E, Rate of reaction, rate, laws and rate constant difference between, first and second, 1, A, order reaction, half life of zero, 1, A, order reaction, Order of reaction and, order of the, 1, A, 1, A, half life period, reaction, half life of first, 1, A, order reaction, E, collision frequency 1, Theories of rate of, reaction, effect of catalyst, 1, E, LOD - Level of Difficulty, , E - Easy, , A - Average, , Topic 1: Rate of Reaction, Rate Laws and, Rate Constant, 1., , 2., , 3., , The rate constant for a first order reaction is, 4.606 × 10–3 s–1. The time required to reduce 2.0, g of the reactant to 0.2 g is :, [2020], (a) 200 s, (b) 500 s, (c) 1000 s, (d) 100 s, If the rate constant for a first order reaction is k,, the time (t) required for the completion of 99%, of the reaction is given by :, [2019], (a) t = 0.693/k, (b) t = 6.909/k, (c) t = 4.606/k, (d) t = 2.303/k, For the chemical reaction, 2NH (g), N2(g) + 3H2(g) , 3, the correct option is:, [2019], , Qns - No. of Questions, , 1 d [H 2 ], 1 d [ NH3 ], =3 dt, 2, dt, d [N2 ], d [ NH3 ], =2, (b) dt, dt, d [ N 2 ] 1 d [ NH3 ], =, (c) dt, 2 dt, d [H 2 ], d [ NH3 ], =2, (d) 3, dt, dt, The correct difference between first and second, order reactions is that, [2018], (a) The rate of a first-order reaction does not, depend on reactant concentrations, the, rate of a second-order reaction does, depend on reactant concentrations, (a), , 4., , D - Difficult, , -
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EBD_8336, 176, , 5., , 6., , 7., , CHEMISTRY, (b) The half-life of a first-order reaction does, not depend on [A]0, the half-life of a, second-order reaction does depend on, [A]0, (c) The rate of a first-order reaction does, depend on reactant concentrations, the, rate of a second-order reaction does not, depend on reactant concentrations, (d) A first-order reaction can be catalyzed, a, second-order reaction cannot be catalyzed, The rate constant of the reaction A ® B is, 0.6 × 10–3 mole per second. If the concentration, of A is 5 M then concentration of B after 20, minutes is :, [2015 RS], (a) 1.08 M, (b) 3.60 M, (c) 0.36 M, (d) 0.72 M, In a reaction, A + B ® Product, rate is doubled, when the concentration of B is doubled, and, rate increases by a factor of 8 when the, concentrations of both the reactants (A and B), are doubled. Rate law for the reaction can be, written as :, [2012], 2, (a) Rate = k [A][B], (b) Rate = k [A]2 [B]2, (c) Rate = k [A] [B], (d) Rate = k [A]2 [B], The rate of the reaction 2N2O5 ® 4NO2 + O2 can, be written in three ways :, [2011 M], -d[N 2 O5 ], = k [N 2 O5 ], dt, d[NO2 ], = k ¢ [N 2 O5 ], dt, , (d), , 9., , 10., , 8., , mol L–1s–1, The rate of the reaction, 2NO + Cl2 ¾¾, ® 2NOCl is given by the rate, equation rate = k [NO]2 [Cl2], [2010], The value of the rate constant can be increased by:, (a) increasing the concentration of NO., (b) increasing the temperature., (c) increasing the concentration of the Cl 2, (d) doing all of these, For the reaction, N2 + 3H2—® 2NH3, [2009], , d [ NH3 ], , dt, – d [ H2 ], , 11., , = 2 × 10–4 mol L–1 s–1 , the value of, , would be :, dt, (a) 4 × 10–4 mol L–1 s–1, (b) 6 × 10–4 mol L–1 s–1, (c) 1 × 10–4 mol L–1 s–1, (d) 3 × 10–4 mol L–1 s–1, In the reaction, BrO3– (aq) + 5Br - (aq) + 6H +, , d[O2 ], = k" [N2O5 ], dt, , The relationship between k and k ' and between, k and k ¢¢ are :, (a) k ' = 2k ; k ' = k, (b) k ' = 2k ; k ¢¢ = k / 2, (c) k ' = 2k ; k ¢¢ = 2k (d) k ' = k ; k ¢¢ = k, For the reaction [N 2O 5 (g) ® 2NO 2 (g) +, 1/2 O2(g)] the value of rate of disappearance of, N2O5 is given as 6.25 × 10–3 mol L–1s–1. The rate, of formation of NO2 and O2 is given respectively, as :, [2010], (a) 6.25 × 10–3 mol L–1s–1 and 6.25 × 10–3, mol L–1s–1, (b) 1.25 × 10–2 mol L–1s–1 and 3.125 × 10–3, mol L–1s–1, (c) 6.25 × 10–3 mol L–1s–1 and 3.125 × 10–3, mol L–1s–1, , 1.25 × 10–2 mol L–1s–1 and 6.25 × 10–3, , 12., , 13., , [2009], ® 3Br2 (l), , +3H2 O(l), The rate of appearance of bromine (Br2) is related, to rate of disappearance of bromide ions as, following:, d[Br2 ], 5 d[Br – ], = –, (a), dt, 3 dt, d[Br2 ] 5 d[Br – ], (b), =, dt, 3 dt, d[Br2 ] 3 d[Br – ], (c), =, dt, 5 dt, d [Br2 ], 3 d[Br – ], (d), = dt, 5 dt, For the reaction 2 A + B ® 3C + D, which of the following does not express the, reaction rate ?, [2006], d[ B ], d[ D], (a) (b), dt, dt, 1 d[A], 1 d[C], (c) –, (d) 2 dt, 3 dt, Consider the reaction, [2006], N2 (g) + 3H2 (g) ® 2 NH3 (g), d[ NH 3 ], The equality relationship between, and, dt, d[ H 2 ], is, dt
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177, , Chemical Kinetics, d[NH 3 ], 2 d[H 2 ], =dt, 3 dt, d[NH3 ], 3 d[H 2 ], (b) +, =dt, 2 dt, d[NH 3 ], d[H 2 ], =(c), dt, dt, d[NH3 ], 1 d[H 2 ], (d), =dt, 3 dt, d[ B ], ,, rate, of, reaction, is equal to, 3 A ® 2B, dt, [2002], 2 d[ A], (a) - 3 d[ A], (b) 3 dt, 2 dt, d[ A], 1, d, [, A, ], (c) (d) + 2, dt, 3 dt, For the reaction 2 N 2O 5 ® 4NO 2 + O 2 , rate, and rate constant are 1.02 × 10–4 mol lit–1 s–1 and, 3.4×10–5 s–1 respectively then concentration of, N2O5 at that time will be, [2001], (a) 1.732 M, (b) 3 M, (c) 3.4 × 105 M, (d) 1.02 × 10–4 M, In the following reaction, how is the rate of, appearance of the underlined product related to, the rate of disappearance of the underlined, reactant ?, [2000], , (a), , 14., , 15., , 16., , +, , 19., , 20., , 21., , 22., , 23., , BrO3- (aq) + 5Br - (aq) + 6H + (aq), , 17., , ¾¾, ® 3Br2 (l) + 3H 2 O(l), (a) d[Br2 ] = - 5 d[Br ], dt, 3 dt, d[Br2 ], d[Br - ], (b), =dt, dt, d[Br, ], 3, d[Br, ], 2, (c), =, dt, 5 dt, d[Br2 ], 3 d[Br - ], (d), =dt, 5 dt, The rate of reaction depends upon the [1995], (a) volume, (b) force, (c) pressure, (d) conc. of reactants, , 24., , 25., , Topic 2: Order of Reaction and Half Life, Period, 18., , A first order reaction has a rate constant of, 2.303 × 10–3 s–1. The time required for 40 g of, this reactant to reduce to 10 g will be, [NEET Odisha 2019], [Given that log102 = 0.3010], (a) 602 s, (b) 230.3 s, (c) 301 s, (d) 2000 s, , 26., , When initial concentration of the reactant is, doubled, the half-life period of a zero order, reaction, [2018], (a) is halved, (b) is doubled, (c) remains unchanged (d) is tripled, Mechanism of a hypothetical reaction [2017], X2 + Y2 ® 2XY is given below :, (i) X2 ® X + X(fast), (ii) X + Y2 XY + Y (slow), (iii) X + Y ® XY (fast), The overall order of the reaction will be :, (a) 2, (b) 0, (c) 1.5, (d) 1, A first order reaction has a specific reaction rate, of 10–2 s–1. How much time will it take for 20 g of, the reactant to reduce to 5 g ?, [2017], (a) 138.6 sec, (b) 346.5 sec, (c) 693.0 sec, (d) 238.6 sec, The rate of a first -order reaction is 0.04, mol l –1s–1 at 10 seconds and 0.03 mol l –1s–1 at, 20 seconds after initiation of the reaction. The, half-life period of the reaction is, [2016], (a) 24.1 s, (b) 34.1 s, (c) 44.1 s, (d) 54.1 s, When initial concentration of a reactant is, doubled in a reaction, its half-life period is not, affected. The order of the reaction is : [2015], (a) First, (b) Second, (c) More than zero but less than first, (d) Zero, A reaction is 50% completed in 2 hours and 75%, completed in 4 hours. The order of reaction is, [NEET Kar. 2013], (a) 0, (b) 1, (c) 2, (d) 3, For a reaction between A and B the order with, respect to A is 2 and the order with respect to B, is 3. The concentrations of both A and B are, doubled, the rate will increase by a factor of:, [NEET Kar. 2013], (a) 10, (b) 12, (c) 16, (d) 32, In a zero-order reaction for every 10° rise of, temperature, the rate is doubled. If the, temperature is increased from 10°C to 100°C, the, rate of the reaction will become :, [2012], (a) 256 times, (b) 512 times, (c) 64 times, (d) 128 times
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EBD_8336, 178, , 27., , 28., , 29., , CHEMISTRY, Which one of the following statements for the, order of a reaction is incorrect ?, [2011], (a) Order can be determined only experimentally., (b) Order is not influenced by stoichiometric, coefficient of the reactants., (c) Order of reaction is sum of power to the, concentration terms of reactants to express, the rate of reaction., (d) Order of reaction is always whole number., The unit of rate constant for a zero order reaction, is, [2011 M], (a) mol L–1 s–1, (b) L mol–1 s–1, (c) L2 mol–2 s–1, (d) s–1, During the kinetic study of the reaction,, 2A + B ® C + D, following results were obtained:, , Run [A]/mol L–1 [B]/mol L–1, , Initial rate of, formation of, –1, , D/mol L min, I, , 0.1, , 0.1, , –1, , –3, , 6.0 × 10, , –2, , II, , 0.3, , 0.2, , 7.2 × 10, , III, , 0.3, , 0.4, , 2.88 × 10, , IV, , 0.4, , 0.1, , 2.40 × 10, , –1, , –2, , Based on the above data which one of the, following is correct?, [2010], 2, , 30., , 31., , 32., , (a) rate = k [A] [B], , (b) rate = k[A] [B], , (c) rate = k [A]2 [B]2, , (d) rate = k [A] [B]2, , Half life period of a first-order reaction is 1386, seconds. The specific rate constant of the, reaction is:, [2009], –2, –1, –3, (a) 0.5 × 10 s, (b) 0.5 × 10 s–1, –2, –1, (c) 5.0 × 10 s, (d) 5.0 × 10–3 s–1, For the reaction A + B ¾¾, ® products, it is, observed that:, [2009], (1) On doubling the initial concentration of A, only, the rate of reaction is also doubled, and, (2) On doubling the initial concentrations of, both A and B, there is a change by a factor, of 8 in the rate of the reaction., The rate of this reaction is given by:, (a) rate = k [A] [B]2 (b) rate = k [A]2 [B]2, (c) rate = k [A] [B], (d) rate = k [A]2 [B], , 33., , The bromination of acetone that occurs in acid, solution is represented by this equation. [2008], CH3COCH3 (aq) + Br2 (aq) ®, CH3COCH2Br (aq) + H+ (aq) + Br– (aq), These kinetic data were obtained for given, reaction concentrations., Initial Concentrations, M, [CH3 COCH3], [Br2], [H+], 0.30, 0.05, 0.05, 0.30, 0.10, 0.05, 0.30, 0.10, 0.10, 0.40, 0.05, 0.20, Initial rate, disappearance of Br2, Ms–1, 5.7×10–5, 5.7 × 10–5, 1.2 × 10–4, 3.1 × 10–4, Base on these data, the rate equations is:, (a) Rate = k [CH3COCH3][H+], (b) Rate = k [CH = COCH3][Br2], (c) Rate = k [CH3 COCH3] [Br2] [H+]2, (d) Rate = k [CH3COCH3][Br2] [H+], The reaction of hydrogen and iodine, monochloride is given as:, [2007], H 2 (g) + 2ICl(g) ¾¾, ® 2HCl(g) + I2 (g), The reaction is of first order with respect to H2(g), and ICI(g), following mechanisms were proposed., Mechanism A:, H 2 (g) + 2ICl(g) ¾¾, ® 2HCl(g) + I 2 (g), , Mechanism B:, , H 2 (g) + ICl(g) ¾¾, ® HI(g);slow, , 34., , HI(g) + ICl(g) ¾¾, ® HCl(g) + I 2 (g);fast, Which of the above mechanism(s) can be, consistent with the given information about the, reaction?, (a) A and B both, (b) neither A nor B, (c) A only, (d) B only, In a first-order reaction A ® B, if k is rate, constant and inital concentration of the reactant, A is 0.5 M, then the half-life is, [2007], log 2, log 2, (b), (a), k, k 0.5, ln 2, 0.693, (d), (c), k, 0.5k
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179, , Chemical Kinetics, 35., , 36., , 37., , 38., , 39., , 40., , 41., , 42., , If 60% of a first order reaction was completed in, 60 minutes, 50% of the same reaction would be, completed in aproximately, [2007], (a) 45 minutes, (b) 60 minutes, (c) 40 minutes, (d) 50 minutes, (log 4 = 0.60, log 5 = 0.69), For a first order reaction A ¾® B the reaction, rate at reactant concentration of 0.01 M is, found to be 2.0 ´ 10-5 mol L-1 s-1. The half, life period of the reaction is, [2005], (a) 30 s, (b) 220 s, (c) 300 s, (d) 347 s, The rate of reaction between two reactants A, and B decreases by a factor of 4 if the, concentration of reactant B is doubled. The, order of this reaction with respect to reactant, B is:, [2005], (a) 2, (b) -2, (c) 1, (d) -1, The rate of a 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 [2004], (a) 0.383 min, (b) 23.1 min, (c) 8.73 min, (d) 7.53 min, If the rate of the reaction is equal to the rate, constant, the order of the reaction is [2003], (a) 3, (b) 0, (c) 1, (d) 2, The reaction A ® B follows first order kinetics., The time taken for 0.8 mole of A to produce 0.6, mole of B is 1 hour. What is the time taken for, conversion of 0.9 mole of A to produce 0.675, mole of B?, [2003], (a) 2 hours, (b) 1 hour, (c) 0.5 hour, (d) 0.25 hour, 3A ® B + C, it would be a zero order reaction, when, [2002], (a) the rate of reaction is proportional to, square of concentration of A, (b) the rate of reaction remains same at any, concentration of A, (c) the rate remains unchanged at any, concentration of B and C, (d) the rate of reaction doubles if concentration, of B is increased to double, Half life of a first order reaction is 4 s and the, initial concentration of the reactants is 0.12 M., The concentration of the reactant left after 16 s, is, [1999], (a) 0.0075 M, (b) 0.06 M, (c) 0.03 M, (d) 0.015 M, , 43., , 44., , 45., , The plot of concentration of the reactant vs., time for a reaction is a straight line with a negative, slope. The reaction follows a, [1996], (a) zero order rate equation, (b) first order rate equation, (c) second order rate equation, (d) third order rate equation, 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, [1995], (a) 53.72 min, (b) 50.49 min, (c) 48.45 min, (d) 46.45 min, Select the rate law that corresponds to data, shown for the following reaction, [1994], A+B ¾, ¾® products., Exp. [A], [B], Initial rate, 1, 0.012 0.035, 0.1, 2, 0.024 0.070, 0.8, 3, 0.024 0.035, 0.1, 4, 0.012 0.070, 0.8, (a) rate = k [B]3, (b) rate = k [B]4, (c) rate = k [A] [B]3 (d) rate = k [A]2 [B]2, Topic 3: Theories of Rate of Reaction, , 46., , 47., , 48., , 49., , An increase in the concentration of the, reactants of a reaction leads to change in, (a) heat of reaction, [2020], (b) threshold energy, (c) collision frequency, (d) activation energy, For a reaction, activation energy Ea = 0 and the, rate constant at 200 K is 1.6 × 106 s–1. The rate, constant at 400 K will be [NEET Odisha 2019], [Given that gas constant, R = 8.314 J K–1 mol–1], (a) 3.2 × 106 s–1, (b) 3.2 × 104 s–1, 6, –1, (c) 1.6 × 10 s, (d) 1.6 × 103 s–1, The addition of a catalyst during a chemical, reaction alters which of the following quantities?, [2016], (a) Entropy, (b) Internal energy, (c) Enthalpy, (d) Activation energy, The activation energy of a reaction can be, determined from the slope of which of the, following graphs ?, [2015], ln K, l, vs.T, (a), (b) ln K vs., T, T, T, l, (c), (d) ln K vs. T, vs., ln K, T
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EBD_8336, 180, , 50., , 51., , 52., , CHEMISTRY, What is the activation energy for a reaction if its, rate doubles when the temperature is raised from, 20°C to 35°C? (R = 8.314 J mol–1 K–1), [NEET 2013], (a) 269 kJ mol–1, (b) 34.7 kJ mol–1, (c) 15.1 kJ mol–1, (d) 342 kJ mol–1, A reaction having equal energies of activation, for forward and reverse reaction has :, [NEET 2013], (a) DG = 0, (b) DH = 0, (c) DH = DG = DS = 0 (d) DS = 0, Activation energy (E a ) and rate constants, (k1 and k2) of a chemical reaction at two different, temperatures (T1 and T2) are related by :, [2012 M], k, E æ1 1ö, (a) ln 2 = – a ç - ÷, k1, R è T1 T2 ø, (b), , ln, , (c), , ln, , (d), 53., , 54., , 55., , k2, E æ 1 1ö, =- aç - ÷, k1, R è T2 T1 ø, , 56., , 57., , 58., , k2, E æ 1 1ö, =- aç + ÷, k1, R è T2 T1 ø, k, E æ1 1ö, ln 2 = a ç - ÷, k1, R è T1 T2 ø, , For an endothermic reaction, energy of, activation is E a and enthalpy of reaction of, DH(both of these in kJ/mol). Minimum value of, Ea will be., [2010], (a) less than DH, (b) equal to DH, (c) more than DH, (d) equal to zero, The rate constants k1 and k2 for two different, reactions are 1016 . e–2000/T and 1015 . e–1000/T,, respectively. The temperature at which k1 = k2 is :, 2000, [2008], K, (a) 1000 K, (b), 2.303, 1000, K, (c) 2000 K, (d), 2.303, The temperature dependence of rate constant, (k) of a chemical reaction is written in terms of, *, , Arrhenius equation, k = Ae- Ea / RT . Activation, energy ( Ea* ) of the reaction can be calculated, by plotting, [2003], 1, (a) log k vs, (b) k vs T, log T, , 1, 1, (d) log k vs, log T, T, The activation energy for a simple chemical, reaction A ® B is Ea in forward direction. The, activation energy for reverse reaction [2003], (a) Is always double of Ea, (b) Is negative of Ea, (c) Is always less than Ea, (d) Can be less than or more than Ea, When a biochemical reaction is carried out in, laboratory in the absence of enzyme then rate of, reaction obtained is 10–6 times, then activation, energy of reaction in the Presence of enzyme is, 6, (a), [2001], RT, (b) Different from Ea obtained in laboratory, (c) P is required, (d) Can't say anything, Activation energy of a chemical reaction can be, determined by, [1998], (a) evaluating rate constant at standard, temperature, (b) evaluating velocities of reaction at two, different temperatures, (c) evaluating rate constants at two different, temperatures, (d) changing concentration of reactants, In a reversible reaction the energy of activation, of the forward reaction is 50 kcal. The energy of, activation for the reverse reaction will be, (a) < 50 kcal, [1996], (b) either greater than or less than 50 kcal, (c) 50 kcal, (d) > 50 kcal, A chemical reaction is catalyzed by a catalyst X., Hence X, [1995], (a) reduces enthalpy of the reaction, (b) decreases rate constant of the reaction, (c) increases activation energy of the reaction, (d) does not affect equilibrium constant of the, reaction, For an exothermic reaction, the energy of, activation of the reactants is, [1994], (a) equal to the energy of activation of products, (b) less than the energy of activation of products, (c) greater than the energy of activation of, products, (d) Sometimes greater and sometimes less than, that of the products, , (c), , 59., , 60., , 61., , k vs
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181, , Chemical Kinetics, ANSWER KEY, 1, , (b), , 8, , (b), , 15, , (b), , 22, , (a), , 29, , (d), , 36, , (d), , 43, , (a), , 50, , (b), , 57, , (b), , 2, , (c), , 9, , (b), , 16, , (d), , 23, , (a), , 30, , (b), , 37, , (b), , 44, , (a), , 51, , (b), , 58, , (c), , 3, , (c), , 10, , (d), , 17, , (d), , 24, , (b), , 31, , (a), , 38, , (b), , 45, , (a), , 52, , (b, d), , 59, , (b), , 4, , (b), , 11, , (d), , 18, , (a), , 25, , (d), , 32, , (a), , 39, , (b), , 46, , (c), , 53, , (c), , 60, , (d), , 5, , (d), , 12, , (d), , 19, , (b), , 26, , (b), , 33, , (d), , 40, , (b), , 47, , (c), , 54, , (d), , 61, , (b), , 6, , (d), , 13, , (a), , 20, , (c), , 27, , (d), , 34, , (c), , 41, , (b), , 48, , (d), , 55, , (d), , 7, , (b), , 14, , (b), , 21, , (a), , 28, , (a), , 35, , (a), , 42, , (a), , 49, , (b), , 56, , (d), , Hints & Solutions, 1., , (b) First order rate equation is, [A ], 2.303, log 0, k=, t, [ A], Þ 4.606 ´ 10 -3 =, , Þt=, , 2., , 5., , 2.303, 2, log, t, 0.2, , 2.303, 4.606 ´ 10-3, , ´ log10 =, , 103, = 500 sec, 2, , 6., , (c) For a first order reaction,, , [A]o, 2.303, log, [A], k, for 99% completion of the reaction,, 2.303, 100, t=, log, k, 100 - 99, 2.303, t=, log 102, k, 4.606, t=, k, t=, , 3., , 4., , 7., , (c) N 2 (g) + 3H 2 (g) , 2NH3 (g), , - d [ N 2 ] -1 d [ H 2 ] 1 d [ NH 3 ], =, =, dt, 3 dt, 2 dt, (b) (t1/2)1st order = Independent of concentration, 1, (t1/2)2nd order µ, [A]o, Half life for the second order reaction is, t1/2 =, , 1, K ´ [ A]0, , 8., , (d) Rate constant k = 0.6 × 10–3 mole per second., (unit mole per second shows zero order reaction), For a zero order reaction, [A] = [A]0 – kt, and [A0] – [A] = [B] = kt, = 0.6 × 10–3 × 20 × 60 = 0.72 M, (d) According to statement given in the, question, it is clear that, r µ [A]2 and r µ [B], That means, order of reaction with respect to B, is 1 and w.r.t A is 2., Hence, Rate = k[A]2[B]1, (b) Rate of disappearance of reactants = Rate, of appearance of products, 1 d(N 2 O5 ) 1 d(NO2 ) d(O 2 ), =, =, 2, dt, 4, dt, dt, 1, 1, k (N 2 O5 ) = k ¢ (N 2 O5 ) = k ¢¢ (N 2 O5 ), 2, 4, k k¢, =, = k ¢¢, 2 4, k, k¢ = 2k, k ¢¢ =, 2, ® 2 NO 2 (g) + 1/2 O 2 (g), (b) N 2O5 (g) ¾¾, –, , d, 1 d, d, [ N2O5 ] = + 2 dt [ NO2 ] = 2 dt [O2 ], dt, , d, [ NO2 ] = 1.25 ´ 10-2 mol L–1s–1 and, dt, d, [ O2 ] = 3.125 ´ 10-3 mol L–1s–1, dt
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EBD_8336, 182, , 9., , 10., , CHEMISTRY, (b) 2 NO(g) + Cl2(g) 2 NOCl(g), Rate = k [NO]2 [Cl2], The value of rate constant can be increased by, increasing the temperature and is independent, of the initial concerntration of the reactants., (d) Rate of disappearance of H2 = rate of, formation of NH3., –, , 1 d[H 2 ] 1 d[NH3 ], =, 3 dt, 2, dt, , Þ, , 16., , 17., , – d[H 2 ] 3 d[NH3 ] 3, –4, =, = × 2×10, dt, 2, dt, 2, , A + B ¾¾, ® Product ;, , 1 d[Br – ] 1 d[Br2 ], =, 5 dt, 3, dt, d[Br2 ], 3 d[Br – ], Þ, =dt, 5, dt, , 12., , 18., , 2, , = 300.91 s, t1/2, Now, 40 g ¾¾® 20 g ¾¾¾, ® 10 g, So, 40 g substance requires 2 half life periods to, reduce upto 10 g, \ Time taken in reduction = 2 × 300.91 s, = 601.82 ; 602 s, , d[C], will not, 3dt, represent the reaction rate. It should not have, , Rate of reaction =, So,, 14., , 15., , 1 d[NH3 ], 1 d[H 2 ], =2, dt, 3 dt, , d[NH3 ], 2 d[H 2 ], =dt, 3 dt, , (b) 3A ¾, ¾® 2 B, Rate of appearance of B is equal to rate of, disappearance of A., 1 d[B], 1 d[A] d[B], 2 d[A], ==Þ, 2 dt, 3 dt, dt, 3 dt, (b) 2 N 2O 5 ® 4 NO 2 + O 2, From the unit of rate constant, it is clear that the, reaction follow first order kinetics. Hence,, by rate law equation, r = k [N2O5], , 0.693, 0.693, s–1, =, k, 2.303 ´ 10 –3, , t1/2, , –ve sign as it is product; since, , 13., , (a) For a first order reaction, Half life period, t 1 =, , (d) In the given options, -, , 1 d [ C], show the, 3 dt, rate of formation of product C, which will be positive., (a) If we write rate of reaction in terms of, concentration of NH3 and H2, then, , r µ [A][B], , By decreasing volume of gas or increasing pressure, of it, concentration will increases. Hence, (a) and, (c) are also correct options., , (d) Rate of disappearance of Br –, = rate of appearance of Br 2, Þ–, , 1 d[Br - ], 1 d[Br2 ], =+, 5 dt, 3 dt, d[Br2 ], 3 d[Br – ], =–, dt, 5 dt, (d) The rate of a reaction is the speed at which, the reactants are converted into products. It, depends upon the concentration of reactants., e.g for the reaction, –, , = 3×10 –4 mol L–1s –1, , 11., , where r = 1.02 × 10–4, k = 3.4 × 10–5, 1.02 × 10–4 = 3.4 × 10–5 [N2O5], [N2O5] = 3M, (d) Rate of reaction, , [A]o, 2k, \ If [A]o = doubled, t1/2 = doubled, , 19., , (b) (t1/2 )zero =, , 20., , (c) Overall rate = k [X][Y2], … (1), k = rate constant, Assuming step (i) to be reversible, its equilibrium, constant,, [ X ]2, 2, keq =, [ X2 ] Þ [ X ] = keq [ X 2 ] ;, , [ X] =, , 1, 2, keq, , 1, , [ X2 ] 2, , … (2), , From eq (1) and (2), 1, , 1, , Rate = kkeq 2 [ X 2 ] 2 [ Y2 ], Overall order =, , 1, 3, + 1 = = 1.5, 2, 2
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183, , Chemical Kinetics, The overall reaction rate depends on the rate of, the slowest step., i.e., Overall rate = Rate of slowest step, , 21., , (a) Half life for a first order reaction,, 0.693, t1/2 =, K, 0.693, sec ., So, t1/2 =, 10-2, Also, for the reduction of 20 g of reactant to 5 g,, two half lives will be required., \ For 20 g of the reactant to reduce to 5 g,, time taken,, 0.693, sec = 138.6 sec., t=2×, 10-2, æ1ö, N = N0 ç ÷, è 2ø, , (d) order of reaction can be zero, whole number, or fractional., , 28., 29., , 31., , (d) Rate1 = k [A]2[B]3, when concentrations of both A and B are, doubled then, Rate2 = k [2A]2[2B]3 = 32 k [A]2[B]3, \ Rate will increase by a factor of 32., , (a) Rate = k [A]0, Unit of k = mol L–1 sec–1, (d) In case of (II) and (III), keepin g, concentration of [A] constant, when the, concentration of [B] is doubled, the rate, quadruples. Hence, it is second order with, respect to B. In case of I & IV, keeping the, concentration of [B] constant, when the, concentration of [A] is increased four times, rate, also increases four times. Hence, the order with, respect to A is one. hence, Rate = k [A] [B]2, , 2.303, æ 0.04 ö, log ç, K= (, è 0.03 ÷ø, 20 - 10), , 25., , 10°C, , 2O3 3O2, Rate - k[O3]2 [O2]–1, , 30., , (b) For a first order reaction,, t75% = 2 × t50%, , r, , The order w.r.t a reactant can be negative also as in, this case:, , (a) For a first order reaction, ( a - x1 ), 2.303, K = t - t log a - x, ( 2 1) ( 2 ), , 24., , = 29 = 512 times, , 27., , K=, , 23., , =2, , æ 100 -10 ö, çè, ÷, 10 ø, , (b), , n, , 2.303 ´ 0.1249, 10, 0.6932, 2.303 ´ 0.1249, t1/2 =, 10, 0.6932 ´10, t1/2 =, = 24.1 sec, 2.303 ´ 0.1249, 0.693, (a) t1/2 =, k, For first order t 1/2 is independent of initial, concentration of reactant., , 100°C, , 26., , where N = mass after n number of half life, n = number of half life, N0 = Initital mass, , 22., , r, , 32., , (b) For a first order reaction, 0.693, 0.693, ; k=, t1/2 =, = 0.5 × 10–3s–1, k, 1386, (a) When concentration of A is doubled, rate is, doubled. Hence order with respect to A is one., When concentrations of both A and B are, doubled, rate increases by 8 times hence order, with respect to B is 2., \ rate = k [A]1 [B]2, (a) Rewriting the given data for the reaction, H+, , ®, CH 3 COCH 3 (aq) + Br2 (aq) ¾¾¾, , CH3COCH 2 Br(aq) + H + (aq) + Br - (aq), S. Initial concent, No., -ration of, CH 3COCH3, , Initial concentr Initial concentr, -ation of Br2, -ation of H +, in M, in M, , in M, , Rate of, disappearance, of Br2 in MS-1, d, dx, i.e. - [Br2 ]or, dt, dt, , 1, 2, , 0.30, 0.30, , 0.05, 0.10, , 0.05, 0.05, , 5.7 ´ 10-5, , 3, , 0.30, , 0.10, , 0.10, , 1.2 ´ 10-4, , 4, , 0.40, , 0.05, , 0.20, , 3.1 ´ 10-4, , 5.7 ´ 10-5, , This reaction is autocatalyzed and involves, complex calculation for concentration terms.
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EBD_8336, 184, , CHEMISTRY, We can look at the above results in a simple way, to find the dependence of reaction rate (i.e. rate, of disappearance of Br 2)., From data (1) and (2) in which concentration of, CH3COCH3 and H+ remain unchanged and only, the concentration of Br 2 is doubled, there is no, change in rate of reaction. It means the rate of, reaction is independent of concentration of Br 2., Again from (2) and (3) in which (CH3CO CH3), and (Br2) remain constant but H+ increases from, 0.05 M to 0.10 i.e. doubled, the rate of reaction, changes from 5.7×10 –5 to 1.2 × 10 –4, (or 12 × 10–5 ), thus it also becomes almost, doubled. It shows that rate of reaction is directly, proportional to [H+]. From (3) and (4), the rate, should have doubled due to increase in conc of, [H+] from 0.10 M to 0.20 M but the rate has, changed from 1.2× 10–4 to 3.1×10–4. This is due, to change in concentration of CH3 CO CH3 from, 0.30 M to 0.40 M. Thus, the rate is directly, proportional to [CH3 COCH3]., rate = k [CH3COCH3]1[Br2]0[H+]1, , 34., , = k [CH3COCH3][H+]., (d) As the slowest step is the rate determining, step thus the mechanism B will be more, consistent with the given information. Also, because it involve one molecule of H2 and one, molecule of ICl, it can be expressed as :, r = k [H2][ICl], Which shows that the reaction is of first order, w.r.t. both H2 & ICl., (c) For a first order reaction, , 35., , a, 2.303, log10, t, a-x, when t = t½, 2.303, a, log10, k=, t½, a-a/2, 2.303, ln 2, or t½ =, log10 2 =, k, k, (a) For a first order reaction, , 33., , Now,, t½ =, , 36., , -5, , k=, , 2.0 ´ 10, [0.01], , t1/2 =, 37., , = 2 × 10–3, , 0.693, 2 ´ 10, , -3, , = 347 sec., , (b) Rate1= k [A]x[B]y, Rate1, = k [A]x [2B]y, 4, or Rate1 = 4k [A]x[2B]y, From (1) and (2) we get, , ... (1), ... (2), , k [A]x [B] y, = k [A]x[2B]y, 4, [B] y, = [2B] y, 4, y, , or, 38., , 39., , 1 æ 2B ö, 1, =ç, ÷ Þ = 2y or (2)–2 = 2y, 4 è Bø, 4, , y = –2., (b) For a first order reaction, A ® products, r, r = k [A] or k =, [A], -2, 1.5 ´ 10, = 3 × 10–2, Þk =, 0.5, 0.693, 0.693, Further, t1/2 =, =, = 23.1 min., k, 3 ´ 10-2, (b) Q r = k [A]n, if n = 0, , 2.303, a, k=, log, t, a-x, when t = 60 and x = 60%, 2.303, 100, 2.303, 100, log, =, log, = 0.0153, 60, 100 - 60, 60, 40, , 2.303, ´ 0.3010 = 45.31 min., 0.0153, (d) Given [A] = 0.01 M, Rate = 2.0 × 10–5 mol L–1 S–1, For a first order reaction, Rate = k [A], =, , k=, , k=, , 2.303, 100, 2.303, log, =, ´ log 2, 0.0153, 100 - 50 0.0153, , 40., , r = k [A]0, or r = k thus for zero order reactions rate is equal, to the rate constant., (b) A ® B For a first order reaction, given a = 0.8 mol, (a – x) = 0.8 – 0.6 = 0.2
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185, , Chemical Kinetics, , 2.303, -2, -1, ´ (0.301 + 0.824) = 1.29 ´ 10 min ., 200, Further, , 2.303, 0.8, log, or k = 2.303 log 4, 1, 0.2, again a = 0.9, a – x = 0.9 – 0.675 = 0.225, , =, , k=, , 41., , 45., , 2.303, 2.303log 4 =, log 4, t, Hence t = 1 hour, (b) For reaction 3A ¾, ¾® B + C, If it is zero order reaction r = k [A]0, i.e the rate, remains same at any concentration of 'A'., , 46., , For zero order reaction rate of reaction is, independent of reactant concentration., , 42., , (a) t1/2 = 4 s, 16, T, =, =4, n=, t1/2, 4, , T = 16, (Q T = n × t½), , n, , 4, , 0.12, æ 1ö, æ 1ö, A = Ao ç ÷ = 0.12 ´ ç ÷ =, = 0.0075 M, è 2ø, è 2ø, 16, , 43., , 47., , k400, Ea é T2 – T1 ù, =, k200 2.303R êë T1T2 úû, Since, given that Ea = 0, , \ log, , So, k400 = k200, , Zero order, reaction, , 48., , For zero order reaction, , K=, , [C 0 ] - [C], t, , [C0] = –Kt + [C0], Þ slope = –K, interecept = [C0], 44., , (a) Given initial concentration (a) = 2.00 m;, Time taken (t) = 200 min and final concentration, (a – x) = 0.15 m. For a first order reaction, rate constant,, k=, , a, 2.303, 2.303, 2.00, log, =, log, t, a-x, 200, 0.15, , k400, =0, k200, , k400, Þ k =1, 200, , [Co], , Time, , (c) From Arrhenius equation, log, , where Ao = initial concentration &, A = concentration left after time t., (a) For a zero order reaction, concentration, decreases at a contant rate with time. Hence, the, plot is a straight line with a negative slope., , [C], , 0.693, 0.693, =, = 53.49 min ., k, 1.29 ´ 10-2, (a) From data 1 and 3, it is clear that keeping (B), const, When [A] is doubled, rate remains, unaffected. Hence rate is independent of [A]., from 1 and 4, keeping [A] constant, when [B] is, doubled, rate become 8 times. Hence, rate µ [ B]3 ., (c) The number of collisions per second per, unit volume of the reaction mixture is known as, collision frequency (Z) . collision frequency µ, no. of reacting molecules or atoms. Higher the, concentration of reactant molecules higher is, the probability of collision and so the collision, frequency., (t1/2 ) =, , 2.303, 0.9, k=, log, t, 0.225, , 49., , So rate constant at 400, k = 1.6 × 106 s–1, (d) A catalyst provides an alternative route for, the reaction with a lower activation energy., (b) Arrhenius equation, Ea, K =A.e - Ea / RT Þ ln K = ln A –, RT, – Ea, slope =, R, so, activation energy of reaction can be, 1, determined from the slope of ln K vs, T, Arrhenius Equation, -Ea RT, represents the, K = Ae - Ea RT , where e, fraction of molecules
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EBD_8336, 186, , 50., , CHEMISTRY, æ1, 1ö, Ea, k, (b) log 2 =, çè T - T ÷ø, 2.303 R 1, k1, 2, Ea, 1 ù, é 1, log 2 = 2.303 ´ 8.314 ê, ú, 293, 308, ë, û, Ea, 15, 0.3 =, ×, 293 ´ 308, 2.303 ´ 8.314, , 55., , 0.3 ´ 2.303 ´ 8.314 ´ 293 ´ 308, Ea =, ., 15, , = 34673 J mol-1 = 34.7 kJ mol-1, 51., , (b) DH = Ea f - Eab = 0, , 52., , (b, d) According to Arrhenius equation, k 2 Ea æ 1 1 ö, =, k1, R çè T1 T2 ÷ø, , ln, , k2, E æ 1 1ö, =– a ç - ÷, k1, R è T2 T1 ø, , 56., , Ea > DH, , and k2, , \ Slope =, , DH, , (d) Given, k1 = 1016.e, , -, , 2000, T, , -, , 2000, T, , 15, = 10 .e, , 15, or 10 ´ 10 .e, , or 10.e, , -, , 2000, T, , or ln 10 -, , -, , -, , 2000, T, , =e, , -, , 1000, T, , 15, = 10 .e, , 1000, T, , 2000, 1000, =T, T, , -, , - Ea, 2.303 R, , or Ea = –2.303R ´ Slope, (d) The activation energy of reverse reaction, will depend upon whether the forward reaction, is exothermic or endothermic., As DH = Ea (forward reaction) – Ea (backward reaction), For exothermic reaction, DH = –ve, \ –DH = Ea(f) – Ea(b), or Ea( f ) = Ea(b) – DH, \ Ea( f ) < Ea(b), for endothermic reaction, DH = + ve, \ DH = Ea(f ) – Ea(b) or Ea(f ) = DH + Ea(b), \ Ea(f ) > Ea(b)., , 1000, = 1015.e T, , When k1 and k2 are equal at any temperature T,, we have, , 1016.e, , Ea, 2.303R, , 1/T, , Ea, , Reaction coordinate, , 54., , Slope =, , log k, , (c), , Energy, , 53., , ln, , 2000 1000, T, T, 1000, or 2.303 log 10 =, T, or 2.303 × 1 × T=1000, [\ log 10 = 1], 1000, K, or T =, 2.303, (d) k = Ae - Ea / RT, Ea, or log T = log A 2.303 RT, Comparing the above equation with, y = mx + c, 1, y = log k, x =, T, Thus a plot of log10k vs 1/T should be a straight, line, with slope equal to – Ea /2.303R and, intercept equal to log A, , or ln 10 =, , 1000, T, , 57., , (b) The presence of enzyme (catalyst) increases, the speed of reaction by lowering the energy, barrier, i.e. a new path is followed with lower, activation energy.
