33 Years NEET Chapterwise & Topicwise Solved Papers CHEMISTRY 2020 PDF
33 Years NEET Chapterwise & Topicwise Solved Papers CHEMISTRY 2020 PDF
33 Years NEET Chapterwise & Topicwise Solved Papers CHEMISTRY 2020 PDF
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CONTENTS
1. Some Basic Concepts of Chemistry 1-10 7. Equilibrium 70-89
Topic 1 : Significant Figures, Laws of Chemical Topic 1: Law of Mass Action, Equilibrium
Combinations and Mole Concept Constant (Kc and Kp) and its Application
Topic 2: Percent Composition and Empirical Topic 2: Relation between K, Q and G and
Formula Factors Effecting Equilibrium
Topic 3: Stoichiometric Calculations Topic 3: Theories of Acids and Bases, Ionic
2. Structure of Atom 11-20 Product of Water and pH Scale
Topic 4: Ionisation of Weak Acids and Bases
Topic 1 : Atomic Models and Dual Nature of
and Relation between Ka and Kb
Electromagnetic Radiation
Topic 5: Common Ion Effect, Salt Hydrolysis,
Topic 2 : Bohr's model for Hydrogen Atom
Buffer Solutions and Solubility Product
(Emission and Absorption Spectra)
Topic 3: Dual Behaviour of Matter and 8. Redox Reactions 90-93
Heisenberg Uncertainty Principle Topic 1: Oxidation and Reduction Reactions
Topic 4 : Quantum Mechanical Model of Atom Topic 2: Oxidation Number
Topic 3: Disproportionation and Balancing of
3. Classification of Elements and Redox Reactions
Periodicity in Properties 21-27 Topic 4: Electrode Potential and Oxidising,
Topic 1: Modern Periodic Table Reducing Agents
Topic 2: Periodic Trends in Properties of
9. Hydrogen 94-97
Elements
Topic 1: Preparation and Properties of
4. Chemical Bonding and Molecular Hydrogen
Structure28-48 Topic 2: Preparation and Properties of Water
Topic 1: Electrovalent, Covalent and Topic 3: Preparation and Properties of
Co‑ordinate Bonding Hydrogen Peroxide
Topic 2: Octet rule, Resonance and Hydrogen
10. The s-Block Elements 98-104
Bonding Topic 1: Preparation and Properties of Alkali
Topic 3: Dipole Moment and Bond Polarity Metals and their Compounds
Topic 4: VSEPR Theory and Hybridisation Topic 2: Some Important Compounds of
Topic 5: Valence Bond and Molecular Orbital Sodium
Theory Topic 3: Preparation and Properties of Alkaline
5. States of Matter 49-57 Earth Metals and their Compounds
Topic 1: Gas laws and Ideal gas Equation Topic 4: Some Important Compounds of
Topic 2: Kinetic Theory of Gases and Molecular Calcium
Speeds 11. The p-Block Elements
Topic 3 : van der Waal's Equation and (Group 13 & 14) 105-109
liquefaction of Gases Topic 1: Boron Family
Topic 4: Liquid State Topic 2: Carbon Family
6. Thermodynamics 58-69 12. Organic Chemistry - Some Basic
Topic 1: First Law and Basic Fundamentals of Principles and Techniques 110-122
Thermodynamics Topic 1: Classification and Nomenclature of
Topic 2: Laws of Thermochemistry Organic Compounds
Topic 3: Entropy and Second Law of Topic 2: Isomerism in Organic Compounds
Thermodynamics Topic 3: Concept of Reaction Mechanism in
Topic 4: Spontaneity and Gibb's Free Energy Organic Compounds
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13. Hydrocarbons 123-141 22. The d-and f-Block Elements 210-222
Topic 1: Alkanes Topic 1: Characteristics of d-Block Elements
Topic 2: Alkenes Topic 2: Compounds of Transition Metals
Topic 3: Alkynes Topic 3: Lanthanoids and Actinoids
Topic 4: Aromatic Hydrocarbons 23. Coordination Compounds 223-240
14. Environmental Chemistry 142-144 Topic 1: Coordination Number, Nomenclature
Topic 1: Air Pollution and Isomerism of Coordination Compounds
Topic 2: Water and Soil Pollution Topic 2: Magnetic Moment, Valence Bond
Theory and Crystal Field Theory
15. The Solid State 145-151
Topic 3: Organometallic Compounds
Topic 1: Properties and Types of Solids
Topic 2: Crystal Structure of Solids 24. Haloalkanes and Haloarenes 241-254
Topic 3: Cubic System and Bragg's Equation Topic 1: Preparation and Properties of
Topic 4: Imperfection in Solids Haloalkanes
Topic 2: Preparation and Properties of
16. Solutions 152-162
Haloarenes
Topic 1: Solubility and Concentration of
Topic 3: Some Important Polyhalogen
Solutions
Compounds
Topic 2: Vapour Pressure, Laws of Solutions
and Ideal, Non-ideal Solutions 25. Alcohols, Phenols and Ethers 255-268
Topic 3: Colligative Properties and Abnormal Topic 1: Preparation and Properties of Alcohols
Molecular Masses Topic 2: Preparation and Properties of Phenols
Topic 3: Preparation and Properties of Ethers
17. Electrochemistry 163-174
26. Aldehydes, Ketones and
Topic 1: Conductance and Conductivity
Carboxylic Acids 269-290
Topic 2: Electrolysis and Types of Electrolysis
Topic 1: Methods of Preparation of Carbonyl
Topic 3: Cells and Electrode Potential, Nernst
Compounds
Equation
Topic 2: Properties of Carbonyl Compounds
Topic 4: Commercial Cells and Corrosion
Topic 3: Preparation and Properties of
18. Chemical Kinetics 175-187 Carboxylic Acids
Topic 1: Rate of Reaction, Rate Laws and Rate
27. Amines 291-304
Constant
Topic 1: Aliphatic and Aromatic+ Amines
Topic 2: Order of Reaction and Half Life Period
Topic 2: Amides, Cyanides and Isocyanides
Topic 3: Theories of Rate of Reaction Topic 3: Nitrocompounds, Alkyl Nitrites and
19. Surface Chemistry 188-192 Diazonium Salts
Topic 1: Adsorption 28. Biomolecules 305-315
Topic 2: Catalysis and Theories of Catalysis Topic 1: Carbohydrates and Lipids
Topic 3: Colloids and Emulsions Topic 2: Amino Acids and Proteins
20. General Principles and Processes Topic 3: Nucleic Acid and Enzymes
of Isolation of Elements 193-195 Topic 4: Vitamins and Hormones
Topic 1: Occurrence of Metals 29. Polymers 316-322
Topic 2: Metallurgical Processes Topic 1: Classification of Polymers
Topic 3: Purification and Uses of Metals
Topic 2: Preparation and Properties of
21. The p-Block Elements Polymers
(Group 15, 16, 17 and 18) 196-209 Topic 3: Uses of Polymers
Topic 1: Nitrogen Family
30. Chemistry in Everyday Life 323-324
Topic 2: Oxygen Family
Topic 3: Halogen Family 31. Nuclear Chemistry 325-328
Topic 4: Noble Gases
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NEET
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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.
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NEET
(iii)
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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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Some Basic Concepts of Chemistry 1
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2 CHEMISTRY
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Some Basic Concepts of Chemistry 3
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4 CHEMISTRY
43. In an experiment it showed that 10 mL of 0.05 M 50. In Haber process 30 litres of dihydrogen and
solution of chloride required 10 mL of 0.1 M 30 litres of dinitrogen were taken for reaction
solution of AgNO3, which of the following will which yielded only 50% of the expected product.
be the formula of the chloride (X stands for the What will be the composition of gaseous mixture
symbol of the element other than chlorine): under the aforesaid condition in the end? [2003]
[NEET Kar. 2013] (a) 20 litres ammonia, 25 litres nitrogen, 15 litres
(a) X2Cl (b) X2Cl2 hydrogen
(c) XCl2 (d) XCl4 (b) 20 litres ammonia, 20 litres nitrogen, 20 litres
44. 6.02 × 1020 molecules of urea are present in hydrogen
100 mL of its solution. The concentration of (c) 10 litres ammonia, 25 litres nitrogen, 15 litres
solution is : [NEET 2013] hydrogen
(a) 0.01 M (b) 0.001 M (d) 20 litres ammonia, 10 litres nitrogen, 30 litres
(c) 0.1 M (d) 0.02 M hydrogen
51. In the reaction
45. What is the [OH] in the final solution prepared 4 NH3 (g) + 5 O2 (g) ® 4 NO(g) + 6 H2O(l)
by mixing 20.0 mL of 0.050 M HCl with 30.0 mL of When 1 mole of ammonia and 1 mole of O2 are
0.10 M Ba(OH)2? [2009] made to react to completion, [1998]
(a) 0.40 M (b) 0.0050 M (a) 1.0 mole of H2O is produced
(c) 0.12 M (d) 0.10 M (b) 1.0 mole of NO will be produced
46. 10 g of hydrogen and 64 g of oxygen were filled (c) all the oxygen will be consumed
in a steel vessel and exploded. Amount of water (d) all the ammonia will be consumed
produced in this reaction will be: [2009] 52. Liquid benzene (C6H6) burns in oxygen according
(a) 3 mol (b) 4 mol to the equation
(c) 1 mol (d) 2 mol 2C6H6(l) + 15O2(g) ¾® 12CO2(g) + 6H2O(g)
47. How many moles of lead (II) chloride will be
How many litres of O2 at STP are needed to
formed from a reaction between 6.5 g of PbO
complete the combustion of 39 g of liquid
and 3.2 g of HCl ? [2008]
benzene? (Mol. wt. of O2 = 32, C6H6 = 78)
(a) 0.044 (b) 0.333
[1996]
(c) 0.011 (d) 0.029
(a) 74 L (b) 11.2 L
48. Concentrated aqueous sulphuric acid is 98%
(c) 22.4 L (d) 84 L
H2SO4 by mass and has a density of 1.80 g mL– 1.
53. A 5 molar solution of H2SO4 is diluted from
Volume of acid required to make one litre of 0.1M
1 litre to a volume of 10 litres, the normality of
H2SO4 solution is [2007]
the solution will be : [1991]
(a) 16.65 mL (b) 22.20 mL (a) 1 N (b) 0.1 N
(c) 5.55 mL (d) 11.10 mL (c) 5 N (d) 0.5 N
49. The mass of carbon anode consumed (giving 54. One litre hard water contains 12.00 mg Mg2+.
only carbon dioxide) in the production of Mili-equivalents of washing soda required to
270 kg of aluminium metal from bauxite by the remove its hardness is : [1988]
Hall process is (Atomic mass: Al = 27) [2005] (a) 1 (b) 12.16
(a) 270 kg (b) 540 kg (c) 1 × 10–3 (d) 12. 16 × 10–3
(c) 90 kg (d) 180 kg
ANSWER KEY
1 (c) 7 (a) 13 (c) 19 (d) 25 (a) 31 (a) 37 (c) 43 (c) 49 (c)
2 (c) 8 (a) 14 (b) 20 (c) 26 (c) 32 (d) 38 (d) 44 (a) 50 (c)
3 (a) 9 (c) 15 (d) 21 (c) 27 (a) 33 (a) 39 (d) 45 (d) 51 (c)
4 (a) 10 (b) 16 (b) 22 (d) 28 (a) 34 (a) 40 (d) 46 (b) 52 (d)
5 (d) 11 (c) 17 (a) 23 (b) 29 (b) 35 (d) 41 (d) 47 (d) 53 (a)
6 (b) 12 (b) 18 (b) 24 (c) 30 (c) 36 (c) 42 (c) 48 (c) 54 (a)
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Some Basic Concepts of Chemistry 5
M
Molar mass moles
.
ol
u lt m
s m
i pl a s
as y
1
m ly b
yb s
Di
.
(a) Number of Mg atoms = ´ NA ´1
l
tip
v id m a
ym
s o
as m
ul
24
e b ss
m by
ol .
M
ym
e
id
1
iv
o l.
D
(b) Number of O atoms = ´ NA ´ 2 Divide by mol.
32 Mass in mass Number of
1 grams molecules
(c) Number of Li atoms = ´ N A ´ 1 Multiply by mo l.
7 mass
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6 CHEMISTRY
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Some Basic Concepts of Chemistry 7
18. (b) 2g of H2 means one mole of H2, hence (29.2 – 20.2)(1.79 ´ 105 )
contains 6.023 × 1023 molecules. Others have 24. (c)
less than one mole, so have less no. of molecules. 1.37
As the least precise number contains 3
19. (d) Specific volume (volume of 1 g) of significant figures therefore, answers should also
cylindrical virus particle = 6.02 × 10–2 cc/g contains 3 significant figures.
Radius of virus (r) = 7 Å = 7 × 10–8 cm 25. (a) No of moles of nitride ion
Length of virus = 10 × 10–8 cm
4.2
Volume of virus = = = 0.3 mol = 0.3 ´ N A nitride ions.
22 14
pr 2 l = ´ (7 ´ 10 -8 ) 2 ´10 ´10 -8 Valence electrons = 8 × 0.3 NA = 2.4 NA
7
Nitride ion has seven protons in the nucleus and
= 154 × 10–23 cc ten electrons surrounding the nucleus. Therefore
volume total no. of electrons is 10. Number of valence
Wt. of one virus particle =
specific volume electrons is (5 + 3) = 8.
Mol. wt. of virus = Wt. of NA particle 26. (c) According to Avogadro's law "equal
154 ´ 10 -23 volumes of all gases contain equal numbers of
= ´ 6.02 ´ 10 23 molecules under similar conditions of
-2
6.02 ´ 10 temperature and pressure". Thus if 1 L of one
= 15400 g/mol = 15.4 kg/mol gas contains N molecules, 2 L of any gas under
20. (c) BaCO3 ® BaO + CO2 the same conditions will contain 2N molecules.
197 g 19 ´ 10 + 81 ´ 11
27. (a) Average atomic mass =
As 197 g of BaCO3 will release 100
22.4 litre of CO2 at STP = 10.81
\ 1 g of BaCO3 will release 28. (a) 1 mol of CO2 = 44 g of CO2
22.4 4.4
= litre of CO2 4.4 g CO2 = = 0.1 mol CO2
197 44
And 9.85 g of BaCO3 will release carbon dioxide = 6 × 1022 molecules
22 .4 = 2 × 6 × 1022 atoms or 1.2 × 10 23 atoms of oxygen.
= ´ 9.85 = 1.12 litre of CO
197 2 29. (b) 6.02 × 1023 molecules of CO =1mole of CO
21. (c) Given : Percentage of the iron = 0.334%; 6.02 × 1024 CO molecules = 10 moles of CO
Molecular weight of the haemoglobin = 10 g atoms of O = 5 g molecules of O2
= 67200 and atomic weight of iron = 56. We know
that the number of iron atoms 30. (c) C2H4 + 3 O2 ¾ ¾® 2CO + 2H O
2 2
Molecular wt. of haemoglobin ´ % of iron
28 kg 96 kg
= As 28 kg of C2H4 undergo complete combustion
100 ´ Atomic weight of iron
67200 ´ 0.334 by 96 kg of O2
= =4
100 ´ 56 \ 2.8 kg of C2H4 undergo complete combustion
22. (d) Number 161 cm, 0.161 cm and 0.0161cm have by 9.6 kg of O2.
3, 3 and 3 significant figures respectively. 31. (a) Cp / Cv = 1.4 shows that the gas is diatomic.
All non-zero digits are significant and the zeros at 22.4 litre at NTP º 6.02 × 1023 molecules
the beginning of a number are not significant. 11.2 L at NTP = 3.01 × 1023 molecules
No. of atoms in gas = 3.01 × 1023 × 2 atoms
23. (b) Molecular weight of C60H122 = 6.02 × 1023 atoms
= (12 × 60) + 122 = 842.
Cp 2
Therefore, weight of one molecule r= = 1 + , where F = degree of freedom of
Cv F
Molecular weight of C60 H122
= the gas molecules
Avogadro's number For mono atomic gas, F = 3
842 Cp 2 5
= 23
= 1.36 ´ 10-21 g ; 1.4 ´ 10-21 g \ g= = 1 + = = 1.67
6.023 ´ 10 Cv 3 3
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Some Basic Concepts of Chemistry 9
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10 CHEMISTRY
47. (d) Writing the equation for the reaction, we get Again,
PbO + 2HCl ¾¾ ® PbCl2 + H2O No. of gram equivalent of C
207 + 16 2 × 36.5 207 + 71 mass in gram
=
= 223 g = 73g = 278g gram equivalent weight
From this equation we find 223 g of PbO reacts mass
with 73 g of HCl to form 278 g of PbCl2. Þ 30 × 103 =
3
If we carry out the reaction between 3.2 g HCl Þ mass = 90 × 103 g = 90 kg
and 6.5 g PbO.
50. (c) N 2 + 3H 2 ® 2NH 3
Amount of PbO that reacts with 3.2 g HCl 1 vol. 3 vol. 2 vol.
223 10 litre 30 litre 20 litre
= ´ 3.2 =9.77 g.
73 It is given that only 50% of the expected product
Since amount of PbO present is only 6.5 g so is formed hence, only 10 litre of NH3 is formed.
PbO is the limiting reagent. N2 used = 5 litres, left = 30 – 5 = 25 litres
Amount of PbCl 2 formed by 6.5 g of PbO H2 used = 15 litres, left = 30 – 15 = 15 litres
278 51. (c) According to Stoichiometry, they should
= ´ 6.5g react as follow:
223
Number of moles of PbCl2 formed 4 NH 3 + 5O 2 ¾ ¾® 4 NO + 6H 2 O
4 moles 5 moles 4 moles 6 moles
0.8 moles
278 6.5 0.8 moles 1 mole 1.2 moles
= ´ moles = 0.029 moles.
223 278 Thus, for 1 mole of O2 only 0.8 moles of NH3 is
48. (c) Molarity of H2SO4 solution consumed. Hence O2 is consumed completely.
98 ´ 1000 52. (d) 2C6 H 6 + 15O2 (g) ¾¾
®12CO2 (g) + 6 H 2 O(g)
= ´ 1.80 = 18.0
98 ´ 100 2(78) 15(32)
Suppose V mL of this H2SO4 is used to prepare Q 156 g of benzene, required oxygen
1 lit. of 0.1M H2SO4 = 15 × 22.4 litre
\ V ´ 18.0 = 1000 ´ 0.1 \ 1g of benzene, required oxygen
15´ 22.4
1000 ´ 0.1 = litre
or V = = 5.55 mL. 156
18.0 \ 39 g of benzene, required oxygen
49. (c) 2Al 2O 3 + 3C ¾¾ ® Al + 3CO 2
15 ´ 22.4 ´ 39
Gram equivalent of Al2O3 º g equivalent of C = = 84.0 litre
156
27
Now equivalent weight of Al = =9 53. (a) 5 M H2SO4 = 10 N H2SO4,
3 (Q Basicity of H2SO4 = 2)
Equivalent weight of C N1V1 = N2V2,
12 0 +4 10 × 1 = N2 × 10 or N2 = 1 N
= = 3 ( C ® C O2 )
4 54. (a) Mg2+ + Na 2 CO3 ¾¾ ® MgCO3 + 2Na +
3
270 ´ 10 1 g eq. 1g eq.
No. of gram equivalent of Al = 1 g eq. of Mg2+ = 12 g of Mg2+ = 12000 mg
9
= 30 × 103 = 1000 milli eq. of Na 2CO3
Hence, \ 12 mg Mg2+ = 1 milli eq. Na2CO3
No. of gram equivalent of C = 30 × 103
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2 Structure of Atom
Topic 1 : Atomic Models and Dual Nature of 4. The energies E1 and E2 of two radiations are 25
Electromagnetic Radiation eV and 50 eV, respectively. The relation between
their wavelengths i.e., l1 and l2 will be: [2011]
1. Calculate the energy in joule corresponding to (a) l1 = l2 (b) l1 = 2l2
light of wavelength 45 nm :
(Planck’s constant h = 6.63 × 10–34 Js; speed of 1
(c) l1 = 4l2 (d) l1 = l2
light c = 3 × 108 ms–1) [2014] 2
(a) 6.67 × 1015 (b) 6.67 × 1011 5. The energy absorbed by each molecule (A2) of
(c) 4.42 × 10–15 (d) 4.42 × 10–18 a substance is 4.4 × 10–19 J and bond energy
2. According to law of photochemical equivalence per molecule is 4.0 × 10–19 J. The kinetic energy
the energy absorbed (in ergs/mole) is given as of the molecule per atom will be: [2009]
(h = 6.62 × 10–27 ergs, c = 3 × 1010 cm s–1, (a) 2.2 × 10–19 J (b) 2.0 × 10–19 J
NA = 6.02 × 1023 mol–1) [NEET Kar. 2013] (c) 4.0 × 10–20 J (d) 2.0 × 10–20 J
1.196 ´ 1016 1.196 ´ 108 6. The value of Planck's constant is 6.63 × 10–34 Js.
(a) (b) The velocity of light is 3.0 × 108 m s–1. Which value
l l
is closest to the wavelength in nanometers of a
2.859 ´ 105 2.859 ´ 1016
(c) (d) quantum of light with frequency of 8 × 1015 s–1 ?
l l [2003]
3. The value of Planck’s constant is 6.63 × 10–34 Js.
(a) 3 × 107 (b) 2 × 10–25
The speed of light is 3 × 1017 nm s–1.. Which
(c) 5 × 10–18 (d) 4 × 101
value is closest to the wavelength in nanometer
7. If the energy of a photon is given as 3.03 × 10–19 J
of a quantum of light with frequency of
then, the wavelength (l) of the photon is :
6 × 1015 s–1? [NEET 2013]
(a) 6.56 nm (b) 65.6 nm [2000]
(a) 25 (b) 50
(c) 656 nm (d) 0.656 nm
(c) 75 (d) 10
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Structure of Atom 13
22. If r is the radius of the first orbit, the radius of 29. Given : The mass of electron is 9.11 × 10–31 kg
nth orbit of H-atom is given by [1988] Planck constant is 6.626 × 10–34 Js,
(a) rn2 (b) rn the uncertainty involved in the measurement of
(c) r/n (d) r2 n2 velocity within a distance of 0.1 Å is [2006]
23. The spectrum of He is expected to be similar to (a) 5.79 × 107 ms–1 (b) 5.79 × 108 ms–1
that [1988] 5
(c) 5.79 × 10 ms –1 (d) 5.79 × 106 ms–1
(a) H (b) Li+
30. The position of both, an electron and a helium
(c) Na (d) He+
atom is known within 1.0 nm. Further the
Topic 3: Dual Behaviour of Matter and momentum of the electron is known within
Heisenberg Uncertainty Principle 5.0 × 10–26 kg ms–1. The minimum uncertainty
in the measurement of the momentum of the
24. In hydrogen atom, the de-Broglie wavelength helium atom is [1998]
of an electron in the second Bohr orbit is (a) 50 kg ms–1 (b) 80 kg ms–1
[Given that Bohr radius, a0 = 52.9 pm]
(c) 8.0 × 10–26 kg ms–1 (d) 5.0 × 10–26 kg ms–1
[NEET Odisha 2019]
(a) 105.8 pm (b) 211.6 pm 31. The momentum of a particle having a de Broglie
(c) 211.6 p pm (d) 52.9 p pm wavelength of 10–17 metres is [1996]
25. Which one is the wrong statement ? [2017] (Given h = 6.625 × 10–34 Js)
(a) The uncertainty principle is (a) 3.3125 × 10–7 kg ms–1
DE ´ Dt ³ h / 4p (b) 26.5 × 10–7 kg ms–1
(b) Half filled and fully filled orbitals have (c) 6.625 × 10–17 kg ms–1
greater stability due to greater exchange (d) 13.25 × 10–17 kg ms–1
energy, greater symmetry and more 32. Uncertainty in position of an electron
balanced arrangement. (mass = 9.1 × 10–28 g) moving with a velocity of
(c) The energy of 2s orbital is less than the energy 3 × 104 cm/s accurate upto 0.001% will be
of 2p orbital in case of Hydrogen like atoms (use h/4(p) in uncertainty expression where
h h = 6.626 ×10–27 erg-second) [1995]
(d) de-Broglie's wavelength is given by l = ,
mn (a) 1.93 cm (b) 3.84 cm
where m = mass of the particle, n = group (c) 5.76 cm (d) 7.68 cm
velocity of the particle 33. When an electron of charge ‘e’ and mass ‘m’
26. A 0.66 kg ball is moving with a speed of 100 m/s. moves with a velocity ‘v’ about the nuclear
The associated wavelength will be charge ‘Ze’ in circular orbit of radius ‘r’, the
( h = 6.6 ´10 -34
)
Js : [2010] potential energy of the electrons is given by
[1994]
(a) 1.0 ´ 10–32m (b) 6.6 ´ 10–32m
(c) 6.6 ´ 10–34m (d) 1.0 ´ 10–35m (a) Ze2 / r (b) - Ze2 / r
27. The measurement of the electron position if 2
associated with an uncertainty in momentum, (c) Ze2 / r (d) mv / r
which is equal to 1 × 10–18 g cm s– 1 . The 34. Which of the following statements do not form
uncertainty in electron velocity is, [2008] a part of Bohr’s model of hydrogen atom ?
(mass of an electron is 9 × 10– 28 g) [1989]
(a) 1 × 109 cm s–1 (b) 1 × 106 cm s–1 (a) Energy of the electrons in the orbits are
5
(c) 1 × 10 cm s –1 (d) 1 × 1011 cm s–1 quantized
28. If uncertainty in position and momentum are (b) The electron in the orbit nearest the nucleus
equal, then uncertainty in velocity is : [2008] has the lowest energy
(c) Electrons revolve in different orbits around
1 h h the nucleus
(a) (b)
2m p 2p (d) The position and velocity of the electrons
1 h h in the orbit cannot be determined
(c) (d) simultaneously.
m p p
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14 CHEMISTRY
Topic 4 : Quantum Mechanical Model of Atom 42. The correct set of four quantum numbers for the
valence electron of rubidium atom (Z = 37) is
35. Orbital having 3 angular nodes and 3 total nodes is
[2012]
[NEET Odisha 2019]
(a) 5, 1, 1 + 1/2 (b) 6, 0, 0, + 1/2
(a) 6 d (b) 5 p
(c) 3 d (d) 4 f (c) 5, 0, 0, + 1/2 (d) 5, 1, 0, + 1/2
36. 4d, 5p, 5f and 6p orbitals are arranged in the 43. Maximum number of electrons in a subshell with :
order of decreasing energy. The correct option l = 3 and n = 4 is : [2012]
is: [2019] (a) 14 (b) 16
(a) 5f > 6p > 5p > 4d (b) 6p > 5f > 5p > 4d (c) 10 (d) 12
(c) 6p > 5f > 4d > 5p (d) 5f > 6p > 4d > 5p 44. If n = 6, the correct sequence for filling of
37. Which one is a wrong statement? [2018] electrons will be : [2011]
(a) Total orbital angular momentum of electron (a) ns ® (n – 2) f ® (n – 1) d ® np
in 's' orbital is equal to zero (b) ns ® (n – 1) d ® (n – 2) f ® np
(b) An orbital is designated by three quantum (c) ns ® (n – 2) f ® np ® (n – 1) d
numbers while an electron in an atom is (d) ns ® np (n – 1) d ® (n – 2) f
designated by four quantum numbers 45. The total number of atomic orbitals in fourth
(c) The value of m for dz2 is zero energy level of an atom is : [2011]
(d) The electronic configuration of N atom is (a) 8 (b) 16
2p1x 2p1y 2p1z (c) 32 (d) 4
1s2 2s2
46. Which of the following is not permissible
arrangement of electrons in an atom? [2009]
(a) n = 5, l = 3, m = 0, s = + 1/2
38. Two electrons occupying the same orbital are
(b) n = 3, l = 2, m = – 3, s = – 1/2
distinguished by [2016]
(c) n = 3, l = 2, m = – 2, s = – 1/2
(a) Principal quantum number
(d) n = 4, l = 0, m = 0, s = – 1/2
(b) Magnetic quantum number
(c) Azimuthal quantum number 47. Maximum number of electrons in a subshell of
(d) Spin quantum number an atom is determined by the following: [2009]
39. What is the maximum number of orbitals that (a) 2 l + 1 (b) 4 l – 2
can be identified with the following quantum (c) 2 n2 (d) 4 l + 2
numbers? [2014] 48. Consider the following sets of quantum numbers:
n = 3, l = 1, ml = 0 n l m s
(a) 1 (b) 2 (i) 3 0 0 + 1/2
(c) 3 (d) 4 (ii) 2 2 1 + 1/2
40. What is the maximum numbers of electrons that (iii) 4 3 –2 – 1/2
can be associated with the following set of (iv) 1 0 –1 – 1/2
quantum numbers? [NEET 2013] (v) 3 2 3 + 1/2
n = 3, l = 1 and m = –1 Which of the following sets of quantum number
(a) 6 (b) 4 is not possible? [2007]
(c) 2 (d) 10 (a) (i), (ii), (iii) and (iv) (b) (ii), (iv) and (v)
41. The orbital angular momentum of a p-electron (c) (i) and (iii) (d) (ii), (iii) and (iv)
is given as : [2012 M] 49. The orientation of an atomic orbital is governed
h h by [2006]
(a) (b) 3 (a) Spin quantum number
2p 2p
(b) Magnetic quantum number
3h h (c) Principal quantum number
(c) (d) 6.
2p 2p (d) Azimuthal quantum number
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Structure of Atom 15
50. The ions O2– , F– , Na +, Mg2+ and Al 3+ are 58. Which of the following species has four lone
isoelectronic. Their ionic radii show [2003] pairs of electrons ? [1993]
2– –
(a) A decrease from O to F and then increase
from Na+ to Al3+ (a) I (b) O -
(b) A significant increase from O2– to Al3+ (c) Cl– (d) He
(c) A significant decrease from O2– to Al3+ 59. The order of filling of electrons in the orbitals of
(d) An increase from O 2– to F– and then an atom will be [1991]
decrease from Na+ to Al3+ (a) 3d, 4s, 4p, 4d, 5s (b) 4s, 3d, 4p, 5s, 4d
51. Which of the following is isoelectronic?[2002]
(c) 5s, 4p, 3d, 4d, 5s (d) 3d, 4p, 4s, 4d, 5s
(a) CO2, NO2 (b) NO2–, CO2
– 60. For azimuthal quantum number l = 3, the
(c) CN , CO (d) SO2, CO2
maximum number of electrons will be [1991]
52. The following quantum numbers are possible
for how many orbital (s) n = 3, l = 2, m = +2 ? (a) 2 (b) 6
[2001] (c) 0 (d) 14.
(a) 1 (b) 3 61. In a given atom no two electrons can have the
(c) 2 (d) 4 same values for all the four quantum numbers.
53. Set of isoelectronic species is [2000] This is called [1991]
(a) Hund’s Rule
(a) N 2 , CO2 , CN- , O2 (b) N, H 2S, CO
(b) Aufbau principle
(c) N 2 , CO, CN - , O 2+2 (d) Ca , Mg, Cl (c) Uncertainty principle
54. The ion that is isoelectronic with CO is [1997] (d) Pauli’s Exclusion principle.
(a) CN– (b) O2+ 62. An ion has 18 electrons in the outermost shell, it
(c) O2 – (d) N2+ is [1990]
55. The orbitals are called degenerate when (a) Cu+ (b) Th4+
[1996]
(c) Cs+ (d) K+
(a) they have the same wave functions
(b) they have the same wave functions but 63. The total number of electrons that can be
different energies accommodated in all the orbitals having principal
(c) they have different wave functions but quantum number 2 and azimuthal quantum
same energy number 1 is [1990]
(d) they have the same energy (a) 2 (b) 4
56. If electron has spin quantum number + 1/2 and (c) 6 (d) 8
a magnetic quantum number – 1, it cannot be 64. The maximum number of electrons in a subshell
present in [1994] is given by the expression [1989]
(a) d-orbital (b) f-orbital
(a) 4l – 2 (b) 4l + 2
(c) p-orbital (d) s-orbital.
(c) 2l + 2 (d) 2n2
57. For which one of the following sets of four
quantum numbers, an electron will have the 65. Number of unpaired electrons in N2+ is [1989]
highest energy? [1994] (a) 2 (b) 0
n l m s (c) 1 (d) 34
(a) 3 2 1 1/2 66. The number of spherical nodes in 3p orbitals are
(b) 4 2 –1 1/2
(a) one (b) three [1988]
(c) 4 1 0 –1/2
(c) none (d) two
(d) 5 0 0 –1/2
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16 CHEMISTRY
ANSWER KEY
1 (d) 8 (c) 15 (c) 22 (a) 29 (d) 36 (a) 43 (a) 50 (c) 57 (b) 64 (b)
2 (b) 9 (b) 16 (a) 23 (b) 30 (d) 37 (d) 44 (a) 51 (c) 58 (c) 65 (c)
3 (b) 10 (b) 17 (c) 24 (c) 31 (c) 38 (d) 45 (b) 52 (a) 59 (b) 66 (a)
4 (b) 11 (b) 18 (d) 25 (c) 32 (a) 39 (a) 46 (b) 53 (c) 60 (d)
5 (d) 12 (c) 19 (a) 26 (d) 33 (b) 40 (c) 47 (d) 54 (a) 61 (d)
6 (d) 13 (c) 20 (a) 27 (a) 34 (d) 41 (a) 48 (b) 55 (d) 62 (a)
7 (c) 14 (b) 21 (c) 28 (a) 35 (d) 42 (c) 49 (b) 56 (d) 63 (c)
=
( 4.4 ×10 ) – ( 4.0 ×10 )
–19 –19
electrons could also be diffracted by crystals
2 just like light of x-rays.
10. (b) Cathode rays are not electromagnetic waves.
–19
0.4×10 –20 11. (b) Balmer series
= = 2.0 ´ 10 J
2
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Structure of Atom 17
12. (c) The electrons has more negative energy in Therefore the centripetal force is supplied by
lower orbits that in higher orbit and also the electrostatic force of attraction between the
electrons are more tightly bound in the smallest electron and nucleus i.e.
allowed orbit. mn2 Ze2
= 2
Energy of an electron at infinite distance from the r r
nucleus is zero. As an electron approaches the
Ze2
nucleus, the electron attraction increases and hence or mn2 =
the energy of electron decreases and thus becomes r
negative. Thus as the value of n decreases, i.e.
1 1 Ze 2
lower the orbit is, more negative is the energy of or mn2 = = K.E
the electron in it. 2 2 r
now total energy (En) = K.E + P.E
13. (c) Energy of photon obtained from th e in first excited state
transition n = 6 to n = 5 will have least energy.
1 é Ze 2 ù
æ 1 1 ö E = mv 2 + ê - ú
DE = 13.6Z 2 - 2 êë r úû
çè n 2 n 2 ÷ø
1 2
1 Ze2 Ze2
æ Zö
2 =+ -
14. (b) E n = E1 ´ ç ÷ 2 r r
è nø
1 Ze 2
E1 -3.4 eV = -
E 2 = 2 = -328kJ/mol 2 r
2
1 Ze 2
E \ K .E = = +3.4 eV
E 4 = 21 2 r
4
For Bohr orbit, kinetic energy = –Total energy and
E 4 22 1 Potential energy = 2 × Total energy.
Þ = 2 = 2
E2 4 2
E -328 17. (c) Energy of electron in 2nd orbit of Li +2
Þ E4 = 2 = kJ /mol = -82 kj /mol
22 4 Z2
= -13.6 2
15. (c) I.E. = E1 – E ¥ n
2.18 × 10–18 = E1 – 0 - 13.6 ´ (3) 2
E1 = 2.18 × 10–18 J-atom–1 = = –30.6 eV
(2) 2
æ 1 1 ö Energy required = 0 – (–30.6) = 30.6 eV
DE = E1 ç 2 - 2 ÷ 18. (d) Radius of hydrogen atom = 0.530 Å, Number
è n1 n 2 ø of excited state (n) = 2 and atomic number of
æ1 1ö hydrogen atom (Z) = 1. We know that the Bohr
Þ hv = 2.18 × 10–18 ç 2 - 2 ÷
è1 4 ø radius.
n2 (2)2
15 (r ) = ´ Bohr Radius of atom = ´ 0.530
Þ 6.625 × 10–34 × v = 2.18 × 10–18 × Z 1
16 = 4 ´ 0.530 = 2.12 Å
2.18 ´ 10 –18
´ 15 19. (a) State of hydrogen atom (n) = 1
Þ v= -34 (due to ground state)
6.625 ´ 10 ´ 16 Radius of hydrogen atom (r) = 0.53 Å. Atomic
Þ v = 3.08 × 1015 s–1 number of Li (Z) = 3.
16. (a) Suppose the nucleus of hydrogen atom Radius of Li2+ ion
have charge of one proton +e. The electron n2 (1) 2
revolves in an orbit of radius r around it. =r´ = 0.53 ´ = 0.17Å.
Z 3
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18 CHEMISTRY
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Structure of Atom 19
\ Uncertainty in the position of the electron 41. (a) Orbital angular momentum
,27
h 6.626´10 ´7
( Dx ) = < =
h
l(l + 1)
,
4 pmDv 4´ 22´(9.1´10 )´ 0.3
28
2p
=1.93cm
For p orbital l = 1
r
Ze 2
Ze2
33. (b) P.E. = work done = ò - 2 dr = - . h h
r r So, = 2 =
¥
2p 2p
Kinetic and potential energy of atoms results from
the motion of electrons. When electrons are excited 42. (c) Electronic configuration of Rb = [Kr] 5s1
they move to a higher energy orbital farther away Set of quantum numbers, n = 5
from the atom. The further the orbital is from the l = 0, \ s-orbital
nucleus, the higher the potential energy of an
electron at that energy level. When the electron m = 0, s = + 1/2
returns to a low energy state, it releases potential The other possible answer is 5, 0, 0, –1/2 In and
energy in the form of kinetic energy.
orbital any electron can take the value of ± 1/2
34. (d) It is uncertainty principle which was given and the other elctron of the same orbital, takes the
by Heisenberg and not Bohr’s postulate. value with opposite sign.
35. (d) Total number of nodes = (n – 1) 43. (a) (n = 4, l = 3) Þ 4f subshell
3= n–1 Þ n=4 Since, maximum no. of electrons in a subshell
Number of angular nodes = l = 3 Þ f-subshell = 2(2l + 1)
36. (a) 5 f 5+3=8 So, total no. of electron in 4f subshell
6p 6+1=7
= 2 (2 × 3 + 1) = 14 electrons.
5p 5+1=6
44. (a)
4d 4+2=6
5f > 6p > 5p > 4d 1s
37. (d) The correct configuration of 'N' is 2s 2p
1s2 2s 2 2p1x 2p1y 2p1z
3s 3p 3d
4s 4p 4d 4f
38. (d) Two electrons occupying the same orbital
should have opposite spins i.e. they differ in 5s 5p 5d 5f
spin quantum number.
39. (a) Given: n = 3, l = 1, m = 0 6s 6p 6d 6f
Hence orbital is 3p
–1 0 +1 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
hence the number of orbital identified by m = 0
45. (b) Total no. of atomic orbital in a shell = n2.
can be one only.
Given n = 4; hence number of atomic orbitals in
40. (c) n = 3 ® 3rd shell
4th shell will be 16.
l = 1 ® p sub shell.
46. (b) m = 2 l +1, thus for l = 2, m = 5, hence values
m = – 1 is possible for two electrons present in
of m will be – 2, –1, 0, + 1, + 2.
an orbital.
Therefore for l = 2, m cannot have the value – 3.
m = 0 is fixed for Pz orbital but Px and Py orbital 47. (d) The number of subshell is (2 l + 1). The
can take any value among ± 1. maximum number of electrons in the sub shell is
2 (2 l + 1) = (4 l + 2)
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20 CHEMISTRY
48. (b) (ii) is not possible for any value of n because 58. (c) Outer electronic configuration of Cl
2 2 2 1
l varies from 0 to (n – 1) thus for n = 2, l can be = 3s 3 p x 3 p y 3 p z
only 0 and 1. Outer electronic configuration of
(iv) is not possible because for l = 0, m = 0. Cl– =3s 2 3px2 3p y 2 3pz 2
(v) is not possible because for l = 2, m varies
hence, Cl– contains four lone pairs of electrons.
from –2 to +2.
59. (b) The sub-shell with lowest value of (n + l) is
49. (b) Magnetic quantum no. represents the filled up first. When two or more sub-shells have
orientation of atomic orbitals in an atom. For same (n + l) value the subshell with lowest value
example px, py & pz have orientation along of 'n' is filled up first therefore the correct order is
X-axis, Y-axis & Z-axis. orbital 4s 3d 4p 5s 4d
50. (c) Amongst isoelectronic species, ionic radii n+l 4 + 0 3 + 2 4 +1 5 + 0 4 + 2
of anion is more than that of cations. Further value =4 5 5 5 6
size of anion increases with increase in –ve 60. (d) l = 3 means f-subshell. Maximum no. of
charge and size of cation decreases with increase electrons = 4l + 2 = 4 × 3 + 2 = 14
in + ve charge. Hence, ionic radii decreases from 61. (d) This is as per the definition of Pauli’s
Exclusion principle.
O– to Al3+.
62. (a) Cu+ = 29 – 1 = 28 e–
51. (c) Both CN– and CO have 14 electrons.
thus the electronic configuration of Cu+ is
52. (a) Quantum number n = 3, l = 2, m = +2 represent
one of the 3d-orbitals with = 1s 2 2s 2 2p 6 3s 2 3p6 3d10
144244 3
1 18e -
s=±
2 63. (c) n = 2, l = 1 means 2 p – orbital. Electrons that
can be accommodated = 6 as p sub-shell has 3
which is possible only for one orbital.
orbitals and each orbital contains 2 electrons.
53. (c) The molecule which contains same number
64. (b) No. of orbitals in a sub-shell = 2l + 1
of electrons are called isoelectronic. eg.
Þ No. of electrons = 2(2l + 1) = 4l + 2
N2 = CO = CN - = O +2 2 = 14e– 65. (c) N(7) = 1s2 2s2 2p3
54. (a) We know that ions which have the same N 2+ = 1s 2 , 2s 2 2 p1x
number of electrons are called isoelectronic. We
Unpaired electrons = 1.
also know that both CO and CN– have 14
66. (a) No. of radial nodes in 3p-orbital
electrons, therefore these are isoelectronic.
= (n – l – 1)
55. (d) The orbitals which have same energy are
[for p ortbital l = 1]
called degenerate orbitals eg. px , p y , pz .
=3–1–1=1
56. (d) For s-orbital; l = 0
Thus value of m must be zero, which is given as There are two types of nodes, angular nodes and
radial notes. Angular nodes are typically flat plane.
–1 in the question. Hence, the electron can not The quantum number l determines the number of
be present in s-orbital. angular nodes in an orbital. Radial nodes are spheres
57. (b) The sub-shell are 3d, 4d, 4p and 5s, 4d has that occurs as the principal quantum number
highest energy as n + l value is maximum for increases. Total nodes of an orbital is the sum of
angular and radial nodes which is given by
this. N=n–l–1
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3 Classification of Elements
and Periodicity in Properties
Topic 1: Modern Periodic Table 3. The element Z = 114 has been discovered recently.
It will belong to which of the following family/
1. Identify the incorrect match. [2020] group and electronic configuration ? [2017]
Name IUPAC Official (a) Carbon family, [Rn] 5f 14 6d10 7s2 7p2
Name (b) Oxygen family, [Rn] 5f 14 6d10 7s2 7p4
(A) Unnilunium (i) Mendelevium (c) Nitrogen family, [Rn] 5f 14 6d10 7s2 7p6
(B) Unniltrium (ii) Lawrencium (d) Halogen family, [Rn] 5f 14 6d10 7s2 7p5
(C) Unnilhexium (iii) Seaborgium 4. An atom has electronic configuration 1s2 2s2 2p6
(D) Unununnium (iv) Darmstadtium 3s2 3p6 3d3 4s2, you will place it in which group?
(a) (B), (ii) (b) (C), (iii) (a) Fifth (b) Fifteenth [2002]
(c) (D), (iv) (d) (A), (i) (c) Second (d) Third
2. Match the following : [2020] 5. The element, with atomic number 118, will be
(a) alkali (b) noble gas [1996]
Oxide Nature
(c) lanthanide (d) transition element
(A) CO (i) Basic 6. The electronic configuration of an element is
(B) BaO (ii) Neutral
(C) Al2O3 (iii) Acidic 1s 2 2s 2 2 p6 3s 2 3 p3. What is the atomic number
(D) Cl2O7 (iv) Amphoteric of the element, which is just below the above
Which of the following is correct option? element in the periodic table? [1995]
(A) (B) (C) (D) (a) 33 (b) 34
(a) (ii) (i) (iv) (iii) (c) 36 (d) 49
(b) (iii) (iv) (i) (ii) 7. If the atomic number of an element is 33, it will
(c) (iv) (iii) (ii) (i) be placed in the periodic table in the [1993]
(a) First group (b) Third group
(d) (i) (ii) (iii) (iv) (c) Fifth group (d) Seventh group.
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22 CHEMISTRY
Topic 2: Periodic Trends in Properties of Thus process of formation of O2– in gas phase
Elements is unfavourable even though O2– is isoelectronic
with neon. It is due to the fact that [2015 RS]
8. Match the oxide given in column A with its (a) Electron repulsion outweighs the stability
property given in column B [NEET Odisha, 2019] gained by achieving noble gas configuration
Column-A Column-B (b) O– ion has comparatively smaller size than
(i) Na2O (A) Neutral oxygen atom
(ii) Al2O3 (B) Basic (c) Oxygen is more electronegative
(iii) N2O (C) Acidic
(d) Addition of electron in oxygen results in
(iv) Cl2O7 (D) Amphoteric
larger size of the ion.
Which of the following options has all correct
14. Which of the following orders of ionic radii is
pairs?
(a) (i)-(B), (ii)-(D), (iii)-(A), (iv)-(C) correctly represented ? [2014]
(b) (i)-(B), (ii)-(A), (iii)-(D), (iv)-(C) (a) H– > H+ > H (b) Na+ > F– > O2–
(c) (i)-(C), (ii)-(B), (iii)-(A), (iv)-(D) (c) F– > O2– > Na+ (d) Al3+> Mg2+> N3–
(d) (i)-(A), (ii)-(D), (iii)-(B), (iv)-(C) 15. Which one of the following arrangements
9. For the second period elements the correct represents the correct order of least negative to
increasing order of first ionisation enthalpy is : most negative electron gain enthalpy for C, Ca,
(a) Li < Be < B < C < N < O < F < Ne [2019] Al, F and O? [NEET Kar. 2013]
(b) Li < B < Be < C < O < N < F < Ne (a) Ca < Al < C < O < F(b) Al < Ca < O < C < F
(c) Li < B < Be < C < N < O < F < Ne (c) Al < O < C < Ca < F (d) C < F < O < Al < Ca
(d) Li < Be < B < C < O < N < F < Ne 16. Identify the wrong statement in the following:
10. The correct order of atomic radii in group 13 [2012]
elements is [2018] (a) Amongst isoelectronic species, smaller the
(a) B < Al < In < Ga < Tl positive charge on the cation, smaller is
(b) B < Al < Ga < In < Tl the ionic radius.
(c) B < Ga < Al < In < Tl (b) Amongst isoelectronic species, greater the
(d) B < Ga < Al < Tl < In negative charge on the anion, larger is the
11. In which of the following options the order of ionic radius.
arrangement does not agree with the variation (c) Atomic radius of the elements increases as
of property indicated against it ? [2016] one moves down the first group of the
(a) Al3+ < Mg2+ < Na+ < F– (increasing ionic periodic table.
size) (d) Atomic radius of the elements decreases
(b) B < C < N < O (increasing first ionisation as one moves across from left to right in
enthalpy) the 2nd period of the periodic table.
(c) I < Br < F < Cl (increasing electron gain 17. What is the value of electron gain enthalpy of
enthalpy) Na+ if IE1 of Na = 5.1 eV ? [2011M]
(d) Li < Na < K < Rb (increasing metallic radius) (a) –5.1 eV (b) –10.2 eV
12. The species Ar, K+ and Ca2+ contain the same
(c) +2.55 eV (d) +10.2 eV
number of electrons. In which order do their radii
18. Among the elements Ca, Mg, P and Cl, the order
increase ? [2015]
of increasing atomic radii is : [2010]
(a) Ca 2+ < Ar < K + (b) Ca 2+ < K + < Ar (a) Ca < Mg < P < Cl (b) Mg < Ca < Cl < P
(c) K + < Ar < Ca 2+ (d) Ar < K + < Ca 2+ (c) Cl < P < Mg < Ca (d) P < Cl < Ca < Mg
13. The formation of the oxide ion O2–(g), from 19. Which of the following represents the correct
oxygen atom requires first an exothermic and order of increasing electron gain enthalpy with
then an endothermic step as shown below : negative sign for the elements O, S, F and Cl ?
[2005, 2010]
O(g) + e– ® O–(g); Df H = –141 kJ mol–1
(a) Cl < F < O < S (b) O < S < F < Cl
O– (g) + e– ® O2– (g); Df H = +780 kJ mol–1 (c) F < S < O < Cl (d) S < O < Cl < F
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Classification of Elements and Periodicity in Properties 23
20. The correct order of the decreasing ionic radii 29. Which of the following statements is true? [2002]
among the following isoelectronic species are : (a) Silicon exhibits 4 coordination number in
(a) Ca 2+ > K + > S2– > Cl – [2010] its compound
– 2-
(b) Cl > S > Ca > K 2+ + (b) Bond energy of F2 is less than Cl2
(c) S2– > Cl – > K + > Ca 2+ (c) Mn(III) oxidation state is more stable than
Mn(II) in aqueous state
(d) K + > Ca 2+ > Cl – > S 2–
21. Amongst the elements with following electronic (d) Elements of 15th group shows only +3 and
configurations, which one of them may have the +5 oxidation states
highest ionization energy? [2009] 30. Which of the following order is wrong? [2002]
(a) Ne [3s23p2] (b) Ar [3d104s24p3 ] (a) NH3 < PH3 < AsH3 – Acidic
(c) Ne [3s23p1] (d) Ne [3s23p3] (b) Li < Be < B < C – First IP
22. The stability of + 1 oxidation state increases in (c) Al2O3 < MgO < Na2O < K2O – Basic
the sequence: [2009] (d) Li+ < Na+ < K+ < Cs+ – Ionic radius
(a) Tl < In < Ga < Al (b) In < Tl < Ga < Al 31. Correct order of first IP among following
(c) Ga < In < Al < Tl (d) Al < Ga < In < Tl elements Be, B, C, N, O is [2001]
23. Which one of the following ionic species has the (a) B < Be < C < O < N (b) B < Be < C < N < O
greatest proton affinity to form stable compound? (c) Be < B < C < N < O (d) Be < B < C < O < N
(a) NH -2 (b) F– [2007] 32. Of the given electronic configurations for the
elements, which electronic configuration
(c) I– (d) HS–
indicates that there will be abnormally high
24. Which of the following electronic configuration
difference in the second and third ionization
of an atom has the lowest ionisation enthalpy?
energy for the element? [1999]
[2007]
(a) 1s2 2s2 2p6 3s2 (b) 1s2 2s2 2p6 3s1
(a) 1s2 2s2 2p3 (b) 1s2 2s2 2p5 3s1
(c) 1s2 2s2 2p6 3s2 3p1(d) 1s2 2s2 2p6 3s2 3p2
(c) 1s2 2s2 2p6 (d) 1s2 2s2 2p5
33. The first ionization potentials (eV) of Be and B
25. Identify the correct order of the size of the following:
respectively are [1998]
(a) Ca2+ < K+ < Ar < Cl– < S2– [2007]
(a) 8.29, 9.32 (b) 9.32, 9.32
(b) Ar < Ca2+ < K+ < Cl– < S2–
(c) 8.29, 8.29 (d) 9.32, 8.29
(c) Ca2+ < Ar < K+ < Cl– < S2–
34. Which of the following does not represent the
(d) Ca2+ < K+ < Ar < S2– < Cl–
correct order of the properties indicated [1997]
26. Which one of the following oxides is expected
(a) Ni2+ > Cr2+ > Fe2+ > Mn2+ (size)
to exhibit paramagnetic behaviour? [2005]
(b) Sc > Ti > Cr > Mn (size)
(a) CO2 (b) SiO2
(c) Mn2+ > Ni2+ < Co2+ <Fe2+ (unpaired electron)
(c) SO2 (d) ClO2
(d) Fe2+ > Co2+ > Ni2+ > Cu2+ (unpaired electron)
27. Ionic radii are [2004]
35. Which one of the following ions will be the
(a) inversely proportional to effective nuclear
smallest in size? [1996]
charge
(a) Na+ (b) Mg2+
(b) inversely proportional to squar e of
(c) F– (d) O2 –
effective nuclear charge
36. Among the following oxides, the one which is
(c) directly proportional to effective nuclear
most basic is [1994]
charge
(a) ZnO (b) MgO
(d) directly proportional to square of effective
(c) Al2O3 (d) N2O5
nuclear charge
37. One of the characteristic properties of non-
28. Among K, Ca, Fe and Zn, the element which can
metals is that they [1993]
form more than one binary compound with
(a) Are reducing agents
chlorine is [2004]
(b) Form basic oxides
(a) Fe (b) Zn
(c) Form cations by electron gain
(c) K (d) Ca
(d) Are electronegative
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24 CHEMISTRY
38. Which electronic configuration of an element (a) Ga, In, Tl (b) Na, Mg, Al
has abnormally high difference between second (c) N, O, F (d) V, Cr, Mn
and third ionization energy? [1993] 43. Elements of which of the following groups will
(a) 1 s2, 2 s2, 2 p6, 3 s1 form anions most readily ? [1992]
(b) 1 s2, 2 s2, 2 p6, 3 s1 3 p1 (a) Oxygen family (b) Nitrogen family
(c) 1 s2, 2 s2, 2 p6, 3 s2 3 p2 (c) Halogens (d) Alkali metals
(d) 1 s2, 2 s2, 2 p6, 3 s2 44. In the periodic table, with the increase in atomic
39. In the periodic table from left to right in a period, number, the metallic character of an element
the atomic volume [1993] [1989]
(a) Decreases (a) Decreases in a period and increases in a group
(b) Increases (b) Increases in a period and decreases in a group
(c) Remains same (c) Increases both in a period and the group
(d) First decrease then increases (d) Decreases in a period and the group.
40. Na+, Mg++, Al3+ and Si4+ are isoelectronic. The 45. The electronic configuration of four elements are
order of their ionic size is [1993] given below. Which element does not belong to
(a) Na+ > Mg++ < Al3 + < Si4+ the same family as others ? [1989]
(b) Na+ < Mg++ > Al3+ > Si4+ (a) [Xe]4f 145d101s2 (b) [Kr]4d 10 5s 2
(c) Na+ > Mg++ > Al3+ > Si4+
(c) [Ne]3s23p5 (d) [Ar] 3d10 4s2
(d) Na+ < Mg++ > Al3+ < Si4+
46. Pauling’s electronegativity values for elements
41. One would expect proton to have very large
are useful in predicting [1989]
[1993]
(a) Polarity of the molecules
(a) Charge (b) Ionization potential
(b) Position in the E.M.F. series
(c) Hydration energy (d) Radius.
(c) Coordination numbers
42. Which of the following sets has strongest
(d) Dipole moments.
tendency to form anions ? [1993]
ANSWER KEY
1 (c) 6 (a) 11 (b) 16 (a) 21 (d) 26 (d) 31 (a) 36 (b) 41 (c) 46 (a)
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 (a) 20 (c) 25 (a) 30 (b) 35 (b) 40 (c) 45 (c)
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Classification of Elements and Periodicity in Properties 25
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26 CHEMISTRY
17. (a) IE1 of Na = – Electron gain enthalpy of Na + 25. (a) For isoelectronic species, size of anion
= – 5.1 eV. increases as negative charge increases whereas
size of cation decreases with increase in positive
In this question, temperature is to be defined as
absolute zero. This is due to the fact that ionization charge. Further ionic radii of anions is more than
energy and electron affinity are defined at absolute that of cations. Thus the correct order is
zero temperature. Ca2+ < K+ < Ar < Cl– < S2–
5 26. (d) Due to odd number of electrons in ClO2, it
Ionization enthalpy = ionization energy + RT
2 is expected to exhibit paramagnetic behaviour.
5 Cl
Electron gain enthalpy = electron affinity - RT
2
18. (c) 12 Mg P Cl Ca O O
15 17 20 Paramagnetic
160 110 99 197 (pm) 27. (a) Ionic radii are inversely proportional to
So, the order will be: Cl < P < Mg < Ca effective nuclear charge.
19. (b) O < S < F < Cl Ionic radii in the nth orbit is given as
Electron gain enthalpy of given elements are
n2 a0 1
– 141, – 200, – 333 and – 349 kJ mol–1 respectively. rn = or rn µ
Z Z
Due to small size of atom, addition of an electron when n = principal quantum number
is not easy. This is the reason why the magnitude
of electron gain enthalpy of oxygen and fluorine is Z=effective nuclear charge.
less than that of sulphur and chlorine respectively. 28. (a) Among the given options, only Fe shows
20. (c) Among the isoelectronic species, size variable oxidation states so it can form two
increases with the increase in negative charge. chlorides, viz. FeCl2 and FeCl3.
Thus S2– has the highest negative charge and 29. (b) This is because of inter-electronic
hence largest in size followed by Cl–, K+ and Ca2+. replusions between lone pairs.
21. (d) The smaller the atomic size, larger is the B.E. : F–F Cl – Cl
value of ionisation potential. Further the atoms (kJ mol–1) : 158.8
.. ..
242.6
«
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Classification of Elements and Periodicity in Properties 27
remove the electron in 2s-orbital in comparison electronegative non metals and will have the
to 2p orbital. strongest tendency to form anions by gaining
34. (a) In a period on moving from left to right ionic electrons from metal atoms.
radii decreases. 43. (c) Elements of halogen group form anions most
So order of cationic radii is Cr2+ > Mn2+ > Fe2+ > Ni2+ readily.
35. (b) Greater is the positive charge on atom, larger Electron affinity values are high in case of halogen
will be the effective nuclear charge. Hence smaller
because halogens have seven electrons ( ns 2 np 5 )
is the size. in the valence shell, they have a strong tendency
36. (b) N2O5 is strongly acidic, ZnO and Al2 O3 are to acquire the nearest inert gas configuration by
amphoteric, therefore, MgO is most basic. gaining an electron from the metallic atom and form
halide ions easily.
37. (a) Non metals form oxides with oxygen and
thus reduce oxides of metals behaving as 44. (a) Metallic character decreases in a period and
reducing agents. increases in a group.
45. (c) Elements (a), (b) and (d) belong to the same
38. (d) Abnormally high difference between 2nd group since each one of them has two electrons
and 3rd ionization energy means that the element in the s-sub shell. In contrast, element (c) has
has two valence electrons. seven electrons in the valence shell and hence
39. (d) Atomic volume is the volume occupied by does not lie in the same group.
one mole of an element. Within a period from left 46. (a) Paulings electronegativity values for elements
to right, atomic volume first decreases and then are useful in predicting polarity of the molecule.
increases due to increase in nuclear charge and
increase in molar mass. Pauling scale of electronegativity was helpful in
predicting :
40. (c) Amongst isoelectronic ions, the size of the (i) Nature of bond between two atoms
cation decreases as the magnitude of the charge (ii) Stability of bond
increases. By calculating the difference in electro-negativities,
41. (c) Proton (H+) being very small in size would polarity of bond can be calculated.
have very large hydration energy.
42. (c) N, O and F (p-block elements) are highly
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28 CHEMISTRY
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Chemical Bonding and Molecular Structure 29
11. Among the following which compound will 17. In X — H — Y, X and Y both are electronegative
show the highest lattice energy ? [1993] elements
(a) KF (b) NaF (a) Electron density on X will increase and on
(c) CsF (d) RbF H will decrease [2001]
12. Among LiCl, BeCl2 BCl3 and CCl4, the covalent (b) In both electron density will decrease
bond character follows the order [1990] (c) In both electron density will increase
(a) LiCl < BeCl2 > BCl3 > CCl4 (d) Electron density will decrease on X and will
(b) BeCl2 < BCl3 < CCl4 < LiCl increase on H
(c) LiCl < BeCl2 < BCl3 < CCl4 18. Which one of the following molecules will form
(d) LiCl > BeCl2 > BCl3 > CCl4 a linear polymeric structure due to hydrogen
13. Which of the following does not apply to metallic bonding? [2000]
bond ? [1989] (a) NH3 (b) H2O
(a) Overlapping valence orbitals (c) HCl (d) HF
(b) Mobile valency electrons 19. In PO43– ion, the formal charge on each oxygen
atom and P—O bond order respectively are
(c) Delocalized electrons
[1998]
(d) Highly directed bonds.
(a) –0.75, 0.6 (b) – 0.75, 1.0
Topic 2: Octet rule, Resonance and (c) – 0.75, 1.25 (d) –3, 1.25
Hydrogen Bonding 20. The low density of ice compared to water is due
to [1997]
14. Which of the following structures is the most (a) hydrogen-bonding interactions
preferred and hence of lowest energy for SO3? (b) dipole-dipole interactions
[2011 M] (c) dipole-induced dipole interactions
(d) induced dipole-induced dipole interactions
21. The boiling point of p-nitrophenol is higher than
(a) (b) 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
(c) (d)
in p-nitrophenol
(d) p-nitrophenol has a higher molecular
15. What is the dominant intermolecular force or weight than o-nitrophenol.
bond that must be overcome in converting liquid 22. Which one of the following is the correct order
CH3OH to a gas? [2009]
of interactions ? [1993]
(a) Dipole-dipole interaction
(a) Covalent < hydrogen bonding < van der
(b) Covalent bonds
Waals < dipole-dipole
(c) London dispersion force
(d) Hydrogen bonding (b) van der Waals < hydrogen bonding <
16. Which of the following is not a correct dipole-dipole < covalent
statement? [2006] (c) van der Waals < dipole-dipole < hydrogen
(a) The cannonical structures have no real bonding < covalent
existence (d) Dipole-dipole < van der Waals < hydrogen
(b) Every AB 5 molecule does in fact have bonding < covalent.
square pyramidal structure 23. Strongest hydrogen bond is shown by [1992]
(c) Multiple bonds are always shorter than (a) Water
corresponding single bonds (b) Ammonia
(d) The electron-deficient molecules can act (c) Hydrogen fluoride
as Lewis acids (d) Hydrogen sulphide.
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30 CHEMISTRY
24. Which one of the following formulae does not 30. The electronegativity difference between N and
correctly represent the bonding capacities of the F is greater than that between N and H yet the
two atoms involved ? [1990] dipole moment of NH3 (1.5 D) is larger than that
+
of NF3 (0.2D). This is because [2006]
é H ù (a) in NH3 the atomic dipole and bond dipole
ê | ú are in the same direction whereas in NF3
(a) êH — P — H ú
ê |
ú these are in opposite directions
ë H û (b) in NH3 as well as NF3 the atomic dipole
(b) F F and bond dipole are in opposite directions
(c) in NH3 the atomic dipole and bond dipole
O
O are in the opposite directions whereas in
(c) O¬ N NF3 these are in the same direction
O–H (d) in NH3 as well as in NF3 the atomic dipole
O and bond dipole are in the same direction
31. Wh ich of th e followi ng woul d ha ve a
(d) H – C = C permanent dipole moment? [2005]
O–H
(a) SiF4 (b) SF 4
25. Which one shows maximum hydrogen bonding?
(c) XeF4 (d) BF3
(a) H2O (b) H2Se [1990]
32. The correct order of the O–O bond length in O2,
(c) H2S (d) HF.
H2O2 and O3 is [1995, 2005]
Topic 3: Dipole Moment and Bond Polarity (a) O 2 > O 3 > H 2O 2
26. Which of the following set of molecules will (b) O 3 > H 2O 2 > O 2
have zero dipole moment? [2020]
(c) O 2 > H 2O 2 > O 3
(a) Boron trifluoride, hydrogen fluoride,
carbon dioxide, 1,3-dichlorobenzene (d) H 2 O 2 > O 3 > O 2
(b) Nitrogen trifluoride, beryllium difluoride, 33. H2O is dipolar, whereas BeF2 is not. It is because
water, 1,3-dichlorobenzene [2004]
(c) Boron trifluoride, beryllium difluoride, (a) the electronegativity of F is greater than
carbon dioxide, 1,4-dichlorobenzene that of O
(d) Ammonia, beryllium difluoride, water, (b) H2O involves hydrogen bonding whereas
1,4-dichlorobenzene BeF2 is a discrete molecule
27. Which of the following is the correct order of (c) H2O is linear and BeF2 is angular
(d) H2O is angular and BeF2 is linear
dipole moment? [NEET Odisha 2019]
34. The dipole moments of diatomic molecules AB
(a) H2O < NF3 < NH3 < BF3 and CD are 10.41D and 10.27 D, respectively
(b) NH3 < BF3 < NF3 < H2O while their bond distances are 2.82 and 2.67 Å,
(c) BF3 < NF3 < NH3 < H2O respectively. This indicates that [1999]
(a) bonding is 100% ionic in both the molecules
(d) BF3 < NH3 < NF3 < H2O
(b) AB has more ionic bond character than CD
28. Which of the following molecules has the (c) AB has lesser ionic bond character than
maximum dipole moment ? [2014] CD
(a) CO2 (b) CH4 (d) bonding is nearly covalent in both the
(c) NH3 (d) NF3 molecules
29. Which of the following is a polar molecule ? 35. Which of the following bonds will be most polar?
[NEET 2013] [1992]
(a) SF4 (b) SiF4 (a) N – Cl (b) O – F
(c) XeF4 (d) BF3 (c) N – F (d) N – N
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Chemical Bonding and Molecular Structure 31
36. H2O has a non zero dipole moment while BeF2 (b) lone pair - lone pair > bond pair - bond pair
has zero dipole moment because [1989] > lone pair - bond pair
(a) H2O molecule is linear while BeF2 is bent (c) bond pair - bond pair > lone pair - bond
(b) BeF2 molecule is linear while H2O is bent pair > lone pair - lone pair
(c) Fluorine has more electronegativity than (d) lone pair - bond pair > bond pair - bond
oxygen pair > lone pair - lone pair
(d) Beryllium has more electronegativity than 43. Which of the following pairs of ions are
oxygen. isoelectronic and isostructural ? [2015]
Topic 4: VSEPR Theory and Hybridisation (a) ClO3– , CO32– (b) SO32– , NO3–
37. In the structure of ClF3, the number of lone pair
of electrons on central atom ‘Cl’ is [2018]
(c) ClO3– , SO32– (d) CO32– , SO32–
(a) One (b) Two 44. Maximum bond angle at nitrogen is present in
(c) Three (d) Four which of the following ? [2015]
38. Which of the following molecules represents (a) NO 2– (b) NO +2
the order of hybridisation sp2, sp2, sp, sp from
left to right atoms? [2018] (c) NO3– (d) NO 2
(a) HC º C – C º CH 45. In which of the following pairs, both the species
(b) CH2 = CH – C º CH are not isostructural ? [2015 RS]
(c) CH3 – CH = CH – CH3
(d) CH2 = CH – CH = CH2 (a) SiCl4 , PCl+
4
39. Which of the following pairs of compounds is (b) diamond, silicon carbide
isoelectronic and isostructural ? [2017] (c) NH3, PH3
(a) TeI2,XeF2 (b) IBr2- , XeF2 (d) XeF4, XeO4
(c) IF3, XeF2 (d) BeCl2,XeF2 46. Be2+ is isoelectronic with which of the following
40. The species, having bond angles of 120° is :- ions? [2014]
[2017] (a) H+ (b) Li+
(a) CIF3 (b) NCl3 (c) Na+ (d) Mg2+
(c) BCl3 (d) PH3 47. Which one of the following species has plane
41. Consider the molecules CH4, NH3 and H2O. triangular shape ? [2014]
(a) N3 – (b) NO3 –
Which of the given statements is false? [2016]
(a) The H–C–H bond angle in CH4, the H–N–H (c) NO2– (d) CO2
bond angle in NH3, and the H–O–H bond 48. In which of the following pair both the species
angle in H2O are all greater than 90° have sp3 hybridization? [NEET Kar. 2013]
(b) The H–O–H bond angle in H2 O is larger (a) H2S, BF3 (b) SiF4, BeH2
than the H–C–H bond angle in CH4. (c) NF3, H2O (d) NF3, BF3
(c) The H–O–H bond angle in H2O is smaller 49. XeF2 is isostructural with [NEET 2013]
than the H–N–H bond angle in NH3. (a) ICl2– (b) SbCl3
(d) The H–C–H bond angle in CH4 is larger (c) BaCl2 (d) TeF2
than the H–N–H bond angle in NH3. 50. Which of the following species contains three
42. Predict the correct order of electron repulsion bond pairs and one lone pair around the central
among the following : [2016] atom ? [2012]
(a) lone pair- lone pair > lone pair - bond pair > (a) H2O (b) BF3
–
bond pair - bond pair (c) NH2 (d) PCl3
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32 CHEMISTRY
51. Which one of the following pairs is isostructural (a) NO -2 and NH 3 (b) BF3 and NO -2
(i.e., having the same shape and hybridization)? (c) NH -2 and H 2 O (d) BF3 and NH -2
(a) é BCl3 and BrCl3- ù [2012] 58. In which of the following molecules / ions
ë û
-ù
BF3, NO2- , NH 2- and H2O , [2009]
é
(b) ë NH3 and NO3 û
the central atom is sp2 hybridized ?
(c) [ NF3 and BF3 ] (a) NH 2- and H2O (b) NO2- and H2O
é BF4- and NH +4 ù (c) BF3 and NO 2- (d) NO 2- and NH 2-
(d) ë û 59. The correct order of increasing bond angles in
52. Which of the two ions from the list given below the following triatomic species is : [2008]
that have the geometry that is explained by the (a) NO-2 < NO+2 < NO 2
same hybridization of orbitals, NO2–, NO3– , (b) NO-2 < NO2 < NO2+
NH2–, NH4+, SCN– ? [2011]
(a) NO2– and NO3– (b) NH4+ and NO3– (c) NO+2 < NO 2 < NO2-
(c) SCN– and NH2– (d) NO2– and NH2– (d) NO +2 < NO-2 < NO
53. Considering the state of hybridization of carbon 60. In which of the following pairs, the two species
atoms, find out the molecule among the following are isostructural? [2007]
which is linear ? [2011] (a) SO32– and NO3– (b) BF3 an NF3
(a) CH3– CH = CH–CH3 (c) BrO3– and XeO3 (d) SF4 and XeF4
(b) CH3 – C º C – CH3 61. Which of the following is not isostructural with
(c) CH2 = CH – CH2 – C º CH SiCl4? [2006]
(d) CH3 – CH2 – CH2 – CH3 (a) SO42– (b) PO43–
54. In which of the following molecules the central (c) NH4+ (d) SCl4
atom does not have sp3 hybridization? [2010] 62. Which of the following species has a linear shape ?
(a) SO2 (b) NO2+ [2006]
(a) NH +4 (b) CH4 (c) O3 (d) NO2–
(c) SF4 (d) BF4– 63. In which of the following molecules all the bonds
55. Some of the properties of the two species, NO3- are not equal? [2006]
and H3O+ are described below. Which one of (a) BF3 (b) AlF3
them is correct? [2010] (c) NF3 (d) ClF3
(a) Similar in hybridization for the central atom 64. Which of the following molecules has trigonal
with different structures. planar geometry? [2005]
(a) BF3 (b) NH3
(b) Dissimilar in hybridization for the central
(c) PCl3 (d) IF3
atom with different structures.
65. In BrF 3 molecule, the lone pairs occupy
(c) isostructural with same hybridization for
equatorial positions to minimize [2004]
the central atom.
(a) lone pair - bond pair repulsion only
(d) Isostructural with different hybridization (b) bond pair - bond pair repulsion only
for the central atom. (c) lone pair - lone pair repulsion and lone pair
56. In which one of the following species the central - bond pair repulsion
atom has the type of hybridization which is not (d) lone pair - lone pair repulsion only
the same as that present in the other three? 66. In an octahedral structure, the pair of d orbitals
(a) SF4 (b) I3– [2010] involved in d 2 sp3 hybridization is [2004]
(c) SbCl52– (d) PCl5 (a) d d (b) d xz, d 2 2
57. In which of the following pairs of molecules/ x2 - y 2 , z 2 x -y
ions, the central atoms have sp2 hybridization? (c) d d (d) d xy , d yz
[2010] z 2 , xz
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Chemical Bonding and Molecular Structure 33
67. In a regular octahedral molecule, MX6 the number 77. The BCl3 is a planar molecule whereas NCl 3 is
of X - M - X bonds at 180° is [2004] pyramidal because [1995]
(a) three (b) two (a) B-Cl bond is more polar than N-Cl bond
(c) six (d) four (b) N-Cl bond is more covalent than B-Cl bond
68. Which of the following has pp – dp bonding? (c) nitrogen atom is smaller than boron atom
(a) NO3– (b) SO32– [2002] (d) BCl3 has no lone pair but NCl3 has a lone
(c) BO33– (d) CO32– pair of electrons
69. Main axis of a diatomic molecule is z, molecular 78. The distance between the two adjacent carbon
orbital px and py overlap to form which of the atoms is largest in [1994]
following orbital? [2001] (a) benzene (b) ethene
(a) p - molecular orbital (c) butane (d) ethyne
(b) s - molecular orbital 79. Among the following orbital bonds, the angle is
(c) d - molecular orbital minimum between [1994]
(d) No bond will be formed
(a) sp3 bonds (b) px and py orbitals
70. Which of the following two are isostructural?
(c) H – O – H in water (d) sp bonds.
[2001]
80. Which of the following does not have a
(a) NH3, BF3 (b) PCl5, ICl5
tetrahedral structure? [1994]
(c) XeF2, IF2– (d) CO3–2, SO3–2
(a) BH –4 (b) BH3
71. In which of the following, the bond angle is
maximum? [2001] (c) NH +4 (d) H 2 O.
(a) NH3 (b) SCl2 81. Which of the following statements is not correct ?
(c) NH4+ (d) PCl3 [1993]
72. Among the following ions the pp–dp overlap (a) Double bond is shorter than a single bond
could be present in [2000] (b) Sigma bond is weaker than a p (pi) bond
(a) NO -2 (b) NO 3- (c) Double bond is stronger than a single bond
(d) Covalent bond is stronger than hydrogen
(c) PO34- (d) CO 23-
bond.
73. Which one of the following has the pyramidal 82. Which structure is linear ? [1992]
shape? [1999] (a) SO2 (b) CO2
(a) CO32– (b) SO3
(c) BF3 (d) PF3 (c) CO 32- (d) SO 24 -
74. Which of the following molecules is planar? 83. Which one of the following has the shortest
(a) SF4 (b) XeF4 [1998] carbon carbon bond length ? [1992]
(c) NF3 (d) SiF4 (a) Benzene (b) Ethene
75. The AsF5 molecule is trigonal bipyramidal. The (c) Ethyne (d) Ethane
hybrid orbitals used by the As atom for bonding are 84. In compound X, all the bond angles are exactly
d 109°28; X is [1991]
(a) 2 , d 2 , s, p x , p y [1997]
x2 - y z (a) Chloromethane
(b) dxy, s, px, py, pz
(b) Carbon tetrachloride
(c) s, px, py, pz, d (c) Iodoform
z2
(d) d , s, px, py, pz (d) Chloroform.
x2 - y 2
85. Which statement is NOT correct ? [1990]
76. The cylindrical shape of an alkyne is due to the (a) A sigma bond is weaker than a p -bond.
fact that it has [1997] (b) A sigma bond is stronger than a p -bond.
(a) three sigma C – C bonds (c) A double bond is stronger than a single
(b) two sigma C – C and one 'p' C – C bond bond.
(c) three 'p' C – C bonds (d) A double bond is shorter than a single
(d) one sigma C– C and two 'p' C – C bonds bond.
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34 CHEMISTRY
86. In which one of the following molecules the (a) NO (b) CN–
central atom said to adopt sp2 hybridization? (c) CN (d) CN+
(a) BeF2 (b) BF3 [1989] 95. Which of the following pairs of species have
(c) C2H2 (d) NH3 the same bond order ? [2017]
87. Which of the following molecule does not have (a) O2, NO + –
(b) CN , CO
a linear arrangement of atoms ? [1989] (c) N 2 , O 2- (d) CO, NO
(a) H2S (b) C2H2
96. The correct bond order in the following species
(c) BeH2 (d) CO2
is: [2015]
88. Equilateral shape has [1988]
(a) sp hybridisation (b) sp2 hybridisation (a) O 22+ < O 2– < O2+ (b) O +2 < O 2– < O 22+
(c) sp3 hybridisaiton (d) sp3 hybridisation
89. The angle between the overlapping of one (c) O 2– < O +2 < O 22+ (d) O 22+ < O 2+ < O 2–
s-orbital and one p-orbital is [1988] 97. Which of the following options represents the
(a) 180° (b) 120° correct bond order ? [2015]
(c) 109°28' (d) 120° 60' (a) O 2– < O 2 < O 2+ (b) O –2 > O2 < O 2+
Topic 5: Valence Bond and Molecular
Orbital Theory (c) O 2– < O2 > O 2+ (d) O –2 > O2 > O2+
90. Identify a molecule which does not exist. [2020] 98. Which of the following species contains equal
(a) Li2 (b) C2 number of s- and p-bonds : [2015]
(c) O2 (d) He2 (a) XeO4 (b) (CN)2
91. Which of the following is paramagnetic? (c) CH2(CN)2 (d) HCO3–
[NEET Odisha 2019]
99. Decreasing order of stability of O2, O- +
2 ,O 2 and
(a) O2 (b) N2
(c) H2 (d) Li2 O 22- is : [2015 RS]
92. The manganate and permanganate ions are
tetrahedral, due to: [2019]
(a) O+ - 2-
2 > O2 > O2 > O2
(a) The p-bonding involves overlap of p- (b) O 22- > O- +
2 > O2 > O2
orbitals of oxygen with d-orbitals of
manganese (c) O2 > O+ 2- -
2 > O2 > O2
(b) There is no p-bonding
(d) O- 2- +
2 > O2 > O2 > O2
(c) The p-bonding involves overlap of p-orbital
of oxygen with p-orbitals of manganese 100. The hybridization in volved in complex
(d) The p- bonding involves overlap of d- [Ni(CN)4]2–. is (At. No. Ni = 28) [2015 RS]
orbital of oxygen with d-orbitals of (a) dsp 2 (b) sp3
manganese (c) d 2 sp2 (d) d 2 sp3
93. Which of the following diatomic molecular 101. The outer orbitals of C in ethene molecule can
species has only p–bonds accordin g to be considered to be hybridized to give three
Molecular Orbital Theory ? [2019] equivalent sp2 orbitals. The total number of sigma
(a) O2 (b) N2 (s) and pi (p) bonds in ethene molecule is
(c) C2 (d) Be2 [NEET Kar. 2013]
94. Consider the following species : (a) 1 sigma (s) and 2 pi (p) bonds
CN+, CN–, NO and CN (b) 3 sigma (s) and 2 pi (p) bonds
Which one of these will have the highest bond (c) 4 sigma (s) and 1 pi (p) bonds
order? [2018] (d) 5 sigma (s) and 1 pi (p) bonds
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Chemical Bonding and Molecular Structure 35
102. In which of the following ionisation processes 111. Which of the following has the minimum bond
the bond energy increases and the magnetic length ? [2011]
behaviour changes from paramagnetic to (a) O2+ (b) O2 –
diamagnetic? [NEET Kar. 2013] (c) O2 2– (d) O2
(a) N2 ® N2+ (b) O2 ® O2+ 112. Which one of the following species does not
(c) C2 ® C2+ (d) NO ® NO+ exist under normal conditions? [2010]
103. The pair of species that has the same bond order (a) Be+2 (b) Be2
in the following is: [NEET Kar. 2013] (c) B 2 (d) Li 2
(a) O2, B2 (b) CO, NO+ 113. According to MO theory which of the following
(c) NO–, CN– (d) O2, N2 lists ranks the nitrogen species in terms of
104. Which of the following is paramagnetic ? increasing bond order? [2009]
[NEET 2013] (a) N 2–2 < N –
2 < N 2 (b) N 2 < N 2–
2 < N2
–
(a) O -2 (b) CN– (c) N 2– < N 2– – 2–
2 < N 2 (d) N 2 < N 2 < N 2
(c) NO+ (d) CO 114. The angular shape of ozone molecule (O 3 )
105. Which one of the following molecules contains consists of : [2008]
no p bond? [NEET 2013] (a) 1 sigma and 2 pi bonds
(a) H2O (b) SO2 (b) 2 sigma and 2 pi bonds
(c) NO2 (d) CO2 (c) 1 sigma and 1 pi bonds
106. Four diatomic species are listed below. Identify (d) 2 sigma and 1 pi bonds
the correct order in which the bond order is 115. The correct order of C–O bond length among
increasing in them: [2008, 2012 M] CO, CO32 - , CO2 is [2007]
(a) NO < O2-< C 22 - < He +2 2– 2–
(a) CO < CO3 < CO2 (b) CO3 < CO2 < CO
(b) O -2 < NO < C22 - < He 2+ (c) CO < CO2 < CO32– (d) CO2 < CO32– < CO
116. The number of unpaired electrons in a
(c) C22 - < He 2+ < O2- < NO paramagnetic diatomic molecule of an element
(d) He+2 < O2- < NO < C 22 - with atomic number 16 is [2006]
(a) 3 (b) 4
107. During change of O2 to O 2- ion, the electron (c) 1 (d) 2
adds on which one of the following orbitals ? 117. Among the following the pair in which the two
[2012 M] species are not isostructural is [2004]
(a) p* orbital (b) p orbital
(a) SiF4 and SF4 (b) IO3- and XeO 3
(c) s* orbital (d) s orbital
108. The pair of species with the same bond order is : (c) BH -4 and NH +4 (d) PF6- and SF6
O2–
(a) (b) O+2 , NO+ [2012] 118. Which of the following statements is not correct
2 , B2
for sigma and pi-bonds formed between two
(c) NO, CO (d) N2, O2 carbon atoms? [2003]
109. Bond order of 1.5 is shown by : [2012] (a) Sigma-bond determines the direction
+
(a) O 2 (b) O 2 -
between carbon atoms but a pi-bond has
(c) O 22 - (d) O2 no primary effect in this regard
(b) Sigma-bond is stronger than a pi-bond
110. The pairs of species of oxygen and their (c) Bond energies of sigma- and pi-bonds are
magnetic behaviours are noted below. Which of of the order of 264 kJ/mol and 347 kJ/mol,
the following presents the correct description ? respectively
(a) O -2 , O 22 - – Both diamagnetic [2011 M] (d) Free rotation of atoms about a sigma-bond
is allowed but not in case of a pi-bond
(b) O + ,O22 - – Both paramagnetic 119. In NO3– ion, number of bond pair and lone pair
(c) O+2 ,O2 – Both paramagnetic of electrons on nitrogen atom respectively are
(a) 2, 2 (b) 3, 1 [2002]
(d) O,O 22 - – Both paramagnetic (c) 1, 3 (d) 4, 0
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36 CHEMISTRY
120. The relationship between the dissociation 125. The correct order of N–O bond lengths in NO,
energy of N2 and N2+ is : [2000] NO2–, NO3– and N2O4 is [1996]
(a) Dissociation energy of N2+ > dissociation (a) N 2 O 4 > NO -2 > NO 3- > NO
energy of N2 (b) NO > NO 3- > N 2 O 4 > NO -2
(b) Dissociation energy of N2 = dissociation (c) NO 3- > NO -2 > N 2 O 4 > NO
energy of N2+ (d) NO > N 2 O 4 > NO-2 > NO3-
(c) Dissociation energy of N2 > dissociation
126. The ground state electronic configuration of
energy of N2+ valence shell electrons in nitrogen molecule (N2)
(d) Dissociation energy of N2 can either be
lower or higher than the dissociation energy is written as KK s 2 s 2 , s * 2 s 2 , p 2 p x2 , p 2 p 2y s 2 pz2
of N2+ Bond order in nitrogen molecule is [1995]
121. Which one of the following arrangements rep- (a) 0 (b) 1
resents the increasing bond orders of the given (c) 2 (d) 3
species? [1999] 127. Which of the following species is paramagnetic?
(a) NO+ < NO < NO– < O2– (a) O 22- (b) NO [1995]
(b) O2– < NO– < NO <NO+ (c) CO (d) CN–
(c) NO– < O2– < NO < NO+ 128. Mark the incorrect statement in the following
(d) NO < NO+ < O2– < NO– [1994]
122. Amon g the followin g which one is not (a) The bond order in the species O2, O2+ and
paramagnetic? [Atomic numbers : Be = 4, O2– decreases as O +2 > O 2 > O -2
Ne = 10, As = 33, Cl = 17] [1998]
(b) The bond energy in a diatomic molecule
(a) Cl– (b) Be+ (c) Ne+2 (d) As +
always increases when an electron is lost
123. The number of anti-bonding electron pairs in
(c) Electrons in antibonding M.O. contribute
O 2- 2 molecular ion on the basis of molecular to repulsion between two atoms.
orbital theory is, (Atomic number of O is 8) [1998] (d) With increase in bond order, bond length
(a) 5 (b) 2 (c) 3 (d) 4 decreases and bond strength increases.
124. N2 and O2 are converted into monoanions, N2– 129. Linear combination of two hybridized orbitals
and O2– respectively. Which of the following belonging to two atoms and each having one
statements is wrong ? [1997] electron leads to a [1990]
(a) In N2, the N—N bond weakens (a) Sigma bond
(b) In O2, the O—O bond order increases (b) Double bond
(c) In O2, bond length decreases (c) Co-ordinate covalent bond
(d) N2– becomes diamagnetic (d) Pi bond.
ANSWER KEY
1 (b) 14 (d) 27 (c) 40 (c) 53 (b) 66 (a) 79 (b) 92 (a) 105 (a) 118 (c)
2 (c) 15 (d) 28 (c) 41 (b) 54 (c) 67 (a) 80 (b) 93 (c) 106 (d) 119 (d)
3 (b) 16 (b) 29 (a) 42 (a) 55 (b) 68 (b) 81 (b) 94 (b) 107 (a) 120 (c)
4 (a) 17 (a) 30 (a) 43 (c) 56 (c) 69 (a) 82 (b) 95 (b) 108 (a) 121 (b)
5 (a) 18 (d) 31 (b) 44 (b) 57 (b) 70 (c) 83 (c) 96 (c) 109 (b) 122 (a)
6 (a) 19 (c) 32 (d) 45 (d) 58 (c) 71 (c) 84 (b) 97 (a) 110 (c) 123 (d)
7 (c) 20 (a) 33 (d) 46 (b) 59 (b) 72 (c) 85 (a) 98 (a) 111 (a) 124 (b)
8 (b) 21 (c) 34 (c) 47 (b) 60 (c) 73 (d) 86 (b) 99 (a) 112 (b) 125 (c)
9 (d) 22 (b) 35 (c) 48 (c) 61 (d) 74 (b) 87 (a) 100 (a) 113 (a) 126 (d)
10 (a) 23 (c) 36 (b) 49 (a) 62 (b) 75 (c) 88 (b) 101 (d) 114 (d) 127 (b)
11 (b) 24 (d) 37 (b) 50 (d) 63 (d) 76 (d) 89 (a) 102 (d) 115 (c) 128 (b)
12 (c) 25 (d) 38 (b) 51 (d) 64 (a) 77 (d) 90 (d) 103 (b) 116 (d) 129 (a)
13 (d) 26 (c) 39 (b) 52 (a) 65 (c) 78 (c) 91 (a) 104 (a) 117 (a)
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Chemical Bonding and Molecular Structure 37
14 that are required to give simple ethane like Formal charge on S atom
structure. 1
= 6 – 0 – (12) = 0
7. (c) Chemical bonds. 2
Formal charge on O atom
8. (b) The stability of the ionic bond depends upon 1
the lattice energy which is expected to be more = 6 – 4 – (4) = 0
2
between Mg and F due to +2 charge on Mg atom.
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38 CHEMISTRY
15. (d) Due to intermolecular hydrogen bonding Intramolecular Hydrogen bonding in o-nitrophenol
in methanol, it exist as assosiated molecule. is a interaction with in the molecule whereas the
16. (b) Statement (a), (c), (d) are correct. Statement intermolecular H-bonding in p-nitrophenol is a
(b) is incorrect statement. interaction between the molecules which results
AB5 may have two structures as follows : into higher boiling point. Intermolecular hydrogen
bonding also results in a stronger driving force for
B
cystal formation in other molecules generating
higher melting temperature e.g. p-hydroxy benzoic
B B acid.
A
B B 22. (b) The strength of the interactions follows the
A order: van der Waal’s < hydrogen – bonding <
B B dipole-dipole < covalent.
B B 23. (c) Hydrogen fluoride shows str ongest
B hydrogen bonding due to high electronegativity
Square Pyramidal Trigonal Bipyramidal of fluorine.
O
Individual cannonical structures do not exist. The ||
molecule as such has a single structure, which is
the resonance hybrid of the cannonical forms. 24. (d) H - C = C* - O - H
The star marked carbon has a valency of 5 and
17. (a) In X — H - - - Y, X and Y both are hence this formula is not correct.
electronegative elements (i.e attracts the 25. (d) As F is most electronegative thus HF shows
electron pair) then electron density on X will maximum strength of hydrogen bond.
increase and on H will decrease. With the increase of electronegativity and decrease
18. (d) F—H-----F—H-----F—H-----F in size of the atom to which hydrogen is covalently
HF form linear polymer structure due to linked, the strength of hydrogen bond increases.
hydrogen bonding.
19. (c) Bond order between P – O 26. (c) BF3, BeF2, CO2 and 1, 4-dichlorobenzene
all are symmetrical molecules.
no. of bonds in all possible direction 5
= = = 1.25 F
total no. of resonating structures 4
µ=0
O O
– B
–
F F
O P O–
–
O P O
Be µ=0
O– O– F F
O– O–
O C O µ=0
–
O P O– –
O P O
O O– Cl Cl µ = 0
3
or Formal charge on oxygen = - = -0.75 27. (c) Dipole moment of a molecule is the vector
4
sum of dipoles of bonds. So based on molecular
20. (a) We know that due to polar nature, water geometry of following molecules,
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.
21. (c) The b.p. of p-nitrophenol is higher than
that of o-nitrophenol because in p-nitrophenol Three equal vectors Vectors not alligned in
there is intermolecular H-bonding but in at 120° has resultant the same direction of
o-nitrophenol it is intramolecular H-bonding. dipole moment = 0, so lone pair, so less dipole
non-polar molecule m o m en t
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Chemical Bonding and Molecular Structure 39
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40 CHEMISTRY
H H H P
10
H H H 104.3° H Cl
Cl Cl
Tetrahedral; Trigonal Bent
pyramidal
51. (d) BF4- hybridisation sp3, tetrahedral structure.
The geometry of H2O should have been tetrahedral
if there are all bond pairs. But due to presence of NH +4 hybridisation sp3, tetrahedral structure.
two lone pairs the shape is distorted tetrahedral.
Hence bond angle reduced to 104.5° from 109.5°. 52. (a)
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Chemical Bonding and Molecular Structure 41
3+3
57. (b) BF3 : = 3 means sp2 hybridisation Cl F
2
5 +1
NO-2 : = 3 means sp2 hybridisation F
2
Here two bonds are in equatorial plane & one
58. (c) We find that in the given molecules
bond is in axial plane.
hybridisation is
64. (a) BF3 is sp2 hybridised. So, it is trigonal
BF3 ® sp2
planar. NH3, PCl3 has sp3 hybridisation hence
NO -2 ® sp2 has trigonal pyramidal shape, IF3, has sp3d
NH 2- ® sp3 hydridization and is T-shaped.
65. (c) The possible structures are:
H2O ® sp3
F
59. (b) From the structure of three species we can F F
determine the number of lone pair electrons on Br F Br F Br
F F
central atom (i.e. N atom) and thus the bond F
F
angle.
I II III
90° lp – lp
...O... ..
N N +
O..
O¬ N=O
O O NO+2 repulsion 0 0 1
- 90° lp – bp
NO 2 NO 2
We know that higher the number of lone pair of repulsion 4 6 3
electron on central atom, greater is the lp – lp 90° bp – bp
repulsion between Nitrogen and oxygen atoms. repulsion 2 0 2
Thus smaller is bond angle. The first structure is minimizing the lp – lp
The correct order of bond angle is repulsion if we compare it with III. The same is
minimizing the lp – bp repulsions if we compare
+
NO2- < NO2 < NO2 it with II. Also it can be noted that, I structure is
not minimizing the bp – bp repulsions.
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42 CHEMISTRY
F F
2 2 6 2 1 1 1
16 S = 1s 2 s 2 p 3s 3 p x 3 p y 3 p z 3d1xy
1442443 { The shape is Xe Square planar shape.
3
sp hybridisation unhybride
F F
In 'S' unhybride d-orbital is present, which will 75. (c) The electronic configuration of As is
involved in p bond formation with oxygen atom.
2 2 6 1 1 1 1 1
8O = 1s 2 2 s 2 2 p 2x 2 p1y 2 p1z As = 1s 2 s 2 p 3s 3 p x 3 p y 3 p z 3d
144424443
In oxygen two unpaired p-orbital is present in ¯ sp3d hybridisation
these one is involved in s bond formation while
other is used in p bond formation So, the hybrid orbitals used by As atom in AsF5
molecule are s, px, py, pz, dz2.
Thus in SO 32 - , pp and dp orbitals are involved
In the formation of a stable trigonal pyramid, the
for pp - d p bonding. 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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Chemical Bonding and Molecular Structure 43
76. (d) In alkynes the hybridisation is sp i.e each 85. (a) A s - bond is stronger than a p -bond
carbon atom undergoes sp hybridisation to form hence option (a) is not correct.
two sp-hybrid orbitals. The two 2p-orbitals
Sigma (s) bonds are formed by head on overlap of
remain unhybridised. Hybrid orbitals form one unhybridised s–s, p–p or s–p orbitals and
sigma and two unhybridised orbitals form p- hybridised orbitals (sp, sp2, sp3, sp3d and sp3d 2)
bonds. 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 . . C
86. (b) BF3 involves sp2-hybridization.
87. (a) For linear arrangement of atoms the
hybridisation should be sp(linear shape, 180°
. . . .
angle). Only H2 S has sp3 -hybridization and
Hence two p bond and one sigma bond between hence has angular shape while C2H2, BeH2 and
C — C lead to cylindrical shape. CO2 all involve sp-hybridization and hence have
77. (d) As there is no lone pair on boron in BCl 3 linear arrangement of atoms.
therefore no repulsion takes place. But there is 88. (b) Equilateral or triangular planar shape
a lone pair on nitrogen in NCl 3. Therefore involves sp2 hybridization.
repulsion takes place. Thus, BCl3 is planar 89. (a)
molecule but NCl3 is a pyramidal molecule.
78. (c) The C–C bond distance decreases as the s-orbital p-orbital
multiplicity of the bond increases. Thus, bond The overlap between s- and p-orbitals occurs
distance decreases in the order: butane (1.54 Å) along internuclear axis and hence the angle is 180°.
> benzene (1.39 Å) > ethene (1.34) Å > ethyne 90. (d) For He2 molecule, Electronic configuration
(1.20 Å). Thus in butane, C – C bond distance is is s1s 2 , s *1s 2
the largest.
79. (b) The angle between the bonds formed by 1 1
Bond order = ( N b - N a ) = (2 - 2) = 0
px and py orbitals is the minimum i.e. 90°. 2 2
80. (b) BH3 has sp2 hybridization and hence does Since, bond order of He2 is zero, so it does not
not have tetrahedral structure while all others exist.
have tetrahedral structures. 91. (a) Molecular orbital configuration of O2 is
81. (b) Sigma bond is stronger than p-bond. The given as :
electrons in the p bond are loosely held. The O2 (16 e–) : s1s2 s*1s2 s2s2 s*2s2 s2p2z
bond is easily broken and is more reactive than
s -bond. Energy released during sigma bond p 2 p 2x = p 2 p 2y p * 2 p1x = p * 2 p1y
formation is always more than p bond because So, in O2 molecule, there are two (2) unpaired
of greater extent of overlapping. electrons, so, it is a “paramagnetic”
82. (b) CO 2 has sp-hybridization and is linear. SO2 substance in nature.
and CO 32- are planar (sp2 ) while SO 24 - is 92. (a) O O–
tetrahedral (sp3).
Mn Mn
83. (c) The bond length decreases in the order
sp3 > sp2 > sp. O – O
O– O O O
Because of the triple bond, the carbon-carbon
Manganate ion Permanganate ion
bond distance in ethyne is shortest.
84. (b) All compounds have sp3 hybridisation, but 93. (c) Only p bond is present in C2 molecule.
in CCl4, all bonded atoms are same. Hence, the s1s2 s*1s2 s2s2 s*2s2 p2p2x = p2p2y
bond angle will be exactly 109°28°
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44 CHEMISTRY
94. (b) NO : (s1s)2, (s*1s)2, (s2s)2,(s*2s)2,(s2pz)2, O 2– (Super oxide ion): Total number of electrons
(p2px)2 = (p2py)2,(p*2px)1 = (p*2py)0 (16 + 1) = 17 .
10 - 5 Electronic configuration
B.O. = = 2.5
2
CN : (s1s) , (s*1s)2, (s2s)2,(s*2s)2,
– 2 s1s 2 < s*1s 2 < s 2s 2 < s* 2s 2 < s 2 p x2
(p2px)2 = (p2py)2, (s2pz)2 < p 2 p 2y = p 2 pz2 < p* 2 p y2 = p* 2 p1z
10 - 4 (Nb - Na ) 10 - 7 3 1
B.O. = =3 Bond order = = = =1
2 2 2 2 2
CN : (s1s)2, (s*1s)2, (s2s)2,(s*2s)2,
O+22 ion: Total number of electrons
(p2px)2 = (p2py)2,(s2pz)1
= (16 – 2) = 14 Electronic configuration
9-4
B.O. = = 2.5 s1s2 < s*1s2 < s2s2 < s*2s2 < s2px2 < p2py2
2 = p2pz2
CN : (s1s) , (s*1s)2, (s2s)2,(s*2s)2,
+ 2
(p2px)2 = (p2py)2 (N b – Na ) 10 – 4 6
Bond order = = = =3
2 2 2
8-4
B.O. = =2 So bond order: O2– < O2+ < O22+
2
97. (a) Oxygen molecule (O2) – Total number of
Hence, option (2) should be the right
answer. electrons = 16 and electronic configuration is
95. (b) CN– and CO have same no. of electrons and s1s 2 < s*1s 2 < s 2s 2 < s* 2s 2 < s 2 p x2
have same bond order equal to 3.
Short cut trick to calculate the bond order: < p 2 p 2y = p 2 pz2 < p* 2 p1y = p* 2 p1z
N2 has 14 electrons and its bond order is 3. We
have to remember this concept that every electron N b - N a 10 - 6 4
added or subtracted to 14, reduces the bond order Bond order = = = =2
by 0.5. For example 2 2 2
CN– Þ no. of electrons = 6 + 7 + 1 = 14
\ bond order = 3 O +2 ion - Total number of electrons (16 – 1) = 15.
CO Þ no. of electrons = 6 + 8 = 14 Electronic configuration
\ bond order = 3
NO Þ no. of electrons = 7 + 8 = 15
s1s 2 < s*1s 2 < s 2s 2 < s* 2s 2 < s 2 p x2
\ bond order = 3 – 0.5 = 2.5
NO+ Þ no. of electrons = 7 + 8 – 1 = 14
\ bond order = 3 < p 2 p 2y = p2 pz2 < p* 2 p1y
O2– Þ no. of electrons = 8 + 8 + 1 = 17
\ bond order = 3 – 1.5 N b - Na 10 - 5 5 1
Please note that this method will work for any Bond order = = = =2
species that have electrons between 10 and 18. 2 2 2 2
96. (c) O +2 ion - Total number of electrons O2– (Super oxide ion) Total number of electrons
(16 – 1) = 15. (16 + 1) = 17 . Electronic configuration
Electronic configuration
σ1s2 < σ*1s2 < σ2s2 < σ* 2s 2 < σ2p2x
s1s 2 < s*1s 2 < s 2 s 2 < s * 2 s 2 < s 2 p x2
< p 2 p 2y = p 2 pz2 < p * 2 p1y < π2p2y = π2p2z < π* 2p 2y = π* 2p1z
N b - N a 10 - 5 5 1 (Nb - Na ) 10 - 7 3 1
Bond order = = = =2 Bond order = = = =1
2 2 2 2 2 2 2 2
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Chemical Bonding and Molecular Structure 45
O (c) ® C2+
C2 ¾¾
98. (a) Xe B.O. 2 2.5
Bond energy decreases
O O O
Magnetic behaviour changes from diamagnetic
Number of s bonds = 4 to paramagnetic
Number of p bonds = 4 (d) NO ¾¾ ® NO+
99. (a) According to molecular orbital theory as
bond order decreases stability of the molecule B.O. 2 2.5
decreases bond energy increases
Magnetic behaviour changes from paramagnetic
1
Bond order = (N b – Na ) to diamagnetic
2 103. (b) No. of electrons in CO = 6 + 8 = 14
1 No. of electrons in NO+ = 7 + 8 – 1 = 14
Bond order for O2+ = (10- 5) = 2.5
2 \ CO and NO+ are isoelectronic species.
1 Isoelectronic species have identical bond order.
Bond order for O 2 = (10 - 6) = 2
2 104. (a) Molecular orbital configuration of O-2 is
1
Bond order for O -2 = (10 - 7) = 1.5
2 O-2 (17) = s1s2, s*1s2, s2s2, s*2s2,
2- 1 s2pz2, p2px2 = p2py2, p*2px2 = p*2py1
Bond order for O2 = (10 - 8) = 1.0
2 O
105. (a) s
hence the correct order is H s H
O+2 > O2 > > O 2–
O2– 2 s s s s
O ¬ S=O O ¬ N =O O = C = O
100. (a) Ni2+ = [Ar] 3d 8, 4s0 p p p p
In the presence of strong ligand CN–, pairing of 106. (d) Calculating the bond order of various
electrons will occur: species.
2+
Ni = G.S. O2- : KK s 2s 2 s* 2 s 2 s2 p 2z
3d 4s 4p
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46 CHEMISTRY
N - N a 10 - 5 5 1 N 2– = s1s 2 s *1s 2 s 2 s 2 s * 2 s 2
Bond order = b = = =2
2 2 2 2 ìp 2 p 2y ìp * 2 p1y
( )
O 2- = s1s 2 s*1s 2 s 2s 2 s* 2s 2 s 2 Pz2
ï
í
2
s 2 p x2
ï
í
0
îïp 2 p z îïp * 2 p z
p2 px2 = p2 p 2y p* 2 px2 p* 2 p1y
10 - 5
Bond order = = 2.5
( N - N a ) 10 - 7 3 1 2
Bond order = b = = =1
2 2 2 2 ìïp 2 p 2y
( ) = s1s s 1s
O22- 2 * 2 2
s 2s s 2s * 2
s 2 pz2 p2 px2 N 2 = s1s 2 s *1s 2 s2 s 2 s * 2 s 2 í
îïp 2 p z
2
, s2 p x2
= p2 p 2y p* 2 px2 = p* 2 p y2 10 – 4
Bond order = =3
Nb - N a 10 - 8 2 2
Bond order = = =1 The correct order is = N 2– –
2 2 2 \ 2 < N2 < N2
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Chemical Bonding and Molecular Structure 47
..
..
O O s-bond formation. 2s electrons are used for
– O .. .. .. O . .O..– coordinate bond formation. Thus there is no
..
.. ..
O lone pair on nitrogen and four bond pairs are
.. ..
present.
In it we find 2 s and 1p bond, i.e., option (d) is
O
correct.
115. (c) All these structures exhibits resonance
and can be represented by the following
resonating structures. N
+ ..
(i) : C º O : ¬¾
® : C = O: O O
– –
120. (c) Dissociation energy of any molecules
O O O depends upon bond order. Bond order in N2
(ii) C C C molecule is 3 while bond order in N +2 is 2.5.
– – – –
O O O O O O Further we know that more the Bond order, more
is the stability and more is the BDE.
+ ..
-
(iii) : O
..
=C=O
..
: ¬¾
®:O -C º O
..
: 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
-
¬¾
® :O
..
º C-O:
1
Bond order of NO+ = (N b - N a )
More is the single bond character. More will be 2
the bond length. Hence, the corret order is :
1 1
CO < CO2 < CO32– = (10 - 4) = ´ 6 = 3
2 2
116. (d) Electronic configuration of the molecule
according to molecular orbital theory, is 1
Similarly, Bond order of NO = (10 - 5)
s1s2s*1s2s2s2s*2s2s2pz2 (p2px2 = s2py2) 2
(p*2px2 = p2py2) s*2pz2s3s2s*3s2s3pz2 1
= (5) = 2.5
(p3px2 = p3py2) (p*3px1 = 3py1) 2
Last two electrons are unpaired. So no. of 1 1
unpaired electron is 2. Bond order of NO– = (10 - 6) = (4) = 2
2 2
117. (a) SiF4 has symmetrical tetrahedral shape which
is due to sp3 hybridisation of silicon atom in its 1 1
excited state while SF4 has distorted tetrahedral Bond order of O -2 = (10 - 7) = (3) = 1.5
2 2
or sea-saw geometry which arises due to sp3d By above calculation, we get
hybridisation of sulphur atom and one lone pair Decreasing bond order
of e–s in one of the equatorial hybrid orbital.
118. (c) As sigma bond is stronger than the p (pi) NO+ > NO > NO– > O -2
bond, so it must be having higher bond energy 122. (a) Paramagnetic character is based upon
than p (pi) bond. 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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48 CHEMISTRY
: :
O
×
×
×
2 * 2 2 * 2
s1s , s 1s , s2s , s 2s , s2 px2 , p2 p 2y In NO, the presence of unpaired electron is clear.
p2 pz2 , p* 2 px2 p* 2 p 2y Therefore, it is paramagnetic.
Anti bonding electron = 8 (4 pairs) 128. (b) The removal of an electron from a diatomic
124. (b) We know that in O2 bond, the order is 2 molecule may increase or decrease the bond
and in O2– bond, the order is 1.5. Therefore, the order.
wrong statements is (b). Removal of an electron from bonding orbital results
125. (c) The N–O bond length decreases in the order in decrease of bond order, hence reduces bond
O strength while removal of an electron from
antibonding orbital results into increase in bond
—
—
— N—N —
> O— — O > N—
—O
O
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States of Matter 49
5 States of Matter
Topic 1: Gas laws and Ideal gas Equation 27°C in identical conditions. The ratio of the
volumes of gases H2 : O2 : methane would be :
1. A mixture of N2 and Ar gases in a cylinder (a) 8 : 16 : 1 (b) 16 : 8 : 1 [2014]
contains 7 g of N2 and 8 g of Ar. If the total (c) 16 : 1 : 2 (d) 8 : 1 : 2
pressure of the mixture of the gases in the 5. Dipole-induced dipole interactions are present
cylinder is 27 bar, the partial pressure of N2 is: in which of the following pairs : [NEET 2013]
[Use atomic masses (in g mol–1) : N = 14, Ar = 40] (a) Cl2 and CCl4 (b) HCl and He atoms
(a) 12 bar (b) 15 bar [2020] (c) SiF4 and He atoms (d) H2O and alcohol
(c) 18 bar (d) 9 bar 6. 50 mL of each gas A and of gas B takes 150 and
2. The volume occupied by 1.8 g of water vapour 200 seconds respectively for effusing through a
at 374°C and 1 bar pressure will be pin hole under the similar condition. If molecular
[Use R = 0.083 bar LK–1 mol–1] mass of gas B is 36, the molecular mass of gas A
[NEET Odisha 2019] will be : [2012]
(a) 5.37 L (b) 96.66 L (a) 96 (b) 128
(c) 55.87 L (d) 3.10 L (c) 32 (d) 64
3. Equal moles of hydrogen and oxygen gases are 7. A certain gas takes three times as long to effuse
placed in a container with a pin-hole through out as helium. Its molecular mass will be :
which both can escape. What fraction of the (a) 27 u (b) 36 u [2012 M]
oxygen escapes in the time required for one-half (c) 64 u (d) 9 u
of the hydrogen to escape ? [2016] 8. Two gases A and B having the same volume
(a) 1/8 (b) 1/4 diffuse through a porous partition in 20 and 10
(c) 3/8 (d) 1/2 seconds respectively. The molecular mass of A
4. Equal masses of H2,O2 and methane have been is 49 u. Molecular mass of B will be : [2011]
taken in a container of volume V at temperature
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50 CHEMISTRY
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States of Matter 51
23. Internal energy and pressure of a gas per unit (c) Z < 1 and attractive forces are dominant
volume are related as : [1993] (d) Z < 1 and repulsive forces are dominant
3 31. The correction factor ‘a’ to the ideal gas equation
(a) P = 2 E (b) P = E
3 2 corresponds to [2018]
1 (a) Density of the gas molecules
(c) P = E (d) P = 2 E (b) Volume of the gas molecules
2
24. The ratio among most probable velocity, mean (c) Forces of attraction between the gas molecules
velocity and root mean square velocity is given (d) Electric field present between the gas molecules
by [1993] 32. Given van der Waals constants for NH3, H2, O2
and CO2 are respectively 4.17, 0.244, 1.36 and
(a) 1 : 2 : 3 (b) 1 : 2. 3 3.59, which one of the following gases is most
(c) 2 : 3 : 8/ p (d) 2 : 8/p : 3 easily liquefied? [2018]
25. A closed flask contains water in all its three (a) NH3 (b) H2
states solid, liquid and vapour at 0°C. In this (c) CO2 (d) O2
situation, the average kinetic energy of water 33. A gas such as carbon monoxide would be most
molecules will be [1992] likely to obey the ideal gas law at : [2015 RS]
(a) the greatest in all the three states (a) high temperatures and low pressures.
(b) the greatest in vapour state (b) low temperatures and high pressures.
(c) the greatest in the liquid state (c) high temperatures and low pressures.
(d) the greatest in the solid state (d) low temperatures and low pressures.
26. In a closed flask of 5 litres, 1.0 g of H2 is heated 34. Maximum deviation from ideal gas is expected
from 300 to 600 K. Which statement is not correct? from : [NEET 2013]
(a) Pressure of the gas increases [1991] (a) N2(g) (b) CH4(g)
(b) The rate of collision increases (c) NH3 (g) (d) H2(g)
(c) The number of moles of gas increases 35. What is the density of N2 gas at 227°C and
(d) The energy of gaseous molecules increases 5.00 atm pressure? (R = 0.0821 atm K–1 mol–1)
27. The root mean square speeds at STP for the [NEET Kar. 2013]
gases H2, N2, O2 and HBr are in the order : [1991] (a) 0.29 g/ml (b) 1.40 g/ml
(a) H2< N2< O2 < HBr (b) HBr < O2 < N2 < H2 (c) 2.81 g/ml (d) 3.41 g/ml
(c) H2 < N2 = O2< HBr (d) HBr < O2 < H2 < N2. 36. For real gases, van der Waals equation is written as
28. Root mean square velocity of a gas molecule is æ an2 ö
proportional to [1990] çè p + 2 ÷ø (V - nb) = nRT
V
(a) m1/2 (b) m0
where ‘a’ and ‘b’ are van der Waals constants.
(c) m–1/2 (d) m
Two sets of gases are :
29. Absolute zero is defined as the temperature
(I) O2, CO2, H2 and He (II) CH4, O2 and H2
(a) at which all molecular motion ceases
The gases given in set-I in increasing order of
(b) at which liquid helium boils [1990]
‘b’ and gases given in set-II in decreasing order
(c) at which ether boils
of ‘a’, are arranged below. Select the correct
(d) all of the above
order from the following : [2012 M]
Topic 3 : van der Waal's Equation and
(a) (I) He < H2 < CO2 < O2 (II) CH4 > H2 > O2
liquefaction of Gases
(b) (I) O2 < He < H2 < CO2 (II) H2 > O2 > CH4
30. A gas at 350 K and 15 bar has molar volume 20 (c) (I) H2 < He < O2 < CO2 (II) CH4 > O2 > H2
percent smaller than that for an ideal gas under the (d) (I) H2 < O2 < He < CO2 (II) O2 > CH4 > H2
same conditions. The correct option about the gas 37. A gaseous mixture was prepared by taking equal
and its compressibility factor (Z) is: [2019] mole of CO and N2. If the total pressure of the
(a) Z > 1 and attractive forces are dominant mixture was found 1 atmosphere, the partial
(b) Z > 1 and repulsive forces are dominant pressure of the nitrogen (N2) in the mixture is :
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52 CHEMISTRY
(a) 0.5 atm (b) 0.8 atm [2011] (b) There is no interaction between the molecules
(c) 0.9 atm (d) 1 atm (c) All molecules of the gas move with same speed
38. The pressure exerted by 6.0 g of methane gas in (d) At a given temperature, PV is proportional
a 0.03 m3 vessel at 129°C is (Atomic masses : to the amount of the gas
C = 12.01, H = 1.01 and R = 8.314 JK–1 mol –1) 45. An ideal gas can’t be liquefied because [1992]
(a) 31684 Pa (b) 215216 Pa [2010] (a) its critical temperature is always above 0°C
(c) 13409 Pa (d) 41648 Pa (b) its molecules are relatively smaller in size
39. van der Waal's real gas, act as an ideal gas, at (c) it solidifies before becoming a liquid
which conditions? [2002] (d) forces operated between its molecules are
(a) High temperature, low pressure negligible
(b) Low temperature, high pressure 46. Select one correct statement. In the gas equation,
(c) High temperature, high pressure PV = nRT [1992]
(d) Low temperature, low pressure (a) n is the number of molecules of a gas
40. Cyclopropane and oxygen at partial pressures (b) V denotes volume of one mole of the gas
170 torr and 570 torr respectively are mixed in a (c) n moles of the gas have a volume V
gas cylinder. What is the ratio of the number of (d) P is the pressure of the gas when only one
moles of cyclopropane to the number of moles mole of gas is present.
of oxygen (nC3H6/nO2)? [1996] 47. A gas is said to behave like an ideal gas when
170 ´ 42 the relation PV/T = constant. When do you
(a) = 0.39 expect a real gas to behave like an ideal gas ?
570 ´ 32
170 æ 170 570 ö (a) When the temperature is low [1991]
(b) ç + ÷ » 0.19 (b) When both the temperature and pressure
42 è 42 32 ø
are low
170 (c) When both the temperature and pressure
(c) = 0.23
740 are high
170 (d) When the temperature is high and pressure
(d) = 0.30
570 is low
41. At which one of the following temperature - 48. In van der Waal's equation of state for a non-
pressure conditions the deviation of a gas from ideal gas, the term that accounts for
ideal behaviour is expected to be minimum? [1996] intermolecular forces is : [1990]
(a) 350 K and 3 atm. (b) 550 K and 1 atm. (a) (V – b) (b) (RT)–1
(c) 250 K and 4 atm. (d) 450 K and 2 atm.
æ a ö
42. Under what conditions will a pure sample of an (c) çè P + 2 ÷ø (d) RT
ideal gas not only exhibit a pressure of 1 atm but V
also a concentration of 1 mole litre–1 ? 49. If P, V, M, T and R are pressure, Volume, molar
(R = 0.082 litre atm mol–1deg–1) [1993] mass, temperature and gas constant respectively,
(a) At STP then for an ideal gas, the density is given by
(b) When V = 22.4 litres [1989]
(c) When T = 12 K RT P
(a) (b)
(d) Impossible under any conditions PM RT
43. When is deviation more in the behaviour of a M PM
(c) (d)
gas from the ideal gas equation PV = nRT ? [1993] V RT
(a) At high temperature and low pressure 50. Correct gas equation is : [1989]
(b) At low temperature and high pressure
(c) At high temperature and high pressure V1T2 V2T1 PV
1 1 = T1
(a) = (b)
(d) At low temperature and low high pressure P1 P2 P2V2 T2
44. Which is not true in case of an ideal gas ? [1992] PT
1 2 = P2V2 V1V2
= P1P2
(a) It cannot be converted into a liquid (c) (d)
V1 T2 T1T2
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States of Matter 53
Topic 4: Liquid State (a) In fixed ratio with that in the lower layer
(b) Same as the lower layer
51. The surface tension of which of the following
(c) Lower than the lower layer
liquid is maximum? [2005]
(d) Higher than the lower layer.
(a) C2H5OH (b) CH3OH
53. A liquid can exist only : [1994]
(c) H2O (d) C6H6
(a) between triple point and critical temperature
52. In a pair of immiscible liquids, a common solute
(b) at any temperature above the melting point
dissolves in both and the equilibrium is reached.
(c) between melting point and critical
Then the concentration of the solute in upper
temperature
layer is [1994]
(d) between boiling and melting temperature.
ANSWER KEY
1 (b) 7 (b) 13 (b) 19 (c) 25 (b) 31 (c) 37 (a) 43 (b) 49 (d)
2 (a) 8 (b) 14 (a) 20 (c) 26 (c) 32 (a) 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)
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54 CHEMISTRY
rA MB weight of H 2 8 ´ 2 1
= = =
8. (b) weight of O 2 32 ´ 1 2
rB MA
13. (b) Given initial volume (V1) = 500 mL ; Initial
V temperature (T1) = 27ºC = 300 K and final
20 = MB 1 MB
Þ = temperature (T2) = –5ºC = 268 K.
V 49 2 49 From Charle’s law :
10
1 V1 V2 500 V2
MB = ´ 49 = 12.25u = or =
4 T1 T2 300 268
9. (a) Given Where V2 = New volume of gas
P1 = 1.5 bar, T1 = 273 + 15 = 288 K, V1 = V 500
P2 = 1.0 bar, T2 = 273 + 25 = 298K, V2 = ? V2 = ´ 268 = 446.66 ml.
300
PV
1 1 = P2V2 14. (a) Given initial volume (V1) = 600 c.c.; Initial
T1 T2 pressure (P 1 ) = 750 mm and final volume
(V2) = 500 c.c. According to Boyle’s law,
1.5 ´ V 1 ´ V2 P1V1 = P2V2
=
288 298 Þ 750 × 600 = P2 × 500
V2 = 1.55 V i.e., volume of bubble will be almost
1.6 time to initial volume of bubble. 750 ´ 600
or P2 = = 900 mm .
10. (a) Rate of diffusion depend upon molecular 500
weight Therefore increase in pressure = (900 – 750)
= 150 mm.
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States of Matter 55
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States of Matter 57
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58 CHEMISTRY
6 Thermodynamics
Trend Analysis with Important Topics & Sub-Topics
second law of
Entropy and second 1 A
thermodynamics
law of thermodynamics
entropy 1 E
Spontaneity and Gibb's spontaniety 1 A 1 E
free energy
LOD - Level of Difficulty E - Easy A - Average D - Difficult Qns - No. of Questions
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Thermodynamics 59
5. A gas is allowed to expand in a well insulated 13. Adiabatic expansions of an ideal gas is
container against a constant external pressure accompanied by [1999]
of 2.5 atm from an initial volume of 2.50 L to a (a) decrease in DE
final volume of 4.50 L. The change in internal (b) increase in temperature
energy DU of the gas in joules will be:- [2017]
(c) decrease in DS
(a) – 500 J (b) – 505 J
(c) + 505 J (d) 1136.25 J (d) no change in any one of the above properties
6. The heat of combustion of carbon to CO2 is Topic 2: Laws of Thermochemistry
–393.5 kJ/mol. The heat released upon formation
of 35.2 g of CO2 from carbon and oxygen gas is 14. The bond dissociation energies of X2, Y2 and
[2015 RS] XY are in the ratio of 1 : 0.5 : 1. DH for the formation
(a) –315 kJ (b) +315 kJ of XY is –200 kJ mol–1. The bond dissociation
(c) –630 kJ (d) –3.15 kJ energy of X2 will be [2018]
7. When 5 litres of a gas mixture of methane and
(a) 200 kJ mol–1 (b) 100 kJ mol–1
propane is perfectly combusted at 0°C and 1
atmosphere, 16 litre of oxygen at the same (c) 400 kJ mol–1 (d) 800 kJ mol–1
temperature and pressure is consumed. The 15. Three thermochemical equations are given below:
amount of heat released from this combustion in (i) C(graphite) + O2(g) ® CO2(g);
kJ (DHcomb (CH4) = 890 kJ mol–1, DHcomb (C3H8) = DrH° = x kJ mol–1
2220 kJ mol–1) is [NEET Kar. 2013] 1
(a) 32 (b) 38 (ii) C(graphite) + O2(g) ® CO(g);
2
(c) 317 (d) 477 DrH° = y kJ mol–1
8. Which of the following is correct option for free
expansion of an ideal gas under adiabatic 1
(iii) CO(g) + O2(g) ® CO2(g);
condition ? [2011] 2
(a) q = 0, DT ¹ 0, w = 0 DrH° = z kJ mol–1
(b) q ¹ 0, DT = 0, w = 0 Based on the above equations, find out which
(c) q = 0, DT = 0, w = 0 of the relationship given below is correct?
(d) q = 0, DT < 0, w ¹ 0 [NEET Kar. 2013]
9. Three moles of an ideal gas expanded (a) x = y – z (b) z = x + y
spontaneously into vacuum. The work done will
(c) x = y + z (d) y = 2z – x
be : [2010]
(a) Zero (b) Infinite 16. Standard enthalpy of vapourisation Dvap H° for
(c) 3 Joules (d) 9 Joules water at 100°C is 40.66 kJ mol–1. The internal
10. Which of the following are not state functions ? energy of vaporisation of water at 100°C
(I) q + w (II) q [2008] (in kJ mol–1) is : [2012]
(III) w (IV) H - TS (a) + 37.56 (b) – 43.76
(a) (I) and (IV) (b) (II), (III) and (IV) (c) + 43.76 (d) + 40.66
(c) (I), (II) and (III) (d) (II) and (III) (Assume water vapour to behave like an ideal gas).
11. In a closed insulated container, a liquid is stirred 17. Equal volumes of two monoatomic gases, A and
with a paddle to increase the temperature, which B, at same temperature and pressure are mixed.
of the following is true? [2002]
The ratio of specific heats (Cp/Cv) of the mixture
(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 will be : [2012 M]
12. When 1 mol of a gas is heated at constant (a) 0.83 (b) 1.50
volume, temperature is raised from 298 to 308 K. (c) 3.3 (d) 1.67
If heat supplied to the gas is 500 J, then which 18. Enthalpy change for the reaction, [2011]
statement is correct ? [2001] 4H(g) ¾¾ ® 2H 2 (g) is – 869.6 kJ.
(a) q = w = 500 J, DU = 0 The dissociation energy of H–H bond is :
(b) q = DU = 500 J, w = 0 (a) – 434.8 kJ (b) – 869.6 kJ
(c) q = –w = 500 J, DU = 0
(c) + 434.8 kJ (d) + 217.4 kJ
(d) DU = 0, q = w = –500 J
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Thermodynamics 61
31. The densities of graphite and diamond at 298 K (a) DH < DE (b) DH = DS
are 2.25 and 3.31 g cm–3, respectively. If the (c) DH = DE (d) DH = DG
standard free energy difference (DGº) is equal 40. One mole of an ideal gas at 300 K is expanded
to 1895 J mol–1, the pressure at which graphite isothermally from an initial volume of 1 litre to
will be transformed into diamond at 298 K is 10 litres. The DE for this process is
[2003] (R = 2 cal. mol–1 K–1) [1998]
(a) 9.92 × 105 Pa (b) 9.92 × 108 Pa (a) 163.7 cal (b) zero
(c) 9.92 × 107 Pa (d) 9.92 × 106 Pa (c) 1381.1 cal (d) 9 lit. atm
32. For which one of the following equations is
DHºreact equal to DHf º for the product? [2003] 41. Given that C + O2 ® CO 2 : DHº = - x kJ
(a) 2 CO(g) + O 2 (g) ® 2 CO 2 (g) 2CO + O2 ® 2CO 2 : DHº = - y kJ
(b) N 2 (g) + O3 (g) ® N 2 O 3 (g) the enthalpy of formation of carbon monoxide
(c) CH 4 ( g ) + 2Cl ( g ) ® CH 2 Cl 2 l + 2HCl ( g ) will be [1997]
()
2x - y y - 2x
(d) Xe(g) + 2F2 (g) ® XeF4 (g) (a) (b)
33. For the reaction 2 2
C3 H8 ( g ) + 5O2 ( g ) ® 3CO2 ( g ) + 4 H 2O(l ) (c) 2x – y (d) y – 2x
at constant temperature, DH – DE is [2003] 42. Hydrogen has an ionisation energy of 1311 kJ
(a) – RT (b) + RT mol –1 and for chlorine it is 1256 kJ mol –1.
(c) – 3 RT (d) + 3 RT Hydrogen forms H+ (aq) ions but chlorine does
34. Heat of combustion DHº for C (s), H2 (g) and not form Cl+ (aq) ions because [1996]
CH4 (g) are –94, –68 and –213 kcal/mol, then (a) H+ has lower hydration enthalpy
(b) Cl+ has lower hydration enthalpy
DHº for C(s) + 2H 2 (g) ® CH 4 (g) is [2002]
(c) Cl has high electron affinity
(a) –17 kcal (b) – 111 kcal (d) Cl has high electronegativity
(c) –170 kcal (d) –85 kcal
1 43. If enthalpies of formation of C2 H 4( g ) , CO2(g)
35. Enthalpy of CH 4 + O 2 ® CH 3 OH is
2 and H 2 O(l) at 25°C and 1atm pressure are 52,
negative. If enthalpy of combustion of CH4 and
CH3OH are x and y respectively, then which – 394 and – 286 kJ/mol respectively, the
relation is correct [2001] enthalpy of combustion of C2H4 is equal to
(a) x > y (b) x < y [1995]
(c) x = y (d) x ³ y (a) – 141.2 kJ/mol (b) – 1412 kJ/mol
36. What is the enthalpy change for, (c) + 14.2 kJ/mol (d) + 1412 kJ/mol
44. Equal volumes of molar hydrochloric acid and
2H 2O 2 (l) ® 2 H 2 O(l) + O 2 (g) if heat of sulphuric acid are neutralized by dil. NaOH
formation of H2O2 (l) and H2O (l) are –188 and solution and x kcal and y kcal of heat are liberated
–286 kJ/mol respectively? [2001] respectively. Which of the following is true ?
(a) –196 kJ/mol (b) + 948 kJ/mol [1994]
(c) + 196 kJ/mol (d) –948 kJ/mol 1
37. The values of heat of formation of SO2 and SO3 (a) x = y (b) x = y
are –298.2 kJ and –98.2 kJ. The heat of formation 2
(c) x = 2y (d) None of these
of the reaction [2000] 45. For the reaction [1994]
SO 2 + (1 / 2) O 2 ® SO 3 will be N 2 + 3H 2 2NH3, D H = ?
(a) –200 kJ (b) –356.2 kJ
(c) + 200 kJ (d) – 396.2 kJ (a) D E + 2RT (b) D E –2RT
38. For a cyclic process, which of the following is (c) D H = RT (d) D E – RT.
not true? [1999] 46. During isothermal expansion of an ideal gas, its
(a) DH = 0 (b) DE = 0 [1991, 94]
(c) DG = 0 (d) Total W = 0 (a) internal energy increases
39. For a reaction in which all reactants and (b) enthalpy decreases
products are liquids, which one of the following (c) enthalpy remains unaffected
equations is most applicable ? [1999] (d) enthalpy reduces to zero.
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47. If DH is the change in enthalpy and DE , the 54. What is the entropy change (in JK–1 mol–1) when
change in internal energy accompanying a one mole of ice is converted into water at 0º C?
gaseous reaction, then [1990] (The enthalpy change for the conversion of ice
to liquid water is 6.0 kJ mol–1 at 0ºC) [2003]
(a) DH is always greater than DE , (a) 21.98 (b) 20.13
(b) DH < D E only if the number of moles of (c) 2.013 (d) 2.198
the products is greater than the number of 55. 2 mole of an ideal gas at 27ºC temperature is
moles of the reactants expanded reversibly from 2 lit to 20 lit. Find the
(c) D H is always less than D E entropy change (R = 2 cal/mol K) [2002]
(d) D H < D E only if the number of moles of (a) 92.1 (b) 0
products is less than the number of moles (c) 4 (d) 9.2
of the reactants. 56. Unit of entropy is [2002]
Topic 3: Entropy and Second Law of (a) JK–1 mol–1 (b) J mol–1
Thermodynamics (c) J–1 K–1 mol–1 (d) JK mol–1
57. The entropy change in the fusion of one mole
48. For the reaction, 2Cl(g) ¾® Cl2(g), the correct of a solid melting at 27ºC (Latent heat of fusion,
option is : [2020] 2930 J mol–1) is : [2000]
(a) DrH > 0 and DrS < 0 (a) 9.77 J K mol
–1 –1 (b) 10.73 J K mol–1
–1
(b) DrH < 0 and DrS > 0 (c) 2930 J K mol–1 –1 (d) 108.5 J K–1 mol–1
(c) DrH < 0 and DrS < 0 58. Identify the correct statement regarding
(d) DrH > 0 and DrS > 0 entropy: [1998]
(a) At absolute zero of temperature, entropy
49. In which case change in entropy is negative ? of a perfectly crystalline substance is taken
[2019] to be zero
(a) Evaporation of water (b) At absolute zero of temperature, the
(b) Expansion of a gas at constant temperature entropy of a perfectly crystalline substance
(c) Sublimation of solid to gas is +ve
(d) 2H(g) ® H2(g) (c) At absolute zero of temperature, the
50. The enthalpy of fusion of water is 1.435 kcal/mol. entropy of all crystalline substances is
The molar entropy change for the melting of ice taken to be zero
at 0°C is : [2012] (d) At 0ºC, the entropy of a perfectly
crystalline substance is taken to be zero
(a) 10.52 cal / (mol K) (b) 21.04 cal / (mol K)
59. According to the third law of thermodynamics
(c) 5.260 cal / (mol K) (d) 0.526 cal / (mol K)
which one of the following quantities for a
51. If the enthalpy change for the transition of liquid perfectly crystalline solid is zero at absolute
water to steam is 30 kJ mol–1 at 27ºC, the entropy zero? [1996]
change for the process would be : [2011] (a) Free energy (b) Entropy
(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 (c) Enthalpy (d) Internal energy
52. Standard entropies of X2, Y2 and XY3 are 60, 40 60. Given the following entropy values (in J K–1 mol–1)
and 50 JK–1mol–1 respectively. For the reaction at 298 K and 1 atm :H2 (g) : 130.6, Cl2 (g) : 223.0,
1 3 HCl (g) : 186.7. The entropy change (in J K–1
X 2 + Y2 XY3 , DH = – 30 kJ mol–1) for the reaction
2 2
to be at equilibrium, the temperature should be: H 2 (g) + Cl2 (g) ¾¾
® 2HCl(g) is [1996]
[2010] (a) +540.3 (b) +727.0
(a) 750 K (b) 1000 K (c) –166.9 (d) +19.8
(c) 1250 K (d) 500 K 61. A chemical reaction will be spontaneous if it is
53. For the gas phase reaction, [2008] accompanied by a decrease of [1994]
PCl5(g) PCl3(g) + Cl2(g) (a) entropy of the system
which of the following conditions are correct ? (b) enthalpy of the system
(a) DH = 0 and DS < 0 (b) DH > 0 and DS > 0 (c) internal energy of the system
(c) DH < 0 and DS < 0 (d) DH > 0 and DS < 0 (d) free energy of the system
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Thermodynamics 67
The state of N2O3 is liquid in NTP conditions. 38. (d) For a cyclic process
Thus, this reaction is not equal to DHf° for the DE = 0, DH = 0 & DG = 0. As all depend upon
product. final state and initial state,W doesn’t depend
(d) Xe(g) + 2F2(g) ¾¾ ® XeF4(g) on path followed.
For this reaction DHr° is equal to DHf° of XeF4(g). 39. (c) As all reactant and product are liquid
33. (c) DH = DE + DnRT Dn ( g ) = 0
Dn = 3 – (1 + 5)
= 3 – 6 = –3 DH = DE - DnRT
DH - DE = (-3RT ) DH = DE (Q Dn = 0)
40. (b) For an isothermal process DE = 0
34. (a) C(s) + O2 (g) ® CO2 (g)
41. (b) Given C + O 2 = CO 2 , DH º = - x kJ ....(a)
DH = -94 kcal/mole ..(i)
1 2CO 2 = 2CO + O 2 DH º = + y kJ … (b)
H 2 (g) + O2 ® H 2 O(g)
2 or CO2 = CO + 1/ 2 O2 , DH º = + y / 2 kJ ...(c)
DH = -68 kcal/mole ..(ii) From eq. no. (a) and (c)
CH 4 + 2O2 ® CO2 + 2H 2O, 1 y - 2x
DH = -213 k cal/mole ...(iii) C + O 2 = CO, DH º = y / 2 - x = kJ
2 2
C(s) + 2H 2 ® CH 4 (g), DH = ? ...(iv) 42. (b) Hydration energy of Cl+ is very less than
Eqn. (iv) can be obtained by H+, hence it doesn’t form Cl+ (aq) ion.
eq. (i) + eq. (ii) × 2 – eq.(iii) 43. (b) Enthalpy of formation of C 2 H 4 , CO2 and
C(s) + O2 (g) ® CO 2 (g) H 2 O ar e 52, – 394 an d – 286 kJ/ mol
2H 2 (g) + O2 (g) ® 2H 2O(g) respectively. (Given)
CO2 (g) + 2H 2 O(g) ® CH 4 (g) + 2O 2 (g) The reaction is
C(s) + 2H 2 (g) ® CH 4 (g) C 2 H 4 + 3O 2 ® 2CO 2 + 2H 2O.
So, DH CH 4 = -94 + 2( -68) - ( -213) change in enthalpy,
= –94 – 136 + 213 = –17 k cal/mole
( DH ) = DH products - DH reactants
35. (a) The enthalpy of combustion is always
negative. = 2 ´ (-394) + 2 ´ (-286) - (52 + 0)
CH4 + 2O2 ¾¾ ® CO2 + 2H2O; DH1 = –x kJ = – 1412 kJ/ mol.
3 44. (b) 1 M H2SO4 = 2g eq. of HCl
CH3OH + O2 ¾¾ ® CO2 + 2H2O; D H2 = - ykJ
2 1
1 Hence y = 2x or x = y.
CH4 + O2 ¾¾ ®CH2OH; + DH = DH1- DH2 2
2
It is given that DH = is negative. 45. (b) D ng = 2 – 4 = – 2, D H = D E – 2RT.
Thus, –x –(–y) = – ve 46. (c) During isothermal expansion of ideal gas,
y – x = –ve D T = 0. Now H = E + PV
Hence, x > y.
Q DH = DE + D( PV )
36. (a) 2H 2 O 2 (l) ¾¾ ® 2H 2 O(l) + O 2 (g) DH = ?
\ D H = D E + D (nRT);
DH = [2 ´ DH f of H 2 O ( l ) + ( DH f of O 2 ) Thus, if D T = 0., DH = DE
-(2 ´ DH f of H 2O2(l) )] i.e., remain unaffected.
47. (d) As DH = DE + DngRT
= [(2 ´ ( -286)) + (0) - (2 ´ (-188))]
if np < nr; D ng = np – nr = – ve.
= [-572 + 376] = -196 kJ / mole
Hence D H < D E.
1
37. (c) SO 2 + O2 ¾
¾® SO 3
48. (c) We know that, Cl 2 (g) ¾¾ ® 2Cl (g) is
2
DH = DH fo (SO ) - DH fo (SO ) endothermic reaction because it required energy
3 2 to break bond.
= –98.2 + 298.2 = 200 kJ/Mole
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68 CHEMISTRY
= 2 × 2 × T × 2.303 × 1 = 9.2 T
So reverse reaction, 2Cl (g) ¾¾ ® Cl2 (g) will
be q 9.2T
Entropy change, DS = = = 9.2 cal.
exothermic, D r H < 0 . T T
q
Also, two gaseous atom combine together to 56. (a) DS =
form 1 gaseous molecule. T
q ¾¾ ® required heat per mole
49. (d) In 2H(g) ¾ ® H 2 (g), no. of moles T ¾¾ ® constant absolute temperature
decreases, therefore entropy decreases. Unit of entropy is JK–1 mol–1
Latent heat of fusion DH y
DH 1.435 ´ 103 57. (a) DS = =
50. (c) DS = = Melting point T
T 273
= 5.260 cal / mol - K 2930
51. (d) Given D H = 30 kJ mol–1 T = 273 + 27 = 300 K = J K–1 mol–1 = 9.77 J K–1 mol–1
300
58. (a) We know from the third law of
DHT 3 ´ 104
DST = = J mol–1 thermodynamics, the entropy of a perfectly
T 300 crystalline substance at absolute zero
= 100 J mol–1 K–1 temperature is taken to be zero.
1 3 59. (b) Entropy is zero for perfectly crystaline solid
52. (a) ΔS for the reaction X 2 + Y2 XY3 at absolute zero.
2 2
ΔS = 50 – (30 + 60) = – 40 J
Entropy states the randomness or disorderness of
For equilibrium D G = 0 = D H – T D S the system. At absolute zero, the movement of
H - 30000
D
molecules of the system or randomenss of the
T = = = 750 K system is zero, hence entropy is also zero.
S
D - 40
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7 Equilibrium
Topic 1: Law of Mass Action, Equilibrium The equilibrium constant (K) of the reaction :
Constant (Kc and Kp) and its Application 5 K
2NH3 + O 2 2NO + 3H2O, will be;
2
1. The equilibrium constant of the following are :
N2 + 3H2 2NH3 K1 [2017] (a) K 2 K33 / K1 (b) K2K 3/K 1
N2 + O2 2NO K2 (c) K 23 K3 / K1 (d) K1K33 / K 2
1
H 2 + O 2 ® H 2O K3
2
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12. The values of Kp1 and Kp2 for the reactions 16. K1 and K2 are equilibrium constant for reactions
Y + Z
X ...(a) [2008] (i) and (ii)
2 NO (g) .............(i)
N2(g) + O2 (g)
2 B
and A ...(b)
1 1
are in the ratio of 9 : 1. If degree of dissociation
NO(g) N 2 (g) + O 2 (g) ......(ii)
2 2
of X and A be equal, then total pressure at
Then, [1989, 94, 2005]
equilibrium for (a) and (b) are in the ratio :
(a) 3 : 1 (b) 1 : 9 2
æ 1 ö
(c) 36 : 1 (d) 1 : 1 (a) K1 = ç (b) K1 = K22
è K 2 ÷ø
13. The value of equilibrium constant of the reaction
1 1 1
HI (g ) H 2 (g) + I 2 is 8.0 [2008] (c) K1 =
K2
(d) K1 = (K2)0
2 2
The equilibrium constant of the reaction 17. Value of KP in the reaction
2HI(g) will be:
H 2 (g) + I 2 (g) MgCO3 (s ) ® MgO( s) + CO2 ( g ) is [2000]
1 1 (a) KP = PCO
(a) (b) 2
16 64 P ´ PMgO
(b) K P = PCO ´ CO2
1 2 PMgCO3
(c) 16 (d)
8
14. The following equilibrium constants are given: PCO2 ´ PMgO
(c) KP =
2NH3 ; K1
N 2 + 3H 2 [2003, 2007] PMgCO3
2NO; K 2
N 2 + O 2 PMgCO3
(d) KP =
1 PCO2 ´ PMgO
H 2O; K3
H 2 + O2
2 18. For dibasic acid correct order is [2000]
The equilibrium constant for the oxidation of 2
(a) Ka1 < Ka 2 (b) K a1 > K a2
moles NH3 by oxygen to give NO is
(c) Ka1 = Ka2 (d) not certain
K 2 K32 K 22 K3
(a) (b) 19. If K1 and K2 are the respective equilibrium
K1 K1 constants for the two reactions
XeF6(g) + H2O(g) XeOF (g) + 2HF(g)
K1 K 2 K 2 K33 4
(c) (d) XeO4(g) + XeF6(g) XeOF (g) + XeO F (g)
K3 K1 4 3 2
the equilibrium constant of the reaction
15. For the reaction [2006]
XeO4(g) + 2HF(g) XeO F (g) + H O(g)
CO (g) + 2H O(l),
CH 4 (g) + 2O 2 (g) 3 2 2
2 2 will be [1998]
DrH = –170.8 k J mol–1 (a) K1/(K2)2 (b) K1 . K2
Which of the following statements is not true ? (c) K1/K2 (d) K2/K1
(a) The equilibrium constant for the reaction
20. If a is the fraction of HI dissociated at
[CO 2 ] H
is given by K p = equilibrium in the reaction, 2 HI (g) 2
[CH 4 ][O2 ]2 (g) + I2 (g), starting with 2 moles of HI, the total
(b) Addition of CH4(g) or O2(g) at equilibrium number of moles of reactants and products at
will cause a shift to the right equilibrium are [1996]
(c) The reaction is exothermic (a) 2 + 2a (b) 2
(d) At equilibrium, the concentrations of (c) 1 + a (d) 2 – a
CO2(g) and H2O(l) are not equal
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Equilibrium 73
21. The rate constant for forward and backward 26. For the reversible reaction, [2014]
reaction of hydrolysis of ester are 1.1 ´ 10 -2 and
N2(g) + 3H2(g) 2NH3(g) + heat
1 .5 ´ 10 -3 per minute respectively. Equilibrium The equilibrium shifts in forward direction
constant for the reaction [1995] (a) By increasing the concentration of NH3(g)
+ (b) By decreasing the pressure
CH COOC H + H
3 2 5
(c) By decreasing concentration of N2(g) and
CH3COOH + C 2 H5OH is H2(g)
(a) 4.33 (b) 5.33 (d) By increasing pressure and decreasing
(c) 6.33 (d) 7.33 temperature.
Topic 2: Relation between K, Q and G and 27. For a given exothermic reaction, Kp and KP¢ are
Factors Effecting Equilibrium the equilibrium constants at temperatures T1 and
T2, respectively. Assuming that heat of reaction
22. Hydrolysis of sucrose is given by the following is constant in temperature range between T1 and
reaction. [2020] T2, it is readily observed that: [2014]
Sucrose + H2O Glucose + Fructose (a) Kp > K P¢ (b) Kp < K P¢
If the equilibrium constant (KC) is 2 × 1013 at
– 1
300 K, th e value of DrG at the same (c) Kp = K P¢ (d) Kp =
K ¢p
temperature will be :
28. The dissociation constants for acetic acid and
(a) 8.314 J mol–1K–1 × 300 K × ln(2 × 1013) HCN at 25°C are 1.5 × 10–5 and 4.5 × 10–10
(b) 8.314 J mol–1K–1 × 300 K × ln(3 × 1013) respectively. The equilibrium constant for the
(c) –8.314 J mol–1K–1 × 300 K × ln(4 × 1013) equilibrium [2009]
(d) –8.314 J mol–1K–1 × 300 K × ln(2 × 1013) CN– + CH3COOH HCN + CH COO–
3
23. Which one of the following conditions will favour would be:
maximum formation of the product in the (a) 3.0 × 10– 5 (b) 3.0 × 10– 4
(c) 3.0 × 10 4 (d) 3.0 × 105
reaction, [2018]
X 2 (g) D r H = – X kJ :
A2 (g) + B2 (g) 29. The reaction quotient (Q) for the reaction
2NH (g)
N2(g) + 3H2(g)
(a) Low temperature and high pressure 3
(b) Low temperature and low pressure [NH 3 ]2
is given by Q = . The reaction will
(c) High temperature and low pressure
[N 2 ][H 2 ]3
(d) High temperature and high pressure
24. Consider the following liquid - vapour proceed from right to left if [2003]
equilibrium. [2016] (a) Q = 0 (b) Q = Kc
Vapour
Liquid (c) Q < Kc (d) Q > Kc
where Kc is the equilibrium constant
Which of the following relations is correct ?
30. For the reaction
dlnG DH v dlnP DH v 2BaO(s) + O (g);
(a) = (b) = 2BaO2(s) 2
2 2 dT RT
dT RT DH = +ve. In equilibrium condition, pressure of
dlnP -DH v dlnP -DH v O2 is dependent on [2002]
(c) = (d) =
dT 2
T 2 dT RT 2 (a) mass of BaO2
(b) mass of BaO
25. Which of the following statements is correct for
(c) temperature of equilibrium
a reversible process in a state of equilibrium ?
(d) mass of BaO2 and BaO both
[2015]
31. In a two-step exothermic reaction
(a) DG = 2.30 RT log K
(b) DGº = –2.30 RT log K
A2(g) + B2(g)
3C(g)
D(g)
Step 1 Step 2
(c) DGº = 2.30 RT log K Steps 1 and 2 are favoured respectively by
(d) DG = –2.30 RT log K [1997]
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74 CHEMISTRY
(a) high pressure, high temperature and low 38. Conjugate base for Brönsted acids H2O and HF
pressure, low temperature are : [2019]
(b) high pressure, low temperature and low (a) OH– and H2F+, respectively
pressure, high temperature (b) H3O+ and F–, respectively
(c) low pressure, high temperature and high
(c) OH– and F–, respectively
pressure, high temperature
(d) low pressure, low temperature and high (d) H3O+ and H2F+, respectively
pressure, low temperature 39. Following solutions were prepared by mixing
32. The equilibrium constant for the reaction different volumes of NaOH and HCl of different
2A at 500 K and 700 K are 1 × 10–10
A2 concentrations : [2018]
and 1 × 10–5 respectively. The given reaction is M M
[1996] a. 60 mL HCl + 40 mL NaOH
10 10
(a) exothermic (b) slow M M
(c) endothermic (d) fast b. 55 mL HCl + 45 mL NaOH
33. Stan dar d Gibb’s free energy change for 10 10
isomerization reaction [1995] M M
c. 75 mL HCl + 25 mL NaOH
trans-2-pentene
cis-2- pentene 5 5
is – 3.67 kJ/mol at 400 K. If more trans-2-pentene M M
is added to the reaction vessel, then d. 100 mL HCl + 100 mL NaOH
10 10
(a) more cis-2-pentene is formed pH of which one of them will be equal to 1?
(b) equilibrium remains unaffected (a) b (b) a
(c) additional trans-2-pentene is formed (c) c (d) d
(d) equilibrium is shifted in forward direction 40. What is the pH of the resulting solution when
34. According to Le-chatelier’s principle, adding heat equal volumes of 0.1 M NaOH and 0.01 M HCl
liquid equilibrium will cause the
to a solid are mixed ? [2015 RS]
(a) temperature to increase [1993] (a) 12.65 (b) 2.0
(b) temperature to decrease (c) 7.0 (d) 1.04
(c) amount of liquid to decrease 41. Which of the following salts will give highest
(d) amount of solid to decrease.
pH in water ? [2014]
35. Which one of the following information can be
(a) KCl (b) NaCl
obtained on the basis of Le Chatelier principle?
(c) Na2CO3 (d) CuSO4
[1992]
42. Which of these is least likely to act as Lewis
(a) Dissociation constant of a weak acid
(b) Entropy change in a reaction base? [NEET 2013]
(c) Equilibrium constant of a chemical reaction (a) F– (b) BF3
(d) Shift in equilibrium position on changing (c) PF3 (d) CO
value of a constraint. 43. Which of the following is least likely to behave
as Lewis base ? [2011]
Topic 3: Theories of Acids and Bases, Ionic (a) H2O (b) NH3
Product of Water and pH Scale (c) BF3 (d) OH–
36. The pH of 0.01 M NaOH (aq) solution will be 44. Which one of the following molecular hydrides
[NEET Odisha 2019] acts as a Lewis acid? [2010]
(a) 9 (b) 7.01 (a) NH3 (b) H2O
(c) 2 (d) 12 (c) B2H6 (d) CH4
37. Which of the following cannot act both as 45. Which of the following molecules acts as a Lewis
Bronsted acid and as Bronsted base? acid ? [2009]
[NEET Odisha 2019] (a) (CH3)2 O (b) (CH3)3 P
(a) HSO4– (b) HCO3–
(c) (CH3)3 N (d) (CH3)3 B
(c) NH3 (d) HCl
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Equilibrium 75
46. The ionization constant of ammonium hydroxide (a) RCOOH > HC º CH > HOH > ROH
is 1.77 × 10–5 at 298 K. Hydrolysis constant of (b) RCOOH > ROH > HOH > HC º CH
ammonium chloride is: [2009] (c) RCOOH > HOH > ROH > HC º CH
(a) 6.50 × 10– 12 (b) 5.65 × 10–13 (d) RCOOH > HOH > HC º CH > ROH
(c) 5.65 × 10–12 (d) 5.65 × 10–10 53. Which one of the following compounds is not a
47. Equal volumes of three acid solutions of pH 3, 4 protonic acid? [2003]
and 5 are mixed in a vessel. What will be the H+ (a) SO2 (OH)2 (b) B (OH)3
ion concentration in the mixture ? [2008] (c) PO (OH)3 (d) SO (OH)2
(a) 1.11 × 10–4 M (b) 3.7 × 10–4 M 54. In HS–, I–, RNH2 and NH3, order of proton
(c) 3.7 × 10– 3 M (d) 1.11× 10–3 M accepting tendency will be [2001]
48. Calculate the pOH of a solution at 25°C that (a) I– > NH3 > RNH2 > HS–
contains 1× 10– 10 M of hydronium ions, i.e. (b) HS– > RNH2 > NH3 > I–
H3O+. [2007] (c) RNH2 > NH3 > HS– > I–
(a) 4.000 (b) 9.0000 (d) NH3 > RNH2 > HS– > I–
(c) 1.000 (d) 7.000 55. A base when dissolved in water yields a solution
with a hydroxyl ion concentration of 0.05 mol litre–1.
49. The hydrogen ion concentration of a 10–8 M
The solution is [2000]
HCl aqueous solution at 298 K (Kw = 10–14) is
(a) basic (b) acidic
[2006] (c) neutral (d) either 'b' or 'c'
(a) 11 × 10 M–8 (b) 9.525 × 10 M –8
56. Conjugate acid of NH -2 is : [2000]
(c) 1.0 × 10–8 M (d) 1.0 × 10–6 M (a) NH4+ (b) NH3
50. What is the correct relationship between the (c) NH2 (d) NH
pHs of isomolar solutions of sodium oxide 57. Which of the following statements about pH
(pH 1 ), sodium sulph ide (pH 2 ), sodium and H+ ion concentration is incorrect? [2000]
selenide (pH3) and sodium telluride (pH4)? (a) Addition of one drop of concentrated HCl in
[2005] NH4OH solution decreases pH of the solution.
(a) pH1 > pH2 > pH3 > pH4 (b) A solution of the mixture of one equivalent of
(b) pH1 > pH2 » pH3 > pH4 each of CH3COOH and NaOH has a pH of 7
(c) pH1 < pH2 < pH3 < pH4 (c) pH of pure neutral water is not zero
(d) pH1 < pH2 < pH3 » pH4 (d) A cold and concentrated H2SO4 has lower
51. The rapid change of pH near the stoichiometric H+ ion concentration than a dilute solution
point of an acid-base titration is the basis of of H2SO4
indicator detection. pH of the solution is related 58. Among boron trifluoride, stannic chloride and
to ratio of the concentrations of the conjugate stannous chloride, Lewis acid is represented by
acid (HIn) and base (In–) forms of the indicator [1999]
by the expression [2004] (a) only stannic chloride
(b) boron trifluoride and stannic chloride
[ In - ]
(a) log = pK In - pH (c) boron trifluoride and stannous chloride
[ HIn ] (d) only boron trifluoride
[ HIn ] 59. What is the H+ ion concentration of a solution
(b) log = pK In - pH
[ In - ] prepared by dissolving 4 g of NaOH (Atomic
[ HIn ] weight of Na = 23 amu) in 1000 mL? [1999]
(c) log = pH - pK In (a) 10–10 M (b) 10–4 M
[ In - ]
(c) 10–1 M (d) 10–13 M
[ In - ]
(d) log = pH - pK In 60. The pH value of a 10 M solution of HCl is [1995]
[ HIn ] (a) less than 0 (b) equal to 0
52. Which one of the following orders of acid (c) equal to 1 (d) equal to 2
strength is correct? [2003]
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76 CHEMISTRY
Topic 4: Ionisation of Weak Acids and 68. The molar solubility of CaF2(Ksp = 5.3 × 10–11)
Bases and Relation between K a and K b in 0.1 M solution of NaF will be
61. At 100°C the Kw of water is 55 times its value at [NEET Odisha 2019]
–10
(a) 5.3 × 10 mol L –1
25°C. What will be the pH of neutral solution?