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187, , Chemical Kinetics, If reaction is exothermic,, , Energy, , ET, E'T, Ea, , DH = - ve Ea(b) > Ea(f), If reaction is endothermic, Ea, , Products, 1, , 60., Reactants + catalyst, Progress of reaction, , 58., , Here ET is the threshold energy., Ea and Ea is energy of activation of reaction in, 1, absence and presence of catalyst respectively., (c) We know that the activation energy of, chemical reaction is given by formula =, , 61., , DH = + ve Ea(b) < Ea(f), (d) A catalyst affects equally both forward and, backward reactions, therefore it does not affect, equilibrium constant of reaction., (b) Ea (Forward) + DH = Ea (backword), For Exothermic reaction, DH = –ve and, \ activation energy of reactant is less than the, energy of activation of products., Threshold energy, , Ea éT2 - T1 ù, k, ê, ú, where k is the rate, log 2 =, 1, k1 2.303R êë T1T2 ú, û, , 59., , constant at temperature T1 and k2 is the rate, constant at temperature T2 and E a is the, activation energy. Therefore, activation energy, of chemical reaction is determined by evaluating, rate constant at two different temperatures., (b) DH = Ea(f) - Ea(b), Thus, energy of activation for reverse reaction, depend upon whether reaction is exothermic or, endothermic, , E, A, , Ea(FR), , DH, , Progress of Reaction, , E¢a(B.R.), , B
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EBD_8336, 188, , CHEMISTRY, , 19, , Surface Chemistry, , Trend Analysis with Important Topics & Sub-Topics, 2020, Topic Name, Adsorption, Catalysis and theoreies, of catalysis, , Sub-Topic, characteristics of, adsorption, , 2019, , 1, 1, , 1, , coagulation, E - Easy, , Topic 1: Adsorption, , 2., , 3., , 2016, , The correct option representing a Freundlich, adsorption isotherm is [NEET Odisha 2019], x, x, (a), = kp–1, (b), = kp0.3, m, m, x, x, = kp2.5, (d), = kp–0.5, (c), m, m, Which one of the following characteristics is, associated with adsorption ?, [2016], (a) DG is negative but DH and DS are positive, (b) DG, DH and DS all are negative, (c) DG and DH are negative but DS is positive, (d) DG and DS are negative but DH is positive, Which of the following statements is correct for, the spontaneous adsorption of a gas? [2014], (a) DS is negative and, therefore, DH should, be highly positive, (b) DS is negative and therefore, DH should, be highly negative, (c) DS is positive and, therefore, DH should, be negative, (d) DS is positive and, therefore, DH should, also be highly positive, , 1, , E, , 1, , E, , E, , E, , Colloids and emulsions colloidal solutions, , 1., , 2017, , catalysis, zeta potential, , LOD - Level of Difficulty, , 2018, , QNS. LOD QNS. LOD QNS. LOD QNS. LOD QNS. LOD, , A, 1, , A - Average, , 4., , 5., , 6., , E, , D - Difficult, , Qns - No. of Questions, , In Freundlich Adsorption isotherm, the value of, 1/n is, [2012], (a) between 0 and 1 in all cases, (b) between 2 and 4 in all cases, (c) 1 in case of physical adsorption, (d) 1 in case of chemisorption, If x is amount of adsorbate and m is amount of, adsorbent, which of the following relations is, not related to adsorption process ?, [2011], (a) x / m = f (p) at constant T., (b) x / m = f (T) at constant p., (c) p = f (T) at constant (x / m)., x, = p´ T, (d), m, The Langmuir adsorption isotherm is deduced, using the assumption:, [2007], (a) the adsorption sites are equivalent in their, ability to adsorb the particles, (b) the heat of adsorption varies with coverage, (c) the adsorbed molecules interact with each other, (d) the adsorption takes place in multilayers.
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189, , Surface Chemistry, 7., , 8., , 9., , 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 whole number), [1994, 2006], (a) k, (b) log k, 1, (c) n, (d), n, Accor ding to the adsorption theor y of, catalysis, the speed of the reaction increases, because, [2003], (a) adsorption lowers the activation energy of, the reaction, (b) the concentration of reactant molecules at, the active centres of the catalyst becomes, high due to strong adsorption, (c) in the process of adsorption, the activation, energy of the molecules becomes large, (d) adsorption produces heat which increases, the speed of the reaction, Which is not correct regarding the adsorption, of a gas on surface of solid?, [2001], (a) On increasing temperature, adsorption, increases continuously, (b) Enthalpy and entropy changes are –ve, (c) Adsorption is more for some specific, substances, (d) This phenomenon is reversible, , Topic 2: Catalysis and Theories of Catalysis, 10. Which one of the following statements is not, correct?, [2017], (a) The value of equilibrium constant is, changed in the presence of a catalyst in, the reaction at equilibrium, (b) Enzymes catalyse mainly bio-chemical, reactions, (c) Coenzymes increase the catalytic activity, of enzyme, (d) Catalyst does not initiate any reaction, , 12., , (b) 50 mL of 1 M AgNO3 + 50 mL of 2 M KI, (c) 50 mL of 2 M AgNO3 + 50 mL of 1.5 M KI, (d) 50 mL of 0.1 M AgNO3 + 50 mL of 0.l M KI, 13., , 14., , 15., , 16., , 17., , Topic 3: Colloids and Emulsions, 11., , Measurin g Zeta potential is useful in, determining which property of colloidal, solution?, [2020], (a) Solubility, (b) Stability of the colloidal particles, (c) Size of the colloidal particles, (d) Viscosity, , Which mixture of the solutions will lead to the, formation of negatively charged colloidal, [AgI]I– sol. ?, [2019], (a) 50 mL of 1 M AgNO3 + 50 mL of 1.5 M KI, , 18., , On which of the following properties does the, coagulating power of an ion depend? [2018], (a) The magnitude of the charge on the ion, alone, (b) Size of the ion alone, (c) The sign of charge on the ion alone, (d) Both magnitude and sign of the charge, on the ion, Fog is colloidal solution of, [2016], (a) Liquid in gas, (b) Gas in liquid, (c) Solid in gas, (d) Gas in gas, Which property of colloidal solution is, independent of charge on the colloidal particles:[2014, 2015], (a) Electrophoresis, (b) Electro-osmosis, (c) Tyndall effect, (d) Coagulation, The protecting power of lyophilic colloidal sol, is expressed in terms of, [2012], (a) Coagulation value, (b) Gold number, (c) Critical miscelle concentration, (d) Oxidation number, Which one of the following forms micelles in, aqueous solution above certain concentration?, [2005], (a) Dodecyl trimethyl ammonium chloride, (b) Glucose, (c) Urea, (d) Pyridinium chloride, Which of the following forms cationic micelles, above certain concentration?, [2004], (a) Sodium dodecyl sulphate, (b) Sodium acetate, (c) Urea, (d) Cetyl trimethyl ammonium bromide
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EBD_8336, 190, , 19., , CHEMISTRY, Position of non-polar and polar part in micelle, is, [2002], (a) polar at outer surface and non-polar at inner, surface, (b) polar at inner surface and non-polar at outer, surface, (c) distributed all over the surface, (d) present in the surface only, Which is used for ending charge on colloidal, solution?, [2000], (a) Electrons, (b) Electrolytes, (c) Positively charged ions, (d) Compounds, Hardy-Schulze rule explains the effect of, electrolytes on the coagulation of colloidal, solution. According to this rule, coagulation, power of cations follow the order, [1999], (a) Ba+2 > Na+ > Al+3 (b) Al+3 > Na+ > Ba+2, (c) Al+3 > Ba+2 > Na+ (d) Ba+2 > Al+3 > Na+, , 20., , 21., , 22., , 23., , 24., , At the Critical Micelle Concentration (CMC) the, surfactant molecules, [1998], (a) decompose, (b) dissociate, (c) associate, (d) become completely soluble, The ability of an ion to bring about coagulation, of a given colloid depends upon, [1997], (a) its size, (b) the magnitude of its charge, (c) the sign of the charge alone, (d) both magnitude and sign of its charge, During dialysis, [1996], (a) only solvent molecules can diffuse, (b) solvent molecules, ions and colloidal, particles can diffuse, (c) all kinds of particles can diffuse through, the semi-permeable membrane, (d) solvent molecules and ions can diffuse, , ANSWER KEY, 1, , (b), , 4, , (a), , 7, , (d), , 10, , (a), , 13, , (d), , 16, , (b), , 19, , (a), , 22, , (c), , 2, , (b), , 5, , (d), , 8, , (a), , 11, , (b), , 14, , (a), , 17, , (a), , 20, , (b), , 23, , (d), , 3, , (b), , 6, , (a), , 9, , (a), , 12, , (b), , 15, , (c), , 18, , (d), , 21, , (c), , 24, , (d), , Hints & Solutions, 1., , (b) According to Freundlich isotherm, , x, = kp1/n ; where (n > 1), m, x, m, x, 1, = kp0.3 as, = 0.3, So,, m, n, So, answer is (b)., , p, , 1, x, = 0, = constant, adsorption, n, m, independent of pressure., , When, , 2., , is, , 1, x, x, = 1, = kp, µ p, adsorption varies, n, m, m, directly with pressure., , When, , 3., , (b) Adsorption is spontaneous process,, therefore change in the free energy (DG) for the, process is negative., According to Gibbs Helmholtz eqn., DG = DH – TDS, DS is negative because adhering of gas, molecules to the surface lowers the randomness., \ DG can be –ve only when DH is –ve., (b) For adsorption DS < 0 and for a, spontaneous change DG = – ve, hence DH should be highly negative which is, clear from the equation, DG = DH – TDS, = – DH – T(– DS) = – DH + TDS, So if DH is highly negative, DG will also be (– ve)
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191, , Surface Chemistry, 4., , (a) According to Freundlich Adsorption, isotherm, , 8., , 1, , x, = KP n, m, at low pressure, \, , x, µ P1, m, , at high pressure, , 5., , 6., , 7., , 9., , 1, =1, n, 1, =0, n, , 10., , x, = constant, m, i.e. the value of n varies between 0 to 1, x, (d), = f (P) at constant T and, m, , x, = f (T) at constant T, m, Thus, it can be stated that for constant value of, x/m, pressure and temperature can take different, values or, P = f (T) at constant (x/m)., x, ¹ P×T, But, m, (a) Langmuir adsorption isotherm is based on, the assumption that every adsorption site is, equivalent and the ability of a particle to bind, there is independent of whether or not nearby, sites are occupied., (d) According to Freundlich adsorption, isotherm., At intermediate pressure, extent of adsorption, x, 1, x, = kp1/n or log = log k + log p;, m, n, m, , 11., , 12., , 13., , AgNO3 + KI ¾¾, ® AgI + KNO 3, , A solution of AgNO3 and KI will form a, negatively charged colloidal sol, [AgI]I–, only, when KI is present in excess (i.e., KI behaves, as a solvent)., Millimole of KI is maximum in option (2), (50 × 2 = 100), (d) According to Hardy Schulze rule,, coagulating power of an ion depends on both, magnitude and sign of the charge on the ion., Greater the valence of the flocculating ion added,, greater is the coagulating power., , 14., 15., 16., , log k, log p, , x, plot of log, vs log p is linear with slope, m, =1, n, , (b), , Negatively, charged colloid, , Slope = 1/n, , log x, m, , (a) A catalyst lowers the activation energy of, the reaction., 1, Rate of reaction µ, activation energy, (a) On increasing temperature, adsorption of, a gas on surface of solid decreases. Solid adsorb, greater amount of substances at lower, temperature., (a) A catalyst speeds up both forward and, backward reaction with the same rate., So, equilibrium constant is not affected by the, presence of a catalyst at any given temperature., (b) In colloidal solution, the potential, difference between the fixed layer and the, diffused layer of opposite charge is known as, Zeta potential., Greater the Zeta potential more will be the, stability of colloidal particle., , 17., , (a) Fog is a colloidal system having dispersed, phase as liquid and dispersion medium as gas., (c) Tyndall effect is an optical property, and it, is independent of charge on colloidal particles., (b) The lyophobic sols are less stable than, lyophilic sols. The lyophilic sols are thus used, to protect the lyophobic sols. This property of, lyophilic sols is known as protective action of, lyophilic sols., Which can be represented by gold number., (a) Micelle formation is shown by surfactants, detergents (Dodecyl trimethyl ammonium, chloride) in their aqueous solutions.
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EBD_8336, 192, , 18., , CHEMISTRY, (d) Cetyl trimethyl ammonium bromide,, , 23., , [C16 H 33 (CH 3 )3 N + Br - ] is a cationic micelle, , o- Polarhead, n- Non-polar tail, (micelle), , 19., , (a), , 20., , (b) On addition of electrolyte, charge of, colloidal particles will get neutralized and, coagulation will occur., (c) According to this law the coagulating, effect of an ion on dispersed phase of opposite, charge increases with the valency of the ion., The precipitating power of Al3+ , Ba++, Na+ ions, is in order Al3+ > Ba2+ > Na+., (c) The critical micelle concentration is the, lowest concentration at which micelle formation, appears. When surfactants are present above, that CMC, they can act as emulsifiers that will, solubilise a compound (oil, dirt, etc.) which is, normally insoluble in the solvent being used., , 21., , 22., , (d) According to the Hardy Schulze rule, the, coagulating effect of an ion on dispersed phase, of opposite charge increases with the valency, of the ion. Therefore, more the charge on, oppositely charged ion, higher is the coagulation, value., Positive ion can coagulate negatively charged, colloid and vice-versa. So, the ability depends on, both magnitude and sign of its charge but the power, of coagulation depends only on the magnitude of, charge., , 24., , (d) Dialysis is a process of removing a, dissolved substance from a colloidal solution, by means of diffusion thr ough suitable, membrane. Colloidal particles cannot pass, through animal membrane. Only solvent, molecules and ions (in case of electrodialysis), can diffuse.
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193, , General Principles and Processes of Isolation of Elements, , 20, , General Principles, and Processes of, Isolation of Elements, , Trend Analysis with Important Topics & Sub-Topics, 2020, Topic Name, Occurrence of metals, Metallurgical, processes, LOD - Level of Difficulty, , 2019, , E - Easy, , A - Average, , Topic 1: Occurrence of Metals, 1., , 2., , 3., , 4., , 2018, , 2017, , 2016, , Sub-Topic, QNS. LOD QNS. LOD QNS. LOD QNS. LOD QNS. LOD, occurrence of, 1, E, metals, metallurgical, 1, A, 1, E, 1, A, processes, Ellingham diagram, 1, A, , Which one is malachite from the following ?, [2019], (a) CuFeS2, (n) Cu(OH)2, (c) Fe3O4, (d) CuCO3.Cu(OH)2, "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 ? [2015], (i) Metal nitrates are highly unstable., (ii) Metal nitrates are highly soluble in water., (a) (i) and (ii) are false, (b) (i) is false but (ii) is true, (c) (i) is true but (ii) is false, (d) (i) and (ii) are true, Which one of the following is a mineral of iron?, [2012], (a) Malachite, (b) Cassiterite, (c) Pyrolusite, (d) Magnetite, Identify the alloy containing a non-metal as a, constituent in it., [2012], (a) Invar, (b) Steel, (c) Bell metal, (d) Bronze, , D - Difficult, , Qns - No. of Questions, , Topic 2: Metallurgical Processes, 5., , 6., , 7., , Identify the correct statement from the following:, [2020], (a) Blister copper has blistered appearance, due to evolution of CO2., (b) Vapour phase refining is carried out for, Nickel by Van Arkel method., (c) Pig iron can be moulded into a variety of, shapes., (d) Wrought iron is impure iron with 4%, carbon., Identify the incorrect statement., [NEET Odisha 2019], (a) Gangue is an ore contaminated with, undesired materials., (b) The scientific and technological process, used for isolation of the metal from its ore is, known as metallurgy., (c) Minerals are naturally occurring chemical, substances in the earth’s crust., (d) Ores are minerals that may contain a metal., Considering Ellingham diagram, which of the, following metals can be used to reduce, alumina?, [2018], (a) Fe, (c) Zn, (c) Cu, (d) Mg
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EBD_8336, 194, , CHEMISTRY, Extraction of gold and silver involves leaching, with CN– ion. Silver is later recovered by [2017], (a) distillation, (b) zone refining, (c) displacement with Zn, (d) liquation, Match items of Column I with the items of, Column II and asign the correct code : [2016], , 8., , 9., , Column-I, (A) Cyanide process, (B) Froth flotation, process, (C) Electrolytic, reduction, (D) Zone refining, , 10., , 11., , 12., , 13., , (a) The DGfo of the sulphide is greater than, those for CS2 and H2S., , Column-II, (i) Ultrapure Ge, (ii) Dressing of ZnS, (iii) Extraction of Al, , 14., , (iv) Extraction of Au, (v) Purification of Ni, , Code :, (A) (B) (C), (D), (a) (iv) (ii), (iii), (i), (b) (ii), (iii) (i), (v), (c) (i), (ii), (iii), (iv), (d) (iii) (iv) (v), (i), In the extraction of copper from its sulphide ore,, the metal finally obtained by the reduction of, cuprous oxide with :, [2012, 2015 RS], (a) iron (II) sulphide (b) carbon monoxide, (c) copper (I) sulphide (d) sulphur dioxide, Which of the following elements is present as, the impurity to the maximum extent in the pig iron ?, (a) Manganese, (b) Carbon [2011], (c) Silicon, (d) Phosphorus, The following reactions take place in the blast, furnace in the preparation of impure iron. Identify, the reaction pertaining to the formation of the, slag., [2011 M], (a) Fe2O3(s) + 3 CO(g) ®2 Fe (l) + 3 CO2 (g), (b) CaCO3 (s) ®CaO (s) + CO2 (g), (c) CaO (s) + SiO2(s) ® CaSiO3 (s), (d) 2C(s) + O2 (g) ®2 CO(g), , Which of the following statements, about the, advantage of roasting of sulphide ore before, reduction is not true?, [2007], , (b) The DGfo is negative for roasting of, sulphide ore to oxide., (c) Roasting of the sulphide to the oxide is, thermodynamically feasible., (d) Carbon and hydrogen are suitable reducing, agents for metal sulphides., Sulphide ores of metals are usually concentrated, by froth flotation process. Which one of the, following sulphide ores offer an exception and, concentrated by chemical leaching?, [2007], (a) Galena, (b) Copper pyrite, (c) Sphalerite, (d) Argentite, , Topic 3: Purification and Uses of Metals, 15. The metal oxide which cannot be reduced to, metal by carbon is, [NEET Kar. 2013], (a) Fe2O3, (b) Al2O3, (c) PbO, (d) ZnO, 16. Which of the following pairs of metals is purified, by van Arkel method ?, [2011], (a) Ga and In, (b) Zr and Ti, (c) Ag and Au, (d) Ni and Fe, 17. The method of zone refining of metals is based, on the principle of, [2003], (a) Greater solubility of the impurity in the, molten state than in the solid, (b) Greater mobility of the pure metal than that, of the impurity, (c) Higher melting point of the impurity than, that of the pure metal, (d) Greater noble character of the solid metal, than that of the impurity, 18. Method used for obtaining highly pure silicon, used as a semiconductor material, is [1994, 96], (a) Oxidation, (b) Electrochemical, (c) Crystallization, (d) Zone refining, , ANS WER KEY, 1, , (d), , 2, , (b), , 3, , (d), , 4, , (b), , 5, , (c), , 6, , (a), , 7, , (d), , 9, , (a), , 11, , (b), , 13, , (d), , 15, , (b), , 17, , (a), , 8, , (c), , 10, , (c), , 12, , (c), , 14, , (d), , 16, , (b), , 18, , (d)
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195, , General Principles and Processes of Isolation of Elements, , Hints & Solutions, 1., 2., 3., 4., , 5., , 6., 7., , 8., , (d) Molachite is a green copper carborate, hydroxide mineral with chemical composition, CuCO3.C4(OH)2., (b) Metal nitrates are highly soluble in water, and are very stable for e.g. NaNO3 and KNO3., (d) [Fe3O4 Þ Magnetite], [CuCO3 Cu(OH)2 – Malachite], [Pyrolusite – MnO2] [Cassiterite – SnO2]., (b) Invar is a nickel iron alloy, Bell metal is an, alloy of about 80% copper and 20% tin, Bronze, is also an alloy of copper and tin., Steel : It always have few % of carbon., (c) The iron obtained from blast furnace, contains about 4% carbon and smaller amount, of impurities like S, P, Si, Mn. This form of iron is, known as pig iron. It can be moulded into variety, of shapes. Blister copper has blistered appearence, due to evolution of SO2.van Arkel method is used, for refining of Zirconium or Titatanium., (a) Gangue is the commercially worthless, material which is contaminating the ore., (d) Mg has more – DG value than alumina. So, it will be in the lower part of Ellingham diagram., Metals which have more – DG value can reduce, those metal oxides which have less – DG value., (c) Zn being more reactive than Ag and Au,, displaces them., , 11., 12., , 13., , 14., , 15., , Leaching, 4Ag + 8NaCN + 2H2O + O2 ¾¾¾¾, ®, 4Na[Ag(CN)2] + 4NaOH, , In Ellingham diagram, place of aluminium oxidation, is much below than that of carbon. Thus, carbon, cannot reduce Al2O3 to Aluminium metal., , Soluble Sodium dicyanoargentate (I), , Soluble cyanide compound can be treated with Zn, to give metal by displacement., Displacement, , 2Na[Ag(CN)2] + Zn ¾¾¾¾¾¾, ®, Na2[Zn(CN)4] + 2Ag¯, , 9., , 10., , (a) Highly electropositive metals like Al, K, Na, etc. are extracted by the electrolytic reduction., •, zone refining method is used for obtaining, metals of high purity e.g. Ge., •, Froth flotation process is suitable for, sulphide ores., •, Cyanide process is used for the extraction, of gold., (c) 2 Cu2S (s) + 3O2 (g) ® 2Cu2O(s) + 2SO2(g), The unchanged Cu2S, mixed with Cu2O and, heated strongly in absence of air, , Cu2S + 2Cu2O ® 6Cu + SO2, (b) Pig iron or cast iron contains 3 – 5% carbon, and varying amounts of Mn, Si, P and S which, makes the iron hard and brittle., (c) In blast furnace at about 1270 K, calcium, carbonate is almost completely decomposed to, give CaO which acts as a flux and combines with, SiO2 present as impurity (gangue) in the ore to, form calcium silicate (fusible slag), CaO(s) (basic flux) + SiO2 (s) (acidic flux), ® CaSiO3 (s) (slag), ¾¾, (d) The sulphide ore is roasted to oxide before, reduction because the DG°f of most of the, sulphides are greater than those of CS2 and H2S,, therefore, neither C nor H can reduce metal, sulphide to metal. Further, the standard free, energies of formation of oxide are much less than, those of SO2. Hence, oxidation of metal sulphides, to metal oxide is thermodynamically favourable., (d) Leaching is the selective dissolution of the, desired mineral leaving behind the impurities in, a suitable dissolving agent, e.g. Argentitie or, Silver glance, Ag2S is an ore of silver. Silver is, extracted from argentite by the mac-Arthur and, Forest process (leaching process)., Ag 2S + 4NaCN ® 2Na[Ag ( CN )2 ] + Na 2S, (b) Al2O3 cannot be reduced by carbon., , 16., , (b) Zr and Ti are purified by van Arkel method., ® ZrI 4 (g), Zr(s) + 2I 2 (g) ¾¾, On the hot, filament, , ® Zr(s) + 2I 2 (g), ZrI4 (g) ¾¾¾ ¾¾, , 17. (a) Zone refining is based on the difference in, solubility of impurities in molten and solid state, of the metal. This method is used for obtaining, metals of very high purity., 18. (d) Si obtained by reduction of SiCl4 with H2 is, further purified by zone refining method to get, Si of very high purity., Silicon is purified by zone-refining process because, the impurities present in it are more soluble in the, liquid phase than in the solid phase.
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197, , The p-Block Elements (Group 15, 16, 17 and 18), 3., , 4., , 5., , 6., , 7., , 8., , 9., , 10., , Which of the following oxoacids of phosphorus, has strongest reducing property?, [NEET Odisha 2019], (a) H3PO4, (b) H4P2O7, (c) H3PO3, (d) H3PO2, The correct order of N-compounds in its, decreasing order of oxidation states is [2018], (a) HNO3, NO, N2, NH4Cl, (b) HNO3, NO, NH4Cl, N2, (c) NH4Cl, N2, NO, HNO3, (d) HNO3, NH4Cl, NO, N2, Which is the correct statement for the given, acids?, [2016], (a) Phosphinic acid is a diprotic acid while, phosphonic acid is a monoprotic acid, (b) Phosphinic acid is a monoprotic acid while, phosphonic acid is a diprotic acid, (c) Both are triprotic acids, (d) Both are diprotic acids, The product obtained as a result of a reaction of, nitrogen with CaC2 is, [2016], (a) CaCN2, (b) CaCN, (c) CaCN3, (d) Ca2CN, Strong reducing behaviour of H3PO2 is due to, [2015 RS], (a) presence of one –OH group and two P–H, bonds, (b) high electron gain enthalpy of phosphorus, (c) high oxidation state of phosphorus, (d) presence of two –OH groups and one P–H bond., In which of the following compounds, nitrogen, exhibits highest oxidation state ?, [2012], (a) N2H4, (b) NH3, (c) N3H, (d) NH2OH, Which of the following statements is not valid, for oxoacids of phosphorus?, [2012], (a) Orthophosphoric acid is used in the, manufacture of triple superphosphate., (b) Hypophosphorous acid is a diprotic acid., (c) All oxoacids contain tetrahedral four, coordinated phosphorus., (d) All oxoacids contain atleast one P = O and, one P — OH group., Oxidation states of P in H4 P2O5 , H4 P2O6 , and, H4 P2O7 , are respectively:, [2010], (a) + 3, + 5, + 4, (b) + 5, + 3, + 4, (c) + 5, + 4, + 3, (d) + 3, + 4, + 5, , 11., , 12., , 13., , 14., , 15., , 16., , How many bridging oxygen atoms are present, in P4O10?, [2010], (a) 5, (b) 6, (c) 4, (d) 2, Nitrogen forms N2, but phosphorus is converted, into P4 from P, the reason is, [2001], (a) Triple bond is present between phosphorus, atom, (b) pp – pp bonding is strong, (c) pp – pp bonding is weak, (d) Multiple bond is formed easily, Which of the following oxy-acids has the, maximum number of hydrogens directly attached, to phosphorus?, [1999], (a) H4P2O7, (b) H3PO2, (c) H3PO3, (d) H3PO4, Repeated use of which one of the following, fertilizers would increase the acidity of the soil?, [1998], (a) Urea, (b) Superphosphate of lime, (c) Ammonium sulphate, (d) Potassium nitrate, Which of the following species has the highest, dipole moment ?, [1997], (a) NH3, (b) PH3, (c) AsH3, (d) SbH3, The structural formula of hypophosphorous, acid is, [1997], , O, , O, (a), , H, , P, H, , OH, , (b) H, , OH, , O, (c), , 17., 18., , HO, , P, OH, , O, , P, OH, , (d) H, , P, , OOH, OH, OH, Brown ring test is used to detect, [1994], (a) Iodine, (b) Nitrate, (c) Iron, (d) Bromide, Which of the following fertilizers has the highest, nitrogen percentage ?, [1993], (a) Ammonium sulphate, (b) Calcium cyanamide, (c) Urea, (d) Ammonium nitrate
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EBD_8336, 198, , 19., , 20., , 21., , 22., , 23., , 24., , 25., , 26., , 27., , 28., , CHEMISTRY, Which one of the following substance is used, in the laboratory for fast drying of neutral, gases?, [1992], (a) Phosphorus pentoxide, (b) Active charcoal, (c) Anhydrous calcium chloride, (d) Na3PO4., Number of electrons shared in the formation of, nitrogen molecule is, [1992], (a) 6, (b) 10, (c) 2, (d) 8, Sugarcane on reaction with nitric acid gives, [1992], (a) CO2 and SO2, (b) (COOH)2, (c) 2 HCOOH (two moles), (d) No reaction., Nitrogen is relatively inactive elemen t, because, [1992], (a) Its atom has a stable electronic configuration, (b) It has low atomic radius, (c) Its electronegativity is fairly high, (d) Dissociation energy of its molecule is fairly, high, H3PO2 is the molecular formula of an acid of, phosphorus. Its name and basicity respectively, are, [1992], (a) Phosphorus acid and two, (b) Hypophosphorous acid and two, (c) Hypophosphorous acid and one, (d) Hypophosphoric acid and two, Aqueous solution of ammonia consists of, [1991], (a) H+, (b) OH–, (d) NH4+and OH–, (c) NH4+, P2O5 is heated with water to give, [1991], (a) Hypophosphorous acid, (b) Phosphorous acid, (c) Hypophosphoric acid, (d) Orthophosphoric acid, Basicity of orthophosphoric acid is, [1991], (a) 2, (b) 3, (c) 4, (d) 5, PCl3 reacts with water to form, [1991], (a) PH3, (b) H3PO3, HCl, (c) POCl3, (d) H3PO4, PH4I + NaOH forms, [1991], (a) PH3, (b) NH3, (c) P4O6, (d) P4O10, , 29., , 30., , 31., , 32., , 33., , 34., , 35., , 36., , 37., , 38., , Pure nitrogen is prepared in the laboratory by, heating a mixture of, [1991], (a) NH4OH + NaCl, (b) NH4 NO3 + NaCl, (c) NH4 Cl + NaOH, (d) NH4 Cl + NaNO2., Which of the following statements is not correct, for nitrogen ?, [1990], (a) Its electronegativity is very high, (b) d-orbitals are available for bonding, (c) It is a typical non-metal, (d) Its molecular size is small, Of the following hybrides which one has the, lowest boiling point ?, [1989], (a) AsH 3, (b) SbH 3, (c) PH 3, (d) NH 3, Which of the following metal evolves hydrogen, on reacting with cold dilute HNO3 ?, [1989], (a) Mg, (b) Al, (c) Fe, (d) Cu., Which one of the following compounds does, not exist ?, [1989], (a) NCl5, (b) AsF5, (c) SbCl5, (d) PF5, Each of the following is true about white and, red phosphorus except that they, [1989], (a) Are both soluble in CS2, (b) Can be oxidised by heating in air, (c) Consist of the same kind of atoms, (d) Can be converted into one another, When orthophosphoric acid is heated to 600°C,, the product formed is, [1989], (a) PH3, (b) P2O5, (c) H3PO3, (d) HPO3, Which of the following is a nitric acid anhydride?, [1988], (a) NO, (b) NO2, (c) N2O5, (d) N2O3., Topic 2: Oxygen Family, Which of the following oxoacid of sulphur has, – O – O – linkage?, [2020], (a) H2SO4, sulphuric acid, (b) H2S2O8, peroxodisulphuric acid, (c) H2S2O7, pyrosulphuric acid, (d) H2SO3, sulphurous acid, Identify the correct formula of ‘oleum’ from the, following., [NEET Odisha 2019], (a) H2S2O8, (c) H2S2O7, (c) H2SO3, (d) H2SO4
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199, , The p-Block Elements (Group 15, 16, 17 and 18), 39., , 40., , 41., , 42., , 43., , 44., , 45., , Which is the correct thermal stability order for, H2E (E = O, S, Se, Te and Po) ?, [2019], (a) H2S < H2O < H2Se < H2Te < H2Po, (b) H2O < H2S < H2Se < H2Te < H2Po, (c) H2Po < H2Te < H2Se < H2S < H2O, (d) H2Se < H2Te < H2Po < H2O < H2S, In which pair of ions both the species contain, S – S bond?, [2017], 2, 2, 2, (b) S2 O7 ,S2 O82 (a) S4 O6 ,S2 O3, (c) S4 O62- ,S2 O72 (d) S2 O72- ,S2 O32 Nitrogen dioxide and sulphur dioxide have some, properties in common. Which property is shown, by one of these compounds, but not by the, other?, [2015], (a) is a reducing agent, (b) is soluble in water, (c) is used as a food-preservative, (d) forms 'acid-rain', Which of the statements given below is, incorrect?, [2015 RS], (a) Cl2O7 is an anhydride of perchloric acid, (b) O3 molecule is bent, (c) ONF is isoelectronic with O2N–., (d) OF2 is an oxide of fluorine, Acidity of diprotic acids in aqueous solutions, increases in the order :, [2014], (a) H2S < H2Se < H2Te, (b) H2Se < H2S < H2Te, (c) H2Te < H2S < H2Se, (d) H2Se < H2Te < H2S, Which of the following does not give oxygen, on heating?, [NEET 2013], (a) Zn(ClO3)2, (b) K2Cr2O7, (c) (NH4)2Cr2O7, (d) KClO3, Sulphur trioxide can be obtained by which of, the following reaction :, [2012], , 47., , 48., , 49., , 50., , 51., , 52., , Δ, , (a), , CaSO 4 + C ¾¾, ®, , (b), , Fe 2 ( SO4 )3 ¾¾, ®, , (c), , S + H 2SO 4 ¾¾, ®, , Δ, , Δ, , 53., Δ, , 46., , (d) H 2SO 4 + PCI5 ¾¾, ®, Which one of the following compounds is a, peroxide ?, [2010], (a), , KO 2, , (b) BaO2, , (c), , MnO 2, , (d) NO 2, , 54., , Match List - I (substances) with List - II (processes), employed in the manufacture of the substances, and select the correct option., [2010], List - I, List - II, Substances, Processes, 1. Sulphuric acid, (i) Haber’s process, 2. Steel, (ii) Bessemer’s, process, 3. Sodium hydroxide (iii) Leblanc process, 4. Ammonia, (iv) Contact process, Code :, 1, 2, 3, 4, (a) (iv) (ii), (iii) (i), (b) (i), (iv) (ii), (iii), (c) (i), (ii), (iii) (iv), (d) (iv) (iii) (ii), (i), Which of the following is the most basic oxide?, [2006], (a) Sb2O3, (b) Bi2O3, (c) SeO2, (d) Al2O3, During its reactions, ozone, [1999], (a) can only combine with hydrogen atoms, (b) accepts electrons, (c) loses electrons, (d) shows the role of electrons to be irrelevant, Which of the following oxides will be the least, acidic?, [1996], (a) As4O 6, (b) As 4O10, (c) P4O10, (d) P4O6, Oxidation of thiosulphate by iodine gives, [1996], (a) tetrathionate ion (b) sulphide ion, (c) sulphate ion, (d) sulphite ion, About 20 km above the earth, there is an ozone, layer. Which one of the following statements, about ozone and ozone layer is true? [1995], (a) ozone has a triatomic linear molecule, (b) it is harmful as it stops useful radiation, (c) it is beneficial to us as it stops U.V radiation, (d) conversion of O3 to O2 is an endothermic, reaction, By passing H 2S gas in acidified KMnO 4, solution, we get, [1995], (a) S, (b) K2S, (c) MnO2, (d) K2SO3, Polyanion formation is maximum in, [1994], (a) Nitrogen, (b) Oxygen, (c) Sulphur, (d) Boron