(log 55 = 1.74) [NEET Kar. 2013] (b) 5.3 × 10–11 mol L–1
(a) 6.13 (b) 7.00 (c) 5.3 × 10–8 mol L–1
(c) 7.87 (d) 5.13 (d) 5.3 × 10–9 mol L–1
62. Accumulation of lactic acid (HC3H5 O3 ), a 69. pH of a saturated solution of Ca(OH)2 is 9. The
monobasic acid in tissues leads to pain and a solubility product (Ksp) of Ca(OH)2 is: [2019]
feeling of fatigue. In a 0.10 M aqueous solution,
lactic acid is 3.7% dissociated. The value of (a) 0.5 10–15 (b) 0.25 10–10
(c) 0.125 10 –15 (d) 0.5 10–10
dissociation constant, Ka, for this acid will be:
[NEET Kar. 2013] 70. The solubility of BaSO4 in water is
(a) 2.8 × 10–4 (b) 1.4 × 10–5 2.42 × 10–3 gL–1 at 298 K. The value of its
(c) 1.4 × 10–4 (d) 3.7 × 10–4 solubility product (Ksp) will be
63. A weak acid, HA, has a Ka of 1.00 × 10–5. If (Given molar mass of BaSO4 = 233 g mol–1)
0.100 mol of this acid is dissolved in one litre of
(a) 1.08 × 10–10 mol2L–2 [2018]
water, the percentage of acid dissociated at –12 2 –2
equilbrium is closest to [2007] (b) 1.08 × 10 mol L
(a) 1.00% (b) 99.9% (c) 1.08 × 10–8 mol2L–2
(c) 0.100% (d) 99.0% (d) 1.08 × 10–14 mol2L–2
64. At 25°C, the dissociation constant of a base, 71. Concentration of the Ag+ ions in a saturated
BOH, is 1.0 ´ 10-12. The concentration of solution of Ag2C2O4 is 2.2 × 10–4 mol L–1.
hydroxyl ions in 0.01 M aqueous solution of Solubility product of Ag2C2O4 is :- [2017]
the base would be [2005]
(a) 2.66 × 10–12 (b) 4.5 × 10–11
(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 (c) 5.3 × 10–12 (d) 2.42 × 10–8
65. Ionisation constant of CH3COOH is 1.7 × 10–5. 72. MY and NY3, two nearly insoluble salts, have
If concentration of H+ ions is 3.4 × 10–4M, then the same Ksp values of 6.2 × 10–13 at room
find out initial concentration of CH3COOH temperature. Which statement would be true in
molecules [2001] regard to MY and NY3 ? [2016]
(a) 3.4 × 10–4M (b) 3.4 × 10–3M (a) The molar solubilities of MY and NY3 in
(c) 6.8 × 10–3M (d) 6.8 × 10–4M water are identical.
66. Aqueous solution of acetic acid contains (b) The molar solubility of MY in water is less
[1991]
than that of NY3
(a) CH3COO– and H+
(b) CH3COO–, H3O+ and CH3COOH (c) The salts MY and NY3 are more soluble in
(c) CH3COO–, H3O+and H+ 0.5 M KY than in pure water.
(d) CH3COOH, CH3COO– and H+ (d) The addition of the salt of KY to solution
of MY and NY3 will have no effect on their
Topic 5: Common Ion Effect, Salt Hydrolysis, solubilities.
Buffer Solutions and Solubility Product
73. Consider the nitration of benzene using mixed
67. Find out the solubility of Ni(OH)2 in 0.1 M conc of H2SO4 and HNO3. If a large amount of
NaOH. Given that the ionic product of Ni(OH)2 KHSO4 is added to the mixture, the rate of
is 2 × 10–15 [2020] nitration will be [2016]
(a) 2 × 10–8 M (b) 1 × 10–13 M (a) faster (b) slower
(c) 1 × 108 M (d) 2 × 10–13 M (c) unchanged (d) doubled
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Equilibrium 77
74. The Ksp of Ag2 CrO4, AgCl, AgBr and AgI (a) BaCl2 (b) AlCl3
are respectively, 1.1 × 10 –12 , 1.8 × 10–10 , (c) LiCl (d) BeCl2
5.0 × 10–13, 8.3 × 10–17. Which one of the 82. Buffer solutions have constant acidity and
following salts will precipitate last if AgNO3 alkalinity because [2012]
solution is added to the solution containing equal (a) these give unionised acid or base on
moles of NaCl, NaBr, NaI and Na2CrO4?[2015] reaction with added acid or alkali.
(b) acids and alkalies in these solutions are
(a) AgCl (b) AgBr
shielded from attack by other ions.
(c) Ag2CrO4 (d) AgI
(c) they have large excess of H+ or OH– ions
75. Which one of the following pairs of solution is
(d) they have fixed value of pH
not an acidic buffer ? [2015 RS]
83. A buffer solution is prepared in which the
(a) HClO4 and NaClO4
concentration of NH3 is 0.30 M and the
(b) CH3COOH and CH3 COONa
concentration of NH4+ is 0.20 M. If the equilibrium
(c) H2CO3 and Na2CO3
constant, Kb for NH3 equals 1.8 × 10–5, what is
(d) H3PO4 and Na3PO4
the pH of this solution ? (log 2.7 = 0.433).
76. Using the Gibbs energy change, DG° = + 63.3kJ,
[2011]
for the following reaction, [2014]
2Ag+ (aq) + CO 2–(aq) (a) 9.08 (b) 9.43
Ag2CO3 3 (c) 11.72 (d) 8.73
the Ksp of Ag2CO3(s) in water at 25°C is:-
84. In qualitative analysis, the metals of Group I can
(R = 8.314 J K–1 mol–1) be separated from other ions by precipitating
(a) 3.2 × 10–26 (b) 8.0 × 10–12 them as chloride salts. A solution initially
(c) 2.9 × 10 –3 (d) 7.9 × 10–2 contains Ag+ and Pb2+ at a concentration of
77. Identify the correct order of solubility in aqueous 0.10 M. Aqueous HCl is added to this solution
medium: [NEET 2013] until the Cl– concentration is 0.10 M. What will
(a) ZnS > Na2S > CuS (b) Na2S > CuS > ZnS the concentrations of Ag+ and Pb2+ be at
(c) Na2S > ZnS > CuS (d) CuS > ZnS > Na2S equilibrium?
78. The values of Ksp of CaCO3 and CaC2O4 are (Ksp for AgCl = 1.8 × 10–10,
4.7 × 10–9 and 1.3 × 10–9 respectively at 25°C. If Ksp for PbCl2 = 1.7 × 10–5) [2011M]
(a) [Ag+] = 1.8 × 10–7 M ; [Pb2+] = 1.7 × 10–6 M
the mixture of these two is washed with water,
(b) [Ag+] = 1.8 × 10–11 M ; [Pb2+] = 8.5 × 10–5 M
what is the concentration of Ca 2+ ions in water? (c) [Ag+] = 1.8 × 10–9 M ; [Pb2+] = 1.7 × 10–3 M
[NEET Kar. 2013] (d) [Ag+] = 1.8 × 10–11 M ; [Pb2+] = 8.5 × 10–4 M
(a) 7.746 × 10–5 M (b) 5.831 × 10–5 M 85. If pH of a saturated solution of Ba(OH)2 is 12,
(c) 6.856 × 10–5 M (d) 3.606 × 10–5 M the value of its K(sp) is : [2010]
79. The dissociation constant of a weak acid is
1 × 10– 4. In order to prepare a buffer solution (a) 4.00 × 10–6 M3 (b) 4.00 × 10–7 M3
with a pH = 5 the [Salt]/[Acid] ratio should be (c) 5.00 × 10–6 M3 (d) 5.00 × 10–7 M3
[NEET Kar. 2013] 86. What is [H+] in mol/L of a solution that is 0.20 M
(a) 1 : 10 (b) 4 : 5 in CH3COONa and 0.10 M in CH3COOH? Ka for
(c) 10 : 1 (d) 5 : 4
CH3COOH = 1.8 × 10-5 . [2010]
80. pH of a saturated solution of Ba(OH)2 is 12.
The value of solubility product (Ksp) of Ba (OH)2 (a) 3.5 × 10–4 (b) 1.1 × 10–5
is : [2012] (c) 1.8 × 10–5 (d) 9.0 × 10–6
(a) 3.3 × 10– 7 (b) 5.0 × 10–7 87. In a buffer solution containing equal
(c) 4.0 × 10–6 (d) 5.0 × 10–6 concentration of B– and HB, the Kb for B–
is 10–10. The pH of buffer solution is : [2010]
81. Equimolar solutions of the following substances
(a) 10 (b) 7
were prepared separately. Which one of these
(c) 6 (d) 4
will record the highest pH value ? [2012]
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78 CHEMISTRY
88. Equimolar solutions of the following were 96. Solubility of a M2S salt is 3.5 × 10–6, then its
prepared in water separately. Which one of the solubility product will be [2001]
solutions will record the highest pH ? [2008] (a) 1.7 × 10–16 (b) 1.7 × 10–6
(a) SrCl2 (b) BaCl2 (c) 1.7 × 10–18 (d) 1.7 × 10–12
(c) MgCl2 (d) CaCl2
97. The solubility product of a sparingly soluble
89. Which of the following pairs constitutes a
salt BA2 is 4 × 10–12. The solubility of BA2 is
buffer? [2006]
[1999]
(a) NaOH and NaCl
(b) HNO3 and NH4NO3 (a) 4 × 10–4 (b) 4 × 10–12
(c) HCl and KCl (c) 4 × 10–3 (d) 1 × 10–4
(d) HNO2 and NaNO2 98. The solubility products of CuS, Ag2S and HgS
90. H2S gas when passed through a solution of are 10 –31 , 10 –44 , 10 –54 respectively. The
cations containing HCl, precipitates the solubilities of these sulphides are in the order
cati ons of secon d group of qualitative [1997]
analysis but not those belonging to the fourth (a) Ag2S > HgS > CuS (b) Ag2S > CuS > HgS
group. It is because [2005]
(c) HgS > Ag2S > CuS (d) CuS > Ag2S > HgS
(a) presence of HCl decreases the sulphide
ion concentration. 99. A physician wishes to prepare a buffer solution
(b) solubility product of group II sulphides of pH = 3.58 that efficiently resists a change in
is more than that of group IV sulphides. pH yet contains only small concentrations of
(c) presence of HCl increases the sulphide the buffering agents. Which one of the following
ion concentration. weak acids together with its sodium salt would
(d) sulphides of group IV cations are be the best to use ? [1997]
unstable in HCl. (a) m-chlorobenzoic acid (pKa = 3.98)
91. The solubility product of a sparingly soluble (b) p-chlorocinnamic acid (pKa = 4.41)
salt AX2 is 3.2 × 10–11. Its solubility ( in moles/ (c) 2, 5-dihydroxy benzoic acid (pKa = 2.97)
litre) is [2004] (d) Acetoacetic acid (pKa = 3.58)
(a) 5.6 × 10 –6 (b) 3.1 × 10 –4
100. The pH value of blood does not appreciably
(c) 2 × 10–4 (d) 4 × 10–4 change by a small addition of an acid or a base,
92. The solubility product of AgI at 25ºC is because the blood [1995]
1.0 × 10–16 mol 2 L–2 . The solubiliy of AgI in (a) is a body fluid
10–4 N solution of KI at 25ºC is approximately
(b) can be easily coagulated
(in mol L–1 ) [2003]
(c) contains iron as a part of the molecule
(a) 1.0 × 10–8 (b) 1.0 × 10–16
(d) contains serum protein which acts as buffer
(c) 1.0 × 10–12 (d) 1.0 × 10–10
101. Which of the following is most soluble ? [1994]
93. Solubility of MX2-type eletrolytes is 0.5 × 10–4
mol/lit, then find out Ksp of electrolytes [2002] (a) Bi2 S3 (Ksp = 1 × 10–17)
(a) 5 × 10–12 (b) 25 × 10–10 (b) MnS (Ksp = 7 × 10–16)
(c) 1 × 10 –13 (d) 5 × 10–13 (c) CuS (Ksp = 8 × 10–37)
94. Which has the highest value of pH? [2002] (d) Ag2 S (Ksp = 6 × 10–51).
(a) CH3COOK (b) Na2CO3 102. 0.1 M solution of which one of these substances
(c) NH4Cl (d) NaNO3 will be basic ? [1992]
(a) Sodium borate
95. Solution of 0.1 N NH4OH and 0.1 N NH4Cl has
pH 9.25. Then find out pKb of NH4OH [2002] (b) Ammonium chloride
(a) 9.25 (b) 4.75 (c) Calcium nitrate
(c) 3.75 (d) 8.25 (d) Sodium sulphate.
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Equilibrium 79
103. In which of the following solvents, AgBr will 104. The compound whose aqueous solution has the
have the highest solubility ? [1992] highest pH is [1988]
(a) 10-3 M NaBr (b) 10 -3 M NH 4 OH (a) NaCl (b) NaHCO3
(c) Na2CO3 (d) NH4Cl.
(c) Pure water (d) 10 -3 M HBr
ANSWER KEY
1 (a) 12 (c) 23 (a) 34 (d) 45 (d) 56 (b) 67 (d) 78 (a) 89 (d) 100 (d)
2 (d) 13 (b) 24 (d) 35 (d) 46 (d) 57 (b) 68 (d) 79 (c) 90 (a) 101 (a)
3 (b) 14 (d) 25 (b) 36 (d) 47 (b) 58 (c) 69 (a) 80 (b) 91 (c) 102 (a)
4 (a) 15 (a) 26 (d) 37 (d) 48 (a) 59 (d) 70 (a) 81 (a) 92 (c) 103 (b)
5 (c) 16 (a) 27 (a) 38 (c) 49 (a) 60 (a) 71 (c) 82 (a) 93 (d) 104 (c)
6 (c) 17 (a) 28 (c) 39 (c) 50 (a) 61 (a) 72 (b) 83 (b) 94 (b)
7 (c) 18 (b) 29 (d) 40 (a) 51 (d) 62 (c) 73 (b) 84 (c) 95 (b)
8 (d) 19 (d) 30 (c) 41 (c) 52 (c) 63 (a) 74 (c) 85 (d) 96 (a)
9 (c) 20 (b) 31 (d) 42 (b) 53 (b) 64 (d) 75 (a) 86 (d) 97 (d)
10 (c) 21 (d) 32 (b) 43 (c) 54 (c) 65 (c) 76 (b) 87 (d) 98 (b)
11 (b) 22 (d) 33 (a) 44 (c) 55 (a) 66 (b) 77 (c) 88 (b) 99 (d)
2NH3 ; K1 =
(i) N 2 + 3H 2
[ NH3 ] 2
volume of container at this stage.
[ N 2 ][ H 2 ]3 V=
nRT
…(i)
P
2NO; K 2 =
[ NO ]2 Since container is sealed and reaction was not
(ii) N 2 + O 2
[ N 2 ][ O 2 ] earlier at equilibrium.
\ n = constant.
1
(iii) H 2 + O 2 ¾¾
® H 2 O; K 3 =
[ H 2O ] PV 0.4 ´ 20
n= = …(ii)
2 [ H 2 ][ O 2 ]1/2 RT RT
Applying (II + 3 × III – I) we will get Put equation (ii) in equation (i)
5 K é 0.4 ´ 20 ù RT
2NH3 + O 2 2NO + 3H 2O; V= ê = 5L
2 ë RT úû 1.6
3. (b) Equilibrium constant for reaction:
K=
[ NO]2
´
[ H 2O ]
3
[ NH3 ]
2
[Product]
[ N2 ][ O 2 ] [ H 2 ]3 ´ [ O2 ]3 / 2 [ N 2 ][ H 2 ]3 K = 1.6 × 1012 =
[Reactant]
\ K = K2 × K33 / K1 The value of K is very high so the system will
2. (d) Max. pressure of CO2 = Pressure of CO2 at contain mostly products at equilibrium.
equilibrium 4. (a) N2(g) + O2(g) 2NO(g)
For reaction, [NO]2
K=
SrO(s) + CO2(g)
SrCO3(s) [N 2 ][O2 ]
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Equilibrium 81
x 1 P é KP 9ù
PB = ´P or 9 = . 1 ê\ 1 = ú
2 2 4 P2 êë KP2 1 úû
x P1 36
( PAB )2 ( PB2 ) ( x )2 ´ P 2 .P ´
2
or
P2 1
= or P1 : P2 = 36 : 1
Now, K P = =
( PAB2 ) 2 (1 - x) 2 ´ P 2 13. (b) Given : Equilibrium constant (K1) for the
reaction:
x 3 .P3
= [Q 1 – x ; 1 ] K1
1 1
2 ´ 1 ´ P2 HI(g) H 2 (g) + I 2 (g); K1 = 8; … (i)
1 2 2
x3 .P 2.K p æ 2K p ö 3 To find equilibrium constant for the following
= or x3 = or x = ç reaction:
2 P è P ÷ø
H 2 (g) + I2 (g) 2HI(g); K2 = ? .....(ii)
12. (c) Given reaction are
multiply (i) by 2, we get
Y + Z
X ..... (i)
2HI(g) H 2 (g) + I2 (g);
2B
and A ......(ii)
Let the total pressure for reaction (i) and (ii) be K1 = 82 = 64. … (iii)
P1 and P2 respectively, then Now reverse equation (iii), we get
1
KP
1 =9
H 2 (g) + I2 (g) 2HI(g); K = .....(iv)
(given) 64
KP 1 Equation (iv) is the same as the required equation
2
After dissociation, 1
(ii), thus K2 for equation (ii) is i.e. option (b)
Y + Z
X 64
(1– a) a a is correct.
At equilibrium
[Let 1 mole of X dissociate with a as degree of When the equation for an equilibrium is multiplied
by a factor, the equilibrium constant must be raised
dissociation] to the power equal to the factor.
Total number of moles = 1– a + a + a = (1+ a) For a reversible reaction, the equilibrium constant
æ1- a ö of the backward reaction is inverse of the
æ a ö
Thus PX = ç .P ;P = P; equilibrium constant for the forward reaction.
è 1 + a ÷ø 1 Y çè 1 + a ÷ø 1
14. (d) Given,
æ a ö 2NH 3 ; K1
PZ = ç .P N 2 + 3H 2 ....(i)
è 1 + a ÷ø 1
æ a ö a 2NO; K2
N2 + O2 ....(ii)
\ KP = ç .P ´ .
1 è 1 + a ÷ø 1 (1 + a ) 1
H2 + H 2 O; K 3
O 2 ....(iii)
æ1- aö 2
P1 / ç .P ....... (i)
è 1 + a ÷ø 1 We have to calculate
Similarly for A 2B 4NH 3 + 5O2 ¾¾ ® 4NO + 6H 2O; K = ?
At equilibrium (1– a) 2a 5
or 2NH 3 + O2 ¾¾ ® 2NO + 3H 2 O
We have, 2
2 [NO]2 [H 2O]3
æ 2aP2 ö æ 1 - a ö For this equation, K =
KP = ç / P ........(ii)
2 è 1 + a ÷ø çè 1 + a ÷ø 2 [NH3 ]2 [O 2 ]5 / 2
Dividing (i) by (ii), we get [NH3 ]2 [NO]2
but K1 = , K2 =
K P1 a 2 .P1 KP 1 P [N 2 ] [H 2 ]3 [N 2 ] [O2 ]
= 2 or 1 = . 1
KP 4a . P2 KP 4 P2
2 2
& K3 =
[H 2O] 3 [H 2 O]3
or K =
[H 2 ] [O2 ]½ 3
[H 2 ]3 [O 2 ]3/2
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82 CHEMISTRY
K 2 . K33 K=
[XeO3F2 ][H 2O]
Now operate,
K1 [XeO4 ][HF]2
[NO]2 [H 2O]3 [N ] [H 2 ]3 \ From eq. no. (a) and (b)
= ´ . 2 K = K 2 / K1
[N 2 ] [O2 ] [H 2 ]3 [O2 ]3/ 2 [NH3 ]2 20. (b) According to equation
[NO]2 [H 2 O]3 2HI H2 + I2
= =K
[NH3 ]2 [O2 ]5 / 2 At t = 0 (2 moles) 0 0
At equilibrium (2 – 2a) moles a mole a mole
K 2 . K 33 Total moles at equilibrium = 2 – 2a + a + a = 2
\ K=
K1 mole
15. (a) First option is incorrect as the value of KP 21. (d) Rate constant of forward reaction (Kf )
given is wrong. It should have been = 1.1 × 10–2 and rate constant of backward
PCO2 reaction (Kb) = 1.5 × 10–3 per minute. Equilibrium
KP = constant (Kc)
PCH4 ´ [PO2 ]2
16. (a) For reaction (i) K f 1.1 ´ 10 -2
= = = 7.33
K b 1.5 ´ 10 -3
[NO]2
K1 = 22. (d) DG = DG° + RT ln Q
[N 2 ][O2 ]
At equilibrium DG = 0, Q = Keq
and for reaction (ii)
So, DrG° = –RT ln Keq
[N 2 ]½ [O2 ]½ 1
DrG° = –8.314 J mol–1 K–1 × 300 K × ln(2 × 1013)
K2 = therefore K1 = 2
[NO] K2 23. (a) A2(g) + B2(g) X (g); DH = –X kJ
2
17. (a) MgCO3 (s) ® MgO(s) + CO 2 (g) On increasing pressure equilibrium shifts in a
MgO & MgCO3 are solid and they do not exert direction where number of moles decreases i.e.
any pressure and hence only pressure exerted forward direction.
is by CO2. Therefore KP = PCO2 On decreasing temperature, equilibrium shifts
18. (b) In polyprotic acids the loss of second in exothermic direction i.e., forward direction.
proton occurs much less readily than the first. So, high pressure and low temperature favours
Usually the Ka values for successive loss of maximum formation of product.
protons from these acids differ by at least a factor 24. (d) Clausius – Clapeyron's equation
of 10–3 i.e., K a1 > K a2 d ln P DH v
=
dT RT 2
H 2X ( 1)
H+ + HX- K a 25. (b) DG° = –2.30RT log K
because at equilibrium DG = 0
H+ + X 2- ( K a )
HX- 2 26. (d) Given reaction is exothermic reaction.
19. (d) For the reaction Hence according to Le-Chatelier's principle low
XeOF (g) + 2HF(g)
XeF6(g) + H2O(g) temperature favours the forward reaction and
4
on increasing pressure equilibrium will shift,
[XeOF4 ][HF]2
K1 = ....(a) towards lesser number of moles i.e. forward
[XeF6 ][H 2 O] direction.
and for the reaction 27. (a) In exothermic reactions on increasing
XeOF (g) + HeO F (g)
XeO4(g) + XeF6(g) temperature value of Kp decreases
4 3 2
[XeOF4 ][XeO 3 F2 ] \ Kp > Kp¢ (Assuming T1 < T2)
K2 = ....(b)
[XeO 4 ][XeF6 ]
28. (c) Given, CH3COOH CH3COO– + H+ ;
For reaction :
K a1 , = 1.5 × 10– 5 ....(i)
XeO 4 (g) + 2HF(g) ® XeO3F2(g) + H 2 O(g)
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Equilibrium 83
HCN H++ CN–; K = 4.5 × 10–10 34. Liquid
(d) Solid
a2
It is an endothermic process. So when
+ –
or H + CN HCN; temperature is raised, more liquid is formed.
1 1 Hence adding heat will shift the equilbrium in
K a' 2 = = ...(ii)
Ka2 4.5 ´ 10 –10 the forward direction.
35. (d) According to Le-chatelier's principle
\ From (i) and (ii), we find that the equilibrium
whenever a constraint is applied to a system in
constant (Ka) for the reaction,
equilibrium, the system tends to readjust so as
CN– + CH3COOH CH COO– + HCN, is
3 to nullify the effect of the constraint.
K a = K a1 ´ K a' 2 36. (d) [OH–] = 0.01 M = 10–2 M
pOH = –log[OH–] = –log(10–2) = 2
1.5×10 –5 1 pH = 14 – pOH = 12
= ´ 105 = 3.33 ´ 10 4
=
4.5×10 –103 37. (d) HCl cannot accept H+ therefore cannot act
29. (d) For reaction to proceed from right to left as Bronsted base.
Q > Kc 38. (c) When a proton is removed from an acid,
æbackward ö æforward ö i.e the reaction will be fast we obtain its conjugate base.
èrate ø èrate ø H + + OH -
H 2 O
in backward direction i.e rb > rf . H + + F -
HF
When Q = Kc, the system is at equilibrium 1
39. (c) Meq. of HCl = 75 ´ ´ 1 = 15
Q < Kc, reaction shift to the right from left. 5
Q > Kc, reaction proceed from right to left. 1
Meq. of NaOH = 25 ´ ´ 1= 5
30. (c) For the reaction 5
BaO(s) + O2 (g); DH = +ve.
BaO 2 (s) Meq. of HCl in resulting solution = 10
At equilibrium K p = PO2 10 1
Molarity of [H+] in resulting mixture = =
[For solid and liquids concentration term is taken 100 10
as unity] +
é1ù
pH = –log[H ] = –log ê ú = 1.0
Hence, the value of equilibrium constant ë 10 û
depends only upon partial pressure of O2 . 40. (a) g eq of NaOH = 0.1 × V = 0.1V
Further on increasing temperature, formation of g eq of HCl = 0.01 × V = 0.01V
O2 increases as this is an endothermic reaction. g eq of NaOH > g eq. HCl
31.
(d) A2(g)+ B2(g)
3C(g)
D(g) hence, resultant solution should be basic, hence
Step 1 Step 2 from the eqn
since the steps 1 and 2 are exothermic hence M1V1 – M2V2 = MV
low temprature will favour both the reactions. In 0.1V – 0.01V = MV
step - 1, moles are increasing hence low pressure
0.09
will favour it. In step 2, moles are decreasing, M= = 0.045 = 4.5 × 10–2
hence high pressure will favour it. 2
32. (b) A2 2A Equilibrium constant is given by Now, pOH = – log [OH–]
= – log 4.5 × 10–2 = 1.34
[ A]2 Q pH + pOH = 14
Kc = \ pH = 14 – 1.34 = 12.65
[ A2 ]
41. (c) Na2CO3 is a salt of strong base (NaOH)
Since the value given is very small, hence conc.
and weak acid (H2CO3). On hydrolysis this salt
of products is less. It means the reaction is slow.
33. (a) If more trans-2-pentene is added, then its will produce strongly basic solution. i.e. pH will
concentration in right hand side will increase. be highest (pH > 7) for this sotluion. Others are
But in order to maintain the K constant, combination of
concentration of cis-2-pentene will also increase.
Therefore more cis-2-pentene will be formed.
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84 CHEMISTRY
KCl = Strong acid + Strong base 50. (a) The solution formed from isomolar
® neutral solution (pH » 7) solutions of sodium oxide, sodium sulphide,
NaCl = Strong acid + Strong base sodium selenide and sodium telluride are H2O,
® neutral solution (pH » 7) H2S, H2Se & H2Te respectively. As the acidic
CuSO4 = Strong acid + wake base strengh increases from H2O to H2Te thus pH
® Acidic solution (pH < 7) decreases and hence the correct order of pHs is
42. (b) BF3 acts as Lewis acid. pH1 > pH2 > pH3 > pH4.
51. (d) For an acid-base indicator
43. (c) BF3 behaves as lewis acid as it is an
H+ + In–
HIn
electron deficient species.
44. (c) Boron in B2H6 is electron deficient.
[H + ][In - ] [HIn]
45. (d) (CH3)3 B – is an electron deficient, thus \ K In = or [H + ] = K In ´
behave as a lewis acid.
[HIn] [In - ]
46. (d) Ammonium chloride is a salt of weak base [HIn]
and strong acid. In this case, hydrolysis or log H + = log K In + log
[In - ]
constant, Kh can be calculated as
Taking negative on both sides
K 1 ´ 10 -14
Kh = w = = 5.65 ´ 10 -10 [HIn]
- log[H + ] = - log K In - log -
K b 1.77 ´ 10 -5
[In ]
47. (b) [H3O]+ for a solution having pH = 3 is given [In - ]
by or we can write pH = pK In + log
[HIn ]
[H3O]+ = 1×10–3 moles/litre [\ [H3O]+ = 10–pH]
Similarly for solution having pH = 4, [In - ]
or log = pH - pK In
[H3O]+ = 1 × 10–4 moles/ litre and for pH=5 [HIn]
[H3O+] = 1×10–5 moles/ litre 52. (c) The higher is the tendency to donate
Let the volume of each solution in mixture be 1L, proton, stronger is the acid. Thus, the correct
then total volume of mixture solution order is R – COOH > HOH > R – OH > CH º CH
= (1 + 1 + 1) L =3L depending upon the rate of donation of proton.
Total [H3O]+ ion present in mixture solution
= (10–3 + 10–4 + 10–5) moles The stability order of conjugate base is
Then [H3 O]+ ion concentration of mixture RCOOQ > H - OQ > R – OQ > HC º CQ
solution
53. (b) B(OH)3 does not provide H+ ions in water
-3 -4 -5
10 + 10 + 10 0.00111 instead it accepts OH– ion and hence it is Lewis
= M= M acid
3 3
= 0.00037 M = 3.7 ×10–4 M. [B(OH) ]- + H +
B(OH )3 + H 2O 4
48. (a) Given [H3O+] = 1 × 10–10 M 54. (c) Strong base has higher tendency to accept
at 25ºC [H3O+] [OH–] = 10–14 the proton. Increasing order of base and hence
10 -14 the order of accepting tendency of proton is
\ [OH - ] = = 10-4
10 -10 I- < HS- < NH3 < RNH 2
55. (a) Given : Hydroxyl ion concentration
Now, [OH - ] = 10 - pOH = 10–4
[OH–] = 0.05 mol L–1. We know that the
\ pOH = 4
49. (a) For a solution of 10–8 M HCl, [H+] = 10–8 [H + ][OH - ] = 1´10 -14
[H+] of water = 10–7 -14
or [H + ] = 1´ 10
–1
Total [H+] = 10–7 + 10–8 = 10 × 10–8 + 10–8 = 2 ´ 10 -13 mol L
0.05
10–8 (10 + 1) = 11 × 10–8
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Equilibrium 85
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86 CHEMISTRY
66. (b) An aqueous solution of acetic acid \ Ksp = (2.2 × 10–4)2 (1.1 × 10–4)
dissociates as = 5.324 × 10–12
CH3COOH + H2O CH COO– + H O+ 72. | M+ + Y–
(b) MY
3 3
67. (d) Ni(OH)2 Ni2+ + 2OH–
2
Ksp = s = 6.2 × 10–13
s s 2s
NaOH ¾¾ ® Na + OH–
s = 6.2´ 10- 13
Total [OH–] = 2s + 0.1 » 0.1 s = 7.87 × 10–7 mol L–1
| N3+ + 3Y –
NY3
Ionic product = [Ni]2+[OH]2
2 × 10–15 = s(0.1)2 Ksp = s × (3s)3 = 27s4 = 6.2 × 10–13
1/4
s = 2 × 10–13 æ 6.2 ´ 10-13 ö
Solubility of Ni(OH)2 = 2 × 10–13 M s= ç ÷
è 27 ø
68. (d) CaF2 Ca2+ + 2F– s = 3.89 × 10 mol L–1
–4
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Equilibrium 87
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88 CHEMISTRY
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Equilibrium 89
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90 CHEMISTRY
8 Redox Reactions
Trend Analysis with Important Topics & Sub-Topics
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Redox Reactions 91
9. The oxidation number of phosphorus in pyro- 15. A mixture of potassium chlorate, oxalic acid and
phosphoric acid is [1999] sulphuric acid is heated. During the reaction
(a) +3 (b) +1 which element undergoes maximum change in
(c) +4 (d) +5 the oxidation number ? [2012]
10. The oxidation number of chromium in potassium (a) S (b) H
dichromate is [1988, 1995] (c) Cl (d) C
(a) + 6 (b) – 5
16. When Cl2 gas reacts with hot and concentrated
(c) – 2 (d) + 2
sodium hydroxide solution, the oxidation number
11. Phosphorus has the oxidation state of + 3 in
of chlorine changes from : [2012]
(a) Phosphorous acid [1994]
(b) Orthophosphoric acid (a) zero to +1 and zero to –5
(c) Hypophosphorous acid (b) zero to –1 and zero to +5
(d) Metaphosphoric acid. (c) zero to –1 and zero to +3
Topic 3: Disproportionation and Balancing (d) zero to +1 and zero to –3
of Redox Reactions 17. The following redox reaction is balanced by
which set of coefficients ? [1999]
12. Which of the following reactions are dispro-
portionation reaction? [2019] aZn + bNO 3- + cH + ® dNH +4 + eH 2 O + fZn 2+
(a) 2Cu+ ® Cu2+ + Cu a b c d e f
(b) 3MnO42– + 4H+ ® 2MnO4– + MnO2 + 2H2O (a) 1 1 10 1 3 1
D
(b) 2 2 10 2 3 2
(c) 2KMnO4 ¾¾® K2MnO4 + MnO2 + O2 (c) 4 2 10 1 3 4
D (d) 4 1 10 1 3 4
(d) 2MnO4–+ 3Mn2+ + 2H2O ¾¾® 5MnO2 + 4H+
18. In which of the following reactions, there is no
Select the correct option from the following: change in valency ? [1994]
(a) (a) and (b) only (b) (a), (b) and (c)
(a) 4 KClO3 ¾ ¾® 3KClO4 + KCl
(c) (a), (c) and (d) (d) (a) and (d) only
13. For the redox reaction [2018] (b) SO2 + 2H2S ¾ ¾® 2H2O + 3S
+ 2+ (c) BaO2 + H2SO4 ¾ ¾® BaSO4 + H2O2
MnO4– + C2 O2–4 + H ® Mn + CO2 + H 2 O
(d) 3 BaO + O2 ¾ ¾® 2 BaO2.
The correct coefficients of the reactants for the 19. Which substance serves as a reducing agent in
balanced equation are the following reaction ? [1994]
MnO4– C2 O2–4 H+
14H+ + Cr2 O 27 - + 3Ni ® 2Cr3+ + 7H2O + 3Ni2+
(a) 16 5 2
(a) H2O (b) Ni
(b) 2 5 16
(c) 5 16 2 (c) H+ (d) Cr2 O 27-
(d) 2 16 5
14. Consider the change in oxidation state of Topic 4: Electrode Potential and Oxidising,
bromine corresponding to different emf values Reducing Agents
as shown in the diagram below : [2018] 20. The standard electrode potential (E°) values of
BrO4– 1.82 V 1.5 V Al3+/Al, Ag+/Ag, K+/K and Cr3+/Cr are –1.66 V,
¾¾¾® BrO3– ¾¾¾® HBrO
0.80 V, –2.93V and –0.74 V, respectively. The
¾¾
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92 CHEMISTRY
21. Standard reduction potentials of the half The strongest oxidising and reducing agents
reactions are given below : respectively are : [2012 M]
F2(g) + 2e– ® 2F– (aq); E° = + 2.85 V (a) F2 and I– (b) Br2 and Cl–
Cl2(g) + 2e– ® 2Cl–(aq); E° = + 1.36 V (c) Cl2 and Br– (d) Cl2 and I2
22. The oxide, which cannot act as a reducing agent,
Br2(l) + 2e– ® 2Br–(aq); E° = + 1.06 V is [1995]
I2(s) + 2e– ® 2I–(aq); E° = + 0.53 V (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 ) 22 (c )
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 )
Light
S2 O 62- ® S is in + 5 oxidation state
(c) N 2 + O2 ¾¾¾® 2 N O (redox reaction)
0 0 +2 -2 The structure of S 2O 42– and S 2 O 6 2– are
here oxidation of N2 & reduction of O2 is taking symmetrical. Thus, both sulphur atoms are in same
place oxidation state. This is not the case with S2O32–
D or S4O62– ions.
(d) H 2O(l) ¾¾® H 2 O(g) (not redox reaction)
3. (a) Losing of electron is called oxidation. 8. (b) Oxidation number of a compound must be
-4 +4 0. Using the values for A, B and C in the four
4. (b) CH 4 (g) + 4Cl2 (g) ® CCl 4 (l ) + 4HCl (g)
options we find that A3(BC4)2 is the answer.
Change in oxidation state of carbon is –4 to +4. Check : (+2)3 + [(+5) + 4(–2)]2 = 6 + (5 – 8)2 = 0
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Redox Reactions 93
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94 CHEMISTRY
9 Hydrogen
Topic 1: Preparation and Properties of 4. The hydride ion, H–, is a stronger base than the
Hydrogen hydroxide ion, OH–. Which one of the following
reactions will occur if sodium hydride (NaH) is
1. Which of the following statements about
dissolved in water? [1997]
hydrogen is incorrect ? [2016]
(a) hydrogen has three isotopes of which (a) H - (aq) + H 2 O(l) ® H 3O - (aq)
tritium is the most common. (b) H - (aq) + H 2 O(l) ® OH - (aq) + H 2 (g)
(b) Hydrogen never acts as cation in ionic salts
(c) Hydronium ion, H3 O+ exists freely in (c) H – (aq) + H 2 O(l) ®
solution OH– (aq) + 2H+ (aq) + 2e–
(d) Dihydrogen does not act as a reducing agent (d) H (aq) + H2O(l) ® reaction
–
2. When a substance A reacts with water, it 5. The ionization of hydrogen atom would give rise
produces a combustible gas B and a solution of to [1990]
substance C in water. When another substance
D reacts with this solution of C, it also produces (a) Hydride ion (b) hydronium ion
the same gas B on warming but D can produce (c) Proton (d) hydroxyl ion.
gas B on reaction with dilute sulphuric acid at Topic 2: Preparation and Properties of Water
room temperature. A imparts a deep golden yellow
colour to a smokeless flame of Bunsen burner. 6. The number of hydrogen bonded water
A, B, C and D respectively are [1998] molecules(s) associated with CuSO4·5H2O is
(a) Na , H2, NaOH, Zn [NEET 2019 Odisha]
(b) K, H2, KOH, Al (a) 5 (b) 3
(c) Ca, H2, Ca(OH)2, Sn (c) 1 (d) 2
(d) CaC2, C2H2, Ca(OH)2, Fe 7. The method used to remove temporary hardness
3. Which one of the following pairs of substances of water is: [2019]
on reaction will not evolve H2 gas? [1998] (a) Calgon’s method
(a) Iron and H2SO4 (aqueous) (b) Clark’s method
(b) Iron and steam (c) Ion-exchange method
(c) Copper and HCl (aqueous) (d) Synthetic resins method
(d) Sodium and ethyl alcohol
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Hydrogen 95
8. Some statements about heavy water are given (c) Na 4 [Na 4 (PO4 )5 ]
below:
(a) Heavy water is used as a moderator in (d) Na 4 [Na 2 (PO 4 )6 ]
nuclear reactors.
Topic 3: Preparation and Properties of
(b) Heavy water is more associated than Hydrogen Peroxide
ordinary water.
(c) Heavy water is more effective solvent than 13. (i) H2O2 + O3 ® H2O + 2O2
ordinary water. (ii) H2O2 + Ag2O ® 2Ag + H2O + O2
Which of the above statements are correct? Role of hydrogen peroxide in the above reactions
(a) (a) and (c) (b) (a) and (b)[2010] is respectively - [2014]
(c) (a), (b) and (c) (d) (b) and (c)
(a) Oxidizing in (i) and reducing in (ii)
9. Which of the following groups of ions makes
the water hard ? [1994] (b) Reducing in (i) and oxidizing in (ii)
(a) Sodium and bicarbonate (c) Reducing in (i) and (ii)
(b) Magnesium and chloride (d) Oxidizing in (i) and (ii)
(c) Potassium and sulphate 14. When H2O2 is oxidised, the product is [1999]
(d) Ammonium and chloride. (a) OH– (b) O2
10. The dielectric constant of H2 O is 80. The (c) O2– (d) HO2–
electrostatic force of attraction between Na+ and
15. The volume strength of 1.5 N H2O2 solution is
Cl– will be [1994]
(a) 4.8 L (b) 5.2 L [1996]
(a) reduced to 1/40 in water than in air
(c) 8.4 L (d) 8.8 L
(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. 16. The O – O – H bond angle in H2O2 is [1994]
11. At its melting point, ice is lighter than water (a) 106° (b) 109° 28'
because [1992] (c) 120° (d) 97°
(a) H2O molecules are more closely packed in 17. Which of the following is the true structure of
solid state H2O2 ? [1989]
(b) Ice crystals h ave hollow hexagonal H
arrangement of H2O molecules.
(a) H– O – O – H (b) O O
(c) On melting of ice the H2O molecule shrinks
in size H
(d) Ice froms mostly heavy water on first H H
(c) O=O (d) O O
melting. H H
12. Calgon used as a water softener is [1989] 18. The reaction of H2O2 with sulphur is an example
of ........reaction [1988]
(a) Na 2 [Na 4 (PO3 )6 ] (a) Addition (b) Oxidation
(b) Na 4 [Na 2 (PO3 )6 ] (c) Reduction (d) Redox
ANSWER KEY
1 (a, d) 3 (c) 5 (c) 7 (b) 9 (b) 11 (b) 13 (c) 15 (c) 17 (b)
2 (a) 4 (b) 6 (c) 8 (b) 10 (b) 12 (a) 14 (b) 16 (d) 18 (d)
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96 CHEMISTRY
( ) 1
speed of neutrons and used as moderators.
Protium 1 H is most common. It is an energetic Boiling point of heavy water is more than that
reducing agent. It reduces oxides, chlorides and of ordinary water, so it is more associated. The
sulphides of certain metals and produce free dielectric constant of heavy water is slightly les
metals at ordinary temperature. than that of ordinary water. Hence, ordinary water
CuO + 2H ® Cu + H2O is more effective solvent than heavy water.
9. (b) Temporary hardness is due to presence of
2. (a) 2Na + 2H 2 O ® 2NaOH + H 2 bicarbonates of calcium and magnesium and
'A' 'C' 'B'
permanent hardness is due to the sulphates or
Zn + 2 NaOH ® Na 2 ZnO 2 + H 2 chlorides of both of calcium and magnesium.
'D ' 'C' 'B' 10. (b) Electrostatic forces of attraction are
reduced to 1/80th in water.
Zn + dil. H 2SO 4 ® ZnSO 4 + H 2
'D ' 'B'
Dielectric constant or relative permittivity is the
Na produces golden yellow colour with
factor by which the electric field between the
smokeless flame of Bunsen burner.
charges is decreased relative to vacuum (or air as
3. (c) Fe + dil. H 2SO 4 ® FeSO 4 + H 2 mentioned in this question).
3Fe + 4H 2 O ® Fe 3O 4 + 4H 2 11. (b) In liquid state, the H2O molecules are held
Steam together by intermolecular H-bonds.
Cu + dil. HCl ® No reaction
When water freezes at atmospheric pressure it
Copper does not evolve H2 from acid as it is crystalises in normal hexagonal form. In it, each
below hydrogen in electrochemical series. oxygen atom is surrounded by 4 hydrogen atoms,
2 Na + C 2 H 5 OH ® 2C 2 H 5 ONa + H 2 two by strong covalent bonds and two by weak
hydrogen bonds. Since hydrogen bonds are longer
than covalent bonds. The molecules of water are
4. (b) H - (aq)+ H 2 O(l) ¾¾
® OH - (aq) + H 2 (g) not closely packed in crystal lattice. There exists a
base 1 acid 1 base 2 acid2
number of hollow spaces in the crystal lattice,
In this reaction H–
acts as bronsted base as it which decreases the density of ice.
accepts one proton (H+) from H2O to form H2. 12. (a) The complex salt of metaphosphoric acid
5. (c) ¾® H + (g ) + e - .
H (g ) ¾ sodium hexametaphosphate (NaPO3)6, is known
6. (c) The actual structure of CuSO4· 5H2O is as calgon. It is represented as Na2[Na4(PO3)6]
[Cu(H2O)4]SO4 · H2O, so only one water molecule Reduction
is associated with the molecule via hydrogen
bond. o –2 o
13. (c) (i) H2–1O2 + O3 H2O + 2O2
7. (b) In this method Ca(OH)2 is used.
Oxidation
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Hydrogen 97
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98 CHEMISTRY
10 The s -Block
Elements
Trend Analysis with Important Topics & Sub-Topics
Topic 1: Preparation and Properties of Alkali 4. Which one of the alkali metals, forms only, the
Metals and their Compounds normal oxide, M2O on heating in air ? [2012]
(a) Rb (b) K
1. The following metal ion activates many (c) Li (d) Na
enzymes, participates in the oxidation of glucose 5. The ease of adsorption of the hydrated alkali
to produce ATP and with Na, is responsible for metal ions on an ion-exchange resins follows the
the transmission of nerve signals. [2020] order : [2012]
(a) Copper (b) Calcium (a) Li+ < K+ < Na+ < Rb+
(c) Potassium (d) Iron (b) Rb+ < K+ <Na+ < Li+
2. Which of the alkali metal chloride (MCl) forms (c) K+ < Na+ < Rb+ < Li+
its dihydrate salt (MCl × 2H2O) easily? (d) Na+ < Li+ < K+ < Rb+
[NEET Odisha 2019 ] 6. In the replacement reaction
(a) KCl (b) LiCl CI + MF CF + MI
(c) CsCl (d) RbCl The reaction will be most favourable if M happens
to be : [2012 M]
3. The function of "Sodium pump" is a biological
(a) Na (b) K
process operating in each and every cell of all
(c) Rb (d) Li
animals. Which of the following biologically
7. The sequence of ionic mobility in aqueous
important ions is also a consituent of this
solution is : [2008]
pump : [2015]
(a) Mg2+ (b) K+ (a) K+ > Na+ > Rb+ > Cs+
(c) Fe2+ (d) Ca2+ (b) Cs+ > Rb+ > K+ > Na+
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The s-Block Elements 99
(c) Rb+ > K+ > Cs+ > Na+ Topic 2: Some Important Compounds of Sodium
(d) Na+ > K+ > Rb+ > Cs+
16. HCl was passed through a solution of CaCl2,
8. The alkali metals form salt-like hydrides by the MgCl 2 and NaCl. Which of the following
direct synthesis at elevated temperature. The compound(s) crystallise(s)? [2020]
thermal stability of these hydrides decreases in
(a) Only NaCl
which of the following orders ? [2008]
(b) Only MgCl2
(a) CsH > RbH > KH > NaH > LiH
(b) KH > NaH > LiH > CsH > RbH (c) NaCl, MgCl2 and CaCl2
(c) NaH > LiH > KH > RbH > CsH (d) Both MgCl2 and CaCl2
(d) LiH > NaH > KH > RbH > CsH 17. Crude sodium chloride obtain ed by
9. The correct order of the mobility of the alkali metal crystallisation of brine solution does not contain
ions in aqueous solutions is [2006]
(a) Na+ > K+ > Rb+ > Li+ [NEET Odisha 2019]
(b) K+ > Rb+ > Na+ > Li+ (a) CaSO4 (b) MgSO4
(c) Rb+ >K+ > Na+ > Li+
(d) Li+ > Na+ > K+ > Rb+ (c) Na2SO4 (d) MgCl2
10. In crystals of which one of the following ionic 18. In Castner-Kellner cell for production of sodium
compounds would you expect maximum distance hydroxide: [NEET Kar. 2013]
between centres of cations and anions? [1998] (a) Brine is electrolyzed with Pt electrodes
(a) LiF (b) CsF (b) Brine is electrolyzed using graphite
(c) CsI (d) LiI electrodes
11. Which of the following metal ions plays an
important role in muscle contraction ? [1994] (c) Molten sodium chloride is electrolysed
(a) K+ (b) Na+ (d) Sodium amalgam is formed at mercury
(c) Mg2+ (d) Ca2+ cathode
12. Which of the following statement is false ? [1994] 19. Which of the following statements is incorrect?
(a) Strontium decomposes water readily than [2011M]
beryllium
(b) Barium carbonate melts at a higher (a) Pure sodium metal dissolves in liquid
temperature than calcium carbonate ammonia to give blue solution.
(c) Barium hydroxide is more soluble in water (b) NaOH reacts with glass to give sodium silicate
than magnesium hydroxide (c) Aluminium reacts with excess NaOH to give
(d) Beryllium hydroxide is more basic than Al(OH)3
barium hydroxide.
(d) NaHCO3 on heating gives Na2CO3
13. Which of the following has largest size ?[1993]
(a) Na (b) Na+ 20. Which of the following oxides is not expected to
(c) Na – (d) Can’t be predicted react with sodium hydroxide? [2009]
14. Compared with the alkaline earth metals, the alkali (a) CaO (b) SiO2
metals exhibit [1990] (c) BeO (d) B2O3
(a) Smaller ionic radii
(b) Highest boiling points 21. In which of the following processes, fused
(c) Greater hardness sodium hydroxide is electrolysed at a 330ºC
(d) Lower ionization energies. temperature for extraction of sodium? [2000]
15. Which one of the following properties of alkali (a) Castner's process (b) Down's process
metals increases in magnitude as the atomic (c) Cyanide process (d) Both 'b' and 'c'
number rises ? [1989] 22. Aqueous solution of sodium carbonate absorbs
(a) Ionic radius NO and NO2 to give [1996]
(b) Melting point
(a) CO2 + NaNO3 (b) CO2 + NaNO2
(c) Electronegativity
(d) First ionization energy. (c) NaNO2 + CO (d) NaNO3 + CO
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100 CHEMISTRY
23. Which of the following is known as fusion 32. Which of the following alkaline earth metal
mixture? [1994] sulphates has hydration enthalpy higher than
(a) Mixture of Na2CO3 + NaHCO3 the lattice enthalpy? [2010]
(b) Na2CO3.10H2O (a) CaSO 4 (b) BeSO 4
(c) Mixture of K2CO3 + Na2CO3 (c) BaSO 4 (d) SrSO 4
(d) NaHCO3 33. Property of the alkaline earth metals that increases
24. Washing soda has formula [1990] with their atomic number [2010]
(a) Na2CO3.7H2O (b) Na2CO3.10H2O (a) Solubility of their hydroxides in water
(c) Na2CO3.3H2O (d) Na2CO3 (b) Solubility of their sulphates in water
Topic 3: Preparation and Properties of Alkaline (c) Ionization energy
Earth Metals and their Compounds (d) Electronegativity
25. Which of the following is an amphoteric 34. The correct order of increasing thermal stability
hydroxide ? [2019] of K2CO3, MgCO3, CaCO3 and BeCO3 is [2007]
(a) Sr(OH)2 (b) Ca(OH)2 (a) BeCO3< MgCO3 < CaCO3 < K2CO3
(c) Mg(OH)2 (d) Be(OH)2 (b) MgCO3 < BeCO3 < CaCO3 < K2CO3
26. Enzymes that utilize ATP in phosphate transfer (c) K2CO3 < MgCO3 < CaCO3 < BeCO3
require an alkaline earth metal (M) as the cofactor.
M is: [2019] (d) BeCO3 < MgCO3 < K2CO3 < CaCO3
(a) Be (b) Mg 35. In which of the following the hydration energy is
(c) Ca (d) Sr higher than the lattice energy? [2007]
27. Which of the following oxides is most acidic in (a) MgSO4 (b) RaSO4
nature? [2018] (c) SrSO4 (d) BaSO4
(a) MgO (b) BeO 36. Calcium is obtained by the [1997]
(c) CaO (d) BaO (a) electrolysis of solution of calcium chloride
28. Among CaH2, BeH2, BaH2, the order of ionic in water
character is [2018]
(b) electrolysis of molten anhydrous calcium
(a) BeH2 < CaH2 < BaH2
chloride or fused calcium chloride
(b) CaH2 < BeH2 < BaH2
(c) roasting of limestone
(d) BaH2 < BeH2 < CaH2 (d) reduction of calcium chloride with carbon
(d) BeH2 < BaH2 < CaH2 37. For two ionic solids CaO and KI, identify the
29. Which of the following statements is false ? wrong statement amongst the following : [1997]
[2016] (a) The lattice energy of CaO is much large
(a) Mg2+ ions form a complex with ATP than that of KI
(b) Ca2+ ions are important in blood clotting (b) KI is more soluble in water
(c) Ca2+ ions are not important in maintaining (c) KI has higher melting point
the regular beating of the heart.
(d) CaO has higher melting point
(d) Mg2+ ions are important in the green parts
of plants. 38. Which one is the correct statement with reference
30. Solubility of the alkaline earth's metal sulphates to solubility of MgSO4 in water? [1996]
in water decreases in the sequence :- [2015] (a) SO 4 2– ion mainly contributes towards
(a) Ca > Sr > Ba > Mg (b) Sr > Ca > Mg > Ba hydration energy
(c) Ba > Mg > Sr > Ca (d) Mg > Ca > Sr > Ba (b) Sizes of Mg2+ and SO42– are similar
31. Which of the following compounds has the (c) Hydration energy of MgSO4 is higher in
lowest melting point ? [2011] comparison to its lattice energy
(a) CaCl2 (b) CaBr2 (d) Ionic potential (charge/radius ratio) of Mg2+
(c) CaI2 (d) CaF2 is very low
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The s-Block Elements 101
ANSWER KEY
1 (c) 6 (c) 11 (d) 16 (a) 21 (a) 26 (b) 31 (c) 36 (b) 41 (c) 46 (b)
2 (b) 7 (b) 12 (d) 17 (b) 22 (b) 27 (b) 32 (b) 37 (c) 42 (b) 47 (b)
3 (b) 8 (d) 13 (c) 18 (d) 23 (c) 28 (a) 33 (a) 38 (c) 43 (c) 48 (b)
4 (c) 9 (c) 14 (d) 19 (c) 24 (b) 29 (c) 34 (a) 39 (d) 44 (c) 49 (d)
5 (b) 10 (c) 15 (a) 20 (a) 25 (d) 30 (d) 35 (a) 40 (c) 45 (b)
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102 CHEMISTRY
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The s-Block Elements 103
16. (a) When HCl is passed through the solution 28. (a) BeH2 < CaH2 < BaH2
Cl– ion concentration increases. Hence ionic Smaller the size of cation, more will be its
product becomes more than solubility product. polarising power. Hence, BeH2 will be least
Only NaCl is crystallised due to less solubility ionic.
than MgCl2 and CaCl2.
29. (c) Calcium regulates muscle contraction,
17. (b) Crude sodium chloride which is obtained including beating of heart muscle, so that it can
by crystallisation of brine solution contains contract and pump out blood to all our body.
Na2SO4, CaSO4, CaCl2 and MgCl2 are present
30. (d) Solubility of alkaline earth metal sulphates
as impurities in crude.
decreases down the group due to decrease in
18. (d) In castner kellner cell, sodium amalgam is hydration energy.
formed at mercury cathode.
MgSO 4 > CaSO4 > SrSO 4 > BaSO 4
Sodium is discharged from brine at a mercury
Hydration Solubility
cathode in preference to hydrogen because of the energy
high hydrogen overpotential at mercury surface.
This is the reason of using mercury cathode instead 31. (c) Melting point of metal halides decreases
of graphite or platinum in castner-kellner cell. as the size of the halogen increases. The correct
order is
19. (c) 2Al(s) + 2NaOH (aq) + 2H2O (l) ¾¾
® CaF2 > CaCl2 > CaBr2 > CaI2
2NaAlO2 + 3H2
sod. meta aluminate 32. (b) Be 2+ is very small, hence its hydration
20. (a) NaOH is a strong alkali. It combines with enthalpy is greater than its lattice enthalpy. Also,
acidic and amphoteric oxides to form salts. Since hydration energy decreases down the group.
CaO is a basic oxide hence does not react with 33. (a) The magnitude of hydration energy for the
NaOH. hydroxides of alkaline earth metals remains
21. (a) In Castner process, for production of (Na) almost same whereas lattice energy decreases
sodium metal, sodium hydroxide (NaOH) is appreciably down the group. Hence, solubility
electrolysed at 330ºC. increases down the group.
22. (b) Na 2 CO3 + NO + NO 2 ® 2 NaNO2 + CO2 34. (a) As the cation size increases down the
23. (c) Mixture of K2CO3 and Na2CO3 is called as group, the metal carbonates become more ionic
fusion mixture. in nature. Hence, the thermal stability increases
24. (b) Washing soda is Na2CO3. 10H2O. as: BeCO3 < MgCO3 < CaCO3.
25. (d) Amphoteric hydroxide means it can react The ionic character of group 1 carbonates is
with both acid and base. more than that of group 2, thus they possess
more thermal stability. Hence, the correct order
Be ( OH ) 2 + 2HCl ¾¾
® BeCl2 + 2H 2O
of thermal stability : BeCO3 < MgCO3 < CuCO3
Be ( OH ) 2 + 2NaOH ¾¾
® Na 2 éë Be ( OH ) 4 ùû < K2CO3.
26. (b) Enzyme that utilise ATP in phosphate 35. (a) The solubility and the hydration energy of
transfer require an alkaline earth metal (M) sulphates of alkaline earth metals decreases as
Mg as the cofactor. we move down the group from Be to Ba due to
27. (b) In metals, moving down the group, metallic the reason that ionic size increases down the
character increases, so basic nature group. The lattice energy remains constant
increases hence most acidic will be BeO. because sulphate ion is so large, so that small
BeO < MgO < CaO < BaO change in cationic sizes do not make any
¾¾¾¾¾¾¾¾¾¾
® difference. Thus the order will be:
increasingbasic character
BeSO4 > MgSO4 > CaSO4 > SrSO4 > BaSO4
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104 CHEMISTRY
36. (b) Calcium is obtained by electrolysis of a 44. (c) Carbonates becomes more thermally stable
fused mass consisting six parts calcium chloride down the group, therefore MgCO3 will leave
and one part calcium fluoride at about 700°C in CO2 easily.
an electrolytic cell. 45. (b) Active ingredient in bleaching powder for
37. (c) CaO has higher lattice energy because bleaching action is Ca (OCl)2.
of higher charge on Ca 2+ and O2– , which
Bleaching powder is a mixture of calcium
results in higher attraction. KI is more soluble
hypochlorite Ca (OCl) 2 , dibasic calcium
in water because of low lattice energy and
hypochlorite Ca (OCl)2. 2 Ca (OH)2 and dibasic
higher hydration energy. Clearly (c) is wrong
calcium chloride CaCl2 . 2Ca(OH)2.
because CaO has higher melting point as
compared to KI. 1
46. (b) (A) Plaster of paris = CaSO4. H2O
38. (c) MgSO4 is the only alkaline earth metal 2
sulphate which is soluble in water and for (B) Epsomite = MgSO4.7H2O
solubility hydration energy should be greater (C) Kieserite = MgSO4.H2O
than lattice energy. (D) Gypsum = CaSO4.2H2O
39. (d) Sodium is obtained by electrolytic
reduction of its chloride. Melting point of Plaster of paris can be more accurately written as
(CaSO4)2 . H2O
chloride of sodium is high (803°C) so in order
D
to lower its melting point(600°C), calcium 47. (b) CaCO3 (s) ¾¾® CO2 (g) + CaO(s)
chloride is added to it. (A) colourless residue
40. (c) Ca and CaH2 both react with water to form
H2 gas, ® Ca (OH )2
CaO(s) + H 2 O ¾ ¾
(B)
Ca + 2 H 2 O ¾
¾® Ca ( OH ) 2 + H 2
Ca (OH )2 + 2CO2 + H2O ¾ ¾ ® Ca (HCO3 )2
CaH 2 + 2H 2 O ¾¾® Ca (OH ) 2 + 2 H 2 (C)
whereas Ñ
Ca (HCO3 )2(s) ¾¾
® CaCO3(s) + CO2(g) + H2 O
K gives H2 while KO2 gives O2 and H2O2 (A)
(C)
2K + 2 H2O ¾
¾® 2KOH + H2
48. (b) heat
X CO2 + Residue
2KO2 + 2H 2 O ¾¾
® 2KOH + O2 + H 2 O2 Solid
H2 O
2KO 2 + 2H 2 O ¾¾ ® 2KOH + O 2 + H 2 O 2 boil
¬¾ ¾ ¾¾2¾ Y
excess CO
Similarly, Na gives H2 while Na2O2 gives H2O2 Z
Clear solution
2Na + 2H 2O ¾¾ ® 2NaOH + H 2
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 heat
CaCO3 ¾ ¾ ¾® CO 2 + CaO
Ba + 2 H2O ¾ ¾® Ba (OH)2 + H2
heat ' X '
Residue
BaO2 + 2H2O ¾ ¾® Ba (OH)2 + H2 O2
H 2O
41. (c) Atomic size of K+ > Ca2+ > Mg2+ and that
of Cl– > F–. Therefore, Mg2+/Cl– ratio has the excess
Ca(HCO3 )2 ¬¾ ¾¾ Ca(OH)2
minimum value. 'Z ' CO2
'Y '
42. (b) 20Ca = 1s22s22p63s23p64s2 = [Ar] 4s2
43. (c) Within a period, the atomic size decreases 49. (d) Gypsum is CaSO4. 2H2O and plaster of
from left to right.Further atomic size increases Paris is (CaSO 4 ) 2 .H 2O. Therefore, gypsum
down the group. Hence the correct order is contains a lower percentage of calcium than
i.e. Na > Mg > Li > Be. plaster of Paris.
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The p-Block Elements (Group 13 & 14) 105
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The p-Block Elements (Group 13 & 14) 107
19. Which of these is not a monomer for a high (c) Zeolites are aluminosilicates having three
molecular mass silicone polymer?[NEET 2013] dimensional network
(a) Me2SiCl2 (b) Me3SiCl (d) Some of the SiO44- units are replaced by
(c) PhSiCl3 (d) MeSiCl3 AlO 54- and AlO 96- ions in zeolites
20. The basic structural unit of silicates is : 25. Glass reacts with HF to produce [2000]
[NEET 2013] (a) SiF4 (b) H2SiF6
(a) SiO4 4– (b) SiO32– (c) H2SiO3 (d) Na3AlF6
(c) SiO4 2– (d) SiO 26. In graphite, electrons are [1993, 1994]
21. Which statement is wrong? [NEET Kar. 2013] (a) Localised on every third C-atom
(a) Feldspars are not aluminosilicates (b) Present in anti-bonding orbital
(b) Beryl is an example of cyclic silicate (c) Localised on each C-atom
(c) Mg2SiO4 is orthosilicate (d) Spread out between the structure
(d) Basic structural unit in silicates is the SiO4
27. Which of the following types of forces bind
tetrahedron
together the carbon atoms in diamond ?[1992]
22. Name the type of the structure of silicate in which
one oxygen atom of [SiO4]4– is shared ?[2011] (a) Ionic (b) Covalent
(a) Linear chain silicate (c) Dipolar (d) van der Waals.
(b) Sheet silicate 28. Water gas is produced by [1992]
(c) Pyrosilicate (a) Passing steam through a red hot coke bed
(d) Three dimensional (b) Saturating hydrogen with moisture
23. The straight chain polymer is formed by [2009] (c) Mixing oxygen and hydrogen in the ratio
(a) hydrolysis of CH3 SiCl 3 followed by of 1 : 2
condensation polymerisation (d) Heating a mixture of CO2 and CH4 in
(b) hydrolysis of (CH 3 ) 4 Si by addition petroleum refineries.
polymerisation 29. Glass is a [1991]
(c) hydrolysis of (CH3)2SiCl 2 followed by (a) Liquid
condensation polymerisation (b) Solid
(d) hydrolysis of (CH3 )3 SiCl followed by (c) Supercooled liquid
condensation polymerisation (d) Transparent organic polymer
24. Which one of the following statements about 30. The substance used as a smoke screen in
the zeolites is false ? [2004] warfare is [1989]
(a) They are used as cation exchangers (a) SiCl4 (b) PH3
(b) They have open structure which enables (c) PCl5 (d) Acetylene
them to take up small molecules
ANSWER KEY
1 (d) 4 (b) 7 (b) 10 (a) 13 (b) 16 (a) 19 (b) 22 (c) 25 (b) 28 (a)
2 (b) 5 (d) 8 (b) 11 (a) 14 (c) 17 (d) 20 (a) 23 (c) 26 (d) 29 (c)
3 (d) 6 (c) 9 (b) 12 (d) 15 (d) 18 (d) 21 (a) 24 (c) 27 (b) 30 (a)
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108 CHEMISTRY
8. (b) Al2O3 can be converted to anhydrous AlCl3 15. (d) Silicones are used in cosmetic surgery.
by heating a mixture of Al2O3 and carbon in dry 16. (a) PbF 4 is ionic in nature due to high
Cl2 gas. electronegativity difference.
1000°C
Al2O3 + 3C + 3Cl2 2AlCl3 + 3CO 17. (d) [SiCl6]2– does not exist because six large
vapours
Solid anhydrous chloride ions cannot be accommodated around
cooled
aluminium
chloride Si4+, due to its small size.
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The p-Block Elements (Group 13 & 14) 109
18. (d) Due to inert pair effect, Pb(II) is more stable CH3 CH3
than Pb(IV)
Sn(IV) is more stable than Sn(II) HO — Si — OH + H O —Si — OH ¾ ®
\ Pb(IV) is easily reduced to Pb(II) and can
CH3 CH3
acts as an oxidising agent whereas Sn(II) is easily
oxidised to Sn(IV) and can acts as a reducing CH3 CH3
agent.
HO — Si — O — Si — OH
Inertness of ns2 electrons of the valence shell to
participate in bonding on moving down the group CH3 CH3
in heavier p-block elements is called inert pair effect.
It occurs due to poor shielding of the ns2 electrons 24. (c) Zeolites are crystaline solid structures
of the valence shell by the intervening d and f made up of silicon, aluminium and oxygen making
electrons. a framework with cavities.
19. (b) Since Me3SiCl contains only one Cl, Zeolites have SiO4– 5–
4 and AlO4 tetrahedrons linked
therefore it can’t form high molecular mass together in a 3-D open structure in which 4 or 6
silicon polymer. It can form only dimer. membered rings predominate. Due to open
structure they have cavities and can take up water
20. (a) SiO44– is basic structural unit of silicates.
and other small molecules.
21. (a) Feldspars are 3-dimensional alumino-
silicates. 25. (b) 6 HF + SiO 2 ® H 2SiF6 + 2 H 2 O
26. (d) In graphite, each carbon is sp2 -hybridized
22. (c) O– O– and the single occupied unh ybridized
p-orbital of C-atoms overlap side ways to give
p -electron cloud which is delocalized and thus
Si Si
– the electrons are spread out between the
O O O– structure.
O
– O– 27. (b) In diamond, each carbon atom is sp 3
hybridized and thus, forms covalent bonds with
Pyrosilicate (Si 2 O 7 ) 6–
four other carbon atoms lying at the corners of a
23. (c) Hydrolysis of substituted chlorosilanes regular tetrahedron.
yield corresponding silanols which undergo 28. (a) Water gas is made by blowing steam
polymerisation. through the layer of incandescent coal.
Cl H OH – 2HCl H 2O + C ¾¾
® H 2 + CO
CH3 Steam Red hot water gas
Si + ¾¾®
CH3 Cl H OH 29. (c) Glass is a super cooled liquid.
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110 CHEMISTRY
12 Organic Chemistry -
Some Basic Principles
and Techniques
Trend Analysis with Important Topics & Sub-Topics
H–C COOH
(c)
OH
(a) 5-formylhex-2-en-3-one OH
(b) 5-methyl-4-oxohex-2-en-5-al
(c) 3-keto-2-methylhex-5-enal COOH
(d)
(d) 3-keto-2-methylhex-4-enal
2. Structure of the compound whose IUPAC name
is 3-ethyl-2-hydroxy-4-methylhex-3-en-5-ynoic 3. The structure of isobutyl group in an organic
acid is : [NEET 2013] compound is : [NEET 2013]
OH
(a) CH3 - CH - CH 2 - CH3
½
COOH
(a) (b) CH3 - CH 2 - CH2 - CH 2 -
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114 CHEMISTRY
Topic 3: Concept of Reaction Mechanism in 43. Which of the following carbocations is expected
Organic Compounds to be most stable? [2018]
NO2 NO2
39. A tertiary butyl carbocation is more stable than
a secondary butyl carbocation because of which Å
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Organic Chemistry - Some Basic Principles and Techniques 115
49. Consider the following compounds [2015] 54. Homolytic fission of the following alkanes forms
CH3 Ph free radicals CH3 – CH3, CH3 – CH2 – CH3,
(CH3)2 CH – CH3, CH3 – CH2 – CH (CH3)2.
CH3—C—CH— Ph—C—PH
CH3 Increasing order of stability of the radicals is
CH3 [NEET Kar. 2013]
g g
(I) (II) (III) (a) (CH3)3 C < (CH3)2 C – CH2CH3 <
g g
Hyperconjugation occurs in : CH3 – C H – CH3 < CH3 – C H2
(a) II only (b) III only g g
(b) (CH3)2 C – CH2CH3 < CH3 – C H – CH3 <
(c) I and III (d) I only g g
CH3 – C H2 < (CH3)3 C
50. Which of the following is the most correct g g
electron displacement for a nucleophilic reaction (c) CH3 – C H2 < CH3 – C H – CH3 <
g g
to take place? [2015] (CH3)2 C – CH2 – CH3 < (CH3)3 C
g g
H H2 (d) CH3 – C H2 < CH3 – C H – CH3 < (CH3)3
(a) g g
H3C—C = C – C – Cl C < (CH3)2 C – CH2CH3
H
H H2 55. What is the hybridisation state of benzyl
(b) H 3C—C = C – C – Cl +
H carbonium ion CH2?
H H2
(c) H 3C—C = C – C – Cl [NEET Kar. 2013]
H (a) sp3 (b) sp2
H H2 (c) spd2 (d) sp2 d
(d) H 3C—C = C – C – Cl 56. Nitrogen detection in an organic compound is
H
carried out by Lassaigne’s test. The blue colour
51. In Duma's method for estimation of nitrogen,
formed corresponds to which of the following
0.25 g of an organic compound gave 40 mL of
formulae? [NEET Kar. 2013]
nitrogen collected at 300 K temperature and 725
(a) Fe3[Fe(CN)6]3 (b) Fe3[Fe(CN)6]2
mm pressure. If the aqueous tension at 300 K is
(c) Fe4[Fe(CN)6]3 (d) Fe4[Fe(CN)6]2
25 mm, the percentage of nitrogen in the
57. The correct order of increasing bond length of
compound is : [2015]
C – H, C – O, C – C and C = C is : [2011]
(a) 18.20 (b) 16.76 (a) C – H < C = C < C – O < C – C
(c) 15.76 (d) 17.36 (b) C – C < C = C < C – O < C – H
52. In the Kjeldahl’s method for estimation of (c) C – O < C – H < C – C < C = C
nitrogen present in a soil sample, ammonia (d) C – H < C – O < C – C < C = C
evolved from 0.75 g of sample neutralized 10 mL 58. In Duma's method of estimation of nitrogen
of 1 M H2SO4. The percentage of nitrogen in 0.35 g of an organic compound gave 55 mL of
the soil is : [2014] nitrogen collected at 300 K temperature and
(a) 37.33 (b) 45.33 715 mm pressure. The percentage composition
(c) 35.33 (d) 43.33 of nitrogen in the compound would be :
53. Arrange the following in increasing order of (Aqueous tension at 300 K = 15 mm) [2011]
stability [NEET Kar. 2013] (a) 15.45 (b) 16.45
Å Å (c) 17.45 (d) 14.45
(A) (CH3 )2 C - CH2CH3 (B) (CH3 )3 - C
59. The Lassaigne’s extract is boiled with conc.
Å Å
(C) (CH3 )2 - CH (D) CH3 - CH 2 HNO3 while testing for halogens. By doing so it
[2011]
Å
(E) CH 3 (a) decomposes Na2S and NaCN, if formed.
(b) helps in the precipitation of AgCl.
(a) E < D < C < B < A (b) E < D < C < A < B (c) increases the solubility product of AgCl.
(c) D < E < C < A < B (d) A < E < D < C < B (d) increases the concentration of NO3– ions.
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116 CHEMISTRY
60. The stability of carbanions in the following : 65. Which one of the following is a technique most
suitable for purification of cyclohexanone from
(a) RC º C (b) a mixture containing benzoic acid, isoamyl
alcohol, cyclohexane and cyclohexanone? [1997]
(a) Crystallization
(c) R 2C = CH (d) R 3C - CH 2 (b) Sublimation
is in the order of : [2008] (c) IR spectroscopy
(a) (a) > (b) > (c) > (d) (b) (b) > (c) > (d) > (a) (d) Gas chromatography
(c) (d) > (b) > (c) > (a) (d) (a) > (c) > (b) > (d) 66. Lassaigne’s test for the detection of nitrogen
61. Which amongst the following is the most stable fails in [1994]
carbocation? [2005] (a) NH2CONHNH2.HCl (b) NH2NH2.HCl
Å Å (c) NH2CONH2 (d) C6H5NHNH2.HCl
(a) CH3 (b) CH3 CH 2 67. A is a lighter phenol and B is an aromatic
carboxylic acid. Separation of a mixture of A and
CH3 B can be carried out easily by using a solution
|
Å
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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118 CHEMISTRY
18. (d) Huckel’s rule states that for aromaticity Hence it is homocyclic (as the ring system is made
there must be (4n + 2) p electrons present in a of one type of atoms, i.e. carbon) but not aromatic.
compound, where n is an integer. As it does not follow (4n +2)p electron rule of
aromaticity.
19. (c) Urea is the first organic compound
synthesized in the laboratory by wohler.
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120 CHEMISTRY
+ PV
1 1 = P2V2
¬¾® T1 T2
+
H H V2 (Volume of nitrogen at STP)
Y Y
(Less stable due to more e– withdrawing effect of –NO2) 273 ´ 700 ´ 40
NO2 = = 33.52 mL
NO2 NO2 300 ´ 760
+ + Percentage of nitrogen
¬¾® ¬¾®
H H H 28× volume of N 2 at STP ×100
Y + Y Y = 22400 × wt. of organic substance
–
(More stable due to less e withdrawing effect of –
NO 2) greater no. of resonating structures.
28 ´ 33.52 ´ 100
44. (a) –I effect increases on increasing = = 16.76%
electronegativity of atom. So, correct order of 22400 ´ 0.25
–I effect is –NH2 < – OR < – F. 52. (a) 10 mL, 1 M H2SO4 = 20 mL, 1 M NH3
45. (c) The o-isomer is steam volatile due to intra- Q wt of N in one mole NH3 = 14
molecular H-bonding. The p-isomer is not steam
volatile due to intermolecular H-bonding or \ 20 × 10– 3 mol NH3 ¾®
association of molecules. Thus, both can be 20 × 10– 3 × 14 nitrogen
separated by steam distillation. \ 0.75 g of sample contains
46. (c) Electrophite is a electron deficient species
and can accpet pair of electrons from nucleophite 14 ´ 20 ´ 10 - 3
= ´ 100 = 37.33%
47. (d) CH3—C º C 0.75
No.of s bp - 1 ù
53. (b) Greater the number of e– donating alkyl
lp –1 û 2 & hybridisation is sp groups (+I effect), greater will be the stability of
carbocations.
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122 CHEMISTRY
This test fails in case of diazo compounds, dinitro is basically used for estimating nitrogen in food
compounds and compounds containing nitrogen fertilizers and agricultural products.
in the ring.
Kjeldahl's method is not used in case of nitro, azo
67. (d) Carboxylic acids dissolve in NaHCO 3 and and azoxy compound.
evolve CO2 gas but phenols do not.
NaHCO 71. (d) Nitrogen, sulphur and halogens are tested
RCOOH ¾¾¾¾ 3
® RCOONa + H 2O + CO2
in an organic compound by Lassaigne's test.