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EBD_8336, 200, , 55., , 56., , 57., , 58., , 59., , 60., , 61., , 62., , CHEMISTRY, The acid which has a peroxy linkage is [1994], (a) Sulphurous acid (b) Pyrosulphuric acid, (c) Dithionic acid, (d) Caro’s acid, Which would quickly absorb oxygen?, [1991, 92], (a) Alkaline solution of pyrogallol, (b) Conc. H2SO4, (c) Lime water, (d) Alkaline solution of CuSO4, Oleum is, [1991], (a) Castor Oil, (b) Oil of vitriol, (c) Fuming H2SO4, (d) None of them, Oxygen will directly react with each of the, following elements except, [1989], (a) P, (b) Cl, (c) Na, (d) S., The gases respectively absorbed by alkaline, pyrogallol and oil of cinnamon are, [1989], (a) O3, CH4, (b) O2, O3, (c) SO2, CH4, (d) N2O, O3., It is possible to obtain oxygen from air by, fractional distillation because, [1989], (a) oxygen is in a different group of the periodic, table from nitrogen, (b) oxygen is more reactive than nitrogen, (c) oxygen has higher b.p. than nitrogen, (d) oxygen has a lower density than nitrogen., Hypo is used in photography to, [1988], (a) reduce AgBr grains to metallic silver, (b) convert metallic silver to silver salt, (c) remove undecomposed silver bromide as a, soluble complex, (d) remove reduced silver, Topic 3: Halogen Family, Match the following:, [2019], (a) Pure nitrogen, (i) Chlorine, (b) Haber process, (ii) Sulphuric acid, (c) Contact proces, (iii) Ammonia, (d) Deacon’s process (iv) Sodium azide or, Barium azide, Which of the following is the correct option ?, (a), (b), (c), (d), (a) (i), (ii), (iii), (iv), (b) (ii), (iv) (i), (iii), (c) (iii), (iv) (ii), (i), (d) (iv) (iii), (ii), (i), , 63., , 64., , 65., , 66, , 67., , Identify the incorrect statement related to PCI5, from the following:, [2019], (a) Three equatorial P – Cl bonds make an, angle of 120° with each other, (b) Two axial P – Cl bonds make an angle of, 180° with each other, (c) Axial P – Cl bonds are longer than, equatorial P – Cl bonds, (d) PC15 molecule is non-reactive, Which of the following statements is not true, for halogens?, [2018], (a) All form monobasic oxyacids, (b) All are oxidizing agents, (c) Chlorine has the highest electron-gain, enthalpy, (d) All but fluorine shows positive oxidation, states, Match the interhalogen compounds of column-I, with the geometry in column II and assign the, correct code., [2017], Column-I, Column-II, 1. XX', (i) T-shape, 2. XX'3, (ii) Pentagonal bipyramidal, 3. XX'5, (iii) Linear, 4. XX'7, (iv) Square-pyramidal, (v) Tetrahedral, Code :, 1, 2, 3, 4, (a) (iii) (i), (iv) (ii), (b) (v), (iv) (iii) (ii), (c) (iv) (iii) (ii), (i), (d) (iii) (iv) (i), (ii), Among the following, the correct order of acidity, is, [2016], (a) HClO3 < HClO4 < HClO2 < HClO, (b) HClO < HClO2 < HClO3 < HClO4, (c) HClO2 < HClO < HClO3 < HClO4, (d) HClO4 < HClO2 < HClO < HClO3, Which one of the following orders is correct for, the bond dissociation enthalpy of halogen, molecules?, [2016], (a) I2 > Br2 > Cl2 > F2 (b) Cl2 > Br2 > F2 > I2, (c) Br2 > I2 > F2 > Cl2 (d) F2 > Cl2 > Br2 > I2
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EBD_8336, 202, , 81., , 82., , 83., , 84., , 85., , 86., , 87., , 88., , 89., , CHEMISTRY, F–, , Cl–, , Regarding, and, which of the following, statements is/are correct?, [1996], (i) Cl– can give up an electron more easily than, F–, (ii) Cl– is a better reducing agent than F–, (iii) Cl– is smaller in size than F–, (iv) F– can be oxidized more readily than Cl–, (a) (i) and (ii), (b) (i), (ii) and (iv), (c) (iii) and (iv), (d) only (i), A certain compound (X) when treated with, copper sulphate solution yields a brown, precipitate. On adding hypo solution, the, precipitate turns white. The compound is [1994], (a) K2CO3, (b) KI, (c) KBr, (d) K3PO4, HI can be prepared by all the following methods,, except, [1994], (a) PI 3 + H 2 O, (b) KI + H 2SO 4, Pt, (c) H 2 + I 2 ¾¾®, (d) I 2 + H 2S, Which among the following is paramagnetic?, [1994], (a) Cl 2 O, (b) ClO 2, (c) Cl 2O 7, (d) Cl 2O 6, Which one of the following oxides of chlorine is, obtained by passing dry chlorine over silver, chlorate at 90°C ?, [1994], (a) Cl2O, (b) ClO3, (c) ClO2, (d) ClO4, The formula for calcium chlorite is, [1994], (a) Ca(ClO 4 )2, (b) Ca(ClO3 )2, (c) CaClO 2, (d) Ca(ClO 2 ) 2, A solution of potassium bromide is treated with, each of the following. Which one would liberate, bromine ?, [1993], (a) Hydrogen iodide (b) Sulphur dioxide, (c) chlorine, (d) Iodine, When chlorine is passed over dry slaked lime at, room temperature, the main reaction product is, [1992], (a) Ca (ClO2 ) 2, (b) CaCl 2, (c) CaOCl 2, (d) Ca (OCl) 2, In the manufacture of bromine from sea water,, the mother liquor containing bromides is treated, with, [1992], (a) Carbon dioxide, (b) Chlorine, (c) Iodine, (d) Sulphur dioxide, , 90., 91., , 92., , 93., , 94., , 95., , 96., , The bleaching action of chlorine is due to [1991], (a) Reduction, (b) Hydrogenation, (c) Chlorination, (d) Oxidation, Bleaching powder reacts with a few drops of, dilute HCl to give, [1989], (a) chlorine, (b) hypochlorous acid, (c) calcium oxide, (d) oxygen, Bleaching powder is obtained by the action of, chlorine gas and, [1988], (a) dilute solution of Ca(OH)2, (b) concentrated solution of Ca(OH)2, (c) dry CaO, (d) dry slaked lime, Topic 4: Noble Gases, Match the compounds given in column I with, the hybridisation and shape given in column II, and mark the correct option., [2016], Column-I, Column-II, 1. XeF6, (i) Distorted octahedral, 2. XeO3, (ii) Square planar, 3. XeOF4, (iii) Pyramidal, 4. XeF4, (iv) Square pyramidal, Code :, 1, 2, 3, 4, (a) (i), (iii) (iv) (ii), (b) (i), (ii), (iv) (iii), (c) (iv) (iii) (i), (ii), (d) (iv) (i), (ii), (iii), Identify the incorrect statement, regarding the, molecule XeO4:, [NEET Kar. 2013], (a) XeO4 molecule is tetrahedral, (b) XeO4 molecule is square planar, (c) There are four pp – dp bonds, (d) There are four sp3 – p, s bonds, Noble gases do not react with other elements, because, [1994], (a) They are mono atomic, (b) They are found in abundance, (c) The size of their atoms is very small, (d) They are completely paired up and stable, electron shells, Which of the following statements is false ?, [1994], (a) Radon is obtained from the decay of radium, (b) Helium is inert gas, (c) Xenon is the most reactive among the rare, gases, (d) The most abundant rare gas found in the
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203, , The p-Block Elements (Group 15, 16, 17 and 18), ANS WER KEY, 1, , (a), , 11, , (b ), , 2, , (b), , 12, , (c), , 3, , (d ), , 13, , (b), , 4, , (a), , 14, , (c), , 5, , (b ), , 15, , (a), , 6, , (a), , 16, , (a), , 7, , (a), , 17, , (b), , 8, , (c), , 18, , (c), , 9, , (b ), , 19, , (a), , 10, , (d ), , 20, , (a), , 21 (b ), 22 (d ), 23 (c), 24 (d ), 25 (d ), , 31, , (c), , 41, , (c), , 51, , (a), , 61, , (c), , 71, , (d), , 81, , (d), , 91, , (a), , 32, , (a), , 42, , (d), , 52, , (c), , 62, , (d ), , 72, , (c), , 82, , (b), , 92, , (d ), , 33, , (a), , 43, , (a), , 53, , (a), , 63, , (d ), , 34, , (a), , 44, , (c), , 54, , (c), , 64, , (d ), , 26 (b ), 27 (b ), 28 (a), , 36, , (c), , 46, , (b), , 56, , (a), , 66, , (b ), , 73 (d) 83, 74 (b,c) 84, 75 (d) 85, 76 (c) 86, , 35, , (d ), , 45, , (b), , 55, , (d), , 65, , (a), , (b), , 93, , (a), , (b), , 94, , (b ), , (c), , 95, , (d ), , 37, , (b ), , 47, , (a), , 57, , (c), , 67, , (b ), , 77, , (b), , 87, , (c), , (d), , 96, , (d ), , 38, , (b ), , 48, , (b), , 58, , (b), , 68, , (b ), , 78, , (a), , 88, , (d), , 29 (d ), 30 (b ), , 39, , (c), , 49, , (a), , 59, , (b), , 69, , (b ), , 79, , (a), , 89, , (b), , 40, , (a), , 50, , (a), , 60, , (c), , 70, , (b ), , 80, , (d), , 90, , (d), , Hints & Solutions, 1., (a), NH2CONH2 + H2O, , 5., , (NH4)2CO3, (A), , D, NH3 (g) + CO3 (g) + H3O (l), , H, , (B), Cu 2 + (aq), , 2., , (C), [Blue coloured, solution], , (b), (X), , (Y), ( rotten fish smell ), , 6., 7., , 2PH3 + 3CuSO 4 ¾¾, ® Cu 3 P2 + 3H 2SO4, (Y), , 3., , 4., , (a), , 0 -3, , HNO3 , NO, N2 , NH4Cl, , Nitrogen in its elemental form has zero oxidation, state., , P, , 4 AgNO3 + 2H2O + H3PO2 —®, , 4AgNO3 + 2H2O + H3PO2 ¾¾, ®, 4Ag + 4HNO3 + H3PO4, +2, , (Monoprotic), , H3PO2 acid is good reducing agent as it contains, two P–H bonds and reduces, for example, AgNO3, to metallic silver., , (d) Hypophosphorous acid is a good reducing, agent as it contains two P–H bonds., , +5, , OH, , (Diprotic acid), OH, OH, (a) CaC2 + N2 ® CaCN2 + C, (a) The acids which contain P-H bond have, strong reducing properties., H, , Ca 3 P2 + 6H 2O ¾¾, ® 3Ca(OH)2 + 2PH 3, , P, , H, Phosphonic acid as shown in structure has two, P–OH bonds, thus, it is dibasic or diprotic, O, , 2+, , NH3 (g) ¾¾¾¾® [Cu(NH 3 ) 4 ], (B), , (b) Phosphinic acid as shown in structure below, has one P—OH bond, thus, it is monobasic or, monoprotic, O, , 4Ag + 4HNO3 + H3PO4, , 8., , (c) Compound, N2H4, NH3, N3H, NH2OH, , =, =, =, =, , Oxidation number of, nitrogen, –2, –3, –1/3, –1
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EBD_8336, 204, , CHEMISTRY, 13., , (b), , O, 9., , (b), , H, , OH, , P, , H Hypophosphorous acid, , (a), , Þ O, , H4P2O7, , P, , OH, O, , OH, , O, , H, , P, , O, , OH, , Pyrophosphoric acid, , O, , (H3PO2) is a monobasic acid. i.e., it has only, one ionisable hydrogen atom or one OH is, O, , (b), , Þ OH, , H3PO2, , P, , H, , H, , Hypophosphorous acid, , present H, , P, , OH, , O, , (c), , H, , Þ HO – P – OH, , H3PO3, , H, , 10., , +3, , Phosphorous acid, , +3, , O, , (d) HO - P - O - P - OH = H 4 P2O5, |, , (d), , |, , OH, , OH, , H3PO4, , Þ HO – P – OH, , orthophosphoric acid, , O, , 14., , O, , ||, , ||, , HO - P - P - OH = H 4 P2 O6, +4 |, , |, , +4, , OH OH, O, , 15., O, , ||, , ||, , HO - P - O - P - OH = H 4 P2 O7, +5 |, , +5, , |, , OH, , OH, , 11., , 16., , O, , (b), bridging, , bridging, , P, , 17., , O, , NaNO3 (aq) + H 2SO4 (aq), , P, O, bridging, , P, O, , P, O, , 12., , (c) Ammonium sulphate is a salt of weak base, and strong acid, so it produces acidity. Hence, aqueous solution of ammonium sulphate, increases the acidity of soil., (a) Order of dipole moment decreases as, NH3 > PH3 > AsH3 > SbH3, (Based upon electronegativity), (a) We know that empirical formula of, hypophosphorus acid is H3PO2. In this only, one ionisable hydrogen atom is present i.e. it is, monobasic. Therefore option (a) is correct, structural formula of it., (b) Brown ring test is done for the confirmation, of NO 3- ions., , O, O, , OH, , O, , ¾¾, ® NaHSO4 (aq) + HNO3 (aq), 6FeSO4 + 2HNO3 +3H 2SO4, , O, O, , ¾¾, ® 3Fe 2 (SO 4 )3 +2NO+4H 2O, , bridging, oxygen, , i.e. 6-bridging oxygen., (c) Nitrogen form N 2 (i. e. N º N) but, phosphorus form P4, because in P2, pp — pp, bonding is present which is a weaker bonding., , ®, FeSO4 + NO ¾¾, , 18., , [Fe(NO)]SO4, Ferrous nitroso-sulphate, (Brown ring), , (c) Urea (46.6%N). % of N in other compound, are : ( NH 4 ) 2 SO 4 = 21.2%;, CaCN 2 = 35.0% and NH 4 NO3 = 35.0%
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205, , The p-Block Elements (Group 15, 16, 17 and 18), , :, , 20., , (a) Phosphorus pentoxide has great affinity for, water. It forms ortho phosphoric acid on, absorbing water, i.e. P4 O10 + 6H 2 O ¾¾, ® 4H 3 PO 4, It is thus used as a powerful dehydrating or, drying agent., (a) Nitrogen molecule is diatomic containing a, triple bond between two N atoms, N º N ., Therefore, nitrogen molecule is formed by, sharing six electrons., (b) Cane sugar is oxidised to oxalic acid, , 21., , (2HNO 3 ¾¾, ® H 2 O + 2NO 2 + O)18, C12 H 22 O11 + 18[O] ¾, ¾® 6 (COOH) 2, Cane sugar, , From HNO3, , 27., , (b) PCl3 + 3H 2 O ® H 3PO 3 + 3HCl, , 28., 29., , (a) PH 4 I + NaOH ® NaI + PH 3 + H 2 O, (d) Pure nitrogen in the lab can be obtained by, heating ammonium nitrate. Ammonium nitrate is, not a stable compound it dissociate to give, nitrogen., Heat, , :, , 19., , NH 4 Cl + NaNO 2 ¾¾¾® NH 4 NO 2, Heat, , ¾¾¾® N 2 + 2 H 2 O., , 30., 31., , Oxalic acid, , + 5H 2O., , 22., , 23., , 24., , ®, C12H22O11 + 36HNO3 ¾¾, 6(COOH)2 + 36NO2 + 23H2O, (d) N2 molecule contains triple bond between, N atoms having very high dissociation energy, (946 kJ mol–1) due to which it is relatively, inactive., (c) H3PO2 is named as hypophosphorous acid., As it contains only one P – OH group, its basicity, is one., (d) Aqueous solution of ammonia is obtained, by passing NH3 in H2O which gives NH4+ and, OH– ions., , NH 4+, , NH 3 + H 2 O, , 25., , 2H2O, , 26., , 34., 35., , 2H 2O, , 36., , 37., , +, , 3–, , +, , 2HPO3, metaphosphoric, acid, , O, , (b) HO – S – O – O – S – OH, , O, , Peroxosulphuric acid is also known as Marshall's, acid and is one of the most powerful peroxy acid, oxidant., , 38., 39., , –, , H + H2PO4, , 3H + PO4, , -2H 2O, , (c) N 2 O 5 + H 2 O ¾¾® 2 HNO 3, , O, , 2H4P2O7, , +, , 600°C, , ¾¾¾¾, ®, , Peroxodisulphuric acid, , (b) It is a tribasic acid as all the three hydrogen, atoms are ionisable. It forms three series of salts., H3PO4, , 2H 3 PO 4, , O, , Pyrophosphoric, acid, , orthophosphoric, acid, , (d), , orthophosphoric, acid, , 4HPO3, , Metaphosphoric, acid, 2H2O, , 4H3PO4, , 33., , -, , + OH, (d) P2O5 have great affinity for water. The final, product is orthophosphoric acid., , P4O10, , 32., , (b) In case of nitrogen, d-orbitals are not, available., (c) NH3 undergoes H-bonding and hence has, the highest b.p. Among the remaining hydrides, i.e. PH3, AsH3 and SbH3 the b.p. increases as, the size of the element increases and hence the, magnitude of the van der Waal’s forces of, attraction increases. Thus, PH3 has the lowest, b.p., (a) Magnesium and manganese are the metals, that produce hydrogen with dilute nitric acid, ¾® Mg (NO3)2 + H2, Mg + 2HNO3 ¾, (a) As Nitrogen does not have d-orbital in its, valence shell, it cannot form NCl5., (a) Red phosphorus is not soluble in CS2, only, white P is soluble., , 2–, , 2H + HPO4, , (b) Oleum is H2S2O7., (c) On going down the group, bond, dissociation enthalpy of the hydrides of oxygen, family decreases. Therefore, thermal stability, also decreases.
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EBD_8336, 206, , CHEMISTRY, , 40., , 41., , 42., , 43., , O, , O, , (a) O – S – S – S – S – O, O, O, , S, S, , O, , O, , S4 O6 2 S2 O32 (c) SO2 is widely used in food and drinks, industries for its property as a preservative and, antioxidant while NO2 is not used as food, preservative., (d) OF2; among the following O and F, F is more, electronegative than oxygen., So OF2 cannot be called oxide because in that, case fluorine is in +1 oxidation state which is, not possible, so OF2 is called oxygen difluoride., (a) The weakning of M—H bond with increase, in size of M (where M = S, Se, Te) explains the, acid character of hydrides., On moving down the group, atomic size increases, hence, bond length increases and hence, removal, tendency of H also increases., , 44., , D, , (c) (NH4)2Cr2O7 ¾¾® N2 + Cr 2O3 + 4H2O, , 49., 50., , +5, , 52., , (c) Ozone layer is beneficial to us, because it, stops harmful ultraviolet radiations to reach the, earth., , 53., , (a) 2 KMnO 4 + 5H 2 S + 3H 2 SO 4 ¾¾®, , 54., , K 2 SO 4 + 2MnSO 4 + 5S + 8H 2 O., Thus, in this reaction sulphur (S) is produced., (c) Due to greater tendency for catenation,, sulphur shows property of polyanion formation, to a greater extent. For example, in polysulphides, , Tetrathion ate, , D, , such as S32 - , S24 - , S52 55., , Δ, , (b) Fe 2 (SO 4 )3 ¾¾, ® Fe2 O3 + SO3, , (d) Caro’s acid is H 2SO 5 which contains one, S – O – O – H peroxy linkage. It is also known as, permonosulphuric acid., O, ||, H – O – O – S – OH, ||, O, , O, , 46., , (b) Ba, O, , 47., , 48., , (a), (1) Sulphuric acid, (2) Steel, , (iv) Contact process, (ii) Bessemer’s, process, (3) Sodium hydroxide (iii) Leblanc process, (4) Ammonia, (i) Haber’s process, (b) More the oxidation state of the central atom, (metal), more is its acidity. Hence, SeO2 (O. S. of, Se = +4) is acidic. Further for a given O.S., the, basic character of the oxides increases with the, , +3, , (a) 2 S2 O 3-2 + I 2 ® S 4O 6-2 + 2I -, , 2 KClO3 ¾¾® 2KCl + 3O2, , 45., , +5, , 51., , D, , D, , +3, , P4O10 > P4O6 > As 4O10 > As 4O6, , Zn(ClO3)2 ¾¾® ZnCl2 + 3O2, , 4K2Cr2O7 ¾¾® 4K2CrO4 + 2Cr2O3 + 3O2, , increasing size of the central atom. Thus Al2O3, and Sb2O3 are amphoteric and Bi2O3 is basic., (a) Since ozone can easily lose oxygen atom, (nascent oxygen), it acts as a powerful oxidising, agent, and hence, reacts with hydrogen atoms., (a) As the O.N of the central atom of the, compounds increases, acidic strength of that, compound also increases and on moving from, top to bottom in groups acidic strength of oxides, also decrease due to decreasing, electronegativity in groups., , Caro's acid, , 56., , (a) Upon oxidation, pyrogallol forms hydroxyquinone and many other higher molecular mass, products., OH, HO, , OH, , HO, O (air), , O, ||, , O, , ||, , O, –, , 2, ¾¾¾®, , pyrogallol, , 1-hydroxy,, 2-benzoquinone
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207, , The p-Block Elements (Group 15, 16, 17 and 18), 66., , (b) Acidic strength increases as the oxidation, number of central atom increases., HClO < HClO2 < HClO3 < HClO4, , 67., , (b) Bond dissociation enthalpy decreases as, the bond distance increases from F2 to I2. This, is due to increase in the size of the atom, on, moving from F to I., F – F bond dissociation enthalpy is smaller then, Cl – Cl and even smaller than Br – Br. This is, because F atom is very small and hence the three, lone pairs of electrons on each F atom repel the, bond pair holding the F-atoms in F2 molecules., The increasing order of bond dissociation, enthalphy is, I2 < F2 < Br2 < Cl2, (b) The H-bonding is present in HF due to high, electronegativity of fluorine atom. While Hbonding is not present in HI, HBr and HCl., (b) HClO4 is the strongest acid amongst all, because the oxidation state of Cl is maximum, (+7)., (b) If acidic nature is high, Ka is high and pKa, is low, H2O, H2S, Ka, 1.8 × 10– 6, 1.3 × 10–7, H2Se, H2Te, Ka, 1.3 × 10–4, 2.3 × 10–3, since pKa = – log Ka, Hence the order of pKa will be, H2O > H2S > H2Se > H2Te, (d) (i) As the number of lone pair of electrons, increases, bond angle decreases due to repulsion, between lp – lp., (ii) As the electronegativity of the central atom, increases, bond angle increasess. Hence, the, correct order of bond angle:, ., :O:, : Cl :, Cl :, <, <, Cl, Cl, O, O, O, O, , Hydroxy groups are strong activators of aromatic, systems, hence pyrogallol is reactive enough to, react with oxygen in air., , 57., , (c) Oleum is H 2S2 O 7 ( H 2SO 4 + SO 3 ) which, , 58., , is obtained by dissolving SO 3 in H2SO4 and is, called fuming sulphuric acid., (b) S + O 2 ¾¾, ® SO 2 (burns with blue light), 4Na + O 2 ¾¾, ® 2NaO, , (burns with yellow light), P4 + 3O 2 ¾¾, ® P4 O 6, , P4 + 5O 2 ¾¾, ® P4 O10, , 59., 60., , Cl + O 2 ¾¾, ® No reaction, Chlorine does not react directly with oxygen., (b) Alkaline pyrogallol absorbs O2 and oil of, cinnamon absorbs O3., (c) Oxygen has higher b.pt. than nitrogen, therefore it can be obtained from air by fractional, distillation., Air is liquified by making use of the joule-Thomson, effect (cooling by expansion of the gas). Water, vapour and CO 2 are removed by solidification., The remaining constituents of liquid air i.e., liquid, oxygen and liquid nitrogen are separated by means, of fractional distillation., , 61., , (c) Undecomposed AgBr forms a soluble, complex with hypo, AgBr + 2 Na 2S 2 O3 ® Na 3[ Ag (S 2 O 3 ) 2 ] + NaBr, , 62., , (d), , 63., , (d) PCl5 is very reactive due to the presence of, weak axial bonds. It is used in the synthesis of, various organic compounds., (d) Due to high electronegativity and small, size, F forms only one oxoacid, HOF known as, fluoric (I) acid., , soluble complex, , 64., , Flourine has –1 oxidation state in most of its, compound. Oxidation number of F is +1 in HOF., , 65., , (a) XX' ® Linear (e.g. ClF, BrF), XX3' ® T-Shape (e.g. ClF3, BrF3), XX5' ® Square pyramidal (e.g. BrF5 IF5), XX7' ® Pentagonal bipyramidal (e.g. IF7), , 68., 69., 70., , 71., , +1, , two lone pairs, , 72., 73., , +3, , two lone pairs, , +5, , +7, , one lone pair, and one electron, , (c) MI > MBr > MCl > MF. As the size of the, anion decreases covalency decreases., (d) Standard reduction potential of halogens, are positive and decreases from fluorine to, iodine. Therefore, halogens act as strong, oxidising agent and their oxidising power, decreases from fluorine to iodine.
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EBD_8336, 208, , 74., , 75., , CHEMISTRY, (b and c), (a) The oxidising power of halogen follow the, order F2 > Cl2 > Br2 > I2., (b) The correct order of electron gain enthalpy, of halogens is Cl2 > F2 > Br2 > I2. The low value, of F2 than Cl2 is due to its small size., (c) The correct order of bond dissociation, energies of halogens is, Cl2 > Br2 > F2 > I2., (d) It is the correct order of electronegativity, values of halogens., F2 > Cl2 > Br2 > I2, , Br–, , & ion act as reducing agents and their, reducing nature is in increasing order, , 82., , +3, , +5, , Cl – Br – I –, ¾¾¾¾¾¾¾¾¾¾¾®, Reducing nature increases, (b) KI reacts with CuSO4 solution to produce, cuprous iodide (white precipitate) and I2 (which, gives brown colour). Iodine reacts with hypo, (Na2S2O3.5H2O) solution. Decolourisation of, solution shows the appearance of white, precipitate., , 2CuSO4 + 4KI ® 2K 2SO4 +, , (d) HO Cl < HO Cl O < HO Cl O 2 < HO Cl O3, +1, , I–, , +7, , 2Na 2S2 O3 + I2 ¾¾, ® Na 2S4 O6 + 2NaI, Sod. tetra, thionate, (colourless), , As the oxidation number of the central atom, increases, strength of acid also increases., , 76., , (c) Bond dissociation energy of fluorine is less, because of its small size and repulsion between, electrons of two atoms. So option (c) is wrong, order. The correct order is, [Cl2 > Br2 > F2 > I2], 0, , 77., , 78., , 79., , 80., , 81., , +1, , 83., , of the size is I - > I > I +, (d) The halide ions act as reducing agents . F–, ion does not show any reducing nature but Cl–,, , (b) HI cannot be prepared by heating iodides, with concentrated H2SO4. The reaction between, KI & H2SO4 is as follows:, 8KI + 5H 2 SO 4 ¾¾, ® 4K 2 SO 4 + 4I 2 + 4H 2 O + H 2 S, , 84., , -1, , ® HOBr+ HBr, (b) H 2O + B r2 ¾¾, Thus, here oxidation number of Br increases from, 0 to +1 and also decreases from 0 to –1. Thus, it, is oxidised as well as reduced., (a) F is more electronegative than Cl therefore, HF bond is stronger than HCl and hence proton, is not given off easily. Hence HF is the weakest, hydrohalic acid., (a) Brown bromine vapour will not react with, (a) Pieces of marble (CaCO3)., (b) Animal charcoal powder can adsorb it., (c) In carbon tetrachloride solution it will, dissolve., (d) In carbon disulphide solution, bromine will, dissolve and thus, intensity of brown colour, will decrease., (d) We know that positive ion is always smaller, and negative ion is always larger than the, corresponding atom. Therefore the correct order, , 2CuI +, I2, Cuprous iodide (Brown colour, (White ppt.), in solution), , 85., , (b) ClO2 contains 7 + 12 i.e. 19 electrons, (valence) which is an odd number, i.e. there is, (are) free electron(s). Hence it is paramagnetic, in nature., (c) Pure ClO2 is obtained by passing dry Cl2, over AgClO3 at 90°C., 90°C, , 2AgClO 3 + Cl 2 (dry ) ¾¾¾®, , 2AgCl + 2ClO 2 + O 2, 86., , (d) Calcium chlorite is Ca (ClO2 ) 2, , 87., , (c) A stronger oxidising agent (Cl 2 ) displaces, a weaker oxidising agent (Br2 ) from its salt, solution., 2KBr + Cl 2 ® 2KCl + Br2, (d), , 88., 89., , 2 Ca(OH) 2 + 2Cl 2 ® Ca(OCl) 2 + 2H 2O + CaCl2, (b) Bromide in the mother liquor is oxidised to, Br2 by Cl 2 which is a stronger oxidising agent., 2Br - + Cl2 ® Br2 + 2Cl -, , (sea water), , 90., , (d) Bleaching action of chlorine is due to, oxidation in presence of moisture. It is permanent, H 2O+Cl2 ® 2HCl+[O], Colouring matter + [O ] ® colourless matter
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209, , The p-Block Elements (Group 15, 16, 17 and 18), 91., 92., , (a) CaOCl 2 + 2HCl ® CaCl 2 + H 2 O + Cl 2, Bleaching powder, , 94., , (d) Cl 2 gas reacts with dry slaked lime,, , (Tetrahedral sp 3 ), , Ca (OH) 2 at 40°C to give bleaching powder, , Xe : Ground state, [Kr] 4d 105s25p65d 0, Xe : Excited state, , Ca (OH) 2 + Cl 2 ® CaOCl 2 + H 2O, 93., , (a) XeF6, , XeO3, , F, , F, F, , Xe, , O, , distorted octahedral, , O, F, , Xe, , 95., , XeF 4, F, , F, F, , F, Xe, , F, , F, 2lp, , Square pyramidal, , Square planar, , 14243, , Form 4 pp-dp bond, with oxygen, , 4 sp3 hybridised orbitals form 4 sigma bond with, oxygen., , O, , Pyramidal, , XeOF 4, , F, , sp3 hybridised, orbital, , O, , F, , F, , 1442443, , Xe, , F, , (b), , 96., , (d) On account of highly stable ns 2 np 6, configuration in the valence shell. These, elements have no tendency either to lose, gain, or share electrons with atoms of other elements, i.e., their combining capacity or valency is zero., Further all the orbitals in the atoms of these, elements are doubly occupied i.e electrons are, not available for sharing., (d) The most abundant rare gas found in the, atmosphere is argon and not helium.
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EBD_8336, 210, , CHEMISTRY, , 22, , The d-and f-Block, Elements, , Trend Analysis with Important Topics & Sub-Topics, 2020, Topic Name, , Characteristics of dblock elements, , Compounds of, transition metals, , Lanthanoids and, actinoids, LOD - Level of Difficulty, , 1., , 2019, , 2018, , 2017, , 2016, , Sub-Topic, QNS. LOD QNS. LOD QNS. LOD QNS. LOD QNS. LOD, d-d transition and, 1, A, paramagnetism, catalytic, activities/oxidation, 1, A, states/reducing, properties, properties of, 1, E, KMnO4, properties of, 1, E, K2 Cr2 O7, reactions of copper, 1, E, characteristics of, 1, E, actinoids, electronic, 1, A, configuration, E - Easy, , Topic 1: Characteristics of d-Block Elements, Identify the incorrect statement., [2020], (a) The transition metals and their compounds, are known for their catalytic activity due to, their ability to adopt multiple oxidation, states and to form complexes., (b) Interstitial compounds are those that are, formed when small atoms like H, C or N are, trapped inside the crystal lattices of metals., (c) The oxidation states of chromium in CrO24 and Cr2 O72 - are not the same., (d) Cr2+ (d4) is a stronger reducing agent than, Fe2+ (d6) in water., , A - Average, , 2., , D - Difficult, , Qns - No. of Questions, , Match the catalyst with the process, [NEET Odisha 2019], Catalyst, Process, (i) V2O5, (a) The oxidation of, ethyne to ethanal, (ii) TiCl4 + Al(CH3)3 (b) Polymerisation of, alkynes, (iii) PdCl2, (c) Oxidation of SO2, in the manufacture, of H2SO4, (iv) Nickel complexes (d) Polymerisation of, ethylene, Which of the following is the correct option?, (a) (i)-(c), (ii)-(a), (iii)-(d), (iv)-(b), (b) (i)-(c), (ii)-(d), (iii)-(a), (iv)-(b), (c) (i)-(a), (ii)-(b), (iii)-(c), (iv)-(d), (d) (i)-(a), (ii)-(c), (iii)-(b), (iv)-(d)
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211, , The d-and f-Block Elements, 3., , 4., , Which one of the following ions exhibits d-d, transition and paramagnetism as well? [2018], (a), , CrO2–, 4, , (b) Cr2 O72–, , (c), , MnO 2–, 4, , (d) MnO4–, , 10., , Magnetic moment 2.84 B.M. is given by :(At. nos, Ni = 28, Ti = 22, Cr = 24, Co = 27) [2015], (a) Ti3+, (b) Cr2+, (c) Co2+, , 5., , (d) Ni2+, , The number of d-electrons in Fe2+ (Z = 26) is, not equal to the number of electrons in which, one of the following?, [2015], (a) p-electrons in Cl (Z = 17), , 11., , (b) d-electrons in Fe (Z = 26), (c) p-electrons in Ne (Z = 10), 6., , (d) s-electrons in Mg (Z = 12), Which is the correct order of increasing energy, of the listed orbitals in the atom of titanium ?, [2015 RS], , 7., , 8., , (a) 3s 4s 3p 3d, , (b) 4s 3s 3p 3d, , (c) 3s 3p 3d 4s, , (d) 3s 3p 4s 3d, , 12., , (, , ), , Magnetic moment 2.83 BM is given by which of, the following ions ?, , °, one of them the standard potential EM2+ /M, , (At. nos. Ti = 22, Cr = 24, Mn = 25, Ni = 28):[2014], , value has a positive sign?, [2012 M], (a) Co (Z = 27), (b) Ni (Z = 28), (c) Cu (Z = 29), (d) Fe (Z = 26), The catalytic activity of transition metals and, their compounds is ascribed mainly to : [2012 M], (a) their magnetic behaviour, (b) their unfilled d-orbitals, (c) their ability to adopt variable oxidation state, (d) their chemical reactivity, 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?, [2011], (a) Mn > Fe > Cr > Co, (b) Fe > Mn > Co > Cr, (c) Co > Mn > Fe > Cr, (d) Cr > Mn > Co > Fe, , (a) Ti3+, , (b) Ni2+, , (c) Cr3+, , (d) Mn 2+, , Reason of lanthanoid contraction is:-, , 13., [2014], , (a) Negligible screening effect of ‘f ’ orbitals, (b) Increasing nuclear charge, (c) Decreasing nuclear charge, (d) Decreasing screening effect, 9., , Sc (Z = 21) is a transition element but Zn (Z = 30), is not because, [NEET Kar. 2013], (a) both Sc and Zn do not exhibit variable, oxidation states, (b) both Sc3+ and Zn2+ ions are colourless and, form white compounds, (c) in case of Sc, 3d orbitals are partially filled, but in Zn these are completely filled, (d) last electron is assumed to be added to 4s, level in case of Zn, Which one of the following does not correctly, represent the correct order of the property, indicated against it?, [2012 M], (a) Ti < V < Cr < Mn : increasing number of, oxidation states, (b) Ti3+ < V3+ < Cr3+ < Mn3+ : increasing, magnetic moment, (c) Ti < V < Cr < Mn : increasing melting, points, (d) Ti < V < Mn < Cr : increasing 2nd ionization, enthalpy, Four successive members of the first series of, the transition metals are listed below. For which, , Which of the following lanthanoid ions is, diamagnetic ?, (At nos. Ce = 58, Sm = 62, Eu = 63, Yb = 70), [NEET 2013], (a), , Sm2+, , Eu2+, , (b), , (c), , Yb2+, , (d) Ce2+, , 14.
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EBD_8336, 212, , 15., , CHEMISTRY, , (c) Lu3+ (Z = 71), 16., , (d) Sc3+ (Z = 21), , (c) Mn 3+, , (a), , Ni3+, , (c), , Fe3+, , (b), , Mn 3+, , (d), , Co3+, , Which of the following pairs has the same size?, (a), , (c) Zr4+, Hf 4+, 18., , 19., , 20., , (b), , 24., , Zr 4+, Ti4+, , (d) Zn 2+, Hf4+, , Which of the following oxidation states is the, most common among the lanthanoids? [2010], (a) 3, , (b) 4, , (c) 2, , (d) 5, , 25., , Which one of the elements with the following, outer orbital configurations may exhibit the, largest number of oxidation states?, [2009], (a) 3d 54s1, , (b) 3d 54s2, , (c) 3d 24s2, , (d) 3d 34s2, , Out of TiF62– , CoF63– , Cu 2 Cl 2 and NiCl2–, 4, , 22., , In which of the following pairs are both the ions, coloured in aqueous solutions ?, [2006], (a) Sc3+, Ti3+, , (b) Sc3+, Co2+, , (c) Ni2+, Cu+, , (d) Ni2+, Ti3+, , The aqueous solution containing which one, of the following ions will be colourless?, (Atomic number: Sc = 21, Fe = 26, Ti = 22, Mn, = 25), [2005], (a) Sc3+, , (b) Fe2+, , (c) Ti 3+, , (d) Mn 2+, , Four successive members of the first row, transition elements are listed below with their, atomic numbers. Which one of them is, expected to have the highest third ionization, enthalpy?, [2005], (a) Vanadium (Z = 23), (b) Chromium (Z = 24), (c) Manganese (Z = 25), (d) Iron (Z = 26), , 26., , (Z of Ti = 22, Co = 27, Cu = 29, Ni = 28), the, colourless species are:, [2009], , 21., , (d) Cr3+, , (At. no. : Sc = 21, Ti = 22, Ni = 28, Cu = 29, Co = 27), , [2010], Fe2+, Ni2+, , (b) Ti3+, , (At.No. Ti = 22, V = 23, Cr = 24, Mn = 25), 23., , Which one of the following ions has electronic, configuration [Ar] 3d6 ?, [2010], , (At. Nos. Mn = 25, Fe = 26, Co = 27, Ni = 28), 17., , (a), , Which of the following ions will exhibit colour, in aqueous solutions?, [2010], 3+, 3+, (a) La (Z = 57), (b) Ti (Z = 22), , V3+, , Among the following series of transition metal, ions, the one where all metal ions have 3d2, electronic configuration is (At. nos. Ti = 22;, V = 23; Cr = 24; Mn = 25), [2004], , (a) Cu2Cl2 and NiCl2–, 4, , (a), , Ti 3+ , V 2+ , Cr 3+ , Mn 4+, , (b), , TiF62– ,and Cu2Cl2, , (b), , Ti + , V 4+ , Cr 6+ , Mn 7 +, , (c), , CoF63– ,and NiCl2–, 4, , (c), , Ti 4+ , V 3+ , Cr 2+ , Mn 3+, , (d), , 3–, TiF62– , and CoF6 ,, , (d), , Ti 2+ , V 3+ , Cr 4+ , Mn 5+, , The correct order of decreasing second, ionisation enthalpy of Ti (22), V(23), Cr(24) and, Mn (25) is :, [2008], (a) Cr > Mn > V > Ti, , (b) V > Mn > Cr > Ti, , (c) Mn > Cr > Ti > V, , (d) Ti > V > Cr > Mn, , Which one of the following ions is the most, stable in aqueous solution?, [2007], , 27., , The basic character of the transition metal, monoxides follows the order, (Atomic No.,Ti = 22, V = 23, Cr = 24, Fe = 26), [2003], (a) TiO > VO > CrO > FeO, (b) VO > CrO > TiO > FeO, (c) CrO > VO > FeO > TiO, (d) TiO > FeO > VO > CrO
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213, , The d-and f-Block Elements, 28., , Which one of the following characteristics of, the transition metals is associated with their, catalytic activity?, [2003], (a) Variable oxidation states, (b) High enthalpy of atomization, , 30., , 37., , Which of the following shows maximum number, of oxidation states?, [2002], (a) Cr, (b) Fe, , [2000], , (c) Iron (Z = 26), (d) Titanium (Z=22), 31., , Which of the following forms colourless, compound?, [2000], (a) Sc3+, (b) V3+, (c) Ti3+, , 32., , 41., , (d) CO2, , When copper is heated with conc. HNO3 it, produces, [2016], (a) Cu(NO3)2 and NO2, (b) Cu (NO3)2 and NO, (c) Cu(NO3)2, NO and NO2, (d) Cu(NO3)2 and N2O, , 42., , Which of the following processes does not, involve oxidation of iron ?, [2015], (a) Decolourization of blue CuSO4 solutution, by iron, (b) Formation of Fe(CO)5 from Fe, (c) Liberation of H2 from steam by iron at high, temperature, , 1s 2 , 2s 2 2 p6 , 3s 2 3 p 6 , 4s 2 3d 9, , (d) 1s 2 , 2s 2 2 p6 ,3s 2 3 p 6 , 4s 2 4 p 6 3d 3, , (c) P2O5, , Which one of the following statements is correct, when SO2 is passed through acidified K2Cr2O7, solution ?, [2016], , (d) Green Cr2(SO4)3 is formed, , (b) + 3, + 4, (d) + 2, + 3, + 4, , 1s 2 , 2s 2 2 p 6 ,3 p 2 3 p6 , 4s 2 4 p 6 ,5s 2 5 p1, , (b) NO2, , (c) SO2 is reduced, , (b) 1s 2 , 2s 2 2 p6 , 3s 2 3 p6 3d 10 , 4 s1, (c), , (a) SO2, , (b) The solution is decolourized, , The electronic configuration of Cu (atomic, number 29) is, [1991], (a), , Name the gas that can readily decolourise, acidified KMnO4 solution :, [2017], , (a) The solution turns blue, , (d) Cr3+, , The common oxidation states of Ti are [1994], (a) + 2, + 3, (c) – 3, – 4, , 35., , 40., , (c) Ge, (d) Am, Which one of the following ionic species will, impart colour to an aqueous solution? [1998], (a) Ti4+, (b) Cu+, (c) Zn2+, , 34., , 39., , (d) Cr3+, , (b) La, , 5, , (c) [ Ne]3s 3 p, (d) [ Ar ]3d10 4s 2, Which one of the following is an ore of silver ?, [1988], (a) Argentite, (b) Stibnite, (c) Haematite, (d) Bauxite, Which of the following metals corrodes readily, in moist air ?, [1988], (a) Gold, (b) Silver, (c) Nickel, (d) Iron, Topic 2: Compounds of Transition Metals, , Which one of the following elements shows, maximum number of different oxidation states in, its compounds?, [1998], (a) Eu, , 33., , 38., , (c) Mn, (d) V, Of the following transition metals, the maximum, numbers of oxidation states are exhibited by:, (a) Chromiun (Z = 24), (b) Manganese (Z = 25), , The electronic configurations of four elements, are given below. Which element does not belong, to the same family as others ?, [1989], (a) [ Xe]4 f 14 5d 10 6s 2 (b) [ Kr ]4d10 5s 2, 2, , (c) Parmagnetic behaviour, (d) Colour of hydrated ions, 29., , 36., , (d) Rusting of iron sheets, 43., , Assuming complete ionization, same moles of, which of the following compounds will require
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EBD_8336, 214, , CHEMISTRY, the least amount of acidified KMnO 4 for, complete oxidation, [2015 RS], , 44., , 45., , 46., , 47., , (a) FeSO4, , (b) FeSO3, , (c) FeC2O4, , (d) Fe(NO2)2, , 51., , 52., , The pair of compounds that can exist together is:, [2014], (a) FeCl3, SnCl2, (b) HgCl2, SnCl2, (c) FeCl2, SnCl2, (d) FeCl3, KI, In acidic medium, H2O2 changes Cr2O7–2 to, CrO5 which has two (–O–O) bonds. Oxidation, state of Cr in CrO5 is:[2014], (a) + 5, (b) + 3, (c) + 6, (d) – 10, The reaction of aqueous KMnO4 with H2O2 in, acidic conditions gives:, [2014], (a) Mn4+ and O2, (b) Mn2+ and O2, (c) Mn2+ and O3, (d) Mn4+ and MnO2, KMnO4 can be prepared from K2MnO4 as per, the reaction:, , 53., , 54., , 55., , , 3MnO 24 - + 2H 2O , , 48., , 49., , 50., , 2MnO -4 + MnO2 + 4OH The reaction can go to completion by removing, OH– ions by adding., [NEET 2013], (a) KOH, (b) CO2, (c) SO2, (d) HCl, Which of the statements is not true? [2012], (a) On passing H2S through acidified K2Cr2O7, solution, a milky colour is observed., (b) Na2Cr2O7 is preferred over K2Cr2O7 in, volumetric analysis., (c) K2Cr2O7 solution in acidic medium is orange., (d) K2Cr 2 O7 solution becomes yellow on, increasing the pH beyond 7., Acidified K2Cr2O7 solution turns green when, Na 2 SO 3 is added to it. This is due to the, formation of :, [2011], (a) Cr2(SO4)3, (b) CrO42–, (c) Cr2(SO3)3, (d) CrSO4, Copper sulphate dissolves in excess of KCN to, give, [2006], (a) [Cu(CN)4]3–, (b) [Cu(CN)4]2–, (c) Cu(CN)2, (d) CuCN, , 56., , 57., , 58., 59., , 60., , German silver is an alloy of, [2000], (a) Fe, Cr, Ni, (b) Cu, Zn, Ag, (c) Cu, Zn, Ni, (d) Cu, Sn, Al, Which of the following combines with Fe (II), ions to form a brown complex?, [2000], (a) NO, (b) N2O, (c) N2O3, (d) N2O5, On heating chromite (FeCr 2O4) with Na2CO3 in, air, which of the following product is obtained?, [1999], (a) Na2Cr2O7, (b) FeO, (c) Fe3O4, (d) Na2CrO4, The addition of excess of aqueous HNO3 to a, solution containing [Cu(NH3)4]2+ produces, [1999], (a) Cu+, (b) [Cu(H2O)4]2+, (c) Cu(OH)2, (d) Cu(NO3)2, An acidic solution of 'X' does not give precipitate, on passing H2 S through it. 'X' gives white, precipitate when NH4OH is added to it. The white, precipitate dissolves in excess of NaOH solution., Pure 'X' fumes in air and dense white fumes are, obtained when a glass rod dipped in NH4OH is, put in the fumes. Compound 'X' can be [1999], (a) ZnCl2, (b) FeCl3, (c) AlCl3, (d) SnCl2, Which one of the following elements constitutes, a major impurity in pig iron ?, [1998], (a) Silicon, (b) Oxygen, (c) Sulphur, (d) Graphite, K2Cr2O7 on heating with aqueous NaOH gives, [1997], (a) CrO 24(b) Cr(OH)3, (c) Cr2O 72(d) Cr(OH)2, CrO3 dissolves in aqueous NaOH to give [1997], (b) CrO42–, (a) Cr2O72–, (c) Cr(OH)3, (d) Cr(OH)2, Cuprous compounds such as CuCl, CuCN and, CuSCN are the only salts stable in water due to, [1996], (a) high hydration energy of Cu+ ions, (b) their inherent tendency to not, disproportionate, (c) diamagnetic nature, (d) insolubility in water, Stainless steel contains iron and, [1995], (a) Cr + Ni, (b) Cr + Zn, (c) Zn + Pb, (d) Fe +Cr + Ni
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215, , The d-and f-Block Elements, 61., , The most durable metal plating on iron to protect, against corrosion is, [1994], (a) Nickel plating, (b) Tin plating, (c) Copper plating, (d) Zinc plating, , 62., , When ( NH 4 ) 2 Cr2O 7 is heated, the gas, evolved is, (a) N2, (c) O2, , 63., , 64., , 65., , 67., , 68., , 69., , 70., , (b) NO2, (d) N2O, , When CuSO 4 is electrolysed using platinum, electrodes,, [1993], (a) Copper is liberated at cathode, sulphur at, anode, (b) Copper is liberated at cathode, oxygen at, anode, (c) Sulphur is liberated at cathode, oxygen at, anode, (d) Oxygen is liberated at cathode, copper at, anode, Cinnabar is an ore of, [1991], (a) Hg, (b) Cu, (c) Pb, (d) Zn, The composition of ‘golden spangles’ is, [1990], (a) PbCrO4, (b) PbI2, (c), , 66., , [1994], , As 2S3, , Topic 3: Lanthanoids and Actinoids, 71., , 72., , (d) BaCrO4, , Prussian blue is formed when, [1989], (a) Ferrous sulphate reacts with FeCl3, (b) Ferric sulphate reacts with K4[Fe(CN)6], (c) Ferrous ammonium sulphate reacts with FeCl3, (d) Ammonium sulphate reacts with FeCl3, Photographic films and plates have an essential, ingredient of, [1989], (a) Silver nitrate, (b) Silver bromide, (c) Sodium chloride (d) Oleic acid, Nitriding is the process of surface hardening of, steel by treating it in an atmosphere of [1989], (a) NH3, (b) O3, (c) N2, (d) H2S, While extracting an element from its ore, the ore, is ground and leached with dil. potassium, , cyanide solution to form the soluble product, potassium argento cyanide. The element is, [1989], (a) Lead, (b) Chromium, (c) Manganese, (d) Silver, A blue colouration is not obtained when, [1989], (a) Ammonium hydroxide dissolves in copper, sulphate, (b) Copper sulphate solution reacts with, K4[Fe(CN)6], (c) Ferric chloride reacts with sod. ferrocyanide, (d) Anhydrous CuSO4 is dissolved in water, , 73., , 74., , The reason for greater range of oxidation states, in actinoids is attributed to :[2017], (a) actinoid contraction, (b) 5f, 6d and 7s levels having comparable, energies, (c) 4f and 5d levels being close in energies, (d) the redioactive nature of actinoids, The electronic configurations of Eu(Atomic No., 63), Gd(Atomic No. 64) and Tb (Atomic No. 65), are, [2016], (a) [Xe]4f76s2, [Xe]4f 8 6s2 and [Xe]4f 85d16s2, (b) [Xe]4f 7 5d1 6s 2 , [Xe]4f 7 5d1 6s2 and, [Xe]4f 96s2, (c) [Xe]4f 6 5d 1 6s 2 , [Xe]4f 7 5d 1 6s 2 and, [Xe]4f 85d16s2, (d) [Xe]4f 76s2, [Xe]4f 75d16s2 and [Xe]4f 96s2, Because of lanthanoid contraction, which of the, following pairs of elements have nearly same, atomic radii ? (Numbers in the parenthesis are, atomic numbers)., [2015], (a) Zr (40) and Nb (41) (b) Zr (40) and Hf (72), (c) Zr (40) and Ta (73) (d) Ti (22) and Zr (40), Gadolinium belongs to 4f series. It's atomic, number is 64. Which of the following is the, correct electronic configuration of gadolinium ?, [1997, NEET Kar. 2013, 2015 RS], (a) [Xe]4f 86d2, (c), , [Xe] 4f 75d16s2, , (b) [Xe]4f 95s1, (d) [Xe] 4f 65d26s2
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EBD_8336, 216, , 75., , 76., , 77., , 78., , CHEMISTRY, Which of the following exhibit only + 3 oxidation, state ?, [2012 M], (a) U, (b) Th, (c) Ac, (d) Pa, Identify the incorrect statement among the, following:, [2007], (a) Lanthanoid contraction is the accumulation, of successive shrinkages., (b) As a result of lanthanoid contraction, the, properties of 4d series of the transition, elements have no similarities with the 5d, series of elements., (c) Shielding power of 4f electrons is quite, weak., (d) There is a decrease in the radii of the atoms, or ions as one proceeds from La to Lu., The main reason for larger number of, oxidation states exhibited by the actinoids, than the corresponding lanthanoids, is, [2005, 2006], (a) more energy difference between 5f and, 6d orbitals than between 4f and 5d, orbitals., (b) lesser energy difference between 5f and, 6d orbitals than between 4f and 5d, orbitals., (c) larger atomic size of actinoids than the, lanthanoids., (d) greater reactive nature of the actinoids, than the lanthanoids., Lanthanoids are, [2004], (a) 14 elements in the sixth period, (atomic no. = 90 to 103) that are filling 4f, sublevel, (b) 14 elements in the seventh period, (atomic no. = 90 to 103) that are filling 5f, sublevel, (c) 14 elements in the sixth period, (atomic no. = 58 to 71) that are filling 4f, sublevel, (d) 14 elements in the seventh period, (atomic no. = 58 to 71) that are filling 4f sub, level, , 79., , The correct order of ionic radii of Y3+, La3+, Eu3+, and Lu3+ is, [2003], (b), , La 3+ < Eu 3+ < Lu 3+ < Y 3+, Y 3+ < La 3+ < Eu 3+ < Lu 3+, , (c), , Y 3+ < Lu 3+ < Eu 3+ < La 3+, , (a), , 80., , (d) Lu 3+ < Eu 3+ < La 3+ < Y 3+, (Atomic nos. Y =39, La = 57, Eu = 63, Lu = 71), General electronic configuration of lanthanides, is, [1991, 2002], (a) (n – 2) f 1 –14 (n –1) s2p6d 0 – 1 ns2, (b) (n – 2) f10 –14 (n –1) d 0 – 1 ns2, (c) (n – 2) f 0 –14 (n –1) d10 ns2, (d) (n – 2) d0 –1 (n –1) f 1 – 14 ns2, , 81., , 82., , Which of the following statement is not correct?, [2001], (a) La (OH)3 is less basic than Li(OH)3, (b) La is actually an element of transition series, rather lanthanides, (c) Atomic radius of Zr and Hf are same, because of lanthanide contraction, (d) In lanthanide series ionic radius of Ln +3, ions decreases, The lanthanide contraction is responsible for, the fact that, [1997], (a) Zr and Y have about the same radius, (b) Zr and Nb have similar oxidation state, (c) Zr and Hf have about the same radius, (d) Zr and Zn have the same oxidation states, (Atomic numbers : Zr = 40, Y = 39, Nb = 41,, Hf = 72, Zn = 30), , 83., , Actinides, , [1994], , (a) Are all synthetic elements, (b) Include element 104, (c) Have any short lived isotopes, (d) Have variable valency, 84., , Among the lanthanides the one obtained by, synthetic method is, [1994], (a) Lu, , (b) Pm, , (c) Pr, , (d) Gd
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217, , The d-and f-Block Elements, , ANSWER KEY, 1, , (c) 10, , (c) 19, , (b) 28, , (a) 37, , (a) 46, , (b) 55, , (a) 64, , (a) 73, , (b) 82, , (c), , 2, , (b) 11, , (c) 20, , (b) 29, , (c) 38, , (d) 47, , (b) 56, , (d) 65, , (b) 74, , (c) 83, , (d), (b), , 3, , (c) 12, , (c) 21, , (a) 30, , (b) 39, , (a) 48, , (b) 57, , (a) 66, , (b) 75, , (c) 84, , 4, , (d) 13, , (c) 22, , (d) 31, , (a) 40, , (d) 49, , (a) 58, , (b) 67, , (b) 76, , (b), , 5, , (a) 14, , (a) 23, , (d) 32, , (d) 41, , (a) 50, , (a) 59, , (d) 68, , (a) 77, , (b), , 6, , (d) 15, , (b) 24, , (a) 33, , (d) 42, , (b) 51, , (c) 60, , (d) 69, , (d) 78, , (c), , 7, , (b) 16, , (d) 25, , (c) 34, , (b) 43, , (a) 52, , (a) 61, , (d) 70, , (b) 79, , (c), , 8, , (a) 17, , (c) 26, , (d) 35, , (b) 44, , (c) 53, , (d) 62, , (a) 71, , (b) 80, , (a), , 9, , (c) 18, , (a) 27, , (a) 36, , (c) 45, , (c) 54, , (b) 63, , (b) 72, , (d) 81, , (a), , Hints & Solutions, 1., , (c) Oxidation state of Cr in CrO 24- and Cr2O72is + 6 i.e. oxidation states are same., , 2., , (b) (i), , 3., , V2O5: Used in the oxidation of SO2 in, the manufacture of H2SO4 by contact, process., (ii) TiCl 4 + Al(CH 3) 3 : Used in the, polymerization of ethylene., (iii) PdCl2: Used in the oxidation of ethyne, to ethanal., (iv) Nickel complexes: Used in the, polymerization of alkynes., , (c), , CrO2–, 4, , Cr6+ diamagnetic, , Cr2O2–, 7, , Cr6+ diamagnetic, , MnO4–, , Mn7+ diamagnetic, , MnO 2–, 4, , Mn6+ paramagnetic, 3d, , 4., , For Ni+2 = 4s0 3d 8, Number of unpaired electrons (n) = 2, Hence, Ni2+ gives magnetic moment of 2.84 B.M., 5., , 6., , Thus, unpaired electron is present, so d–d, transition is possible., For tetrahedral geometry:, t2g, , t2g, , eg, , eg, , Before, transition, , After, transition, , (a) Fe+2 = 3d 6 (number of ‘d’ electrons = 6), In Cl = 1s2 2s2 2p6 3s2 3p5, total p electrons = 11, which are not equal to, number of ‘d’ electrons in Fe+2, p-electrons in Ne = 1s2 2s2 2p6 = 6, s-electrons in Mg = 1s2 2s2 2p6 3s2 = 6, (d) (n + l) rule can be used. Titanium is a multi, electron system, (n + l) 3s < 3p < 4s < 3d, ¯, ¯, ¯, ¯, (3 + 0) (3 + 1) (4 + 0) (3 + 2), ||, ||, ||, ||, 3, 4, 4, 5, If (n + l) values are same, then value of "n" has to, be considered. The orbital with lower n value is, filled first., , 7., , 3d, , (d) Magnetic moment = 2.84 B.M. This, indicates that 2 unpaired electrons are present., , (b) Magnetic moment, m = n(n + 2), 2.83 = n(n + 2), On solving n = 2, Ni2+ has two unpaired electrons.
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EBD_8336, 218, , 8., , CHEMISTRY, (a) The shape of f-orbitals is very much, diffused and they have poor shielding effect., The effective nuclear charge increases due to, imperfect shielding of one electron by another in, the same subshell which causes the contraction in, the size of electron charge cloud. This contraction, in size is known as lanthanoid contraction., , 9., , 10., , 11., , 12., , (c) Sm2+(Z = 62), [Xe]4f 6 6s2 – 6 unpaired e–, Eu2+(Z = 63), [Xe]4f7 6s2 – 7 unpaired e–, Yb2+(Z = 70), [Xe]4f14 6s2 – 0 unpaired e–, Ce2+(Z = 58), [Xe]4f1 5d1 6s2 – 2 unpaired e–, Only Yb2+ is diamagnetic., (c) A transition element must have incomplete, d-subshell. Zinc have completely filled d, subshell having 3d10 configuration. Hence do, not show properties of transition elements to, any appreciable extent except for their ability to, form complexes., (c) The melting points of the transition, elements first rise to a maximum and then fall as, the atomic number increases, manganese have, abnormally low melting points., = 0.34 volt,, (c) E° +2, Cu, , 15., , 16., , 17., 18., , 19., , 20., , In CoF63– –, Co is in + 3 O.S ; 3d5 = coloured, In Cu2Cl2– Cu is in +1 O.S. ; 3d10 – colourless, 8, In NiCl 2–, 4 – Ni is in + 2 O.S ; 3d – coloured, , The colour exhibited by transition metal ions is, due to the presence of unpaired electrons in, d-orbitals which permits the d - d excitation of, electrons., , /Cu, , Other has – ve E°R.P., E° ++, = – 0.28, Co /Co, E° ++ = – 0.25, Ni / Ni, E° ++, Fe /Fe = – 0.44, , 13., , (c) The transition metals and their compounds, are used as catalysts because of the variable, oxidation states. Due to this, they easily absorb, and re-emit wide range of energy to provide the, necessary activation energy., , 14., , (a), , Cr2+ =, Mn2+=, Fe2+ =, Co2+=, , Mn 2+ is most stable due to half-filled, configuration. Hence, only option (a) can be the, correct answer., , (b) La3+ : 54 e– = [Xe], To3+ : 19 e– = [Ar] 3 d 1 (Colour), Lu3+ : 68 e– = [Xe] 4 f 14, Sc3+ : 18 e– = [Ar], (d) Ni3+ : [Ar] 3d 7, Mn3+ : [Ar] 3d 4, Fe3+ : [Ar] 3d 5, Co3+ : [Ar] 3d 6, (c) Due to lanthanide contraction, the size of, Zr and Hf (atom and ions) become nearly similar, (a) +3 oxidation state is most common for, lanthanoids. But occasionally +2 and +4 states, are also obtained due to extra stability of, empty, half-filled or filled f-subshell., 4s, 3d, (b) Mn - 3d5 4s2, The no. of various oxidation states possible are, + 2, + 3, + 4, (+ 5), + 6, + 7., (b) In TiF62– ,– Ti is in + 4 O.S. ; 3d 0 = colourless, , 21., , (a) Ti ; Z (22) is 1s22s22p63s23p64s23d2, V ; Z (23) is 1s22s22p63s23p64s23d3, Cr ; Z (24) is 1s22s22p63s23p63d54s1, Mn ; Z (25) is 1s22s22p63s23d54s2, The second electron in all the cases (except Cr), is taken out from 4s-orbital and for Cr it is taken, from half-filled 3d-orbital. The force required for, removal of second electron will be more for Mn, than others (except for Cr) due to having more, positive charge. Based on this, we find the, correct order Mn > V > Ti., i.e. Cr > Mn > V > Ti., , 22., , (d) V =, Ti3+=, Mn3+ =, Cr3+=, , 3+
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EBD_8336, 222, , CHEMISTRY, 82., , (c) In lanthanide series there is a regular, decrease in the atomic as well as ionic radii of, trivalent ions (M3+) as the atomic number, increases. Although the atomic radii do show, some irregularities but ionic radii decreases from, La(103 pm) to Lu (86pm)., , (c) We know that regular decrease in the size, of the atoms and ions is called lanthanide, contraction. In vertical column of transition, elements there is a very small change in size and, some times size is found to be same from second, member to third member.The similarity in size of, the atoms of Zr and Hf is evident due to the fact, of lanthanide contraction. Therefore, Zr and Hf, both have same radius 160 pm., , 83., , (a) The Lanthanides are transition metals from, atomic numbers 58 (Ce) to 71(Lu)., , (d) Actinides have variable valency due to, very small difference in energies of 5f, 6d and 7s, orbitals. Actinides are the elements from atomic, number 89 to 103., , 84., , (b) Pm is obtained by synthetic method., , 77., , (b) Lesser energy difference between 5f and, 6d orbitals than between 4f and 5d orbitals result, in larger no. of oxidation state., , 78., , (c) Lanthanides are 4 f-series elements starting, from cerium (Z = 58) to lutetium ( Z = 71). These, are placed in the sixth period and in third group., , 79., , 80., , Hence the electronic configuration becomes :, (n –2) f, 81., , 1– 14, , (n –, , 1) s2p6 d0 – 1, , ns2., , (a) La (OH)3 is more basic than Li (OH)3. In, lanthanides, the basic character of hydroxides, decreases as the ionic radius decreases., , Promethium is not present in nature. It is synthetic, radioactive lanthanoid.
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223, , Coordination Compounds, , 23, , Coordination, Compounds, , Trend Analysis with Important Topics & Sub-Topics, 2020, Topic Name, , Sub-Topic, , 2019, , 2018, , isomerism, Coordination number,, nomenclature and, isomerism of, primary/secondary, coordination, valency, compounds, , 1, , ligand strength, , 1, , E, , magnetic moment, , 1, , A, , Magnetic, cystal field theory, moment,valence bond, complex, formation, theory and crystal field, geometry/hybridisa, theory, tion and magnetic, behaviour, LOD - Level of Difficulty, , E - Easy, , The type of isomerism shown by the complex, [CoCl2(en)2] is, [2018], , A - Average, , (a) 3 AgCl, 1 AgCl, 2 AgCl, (b) 3 AgCl, 2 AgCl, 1 AgCl, , D, , A, , A, , D - Difficult, , 1, , D, , 1, , A, , 1, , A, , Qns - No. of Questions, , (d) 1 AgCl, 3 AgCl, 2 AgCl, 3., , (c) Linkage isomerism, The correct order of the stoichiometries of AgCl, formed when AgNO3 in excess is treated with, the complexes : CoCl3.6NH3, CoCl 3.5NH3,, CoCl3.4NH3 respectively is :[2017], , 1, , (c) 2 AgCl, 3 AgCl, 1 AgCl, , (b) Coordination isomerism, , 2., , 2016, , A, , A, , 1, , (a) Geometrical isomerism, , (d) Ionization isomerism, , 1, 1, , Topic 1: Coordination Number, Nomenclature, and Isomerism of Coordination Compounds, 1., , 2017, , QNS. LOD QNS. LOD QNS. LOD QNS. LOD QNS. LOD, , 4., , 5., , Cobalt (III) chloride forms several octahedral, complexes with ammonia. Which of the, following will not give test of chloride ions with, silver nitrate at 25ºC ?, [2015], (a) CoCl3·4NH3, , (b) CoCl3·5NH3, , (c) CoCl3·6NH3, , (d) CoCl3·3NH3, , The sum of coordination number and oxidation, number of the metal M in the complex, [M(en)2(C2O4)]Cl (where en is ethylenediamine) is:, (a) 9, , (b) 6, , (c) 7, , (d) 8, , [2015 RS], , The name of complex ion, [Fe(CN)6]3– is :
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EBD_8336, 224, , CHEMISTRY, (a) Hexacyanoiron (III) ion, , [2015 RS], , with the composition of [Co(NH3 )4 Cl2 ]+ is due, , (b) Hexacyanitoferrate (III) ion, , to :, , (c) Tricyanoferrate (III) ion, (d) Hexacyanidoferrate (III) ion, 6., , 7., , 8., , (b) geometrical isomerism, , Number of possible isomers for the complex, [Co(en)2Cl2]Cl will be (en = ethylenediamine), (a) 2, , (b) 1, , (c) 3, , (d) 4, , [2015 RS], , (c) coordination isomerism, (d) ionization isomerism, 13., , An excess of AgNO3 is added to 100 mL of a, 0.01 M solution of dichlorotetraaquachromium, (III) chloride. The number of moles of AgCl, precipitated would be :, [NEET 2013], (a) 0.002, , (b) 0.003, , (c) 0.01, , (d) 0.001, , The correct IUPAC name for [CrF2(en)2]Cl is:, [NEET Kar. 2013], (a) Chlorodifluoridobis (ethylene diamine), chromium (III), , 14., , (b) Chlorodifluoridoethylenediaminechromium (III) chloride, , 10., , 11., , 12., , (b) Optical isomer, , (c) cis-isomer, , (d) Position isomer, , The complexes [Co(NH3)6] [Cr(CN)6] and, [Cr(NH3)6] [Co(CN)6] are the examples of which, type of isomerism?, [2011], (a) Linkage isomerism, (b) Ionization isomerism, (c) Coordination isomerism, (d) Geometrical isomerism, The complex, [Pt(Py)(NH3)BrCl] will have how, many geometrical isomers ?, [2011], (a) 3, (b) 4, (c) 0, (d) 2, The existence of two different coloured complexes, , [ Ni(en)3 ]2+, , (b), , é Ni ( NH3 ) ( H 2 O ) ù, 4, 2û, ë, , (c), , é Pt ( NH3 ) Cl 2 ù, 2, ë, û, , (d), , é Ni ( NH3 ) Cl 2 ù, 2, ë, û, , 2+, , Which of the following does not show optical, isomerism?, [2009], , (c) [Co (en)3]3+, (d) [Co (en)2Cl2]+, (en = ethylenediamine), 15., , [NEET Kar. 2013], (a) Linkage isomer, , (a), , (b) [Co (en) Cl2 (NH3)2]+, , (d) Difluorobis-(ethylene diamine) chromium, (III) chloride, In a particular isomer of [Co(NH3)4Cl2]0, the, Cl-Co-Cl angle is 90°, the isomer is known as, , Which one of the following complexes is not, expected to exhibit isomerism?, [2010], , (a) [Co(NH3)3Cl3]0, , (c) Difluoridobis (ethylene diamine) chromium, (III) chloride, , 9., , [2010], , (a) linkage isomerism, , Which of the following will give a pair of, enantiomorphs?, [2007], (a) [Cr(NH3)6][Co(CN)6], (b) [Co(en)2Cl2]Cl, (c) [Pt(NH3)4] [PtCl6], (d) [Co(NH3)4Cl2]NO2. (en =NH2CH2CH2NH2), , 16., , [Co(NH3)4 (NO2)2] Cl exhibits, , [2006], , (a) linkage isomerism, ionization isomerism, and geometrical isomerism, (b) ionization isomerism, geometrical isomerism, and optical isomerism, (c) linkage isomerism, geometrical isomerism, and optical isomerism, (d) linkage isomerism, ionization isomerism, and optical isomerism
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225, , Coordination Compounds, 17., , (en = ethylenediamine), , [2005], , Which of the following will give maximum, number of isomers?, [2001], , (b) trans-[Pt(NH3)2Cl 2], , (a) [Ni (C2O4) (en)2]2– (b) [Ni (en) (NH3)4]2+, , (c) cis-[Co(en)2Cl 2 ], , (c) [Cr (SCN)2 (NH3)4]+ (d) [Co (NH3)4 Cl2], , 23., , 24., , Which of the following is considered to be an, anticancer species ?, [2004], (a), , Cl, , CH2, Pt, , Cl, , (c), , CH2, , Cl, , (b), , Cl, , H3N, , Cl, , Cl, , H3N, , 26., , Pt, Cl, , Cl, , 25., , Pt, , (d), , Pt, , Cl, , Cl, , Cl, , H3N, , 19., , (d) [Co(NH3)3 (H2O)3]Cl3, , (a) cis-[Pt(NH3)2 Cl 2], , (d) trans-[Co(en)2Cl 2 ], 18., , (c) [Co(NH2)3 (H2O)2 Cl], , Which one of the following is expected to, exhibit optical isomerism?, , Which one of th e following octahedral, complexes will not show geometric isomerism?, , 21., , (d) [MA4B2], , According to IUPAC nomenclature sodium, nitroprusside is named as, [2003], (b) Sodium nitroferrocyanide, (c) Sodium nitroferrocyanide, (d) Sodium pentacyanonitrosyl ferrate (II), The hypothetical complex chloridodiaquatriammine, cobalt (III) chloride can be represented as [2002], (a) [CoCl(NH3)3 (H2O)2]Cl2, (b) [Co(NH3)3 (H2O)Cl3], , (c) 2, , (d) 6, , [2001], , Which one of the following complexes will have, four different isomers ?, [2000], [ Co (en ) 2 Cl 2 ]Cl, , [Co(PPh 3 )2 Cl2 ]Cl, , In which of the following compounds does iron, exhibit zero oxidation state?, [1999], (a) [Fe(H2O)6] (NO3)3 (b) K3[Fe(CN)6], , 28., , The total number of possible isomers for the, complex compound [CuII (NH3)4] [PtII Cl4], , (c) K4[Fe(CN)6], , [2003], , (a) Sodium pentacyanonitrosyl ferrate (III), , 22., , (b) 4, , 27., , (d) tris-(ethylendiamine) cobalt (III) bromide, , (c) [MA3B3], , (a) 3, , (d) [ Co (en ) 3 ]Cl 3, , (c) trans-dicyanobis (ethylenediamine), chromium (III) chloride, , (b) [MA2B4], , Oxidation number of Ni in [Ni(C2O4)3]4– is, , (c), , (b) diamminedichloroplatinum(II), , (a) [MA5B], , (b) [Co(NH3)6] Cl3, (d) [Ni(CO)4], , (b) [Co(en)(NH3 )2 Cl2 ]Cl, , Which of the following coordination compounds, would exhibit optical isomerism?, [2004], , (A and B are monodentate ligands), , (a) K4[Fe(CN)6], (c) [Cu(NH3)4]Cl2, , (a), , NH3, , (a) pentaamminenitrocobalt(III) iodide, , 20., , Which of the following will exhibit maximum, ionic conductivity?, [2001], , 29., , (d) [Fe(CO)5], , (a) 3, , (b) 6, , (c) 5, , (d) 4, , [1998], , IUPAC name of [Pt(NH3)3 (Br) (NO2) Cl] Cl is, [1998], (a) Triamminechlorobromonitroplatinum (IV), chloride, (b) Triamminebromonitrochloroplatinum (IV), chloride, (c) Triamminebromochloronitroplatinum (IV), chloride, (d) Triamminenitrochlorobromoplatinum (IV), chloride
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EBD_8336, 226, , 30., , CHEMISTRY, A coordination complex compound of cobalt has, the molecular formula containing five ammonia, 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, [1998], , (a), , (b) [ Co ( H 2 O ) 2 ( NH 3 ) 2 Cl 2 ] +, (c), , 37., , (d) [Co (NH3)5] [(NO2)2Cl2], The formula for the complex, dichlorobis (urea), copper (II) is, [1997], , 38., , 34., , 35., , (d) Potash alum, , Which of the following is the correct order of, increasing field strength of ligands to form, coordination compounds?, [2020], (b) F– < SCN–< C2 O 24 - < CN–, (c) CN– < C2 O 24 - < SCN– < F–, , The number of geometrical isomers of the, complex [Co(NO2)3 (NH3)3] is, [1997], (b) 3, , (c) 4, , (d) zero, , Among the following, the compound that is both, paramagnetic and coloured, is, [1996], (a) KMnO4, , (b) CuF2, , (c) K4[Fe(CN)6], , (d) K2Cr2O7, , The number of geometrical isomers for, [Pt (NH3)2 Cl2] is, [1995], (a) 2, , (b) 1, , (c) 3, , (d) 4, , K 3[Al(C 2O 4 )3 ] is called, , (d) SCN– < F– < C2 O 24 - < CN–, 39., , 40., , 41., , [1994], , (a) Potassium alumino oxalate, (b) Potassium trioxalatoaluminate (III), (c) Potassium aluminium (III) oxalate, (d) Potassium trioxalato aluminate (VI), 36., , (c) Hypo, , (b) [Cu{O = C (NH2)2}Cl]Cl, (d) [CuCl2] [{O = C (NH2)2}]2, , 33., , (b) K 4 [Fe(CN) 6 ], , (a) SCN– < F– < CN– < C2 O 24 -, , (a) 2, , [1989], , (a) Bleaching powder, , (a) [Cu{O = C (NH2)2}] Cl2, (c) [CuCl2 {O = C(NH2)2}2], 32., , An example of double salt is, , Topic 2: Magnetic Moment, Valence Bond, Theory and Crystal Field Theory, , (c) [Co (NH3)5 (NO2)] Cl2, 31., , [Cr(H 2 O) 2 Cl 2 ]+, , (d) [Co(CN) 5 NC], , (a) [Co(NH3)4 (NO2) Cl] [(NH3) Cl], (b) [Co (NH3)5 Cl] [Cl (NO2)], , [Co( NH 3 ) 6 ]3+, , Among the following complexes, optical activity, is possible in, [1994], , 42., , The calculated spin only magnetic moment of, Cr2+ ion is, [2020], (a) 4.90 BM, (b) 5.92 BM, (c) 2.84 BM, (d) 3.87 BM, The Crystal Field Stabilisation Energy (CFSE), for [CoCl6]4– is 18000 cm–1. The CFSE for, [CoCl4]2– will be, [NEET Odisha 2019], –1, (a) 8000 cm, (c) 6000 cm–1, –1, (c) 16000 cm, (d) 18000 cm–1, What is the correct electronic configuration of, the central atom in K4[Fe(CN)6] based on crystal, field theory ?, [2019], (a), , t24g eg2, , (b) t26g eg0, , (c), , e3g t23g, , (d) eg4 t22g, , The geometry and magnetic behaviour of the, complex [Ni(CO)4] are, [2018], (a) Square planar geometry and diamagnetic, (b) Tetrahedral geometry and diamagnetic, (c) Tetrahedral geometry and paramagnetic, (d) Square planar geometry and paramagnetic
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227, , Coordination Compounds, 43., , a., , Co3+, , i., , 8 BM, , b., , Cr3+, , ii., , 35 BM, , c., , Fe3+, , iii., , 3 BM, , d., , Ni2+, , iv., , 24 BM, , v., , 15 BM, , a, , 44., , 45., , (c) [Co(CN)6]3– has no unpaired electrons and, will be in a high-spin configuration., , Match the metal ions given in Column I with, the spin magnetic moments of the ions given in, Column II and assign the correct code : [2018], Column I, Column II, , b, , c, , d, , (a) iv, , v, , ii, , i, , (b) i, , ii, , iii, , iv, , (c) iii, , v, , i, , ii, , (d) iv, , i, , ii, , iii, , (d) [Co(CN)6]3– has no unpaired electrons and, will be in a low-spin configuration., 48., , 49., , (b) HgI42 - , I3- [2017], , (c) Hg2I2, I–, , (d) HgI2 , I3-, , Correct increasing order for the wavelengths of, absorption in the visible region the complexes, of Co3+ is :[2017], , (c) [Co(H2O)6]2+, , (d) [Co(H2O)6]3+, , Which of the following complexes is used as an, anti-cancer agent:, [2014], (c) cis-K2[PtCl2Br2], , 50., , (c), 51., , [CoCl6]4–, , Which is diamagnetic?, (a), (c), , 52., , (d) Na2CoCl4, , A magnetic moment of 1.73 BM will be shown, by one among the following : [NEET 2013], (a) [Ni(CN)4]2–, , [Fe(CN)6]3–, [Ni(CN)4]2–, , (b) [TiCl4], (d) [Cu(NH3)4]2+, [NEET Kar. 2013], (b) [Co(F6)]3–, (d) [NiCl4]2–, , The anion of acetylacetone (acac) forms, Co(acac)3 chelate with Co3+. The rings of the, chelate are, [NEET Kar. 2013], (a) three membered, , (b) five membered, , (b) [Co(H2O)6]3+, [Co(NH3)6]3+, [Co(en)3]3+, , (c) four membered, , (d) six membered, , 53., , (d) [Co(en)3]3+, [Co(NH3)6]3+, [Co(H2O)6]3+, Pick out the correct statement with respect to, [Mn(CN)6]3[2017], (a) It is sp3d2 hybridised and tetrahedral, (b) It is d2sp3 hybridised and octahedral, (c) It is dsp2 hybridised and square planar, , 54., , Which of these statements about [Co(CN)6]3–, is true ?, [2015], (a), , [Co(CN)6]3–, , has four unpaired electrons, and will be in a low-spin configuration., , (b) [Co(CN)6]3– has four unpaired electrons, and will be in a high spin configuration., , Which among the following is a paramagnetic, complex?, [NEET Kar. 2013], (a) [Mo(CO)6], , (b) [Co(NH3)6]3+, , (c) [Pt(en)Cl2], , (d) [CoBr4]2–, , (At. No. of Mo = 42, Pt = 78), , (d) It is sp3d2 hybridised and octahedral, 47., , (b) [Fe(H2O)6]3+, , (a) [Co(H2O)6]3+, [Co(en)3]3+, [Co(NH3)6]3+, (c) [Co(NH3)6]3+, [Co(en)3]3+, [Co(H2O)6]3+, 46., , (a) [Mn(H2O)6]3+, , (a) mer-[Co(NH3)3Cl3] (b) cis-[PtCl2(NH3)2], , HgCl2 and I2 both when dissolved in water, containing I – ions, the pair of species formed is:, (a) HgI2, I–, , Among the following complexes the one which, shows zero crystal field stabilization energy, (CFSE):, [2014], , 55., , Which one of the following is an outer orbital, complex and exhibits paramagnetic behaviour ?, [2012], (a) [Ni(NH3)6]2+, , (b) [Zn(NH3)6)]2+, , (c) [Cr(NH3)6]3+, , (d) [Co(NH3)6]3+, , Red precipitate is obtained when ethanol, solution of dimethylglyoxime is added to, ammoniacal Ni(II). Which of the following, statements is not true ?, [2012 M]
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EBD_8336, 228, , CHEMISTRY, (a) Red complex has a square planar geometry., , 61., , (b) Complex has symmetrical H-bonding, , d 4 octahedral complex is:, , (c) Red complex has a tetrahedral geometry., , (a) – 1.8 D 0, , (b) – 1.6 D 0 + P, , (c) – 1.2 D 0, , (d) – 0.6 D 0, , (d) Dimethylglyoxime functions as bidentate, ligand., , dimethylglyoxime =, , 56., , H3C, , C, , N, , H3C, , C, , N, , OH, , 62., , (a), , (b), , -12, D o + 3P, 5, , (c), , -2, D o + 2P, 5, , (d), , -2, Do + P, 5, , (c) [Zn (NH3)6]2 +, (d) [Sc (H2O)3 (NH3)3]3+, (At. no. Zn = 30, Sc = 21, Ti = 22, Cr = 24), 63., , 58., , 59., , 60., , (b) [Ni(CN)4]2–, , (c) [CuCl4]2–, , (d) [CoF6]3–, , The d-electron configurations of Cr 2+, Mn2+,, Fe2+ and Co2+ are d 4, d 5, d 6 and d 7, respectively., Which one of the following will exhibit minimum, paramagnetic behaviour?, [2011], , Which of the following complexes exhibits the, highest paramagnetic behaviour ?, [2008], (a) [V(gly)2(OH)2(NH3)2]+, (b) [Fe(en)(bpy)(NH3)2]2+, (c) [Co(ox)2(OH)2]–, , Of the following complex ions, which is, diamagnetic in nature ?, [2011], (a) [NiCl4]2–, , Which of the following complex ions is expected, to absorb visible light?, [2009], (b) [Cr (NH3)6]3 +, , Low spin complex of d 6-cation in an octahedral, field will have the following energy : [2012 M], , -12, Do + P, 5, , [2010], , (a) [Ti (en)2(NH3)2]4 +, , OH, , (D o= Crystal Field Splitting Energy in an, octahedral field, P = Electron pairing energy), 57., , Crystal field stabilization energy for high spin, , (d), , [Ti(NH3)6]3+, , where gly = glycine, en = ethylenediamine and, bpy = bipyridyl moities), (At.nosTi = 22, V = 23, Fe = 26, Co = 27), , (a) [Mn(H2O)6]2+, , (b) [Fe(H2O)6]2+, , In which of the following coordination entities, the magnitude Do (CFSE in octahedral field) will, be maximum?, [2008], , (c) [Co(H2O)6]2+, , (d) [Cr(H2O)6]2+, , (a) [Co(H2O)6]3+, , (b) [Co(NH3)6]3+, , (c) [Co(CN)6]3–, , (d) [Co (C2O4)3]3–, , (At, nos. Cr = 24, Mn = 25, Fe = 26, Co = 27), Which of the following complex compounds will, exhibit highest paramagnetic behaviour?, [2011M], (At. No. : Ti = 22, Cr = 24, Co = 27, Zn = 30), (a) [Ti (NH3)6]3+, , (b) [Cr (NH3)6]3+, , (c) [Co (NH3)6]3+, , (d) [Zn (NH3)6]2+, , Which of the following complex ion is not, expected to absorb visible light ?, [2010], (a), (c), , 2-, , [ Ni(CN)4 ], [ Fe(H2 O)6 ]2+, , (b), (d), , [Cr(NH3 )6 ], [ Ni(H 2O)6 ]2+, 3+, , 64., , (At. No. Co = 27), 65., , The d electron configurations of Cr2+, Mn2+, Fe2+, and Ni2+ are 3d 4, 3d 5, 3d 6 and 3d 8 respectively., Which one of the following aqua complexes will, exhibit the minimum paramagnetic behaviour?, [2007], (a), , [Fe(H2O)6]2+, , [Ni(H2O)6]2+, , (b), , (c), , [Cr(H2O)6]2+, , (d) [Mn(H2O)6]2+, , (At. No. Cr = 24, Mn = 25, Fe = 26, Ni = 28)
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229, , Coordination Compounds, 66., , [Cr(H2O)6]Cl3 (at no. of Cr = 24) has a magnetic, moment of 3.83 B. M. The correct distribution of, 3d electrons in the chromium of the complex is, , (, , (a) 3d xy1, 3d x 2 - y 2, , ) , 3d, 1, , 1, yz, , 73., , [2006], , (b) 3dxy1, 3dyz1, 3dxz1, (c) 3dxy1, 3dyz1, 3d1z 2, 67., , (d) (3dx2 – y2)1, 3d z 2 , 3dxz1, , 74., , Which one of the following is an inner orbital, complex as well as diamagnetic in behaviour?, (Atomic number: Zn = 30, Cr = 24, Co = 27,, Ni = 28), [2005], 2+, 3+, (a) [Zn(NH3)6], (b) [Cr(NH3)6], , 75., , (c) [Co(NH3)6]3+, 68., , 69., , (d) [Ni(NH3)6]2+, , Among [ Ni( CO ) 4 ], [ Ni( CN ) 4 ] 2 - , [ NiCl 4 ] 2 species, the hybridization states of the Ni atom, are, respectively (At. No. of Ni = 28), [2004], (a), , sp 3 , dsp2 , dsp 2, , (b) sp3 , dsp 2 , sp3, , (c), , sp3 , sp 3 , dsp 2, , (d) dsp 2 , sp 3 , sp 3, , CN– is a strong field ligand. This is due to the, fact that, [2004], , 76., , 77., , (a) it carries negative charge, (b) it is a pseudohalide, , 70., , (c) it can accept electrons from metal species, (d) it forms high spin complexes with metal, species, Considering H2O as a weak field ligand, the, number of unpaired electrons in [ Mn ( H 2 O ) 6 ]2 +, [2004], will be (At. no. of Mn = 25), (a) three, (c) two, , 71., , 72., , (b) five, (d) four, , The number of unpaired electrons in the complex, ion [CoF6]3– is (Atomic no.: Co = 27) [2003], (a) Zero, (b) 2, (c) 3, (d) 4, Atomic number of Cr and Fe are respectively 25, and 26, which of the following is paramagnetic?, [2002], , 78., , 79., , (a) [Cr(CO)6], (b) [Fe(CO)5], –4, (c) [Fe(CN)6], (d) [Cr(NH3)6]+3, CuSO4 when reacts with KCN forms CuCN,, which is insoluble in water. It is soluble in excess, of KCN due to formation of the following complex, [2002], (a) K2[Cu(CN)4], (b) K3[Cu(CN)4], (c) CuCN2, (d) Cu[K Cu(CN)4], Which statement is incorrect?, [2001], (a) [Ni(CO)4] – Tetrahedral, paramagnetic, (b) [Ni(CN)4]2– – Square planar, diamagnetic, (c) [Ni(CO)4] – Tetrahedral, diamagnetic, (d) [NiCl4]2– – Tetrahedral, paramagnetic, Which one of the following will show, paramagnetism corresponding to 2 unpaired, electrons?(Atomic numbers : Ni = 28, Fe = 26), (a) [Fe F6]3–, (b) [Ni Cl4]2– [1999], 3–, (c) [Fe (CN)6], (d) [Ni (CN)4]2–, The number of unpaired electrons in the complex, [Cr(NH3)6]Br3 is (Atomic number Cr = 24), (a) 4, (b) 1, [1999], (c) 2, (d) 3, Which of the following statements is correct ?, (Atomic number of Ni = 28), [1997], (a) [Ni(CO)4] is diamagnetic and [NiCl4]2– and, [Ni(CN)4]2– are paramagnetic, (b) [Ni(CO)4] and [Ni(CN)4]2– are diamagnetic, and [NiCl4]2– is paramagnetic, (c) [Ni(CO)4] and [NiCl4]2–are diamagnetic and, [Ni(CN)4]2– is paramagnetic, (d) [NiCl4]2– and [Ni(CN)4]2– are diamagnetic, and [Ni(CO)4] is paramagnetic, Which of the following is common donor atom, in ligands?, [1995], (a) arsenic, (b) nitrogen, (c) oxygen, (d) both 'b' and 'c', The complex ion [Co(NH3)6]3+ is formed by sp3d 2, hybridisation. Hence, the ion should possess, (a) Octahedral geometry, [1990], (b) Tetrahedral geometry, (c) Square planar geometry, (d) Tetragonal geometry.