68. (c) Higher the +I effect greater is stability of the
species. Thus,
+ + + The organic compound is fused with sodium metal
(CH3 )3 C + > (CH3 )2 C H > C6 H 5 C H 2 > CH 3 C H 2 as to convert these elements into ionisable inorganic
substances,
Also, primary benzyl carbocation have almost
the same stability as 2°-alkyl carbocations. Na + C + N ¾¾
® NaCN
69. (b) Sodium cyanide (Na + C + N ® NaCN). 2Na + S ¾¾
® Na 2S
(Lassaigne's test) 2Na + X 2 ¾¾
® 2NaX
70. (a) Kjeldahl's method is suitable for estimating The cyanide, sulphide or halide ions can be
nitrogen in those compounds in which nitrogen confirmed in aqueous solution by usual test.
is linked to carbon and hydrogen. This method
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13 Hydrocarbons
Topic 1: Alkanes (c) Both bond angles and bond length remains
same
1. The alkane that gives only one mono-chloro (d) Bond angle remains same but bond length
product on chlorination with Cl2 in presence of changes
diffused sunlight is [NEET Odisha 2019] 4. The correct statement regarding the comparison
(a) Isopentane of staggered and eclipsed conformation of
ethane, is [2016]
(b) 2, 2-dimethylbutane
(a) The staggered conformation of ethane is
(c) neopentane
less stable than eclipsed conformation,
(d) n-pentane because staggered conformation has
2. Hydrocarbon (A) reacts with bromine by torsional strain
substitution to form an alkyl bromide which by (b) The eclipsed conformation of ethane is
Wurtz r eaction is converted to gaseous more stable than staggered conformation,
hydrocarbon containing less than four carbon because eclipsed conformation has no
atoms. (A) is [2018] torsional strain
(a) CH º CH (b) CH2 = CH2 (c) The eclipsed conformation of ethane is
(c) CH4 (d) CH3 – CH3 more stable than staggered conformation
3. With respect to the conformers of ethane, which even though the eclipsed conformation has
of the following statements is true ? [2017] torsional strain
(a) Bond angle changes but bond length (d) The staggered conformation of ethane is
remains same more stable than eclipsed conformation,
because staggered conformation has no
(b) Both bond angle and bond length change
torsional strain.
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124 CHEMISTRY
5. Given OH
OH
Br Br H
H Br (c)
CH3 H
CH3 and H
H CH3 CH3
H H H
Br
OH
I II H
I and II are [NEET Kar. 2013] (d)
(a) A pair of optical isomers
(b) Identical HO
(c) A pair of conformers H H H
(d) A pair of geometrical isomers 8. Liquid hydrocarbons can be converted to a
6. In the following the most stable conformation mixture of gaseous hydrocarbons by : [2010]
of n-butane is: [2010] (a) Oxidation
CH3 CH3 (b) Cracking
H CH3 H H (c) Distillation under reduced pressure
(d) Hydrolysis
(a) (b) 9. A compound of molecular formula of C7H16
shows optical isomerism, compound will be
H H H H
(a) 2, 3-Dimethylpentane [2001]
H CH3 (b) 2,2-Dimethylbutane
CH3 CH3 (c) 2-Methylhexane
CH3 H (d) None of the above
10. The reaction of ethyl magnesium bromide with
(c) (d) water would give [1999]
H H (a) Ethane (b) Ethyl alcohol
H H H
HH H3C (c) Ethyl bromide (d) Ethyl ether
7. Which of the following conformers for ethylene 11. In commercial gasolines the type of
glycol is most stable? [2010] hydrocarbons which are more desirable, is
(a) branched hydrocarbons [1997]
OH (b) straight-chain hydrocarbons
H OH (c) aromatic hydrocarbons such as toluene
(d) linear unsaturated hydrocarbons
(a) 12. The most stable conformation of n-butane is
[1997]
(a) skew boat (b) gauche
H H (c) staggered-anti (d) eclipsed
H 13. Which one of the following reactions is
expected to readily give a hydrocarbon product
OH in good yields ? [1997]
H H Electrolyt ic
(a) RCOOK ¾¾ ¾ ¾¾®
oxidation
(b) (b) - + Br2
RCOO Ag ¾¾
¾®
Cl
H H (c) CH 3 CH 3 ¾¾
¾2®
hu
OH C 2 H 5OH
(d) (CH 3 ) 3 CCl ¾¾ ¾ ¾®
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Hydrocarbons 125
–
chain initiating step involves the formation of (a) Cl – CH2 – CH2 – CH
–
(a) Chlorine free radical [1994] CH3
(b) Hydrogen chloride
CH2Cl
(c) Methyl radical
–
(b) H3C – CH2 – CH – CH3
(d) Chloromethyl radical.
15. Reactivity of hydrogen atoms attached to
CH3
different carbon atoms in alkanes has the order
–
(c) H3C – CH2 – C – CH3
[1993]
–
(a) Tertiary > Primary > Secondary Cl
(b) Primary > Secondary > Tertiary CH3
–
(c) Both (a) and (b) (d) H3C – CH – CH
–
(d) Tertiary > Secondary > Primary Cl CH3
18. 2,3-Dimethyl-2-butene can be prepared by
Topic 2: Alkenes
heating which of the following compounds with
16. An alkene on ozonolysis gives methanal as one a strong acid ? [2015 RS]
of the product. Its structure is [2020] (a) (CH3)2 CH – CH – CH = CH2
|
CH3
(b) (CH3)3 C – CH = CH2
(c) (CH3)2C = CH – CH2 – CH3
(a) (d) (CH3)2CH – CH2 – CH = CH2
19. In the reaction with HCl, an alkene reacts in
accordance with the Markovnikov's rule, to give
a product 1-chloro-1-methylcyclohexane. The
possible alkene is : [2015 RS]
CH2 CH3
(b)
(a) (b)
CH3
(c)
(c) (a) and (b) (d)
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126 CHEMISTRY
(a) C CH CH3
CH3
OH CH3
(c)
CH3 CH3
CH3
H2C CH3 H
(a) 8 (b) 12 (d) C CH2 CH2
(c) 16 (d) 4
CH3 OH
22. The reaction of C6H5CH = CHCH3 with HBr
produces: [2015] 24. In the following reactions, [2011]
(a) C6 H5CH 2CHCH3
| CH3
Br (i) CH3–CH–CH–CH3
H + /Heat
A + B
Major Minor
(b) C6 H5CH 2CH 2CH 2Br OH products products
HBr,dark
CH=CHCH3 (ii) A ¾¾¾¾¾¾¾¾® C + D
in absenceof peroxide
A ¾¾¾¾¾¾¾¾® C + D
(c) in absenceof peroxide æMajor ö æMinor ö
èproduct ø èproduct ø
+ Br
H 2O/H CH3
C CH —
— CH2
(b) CH3– C = CH–CH3
CH3 CH3
A B 3 3 and CH 3– C – CH2– CH3
Minor Product + Major Product Br
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Hydrocarbons 127
CH3 R R R H
(c) CH3– C = CH – CH3 (a) (b)
H H R H
CH 3 R R R R
3 3 and CH3 – CH – CH – CH 3 (c) (d)
R H R R
Br 30. Reaction of HBr with propene in the presence of
CH3
peroxide gives [2004]
(d) CH2 = C – CH2 – CH3 3 2 3 (a) isopropyl bromide (b) 3-bromo propane
CH3 (c) allyl bromide (d) n-propyl bromide
2 2 3 and CH3– C – CH2 – CH3
CH 3
Br |
25. The IUPAC name of the compound having the 31. The compound CH 3— C = CH — CH 3
formula CH º C – CH = CH2 is : [2009] on reaction with NaIO4 in the presence of
(a) 1-butyn-3-ene (b) but-1-yne-3-ene KMnO4 gives [2003]
(c) 1-butene-3-yne (d) 3-butene-1-yne (a) CH3CHO + CO2
26. Which of the following compounds will exhibit (b) CH3COCH3
cis-trans (geometrical) isomerism? [2009] (c) CH3COCH3 + CH3COOH
(d) CH3COCH3 + CH3CHO
(a) Butanol (b) 2-Butyne
32. Geometrical isomers differ in [2002]
(c) 2-Butenol (d) 2-Butene (a) position of functional group
27. H 3C - CH - CH = CH 2 + HBr ® A (b) position of atoms
|
(c) spatial arrangement of atoms
CH 3
(d) length of carbon chain
A (predominantly) is : [2008] 33. Which alkene on ozonolysis gives CH3CH2CHO
(a) CH 3 - CH - CH 2 - CH 2 Br and CH3CCH3 [2001]
|
||
CH 3 O
Br
| CH3
(b) CH3 - C- CH 2 - CH3 (a) CH3CH2CH = C
| CH3
CH3 (b) CH3CH2CH = CHCH2CH3
(c) CH3 - CH- CH - CH 3 (c) CH3 CH2 CH = CH CH3
| | (d) CH 3 - C = CHCH 3
Br CH3 |
CH3
(d) CH 3 - CH - CH - CH 3
| | 34. In preparation of alkene from alcohol using Al2O3
CH 3 Br which is effective factor? [2001]
28. Which of the compounds with molecular formula (a) Temperature
C5H10 yields acetone on ozonolysis? (b) Concentration
[2007] (c) Surface area of Al2O3
(a) 3-methyl-1-butene (b) cyclopentane (d) Porosity of Al2O3
(c) 2-methyl-1-butene (d) 2-methyl-2-butene. 35. Which of the following reagents convert
propene to 1-propanol? [2000]
29. Which one of the following alkenes will react (a) H2O, H2SO4
faster with H2 under catalytic hydrogenation (b) aqueous KOH
conditions? [2005] (c) MgSO4, NaBH4/H2O
(R = Alkyl Substituent) (d) B2H6, H2O2, OH–
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128 CHEMISTRY
36. Correct order of stability is : [2000] 42. When 3, 3-dimethyl –2-butanol is heated with
(a) cis -2- butene > 1-butene > trans -2-butene H2SO4, the major product obtained is [1995]
(b) trans-2-butene > cis-2-butene > 1-butene (a) 2, 3-dimethyl –2-butene
(c) 1-butene > cis-2-butene > trans-2- butene
(b) 3, 3-dimethyl –1- butene
(d) cis-2-butene > trans-2-butene > 1-butene
37. The correct structure of trans-2 hexenal is [1999] (c) 2, 3-dimethyl –1- butene
CHO (d) cis & trans isomers of 2, 3-dimethyl – 2-butene
(a) 43. The alkene R – CH = CH2 reacts readily with
(b) CHO B2H6 and formed the product B which on
(c) CHO (d)
CHO oxidation with alkaline hydrogen peroxides
38. A hydrocarbon ‘A’ on chlorination gives ‘B’ produces [1995]
which on heating with alcoholic potassium (a) R – CH2 – CHO (b) R– CH2 – CH2 – OH
hydroxide changes into another hydrocarbon
‘C’. The latter decolourises Baeyer's reagent and (c) R-C =O (d) R - CH - C H 2
on ozonolysis forms formaldehyde only. ‘A’ is | | |
(a) Ethane (b) Butane [1998] CH3 OH OH
(c) Methane (d) Ethene
39. In reaction sequence [1997] 44. Which of the following compounds has the
lowest boiling point ? [1994]
CH2OH
Hypochlorous R
CH 2 = CH 2 ¾¾ ¾¾¾ ¾® M ¾¾® | (a) CH3CH2CH2CH2CH3
acid CH2OH (b) CH3CH = CHCH2CH3
molecule 'M' and reagent 'R' respectively are
(c) CH3CH = CH – CH = CH2
(a) CH3CH2Cl and NaOH
(d) CH3CH2CH2CH3
(b) CH3CH2OH and H2SO4
(c) CH2Cl – CH2OH and aqueous NaHCO3 45. When hydrochloric acid gas is treated with
propene in presence of benzoyl peroxide, it gives
(d) CH2 — CH2 and heat
[1993]
O (a) 2-Chloropropane (b) Allyl chloride
40. In the presence of platinum catalyst,
hydrocarbon A adds hydrogen to form n-hexane. (c) No reaction (d) n-Propyl chloride.
When hydrogen bromide is added to A instead 46. The restricted rotation about carbon carbon
of hydrogen, only a single bromo compound is double bond in 2-butene is due to [1993]
formed. Which of the following is A? [1996] 2
(a) Overlap of one s- and sp - hybridized
(a) CH 3 — CH 2 — CH = CH — CH 2 — CH 3 orbitals
(b) CH 3 — CH 2 — CH 2 — CH = CH — CH 3 (b) Overlap of two sp2 - hybridized orbitals
(c) CH 3 — CH = CH — CH 2 — CH 2 — CH 3 (c) Overlap of one p- and one sp2 - hybridized
(d) CH 2 = CH — CH 2 — CH 2 — CH 2 — CH 3 orbitals
41. Which of the following will not show cis-trans (d) Sideways overlap of two p- orbitals.
isomerism? [1996]
47. Which one of the following can exhibit cis-trans
(a) CH 3 — CH = CH — CH 3 isomerism ? [1989]
(b) CH 3 — CH 2 — CH = CH — CH 2 — CH 3 (a) CH 3 - CHCl - COOH
(c) CH — CH = CH — CH — CH
3 2 3
| (b) H - C º C - Cl
CH3
(c) ClCH = CHCl
(d) CH 3— C H — CH = CH — CH 2 — CH 3
| (d) ClCH 2 - CH 2Cl
CH 3
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Hydrocarbons 129
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130 CHEMISTRY
58. Predict the product C obtained in the following (a) Bromine in carbon tetrachloride
reaction of butyne-1. [2007] (b) Bromine in acetic acid
HI (c) Alk. KMnO4
CH 3CH 2 - C º CH + HCl ¾¾
® B ¾¾® C
I (d) Ammoniacal silver nitrate.
| 64. Acetylenic hydrogens are acidic because [1989]
(a) CH3 - CH 2 - CH 2 - C - H (a) Sigma electron density of C – H bond in
| acetylene is nearer to carbon, which has
Cl
50% s-character
I
| (b) Acetylene has only open hydrogen in each
(b) CH 3 - CH 2 - CH - CH 2 Cl carbon
I (c) Acetylene contains least number of
| hydrogens among the possible
(c) CH3CH 2 - C - CH3
| hydrocarbons having two carbons
Cl (d) Acetylene belongs to the class of alkynes
(d) CH3 - CH - CH 2CH 2 I with molecular formula, CnH2n – 2.
|
Cl Topic 4: Aromatic Hydrocarbons
59. Products of the following reaction: [2005]
65. Given:
(1) O CH3 H3C CH2 H2C
CH3CºC·CH2 CH3 ¾¾¾¾¾¾3
(2) Hydrolysis
® ...... are: H3C CH2
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Hydrocarbons 131
69. Some meta-directing substituents in aromatic 75. Which of the following species is not
substitution are given. Which one is most electrophilic in nature? [2010]
deactivating? [NEET 2013] Å Å
(a) –SO3H (b) –COOH (a) NO 2 (b) Cl
(c) –NO2 (d) –C º N Å
70. Which of the following compounds will not (c) BH3 (d) H3O
undergo Friedal-Craft’s reaction easily : 76. Which one of the following is most reactive
[NEET 2013] towards electrophilic attack ? [2008]
(a) Xylene (b) Nitrobenzene NO2
CH2OH
(c) Toluene (d) Cumene
71. Among the following compounds the one that (a) (b)
is most reactive towards electrophilic nitration is:
[2012] Cl
(a) Benzoic acid (b) Nitrobenzene
(c) Toluene (d) Benzene OH
72. Which one of the following is most reactive (c) (d)
towards electrophilic reagent ? [2011]
CH3 77. The order of decreasing reactivity towards an
(a) electrophilic reagent, for the following would be
OCH3 [2007]
CH3 (i) benzene (ii) toluene
(b) (iii) chlorobenzene (iv) phenol
(a) (ii) > (iv) > (i) > (iii) (b) (iv) > (iii) > (ii) > (i)
OH
(c) (iv) > (ii) > (i) > (iii) (d) (i) > (ii) > (iii) > (iv)
CH3
78. Using anhydrous AlCl3 as catalyst, which one
(c) of the following reactions produces
NHCOCH3 ethylbenzene (PhEt)? [2004]
CH3
(a) H 3 C - CH 2 OH + C 6 H 6
(d)
CH2OH (b) CH 3 - CH = CH 2 + C6 H 6
73. In the following reaction, C 6 H 5 CH 2 Br
1.Mg, Ether
(c) H 2 C = CH 2 + C 6 H 6
¾¾¾¾¾
+
® X, the product ‘X’ is [2010]
2.H 3O (d) H 3C - CH 3 + C 6 H 6
(a) C6H5CH2CH2C6H5
79. Which one of the following is a free-radical
(b) C6H5CH2OCH2C6H5 substitution reaction? [2003]
(c) C6H5CH2OH
(d) C6H5CH3 (a) CH 3CHO + HCN ¾
¾® CH 3CH ( OH ) CN
74. Which one is most reactive towards electrophilic CH3 CH2Cl
Boiling
reagent? [2010] (b) +Cl2 ¾¾®
CH3 CH3
CH 3
OH CH2OH (c) Anh. AlCl
+CH 3Cl ¾¾¾®
3
(a) (b)
CH2Cl
CH3 CH3 (d) + AgNO2
NHCOCH3 OCH3 CH2NO2
(c) (d) ¾¾®
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132 CHEMISTRY
ANSWER KEY
1 (c) 10 (a) 19 (c) 28 (d) 37 (a) 46 (d) 55 (a) 64 (a) 73 (d) 82 (a)
2 (c) 11 (a) 20 (d) 29 (a) 38 (a) 47 (c) 56 (c) 65 (a) 74 (d) 83 (d)
3 (c) 12 (c) 21 (a) 30 (d) 39 (c) 48 (b) 57 (d) 66 (b) 75 (d) 84 (c)
4 (d) 13 (a) 22 (d) 31 (c) 40 (a) 49 (b) 58 (c) 67 (d) 76 (c) 85 (d)
5 (c) 14 (a) 23 (a) 32 (c) 41 (c) 50 (a) 59 (b) 68 (a) 77 (c) 86 (c)
6 (b) 15 (d) 24 (b) 33 (a) 42 (a) 51 (c) 60 (a) 69 (c) 78 (c) 87 (c)
7 (a) 16 (b) 25 (c) 34 (a) 43 (b) 52 (a) 61 (c) 70 (b) 79 (b) 88 (c)
8 (b) 17 (c) 26 (d) 35 (d) 44 (d) 53 (b) 62 (c) 71 (c) 80 (a) 89 (d)
9 (a) 18 (b) 27 (b) 36 (b) 45 (a) 54 (d) 63 (d) 72 (b) 81 (c)
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Hydrocarbons 133
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134 CHEMISTRY
At Cathode 2K + + 2e – ® 2K CH3
CH3 CH3 CH3
2K + H 2O ® 2KOH + H 2 C=C ¾® CH3 – C – C
CH3 CH3 CH3
hv H
14. (a) Cl2 ¾¾¾¾¾¾¾¾ ® 2 Cl• 2, 3 - dimethyl – 2 - butene (Stable 3° carbocation)
Chain initiation step
15. (d) The reactivity of H-atoms depends upon
The driving force of the reaction is the stability of
the stability of free radicals, follows the order the carbocation. Tertiary carbocation is more stable
Tertiary > secondary > primary. than secondary carbocation which is achieved by
Therefore, reactivity of H-atoms follows the 1, 2-rearrangement
same order, i.e., tertiary > secondary > primary. 19. (c) 1-chloro-1-methylcyclohexane.
R R H H CH3
| | | | Cl
R – C• > R - C • > R - C • > H - C •
| | | |
R H H H
Tertiary Secondary primary free methyl
free radical free radical radical free radical CH2 CH3 CH3
16. (b) Cl
–
+ H – Cl + Cl
O
CH2 – CH = CH2 CH2 – C – H
CH3 CH3 CH3
O
(i) O 3 Cl
+ H H
(ii) Zn.H 2O H–C–H
+ H – Cl + Cl
Methanal
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Hydrocarbons 135
HBr, dark
Å (ii) CH3— C = CH — CH 3 ¾¾¾®
in absence
C6H5—CH—CH2—CH3 (A) of peroxide
(Benzyl carbocation)
(CH3) 2— CH — CH — CH 3 +
¾®
–
Br
Br
C6H5—CH—CH2—CH3 (Minor)
CH 3
Br
CH3 — C — CH2 – CH3
CH3
H
+
Br
23. (a) C CH CH2 (Major)
This reaction is governed by Markownikoff’s
CH3 rule according to which when an unsymmetrical
CH3 CH3 reagent e.g. HBr adds to an unsymmetrical
OH
– alkene, then the negative part of the reagent is
C CH CH3 ¾¾® C CH CH3 added to that carbon atom of the double bond
Å
which bears the least number of hydrogen atom.
CH3 H3C OH Thus, in above case 2-methyl-2-bromobutane
¾¾®
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136 CHEMISTRY
CH3 (2°)
suggested by Markownikoff. This is termed
Of the two possibilities, 2° carbocation is more anti-Markownikoff or peroxide effect.
stable so the product of the reaction is expected CH3
to be predominantly formed by 2° carbocation |
4® NaIO
31. (c) CH3 - C = CH - CH3 ¾¾¾
i.e. CH 3 - CH - CH - CH 3 KMnO4
| | O
CH 3 Br ||
CH 3 - C - CH 3 + CH 3COOH
i.e. 2– Bromo-3-Methylbutane 32. (c) Geometrical isomers differ in spatial
arrangement of atoms.
Some electrophilic addition reaction form products 33. (a)
that are clearly not the result of the addition of
electrophile to the sp 2 carbon bonded with the H CH3 H O CH3
| | O
most hydrogen and the addition of a nucleophile 3
to the other sp2 carbon. CH 3 – CH2– C = C CH3 – CH2– C C
The unexpected product resu lts from a | | |
rearrangement of carbocation intermediate. Please CH3 O–O CH3
O
note that all carbocation do not rearrange. ||
CH3– C – CH3+ CH3 CH2CHO
(–H 2O)
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Hydrocarbons 137
HBr
R — CH = CH 2 + H 2O n-hexane |
CH3
At 220º – 250ºC it forms ether. CH3 - CH 2 - CH - CH2 - CH2 - CH3
|
Br
35. (d) We know that 41. (c) CH3
B H CH CH CH2 CH3
6CH3 - CH = CH 2 ¾¾¾®
2 6
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138 CHEMISTRY
52. (a)
Peroxide effect is observed only in case of NaNH
Å
HBr. HC º CH ¾¾¾¾
2 ® HC º CNa
liq. NH 3
Å
Rotation around p bond is not possible. If any H C -CH – Br
¬¾¾¾¾¾
3 2 ¾ H3C – CH 2 – C º C N a
attempt is made to rotate one of the carbon atoms,
the lobes of p-orbital will no longer remain H3C – CH 2 – C º C – CH 2 – CH3
coplanar i.e no parallel overlap will be possible 3-Hexyne (Y)
and thus p-bond will break . This is known as
concept of restricted rotation. In other words 53. (b) The combustion reaction of ethylene is
the presence of p-bonds fix the position of two
5
carbon atoms. C 2 H 2 + O2 ® 2CO2 + H 2 O
2
47. (c) Such isomers, which possess the same
Both HC CH and CO2 have sp hybridization.
molecular and structural formula but differ in
the arrangement of atoms around the double O
bonded carbon atoms are known as geometrical P
Hg +2 / H2SO 4
isomers. 54. (d) CH º CH ¾¾¾¾¾¾
® CH3 - C - H
H - C - Cl H - C - Cl CH3CHO does not give Victor Meyer test.
|| || 55. (a) 1-Butyne and 2-butyne are distinguished
H - C - Cl Cl - C - H by NaNH2 because 1-Butyne react with NaNH2
(cis) (trans) due to presence of terminal hydrogen.
48. (b) CH3 CH2 C CH + NaNH2
1– Butyne
3 h CH3CH 2 C CNa + NH 3
3
i 56. (c) When double and triple, both bonds are
¾¾¾®
present, then ‘lowest set of locants’ rule will
decide the IUPAC name.
3 1 2 3 4 5 (incorrect)
C H3 - C H = C H - C º C H
Number of s 5 4 3 2 1 (correct)
bonds = 21
49. (b) Alkynes can be reduced to cis-alkenes Hence, the correct name will be pent-3-en-1-yne.
with the use of Lindlar’s catalyst 57. (d) The amount of s-character in various hybrid
50. (a) Correct order is orbitals is as follows.
H - C º C - H > H 3C - C º C - H > H 2 C sp = 50%, sp2 = 33% and sp3 = 25%
( Two acidic
hydrogens ) ( One acidic
hydrogen ) 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
= CH 2 > CH3 - CH3
turn has greater s-character of the C – H bond in
51. (c) Hydration of alkynes give ketones. alkanes. Thus, owing to a high s-character of the
C – H bond in alkynes, the electrons constituting
OH this bond are more strongly held by the carbon
H3 C - C º CH ¾¾
® H3C - C = CH 2 nucleus with the result the hydrogen present on
(A) such a carbon atom can be easily removed as
O
proton. The acidic nature of three types of C – H
Tautomerism
H3 C - C - CH3
¾¾¾¾® bonds follows the following order
(B) º C - H > = C - H > - C - H.
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Hydrocarbons 139
Further, as we know that conjugate base of a 62. (c) On heating ethylene chloride (1, 1 dichloro
strong acid is a weak base, hence the correct ethane) with alcoholic KOH followed by
order of basicity is sodamide, alkyne is obtained
(- ) (- ) (-) R - CH 2 - CCl 2 - R
H - C º C < CH 2 = CH < CH 2 - CH3
alc.KOH
¾¾¾¾® R - CH = CCl - R
58. (c) CH3 - CH 2 - C º CH + HCl
NaNH
¾¾¾¾ 2®R - C º C - R
® CH3 - CH 2 - C = CH 2
¾¾ 63. (d) Br2 in CCl4 (a), Br2 in CH3 COOH (b) and
|
Cl alk. KMnO4 (c) will react with all unsaturated
compounds, i.e., 1, 3 and 4 while ammoniacal
I AgNO3 (d) reacts only with terminal alkynes,
| i.e., 3 and h ence compound 3 can be
HI
¾¾® CH3 - CH 2 - C - CH3
| distinguished from 1, 2 and 4 by. ammoniacal
Cl AgNO3 (d).
64. (a) The acidity of acetylene or 1–alkynes can
According to Markownikoff’s rule which states be explained on the basis of molecular orbital
that when an unsymmetrical alkene undergo concept according to which formation of
hydrohalogenation, the negative part goes to C–H bond in acetylene involves sp-hybridised
the C-atom which contain lesser no. of H-atom. carbon atom.
O
59. (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 s-
O character will be correspondingly more closer to
2H O the nucleus.
CH3– C – C – CH2 – CH3 ¾®
2
Thus, owing to high s character of the C—H bond
O O 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
(O ) acts as more electronegative species than the sp2
CH 3 - C O + H 2O2 ¾¾
¾®
| and sp 3 with the result the hydrogen present on
CH3CH 2CO such a carbon atom (º C—H) can be easily removed
Methylethy lglyoxal
as a proton.
65. (a) Enthalpy of hydrogenation
CH3COOH + CH3CH 2COOH
Acetic acid Propionic acid 1
µ
The glyoxal formed as an intermediate is stability of alkene
oxidised by H2O2 to give the acids.` \ III > II > I
60. (a) Only C2H2 (acetylene) has acidic H-atoms 66. (b) 2C6 H 6 (g) + 9O 2 (g) ¾¾¾®
2 5 VO
and hence reacts with NaNH2 to form sodium 410°C
salt, i.e., O
HC º CH + NaNH2 ¾ ¾® HC º CNa + NH3. CH—C
61. (c) Reduction of alkynes with Na/liq. NH3 2 O + 4CO2(g) + 4H2O(g)
gives trans-alkenes. This reaction is called Birch CH—C
reduction
CH 3 - C º C - CH 3 + H 2 O
Maleic anhydride
H3C H 67. (d) Presence of 6 p orbitals, each containing
Na / liq. NH
¾¾¾¾¾¾
3® C=C one unpaired electron, in a six membered cyclic
H CH3 structure is in accordance with Huckel rule of
trans-Butene or trans-2-Butene aromaticity.
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140 CHEMISTRY
68. (a) Huckel rule is not obeyed. It has only four 76. (c) Out of the given compounds the most
electrons. Further it does not have continous reactive towards nucleophilic attack is
conjugation. OH
In thiophene, one lone pair of electrons are involved
in delocalisation.
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Hydrocarbons 141
81. (c) 85. (d) –OCH3 activates the benzene ring. –NO2
CH3 deactivates the ring. Hence, the reaction of the
given compounds with electrophiles is in the
anhydrous AlCl
+ CH3Cl ¾¾¾¾¾®
3
+ HCl order, I > II > III.
Å
82. (a) CH3 and NH2 and OH are electron donating 86. ¾® CH ·3 + AlCl4–
(c) CH3Cl + AlCl3 ¾
groups, whereas NO2 is an electron withdrawing
Electrophile
group and leaves the benzene ring deactivated.
Due to stronger electron attracting (–I effect) 87. (c) Chlorination of methane proceeds via free
effect of NO2 group, C6 H5NO2 shows least radical mechanism. Conversion of methyl
reactivity towards electrophilic substitution. chloride to methyl alcohol proceeds via
83. (d) –Cl atom shows o/p-directive influence nucleophilic substitution. Formation of ethylene
but deactivate the benzene ring, while from ethyl alcohol proceeds via dehydration
[–OH/–CH3] groups show o/p influence and reaction. Nitration of benzene is electrophilic
activate the benzene ring but –OH group is more substitution reaction.
activating than –CH3. 88. (c) Benzene do not show addition reaction like
Hence order of electrophilic substitution other un saturated hydrocarbons. Due to
resonance all the C – C bonds have the same
OH CH3 Cl 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
84. (c) In arenes, p electrons are delocalised, hence has much higher electron density in the ring than
arenes do not undergo addition reactions easily. benzene, Nitrobenzene and benzoic acid.
Aromatic compounds (Arenes) are highly stable Nitro and carboxylic group show- I-effect and
and show resonance eg. Benzene is the simplest hence deactivate the ring towards electrophillic
example. substutution .
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142 CHEMISTRY
14 Environmental
Chemistry
Trend Analysis with Important Topics & Sub-Topics
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Environmental Chemistry 143
(a) CO2 (b) Water gas (c) Fluoride deficiency in drinking water is
(c) Petroleum gas (d) NO2 harmful. Soluble fluoride is often used to
8. Which of the following is a sink for CO? [2017] bring its concentration upto 1 ppm.
(a) Microorganism present in the soil (d) When the pH of rain water is higher than 6.5,
it is called acid rain.
(b) Oceans
11. Which one of the following statement is not
(c) Plants
true ? [2011]
(d) Haemoglobin
(a) pH of drinking water should be between
9. Roasting of sulphides give the gas X as a by 5.5 – 9.5.
product. This is colorless gas with choking smell
(b) Concentration of DO below 6 ppm is good
of burnt sulphur and caused great damage to
for the growth of fish.
respiratory organs as a result of acid rain. Its
aqueous solution is acidic, acts as a reducing (c) Clean water would have a BOD value of less
agent and its acid has never been isolated. The than 5 ppm.
gas X is : [NEET 2013] (d) Oxides of sulphur, nitrogen and carbon are
(a) SO2 (b) CO2 the most widespread air pollutant.
(c) SO3 (d) H2S 12. Green chemistry means such reactions which
10. Which one of the following statements is not (a) produce colour during reactions [2008]
true? [NEET Kar. 2013]
(b) reduce the use and production of hazardous
(a) Dissolved oxygen (DO) in cold water can chemicals
reach a concentration upto 10 ppm.
(c) are related to the depletion of ozone layer
(b) Clean water would have a BOD value of 5
(d) study the reactions in plants
ppm.
ANSWER KEY
1 (d) 2 (a) 3 (d) 4 (d) 5 (a) 6 (c) 7 (a) 8 (a) 9 (a) 10 (d)
11 (b) 12 (b)
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144 CHEMISTRY
It is oxidising in nature and causes irritation to
Microorganism present in soil convert CO to CO2,
eyes, lungs, nose, asthamatic attack and damage thus it's sink.
plants.
5. (a) Green house gases such as CO2, ozone, 9. (a) Based on the features given, gas must be SO2.
methane, the chlorofluoro carbon compounds 10. (d) Acid rain is the rain water containing
and water vapour form a thick cover around the sulphuric acid and nitric acid which are formed
earth which prevents the IR rays emitted by the from the oxides of sulphur and nitrogen present
earth to escape. It gradually leads to increase in in the air as pollutants. Rain water has a pH
temperature of atmosphere. below 5.6.
11. (b) The growth of fishes get hindered if the
Water vapour is the largest contributor to the earth's concentration of D.O. is below 6 ppm.
greenhouse effect. However water vapour does not 12. (b) Green chemistry are the reaction, which
control the Earth's temperature. reduces the use of hazardous chemicals.
6. (c) CO and oxides of Nitrogen are poisonous Green chemistry is the programme of developing
gases present in automobile exhaust gases. new chemical products and chemical processes or
7. (a) Liquified CO2 (carbon dioxide) with a making improvements in the already existing
compounds and processes so as to make them less
suitable detergent is used in dry cleaning.
harmful to human health and environment. This
8. (a) Microorganisms present in the soil is a sink means the same as to reduce the use and production
for CO. of hazardous chemicals.
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The Solid State 145
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146 CHEMISTRY
9. For orthorhombic system, axial ratios are is 2.72 g cm–3. The molar mass of the metal is :
a ¹ b ¹ c and the axial angles are [1991] (NA, Avogadro’s constant = 6.02 × 1023 mol-1)
(a) a = b = g ¹ 90° [NEET 2013]
(b) a = b = g = 90° (a) 30 g mol–1 (b) 27 g mol–1
(c) a = g = 90°, b ¹ 90° (c) 20 g mol–1 (d) 40 g mol–1
(d) a ¹ b ¹ g = 90° 16. Structure of a mixed oxide is cubic close-packed
(c.c.p). The cubic unit cell of mixed oxide is
Topic 3: Cubic System and Bragg's Equation composed of oxide ions. One fourth of the
10. An element has a body centered cubic (bcc) tetrahedral voids are occupied by divalent metal
A and the octahedral voids are occupied by a
structure with a cell edge of 288 pm. The atomic
monovalent metal B. The formula of the oxide is :
radius is: [2020]
[2012 M]
2 4 (a) ABO2 (b) A2BO2
(a) ´ 288pm (b) ´ 288pm
4 3 (c) A2B3O4 (d) AB2O2
4 17. The number of octahedral void(s) per atom
´ 288pm 3 present in a cubic close-packed structure is :
(c) (d) ´ 288pm
2 4 [2012]
11. A compound is formed by cation C and anion A. (a) 1 (b) 3
The anions form hexagonal close packed (hcp) (c) 2 (d) 4
lattice and the cations occupy 75% of octahedral 18. A metal crystallizes with a face-centered cubic
voids. The formula of the compound is: lattice. The edge length of the unit cell is 408
[2019] pm. The diameter of the metal atom is : [2012]
(a) C2A 3 (b) C3A 2 (a) 288 pm (b) 408 pm
(c) C3A 4 (d) C4A 3 (c) 144 pm (d) 204 pm
12. Iron exhibits bcc structure at room temperature. 19. AB crystallizes in a body centred cubic lattice
Above 900°C, it transforms to fcc structure. The with edge length ‘a’ equal to 387 pm. The
ratio of density of iron at room temperature to distance between two oppositely charged ions
that at 900°C (assuming molar mass and atomic in the lattice is : [2010]
radii of iron remains constant with temperature) (a) 335 pm (b) 250 pm
is [2018] (c) 200 pm (d) 300 pm
20. Copper crystallises in a face-centred cubic lattice
3 4 3 with a unit cell length of 361 pm. What is the
(a) (b)
2 3 2 radius of copper atom in pm?[2009]
1 3 3 (a) 157 (b) 181
(c) (d) (c) 108 (d) 128
2 4 2 21. Lithium metal crystallises in a body centred cubic
13. Which of the following statements about the crystal. If the length of the side of the unit cell of
interstitial compounds is incorrect ? lithium is 351 pm, the atomic radius of the lithium
[NEET 2013] will be: [2009]
(a) They are chemically reactive. (a) 151.8 pm (b) 75.5 pm
(b) They are much harder then the pure metal. (c) 300.5 pm (d) 240.8 pm
(c) They have higher melting points than the 22. Percentage of free space in a body centred cubic
pure metal. unit cell is : [2008]
(d) They retain metallic conductivity. (a) 30% (b) 32%
14. The number of carbon atoms per unit cell of (c) 34% (d) 28%
diamond unit cell is : [NEET 2013] 23. If ‘a’ stands for the edge length of the cubic
(a) 8 (b) 6 systems : simple cubic, body centred cubic and
(c) 1 (d) 4 face centred cubic, then the ratio of radii of the
15. A metal has an fcc lattice. The edge length of spheres in these systems will be respectively,
the unit cell is 404 pm. The density of the metal [2008]
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The Solid State 147
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148 CHEMISTRY
39. When electrons are trapped into the crystal in 40. On doping Ge metal with a little of In or Ga, one
anion vacancy, the defect is known as [1994] gets [1993]
(a) Schottky defect (a) p-type semi conductor
(b) Frenkel defect (b) n-type semi conductor
(c) Stoichiometric defect (c) insulator
(d) F-centres (d) rectifier
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The Solid State 149
13. (a) Interstitial compounds are chemically inert. a
charged ions = 3
In interstitial compounds small atoms like H, B &
2
387 ´ 1.732
C enter into the void sites between the packed = ; 335pm
atoms of crystalline metal. They retain metallic 2
conductivity and are chemically inert. 20. (d) Since Cu metal crystallises in a face centred
cubic lattice
14. (a) Diamond is like ZnS. In diamond cubic unit a
cell, there are eight corner atoms, six face \ r=
centered atoms and four more atoms inside the 2 2
361
structure (tetrahedral voids). r= = 127.6 ; 128pm
Total no. of atoms present per unit cell 2 2
1 1 21. (a) Since lithium metal crystallises in a body
= 8´ + 6´ + 4 = 8 centred cubic crystal
8 2
(corners) (face (inside a 3 351 ´ 1.732
centered) body) \ r= = =151.98 pm
4 4
15. (b) Density is given by 22. (b) Percentage of occupied space in a bcc is 68%.
Z ´M \ Percentage of free space in a bcc is
d= ; where Z = number of formula units
N Aa 3 (100 – 68) = 32%
present in unit cell, which is 4 for fcc 23. (a) Following generalization can be easily
a = edge length of unit cell. M = Molecular mass derived for various types of lattice arrangements
4´ M in cubic cells between the edge length (a) of the
2.72 = cell and r the radius of the sphere.
6.02 ´ 10 ´ (404 ´ 10 -10 )3
23
a
(Q 1pm = 10-10 cm) For simple cubic : a = 2r or r =
2
2.72 ´ 6.02 ´ (404)3 For body centred cubic :
M= = 26.99 = 27 g mol–1
4 ´ 107 4 3
a= r or r = a
1 3 4
16. (d) No. of A2 + ions = ´ 8 = 2 For face centred cubic :
4
No. of B+ ions = 4 × 1 = 4 1
a = 2 2 r or r = a
No. of atoms in ccp, 2 2
1 1 Thus the ratio of radii of spheres will be simple :
O2 - = 8 ´ + 6 ´ = 4
8 2 bcc : fcc
Hence the formula of the oxide will be a 3 1
A2B4O4 or AB2O2. = : a: a
2 4 2 2
17. (a) Number of octahedral voids in ccp is equal 24. (d) Number of atoms per unit cell = 1
to effective number of atoms. Atoms touch each other along edges. Hence
In ccp, effective number of atoms are 4 so, 4 a
octahedral voids. r=
2
So, 1 octahedral voids per atom. (r = radius of atom and a = edge length)
18. (a) For fcc structure, 2 a = 4r 4 3
pr
p
2 ´ 408 Therefore % fraction = 3 = = 0.52
= 2r (2r = Diameter) (2r ) 3 6
2
Diameter = 288.5 ; 288 pm 25. (c) In body centred cubic lattice one molecule
of CsBr is within one unit cell.
19. (a) For bcc lattice, diagonal = a 3 .
Atomic mass of unit cell = 133 + 80 = 213 a.m.u
The distance between the two oppositely
Volume of cell = (436.6 × 10–10)3 cm3
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150 CHEMISTRY
Z ´ at.wt. therefore number of atoms per unit cell = 1. Thus
Density =
Av.no. ´ vol.of unit cell formula of the compound is AB.
1´ 213 29. (d) Given : Order of Bragg diffraction (n) = 2;
Density = 23 3 -30 Wavelength (l) = 1 Å and angle (q) = 60º. We
6.02 ´10 ´ (436.6) ´ 10
know from the Bragg’s equation nl = 2d sin q
213 ´ 107 or 2 × 1 = 2d sin 60º
= = 4.25 g / cm3
3
6.02 ´ (436.6) 3 2
Þ 2 ´ 1 = 2.d . Þ d= = 1.15 Å
In the density formula, Z is number of particles 2 3
present in a unit cell and M is molar mass of that
(where d = difference between the scattering
particle. For any diatomic molecule like CsBr, Z =
2 but for any mono atomic molecule like Li, Z = 1
planes)
for bcc lattice. 30. (c) A body-centred cubic system consists of
eight atoms at the corners plus one atom at the
26. (c) centre of cube.
31. (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
32. (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.
33. (c) The no. of atoms is a unit cell may be
calculated by the the formula
An isolated fcc cell is shown here. Each face n n n f ne
Z= c+ b+ +
of the cell is common to two adjacent cells. 8 1 2 4
Therefore, each face centre atom contributes where nc = no. of atom at the corner nb = no. of
only half of its volume and mass to one cell. atoms at body centre nf = no. of atoms at face
Arranging six cells each sharing the remaining centre, ne = no. of atoms at edge centre.
half of the face centred atoms, constitutes An fcc crystal contains
fcc cubic lattice. e.g., Cu and Al. 8 6
27. (a) For the given cubic structure, = + = 4 atoms in a unit cell.
8 2
34. (b) Ni0.98O = (Ni2+)x (Ni3+)0.98–x (O2–)1
No. of X atoms at the corners = 8 ´ 1 = 1
8 Net charge = 0
1 [x × 2] + [(0.98 – x) × 3] + [–2 × 1] = 0
No. of Y atoms at the face-centres = 6 ´ = 3 x = 0.94
2
Fraction of nickel existing as
\ Formula of the compound = XY3
0.94
28. (a) Given: Atoms are present at the corners of Ni2+ = = 0.959 » 0.96
the cube = A and atoms present at body 0.98
35. (d) The semiconductors formed by the
centre = B. We know that a cubic unit cell has 8
introduction of impurity atoms containing one
corners. Therefore contribution of each atom at
1 elecron less than the parent atoms of insulators
the corner = . Since number of atoms per unit are termed as p-type semiconductors. Therefore,
8
1 silicon containing 14 electrons has to be doped
cell is 8, therefore total contribution = 8 ´ = 1 . with boron containing 13 electrons to give a
8
We also know the one atom is in the body centre, p-type semi-conductor.
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The Solid State 151
36. (b) For each Sr 2+
ion added, one Na+
ion is from their lattice sites so that the electrical
removed to maintain the electrical neutrality. neutrality is maintained. The defect is called
Hence concentration of cation vacancies Schottky defect.
= mole % of SrCl2 added = 10–4 mole % 39. (d) When electrons are trapped in anion
10 -4 vacancies, these are called F-centres.
= ´ 6.023 ´ 1023 = 6.023 ´ 1017
100
37. (d) When the crystal is irradiated with white +ve –ve
light, the trapped electrons in a hole absorbs e– ion ion
photon for excitation from ground state to excited
state. This gives colour to the compund. F- centre in crystal
Alkali metal halide may have excess metal ion if a 40. (a) p-type of semiconductors are produced
negative ion is absent from its lattice site, leaving a by adding impurity containing less electrons
hole, which is occupied by electron to maintain (i.e. atoms of group 13).
electrical neutrality. The 'holes' occupied by Ge belongs to Group 14 and In to Group 13.
electrons are called F-centres or colour centres.
Hence, on doping, p-type semiconductor is
38. (d) If in an ionic crystal of the type A+. B–, obtained.
equal number of cations and anions are missing
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16 Solutions
Trend Analysis with Important Topics & Sub-Topics
(a) Molarity
Topic 1: Solubility and Concentration of
Solutions (b) Mole fraction
(c) Weight percentage
1. The density of 2 M aqueous solution of NaOH (d) Molality
is 1.28 g/cm3. The molality of the solution is 4. What is the mole fraction of the solute in a 1.00
[Given that molecular mass of NaOH = 40 g mol–1] m aqueous solution ? [2011, 2015 RS]
[NEET Odisha 2019 ] (a) 0.177 (b) 1.770
(a) 1.32 m (b) 1.20 m (c) 0.0354 (d) 0.0177
(c) 1.56 m (d) 1.67 m 5. How many grams of concentrated nitric acid
2. If molality of the dilute solutions is doubled, the solution should be used to prepare 250 mL of
value of molal depression constant (Kf ) will be:- 2.0M HNO3 ? The concentrated acid is 70%
[2017] HNO3 [NEET 2013]
(a) halved (b) tripled (a) 90.0 g conc. HNO3
(c) unchanged (d) doubled (b) 70.0 g conc. HNO3
3. Which of the following is dependent on (c) 54.0 g conc. HNO3
temperature? [2017] (d) 45.0 g conc. HNO3
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154 CHEMISTRY
19. At 100 °C the vapour pressure of a solution of 26. Formation of a solution from two components
6.5 g of a solute in 100 g water is 732 mm. If can be considered as [2003]
Kb = 0.52, the boiling point of this solution will (i) Pure solvent ® separated solvent
be [2016] molecules, DH1
(a) 101 °C (b) 100 °C (ii) Pure solute ® separated solute
(c) 102 °C (d) 103 °C molecules, DH2
20. Which one is not equal to zero for an ideal (iii) Separted solvent ® Solution, DH3
solution: [2015] & solute molecules
(a) DSmix Solution so formed will be ideal if
(b) DVmix
(c) DP = Pobserved - PRaoult (a) DH soln = DH 3 - DH1 - DH 2
(d) DHmix (b) DH soln = DH1 + DH 2 + DH 3
21. PA and PB are the vapour pressure of pure liquid (c) D H soln = D H 1 + D H 2 - D H 3
components, A and B, respectively of an ideal (d) DH soln = DH1 - DH 2 - DH 3
binary solution. If X A represents the mole 27. A solution containing components A and B
fraction of component A, the total pressure of follows Raoult's law when [2002]
the solution will be [2012] (a) A – B attraction force is greater than A – A
(a) pa + xa (pb – pa) (b) pa + xa (pa – pb) and B – B
(c) pb + xa (pb – pa) (d) pb + xa (pa – pb) (b) A – B attraction force is less than A – A and
22. Vapour pressur e of chloroform (CHCl 3 ) B–B
and dichloromethane (CH2Cl2 ) at 25 ºC are (c) A – B attraction force remains same as
200 mm Hg and 41.5 mm Hg respectively. Vapour A–A and B –B
pressure of the solution obtained by mixing (d) Volume of solution is different from sum of
25.5 g of CHCl3 and 40 g of CH2Cl2 at the same volume of solute and solvent
temperature will be : (Molecular mass of CHCl3 28. The beans are cooked earlier in pressure cooker,
= 119.5 u and molecular mass of CH2Cl2 = 85 u). because [2001]
[2012 M] (a) Boiling point increases with increasing
(a) 173.9 mm Hg (b) 615.0 mm Hg pressure
(c) 347.9 mm Hg (d) 285.5 mm Hg (b) Boiling point decreases with increasing
23. An aqueous solution is 1.00 molal in KI. Which pressure
change will cause the vapour pressure of the (c) Internal energy is not lost while cooking in
solution to increase? [2010] pressure cooker
(a) Addition of NaCI (d) Extra pressure of pressure cooker, softens
(b) Addition of Na 2SO4 the beans
(c) Addition of 1.00 molal KI 29. The vapour pressure of a solvent decreased by
(d) Addition of water 10 mm of mercury when a non-volatile solute
24. A solution of acetone in ethanol [2006] was added to the solvent. The mole fraction of
(a) shows a positive deviation from Raoult’s law the solute in the solution is 0.2. What should be
(b) behaves like a non ideal solution the mole fraction of the solvent if the decrease
(c) obeys Raoult’s law in the vapour pressure is to be 20 mm of mercury?
(d) shows a negative deviation from Raoult’s law [1998]
25. The vapour pressure of two liquids ‘P’ and (a) 0.8 (b) 0.6
‘Q’ are 80 and 60 torr, respectively. The total (c) 0.4 (d) 0.2
vapour pressure of solution obtained by 30. The vapour pressure at a given temperature of
an ideal solution containing 0.2 mol of a non-
mixing 3 mole of ‘P’ and 2 mole of ‘Q’ would be
volatile solute and 0.8 mol of solvent is 60 mm of
[2005]
Hg. The vapour pressure of the pure solvent at
(a) 72 torr (b) 140 torr
the same temperature is [1996]
(c) 68 torr (d) 20 torr
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Solutions 155
(a) 150 mm of Hg (b) 60 mm of Hg (a) 0.80K (b) 0.40 K [2020]
(c) 75 mm of Hg (d) 120 mm of Hg (c) 0.60 K (d) 0.20 K
31. Vapour pressure of benzene at 30 °C is 121.8 mm. 37. Which one of the following electrolytes has the
When 15 g of a non volatile solute is dissolved same value of van't Hoff's factor (i) as that of
in 250 g of benzene, its vapour pressure the Al2(SO4)3 (if all are 100% ionised) ? [2015]
decreased to 120.2 mm. The molecular weight of (a) K3[Fe(CN)6] (b) Al(NO3)3
the solute (Mo. wt. of solvent = 78) [1995]
(c) K4[Fe(CN)6] (d) K2SO4
(a) 356.2 (b) 456.8
38. The boiling point of 0.2 mol kg–1 solution of X
(c) 530.1 (d) 656.7
in water is greater than equimolal solution of Y
32. According to Raoult's law, relative lowering of
in water. Which one of the following statements
vapour pressure for a solution is equal to
is true in this case ? [2015]
[1995]
(a) moles of solute (a) Molecular mass of X is greater than the
molecular mass of Y.
(b) moles of solvent
(c) mole fraction of solute (b) Molecular mass of X is less than the
(d) mole fraction of solvent molecular mass of Y.
33. The relative lowering of the vapour pressure is (c) Y is undergoing dissociation in water while
equal to the ratio between the number of X undergoes no change.
[1991] (d) X is undergoing dissociation in water.
(a) solute molecules to the solvent molecules 39. Of the following 0.10 m aqueous solutions,
(b) solute molecules to the total molecules in which one will exhibit the largest freezing point
the solution depression? [2014]
(c) solvent molecules to the total molecules in (a) KCl (b) C6H12O6
the solution (c) Al2(SO4)3 (d) K2SO4
(d) solvent molecules to the total number of
40. The freezing point depression constant for water
ions of the solute.
is – 1.86 ºC m–1. If 5.00 g Na2SO4 is dissolved in
34. An ideal solution is formed when its components 45.0 g H2O, the freezing point is changed by
[1988]
– 3.82 ºC. Calculate the van’t Hoff factor for
(a) have no volume change on mixing Na2SO4. [2011]
(b) have no enthalpy change on mixing
(a) 2.05 (b) 2.63
(c) have both the above characteristics
(c) 3.11 (d) 0.381
(d) have high solubility.
41. The van’t Hoff factor i for a compound which
35. All form ideal solution except [1988]
undergoes dissociation in one solvent and
(a) C6H6 and C6H5 CH3 association in other solvent is respectively :
(b) C2H6 and C2H5I [2011]
(c) C6H5Cl and C6H5 Br
(a) less than one and greater than one.
(d) C2H5 I and C2H5 OH.
(b) less than one and less than one.
Topic 3: Colligative Properties and (c) greater than one and less than one.
Abnormal Molecular Masses (d) greater than one and greater than one.
36. The freezing point depression constant (Kf) of 42. A 0.1 molal aqueous solution of a weak acid is
benzene is 5.12 K kg mol–1. The freezing point 30% ionized. If K f for water is 1.86 °C/m, the
depression for the solution of molality 0.078 m freezing point of the solution will be : [2011 M]
containing a non-electrolyte solute in benzene (a) – 0.18 °C (b) – 0.54 °C
is (rounded off upto two decimal places) : (c) – 0.36 °C (d) – 0.24 °C
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156 CHEMISTRY
43. 200 mL of an aqueous solution of a protein 49. 1.00 g of a non-electrolyte solute (molar mass
contains its 1.26 g. The osmotic pressure of this 250 g mol–1) was dissolved in 51.2 g of benzene.
solution at 300 K is found to be 2.57 × 10–3 If the freezing point depression constant, Kf of
bar. The molar mass of protein will be benzene is 5.12 K kg mol–1, the freezing point of
(R = 0.083 L bar mol–1 K–1) [2011 M] benzene will be lowered by [2006]
(a) 51022 g mol–1 (b) 122044 g mol–1 (a) 0.3 K (b) 0.5 K
(c) 31011 g mol–1 (d) 61038 g mol–1 (c) 0.4 K (d) 0.2
44. A solution of sucrose (molar mass = 342 g mol–1) 50. A solution of urea (mol. mass 56 g mol-1) boils
has been prepared by dissolving 68.5 g of at 100.18 °C at the atmospheric pressure. If Kf
and Kb for water are 1.86 and 0.512 K kg mol-1
sucrose in 1000 g of water. The freezing point of
respectively, the above solution will freeze at
the solution obtained will be (Kf for water
= 1.86 K kg mol–1). [2010] [2005]
(a) – 0.372 °C (b) – 0.520 °C (a) 0.654 °C (b) - 0.654 °C
(c) 6.54 °C (d) - 6.54 °C
(d) + 0.372 °C (d) – 0.570 °C
51. Camphor is often used in molecular mass
45. A 0.0020 m aqueous solution of an ionic
determination because [2004]
compound Co(NH 3 ) 5 (NO 2 )Cl freezes at
– 0.00732 °C. Number of moles of ions which (a) it is readily available
1 mol of ionic compound produces on being (b) it has a very high cryoscopic constant
dissolved in water will be (Kf = – 1.86 °C/m) (c) it is volatile
[2009] (d) it is solvent for organic substances
(a) 3 (b) 4 52. A solution contains non-volatile solute of
(c) 1 (d) 2 molecular mass M2. Which of the following can
46. 0.5 molal aqueous solution of a weak acid (HX) be used to calculate the molecular mass of solute
is 20% ionised. If Kf for water is 1.86 K kg in terms of osmotic pressure ? [2002]
mol–1 ,the lowering in freezing point of the
æm ö æ m ö RT
solution is [2007] (a) M 2 = ç 2 ÷ VRT (b) M 2 = ç 2 ÷
è p ø èV ø p
(a) 0.56 K (b) 1.12 K
(c) – 0.56 K (d) – 1.12 K æm ö æm ö p
(c) M 2 = ç 2 ÷ pRT (d) M2 = ç 2 ÷
47. During osmosis, flow of water through a èV ø è V ø RT
semipermeable membrane is [2006] 53. Which of the following colligative property can
(a) from both sides of semipermeable provide molar mass of proteins (or polymers or
membrane with equal flow rates colloids) with greatest precision ? [2000]
(b) from both sides of semipermeable (a) Osmotic pressure
membrane with unequal flow rates (b) Elevation of boiling point
(c) from solution having lower concentration only (c) Depression of freezing point
(d) from solution having higher concentration (d) Relative lowering of vapour pressure
only 54. Which of the following 0.10 m aqueous solutions
48. A solution containing 10 g dm –3 of urea will have the lowest freezing point ? [1997]
(molecular mass = 60 g mol–1) is isotonic with a (a) Al2(SO4)3 (b) C6H12O6
5% solution of a non-volatile solute. The
molecular mass of this non-volatile solute is (c) KCl (d) C12H22O11
[2006] 55. At 25°C, the highest osmotic pressure is
exhibited by 0.1 M solution of [1994]
(a) 300 g mol–1 (b) 350 g mol–1
(a) CaCl2 (b) KCl
(c) 200 g mol–1 (d) 250 g mol–1
(c) Glucose (d) Urea.
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Solutions 157
56. Which one of the following salts will have the (b) Glucose will flow towards urea solution
same value of van’t Hoff factor (i) as that of (c) urea will flow towards glucose solution
K4[Fe (CN)6]. [1994] (d) water will flow from urea solution to glucose
(a) Al2(SO4)3 (b) NaCl 59. Which of the following aqueous solution has
(c) Al (NO3)3 (d) Na2SO4. minimum freezing point ? [1991]
57. Which one is a colligative property ? [1992] (a) 0.01 m NaCl (b) 0.005 m C2H5OH
(a) boiling point (b) vapour pressure (c) 0.005 m MgI2 (d) 0.005 m MgSO4.
(c) osmotic pressure (d) freezing point 60. Blood cells retain their normal shape in solution
58. If 0.1 M solution of glucose and 0.1 M solution which are [1991]
of urea are placed on two sides of the semi- (a) hypotonic to blood
permeable membrane to equal heights, then it (b) isotonic to blood
will be correct to say that [1992]
(c) hypertonic to blood
(a) There will be no net movement across the
membrane (d) equinormal to blood.
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158 CHEMISTRY
wt ´ 1000 \ 1000 mL of solution contain cane sugar
5. (d) Molarity (M) =
mol. wt. ´ vol (mL) 5
= ´ 1000 = 50 g/L
wt. 1000 100
2= ×
63 250 1 ´ 1000
63 Similarily 1% solution contains =
wt. = g 100
2 = 10 g/L of X.
100
wt. of 70% acid = ´ 31.5 = 45 g 50 g / L
70 (p1) = C × R × T = ´R´T
6. (b) Concentration of 342
Osmotic pressure of 1% solution of susbtance
25.3 1000
Na 2CO3 = ´ = 0.955 M
106 250 10 g / L
(p2) = ´R´T
M
[Na + ] = 2 × 0.955 = 1.91 M As both are isotonic,
é CO32– ù = 0.955 M So p1 = p2
ë û 50 10
7. (b) One molal solution means one mole solute or ´R´T = ´R´T
342 M
present in 1 kg (1000 g) solvent
342
i.e., mole of solute = 1 \ M (mol. wt. of X) = = 68.4
5
1000g 1000 12. (c) Percentage volume of oxygen = 21%.
Mole of solvent (H2O) = =
18g 18 So, 100 mL of air contains = 21mL of O2
1 \ Volume of oxygen in one litre of air
Mole fraction of solute =
æ 1000 ö 21
ç1 + ÷ = ´ 1000 = 210mL.
è 18 ø 100
18 210
= = 0.01768 ; 0.018. Therefore, no. of moles = = 0.0093
1018 22400
8. (c) From molarity equation
(Q volume of 1 mole of gas at S.T.P. is 22400 mL)
M1V1 + M2V2 = MV
13. (b) The molality involves weights of the
1 × 2.5 + 0.5 × 3 = M × 5.5 solute and the solvent. Since weight does not
4 change with the temperature, therefore molality
M= = 0.727 ; 0.73M
5.5 does not depend upon the temperature.
9. (b) Density = 1.17 g/cc 14. (d) Hydron bond of ethanol gets weakened
Mass by addition of acetone. Thus the mixture of
As d = ethanol and acetone show positive deviation
Volume
from Raoult's law.
volume = 1cc \ mass = 1.17g
15. (b) In case of positive deviation from ideal
No. of moles of solute behaviour A-B interactions are weaker than A-
Now molarity =
Volume of solution in litre A and B-B interactions.
16. (d) Mole fraction of water vapour = 0.02
1.17 ´ 1000 1170 \ Mole fraction of dry air = 1 – 0.02 = 0.98
= = =32.05M
36.5 ´ 1 36.5 Total pressure of saturated air = 1.2 atm
10. (d) The mole fraction can never be equal to –2 Acc. to Dalton’s law of partial pressure,
or 2. It is always between 0 and 1 i.e., 0 < x < 1. PDry air = PTotal xDry air = 1.2 × 0.98 = 1.176 atm
11. (c) 5% cane sugar solution means 100ml of 17. (a) The solutions which show a large negative
solution contain cane sugar = 5g deviation from Raoult’s law form maximum
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Solutions 159
boiling azeotrope. Nitric acid and water forms a 21. (d) P = PAXA + PBXB = PAXA + PB (1 – XA)
maximum boiling azeotrope with a boiling point (for binary sol. XA + XB = 1)
of 393.5 K.
Þ PAXA + PB – PBXA
18. (a) Let us consider that A is benzene and B is
toluene Þ PB + XA (PA – PB)
1 : 1 molar mixture of A and B 25.5
1 1 22. (N) n CHCl3 = = 0.213
\ xA = and xB = 119.5
2 2 40
0 0 n CH 2Cl2 = = 0.47
Total pressure of solution (P) = PA xA + PB xB 85
1 1 PT = PA° X A + PB° X B
P = 12.8 × + 3.85 × = 8.325 kPa
2 2 0.213 0.47
1 = 200 ´ + 41.5 ´
0 12.8 ´ 0.683 0.683
yA = PA xA = 2 = 0.768 = 62.37 + 28.55 = 90.92
P 8.325 23. (d) When the aqueous solution of one molal
\ yB = 1 – yA = 1 – 0.768 = 0.232
KI is diluted with water, concentration decreases,
so, the vapour will contain higher percentage of therefore the vapour pressure of the resulting
benzene. solution increases.
24. (a) A solution of acetone in ethanol shows
As the molar composition is same and benzene positive deviation from Raoult's law. It is because
has higher vapour pressure, the vapour will contain
higher percentage of benzene. ethanol molecules are strongly hydrogen
bonded. When acetone is added, these
æ P° - Ps ö n W1 M 2 molecules break the hydrogen bonds and
19. (a) = = ´
èç P° ø÷ N M1 W2 ethanol becomes more volatile. Therefore, its
vapour pressure is increased.
Where, W1 = wt of solute
W2 = wt of solvent 25. (a) Given V.PP = 80 torr
M1 = Mass of solute V.PQ = 60 torr
M2 = Mass of solvent Ptotal = V·PP × XP + V·PQ × XQ
at 100 °C, P° = 760 mm é 3 2ù
= ê80 ´ + 60 ´ ú = 16 × 3 + 12 × 2
760 - 732 6.5 ´ 18 ë 5 5û
=
760 M1 ´ 100 Ptotal = 48 + 24 = 72 torr
M1 = 31.75 g mol–1 26. (b) For an ideal solution, DHmixing = 0
DH = DH1 + DH2 + DH3 (Accroding to Hess's
W1 ´ 1000 law)
DTb = m × Kb = × Kb
M 1 ´ W2
for ideal solutions there is no change in magnitude
0.52 ´ 6.5 ´1000 of the attractive forces in the two components
DTb = = 1.06 °C
31.75 ´100 present.
\ boiling point of solution 27. (c) These two components A and B follows
the condition of Raoult’s law if the force of
= 100 °C + 1.06 °C = 101 °C
attraction between A and B is equal to the force
20. (a) For an ideal solution DSmix > 0 of attraction between A and A or B and B.
28. (a) The beans are cooked earlier in pressure
Entropy will always increase upon mixing two
compounds as disorder increases.
cooker because boiling point increases with
increasing pressure.
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29. (b) According to Raoult's law 33. (b) According to Raoult's law, the relative
p° - p n lowering in vapour pressure of a dilute solution
= (mole fraction of solute) is equal to the mole fraction of the solute present
p° n+ N
in the solution.
10
= 0.2 \ p° = 50 mm p° - p n
p° = mole fraction of solute =
p° n+ N
For other solution of same solvent 34. (c) For ideal solution,
20 n DVmixing = 0 and DH mixing = 0 .
= (mole fraction of solute)
p° n + N
35. (d) Ethyl alcohol forms hydrogen bonding
20
Þ = mole fraction of solute with itself, hence it will not form ideal solution
50
with C2H5I.
Þ Mole fraction of solute = 0.4
Hence, mole fraction of solvent = 1 – 0.4 = 0.6 36. (b) DTf = K f m = 5.12 × 0.078 = 0.399 K = 0.40 K
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Solutions 161
50 K + + Cl-
KCl
volatile solute p2 = × .0821 × T
M
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162 CHEMISTRY
C6H12O6 and C12H22O11 are not ionised so they 58. (a) As both the solutions are isotonic hence
have single particle. there will be no net movement of the solvent
through the semipermeable membrane between
Hence, Al 2 (SO4 )3 have maximum value of
two solutions.
depression in F.P. or lowest F.P.
59. (a) DTf = i × Kf × m
55. (a) Conc. of particles in CaCl2 sol. will be max. Van't Hoff factor, i = 2 for NaCl,
as i = 3 is max. So it exhibits highest osmotic hence DTf = 0.02 Kf which is maximum in the
pressure.
present case.
Glucose and urea do not dissociate into ions, as
Hence DTf is maximum or freezing point is
they are non-electrolytes. minimum.
60. (b) Blood cells neither swell nor shrink in
56. (a) K 4 [Fe(CN) 6 ] and Al 2 (SO 4 )3 both isotonic solution. As isotonic solution have
dissociates to give 5 ions or i = 5 equal concentration therefore, there is no flow
4 K + + [ Fe(CN ) ]- of solvent.
K 4 [Fe(CN )6 ] 6
and Blood cell shrink when placed in hypertonic
2Al3+ + 3SO 42-
Al 2 (SO 4 )3 solution and swells when placed in hypotonic
solution.
57. (c) Osmotic pressure, elevation in boiling point,
lowering of vapour pressure and depression in
freezing point are colligative properties.
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Electrochemistry 163
17 Electrochemistry
Trend Analysis with Important Topics & Sub-Topics
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164 CHEMISTRY
6. Molar conductivities (L°m) at infinite dilution of (c) infinite dilution, each ion makes definite
NaCl, HCl and CH3COONa are 126.4, 425.9 and contribution to conductance of an
91.0 S cm2 mol–1 respectively. L°m for CH3COOH electrolyte whatever be the nature of the
will be : [2012 M] other ion of the electrolyte.
(a) 425.5 S cm2 mol–1 (b) 180.5 S cm2 mol–1 (d) infinite dilution, each ion makes definite
(c) 290.8 S cm2 mol–1 (d) 390.5 S cm2 mol–1 contriubtion to equivalent conductance of
7. An increase in equivalent conductance of a an electrolyte, whatever be the nature of
strong electrolyte with dilution is mainly due to: the other ion of the electrolyte.
[2010] 11. The ionic conductance of Ba 2+ and Cl – are
(a) increase in ionic mobility of ions respectively 127 and 76 ohm–1 cm2 at infinite
(b) 100% ionisation of electrolyte at normal dilution. The equivalent conductance
dilution (in ohm–1 cm2) of BaCl2 at infinite dilution will be :
(c) increase in both i.e. number of ions and (a) 139.5 (b) 203 [2000]
ionic mobility of ions (c) 279 (d) 101.5
(d) increase in number of ions 12. Specific conductance of a 0.1 N KCl solution at
23ºC is 0.012 ohm–1 cm–1. Resistance of cell
8. Which of the following expressions correctly
containing the solution at same temperature was
represents the equivalent conductance at infinite
found to be 55 ohm. The cell constant is [2000]
dilution of Al2 (SO4)3,? Given that L °Al3+ and
(a) 0.918 cm–1 (b) 0.66 cm–1
(c) 1.142 cm –1 (d) 1.12 cm–1
L°SO2 - are the equivalent conductances at
4 13. Equivalent conductances of NaCl, HCl and
infinite dilution of the respective ions. [2010] CH3COONa at infinite dilution are 126.45, 426.16
1 ° 1 and 91 ohm–1 cm2 respectively. The equivalent
(a) L
3 Al3+
+ L°
2 SO 24- conductance of CH3COOH at infinite dilution
would be [1997]
(b) 2L ° + 3L° (a) 101.38 ohm–1 cm2 (b) 253.62 ohm–1 cm2
Al3+ SO 24- (c) 390.71 ohm–1 cm2 (d) 678.90 ohm–1 cm2
14. If 0.01 M solution of an electrolyte has a
(c) L° + L°
Al3+ SO24 - resistance of 40 ohms in a cell having a cell
constant of 0.4 cm–1, then its molar conductance
æL ° + L ° 2- ö ´ 6
(d) è Al3+ SO4 ø in ohm–1 cm2 mol–1 is [1997]
(a) 102 (b) 104
M (c) 10 (d) 103
9. The equivalent conductance of solution of
32 15. On heating one end of a piece of a metal, the
a weak monobasic acid is 8.0 mho cm2 and at other end becomes hot because of [1995]
infinite dilution is 400 mho cm2. The dissociation (a) resistance of the metal
constant of this acid is: [2009] (b) mobility of atoms in the metal
(a) 1.25 × 10–6 (b) 6.25 × 10–4 (c) energised electrons moving to the other end
(c) 1.25 × 10–4 (d) 1.25 × 10–5 (d) minor perturbation in the energy of atoms
10. Kohlrausch’s law states that at : [2008] 16. If 0.5 A current is passed through acidified silver
(a) finite dilution, each ion makes definite nitrate solution for 100 minutes. The mass of
contribution to equivalent conductance of silver deposited on cathode, is (eq.wt.of silver
an electrolyte, whatever be the nature of nitrate = 108) [1995]
the other ion of the electrolyte. (a) 2.3523 g (b) 3.3575 g
(b) infinite dilution each ion makes definite (c) 5.3578 g (d) 6.3575 g
contribution to equivalent conductance of 17. Which of the following is an insulator ?[1992]
an electrolyte depending on the nature of (a) Graphite (b) Aluminium
the other ion of the electrolyte. (c) Diamond (d) Silicon.
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Electrochemistry 165
volume of hydrogen produced at STP from (c) 1.0 × l010 (d) 1.0 × l030
H+ ions in solution by the same quantity of 30. For the cell reaction [2019]
electric charge will be [2005] 2Fe3+(aq) + 2I– (aq) ® 2Fe2+ (aq) + I2(aq)
(a) 44.8 L (b) 22.4 L 0.24 V at 298 K. The standard Gibbs energy (D,
(c) 11.2 L (d) 5.6 L G°) of the cell reaction is:
25. In electrolysis of NaCl, when Pt electrode is
[Given that Faraday constant F = 96500 C mol–1]
taken, then H2 is liberated at cathode while with
Hg cathode, it forms sodium amalgam. This is (a) –46.32 kJ mol–1 (b) –23.16 kJ mol–1
because [2002] (c) 46.32 kJ mol–1 (d) 23.16 kJ mol–1
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166 CHEMISTRY
31. In the electrochemical cell :- [2003, 2017] (a) 0.500 V (b) 0.325 V
Zn | ZnSO4 (0.01M) | | CuSO4 (1.0 M) | Cu, the emf (c) 0.650 V (d) 0.150 V
of this Daniel cell is E1. When the concentration 38. Standard electrode potential for Sn4+ / Sn2+ couple
of ZnSO4 is changed to 1.0M and that of CuSO4 is + 0.15 V and that for the Cr3+ / Cr couple is
changed to 0.01M, the emf changes to E2. From – 0.74 V. These two couples in their standard state
the followings, which one is the relationship are connected to make a cell. The cell potential
RT will be : [2011]
between E1 and E2? (Given, = 0.059) (a) + 1.19 V (b) + 0.89 V
F
(a) E1 < E2 (b) E1 > E2 (c) + 0.18 V (d) + 1.83 V
(c) E2 = 0 ¹ E1 (d) E1 = E2 39. If the E°cell for a given reaction has a negative
32. The pressure of H2 required to make the potential value, then which of the following gives the
of H2-electrode zero in pure water at 298 K is : correct relationships for the values of DG° and
[2016] Keq ? [2011]
–14 –12 (a) DG° > 0 ; Keq > 1 (b) DG° < 0 ; Keq > 1
(a) 10 atm (b) 10 atm
(c) DG° < 0 ; Keq < 1 (d) DG° > 0 ; Keq < 1
(c) 10–10 atm (d) 10–4 atm
40. A solution contains Fe2+, Fe3+ and I– ions. This
33. A device that converts energy of combustion of
solution was treated with iodine at 35°C. E° for
fuels like hydrogen and methane, directly into
Fe3+ / Fe2+ is + 0.77 V and E° for I2/2I– = 0.536 V.
electrical energy is known as : [2015]
The favourable redox reaction is : [2011 M]
(a) Electrolytic cell (b) Dynamo
(a) I2 will be reduced to I–
(c) Ni-Cd cell (d) Fuel Cell (b) There will be no redox reaction
34. A hydrogen gas electrode is made by dipping (c) I– will be oxidised to I2
platinum wire in a solution of HCl of pH = 10 and (d) Fe2+ will be oxidised to Fe3+
by passing hydrogen gas around the platinum 41. For the reduction of silver ions with copper metal,
wire at one atm pressure. The oxidation potential the standard cell potential was found to be
of electrode would be ? [NEET 2013] + 0.46 V at 25°C. The value of standard
(a) 0.59 V (b) 0.118 V Gibbs energy, DG0 will be (F = 96500 C mol–1)
(c) 1.18 V (d) 0.059 V (a) – 89.0 kJ (b) – 89.0 J [2010]
35. Consider the half-cell reduction reaction : (c) – 44.5 kJ (d) – 98.0 kJ
Mn2+ + 2e– ® Mn, E° = –1.18 V 42. Consider the following relations for emf of a
Mn2+ ® Mn3+ + e–, E° = –1.51 V electrochemical cell: [2010]
The E° for the reaction 3Mn 2+ ® Mn0 + 2Mn3+, (i) emf of cell = (Oxidation potential of anode)
and possibility of the forward reaction are, – (Reduction potential of cathode)
respectively [NEET Kar. 2013] (ii) emf of cell = (Oxidation potential of anode)
+ (Reduction potential of cathode)
(a) – 2.69 V and no (b) – 4.18 V and yes (iii) emf of cell = (Reduction potential of anode)
(c) + 0.33 V and yes (d) + 2.69 V and no + (Reduction potential of cathode)
36. Standard electrode potential of three metals X, Y (iv) emf of cell = (Oxidation potential of anode)
and Z are – 1.2 V, + 0.5 V and – 3.0 V, respectively. – (Oxidation potential of cathode)
The reducing power of these metals will be : Which of the above relations are correct?