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EBD_8336, 230, , CHEMISTRY, 85., , Topic 3: Organometallic Compounds, 80., , 81., , Iron carbonyl, [Fe(CO)5] is, , [2018], , (a) Tetranuclear, , (b) Mononuclear, , (c) Dinuclear, , (d) Trinuclear, , An example of a sigma bonded organometallic, compound is :, [2017], , Among the following, which is not the p-bonded, organometallic compound?, [2003], (a), , (CH 3 ) 4 Sn, , (b), , K [PtCl3 (h2 - C 2 H 4 )], , (c), , Fe (h5 - C5 H 5 ) 2, , (d), , Cr (h6 - C 6 H 6 ) 2, , (a) Grignard's reagent (b) Ferrocene, (c) Cobaltocene, 82., , (d) Ruthenocene, , Which of the following has longest C–O bond, length? (Free C–O bond length in CO is 1.128Å), , 86., , [2016], (a) [Ni(CO)4], (c), 83., , 84., , [Fe(CO)4]2–, , Which of the following organometallic, compound is s and p bonded?, [2001], (a) [Fe (h5 – C5H5)2], , (b) [Co(CO)4]–, , (b) Fe (CH3)3, , (d) [Mn(CO)6]+, , (c) K [PtCl3(h2 – C2H4)], , Which of the following carbonyls will have the, strongest C – O bond ?, [2011 M], , (d) [Co(CO)5 NH3]2+, , (a) [Mn (CO)6]+, , (b) [Cr (CO)6 ], , 87., , Which of the following may be considered to be, an organometallic compound?, [1996], , (c) [V (CO)6]–, , (d) [Fe (CO)5], , (a) Nickel tetracarbonyl, , Which of the following does not have a metalcarbon bond?, [2004], , (b) Chlorophyll, , (a), , Al(OC 2 H 5 )3, , (d) [Co (en)3] Cl3, , (c), , K[Pt (C 2 H 4 )Cl 3 ] (d), , (b) C 2 H 5MgBr, , (c) K3 [Fe (C2O4)3], , [ Ni(CO) 4 ], ANSWER KEY, , 1, , (b) 10, , (c) 19 (d) 28 (d) 37, , 2, , (a) 11, , (a) 20 (a) 29 (c) 38, , 3, , (d) 12, , (b) 21 (a) 30 (c) 39, , 4, , (a) 13, , (d) 22 (a) 31 (c) 40, , (d) 46 (b) 55 (c) 64, (d) 47 (d) 56 (b) 65, , (c) 73 (b) 82 (c), (b) 74 (a) 83 (a), , (a) 48 (b) 57 (b) 66, (a) 49 (b) 58 (c) 67, , (b) 75 (b) 84 (a), (c) 76 (d) 85 (a), , 5, , (d) 14, , (a) 23 (c) 32 (a) 41 (b), , 6, , (c) 15, , (b) 24 (a) 33 (b) 42, (a) 25 (c) 34 (a) 43, , (b) 51 (c) 60 (a) 69, (a) 52 (d) 61 (d) 70, , 7, , (d) 16, , 8, 9, , (c) 17, (c) 18, , 50 (d) 59 (b) 68, , (b) 77 (b) 86 (d), (b) 78 (d) 87 (a), (b) 79 (a), , (c) 26 (b) 35 (b) 44, (c) 27 (d) 36 (b) 45, , (b) 53 (d) 62 (b) 71, (d) 54 (a) 63 (c) 72, , (d) 80 (b), (d) 81 (a)
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233, , Coordination Compounds, (c), , en, , +, , en, Cl, , +, , Co, Cl, , Cl, en, Cis-d-isomer, , N, , N, , 3+, , en – N, , N, , [Co( NH 3 ) 6 ]Cl 3 ® [Co( NH 3 ) 6 ]3+ + 3 Cl - ® 4 ions, , en, N, , Co, , en, N, , [ Ni(CO) 4 ] ® No ions, 25., 26., , 3+, , N, , N, en, N, , en, N, d form, , en, , [Cu ( NH 3 ) 4 ]Cl 2 ® [Cu ( NH 3 ) 4 ]2+ + 2Cl - ® 3 ions, , Enantiomers, , l form, , (ii) Optical isomers, , (a) MA3 B3 – 2 geometrical isomers, , Cl, en CO, NH3, , The complexes of general formula Ma6 and Ma5b, octahedral geometry do not show geometrical, isomerism., , (a) IUPAC name of sodium nitroprusside, Na 2 [Fe(CN) 5 NO] is sodium pentacyanonitrosoyl ferrate (III) because in it NO is neutral, ligand. Hence, 2×O.N. of Na + O.N. of Fe + 5×O.N. of CN, 1×O.N. of NO = 0, 2×(+1) + O.N. of Fe + 5 ×(–1) +1×0 = 0, O.N. of Fe = 5 – 2 = +3, Hence, ferrate (III), (a) Chloridodiaquatriammine cobalt (III), chloride is [CoCl( NH 3 )3 (H 2O) 2 ]Cl 2, , Plane of symmetry, , trans (for NH 3), , MA4 B2 – 2 geometrical isomers, , 22., , = x + 3( -2) = -4 Þ x = -4 + 6 = 2, (b) Complex [Co(en)(NH3)2Cl2]Cl will have four, different isomers., (i) Geometrical isomers, Cl, NH3, |, en – CO, |, NH3, Cl, NH3, Cl, |, en – CO, |, Cl, NH3, , MA2 B4 – 2 geometrical isomers, , 21., , 4-, , trans (for Cl), , The two optically active isomers are collectively, called enantiomers., 20., , (c) O.N. of Ni in [ N i(C 2 O 4 ) 3 ], , |, , N, Co, , (a) The complex ion which give maximum ions, in solution exhibit maximum ionic conductivity., K 4 [ Fe(CN) 6 ] ® [ Fe (CN) 6 ]-4 + 4K + ®5 ions, , Cis-l-isomer, , Trans-form of [M(AA) 2 a 2]n± does not show, optical isomerism., (c) Diaminodichloroplatinum (II) commonly, known as cis - platin is found to have anticancer, property., (d) The optical isomers are pair of molecules, which are non superimposable mirror images of, each other., N, , 24., en, , Mirror, , 19., , (c) [Cr(SCN ) 2 ( NH 3 ) 4 ]+ shows linkage and, geometrical isomerism. Hence, produces maximum, no. of isomers., , Cl, , Co, , 18., , 23., , |, , 17., , Cl, NH3, , cis-isomer, , 27., 28., , Cl, Cl, H3N, , CO, , en, , NH3, , d- and l -forms, , (no element of symmetry), Hence, the compound has 4 different isomers, (d) In [Fe(CO)5] iron exist in zero oxidation state., (d) The total number of isomers for the complex, compound, [Cu II (NH3 )4 ][Pt II Cl 4 ] is four.., , These four isomers are, [Cu ( NH 3 )3 Cl] [Pt ( NH 3 )Cl 3 ],, [Pt(NH3 )3 Cl] [Cu(NH3 )Cl3 ] ,
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235, , Coordination Compounds, 43., , Co3+ = [Ar], , 3d 6 , unpaired e–(n) = 4, , As, coordination number of Mn = 6, so it will, form an octahedral complex., \ [Mn(CN)6]3– =, , (a), Spin magnetic moment, 4(4 + 2) =, , =, , 24 B.M., , Cr3+ = [Ar] 3d 3 , unpaired e–(n) = 3, Spin magnetic moment, , 3d, , 3(3 + 2) = 15 B.M., , =, , Fe3+ = [Ar] 3d 5 , unpaired e–(n) = 5, Spin magnetic moment, , 47., , Ni2+ = [Ar], , Red ppt, , HgI2 + 2I- ® [ HgI 4 ], 45., , (b) Due to d 5 configuration CFSE is zero., , 49., , (b) cis-platin, , 50., , (d) [Cu(NH3)4]2+ hybridisation dsp2, Cu+2 – 3d9 has one unpaired e-, , So, magnetic moment, m=, , soluble, , 46., , 51., , 52., , C, , 53., 4s, 4p, d 2 sp3, , \ It is diamagnetic., (d) Structure of Co(acac)3:, , C, , The presence of a strong field ligand CN– causes, pairing of electrons., , 3d, , (c) Ni+2 ® 3d 8 =, CN– is a strong ligand and causes pairing of 3d, electrons of Ni2+., , (b) In the complex [Mn(CN)6]3–, O.S. of Mn is + 3, E.C. of Mn+3 ® 3d 4, 3d, 4s, 4p, , Þ, , n ( n + 2 ) = 1(1 + 2 ), , = 3 = 1.73, , Hence, the increasing order of wavelengths of, absorption is:, [Co(en)3]3+ < [Co(NH3)6]3+ < [Co(H2O)6]3+, Higher value of D0 means higher stability, which, means it will absorb high energy photons. High, energy photons have low value of wavelength., , (d) In [Co(CN)6]–3, O.N. of Co is +3, , 48., , 2-, , (d) The order of strength of ligand is, en > NH3 > H2O, For strong ligand, the value of D0 for the complex, is higher., , 3, , \ Pairing of electrons occurs so in this complex, no unpaired electron is present and it is low spin, complex., , 2(2 + 2) = 8 B.M., , (b) In a solution containing HgCl2, I2 and I–,, both HgCl2 and I2 compete for I–., Since formation constant of [HgI4]2– is very, large (1.9 × 10 30 ) as compared with I 3–, (Kf = 700)., \ I– will preferentially combine with HgCl2., HgCl2 + 2I– ® HgI2 ¯ + 2Cl–, , ´´ ´´ ´´, 4p, , CN– is a strong field ligand, , 3d 8 , unpaired e–(n) = 2, , Spin magnetic moment, =, , ´´, 4s, d2s, , \ Co+3 = 3d 6 4s 0, , 5(5 + 2) = 35 B.M., , =, , 44, , ´´ ´´, , [Ar], , O, Co O, , O, O, , 1, , 6, , O, , 5, , C, , C, C, , O2, C, , 3, , 4, , Hence, the rings of chelate are six membered ring., (d) Co2+ Þ [Ar]3d74s0, here, Br– is a weak field, ligand so will not cause pairing of d-electrons, in Co2+., \ [CoBr 4 ] 2– will exhibit paramagnetic, behaviour due to unpaired electrons.
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237, , Coordination Compounds, (c) [Co(NH3)6]3+ : In this complex, cobalt ion is, in + 3 oxidation state with 3d 6 configuration., Co3+,[Ar]3d 6, , 63., , 3d, , 4s, , 4p, , Zn= [Ar] 3d 10 4s2;, , Zn2+= 3d10, , Sc = [Ar] 3d 1 4s2;, , Sc3+ = 3d 0, , (c) More the number of unpaired electrons present, in a complex more is its paramagnetic behaviour., x, , -1, , -1, , 0, , (a) [V ( gly ) (OH) (NH3 ) ]+, 2, 2, 2, [Co(NH3)6]3+, , x + 2(–1) + 2(–1) + 2(0) = +1, Þ x = +5, V5+ = [Ar]; no unpaired electron, , NH3 NH3 NH3 NH3 NH3 NH3, , x, , 0, , Fe2+ = [Ar] 3d 6, , ion, , Zn2+ ion, , 0, , x = +2, , (d) In this complex Zn exists as, Zn2+, , 0, , (b) [Fe(en)(bpy )(NH3 ) ]2 +, 2, , (inner orbital or d 2sp3 hybrid orbital low spin, complex) diamagnetic, , But due to presence of strong field ligands, en,, bpy and NH3; the electrons will try to pair up., Thus, the complex will contain no unpaired, electron., , : 3d10 4s0, , Zn2+ ion in [Zn(NH3)4]2+, , e, , -2, , –1, , (c) [Co (Ox ) (OH) ]–, 2, 2, x + 2(–2) + 2(–1) = –1, Þ x = +5, , NH3 NH3 NH3 NH3, , Co5+ = [Ar] 3d 4, , Due to presence of paired electrons, complex is, diamagnetic in nature., 60., , (a) [ Ni(CN)4 ], , [Cr(NH3 )6 ], , 3+, , 2-, , Thus, 4 unpaired electrons., x, , : Number of unpaired electrons = 0, , x = +3, Ti3+ = [Ar] 3d1, Thus, 1 unpaired electron., Hence, option (c) is the correct answer., , : Number of unpaired electrons = 3, , [Fe(H 2O)6 ]2+ : Number of unpaired electrons = 4, , [ Ni(H2O)6 ]2+ : Number of unpaired electrons = 2, 61., , –, , t2g, CFSE = [0.6 × 1] + [–0.4 × 3] = – 0.6 D 0, 62., , To find unpaired electrons let us calculate the, oxidation states of elements in each complex and, then write the electronic configuration for that, oxidation state to find the number of unpaired, electrons in it., , (d) d 4 in high spin octahedral complex, , eg, , (b) Since Cr 3+ in the complex has unpaired, electrons in the d orbital, hence will be coloured, Ti = [Ar]3d 2 4s2 ;, , Ti4 + = 3d 0, , Cr = [Ar] 3d 5 4s1;, , Cr3+ = 3d 3, , 0, , (d) [Ti (NH3 ) ]3+, 6, , 64., , (c) Out of the given ligands water, ammonia,, cyanide and oxalate, maximum splitting will occur, in case of cyanide (CN–) i.e. the magnitude of Do, will be maximum in case of [Co(CN)6]3–.
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239, , Coordination Compounds, 69., , CN–, , [Fe(CN)6]–4, , (b), is a strong field ligand as it is a, psuedohalide ion., , ×, ×, , Psuedohalide ions are strong coordinating ligands, and hence have the tendency to form s-bond (from, the pseudo halide to the metal) and p-bond (from, the metal to pseudo halide)., , 70., , 71., , 3d, , ×, ×, , × × ×, × × ×, , 4s, , 4p, , ×, ×, , × × ×, × × ×, , 4s, , 4p, , ––– diamagnetic, [Cr(NH3)6]+3, , × ×, × ×, , (b) Since H 2 O is a weak ligand, it will not cause, pairing of electrons in the metal ion Mn 2+. Thus, electronic configuration of the metal (Mn2+) in, the complex will be, , 3d, , ––– Paramagnetic, 73., , Mn 2+ :1s 2 2s 2 p6 3s 2 3 p 6 3d 5, , ×, ×, , i.e. 5 unpaired electrons., (d) Co here is in +3 oxidation state, , (b) Copper sulphate react with KCN to give, white ppt of Cu(CN)2 and cyanogen gas. The, insoluble copper cyanide dissolve in excess of, KCN and give soluble potassium cuprocyanide, CuSO 4 + 2 KCN ¾, ¾® K 2 SO 4 + Cu (CN ) 2, , Co, , ® 2CuCN + CN - CN, 2Cu(CN) 2 ¾¾, insoluble, , Co3+, , ® K 3 [Cu(CN) 4 ], CuCN + 3KCN ¾¾, Soluble, , = 4 and sp3d 2 hybridisation, , Unpaired electrons, and octahedral shape., 72., , cyanogen, , 74., , (d) Cr 3+ has 4sº 3d 3 electronic configuration, with 3 unpaired electrons, hence paramagnetic., In other cases, pairing of d-electrons take place, in presence of strong field ligands such as CO, or CN–., , (a) (CO) carbonyl group being a strong ligand, paired all electrons present in d orbital of Ni., In [Ni(CO)4 ] complex, Ni will have 3d10, configuration. It has tetrahedral geometry but, diamagnetic as there are no unpaired electrons., , In Cr(CO)6 molecule 12 electrons are contributed, by CO group and it contain no odd electron, , 75., , (b) As in [NiCl4]–2, chloride ion being a weak, ligand is not able to pair up the d–e–s., , [Cr(CO)6], , 76., , (d) In [Cr (NH3)6]Br3, Cr is in +3 oxidation state etc., , ×, ×, , ×, ×, , 3d, , ×, ×, , ×, ×, , 4s, , × ×, × ×, 4p, , 3d, , 3d, , ––– diamagnetic, , 4p, , 24, , d2sp 3 hybridized, , ––– diamagnetic, [Fe(CO)5], ×, ×, , 4s, , Cr, , 3d, ×, ×, 4s, , ×, ×, , ×, ×, 4p, , ×, ×, , 4s, , 4p, , +3, , Cr, , Its ion is octahedral in nature. Due to the, presence of three unpaired electrons it is, paramagnetic.
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EBD_8336, 240, , 77., , CHEMISTRY, (b), , Atom/Ion, Complex, 3d, Ni, , 2+, , No. of, Magnetic, unpaired, nature, electrons, , Configuration, , 4s, , 4p, , 8, , (d ), , [NiCl4], , 2, , 2–, , 2, , sp3, , 2–, , 0, , [Ni(CN)4], , Rearrangement, , dsp, , 2, , 8 2, , 2, , Ni (d s ), [Ni(CO)4], , 0, Rearrangement, , 78., , 79., 80., , 81., 82., , (d) In the formation of a coordinate bond, the, ligands donate a pair of electrons to the metal, atom. Futher nitrogen and oxygen has great, tendency to donate the pair of electrons in most, of the compounds. Therefore, both nitrogen, and oxygen are common donor atoms in ligands., (a) According to VSEPR theory, a molecule with, six bond pairs must be octahedral., (b) [Fe(CO)5], EAN = Z – O.N. + 2(C.N.), = 26 – 0 + 2(5), = 26 + 10, = 36, Only one central metal atom/ion is present and, it follows EAN rule, so it is mononuclear., (a) Grignard's reagent (RMgX) is a s-bonded, organometallic compound., (c) [Fe(CO)4]2–, Since metal atom is carrying maximum –ve, charge therefore, it would show maximum, synergic bonding as a resultant C—O bond, length would be maximum., M–C p bond in metal carbonyl which is formed, by the donatio of an electron pair from a filled dorbital of metal into the vacant anti bonding porbtial of CO, strengthen the M–C s bond. This is, called synergic effect., , sp3, , 83., , 84., , 85., , 86., , 87., , (a) As + ve charge on the central metal atom, increases, the less readily the metal can donate, electron density into the p* orbitals of CO ligand, to weaken the C – O bond. Hence, the C – O, bond would be strongest in [Mn(CO)6]+., (a) Triethoxyaluminium has no Al – C linkage, O - CH 2 CH 3, Al, O - CH 2 CH 3, O - CH 2 CH 3, (a) The number of carbon atom found in p bonded, organometallic compounds is indicated by the greek, latter 'h' with a number. The prefixes h2, h5 and h6, indicate that 2, 5 and 6 carbon atom are bound to, the metal in the compound.(CH3)4Sn does not, involve any pi (p) bond formation. (CH3)4Sn is a s, bonded organometallic compound., (d) In metal carbonyl complexes, s and p both, bonds are present because of synergic bonding., Thus, [Co(CO)5(NH3)]2+ has s and p bonding., (a) Organometallic compounds are those, compounds in which there is a bond which, involve metal-carbon bond. In [Ni(CO)4] there, are bond involving Ni – C., In chlorophyll, there are 4 bonds between Mg – N.
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EBD_8336, 242, , 6., , 7., , CHEMISTRY, (a) CH3 – CH2 – CH2Br, , Which of the following acids does not exhibit, optical isomerism ?, [2012], (a) Maleic acid, (b) a -amino acids, (c) Lactic acid, (d) Tartaric acid, Consider the reactions :, [2011 M], (i), , (b), , C H OH, , 2 5 ¾¾, ® (CH 3 ) 2, (CH 3 ) 2 CH - CH 2 Br ¾¾¾, , (CH3)2 CH - CH2OC2H5 + HBr, (ii), , (CH3 )2 CH - CH 2 Br, , C 2 H 5O ®, ¾¾¾¾¾, , (CH 3 )2, , (CH3)2 CH - CH2OC2H5 + Br, , 8., , (d) CH3CH2Br, 11., , (b) SN1 and SN1, (d) SN2 and SN1, , 12., , CH3 – CH = CH2 + H2O ¾®, CH3 – CH – CH3, , (c) 8, , (d) 2, , If there is no rotation of plane polarised light by, a compound in a specific solvent, though to be, chiral, it may mean that, [2007], , (c) the compound may be a racemic mixture, , OH, , CH3 – CH 2–CH–CH 2Br ¾®, , (b) 6, , (b) there is no compound in the solvent, , (b) RCHO + R ¢MgX ¾® R – CH – R¢, , (c), , (a) 4, , (a) the compound is certainly meso, , OH, , CH 3, , How many stereoisomers does this molecule, have ?, [2008], CH3CH = CHCH 2CHBrCH 3, , Which one is a nucleophilic substitution, reaction among the following ?, [2011], (a), , (d) the compound is certainly achiral., 13., , CH3, , CH3 – CHCl – CH2 – CH3 has a chiral centre., Which one of the following represents its, R-configuration?, [2007], , (a), , C2 H 5, |, H - C - CH3, |, Cl, , (c), , CH3, |, H - C - Cl, |, C2 H5, , CH 3– CH 2–CH–CH 2NH 2, , 9., , 10., , (d) CH3CHO + HCN ¾¾, ® CH3CH (OH) CN, Which of the following reactions is an example, of nucleophilic substitution reaction? [2009], (a) 2 RX + 2 Na ® R – R + 2 NaX, (b) RX + H2 ® RH + HX, (c) RX + Mg ® RMgX, (d) RX + KOH ® ROH + KX, In an SN2 substitution reaction of the type, -, , -, , [2008], ® R - Cl + Br, R - Br + Cl ¾¾¾, which one of the following has the highest, reactivity rate ?, DMF, , CH3, |, CH3 - C - CH 2 Br, |, CH, 3, , -, , The mechanisms of reactions (i) and (ii) are, respectively :, (a) SN1 and SN2, (c) SN2 and SN2, , (c), , CH3 - CH - CH 2 Br, |, CH3, , 14., , (b), , C2 H5, |, Cl - C - CH3, |, H, , (d), , C2 H 5, |, H3C - C - Cl, |, H, , For (i) I–, (ii) Cl–, (iii) Br–, the increasing order of, nucleophilicity would be, [2007], –, –, –, –, –, (a) Cl < Br < I, (b) I < Cl < Br–, (c) Br– < Cl– < I–, , (d) I– < Br– < Cl–
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243, , Haloalkanes and Haloarenes, 15., , 16., , Which of the following is not chiral ?, (a) 2, 3- Dibromopentane, (b) 3-Bromopentane, (c) 2-Hydroxypropanoic acid, (d) 2-Butanol, The chirality of the compound, , [2006], (c), , H3C, , 17., , 18., , 20., , H is, Cl, , H, (d), HO, , (CH 3 )3 C - Cl, , (b), , (c), , CH 3CH 2Cl, , (d) CH 2 = CHCH 2 Cl, , 2®Y, CH 3 CH 2 Cl ¾¾ ¾, ¾® X ¾¾ ¾, ¾, , H, H, , CH3, , CH3, , CH3, , CH3, , H, HO, , OH, HO, and, H, H, CH3, , CH3, , NaCN, , Ni / H, , Acetic, , Y ¾¾ ¾, ¾® Z, anhydride, , Z in the above reaction sequence is, , [2002], , (a) CH3CH2CH2NHCOCH3, (b) CH3CH2CH2NH2, (d) CH3CH2CH2CONHCOCH3, 23., , When CH3CH2CHCl2 is treated with NaNH2,, the product formed is, [2002], (a) CH3 — CH = CH2, (b), , CH 3 — C º CH, , (c), , CH3CH2CH, , (d) CH3CH2CH, 24., , CH3, , H, HO, and, HO, OH, , CH3, , 22., , CH 2 = CHCl, , CH3, , OH, OH, , Reactivityorder of halides for dehydrohalogenation, is, [2002], (a) R – F > R – Cl > R – Br > R –I, (b) R –I > R – Br > R – Cl > R – F, (c) R –I > R – Cl > R – Br > R – F, (d) R – F > R –I > R – Br > R – Cl, , Which of the following pairs of compounds are, enantiomers?, [2003], , H, HO, , CH3, , (c) CH3CH2CH2CONHCH3, , (a), , (b), , OH, H, and, H, H, , 21., , , is C13H12., , How many structural isomers are possible when, one of the hydrogens is replaced by a chlorine, atom?, [2004], (a) 6, (b) 4, (c) 8, (d) 7, Which of the following is least reactive in a, nucleophilic substitution reaction., [2004], , HO, H, , CH3, , CH3, [2005], , (a) R, (b) S, (c) E, (d) Z, Which of the following undergoes nucleophilic, substitution exclusively by SN1 mechanism?, [2005], (a) Ethyl chloride, (b) Isopropyl chloride, (c) Chlorobenzene, (d) Benzyl chloride, The molecular formula of diphenylmethane,, , (a), , H, H, , OH, HO, and, HO, H, , CH3, , CH2, , 19., , CH3, , CH3, , CH3, , Br, , C, , H, HO, , NH2, , NH2, , Cl, , NH2, , An organic compound A (C4H9Cl) on reaction, with Na/diethyl ether gives a hydrocarbon which, on monochlorination gives only one chloro, derivative, then A is, [2001], (a) tert-butyl chloride, (b) sec-butyl chloride, (c) isobutyl chloride, (d) n-butyl chloride
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EBD_8336, 246, , 43., , CHEMISTRY, Which of the following compounds undergoes, nucleophilic substitution reaction most easily ?, , (I), , [2011 M], 47., , Cl, , Cl, (b), , (a), , NO2, CH3, , 48., , Cl, Cl, , (c), , (d) III < II < I < IV, , Benzene reacts with CH3Cl in the presence of, anhydrous AlCl3 to form:, [2009], (a) chlorobenzene, , (b) benzylchloride, , (c) xylene, , (d) toluene, , The replacement of chlorine of chlorobenzene, to give phenol requires drastic conditions, but, the chlorine of 2,4-dinitrochlorobenzene is, readily replaced since,, [1997], , (c) nitro groups donate electrons at meta, position, , The reaction of toluene with Cl2 in presence of, FeCl3 gives 'X' and reaction in presence of light, gives ‘Y’. Thus, ‘X’ and ‘Y’ are :, [2010], , (d) nitro groups withdraw electrons from, ortho/para positions of the aromatic ring, , (a) X = Benzal chloride, Y = o – Chlorotoluene, 49., , (c) X = o –and p – Chlorotoluene,, , Benzene reacts with n-propyl chloride in the, presence of anhydrous AlCl3 to give [1993], (a) 3 – Propyl – 1 – chlorobenzene, , Y = Trichloromethyl – benzene, , (b) n-Propylbenzene, , (d) X = Benzyl chloride, Y = m – Chlorotoluene, , 46., , (c) IV < III < I < II, , (b) nitro groups withdraw electrons from the, meta position of the aromatic ring, , (b) X = m – Chlorotoluene, Y = p –Chlorotoluene, , 45., , (IV), , (b) II < III < I < IV, , (a) nitro groups make the aromatic ring electron, rich at ortho/para positions, , (d), OCH3, , 44., , (III), , (II), , (a) I < II < IV < III, , (c) No reaction, , Which one is most reactive towards SN 1 reaction ?, , (d) Isopropylbenzene., , (a), , C6 H5CH(C6 H5 )Br, , (b), , C6 H5CH(CH3 )Br, , Chlorobenzene reacts with Mg in dry ether to, give a compound (A) which further reacts with, ethanol to yield, [1993], , (c), , C6 H5C(CH3 )(C6 H5 )Br, , (a) Phenol, , (b) Benzene, , (d), , C6 H5CH 2 Br, , (c) Ethylbenzene, , (d) Phenyl ether., , [2010], , The correct order of increasing reactivity of, C – X bond towards nucleophile in the following, compounds is:, [2010], , X, , X, , 50., , 51., , Which chloro derivative of benzene among the, following would undergo hydrolysis most, readily with aqueous sodium hydroxide to, furnish the corresponding hydroxy derivative?, [1989], , NO2, , NO2, , (CH3)3 C – X, (CH3)2CH – X, NO2, , (a), , O2 N, , Cl, NO2
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247, , Haloalkanes and Haloarenes, (b) O2N, , Cl, , Cl, , (d) Cl, (c), , Cl, , CH2Cl, , Cl, , Me2N, , 53., , (d) C6H5Cl., Topic 3: Some Important Polyhalogen, Compo unds, 52., , C, , Trichloroacetaldehyde, CCl3CHO reacts with, chlorobenzene in presence of sulphuric acid, and produces:, [2009], , 54., , Which of the following is responsible for, depletion of the ozone layer in the upper strata, of the atmosphere?, [2004], (a) Polyhalogens, , (b) Ferrocene, , (c) Fullerenes, , (d) Freons, , Chloropicrin is obtained by the reaction of, [2004], (a) steam on carbon tetrachloride, , Cl, , (b) nitric acid on chlorobenzene, (c) chlorine on picric acid, (d) nitric acid on chloroform, (a), , Cl, , C, , 55., , Cl, , H, , (a) Phosgene, (b) Calcium hypochlorite, , OH, (b) Cl, , Industrial preparation of chloroform employs, acetone and, [1993], , C, , (c) Chlorine gas, (d) Sodium chloride., , Cl, 56., , Cl, , Phosgene is a common name for, , [1988], , (a) phosphoryl chloride, (c), , Cl, , CH, , (b) Thionyl chloride, , Cl, , (c) Carbon dioxide and phosphine, , CCl3, , (d) Carbonyl chloride., , ANSWER KEY, 1, 2, 3, 4, 5, 6, , (d) 7, (b) 8, (b) 9, (c) 10, (N) 11, (a) 12, , (a), (c), (d), (d), (a), (a), , 13, 14, 15, 16, 17, 18, , (b), (a), (b), (a), (d), (b), , 19, 20, 21, 22, 23, 24, , (b), (b), (b), (a), (b), (a), , 25, 26, 27, 28, 29, 30, , (d), (c), (c), (a), (c), (d), , 31, 32, 33, 34, 35, 36, , (b), (b), (c), (a), (b), (c), , 37, 38, 39, 40, 41, 42, , (b), (a), (b), (b), (b), (c), , 43, 44, 45, 46, 47, 48, , (a), (c), (c), (a), (d), (d), , 49, 50, 51, 52, 53, 54, , (d) 55, (b) 56, (a), (c), (d), (d), , (b), (d)
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249, , Haloalkanes and Haloarenes, 9., , 10., , (d) In nucleophilic substitution, a nucleophile, provides an electron pair to the substrate and, the leaving group departs with an electron pair., Nu: + R—X ¾® R—Nu + X:, (d) SN2 mechanism is followed in case of, primary and secondary halides i.e., SN2 reaction, is favoured by small groups on the carbon atom, attached to halogens so reactivity order is:, , CH3 CH2 Br > CH3 CH2 CH2 Br >, , 2, , C2H5, 3, , 13. (b), , 1, , H, , 4, R-configuration, , 14. (a) Nucleophilicity increases down the periodic, table : I - > Br - > Cl - > F-, , CH3, |, CH3– CH – CH2Br > CH3 – C – Br, |, |, CH3, CH3, , Nucleophilicity governs the kinetics of a reaction, while basicitydetermines its thermodynamics., Since, electronegativity decreases down the group, (or electropositivity increases), nucleophilicity, also increases down the group. But the order of, basicity is:, , i.e. option (d) is correct., , I– < Br– < Cl– < F–, , 11. (a) In the molecule, CH3 CH = CH CH 2 CH- CH 3,, |, Br, , the number of stereoisomers is given by sum of, geometrical isomers (because of presence of C = C), and optical isomers (because of presence of chiral, carbon atom)., Number of geometrical isomers = 2 (one C = C is, present)., Number of optical isomers = 2 (one chiral carbon, atom)., Total number of stereoisomers = 2 + 2 = 4, 12. (a) Compounds which do not show optical, activity inspite of the presence of chiral carbon, atoms are called meso-compounds., The absence of optical activity in these compound, is due to the presence of a plane of symmetry in, their molecules. e.g. meso-tartaric acid is optically, inactive., , COOH, , 15. (b), | | | |, |, (a) C – C*– C*– C – C –, | | |, | |, Br Br, |, , |, , |, , |, , |, , |, , |, , |, |, Br, , |, , OH, , H, , OH, , (Plane of symmetry), , [Chiral], , (b) – C – C – C – C – C –, , [Not chiral], , OH, | |, *, (c) – C – C – COOH, | |, , [Chiral], , OH, | |, | |, *, (d) – C – C – C – C –, | | |, |, , [Chiral], , * marked are chiral carbons., 16. (a) Clockwise rotation., 1 Br, H3C, , H, , COOH, , Cl — C — CH3, , C, 3, , H 4, Cl 2, , Hence configuration is R., If the eye travel in a clockwise direction, the, configuration is (R) as the order of priority is, Br > Cl > CH3 > H.
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EBD_8336, 250, , CHEMISTRY, , 17. (d) SN1 reaction is favoured by heavy groups, on the carbon atom attached to halogen i.e, Benzyl > allyl > tertiary > secondary > primary, > alkyl halides, Å, , CH2Cl, , OH, , H, HO, , CH2, , – Cl, , H, , Obtained from SN1 path., This molecule is resonance stabilised., 18. (b) In diphenylmethane monochlorination at, following positions will produce structured, isomers, , and, , HO, , H, OH, , H, , CH3, , ¾®, (Benzyl cation), , CH3, , CH3, , CH3, , 21. (b) The order of atomic size of halogens, decrease in the order I > Br > Cl > F, i.e on moving down a group atomic size increases., Further the bond length of C-X bond decreases, in the order, C – I > C – Br > C – Cl > C – F, and hence the bond dissociation energy, decreases in the order, , CH2, , R – F > R – Cl > R – Br > R – I, , 1, , 4, , 2 3, Cl, , hence R – I being a weakest bond break most, easily. Hence R – I is most reactive., 22. (a), , CH2, , Ni/H, , NaCN, , 2®, CH3CH 2 Cl ¾¾¾¾, ® CH3CH 2CN ¾¾¾¾, , (1), , (X), , CH2, , CH 3 CH 2 CH 2 NH 2 ¾¾¾¾¾¾, ®, (CH 3CO) 2 O, , (Y), , Cl, , (2), , CH3CH 2 CH 2 NHCOCH 3, (Z), , CH2, , NaNH 2, 23. (b) CH 3 — CH 2 — CHCl 2 ¾¾, ¾¾®, , Cl, , D, , (3), Cl, , NaNH, , 2 ® CH — C º CH, CH 3 — CH = CHCl ¾¾ ¾¾, 3, , D, , CH2, (4), , Final Product, , CH3, , CH3, , CH3 – C – Cl + 2Na + Cl – C – CH3, CH3, , CH3, , t-Butyl chloride, , CH3CH3, , CH3CH3, CH3 – C – C –CH2Cl, CH3CH3, , ¾¾®, , _, , H 2C – CH=Cl +, H 2 C==CH –Cl, 20. (b) Compound which are mirror image of each, other and are not superimposable are termed as, enantiomers., , 24. (a), , Wurtz, ¾¾®, Rxn, , 19. (b) H 2 C = CHCl is capable of showing, resonance which develops a partial double bond,, character to C–Cl bond, thereby making it less, reactive towards nucleophilic substitution., , CH3– C – C – CH3, , Mono, Clorination, , CH3 CH3
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251, , Haloalkanes and Haloarenes, hn, 25. (d) Cl2 ¾¾®, Cl + Cl, •, , •, , Cl / hn, , CH3 — CH2 — CH2 — CH3¾¾2¾¾®, CH3, , CH3, , H — C — Cl + Cl — C — H, , C2H5, , C2H5, , d, , l, , Racemic mixture, 50% d form + 50% l form, , Chlorination of alkane takes place via free radical, formation. Cl· may attack on either side and gives, a racemic mixture., , 26. (c) Chiral molecules are those molecules which, have atleast one asymmetric carbon atom (a, carbon atom attached to 4 different groups). This, is true in case of 3-methylpentanoic acid., H, |, , CH3, , 27. (c) Potassium ethoxide is a strong base, and 2bromopentane is a 2º bromide, so elimination, reaction predominates, OC H -, , 2 ¾, 5®, CH 3 CH (Br )CH 2 CH 2 CH 3 ¾¾ ¾, , CH3CH = CHCH 2CH3 + CH 2 = CHCH2CH 2CH3, Pentene -1(min or) cis, , Since, trans- alkene is more stable than cis, thus, trans-pentene -2 is the main product., 28. (a) Due to absence of a chiral, C-atom. D — CH 2 — CH 2 — CH 2 Cl, molecule is not a chiral molecule., 29. (c) The compound is diethyl ether (CH3CH2)2O, which is resistant to nucleophilic attack by, hydroxyl ion due to absence of double or triple, bond, whereas all other compounds given are, unsaturated., O, ether, , ||, , CH3 - C - OCH3, Methyl acetate, , O, CH3 - C º N, Acetonitrile, , OH, , –, , C6H13, , C6H13, , H, , H C Br ¾® HO -----C -----Br ® HO, CH3, (–) 2-Bromooctane, , C H + Br–, , CH3, (+) 2–octanol, , CH3, Transition state, , In this reaction Walden Inversion takes place so, it is an example of SN2-reaction., 31. (b) H 3C — CH 2 — *CH — CH 2 — CH 2 — CH3, |, , CH3, 3-Methylhexane, , Due to presence of four different groups on, carbon,(C*) it is chiral, 32. (b) Number of active hydrogen in a compound, corresponds to the number of moles of CH4, evolved per mole of the compound., , 3, -OH, - NH2 , - SH, - OH or - C º CH ¾¾¾¾, ® CH 4 (2CH 4 from, , |, , C2 H5 OC2 H5, , C6H13, , CH MgI, , C2H5 – C - CH2 COOH, , Pentene - 2(major) trans, , 30. (d) When (–) 2-bromooctane is allowed to react, with sodium hydroxide under given conditions,, where second order kinetics is followed, the, product obtained is (+) 2-octanol., , ||, , CH3 - C - NH 2, acetamide, , ® CH 4 ( 2CH 4 from - NH 2 ), , 33. (c) Racemic mixtures are resolved by the, formation of Diastereomers., Enantiomers have same physical and chemical, properties except rotation of plane polarized light., Diasteromers have different physical and chemical, properties., , 34. (a) Resolution., 35. (b) At high temp. i.e., 400°C substitution occurs, in preference to addition., Cl , 400°C, , CH 3 CH = CH 2 ¾¾2¾ ¾¾®, - HCl, , ClCH 2 CH = CH 2, H, |, , 36. (c) H3C - CH 2 - C*- CH3, |, Cl, The compound containing a chiral carbon atom, i.e., (a carbon atom which is attached to four, different substituents is known as a chiral carbon, atom) is optically active.
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EBD_8336, 252, , CHEMISTRY, , 37. (b) Aryl halides do not give substitution, reactions due to partial double bond, character between carbon and halogen., CH3, CCl3, 38., , Br, , 3Cl, , (a), , 2, ¾¾®, D, , 2, ¾¾®, Fe, , CCl3, , CH3, , Br, , 39., , Zn, ¾¾®, HCl, , Br, m-Bromotoluene, , (b) (1) CH3CH2CH2–Br + KOH, ® CH3CH=CH2 + KBr + H2O, This is dehydrohalogenation reaction which is, an example of elimination reaction., (2), , H3C, , C, , Biphenyl compounds are devoid of individual, chiral carbon atoms, but o-substituted biphenyl, show optical isomerism. This is due to the fact, that, both the rings are not in the same plane because, of steric hindrance of the groups present in oposition. Also, these groups restrict the rotation, around the single bond and hence they must exist, in two non-super imposable mirror images of each, other., , 41., 42., , (b), (c) Methyl group is ortho para directing but, due to steric hindrance effect, generated by two, CH3 groups substitution will not take place on, position (I). Hence, only two products are, possible., CH3, , H3C, CH3, CH3, + KOH ®, C, + KBr, , Br, , OH, , I, , III, , CH3, , II, , i.e., , CH3, , Replacement of Br – by OH– is substitution, reaction thus, it is a nucleophilic substitution, reaction., Br, (3), + Br2 ®, Br, Above reaction involves addition of Br 2 across, double bond. Thus, it is called addition reaction., , CH3, Br2, , ¾¾®, FeBr3, , CH3, , Br, , Br, (A), CH3, , CH3, , +, CH3, , 40. (b), In option (a), there is no chiral carbon., In option (c), there exist a center of symmetry., In option (d), there exist a plane of symmetry., In option (d), both the rings are not in the same, plane and the rotation is restricted around the, single bond., , Br Br, , I I, , Restricted rotation, , 43. (a) To compare the rates of substitution, in, chlorobenzene, electron-withdrawing group, and, chlorobenzene having electron-releasing group,, we compare the structures carbanion I (from, chlorobenzene), II (from chlorobenzene, containing electron-withdrawing group) and III, (from chlorobenzene containing electronreleasing group)., Cl, Z, Z, Cl, –, , –, , I, , II, , G
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EBD_8336, 254, , CHEMISTRY, produce free atomic chlorine which results in, decomposition of ozone., , Anhyd., ®, 49. (d) C 6H6 + CH 3CH 2CH 2CI ¾¾¾¾, , AlCl3, , CH 3, |, C6 H5 - CH - CH 3, Isopropyl benzene, , 54. (d) Chloropicrin is nitrochloroform. It is obtained, by the nitration of chloroform with HNO3., , Mg, , ® C6 H5 MgCl, 50. (b) C6 H5 Cl ¾¾¾¾, dry ether, , HNO, , 3 ® O NCCl, HCCl 3 ¾¾ ¾, ¾, 2, 3, , (A), ® C6 H 6 + CH3CH 2OMgCl, ¾¾¾¾¾¾, , 51. (a) Cl in 2, 4, 6-trinitrochlorobenzene is activated, by three NO2 groups at o- and p-positions and, hence, undergoes hydrolysis most readily., (c), Cl, Cl 3C–C=O +, , H, , Cl, , conc. H SO, , 2, 4, ¾¾¾¾® Cl2C–CH, , Cl, , Chloropicrin, , Chloroform, , CH 3CH 2OH, , 52., , Chlorofluorocarbons are chemically inert and hence, do not react with any substance with which they, come in contact and therefore, float through the, atmosphere and as a result enter the stratosphere., , Cl, , Cl, DDT, , Chloropicrin is poisonous and used as an, insecticide and a war gas., 55. (b) By distilling ethanol or acetone with a paste, of bleach in g powder (laboratory and, commercial method)., CaOCl 2 + H 2 O ¾, ¾® Ca (OH ) 2 + Cl 2, , Cl2, so obtained acts as a mild oxidising as well, as chlorinating agent, (i) CH 3COCH 3 + 3Cl 2, ¾¾® CCl 3 .CO.CH 3 + 3HCl, (Chlorination), , 53. (d) Chlorofluorocarbons, e.g. CF2Cl2, CHF2Cl2,, HCF2 CHCl 2 . These are non-inflammable,, colourless and stable upto 550ºC. These are, emitted as propellants in aerosol spray,, refrigerators, fire fighting reagents etc. They, absorb UV-rays in the stratosphere and, , (ii) 2CCl3 .CO.CH 3 + Ca (OH) 2 ¾¾®, 2CHCl 3 + (CH 3COO) 2 Ca, (Hydrolysis), , 56. (d) Carbonyl chloride (COCl2).
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EBD_8336, 256, , 5., , 6., , CHEMISTRY, (a) (I), (III) and (IV) only, (b) (I) and (II) only, (c) (IV) only, (d) (III) and (IV) only, Number of isomeric alcohols of molecular, formula C6H14O which give positive iodoform, test is, [NEET Kar. 2013], (a) two, (b) three, (c) four, (d) five, In the following sequence of reactions [2012], KCN, , (a), (b), (c), 12., , H O+, , 3 ®, CH3 - Br ¾¾¾® A ¾¾¾¾, , 13., , LiAlH 4, ® C, the end product (C) is :, B ¾¾¾¾, ether, , 7., , 8., , (a) Acetone, (b) Methane, (c) Acetaldehyde, (d) Ethyl alcohol, When glycerol is treated with excess of HI, it, produces:, [2010], (a) glycerol triiodide (b) 2–iodopropane, (c) allyl iodide, (d) propene, Consider the following reaction,, [2009], PBr, , alc.KOH, , 3 ® X ¾¾¾¾® Y, ethanol ¾¾¾, , 14., , (i) H 2SO 4 room temperature, ® Z;, ¾¾¾¾¾¾¾¾¾¾¾¾¾, (ii) H 2O, heat, , 9., , 10., , 11., , the product Z is:, (a) CH3CH2 – O – CH2 – CH3, (b) CH3 – CH2 – O – SO3H, (c) CH3CH2OH, (d) CH2 = CH2, H2COH · CH2OH on heating with periodic acid, gives:, [2009], CHO, (a) 2 HCOOH, (b) |, CHO, H, (d) 2 CO2, (c) 2, C=O, H, Ethylene oxide when treated with Grignard, reagent yields, [2006], (a) Tertiary alcohol, (b) Cyclopropyl alcohol, (c) Primary alcohol, (d) Secondary alcohol, Which of the following will not form a yellow, precipitate on heating with an alkaline solution, of iodine?, [2004], , 15., , CH 3CH (OH)CH 3, CH 3CH 2 CH(OH )CH 3, CH 3OH, , (d) CH 3CH 2OH, n-Propyl alcohol and isopropyl alcohol can be, chemically distinguished by which reagent?, (a) PCl5, [2002], (b) Reduction, (c) Oxidation with potassium dichromate, (d) Ozonolysis, Which of the following is correct ?, [2001], (a) On reduction of any aldehyde, secondary, alcohol is formed, (b) Reaction of vegetable oil with H2SO4 gives, glycerine, (c) Sucrose on reaction with NaCl gives invert, sugar, (d) Alcoholic iodine gives iodoform with, NaOH, Propan - 1- ol may be prepared by the reaction, of propene with, [2000], (a) H3BO3, (b) H2SO4/H2O, (c) B2H6, NaOH–H2O2, O, ||, (d) CH 3 C - O - O - H, Reaction of CH — CH with RMgX leads to, 2, , formation of, (a) RCHOHR, , O, , 2, , [1998], (b) RCHOHCH3, R, , CHCH2OH, R, The stablest among the following is, [1994], (a) CH3CH(OH)2, (b) ClCH2CH(OH)2, (c) (CH3)2 C (OH)2, (d) CCl3 CH (OH)2., Which one of the following on oxidation gives a, ketone ?, [1993], (a) Primary alcohol, (b) Secondary alcohol, (c) Tertiary alcohol (d) All of these, What is formed when a primary alcohol, undergoes catalytic dehydrogenation ? [1993], (a) Aldehyde, (b) Ketone, (c) Alkene, (d) Acid, , (c) RCH2CH2OH, , 16., , 17., , 18., , (d)
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259, , Alcohols, Phenols and Ethers, H3C, , (c), , H 3C, , C, , O, , O, , H, , 40., , CH3, (d), , 35., , H3C, , C, , OH, , O O, OH, , (c), , CH2OH, , (b), , OH, , I, , 41., , OH, , OH, , (d), 36., , 37., , CH, 42., , Among the following four compounds [2010], (i) phenol, (ii) methylphenol, (iii) meta-nitrophenol (iv) para-nitrophenol, the acidity order is :, (a) ii > i > iii > iv, (b) iv > iii > i > ii, (c) iii > iv > i > ii, (d) i > iv > iii > ii, Consider the following reaction:, [2009], Zn dust, , 38., , (c), , 39., , CH3, , NO2, , OH, , (d), CH3, NO2, When phenol is treated with CHCl3 and NaOH,, the product formed is, [2002], , II, , OH, , III, , OH, , OH, , NO2, CH3, I, , CH Cl, Anhydrous AlCl3, , 4®Z, ¾¾¾¾¾¾¾, The product Z is:, (a) benzaldehyde, (b) benzoic acid, (c) benzene, (d) toluene, Which one of the following compounds is most, acidic?, [2005], OH, (a) Cl–CH2–CH2–OH, (b), , OH, , OH, , (b) I > III > II, (a) III > I > II, (c) II > III > I, (d) I > II > III, The ionization constant of phenol is higher than, that of ethanol because :, [2000], (a) Phenoxide ion is bulkier than ethoxide, (b) Phenoxide ion is stronger base than, ethoxide, (c) Phenoxide ion is stabilized through, delocalization, (d) Phenoxide ion is less stable than ethoxide, Consider the following phenols :, OH, , 3, ® X ¾¾¾¾¾¾¾, ®Y, Phenol ¾¾¾¾, , Alkaline KMnO, , OH, , H, , Which one of the following compounds has the, most acidic nature?, [2010], (a), , (a) Benzaldehyde, (b) Salicylaldehyde, (c) Salicylic acid, (d) Benzoic acid, The correct order of acidic strength for following, compounds will be, [2001], , 43., , 44., , NO2, II, , III, , IV, , The decreasing order of acidity of the above, phenols is, [1999], (a) III > IV > II > I, (b) II > I > IV > III, (c) I > IV > II > III, (d) III > IV > I > II, 1-Phenylethanol can be prepared by the reaction, of benzaldehyde with, [1997], (a) Ethyl iodide and magnesium, (b) Methyl iodide and magnesium, (c) Methyl bromide and aluminium bromide, (d) Methyl bromide, Correct increasing order of acidity is as follows:, [1994], (a) H2O, C2H2, H2CO3, Phenol, (b) C2H2, H2O, H2CO3, Phenol, (c) Phenol, C2H2, H2CO3, H2O, (d) C2H2, H2O, Phenol, H2CO3
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EBD_8336, 262, , 1, 2, 3, 4, 5, 6, 7, , CHEMISTRY, , (d ), (c), (b ), (a), (c), (d ), (b ), , 8, 9, 10, 11, 12, 13, 14, , 15, 16, 17, 18, 19, 20, 21, , (c), (c), (c), (c), (c), (d ), (c), , (c), (d ), (b ), (a), (d ), (b ), (d ), , ANSW E R K E Y, (c) 2 9 (a) 3 6, (a) 3 0 (d ) 3 7, (b ) 3 1 (b ) 3 8, (c) 3 2 (b ) 3 9, (d ) 3 3 (c) 4 0, (c) 3 4 (c) 4 1, (c) 3 5 (b ) 4 2, , 22, 23, 24, 25, 26, 27, 28, , 43, 44, 45, 46, 47, 48, 49, , (b ), (b ), (c), (b ), (a), (c), (a), , (b ), (d ), (d ), (d ), (c), (b ), (d ), , 50, 51, 52, 53, 54, 55, 56, , (b ), (d ), (b ), (d ), (a), (d ), (a), , 57, 58, 59, 60, 61, , (b ), (c), (c), (a), (a), , Hints & Solutions, 1., , 2., , Å, , OH, O, |, P, Cu, ®, R, —, C, —R, (d) R — CH — R ¾¾¾, 300°C, (Secondary, Alcohol), , R — O— ZnCl2 ® R Å + [HOZnCl2 ]d, |, , H, (I), , (Ketone), , –, Na, (c) C2H5OH ¾¾®, C2H5O Na+, , (A), , Carbocation is formed as intermediate in the, SN mechanism which these reaction undergoes., 1, (III) Tertiary carbocation easily formed due to, the stability., , (B), , PCl5, , OH, , C2H5Cl, , |, , (C), , CH3 — C — CH3 ¾¾, ®, , SN2, , –, C2H5O Na+ + C2H5Cl ¾¾® C2H5OC2H5, , (B), , 3., , |, , CH3, , (C), , Å, , So, the correct option is (c), (b) Four primary alcohols of C 5H11 OH are, possible. These are:, (i) CH 3CH 2 CH 2 CH 2 CH 2 OH, (ii) CH3CH 2—CH—CH2OH, , CH3 — C — CH 3 + H 2 O, |, , CH3, (IV) In the presence of ZnCl2, 2° carbocation is, formed from (CH3 ) 2 — C— OH, |, , CH3, , Å, , i.e., CH3 — C H— CH3, , (iii) CH3—CH—CH 2CH2OH, , ZnCl2 is a lewis acid and interact with alcohol., In the absence of ZnCl2 formation of primary, carbocation is difficult., , CH3, , CH3, 5., , (iv) CH3—C—CH2OH, CH3, 4., , (a), Å, , CH 3 — CH 2OH + ZnCl 2 ® R — O— ZnCl 2, |, , (R = CH3 — CH 2 —), , H, , H, (I), , (c) Compound containing CH3 CH(OH) or, CH3CO–group give positive iodoform test., OH, |, CH3 CH— CH 2 CH 2 CH 2 CH 3, OH, |, CH3 — CH — CH CH 2CH 3 ,, |, CH3
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265, , Alcohols, Phenols and Ethers, Trans isomer can not react with acetone to form, cyclic ketal., , 30., , (d), , 34., , OH, , d-, , CHCl3 + NaOH ƒ : CCl3 + H 2 O, d-, , –, , –, , O, Cl, , CH3, , Cl, , C, –, , ¾®, , CH3, , Cl + H O ¾, 2, , ¾®, , Cl, , CH 3CH 2CH 2Cl, , d-, , O, , + :C, , AlCl3, , Zn, D, , : CCl3 ¾¾, ® : CCl2 + Cl, , –, , – OH, , C, , O, , O, , H, , CH3, , CH, , O2, , –, , O, C=O, , CCl2, , –, OH, – 2HCl, , ¾ ¾®, , H, , 35., , H, , Therefore, functional group – CHO is introduced., (b), , (b) Phenol is most acidic because its conjugate, base is stabilised due to resonance, while the, rest three compounds are alcohols, hence, their, corrosponding conjugate bases do not exhibit, resonance., OH, , OH, , CH3, , –, , O, , 31., , (c), , 36., , O—H, Å, , Å, , +H ¾®, , (b), , >, NO2, NO2, , H, , ¯, H—CH2, H, , Å, , (d), , (iv), , ¾(c), ¾, , OH, , OH, , ¾¾®, , ¾, , ®, , ), (a, , (iii), , (– I and – M effects, (only – I effect), both increase acidity), , CH3, , H H, , ¾, , OH, , ®, , >, , >, , CH3, , (i), , 32., , The intermediate is a 3° carbocation, which is, very stable. Hence, rearrangement will not take, place and product mentioned in option (b) will, not form., (b) Phenols react with alkyl halides in alkaline, medium to form ethers. Therefore,, , OCH3, , OH, (i) NaOH, , ¾¾¾®, , (ii), (+ I effect of CH3 group decreases, acidity), \ Correct choice : (b), OH, 37., , CH Cl, , Zn, , ¾®, , (b), , ¾ ¾3 ®, , Phenol, , X, , CH3, , (ii) CH3I, , 33., , (c) o-nitrophenol will not be soluble in NaHCO3., Due to intramolecular hydrogen bonding,, hydrogen on OH is strongly bound. So, it cannot, be have as an acid and can not react with sodium, bicarbonate., , anhy, AlCl3, , COOH, alc., , ¾¾®, KMnO 4, , Y, , Z
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EBD_8336, 266, , CHEMISTRY, , 38. (c) Phenols are more acidic than alcohols as, they are resonance stabilised whereas alcohols, are not., Further –NO2 is an electron withdrawing group, which increases acidic character and facilitates, release of proton, whereas –CH3 is an electron, donating group which decrea ses acidic, character, thus removal of H+ becomes difficult., CHO, , CHCl, , 3, ¾¾®, , 39. (b), , Salicylaldehyde, , If CCI4 is used in place of CHCl3, then the, product is salicylic acid., 40. (a) Electron releasing group (–CH3)decreases, acidity while electron withdrawing group, (–NO2)increases acidity., 41. (c) The acidic nature of phenol is due to the, formation of stable phenoxide ion in solution, C 6 H5 O, C6 H 5OH + H 2 O , , -, , +, , Phenoxide ion, , H 3O, , +, , The phenoxide ion is stable due to resonance., O, , –, , O, , O, , ••, ••, , O–, , OH, d, , d, , –, , d, , –, , ¾, , ¾, , CH 3 MgI, , ¾ ®, , C O ¾®, CH O MgI, , CH3, , H O, , 2 ¾, CHOH ¬¾¾, , CH3, , - Mg ( OH ) I, , (1-Phenyl ethanol), , NaOH, , (Riemer Tiemann reaction), , Dry ether, , C6H5, CH3MgI +, H, C6H5, , C6H5, , OH, , OH, , 43. (b) CH 3 I + Mg ¾, , O, ••, , –, , The negative charge is delocalized in the, benzene ring which is a stabilizing factor in the, phenoxide ion and increase acidity of phenol,, wheras no resonance is possible in alkoxide ions, (RO–) derived from alcohol. The negative charge, is localized on oxygen atom. Thus, alcohols are, less acidic., 42. (a) Electron withdrawing group (–NO 2 ), increases the acidity while electron releasing, group (–CH3, –H) decreases acidity. Also effect, will be more if functional group is present at para, position than ortho and meta position., , 44. (d) Such questions can be solved by considering, the relative basic character of their conjugated bases, which for H2O, C2H2, H2CO3 and C6H5OH are, -, , OH, HC º C : , HCO 3 , C 6 H 5 O, More the possibility for the dispersal of the, negative charge, weaker will be the base. Thus, the relative basic character of the four bases is, HCO 3 < C 6 H 5 O < OH < HC º C, -, , -, , Equivalent, , -, , -, , -, , Non-equivalent, , -, , -, , Oxygen can accommodate, , resonating structures, , (–) charge easily, , Due to resonance, , Thus, the acidic character of the four, corresponding acids wil be, HC º CH < H 2O < C6 H5OH < H 2CO3, 45. (d) Electron-donating groups (– OCH3, – CH3, etc.) tend to decrease and electron withdrawing, groups (–NO2) tend to increase the acidic, character of phenols. Since –OCH3 is a more, powerful electron-donating group than –CH3, group, therefore, p-methylphenol is slightly, more acidic than p-methoxyphenol while, p-nitrophenol is the strongest acid. Thus,, op t i o n ( d) , i . e . p - m e t h ox yp h en ol ,, p-methylphenol, p-nitrophenol is correct., 46. (d) Wit h Br 2 wa ter, phen ol gi ves 2, 4,, 6- tribromophenol., OH, , + 3Br2 (excess), , H2O, , Br, , OH, , ¾®, , Br, + 3HBr, , Br, 2, 4, 6 Tribromophenol
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271, , Aldehydes, Ketones, and Carboxylic acids, 13., , Of the following, which is the product formed when, cyclohexanone undergoes aldol condensation, followed by heating ?, [2017], , 16., , An organic compound 'X' having molecular, formula C5H10O yields phenyl hydrazone and, gives negative response to the Iodoform test, and Tollen's test. It produces n-pentane on, reduction. 'X' could be :[2015], (a) 2-pentanone, (b) 3-pentanone, (c) n-amyl alcohol, (d) pentanal, , 17., , Treatment of cyclopentanone, , (a), O, (b), , OH, (c), OH, , OH, , 18., , OH, , methyl lithium gives which of the following, species?, [2015], (a) Cyclopentanonyl cation, (b) Cyclopentanonyl radical, (c) Cyclopentanonyl biradical, (d) Cyclopentanonyl anion, The enolic form of ethyl acetoacetate as below, has:, [2015], H3C, , (d), , C, OH, , OH, 14., , 15., , The correct statement regarding a carbonyl, compoun d with a hydrogen atom on its, alphacarbon, is :, [2016], (a) a carbonyl compound with a hydrogen atom, on its alpha-carbon never equilibrates with, its corresponding enol., (b) a carbonyl compound with a hydrgen atom, on its alpha-carbon rapidly equilibrates, with its corresponding enol and this, process is known as aldehyde-ketone, equilibration., (c) a carbonyl compound with a hydrogen atom, on its alpha-carbon rapidly equilibrates, with its corresponding enol and this, process is known as carbonylation., (d) a carbonyl compound with a hydrogen atom, on its alpha-carbon rapidly equilibrates with its, corresponding enol and this process is known, as keto-enol tautomerism., The product formed by the reaction of an, aldehyde with a primary amine is, [2016], (a) Schiff base, (b) Ketone, (c) Carboxylic acid, (d) Aromatic acid, , 19., , 20., , =O with, , H, C, , C, , O, , H3C, , OC2H5, , C, , H2, C, , C, , O, , OC2H5, , O, , (a) 16 sigma bonds and 1 pi - bond, (b) 9 sigma bonds and 2 pi - bonds, (c) 9 sigma bonds and 1 pi - bond, (d) 18 sigma bonds and 2 pi - bonds, Reaction of a carbonyl compound with one of, the following reagents involves nucleophilic, addition followed by elimination of water. The, reagent is :, [2015 RS], (a) a Grignard reagent, (b) hydrazine in presence of feebly acidic solution, (c) hydrocyanic acid, (d) sodium hydrogen sulphite, Which one is most reactive towards Nucleophilic, addition reaction?, [2014], CHO, , (a), , CHO, , CHO, , (c), , COCH3, , (b), , (d), CH3, , NO2
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EBD_8336, 272, , 21., , CHEMISTRY, The order of stability of the following tautomeric, compounds is :, [NEET 2013], OH, , (c), , O, , |, , ||, , CH 2 = C - CH 2 - C - CH 3, , (d) (CH3)2C, , I, , 24., , O, O, ||, ||, , CH3 - C- CH 2 - C- CH 3, II, , OH, , O, , |, , 25., , ||, , CH 3 - C = CH - C - CH 3, , III, , 22., , (a) III > II > I, (b) II > I > III, (c) II > III > I, (d) I > II > III, Predict the product in the given reaction. [2012], , CHO, 26., , 50 % KOH, ¾¾¾®, , Cl, , Cl, , CH2OH, , –, , COO, , +, , Cl, , 23., , OH, OH, , OH, (c), , 27., , Cl, CH2OH, +, , (b), , (d), , CH2COO–, , CH2OH, +, , (a), , Cl, , CH2OH, , +, , 28., , COO–, , H3C, , CH3CH2CH2, , C, O, , CH3, , (b) CH3CH2CH2, , C, , CH 2CH 2CH 3, , 29., , OH, OC2H5, OC2H5, , OC2H5, , CH 3 CHO and C 6 H5 CH 2 CHO can be, distinguished chemically by :, [2012], (a) Benedict test, (b) Iodoform test, (c) Tollen’s reagent test (d) Fehling solution test, Consider the reaction :, RCHO + NH2NH2 ® RCH = N – NH2, What sort of reaction is it ?, [2012 M], (a) Electrophilic addition – elimination reaction, (b) Free radical addition – elimination reaction, (c) Electrophilic substitution – elimination, reaction, (d) Nucleophilic addition – elimination reaction, Which of the following compounds will give a, yellow precipitate with iodine and alkali?, [2012 M], (a) Acetophenone, (c) Methyl acetate, (b) Acetamide, (d) 2-Hydroxypropane, Clemmensen reduction of a ketone is carried out in, the presence of which of the following? [2011], (a) Glycol with KOH, (b) Zn-Hg with HCl, (c) Li Al H4, (d) H2 and Pt as catalyst, The order of reactivity of phenyl magnesium, bromide (PhMgBr) with the following compounds, [2011 M], , H, , OH, OH, Acetone is treated with excess of ethanol in the, presence of hydrochloric acid. The product, obtained is :, [2012], O, (a), , (CH3)2C, , C=O,, , H3C, H3C, , C=O and, , Ph, , C=O, , Ph, , I, II, III, (a) III > II > I, (b) II > I > III, (c) I > III > II, (d) I > II > III, Following compounds are given:, (1) CH3CH2OH, (2) CH3COCH3, (3), , CH 3 - CHOH, CH 3, , (4) CH3OH
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273, , Aldehydes, Ketones, and Carboxylic acids, , 30., , Which of the above compound(s), on being, warmed with iodine solution and NaOH, will give, iodoform?, [2010], (a) (1) and (2), (b) (1), (3) and (4), (c) only (2), (d) (1), (2) and (3), Which one of the following compounds will be, most readily dehydrated?, [2010], OH, (a), , H3C, , 33., , 34., , O, , O, (b), , H3C, , (c), , OH, , 35., , O, (c), , H3 C, , OH, O, , (d), 31., , Reduction of aldehydes and ketones into, hydrocarbons using zinc amalgam and conc. HCl, is called, [2007], (a) Cope reduction, (b) Dow reduction, (c) Wolff-Kishner reduction, (d) Clemmensen reduction., Which one of the following on treatment with, 50% aqueous sodium hydroxide yields the, corresponding alcohol and acid?, [2007], (a) C6H5CHO, (b) CH3CH2CH2CHO, , OH, , H3C, , 36., , Acetophenone when reacted with a base,, C2H5ONa, yields a stable compound which has, the structure., [2008], , O, ||, CH 3 - C - CH 3, , (d) C6H5CH2CHO, , The product formed in Aldol condensation is, [2007], (a) A beta-hydroxy aldehyde or a beta-hydroxy, ketone, (b) An alpha-hydroxy aldehyde or ketone, (c) An alpha, beta unsaturated ester, (d) A beta-hydroxy acid, Consider the following compounds., [2007], (i) C6H5COCl, (ii), , O2N, , COCl, , (iii) H3C, , COCl, , C = CH – C, (a), , (b), , CH3, , O, , CH – CH2C, CH3, , O, , (iv) OHC, , CH3CH3, , (c), , C–C, OH OH, , (d), , 32., , CH – CH, , OH OH, A strong base can abstract an a -hydrogen, from :, [2008], (a) alkene, (b) amine, (c) ketone, (d) alkane, , 37., , COCl, , The correct decreasing order of their reactivity, towards hydrolysis is, (a) (i) > (ii) > (iii) > (iv) (b) (iv) > (ii) > (i) > (iii), (c) (ii) > (iv) > (i) > (iii) (d) (ii) > (iv) > (iii) > (i), A carbonyl compound reacts with hydrogen, cyanide to form cyanohydrin which on, hydrolysis forms a racemic mixture of a-hydroxy, acid. The carbonyl compound is, [2006], (a) Acetone, (b) Diethyl ketone, (c) Formaldehyde, (d) Acetaldehyde
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EBD_8336, 274, , 38., , CHEMISTRY, Nucleophilic addition reaction will be most, favoured in, [2006], (a) (CH3)2C = O, (b) CH3CH2CHO, (c) CH3CHO, O, ||, , A and B in the following reactions are [2003], , R–C–R', , HCN/, ¾¾®, KCN, , O, , OH, B, ¾®, R, –, C, A, CH2NH2, R', , 42., , CN, (a), , OH, OH, , , B = LiAlH4, , A = RR'C, , , B = NH3, COOH, CN, (c) A = RR'C, , B = H 3O Å, OH, (d) A = RR'CH2CN, B = NaOH, When m-chlorobenzaldehyde is treated with 50%, KOH solution, the product(s) obtained is (are), [2003], (b), , 40., , A = RR'C, , OH OH, , 43., , 44., , OH, , OH, –, , COO, , CH2OH, , +, OH, COO–, +, Cl, , (d), Cl, , 45., , OH, CH2OH, , Cl, OH, , OH, , CH, , CH, , CH2 =CH — CH= O, , (b), , CH 2 = CH—CH=O, , (c), , CH 2 = CH — CH= O, , Cl, , δ–, , δ+, , δ+, , δ-, , δ-, , δ+, , CH 2 = C H — CH = O, , During reduction of aldehydes with hydrazine, and potassium hydroxide, the first is the, formation of :, [2000], (a) R — CH == N — NH2, (b) R — C ºº N, (c), , 46., , δ+, , (a), , (d), , (b), , (c), , ¾¾ ¾® CH 3 CH (OH )COOH, an asymmetric centre is generated. The acid, obtained would be, [2003], (a) 20 % D + 80 % L-isomer, (b) D-isomer, (c) L-isomer, (d) 50% D + 50% L-isomer, Which of the following is correct?, [2001], (a) Diastase is an enzyme, (b) Acetophenone is an ether, (c) Cycloheptane is an aromatic compound, (d) All the above, Which of the following is incorrect?, [2001], (a) NaHSO3 is used in detection of carbonyl, compound, (b) FeCl3 is used in detection of phenolic group, (c) Tollens’ reagent is used in detection of, unsaturation, (d) Fehling solution is used in detection of, glucose, Polarization of electrons in acrolein may be, written as:, [2000], δ–, , CH CH, , (a), , In the reaction, CH 3CHO + HCN ® CH 3CH (OH) CN, H.OH, , CH 3 — CH 2 — CH 2C — CH 3, , (d), 39., , 41., , R — C — NH 2, ||, , O, (d) R — CH == NH, Iodoform test is not given by, [1999], (a) 2-Pentanone, (b) Ethanol, (c) Ethanal, (d) 3-Pentanone
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EBD_8336, 276, , 60., , CHEMISTRY, O, , HCl, -3H 2O, , 3CH3COCH3 ¾¾¾¾, ®, (A), , NH, , (b), , (CH 3 )2 C = CH - CO - CH = C(CH 3 ) 2, (B), , 61., , 62., , This polymer (B) is obtained when acetone is, saturated with hydrogen chloride gas, B can be, (a) phorone, (b) formose [1989], (c) diacetone alcohol (d) mesityl oxide., If formaldehyde and KOH are heated, then we, get, [1988], (a) Methane, (b) Methyl alcohol, (c) Ethyl formate, (d) Acetylene., Formalin is an aqueous solution of, [1988], (a) Fluorescein, (b) Formic acid, (c) Formaldehyde, (d) Furfuraldehyde., , O, COOH, (c), , NH2, NH2, (d), NH2, 65., , Topic 3: Preparation and Properties of, Carboxylic Acids, 63., , The reaction that does not give benzoic acid as, the major product is, [NEET Odisha 2019], 66., , CH2OH, KMnO /H, , (a), , +, , 4, ¾¾¾¾¾®, , CH2OH, 2 2 7, ¾¾¾¾®, , COCH3, , (c), , OCOCH 3, (i) NaOCl, , (b), , ¾¾¾¾¾, +®, , CH2OH, , 64., , O 2N, , (ii) H 3 O, , (d), , (c), OCOCH3, , COOH, , (d), , strong heating, , + NH3 ¾¾¾¾¾ ®, , 67., , COOH, , COOH, [2019], , CONH, , H 3CO, , OCOCH3, , PCC, Pyridinium, chlorochromate, , ¾¾¾¾¾¾®, , The major product of the following reaction is :, , (a), , OCOCH3, , (a), , K Cr O, , (b), , Carboxylic acids have higher boiling points than, aldehydes, ketones and even alcohols of, comparable molecular mass. It is due to their, [2018], (a) Formation of intramolecular H-bonding, (b) Formation of carboxylate ion, (c) Formation of intermolecular H-bonding, (d) More extensive association of carboxylic, acid via van der Waals force of attraction, Which one of the following esters gets, hydrolysed most easily under alkaline conditions?, [2015 RS], , Cl, The correct order of decreasing acid strength of, trichloroacetic acid (A), trifluoroacetic acid (B),, acetic acid (C) and formic acid (D) is : [2012], (a) B > A > D > C, (b) B > D > C > A, (c) A > B > C > D, (d) A > C > B > D
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277, , Aldehydes, Ketones, and Carboxylic acids, 68., , 69., , Match the compounds given in List-I with, List-II and select the suitable option using the, code given below :, [2011 M], List I, List-II, (A) Benzaldehyde, (i) Phenolphthalein, (B) Phthalic anhydride (ii) Benzoin, condensation, (C) Phenyl benzoate (iii) Oil of wintergreen, (D) Methyl salicylate (iv) Fries, rearrangement, Code :, (A), (B), (C), (D), (a) (iv), (i), (iii), (ii), (b) (iv), (ii), (iii), (i), (c) (ii), (iii), (iv), (i), (d) (ii), (i), (iv), (iii), An organic compound ‘A’ on treatment with NH3, gives ‘B’ which on heating gives ‘C’, ‘C’ when, treated with Br2 in the presence of KOH produces, ethylamine. Compound ‘A’ is:, [2011 M], (a) CH3COOH, (b) CH3 CH2 CH2 COOH, (c), , COOH, (c), OCH2CH3, COOC2H5, , (d), 71., , 72., , 73., , Br, |, (a) H– C – CH2COOH, |, Br, (b) CH2Br – CH2 – COBr, , CH3 – CHCOOH, CH3, , 70., , (d) CH3CH2COOH, In a set of reactions, ethylbenzene yielded a, product D., [2010], KMnO4, Br2, CH2CH3 ¾¾¾¾, ® B ¾¾¾®, KOH, , FeCl3, , C H OH, , 2 5, C ¾¾¾¾¾, ®D, +, H, , CH2 – CH – COOC 2H5, , (a), , Br, , Br, (b), , Br, CH2 COOC2H5, , Br, Given are cyclohexanol (I) acetic acid (II),, 2, 4, 6 – trinitrophenol (III) and phenol (IV). In, these, the order of decreasing acidic character, will be :, [2010], (a) III > II > IV > I, (b) II > III > I > IV, (c) II > III > IV > I, (d) III > IV > II > I, Among the given compounds, the most, susceptible to nucleophilic attack at the carbonyl, group is:, [2010], (a) CH3COOCH3, (b) CH3CONH2, (b) CH3COOCOCH3 (d) CH3COCl, Propionic acid with Br 2|P yields a dibromo, product. Its structure would be:, [2009], , 74., , Br, |, (c) CH3– C – COOH, |, Br, (d) CH2 Br – CHBr – COOH, The relative reactivities of acyl compounds, towards nucleophilic substitution are in the order, of :, [2008], (a) Acyl chloride > Acid anhydride > Ester, > Amide, (b) Ester > Acyl chloride > Amide, > Acid anhydride, (c) Acid anhydride > Amide > Ester, > Acyl chloride, (d) Acyl chloride > Ester > Acid anhydride, > Amide
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279, , Aldehydes, Ketones, and Carboxylic acids, 84., , 85., , 86., , 87., , Aspirin is an acetylation product of, [1998], (a) o-Hydroxybenzoic acid, (b) o-Dihydroxybenzene, (c) m-Hydroxybenzoic acid, (d) p-Dihydroxybenzene, An ester (A) with molecular fomula, C9H10O2, was treated with excess of CH3MgBr and the, complex so formed was treated with H2SO4 to, give an olefin (B). Ozonolysis of (B) gave a ketone, with molecular formula C8H8O which shows +ve, iodoform test. The structure of (A) is [1998], (a) C6H5COOC2H5, (b) C2H5COOC6H5, (c) H3COCH2COC6H5, (d) p – H3CO – C6 H 4 – COCH3, Which one of the following esters cannot, undergo Claisen self-condensation?, [1998], (a) CH 3 - CH 2 - CH 2 - CH 2 - COOC 2 H 5, (b) C6H5COOC2H5, (c) C6H5CH2COOC2H5, (d) C6H11CH2COOC2H5, Consider the following transformations :, heat, , I, , CaCO 3, 2, ¾® B ¾¾®, ¾® A ¾¾, CH3COOH¾¾ ¾, C, , 91., , 92., , 93., , 94., , 95., , NaOH, , The molecular formula of C is, OH, (a), , |, , CH 3 - C - CH 3, |, , [1996], , (b) ICH2 — COCH3, , 96., , I, , 88., , 89., , 90., , (c) CHI3, (d) CH3I, Formic acid is obtained when, [1994], (a) Calcium acetate is heated with conc. H2SO4, (b) Calcium formate is heated with calcium acetate, (c) Glycerol is heated with oxalic acid at 373 K, (d) Acetaldehyde is oxidised with K2Cr2O7 and, H2SO4., The preparation of ethyl acetoacetate involves, (a) Wittig reaction, [1994], (b) Cannizzaro’s reaction, (c) Reformatsky reaction, (d) Claisen condensation., Schotten-Baumann reaction is a reaction of, phenols with, [1994], (a) Benzoyl chloride and sodium hydroxide, (b) Acetyl chloride and sodium hydroxide, (c) Salicylic acid and conc. H2SO4, (d) Acetyl chloride and conc H2SO4, , 97., , 98., , 99., , An ester is boiled with KOH. The product is cooled, and acidified with concentrated HCl. A white, crystalline acid separates. The ester is, [1994], (a) Methyl acetate, (b) Ethyl acetate, (c) Ethyl formate, (d) Ethyl benzoate, Sodium formate on heating yields, [1993], (a) Oxalic acid and H2, (b) Sodium oxalate and H2, (c) CO2 and NaOH, (d) Sodium oxalate., Among acetic acid, phenol and n-hexanol, which, of the following compounds will react with, NaHCO3 solution to give sodium salt and carbon, dioxide?, [1993], (a) Acetic acid, (b) n-Hexanol, (c) acetic acid and phenol, (d) Phenol., (CH3)2 C = CHCOCH3 can be oxidized to, (CH3)2C = CHCOOH by, [1993], (a) Chromic acid, (b) NaOI, (c) Cu at 300°C, (d) KMnO4., In which of the following, the number of carbon, atoms does not remain same when carboxylic, acid is obtained by oxidation, [1992], (a) CH 3COCH 3, (b) CCl 3CH 2CHO, (c) CH 3CH 2 CH 2OH (d) CH 3CH 2CHO., Benzoic acid gives benzene on being heated with, X and phenol gives benzene on being heated, with Y. Therefore X and Y are respectively [1992], (a) Soda-lime and copper, (b) Zn dust and NaOH, (c) Zn dust and soda-lime, (d) Soda-lime and zinc dust., The compound formed when malonic ester is, heated with urea is, [1989], (a) Cinnamic acid, (b) Butyric acid, (c) Barbituric acid, (d) Crotonic acid., Which of the following is the correct decreasing, order of acidic strength of, [1988], (i) methanoic acid, (ii) ethanoic acid, (iii) propanoic acid, (iv) butanoic acid., (a) (i) > (ii) > (iii) > (iv) (b) (ii) > (iii) > (ii) > (i), (c) (i) > (iv) > (iii) > (ii) (d) (iv) > (i) > (iii) > (ii), Among the following the strongest acid is [1988], (a) CH3COOH, (b) CH2ClCH2COOH, (c) CH2ClCOOH, (d) CH3CH2COOH
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EBD_8336, 280, , CHEMISTRY, ANSWER KEY, , 1, , (b) 11 (c) 21, , (a), , 31, , 41, , (a), , (d), , 51, , (c), , 61, , (b), , 71, , (a), , 81, , (b), , 91, , (d), , 2, , (c), , 12 (b) 22, , (c), , 32, , (c), , 42, , (a), , 52, , (b), , 62, , (c), , 72, , (d), , 82, , (c), , 92, , (b), , 3, , (a), , 13 (a) 23, , (d), , 33, , (d), , 43, , (c), , 53, , (b), , 63, , (d), , 73, , (c), , 83, , (b), , 93, , (a), , 4, , (b) 14 (d) 24, , (b), , 34, , (a), , 44, , (c), , 54, , (d), , 64, , (b), , 74, , (a), , 84, , (a), , 94, , (b), , 5, , (a), , 15 (a) 25, , (d), , 35, , (a), , 45, , (a), , 55, , (d), , 65, , (c), , 75, , (c), , 85, , (a), , 95, , (a), , 16 (b) 26 (a, d) 36, , 6, , (c), , (c), , 46, , (d), , 56, , (b), , 66, , (a, , 76, , (a), , 86, , (b), , 96, , (d), , 7, , (b) 17 (d) 27, , (b), , 37, , (d), , 47, , (a), , 57, , (d), , 67, , (a), , 77, , (c), , 87, , (c), , 97, , (c), , 8, , (c), , 18 (d) 28, , (d), , 38, , (c), , 48, , (c), , 58, , (a), , 68, , (d), , 78, , (d), , 88, , (c), , 98, , (a), , 9, , (b) 19 (b) 29, , (d), , 39, , (a), , 49, , (b), , 59, , (d), , 69, , (d), , 79, , (a), , 89, , (d), , 99, , (c), , 10, , (b) 20 (d) 30, , (d), , 40, , (c), , 50, , (c), , 60, , (a), , 70, , (d), , 80, , (a), , 90, , (a), , Hints & Solutions, 1., , OH, , (b), CH3, , CHCl2, Cl 2/hv, , O, ||, , CH – OH, , (b), , NaOH, , H C H ¾¾¾¾, ® CH3OH + HCOONa, Cannizzaro reaction, , H2 O, 373 K, , (c) 2CH3Cl, , (X), , Na, ¾¾¾¾®, dry ether, , CH3 – CH3, , Wurtz reaction, , –H 2 O, , COCH3, , CHO, CH COCl, anh.AlCl3, , 3, ¾¾¾¾¾, ®, , (d), , Friedel-Craft acylation, , In presence of sunlight chlorination takes place at, the side chain by free radical mechanism., , 2., , 5., , (c) Zn/Hg and conc. HCl reduce carboxyl, group to meth ylene group (Clemmensen, reduction)., O, , C, , O, , Cl, , 3., , (a), , 4., , It is Rosenmund reaction., (b), OH, OH, (a), , H2, ¾¾¾®, Pd–BaSO4, , CHCl3, ¾¾¾¾, ®, NaOH, , CH, , Note that, no new C–C bond is formed in Cannizzaro, reaction., (a) Among the given options, only (a) can be, oxidized to ketone., OH, O, |, ||, oxidation, H3C – CH – CH3 ¾¾¾¾® H3C – C – CH3, 2-hydroxypropane, , Acetone, , Carbonyl compounds (aldehydes and ketones) are, obtained by the oxidation of 1° and 2° alcohols, respectively., , CHO, , Riemer-Tiemann reaction, , 6., , H, , 2, ¾, R — C — H + HCl, (c) R — C — Cl ¾¾®, , O, , Pd – BaSO4, , O, ‘P’
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EBD_8336, 282, , CHEMISTRY, 21., , On reduction, it gives n-pentane, reduction, , H3C—CH2—C—CH2—CH3 ¾¾¾¾ ¾®, Zn-Hg/HCl, , or NH 2-NH 2/OH, O, CH 3—CH 2—CH2—CH 2—CH 3, , –, , n-pentane, , 17., , (d), , –, , O, , O, –, , CH3, , Å, , 22., , + CH3 Li ®, Cylopentanoyl anion, , Reactions involving methyl lithium require and an, hydrous condtions, because the compound is highly, reactive towards water. Methyl lithium is stable, as a solution in ether., , 18., , H, , s, , s, , C, , s, , H, , H, s, , C, , s, , s, p, , C, , s, , C, , O, H, , H, , Os C, , C, , s, , Os H, , s, p, , s, , s, , s, , H, O, , (b), , ||, , s, , s, , H, , H, , 1, , C H OH, , 2 5, CH3, CH3 ¾¾¾®, , CH3, , C, OH, , OC2H5, C H OH, , 2 5, ¾¾¾®, CH3, , C, , CH3, , OC2H5, Acetal, , 24., , (b) CH 3 CHO gives Iodoform test but, C6H5CH2CHO does not give Iodoform test due, to absence of methyl group., , 25., , (d) R – CH = O + H2N – NH2 ®, R – CH = N – NH2, Such reactions take place in slightly acidic, medium and involve nucleophilic addition of the, ammonia derivative., (a, d) The compounds with CH3 – C – group, ||, O, or CH3 — CH — group give iodoform test., |, OH, , CHO, , NO 2, , C, , Hemiacetal, , Elimination, , is most reactive towards nucleophilic, addition reaction., , (d) Anhydrous alcohols add to the carbonyl, group of aldehydes in the presence of anhydrous, hydrogen chloride to form acetals via hemiacetals., OC2H5, , O, , H2O, ¾¾¾¾¾, ® R—C—R¢, , NH—NH2, (Addition), While in all other cases no elimination take place., (d) Any substituent in the carbonyl compound, that increases the positive charge on the carbonyl, carbon will increase reactivity towards nucleophilic, addition., —NO2 shows –M effect hence, , Cl, , Cl, , Cl, , CH3, , R — C— R + NH 2 — NH 2 ¾¾¾, ®, NH2, N, OH, , R—C—R, , +, , 50 % KOH, , ¾¾¾®, , HÅ, , 1, , CH2–OH, , C–O, , CHO, , 23., , s, , s, , 20., , O, , (d) Enolic form of ethyl acetoacetate has 18, sigma and 2 pi-bonds as shown below:, H, , 19., , (a) Enolic form predominates in compounds, containing two carbonyl groups separated by a, – CH2 group. This is due to following two factors., (i) Presence of conjugation which increases, stability., (ii) Formation of intramolecular hydrogen, bond between enolic hydroxyl group and, second carbonyl group which leads to, stablisation of the molecule. Hence, the correct answer is III > II > I., (c) Cannizzaro reaction - when an aldehyde, containing no a – H undergo reaction in presence, of 50% KOH. It disproportionates to form a molecule, of carboxylic acid and a molecule of alcohol., , 26.