[2011] (a) (ii) and (iv) (b) (iii) and (i)
(a) Y > Z > X (b) X > Y > Z (c) (i) and (ii) (d) (iii) and (iv)
43. Given: [2009]
(c) Z > X > Y (d) X > Y > Z
(i) Cu2+ + 2e– ® Cu, Eo = 0.337 V
37. The electrode potentials for [2011] (ii) Cu2+ + e– ® Cu+, Eo = 0.153 V
Cu2+(aq) + e– ¾¾ ® Cu+(aq) Electrode potential, Eo for the reaction,
and Cu+(aq) + e– ¾¾ ® Cu(s) Cu + + e– ® Cu, will be :
are + 0.15 V and + 0.50V, respectively. The value (a) 0.90 V (b) 0.30 V
of E°Cu 2+ /Cu will be : (c) 0.38 V (d) 0.52 V
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Electrochemistry 167
44. On the basis of the following E° values, the 51. Cu(aq) is unstable in solution and undergoes
strongest oxidizing agent is : [2008] simultaneous oxidation and reduction according
[Fe(CN)6]4– ® [Fe(CN)6]3– + e–; E° = – 0.35 V to the reaction : [2000]
Fe2+ ® Fe3+ + e– E° = – 0.77 V 2Cu + (aq)
Cu 2+ (aq) + Cu( s )
(a) [Fe(CN)6]4– (b) Fe2+ choose correct Eº for above reaction if Eº
(c) Fe3+ (d) [Fe(CN)6]3– Cu2+/Cu = 0.34 V and Eº Cu2+/Cu+ = 0.15 V
45. The equilibrium constant of the reaction: (a) –0.38 V (b) +0.49 V
Cu(s) + 2Ag + (aq) ¾¾ ® Cu 2+ (aq) + 2Ag(s) ; (c) +0.38 V (d) –0.19 V
52. What is the Eºcell for the reaction
E° = 0.46 V at 298 K is [2007]
(a) 2.0 × 1010 (b) 4.0 × 1010 Cu 2+ (aq) + Sn 2+ (aq) +
Cu(s) + Sn 4+ (aq)
(c) 4.0 × 1015 (d) 2.4 × 1010 at 25ºC if the equilibrium constant for the reaction
is 1 × 106? [1999]
46. If E ° 2+ = - 0.441V and E ° 3+ 2+
Fe /Fe Fe /Fe (a) 0.5328 V (b) 0.3552 V
= + 0.771V, the standard EMF of the reaction
(c) 0.1773 V (d) 0.7104 V
Fe + 2Fe3+ ® 3Fe2+ will be [2006]
(a) 1.653 V (b) 1.212 V 53. For the cell reaction, [1998]
(c) 0.111 V (d) 0.330 V Cu2+ (C1, aq) + Zn(s) = Zn 2+ (C2, aq) + Cu(s) of
47. A hypothetical electrochemical cell is shown below an electrochemical cell, the change in free energy,
DG, at a given temperature is a function of
A|A + (xM)||B+ (yM)|B [2006] (a) ln (C1) (b) ln (C2/C1)
The emf measured is +0.20 V. The cell reaction is (c) ln (C2) (d) ln (C1 + C2)
(a) A+ + e–® A; B++ e– ® B 54. Without losing its concentration ZnCl2 solution
(b) The cell reaction cannot be predicted cannot be kept in contact with [1998]
(c) A + B+ ® A+ + B (a) Au (b) Al
(d) A+ + B ® A + B+ (c) Pb (d) Ag
48. The standard e.m.f. of a galvanic cell involving
55. Eº for the cell, Zn Zn 2+ (aq) Cu 2+ (aq) Cu is
cell reaction with n = 2 is found to be 0.295 V
1.10 V at 25ºC. The equilibrium constant for the
at 25°C. The equilibrium constant of the
cell reaction:
reaction would be 2+ 2+
(Given F = 96500 C mol–1; R = 8.314JK–1mol–1) Zn + Cu (aq) Cu + Zn (aq) ,
[2004] is of the order of [1997]
(a) 2.0 × 1011 (b) 4.0 × 1012 (a) 10–18 (b) 10–37
(c) 1.0 × 102 (d) 1.0 × 1010 (c) 1018 (d) 1037
49. On the basis of the information available from 56. Standard potentials (Eº) for some half-reactions
the reaction are given below : [1997]
4 2 4 + - 2 +
(1) Sn + 2e ® Sn ; E º = +0.15 V
Al + O 2 ® Al 2 O 3 , DG = – 827 kJ mol–1 of
3 3 (2) 2 Hg2 + + 2e - ® Hg 2 2+ ; E º = +0.92 V
O2. The minimum e.m.f required to carry out an
electrolysis of Al2O3 is (F = 96500 C mol–1) (3) PbO 2 + 4H + + 2e - ® Pb 2+ + 2H 2O ;
(a) 8.56 V (b) 2.14 V [2003] E° + 1.45 V
(c) 4.28 V (d) 6.42 V Based on the above, which one of the following
50. Which reaction is not feasible? [2002] statements is correct ?
(a) 2KI + Br2 ® 2KBr + I2 (a) Sn4+ is a stronger oxidising agent than Pb4+
(b) 2KBr + I 2 ® 2KI + Br2 (b) Sn2+ is a stronger reducing agent than
(c) 2KBr + Cl 2 ® 2KCl + Br2 Hg22+
(c) Hg2+ is a stronger oxidising agent than Pb4+
(d) 2H 2O + 2F2 ® 4HF + O 2
(d) Pb2+ is a stronger reducing agent than Sn2+
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168 CHEMISTRY
57. The correct relationship between free energy and The cell potential will be : [NEET 2013]
equilibrium constant K of a reaction is [1996] (a) 0.42 V (b) 0.84 V
(a) DG° = - RT ln K (b) DG = RT ln K (c) 1.34 V (d) 1.10 V
(c) DG° = RT ln K (d) DG = - RT ln K 61. Standard free energies of formation (in kJ/mol)
58. An electrochemical cell is set up as: Pt; H2 at 298 K are – 237.2, – 394.4 and – 8.2 for H2O(l),
(1atm) | HCl(0.1 M) || CH3COOH (0.1 M)| H2 CO2(g) and pentane (g), respectively. The value
(1atm); Pt. The e.m.f of this cell will not be zero, of E°cell for the pentane-oxygen fuel cell is :
because [1995] (a) 1.968 V (b) 2.0968 V [2008]
(a) the temperature is constant (c) 1.0968 V (d) 0.0968 V
(b) e.m.f depends on molarities of acids used 62. The efficiency of a fuel cell is given by [2007]
(c) acids used in two compartments are different DG DG
(a) (b)
(d) pH of 0.1 M HCl and 0.1 M CH3COOH is DS DH
not same DS DH
(c) (d)
59. The standard reduction potentials at 25°C of DG DG
Li+ / Li, Ba2+ / Ba, Na+ / Na and Mg2+ /Mg are 63. In the silver plating of copper, K[Ag(CN)2] is
– 3.03, – 2.73, – 2.71 and – 2.37 respectively. used instead of AgNO3. The reason is [2002]
Which one of the following is the strongest (a) A thin layer of Ag is formed on Cu
oxidising agent? [1994] (b) More voltage is required
(a) Na+ (b) Li+ (c) Ag+ ions are completely removed from
solution
(c) Ba2+ (d) Mg2+
(d) Less availability of Ag+ ions, as Cu cannot
Topic 4: Commercial Cells and Corrosion displace Ag from [Ag(CN)2]– ion
60. A button cell used in watches functions as 64. The most convenient method to protect the
following bottom of ship made of iron is [2001]
Zn(s) + Ag2O(s) + H2O(l) (a) Coating it with red lead oxide
(b) White tin plating
2Ag(s) + Zn2+(aq) + 2OH–(aq)
(c) Connecting it with Mg block
If half cell potentials are : (d) Connecting it with Pb block
Zn2+(aq) + 2e– ® Zn(s); Eo = – 0.76 V 65. The most durable metal plating on iron to protect
Ag2O(s) + H2O (l) + 2e– ® 2Ag(s) + against corrosion is [1994]
2OH–(aq); Eo = 0.34 V (a) nickel plating (b) copper plating
(c) tin plating (d) zinc plating.
ANSWER KEY
1 (a) 8 (c) 15 (c) 22 (c) 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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170 CHEMISTRY
17. (c) Diamond is an insulator, because of no free
At infinite dilution, if molar conductances of cation, electrons.
anion and electrolyte are l¥ ¥
+ m , l - m and
l ¥m 18. (a) During the electrolysis of dil. sulphuric acid
respectively for the electrolyte Ax B y, th en using Pt electrodes following reaction occurs.
xAy+ + yBx–
AxBy At cathode : 4H+ (aq) + 4e– ¾® 2H2 (g)
¥ 1 ¥ 1 At anode : 2H2O (l) ¾® O2 (g) + 4H+ (aq) + 4e–
l eq = l+ m + l¥ - m and
y x 19. (d) 1 equivalent of any substance is deposited
l¥ ¥ ¥
m = xl + m + yl - m by 1 F of charge.
12. (b) Given specific conductance of the solution 20 g calcium contains,
(k) = 0.012 ohm–1 cm–1 and resistance (R) = 55 Given mass
ohm. We know that cell constant = Specific Number of equivalents =
Equivalent mass
conductance × Observed resistance = 0.012 × (Equivalent mass of Ca
55 = 0.66 cm–1.
Atomic mass 40
13. (c) By Kohlraush's law, Ù°eq NaCl = 126.45 = = = 20 )
Valency 2
l° + l° = 126.45 ....(1) 20
Na + Cl - = =1
l° + + l° = 426.16 20
H Cl -
...(2)
So, 1 Faraday of charge is required to deposit 1
l° + l° = 91 ....(3) equivalent of Ca.
CH3OO – Na +
( +6) ( +7)
on adding (2) and (3) then subtract (1) from it 20. (c) Mn O 24 - ¾® MnO4- + e -
l° °
- + l + = 517.16 - 126.45 0.1 mole
CH3COO H
Quantity of electricity required = 0.1F
Ù°(CH3COOH) = 390.71ohm -1 cm 2 = 0.1 × 96500 = 9650 C
14. (d) Given molarity = 0.01 M 21. (d) wO = nO × 32
2 2
Resistance = 40 ohm; 5600
l wO = ´ 32 = 8g = 1 equivalent of O2
-1 2 22400
Cell constant = 0.4cm
A = 1 equivalent of Ag = 108
Specific conductivity ( k) 22. (c) Applying,
cell constant 0.4 Eit
= = = 0.01 ohm -1 cm -1 w = Zit =
resistance 40 96500
1000k Equivalent weight of cobalt (II) = 59/2
Molar conductance ( Ù m ) =
Molarity I = 10 A
1000 ´ .01 Time (t) = 109 min = 109 × 60 sec
= = 103 ohm -1cm 2 mol -1
.01 Substituting these values we get,
15. (c) When one end of a metal is heated, the 59 ´ 10 ´ 109 ´ 60
free electrons are energised and move to the w= = 20.0
2 ´ 96500
other end. It heats up the other end of the metal.
23. (a) As Q = i × t
16. (b) Given current (i) = 0.5 A;
Time (t) = 100 minutes × 60 = 6000 sec \ Q = 4.0 × 104 × 6 × 60 × 60 C
Equivalent weight of silver nitrate (E) = 108. = 8.64 × 10 8 C
According to Faraday's first law of electrolysis Now since 96500 C liberates 9 g of Al
Eit 108 ´ 0.5 ´ 6000 8.64 × 108 C liberates 9
´ 8.64 ´ 108 g Al
W= = = 3.3575 g.
96500 96500 96500
= 8.1 × 104 g of Al
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Electrochemistry 171
24. (d) No. of gram equivalent of H+ = No. of eq. of 2SO42–(aq) ¾¾
® S2O82–(aq) + 2e–, E° = +1.96 V
4.5 For dilute sulphuric acid, first reaction is
Al3+ = = 0.5
9 preferred but for concentrated acid, second
27 reaction is preferred.
Eq. wt of Al3+ = = 9
3 ° =
2.303RT
29. (c) E cell log K
No. of gm eq. of H+ = no. of mole of H+ nF
Hence, Mass of H+ = 0.5 × 1g = 0.5 g ° = 0.59 V , n = 1
Given : E cell
We know that, 2g H2 at STP = 22.4 L
0.059
22.4 0.59 = log K
\ 0.5 g H2 at STP = ´ 0.5 = 5.6 L 1
2 0.59
= log K
25. (b) If mercury is used as cathode, H+ ions are 0.059
not discharged at mercury cathode because 10 = log k
mercury has a high over voltage for hydrogen. K = 1010
30. (a) DG = –nFE°
In electrolysis of NaCl, following ionisations take = –2 × 96500 × 0.24 = – 46320 J/mol
place: = – 46.32 kJ/mol
NaCl Na+ + Cl– 31. (b) For cell,
H2O H+ + OH– Zn|ZnSO4(0.01 M) || CuSO4(1.0 M)|Cu
Na+ and H+ ions move towards cathode.
However, only H+ ions are discharged more readily 2.303RT log [ Zn 2+ ]
than Na+ ions because in electrochemical series, Ecell = E °cell -
hydrogen is lower than sodium.
nF [Cu 2+ ]
At cathode : 2H+ + 2e– ¾® H2(g) 2.303RT ( 0.01)
If mercury is used as cathode, then Na+ ions are \ E1 = E °cell - ´ log
2´ F 1
discharged at cathode in preference to H+ ions,
yielding sodium, which dissolves in mercury to
When concentrations are changed for ZnSO4
form sodium amalgam. and CuSO4, we can write
At cathode : Na+ + e– ¾® Na(s) 2.303RT 1
E2 = E °cell - ´ log
Na(s) + Hg(s) ¾® Na – Hg(s) 2F 0.01
26. (b) The reduction potential is higher in case of \ E1 > E2
E°(Fe3+/Fe2+) = +0.77V. Thus, Fe3+ will readily 32. (a) 2H+(aq) + 2e– ® H2(g)
reduce to Fe2+ and quantity of Fe3+ will decrease. P
\ E = E0 – 0.0591 log H 2
W q 2 [ H + ]2
27. (a) By Faraday's Ist Law, =
E 96500 PH 2
(where q = it = total electric charge) 0 = 0 – 0.0295 log
we know that no of equivalent (10 -7 )2
PH 2
W it 1 ´ 965 1 =1
= = = = (10-7 )2
E 96500 96500 100
(where i = 1 A, t = 16 × 60 + 5 = 965 sec.) PH 2 = 10–14 atm
Since, we know that 33. (d) A device that converts energy of
1 combustion of fuels, directly into electrical
no. of equivalent 100 energy is known as fuel cell.
Normality = = = 0.01 N 34. (a) H2 ¾¾ ® 2H+ + 2e–
Volume (in litre) 1
1 atm 10-10
28. (b) During electrolysis of sulphuric acid, the
following processes are possible at cathode: E 0.059 (10-10 )2
+ =0– log
H 2 /H
2H2O(l) ¾¾ ® O2(g) + 4H+(aq) + 4e–, 2 1
E° = +1.23 V E + = +0.59 V
H 2 /H
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35. (a) DE° = E°red + E°oxd = –1.81 – 1.51 = –2.69 40. (c) Given Fe+3/Fe2+ = + 0.77 V
Since DE° is negative and I2 / 2I– = 0.536V
\ DG = –nFE°, DG will have positive value so, Reduction potential of Fe3+ / Fe2+ > I2 / I–, so
forward reaction is not possible. Fe3+ will be reduced and I– will be oxidised.
+
The value of electrode potential of a half-cell 41. (a) Cu + 2 Ag (aq) ® Cu 2+
¾¾ (aq) + 2 Ag(s)
reaction does not depend on stoichiometric co-
efficient. e.g. Here, n = 2 , 0 = + 0.46 V
Ecell
Mn2+ ¾® Mn3+ + e– ; E° = –1.51V
2Mn2+ ¾® 2Mn3+ + 2e– ; E° = – 1.51V ΔGo = –nE°F = – 2×0.46×96500 kJ ; – 89 kJ
1000
36. (c) As th e value of standard reduction
potential decreases the reducing power increases 42. (a) Ecell = Ecathode – Eanode
i.e., Þ Ecell = Ered/cathode + Eoxi/anode
Z > X > Y
( -3.0) ( -1.2) ( +0.5) Þ Ecell = - Eoxi/cathode + Eoxi/anode
Hence, option (a) is correct.
37. (b) Cu 2+ + 1e - ® Cu +
43. (d) Cu2+ + 2e– ® Cu; DGo = – nEo F
E1o = 0.15V ; DG1o = -n1E1o F
= – 2 × F × 0.337 = – 0.674 F ....(i)
Cu + + 1e - ® Cu Cu+ ® Cu2+ + e– ; DGo = – nEo F
E2o = 0.50V ; DG2o = - n2 E2o F = – 1 × F × – 0.153 = 0.153 F ....(ii)
On adding eqn (i) & (ii)
Cu 2+ + 2e - ® Cu Cu+ + e– ® Cu ;
DG° = DG°1 + DG°2 DGo = – 0.521 F = – nE°F;
– nE° F = –1 n1 E1o F + (–1) n2 E2o F Here n = 1 \ E° = + 0.52 V
n1E1o + n2 E2o 0.15 ´ 1 + 0.50 ´ 1 44. (c) From the given data we fin d Fe 3+
E° = =
Þ 0.325 V
n 2 (E °
(Fe3+ /Fe 2 + ) )
= +0.77V is strongest oxidizing
agent.
When exchange of electrons (n) is different for all
three reactions, we can not use the formula for More the positive value of reduction potential,
standard electrode potential (Ecell). e.g. more is the tendency to get reduced (better
Cu2+ + e– ¾¾ ® Cu+ (n = 1) oxidising agent).
Cu+ + e– ¾¾ ® Cu (n = 1) 0.0591
o
Cu2+ + 2e– ¾¾ ® Cu (n = 2) 45. (c) As Ecell = log K c
n
E° ¹ E° + E° 0.0591
Cu 2+ /Cu Cu 2+ /Cu + Cu + /Cu
\ 0.46 = log K c
38. (b) The couple for which SRP value is high, 2
will act as cathode and the other couple will act 2 ´ 0.46
as an anode. \ log K c = = 15.57
0.0591
o o o
Ecell = Ecathode – Eanode or Kc = Antilog 15.57 = 3.7 × 1015 » 4 × 1015
= 0.15 V –(–0.74 V) = +0.89 V 46. (b)
39. (d) Standard Gibbs free energy is given as Fe ¾® Fe 2 + + 2e– [Anode] E = –0.441V
DG° = – nE° F [Fe 3+
+ e – ¾® Fe 2 + ] ´ 2 [Cathode] E = +0.771V
If E°cell < 0 i.e. – ve
DG° > 0 Fe + 2Fe 3+ ¾® 3Fe 2+
o o o
Further DG° = – RT ln Keq Ecell = Ecathode - EAnode
DG° > 0 when Keq < 1 = 0.771 – (–0.441) = 0.771 + 0.441 = 1.212 V
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Electrochemistry 173
47. (c) The cell reacton is as follows :
53. (b) DG = - nE ° F
+
A® A + e– RT C2
+ For concentration cell, E = ln
B + e® B nF C1
+ +
Adding A + B ® A + B In it R, T, n and F are constant
So E is based upon ln C2 / C1
0.0591
48. (d) E° = log10 K RT
n Now DG = -nEF = - nF ´ ln C2 / C1
Here, n = 2, E ° = 0.295 nF
= –RT ln C2/C1
2 ´ 0.295 At constant temperature DG is based upon
\ log10 K = = 10 or K = 1 × 1010
0.0591 lnC2/C1.
49. (b) DG = –nEF, Al ® Al 3+ + 3e– 54. (b) Without losing its concentration ZnCl2
For 1 mol of Al, n = 3 solution cannot be kept in contact with Al
4 4 because Al is more reactive than Zn due to high
\ for mol of Al, n = 3 ´ = 4
3 3 electrode (reduction) potential.
According to question,
55. (d) Given for the reaction
827 ´1000 = 4 ´ E ´ 96500
Zn(s) + Cu 2+ (aq)
Cu(s) + Zn 2+ (aq),
827 ´ 1000
E= = 2.14V Eº = +1.10 V.
4 ´ 96500
At equilibrium
50. (b) 2KBr + I 2 ¾¾ ® 2KI + Br2
° 0.0591
reaction is not possible because Br – ion is not \ Ecell = log10 K eq
n
oxidised to Br2 with I2 due to higher electrode here (n ® number of exchange of electrons)
(oxidation) potential of I2 than bromine.
0.0591 2.20
51. (c) 2 Cu + ¾¾
® Cu +2 + Cu or 1.10 = log10 Keq = log10 K eq
2 0.059
2 e - + Cu +2 ¾¾
® Cu ; E1º = 0.34V ; ...(i) = 37.22 or Keq = 1.66 × 1037
e - + Cu +2 ¾¾
® Cu + ; E2º = 0.15V ; ...(ii) 56. (b) Sn 2+ ¾¾
® Sn 4+ + 2e - ; E o = -0.15 V
Cu + + e - ¾¾
® Cu; E3º = ? ... (iii) Hg 22+ ¾¾
® 2 Hg 2+ + 2e – ; E o = -0.92 V
Now, DG1º = -nFE1º = -2 ´ 0.34 F Thus, Sn2+ can more easily get oxidized than
Hg22+ or Sn2+ is stronger reducing agent than
DG2º = -1 ´ 0.15F , DG3º = -1 ´ E3º F
Hg22+.
Again, DG1º = DG2º + DG3º 57. (a) The relation between free energy change
and equilibrium constant is given by Nernst
Þ -0.68F = -0.15F - E3º F
equation
Þ E3º = 0.68 - 0.15 = 0.53V RT
Ecell = E º - ln Q
º
Ecell º
= Ecathode (Cu + / Cu ) nF
At equilibrium, E cell = 0 and Q = KC
º
- Eanode (Cu +2 / Cu + )
RT
= 0.53 – 0.15 = 0.38 V. \ Eº = ln K … (i)
nF
52. ° = 2.303 RT log K = .0591 log K
(c) Ecell Again DGº = – nFEº … (ii)
eq eq
nF n
put in (i)
.0591
= log 106 = .0591 × 3 = 0.1773V -DG º RT
2 = ln K ; DG º = - RT ln K
nF nF
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174 CHEMISTRY
58. (d) For a concentration cell having different 3387000 3387000
0
concentrations of ions. or Ecell = =
32 ´ 96500 3088000
0.0591 C
E= log 1 3387
n C2 or V = 1.0968 V
If all the concentrations are identical then 3088
obviously the cell voltage is zero. But as the pH DG
62. (b) Efficiency of a fuel cell ( h) =
of 0.1 M HCl (strong acid) & pH of 0.1M DH
CH3COOH(weak acid) is not same, therefore
DH is the required energy (input) and DG is the
the cell voltage will not be zero. useful energy obtained (output).
59. (d) Higher the reduction potential, stronger is
the oxidising agent. 63. (d) In th e silver platin g of copper,
60. (d) E°Cell = E°OP + E°RP = 0.76 + 0.34 = 1.10 V K[Ag(CN) 2 ] is used in stead of AgNO 3 .
61. (c) Writing the equation for pentane-oxygen Copper being more electropositive readily
fuel cell at respective electrodes and over all precipitate silver from their salt solution
reaction, we get Cu + 2AgNO 3 ¾¾ ® Cu(NO3 ) 2 + Ag
At Anode:
whereas in K[Ag (CN)2] solution a complex anion
C 5 H12 + 10H 2 O ® 5CO 2 + 32H + + 32e - [Ag(CN)2]– is formed and hence Ag+ are less
(pentane)
available in the solution and therefore copper
At Cathode:
cannot displace Ag from its complex ion.
8O 2 + 32H + + 32e- ® 16H 2 O 64. (c) For bottom of ship to be protected, it is
Over all :C5 H12 + 8O 2 ® 5CO2 + 6H 2 O connected with more reactive metal than iron
like magnesium. This technique is called cathodic
Calculation of DG° for the above reaction
protection.
DG° = [5×(–394.4) + 6× (–237.2)] – [–8.2] 65. (d) This is because zinc has higher oxidation
= – 1972.0 – 1423.2 + 8.2 = – 3387.0 kJ potential than Ni, Cu and Sn.
= – 3387000 Joules.
From the overall equation we find n = 32 The process of coating of iron surface with zinc is
0 known as galvanization. Galvanized iron sheets
Using the relation, DG° = – nFEcell and maintain their lustrue due to the formation of
substituting various values, we get protective layer of basic zinc carbonate.
0
– 3387000 = –32×96500× Ecell (F = 96500C)
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Chemical Kinetics 175
18 Chemical Kinetics
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176 CHEMISTRY
(b) The half-life of a first-order reaction does (d) 1.25 × 10–2 mol L–1s–1 and 6.25 × 10–3
not depend on [A]0, the half-life of a mol L–1s–1
second-order reaction does depend on 9. The rate of the reaction
[A]0 2NO + Cl2 ¾¾ ® 2NOCl is given by the rate
(c) The rate of a first-order reaction does equation rate = k [NO]2 [Cl2] [2010]
depend on reactant concentrations, the The value of the rate constant can be increased by:
rate of a second-order reaction does not (a) increasing the concentration of NO.
depend on reactant concentrations (b) increasing the temperature.
(d) A first-order reaction can be catalyzed, a (c) increasing the concentration of the Cl 2
second-order reaction cannot be catalyzed (d) doing all of these
5. The rate constant of the reaction A ® B is 10. For the reaction, N2 + 3H2—® 2NH3, [2009]
0.6 × 10–3 mole per second. If the concentration
d [ NH3 ]
of A is 5 M then concentration of B after 20 = 2 × 10–4 mol L–1 s–1 , the value of
minutes is : [2015 RS] dt
(a) 1.08 M (b) 3.60 M – d [ H2 ]
would be :
(c) 0.36 M (d) 0.72 M dt
6. In a reaction, A + B ® Product, rate is doubled (a) 4 × 10–4 mol L–1 s–1
when the concentration of B is doubled, and (b) 6 × 10–4 mol L–1 s–1
rate increases by a factor of 8 when the (c) 1 × 10–4 mol L–1 s–1
concentrations of both the reactants (A and B) (d) 3 × 10–4 mol L–1 s–1
are doubled. Rate law for the reaction can be 11. In the reaction [2009]
written as : [2012]
2 BrO3– (aq) + 5Br - (aq) + 6H + ® 3Br2 (l)
(a) Rate = k [A][B] (b) Rate = k [A]2 [B]2
(c) Rate = k [A] [B] (d) Rate = k [A]2 [B] +3H2 O(l)
7. The rate of the reaction 2N2O5 ® 4NO2 + O2 can The rate of appearance of bromine (Br2) is related
be written in three ways : [2011 M] to rate of disappearance of bromide ions as
-d[N 2 O5 ] following:
= k [N 2 O5 ] d[Br2 ] 5 d[Br – ]
dt (a) = –
d[NO2 ] dt 3 dt
= k ¢ [N 2 O5 ] d[Br2 ] 5 d[Br – ]
dt (b) =
d[O2 ] dt 3 dt
= k" [N2O5 ] d[Br2 ] 3 d[Br – ]
dt (c) =
The relationship between k and k ' and between dt 5 dt
k and k ¢¢ are : d [Br2 ] 3 d[Br – ]
(d) = -
(a) k ' = 2k ; k ' = k (b) k ' = 2k ; k ¢¢ = k / 2 dt 5 dt
(c) k ' = 2k ; k ¢¢ = 2k (d) k ' = k ; k ¢¢ = k 12. For the reaction 2 A + B ® 3C + D
8. For the reaction [N 2O 5 (g) ® 2NO 2 (g) + which of the following does not express the
1/2 O2(g)] the value of rate of disappearance of reaction rate ? [2006]
N2O5 is given as 6.25 × 10–3 mol L–1s–1. The rate d[ B ] d[ D]
(a) - (b)
of formation of NO2 and O2 is given respectively dt dt
as : [2010] 1 d[A] 1 d[C]
(c) – (d) -
(a) 6.25 × 10–3 mol L–1s–1 and 6.25 × 10–3 2 dt 3 dt
mol L–1s–1 13. Consider the reaction [2006]
(b) 1.25 × 10–2 mol L–1s–1 and 3.125 × 10–3 N2 (g) + 3H2 (g) ® 2 NH3 (g)
mol L–1s–1 d[ NH 3 ]
(c) 6.25 × 10–3 mol L–1s–1 and 3.125 × 10–3 The equality relationship between and
dt
mol L–1s–1 d[ H 2 ]
- is
dt
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178 CHEMISTRY
27. Which one of the following statements for the 32. The bromination of acetone that occurs in acid
order of a reaction is incorrect ? [2011] solution is represented by this equation. [2008]
(a) Order can be determined only experimentally. CH3COCH3 (aq) + Br2 (aq) ®
(b) Order is not influenced by stoichiometric CH3COCH2Br (aq) + H+ (aq) + Br– (aq)
coefficient of the reactants. These kinetic data were obtained for given
(c) Order of reaction is sum of power to the reaction concentrations.
concentration terms of reactants to express Initial Concentrations, M
the rate of reaction.
[CH3 COCH3] [Br2] [H+]
(d) Order of reaction is always whole number.
0.30 0.05 0.05
28. The unit of rate constant for a zero order reaction
is [2011 M] 0.30 0.10 0.05
(a) mol L–1 s–1 (b) L mol–1 s–1 0.30 0.10 0.10
(c) L2 mol–2 s–1 (d) s–1 0.40 0.05 0.20
29. During the kinetic study of the reaction, Initial rate, disappearance of Br2, Ms–1
2A + B ® C + D, following results were obtained: 5.7×10–5
5.7 × 10–5
Run [A]/mol L–1 [B]/mol L–1 Initial rate of
1.2 × 10–4
formation of
–1 –1 3.1 × 10–4
D/mol L min
–3
Base on these data, the rate equations is:
I 0.1 0.1 6.0 × 10 (a) Rate = k [CH3COCH3][H+]
–2
II 0.3 0.2 7.2 × 10 (b) Rate = k [CH = COCH3][Br2]
–1
III 0.3 0.4 2.88 × 10 (c) Rate = k [CH3 COCH3] [Br2] [H+]2
IV 0.4 0.1 2.40 × 10
–2
(d) Rate = k [CH3COCH3][Br2] [H+]
33. The reaction of hydrogen and iodine
Based on the above data which one of the
monochloride is given as: [2007]
following is correct? [2010]
2
H 2 (g) + 2ICl(g) ¾¾ ® 2HCl(g) + I2 (g)
(a) rate = k [A] [B] (b) rate = k[A] [B] The reaction is of first order with respect to H2(g)
and ICI(g), following mechanisms were proposed.
(c) rate = k [A]2 [B]2 (d) rate = k [A] [B]2
Mechanism A:
30. Half life period of a first-order reaction is 1386
H 2 (g) + 2ICl(g) ¾¾
® 2HCl(g) + I 2 (g)
seconds. The specific rate constant of the
Mechanism B:
reaction is: [2009]
–2 –1 H 2 (g) + ICl(g) ¾¾
® HI(g);slow
(a) 0.5 × 10 s (b) 0.5 × 10 s–1
–3
(c) 5.0 × 10 s–2 –1 (d) 5.0 × 10–3 s–1 HI(g) + ICl(g) ¾¾ ® HCl(g) + I 2 (g);fast
31. For the reaction A + B ¾¾ ® products, it is Which of the above mechanism(s) can be
consistent with the given information about the
observed that: [2009]
reaction?
(1) On doubling the initial concentration of A
(a) A and B both (b) neither A nor B
only, the rate of reaction is also doubled
(c) A only (d) B only
and
34. In a first-order reaction A ® B, if k is rate
(2) On doubling the initial concentrations of
constant and inital concentration of the reactant
both A and B, there is a change by a factor A is 0.5 M, then the half-life is [2007]
of 8 in the rate of the reaction.
log 2 log 2
The rate of this reaction is given by: (a) (b)
k k 0.5
(a) rate = k [A] [B]2 (b) rate = k [A]2 [B]2
ln 2 0.693
(c) rate = k [A] [B] (d) rate = k [A]2 [B] (c) (d)
k 0.5k
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Chemical Kinetics 179
35. If 60% of a first order reaction was completed in 43. The plot of concentration of the reactant vs.
60 minutes, 50% of the same reaction would be time for a reaction is a straight line with a negative
completed in aproximately [2007] slope. The reaction follows a [1996]
(a) 45 minutes (b) 60 minutes (a) zero order rate equation
(c) 40 minutes (d) 50 minutes (b) first order rate equation
(log 4 = 0.60, log 5 = 0.69) (c) second order rate equation
36. For a first order reaction A ¾® B the reaction (d) third order rate equation
rate at reactant concentration of 0.01 M is 44. A substance 'A' decomposes by a first order
found to be 2.0 ´ 10-5 mol L-1 s-1. The half reaction starting initially with [A] = 2.00 m and
life period of the reaction is [2005] after 200 min, [A] becomes 0.15 m. For this
(a) 30 s (b) 220 s reaction t1/2 is [1995]
(c) 300 s (d) 347 s (a) 53.72 min (b) 50.49 min
37. The rate of reaction between two reactants A (c) 48.45 min (d) 46.45 min
and B decreases by a factor of 4 if the 45. Select the rate law that corresponds to data
concentration of reactant B is doubled. The shown for the following reaction [1994]
order of this reaction with respect to reactant A+B ¾ ¾® products.
B is: [2005] Exp. [A] [B] Initial rate
(a) 2 (b) -2 1 0.012 0.035 0.1
(c) 1 (d) -1 2 0.024 0.070 0.8
38. The rate of a first order reaction is 1.5 × 10–2 3 0.024 0.035 0.1
mol L–1 min–1 at 0.5 M concentration of the 4 0.012 0.070 0.8
reactant. The half life of the reaction is [2004]
(a) rate = k [B]3 (b) rate = k [B]4
(a) 0.383 min (b) 23.1 min
(c) 8.73 min (d) 7.53 min (c) rate = k [A] [B]3 (d) rate = k [A]2 [B]2
39. If the rate of the reaction is equal to the rate Topic 3: Theories of Rate of Reaction
constant, the order of the reaction is [2003]
(a) 3 (b) 0 46. An increase in the concentration of the
(c) 1 (d) 2 reactants of a reaction leads to change in
40. The reaction A ® B follows first order kinetics. (a) heat of reaction [2020]
The time taken for 0.8 mole of A to produce 0.6 (b) threshold energy
mole of B is 1 hour. What is the time taken for (c) collision frequency
conversion of 0.9 mole of A to produce 0.675 (d) activation energy
mole of B? [2003] 47. For a reaction, activation energy Ea = 0 and the
(a) 2 hours (b) 1 hour rate constant at 200 K is 1.6 × 106 s–1. The rate
(c) 0.5 hour (d) 0.25 hour constant at 400 K will be [NEET Odisha 2019]
41. 3A ® B + C, it would be a zero order reaction [Given that gas constant, R = 8.314 J K–1 mol–1]
when [2002] (a) 3.2 × 106 s–1 (b) 3.2 × 104 s–1
(a) the rate of reaction is proportional to (c) 1.6 × 10 s 6 –1 (d) 1.6 × 103 s–1
square of concentration of A 48. The addition of a catalyst during a chemical
(b) the rate of reaction remains same at any reaction alters which of the following quantities?
concentration of A [2016]
(c) the rate remains unchanged at any (a) Entropy (b) Internal energy
concentration of B and C (c) Enthalpy (d) Activation energy
(d) the rate of reaction doubles if concentration 49. The activation energy of a reaction can be
of B is increased to double determined from the slope of which of the
42. Half life of a first order reaction is 4 s and the following graphs ? [2015]
initial concentration of the reactants is 0.12 M. ln K l
The concentration of the reactant left after 16 s (a) vs.T (b) ln K vs.
is [1999] T T
(a) 0.0075 M (b) 0.06 M (c) T l (d) ln K vs. T
(c) 0.03 M (d) 0.015 M vs.
ln K T
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180 CHEMISTRY
50. What is the activation energy for a reaction if its 1 1
(c) k vs (d) log k vs
rate doubles when the temperature is raised from log T T
20°C to 35°C? (R = 8.314 J mol–1 K–1) 56. The activation energy for a simple chemical
[NEET 2013] reaction A ® B is Ea in forward direction. The
(a) 269 kJ mol–1 (b) 34.7 kJ mol–1 activation energy for reverse reaction [2003]
(c) 15.1 kJ mol–1 (d) 342 kJ mol–1
(a) Is always double of Ea
51. A reaction having equal energies of activation
(b) Is negative of Ea
for forward and reverse reaction has :
(c) Is always less than Ea
[NEET 2013]
(a) DG = 0 (b) DH = 0 (d) Can be less than or more than Ea
(c) DH = DG = DS = 0 (d) DS = 0 57. When a biochemical reaction is carried out in
52. Activation energy (E a ) and rate constants laboratory in the absence of enzyme then rate of
(k1 and k2) of a chemical reaction at two different reaction obtained is 10–6 times, then activation
temperatures (T1 and T2) are related by : energy of reaction in the Presence of enzyme is
[2012 M] 6
(a) [2001]
k E æ1 1ö RT
(a) ln 2 = – a ç - ÷ (b) Different from Ea obtained in laboratory
k1 R è T1 T2 ø
(c) P is required
k2 E æ 1 1ö (d) Can't say anything
(b) ln =- aç - ÷
k1 R è T2 T1 ø 58. Activation energy of a chemical reaction can be
determined by [1998]
k2 E æ 1 1ö (a) evaluating rate constant at standard
(c) ln =- aç + ÷
k1 R è T2 T1 ø temperature
k E æ1 1ö (b) evaluating velocities of reaction at two
(d) ln 2 = a ç - ÷ different temperatures
k1 R è T1 T2 ø
(c) evaluating rate constants at two different
53. For an endothermic reaction, energy of temperatures
activation is E a and enthalpy of reaction of (d) changing concentration of reactants
DH(both of these in kJ/mol). Minimum value of 59. In a reversible reaction the energy of activation
Ea will be. [2010] of the forward reaction is 50 kcal. The energy of
(a) less than DH (b) equal to DH activation for the reverse reaction will be
(c) more than DH (d) equal to zero (a) < 50 kcal [1996]
54. The rate constants k1 and k2 for two different (b) either greater than or less than 50 kcal
reactions are 1016 . e–2000/T and 1015 . e–1000/T, (c) 50 kcal
respectively. The temperature at which k1 = k2 is : (d) > 50 kcal
60. A chemical reaction is catalyzed by a catalyst X.
2000 [2008] Hence X [1995]
(a) 1000 K (b) K
2.303 (a) reduces enthalpy of the reaction
1000 (b) decreases rate constant of the reaction
(c) 2000 K (d) K (c) increases activation energy of the reaction
2.303
55. The temperature dependence of rate constant (d) does not affect equilibrium constant of the
(k) of a chemical reaction is written in terms of reaction
* 61. For an exothermic reaction, the energy of
Arrhenius equation, k = Ae- Ea / RT . Activation activation of the reactants is [1994]
energy ( Ea* ) of the reaction can be calculated (a) equal to the energy of activation of products
by plotting [2003] (b) less than the energy of activation of products
1 (c) greater than the energy of activation of
(a) log k vs (b) k vs T products
log T
(d) Sometimes greater and sometimes less than
that of the products
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Chemical Kinetics 181
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)
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182 CHEMISTRY
9. (b) 2 NO(g) + Cl2(g) 2 NOCl(g) where r = 1.02 × 10–4, k = 3.4 × 10–5
1.02 × 10–4 = 3.4 × 10–5 [N2O5]
Rate = k [NO]2 [Cl2] [N2O5] = 3M
The value of rate constant can be increased by 16. (d) Rate of reaction
increasing the temperature and is independent
of the initial concerntration of the reactants. 1 d[Br - ] 1 d[Br2 ]
– =+
10. (d) Rate of disappearance of H2 = rate of 5 dt 3 dt
formation of NH3. d[Br2 ] 3 d[Br – ]
=–
1 d[H 2 ] 1 d[NH3 ] dt 5 dt
– = 17. (d) The rate of a reaction is the speed at which
3 dt 2 dt the reactants are converted into products. It
– d[H 2 ] 3 d[NH3 ] 3 –4
depends upon the concentration of reactants.
Þ = = × 2×10 e.g for the reaction
dt 2 dt 2
A + B ¾¾
® Product ; r µ [A][B]
= 3×10 –4 mol L–1s –1
By decreasing volume of gas or increasing pressure
11. (d) Rate of disappearance of Br – of it, concentration will increases. Hence, (a) and
= rate of appearance of Br 2 (c) are also correct options.
1 d[Br – ] 1 d[Br2 ] 18. (a) For a first order reaction
Þ– =
5 dt 3 dt 0.693 0.693
Half life period, t 1 = = s–1
d[Br2 ] 3 d[Br – ] k 2.303 ´ 10 –3
Þ =- 2
dt 5 dt = 300.91 s
t1/2 t1/2
d[C] Now, 40 g ¾¾® 20 g ¾¾¾ ® 10 g
12. (d) In the given options, - will not
3dt So, 40 g substance requires 2 half life periods to
represent the reaction rate. It should not have reduce upto 10 g
\ Time taken in reduction = 2 × 300.91 s
1 d [ C]
–ve sign as it is product; since show the = 601.82 ; 602 s
3 dt
rate of formation of product C, which will be positive. [A]o
19. (b) (t1/2 )zero =
13. (a) If we write rate of reaction in terms of 2k
concentration of NH3 and H2, then \ If [A]o = doubled, t1/2 = doubled
1 d[NH3 ] 1 d[H 2 ] 20. (c) Overall rate = k [X][Y2] … (1)
Rate of reaction = =-
2 dt 3 dt k = rate constant
Assuming step (i) to be reversible, its equilibrium
d[NH3 ] 2 d[H 2 ]
So, =- constant,
dt 3 dt
[ X ]2
[ X2 ] Þ [ X ] = keq [ X 2 ] ;
14. (b) 3A ¾ ¾® 2 B keq = 2
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Chemical Kinetics 183
r æ 100 -10 ö
The overall reaction rate depends on the rate of 100°C çè
10 ø
÷
the slowest step. 26. (b) =2 = 29 = 512 times
r
10°C
i.e., Overall rate = Rate of slowest step
27. (d) order of reaction can be zero, whole number
21. (a) Half life for a first order reaction, or fractional.
0.693
t1/2 = The order w.r.t a reactant can be negative also as in
K this case:
0.693
So, t1/2 = sec . 2O3 3O2
10-2 Rate - k[O3]2 [O2]–1
Also, for the reduction of 20 g of reactant to 5 g,
two half lives will be required. 28. (a) Rate = k [A]0
\ For 20 g of the reactant to reduce to 5 g, Unit of k = mol L–1 sec–1
time taken, 29. (d) In case of (II) and (III), keepin g
0.693 concentration of [A] constant, when the
t=2× sec = 138.6 sec. concentration of [B] is doubled, the rate
10-2
quadruples. Hence, it is second order with
n
æ1ö respect to B. In case of I & IV, keeping the
N = N0 ç ÷
è 2ø concentration of [B] constant, when the
where N = mass after n number of half life concentration of [A] is increased four times, rate
n = number of half life also increases four times. Hence, the order with
N0 = Initital mass respect to A is one. hence
22. (a) For a first order reaction Rate = k [A] [B]2
2.303 ( a - x1 ) 30. (b) For a first order reaction
K = t - t log a - x
( 2 1) ( 2 ) 0.693 0.693
t1/2 = ; k= = 0.5 × 10–3s–1
2.303 æ 0.04 ö k 1386
log ç
K= (
20 - 10) è 0.03 ÷ø 31. (a) When concentration of A is doubled, rate is
doubled. Hence order with respect to A is one.
2.303 ´ 0.1249 When concentrations of both A and B are
K=
10 doubled, rate increases by 8 times hence order
0.6932 2.303 ´ 0.1249 with respect to B is 2.
t1/2 = 10 \ rate = k [A]1 [B]2
0.6932 ´10 32. (a) Rewriting the given data for the reaction
t1/2 = = 24.1 sec
2.303 ´ 0.1249 H+
CH 3 COCH 3 (aq) + Br2 (aq) ¾¾¾
®
0.693
23. (a) t1/2 = CH3COCH 2 Br(aq) + H + (aq) + Br - (aq)
k
For first order t 1/2 is independent of initial S. Initial concent Initial concentr Initial concentr Rate of
No. -ration of -ation of Br2 -ation of H + disappearance
concentration of reactant. CH 3COCH3 in M in M of Br2 in MS-1
in M d dx
24. (b) For a first order reaction, i.e. - [Br2 ]or
dt dt
t75% = 2 × t50%
1 0.30 0.05 0.05 5.7 ´ 10-5
25. (d) Rate1 = k [A]2[B]3 2 0.30 0.10 0.05 5.7 ´ 10-5
when concentrations of both A and B are 3 0.30 0.10 0.10 1.2 ´ 10-4
doubled then 4 0.40 0.05 0.20 3.1 ´ 10-4
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184 CHEMISTRY
We can look at the above results in a simple way Now,
to find the dependence of reaction rate (i.e. rate
2.303 100 2.303
of disappearance of Br 2). t½ = log = ´ log 2
From data (1) and (2) in which concentration of 0.0153 100 - 50 0.0153
CH3COCH3 and H+ remain unchanged and only 2.303
= ´ 0.3010 = 45.31 min.
the concentration of Br 2 is doubled, there is no 0.0153
change in rate of reaction. It means the rate of 36. (d) Given [A] = 0.01 M
reaction is independent of concentration of Br 2. Rate = 2.0 × 10–5 mol L–1 S–1
Again from (2) and (3) in which (CH3CO CH3) For a first order reaction
and (Br2) remain constant but H+ increases from Rate = k [A]
0.05 M to 0.10 i.e. doubled, the rate of reaction -5
changes from 5.7×10 –5 to 1.2 × 10 –4 2.0 ´ 10
k= = 2 × 10–3
(or 12 × 10–5 ), thus it also becomes almost [0.01]
doubled. It shows that rate of reaction is directly 0.693
proportional to [H+]. From (3) and (4), the rate t1/2 = -3 = 347 sec.
2 ´ 10
should have doubled due to increase in conc of
[H+] from 0.10 M to 0.20 M but the rate has 37. (b) Rate1= k [A]x[B]y ... (1)
changed from 1.2× 10–4 to 3.1×10–4. This is due Rate1
to change in concentration of CH3 CO CH3 from = k [A]x [2B]y ... (2)
4
0.30 M to 0.40 M. Thus, the rate is directly
proportional to [CH3 COCH3]. or Rate1 = 4k [A]x[2B]y
From (1) and (2) we get
rate = k [CH3COCH3]1[Br2]0[H+]1
= k [CH3COCH3][H+]. k [A]x [B] y
= k [A]x[2B]y
33. (d) As the slowest step is the rate determining 4
step thus the mechanism B will be more [B] y
= [2B] y
consistent with the given information. Also 4
because it involve one molecule of H2 and one y
molecule of ICl, it can be expressed as : 1 æ 2B ö 1
or =ç ÷ Þ = 2y or (2)–2 = 2y
r = k [H2][ICl] 4 è Bø 4
Which shows that the reaction is of first order y = –2.
w.r.t. both H2 & ICl. 38. (b) For a first order reaction, A ® products
34. (c) For a first order reaction r
2.303 a r = k [A] or k =
k= log10 [A]
t a-x 1.5 ´ 10 -2
when t = t½ Þk = = 3 × 10–2
0.5
2.303 a 0.693 0.693
k= log10 Further, t1/2 = = = 23.1 min.
t½ a-a/2 k 3 ´ 10-2
2.303 ln 2
or t½ = log10 2 = 39. (b) Q r = k [A]n
k k
35. (a) For a first order reaction if n = 0
2.303 a r = k [A]0
k= log or r = k thus for zero order reactions rate is equal
t a-x
when t = 60 and x = 60% to the rate constant.
2.303 100 2.303 100 40. (b) A ® B For a first order reaction
k= log = log = 0.0153 given a = 0.8 mol, (a – x) = 0.8 – 0.6 = 0.2
60 100 - 60 60 40
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Chemical Kinetics 185
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186 CHEMISTRY
2000 1000
k Ea æ1 1ö or ln 10 = -
50. (b) log 2 = çè T - T ÷ø T T
k1 2.303 R 1 2 1000
or 2.303 log 10 =
Ea é 1 1 ù T
log 2 = 2.303 ´ 8.314 ê - ú or 2.303 × 1 × T=1000 [\ log 10 = 1]
ë 293 308 û
1000
Ea 15 or T = K
0.3 = × 2.303
2.303 ´ 8.314 293 ´ 308
55. (d) k = Ae - Ea / RT
0.3 ´ 2.303 ´ 8.314 ´ 293 ´ 308 Ea
Ea = . or log T = log A -
15 2.303 RT
= 34673 J mol-1 = 34.7 kJ mol-1 Comparing the above equation with
y = mx + c
51. (b) DH = Ea f - Eab = 0
1
y = log k, x =
52. (b, d) According to Arrhenius equation T
k 2 Ea æ 1 1 ö Thus a plot of log10k vs 1/T should be a straight
ln = - line, with slope equal to – Ea /2.303R and
k1 R çè T1 T2 ÷ø
intercept equal to log A
k2 E æ 1 1ö
ln =– a ç - ÷ Ea
k1 R è T2 T1 ø Slope =
log k 2.303R
53. (c)
Ea 1/T
Energy
- Ea
DH \ Slope =
2.303 R
or Ea = –2.303R ´ Slope
Reaction coordinate 56. (d) The activation energy of reverse reaction
Ea > DH will depend upon whether the forward reaction
2000 is exothermic or endothermic.
-
54. (d) Given, k1 = 1016.e T
As DH = Ea (forward reaction) – Ea (backward reaction)
-
1000 For exothermic reaction
and k2 = 1015.e T DH = –ve
When k1 and k2 are equal at any temperature T, \ –DH = Ea(f) – Ea(b)
we have or Ea( f ) = Ea(b) – DH
2000 1000 \ Ea( f ) < Ea(b)
- -
1016.e T 15
= 10 .e T for endothermic reaction
DH = + ve
2000 1000
15
-
15
- \ DH = Ea(f ) – Ea(b) or Ea(f ) = DH + Ea(b)
or 10 ´ 10 .e T
= 10 .e T
\ Ea(f ) > Ea(b).
2000 1000
- -
T T 57. (b) The presence of enzyme (catalyst) increases
or 10.e =e
the speed of reaction by lowering the energy
2000 1000 barrier, i.e. a new path is followed with lower
or ln 10 - =-
T T activation energy.
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Chemical Kinetics 187
If reaction is exothermic,
ET
E'T DH = - ve Ea(b) > Ea(f)
Ea
If reaction is endothermic
Products
DH = + ve Ea(b) < Ea(f)
Energy
Ea
1
60. (d) A catalyst affects equally both forward and
Reactants + catalyst backward reactions, therefore it does not affect
Progress of reaction
equilibrium constant of reaction.
Here ET is the threshold energy. 61. (b) Ea (Forward) + DH = Ea (backword)
Ea and Ea is energy of activation of reaction in For Exothermic reaction, DH = –ve and
1
absence and presence of catalyst respectively. \ activation energy of reactant is less than the
58. (c) We know that the activation energy of energy of activation of products.
chemical reaction is given by formula =
Threshold energy
k Ea éT2 - T1 ù
log 2 = ê ú, where k is the rate
k1 2.303R êë T1T2 ú
û
1
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188 CHEMISTRY
19 Surface Chemistry
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Surface Chemistry 189
7. For adsorption of a gas on a solid, the plot of 12. Which mixture of the solutions will lead to the
log x/m vs log p is linear with slope equal to formation of negatively charged colloidal
(n being whole number) [1994, 2006] [AgI]I– sol. ? [2019]
(a) k (b) log k (a) 50 mL of 1 M AgNO3 + 50 mL of 1.5 M KI
1 (b) 50 mL of 1 M AgNO3 + 50 mL of 2 M KI
(c) n (d)
n
(c) 50 mL of 2 M AgNO3 + 50 mL of 1.5 M KI
8. Accor ding to the adsorption theor y of
catalysis, the speed of the reaction increases (d) 50 mL of 0.1 M AgNO3 + 50 mL of 0.l M KI
because [2003] 13. On which of the following properties does the
(a) adsorption lowers the activation energy of coagulating power of an ion depend? [2018]
the reaction (a) The magnitude of the charge on the ion
(b) the concentration of reactant molecules at alone
the active centres of the catalyst becomes (b) Size of the ion alone
high due to strong adsorption
(c) The sign of charge on the ion alone
(c) in the process of adsorption, the activation
(d) Both magnitude and sign of the charge
energy of the molecules becomes large
on the ion
(d) adsorption produces heat which increases
the speed of the reaction 14. Fog is colloidal solution of [2016]
9. Which is not correct regarding the adsorption (a) Liquid in gas (b) Gas in liquid
of a gas on surface of solid? [2001] (c) Solid in gas (d) Gas in gas
(a) On increasing temperature, adsorption 15. Which property of colloidal solution is
increases continuously independent of charge on the colloidal particles:-
(b) Enthalpy and entropy changes are –ve [2014, 2015]
(c) Adsorption is more for some specific (a) Electrophoresis (b) Electro-osmosis
substances (c) Tyndall effect (d) Coagulation
(d) This phenomenon is reversible
16. The protecting power of lyophilic colloidal sol
Topic 2: Catalysis and Theories of Catalysis is expressed in terms of [2012]
10. Which one of the following statements is not (a) Coagulation value
correct? [2017] (b) Gold number
(a) The value of equilibrium constant is (c) Critical miscelle concentration
changed in the presence of a catalyst in (d) Oxidation number
the reaction at equilibrium 17. Which one of the following forms micelles in
(b) Enzymes catalyse mainly bio-chemical aqueous solution above certain concentration?
reactions [2005]
(c) Coenzymes increase the catalytic activity
(a) Dodecyl trimethyl ammonium chloride
of enzyme
(d) Catalyst does not initiate any reaction (b) Glucose
(c) Urea
Topic 3: Colloids and Emulsions (d) Pyridinium chloride
11. Measurin g Zeta potential is useful in 18. Which of the following forms cationic micelles
determining which property of colloidal above certain concentration? [2004]
solution? [2020] (a) Sodium dodecyl sulphate
(a) Solubility (b) Sodium acetate
(b) Stability of the colloidal particles (c) Urea
(c) Size of the colloidal particles
(d) Cetyl trimethyl ammonium bromide
(d) Viscosity
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190 CHEMISTRY
19. Position of non-polar and polar part in micelle 22. At the Critical Micelle Concentration (CMC) the
is [2002] surfactant molecules [1998]
(a) polar at outer surface and non-polar at inner (a) decompose
surface (b) dissociate
(b) polar at inner surface and non-polar at outer (c) associate
surface (d) become completely soluble
(c) distributed all over the surface 23. The ability of an ion to bring about coagulation
(d) present in the surface only of a given colloid depends upon [1997]
20. Which is used for ending charge on colloidal (a) its size
solution? [2000]
(b) the magnitude of its charge
(a) Electrons
(c) the sign of the charge alone
(b) Electrolytes
(d) both magnitude and sign of its charge
(c) Positively charged ions
24. During dialysis [1996]
(d) Compounds
(a) only solvent molecules can diffuse
21. Hardy-Schulze rule explains the effect of
electrolytes on the coagulation of colloidal (b) solvent molecules, ions and colloidal
solution. According to this rule, coagulation particles can diffuse
power of cations follow the order [1999] (c) all kinds of particles can diffuse through
(a) Ba+2 > Na+ > Al+3 (b) Al+3 > Na+ > Ba+2 the semi-permeable membrane
(c) Al+3 > Ba+2 > Na+ (d) Ba+2 > Al+3 > Na+ (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)
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Surface Chemistry 191
4. (a) According to Freundlich Adsorption 8. (a) A catalyst lowers the activation energy of
isotherm the reaction.
1 1
x Rate of reaction µ
= KP n activation energy
m
1 9. (a) On increasing temperature, adsorption of
at low pressure =1 a gas on surface of solid decreases. Solid adsorb
n
x greater amount of substances at lower
\ µ P1
m temperature.
1 10. (a) A catalyst speeds up both forward and
at high pressure =0
n backward reaction with the same rate.
x So, equilibrium constant is not affected by the
= constant
m presence of a catalyst at any given temperature.
i.e. the value of n varies between 0 to 1 11. (b) In colloidal solution, the potential
x difference between the fixed layer and the
5. (d) = f (P) at constant T and
m diffused layer of opposite charge is known as
x Zeta potential.
= f (T) at constant T Greater the Zeta potential more will be the
m
stability of colloidal particle.
Thus, it can be stated that for constant value of
x/m, pressure and temperature can take different 12. (b) AgNO3 + KI ¾¾
® AgI + KNO 3
values or Negatively
charged colloid
P = f (T) at constant (x/m). A solution of AgNO3 and KI will form a
x negatively charged colloidal sol, [AgI]I–, only
But ¹ P×T
m when KI is present in excess (i.e., KI behaves
6. (a) Langmuir adsorption isotherm is based on as a solvent).
the assumption that every adsorption site is Millimole of KI is maximum in option (2)
equivalent and the ability of a particle to bind (50 × 2 = 100)
there is independent of whether or not nearby 13. (d) According to Hardy Schulze rule,
sites are occupied. coagulating power of an ion depends on both
7. (d) According to Freundlich adsorption magnitude and sign of the charge on the ion.
isotherm. Greater the valence of the flocculating ion added,
At intermediate pressure, extent of adsorption greater is the coagulating power.
x x 1
= kp1/n or log = log k + log p;
m m n 14. (a) Fog is a colloidal system having dispersed
phase as liquid and dispersion medium as gas.
15. (c) Tyndall effect is an optical property, and it
Slope = 1/n is independent of charge on colloidal particles.
log x 16. (b) The lyophobic sols are less stable than
m
lyophilic sols. The lyophilic sols are thus used
log k
to protect the lyophobic sols. This property of
lyophilic sols is known as protective action of
log p lyophilic sols.
x Which can be represented by gold number.
plot of log vs log p is linear with slope
m 17. (a) Micelle formation is shown by surfactants
detergents (Dodecyl trimethyl ammonium
=1
n chloride) in their aqueous solutions.
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192 CHEMISTRY
18. (d) Cetyl trimethyl ammonium bromide, 23. (d) According to the Hardy Schulze rule, the
coagulating effect of an ion on dispersed phase
[C16 H 33 (CH 3 )3 N + Br - ] is a cationic micelle
of opposite charge increases with the valency
of the ion. Therefore, more the charge on
19. (a) o- Polarhead oppositely charged ion, higher is the coagulation
n- Non-polar tail value.
(micelle)
Positive ion can coagulate negatively charged
20. (b) On addition of electrolyte, charge of colloid and vice-versa. So, the ability depends on
colloidal particles will get neutralized and both magnitude and sign of its charge but the power
coagulation will occur. of coagulation depends only on the magnitude of
21. (c) According to this law the coagulating charge.
effect of an ion on dispersed phase of opposite
24. (d) Dialysis is a process of removing a
charge increases with the valency of the ion.
dissolved substance from a colloidal solution
The precipitating power of Al3+ , Ba++, Na+ ions
by means of diffusion thr ough suitable
is in order Al3+ > Ba2+ > Na+.
membrane. Colloidal particles cannot pass
22. (c) The critical micelle concentration is the through animal membrane. Only solvent
lowest concentration at which micelle formation molecules and ions (in case of electrodialysis)
appears. When surfactants are present above can diffuse.
that CMC, they can act as emulsifiers that will
solubilise a compound (oil, dirt, etc.) which is
normally insoluble in the solvent being used.
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General Principles and Processes of Isolation of Elements 193
20
General Principles
and Processes of
Isolation of Elements
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194 CHEMISTRY
8. Extraction of gold and silver involves leaching 13. Which of the following statements, about the
with CN– ion. Silver is later recovered by [2017] advantage of roasting of sulphide ore before
(a) distillation reduction is not true? [2007]
(b) zone refining (a) The DGfo of the sulphide is greater than
(c) displacement with Zn those for CS2 and H2S.
(d) liquation
9. Match items of Column I with the items of (b) The DGfo is negative for roasting of
Column II and asign the correct code : [2016] sulphide ore to oxide.
(c) Roasting of the sulphide to the oxide is
Column-I Column-II
thermodynamically feasible.
(A) Cyanide process (i) Ultrapure Ge
(d) Carbon and hydrogen are suitable reducing
(B) Froth flotation (ii) Dressing of ZnS agents for metal sulphides.
process 14. Sulphide ores of metals are usually concentrated
(C) Electrolytic (iii) Extraction of Al by froth flotation process. Which one of the
reduction following sulphide ores offer an exception and
(D) Zone refining (iv) Extraction of Au concentrated by chemical leaching? [2007]
(v) Purification of Ni (a) Galena (b) Copper pyrite
Code : (c) Sphalerite (d) Argentite
(A) (B) (C) (D) Topic 3: Purification and Uses of Metals
(a) (iv) (ii) (iii) (i)
(b) (ii) (iii) (i) (v) 15. The metal oxide which cannot be reduced to
metal by carbon is [NEET Kar. 2013]
(c) (i) (ii) (iii) (iv)
(a) Fe2O3 (b) Al2O3
(d) (iii) (iv) (v) (i)
(c) PbO (d) ZnO
10. In the extraction of copper from its sulphide ore,
16. Which of the following pairs of metals is purified
the metal finally obtained by the reduction of
by van Arkel method ? [2011]
cuprous oxide with : [2012, 2015 RS]
(a) Ga and In (b) Zr and Ti
(a) iron (II) sulphide (b) carbon monoxide
(c) Ag and Au (d) Ni and Fe
(c) copper (I) sulphide (d) sulphur dioxide
17. The method of zone refining of metals is based
11. Which of the following elements is present as
on the principle of [2003]
the impurity to the maximum extent in the pig iron ?
(a) Greater solubility of the impurity in the
(a) Manganese (b) Carbon [2011] molten state than in the solid
(c) Silicon (d) Phosphorus (b) Greater mobility of the pure metal than that
12. The following reactions take place in the blast of the impurity
furnace in the preparation of impure iron. Identify (c) Higher melting point of the impurity than
the reaction pertaining to the formation of the that of the pure metal
slag. [2011 M] (d) Greater noble character of the solid metal
(a) Fe2O3(s) + 3 CO(g) ®2 Fe (l) + 3 CO2 (g) than that of the impurity
(b) CaCO3 (s) ®CaO (s) + CO2 (g) 18. Method used for obtaining highly pure silicon
(c) CaO (s) + SiO2(s) ® CaSiO3 (s) used as a semiconductor material, is [1994, 96]
(d) 2C(s) + O2 (g) ®2 CO(g) (a) Oxidation (b) Electrochemical
(c) Crystallization (d) Zone refining
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196 CHEMISTRY
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The p-Block Elements (Group 15, 16, 17 and 18) 197
3. Which of the following oxoacids of phosphorus 11. How many bridging oxygen atoms are present
has strongest reducing property? in P4O10? [2010]
[NEET Odisha 2019] (a) 5 (b) 6
(a) H3PO4 (b) H4P2O7 (c) 4 (d) 2
(c) H3PO3 (d) H3PO2 12. Nitrogen forms N2, but phosphorus is converted
4. The correct order of N-compounds in its into P4 from P, the reason is [2001]
decreasing order of oxidation states is [2018] (a) Triple bond is present between phosphorus
(a) HNO3, NO, N2, NH4Cl atom
(b) pp – pp bonding is strong
(b) HNO3, NO, NH4Cl, N2
(c) pp – pp bonding is weak
(c) NH4Cl, N2, NO, HNO3
(d) Multiple bond is formed easily
(d) HNO3, NH4Cl, NO, N2 13. Which of the following oxy-acids has the
5. Which is the correct statement for the given maximum number of hydrogens directly attached
acids? [2016] to phosphorus? [1999]
(a) Phosphinic acid is a diprotic acid while (a) H4P2O7 (b) H3PO2
phosphonic acid is a monoprotic acid
(c) H3PO3 (d) H3PO4
(b) Phosphinic acid is a monoprotic acid while
14. Repeated use of which one of the following
phosphonic acid is a diprotic acid
fertilizers would increase the acidity of the soil?
(c) Both are triprotic acids
[1998]
(d) Both are diprotic acids
(a) Urea
6. The product obtained as a result of a reaction of
(b) Superphosphate of lime
nitrogen with CaC2 is [2016]
(c) Ammonium sulphate
(a) CaCN2 (b) CaCN
(d) Potassium nitrate
(c) CaCN3 (d) Ca2CN
15. Which of the following species has the highest
7. Strong reducing behaviour of H3PO2 is due to
dipole moment ? [1997]
[2015 RS]
(a) NH3 (b) PH3
(a) presence of one –OH group and two P–H (c) AsH3 (d) SbH3
bonds 16. The structural formula of hypophosphorous
(b) high electron gain enthalpy of phosphorus acid is [1997]
(c) high oxidation state of phosphorus
O O
(d) presence of two –OH groups and one P–H bond.
8. In which of the following compounds, nitrogen P P
(a) H (b) H
exhibits highest oxidation state ? [2012] OH OH
H OH
(a) N2H4 (b) NH3
(c) N3H (d) NH2OH O O
9. Which of the following statements is not valid P P
for oxoacids of phosphorus? [2012] (c) HO (d) H
OH OOH
(a) Orthophosphoric acid is used in the OH OH
manufacture of triple superphosphate. 17. Brown ring test is used to detect [1994]
(b) Hypophosphorous acid is a diprotic acid. (a) Iodine (b) Nitrate
(c) All oxoacids contain tetrahedral four (c) Iron (d) Bromide
coordinated phosphorus. 18. Which of the following fertilizers has the highest
(d) All oxoacids contain atleast one P = O and nitrogen percentage ? [1993]
one P — OH group. (a) Ammonium sulphate
10. Oxidation states of P in H4 P2O5 , H4 P2O6 , and (b) Calcium cyanamide
H4 P2O7 , are respectively: [2010] (c) Urea
(a) + 3, + 5, + 4 (b) + 5, + 3, + 4 (d) Ammonium nitrate
(c) + 5, + 4, + 3 (d) + 3, + 4, + 5
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198 CHEMISTRY
19. Which one of the following substance is used 29. Pure nitrogen is prepared in the laboratory by
in the laboratory for fast drying of neutral heating a mixture of [1991]
gases? [1992] (a) NH4OH + NaCl (b) NH4 NO3 + NaCl
(a) Phosphorus pentoxide (c) NH4 Cl + NaOH (d) NH4 Cl + NaNO2.
(b) Active charcoal 30. Which of the following statements is not correct
(c) Anhydrous calcium chloride for nitrogen ? [1990]
(d) Na3PO4.
(a) Its electronegativity is very high
20. Number of electrons shared in the formation of
(b) d-orbitals are available for bonding
nitrogen molecule is [1992]
(c) It is a typical non-metal
(a) 6 (b) 10
(c) 2 (d) 8 (d) Its molecular size is small
31. Of the following hybrides which one has the
21. Sugarcane on reaction with nitric acid gives lowest boiling point ? [1989]
[1992]
(a) CO2 and SO2 (a) AsH 3 (b) SbH 3
(b) (COOH)2 (c) PH 3 (d) NH 3
(c) 2 HCOOH (two moles) 32. Which of the following metal evolves hydrogen
(d) No reaction. on reacting with cold dilute HNO3 ? [1989]
22. Nitrogen is relatively inactive elemen t (a) Mg (b) Al
because [1992] (c) Fe (d) Cu.
(a) Its atom has a stable electronic configuration 33. Which one of the following compounds does
(b) It has low atomic radius not exist ? [1989]
(c) Its electronegativity is fairly high (a) NCl5 (b) AsF5
(d) Dissociation energy of its molecule is fairly (c) SbCl5 (d) PF5
high 34. Each of the following is true about white and
23. H3PO2 is the molecular formula of an acid of red phosphorus except that they [1989]
phosphorus. Its name and basicity respectively (a) Are both soluble in CS2
are [1992]
(b) Can be oxidised by heating in air
(a) Phosphorus acid and two
(c) Consist of the same kind of atoms
(b) Hypophosphorous acid and two
(c) Hypophosphorous acid and one (d) Can be converted into one another
35. When orthophosphoric acid is heated to 600°C,
(d) Hypophosphoric acid and two
the product formed is [1989]
24. Aqueous solution of ammonia consists of
(a) PH3 (b) P2O5
[1991]
(c) H3PO3 (d) HPO3
(a) H+ (b) OH–
36. Which of the following is a nitric acid anhydride?
(c) NH4+ (d) NH4+and OH–
[1988]
25. P2O5 is heated with water to give [1991]
(a) NO (b) NO2
(a) Hypophosphorous acid
(b) Phosphorous acid (c) N2O5 (d) N2O3.
(c) Hypophosphoric acid Topic 2: Oxygen Family
(d) Orthophosphoric acid
37. Which of the following oxoacid of sulphur has
26. Basicity of orthophosphoric acid is [1991]
– O – O – linkage? [2020]
(a) 2 (b) 3 (a) H2SO4, sulphuric acid
(c) 4 (d) 5 (b) H2S2O8, peroxodisulphuric acid
27. PCl3 reacts with water to form [1991] (c) H2S2O7, pyrosulphuric acid
(a) PH3 (b) H3PO3, HCl (d) H2SO3, sulphurous acid
(c) POCl3 (d) H3PO4 38. Identify the correct formula of ‘oleum’ from the
28. PH4I + NaOH forms [1991] following. [NEET Odisha 2019]
(a) PH3 (b) NH3 (a) H2S2O8 (c) H2S2O7
(c) P4O6 (d) P4O10 (c) H2SO3 (d) H2SO4
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The p-Block Elements (Group 15, 16, 17 and 18) 199
39. Which is the correct thermal stability order for 47. Match List - I (substances) with List - II (processes)
H2E (E = O, S, Se, Te and Po) ? [2019] employed in the manufacture of the substances
(a) H2S < H2O < H2Se < H2Te < H2Po and select the correct option. [2010]
(b) H2O < H2S < H2Se < H2Te < H2Po List - I List - II
(c) H2Po < H2Te < H2Se < H2S < H2O Substances Processes
(d) H2Se < H2Te < H2Po < H2O < H2S 1. Sulphuric acid (i) Haber’s process
40. In which pair of ions both the species contain 2. Steel (ii) Bessemer’s
S – S bond? [2017] process
2 -
(a) S4 O6 ,S2 O3 2 - (b) S2 O7 ,S2 O82 -
2 -
3. Sodium hydroxide (iii) Leblanc process
(c) S4 O62- ,S2 O72 - (d) S2 O72- ,S2 O32 - 4. Ammonia (iv) Contact process
41. Nitrogen dioxide and sulphur dioxide have some Code :
properties in common. Which property is shown 1 2 3 4
by one of these compounds, but not by the (a) (iv) (ii) (iii) (i)
other? [2015] (b) (i) (iv) (ii) (iii)
(a) is a reducing agent (c) (i) (ii) (iii) (iv)
(b) is soluble in water (d) (iv) (iii) (ii) (i)
(c) is used as a food-preservative 48. Which of the following is the most basic oxide?
(d) forms 'acid-rain' [2006]
42. Which of the statements given below is (a) Sb2O3 (b) Bi2O3
incorrect? [2015 RS] (c) SeO2 (d) Al2O3
(a) Cl2O7 is an anhydride of perchloric acid 49. During its reactions, ozone [1999]
(b) O3 molecule is bent (a) can only combine with hydrogen atoms
(c) ONF is isoelectronic with O2N–. (b) accepts electrons
(d) OF2 is an oxide of fluorine (c) loses electrons
43. Acidity of diprotic acids in aqueous solutions (d) shows the role of electrons to be irrelevant
increases in the order : [2014] 50. Which of the following oxides will be the least
(a) H2S < H2Se < H2Te acidic? [1996]
(b) H2Se < H2S < H2Te (a) As4O 6 (b) As 4O10
(c) H2Te < H2S < H2Se (c) P4O10 (d) P4O6
(d) H2Se < H2Te < H2S 51. Oxidation of thiosulphate by iodine gives
44. Which of the following does not give oxygen [1996]
(a) tetrathionate ion (b) sulphide ion
on heating? [NEET 2013]
(c) sulphate ion (d) sulphite ion
(a) Zn(ClO3)2 (b) K2Cr2O7
52. About 20 km above the earth, there is an ozone
(c) (NH4)2Cr2O7 (d) KClO3 layer. Which one of the following statements
45. Sulphur trioxide can be obtained by which of about ozone and ozone layer is true? [1995]
the following reaction : [2012] (a) ozone has a triatomic linear molecule
(a) CaSO 4 + C ¾¾
®
Δ (b) it is harmful as it stops useful radiation
(c) it is beneficial to us as it stops U.V radiation
Fe 2 ( SO4 )3 ¾¾
Δ
(b) ® (d) conversion of O3 to O2 is an endothermic
reaction
Δ
(c) S + H 2SO 4 ¾¾
® 53. By passing H 2S gas in acidified KMnO 4
Δ solution, we get [1995]
(d) H 2SO 4 + PCI5 ¾¾ ®
46. Which one of the following compounds is a (a) S (b) K2S
peroxide ? [2010] (c) MnO2 (d) K2SO3
54. Polyanion formation is maximum in [1994]
(a) KO 2 (b) BaO2 (a) Nitrogen (b) Oxygen
(c) MnO 2 (d) NO 2 (c) Sulphur (d) Boron
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200 CHEMISTRY
55. The acid which has a peroxy linkage is [1994] 63. Identify the incorrect statement related to PCI5
(a) Sulphurous acid (b) Pyrosulphuric acid from the following: [2019]
(c) Dithionic acid (d) Caro’s acid (a) Three equatorial P – Cl bonds make an
56. Which would quickly absorb oxygen? angle of 120° with each other
[1991, 92] (b) Two axial P – Cl bonds make an angle of
(a) Alkaline solution of pyrogallol 180° with each other
(b) Conc. H2SO4
(c) Axial P – Cl bonds are longer than
(c) Lime water
equatorial P – Cl bonds
(d) Alkaline solution of CuSO4
57. Oleum is [1991] (d) PC15 molecule is non-reactive
(a) Castor Oil (b) Oil of vitriol 64. Which of the following statements is not true
(c) Fuming H2SO4 (d) None of them for halogens? [2018]
58. Oxygen will directly react with each of the (a) All form monobasic oxyacids
following elements except [1989] (b) All are oxidizing agents
(a) P (b) Cl (c) Chlorine has the highest electron-gain
(c) Na (d) S. enthalpy
59. The gases respectively absorbed by alkaline (d) All but fluorine shows positive oxidation
pyrogallol and oil of cinnamon are [1989] states
(a) O3, CH4 (b) O2, O3 65. Match the interhalogen compounds of column-I
(c) SO2, CH4 (d) N2O, O3. with the geometry in column II and assign the
60. It is possible to obtain oxygen from air by correct code. [2017]
fractional distillation because [1989] Column-I Column-II
(a) oxygen is in a different group of the periodic 1. XX' (i) T-shape
table from nitrogen
2. XX'3 (ii) Pentagonal bipyramidal
(b) oxygen is more reactive than nitrogen
3. XX'5 (iii) Linear
(c) oxygen has higher b.p. than nitrogen
4. XX'7 (iv) Square-pyramidal
(d) oxygen has a lower density than nitrogen.
61. Hypo is used in photography to [1988] (v) Tetrahedral
(a) reduce AgBr grains to metallic silver Code :
(b) convert metallic silver to silver salt 1 2 3 4
(c) remove undecomposed silver bromide as a (a) (iii) (i) (iv) (ii)
soluble complex (b) (v) (iv) (iii) (ii)
(d) remove reduced silver
(c) (iv) (iii) (ii) (i)
Topic 3: Halogen Family (d) (iii) (iv) (i) (ii)
62. Match the following: [2019] 66 Among the following, the correct order of acidity
(a) Pure nitrogen (i) Chlorine is [2016]
(b) Haber process (ii) Sulphuric acid (a) HClO3 < HClO4 < HClO2 < HClO
(c) Contact proces (iii) Ammonia (b) HClO < HClO2 < HClO3 < HClO4
(d) Deacon’s process (iv) Sodium azide or (c) HClO2 < HClO < HClO3 < HClO4
Barium azide
(d) HClO4 < HClO2 < HClO < HClO3
Which of the following is the correct option ?
67. Which one of the following orders is correct for
(a) (b) (c) (d)
the bond dissociation enthalpy of halogen
(a) (i) (ii) (iii) (iv)
molecules? [2016]
(b) (ii) (iv) (i) (iii)
(a) I2 > Br2 > Cl2 > F2 (b) Cl2 > Br2 > F2 > I2
(c) (iii) (iv) (ii) (i)
(c) Br2 > I2 > F2 > Cl2 (d) F2 > Cl2 > Br2 > I2
(d) (iv) (iii) (ii) (i)
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The p-Block Elements (Group 15, 16, 17 and 18) 201
68. The variation of the boiling point of the 74. Which one of the following arrangements does
hydrogen halides is in the order HF > HI > HBr > not give the correct picture of the trends
HCl. [2015 RS] indicated against it ? [2000, 2008]
What explains the higher boiling point of (a) F2 > Cl2 > Br2 > I2 : Oxidizing power
hydrogen fluoride? (b) F2 > Cl2 > Br2 > I2 : Electron gain enthalpy
(c) F2 > Cl2 > Br2 > I2 : Bond dissociation
(a) The electronegativity of fluorine is much
energy
higher than for other elements in the group. (d) F2 > Cl2 > Br2 > I2 : Electronegativity.
(b) There is strong hydrogen bonding between 75. Which one of the following orders correctly
HF molecules represents the increasing acid strengths of the
(c) The bond energy of HF molecules is greater given acids? [2005, 2007]
than in other hydrogen halides. (a) HOClO < HOCl < HOClO3 < HOClO2
(d) The effect of nuclear shielding is much (b) HOClO2 < HOClO3 < HOClO < HOCl
reduced in fluorine which polarises the HF (c) HOClO3 < HOClO2 < HOClO < HOCl
(d) HOCl < HOClO < HOClO2 < HOClO3
molecule.
76. Which one of the following orders is not in
69. Which is the strongest acid in the following : accordance with the property stated against is ?
[NEET 2013] [2006]
(a) HClO3 (b) HClO4 (a) HI > HBr > HCl > HF : Acidic property in
(c) H2SO3 (d) H2SO4 water
70. In which of the following arrangements the given (b) F2 > Cl2 > Br2 > I2 : Electronegativity
sequence is not strictly according to the property (c) F2 > Cl2 > Br2 > I2 : Bond dissociation
indicated against it ? [2012 M] energy
(d) F2 > Cl2 > Br2 > I2 : Oxidising power
(a) HF < HCl < HBr < HI : increasing acidic 77. Which is the best description of the behaviour
strength of bromine in the reaction given below? [2004]
(b) H2O < H2S < H2Se < H2Te : increasing H 2 O + Br2 ® HOBr + HBr
pKa values (a) Proton acceptor only
(c) NH3 < PH3 < AsH3 < SbH3 : increasing (b) Both oxidized and reduced
acidic character (c) Oxidized only
(d) CO2 < SiO2 < SnO2 < PbO2 : increasing (d) Reduced only
oxidising power 78. Which of the following statements is not true ?
71. The correct order of increasing bond angles in (a) HF is a stronger acid than HCl [2003]
the following species are : [2010] (b) Among halide ions, iodide is the most
(a) Cl 2 O < ClO 2 < ClO 2– powerful reducing agent
(c) Fluorine is the only halogen that does not
(b) ClO 2 < Cl 2 O < ClO 2– show a variable oxidation state
(c) Cl 2 O < ClO 2– < ClO 2 (d) HOCl is a stronger acid than HOBr
(d) ClO 2– < Cl 2 O < ClO 2 79. A one litre flask is full of brown bromine vapour.
72. In the case of alkali metals, the covalent character The intensity of brown colour of vapour will not
decreases in the order: [2009] decrease appreciably on adding to the flask
(a) MF > MCl > MBr > MI some [1998]
(b) MF > MCl > MI > MBr (a) pieces of marble
(c) MI > MBr > MCl > MF (b) animal charcoal powder
(c) carbon tetrachloride
(d) MCl > MI > MBr > MF
(d) carbon disulphide
73. Among the following which is the strongest
80. Which one is the correct order of the size of
oxidising agent? [2009]
iodine species? [1997]
(a) Br2 (b) I2
(a) I > I+ > I– (b) I > I– > I+
(c) Cl2 (d) F2
(c) I+ > I– > I (d) I– > I > I+
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202 CHEMISTRY
81. Regarding F– and Cl– which of the following 90. The bleaching action of chlorine is due to [1991]
statements is/are correct? [1996] (a) Reduction (b) Hydrogenation
(i) Cl– can give up an electron more easily than (c) Chlorination (d) Oxidation
F– 91. Bleaching powder reacts with a few drops of
(ii) Cl– is a better reducing agent than F– dilute HCl to give [1989]
(iii) Cl– is smaller in size than F– (a) chlorine
(b) hypochlorous acid
(iv) F– can be oxidized more readily than Cl–
(c) calcium oxide
(a) (i) and (ii) (b) (i), (ii) and (iv) (d) oxygen
(c) (iii) and (iv) (d) only (i) 92. Bleaching powder is obtained by the action of
82. A certain compound (X) when treated with chlorine gas and [1988]
copper sulphate solution yields a brown (a) dilute solution of Ca(OH)2
precipitate. On adding hypo solution, the (b) concentrated solution of Ca(OH)2
precipitate turns white. The compound is [1994] (c) dry CaO
(a) K2CO3 (b) KI (d) dry slaked lime
(c) KBr (d) K3PO4
Topic 4: Noble Gases
83. HI can be prepared by all the following methods,
93. Match the compounds given in column I with
except [1994]
the hybridisation and shape given in column II
(a) PI 3 + H 2 O (b) KI + H 2SO 4 and mark the correct option. [2016]
(c) H 2 + I 2 ¾¾® Pt (d) I 2 + H 2S Column-I Column-II
84. Which among the following is paramagnetic? 1. XeF6 (i) Distorted octahedral
[1994] 2. XeO3 (ii) Square planar
(a) Cl 2 O (b) ClO 2 3. XeOF4 (iii) Pyramidal
4. XeF4 (iv) Square pyramidal
(c) Cl 2O 7 (d) Cl 2O 6 Code :
85. Which one of the following oxides of chlorine is 1 2 3 4
obtained by passing dry chlorine over silver (a) (i) (iii) (iv) (ii)
chlorate at 90°C ? [1994] (b) (i) (ii) (iv) (iii)
(a) Cl2O (b) ClO3 (c) (iv) (iii) (i) (ii)
(c) ClO2 (d) ClO4 (d) (iv) (i) (ii) (iii)
86. The formula for calcium chlorite is [1994] 94. Identify the incorrect statement, regarding the
molecule XeO4: [NEET Kar. 2013]
(a) Ca(ClO 4 )2 (b) Ca(ClO3 )2 (a) XeO4 molecule is tetrahedral
(c) CaClO 2 (d) Ca(ClO 2 ) 2 (b) XeO4 molecule is square planar
87. A solution of potassium bromide is treated with (c) There are four pp – dp bonds
each of the following. Which one would liberate (d) There are four sp3 – p, s bonds
95. Noble gases do not react with other elements
bromine ? [1993]
because [1994]
(a) Hydrogen iodide (b) Sulphur dioxide (a) They are mono atomic
(c) chlorine (d) Iodine (b) They are found in abundance
88. When chlorine is passed over dry slaked lime at (c) The size of their atoms is very small
room temperature, the main reaction product is (d) They are completely paired up and stable
[1992] electron shells
(a) Ca (ClO2 ) 2 (b) CaCl 2 96. Which of the following statements is false ?
[1994]
(c) CaOCl 2 (d) Ca (OCl) 2 (a) Radon is obtained from the decay of radium
89. In the manufacture of bromine from sea water, (b) Helium is inert gas
the mother liquor containing bromides is treated (c) Xenon is the most reactive among the rare
with [1992] gases
(a) Carbon dioxide (b) Chlorine (d) The most abundant rare gas found in the
(c) Iodine (d) Sulphur dioxide
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The p-Block Elements (Group 15, 16, 17 and 18) 203
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204 CHEMISTRY
13. (b)
O OH OH
O OH OH
H Pyrophosphoric acid
O
(H3PO2) is a monobasic acid. i.e., it has only
one ionisable hydrogen atom or one OH is P
(b) H3PO2 Þ OH H
O H
Hypophosphorous acid
present H P OH
O
H (c) H3PO3 Þ HO – P – OH
H
Phosphorous acid
+3 +3
O
10. (d) HO - P - O - P - OH = H 4 P2O5
| | (d) H3PO4 Þ HO – P – OH
OH OH
OH
orthophosphoric acid
14. (c) Ammonium sulphate is a salt of weak base
O O
|| || and strong acid, so it produces acidity. Hence
HO - P - P - OH = H 4 P2 O6 aqueous solution of ammonium sulphate
+4 | | +4
increases the acidity of soil.
OH OH
15. (a) Order of dipole moment decreases as
NH3 > PH3 > AsH3 > SbH3
O O (Based upon electronegativity)
|| ||
16. (a) We know that empirical formula of
HO - P - O - P - OH = H 4 P2 O7
+5 | | +5 hypophosphorus acid is H3PO2. In this only
OH OH one ionisable hydrogen atom is present i.e. it is
monobasic. Therefore option (a) is correct
11. (b) O structural formula of it.
bridging P bridging 17. (b) Brown ring test is done for the confirmation
O of NO 3- ions.
O
O
NaNO3 (aq) + H 2SO4 (aq)
P O
¾¾
® NaHSO4 (aq) + HNO3 (aq)
O P
O 6FeSO4 + 2HNO3 +3H 2SO4
O P O ¾¾
® 3Fe 2 (SO 4 )3 +2NO+4H 2O
bridging
O bridging FeSO4 + NO ¾¾
® [Fe(NO)]SO4
oxygen
Ferrous nitroso-sulphate
i.e. 6-bridging oxygen. (Brown ring)
12. (c) Nitrogen form N 2 (i. e. N º N) but 18. (c) Urea (46.6%N). % of N in other compound
phosphorus form P4, because in P2, pp — pp are : ( NH 4 ) 2 SO 4 = 21.2%;
bonding is present which is a weaker bonding.
CaCN 2 = 35.0% and NH 4 NO3 = 35.0%
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The p-Block Elements (Group 15, 16, 17 and 18) 205
19. (a) Phosphorus pentoxide has great affinity for 27. (b) PCl3 + 3H 2 O ® H 3PO 3 + 3HCl
water. It forms ortho phosphoric acid on
absorbing water 28. (a) PH 4 I + NaOH ® NaI + PH 3 + H 2 O
i.e. P4 O10 + 6H 2 O ¾¾ ® 4H 3 PO 4 29. (d) Pure nitrogen in the lab can be obtained by
It is thus used as a powerful dehydrating or heating ammonium nitrate. Ammonium nitrate is
drying agent. not a stable compound it dissociate to give
20. (a) Nitrogen molecule is diatomic containing a nitrogen.
triple bond between two N atoms, N º N . Heat
:
:
NH 4 Cl + NaNO 2 ¾¾¾® NH 4 NO 2
Therefore, nitrogen molecule is formed by
sharing six electrons. Heat
¾¾¾® N 2 + 2 H 2 O.
21. (b) Cane sugar is oxidised to oxalic acid
30. (b) In case of nitrogen, d-orbitals are not
(2HNO 3 ¾¾ ® H 2 O + 2NO 2 + O)18 available.
C12 H 22 O11 + 18[O] ¾ ¾® 6 (COOH) 2 31. (c) NH3 undergoes H-bonding and hence has
Cane sugar From HNO3 Oxalic acid the highest b.p. Among the remaining hydrides
+ 5H 2O. i.e. PH3, AsH3 and SbH3 the b.p. increases as
the size of the element increases and hence the
C12H22O11 + 36HNO3 ¾¾ ® magnitude of the van der Waal’s forces of
6(COOH)2 + 36NO2 + 23H2O attraction increases. Thus, PH3 has the lowest
22. (d) N2 molecule contains triple bond between b.p.
N atoms having very high dissociation energy 32. (a) Magnesium and manganese are the metals
(946 kJ mol–1) due to which it is relatively that produce hydrogen with dilute nitric acid
inactive. Mg + 2HNO3 ¾ ¾® Mg (NO3)2 + H2
23. (c) H3PO2 is named as hypophosphorous acid. 33. (a) As Nitrogen does not have d-orbital in its
As it contains only one P – OH group, its basicity valence shell, it cannot form NCl5.
is one. 34. (a) Red phosphorus is not soluble in CS2, only
24. (d) Aqueous solution of ammonia is obtained white P is soluble.
by passing NH3 in H2O which gives NH4+ and
OH– ions. 600°C
35. (d) 2H 3 PO 4 ¾¾¾¾
® 2HPO3
orthophosphoric -2H 2O metaphosphoric
NH 3 + H 2 O NH 4+
+ OH -
acid acid
25. (d) P2O5 have great affinity for water. The final
36. (c) N 2 O 5 + H 2 O ¾¾® 2 HNO 3
product is orthophosphoric acid.
2H2O O O
P4O10 4HPO3
Metaphosphoric
acid 37. (b) HO – S – O – O – S – OH
2H2O O O
2H 2O Peroxodisulphuric acid
4H3PO4 2H4P2O7
orthophosphoric Pyrophosphoric
acid acid Peroxosulphuric acid is also known as Marshall's
26. (b) It is a tribasic acid as all the three hydrogen acid and is one of the most powerful peroxy acid
oxidant.
atoms are ionisable. It forms three series of salts.
+ – 38. (b) Oleum is H2S2O7.
H3PO4 H + H2PO4
39. (c) On going down the group, bond
dissociation enthalpy of the hydrides of oxygen
+ 3– + 2– family decreases. Therefore, thermal stability
3H + PO4 2H + HPO4
also decreases.
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206 CHEMISTRY
O O O increasing size of the central atom. Thus Al2O3
– and Sb2O3 are amphoteric and Bi2O3 is basic.
40. (a) O – S – S – S – S – O S
49. (a) Since ozone can easily lose oxygen atom
O O S O
O (nascent oxygen), it acts as a powerful oxidising
S4 O6 2 - S2 O32 - agent, and hence, reacts with hydrogen atoms.
41. (c) SO2 is widely used in food and drinks 50. (a) As the O.N of the central atom of the
industries for its property as a preservative and compounds increases, acidic strength of that
compound also increases and on moving from
antioxidant while NO2 is not used as food
top to bottom in groups acidic strength of oxides
preservative.
also decrease due to decreasing
42. (d) OF2; among the following O and F, F is more electronegativity in groups.
electronegative than oxygen. +5 +3 +5 +3
So OF2 cannot be called oxide because in that P4O10 > P4O6 > As 4O10 > As 4O6
case fluorine is in +1 oxidation state which is
not possible, so OF2 is called oxygen difluoride. 51. (a) 2 S2 O 3-2 + I 2 ® S 4O 6-2 + 2I -
Tetrathion ate
43. (a) The weakning of M—H bond with increase
52. (c) Ozone layer is beneficial to us, because it
in size of M (where M = S, Se, Te) explains the
stops harmful ultraviolet radiations to reach the
acid character of hydrides.
earth.
On moving down the group, atomic size increases 53. (a) 2 KMnO 4 + 5H 2 S + 3H 2 SO 4 ¾¾®
hence, bond length increases and hence, removal
tendency of H also increases. K 2 SO 4 + 2MnSO 4 + 5S + 8H 2 O.
Thus, in this reaction sulphur (S) is produced.
D
44. (c) (NH4)2Cr2O7 ¾¾® N2 + Cr 2O3 + 4H2O 54. (c) Due to greater tendency for catenation,
D sulphur shows property of polyanion formation
Zn(ClO3)2 ¾¾® ZnCl2 + 3O2 to a greater extent. For example, in polysulphides
D
2 KClO3 ¾¾® 2KCl + 3O2 such as S32 - , S24 - , S52 -
D
4K2Cr2O7 ¾¾® 4K2CrO4 + 2Cr2O3 + 3O2 55. (d) Caro’s acid is H 2SO 5 which contains one
S – O – O – H peroxy linkage. It is also known as
Δ permonosulphuric acid.
45. (b) Fe 2 (SO 4 )3 ¾¾
® Fe2 O3 + SO3
O
O ||
H – O – O – S – OH
46. (b) Ba ||
O
Caro's acid
O 56. (a) Upon oxidation, pyrogallol forms hydroxy-
47. (a) quinone and many other higher molecular mass
(1) Sulphuric acid (iv) Contact process products.
(2) Steel (ii) Bessemer’s
process OH O
HO OH HO ||
(3) Sodium hydroxide (iii) Leblanc process O
||
O (air)
(4) Ammonia (i) Haber’s process ¾¾¾®
2
48. (b) More the oxidation state of the central atom 1-hydroxy,
pyrogallol
(metal), more is its acidity. Hence, SeO2 (O. S. of 2-benzoquinone
Se = +4) is acidic. Further for a given O.S., the
basic character of the oxides increases with the
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The p-Block Elements (Group 15, 16, 17 and 18) 207
66. (b) Acidic strength increases as the oxidation
Hydroxy groups are strong activators of aromatic number of central atom increases.
systems, hence pyrogallol is reactive enough to HClO < HClO2 < HClO3 < HClO4
react with oxygen in air.
+1 +3 +5 +7
57. (c) Oleum is H 2S2 O 7 ( H 2SO 4 + SO 3 ) which 67. (b) Bond dissociation enthalpy decreases as
the bond distance increases from F2 to I2. This
is obtained by dissolving SO 3 in H2SO4 and is
is due to increase in the size of the atom, on
called fuming sulphuric acid. moving from F to I.
58. (b) S + O 2 ¾¾ ® SO 2 (burns with blue light) F – F bond dissociation enthalpy is smaller then
4Na + O 2 ¾¾
® 2NaO Cl – Cl and even smaller than Br – Br. This is
because F atom is very small and hence the three
(burns with yellow light)
lone pairs of electrons on each F atom repel the
P4 + 3O 2 ¾¾
® P4 O 6 bond pair holding the F-atoms in F2 molecules.
P4 + 5O 2 ¾¾
® P4 O10 The increasing order of bond dissociation
Cl + O 2 ¾¾ ® No reaction enthalphy is
I2 < F2 < Br2 < Cl2
Chlorine does not react directly with oxygen.
68. (b) The H-bonding is present in HF due to high
59. (b) Alkaline pyrogallol absorbs O2 and oil of
electronegativity of fluorine atom. While H-
cinnamon absorbs O3.
bonding is not present in HI, HBr and HCl.
60. (c) Oxygen has higher b.pt. than nitrogen 69. (b) HClO4 is the strongest acid amongst all
therefore it can be obtained from air by fractional because the oxidation state of Cl is maximum
distillation. (+7).
Air is liquified by making use of the joule-Thomson 70. (b) If acidic nature is high, Ka is high and pKa
effect (cooling by expansion of the gas). Water is low
vapour and CO 2 are removed by solidification. H2O H2S
The remaining constituents of liquid air i.e., liquid Ka 1.8 × 10– 6 1.3 × 10–7
oxygen and liquid nitrogen are separated by means
H2Se H2Te
of fractional distillation.
Ka 1.3 × 10–4 2.3 × 10–3
61. (c) Undecomposed AgBr forms a soluble since pKa = – log Ka
complex with hypo Hence the order of pKa will be
AgBr + 2 Na 2S 2 O3 ® Na 3[ Ag (S 2 O 3 ) 2 ] + NaBr H2O > H2S > H2Se > H2Te
soluble complex 71. (d) (i) As the number of lone pair of electrons
62. (d) increases, bond angle decreases due to repulsion
63. (d) PCl5 is very reactive due to the presence of between lp – lp.
weak axial bonds. It is used in the synthesis of (ii) As the electronegativity of the central atom
various organic compounds. increases, bond angle increasess. Hence, the
64. (d) Due to high electronegativity and small correct order of bond angle:
.
size, F forms only one oxoacid, HOF known as : Cl : :O: Cl :
fluoric (I) acid. < <
O O Cl Cl O O
Flourine has –1 oxidation state in most of its two lone pairs two lone pairs one lone pair
compound. Oxidation number of F is +1 in HOF. and one electron
72. (c) MI > MBr > MCl > MF. As the size of the
anion decreases covalency decreases.
65. (a) XX' ® Linear (e.g. ClF, BrF)
73. (d) Standard reduction potential of halogens
XX3' ® T-Shape (e.g. ClF3, BrF3) are positive and decreases from fluorine to
XX5' ® Square pyramidal (e.g. BrF5 IF5) iodine. Therefore, halogens act as strong
XX7' ® Pentagonal bipyramidal (e.g. IF7) oxidising agent and their oxidising power
decreases from fluorine to iodine.
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208 CHEMISTRY
74. (b and c) Br– I–
& ion act as reducing agents and their
(a) The oxidising power of halogen follow the reducing nature is in increasing order
order F2 > Cl2 > Br2 > I2.
Cl – Br – I –
(b) The correct order of electron gain enthalpy ¾¾¾¾¾¾¾¾¾¾¾®
Reducing nature increases
of halogens is Cl2 > F2 > Br2 > I2. The low value
of F2 than Cl2 is due to its small size. 82. (b) KI reacts with CuSO4 solution to produce
(c) The correct order of bond dissociation cuprous iodide (white precipitate) and I2 (which
gives brown colour). Iodine reacts with hypo
energies of halogens is
(Na2S2O3.5H2O) solution. Decolourisation of
Cl2 > Br2 > F2 > I2. solution shows the appearance of white
(d) It is the correct order of electronegativity precipitate.
values of halogens. 2CuSO4 + 4KI ® 2K 2SO4 + 2CuI + I2
F2 > Cl2 > Br2 > I2 Cuprous iodide (Brown colour
(White ppt.) in solution)
75. (d) HO Cl < HO Cl O < HO Cl O 2 < HO Cl O3
+1 +3 +5 +7
2Na 2S2 O3 + I2 ¾¾
® Na 2S4 O6 + 2NaI
Sod. tetra
As the oxidation number of the central atom thionate
increases, strength of acid also increases. (colourless)
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The p-Block Elements (Group 15, 16, 17 and 18) 209
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210 CHEMISTRY
Topic 1: Characteristics of d-Block Elements 2. Match the catalyst with the process
1. Identify the incorrect statement. [2020] [NEET Odisha 2019]
(a) The transition metals and their compounds Catalyst Process
are known for their catalytic activity due to (i) V2O5 (a) The oxidation of
their ability to adopt multiple oxidation ethyne to ethanal
states and to form complexes. (ii) TiCl4 + Al(CH3)3 (b) Polymerisation of
alkynes
(b) Interstitial compounds are those that are (iii) PdCl2 (c) Oxidation of SO2
formed when small atoms like H, C or N are in the manufacture
trapped inside the crystal lattices of metals. of H2SO4
(iv) Nickel complexes (d) Polymerisation of
(c) The oxidation states of chromium in CrO24 - ethylene
Which of the following is the correct option?
and Cr2 O72 - are not the same. (a) (i)-(c), (ii)-(a), (iii)-(d), (iv)-(b)
(d) Cr2+ (d4) is a stronger reducing agent than (b) (i)-(c), (ii)-(d), (iii)-(a), (iv)-(b)
Fe2+ (d6) in water. (c) (i)-(a), (ii)-(b), (iii)-(c), (iv)-(d)
(d) (i)-(a), (ii)-(c), (iii)-(b), (iv)-(d)
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The d-and f-Block Elements 211
3. Which one of the following ions exhibits d-d 10. Sc (Z = 21) is a transition element but Zn (Z = 30)
transition and paramagnetism as well? [2018] is not because [NEET Kar. 2013]
(a) both Sc and Zn do not exhibit variable
(a) CrO2–
4 (b) Cr2 O72–
oxidation states
(c) MnO 2– –
(d) MnO4 (b) both Sc3+ and Zn2+ ions are colourless and
4
form white compounds
4. Magnetic moment 2.84 B.M. is given by :- (c) in case of Sc, 3d orbitals are partially filled
(At. nos, Ni = 28, Ti = 22, Cr = 24, Co = 27) [2015] but in Zn these are completely filled
(a) Ti3+ (b) Cr2+ (d) last electron is assumed to be added to 4s
(c) Co2+ (d) Ni2+ level in case of Zn
5. The number of d-electrons in Fe2+ (Z = 26) is 11. Which one of the following does not correctly
not equal to the number of electrons in which represent the correct order of the property
one of the following? [2015] indicated against it? [2012 M]
(a) p-electrons in Cl (Z = 17) (a) Ti < V < Cr < Mn : increasing number of
oxidation states
(b) d-electrons in Fe (Z = 26)
(b) Ti3+ < V3+ < Cr3+ < Mn3+ : increasing
(c) p-electrons in Ne (Z = 10)
magnetic moment
(d) s-electrons in Mg (Z = 12)
(c) Ti < V < Cr < Mn : increasing melting
6. Which is the correct order of increasing energy points
of the listed orbitals in the atom of titanium ?
(d) Ti < V < Mn < Cr : increasing 2nd ionization
[2015 RS] enthalpy
(a) 3s 4s 3p 3d (b) 4s 3s 3p 3d 12. Four successive members of the first series of
(c) 3s 3p 3d 4s (d) 3s 3p 4s 3d the transition metals are listed below. For which
7. 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):- value has a positive sign? [2012 M]
[2014] (a) Co (Z = 27) (b) Ni (Z = 28)
(a) Ti3+ (b) Ni2+ (c) Cu (Z = 29) (d) Fe (Z = 26)
(c) Cr3+ (d) Mn 2+ 13. The catalytic activity of transition metals and
their compounds is ascribed mainly to : [2012 M]
8. Reason of lanthanoid contraction is:- [2014]
(a) their magnetic behaviour
(a) Negligible screening effect of ‘f ’ orbitals
(b) their unfilled d-orbitals
(b) Increasing nuclear charge
(c) their ability to adopt variable oxidation state
(c) Decreasing nuclear charge
(d) their chemical reactivity
(d) Decreasing screening effect
14. For the four successive transition elements (Cr,
9. Which of the following lanthanoid ions is Mn, Fe and Co), the stability of +2 oxidation
diamagnetic ? state will be there in which of the following order?
(At nos. Ce = 58, Sm = 62, Eu = 63, Yb = 70) [2011]
[NEET 2013] (a) Mn > Fe > Cr > Co
(a) Sm2+ (b) Eu2+ (b) Fe > Mn > Co > Cr
(c) Yb2+ (d) Ce2+ (c) Co > Mn > Fe > Cr
(d) Cr > Mn > Co > Fe
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212 CHEMISTRY
15. Which of the following ions will exhibit colour (a) V3+ (b) Ti3+
in aqueous solutions? [2010] (c) Mn 3+ (d) Cr3+
3+
(a) La (Z = 57) 3+
(b) Ti (Z = 22) (At.No. Ti = 22, V = 23, Cr = 24, Mn = 25)
(c) Lu3+ (Z = 71) (d) Sc3+ (Z = 21) 23. In which of the following pairs are both the ions
16. Which one of the following ions has electronic coloured in aqueous solutions ? [2006]
configuration [Ar] 3d6 ? [2010] (a) Sc3+, Ti3+ (b) Sc3+, Co2+
(a) Ni3+ (b) Mn 3+ (c) Ni2+, Cu+ (d) Ni2+, Ti3+
(c) Fe3+ (d) Co3+ (At. no. : Sc = 21, Ti = 22, Ni = 28, Cu = 29, Co = 27)
(At. Nos. Mn = 25, Fe = 26, Co = 27, Ni = 28) 24. The aqueous solution containing which one
17. Which of the following pairs has the same size? of the following ions will be colourless?
[2010] (Atomic number: Sc = 21, Fe = 26, Ti = 22, Mn
(a) Fe2+, Ni2+ (b) Zr 4+, Ti4+ = 25) [2005]
(d) 3–
TiF62– , and CoF6 ,
(d) Ti 2+ , V 3+ , Cr 4+ , Mn 5+
27. The basic character of the transition metal
21. The correct order of decreasing second
monoxides follows the order
ionisation enthalpy of Ti (22), V(23), Cr(24) and
Mn (25) is : [2008] (Atomic No.,Ti = 22, V = 23, Cr = 24, Fe = 26)
[2003]
(a) Cr > Mn > V > Ti (b) V > Mn > Cr > Ti
(a) TiO > VO > CrO > FeO
(c) Mn > Cr > Ti > V (d) Ti > V > Cr > Mn (b) VO > CrO > TiO > FeO
22. Which one of the following ions is the most (c) CrO > VO > FeO > TiO
stable in aqueous solution? [2007] (d) TiO > FeO > VO > CrO
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The d-and f-Block Elements 213
28. Which one of the following characteristics of 36. The electronic configurations of four elements
the transition metals is associated with their are given below. Which element does not belong
catalytic activity? [2003] to the same family as others ? [1989]
(a) Variable oxidation states (a) [ Xe]4 f 14 5d 10 6s 2 (b) [ Kr ]4d10 5s 2
2 5
(b) High enthalpy of atomization (c) [ Ne]3s 3 p (d) [ Ar ]3d10 4s 2
(c) Parmagnetic behaviour 37. Which one of the following is an ore of silver ?
[1988]
(d) Colour of hydrated ions (a) Argentite (b) Stibnite
29. Which of the following shows maximum number (c) Haematite (d) Bauxite
of oxidation states? [2002] 38. Which of the following metals corrodes readily
in moist air ? [1988]
(a) Cr (b) Fe
(a) Gold (b) Silver
(c) Mn (d) V (c) Nickel (d) Iron
30. Of the following transition metals, the maximum
Topic 2: Compounds of Transition Metals
numbers of oxidation states are exhibited by:
(a) Chromiun (Z = 24) [2000] 39. Name the gas that can readily decolourise
acidified KMnO4 solution : [2017]
(b) Manganese (Z = 25)
(a) SO2 (b) NO2
(c) Iron (Z = 26)
(c) P2O5 (d) CO2
(d) Titanium (Z=22)
40. Which one of the following statements is correct
31. Which of the following forms colourless
compound? [2000] when SO2 is passed through acidified K2Cr2O7
(a) Sc3+ (b) V3+ solution ? [2016]
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214 CHEMISTRY
the least amount of acidified KMnO 4 for 51. German silver is an alloy of [2000]
complete oxidation [2015 RS] (a) Fe, Cr, Ni (b) Cu, Zn, Ag
(a) FeSO4 (b) FeSO3 (c) Cu, Zn, Ni (d) Cu, Sn, Al
52. Which of the following combines with Fe (II)
(c) FeC2O4 (d) Fe(NO2)2 ions to form a brown complex? [2000]
44. The pair of compounds that can exist together is: (a) NO (b) N2O
[2014] (c) N2O3 (d) N2O5
53. On heating chromite (FeCr 2O4) with Na2CO3 in
(a) FeCl3, SnCl2 (b) HgCl2, SnCl2
air, which of the following product is obtained?
(c) FeCl2, SnCl2 (d) FeCl3, KI [1999]
45. In acidic medium, H2O2 changes Cr2O7–2 to (a) Na2Cr2O7 (b) FeO
CrO5 which has two (–O–O) bonds. Oxidation (c) Fe3O4 (d) Na2CrO4
state of Cr in CrO5 is:- [2014] 54. The addition of excess of aqueous HNO3 to a
(a) + 5 (b) + 3 solution containing [Cu(NH3)4]2+ produces
(c) + 6 (d) – 10 [1999]
46. The reaction of aqueous KMnO4 with H2O2 in (a) Cu+ (b) [Cu(H2O)4]2+
acidic conditions gives: [2014] (c) Cu(OH)2 (d) Cu(NO3)2
(a) Mn4+ and O2 (b) Mn2+ and O2 55. An acidic solution of 'X' does not give precipitate
(c) Mn2+ and O3 (d) Mn4+ and MnO2 on passing H2 S through it. 'X' gives white
precipitate when NH4OH is added to it. The white
47. KMnO4 can be prepared from K2MnO4 as per precipitate dissolves in excess of NaOH solution.
the reaction: Pure 'X' fumes in air and dense white fumes are
3MnO 24 - + 2H 2O
obtained when a glass rod dipped in NH4OH is
put in the fumes. Compound 'X' can be [1999]
2MnO -4 + MnO2 + 4OH -
(a) ZnCl2 (b) FeCl3
The reaction can go to completion by removing (c) AlCl3 (d) SnCl2
OH– ions by adding. [NEET 2013] 56. Which one of the following elements constitutes
(a) KOH (b) CO2 a major impurity in pig iron ? [1998]
(c) SO2 (d) HCl (a) Silicon (b) Oxygen
48. Which of the statements is not true? [2012] (c) Sulphur (d) Graphite
(a) On passing H2S through acidified K2Cr2O7 57. K2Cr2O7 on heating with aqueous NaOH gives
solution, a milky colour is observed. [1997]
(b) Na2Cr2O7 is preferred over K2Cr2O7 in (a) CrO 24- (b) Cr(OH)3
volumetric analysis. (c) Cr2O 72- (d) Cr(OH)2
58. CrO3 dissolves in aqueous NaOH to give [1997]
(c) K2Cr2O7 solution in acidic medium is orange.
(a) Cr2O72– (b) CrO42–
(d) K2Cr 2 O7 solution becomes yellow on
(c) Cr(OH)3 (d) Cr(OH)2
increasing the pH beyond 7.
59. Cuprous compounds such as CuCl, CuCN and
49. Acidified K2Cr2O7 solution turns green when CuSCN are the only salts stable in water due to
Na 2 SO 3 is added to it. This is due to the [1996]
formation of : [2011] (a) high hydration energy of Cu+ ions
(a) Cr2(SO4)3 (b) CrO42– (b) their inherent tendency to not
(c) Cr2(SO3)3 (d) CrSO4 disproportionate
50. Copper sulphate dissolves in excess of KCN to (c) diamagnetic nature
give [2006] (d) insolubility in water
(a) [Cu(CN)4]3– (b) [Cu(CN)4]2– 60. Stainless steel contains iron and [1995]
(c) Cu(CN)2 (d) CuCN (a) Cr + Ni (b) Cr + Zn
(c) Zn + Pb (d) Fe +Cr + Ni
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The d-and f-Block Elements 215
61. The most durable metal plating on iron to protect cyanide solution to form the soluble product
against corrosion is [1994] potassium argento cyanide. The element is
(a) Nickel plating (b) Tin plating [1989]
(c) Copper plating (d) Zinc plating (a) Lead (b) Chromium
(c) Manganese (d) Silver
62. When ( NH 4 ) 2 Cr2O 7 is heated, the gas
70. A blue colouration is not obtained when
evolved is [1994]
[1989]
(a) N2 (b) NO2
(a) Ammonium hydroxide dissolves in copper
(c) O2 (d) N2O sulphate
63. When CuSO 4 is electrolysed using platinum (b) Copper sulphate solution reacts with
electrodes, [1993] K4[Fe(CN)6]
(a) Copper is liberated at cathode, sulphur at (c) Ferric chloride reacts with sod. ferrocyanide
anode (d) Anhydrous CuSO4 is dissolved in water
(b) Copper is liberated at cathode, oxygen at Topic 3: Lanthanoids and Actinoids
anode
71. The reason for greater range of oxidation states
(c) Sulphur is liberated at cathode, oxygen at
in actinoids is attributed to :- [2017]
anode
(a) actinoid contraction
(d) Oxygen is liberated at cathode, copper at
anode (b) 5f, 6d and 7s levels having comparable
energies
64. Cinnabar is an ore of [1991]
(c) 4f and 5d levels being close in energies
(a) Hg (b) Cu
(c) Pb (d) Zn (d) the redioactive nature of actinoids
65. The composition of ‘golden spangles’ is 72. The electronic configurations of Eu(Atomic No.
63), Gd(Atomic No. 64) and Tb (Atomic No. 65)
[1990]
are [2016]
(a) PbCrO4 (b) PbI2
(a) [Xe]4f76s2, [Xe]4f 8 6s2 and [Xe]4f 85d16s2
(c) As 2S3 (d) BaCrO4 (b) [Xe]4f 7 5d1 6s 2 , [Xe]4f 7 5d1 6s2 and
66. Prussian blue is formed when [1989] [Xe]4f 96s2
(a) Ferrous sulphate reacts with FeCl3 (c) [Xe]4f 6 5d 1 6s 2 , [Xe]4f 7 5d 1 6s 2 and
(b) Ferric sulphate reacts with K4[Fe(CN)6] [Xe]4f 85d16s2
(c) Ferrous ammonium sulphate reacts with FeCl3 (d) [Xe]4f 76s2, [Xe]4f 75d16s2 and [Xe]4f 96s2
(d) Ammonium sulphate reacts with FeCl3 73. Because of lanthanoid contraction, which of the
67. Photographic films and plates have an essential following pairs of elements have nearly same
ingredient of [1989] atomic radii ? (Numbers in the parenthesis are
atomic numbers). [2015]
(a) Silver nitrate (b) Silver bromide
(a) Zr (40) and Nb (41) (b) Zr (40) and Hf (72)
(c) Sodium chloride (d) Oleic acid
68. Nitriding is the process of surface hardening of (c) Zr (40) and Ta (73) (d) Ti (22) and Zr (40)
steel by treating it in an atmosphere of [1989] 74. Gadolinium belongs to 4f series. It's atomic
(a) NH3 (b) O3 number is 64. Which of the following is the
correct electronic configuration of gadolinium ?
(c) N2 (d) H2S
[1997, NEET Kar. 2013, 2015 RS]
69. While extracting an element from its ore, the ore
is ground and leached with dil. potassium (a) [Xe]4f 86d2 (b) [Xe]4f 95s1
(c) [Xe] 4f 75d16s2 (d) [Xe] 4f 65d26s2
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216 CHEMISTRY
75. Which of the following exhibit only + 3 oxidation 79. The correct order of ionic radii of Y3+, La3+, Eu3+
state ? [2012 M] and Lu3+ is [2003]
(a) U (b) Th (a) La 3+ < Eu 3+ < Lu 3+ < Y 3+
(c) Ac (d) Pa
(b) Y 3+ < La 3+ < Eu 3+ < Lu 3+
76. Identify the incorrect statement among the
following: [2007]
(c) Y 3+ < Lu 3+ < Eu 3+ < La 3+
(a) Lanthanoid contraction is the accumulation (d) Lu 3+ < Eu 3+ < La 3+ < Y 3+
of successive shrinkages. (Atomic nos. Y =39, La = 57, Eu = 63, Lu = 71)
(b) As a result of lanthanoid contraction, the 80. General electronic configuration of lanthanides
properties of 4d series of the transition is [1991, 2002]
elements have no similarities with the 5d (a) (n – 2) f 1 –14 (n –1) s2p6d 0 – 1 ns2
series of elements. (b) (n – 2) f10 –14 (n –1) d 0 – 1 ns2
(c) Shielding power of 4f electrons is quite (c) (n – 2) f 0 –14 (n –1) d10 ns2
weak.
(d) (n – 2) d0 –1 (n –1) f 1 – 14 ns2
(d) There is a decrease in the radii of the atoms
or ions as one proceeds from La to Lu. 81. Which of the following statement is not correct?
77. The main reason for larger number of [2001]
oxidation states exhibited by the actinoids (a) La (OH)3 is less basic than Li(OH)3
than the corresponding lanthanoids, is (b) La is actually an element of transition series
[2005, 2006] rather lanthanides
(a) more energy difference between 5f and (c) Atomic radius of Zr and Hf are same
because of lanthanide contraction
6d orbitals than between 4f and 5d
(d) In lanthanide series ionic radius of Ln +3
orbitals.
ions decreases
(b) lesser energy difference between 5f and
6d orbitals than between 4f and 5d 82. The lanthanide contraction is responsible for
orbitals. the fact that [1997]
(c) larger atomic size of actinoids than the (a) Zr and Y have about the same radius
lanthanoids. (b) Zr and Nb have similar oxidation state
(d) greater reactive nature of the actinoids (c) Zr and Hf have about the same radius
than the lanthanoids.
(d) Zr and Zn have the same oxidation states
78. Lanthanoids are [2004]
(Atomic numbers : Zr = 40, Y = 39, Nb = 41,
(a) 14 elements in the sixth period
(atomic no. = 90 to 103) that are filling 4f Hf = 72, Zn = 30)
sublevel 83. Actinides [1994]
(b) 14 elements in the seventh period (a) Are all synthetic elements
(atomic no. = 90 to 103) that are filling 5f
(b) Include element 104
sublevel
(c) Have any short lived isotopes
(c) 14 elements in the sixth period
(atomic no. = 58 to 71) that are filling 4f (d) Have variable valency
sublevel 84. Among the lanthanides the one obtained by
(d) 14 elements in the seventh period synthetic method is [1994]
(atomic no. = 58 to 71) that are filling 4f sub (a) Lu (b) Pm
level
(c) Pr (d) Gd
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The d-and f-Block Elements 217
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)
3 (c) 12 (c) 21 (a) 30 (b) 39 (a) 48 (b) 57 (a) 66 (b) 75 (c) 84 (b)
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)
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218 CHEMISTRY
8. (a) The shape of f-orbitals is very much 15. (b) La3+ : 54 e– = [Xe]
diffused and they have poor shielding effect. To3+ : 19 e– = [Ar] 3 d 1 (Colour)
The effective nuclear charge increases due to Lu3+ : 68 e– = [Xe] 4 f 14
imperfect shielding of one electron by another in Sc3+ : 18 e– = [Ar]
the same subshell which causes the contraction in
the size of electron charge cloud. This contraction 16. (d) Ni3+ : [Ar] 3d 7
in size is known as lanthanoid contraction. Mn3+ : [Ar] 3d 4
9. (c) Sm2+(Z = 62) Fe3+ : [Ar] 3d 5
[Xe]4f 6 6s2 – 6 unpaired e– Co3+ : [Ar] 3d 6
Eu2+(Z = 63) 17. (c) Due to lanthanide contraction, the size of
[Xe]4f7 6s2 – 7 unpaired e– Zr and Hf (atom and ions) become nearly similar
Yb2+(Z = 70) 18. (a) +3 oxidation state is most common for
[Xe]4f14 6s2 – 0 unpaired e– lanthanoids. But occasionally +2 and +4 states
Ce2+(Z = 58) are also obtained due to extra stability of
[Xe]4f1 5d1 6s2 – 2 unpaired e– empty, half-filled or filled f-subshell.
Only Yb2+ is diamagnetic. 3d 4s
10. (c) A transition element must have incomplete 19. (b) Mn - 3d5 4s2
d-subshell. Zinc have completely filled d The no. of various oxidation states possible are
subshell having 3d10 configuration. Hence do + 2, + 3, + 4, (+ 5), + 6, + 7.
not show properties of transition elements to 20. (b) In TiF62– ,– Ti is in + 4 O.S. ; 3d 0 = colourless
any appreciable extent except for their ability to
form complexes. In CoF63– –, Co is in + 3 O.S ; 3d5 = coloured
11. (c) The melting points of the transition In Cu2Cl2– Cu is in +1 O.S. ; 3d10 – colourless
elements first rise to a maximum and then fall as
the atomic number increases, manganese have In NiCl 2– 8
4 – Ni is in + 2 O.S ; 3d – coloured
abnormally low melting points. The colour exhibited by transition metal ions is
12. (c) E° +2 = 0.34 volt, due to the presence of unpaired electrons in
Cu /Cu d-orbitals which permits the d - d excitation of
Other has – ve E°R.P. electrons.
E° ++ = – 0.28
Co /Co 21. (a) Ti ; Z (22) is 1s22s22p63s23p64s23d2
E° ++ = – 0.25 V ; Z (23) is 1s22s22p63s23p64s23d3
Ni / Ni
E° ++ Cr ; Z (24) is 1s22s22p63s23p63d54s1
Fe /Fe = – 0.44
Mn ; Z (25) is 1s22s22p63s23d54s2
13. (c) The transition metals and their compounds The second electron in all the cases (except Cr)
are used as catalysts because of the variable is taken out from 4s-orbital and for Cr it is taken
oxidation states. Due to this, they easily absorb
from half-filled 3d-orbital. The force required for
and re-emit wide range of energy to provide the
removal of second electron will be more for Mn
necessary activation energy.
than others (except for Cr) due to having more
14. (a) Cr2+ = positive charge. Based on this, we find the
2+ correct order Mn > V > Ti.
Mn =
2+ i.e. Cr > Mn > V > Ti.
Fe = 3+
22. (d) V =
Co2+=
Ti3+=
Mn 2+ is most stable due to half-filled
Mn3+ =
configuration. Hence, only option (a) can be the
correct answer. Cr3+=
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The d-and f-Block Elements 219
Cr3+ will be most stable in aqueous solution 27. (a) The basic character of the transition metal
according to crystal field theory. It has t
3 monoxide is TiO > VO > CrO > FeO because
2g
configuration, which is most stable among the basic character of oxides decrease with increase
given options. in atomic number.
23. (d) Sc3+ : 1s2, 2s2p6, 3s2p6d0, 4s0; no unpaired Oxides of transitional metals in low oxidation state
electron. i.e., + 2 and + 3 are generally basic except Cr2O3.
Cu+ : 1s2, 2s2p6, 3s2p6d10, 4s0; no unpaired
electron. 28. (a) Transition metals have the ability to utilize
Ni 2+: 1s2, 2s2p6 , 3s2 p6 d 8, 4s0; 2 unpaired (n –1)d -orbitals. Thus variable oxidation states
electron present. enables the transition element to associate with
Ti3+ : 1s2, 2s2p6, 2s2p6d1, 4s0; 1 unpaired electron the reactants in different forms.
present 29. (c) Mn : [Ar] 3d5 4s2
Co2+ : 1s2 , 2s2 p6 , 3s2 p6d7 , 4s0 ; 3 unpaired Shows +2, +3, +4, +5, +6 & +7 oxidation states
electron present 30. (b) Manganese shows max. no. of oxidation
So, from the given option, the only correct states,
combination is Ni2+ and Ti3+ +2, +3, + 4, + 5, + 6, + 7
24. (a) Sc3+® 3d0 4s0 Other metals shows the following oxidation
Fe2+® 3d6 4s0 states
Cr = + 2, +3, + 4, + 5, + 6
4 unpaired e– Fe = + 2, + 3
Ti3+® 3d14s0 Ti = + 2, + 3, + 4
1 unpaired e– 31. (a) Sc +3 ® [Ar]18 3d 0 ; V +3 ® [Ar]18 3d 2
Mn2+ ® 3d5 4s0
Ti +3 ® [Ar]18 3d1 ; Cr +3 ® [Ar]18 3d 3
5 unpaired e– Scandium ion doesn’t contain any unpaired
3+
In Sc there is no unpaired electron. So the electron in d orbitals hence it forms colourless
aqueous solution of Sc3+ will be colourless. compound.
25. (c) For third ionization enthalpy electronic 32. (d) Eu – + 2, + 3
configuration of La – +3
3d 4s
2+ – 4s0 3d3 Gd – + 3
23V Am = +2, +3, +4, +5, +6
2+ – 4s0 3d4 33. (d) We know that chromium (III) salts dissolve
24Cr
2+ 0 5
in water to give violet solution. The violet colour
25Mn – 4s 3d is due to the hydrated chromium (III) in
2+ – 4s0 3d6
26Fe [Cr(H2O)6]3+.
Mn has most stable configuration due to half 34. (b) Ti (22) = [Ar] 3d2 4s2
filled d-orbital. Hence, 3rd ionization energy will minimum oxidation state is +2 and maximum
be highest for Mn. oxidation state is +4.
26. (d) The electronic configuration of different Ti4+ = [Ar]
species given in the option (d) are : Hence, it is most stable state of Ti.
Ti 2+ :1s 2 2s 2 2p 6 3s 2 3p 6 3d 2 Ti3+ = [Ar]3d1
It acts as a good reducing agent as it can readily
V3+ :1s 2 2s 2 2p6 3s 2 3p6 3d 2 oxidized to Ti4+ ion.
Ti2+ = [Ar]3d2
Cr 4+ :1s 2 2s 2 2p 6 3s 2 3p 6 3d 2
It is the most unstable form.
Mn5+ :1s 2 2s 2 2p6 3s 2 3p 6 3d 2 Thus, +3 and +4 are the common oxidation states.
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220 CHEMISTRY
The minimum oxidation state of transition metals 44. (c) Both are reducing agents.
is equal to the number of electrons in 4s shell and 45. (c) The structure of CrO5 is
the maximum oxidation state is equal to the sum of
O O
the 4s and 3d electrons.
Cr
35. (b) Electronic configuration of Cu (29) is 1s2
2s 2 2p6 3s2 3p6 4s1 3d10 and not 1s2, 2s2 2p6 3s2 O O
3p6 4s2, 3d9 due to extra stability of fully filled O
orbitals. Hence, CrO5 has two proxy linkage.
36. (c) [Ne]3s23p5 is the electronic configuration Now suppose the oxi. no. of Cr is x then
of a p-block element whereas other x + (– 1 × 4) + (– 2) = 0
configurations are those of d-block elements \ x=+6
37. (a) Argentite or silver glance (Ag2S) is an ore 46. (b) 2 KMnO4 + 3H2SO4 + 5H2O2 ¾®
of Ag.
38. (d) In moist air, corrosion of iron takes place in K2SO4 + 2MnSO4 + 8H2O + 5O2
i.e. Mn2+ ion and O2.
the form of rust (Fe 2O 3 + Fe(OH)3 ) . As iron is
most reactive among all. 47. (b) HCl and SO2 are reducing agents and can
39. (a) Potassium permanganate has a purple colour. reduce MnO4–. CO2 which is neither oxidising
When sulphur dioxide reacts with potassium and nor reducing will provide only acidic
permanganate, the solution decolourises. medium. It can shift reaction in forward direction
5SO2 + 2KMnO4 + 2H2O ® 2H2SO4 + 2MnSO4 and reaction can go to completion.
+ K2SO4 48. (b) Na2Cr 2O7 is hygroscopic in nature and
40. (d) K2Cr2O7 + SO2 + H2SO4 therefore accurate weighing is not possible in
® K2SO4 + Cr 2(SO4)3 + H2O normal atmospheric conditions. A hygroscopic
green colour substance absorbs moisture from atmosphere
41. (a) Less active metals like Cu, Ag etc. react and this could lead to inaccuracies in weight.
with conc. acid and form nitrate and nitrogen
dioxide. 49. (a) The green colour appears due to the
42. (b) Formation of Fe(CO)5 from Fe involves no formation of Cr3+ion
change in oxidation state of iron. 2– 2– 2–
Cr2O7 + 3SO3 + 8H+ ¾® 3SO4 + 2Cr3+ + 4H2O
43. (a) Considering the same moles of compounds 50. (a) Copper sulphate on treatment with excess
Fe2+ SO2–
4 only Fe
2+ is oxidised by KMnO .
4 of KCN forms complex K 3 [Cu(CN) 4 ] or
( +6) [Cu(CN)4]3–
SO2–
4 in which sulphur is in highest oxidation
state cannot be oxidised. CuSO 4 + 2KCN ¾
¾® Cu ( CN ) 2 + K 2SO 4
Fe 2+ SO32– ¾¾ ® Fe3+ 2 Cu ( CN ) 2 ¾
¾® Cu 2 (CN ) 2 + (CN ) 2
(+4) (+6)
2–
SO3 —® SO4 + 2e
2– –
Cu2(CN)2 + 3KCN ¾® K3[Cu(CN)4] + CuCN
Sulphur can be oxidised to (+6) i.e to SO 2– 51. (c) German Silver Composition : Cu = 56%,
4
Zn = 24%, Ni = 20%
In FeC2 O4 ¾¾ ® Fe3+
Carbon can be oxidised to (+IV) i.e to CO2 52. (a) We know that when nitrogen oxide (NO)
combines with Fe (II) ions, a brown complex is
Fe(NO 2 ) 2 ¾¾® Fe3+
formed. This reaction is called brown ring test
(+3)
+ 2NO2– Nitrogen can be 53. (d) By heating chromite with Na2 CO3, Na2CrO4
oxidised to (+V) is obtained
¯ – – state
2NO3 + 2e 4FeCr2O4 + 8Na2CO3 + 7O2 ¾®
8Na2CrO4 + 2Fe2O3 + 8CO2
(+5)
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The d-and f-Block Elements 221
54. (b) When excess of aq. HNO3 to [C4(H3)4]2+ 63. (b) CuSO 4 Cu 2+ + SO 24 - ;
produces [(4(H2O)4]2+. H + + OH -
H 2 O
Following equillibrium exist in the given aqueous At cathode : Cu2+ + 2e– ¾® Cu
solution At anode : 4OH– ¾® 2H2O + O2 + 4e–
[Cu(H2O)4]2+(aq) + 4NH3(aq) 64. (a) Cinnabar (HgS) is an ore of Hg.
[Cu(NH3)4]2+(aq) + 4H2O(l) 65. (b) PbI2 is yellow and is called golden spangles.
` When HNO3 is added to this solution: 66. (b) Potassium ferrocyanide solution is added
(excess) HNO3 (aq) +NH3 (aq) ¾® to Fe3+ ions in solution to give deep blue solution
NH +4 (aq) + NO 3– (aq) or precipitate.
Thus, amount of NH3(aq) will reduce drastically. 2Fe2(SO4)3 + 3K4[Fe(CN)6] ¾®
With a simple application of Le Chatelier's
principle, the equillibrium will shift towards left Fe4[Fe(CN)6]3 + 6K2SO4
in the first reaction. Hence, [Cu(H2O)4]2+ will form. Prussian blue
67. (b) AgBr is highly photosensitive and is used
NH 4 OH
55. (a) X ¾¾ ¾¾® White ppt in photographic films and plates.
68. (a) When steel is heated in presence of NH3,
excess
¾¾ ¾
¾® Acidic solution (soluble ) iron nitride on the surface of steel is formed which
NaOH ( No. ppt with H 2S) imparts a hard coating. This process is called
Given reactions (white precipitate with H2S in nitriding.
presence of NH4OH) indicate that 'X' should be 69. (d) Cyanide process is used in the metallurgy of
ZnCl2 which explains all given reactions. Ag.
ZnCl2 + 2H2O ¾® Zn(OH)2 + 2HCl 2Ag 2S + 8 NaCN + O 2 + 2 H 2 O ¾
¾®
White fumes
4 Na[Ag(CN ) 2 ] + 4 NaOH + 2S
NH4OH + HCl ¾¾¾® NH4Cl
White fumes
- H 2O
Dense white fumes
2 Na[ Ag (CN ) 2 ] + Zn ¾¾®
ZnCl 2 + 2 NaOH ® Zn(OH) 2 + 2NaCl Na 2 [ Zn (CN ) 4 ] + 2 Ag ¯
2 NaOH
Zn (OH ) 2 ¾¾ ¾¾® Na 2 ZnO 2 + 2H 2 O 70. (b) 2CuSO4 + K4[Fe(CN)6] ¾®
Excess Cu2[Fe(CN)6] + 2K2SO4
56. (d) Pig iron is the most impure form of iron Chocolate ppt.
and contain highest proportion of carbon
(2.5 – 4%). 71. (b) Minimum or comparable energy gap
57. (a) K2Cr2O7 + 2NaOH ¾® between 5f, 6d and 7s subshell makes electron
K2CrO4 + Na2CrO4 + H2O excitation easier, hence there is a greater range
of oxidation states in actinoids.
Hence CrO42– ion is obtained.
72. (d) Eu (63) = [Xe] 4f 7 6s2
58. (b) CrO3 + 2NaOH ® Na 2CrO 4 + H 2 O
Gd (64) = [Xe] 4f 7 5d1 6s2
59. (d) Although Cu2+ state is more stable than Tb (65) = [Xe] 4f 9 6s2
Cu+, but Cu+ ion can be stabilized by formation 73. (b) Zr (40) and Hf (72) are in the same group
of insoluble substances such as CuCl, CuCN (4). Because of lanthanoid contraction, they have
and CuSCN. Cu2+ is more stable than Cu+ nearly same atomic radii.
because of its high hydration energy. 74. (c) Gd (64) = [Xe]4f 7 5d1 6s2
60. (d) Stainless steel contains 73% Fe, 18% Cr
75. (c) Ac (89) = [Rn] [6d1] [7s2]
and 8% Ni.
76. (b) Due to lanthanoid contraction the
61. (d) Galvanisation is the process of deposition
properties of 4d series of the transition element
of zinc metal on the surface of Fe to prevent it have similarities with the 5d series of elements.
from rusting. Zinc forms a protective layer of basic
carbonate (ZnCO3. Zn(OH)2) on it. There is a steady decrease in the radii as the atomic
number of the lanthanide elements increases. For
Heat every additional proton added in nucleus the
62. (a) (NH 4 ) 2 Cr2 O7 ¾¾¾® Cr2O3 + N 2 + 4H 2O
corresponding electron goes to 4f subshell.
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222 CHEMISTRY
77. (b) Lesser energy difference between 5f and 82. (c) We know that regular decrease in the size
6d orbitals than between 4f and 5d orbitals result of the atoms and ions is called lanthanide
in larger no. of oxidation state. contraction. In vertical column of transition
78. (c) Lanthanides are 4 f-series elements starting elements there is a very small change in size and
from cerium (Z = 58) to lutetium ( Z = 71). These some times size is found to be same from second
are placed in the sixth period and in third group. member to third member.The similarity in size of
the atoms of Zr and Hf is evident due to the fact
79. (c) In lanthanide series there is a regular
of lanthanide contraction. Therefore, Zr and Hf
decrease in the atomic as well as ionic radii of
both have same radius 160 pm.
trivalent ions (M3+) as the atomic number
increases. Although the atomic radii do show 83. (d) Actinides have variable valency due to
some irregularities but ionic radii decreases from very small difference in energies of 5f, 6d and 7s
La(103 pm) to Lu (86pm). orbitals. Actinides are the elements from atomic
number 89 to 103.
80. (a) The Lanthanides are transition metals from
atomic numbers 58 (Ce) to 71(Lu). 84. (b) Pm is obtained by synthetic method.
Hence the electronic configuration becomes : Promethium is not present in nature. It is synthetic
1– 14
radioactive lanthanoid.
(n –2) f (n – 1) s2p6 d0 – 1 ns2.
81. (a) La (OH)3 is more basic than Li (OH)3. In
lanthanides, the basic character of hydroxides
decreases as the ionic radius decreases.
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Coordination Compounds 223
23 Coordination
Compounds
Trend Analysis with Important Topics & Sub-Topics
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224 CHEMISTRY
(a) Hexacyanoiron (III) ion [2015 RS]
with the composition of [Co(NH3 )4 Cl2 ]+ is due
(b) Hexacyanitoferrate (III) ion
to : [2010]
(c) Tricyanoferrate (III) ion
(a) linkage isomerism
(d) Hexacyanidoferrate (III) ion
(b) geometrical isomerism
6. Number of possible isomers for the complex
(c) coordination isomerism
[Co(en)2Cl2]Cl will be (en = ethylenediamine)
(d) ionization isomerism
(a) 2 (b) 1 [2015 RS]
13. Which one of the following complexes is not
(c) 3 (d) 4
expected to exhibit isomerism? [2010]
7. An excess of AgNO3 is added to 100 mL of a
0.01 M solution of dichlorotetraaquachromium (a) [ Ni(en)3 ]2+
(III) chloride. The number of moles of AgCl
precipitated would be : [NEET 2013] 2+
(b) é Ni ( NH3 ) ( H 2 O ) ù
(a) 0.002 (b) 0.003 ë 4 2û
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Coordination Compounds 225
17. Which one of the following is expected to (c) [Co(NH2)3 (H2O)2 Cl]
exhibit optical isomerism? (d) [Co(NH3)3 (H2O)3]Cl3
(en = ethylenediamine) [2005] 23. Which of the following will give maximum
(a) cis-[Pt(NH3)2 Cl 2] 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]
(d) trans-[Co(en)2Cl 2 ] 24. Which of the following will exhibit maximum
ionic conductivity? [2001]
18. Which of the following is considered to be an
anticancer species ? [2004] (a) K4[Fe(CN)6] (b) [Co(NH3)6] Cl3
(c) [Cu(NH3)4]Cl2 (d) [Ni(CO)4]
Cl CH2 Cl Cl 25. Oxidation number of Ni in [Ni(C2O4)3]4– is
(a) Pt CH2 (b) Pt
(a) 3 (b) 4 [2001]
Cl Cl Cl Cl
(c) 2 (d) 6
Cl 26. Which one of the following complexes will have
H3N H3N Cl
four different isomers ? [2000]
(c) Pt (d) Pt
H3N Cl Cl NH3 (a) [ Co (en ) 2 Cl 2 ]Cl
19. Which of the following coordination compounds (b) [Co(en)(NH3 )2 Cl2 ]Cl
would exhibit optical isomerism? [2004]
(c) [Co(PPh 3 )2 Cl2 ]Cl
(a) pentaamminenitrocobalt(III) iodide
(b) diamminedichloroplatinum(II) (d) [ Co (en ) 3 ]Cl 3
(c) trans-dicyanobis (ethylenediamine) 27. In which of the following compounds does iron
chromium (III) chloride exhibit zero oxidation state? [1999]
(d) tris-(ethylendiamine) cobalt (III) bromide (a) [Fe(H2O)6] (NO3)3 (b) K3[Fe(CN)6]
20. Which one of th e following octahedral (c) K4[Fe(CN)6] (d) [Fe(CO)5]
complexes will not show geometric isomerism? 28. The total number of possible isomers for the
(A and B are monodentate ligands) [2003] complex compound [CuII (NH3)4] [PtII Cl4]
(a) [MA5B] (b) [MA2B4] (a) 3 (b) 6 [1998]
(c) [MA3B3] (d) [MA4B2] (c) 5 (d) 4
21. According to IUPAC nomenclature sodium 29. IUPAC name of [Pt(NH3)3 (Br) (NO2) Cl] Cl is
nitroprusside is named as [2003] [1998]
(a) Sodium pentacyanonitrosyl ferrate (III) (a) Triamminechlorobromonitroplatinum (IV)
(b) Sodium nitroferrocyanide chloride
(c) Sodium nitroferrocyanide (b) Triamminebromonitrochloroplatinum (IV)
chloride
(d) Sodium pentacyanonitrosyl ferrate (II)
(c) Triamminebromochloronitroplatinum (IV)
22. The hypothetical complex chloridodiaquatriammine
chloride
cobalt (III) chloride can be represented as [2002]
(d) Triamminenitrochlorobromoplatinum (IV)
(a) [CoCl(NH3)3 (H2O)2]Cl2
chloride
(b) [Co(NH3)3 (H2O)Cl3]
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226 CHEMISTRY
30. A coordination complex compound of cobalt has (a) [Co( NH 3 ) 6 ]3+
the molecular formula containing five ammonia
molecules, one nitro group and two chlorine (b) [ Co ( H 2 O ) 2 ( NH 3 ) 2 Cl 2 ] +
atoms for one cobalt atom. One mole of this
compound produces three mole ions in an (c) [Cr(H 2 O) 2 Cl 2 ]+
aqueous solution. On reacting this solution with
excess of AgNO3 solution, we get two moles of (d) [Co(CN) 5 NC]
AgCl precipitate. The ionic formula for this 37. An example of double salt is [1989]
complex would be [1998]
(a) Bleaching powder (b) K 4 [Fe(CN) 6 ]
(a) [Co(NH3)4 (NO2) Cl] [(NH3) Cl]
(c) Hypo (d) Potash alum
(b) [Co (NH3)5 Cl] [Cl (NO2)]
Topic 2: Magnetic Moment, Valence Bond
(c) [Co (NH3)5 (NO2)] Cl2 Theory and Crystal Field Theory
(d) [Co (NH3)5] [(NO2)2Cl2]
38. Which of the following is the correct order of
31. The formula for the complex, dichlorobis (urea) increasing field strength of ligands to form
copper (II) is [1997] coordination compounds? [2020]
(a) [Cu{O = C (NH2)2}] Cl2 (a) SCN– < F– < CN– < C2 O 24 -
(b) [Cu{O = C (NH2)2}Cl]Cl
(b) F– < SCN–< C2 O 24 - < CN–
(c) [CuCl2 {O = C(NH2)2}2]
(c) CN– < C2 O 24 - < SCN– < F–
(d) [CuCl2] [{O = C (NH2)2}]2
32. The number of geometrical isomers of the (d) SCN– < F– < C2 O 24 - < CN–
complex [Co(NO2)3 (NH3)3] is [1997] 39. The calculated spin only magnetic moment of
(a) 2 (b) 3 Cr2+ ion is [2020]
(c) 4 (d) zero (a) 4.90 BM (b) 5.92 BM
(c) 2.84 BM (d) 3.87 BM
33. Among the following, the compound that is both
paramagnetic and coloured, is [1996] 40. The Crystal Field Stabilisation Energy (CFSE)
for [CoCl6]4– is 18000 cm–1. The CFSE for
(a) KMnO4 (b) CuF2
[CoCl4]2– will be [NEET Odisha 2019]
(c) K4[Fe(CN)6] (d) K2Cr2O7 –1
(a) 8000 cm (c) 6000 cm–1
34. The number of geometrical isomers for –1
(c) 16000 cm (d) 18000 cm–1
[Pt (NH3)2 Cl2] is [1995]
41. What is the correct electronic configuration of
(a) 2 (b) 1 the central atom in K4[Fe(CN)6] based on crystal
(c) 3 (d) 4 field theory ? [2019]
35. K 3[Al(C 2O 4 )3 ] is called [1994] (a) t24g eg2 (b) t26g eg0
(a) Potassium alumino oxalate (c) e3g t23g (d) eg4 t22g
(b) Potassium trioxalatoaluminate (III) 42. The geometry and magnetic behaviour of the
(c) Potassium aluminium (III) oxalate complex [Ni(CO)4] are [2018]
(d) Potassium trioxalato aluminate (VI) (a) Square planar geometry and diamagnetic
36. Among the following complexes, optical activity (b) Tetrahedral geometry and diamagnetic
is possible in [1994] (c) Tetrahedral geometry and paramagnetic
(d) Square planar geometry and paramagnetic
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Coordination Compounds 227
43. Match the metal ions given in Column I with (c) [Co(CN)6]3– has no unpaired electrons and
the spin magnetic moments of the ions given in will be in a high-spin configuration.
Column II and assign the correct code : [2018] (d) [Co(CN)6]3– has no unpaired electrons and
Column I Column II will be in a low-spin configuration.
a. Co3+ i. 8 BM 48. Among the following complexes the one which
b. Cr3+ ii. 35 BM shows zero crystal field stabilization energy
(CFSE): [2014]
c. Fe3+ iii. 3 BM
(a) [Mn(H2O)6]3+ (b) [Fe(H2O)6]3+
d. Ni2+ iv. 24 BM
(c) [Co(H2O)6]2+ (d) [Co(H2O)6]3+
v. 15 BM 49. Which of the following complexes is used as an
a b c d anti-cancer agent: [2014]
(a) iv v ii i (a) mer-[Co(NH3)3Cl3] (b) cis-[PtCl2(NH3)2]
(b) i ii iii iv (c) cis-K2[PtCl2Br2] (d) Na2CoCl4
(c) iii v i ii 50. A magnetic moment of 1.73 BM will be shown
(d) iv i ii iii by one among the following : [NEET 2013]
44. HgCl2 and I2 both when dissolved in water (a) [Ni(CN)4]2– (b) [TiCl4]
containing I – ions, the pair of species formed is: (c) [CoCl6]4– (d) [Cu(NH3)4]2+
(a) HgI2, I– (b) HgI42 - , I3- [2017] 51. Which is diamagnetic? [NEET Kar. 2013]
(a) [Fe(CN)6]3– (b) [Co(F6)]3–
(c) Hg2I2, I– (d) HgI2 , I3- (c) [Ni(CN)4]2– (d) [NiCl4]2–
45. Correct increasing order for the wavelengths of 52. The anion of acetylacetone (acac) forms
absorption in the visible region the complexes Co(acac)3 chelate with Co3+. The rings of the
of Co3+ is :- [2017] chelate are [NEET Kar. 2013]
(a) [Co(H2O)6]3+, [Co(en)3]3+, [Co(NH3)6]3+ (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
(c) [Co(NH3)6]3+, [Co(en)3]3+, [Co(H2O)6]3+ 53. Which among the following is a paramagnetic
(d) [Co(en)3]3+, [Co(NH3)6]3+, [Co(H2O)6]3+ complex? [NEET Kar. 2013]
46. Pick out the correct statement with respect to (a) [Mo(CO)6] (b) [Co(NH3)6]3+
[Mn(CN)6]3- [2017] (c) [Pt(en)Cl2] (d) [CoBr4]2–
(a) It is sp3d2 hybridised and tetrahedral (At. No. of Mo = 42, Pt = 78)
(b) It is d2sp3 hybridised and octahedral 54. Which one of the following is an outer orbital
(c) It is dsp2 hybridised and square planar complex and exhibits paramagnetic behaviour ?
(d) It is sp3d2 hybridised and octahedral [2012]
47. Which of these statements about [Co(CN)6]3– (a) [Ni(NH3)6]2+ (b) [Zn(NH3)6)]2+
is true ? [2015] (c) [Cr(NH3)6]3+ (d) [Co(NH3)6]3+
(a) [Co(CN)6]3– has four unpaired electrons
55. Red precipitate is obtained when ethanol
and will be in a low-spin configuration.
solution of dimethylglyoxime is added to
(b) [Co(CN)6]3– has four unpaired electrons ammoniacal Ni(II). Which of the following
and will be in a high spin configuration. statements is not true ? [2012 M]
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228 CHEMISTRY
(a) Red complex has a square planar geometry. 61. Crystal field stabilization energy for high spin
(b) Complex has symmetrical H-bonding d 4 octahedral complex is: [2010]
(c) Red complex has a tetrahedral geometry. (a) – 1.8 D 0 (b) – 1.6 D 0 + P
(d) Dimethylglyoxime functions as bidentate
ligand. (c) – 1.2 D 0 (d) – 0.6 D 0
62. Which of the following complex ions is expected
OH
H3C C N to absorb visible light? [2009]
dimethylglyoxime =
H3C C N (a) [Ti (en)2(NH3)2]4 +
OH
(b) [Cr (NH3)6]3 +
56. Low spin complex of d 6-cation in an octahedral (c) [Zn (NH3)6]2 +
field will have the following energy : [2012 M]
(d) [Sc (H2O)3 (NH3)3]3+
-12 -12
(a) Do + P (b) D o + 3P (At. no. Zn = 30, Sc = 21, Ti = 22, Cr = 24)
5 5
63. Which of the following complexes exhibits the
-2 -2
(c) D o + 2P (d) Do + P highest paramagnetic behaviour ? [2008]
5 5
(a) [V(gly)2(OH)2(NH3)2]+
(D o= Crystal Field Splitting Energy in an
octahedral field, P = Electron pairing energy) (b) [Fe(en)(bpy)(NH3)2]2+
57. Of the following complex ions, which is (c) [Co(ox)2(OH)2]–
diamagnetic in nature ? [2011]
(d) [Ti(NH3)6]3+
(a) [NiCl4]2– (b) [Ni(CN)4]2–
where gly = glycine, en = ethylenediamine and
(c) [CuCl4]2– (d) [CoF6]3–
bpy = bipyridyl moities)
58. The d-electron configurations of Cr 2+, Mn2+,
Fe2+ and Co2+ are d 4, d 5, d 6 and d 7, respectively. (At.nosTi = 22, V = 23, Fe = 26, Co = 27)
Which one of the following will exhibit minimum 64. In which of the following coordination entities
paramagnetic behaviour? [2011] the magnitude Do (CFSE in octahedral field) will
(a) [Mn(H2O)6]2+ (b) [Fe(H2O)6]2+ be maximum? [2008]
(c) [Co(H2O)6]2+ (d) [Cr(H2O)6]2+ (a) [Co(H2O)6]3+ (b) [Co(NH3)6]3+
(At, nos. Cr = 24, Mn = 25, Fe = 26, Co = 27) (c) [Co(CN)6]3– (d) [Co (C2O4)3]3–
59. Which of the following complex compounds will
(At. No. Co = 27)
exhibit highest paramagnetic behaviour?
[2011M] 65. The d electron configurations of Cr2+, Mn2+, Fe2+
(At. No. : Ti = 22, Cr = 24, Co = 27, Zn = 30) and Ni2+ are 3d 4, 3d 5, 3d 6 and 3d 8 respectively.
(a) [Ti (NH3)6]3+ (b) [Cr (NH3)6]3+ Which one of the following aqua complexes will
exhibit the minimum paramagnetic behaviour?
(c) [Co (NH3)6]3+ (d) [Zn (NH3)6]2+
60. Which of the following complex ion is not [2007]
expected to absorb visible light ? [2010] (a) [Fe(H2O)6]2+ (b) [Ni(H2O)6]2+
(a) [ Ni(CN)4 ] 2-
(b) [Cr(NH3 )6 ] 3+ (c) [Cr(H2O)6]2+ (d) [Mn(H2O)6]2+
(c) [ Fe(H2 O)6 ]2+ (d) [ Ni(H 2O)6 ]2+ (At. No. Cr = 24, Mn = 25, Fe = 26, Ni = 28)
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Coordination Compounds 229
66. [Cr(H2O)6]Cl3 (at no. of Cr = 24) has a magnetic (a) [Cr(CO)6] (b) [Fe(CO)5]
moment of 3.83 B. M. The correct distribution of (c) [Fe(CN)6] –4 (d) [Cr(NH3)6]+3
3d electrons in the chromium of the complex is 73. CuSO4 when reacts with KCN forms CuCN,
which is insoluble in water. It is soluble in excess
( ) , 3d [2006]
1
(a) 3d xy1, 3d x 2 - y 2 yz
1 of KCN due to formation of the following complex
(b) 3dxy1, 3dyz1, 3dxz1 [2002]
(a) K2[Cu(CN)4] (b) K3[Cu(CN)4]
(c) 3dxy1, 3dyz1, 3d1z 2
(c) CuCN2 (d) Cu[K Cu(CN)4]
(d) (3dx2 – y2)1, 3d z 2 , 3dxz1 74. Which statement is incorrect? [2001]
67. Which one of the following is an inner orbital (a) [Ni(CO)4] – Tetrahedral, paramagnetic
complex as well as diamagnetic in behaviour? (b) [Ni(CN)4]2– – Square planar, diamagnetic
(Atomic number: Zn = 30, Cr = 24, Co = 27, (c) [Ni(CO)4] – Tetrahedral, diamagnetic
Ni = 28) [2005] (d) [NiCl4]2– – Tetrahedral, paramagnetic
(a) [Zn(NH3)6] 2+ (b) [Cr(NH3)6] 3+
75. Which one of the following will show
(c) [Co(NH3)6]3+ (d) [Ni(NH3)6]2+ paramagnetism corresponding to 2 unpaired
electrons?(Atomic numbers : Ni = 28, Fe = 26)
68. Among [ Ni( CO ) 4 ], [ Ni( CN ) 4 ] 2 - , [ NiCl 4 ] 2 -
(a) [Fe F6]3– (b) [Ni Cl4]2– [1999]
species, the hybridization states of the Ni atom 3–
are, respectively (At. No. of Ni = 28) [2004] (c) [Fe (CN)6] (d) [Ni (CN)4]2–
76. The number of unpaired electrons in the complex
(a) sp 3 , dsp2 , dsp 2 (b) sp3 , dsp 2 , sp3 [Cr(NH3)6]Br3 is (Atomic number Cr = 24)
(a) 4 (b) 1 [1999]
(c) sp3 , sp 3 , dsp 2 (d) dsp 2 , sp 3 , sp 3
(c) 2 (d) 3
69. CN– is a strong field ligand. This is due to the 77. Which of the following statements is correct ?
fact that [2004]
(Atomic number of Ni = 28) [1997]
(a) it carries negative charge
(a) [Ni(CO)4] is diamagnetic and [NiCl4]2– and
(b) it is a pseudohalide [Ni(CN)4]2– are paramagnetic
(c) it can accept electrons from metal species (b) [Ni(CO)4] and [Ni(CN)4]2– are diamagnetic
(d) it forms high spin complexes with metal and [NiCl4]2– is paramagnetic
species (c) [Ni(CO)4] and [NiCl4]2–are diamagnetic and
70. Considering H2O as a weak field ligand, the [Ni(CN)4]2– is paramagnetic
number of unpaired electrons in [ Mn ( H 2 O ) 6 ]2 + (d) [NiCl4]2– and [Ni(CN)4]2– are diamagnetic
and [Ni(CO)4] is paramagnetic
will be (At. no. of Mn = 25) [2004]
78. Which of the following is common donor atom
(a) three (b) five in ligands? [1995]
(c) two (d) four (a) arsenic (b) nitrogen
71. The number of unpaired electrons in the complex (c) oxygen (d) both 'b' and 'c'
ion [CoF6]3– is (Atomic no.: Co = 27) [2003] 79. The complex ion [Co(NH3)6]3+ is formed by sp3d 2
(a) Zero (b) 2 hybridisation. Hence, the ion should possess
(c) 3 (d) 4 (a) Octahedral geometry [1990]
72. Atomic number of Cr and Fe are respectively 25 (b) Tetrahedral geometry
and 26, which of the following is paramagnetic? (c) Square planar geometry
(d) Tetragonal geometry.