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EBD_8336, 284, , CHEMISTRY, , 36. (c) The degree of hydrolysis increases as the, magnitude of positive charge on carbonyl group, increases. Electron withdrawing group increases, the positive charge and electron releasing group, decreases the positive charge. Among these NO2, &-CHO are electron withdrawing group from, which-NO2 exhibit more –I effect than –CHO. On, the other hand-CH3 is a electron releasing group, therefore the order of reactivity towards, hydrolysis is, COCl, , COCl, >, , COCl, , COCl, , >, , NO2, , O, , OH, |, , HCN, , CH3 - C - H ¾¾¾® CH3 - C - CN, |, , H, , OH, |, , hydrolysis, , ¾¾¾¾¾, ® CH3 - C*- H, |, , COOH, 38. (c) Aldehydes are more reactive than ketones due, to +I effect of –CH3 group. There are two –CH3, group in acetone which reduces +ve charge density, on carbon atom of carbonyl group. More hindered, carbonyl group too becomes less reactive. So, in, the given case CH3CHO is the right choice., OH, , O, ||, , |, , HCN, KCN, , |, , R ¢ (A), , OH, |, , Reduction by, , ¾¾¾¾¾¾, ® R– C – CH 2 NH 2, LiAlH 4 (B), , |, , R¢, 40. (c) It is a simple Cannizzaro reaction., –, , COO, , CHO, 50% KOH, , 2, Cl, , |, , CH2OH, , CN, H, , H, , |, , |, , H + / H 2O, , ¾¾ ¾ ¾, ¾® OH - C - CH 3 + CH 3 - C - OH, |, , |, , COOH, , 42., , 43., , COOH, , 50% L - Lactic, , 50% D-Lactic, , acid, , acid, , (a) Acetophenone is a ketone, cyclopentane, doesn’t contain (4n + 2)p electron hence is not, aromatic. Diastase is an enzyme used in the, preparation of Maltose (Malt sugar,, C12H22O11) through hydrolysis of starch., (c) Tollen's reagent is used to detect aldehydic, group. Tollen's reagent is an ammonical solution, of silver nitrate. When aldehyde is added to, Tollen's reagent, silver oxide is reduced to metallic, silver which deposits as mirror., RCHO + Ag2O ¾¾, ® RCOOH + 2Ag, , 44. (c) In CH 2 = CH — CHO due to – M effect of, — CHO group, polarization of electron takes, place as follows:, +, –, CH 2 = CH — C = O « CH 2 — CH = C — O, |, , |, , H, H, Hence, partial polarization is represented as, d+, , d-, , C H 2 = CH — CH = O, 45. (a), , O, R C H + NH2 NH2 ¾® R C N, H, Aldehyde, , 39. (a) R– C –R ¢ ¾¾¾® R — C — CN, , |, , 41. (d) CH 3 - C = O + HCN ® CH 3 - C - OH, , CH3, , 37. (d) Out of given compound only acetaldehyde, can form optical active hydroxy acid as it contains, one asymmetric carbon atom as marked below :, ||, , |, , >, , CHO, , H, , H, , Hydrazine, , NH2, , Aldehyde hydrazone, , 46. (d) Iodoform test is exhibited by ethyl alcohol, acetaldehyde, acetone, methyl ketones and those, alcohols which possess CH3CH(OH)-group. As, 3-pentanone does not contain CH3 CO-group, therefore it does not give iodoform test., 47. (a), OH, O, C, , CH, , C CH2, , C CH3, , H SO /HgSO 4, , +, Cl, , 2 4, ¾¾¾¾¾®, , Cl, , enol form, , Keto form
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287, , Aldehydes, Ketones, and Carboxylic acids, (b), , O, , O, +, , H, , ¾®, , O, O, , H, , OH, 71. (a) N6O2, , O, , NO2 > CH3COOH >, , H, , NO2, (II), , OH, , OH, , OH, , OH, , OH, , Phenolphthalein, , R, , (IV), , R, , AlCl, , 3, ¾¾ ®, , R, Fries rearrangement, , O, , COOCH3, OH, (d), 69., , (Oil of wintergreen), , (d) Since, C when heated with Br2 in presence, of KOH produces ethylamine, hence it must be, propanamide and hence the organic compound, (A) will be propanoic acid. The reactions follows:, , 72., 73., , NH, , 3, CH3 - CH 2 - COOH ¾¾¾, ®, , (A), , - +, , Br /P, , (B), , Hell-volhard-zelinsky reaction:, Carboxylic acids having an a-hydrogen are, halogenated at the a-position on treatment with, chlorine or bromine in the presence of small amount, of red phosphoms to give a-halocarboxylic acids., The reaction fails to accomplish the fluorination, and iodination of carboxylic acids., , KOH + Br, Hoffmann, bromamide, reaction, , 2, CH 3 - CH 2 - CONH 2 ¾¾¾¾¾, ®, , CH3 - CH 2 - NH 2, (Ethylamine), , COOH, , CH2 – CH3, 70., , 74., , ( KMnO4 )®, ¾¾¾¾¾, , (d), , KOH, , (B), , COOH, , COOC2H5, C H OH, , Br, FeCl3, , 2 5, ¾¾¾¾®, +, , 2, ¾®, , (C), , Br, , (I), , Explanation: Presence of three — NO2 groups, in o–, p– positions to phenolic groups (in III), makes phenol strongly acidic because its, corresponding phenate ion (conjugate base) is, highly stabilised due to resonance. Conjugate, base of CH3 COOH, II (i.e. CH3 COO – ) is, resonance hybrid of two equivalent structures., The conjugate base of phenol, IV is stabilized, due to resonance (note that here all resonating, structures are not equivalent). The conjugate, base of cyclohexanol, I does not exhibit, resonance, hence not formed., (d) Cl– is the weakest base and hence better, leaving group., (c) This reaction is an example of Hell - Volhard, Zelinsky reaction. In this reaction acids, containing a – H on treatment with X2 /P give, di-halo substituted acid., , 2 ® CH - CBr - COOH, CH3 – CH 2 COOH ¾¾¾, 3, 2, , D, , ®, CH3 - CH 2 - CO O N H 4 ¾¾, , (C), , OH, >, , (c) The fries rearrangement enables the, preparation of acylphenols. The reaction is, catalysed by Bronsted or lewis acid such as, AlCl3., O, OH, OH O, O, , (III), , H, , (D), , Br, , (a) The more the basic character of the leaving, group, the lesser is the reactivity. The basic, character follows the order:, NH2– > OR– > RCOO – > Cl–, Hence, the relative reactivities of acyl, compounds towards nucleophilic substitution, follow the order Acyl halides > Acid anhydride, > Ester > Amide.
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EBD_8336, 292, , 4., , CHEMISTRY, Nitration of aniline in strong acidic medium also, gives m-nitroaniline because, [2018], (a) Inspite of substituents nitro group always, goes to only m-position., , (d) Arylamines are generally more basic than, alkylamines, because the n itrogen atom, in arylamines is sp-hybridized., 8., , (b) In electrophilic substitution reactions,, amino group is meta directive., (c) In acidic (strong) medium aniline is present, as anilinium ion., , 9., , (d) In absence of substituents, nitro group, always goes to m-position., 5., , The correct increasing order of basic strength, for the following compounds is :, [2017], , (I), , NH2, , NH2, , NH2, (II), , (a) III < I < II, , (b) III < II < I, , (c) II < I < III, , (d) II < III < I, , (a) Hoffmann hypobromamide reaction, , (b) Azobenzene, , (c) Aniline, , (d) p-Aminophenol, , The number of structural isomers possible from, the molecular formula C3H9N is :, [2015 RS], (a) 4, , (b) 5, , (c) 2, , (d) 3, , Method by which aniline cannot be prepared is:, [2015 RS], , (b) degradation of benzamide with bromine in, alkaline solution, , CH3, , Which of the following reactions is appropriate, for converting acetamide to methanamine? [2017], , (a) Azoxybenzene, , (a) hydrolysis of phenylisocyanide with acidic, solution, , (III), NO2, , 6., , 10., , The electrolytic reduction of nitrobenzene in, strongly acidic medium produces :[2015], , (c) reduction of nitrobenzene with H2/Pd in, ethanol, (d) potassium salt of phthalimide treated with, chlorobenzene followed by hydrolysis with, aqueous NaOH solution., 11., , Some reactions of amines are given. Which one, is not correct ?, [NEET Kar. 2013], (a) (CH3)2NH + NaNO2 + HCl ®, , (b) Stephens reaction, , (CH3)2 N – N = O, , (c) Gabriels phthalimide synthesis, (d) Carbylamine reaction, 7., , (b) (CH3)2N –, , The correct statement regarding the basicity of, arylamines is, [2016], , (CH3)2N –, , (a) Arylamines are generally less basic than, alkylamines because the nitrogen lone-pair, electrons are delocalized by interaction, with the aromatic ring p electron system., (b) Arylamines are generally more basic than, alkylamines because the nitrogen lone-pair, electrons are not delocalized by interaction, with the aromatic ring p electron system., (c) Arylamines are generally more basic than, alkylamines because of aryl group., , + NaNO2 + HCl ®, – N = NCl, , (c) CH3CH2NH2 + HNO2 ® CH3CH2OH + N2, (d) CH3NH2 + C6H5SO2Cl ®, CH3NHSO2C6H5., 12., , In a set of reactions m-bromobenzoic acid gave, a product D. Identify the product D., [2011], COOH, SOCl, , NH, , NaOH, 3, 2, ¾¾®, D, C ¾¾®, B ¾¾®, Br, , Br, , 2
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293, , Amines, SO2NH2, , (a), , (4) CH3CH(OH)CH3, , CONH2, , Options :, (1), (2) (3), (a) (iv) (ii) (iii), (b) (ii), (i), (iv), (c) (iii), (ii) (i), (d) (ii), (iii) (i), Predict the product:, , NH2, NH2, , (c), , (d), Br, , Br, , Which of the following compounds is most basic?, [2011 M], (a), , (iv) with Lucas, reagent, cloudiness, appears, after 5 minutes, , (b), , Br, , 13., , COOH, , O2N, , NH2, , (b), , 16., , (4), (i), (iii), (iv), (iv), [2009], , NHCH3+NaNO2+ HCl ¾® Product, , CH2NH2, , CH3, (c), , N – COCH3, , (d), , NH2, , (a), , H, , 14., , Which of the following statements about primary, amines is ‘False’ ?, [2010], (a) Alkyl amines are stronger bases than aryl, amines, (b) Alkyl amines react with nitrous acid to, produce alcohols, , (b), , (c), , CH3C º CH, , (ii) with KOH, (alcohol) and, CHCl3 produces, bad smell, , (3) CH3CH2COOCH3, , (iii) gives white, ppt. with, ammonical, AgNO3, , CH3, , N, , CH3, , (d), 17., , (1) CH3CH2CH2CH2NH2 (i) alkaline, hydrolysis, (2), , +, , OH, , Match the compounds given in List - I with their, characteristic reactions given in List - II. Select, the correct option., [2010], List - II, Reactions, , NO, , NO, , (d) Alkyl amines are stronger bases than ammonia, , List - I, Compounds, , NHCH3, , NHCH3, , (c) Aryl amines react with nitrous acid to, produce phenols, 15., , NO2, , N, , 18., , N, , N=O, , Which of the following is more basic than, aniline?, [2006], (a) Triphenylamine, (b) p-Nitroaniline, (c) Benzylamine, (d) Diphenylamine, The major organic product formed from the, following reaction :, [2005], , O, , (i) CH NH, , 3, 2, ¾ ¾ ¾ ¾ ¾ ¾ ® ......, , (ii) LiAlH 4, (iii) H 2O, , is
Page 314 :
EBD_8336, 294, , CHEMISTRY, (a), , 23., , O – NHCH3, , (a) an alkyl cyanide, , (b), 24., , H, H, NCH3, OH, , 25., , H, , (d), , (b) a nitro compound, , (c) an alkyl isocyanide (d) an amide, , NCH3, , (c), , The compound obtained by heating a mixture, of a primary amine and chloroform with ethanolic, potassium hydroxide (KOH) is, [1997], , NCH3, , When aniline reacts with oil of bitter almonds, (C6 H5CHO), condensation takes place and benzal, derivative is formed. This is known as, [1995], (a) Million's base, , (b) Schiff's reagent, , (c) Schiff's base, , (d) Benedict's reagent, , What is the decreasing order of basicity of, primary, secondary and tertiary ethylamines and, NH3 ?, [1994], (a), , NH 3 > C 2 H 5 NH 2 > (C 2 H 5 ) 2 NH >, , OH, , 19., , (C 2 H 5 )3 N, , Electrolytic reduction of nitrobenzene in weakly, acidic medium gives, [2005], , (b) (C2 H 5 )3 N > (C2 H 5 )2 NH >, C2 H5 NH 2 > NH 3, , (a) N-Phenylhydroxylamine, (c), , (b) Nitrosobenzene, , (C2 H 5 )2 NH > C2 H 5 NH 2 >, (C 2 H 5 )3 N > NH 3, , (c) Aniline, (d) p-Hydroxyaniline, 20., , (a) Toluene, (c) Glycerol, 21., , 22., , (d), , The consituent of the powerful explosive RDX, is formed during the nitration of, [2000], , C 2 H 5 NH 2 > NH 3 ., 26., , (b) Phenol, , (C 2 H 5 ) 2 NH > (C 2 H 5 )3 N >, , Mark the correct statement, , [1988], , (a) Methylamine is slightly acidic, , (d) Urotropine, , Which of the following is most basic in nature?, , (b) Methylamine is less basic than ammonia, , (a) NH3, , (b) CH3NH2 [2000], , (c) Methylamine is a stronger base than ammonia, , (c) (CH3)2NH, , (d) C6H5NHCH3, , Aniline is reacted with bromine water and the, resulting product is treated with an aqueous, solution of sodium nitrite in presence of dilute, hydrochloric acid. The compound so formed is, converted into a tetrafluoroborate which is, subsequently heated dry. The final product is, (a) 1,3, 5-Tribromobenzene, , [1998], , (d) Methylamine forms salts with alkalies., Topic 2: Amides, Cyanides and Isocyanides, 27., , The following reaction, NH2, + Cl, O, , H, N, , (b) p-Bromofluorobenzene, (c) p-Bromoaniline, (d) 2,4, 6-Tribromofluorobenzene, , NaOH, ¾¾¾®, , O, is known by the name :, , O, , [2015 RS]
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EBD_8336, 298, , CHEMISTRY, (b) CH3, , 51., , N=N, , NH2, , (c) (CH3)2N, , N=N, , (d) (CH3)2N, , NH, CHCl +KOH, , reduction, , [2000], , 3, [A] ¾¾¾¾® [B] ¾¾¾¾¾¾, ®, , reduction, , [C] ¾¾¾¾® N-Methylaniline, A is, , (a) Formaldehyde, (b) Trichloromethane, (c) Nitrobenzene, (d) Toluene, 52. Which of the following reagents will convert, p-methylbenzenediazonium chloride into, p-cresol?, (a) Cu powder, (b) H2O, [1999], (c) H3PO2, (d) C6H5OH, 53. Diazo coupling is useful to prepare some [1994], (a) Pesticides, (b) dyes, (c) proteins, (d) vitamins, , ANSW E R KE Y, 1, 2, 3, 4, 5, 6, , (d), (b), (a), (c), (c), (a), , 7, 8, 9, 10, 11, 12, , (a), (d ), (a), (d ), (b ), (c), , 13, 14, 15, 16, 17, 18, , (b ), (c), (d ), (d ), (c), (b ), , (c), (d ), (c), (d ), (c), (c), , 19, 20, 21, 22, 23, 24, , 25, 26, 27, 28, 29, 30, , (d ), (c), (d ), (b ), (a), (a), , 31, 32, 33, 34, 35, 36, , (a), (d ), (b ), (c), (c), (d ), , 37, 38, 39, 40, 41, 42, , (d ), (b ), (b ), (a), (c), (d ), , 43, 44, 45, 46, 47, 48, , (b ), (b ), (a), (a), (a), (b ), , 49, 50, 51, 52, 53, , (d ), (c), (c), (b ), (b ), , Hints & Solutions, 1., , (d) Aliphatic and aromatic primary amines give, carbylamine reaction. Since aniline is primary, aromatic amine, it gives carbylamine test., , 2., , (b) Hinsberg’s reagent is used to distinguish, 1°, 2° and 3° amine. Primary amines react with, Hinsberg’s reagent to give precipitate. which is, soluble in alkali while secondary amines give a, product which is insoluble in alkali. Tertiary, amines do not react with Hinsberg’s reagent., , 3., , 5., , Order of basic strength is, NH2, , NH2, , NH2, , <, , <, , NO2, , CH3, O, , 6., , (c), , (c) – NO2 group has strong – R effect and –, CH3 shows +R effect., \, , (a) Account for the inductive effect, solvation, effect (H-bonding.) and steric hinderance for, basic character in aqueous solutions, (CH3)2 NH > CH3 NH2 > (CH3)3 N, , 4., , In acidic medium aniline is protonated to form, anilinium ion which is meta-directing., , D, (a) CH3 - C - NH 2 + Br2 + 4NaOH ¾¾, ®, acetamide, , +, , NH2, , NH3, H+, ¾¾¾¾®, nitrating, mixture, , CH3 – NH2 + 2NaBr + Na2CO3 + 3H2O, , +, , NH3, , methanamine, , +, , It is called Hoffmann Bromamide reaction., , NO2, ¾¾¾®, , NO2
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305, , Biomolecules, , 28, , Biomolecules, , Trend Analysis with Important Topics & Sub-Topics, 2020, Topic Name, , Amino acids and, proteins, , Sub-Topic, hydrolysis of, sucrose, reducing/non, reducing sugars, aminoacids/, proteins, , Nucleic acid and, enzymes, , structure of, DNA/RNA, , LOD - Level of Difficulty, , E - Easy, , Carbohydrates and, lipids, , 2019, , 2., , 3., , 1, , 1, , Sucrose on hydrolysis gives:, [2020], (a) a-D-Glucose + b-D-Glucose, (b) a-D-Glucose + b-D-Fructose, (c) a-D-Fructose + b-D-Fructose, (d) a-D-Glucose + b-D-Fructose, Which one given below is a non-reducing sugar?, (a) Maltose, (b) Lactose [2016], (c) Glucose, (d) Sucrose, D (+) glucose reacts with hydroxylamine and, yields an oxime. The structure of the oxime would, be :, [2014], (a), , CH = NOH, H – C – OH, , HO – C – H, HO – C – H, , (b), , 2017, , 2016, , QNS. LOD QNS. LOD QNS. LOD QNS. LOD QNS. LOD, E, , E, , CH2OH, , CH2OH, , E, , 1, , A, , 1, , E, , 1, , A, , (c), , CH = NOH, , (d), , HO – C – H, H – C – OH, HO – C – H, , 4., , HO – C – H, , H – C – OH, , 1, , Qns - No. of Questions, , CH = NOH, , H – C – OH, , E, , E, , D - Difficult, , HO – C – H, , H – C – OH, , 1, , 1, , A - Average, , Topic 1: Carbohydrates and Lipids, 1., , 2018, , 5., , CH = NOH, H – C – OH, , HO – C – H, H – C – OH, , H – C – OH, , H – C – OH, , CH2OH, , CH2OH, , Which one of the following sets of monosaccharides forms sucrose?, [2012], (a) a–D-Galactopyranose and, a–D-Glucopyranose, (b) a–D-Glucopyranose and, b–D-Fructofuranose, (c) b–D-Glucopyranose and, a–D- Fructofuranose, (d) a–D-Glucopyranose and, b–D-Fructopyranose, Which one of the following statements is not, true regarding (+) Lactose ?, [2011], (a) On hydrolysis (+) Lactose gives equal, amount of D(+) glucose and D(+), galactose.
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EBD_8336, 306, , 6., , 7., , 8., 9., , 10., , 11., , 12., 13., , 14., , CHEMISTRY, (b) (+) Lactose is a b-glycoside formed by the, union of a molecule of D(+) glucose and a, molecule of D(+) galactose., (c) (+) Lactose is a reducing sugar and does, not exhibit mutarotation., (d) (+) Lactose, C12H22 O11 contains 8-OH, groups., Which one of the following does not exhibit the, phenomenon of mutarotation ?, [2010], (a) (+) – Sucrose, (b) (+) – Lactose, (c) (+) – Maltose, (d) (–) – Fructose, Fructose reduces Tollen’s reagent due to [2010], (a) enolisation of fructose followed by, conversion to aldehyde by base, (b) asymmetric carbons, (c) primary alcoholic group, (d) secondary alcoholic group, The cell membranes are mainly composed of, (a) fats, (b) proteins [2005], (c) phospholipids, (d) carbohydrates, Number of chiral carbons in b – D – (+) – glucose, is, [2004], (a) five, (b) six, (c) three, (d) four, Glycolysis is, [2003], (a) conversion of glucose to haem, (b) oxidation of glucose to glutamate, (c) conversion of pyruvate to citrate, (d) oxidation of glucose to pyruvate, Phospholipids are esters of glycerol with [2003], (a) Three phosphate groups, (b) Three carboxylic acid residues, (c) Two carboxylic acid residues and one, phosphate group, (d) One carboxylic acid residue and two, phosphate groups, Cellulose is a polymer of, [2002], (a) Glucose, (b) Fructose, (c) Ribose, (d) Sucrose, Which of the following gives positive Fehling, solution test ?, [2001], (a) Protein, (b) Sucrose, (c) Glucose, (d) Fats, Which is correct statement?, [2001], (a) Starch is a polymer of a-glucose, (b) In cyclic structure of fructose, there are four, carbons and one oxygen atom, (c) Amylose is a component of cellulose, (d) Proteins are composed of only one type of, amino acids, , 15., 16., 17., , 18., , 19., , 20., , 21., , 22., , Mg is present in which of the following : [2000], (a) Starch, (b) Chlorophyll, (c) Both, (d) None, Which of the following is the sweetest sugar?, (a) Sucrose, (b) Glucose [1999], (c) Fructose, (d) Maltose, In cells, the net production of ATP molecules, generated from one glucose molecule is [1999], (a) 46, (b) 32, (c) 36, (d) 40, The number of molecules of ATP produced in, the lipid metabolism of a molecule of palmitic, acid is, [1998], (a) 130, (b) 36, (c) 56, (d) 86, Glucose molecule reacts with 'X' number of, molecules of phenylhydrazine to yield osazone., The value of 'X' is, [1998], (a) four, (b) one, (c) two, (d) three, Sucrose in water is dextro-rotatory, [a]D= + 66.4º., When boiled with dilute HCl, the solution, becomes leavo-rotatory, [a]D= –20º. In this, process, the sucrose molecule breaks into[1996], (a) L-glucose + D-fructose, (b) L-glucose + L-fructose, (c) D-glucose + D-fructose, (d) D-glucose + L-fructose, The a-D glucose and b-D glucose differ from, each other due to difference in carbon atom with, respect to its, [1995], (a) Conformation, (b) Configuration, (c) Number of OH groups, (d) Size of hemiacetal ring, On hydrolysis of starch, we finally get [1991], (a) Glucose, (b) Fructose, (c) Both (a) and (b) (d) Sucrose, Topic 2: Amino Acids and Proteins, , 23., 24., , Which of the following is a basic amino acid ?, (a) Alanine, (b) Tyrosine [2020], (c) Lysine, (d) Serine, Which structure(s) of proteins remain(s) intact, during denaturation process?, [NEET Odisha, 2019], (a) Tertiary structure only, (b) Both secondary and tertiary structures, (c) Primary structure only, (d) Secondary structure only
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307, , Biomolecules, 25., , 26., , 27., , 28., , 29., , 30., , 31., , The non-essential amino acid among the, following is:, [2019], (a) valine, (b) leucine, (c) alanine, (d) lysine, Which of the following compounds can form a, zwitterion?, [2018], (a) Aniline, (b) Acetanilide, (c) Glycine, (d) Benzoic acid, Which of the following statements is not, correct :, [2017], (a) Ovalbumin is a simple food reserve in eggwhite, (b) Blood proteins thrombin and fibrinogen are, involved in blood clotting, (c) Denaturation makes the proteins more active, (d) Insulin maintains sugar level in the blood, of a human body, In a protein molecule various amino acids are, linked together by, [2016], (a) a-glycosidic bond (b) b-glycosidic bond, (c) peptide bond, (d) dative bond, Which of the statements about "Denaturation", given below are correct ?, [2011 M], (a) Denaturation of proteins causes loss of, secondary and tertiary structures of the protein., (b) Denturation leads to the conversion of, double strand of DNA into single strand, (c) Denaturation affects primary strucrture, which gets distorted, (a) (b) and (c), (b) (a) and (c), (c) (a) and (b), (d) (a), (b) and (c), Which functional group participates in, disulphide bond formation in proteins? [2005], (a) Thioester, (b) Thioether, (c) Thiol, (d) Thiolactone, Which one of the following structures, represents the peptide chain?, [2004], O, H, |, ||, |, (a) - N - C - N - C - NH - C - NH || | |, O H, H, H, |, | | | | | | |, (b) - N - C - C- C- C- N - C- C- C|| | | |, | | |, O, H, O, H, H, | | ||, | |, | |, (c) - N - C- C - N - C - C - N - C - C |, | ||, | ||, O, O, , (d), 32., , 33., , 34., , 35., , H, O, H, | | | ||, | | |, | |, - N - C- C- C - N - C- C- N- C - C- C| |, | | |, || | |, H, O, , The correct statement in respect of protein, haemoglobin is that it, [2004], (a) functions as a catalyst for biological reactions, (b) maintains blood sugar level, (c) acts as an oxygen carrier in the blood, (d) forms antibodies and offers resistance to, dieases, The helical structure of protein is stabilized by, [2004], (a) dipeptide bonds (b) hydrogen bonds, (c) ether bonds, (d) peptide bonds, Which is not a true statement?, [2002], (a) a-Carbon of a-amino acid is asymmetric, (b) All proteins are found in L-form, (c) Human body can synthesize all proteins, they need, (d) At pH = 7 both amino and carboxylic, groups exist in ionised form, O, || · ·, For - C - N H - (peptide bond), [2001], Which statement is incorrect about peptide bond?, (a) C — N bond length in proteins is longer, than usual bond length of the C — N bond, (b) Spectroscopic analysis shows planar, structure of the — C— NH — group, ||, , 36., 37., , 38., , O, (c) C — N bond length in proteins is smaller, than usual bond length of the C—N bond, (d) None of the above, The number of essential amino acids in man is, (a) 8, (b) 10, [2000], (c) 18, (d) 20, The dominant cation in the blood plasma, (extracellular fluid) is, [1999], (a) potassium, (b) calcium, (c) magnesium, (d) sodium, The reactions of (a) oxygen and (b) carbon, monoxide with haeme (the prosthetic group of, haemoglobin) give, [1997], (a) only oxygen-haeme complex, (b) only carbon monoxide-haeme complex, (c) both oxygen-haeme and carbon monoxidehaeme complexes but oxygen-haeme, complex is more stable
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EBD_8336, 308, , 39., , 40., , 41., , 42., , CHEMISTRY, (d) both oxygen-haeme and carbon monoxidehaeme complexes but carbon monoxidehaeme complex is more stable, Identify an element amongst the following which, is responsible for oxidation of water to O2 in, biological processes, [1997], (a) Fe, (b) Mn, (c) Mo, (d) Cu, In reference to biological role, Ca2+ ions are, important in, [1996], (a) triggering the contraction of muscles, (b) generating right electrical potential across, cell membrane, (c) hydrolysis of ATP, (d) defence mechanism, A reagent suitable for the determination of, N-terminal residue of a peptide is, [1996], (a) p-Toluenesulphonyl chloride, (b) 2, 4-Dinitrophenylhydrazine, (c) Carboxypeptidase, (d) 2, 4-Dinitrofluorobenzene, Which of the following protein destroy the, antigen when it enters in body cell?, [1995], (a) Antibodies, (b) Insulin, (c) Chromoprotein, (d) Phosphoprotein, , 46., , 47., , 48., , 49., , Enzyme (B), , Topic 3: Nucleic Acid and Enzymes, 43., , 44., , 45., , The correct statement regarding RNA and DNA,, respectively is, [2016], (a) The sugar component in RNA is arabinose, and the sugar component in DNA is, 2'-deoxyribose., (b) The sugar component in RNA is ribose and the, sugar component in DNA is 2'-deoxyribose., (c) The sugar component in RNA is arabinose, (d) The sugar component in RNA is, 2'-deoxyribose and the sugar component, in DNA is arabinose, In DNA, the linkages between different, nitrogenous bases are:, [NEET Kar. 2013], (a) peptide linkage, (b) phosphate linkage, (c) H-bonding, (d) glycosidic linkage, Which one of the following, statements is, incorrect about enzyme catalysis?, [2012], (a) Enzymes are mostly proteinous in nature., (b) Enzyme action is specific., (c) Enzymes are denaturated by ultraviolet, rays and at high temperature., (d) Enzymes are least reactive at optimum, temperature., , The segment of DNA which acts as the, instrumental manual for the synthesis of the, protein is:, [2009], (a) ribose, (b) gene, (c) nucleoside, (d) nucleotide, In DNA, the complimentary bases are:, [1998, 2008], (a) Adenine and thymine; guanine and cytosine, (b) Adenine and thymine ; guanine and uracil, (c) Adenine and guanine; thymine and cytosine, (d) Uracil and adenine; cytosine and guanine, RNA and DNA are chiral molecules, their, chirality is due to, [2007], (a) chiral bases, (b) chiral phosphate ester units, (c) D-sugar component, (d) L-sugar component., During the process of digestion, the proteins, present in food materials are hydrolysed to, amino acids. The two enzymes involved in the, process, Enzyme (A), Proteins ¾¾¾¾¾, ® Polypeptides, , 50., , 51., , 52., , ¾¾¾¾¾® Amino acids, are respectively, [2006], (a) Diastase and Lipase, (b) Pepsin and Trypsin, (c) Invertase and Zymase, (d) Amylase and Maltase, A sequence of how many nucleotides in, messenger RNA makes a codon for an amino, acid?, [2004], (a) Three, (b) Four, (c) One, (d) Two, The enzyme which hydrolyses triglycerides to, fatty acids and glycerol is called, [2004], (a) Maltase, (b) Lipase, (c) Zymase, (d) Pepsin, Chargaff's rule states that in an organism [2003], (a) Amounts of all bases are equal, (b) Amount of adenine (A) is equal to that of, thymine (T) and the amount of guanine (G), is equal to that of cytosine (C), (c) Amount of adenine (A) is equal to that of, guanine (G) and the amount of thymine (T), is equal to that of cytosine (C), (d) Amount of adenine (A) is equal to that of, cytosine (C) and the amount of thymine, (T) is equal to that of guanine (G)
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309, , Biomolecules, 53., , 54., , 55., , Enzymes are made up of, [2002], (a) Edible proteins, (b) Proteins with specific structure, (c) Nitrogen containing carbohydrates, (d) Carbohydrates, Which of the following is correct about Hbonding in nu cleotide?, [2001], (a) A --- A and T --- T (b) G --- T and A --- C, (c) A --- G and T --- C (d) A --- T and G --- C, Which one of the following chemical units is, certainly to be found in an enzyme?, [1997], OH, H, O, O, (b), N—C, (a), HO, HO, O, O, O, , (c), 56., , 57., , 58., , 1, 2, 3, 4, 5, 6, 7, , (d), , O O, , R, , O, , R, , O, , R, , O, Chemically considering, digestion is basically, [1994], (a) Anabolism, (b) Hydrogenation, (c) Hydrolysis, (d) Dehydrogenation., Enzymes take part in a reaction and, [1993], (a) decrease the rate of a chemical reaction, (b) increase the rate of a chemical reaction, (c) both (a) and (b), (d) none of these, The couplings between base units of DNA is, through :, [1992], (a) Hydrogen bonding, (b) Electrostatic bonding, (c) Covalent bonding, (d) van der Waals forces, , (b), (d), (d), (b), (c), (a), (a), , 8, 9, 10, 11, 12, 13, 14, , (c), (a), (d), (c), (a), (c), (a), , 15, 16, 17, 18, 19, 20, 21, , (b), (c), (c), (a), (d), (c), (b), , 22, 23, 24, 25, 26, 27, 28, , (a), (c), (c), (c), (c), (c), (c), , Topic 4: Vitamins and Hormones, 59., , 60., 61., , 62., 63., 64., , 65., 66., , 67., 68., , Which of the following hormones is produced, under the condition of stress which stimulates, glycogenolysis in the liver of human beings?, (a) Thyroxin, (b) Insulin, [2014], (c) Adrenaline, (d) Estradiol, Deficiency of vitamin B1 causes the disease, (a) Convulsions, (b) Beri-Beri [2012], (c) Cheilosis, (d) Sterility, Which of the following hormones contains, iodine?, [2009], (a) Testosterone, (b) Adrenaline, (c) Thyroxine, (d) Insulin, Which one of the following is an amine hormone ?, (a) Thyroxine, (b) Oxypurin [2008], (c) Insulin, (d) Progesterone, Which of the following is water-soluble? [2007], (a) Vitamin E, (b) Vitamin K, (c) Vitamin A, (d) Vitamin B, Which one of the following is a peptide, hormone ?, [2006], (a) Testosterone, (b) Thyroxin, (c) Adrenaline, (d) Glucagon, The human body does not produce, [2006], (a) Vitamins, (b) Hormones, (c) Enzymes, (d) DNA, The hormone that helps in the conversion of, glucose to glycogen is, [2004], (a) Cortisone, (b) Bile acids, (c) Adrenaline, (d) Insulin, Vitamin B12 contains, [2003], (a) Ca(II), (b) Fe(II), (c) Co(III), (d) Zn(II), Which of the following is a steroid hormone?, [1999], (a) Cholesterol, (b) Adrenaline, (c) Thyroxine, (d) Progesterone, , ANSWER KEY, 29 (c) 36 (b), 30 (c) 37 (d), 31 (c) 38 (d), 32 (c) 39 (b), 33 (b) 40 (a), 34 (b) 41 (d), 35 (a) 42 (a), , 43, 44, 45, 46, 47, 48, 49, , (b), (c), (d), (b), (a), (c), (b), , 50, 51, 52, 53, 54, 55, 56, , (a), (b), (b), (b), (d), (b), (c), , 57, 58, 59, 60, 61, 62, 63, , (b), (a), (c), (b), (c), (a), (d), , 64, 65, 66, 67, 68, , (d), (a), (d), (c), (d)
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EBD_8336, 310, , CHEMISTRY, , Hints & Solutions, HO, , Hydrolysis, (b) Sucrose ¾¾¾¾®, a-D-Glucose +, b-D-Fructose, , 1., , CH2OH, O, , H, HO, , O, , H HOH2C, , H, OH, , O, , H, , H, , OH, , H, , O, , OH, , HO, H, , CH2–OH, , D(+)glucose, , 5., , Reducing sugars have hydroxyl group at the, anomeric C1 position. In solution, a carbohydrate, can open up to its aldehyde form. Hence, it can, switch configuration between a– and b– forms., , 6., 7., , 8., , 9., , (a) Sucrose does not have free — CHO or CO, group, hence it does not undergo mutarotation., (a) Fructose, a ketose as the substrate, under, the alkaline medium of Tollen’s reagent, a part of, fructose is transformed into glucose and, mannose, both aldoses. Then these aldoses give, positive silver mirror test., (c) Cell membranes (Plasma membranes) constitutes bilayer of phospholipid with embedded, proteins. In humans, lipids accounts for upto, 79% of cell membrane., (a), HO, , HO, , OH, , O, OH, , CH2, OH, , OH, OH, , a-D-Glucopyranose, , +, HO, , OH, , (c) All reducing sugar shows mutarotation., , H, CH2, , CH2, OH, , The glycoside linkage is between C1 and C2., , (d) Glucose reacts with hydroxyl amine to form, an oxime., CHO, CH=N–OH, OH, OH, H, NH2OH HO, H, H, ¾¾¾, ®, OH, OH, H, OH, OH, H, , HO, , + H2O, , Sucrose is a disaccharide of a–D-Glucopyranose, and b–D-Fructofuranose., , CH2OH, , b-D-Fructose, , CH2–OH, , O, HO, , Sucrose, , (d) Sucrose is non-reducing disaccharide as, the two monosaccharide units are linked through, their respective carbonyl groups., , 4. (b), , OH, , OH, , Only carbonyl group can be reduced as with other, carbons, –OH group is attached., , H, HO, H, H, , CH2, , OH, , Sucrose on hydrolysis brings about a change in the, sign of rotation from dextro to laevo and the product, is called as invert sugar., , 3., , O, , ¾¾, ®, OH, , a-D-Glucose, , 2., , OH, , CH2, , H, , O, OH, , H, , H2C, OH, , b-D-Fructofuranose, , OH, , 1, C, 2, 3, , H, OH, H, , 4, , O, , b – D –(+)– Glucose, , OH, , 5, CH2OH, , Carbon atoms from C 1 to C 5 are chiral
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311, , Biomolecules, 10., , 11., , (d) It is a common pathway for both the aerobic, & anaerobic respiration in which 1 glucose, molecule is converted to 2 molecules of pyruvate., (c) Phospholipids are derivatives of glycerol, in which two of the hydroxyl groups are, esterified with fatty acids while the third is, esterified with some derivative of phosphoric, acid with some alcohol such as choline,, ethanolamine, serine or inositol., R¢¢, O = P – O – CH2, OH, , 13., 14., , 15., 16., 17., , CH – O – C –R¢, , O, (a) We know that cellulose (C 6 H12 O 6 ) n is, the chief constituent of cell walls of plants. It is, the most abundant organic substance found in, nature. It is a polymer of glucose with 3500, repeating units in a chain., (c) Glucose contain aldehyde group. Hence, it, gives positive fehling solution test., (a) Starch is also know as amylum which, occurs in all green plants. A molecule of starch, (C 6 H10 O5 ) n is built of a large number of, a-glucose ring joined through oxygen-atom., (b) Chlorophyll contain Mg., (c) Fructose is the sweetest sugar., (c) C 6 H12 O 6 + 6O 2 + 2 ATP ¾, ¾®, , 6CO 2 + 6H 2O + 38 ATP Molecules, , 18., , 19., , CHOH, warm, , ¾¾®, , (CHOH)3, CH2OH, , Net total number of ATP molecules evolved =, 36 molecules., (a) In the lipid metabolism, when palmitic acid, is oxidised, two carbon fragments are removed, sequentially to form acetyl coenzyme. It enters, the citric acid cycle for production of 130 ATP., (d) Glucose reacts with one molecule of, phenyl hydrazine to give phenyl hydrazone., When warmed with excess of phenylhydrazine,, the secondary alcoholic group adjacent to the, aldehyde group is oxidised by another molecule, of phenylhydrazine to a ketonic group. With this, , (CHOH)3, CH2OH, , Glucose, , Glucose Phenylhydrozone, , H2NNHC6H5, , O, , CH2O – C –R, , 12., , CHOH, , ¾¾®, , O, , ketonic group, the third molecule of, phenylhydrazine condenses to glucosazone., Therefore, the value of X is 3, CHO + H2NNHC6H5, CH = NNHC6H5, , CH = NNHC6H5, , CH = NNHC6H5, H2NNHC6H5, – H2 O, , C = NNHC6H5, , C=O, (CHOH)3, , (CHOH)3, CH2OH, , CH2OH, , Glucosazone, , keto compound of glucose, phenyl hydrazone, , 20., , (c) The hydrolysis of sucrose by boiling with, mineral acid or by enzyme invertase or sucrase, produces a mixture of equal molecules of D(+), glucose and D(–) Fructose., HCl, , ® C6 H12 O6, C12 H 22 O11 + H 2 O ¾¾¾, sucrose, [ a D ]=+66.5º, , 21., , + C6 H12 O6, D - Fructose, D - glu cos e, [ aD ]=+52.5º [a D ]=-92º, 3, 144444244444, Invert sugar,[ a D ]=-20º, , (b) a-D glucose and b-D glucose are the, isomers which differ in the orientation, (configuration) of H and OH groups around C1, atom., H —C1—OH, H —C2—OH, O, HO —C3—H, H —C4—OH, 5, , H —C, , 6, , CH2OH, , a– D - glucose, , 1, , HO —C —H, H —C2—OH, 3, , HO —C —H, H —C4—OH, 5, , H —C, , 6, , CH2OH, , b– D - glucose, , O
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EBD_8336, 312, , 22., , CHEMISTRY, (a) Manufacture - By hydrolysis of starch with, hot dil. mineral acids, , HO 2CCHC H 2S - SCH 2 CHC O 2 H, |, |, NH 2, NH2, , (C6 H10O5 ) n ® (C6 H10O5 )n ®, Starch, , Cystine, , Dextrin, , 31., , H O, , 2 ® 2C H O, C12 H 22 O11 ¾¾¾, 6 12 6, , Maltose, , 23., , (c), , Glucose, , H2N – (CH2)4 – CH – COO–, +, , - C OH + H — N— ¾¾, ® — C– N— + H2 O, , NH3, , |, , 25., , Since it contains more number of –NH2 groups, as compared to –COOH groups, hence it is basic, amino acid., (c) During denaturation primary structure of, proteins remain intact while secondary and, tertiary structures are destroyed., (c) Alanine is non-essential amino acid., , 26., , (c), , , HOOC – CH 2 – NH 2 , Glycine, , 28., , 32., , 33., +, , Zwitter ion, , (c) Due to denaturation of proteins, helix get, uncoiled and protein loses its biological activity., (c) Peptide bond, — C — NH —, ||, O, , 29., , 30., , (c) When the proteins are subjected to the, action of heat, mineral acids or alkali, the water, soluble form of globular protein changes to water, insoluble fibrous protein. This is called, denaturation of proteins. During denaturation,, secondary and tertiary structures of protein, destroyed but primary structures remains intact., (c), , 2R - S - H, Thiol, , R -S-S- R, Disulphide, , Example :, 2HO 2C CHC H 2SH, |, NH2, Cysteine, , [O], [H], , ||, , H, , |, , O H, , Carboxyl group Amine group of, of one amino acid other amino acid, , OOC – CH 2 – NH 3, , 27., , |, , O, , Lysine, , 24., , (c) The bond formed between two amino acids, by the elimination of a water molecule is called, a peptide linkage or bond. The peptide bond is, simply another name for amide bond., , Peptide bond, , The product formed by linking amino acid, molecules through peptide linkages. —CO—, NH—, is called a peptide., (c) Haemoglobin acts as an oxygen carrier in, the blood since it reacts with oxygen to form, unstable oxyhaemoglobin which easily breaks, to give back haemoglobin and oxygen., (b) Hydrogen bonding between different units, is responsible for holding helix in a position., NH group in one unit is linked to carbonyl, oxygen of the third unit by hydrogen bonding., The a-helix structure is formed when the chain of, a-amino acids coils as a right handed screw (called, a-helix) because of the formation of hydrogen, bonds between amide groups of the same peptide, chain., , 34., 35., , (b) All proteins are not found in L-form but, they may be present in form of D or L, (a) Due to resonance C — N bond in protein, acquires double bond character and is smaller, than usual C — N bond., —, O, O, C, , 36., , NH, , C, , NH, , Å, , (b) There are 20 amino acid in man, out of, which 10 amino acids are essential amino acids., Essential amino acids are supplied to our bodies, by food which we take because they cannot be, synthesised in the body. These are (i) valine (ii), leucine (iii) Isoleucine (iv) Phenylalanine (v), Threonine (vi) Methionine (vii) Lysine (viii), Tryptophan (ix) Arginine (x) Histidine.
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313, , Biomolecules, 37., , (d) Na+, Cl– and HCO3– are the electrolytes, present in blood plasma. Na+ and Cl– helps in, maintainence of osmotic pressure and fluid, balance., , 38., , (d), , H+, O2N, , Hb + CO ¾, ¾® HbCO (stable compound), , NO2, , carboxy haemoglobin, , , , , Hb + O 2, , 40., , 41., , (b) Water oxidation is catalyzed by a, manganese containing cofactor. Here, cofactor, is a species which increases the activity of an, enzyme., (a), (a) Ca2+ ions are responsible for triggering, contraction of muscles and functions of, nerves., (b) K+ (and Na + ions) are responsible for, maintaining right electrical potential across, cell membrane., (c) Mg2+ act as a cofactor for hydrolysis of, ATP., , (DNP derivative), , H 2 NCHCOOH + H 2 NCHCOOH, |, |, R2, R3, , HbO 2, , oxyhaemoglobin (unstable)., , 39., , NH CHCOOH, |, +, R1, , (Amino acid mixture), , 42., , (a) When antigens enter in to the body cells, and destroy them, then antibodies being, proteins are synthesised in the body and, combine with antigens and destroy these, antigens by forming inactive complexes., Therefore, antibodies protein destroy antigens., , 43., , (b) Sugar in DNA is 2-deoxyribose whereas, sugar in RNA is ribose., , 44., , (c) The base pairs of the two strands of DNA, are linked together through H-bonds., , (d) 2, 4- Dinitroflurobenzene, also known as, Sanger's reagent, reacts with the H2N - group of, the peptide to from 2, 4- dinitrophenyl (DNP), derivative of the peptide. The DNP derivative of, the peptide is hydrolysed to give DNP derivative, of the single amino acid, , 45., , (d) Enzymes are most reactive at optimum, temperature. The optimum temperature for, enzyme activity is the human body temp., , 46., , (b) The DNA sequence that codes for a, specific protein is called a Gene and thus every, protein in a cell has a corrosponding gene., , H 2 NCHCONH CHCONH CHCOOH +, |, |, |, R3, R2, R1, , 47., , (a) In DNA the complimentary bases are, Adenine and thymine;, Guanine and cytosine., The genetic information for cell is contained in, the sequence of bases A, T, G and C in DNA, molecule., , ¾®, F ¾, , O2N, NO2, , Adenine pair with thymine with two hydrogen, bonds and Guanine pairs with cytosine with three, hydrogen bonds., , 2, 4, Dinitrofluoro benzene, , NHC HCONH C HCONH C HCOOH, , O2N, NO2, , |, R1, , |, R2, , (DNP derivative of polypeptide), , |, R3, , 48., , (c) Each nucleic acid consists of a pentose, sugar a heterocyclic base, and phosphoric acid., The sugar present in DNA is -D (–) -2-deoxy, ribose and the sugar present in RNA is D (–)-
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EBD_8336, 314, , CHEMISTRY, ribose. The chirality of DNA and RNA molecules, are due to the presence of sugar components., , 54., , O, , 5, , HOCH2, , OH, , 4, , 1, , H, , H, 3, , H, , 53., , H, , OH, , 2, , OH, , O, , C1, C2, C3 and C4 are chiral carbons., , O, , 5, , OH, , 4, , 1, , H, , H, 3, , H, , OH, , H, , 55., , 2, , 56., , Pepsin, Proteases, , Trypsin, Chemotrypsin, (Pancreatic juice Intestine), , 50., , (a) The sequence of bases in mRNA are read, in a serial order in groups of three at a time., Each triplet of nucleotides (having a specific, sequence of bases) is known as codon. Each, codon specifies one amino acid., , 57., , 58., , There are four bases A, G, C,T, therefore, 4 3 = 64, triplets or codons are possible., , 51., , (b) Triglycerides are lipids, hence these are, hydrolysed by lipases to glycerol and fatty acids., , 52., , (b), , 59., , A G, = =1, T C, , Amount of A = T and that of G = C., , Deoxyribose-Adenine ... Thymine -Deoxyribose, , The hydrogen bonds are formed between the base, (shown by dotted lines). Because of size and, geometrics of the bases, the only possible pairing, in DNA are between G (Guanine) and C (Cytosine), through three H-bonds and between A (Adenosine), and T (Thymine) through two H-bonds., (b) Peptide bonds are present in enzyme., , O, , Proteins ¾¾¾¾® Polypeptides, , ® Amino acids, ¾¾¾¾¾¾¾¾¾¾, , Deoxyribose-Guanine ... Cytosine -Deoxyribose, , N C, , D(–) -2-deoxy ribose, , 49., , Deoxyribose-Guanine ... Cytosine -Deoxyribose, , O, P OH, O, P OH, O, P OH, , H, , H, , C1, C3 and C4 are chiral carbons., (b) Pepsin and Trypsin are two enzymes, involved in the process (hydrolysis of proteins), , Deoxyribose-Adenine ... Thymine -Deoxyribose, , OH P, O, OH P, O, OH P, , D(–)–ribose, , HOCH2, , (b) Enzymes are made up of protein with, specific structure., (d), , 60., , (c) In digestion, large molecules are, hydrolysed to give smaller molecules. For, example, protein gets hydrolysed to amino acids., Thus, chemically considering, digestion is, basically hydrolysis., (b) Enzymes being biocatalyst can increase the, rate of a reaction upto 10 million times. Even, very small amount can accelerate a reaction., (a) DNA consists of two polynucleotide, chains, each chain forms a right handed helical, spiral with ten bases in one turn of the spiral., The two chains coil to double helix and run in, opposite direction held together by hydrogen, bonding., (c) Adrenaline is a hormone produced by, adrenal glands during high stress or exciting, situations. This powerful hormone is part of the, human body’s acute stress response system,, also called the fight or flight response., (b) Beri-Beri.
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315, , Biomolecules, 61., , (c) Thyroxine is the only hormone among the, given choices that contains iodine. Its structure, is as follows:, , Vitamin H also known as Biotin is part of B, complex group of vitamin., , 64., , OH, , I, , I, , 65., , I, O, I, H2N, , 62., 63., , COOH, , (a) Thyroxine is an amino hormone., (d) Vitamin B is water soluble whereas all other, are fat soluble., , 66., 67., 68., , (d) Testosterone and Adrenaline are steroid, hormone, Thyroxin is a non-steroid hormone and, glucagon is a peptide hormone., (a) Vitamins are organic substances which, does not provide energy but are essential for, healthy growth and proper functioning of body., Vitamins are not synthesized inside human body, but they are essential part of our diet., (d) Insulin helps in converting glucose to, glycogen., (c) Vit B12 also called Cyanocobalamin, is, anti-pernicious anaemia vitamin., (d) Progesterone (Gestogens) is a steroid, hormone, which controls the development and, maintainance of pregnancy. Thryoxine and, Adrenaline are Amine hormones.
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EBD_8336, 316, , CHEMISTRY, , 29, , Polymers, , Trend Analysis with Important Topics & Sub-Topics, 2020, Topic Name, Classification of, polymers, Preparation and, properties of polymers, , Sub-Topic, natural polymers, biodegradable, polymers, properties of, polymers, properties of, natural rubber, , LOD - Level of Difficulty, , 1, , 2017, , A - Average, , 5., , Which of the following is a natural polymer?, [2020], , (b) polybutadiene, 6., , (c) poly (Butadiene-acrylonitrile), , A, , D - Difficult, , Qns - No. of Questions, , Of the following which one is classified as, polyester polymer ?, [2011], (a) Terylene, , (b) Bakelite, , (c) Melamine, , (d) Nylone-66, , [NH(CH 2 ) 6 NHCO(CH 2 ) 4 CO]n, , (d) cis-1, 4-polyisoprene, 2., , The biodegradable polymer is:, , [2019], , 4., , (a) addition polymer, , (a) nylon-6, 6, , (b) thermosetting polymer, , (b) nylon 2-nylon 6, , (c) homopolymer, , (c) nylon-6, , (d) copolymer, , Nylon is an example of :, , is a, [2006], , (d) Buna-S, 3., , 2016, , E, 1, , E - Easy, , (a) poly (Butadiene-styrene), , 2018, , E, 1, , Topic 1: Classification of Polymers, 1., , 2019, , QNS. LOD QNS. LOD QNS. LOD QNS. LOD QNS. LOD, , 7., [NEET 2013], , Which one of the following is a chain growth, polymer?, [2004], , (a) Polysaccharide, , (b) Polyamide, , (a) Starch, , (b) Nucleic acid, , (c) Polythene, , (d) Polyester, , (c) Polystyrene, , (d) Protein, , Which one of the following is not a, condensation polymer ?, [2012], (a) Melamine, , (b) Glyptal, , (c) Dacron, , (d) Neoprene, , 8., , In elastomer, intermolecular forces are [1995], (a) strong, (c) nil, , (b) weak, (d) none of these
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317, , Polymers, 14., , Topic 2: Preparation and Properties, of Polymers, 9., , [2014], , Regarding cross-linked or network polymers,, which of the following statements is incorrect?, [2018], , (a) Propylene and para HO—(C6H4)— OH, , (a) They contain covalent bonds between, various linear polymer chains., , (c) Terephthalic acid and ethylene glycol, , (b) They are formed from bi- and tri-functional, monomers., , (b) Benzoic acid and ethanol, (d) Benzoic acid and para HO–(C6H4)—OH, 15., , (c) They contain strong covalents bonds in, their polymer chains., Natural rubber has, , Which is the monomer of Neoprene in the, following ?, [NEET 2013], (a), , (d) Examples are bakelite and melamine., 10., , Which of the following organic compounds, polymerizes to form the polyester Dacron?, , (b) CH 2 = C ¾ CH = CH 2, , (b) all trans-configuration, , Cl, , (c) alternate cis-and trans-configuration, , 12., , (c) CH2 = CH ¾ C º CH, , (d) random cis-and trans-configuration, Biodegradable polymer which can be produced, from glycine and aminocaproic acid is : [2015], (a) PHBV, , (d) CH2 = CH ¾ CH = CH2, 16., , (a) Artificial silk is derived from cellulose., , (c) Nylon 6, 6, (d) Nylon 2- nylon 6, Caprolactum is used for the manufacture of :, (a) Nylon - 6, , (b) Teflon, , (c) Terylene, , (d) Nylon - 6,6, , (b) Nylon-6, 6 is an example of elastomer., (c) The repeat unit in natural rubber is, isoprene., (d) Both starch and cellulose are polymers of, glucose., , Which one of the following is an example of a, thermosetting polymer? [2014], (a), , (b), , (c), , 17., , Which one of the following sets forms the, biodegradable polymer?, [2012 M], (a) CH2 = CH – CN and CH2 = CH – CH = CH2, , ( CH 2 - C = CH - CH 2 ) n, |, Cl, , (b) H2N – CH2 – COOH and, H2N–(CH2)5 – COOH, (c) HO – CH2 – CH2 – OH and, , ( CH 2 - CH ) n, |, Cl, , H, H O, O, | ||, ||, |, ( N - (CH 2 )6 - N - C - (CH 2 )4 - C )n, , OH, (d), , Which of the following statements is false?, [2012], , (b) Buna - N, , [2015 RS], , 13., , OH, , CH2, , ½, , CH 3, , [2016], , (a) all cis-configuration, , 11., , CH 2 = C ¾ CH = CH 2, , HOOC, , (d), , CH2, n, , COOH, , CH = CH2 and, CH2 = CH – CH = CH2
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319, , Polymers, 28. Natural rubber is a polymer of, , CH3, 25., , Monomer of — C — CH2 —, , CH3, , 26., , is, , (a) butadiene, , [2002], , n, , (a) 2-methylpropene, , (b) Styrene, , (c) Propylene, , (d) Ethene, , (b) isoprene, , (c) 2-methylbutadiene (d) Hexa-1, 3-diene, 29. Nylon 6, 6 is a polyamide obtained by the reaction, of, [1996], (a) COOH(CH2)4 COOH + NH2C6H4NH2–(p), (b) COOH(CH2)4 COOH + NH2 (CH2)6 NH2, , Which of the following is not correctly matched?, [2001], O, , (c) COOH (CH2)6 COOH + NH2 (CH2)4 NH2, (d) COOHC6H4 COOH–(p) + NH2 (CH2)6 NH2, , O, , (a) Terylene –OCH2–CH2–C–, , [1999], , 30. Bakelite is prepared by the reaction between, , –C –, , (a) urea and formaldehyde, , n, , [1995], , (b) ethylene glycol, (b) Neoprene, , (c) phenol and formaldehyde, (d) tetramethylene glycol, , — CH 2 — C = CH — CH 2 —, , 31. An example of biopolymer is, , Cl, , n, , (c) Nylon-66, , O, , (a) Teflon, , (b) Neoprene, , (c) Nylon-6, 6, , (d) DNA, , Topic 3: Uses of Polymers, , O, , –NH–(CH2)6–NH –C–(CH2)4–C–O–, , 32. The polymer that is used as a substitute for wool, in making commercial fibres is, [NEET Odisha 2019], , n, , CH3, (d) PMMA, , – CH2–C, COOCH3, , 27. CF2 = CF2 is a unit of, , [1994], , n, , [2000], , (a) Teflon, , (a) Buna-N, , (b) melamine, , (c) nylon-6, 6, , (d) polyacrylonitrile, , 33. Which one of the following is used to make "nonstick" cook-wares ?, [1997], (a) Polystyrene, , (b) Buna – S, , (b) Polyethylene terephthalate, , (c) Bakelite, , (c) Polytetrafluoroethylene, , (d) Polyethene, , (d) Polyvinyl chloride, A NSW E R K E Y, , 1, , (d ), , 5, , (a), , 9, , (c), , 13, , (d ), , 17, , (b ), , 21, , (d ), , 25, , (a), , 29, , (b ), , 2, , (b ), , 6, , (d ), , 10, , (a), , 14, , (c), , 18, , (a), , 22, , (a ), , 26, , (a), , 30, , (c), , 3, , (b ), , 7, , (c), , 11, , (d ), , 15, , (b ), , 19, , (a), , 23, , (b ), , 27, , (a), , 31, , (d ), , 4, , (d ), , 8, , (b ), , 12, , (a), , 16, , (b ), , 20, , (b ), , 24, , (d ), , 28, , (b ), , 32, , (d ), , 33, , (c)
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EBD_8336, 320, , CHEMISTRY, , Hints & Solutions, 1., 2., 3., 4., 5., 6., , 7., , (d) cis-1,4-Polyisoprene is a nautral polymer., (b) nylon 2 – nylon 6, (b) Nylon is a synthetic polyamide polymer., (d) Neoprene is an addition polymer of, isoprene., (a) Polyesters are condensation polymers of a, diabasic acid and a diol. e.g., Terylene., (d) The given compound is a copolymer of, hexamethylene diamine and adipic acid. It is, actually Nylon-6,6., (c) Chain growth polymers involve a series, of reactions, each step of which consumes a, reactive particle and produce another similar, particle. The reactive particles may be free, radicals or ions to which monomers get added, by a chain reaction., , 11., , (d) H2 N—CH2—COOH, Glycine, , + H2 N — (CH2)5—COOH ®, Aminocaproic acid, , (HN—CH2—CO—NH— (CH2)5—CO) —, Nylon-2-nylon-6, , 12., , (a), O, , O, , H, N, , NH, , n, , Nylon–6, (Caprolactum), , 13., , (d) Thermosetting polymers undergo chemical, changes when heated and set to hard mass when, cooled e.g. Bakelite., , Benzoyl Peroxide, nC6H5 – CH = CH2 ¾¾¾¾¾¾¾, ®, , C 6H 5, , (CH, , CH2)n, , Polystyrene, , 8., , (b) Elastomers are the polymers having very, weak intermolecular forces of attraction between, the polymer chain. The weak forces permit the, polymer to be streched., , 9., , (c) Cross-linked or network polymers are, usually formed from bi-functional and trifunctional monomers and contain strong, covalent bonds between various linear polymer, chains like melamine, bakelite etc., , 10., , (a) Natural rubber is found to be a polymer of, cis-isoprene i.e. it is cis-polyisoprene, H2C, , C — CH, |, CH3, Isoprene, , polymerisation, CH2 ¾¾¾¾¾¾, ®, , CH2 CH2, , H3C, , H, , cis-polyisoprene, , Thermoset plastics contain polymers and cross, linked together during the curing process to form, an irreversible chemical bond. Cross linking process, making thermosets plastic ideal for high heat, applications., , 14., , (c), , n HO.CH2CH2OH+n HOOC, Ethylene glycol, , COOH D, , COOH, Terephthalic acid, , O, ||, -( O CH2CH2–O– C, Terylene, , O, ||, C)n
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323, , Chemistry in Everyday Life, , 30, , Chemistry in, Everyday Life, , Trend Analysis with Important Topics & Sub-Topics, 2020, Topic Name, Drugs and their, classification, Cleansing agents, LOD - Level of Difficulty, , 1., , 2., , 3., , 4., , 5., , 6., , 2019, , 2018, , 2017, , 2016, , Sub-Topic, QNS. LOD QNS. LOD QNS. LOD QNS. LOD QNS. LOD, Therapeutic action, 1, E, 1, E, 1, E, of drugs, detergents, 1, A, E - Easy, , Which of the following is a cationic detergent?, (a) Sodium stearate, [2020], (b) Cetyltrimethyl ammonium bromide, (c) Sodium dodecylbenzene sulphonate, (d) Sodium lauryl sulphate, The artificial sweetener stable at cooking, temperature and does not provide calories is, [NEET Odisha 2019], (a) alitame, (b) saccharin, (c) aspartame, (d) sucralose, Among the following, the narrow spectrum, antibiotic is:, [2019], (a) penicillin G, (b) ampicillin, (c) amoxycillin, (d) chloramphenicol, Mixture of chloroxylenol and terpineol acts as:, [2017], (a) antiseptic, (b) antipyretic, (c) antibiotic, (d) analgesic, Which of the following is an analgesic? [2016], (a) Novalgin, (b) Penicillin, (c) Streptomycin, (d) Chloromycetin, Bithionol is generally added to the soaps as an, additive to function as a/an :, [2015], , A - Average, , 7., , 8., , 9., , 10., , D - Difficult, , Qns - No. of Questions, , (a) Dryer, (b) Buffering agent, (c) Antiseptic, (d) Softner, Artificial sweetner which is stable under cold, conditions only is :, [2014], (a) Saccharine, (b) Sucralose, (c) Aspartame, (d) Alitame, Antiseptics and disinfectants either kill or, prevent growth of microorganisms. Identify, which of the following statements is not true:, [NEET 2013], (a) Chlorine and iodine are used as strong, disinfectants., (b) Dilute solutions of Boric acid and, Hydrogen Peroxide are strong antiseptics., (c) Disinfectants harm the living tissues., (d) A 0.2% solution of phenol is an antiseptic, while 1% solution acts as a disinfectant., Dettol is the mixture of [1996, NEET Kar. 2013], (a) Terpineol and Bithionol, (b) Chloroxylenol and Bithionol, (c) Chloroxylenol and Terpineol, (d) Phenol and Iodine, Chloroamphenicol is an :, [2012 M], (a) antifertility drug, (b) antihistaminic, (c) antiseptic and disinfectant, (d) antibiotic-broad spectrum
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EBD_8336, 324, , 11., , 12., , 13., , CHEMISTRY, Which one of the following is employed as, Antihistamine ?, [2011], (a) Chloramphenicol, (b) Diphenyl hydramine, (c) Norothindrone, (d) Omeprazole, Which one of the following is employed as a, tranquilizer drug?, [2010], (a) Promethazine, (b) Valium, (c) Naproxen, (d) Mifepristone, Which one of the following is employed as a, , tranquilizer?, [2009], (a) Naproxen, (b) Tetracycline, (c) Chlorpheninamine(d) Equanil, 14. Which one of the following can possibly be used, as analgesic without causing addiction and, mood modification ?, [1997], (a) Diazepam, (b) Morphine, (c) N-acetyl-para-aminophenol, (d) Tetrahydrocannabinol, , ANSWER KEY, 1, , (b), , 3, , (a), , 5, , (a), , 7, , (c), , 9, , (c) 11, , (b) 13, , (d), , 2, , (d), , 4, , (a), , 6, , (c), , 8, , (b) 10, , (d) 12, , (b) 14, , (c), , Hints & Solutions, 8., , CH3, , 1., , (b) CH3 – (CH2)15 – N – CH3 Br, , –, , CH3, , Cetyltrimethyl ammonium bromide, , 2., , 3., 4., , (b) Dilute solutions of boric acid and, hydrogen peroxide are weak antiseptics., , (d) Sucralose is trichloro derivative of sucrose., It is stable at cooking temperature. It does, not provide calories., (a) Penicillin G, (a) Dettol is a mixture of chloroxylenol and, terpineol which is a very commonly known, antiseptic., , 5., , (a) Novalgin is most widely used as analgesic., Analgesics are pain releiving, , 6., , (c) Bithionol is added to soaps to impart, antiseptic properties., , 7., , (c) Aspartame is stable under cold conditions., , Antiseptics are used for humans, pets, etc. but, disinfectants are used to kill bacteria etc. on floor,, wall surfaces., , 9., , (c) Dettol is a mixture of chloroxylenol and, terpineol., , 10., , (d) Chloroamphenicol is a broad spectrum, antibiotic., , 11. (b), , Diphenyl hydramine also known as, benadryl is an antihistamine., , 12. (b), , Valium is used as a sedative., , 13. (d), , Equanil is an important medicine used in, depression and hypertension., , 14. (c), , We know that N-acetyl-para-aminophenol, (or paracetamol) is an antipyretic which, can also be used as an analgesic to relieve, pains.
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31, , 325, , Nuclear Chemistry, , 1., , Nuclear Chemistry, , 2., , The half life of a substance in a certain enzymecatalysed reaction is 138s. The time required for, the concentration of the substance to fall from, 1.28 mg L–1 to 0.04 mg L–1, is :, [2011], (a) 414 s, (b) 552 s, (c) 690 s, (d) 276 s, A nuclide of an alkaline earth metal undergoes, radioactive decay by emission of the a-particles, in succession. The group of the periodic table, to which the resulting daughter element would, belong is, [2005], (a) Gr. 4, (b) Gr. 6, (c) Gr. 14, (d) Gr. 16, , 3., , The radioactive isotope, , 60, 27 Co, , which is used in, , the treatment of cancer can be made by (n, p), reaction. For this reaction the target nucleus is, (a) 59, 28 Ni, (c), 4., , 5., , 6., , 60, 28 Ni, , (b) 59, 27 Co, (d), , [2004], , 60, 27 Co, , The radioactive isotope, tritium, (13 H ) has a halflife of 12.3 years. If the initial amount of tritium is, 32 mg, how many milligrams of it would remain, after 49.2 years?, [2003], (a) 8 mg, (b) 1 mg, (c) 2 mg, (d) 4 mg, 235 , nucleus absorbs a neutron and, U, 92, disintegrates into 54Xe139, 38Sr94 and x. So what, will be the product x ?, [2002], (a) 3-neutrons, (b) 2-neutrons, (c) a-particle, (d) b-particle, A human body required 0.01M activity of, radioactive substance after 24 hours. Half life of, , radioactive substance is 6 hours. Then injection, of maximum activity of radioactive substance, that can be injected will be, [2001], (a) 0.08 M, (b) 0.04 M, (c) 0.32 M, (d) 0.16 M, If species ba X emits firstly a positron, then two, a and two b and in last one a and finally converted, to species dc Y , so correct relation is, [2001], (a) c = a – 5, d = b – 12, (b) c = a – 6, d = b – 8, (c) c = a – 4, d = b – 12, (d) c = a – 5, d = b – 8, 8. When a radioactive element emits successively, one a-particle and two b-particles, the mass, number of the daughter element, [1999], (a) is reduced by 4 units, (b) remains the same, (c) is reduced by 2 units, (d) is increased by 2 units, 9., Number of neutrons in a parent nucleus X, which, gives 14, nucleus, after two successive b, 7 N, emissions, would be, [1998], (a) 9, (b) 6, (c) 7, (d) 8, 10. Carbon - 14 dating method is based on the fact, that:, [1997], (a) C-14 fraction is same in all objects, (b) C-14 is highly insoluble, (c) Ratio of carbon-14 and carbon-12 is constant, (d) all the above, , 7., , 11., , 235, 1, 92 U + 0 n, , 236, ® 92, U®
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EBD_8336, 326, , CHEMISTRY, , fission products + neutrons + 3.20 × 10–11 J, 235, The energy released when 1 g of 92, U finally, undergoes fission is, [1997], (a) 12.75 × 108 kJ, (b) 16.40 × 107 kJ, (c) 8.20 × 107 kJ, (d) 6.50 × 106 kJ, 24, 12. One microgram of radioactive sodium 11, Na with, a half-life of 15 hours was injected into a living, system for a bio-assay. How long will it take for, the radioactive subtance to fall up to 25% of the, initial value?, [1996], (a) 60 hours, (b) 22.5 hours, (c) 375 hours, (d) 30 hours, 13. Half-life for radioactive 14C is 5760 years. In how, many years, 200 mg of 14C will be reduced to, 25 mg?, [1995], (a) 5760 years, (b) 11520 years, (c) 17280 years, (d) 23040 years, 14. In a radioactive decay, an emitted electron comes, from, [1994], (a) The nucleus of atom, (b) The orbit with principal quantum number 1, (c) the inner orbital of the atom, (d) the outermost orbit of the atom., , 1, 2, , (c), (c), , 3, 4, , (c), (c), , 5, 6, , (a), (d), , 7, 8, , 15. India has the world’s largest deposits of thorium, in the form of, [1994], (a) rutile, (b) magnesite., (c) lignite, (d) monazite., 16. If an isotope of hydrogen has two neutrons in its, atom, its atomic number and atomic mass number, will respectively be, [1992], (a) 2 and 1, (b) 3 and 1, (c) 1 and 1, (d) 1 and 3., 17. The age of most ancient geological formations is, estimated by, [1989], (a) Potassium–argon method, (b) Carbon-14 dating method, (c) Radium-silicon method, (d) Uranium-lead method., 18. Emission of an alpha particle leads to a [1989], (a) Decrease of 2 units in the charge of the atom, (b) Increase of 2 units in the mass of the atom, (c) Decrease of 2 units in the mass of the atom, (d) Increase of 4 units in the mass of the atom., , ANSWER KEY, (a), (a) 11, 9, (a) 10 (c) 12, , (c), (d), , 13, 14, , (c), (a), , 15, 16, , (d), (d), , 17, 18, , (d), (a)
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327, , Nuclear Chemistry, , Hints & Solutions, 1., , 2., , 3., , (c) For a first order reaction, Total time T = no. of half lives (n) × half life (t1/2), n, N æ 1ö, =ç ÷, N0 è 2 ø, where n = no. of half lives, Give N0 (original amount) = 1.28 mg/l, N (amount of substance left after time T ), = 0.04 mgL–1, n, 5, n, n, 1 æ1ö æ1ö, æ1ö, 0.04 æ 1 ö, =ç ÷ ; ç ÷ =ç ÷, \, =ç ÷ ;, è2ø, 1.28 è 2 ø 32 è 2 ø è 2 ø, n =5, T = 5 × 138 = 690, (c) When IIA group element (Ra) emits one, a-particle its group no. decreases by two, unit. i.e., go into zero group (Gr. 16) But as it, i s r a d i oa ct i ve t h us d ue t o successi ve, emission last product is Pb i.e., (Gr.14)., 1, (c) [ X ] +10 n ®60, 27 Co +1 H, Balancing the mass and atomic numbers on both, 1, +10 n ®60, 27 Co +1 H, Thus X should be 60, 28 Ni, (c) Given t1/2 = 12.3 years, Initial amount (N0) = 32 mg, Total time = 49.2 years, , sides, , 4., , 60, 28 X, , No. of half lives (n) =, , Used half life time (n) =, , æ1ö, è2ø, , Initial activity = 0.01 × 16 = 0.16 M, 7., , (a), , b, b, 0, a X® a -1 X + 1e, , –2a, , b -8, 4, a -5 X + 2 He, , –2b, , b -8, a -3 X, , –a, 8., , 1, , ZX, , A, , a, , b, , ¾¾, ® Z X A- 4, , 9., , b, -b, 14, 14, (a) 5 X14 ¾-¾®, 6 Y ¾¾® 7 N, No. of neutrons = Mass number – No. of proton, =14 – 5 = 9, (c) By carbon dating method, , Age of wood =, , + 0 n ® 54 Xe, , + 38 Sr, , 94, , +X, , 5., , (a), , 6., , 92 + 0 = 54 + 38 + a Þ a = 0, 235 + 1 = 139 + 94 + b Þ b = 3 So, X = 3 0n1, i.e 3 neutrons., (d) Remaining activity = 0.01M, after 24 hrs, n, æ 1ö, Remaining activity = Initial activity ´ ç ÷, è 2ø, , 92 U, , b, , ¾¾, ® Z - 2 X A - 4 ¾¾, ® Z - 2 +1 X A - 4, , T, 49.2, =, =4, t1/2 12.3, , 139, , b -12, 4, a -5 X + 2 He, , (a) Mass number is affected by emmision of a, particle while b-particle has negligible mass does, not effect mass number. e.g., , 2.303, æN ö, ´ t½ log ç 0 ÷, è Nø, 0.693, , é ratio of C14 / C12 in living wood ù, ú, ê, 14, 12, êë ratio of C / C in dead wood úû, , 4, , 235, , 4, , 0.01 = Initial activity ´ ç ÷, , So,, , 10., , n, 32, æ1ö, æ 1ö, = 2 mg, Now Nt = N0 ç ÷ = 32ç ÷ =, è 2ø, 16, è2ø, Hence 32 mg becomes 2 mg in 49.2 years, , Total time 24, =, =4, T1/2, 6, , Hence, it is based upon the ratio of C14 and C12., 11., , (c) 1 atom of, , 235, 92 U on, , fission gives energy, , = 3.2 × 10–11 J, 6.023 × 1023 atom (1 mole) on fission gives energy, = 3.2 × 10–11 × 6.023 × 1023 J, 1 g of, =, , 235, 92 U, , on fission gives energy, , 6.023, ´ 3.2 ´ 1012 J = 8.2 × 107 kJ, 235
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EBD_8336, 328, , 12., , CHEMISTRY, (d), , n, N, æ 1ö, N = N 0 ç ÷ ; here, N = 0, è 2ø, 4, , n, , 2, , n, , N0, æ 1ö, æ 1ö, æ 1ö, = N0 ç ÷ Þ ç ÷ = ç ÷ Þ n = 2, è 2ø, è 2ø, è 2ø, 4, Total time (T)= n × t1/2 = 2 × 15 = 30 hrs, , 13., , (c) As we know that, n, , N æ 1ö, =ç ÷, N0 è 2 ø, where N 0 original amount of radioactive, sustacnce, N = Amount of substance remain after n half lives, n, , \, , 25 æ 1 ö, 1 æ 1ö, = ç ÷ or = ç ÷, 200 è 2 ø, 8 è 2ø, 3, , 14., , 15., 16., , 17., , n, , C-14 dating method is used to predict the age of, fossils or dead animals or a fallen tree., , n, , æ 1ö, æ 1ö, çè ÷ø = çè ÷ø ; \ n = 3, 2, 2, now T = n × t½, where T = total time, T = 3 × 5760 years = 17280 years, , (a) When a radioactive elements emits a or b, particle the new element formed may have, unstable nucleus. It may further disintegrate by, emitting a- or b particle forming a new element., This process of integration may continue till end, product formed, is a stable compound., (d) The ore of thorium is monazite., (d) As number of neutron =, Mass number – atomic number, Give number of neutron = 2, \ Mass number will be 3 and atomic number, will be one., (d) Age of geological formations (i.e. predicting, the age of the earth and rocks) is estimated by, U– Pb method, also known as helium dating., , or, , 18., , (a) Emission of a -particle ( 42 He ) leads to, decrease of 2 units of charge. e.g., 92 U, , 238, , ¾¾, ®90 Th 234 + 2 He 4
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