[2002]
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230 CHEMISTRY
Topic 3: Organometallic Compounds 85. Among the following, which is not the p-bonded
organometallic compound? [2003]
80. Iron carbonyl, [Fe(CO)5] is [2018]
(a) Tetranuclear (b) Mononuclear (a) (CH 3 ) 4 Sn
ANSWER KEY
1 (b) 10 (c) 19 (d) 28 (d) 37 (d) 46 (b) 55 (c) 64 (c) 73 (b) 82 (c)
2 (a) 11 (a) 20 (a) 29 (c) 38 (d) 47 (d) 56 (b) 65 (b) 74 (a) 83 (a)
3 (d) 12 (b) 21 (a) 30 (c) 39 (a) 48 (b) 57 (b) 66 (b) 75 (b) 84 (a)
4 (a) 13 (d) 22 (a) 31 (c) 40 (a) 49 (b) 58 (c) 67 (c) 76 (d) 85 (a)
5 (d) 14 (a) 23 (c) 32 (a) 41 (b) 50 (d) 59 (b) 68 (b) 77 (b) 86 (d)
6 (c) 15 (b) 24 (a) 33 (b) 42 (b) 51 (c) 60 (a) 69 (b) 78 (d) 87 (a)
7 (d) 16 (a) 25 (c) 34 (a) 43 (a) 52 (d) 61 (d) 70 (b) 79 (a)
8 (c) 17 (c) 26 (b) 35 (b) 44 (b) 53 (d) 62 (b) 71 (d) 80 (b)
9 (c) 18 (c) 27 (d) 36 (b) 45 (d) 54 (a) 63 (c) 72 (d) 81 (a)
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Coordination Compounds 231
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232 CHEMISTRY
Cl NH3 Cl Cl
M NH3
Cl Mer
fac.
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Coordination Compounds 233
17. (c)
en 23. (c) [Cr(SCN ) 2 ( NH 3 ) 4 ]+ shows linkage and
en + +
Cl Cl geometrical isomerism. Hence, produces maximum
no. of isomers.
Co Co
24. (a) The complex ion which give maximum ions
Cl Cl in solution exhibit maximum ionic conductivity.
en
en K 4 [ Fe(CN) 6 ] ® [ Fe (CN) 6 ]-4 + 4K + ®5 ions
Cis-d-isomer Cis-l-isomer
[Co( NH 3 ) 6 ]Cl 3 ® [Co( NH 3 ) 6 ]3+ + 3 Cl - ® 4 ions
Mirror
Trans-form of [M(AA) 2 a 2]n± does not show [Cu ( NH 3 ) 4 ]Cl 2 ® [Cu ( NH 3 ) 4 ]2+ + 2Cl - ® 3 ions
optical isomerism. [ Ni(CO) 4 ] ® No ions
18. (c) Diaminodichloroplatinum (II) commonly 25. (c) O.N. of Ni in [ N i(C 2 O 4 ) 3 ]
4-
|
Co en Cl
N N
N N trans (for Cl)
N en
en NH3
| Cl
N
d form l form
en – CO Plane of symmetry
| Cl
|
Enantiomers
NH3
The two optically active isomers are collectively trans (for NH 3)
called enantiomers. (ii) Optical isomers
20. (a) MA3 B3 – 2 geometrical isomers
Cl Cl
MA2 B4 – 2 geometrical isomers Cl
Cl
MA4 B2 – 2 geometrical isomers en CO CO en
NH3 H3N
The complexes of general formula Ma6 and Ma5b NH3 NH3
octahedral geometry do not show geometrical cis-isomer d- and l -forms
isomerism. (no element of symmetry)
21. (a) IUPAC name of sodium nitroprusside Hence, the compound has 4 different isomers
Na 2 [Fe(CN) 5 NO] is sodium pentacyanoni- 27. (d) In [Fe(CO)5] iron exist in zero oxidation state.
trosoyl ferrate (III) because in it NO is neutral 28. (d) The total number of isomers for the complex
ligand. Hence compound
2×O.N. of Na + O.N. of Fe + 5×O.N. of CN
[Cu II (NH3 )4 ][Pt II Cl 4 ] is four..
1×O.N. of NO = 0
2×(+1) + O.N. of Fe + 5 ×(–1) +1×0 = 0 These four isomers are
O.N. of Fe = 5 – 2 = +3, Hence, ferrate (III) [Cu ( NH 3 )3 Cl] [Pt ( NH 3 )Cl 3 ],
22. (a) Chloridodiaquatriammine cobalt (III)
[Pt(NH3 )3 Cl] [Cu(NH3 )Cl3 ] ,
chloride is [CoCl( NH 3 )3 (H 2O) 2 ]Cl 2
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234 CHEMISTRY
– – – 2– – 2–
[Pt(NH3 )4 ] [CuCl 4 ] I < Br < SCN < Cl < S < F < OH < C2O4
and [Cu(NH3 )4 ] [ PtCl4 ] . 4–
< H2O < NCS < EDTA < NH 3 < en < CN < CO
The isomer [Cu (NH3)2 Cl2][Pt (NH3)2 Cl2] does So, correct order is
not exist as both the parts are neutral.
SCN - < F- < C2 O42- < CN -
For writing all combination of structural isomers,
we have to keep the same value of primary valence Strong and weak field ligands affect the electronic
and secondary valence of central atom/ion for all configuration of elements of first transition series.
the isomers. Second and third transition series electrons pair
up irrespective of nature of ligand provided pairing
29. (c) We know that IUPAC name of of electrons is allowed.
[Pt(NH3)3 (Br) (NO2)Cl]Cl is triammine-
bromochloronitroplatinum (IV) chloride. 39. (a) Electronic configuration of Cr2+ – [Ar] 3d4
30. (c) As it forms two moles of silver chloride thus n=4
it has two moles of ionisable Cl.
m n = n(n + 2)
[Co(NH3 )5 NO2 ]Cl 2 ® [Co(NH3 )5 NO 2 ]2+ + 2Cl -
2Cl - + 2AgNO3 ® 2AgCl + 2NO 3- \ m n = 4(4 + 2) = 24 BM = 4.9 BM
31. (c) [CuCl2{(O = C(NH2)2}2] 4
40. (a) Q Dt = D0
32. (a) [Co( NO 2 )3 ( NH 3 )3 ] 9
4
Possible geometrical isomers Dt = × 18000 cm–1 = 8000 cm–1
NH3 NH3 9
H3N NH3 O2N NH3 41. (b)
4-
K 4 ëé Fe ( CN )6 ûù ® 4K + + [Fe(CN) 6 ]4 -
¾¾
Co Co
O2N NO2 NH3 NO2 x–6=–4
x=+2
NO2 NO2 Fe: 4s2 3d6
fac mer Fe2+ : 3d6
Total geometrical isomers = 2
33. (b) CuF2 is both paramagnetic and coloured eg
due to unpaired electron in d-orbital (3d 94s0). t2g
34. (a) [Pt (NH3)2 Cl2] is a disubstituted complex In spherical field
and shows only cis-& trans-isomers CN– is a strong field ligand and causes pairing of
H3N Cl electrons.
Cl NH 3
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Coordination Compounds 235
43. (a) Co3+ = [Ar] 3d 6 , unpaired e–(n) = 4 As, coordination number of Mn = 6, so it will
Spin magnetic moment form an octahedral complex.
\ [Mn(CN)6]3– =
= 4(4 + 2) = 24 B.M.
Cr3+ = [Ar] 3d 3 , unpaired e–(n) = 3 [Ar] ´´ ´´ ´´ ´´ ´´ ´´
Spin magnetic moment
3d 4s 4p
= 3(3 + 2) = 15 B.M. d2s 3
Hence, the increasing order of wavelengths of CN– is a strong ligand and causes pairing of 3d
absorption is: electrons of Ni2+.
[Co(en)3]3+ < [Co(NH3)6]3+ < [Co(H2O)6]3+ \ It is diamagnetic.
Higher value of D0 means higher stability, which 52. (d) Structure of Co(acac)3:
means it will absorb high energy photons. High
energy photons have low value of wavelength. C
O 1 O C
46. (b) In the complex [Mn(CN)6]3–, O.S. of Mn is + 3 C
Co O
E.C. of Mn+3 ® 3d 4 O
3d 4s 4p C O2
6 O
5 C C 3
The presence of a strong field ligand CN– causes 4
pairing of electrons. Hence, the rings of chelate are six membered ring.
53. (d) Co2+ Þ [Ar]3d74s0, here, Br– is a weak field
Þ ligand so will not cause pairing of d-electrons
3d 4s 4p in Co2+.
d 2 sp3 \ [CoBr 4 ] 2– will exhibit paramagnetic
behaviour due to unpaired electrons.
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236 CHEMISTRY
54. (a) [Ni(NH3)6 ]2+ accordance with the fact that the ligand involved
here is strong i.e., CN– ion.
6 2
Ni2+ = 3d 8, according to CFT = t 2g eg therefore, Ni2+ (after rearrangement)
hybridisation is sp 3 d 2 & complex is 3d 4s 4p
paramagnetic and outer orbital.
55. (c) Nickel ions are frequently detected by the
formation of red precipitate of the complex of Hence, now dsp2 hybridization involving one
nickel dimethylglyoxime, when heated with 3d, one 4s and two 4p orbitals, takes place leading
dimethylglyoxime. to four dsp2 hybrid orbitals, each of which
CH3 C NOH accepts four electron pairs from CN– ion forming
+ Ni2+ [Ni (CN)4]2– ion.
CH3 C NOH
[Ni (CN)4]2–
Dimethylglyoxime 3d 4s 4p
OH O
×× ×× ×× ××
CH3 C N 2+
N C CH3
Ni
CH3 C N N C CH3 four dsp2 hybrid bonds
Thus, the complex is diamagnetic as it has no
O OH unpaired electron.
Nickel dimethylglyoxime
58. (c) Cr2+, d 4
2+
Ni =
3d 4s 4p 4 unpaired e- s
dmg is a strong field ligand, so electrons will Mn2+, d 5
rearrange as:
2+
5 unpaired e- s
Ni = ×× ×× ×× ××
3d 4s 4p Fe2+, d 6
dsp 2 4 unpaired e- s
Thus, the complex is square plan ar geometry.
Co2+, d7
56. (b) for d 6, t2g6 eg0 (in low spin)
C.F.S.E = – 0.4 × 6D0 + 3P 3 unpaired e- s
Minimum paramagnetic behaviour
12
= - D + 3P = [Co (H2O)6]2+
5 o 59. (b)
[P is pairing energy for one electron] (a) [Ti(NH3)6]3+ : 3d1 configuration and thus has
57. (b) Ni2+ = 3d8 4sº one unpaired electron.
(b) [Cr(NH3)6]3+ : In this complex Cr is in +3
oxidation state.
Since, the coordination number of Ni in this Cr3+ : 3d 3 4so
complex is 4, the configuration of Ni2+ at first
sight shows that the complex is paramagnetic with
two unpaired electron. However, experiments Cr3+ ion [Cr (NH3)6]3+
show that the complex is diamagnetic. This is
possible when the 3d electrons rearrange against NH3NH3 NH3 NH3NH3NH3
the Hund’s rule as shown below. This is in
Thus, the complex is paramagnetic.
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Coordination Compounds 237
(c) [Co(NH3)6]3+ : In this complex, cobalt ion is Zn= [Ar] 3d 10 4s2; Zn2+= 3d10
in + 3 oxidation state with 3d 6 configuration. Sc = [Ar] 3d 1 4s2; Sc3+ = 3d 0
Co3+,[Ar]3d 6 63. (c) More the number of unpaired electrons present
3d 4s 4p 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 0 0
(inner orbital or d 2sp3 hybrid orbital low spin (b) [Fe(en)(bpy )(NH3 ) ]2 +
2
complex) diamagnetic x = +2
(d) In this complex Zn exists as Fe2+ = [Ar] 3d 6
Zn2+ ion But due to presence of strong field ligands, en,
Zn2+ ion : 3d10 4s0 bpy and NH3; the electrons will try to pair up.
Thus, the complex will contain no unpaired
electron.
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 Thus, 4 unpaired electrons.
diamagnetic in nature. x 0
2- (d) [Ti (NH3 ) ]3+
60. (a) [ Ni(CN)4 ] : Number of unpaired electrons = 0 6
x = +3
[Cr(NH3 )6 ] 3+
: Number of unpaired electrons = 3 Ti3+ = [Ar] 3d1
Thus, 1 unpaired electron.
[Fe(H 2O)6 ]2+ : Number of unpaired electrons = 4
Hence, option (c) is the correct answer.
[ Ni(H2O)6 ]2+ : Number of unpaired electrons = 2 To find unpaired electrons let us calculate the
oxidation states of elements in each complex and
61. (d) d 4 in high spin octahedral complex then write the electronic configuration for that
eg – oxidation state to find the number of unpaired
electrons in it.
t2g
64. (c) Out of the given ligands water, ammonia,
CFSE = [0.6 × 1] + [–0.4 × 3] = – 0.6 D 0
cyanide and oxalate, maximum splitting will occur
62. (b) Since Cr 3+ in the complex has unpaired in case of cyanide (CN–) i.e. the magnitude of Do
electrons in the d orbital, hence will be coloured will be maximum in case of [Co(CN)6]3–.
Ti = [Ar]3d 2 4s2 ; Ti4 + = 3d 0
Cr = [Ar] 3d 5 4s1; Cr3+ = 3d 3
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238 CHEMISTRY
[Cr(NH3)6]3+ Cr3+(24 –3 = 21)
In octahedral field the crystal field splitting of d-
orbitals of a metal ion depends upon the field 4p
3d 4s
produced by the ligands.
65. (b) Lesser is the number of unpaired electrons
d 2sp 3 ® (inner octahedral
smaller will be the paramagnetic behaviour. As complex & paramagnetic)
Cr2+, Mn2+, Fe2+ & Ni2+ contains. 2+ 2+
[Ni(NH3)6] Ni (28 – 2 = 26)
Cr2+ (3d 4) =
3d 4s 4p
= 4 unpaired e–.
sp 3d 2 ® (outer octahedral
complex & paramagnetic)
Mn2+ (3d 5) =
[Zn(NH3)6]2+ Zn 2+(30 – 2 = 28)
= 5 unpaired e–.
3d 4s 4p
Fe2+ (3d 6) =
sp3d2 ® (outer octahedral
= 4 unpaired e–. complex & diamagnetic)
dsp 2 Hybridisation
dxy dyz dzx Remember CN– is also a strong ligand
Ni 2+ in [ NiCl 4 ]2 -
So 3dxy1 3dyz1 3dxz1
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Coordination Compounds 239
69. (b) CN– is a strong field ligand as it is a [Fe(CN)6]–4
psuedohalide ion. × × × × × ×
× × × × × ×
Psuedohalide ions are strong coordinating ligands 3d 4s 4p
and hence have the tendency to form s-bond (from
the pseudo halide to the metal) and p-bond (from ––– diamagnetic
the metal to pseudo halide). [Cr(NH3)6]+3
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240 CHEMISTRY
77. (b)
No. of
Atom/Ion Configuration Magnetic
unpaired
Complex nature
electrons
3d 4s 4p
2+ 8
Ni (d ) 2
2–
[NiCl4]
2
sp3
2–
[Ni(CN)4] 0
Rearrangement dsp
2
8 2
Ni (d s ) 2
[Ni(CO)4]
0
Rearrangement
sp3
78. (d) In the formation of a coordinate bond, the 83. (a) As + ve charge on the central metal atom
ligands donate a pair of electrons to the metal increases, the less readily the metal can donate
atom. Futher nitrogen and oxygen has great electron density into the p* orbitals of CO ligand
tendency to donate the pair of electrons in most to weaken the C – O bond. Hence, the C – O
of the compounds. Therefore, both nitrogen bond would be strongest in [Mn(CO)6]+.
and oxygen are common donor atoms in ligands.
79. (a) According to VSEPR theory, a molecule with 84. (a) Triethoxyaluminium has no Al – C linkage
six bond pairs must be octahedral. O - CH 2 CH 3
80. (b) [Fe(CO)5] Al O - CH CH2 3
EAN = Z – O.N. + 2(C.N.) O - CH 2 CH 3
= 26 – 0 + 2(5)
85. (a) The number of carbon atom found in p bonded
= 26 + 10 organometallic compounds is indicated by the greek
= 36 latter 'h' with a number. The prefixes h2, h5 and h6
Only one central metal atom/ion is present and indicate that 2, 5 and 6 carbon atom are bound to
it follows EAN rule, so it is mononuclear. the metal in the compound.(CH3)4Sn does not
81. (a) Grignard's reagent (RMgX) is a s-bonded involve any pi (p) bond formation. (CH3)4Sn is a s
organometallic compound. bonded organometallic compound.
82. (c) [Fe(CO)4]2– 86. (d) In metal carbonyl complexes, s and p both
Since metal atom is carrying maximum –ve
bonds are present because of synergic bonding.
charge therefore, it would show maximum
synergic bonding as a resultant C—O bond Thus, [Co(CO)5(NH3)]2+ has s and p bonding.
length would be maximum. 87. (a) Organometallic compounds are those
compounds in which there is a bond which
M–C p bond in metal carbonyl which is formed involve metal-carbon bond. In [Ni(CO)4] there
by the donatio of an electron pair from a filled d-
orbital of metal into the vacant anti bonding p- are bond involving Ni – C.
orbtial of CO, strengthen the M–C s bond. This is In chlorophyll, there are 4 bonds between Mg – N.
called synergic effect.
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Haloalkanes and Haloarenes 241
24 Haloalkanes and
Haloarenes
Trend Analysis with Important Topics & Sub-Topics
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242 CHEMISTRY
6. Which of the following acids does not exhibit (a) CH3 – CH2 – CH2Br
optical isomerism ? [2012]
(b) CH3 - CH - CH 2 Br
(a) Maleic acid (b) a -amino acids |
(c) Lactic acid (d) Tartaric acid CH3
7. Consider the reactions : [2011 M]
(i) C H OH
2 5 ¾¾
(CH 3 ) 2 CH - CH 2 Br ¾¾¾ ® (CH 3 ) 2 CH3
|
(CH3)2 CH - CH2OC2H5 + HBr CH3 - C - CH 2 Br
(c)
|
C 2 H 5O -
(ii) (CH3 )2 CH - CH 2 Br ¾¾¾¾¾ ® (CH 3 )2 CH
3
(CH3)2 CH - CH2OC2H5 + Br -
(d) CH3CH2Br
The mechanisms of reactions (i) and (ii) are 11. How many stereoisomers does this molecule
respectively : have ? [2008]
(a) SN1 and SN2 (b) SN1 and SN1 CH3CH = CHCH 2CHBrCH 3
(c) SN2 and SN2 (d) SN2 and SN1
(a) 4 (b) 6
8. Which one is a nucleophilic substitution
reaction among the following ? [2011] (c) 8 (d) 2
12. If there is no rotation of plane polarised light by
(a) CH3 – CH = CH2 + H2O ¾®
a compound in a specific solvent, though to be
CH3 – CH – CH3 chiral, it may mean that [2007]
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Haloalkanes and Haloarenes 243
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244 CHEMISTRY
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Haloalkanes and Haloarenes 245
38. The compound C7H8 undergoes the following 41. In which of the following compounds, the C - Cl
reactions: [2018] bond ionisation shall give most stable carbonium
C7 H8 ¾¾¾¾
3Cl / D
2 ® A ¾¾¾¾ Br /Fe
2 Zn /HCl
® B ¾¾¾¾® C
ion? [2015]
The product 'C' is H
(a) m-bromotoluene CH — Cl
(b) o-bromotoluene
(c) p-bromotoluene H3C
(d) 3-bromo-2,4,6-trichlorotoluene C — Cl
(a) H3C (b)
39. For the following reactions : [2016] CH3
(1) CH3CH2CH2Br + KOH ®
CH3CH=CH2+KBr + H2O H H
C — Cl H 3C
H3C CH3 H3C CH3 (c) O 2NH2C H (d) C — Cl
H 3C
(2) + KOH ¾® + KBr
42. What products are formed when the following
Br OH compounds is treated with Br2 in the presence
Br of FeBr 3? [2014]
(3) + Br2 ¾®
CH3
Br
Which of the following statements is correct ?
(a) (1) and (2) are elimination reactions and (3)
is addition reaction CH3
(b) (1) is elimination, (2) is substitution and (3)
is addition reaction CH3 CH3
(c) (1) is elimination, (2) an d (3) are
substitution reactions Br
(a) and
(d) (1) is substitution, (2) and (3) are addition
reactions CH3 CH3
40. Which of the following biphenyls is optically Br
active ? [2016]
CH3 CH3
(a) O2N
Br Br
(b) and
CH3 CH3
I
Br Br CH3 CH3
Br
(b) (c) and
I I CH3 CH3
I Br
I (d) and
CH 3 CH3 Br CH3
(d) Br
CH 3
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246 CHEMISTRY
43. Which of the following compounds undergoes (I) (II) (III) (IV)
nucleophilic substitution reaction most easily ?
(a) I < II < IV < III (b) II < III < I < IV
[2011 M]
(c) IV < III < I < II (d) III < II < I < IV
Cl 47. Benzene reacts with CH3Cl in the presence of
Cl
anhydrous AlCl3 to form: [2009]
(a) (b) (a) chlorobenzene (b) benzylchloride
NO2 (c) xylene (d) toluene
CH3 48. The replacement of chlorine of chlorobenzene
to give phenol requires drastic conditions, but
the chlorine of 2,4-dinitrochlorobenzene is
Cl
readily replaced since, [1997]
Cl
(a) nitro groups make the aromatic ring electron
(c) (d)
rich at ortho/para positions
(b) nitro groups withdraw electrons from the
OCH3
meta position of the aromatic ring
44. The reaction of toluene with Cl2 in presence of (c) nitro groups donate electrons at meta
FeCl3 gives 'X' and reaction in presence of light position
gives ‘Y’. Thus, ‘X’ and ‘Y’ are : [2010] (d) nitro groups withdraw electrons from
(a) X = Benzal chloride, Y = o – Chlorotoluene ortho/para positions of the aromatic ring
(b) X = m – Chlorotoluene, Y = p –Chlorotoluene 49. Benzene reacts with n-propyl chloride in the
presence of anhydrous AlCl3 to give [1993]
(c) X = o –and p – Chlorotoluene,
(a) 3 – Propyl – 1 – chlorobenzene
Y = Trichloromethyl – benzene
(b) n-Propylbenzene
(d) X = Benzyl chloride, Y = m – Chlorotoluene
(c) No reaction
45. Which one is most reactive towards SN 1 reaction ?
(d) Isopropylbenzene.
(a) C6 H5CH(C6 H5 )Br [2010]
50. Chlorobenzene reacts with Mg in dry ether to
(b) C6 H5CH(CH3 )Br 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.
51. Which chloro derivative of benzene among the
46. The correct order of increasing reactivity of
following would undergo hydrolysis most
C – X bond towards nucleophile in the following
readily with aqueous sodium hydroxide to
compounds is: [2010]
furnish the corresponding hydroxy derivative?
X X [1989]
NO2
NO2
(CH3)3 C – X, (CH3)2CH – X
(a) O2 N Cl
NO2
NO2
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Haloalkanes and Haloarenes 247
Cl
(b) O2N Cl
(d) Cl C Cl
ANSWER KEY
1 (d) 7 (a) 13 (b) 19 (b) 25 (d) 31 (b) 37 (b) 43 (a) 49 (d) 55 (b)
2 (b) 8 (c) 14 (a) 20 (b) 26 (c) 32 (b) 38 (a) 44 (c) 50 (b) 56 (d)
3 (b) 9 (d) 15 (b) 21 (b) 27 (c) 33 (c) 39 (b) 45 (c) 51 (a)
4 (c) 10 (d) 16 (a) 22 (a) 28 (a) 34 (a) 40 (b) 46 (a) 52 (c)
5 (N) 11 (a) 17 (d) 23 (b) 29 (c) 35 (b) 41 (b) 47 (d) 53 (d)
6 (a) 12 (a) 18 (b) 24 (a) 30 (d) 36 (c) 42 (c) 48 (d) 54 (d)
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250 CHEMISTRY
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
Cl 23. (b) CH 3 — CH 2 — CHCl 2 ¾¾ ¾¾®
D
(3)
Cl CH 3 — CH = CHCl ¾¾ ¾¾
NaNH
2 ® CH — C º CH
3
D Final Product
CH2
_
H 2 C==CH –Cl H 2C – CH=Cl + CH3CH3 CH3CH3
¾¾®
20. (b) Compound which are mirror image of each CH3 – C – C –CH2Cl Mono
CH3– C – C – CH3
other and are not superimposable are termed as CH3CH3 Clorination CH3 CH3
enantiomers.
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Haloalkanes and Haloarenes 251
hn
25. (d) Cl2 ¾¾® Cl + Cl
• • 30. (d) When (–) 2-bromooctane is allowed to react
with sodium hydroxide under given conditions,
Cl / hn where second order kinetics is followed, the
CH3 — CH2 — CH2 — CH3¾¾2¾¾®
product obtained is (+) 2-octanol.
CH3 CH3 C6H13 H C6H13 C6H13
–
OH
H C Br ¾® HO -----C -----Br ® HO C H + Br–
H — C — Cl + Cl — C — H
CH3 CH3 CH3
C2H5 C2H5 (–) 2-Bromooctane Transition state (+) 2–octanol
d l In this reaction Walden Inversion takes place so
Racemic mixture it is an example of SN2-reaction.
50% d form + 50% l form
31. (b) H 3C — CH 2 — *CH — CH 2 — CH 2 — CH3
|
Chlorination of alkane takes place via free radical
CH3
formation. Cl· may attack on either side and gives
a racemic mixture. 3-Methylhexane
Due to presence of four different groups on
26. (c) Chiral molecules are those molecules which
have atleast one asymmetric carbon atom (a carbon,(C*) it is chiral
carbon atom attached to 4 different groups). This 32. (b) Number of active hydrogen in a compound
is true in case of 3-methylpentanoic acid. corresponds to the number of moles of CH4
H evolved per mole of the compound.
|
CH MgI
C2H5 – C - CH2 COOH -OH, - NH2 , - SH, - OH or - C º CH ¾¾¾¾
3
® CH 4 (2CH 4 from
|
CH3
® CH 4 ( 2CH 4 from - NH 2 )
27. (c) Potassium ethoxide is a strong base, and 2-
bromopentane is a 2º bromide, so elimination 33. (c) Racemic mixtures are resolved by the
formation of Diastereomers.
reaction predominates
OC H - Enantiomers have same physical and chemical
CH 3 CH (Br )CH 2 CH 2 CH 3 ¾¾ ¾
2 ¾
5®
properties except rotation of plane polarized light.
CH3CH = CHCH 2CH3 + CH 2 = CHCH2CH 2CH3 Diasteromers have different physical and chemical
Pentene - 2(major) trans Pentene -1(min or) cis properties.
Since, trans- alkene is more stable than cis, thus 34. (a) Resolution.
trans-pentene -2 is the main product.
28. (a) Due to absence of a chiral 35. (b) At high temp. i.e., 400°C substitution occurs
C-atom. D — CH 2 — CH 2 — CH 2 Cl in preference to addition.
molecule is not a chiral molecule.
Cl , 400°C
29. (c) The compound is diethyl ether (CH3CH2)2O CH 3 CH = CH 2 ¾¾2¾ ¾¾®
which is resistant to nucleophilic attack by - HCl
hydroxyl ion due to absence of double or triple ClCH 2 CH = CH 2
bond, whereas all other compounds given are
unsaturated. H
|
O 36. (c) H3C - CH 2 - C*- CH3
||
|
C2 H5 OC2 H5 CH3 - C - OCH3
Cl
ether Methyl acetate
The compound containing a chiral carbon atom
O i.e., (a carbon atom which is attached to four
||
CH3 - C º N CH3 - C - NH 2 different substituents is known as a chiral carbon
Acetonitrile acetamide atom) is optically active.
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252 CHEMISTRY
Zn 41. (b)
¾¾®
HCl 42. (c) Methyl group is ortho para directing but
Br Br due to steric hindrance effect, generated by two
m-Bromotoluene CH3 groups substitution will not take place on
39. (b) (1) CH3CH2CH2–Br + KOH position (I). Hence, only two products are
® CH3CH=CH2 + KBr + H2O possible.
This is dehydrohalogenation reaction which is
CH3
an example of elimination reaction.
(2) III I
– –
I I I II G
Restricted rotation
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Haloalkanes and Haloarenes 253
G withdraws electrons, neutralises (disperses) – 45. (c) SN1 reactions involve the formation of
ve charge of the ring, stabilises carbanion, carbocations, hence higher the stability of
facilitates SN reaction (activation effect) intermediate carbocation, more will be reactivity
Z Cl of the parent alkyl halide. The tertiary carbocation
formed from (c) is stabilized by two phenyl groups
– and one methyl group, hence most stable.
46. (a) Tertiary alkyl halide is most reactive towards
III G nucleophilic substitution because the
corresponding carbocation (3°) is most stable.
G releases electrons, intensifies –ve charge,
Aryl halide is least reactive due to partial double
destabilizes carbanion, retards SN reaction
bond character of the C – Cl bond.
(deactivation)
NO2 is activating group and CH3 and OCH3 are Presence of — NO 2 groups in ortho and para
deactiving group. positions increases the reactivity of the – Cl
towards nucleophiles.
Hence, the correct order of nucleophilic
substitution reaction is : (CH3)3 – C – X > (CH3)2 – CH – X >
III IV
Cl Cl Cl
X
Cl
X
> > > NO2
>
NO2 CH3 OCH3 NO2
II I
In SN reactions, a carbanion is formed as an or I < II < IV < III
intermediate, so any substituent that increases the
stability of carbanion and hence the transition state CH3
leading to its formation will enhance the S N
reactions. Anhyd.
47. (d) + CH3Cl ¾¾® +HCl
AlCl3
44. (c)
CH3 CH3 CH3 Toluene
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Alcohols, Phenols and Ethers 255
25 Alcohols, Phenols
and Ethers
Trend Analysis with Important Topics & Sub-Topics
substitution reactions 1 D
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Alcohols, Phenols and Ethers 257
CH3 CH3
O CH (c) CH3 OH and I2
CH3
(a)
(d) CH – CH3 and I2
OH
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(b) , ,
(a) (b)
OH
CH(CH3)2
(c) , , (c) (d)
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Alcohols, Phenols and Ethers 259
H3 C (a) Benzaldehyde
C O O H (b) Salicylaldehyde
H 3C (c) Salicylic acid
(c) (d) Benzoic acid
40. The correct order of acidic strength for following
compounds will be [2001]
CH3 OH OH OH
H3C C O O H
(d) OH
CH3 NO2
35. Which one of the following compounds has the I II III
most acidic nature? [2010] (a) III > I > II (b) I > III > II
(c) II > III > I (d) I > II > III
CH2OH OH 41. The ionization constant of phenol is higher than
(a) (b)
that of ethanol because : [2000]
(a) Phenoxide ion is bulkier than ethoxide
OH (b) Phenoxide ion is stronger base than
(c)
ethoxide
OH (c) Phenoxide ion is stabilized through
delocalization
CH (d) Phenoxide ion is less stable than ethoxide
(d) 42. Consider the following phenols :
36. Among the following four compounds [2010] OH OH OH OH
(i) phenol (ii) methylphenol
(iii) meta-nitrophenol (iv) para-nitrophenol
the acidity order is : NO2
(a) ii > i > iii > iv (b) iv > iii > i > ii CH3 NO2
(c) iii > iv > i > ii (d) i > iv > iii > ii I II III IV
37. Consider the following reaction: [2009] The decreasing order of acidity of the above
Zn dust CH Cl phenols is [1999]
Phenol ¾¾¾¾
® X ¾¾¾¾¾¾¾
3 ®Y
Anhydrous AlCl3
(a) III > IV > II > I (b) II > I > IV > III
Alkaline KMnO
¾¾¾¾¾¾¾ 4®Z
(c) I > IV > II > III (d) III > IV > I > II
The product Z is:
43. 1-Phenylethanol can be prepared by the reaction
(a) benzaldehyde (b) benzoic acid
of benzaldehyde with [1997]
(c) benzene (d) toluene
38. Which one of the following compounds is most (a) Ethyl iodide and magnesium
acidic? [2005] (b) Methyl iodide and magnesium
OH (c) Methyl bromide and aluminium bromide
(a) Cl–CH2–CH2–OH (b) (d) Methyl bromide
44. Correct increasing order of acidity is as follows:
OH OH [1994]
(c) (d) (a) H2O, C2H2, H2CO3, Phenol
NO2 CH3 (b) C2H2, H2O, H2CO3, Phenol
39. When phenol is treated with CHCl3 and NaOH, (c) Phenol, C2H2, H2CO3, H2O
the product formed is [2002] (d) C2H2, H2O, Phenol, H2CO3
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260 CHEMISTRY
reaction
(b) + C2H5I OCH3
I Br
(b) and cine substitution reaction
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Alcohols, Phenols and Ethers 261
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262 CHEMISTRY
ANSW E R K E Y
1 (d ) 8 (c) 15 (c) 22 (c) 2 9 (a) 3 6 (b ) 43 (b ) 50 (b ) 57 (b )
2 (c) 9 (c) 16 (d ) 23 (a) 3 0 (d ) 3 7 (b ) 44 (d ) 51 (d ) 58 (c)
3 (b ) 10 (c) 17 (b ) 24 (b ) 3 1 (b ) 3 8 (c) 45 (d ) 52 (b ) 59 (c)
4 (a) 11 (c) 18 (a) 25 (c) 3 2 (b ) 3 9 (b ) 46 (d ) 53 (d ) 60 (a)
5 (c) 12 (c) 19 (d ) 26 (d ) 3 3 (c) 4 0 (a) 47 (c) 54 (a) 61 (a)
6 (d ) 13 (d ) 20 (b ) 27 (c) 3 4 (c) 4 1 (c) 48 (b ) 55 (d )
7 (b ) 14 (c) 21 (d ) 28 (c) 3 5 (b ) 4 2 (a) 49 (d ) 56 (a)
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Alcohols, Phenols and Ethers 263
H 2SO 4
16. (d) Due to –I-effect of the three C–Cl-bonding
HO
HO between Cl and C-atom of the OH group, CCl3
CH3 CH 2 OH ¬¾¾
22 ¾ CH – CH – HSO
3 2 4
heat
CH (OH)2 is most stable.
9. (c) 1, 2 – Diols, when treated with an aqueous
solution of periodic acid give aldehyde or 17. (b) Secondary alcohols on oxidation give
ketones ketones.
CH2OH HIO4 R R
| ¾¾® CH2O +CH2O [O]
CH2OH CHOH ¾¾® C=O
R R
A 1° alcohol on reaction with periodic acid gives Isopropyl Ketone
CH 2O. Since in glycol, both the OH groups are alcohol
primary hence, 2 molecules of CH 2O is formed as
product.
Primary alcohols form aldehydes.
10. (c) Ethylene oxide when treated with Grignard
Reagent gives primary alcohol.
18. (a) 1° Alcohols on catalytic dehydrogenation
CH 2 CH 2 – OMgX
give aldehydes.
| O + R–MgX ¾® | + H 2O
CH 2 CH 2 – R Cu
RCH2OH ¾¾® RCHO + H2
300°C
X 1° alcohol Aldehyde
¾® R – CH2– CH –2 OH + Mg
OH
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264 CHEMISTRY
¾®
approximately 200 – 300 atmospheres. It is then – Cl (a-elimination)
passed over a catalyst (ZnO + Cr 2O3) at 300°C. : CCl2 dichlorocarbene
Methyl alcohol vapours are formed which are (electrophile)
condensed 26. (d) Haloform reaction is shown by compound
ZnO+ Cr O having
CO + 2H 2 ¾¾¾¾¾®
2 3 CH3OH
300°C
Compressed gas Methyl alcohol CH3 – C – or CH 3 – CH –
20. (b) Since 2-methylpropan-2-ol generates 3° || | Group
carbocation, therefore, it reacts fastest with HBr. O OH
NaOI
¾¾¾¾®
Greater the stability of the intermediate CH – CH3 or
carbocation, more reactive is the alcohol. | NaOH + I 2
OH
– +
21. (d) The rates of reaction of different alcohols C – ONa + CHI3
with lucas reagent follows the order. || Yellow
3° alcohol > 2° alcohol > 1° alcohol O ppt.
Since carbocations are formed as 27. (c) Mechanism :
intermediate, more stable the carbocation, higher AlCl
CH 3 – CH 2 – CH 2 – Cl ¾¾¾®
3
will be the reactivity of the parent compound + –
(alcohol). 2-Methylpropan-2-ol generates a 3º CH3 – CH 2 – CH 2 + AlCl4
carbocation, so it will react fastest; other three 1° Carbocation
generates either 1º or 2º carbocations.
CH3 – CH – CH 3
CH3 CH3 CH3 + –
| + | - | ESR H
H Å Br ¬¾¾¾ C 3 – CH – CH3
CH3 - C - OH ¾¾¾ ® CH3 - C ¾¾¾® CH3 - C - Br shift
| | | 2° Carbocation
CH3 CH3 CH3 (P)
2-Methylpropan-2-ol CH3
22. (c) KMnO4 (alkaline) and OsO4 / CH2Cl2 are CH3 – CH – CH3 CH3–C– O– O–H
used for hydroxylation of double bond while O3
O +
/Zn is used for ozonolysis. Therefore, the right ¾¾®
D
2 H3O
option is (c), i.e., D
BH3 in THF Cumene Cumene
3CH 3CH = CH 2 ¾¾¾¾¾¾ ® (CH 3CH 2CH 2 )3 B (P) hydroperoxide
3H2O2
¾¾¾¾® 3CH 3CH 2CH 2OH + H3BO3 OH
NaOH O
1-propanol
23. (a) Lucas reagent is conc. HCl + anhyd. ZnCl2. + CH3 – C– CH3
CH H3C–C– O–O–H
CH3 28. (c) Electron withdrawing – NO2 group has very
O
strong –I and –R effects so, compound 3 will be
24. (b) ¾¾
2¾
® most acidic.
Cumene OH CH3
Cumene
hydroperoxide 29. (a) +O C ¾¾®
–H2O
OH CH3
OH O cis-cyclopenta-1, acetone
|| 2-diol
H+ O CH3
¾¾¾
® + H3C CH3 C
H2 O
O CH3
cyclic ketal
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Alcohols, Phenols and Ethers 265
¾®
Cl – OH
– –
O O
C=O – CCl2 35. (b) Phenol is most acidic because its conjugate
OH
H
¾ ¾®
H
base is stabilised due to resonance, while the
– 2HCl
rest three compounds are alcohols, hence, their
Therefore, functional group – CHO is introduced. corrosponding conjugate bases do not exhibit
resonance.
31. (b)
OH OH
¾
®
(i) (ii)
(+ I effect of CH3 group decreases
acidity)
The intermediate is a 3° carbocation, which is \ Correct choice : (b)
very stable. Hence, rearrangement will not take
place and product mentioned in option (b) will OH
not form.
32. (b) Phenols react with alkyl halides in alkaline Zn CH Cl
medium to form ethers. Therefore, 37. (b) ¾® ¾ ¾3 ®
anhy
AlCl3
OH OCH3 Phenol X
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266 CHEMISTRY
38. (c) Phenols are more acidic than alcohols as Dry ether
43. (b) CH 3 I + Mg ¾ CH 3 MgI
they are resonance stabilised whereas alcohols ¾ ¾ ¾ ®
OH OH C6H5
H O
CHO CHOH ¬¾¾
2 ¾
CHCl CH3 - Mg ( OH ) I
39. (b) ¾¾®
3
(1-Phenyl ethanol)
NaOH
(Riemer Tiemann reaction) Salicylaldehyde 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
If CCI4 is used in place of CHCl3, then the -
41. (c) The acidic nature of phenol is due to the Equivalent Non-equivalent Oxygen can accommodate
formation of stable phenoxide ion in solution resonating structures (–) charge easily
C 6 H5 O
C6 H 5OH + H 2 O -
+ H 3O +
Due to resonance
Phenoxide ion Thus, the acidic character of the four
The phenoxide ion is stable due to resonance. corresponding acids wil be
–
O O O 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
OH O– O group, therefore, p-methylphenol is slightly
d
–
d
–
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 ,
d
–
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Alcohols, Phenols and Ethers 267
O – CH3 OH
When phenol is treated Br2/CS2 monobromination
takes place.
HI
47. (c) Due to strong electron-donating effect of the ¾® + CH3I
OH group, the electron density in phenol is much
higher than that in toluene, benzene and 53. (d)
chlorobenzene and hence phenol is readily
OC3H OC3H OC3H
attacked by the electrophile.
OH OH H
NH
CHCl2 ¾®
2
¾®
48. (b) Br Br
+ CHCl ¾® 3 Benzyne
OH OH OC3H
CH(OH)2 CHO
2KOH N2H
¾¾® ¾¾®
–H O
OC3H a (Less stable)
2
:O–H
HBr/H O
51. (d) In lone pair of e–s present on CH3 CH2 CH ¾¾¾¾®
CH2 (Peroxide
2 2
effect)
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268 CHEMISTRY
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Aldehydes, Ketones, and Carboxylic acids 269
26 Aldehydes, Ketones,
and Carboxylic Acids
Trend Analysis with Important Topics & Sub-Topics
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270 CHEMISTRY
COOH (a) oxidation of primary alcohols
(c) + Zn/Hg and conc. HCl (b) hydrolysis of esters
(c) oxidation of tertiary alcohols
CH3 (d) reaction of acid halides with alcohols
(d) + CrO2Cl 2 in CS2 followed by 9. Pinacolone is [1994]
(a) 2, 3-Dimethyl-2, 3-butanediol
H3O+
(b) 3, 3-Dimethyl-2-butanone
3. Consider the following reaction :
(c) 1-Phenyl-2-Propanone
COCl
H (d) 1,1-Diphenyl-2-ethandiol.
¾¾¾®
2
'A'
Pd-BaSO4
Topic 2: Properties of Carbonyl Compounds
The product ‘A’ is : [2012 M]
10. Reaction between acetone and methyl-
(a) C6H5CHO (b) C6H5OH
magnesium chloride followed by hydrolysis will
(c) C6H5COCH3 (d) C6H5Cl give : [2020]
4. Which of the following reactions will not result (a) Sec. butyl alcohol (b) Tert. butyl alcohol
in the formation of carbon-carbon bonds?
(c) Isobutyl alcohol (d) Isopropyl alcohol
(a) Reimer-Tiemann reaction [2010]
11. Reaction between benzaldehyde and
(b) Cannizaro reaction
acetophenone in presence of dilute NaOH is
(c) Wurtz reaction known as [2020]
(d) Friedel-Crafts acylation (a) Cannizzaro's reaction
5. Which one of the following can be oxidised to
(b) Cross Cannizzaro's reaction
the corresponding carbonyl compound? [2004]
(c) Cross Aldol condensation
(a) 2-hydroxypropane
(d) Aldol condensation
(b) Ortho-nitrophenol
(c) Phenol 12. Consider the reactions :- [2017]
(d) 2-methyl-2 hydroxypropane +
[Ag(NH3)2]
Cu
6. In the following reaction, product 'P' is [2002] X ¾¾®573K
A ¾¾¾¾®
–
Silver mirror
(C2H6O) OH D observed
H
R - C - Cl ¾¾¾
2® P –
OH D
|| Pd - BaSO 4 ¾¾¾¾¾¾¾®
O Y
O
(a) RCH2OH (b) RCOOH NH2 – NH – C – NH2
Z
(c) RCHO (d) RCH3
7. The catalyst used in Rosenmund's reduction is Identify A, X, Y and Z
[2000] (a) A-Methoxymethane, X-Ethanol,
(a) HgSO4 (b) Pd/BaSO4 Y-Ethanoic acid, Z-Semicarbazide.
(c) anhydrous AlCl3 (d) anhydrous ZnCl2 (b) A - Ethanal, X-Ethanol, Y - But - 2-enal,
8. Ketones Z-Semicarbazone
(c) A-Ethanol, X-Acetaldehyde, Y - Butanone,
[ R — C — R1 , where R = R1 = alkyl groups]
|| Z-Hydrazone
O (d) A-Methoxymethane, X-Ethanoic acid,
can be obtained in one step by [1998] Y-Acetate ion, Z-hydrazine.
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Aldehydes, Ketones, and Carboxylic acids 271
13. Of the following, which is the product formed when 16. An organic compound 'X' having molecular
cyclohexanone undergoes aldol condensation formula C5H10O yields phenyl hydrazone and
followed by heating ? [2017] gives negative response to the Iodoform test
and Tollen's test. It produces n-pentane on
(a) reduction. 'X' could be :- [2015]
(a) 2-pentanone (b) 3-pentanone
(c) n-amyl alcohol (d) pentanal
O
17. Treatment of cyclopentanone =O with
(b)
methyl lithium gives which of the following
species? [2015]
OH (a) Cyclopentanonyl cation
(b) Cyclopentanonyl radical
(c) (c) Cyclopentanonyl biradical
(d) Cyclopentanonyl anion
OH OH 18. The enolic form of ethyl acetoacetate as below
has: [2015]
OH H H2
H3C C O H3C C O
(d) C C C C
OH OC2H5 O OC2H5
OH (a) 16 sigma bonds and 1 pi - bond
14. The correct statement regarding a carbonyl (b) 9 sigma bonds and 2 pi - bonds
compoun d with a hydrogen atom on its (c) 9 sigma bonds and 1 pi - bond
alphacarbon, is : [2016] (d) 18 sigma bonds and 2 pi - bonds
(a) a carbonyl compound with a hydrogen atom 19. Reaction of a carbonyl compound with one of
on its alpha-carbon never equilibrates with the following reagents involves nucleophilic
its corresponding enol. addition followed by elimination of water. The
(b) a carbonyl compound with a hydrgen atom reagent is : [2015 RS]
on its alpha-carbon rapidly equilibrates (a) a Grignard reagent
with its corresponding enol and this (b) hydrazine in presence of feebly acidic solution
process is known as aldehyde-ketone (c) hydrocyanic acid
equilibration. (d) sodium hydrogen sulphite
(c) a carbonyl compound with a hydrogen atom 20. Which one is most reactive towards Nucleophilic
on its alpha-carbon rapidly equilibrates addition reaction? [2014]
with its corresponding enol and this
process is known as carbonylation. CHO COCH3
(a) (b)
(d) a carbonyl compound with a hydrogen atom
on its alpha-carbon rapidly equilibrates with its
corresponding enol and this process is known CHO CHO
as keto-enol tautomerism.
15. The product formed by the reaction of an (c) (d)
aldehyde with a primary amine is [2016]
(a) Schiff base (b) Ketone NO2
CH3
(c) Carboxylic acid (d) Aromatic acid
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272 CHEMISTRY
21. The order of stability of the following tautomeric OH
compounds is : [NEET 2013] (c) (CH3)2C
OH O OC2H5
| || OC2H5
CH 2 = C - CH 2 - C - CH 3 (d) (CH3)2C
I OC2H5
O O 24. CH 3 CHO and C 6 H5 CH 2 CHO can be
|| ||
distinguished chemically by : [2012]
CH3 - C- CH 2 - C- CH 3
(a) Benedict test (b) Iodoform test
II
(c) Tollen’s reagent test (d) Fehling solution test
OH O 25. Consider the reaction :
| ||
CH 3 - C = CH - C - CH 3
RCHO + NH2NH2 ® RCH = N – NH2
III What sort of reaction is it ? [2012 M]
(a) III > II > I (b) II > I > III (a) Electrophilic addition – elimination reaction
(c) II > III > I (d) I > II > III (b) Free radical addition – elimination reaction
22. Predict the product in the given reaction. [2012] (c) Electrophilic substitution – elimination
CHO reaction
(d) Nucleophilic addition – elimination reaction
50 % KOH
¾¾¾® 26. Which of the following compounds will give a
yellow precipitate with iodine and alkali?
Cl
[2012 M]
CH2OH CH2COO– (a) Acetophenone (c) Methyl acetate
(a) +
(b) Acetamide (d) 2-Hydroxypropane
Cl Cl 27. Clemmensen reduction of a ketone is carried out in
the presence of which of the following? [2011]
CH2OH OH (a) Glycol with KOH
(b) +
(b) Zn-Hg with HCl
OH (c) Li Al H4
OH
– (d) H2 and Pt as catalyst
CH2OH COO
(c) + 28. The order of reactivity of phenyl magnesium
bromide (PhMgBr) with the following compounds
Cl Cl [2011 M]
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Aldehydes, Ketones, and Carboxylic acids 273
Which of the above compound(s), on being 33. Reduction of aldehydes and ketones into
warmed with iodine solution and NaOH, will give hydrocarbons using zinc amalgam and conc. HCl
iodoform? [2010] is called [2007]
(a) (1) and (2) (b) (1), (3) and (4) (a) Cope reduction
(c) only (2) (d) (1), (2) and (3) (b) Dow reduction
30. Which one of the following compounds will be (c) Wolff-Kishner reduction
most readily dehydrated? [2010]
(d) Clemmensen reduction.
OH 34. Which one of the following on treatment with
50% aqueous sodium hydroxide yields the
(a)
H3C corresponding alcohol and acid? [2007]
O
O (a) C6H5CHO (b) CH3CH2CH2CHO
O
(b) ||
H3C (c) CH 3 - C - CH 3 (d) C6H5CH2CHO
OH
O 35. The product formed in Aldol condensation is
[2007]
(c) (a) A beta-hydroxy aldehyde or a beta-hydroxy
H3 C
OH ketone
(b) An alpha-hydroxy aldehyde or ketone
O OH
(c) An alpha, beta unsaturated ester
(d) A beta-hydroxy acid
(d)
H3C 36. Consider the following compounds. [2007]
31. Acetophenone when reacted with a base, (i) C6H5COCl
C2H5ONa, yields a stable compound which has
the structure. [2008] (ii) O2N COCl
C = CH – C
(a)
CH3 O (iii) H3C COCl
CH – CH2C
(b)
CH3 O (iv) OHC COCl
CH3CH3
The correct decreasing order of their reactivity
C–C towards hydrolysis is
(c)
(a) (i) > (ii) > (iii) > (iv) (b) (iv) > (ii) > (i) > (iii)
OH OH
(c) (ii) > (iv) > (i) > (iii) (d) (ii) > (iv) > (iii) > (i)
CH – CH 37. A carbonyl compound reacts with hydrogen
(d) cyanide to form cyanohydrin which on
OH OH hydrolysis forms a racemic mixture of a-hydroxy
32. A strong base can abstract an a -hydrogen acid. The carbonyl compound is [2006]
from : [2008] (a) Acetone (b) Diethyl ketone
(a) alkene (b) amine (c) Formaldehyde (d) Acetaldehyde
(c) ketone (d) alkane
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38. Nucleophilic addition reaction will be most 41. In the reaction
favoured in [2006] CH 3CHO + HCN ® CH 3CH (OH) CN
(a) (CH3)2C = O H.OH
(b) CH3CH2CHO ¾¾ ¾® CH 3 CH (OH )COOH
(c) CH3CHO an asymmetric centre is generated. The acid
O obtained would be [2003]
||
(d) CH 3 — CH 2 — CH 2C — CH 3 (a) 20 % D + 80 % L-isomer
(b) D-isomer
39. A and B in the following reactions are [2003]
(c) L-isomer
OH (d) 50% D + 50% L-isomer
HCN/ B
R–C–R' ¾¾® A ¾® R– C 42. Which of the following is correct? [2001]
KCN
R' CH2NH2
O (a) Diastase is an enzyme
CN (b) Acetophenone is an ether
(a) A = RR'C , B = LiAlH4 (c) Cycloheptane is an aromatic compound
OH
OH (d) All the above
(b) A = RR'C , B = NH3 43. Which of the following is incorrect? [2001]
COOH (a) NaHSO3 is used in detection of carbonyl
CN compound
(c) A = RR'C , B = H 3O Å (b) FeCl3 is used in detection of phenolic group
OH
(c) Tollens’ reagent is used in detection of
(d) A = RR'CH2CN, B = NaOH
unsaturation
40. When m-chlorobenzaldehyde is treated with 50%
(d) Fehling solution is used in detection of
KOH solution, the product(s) obtained is (are)
glucose
[2003]
44. Polarization of electrons in acrolein may be
OH OH written as: [2000]
δ– δ+
CH CH (a) CH2 =CH — CH= O
(a)
δ+ δ–
(b) CH 2 = CH—CH=O
OH OH
δ+ δ-
COO –
CH2OH (c) CH 2 = CH — CH= O
+ δ- δ+
(d) CH 2 = C H — CH = O
(b)
OH OH 45. During reduction of aldehydes with hydrazine
and potassium hydroxide, the first is the
COO– CH2OH formation of : [2000]
(c) + (a) R — CH == N — NH2
(b) R — C ºº N
Cl Cl (c) R — C — NH 2
||
OH OH O
(d) R — CH == NH
CH CH
(d) 46. Iodoform test is not given by [1999]
(a) 2-Pentanone (b) Ethanol
Cl Cl (c) Ethanal (d) 3-Pentanone
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Aldehydes, Ketones, and Carboxylic acids 275
47. Reaction of phenylacetylene with dil. H2SO4 and 55. Which of the following compound will undergo
HgSO4 gives [1999] self aldol condensation in the presence of cold
(a) Acetophenone (b) 2-Phenylethanol dilute alkali ? [1994]
(c) Phenylacetaldehyde (d) Phenylacetic acid (a) CH2 == CH — CHO (b) CH == C — CHO
48. The cyanohydrin of a compound on hydrolysis (c) C6H5CHO (d) CH3CH2CHO.
gives an optically active a-hydroxy acid. The
56. Acetaldehyde reacts with [1991]
compound is [1999]
(a) Diethyl ketone (b) Formaldehyde (a) Electrophiles only
(c) Acetaldehyde (d) Acetone (b) Nucleophiles only
49. Phenylmethyl ketone can be converted into (c) Free radicals only
ethylbezene in one step by which of the (d) Both electrophiles and nucleophiles.
following reagents? [1999]
57. The reagent (s) which can be used to distinguish
(a) LiAlH4 (b) Zn-Hg/HCl
acetophenone from benzophenone is (are)
(c) NaBH4 (d) CH3MgI
50. (CH3 ) 3 C—CHO does not undergo Aldol (a) 2,4- Dinitrophenylhydrazine [1990]
condensation due to [1996] (b) Aqueous solution of NaHSO3
(a) three electron donating methyl groups (c) Benedict reagent
(b) cleavage taking place between —C— CHO (d) I2 and NaOH.
bond
(c) absence of alpha hydrogen atom in the CH 2 O
molecule 58. O CH 2
(d) bulky (CH3)3 C—group CH 2 O
51. Decreasing order of reactivity towards
nucleophilic addition to carbonyl group among The above shown polymer is obtained when a
cyclopentanone, 3-pentanone and n-pentanal is carbonyl compound is allowed to stand. It is a
[1996] white solid. The polymer is [1989]
(a) 3-pentanone, cyclopentanone, n-pentanal (a) Trioxane (b) Formose
(b) n-pentanal, 3-pentanone, cyclopentanone (c) Paraformaldehyde (d) Metaldehyde.
(c) n-pentanal, cyclopentanone, 3-pentanone
59. CH 3
(d) cyclopentanone, 3-pentanone, n-pentanal
52. Acetone reacts with iodine (I2) to form iodoform
in the presence of [1995]
(a) CaCO3 (b) NaOH CH 3
CH 3
(c) KOH (d) MgCO3
53. Benzaldehyde reacts with ethanoic KCN to give The above compound describes a condensation
(a) C6H5CHOHCN [1994] polymer which can be obtained in two ways :
(b) C6H5CHOHCOC6H5 either treating 3 molecules of acetone
(c) C6H5CHOHCOOH (CH3 COCH3) with conc. H2SO4 or passing
(d) C6H5CHOHCHOHC6H5 propyne (CH3 C º CH) through a red hot tube.
54. Aldehydes and ketones will not form crystalline The polymer is [1989]
derivatives with [1994] (a) Phorone
(a) Sodium bisulphite (b) Mesityl oxide
(b) Phenylhydrazine (c) Deacetonyl alcohol
(c) Semicarbazide hydrochloride
(d) Mesitylene.
(d) Dihydrogen sodium phosphate.
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276 CHEMISTRY
HCl O
60. 3CH3COCH3 ¾¾¾¾
®
-3H 2O
(A)
(CH 3 )2 C = CH - CO - CH = C(CH 3 ) 2 (b) NH
(B)
This polymer (B) is obtained when acetone is O
saturated with hydrogen chloride gas, B can be COOH
(a) phorone (b) formose [1989] (c)
(c) diacetone alcohol (d) mesityl oxide. NH2
61. If formaldehyde and KOH are heated, then we NH2
get [1988] (d)
(a) Methane (b) Methyl alcohol NH2
(c) Ethyl formate (d) Acetylene.
65. Carboxylic acids have higher boiling points than
62. Formalin is an aqueous solution of [1988] aldehydes, ketones and even alcohols of
(a) Fluorescein (b) Formic acid comparable molecular mass. It is due to their
(c) Formaldehyde (d) Furfuraldehyde. [2018]
(a) Formation of intramolecular H-bonding
Topic 3: Preparation and Properties of
Carboxylic Acids (b) Formation of carboxylate ion
(c) Formation of intermolecular H-bonding
63. The reaction that does not give benzoic acid as (d) More extensive association of carboxylic
the major product is [NEET Odisha 2019] acid via van der Waals force of attraction
CH2OH 66. Which one of the following esters gets
KMnO /H + hydrolysed most easily under alkaline conditions?
(a) ¾¾¾¾¾®
4
[2015 RS]
CH2OH OCOCH3
(a)
(b) K Cr O
¾¾¾¾®
2 2 7
O 2N
OCOCH 3
COCH3
(i) NaOCl
(c) ¾¾¾¾¾
+®
(b)
(ii) H 3 O H 3CO
CH2OH PCC
OCOCH3
¾¾¾¾¾¾®
(d) Pyridinium (c)
chlorochromate
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Aldehydes, Ketones, and Carboxylic acids 277
68. Match the compounds given in List-I with
COOH
List-II and select the suitable option using the
code given below : [2011 M]
List I List-II (c)
OCH2CH3
(A) Benzaldehyde (i) Phenolphthalein
(B) Phthalic anhydride (ii) Benzoin COOC2H5
condensation
(C) Phenyl benzoate (iii) Oil of wintergreen (d)
(D) Methyl salicylate (iv) Fries Br
rearrangement 71. Given are cyclohexanol (I) acetic acid (II),
Code : 2, 4, 6 – trinitrophenol (III) and phenol (IV). In
(A) (B) (C) (D) these, the order of decreasing acidic character
will be : [2010]
(a) (iv) (i) (iii) (ii)
(a) III > II > IV > I (b) II > III > I > IV
(b) (iv) (ii) (iii) (i) (c) II > III > IV > I (d) III > IV > II > I
(c) (ii) (iii) (iv) (i) 72. Among the given compounds, the most
(d) (ii) (i) (iv) (iii) susceptible to nucleophilic attack at the carbonyl
69. An organic compound ‘A’ on treatment with NH3 group is: [2010]
gives ‘B’ which on heating gives ‘C’, ‘C’ when (a) CH3COOCH3 (b) CH3CONH2
treated with Br2 in the presence of KOH produces (b) CH3COOCOCH3 (d) CH3COCl
ethylamine. Compound ‘A’ is: [2011 M] 73. Propionic acid with Br 2|P yields a dibromo
(a) CH3COOH product. Its structure would be: [2009]
(b) CH3 CH2 CH2 COOH Br
|
(c) CH3 – CHCOOH (a) H– C – CH2COOH
|
CH3
Br
(d) CH3CH2COOH (b) CH2Br – CH2 – COBr
70. In a set of reactions, ethylbenzene yielded a
product D. [2010] Br
|
KMnO4 Br2
(c) CH3– C – COOH
CH2CH3 ¾¾¾¾ ® B ¾¾¾® |
KOH FeCl3
Br
(d) CH2 Br – CHBr – COOH
C H OH
2 5 74. The relative reactivities of acyl compounds
C ¾¾¾¾¾
+
®D
H towards nucleophilic substitution are in the order
of : [2008]
CH2 – CH – COOC 2H5
(a) (a) Acyl chloride > Acid anhydride > Ester
Br > Amide
Br (b) Ester > Acyl chloride > Amide
> Acid anhydride
(c) Acid anhydride > Amide > Ester
(b)
> Acyl chloride
Br
(d) Acyl chloride > Ester > Acid anhydride
CH2 COOC2H5
> Amide
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278 CHEMISTRY
75. Which of the following represents the correct 79. The –OH group of an alcohol or the –COOH
order of the acidity in the given compounds? group of a carboxylic acid can be replaced by
[2007] –Cl using [2004]
(a) FCH2COOH > CH3COOH > BrCH2COOH (a) Phosphorus pentachloride
> ClCH2COOH (b) Hypochlorous acid
(b) BrCH2COOH > ClCH2COOH > FCH2COOH (c) Chlorine
(d) Hydrochloric acid
> CH3COOH
80. In a set of the given reactions, acetic acid yielded
(c) FCH2COOH > ClCH2COOH > BrCH2COOH
a product C.
> CH3COOH
C H
(d) CH3 COOH > BrCH2COOH > ClCH2COOH CH3COOH + PCl5 ¾¾
® A ¾¾¾®
6 6 B
Anh.AlCl3
> FCH2COOH
C2H 5MgBr
76. Self condensation of two moles of ethyl acetate ¾¾¾¾¾¾ ®C
Ether
in presence of sodium ethoxide yields [2006] Product C would be - [2003]
(a) acetoacetic ester (b) methyl acetoacetate
(c) ethyl propionate (d) ethyl butyrate C2 H 5
|
77. In a set of reactions propionic acid yielded a (a) CH 3 - C (OH)C 6 H 5
compound D. [2006]
SOCl2 NH3 (b) CH3CH(OH)C2H5
CH3CH2COOH ¾¾¾¾ ® B ¾¾¾ ®C
(c) CH3COC6H5
KOH
¾¾¾® D (d) CH3CH(OH)C6H5
Br2
The structure of D would be 81. MgBr
(a) CH3CH2CONH2 (b) CH3CH2NHCH3
(c) CH3CH2NH2 (d) CH3CH2CH2NH2
( i ) CO
78. In a set of reactions acetic acid yielded a ¾¾¾¾
2®
P
product D. ( ii ) H 3O Å
SOCl Benzene
CH3COOH ¾¾¾¾
2
® A ¾¾¾¾¾® B In the above reaction product 'P' is [2002]
Anhy.AlCl3
HCN H.OH
CHO COOH
¾¾¾® C ¾¾¾® D
The structure of D would be: [2005] (a) (b)
COOH
CH2 – C – CH3 OH
(a) O
OH (c) (d) C 6H 5 – C – C 6H5
CN
82. Benzoic acid may be converted to ethyl benzoate
C – CH3 by reaction with : [2000]
(b)
OH (a) Sodium ethoxide (b) Ethyl chloride
(c) Dry HCl—C2H5OH (d) Ethanol
OH 83. Acetaldehyde reacts with semicarbazide and
CH2 – C – CH3 forms semicarbazone. Its structure is [1999]
(c)
CN (a) CH3CH = NNHCON = CHCH3
(b) CH3CH = NNHCONH2
OH (c) CH 3 CH = N — N — CONH 2
|
C – COOH OH
(d)
CH3 (d) CH3CH = N—CONHNH2
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Aldehydes, Ketones, and Carboxylic acids 279
84. Aspirin is an acetylation product of [1998] 91. An ester is boiled with KOH. The product is cooled
(a) o-Hydroxybenzoic acid and acidified with concentrated HCl. A white
(b) o-Dihydroxybenzene crystalline acid separates. The ester is [1994]
(c) m-Hydroxybenzoic acid (a) Methyl acetate (b) Ethyl acetate
(d) p-Dihydroxybenzene (c) Ethyl formate (d) Ethyl benzoate
85. An ester (A) with molecular fomula, C9H10O2 92. Sodium formate on heating yields [1993]
was treated with excess of CH3MgBr and the (a) Oxalic acid and H2
complex so formed was treated with H2SO4 to (b) Sodium oxalate and H2
give an olefin (B). Ozonolysis of (B) gave a ketone (c) CO2 and NaOH
with molecular formula C8H8O which shows +ve (d) Sodium oxalate.
iodoform test. The structure of (A) is [1998] 93. Among acetic acid, phenol and n-hexanol, which
(a) C6H5COOC2H5 of the following compounds will react with
(b) C2H5COOC6H5 NaHCO3 solution to give sodium salt and carbon
(c) H3COCH2COC6H5 dioxide? [1993]
(d) p – H3CO – C6 H 4 – COCH3 (a) Acetic acid
86. Which one of the following esters cannot (b) n-Hexanol
undergo Claisen self-condensation? [1998] (c) acetic acid and phenol
(a) CH 3 - CH 2 - CH 2 - CH 2 - COOC 2 H 5 (d) Phenol.
(b) C6H5COOC2H5 94. (CH3)2 C = CHCOCH3 can be oxidized to
(c) C6H5CH2COOC2H5 (CH3)2C = CHCOOH by [1993]
(d) C6H11CH2COOC2H5 (a) Chromic acid (b) NaOI
87. Consider the following transformations : (c) Cu at 300°C (d) KMnO4.
CaCO 3 heat I 95. In which of the following, the number of carbon
CH3COOH¾¾ ¾ ¾® B ¾¾®
¾® A ¾¾ 2
C atoms does not remain same when carboxylic
NaOH
acid is obtained by oxidation [1992]
The molecular formula of C is [1996]
(a) CH 3COCH 3 (b) CCl 3CH 2CHO
OH
| (c) CH 3CH 2 CH 2OH (d) CH 3CH 2CHO.
(a) CH 3 - C - CH 3 (b) ICH2 — COCH3
| 96. Benzoic acid gives benzene on being heated with
I X and phenol gives benzene on being heated
(c) CHI3 (d) CH3I with Y. Therefore X and Y are respectively [1992]
88. Formic acid is obtained when [1994] (a) Soda-lime and copper
(a) Calcium acetate is heated with conc. H2SO4 (b) Zn dust and NaOH
(b) Calcium formate is heated with calcium acetate (c) Zn dust and soda-lime
(c) Glycerol is heated with oxalic acid at 373 K (d) Soda-lime and zinc dust.
(d) Acetaldehyde is oxidised with K2Cr2O7 and 97. The compound formed when malonic ester is
H2SO4. heated with urea is [1989]
89. The preparation of ethyl acetoacetate involves (a) Cinnamic acid (b) Butyric acid
(a) Wittig reaction [1994] (c) Barbituric acid (d) Crotonic acid.
(b) Cannizzaro’s reaction 98. Which of the following is the correct decreasing
(c) Reformatsky reaction order of acidic strength of [1988]
(d) Claisen condensation. (i) methanoic acid (ii) ethanoic acid
90. Schotten-Baumann reaction is a reaction of (iii) propanoic acid (iv) butanoic acid.
phenols with [1994] (a) (i) > (ii) > (iii) > (iv) (b) (ii) > (iii) > (ii) > (i)
(a) Benzoyl chloride and sodium hydroxide (c) (i) > (iv) > (iii) > (ii) (d) (iv) > (i) > (iii) > (ii)
(b) Acetyl chloride and sodium hydroxide 99. Among the following the strongest acid is [1988]
(c) Salicylic acid and conc. H2SO4 (a) CH3COOH (b) CH2ClCH2COOH
(d) Acetyl chloride and conc H2SO4 (c) CH2ClCOOH (d) CH3CH2COOH
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280 CHEMISTRY
ANSWER KEY
1 (b) 11 (c) 21 (a) 31 (a) 41 (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)
6 (c) 16 (b) 26 (a, d) 36 (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)
Friedel-Craft acylation
In presence of sunlight chlorination takes place at Note that, no new C–C bond is formed in Cannizzaro
the side chain by free radical mechanism. reaction.
5. (a) Among the given options, only (a) can be
2. (c) Zn/Hg and conc. HCl reduce carboxyl oxidized to ketone.
group to meth ylene group (Clemmensen OH O
reduction). | ||
O O oxidation
H3C – CH – CH3 ¾¾¾¾® H3C – C – CH3
C Cl CH 2-hydroxypropane Acetone
3. (a) H2
¾¾¾®
Pd–BaSO4
Carbonyl compounds (aldehydes and ketones) are
It is Rosenmund reaction. obtained by the oxidation of 1° and 2° alcohols
4. (b) respectively.
OH OH
CHO H
CHCl3 6. (c) R — C — Cl ¾¾®
¾2
R — C — H + HCl
(a) ¾¾¾¾ ® Pd – BaSO4
NaOH O O
Riemer-Tiemann reaction ‘P’
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Aldehydes, Ketones, and Carboxylic acids 281
7. (b) Catalyst used in Rosenmund reduction is Pd/ Cu
CH 3 - CH 2 OH ¾¾¾® CH3 - CHO
BaSO4. Rosenmund Reduction is used for 573K
(X) (A)
+
reduction of acid chloride. éë Ag(NH3 )2 ùû OH –
O O ¾¾¾¾¾¾¾® silver mirror observed
D
Pd / BaSO4
R C Cl ¾¾¾ ¾¾® R C H
8. (c) By oxidation of tertiary alcohol with stronger
oxidising agents ketones may be formed along
with carboxylic acid.
4[O]
CH3COCH3+CO2+2H2O
OH
(CH3)3 COH | D
8[O] CH3 - CH - CH 2 - CHO
CH3COOH+2CO2+3H2O 3-Hydroxybutanal
9. (b) Pinacolone is 3,3-dimethyl-2-butanone.
CH3 - CH = CH - CHO
CH3 (Y)
But-2-enal
CH3 – C – C – CH3 O
CH3O CH3 - CH = N - NH - C - NH 2
10. (b) (Z)
d– Semicarbazone
O d– d+ O MgBr 13. (a)
CH3–MgBr O
CH3–C–CH3 ¾¾¾ ¾¾ ® CH3–C–CH3 O
d+ H
Acetone CH3 (i) OH
+H
||
¾®
Hydrolysis (ii) D
OH O
14. (d) Keto-enol tautomerism is possible only in
CH3–C–CH3 those aldehydes and ketones which have at least
one a-hydrogen atom, which can convert the
CH3 ketonic group to the enolic group. e.g.
(tert- Butyl alcohol)
O O
11. (c) || ||
O O CH3 — C — CH2 — C CH3
dil. OH –
C–H+ C – CH3 ¾ ¾¾¾®
Cross Aldol
Ketonic form
reaction OH O
OH H O | ||
CH 3 — C CH — C — CH 3
||
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282 CHEMISTRY
On reduction, it gives n-pentane 21. (a) Enolic form predominates in compounds
reduction containing two carbonyl groups separated by a
H3C—CH2—C—CH2—CH3 ¾¾¾¾ ¾® – CH2 group. This is due to following two factors.
Zn-Hg/HCl
or NH 2-NH 2/OH
– (i) Presence of conjugation which increases
O stability.
CH 3—CH 2—CH2—CH 2—CH 3 (ii) Formation of intramolecular hydrogen
n-pentane
bond between enolic hydroxyl group and
17. (d) – second carbonyl group which leads to
O O
CH3 stablisation of the molecule. Hence, the cor-
–
rect answer is III > II > I.
Å
+ CH3 Li ® 22. (c) Cannizzaro reaction - when an aldehyde
containing no a – H undergo reaction in presence
Cylopentanoyl anion of 50% KOH. It disproportionates to form a molecule
of carboxylic acid and a molecule of alcohol.
Reactions involving methyl lithium require and an O
hydrous condtions, because the compound is highly
CH2–OH
reactive towards water. Methyl lithium is stable CHO C–O
as a solution in ether.
50 % KOH +
18. (d) Enolic form of ethyl acetoacetate has 18 ¾¾¾®
sigma and 2 pi-bonds as shown below: Cl
Cl Cl
H H 23. (d) Anhydrous alcohols add to the carbonyl
s s s group of aldehydes in the presence of anhydrous
s s s
H C C p C s
C p
O hydrogen chloride to form acetals via hemiacetals.
s s H H
s OC2H5
H Os H s s
s
Os C s C H C H OH
s s CH3 C 2 5
CH3 ¾¾¾® CH3 C CH3
H H
O O OH
||
1 HÅ Hemiacetal
19. (b) R — C— R + NH 2 — NH 2 ¾¾¾®
NH2 OC2H5
OH N C H OH
2 5
¾¾¾® CH3 C CH3
1 H2O
R—C—R ¾¾¾¾¾ ® R—C—R¢
Elimination
OC2H5
NH—NH2 Acetal
(Addition) 24. (b) CH 3 CHO gives Iodoform test but
While in all other cases no elimination take place. C6H5CH2CHO does not give Iodoform test due
20. (d) Any substituent in the carbonyl compound to absence of methyl group.
that increases the positive charge on the carbonyl
carbon will increase reactivity towards nucleophilic 25. (d) R – CH = O + H2N – NH2 ®
addition. R – CH = N – NH2
—NO2 shows –M effect hence Such reactions take place in slightly acidic
medium and involve nucleophilic addition of the
CHO ammonia derivative.
26. (a, d) The compounds with CH3 – C – group
is most reactive towards nucleophilic ||
addition reaction. O
or CH3 — CH — group give iodoform test.
NO 2 |
OH
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Aldehydes, Ketones, and Carboxylic acids 283
Hence, Acetophenone Thus, two molecules of acetophenone condense
to form a b-hydroxy ketone which gets dehydrated
CH3 C and 2-Hydroxypropane in the presence of acid upon heating to form a,
b-unsaturated compound.
O i.e., option (a) is correct.
CH3 CH CH3 32. (c) A strong base (OH–) abstract an a – H atom
both give a yellow from ketone which is acidic in nature.
OH
The acidity of a-hydrogen atom in ketone is due
precipitate of CHI3 (iodoform) with iodine and
to resonance stabilization of enolate anoin.
alkali.
27. (b) Clemmensen reduction is
Zn-Hg / conc. HCl
Zn–Hg/HCl 33. (d) C = O ¾¾¾¾¾ ¾¾® CH2 + H2O
C = O ¾¾¾¾® CH 2
Clemmensen reduction
28. (d) The reactivity of the carbonyl group
towards the nucleophilic addition reactions e.g.
depend upon the magnitude of the positive CH3 Zn-Hg / CH3
charge on the carbonyl carbon atom (electronic conc. HCl
C = O ¾¾¾ ¾® CH2 + H2O
factor) and also on the crowding around the
CH3 CH3
carboxyl carbon atom in the transition state
(steric factor). Both these factors predict the Clemmensen reduction is complementary to wolf-
following order kishner reduction, which also convert aldehyde and
H3C H3C Ph ketones to hydrocarbons. Clemmensen reduction
C=O> C=O> C=O carried out in strongly acidic conditions and wolf
H H3C Ph kishner reduction carried out in strongly basic
(due to steric crowding). conditions.
29. (d) Among the given compounds only CH3OH 34. (a) Aldehydes containing no a-hydrogen
does not give iodoform reaction. atom on warming with 50% NaOH or KOH
30. (d) The intermediate is carbocation which is undergo disproportionation i.e. self oxidation -
destabilised by C = O group (present on a-
reduction known as Cannizzaro’s reaction.
carbon to the –OH group) in the first three cases.
In (d), a–hydrogen is more acidic which can be 50% NaOH
removed as water. Moreover, the positive charge 2 C6 H5 CHO + NaOH ¾¾¾¾¾®
on the intermediate carbocation is relatively C6 H5COONa + C6 H5CH 2 OH
away from the C = O group.
O O 35. (a) Aldehydes and ketones having at least one
a-hydrogen atom in presence of dilute alkali give
OH -
31. (a) C6H5 – C + H.CH 2– C ¾¾¾® b-hydroxy aldehyde or b-hydroxy ketone.
O
CH3 C6H5 ||
CH 3 - C + HCH 2 CHO
OH O |
+
H
H Heat Acetaldehyde
C6H5 – C – CH 2– C ¾¾¾¾®
H O 2 OH
|
CH3 dil.NaOH
6 5 ¾¾¾¾¾
® CH3 - C - CH 2 - CHO
O |
H
Aldol
C6H5 – C = CH – C– C6H5
D
¾¾
® CH3 - CH = CH.CHO
CH3 - H 2O Crotonaldehyde
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284 CHEMISTRY
36. (c) The degree of hydrolysis increases as the
magnitude of positive charge on carbonyl group H H
| |
increases. Electron withdrawing group increases 41. (d) CH 3 - C = O + HCN ® CH 3 - C - OH
the positive charge and electron releasing group |
decreases the positive charge. Among these NO2 CN
&-CHO are electron withdrawing group from H H
which-NO2 exhibit more –I effect than –CHO. On | |
H + / H 2O
the other hand-CH3 is a electron releasing group ¾¾ ¾ ¾
¾® OH - C - CH 3 + CH 3 - C - OH
| |
therefore the order of reactivity towards COOH COOH
hydrolysis is 50% L - Lactic 50% D-Lactic
acid acid
COCl COCl COCl COCl 42. (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,
NO2 CHO CH3 C12H22O11) through hydrolysis of starch.
37. (d) Out of given compound only acetaldehyde 43. (c) Tollen's reagent is used to detect aldehydic
can form optical active hydroxy acid as it contains group. Tollen's reagent is an ammonical solution
one asymmetric carbon atom as marked below : of silver nitrate. When aldehyde is added to
Tollen's reagent, silver oxide is reduced to metallic
O OH silver which deposits as mirror.
|| |
HCN
CH3 - C - H ¾¾¾® CH3 - C - CN RCHO + Ag2O ¾¾ ® RCOOH + 2Ag
|
H 44. (c) In CH 2 = CH — CHO due to – M effect of
OH — CHO group, polarization of electron takes
hydrolysis
| place as follows:
¾¾¾¾¾
® CH3 - C*- H + –
| CH 2 = CH — C = O « CH 2 — CH = C — O
| |
COOH H H
38. (c) Aldehydes are more reactive than ketones due Hence, partial polarization is represented as
to +I effect of –CH3 group. There are two –CH3 d+ d-
group in acetone which reduces +ve charge density C H 2 = CH — CH = O
on carbon atom of carbonyl group. More hindered 45. (a)
carbonyl group too becomes less reactive. So, in O
the given case CH3CHO is the right choice.
R C H + NH2 NH2 ¾® R C N NH2
O OH H
|| |
HCN Aldehyde Hydrazine Aldehyde hydrazone
39. (a) R– C –R ¢ ¾¾¾® R — C — CN
KCN | 46. (d) Iodoform test is exhibited by ethyl alcohol
R ¢ (A) acetaldehyde, acetone, methyl ketones and those
OH alcohols which possess CH3CH(OH)-group. As
| 3-pentanone does not contain CH3 CO-group
Reduction by
¾¾¾¾¾¾
® R– C – CH 2 NH 2 therefore it does not give iodoform test.
LiAlH 4 (B) |
R¢ 47. (a)
40. (c) It is a simple Cannizzaro reaction. OH O
– C CH C CH2 C CH3
CHO COO CH2OH
H SO /HgSO 4
2 4
50% KOH ¾¾¾¾¾®
2 +
Cl Cl Cl enol form Keto form
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Aldehydes, Ketones, and Carboxylic acids 285
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286 CHEMISTRY
CH3 CH3 65. (c) Carboxylic acids have higher boiling points
C = O + H2 CHCOCH H2 + O = C than aldehydes, ketones and even alcohols of
CH3 CH3 comparable molecular mass.
CH 3 CH 3 This is due to more extensive association through
HCl
¾¾® C = CH COCH = C intermolecular H-bonding.
–2H 2O
CH 3 CH 3
Phorone O ¼¼¼H O
R C C R
Two molecules of acetone condense to form O H ¼¼¼O
mesityl oxide.
66. (a) Among the substituent attached to the
CH3
C = O + H 2 CHCOCH 3 ¾® benzene ring, –NO2 group is the most electron
CH3 withdrawing, thus withdraws electron density
CH3
from carbonyl carbon thus facilitate the attack
C == CH.CO.CH3
CH3 –
of OH ion.
61. (b) Aldehydes containing no a-hydrogen atom O O
on warming with 50% NaOH or KOH undergo . . || ||
O – C – CH3 O – C – CH3
..
disproportionation i.e. self oxidation - reduction
known as Cannizzaro’s reaction.
50%NaOH
2 HCHO + ¾¾¾¾¾
® HCOONa + CH 3OH
62. (c) Formalin is an aqueous solution (40%) of N N
– – –
formaldehyde. O O O O
63. (d) (–R effect of –NO 2 group)
CH2 CHO 67. (a) CF3 COOH > CCl3 COOH > HCOOH >
OH ¾¾¾¾®
PCC CH3COOH (Ka order)
Pyridinium
chlorochromate The halogenated fatty acids are much stronger
PCC is a mild oxidizing agent which oxidises acids than the parent fatty acid and moreover
primary alcohol to aldehyde. the acidity among the halogenated fatty acid
COOH increases almost proportionely with the increase
64. (b) + NH3 ¾¾® in electronegativity of the halogen present.
COOH Further, formic acid having no alkyl group is more
Å acidic than acetic acid.
COONH4 CONH2
D
Å ¾¾¾¾ ® 68. (d)
-2H 2O
COONH4 CONH2
KCN
(a) 2C6H5CHO ¾¾¾¾¾®
H O, C H OH 2 2 5
O
Strong C OH
Heating
¾¾¾® NH KCN
C6H5 – CH – C – C6H5
–NH3 C H O C H OH
O O
(Benzoin)
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Aldehydes, Ketones, and Carboxylic acids 287
(b) O O OH
71. (a) N6O2
+
O ¾®
H
O NO2 > CH3COOH >
H O H
NO2
(II) (III)
OH OH
OH OH OH OH
Phenolphthalein
(c) The fries rearrangement enables the >
preparation of acylphenols. The reaction is
catalysed by Bronsted or lewis acid such as (IV) (I)
AlCl3. Explanation: Presence of three — NO2 groups
O OH O OH in o–, p– positions to phenolic groups (in III)
makes phenol strongly acidic because its
O R AlCl R corresponding phenate ion (conjugate base) is
¾¾ ®
3
highly stabilised due to resonance. Conjugate
base of CH3 COOH, II (i.e. CH3 COO – ) is
resonance hybrid of two equivalent structures.
R O The conjugate base of phenol, IV is stabilized
Fries rearrangement
due to resonance (note that here all resonating
COOCH3
structures are not equivalent). The conjugate
OH base of cyclohexanol, I does not exhibit
(d) (Oil of wintergreen) resonance, hence not formed.
72. (d) Cl– is the weakest base and hence better
69. (d) Since, C when heated with Br2 in presence leaving group.
of KOH produces ethylamine, hence it must be 73. (c) This reaction is an example of Hell - Volhard
propanamide and hence the organic compound Zelinsky reaction. In this reaction acids
(A) will be propanoic acid. The reactions follows:
NH
containing a – H on treatment with X2 /P give
CH3 - CH 2 - COOH ¾¾¾
3
® di-halo substituted acid.
(A)
- + Br /P
D CH3 – CH 2 COOH ¾¾¾
2 ® CH - CBr - COOH
CH3 - CH 2 - CO O N H 4 ¾¾
® 3 2
(B) Hell-volhard-zelinsky reaction:
KOH + Br Carboxylic acids having an a-hydrogen are
CH 3 - CH 2 - CONH 2 ¾¾¾¾¾
2
®
Hoffmann halogenated at the a-position on treatment with
(C) bromamide chlorine or bromine in the presence of small amount
reaction
of red phosphoms to give a-halocarboxylic acids.
CH3 - CH 2 - NH 2 The reaction fails to accomplish the fluorination
(Ethylamine)
and iodination of carboxylic acids.
CH2 – CH3 COOH
74. (a) The more the basic character of the leaving
( KMnO4 )®
¾¾¾¾¾ group, the lesser is the reactivity. The basic
70. (d) KOH
character follows the order:
(B) NH2– > OR– > RCOO – > Cl–
COOH COOC2H5 Hence, the relative reactivities of acyl
Br C H OH
compounds towards nucleophilic substitution
¾®
2
¾¾¾¾®
2 5
+ follow the order Acyl halides > Acid anhydride
FeCl3 H
Br Br > Ester > Amide.
(C) (D)
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288 CHEMISTRY
75. (c) Electron withdrawing substituents (like 79. (a) Among the given options, only PCl5 can
halogen, – NO2, – C6H5 etc.) would disperse convert an alcoholic group as well as a carboxyl
the negative charge and hence stabilise the group to chloride.
carboxylate ion and thus increase acidity of the PCl
5® RCH Cl
RCH 2 OH ¾¾¾
parent acid. On the other hand, electron-releasing 2
substituents would intensify the negative PCl
5 ® RCOCl
RCOOH ¾¾¾
charge, destabilise the carboxylate ion and thus
decrease acidity of the parent acid. Other than PCl5 or PCl3, SOCl2 also can be used.
Electronegativity decreases in order SOCl2 is preferred because the other two products
F > Cl > Br in this case are gaseous and escape the reaction
and hence –I effect also decreases in the same mixture. This makes the purification of the
order, therefore the correct option is products easier.
RCOOH + SOCl2(l) ® RCOCl + SO2(g) + HCl(g)
[FCH2COOH > ClCH2COOH >
BrCH2COOH > CH3COOH] 80. (a) CH3COOH + PCl5 ® CH3COCl
76. (a) It is an example of Claisen condensation. The COCH3
product is acetoacetic ester.
O O C H
¾¾¾¾¾
6 6
¾
® MgBrC2H 5
|| || Anhyd AlCl3
CH 3 C - OC 2 H 5 + H - CH 2 - C - OC 2 H 5 ¾
¾® Friedel Craft
reaction
O O [B]
|| ||
CH 3 - C - CH 2 - C - OC 2 H 5
(ethyl acetoacetate ) OH
OMgBr
C 2 H5 – C – CH3
77. (c) CH 3 CH 2 COOH ¾SOCl
¾¾ ¾ 2® C 2 H 5 – C – CH3
NH H+
CH 3 CH 2 COCl ¾¾¾
3®
ether
KOH hydrolysis [C]
CH3 CH 2 CONH 2 ¾¾¾® CH3CH 2 NH 2
Br2
81. (b) Grignard reagent forms addition product with
SOCl bubbled carbondioxide which on hydrolysis with
78. (d) CH 3COOH ¾ ¾¾2® CH 3 - COCl HCl yields benzoic acid.
Friedel Craft C6 H 6 / AlCl3
reaction O
MgBr C OMgBr
+
(i) CO H O
¾¾®
2
¾¾®
3
HCN
¬¾¾
¾¾¾®
O
H2O
(hydrolysis of cyano group) C O H
+ Mg(OH)Br
'P'
Benzoic acid
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Aldehydes, Ketones, and Carboxylic acids 289
82. (c) 86. (b) The ester having a-hydrogen atom show
Claisen condensation reaction. We know that
COOH COOC2H5 ethyl benzoate (C6 H5 COOC 2 H5 ) does not
contain a-hydrogen. Therefore, C6H5COOC2H5
HCl does not undergo Claisen self condensation.
+ C2H5OH ¾®
dry
+ H2O
Ethanol 87. (c) CH3COOH + CaCO3 ¾® (CH3COO)2Ca
CH 3
This process is known as esterification. ¾¾¾®
Heat
CH 3
> CO ¾¾¾¾¾
I2 + NaOH
®
O CHI3 + NaI + CH3COONa + 3H2O
83. (b) CH3 —C == O + H2N—NH—C—NH2 88. (c) When glycerol is heated with oxalic acid
following reaction occurs.
H CH2OH HOOC CH2OC.COOH
O | |
100 -110°C
|
(–H2O) CHOH + HOOC ¾¾¾¾¾
® CHOH
- H2O
¾ ¾ ® CH3—CH == N—NH—C—NH2 | oxalic |
acetaldehyde semicarbazone CH 2OH acid CH2OH
CH2 CHO CH2OH
OH | |
- CO H O
COOH ¾¾¾®
2 CHOH ¾¾¾
2 ® CHOH + HCOOH
| | Formic acid
84. (a) + ClCOCH3 CH 2OH CH2OH
89. (d) In Claisen condensation intermolecular
o-hydroxy benzoic acid acetyl chloride
(Salicylic acid) condensation of esters containing a-hydrogen
OCOCH3 atom in presence of strong base form b-keto ester.
COOH CH3COO C2H5 + H.CH2.CO.OC2H5
Pyridine
¾¾¾® ethyl acetate
O
C2H5ONa
Aspirin CH3C. CH2COOC2 H5+ C2H5OH
CH COO C H + H.CH .CO.OC H
85. (a) Ethyl acetoacetate
OMgBr (b-ketoester)
| OH
CH MgBr
C6 H5COOC2 H5 ¾¾¾¾¾
3
® C6 H5 - C - OC2 H5
( A) |
CH 90. (a) aq. NaOH
3 + C6H5COCl
O O.COC6H5
||
- Mg(OC2 H5 )Br Excess
¾¾¾¾¾¾¾
® C6 H5 - C - CH3 ¾¾¾¾
®
CH3MgBr
OMgBr CH3
HO phenyl benzoate
C6 H5 C CH3 ¾®
2
C6 H5 C CH3
The function of NaOH is
CH3 OH (i) To convert phenol to more stronger nucleophile
CH3 PhO –
| (ii) To neutralize the acid formed
Conc. H 2SO 4
¾¾¾¾¾¾
® C 6 H5 — C = CH 2
D 91. (d) Methyl acetate and ethyl acetate on
'B' hydrolysis give CH3COOH which is a liquid.
Ozonolysis
¾¾¾¾¾ ® C 6 H 5 COCH 3 + HCHO Similarly ethyl formate on hydrolysis will give
formic acid which is also a liquid. Only ethyl
3I + 4 NaOH
C6 H 5 COCH 3 ¾¾¾¾¾¾
2
® CHI3 benzoate on hydrolysis will give benzoic acid
which is a solid.
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290 CHEMISTRY
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Amines 291
27 Amines
Trend Analysis with Important Topics & Sub-Topics
NH2
|
NHC2H5 NH3 (d) CH3 — C — CH 2CH 2CH3
|
(c) (d) CH3
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292 CHEMISTRY
4. Nitration of aniline in strong acidic medium also (d) Arylamines are generally more basic than
gives m-nitroaniline because [2018] alkylamines, because the n itrogen atom
(a) Inspite of substituents nitro group always in arylamines is sp-hybridized.
goes to only m-position. 8. The electrolytic reduction of nitrobenzene in
(b) In electrophilic substitution reactions, strongly acidic medium produces :- [2015]
amino group is meta directive. (a) Azoxybenzene (b) Azobenzene
(c) In acidic (strong) medium aniline is present (c) Aniline (d) p-Aminophenol
as anilinium ion. 9. The number of structural isomers possible from
(d) In absence of substituents, nitro group the molecular formula C3H9N is : [2015 RS]
always goes to m-position. (a) 4 (b) 5
5. The correct increasing order of basic strength (c) 2 (d) 3
for the following compounds is : [2017] 10. Method by which aniline cannot be prepared is:
[2015 RS]
NH2 NH2 NH2
(a) hydrolysis of phenylisocyanide with acidic
solution
(I) (II) (III)
(b) degradation of benzamide with bromine in
CH3 alkaline solution
NO2
(c) reduction of nitrobenzene with H2/Pd in
(a) III < I < II (b) III < II < I ethanol
(c) II < I < III (d) II < III < I (d) potassium salt of phthalimide treated with
chlorobenzene followed by hydrolysis with
6. Which of the following reactions is appropriate
aqueous NaOH solution.
for converting acetamide to methanamine? [2017]
11. Some reactions of amines are given. Which one
(a) Hoffmann hypobromamide reaction is not correct ? [NEET Kar. 2013]
(b) Stephens reaction (a) (CH3)2NH + NaNO2 + HCl ®
(c) Gabriels phthalimide synthesis (CH3)2 N – N = O
(d) Carbylamine reaction (b) (CH3)2N – + NaNO2 + HCl ®
7. The correct statement regarding the basicity of
arylamines is [2016] (CH3)2N – – N = NCl
(a) Arylamines are generally less basic than
alkylamines because the nitrogen lone-pair (c) CH3CH2NH2 + HNO2 ® CH3CH2OH + N2
electrons are delocalized by interaction (d) CH3NH2 + C6H5SO2Cl ®
with the aromatic ring p electron system. CH3NHSO2C6H5.
(b) Arylamines are generally more basic than 12. In a set of reactions m-bromobenzoic acid gave
alkylamines because the nitrogen lone-pair a product D. Identify the product D. [2011]
electrons are not delocalized by interaction COOH
with the aromatic ring p electron system.
SOCl NH
NaOH
¾¾®
2
B ¾¾®
3
C ¾¾®
Br D
(c) Arylamines are generally more basic than 2
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Amines 293
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294 CHEMISTRY
O – NHCH3 23. The compound obtained by heating a mixture
(a) of a primary amine and chloroform with ethanolic
potassium hydroxide (KOH) is [1997]
(a) an alkyl cyanide (b) a nitro compound
(b)
(c) an alkyl isocyanide (d) an amide
NCH3
24. When aniline reacts with oil of bitter almonds
H
(C6 H5CHO), condensation takes place and benzal
derivative is formed. This is known as [1995]
H
(a) Million's base (b) Schiff's reagent
NCH3
(c) (c) Schiff's base (d) Benedict's reagent
OH 25. What is the decreasing order of basicity of
primary, secondary and tertiary ethylamines and
H NH3 ? [1994]
NCH3 (a) NH 3 > C 2 H 5 NH 2 > (C 2 H 5 ) 2 NH >
(d)
OH (C 2 H 5 )3 N
19. Electrolytic reduction of nitrobenzene in weakly (b) (C2 H 5 )3 N > (C2 H 5 )2 NH >
acidic medium gives [2005]
C2 H5 NH 2 > NH 3
(a) N-Phenylhydroxylamine
(b) Nitrosobenzene (c) (C2 H 5 )2 NH > C2 H 5 NH 2 >
(c) Aniline (C 2 H 5 )3 N > NH 3
(d) p-Hydroxyaniline (d) (C 2 H 5 ) 2 NH > (C 2 H 5 )3 N >
20. The consituent of the powerful explosive RDX
C 2 H 5 NH 2 > NH 3 .
is formed during the nitration of [2000]
(a) Toluene (b) Phenol 26. Mark the correct statement [1988]
(c) Glycerol (d) Urotropine (a) Methylamine is slightly acidic
21. 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 (d) Methylamine forms salts with alkalies.
22. Aniline is reacted with bromine water and the
Topic 2: Amides, Cyanides and Isocyanides
resulting product is treated with an aqueous
solution of sodium nitrite in presence of dilute 27. The following reaction
hydrochloric acid. The compound so formed is NH2
converted into a tetrafluoroborate which is + Cl
subsequently heated dry. The final product is
O
(a) 1,3, 5-Tribromobenzene [1998]
H
(b) p-Bromofluorobenzene N
NaOH
(c) p-Bromoaniline ¾¾¾®
O
(d) 2,4, 6-Tribromofluorobenzene
O
is known by the name : [2015 RS]
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Amines 295
(a) Friedel-Craft's reaction 32. The final product C, obtained in this reaction
(b) Perkin's reaction NH2
(c) Acetylation reaction Ac O Br H O
¾¾®
2
¾¾¾®
A CH 2
B ¾¾®
2
C
(d) Schotten-Baumen reaction COOH
3
+
H
28. On hydrolysis of a “compound”, two compounds
CH3
are obtained. One of which on treatment with sodium
would be [2003]
nitrite and hydrochloric acid gives a product which
does not respond to iodoform test. The second one NHCOCH3 NH2
reduces Tollen’s reagent and Fehling’s solution. The Br COCH3
“compound” is [NEET Kar. 2013] (a) (b)
(a) CH3 CH2 CH2 CON(CH3)2
CH3 CH3
(b) CH3 CH2 CH2 NC
COCH3 NH2
(c) CH3 CH2 CH2 CN Br Br
(d) CH3 CH2 CH2 ON = O (c) (d)
29. An organic compound (C3H9N) (A), when treated
with nitrous acid, gave an alcohol and N2 gas was CH3 CH3
evolved. (A) on warming with CHCl3 and caustic C N
potash gave (C) which on reduction gave
isopropylmethylamine. Predict the structure of (A). H OÅ
33. + CH3MgBr ¾¾¾¾
3 ®P
CH3
(a) CH NH2 [2012 M] OCH3
CH3
Product 'P' in the above reaction is [2002]
(b) CH3CH2 NH CH3 OH O
(c) CH3 N CH3
CH – CH3 C – CH3
CH3
(a) (b)
(d) CH3CH2 CH2 NH2
OCH3 OCH3
30. Acetamide is treated with the following reagents
separately. Which one of these would yield CHO COOH
methylamine? [2010]
(a) NaOH – Br2
(b) Sodalime (c) (d)
(c) Hot conc. H2SO4 OCH3 OCH3
(d) PCl5 34. Intermediates formed during reaction of
31. Which one of the following on reduction with R C NH 2 with Br2 and KOH are [2001]
lithium aluminium hydride yields a secondary ||
O
amine? [2007]
(a) Methyl isocyanide (a) RNHBr and RCONHBr
(b) Acetamide (b) RNHCOBr and RNCO
(c) Methyl cyanide (c) RCONHBr and RNCO
(d) Nitroethane. (d) RCONBr2
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296 CHEMISTRY
35. An isocyanide is prepared by [1999] 39. Acetamide and ethylamine can be distinguished
(a) Friedel-Crafts reaction by reacting with [1994]
(b) Perkin reaction (a) Aqueous HCl and heat
(c) Carbylamine reaction (b) Aqueous NaOH and heat
(d) Wurtz reaction (c) Acidified KMnO4
36. Consider the following sequence of reactions :
(d) Bromine water.
Reduction HNO
® [B] ¾¾¾¾
Compound[A] ¾¾¾¾¾ 2®
40. For carbylamine reaction, we need hot alcoholic
CH3CH 2OH KOH and [1992]
The compound [A] is [1996] (a) Any primary amine and chloroform
(a) CH3CH2CN (b) CH3NO2 (b) Chloroform and silver powder
(c) CH3NC (d) CH3CN (c) A primary amine and an alkyl halide
37. Aniline is an activated system for electrophilic
(d) A monoalkylamine and trichloromethane.
substitution. The compound formed on heating
aniline with acetic anhydride is [1996] 41. Indicate which nitrogen compound amongst the
following would undergo Hoffmann reaction
NH2 (i.e., reaction with Br2 and strong KOH) to
furnish the primary amine (R – NH2) [1989]
(a) (a) RCONHCH3 (b) RCOONH4
COCH3
(c) RCONH2 (d) R – CO – NHOH.
NH2
Topic 3: Nitrocompounds, Alkyl Nitrites and
Diazonium Salts
(b)
42. In the following reaction, the product (A) [2014]
COCH3 +
NºNCl
–
NH2
NH2
+ H +
is :
(c) ¾¾® (A)
Yellow dye
COCH3
NHCOCH3
(d) (a) N=N–NH
NH2
38. Which is formed, when acetonitrile is hydrolysed
partially with cold concentrated HCl? [1995] (b) N=N
(a) Acetic acid NH2
(b) Acetamide
(c) N=N
(c) Methyl cyanide
(d) Acetic anhydrides
(d) N=N NH2
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Amines 297
43. Which of the following will be most stable 47. Aniline in a set of the following reactions yielded
diazonium salt RN2+X– ? [2014] a coloured product ‘Y’ [2002, 2010]
(a) CH3 N2+X– (b) C6H5N2+X– NH2
A
The structure of ‘Y’ would be ;
¾®
Br Br
CH3
(a) N=N N
ÅN2C I –
CH3
A is : [NEET 2013]
(a) Cu2Cl2 (b) H3PO2 and H2O CH3 CH3
(a) 1, 3 - Dinitrobenzene
CH3 CH3
(b) 1, 4 - Dinitrobenzene
(c) 1, 2, 4 - Trinitrobenzene (d) HN N=N NH
(d) 1, 2 - Dinitrobenzene
48. Nitrobenzene can be prepared from benzene by
46. What is the product obtained in the following using a mixture of conc. HNO3 and conc. H2SO4;
reaction : [2011] in the mixture, nitric acid acts as a/an: [2009]
NO2 (a) acid (b) base
Zn ........... ?
¾¾®
NH Cl (c) catalyst (d) reducing agent
4
49. Aniline in a set of reactions yielded a product D.
NHOH NH2
NaNO CuCN
(a) ¾¾¾®
HCl
2
A ¾¾¾® B
N H HNO
¾¾¾®
2
C ¾¾¾®
2
D
(b) N Ni
The structure of the product D would be: [2005]
O
– (a) C6H5NHOH (b) C6H5NHCH2CH3
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298 CHEMISTRY
(a) Formaldehyde
(b) CH3 N=N NH2 (b) Trichloromethane
(c) Nitrobenzene
(d) Toluene
(c) (CH3)2N N=N 52. Which of the following reagents will convert
p-methylbenzenediazonium chloride into
p-cresol?
(d) (CH3)2N NH (a) Cu powder (b) H2O [1999]
(c) H3PO2 (d) C6H5OH
51. reduction
[A] ¾¾¾¾® [B] ¾¾¾¾¾¾
3 ®
CHCl +KOH [2000] 53. Diazo coupling is useful to prepare some [1994]
reduction (a) Pesticides (b) dyes
[C] ¾¾¾¾® N-Methylaniline, A is
(c) proteins (d) vitamins
ANSW E R KE Y
1 (d) 7 (a) 13 (b ) 19 (c) 25 (d ) 31 (a) 37 (d ) 43 (b ) 49 (d )
2 (b) 8 (d ) 14 (c) 20 (d ) 26 (c) 32 (d ) 38 (b ) 44 (b ) 50 (c)
3 (a) 9 (a) 15 (d ) 21 (c) 27 (d ) 33 (b ) 39 (b ) 45 (a) 51 (c)
4 (c) 10 (d ) 16 (d ) 22 (d ) 28 (b ) 34 (c) 40 (a) 46 (a) 52 (b )
5 (c) 11 (b ) 17 (c) 23 (c) 29 (a) 35 (c) 41 (c) 47 (a) 53 (b )
6 (a) 12 (c) 18 (b ) 24 (c) 30 (a) 36 (d ) 42 (d ) 48 (b )
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Amines 299
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300 CHEMISTRY
17. (c) Benzylamine is more basic than aniline. The 21. (c) (CH3)2NH is most basic because two electron
reason is that in aniline, the lone pair of nitrogen releasing groups are present on Nitrogen. Also
aromatic amines are less basic than aliphatic amines.
is conjugated with benzene ring so it is not
The basic character of amines follow the order
available readily for others. On the other hand
R2NH > RNH2 > C6H5NHCH3 > NH3
in Benzylamine, nitrogen is not directly attached
with ring so lone pairs are not conjugated with ring. NH2 NH2
18. (b) Br Br
NaNO
O NCH3 22. (d) +3Br2 ¾® ¾¾¾® 2
& dil HCl
CH 3NH 2
¾¾¾®
Br
¾¾¾®
LiAlH4/H2O
2, 4, 6 tribromoaniline
[H]
+ –
NHCH3 N2Cl N2BF4
Br Br Br Br
NaBF4
¾¾¾®
(- NaCl)
19. (c) Electrolytic reduction of Nitrobenzene in
weakly acidic medium give aniline
Br Br
NO2 NH2 Diazonium Diazonium
salt tetra fluro borate
Electrolytic reduction
¾¾¾¾¾¾¾®
(weakly acidic medium) F
Aniline Br Br
D
Nitroalkane in strongly acidic medium gives ¾¾®
-N2
p-hydroxyaniline - BF3
Br
NO2 NHOH 2,4,6 tribromofluorobenzene
electrolytic
¾¾¾¾¾¾¾®
reduction in presence 23. (c) CH 3CH 2 NH 2 + CHCl 3 + 3KOH ¾
¾®
of conc. H2 SO4
CH 3CH 2 NC + 3KCl + 3H 2O
(Strongly acidic medium)
In this reaction, bad smelling compound ethyl
NH2 isocyanide (CH 3CH 2 NC) is produced. This
rearrangement equation is known as carbyl amine reaction.
¾¾¾¾®
24. (c) Benzaldehyde reacts with primary aromatic
amine to form schiff's base
OH
p-Hydroxyaniline C6 H 5CH = O + C 6 H 5 NH 2 ¾
¾®
Benzaldehyde Aniline
20. (d) RDX is prepared by treating urotropine with
fuming nitric acid. When the inner bridge system is C 6 H 5 C H = NC 6 H 5 + H 2 O
destroyed by oxidation and the peripheral nitrogen Benzyliden e aniline
atom are nitrated, it forms cyclonite (or RDX). 25. (d) All aliphatic amines are stronger bases than
N N conc. HNO O2N — N N — NO2 NH3 and among different ethylamines order of
N ¾¾¾¾® 3
basictity is 2° > 3° > 1°. Thus, the correct order
N N is (d) (in polar solvent):
Urotropine
NO2 (C 2 H 5 ) 2 NH > (C 2 H 5 )3 N >
RDX
C 2 H 5 NH 2 > NH 3 .
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Amines 301
This anomalous behaviour of tertiary ethyl amine
is due to steric factors i.e., crowding of alkyl NH2
Cl
groups cover nitrogen atom from all sides and +
(d)
thus makes the approach and bonding by a lewis O
acid relatively difficult which results the Schotten-Baumen Reaction
maximum steric strain in tertiary amines. The
electrons are there but the path is blocked NH – C –
resulting the reduction in its basicity. ¾¾®
NaOH O
This comparison is true for amines in a protic 28. (b) Hydrolysis of propyl isocyanide
solvent like water. In gaseous phase, the order of
basicity is: tertiary amine > secondary amine > (CH3CH2CH2NC) gives CH3CH2CH2 NH2 +
primary amine > NH3. HCOOH.
26. (c) Methyl amine is a stronger base than On treatment with NaNO2 and HCl (I) gives
ammonia due to +I effect. The alkyl groups which CH3CH2CH2OH which does not give iodoform
are electron releasing groups increase the test. II (HCOOH) reduces Tollen’s reagent and
electron density around the nitrogen thereby Fehling’s solution.
increasing the availability of the lone pair of H O
electrons to proton or lewis acid and making the CH3CH 2CH 2 NC ¾¾¾
2 ®
Propyl isocyanide
amine more basic
NH3 CH3NH2 CH3CH 2 CH 2 NH 2 + HCOOH
I II
Kb = 1.8 × 10–5 44 × 10–5
27. (d) HNO2
29. (a) CH3 CH NH2 ¾¾®
CH3
Base
Ar OH + HO reduction
C ¾¾¾® CH3 CH NH CH3
¾¾®
R R CH3
O Isopropylmethylamine
OH
+ CH3 – C – Cl ¾® Only primary aliphatic and aromatic amines give
(c)
Acetylation Reaction
carbylamine reaction. Primary amine on reaction
with HNO2 give alcohol
O
NaOH
O – C – CH3 30. (a) CH3CONH2 ¾¾¾¾
® CH3 NH 2
Br2
(Hofmann bromamide reaction)
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302 CHEMISTRY
31. (a) Reduction of alkyl isocyanides in presence
– +
of LiAlH4 yields secondary amines containing K
(iii) R — C — N —Br R — C — N + KBr
methyl as one of the alkyl group.
O O
r LiAlH 4
R - N = C + 4[H] ¾¾¾¾ ® R - NH - CH3 R
2°amine
N ••
••
r LiAlH 4 (iv) O = C R—N=C=O
e.g., CH3 - N = C + 4[H] ¾¾¾¾ ®
•
•
(rearrangement)
CH 3 - NH - CH 3 · ·
(v) R - N =
C=O + 2KOH ® RNH 2 + K 2 CO 3
dimethyl amine
whereas, alkyl cyanides give 1° amine on reduction. 35. (c) Carbylamine reaction :
32. (d) R– NH2+3KOH+CHCl3 ¾¾®
NH2 NHCOCH3 R –N C+3KCl+3H2O
Reduction CH
(CH CO) O
¾
Br
¾¾¾
2
® 36. (d) CH3CN ¾¾¾® 3 CH 2 NH2
¾¾¾¾
3 2
® CH COOH
3 HONO
CH3 CH 2 OH + N 2+ HO2¬
CH3 CH3
(A) 37. (d) Aniline when treated with acetic anhydride
NHCOCH3 NH2 forms acetanilide (nucleophilic substitution)
O
Br +
H /H O Br
¾¾2® NH2 NH – C – CH3
O O
O
CH3 CH3 CH 3–C–O–C–CH 3
¾¾¾¾¾¾® + CH3– C – OH
(B) (C)
CºN
38. (b) Acetonitrile (Methyl cyanide) on treatment
33. (b) + CH3MgBr ¾¾ with conc. HCl give acetamide.
®
Conc .HCl
OCH3 CH 3C º N ¾¾ ¾ ¾
¾® CH 3CONH 2.
Acetonitri le Acetamide
H3C –C = NMgBr H3C –C = NH
39. (b) Acetamide and ethylamine can be
H O+ distinguished by heating with NaOH solution.
¾¾®
3
Acetamide evolves NH3 but ethylamine does not.
OCH3
OCH3 D
COCH3 CH 3CONH 2 + NaOH ¾¾®
H O
+ CH 3COONa + NH 3
¾¾®
3
OCH3 ® No reaction.
CH 3 CH 2 NH 2 + NaOH ¾¾
34. (c) The mechanism of Hoffmann bromamide
40. (a) Any primary amine means both aliphatic as
reaction is
well as aromatic but monoalkylamines means
(i) RCONH 2 + Br2 ® RCONHBr + HBr
only 1° aliphatic amines. Therefore, option (a) is
-
OH
(ii) RCONHBr + HBr ¾¾¾ ® correct while (d) is wrong.
·-·
R — CO N Br + H 2O RNH2 + CHCl3 + 3KOH ¾¾®
RNC + 3KCl + 3H 2 O
isocyanide
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Amines 303
41. (c) Only 1° amides (i.e. RCONH2) in the present 46. (a) When nitro compound is reduced with a
case undergo Hofmann bromamide reaction. neutral reducing agent (Zn dust + NH4Cl) the
RCONH2 + Br2 + 4KOH ¾¾ ® corresponding hydroxyl amine is formed
R - NH 2 + 2KBr + K 2CO3 + 2H 2O Zn dust + NH Cl
C6 H 5 NO 2 + 4[H] ¾¾¾¾¾¾¾
4
®
(Hofmann's bromamide reaction) C6 H 5 NHOH
Phenyl hydroxyl amine
NH 2 47. (a)
42. (d) NH2 N = N – Cl
N 2 Cl ¾¾¾¾¾¾®
+
CH3
H N
NaNO2/HCl CH3
¾¾ ¾¾®
(273–278K)
¾¾¾¾¾®
N=N NH 2 (X)
CH3
43. (b) Arene diazonium salts are most stable N=N N
among the given options because of the CH3
dispersal of +ve charge on the benzene ring (Y)
due to resonance. 48. (b) HONO2 + H2SO4
Base Acid
Resonance structures of arene diazonium
® NO +2 + H 2 O + HSO 4-
¾¾
N º ··
Å Å Å
N N= N N= N Nitric acid acts as a base by accepting a proton.
··
··
·· ·· 49. (d)
Å + –
¬¾® ¬¾® NH2 N2 Cl
NaNO
Å ¾¾®
2
HCl
Å
N º ··
Å Å
··
N = ··
N N NºN Arene diazonium salt
¾¾®
Å CuCN
¬¾® ¬¾® ¬¾® Å
¾®
Br +H3PO2+H2O Br Benzyl a lcohol
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304 CHEMISTRY
51. (c) 53. (b) Azo dyes can be prepared by diazotizing an
NO2 N C aromatic amine and subsequent coupling with a
suitable aromatic phenol or amine
Sn + HCl CHCl +KOH
3
¾¾¾¾®
reduction
¾¾¾¾¾®
Nitrobenzene N =NCl + H N(CH3)2
(A) (B) (C)
NH–CH3
Diazonium chloride N,N-dimethylaniline
Re duction
¬¾ ¾¾ ¾¾
Na/C2H5OH
¾® N=N N(CH3)2
N-methylaniline
p-Dimethyl amino azobenzene
–
N2+ Cl OH p-Dimethyl amino azobenzene is used as a dye
for colouring polishes, wax products and soap.
H O
¾¾¾
2 ®
+ N2+HCl
52. (b)
CH3 CH3
p-cresol
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Biomolecules 305
28 Biomolecules
Trend Analysis with Important Topics & Sub-Topics
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306 CHEMISTRY
(b) (+) Lactose is a b-glycoside formed by the 15. Mg is present in which of the following : [2000]
union of a molecule of D(+) glucose and a (a) Starch (b) Chlorophyll
molecule of D(+) galactose. (c) Both (d) None
(c) (+) Lactose is a reducing sugar and does 16. Which of the following is the sweetest sugar?
not exhibit mutarotation. (a) Sucrose (b) Glucose [1999]
(d) (+) Lactose, C12H22 O11 contains 8-OH (c) Fructose (d) Maltose
groups. 17. In cells, the net production of ATP molecules
6. Which one of the following does not exhibit the generated from one glucose molecule is [1999]
phenomenon of mutarotation ? [2010] (a) 46 (b) 32
(a) (+) – Sucrose (b) (+) – Lactose (c) 36 (d) 40
(c) (+) – Maltose (d) (–) – Fructose 18. The number of molecules of ATP produced in
7. Fructose reduces Tollen’s reagent due to [2010] the lipid metabolism of a molecule of palmitic
(a) enolisation of fructose followed by acid is [1998]
conversion to aldehyde by base (a) 130 (b) 36
(b) asymmetric carbons (c) 56 (d) 86
(c) primary alcoholic group 19. Glucose molecule reacts with 'X' number of
(d) secondary alcoholic group molecules of phenylhydrazine to yield osazone.
8. The cell membranes are mainly composed of The value of 'X' is [1998]
(a) fats (b) proteins [2005] (a) four (b) one
(c) phospholipids (d) carbohydrates (c) two (d) three
9. Number of chiral carbons in b – D – (+) – glucose 20. Sucrose in water is dextro-rotatory, [a]D= + 66.4º.
is [2004] When boiled with dilute HCl, the solution
(a) five (b) six becomes leavo-rotatory, [a]D= –20º. In this
(c) three (d) four process, the sucrose molecule breaks into[1996]
10. Glycolysis is [2003] (a) L-glucose + D-fructose
(a) conversion of glucose to haem (b) L-glucose + L-fructose
(b) oxidation of glucose to glutamate (c) D-glucose + D-fructose
(c) conversion of pyruvate to citrate (d) D-glucose + L-fructose
(d) oxidation of glucose to pyruvate 21. The a-D glucose and b-D glucose differ from
11. Phospholipids are esters of glycerol with [2003] each other due to difference in carbon atom with
(a) Three phosphate groups respect to its [1995]
(b) Three carboxylic acid residues (a) Conformation
(c) Two carboxylic acid residues and one (b) Configuration
phosphate group (c) Number of OH groups
(d) One carboxylic acid residue and two (d) Size of hemiacetal ring
phosphate groups 22. On hydrolysis of starch, we finally get [1991]
12. Cellulose is a polymer of [2002] (a) Glucose (b) Fructose
(a) Glucose (b) Fructose (c) Both (a) and (b) (d) Sucrose
(c) Ribose (d) Sucrose
13. Which of the following gives positive Fehling Topic 2: Amino Acids and Proteins
solution test ? [2001] 23. Which of the following is a basic amino acid ?
(a) Protein (b) Sucrose (a) Alanine (b) Tyrosine [2020]
(c) Glucose (d) Fats (c) Lysine (d) Serine
14. Which is correct statement? [2001] 24. Which structure(s) of proteins remain(s) intact
(a) Starch is a polymer of a-glucose during denaturation process?
(b) In cyclic structure of fructose, there are four [NEET Odisha, 2019]
carbons and one oxygen atom (a) Tertiary structure only
(c) Amylose is a component of cellulose (b) Both secondary and tertiary structures
(d) Proteins are composed of only one type of (c) Primary structure only
amino acids (d) Secondary structure only
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Biomolecules 307
25. The non-essential amino acid among the H O H
following is: [2019] | | | || | | | | |
(a) valine (b) leucine (d) - N - C- C- C - N - C- C- N- C - C- C-
| | | | | || | |
(c) alanine (d) lysine H O
26. Which of the following compounds can form a 32. The correct statement in respect of protein
zwitterion? [2018] haemoglobin is that it [2004]
(a) Aniline (b) Acetanilide (a) functions as a catalyst for biological reactions
(c) Glycine (d) Benzoic acid (b) maintains blood sugar level
27. Which of the following statements is not (c) acts as an oxygen carrier in the blood
correct : [2017] (d) forms antibodies and offers resistance to
(a) Ovalbumin is a simple food reserve in egg- dieases
white 33. The helical structure of protein is stabilized by
(b) Blood proteins thrombin and fibrinogen are [2004]
involved in blood clotting (a) dipeptide bonds (b) hydrogen bonds
(c) Denaturation makes the proteins more active (c) ether bonds (d) peptide bonds
(d) Insulin maintains sugar level in the blood
34. Which is not a true statement? [2002]
of a human body
(a) a-Carbon of a-amino acid is asymmetric
28. In a protein molecule various amino acids are
linked together by [2016] (b) All proteins are found in L-form
(a) a-glycosidic bond (b) b-glycosidic bond (c) Human body can synthesize all proteins
(c) peptide bond (d) dative bond they need
29. Which of the statements about "Denaturation" (d) At pH = 7 both amino and carboxylic
given below are correct ? [2011 M] groups exist in ionised form
(a) Denaturation of proteins causes loss of O
|| · ·
secondary and tertiary structures of the protein. 35. For - C - N H - (peptide bond) [2001]
(b) Denturation leads to the conversion of Which statement is incorrect about peptide bond?
double strand of DNA into single strand
(a) C — N bond length in proteins is longer
(c) Denaturation affects primary strucrture
than usual bond length of the C — N bond
which gets distorted
(b) Spectroscopic analysis shows planar
(a) (b) and (c) (b) (a) and (c)
(c) (a) and (b) (d) (a), (b) and (c) structure of the — C— NH — group
||
30. Which functional group participates in O
disulphide bond formation in proteins? [2005] (c) C — N bond length in proteins is smaller
(a) Thioester (b) Thioether than usual bond length of the C—N bond
(c) Thiol (d) Thiolactone (d) None of the above
31. Which one of the following structures 36. The number of essential amino acids in man is
represents the peptide chain? [2004] (a) 8 (b) 10 [2000]
H O (c) 18 (d) 20
| | || 37. The dominant cation in the blood plasma
(a) - N - C - N - C - NH - C - NH - (extracellular fluid) is [1999]
|| | |
O H (a) potassium (b) calcium
H H (c) magnesium (d) sodium
| | | | | | | | 38. The reactions of (a) oxygen and (b) carbon
(b) - N - C - C- C- C- N - C- C- C- monoxide with haeme (the prosthetic group of
|| | | | | | | haemoglobin) give [1997]
O (a) only oxygen-haeme complex
H H H O (b) only carbon monoxide-haeme complex
| | | | | | ||
(c) both oxygen-haeme and carbon monoxide-
(c) - N - C- C - N - C - C - N - C - C -
| || | || | haeme complexes but oxygen-haeme
O O complex is more stable
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308 CHEMISTRY
(d) both oxygen-haeme and carbon monoxide- 46. The segment of DNA which acts as the
haeme complexes but carbon monoxide- instrumental manual for the synthesis of the
haeme complex is more stable protein is: [2009]
39. Identify an element amongst the following which (a) ribose (b) gene
is responsible for oxidation of water to O2 in (c) nucleoside (d) nucleotide
biological processes [1997] 47. In DNA, the complimentary bases are:
(a) Fe (b) Mn [1998, 2008]
(c) Mo (d) Cu (a) Adenine and thymine; guanine and cytosine
40. In reference to biological role, Ca2+ ions are (b) Adenine and thymine ; guanine and uracil
important in [1996] (c) Adenine and guanine; thymine and cytosine
(a) triggering the contraction of muscles
(d) Uracil and adenine; cytosine and guanine
(b) generating right electrical potential across
48. RNA and DNA are chiral molecules, their
cell membrane
chirality is due to [2007]
(c) hydrolysis of ATP
(d) defence mechanism (a) chiral bases
41. A reagent suitable for the determination of (b) chiral phosphate ester units
N-terminal residue of a peptide is [1996] (c) D-sugar component
(a) p-Toluenesulphonyl chloride (d) L-sugar component.
(b) 2, 4-Dinitrophenylhydrazine 49. During the process of digestion, the proteins
(c) Carboxypeptidase present in food materials are hydrolysed to
(d) 2, 4-Dinitrofluorobenzene amino acids. The two enzymes involved in the
42. Which of the following protein destroy the process
antigen when it enters in body cell? [1995] Proteins ¾¾¾¾¾Enzyme (A)
® Polypeptides
(a) Antibodies (b) Insulin Enzyme (B)
(c) Chromoprotein (d) Phosphoprotein ¾¾¾¾¾® Amino acids
are respectively [2006]
Topic 3: Nucleic Acid and Enzymes (a) Diastase and Lipase
(b) Pepsin and Trypsin
43. The correct statement regarding RNA and DNA, (c) Invertase and Zymase
respectively is [2016] (d) Amylase and Maltase
(a) The sugar component in RNA is arabinose 50. A sequence of how many nucleotides in
and the sugar component in DNA is messenger RNA makes a codon for an amino
2'-deoxyribose. acid? [2004]
(b) The sugar component in RNA is ribose and the (a) Three (b) Four
sugar component in DNA is 2'-deoxyribose. (c) One (d) Two
(c) The sugar component in RNA is arabinose 51. The enzyme which hydrolyses triglycerides to
(d) The sugar component in RNA is fatty acids and glycerol is called [2004]
2'-deoxyribose and the sugar component (a) Maltase (b) Lipase
in DNA is arabinose (c) Zymase (d) Pepsin
44. In DNA, the linkages between different 52. Chargaff's rule states that in an organism [2003]
nitrogenous bases are: [NEET Kar. 2013] (a) Amounts of all bases are equal
(a) peptide linkage (b) phosphate linkage (b) Amount of adenine (A) is equal to that of
(c) H-bonding (d) glycosidic linkage thymine (T) and the amount of guanine (G)
45. Which one of the following, statements is is equal to that of cytosine (C)
incorrect about enzyme catalysis? [2012] (c) Amount of adenine (A) is equal to that of
(a) Enzymes are mostly proteinous in nature. guanine (G) and the amount of thymine (T)
(b) Enzyme action is specific. is equal to that of cytosine (C)
(c) Enzymes are denaturated by ultraviolet
(d) Amount of adenine (A) is equal to that of
rays and at high temperature.
cytosine (C) and the amount of thymine
(d) Enzymes are least reactive at optimum
(T) is equal to that of guanine (G)
temperature.
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Biomolecules 309
53. Enzymes are made up of [2002] Topic 4: Vitamins and Hormones
(a) Edible proteins
(b) Proteins with specific structure 59. Which of the following hormones is produced
(c) Nitrogen containing carbohydrates under the condition of stress which stimulates
(d) Carbohydrates glycogenolysis in the liver of human beings?
54. Which of the following is correct about H- (a) Thyroxin (b) Insulin [2014]
bonding in nu cleotide? [2001] (c) Adrenaline (d) Estradiol
(a) A --- A and T --- T (b) G --- T and A --- C 60. Deficiency of vitamin B1 causes the disease
(c) A --- G and T --- C (d) A --- T and G --- C (a) Convulsions (b) Beri-Beri [2012]
55. Which one of the following chemical units is (c) Cheilosis (d) Sterility
certainly to be found in an enzyme? [1997] 61. Which of the following hormones contains
OH iodine? [2009]
O
H (a) Testosterone (b) Adrenaline
O (c) Thyroxine (d) Insulin
(a) (b) N—C
HO 62. Which one of the following is an amine hormone ?
HO O O (a) Thyroxine (b) Oxypurin [2008]
O (c) Insulin (d) Progesterone
O O R 63. Which of the following is water-soluble? [2007]
(a) Vitamin E (b) Vitamin K
O R (c) Vitamin A (d) Vitamin B
(c) (d) 64. Which one of the following is a peptide
O R
hormone ? [2006]
O (a) Testosterone (b) Thyroxin
56. Chemically considering, digestion is basically (c) Adrenaline (d) Glucagon
[1994] 65. The human body does not produce [2006]
(a) Anabolism (b) Hydrogenation (a) Vitamins (b) Hormones
(c) Hydrolysis (d) Dehydrogenation. (c) Enzymes (d) DNA
57. Enzymes take part in a reaction and [1993] 66. The hormone that helps in the conversion of
(a) decrease the rate of a chemical reaction glucose to glycogen is [2004]
(b) increase the rate of a chemical reaction (a) Cortisone (b) Bile acids
(c) both (a) and (b) (c) Adrenaline (d) Insulin
(d) none of these 67. Vitamin B12 contains [2003]
58. The couplings between base units of DNA is (a) Ca(II) (b) Fe(II)
through : [1992] (c) Co(III) (d) Zn(II)
(a) Hydrogen bonding 68. Which of the following is a steroid hormone?
(b) Electrostatic bonding [1999]
(c) Covalent bonding (a) Cholesterol (b) Adrenaline
(d) van der Waals forces (c) Thyroxine (d) Progesterone
ANSWER KEY
1 (b) 8 (c) 15 (b) 22 (a) 29 (c) 36 (b) 43 (b) 50 (a) 57 (b) 64 (d)
2 (d) 9 (a) 16 (c) 23 (c) 30 (c) 37 (d) 44 (c) 51 (b) 58 (a) 65 (a)
3 (d) 10 (d) 17 (c) 24 (c) 31 (c) 38 (d) 45 (d) 52 (b) 59 (c) 66 (d)
4 (b) 11 (c) 18 (a) 25 (c) 32 (c) 39 (b) 46 (b) 53 (b) 60 (b) 67 (c)
5 (c) 12 (a) 19 (d) 26 (c) 33 (b) 40 (a) 47 (a) 54 (d) 61 (c) 68 (d)
6 (a) 13 (c) 20 (c) 27 (c) 34 (b) 41 (d) 48 (c) 55 (b) 62 (a)
7 (a) 14 (a) 21 (b) 28 (c) 35 (a) 42 (a) 49 (b) 56 (c) 63 (d)
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310 CHEMISTRY
Sucrose on hydrolysis brings about a change in the 5. (c) All reducing sugar shows mutarotation.
sign of rotation from dextro to laevo and the product
is called as invert sugar. Reducing sugars have hydroxyl group at the
anomeric C1 position. In solution, a carbohydrate
2. (d) Sucrose is non-reducing disaccharide as can open up to its aldehyde form. Hence, it can
the two monosaccharide units are linked through switch configuration between a– and b– forms.
their respective carbonyl groups. 6. (a) Sucrose does not have free — CHO or CO
Only carbonyl group can be reduced as with other
group, hence it does not undergo mutarotation.
carbons, –OH group is attached.
7. (a) Fructose, a ketose as the substrate, under
the alkaline medium of Tollen’s reagent, a part of
fructose is transformed into glucose and
3. (d) Glucose reacts with hydroxyl amine to form
mannose, both aldoses. Then these aldoses give
an oxime.
positive silver mirror test.
CHO CH=N–OH
8. (c) Cell membranes (Plasma membranes) con-
H OH H OH stitutes bilayer of phospholipid with embedded
HO H NH2OH HO H proteins. In humans, lipids accounts for upto
¾¾¾ ®
H OH H OH 79% of cell membrane.
H OH H OH 9. (a)
CH2–OH CH2–OH
HO 1
D(+)glucose C H
2
H OH
4. (b) HO 3
CH2 OH HO H
4 O b – D –(+)– Glucose
O CH2 H OH
O
OH + OH 5
OH H
HO H2C
OH
OH CH2OH
OH OH
a-D-Glucopyranose b-D-Fructofuranose Carbon atoms from C 1 to C 5 are chiral
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Biomolecules 311
10. (d) It is a common pathway for both the aerobic ketonic group, the third molecule of
& anaerobic respiration in which 1 glucose phenylhydrazine condenses to glucosazone.
molecule is converted to 2 molecules of pyruvate. Therefore, the value of X is 3
11. (c) Phospholipids are derivatives of glycerol CHO + H2NNHC6H5 CH = NNHC6H5
in which two of the hydroxyl groups are
esterified with fatty acids while the third is CHOH CHOH
esterified with some derivative of phosphoric warm
¾¾®
acid with some alcohol such as choline, (CHOH)3 (CHOH)3
ethanolamine, serine or inositol.
CH2OH CH2OH
R¢¢
Glucose Glucose Phenylhydrozone
O
¾¾®
H2NNHC6H5
O = P – O – CH2 O
OH CH – O – C –R¢ CH = NNHC6H5 CH = NNHC6H5
CH2O – C –R H2NNHC6H5
C = NNHC6H5 C=O
– H2 O
O
12. (a) We know that cellulose (C 6 H12 O 6 ) n is (CHOH)3 (CHOH)3
the chief constituent of cell walls of plants. It is
the most abundant organic substance found in CH2OH CH2OH
nature. It is a polymer of glucose with 3500 Glucosazone keto compound of glucose
phenyl hydrazone
repeating units in a chain.
20. (c) The hydrolysis of sucrose by boiling with
13. (c) Glucose contain aldehyde group. Hence, it
mineral acid or by enzyme invertase or sucrase
gives positive fehling solution test.
produces a mixture of equal molecules of D(+)
14. (a) Starch is also know as amylum which
glucose and D(–) Fructose.
occurs in all green plants. A molecule of starch
(C 6 H10 O5 ) n is built of a large number of HCl
C12 H 22 O11 + H 2 O ¾¾¾
® C6 H12 O6 + C6 H12 O6
a-glucose ring joined through oxygen-atom. sucrose D - glu cos e D - Fructose
15. (b) Chlorophyll contain Mg. [ a D ]=+66.5º [ aD ]=+52.5º [a D ]=-92º
144444244444 3
16. (c) Fructose is the sweetest sugar. Invert sugar,[ a D ]=-20º
17. (c) C 6 H12 O 6 + 6O 2 + 2 ATP ¾ ¾® 21. (b) a-D glucose and b-D glucose are the
6CO 2 + 6H 2O + 38 ATP Molecules isomers which differ in the orientation
Net total number of ATP molecules evolved = (configuration) of H and OH groups around C1
36 molecules. atom.
18. (a) In the lipid metabolism, when palmitic acid
is oxidised, two carbon fragments are removed
H —C1—OH HO —C1—H
sequentially to form acetyl coenzyme. It enters
the citric acid cycle for production of 130 ATP. H —C2—OH H —C2—OH
O O
19. (d) Glucose reacts with one molecule of HO —C3—H HO —C —H
3
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312 CHEMISTRY
22. (a) Manufacture - By hydrolysis of starch with
HO 2CCHC H 2S - SCH 2 CHC O 2 H
hot dil. mineral acids | |
NH 2 NH2
(C6 H10O5 ) n ® (C6 H10O5 )n ® Cystine
Starch Dextrin
31. (c) The bond formed between two amino acids
H O
C12 H 22 O11 ¾¾¾
2 ® 2C H O
6 12 6 by the elimination of a water molecule is called
Maltose Glucose a peptide linkage or bond. The peptide bond is
simply another name for amide bond.
–
23. (c) H2N – (CH2)4 – CH – COO
+ - C OH + H — N— ¾¾
® — C– N— + H2 O
NH3 | | || |
Lysine O H O H
Carboxyl group Amine group of Peptide bond
of one amino acid other amino acid
Since it contains more number of –NH2 groups
as compared to –COOH groups, hence it is basic The product formed by linking amino acid
amino acid. molecules through peptide linkages. —CO—
24. (c) During denaturation primary structure of NH—, is called a peptide.
proteins remain intact while secondary and 32. (c) Haemoglobin acts as an oxygen carrier in
tertiary structures are destroyed. the blood since it reacts with oxygen to form
unstable oxyhaemoglobin which easily breaks
25. (c) Alanine is non-essential amino acid.
to give back haemoglobin and oxygen.
26. (c)
HOOC – CH 2 – NH 2 33. (b) Hydrogen bonding between different units
Glycine is responsible for holding helix in a position.
+
OOC – CH 2 – NH 3 NH group in one unit is linked to carbonyl
Zwitter ion oxygen of the third unit by hydrogen bonding.
27. (c) Due to denaturation of proteins, helix get
uncoiled and protein loses its biological activity. The a-helix structure is formed when the chain of
a-amino acids coils as a right handed screw (called
28. (c) Peptide bond
a-helix) because of the formation of hydrogen
bonds between amide groups of the same peptide
— C — NH —
|| chain.
O 34. (b) All proteins are not found in L-form but
29. (c) When the proteins are subjected to the they may be present in form of D or L
action of heat, mineral acids or alkali, the water 35. (a) Due to resonance C — N bond in protein
soluble form of globular protein changes to water acquires double bond character and is smaller
insoluble fibrous protein. This is called than usual C — N bond.
—
denaturation of proteins. During denaturation, O O
secondary and tertiary structures of protein C NH C NH
Å
destroyed but primary structures remains intact. 36. (b) There are 20 amino acid in man, out of
R -S-S- R which 10 amino acids are essential amino acids.
30. (c) 2R - S - H
Thiol Disulphide
Essential amino acids are supplied to our bodies
Example : by food which we take because they cannot be
[O] synthesised in the body. These are (i) valine (ii)
2HO 2C CHC H 2SH leucine (iii) Isoleucine (iv) Phenylalanine (v)
| [H]
NH2 Threonine (vi) Methionine (vii) Lysine (viii)
Cysteine Tryptophan (ix) Arginine (x) Histidine.
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314 CHEMISTRY
ribose. The chirality of DNA and RNA molecules 53. (b) Enzymes are made up of protein with
are due to the presence of sugar components. specific structure.
54. (d)
5
O
HOCH2 OH O Deoxyribose-Adenine ... Thymine -Deoxyribose O
4 1
OH P P OH
H H O Deoxyribose-Guanine ... Cytosine -Deoxyribose
O
H H OH P Deoxyribose-Guanine ... Cytosine -Deoxyribose
P OH
O O
3 2
OH OH OH P Deoxyribose-Adenine ... Thymine -Deoxyribose
P OH
D(–)–ribose
The hydrogen bonds are formed between the base
C1, C2, C3 and C4 are chiral carbons. (shown by dotted lines). Because of size and
geometrics of the bases, the only possible pairing
O in DNA are between G (Guanine) and C (Cytosine)
5
HOCH2 OH through three H-bonds and between A (Adenosine)
4 1 and T (Thymine) through two H-bonds.
H H 55. (b) Peptide bonds are present in enzyme.
H H
3 2 H
OH H
N C
D(–) -2-deoxy ribose
C1, C3 and C4 are chiral carbons. O
49. (b) Pepsin and Trypsin are two enzymes 56. (c) In digestion, large molecules are
involved in the process (hydrolysis of proteins) hydrolysed to give smaller molecules. For
Pepsin
Proteins ¾¾¾¾® Polypeptides example, protein gets hydrolysed to amino acids.
Proteases
Thus, chemically considering, digestion is
Trypsin basically hydrolysis.
® Amino acids
¾¾¾¾¾¾¾¾¾¾
Chemotrypsin 57. (b) Enzymes being biocatalyst can increase the
(Pancreatic juice Intestine)
rate of a reaction upto 10 million times. Even
50. (a) The sequence of bases in mRNA are read very small amount can accelerate a reaction.
in a serial order in groups of three at a time.
Each triplet of nucleotides (having a specific 58. (a) DNA consists of two polynucleotide
sequence of bases) is known as codon. Each chains, each chain forms a right handed helical
codon specifies one amino acid. spiral with ten bases in one turn of the spiral.
The two chains coil to double helix and run in
There are four bases A, G, C,T, therefore, 4 3 = 64 opposite direction held together by hydrogen
triplets or codons are possible. bonding.
59. (c) Adrenaline is a hormone produced by
51. (b) Triglycerides are lipids, hence these are
adrenal glands during high stress or exciting
hydrolysed by lipases to glycerol and fatty acids.
situations. This powerful hormone is part of the
A G human body’s acute stress response system,
52. (b) = =1
T C also called the fight or flight response.
Amount of A = T and that of G = C. 60. (b) Beri-Beri.
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Biomolecules 315
61. (c) Thyroxine is the only hormone among the Vitamin H also known as Biotin is part of B
given choices that contains iodine. Its structure complex group of vitamin.
is as follows:
64. (d) Testosterone and Adrenaline are steroid
OH hormone, Thyroxin is a non-steroid hormone and
glucagon is a peptide hormone.
I I
65. (a) Vitamins are organic substances which
does not provide energy but are essential for
I healthy growth and proper functioning of body.
O Vitamins are not synthesized inside human body
but they are essential part of our diet.
66. (d) Insulin helps in converting glucose to
I
glycogen.
67. (c) Vit B12 also called Cyanocobalamin, is
H2N COOH
anti-pernicious anaemia vitamin.
68. (d) Progesterone (Gestogens) is a steroid
62. (a) Thyroxine is an amino hormone.
hormone, which controls the development and
63. (d) Vitamin B is water soluble whereas all other maintainance of pregnancy. Thryoxine and
are fat soluble. Adrenaline are Amine hormones.
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316 CHEMISTRY
29 Polymers
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Polymers 317
Topic 2: Preparation and Properties 14. Which of the following organic compounds
of Polymers polymerizes to form the polyester Dacron?
9. Regarding cross-linked or network polymers, [2014]
which of the following statements is incorrect? (a) Propylene and para HO—(C6H4)— OH
[2018] (b) Benzoic acid and ethanol
(a) They contain covalent bonds between (c) Terephthalic acid and ethylene glycol
various linear polymer chains.
(d) Benzoic acid and para HO–(C6H4)—OH
(b) They are formed from bi- and tri-functional
15. Which is the monomer of Neoprene in the
monomers.
following ? [NEET 2013]
(c) They contain strong covalents bonds in
their polymer chains. (a) CH 2 = C ¾ CH = CH 2
(d) Examples are bakelite and melamine. ½
CH 3
10. Natural rubber has [2016]
(a) all cis-configuration (b) CH 2 = C ¾ CH = CH 2
(b) all trans-configuration Cl
(c) alternate cis-and trans-configuration (c) CH2 = CH ¾ C º CH
(d) random cis-and trans-configuration (d) CH2 = CH ¾ CH = CH2
11. Biodegradable polymer which can be produced 16. Which of the following statements is false?
from glycine and aminocaproic acid is : [2015] [2012]
(a) PHBV (b) Buna - N (a) Artificial silk is derived from cellulose.
(c) Nylon 6, 6 (d) Nylon 2- nylon 6
(b) Nylon-6, 6 is an example of elastomer.
12. Caprolactum is used for the manufacture of :
(c) The repeat unit in natural rubber is
[2015 RS] isoprene.
(a) Nylon - 6 (b) Teflon (d) Both starch and cellulose are polymers of
(c) Terylene (d) Nylon - 6,6 glucose.
13. Which one of the following is an example of a 17. Which one of the following sets forms the
thermosetting polymer? [2014] biodegradable polymer? [2012 M]
(a) ( CH 2 - C = CH - CH 2 ) n (a) CH2 = CH – CN and CH2 = CH – CH = CH2
|
Cl (b) H2N – CH2 – COOH and
H2N–(CH2)5 – COOH
(b) ( CH 2 - CH ) n
| (c) HO – CH2 – CH2 – OH and
Cl
H H O O HOOC COOH
| | || ||
(c) ( N - (CH 2 )6 - N - C - (CH 2 )4 - C )n
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318 CHEMISTRY
18. Which of the following structures represents 22. The monomer of the polymer; [2005]
neoprene polymer? [2010] CH3
CH3
(a) –(CH 2 – C = CH – CH 2 –) n |
Å
| ÚÚÚÚÚCH 2 - C - CH 2- C
| is
Cl CH3 CH3
CN
| CH3
(b) –(CH 2 – CH –) n (a) H 2C = C
Cl CH3
|
(c) –( CH 2 – CH –)n (b) CH3CH=CHCH3
(d) –( CH – CH 2 –)n (c) CH3CH = CH2
|
C6 H5 (d) (CH3)2C = C(CH3)2
19. Structures of some common polymers are given. 23. Acrilan is a hard, horny and a high melting
Which one is not correctly presented? [2009] material. Which of the following represents its
(a) Neoprene; structure ? [2003]
æ ö
ç ÷ æ ö
ç -CH 2 - C = CH - CH 2 - CH 2 - ÷ (a) ç —CH 2 — CH — ÷
ç | ÷
ç | ÷
ç ÷ ç Cl ÷
è Cl øn è øn
(b) Terylene;
(b) æç —CH 2 — CH— ö÷
COOCH2 – CH2 – O–)n |
(– OC ç ÷
ç CN ÷
è øn
(c) Nylon 6, 6;
CH3
[ –NH(CH 2 )6 NH CO(CH2 )4 CO - ]n æ | ö
(c) ç —CH 2 — C — ÷
(d) Teflon; (-CF2 - CF2 -)n ç | ÷
ç COOCH3 ÷
20. Which one of the following statement is not true? ç ÷
è øn
[2008]
(a) In vulcanization the formation of sulphur æ ö
(d) ç—CH — CH ÷
bridges between different chains make 2
ç | ÷
rubber harder and stronger. ç COOC2H5 ÷
è øn
(b) Natural rubber has the trans -configuration
at every double bond 24. Which one of the following monomers gives the
(c) Buna-S is a copolymer of butadiene and polymer neoprene on polymerization ? [2003]
styrene (a) CF2 = CF2
(d) Natural rubber is a 1, 4 - polymer of isoprene (b) CH2 = CHCl
21. Which one of the following polymers is prepared (c) CCl2 = CCl2
by condensation polymerisation? [2007]
(a) Teflon (b) Natural rubber Cl
|
(c) Styrene (d) Nylon-6,6 (d) CH 2 = C — CH = CH 2
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Polymers 319
A NSW E R K E Y
1 (d ) 5 (a) 9 (c) 13 (d ) 17 (b ) 21 (d ) 25 (a) 29 (b ) 33 (c)
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 )
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Polymers 321
It is resistant to mineral and organic acids. It is
used for blending with wool to provide better CH3
crease, in safety helmets and aircraft battery 22. (a) nH2C = C
boxes. CH3
Cl
| CH3
15. (b) CH 2 = C- CH = CH 2 (chloroprene) is the CH3
monomer of neoprene. C –H2C C CH2 C CH2–
CH3
16. (b) Nylon-6,6 is an example of first synthetic CH3
fibres produced from simple molecules. It is
23. (b) Acrilan is a polyacrylonitrile ( PAN).
prepared by condensation polymerisation of
adipic acid and hexamethylene diamine. Acrylic fibers are synthetic fibers made from
17. (b) Biodegradable polymer is Nylon-2- acrylonitridle monomer. Popular brand names are
Dralon, Orlon, Drytex, Courtelle, etc.
Nylon-6 which is copolymer of glycine
(H2N – CH2– COOH) and amino caproic acid Cl
(H2N–(CH2)5 – COOH). |
K S O
24. (d) nCH 2 = CH - C = CH 2 ¾¾2 ¾
2¾8®
nH2N – CH2 – COOH + Chloropren e
glycine nH2N – (CH2)5– COOH
amino caproic acid Cl
|
-( CH 2 - CH = C - CH 2 -
)n
O O
Neoprene
–( HN – CH2 – C – HN – (CH2)5 – C –) n
nylon – 2 – nylon – 6 é CH3 ù
ê | ú
25. (a) Monomer of ê — C — CH 2 — ú
18. (a) Neoprene is a polymer of chloroprene ê | ú
(2 – chloro – 1, 3 – butadiene). ëê CH3 ûú n
19. (a) Neoprene is a polymer of chloroprene. polymer is 2-methylpropene.
CH3 H CH2 26. (a) Terylene is prepared by condensing
20. (b) C=C terephthalic acid and ethylene glycol.
– CH2 CH2 CH3
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322 CHEMISTRY
28. (b) Natural rubber is a polymer of isoprene,
OH OH
Polymerisation CH2OH
n CH2=CH–C=CH2
n Polymerisation
+ n
CH3
Isoprene
(2-methyl –CH2–CH=C–CH2– CH2OH
n
butadiene) CH3
Polyisoprene OH OH OH
(Natural rubber)
CH2 CH2 CH2
29. (b)
nHOOC (CH 2 ) 4 COOH + nH 2 N(CH 2 )6 NH 2
Adipic acid Hexamethylene CH2 CH2
diamine
O O CH2
¾¾¾¾
525K
® [– C – (CH2)4 – C –NH–(CH2) 6 – NH –]n CH2 CH2
Polymerisation Nylon 6, 6
OH OH OH
30. (c) Phenol and formaldehyde undergo
(cross linked Polymer Bakelite)
condensation polymerisation under two
different conditions to give a cross linked
polymer called bakelite. 31. (d) DNA is a biopolymer.
32. (d) Polyacrylonitrile is used as a substitute for
OH OH OH wool in making commercial fibre as orlon or
CH2OH acrilan.
–
OH 33. (c) Polytetrafluoroethylene commonly known
+ HCHO +
as teflon is a tough material. This is resistant to
heat and chemicals and have high melting point.
CH2OH
o & p-Hydroxy methyl phenol Therefore it is used for coating the cookware to
(intermediate) make them non-sticky.
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Chemistry in Everyday Life 323
30 Chemistry in
Everyday Life
Trend Analysis with Important Topics & Sub-Topics
1. Which of the following is a cationic detergent? (a) Dryer (b) Buffering agent
(a) Sodium stearate [2020] (c) Antiseptic (d) Softner
7. Artificial sweetner which is stable under cold
(b) Cetyltrimethyl ammonium bromide
conditions only is : [2014]
(c) Sodium dodecylbenzene sulphonate (a) Saccharine (b) Sucralose
(d) Sodium lauryl sulphate (c) Aspartame (d) Alitame
2. The artificial sweetener stable at cooking 8. Antiseptics and disinfectants either kill or
temperature and does not provide calories is prevent growth of microorganisms. Identify
which of the following statements is not true:
[NEET Odisha 2019] [NEET 2013]
(a) alitame (b) saccharin (a) Chlorine and iodine are used as strong
(c) aspartame (d) sucralose disinfectants.
3. Among the following, the narrow spectrum (b) Dilute solutions of Boric acid and
antibiotic is: [2019] Hydrogen Peroxide are strong antiseptics.
(a) penicillin G (b) ampicillin (c) Disinfectants harm the living tissues.
(c) amoxycillin (d) chloramphenicol (d) A 0.2% solution of phenol is an antiseptic
4. Mixture of chloroxylenol and terpineol acts as: while 1% solution acts as a disinfectant.
[2017] 9. Dettol is the mixture of [1996, NEET Kar. 2013]
(a) antiseptic (b) antipyretic (a) Terpineol and Bithionol
(c) antibiotic (d) analgesic (b) Chloroxylenol and Bithionol
5. Which of the following is an analgesic? [2016] (c) Chloroxylenol and Terpineol
(a) Novalgin (b) Penicillin (d) Phenol and Iodine
(c) Streptomycin (d) Chloromycetin 10. Chloroamphenicol is an : [2012 M]
(a) antifertility drug
6. Bithionol is generally added to the soaps as an (b) antihistaminic
additive to function as a/an : [2015] (c) antiseptic and disinfectant
(d) antibiotic-broad spectrum
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324 CHEMISTRY
11. Which one of the following is employed as tranquilizer? [2009]
Antihistamine ? [2011] (a) Naproxen (b) Tetracycline
(a) Chloramphenicol (c) Chlorpheninamine(d) Equanil
(b) Diphenyl hydramine 14. Which one of the following can possibly be used
(c) Norothindrone as analgesic without causing addiction and
(d) Omeprazole mood modification ? [1997]
12. Which one of the following is employed as a (a) Diazepam
tranquilizer drug? [2010] (b) Morphine
(a) Promethazine (b) Valium (c) N-acetyl-para-aminophenol
(c) Naproxen (d) Mifepristone (d) Tetrahydrocannabinol
13. Which one of the following is employed as a
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)
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31
Nuclear Chemistry 325
Nuclear Chemistry
1. The half life of a substance in a certain enzyme- radioactive substance is 6 hours. Then injection
catalysed reaction is 138s. The time required for of maximum activity of radioactive substance
the concentration of the substance to fall from that can be injected will be [2001]
1.28 mg L–1 to 0.04 mg L–1, is : [2011] (a) 0.08 M (b) 0.04 M
(a) 414 s (b) 552 s (c) 0.32 M (d) 0.16 M
(c) 690 s (d) 276 s
7. If species ba X emits firstly a positron, then two
2. A nuclide of an alkaline earth metal undergoes
radioactive decay by emission of the a-particles a and two b and in last one a and finally converted
in succession. The group of the periodic table to species dc Y , so correct relation is [2001]
to which the resulting daughter element would (a) c = a – 5, d = b – 12
belong is [2005] (b) c = a – 6, d = b – 8
(a) Gr. 4 (b) Gr. 6 (c) c = a – 4, d = b – 12
(c) Gr. 14 (d) Gr. 16 (d) c = a – 5, d = b – 8
3. The radioactive isotope 60 which is used in 8. When a radioactive element emits successively
27 Co
the treatment of cancer can be made by (n, p) one a-particle and two b-particles, the mass
reaction. For this reaction the target nucleus is number of the daughter element [1999]
(a) is reduced by 4 units
(a) 59
28 Ni (b) 59
27 Co [2004]
(b) remains the same
(c) 60 (d) 60
28 Ni 27 Co (c) is reduced by 2 units
4. The radioactive isotope, tritium, (13 H ) has a half- (d) is increased by 2 units
9. Number of neutrons in a parent nucleus X, which
life of 12.3 years. If the initial amount of tritium is
gives 14
7 N
nucleus, after two successive b
32 mg, how many milligrams of it would remain
emissions, would be [1998]
after 49.2 years? [2003]
(a) 9 (b) 6
(a) 8 mg (b) 1 mg
(c) 7 (d) 8
(c) 2 mg (d) 4 mg 10. Carbon - 14 dating method is based on the fact
5. U 235 , nucleus absorbs a neutron and
92 that: [1997]
disintegrates into 54Xe139, 38Sr94 and x. So what (a) C-14 fraction is same in all objects
will be the product x ? [2002] (b) C-14 is highly insoluble
(a) 3-neutrons (b) 2-neutrons (c) Ratio of carbon-14 and carbon-12 is constant
(c) a-particle (d) b-particle (d) all the above
6. A human body required 0.01M activity of 11. 235 1 236
92 U + 0 n ® 92 U®
radioactive substance after 24 hours. Half life of
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326 CHEMISTRY
fission products + neutrons + 3.20 × 10–11 J 15. India has the world’s largest deposits of thorium
The energy released when 1 g of 92 235 in the form of [1994]
U finally
undergoes fission is [1997] (a) rutile (b) magnesite.
(a) 12.75 × 108 kJ (b) 16.40 × 107 kJ (c) lignite (d) monazite.
(c) 8.20 × 107 kJ (d) 6.50 × 106 kJ 16. If an isotope of hydrogen has two neutrons in its
12. One microgram of radioactive sodium 11 24
Na with atom, its atomic number and atomic mass number
a half-life of 15 hours was injected into a living will respectively be [1992]
system for a bio-assay. How long will it take for (a) 2 and 1 (b) 3 and 1
the radioactive subtance to fall up to 25% of the
initial value? [1996] (c) 1 and 1 (d) 1 and 3.
(a) 60 hours (b) 22.5 hours 17. The age of most ancient geological formations is
(c) 375 hours (d) 30 hours estimated by [1989]
13. Half-life for radioactive 14C is 5760 years. In how (a) Potassium–argon method
many years, 200 mg of 14C will be reduced to (b) Carbon-14 dating method
25 mg? [1995]
(c) Radium-silicon method
(a) 5760 years (b) 11520 years
(d) Uranium-lead method.
(c) 17280 years (d) 23040 years
14. In a radioactive decay, an emitted electron comes 18. Emission of an alpha particle leads to a [1989]
from [1994] (a) Decrease of 2 units in the charge of the atom
(a) The nucleus of atom (b) Increase of 2 units in the mass of the atom
(b) The orbit with principal quantum number 1 (c) Decrease of 2 units in the mass of the atom
(c) the inner orbital of the atom (d) Increase of 4 units in the mass of the atom.
(d) the outermost orbit of the atom.
ANSWER KEY
1 (c) 3 (c) 5 (a) 7 (a) 9 (a) 11 (c) 13 (c) 15 (d) 17 (d)
2 (c) 4 (c) 6 (d) 8 (a) 10 (c) 12 (d) 14 (a) 16 (d) 18 (a)
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Nuclear Chemistry 327
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