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A Complete Resource Book in Physics for JEE Main 2019
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A Complete Resource Book in for JEE Main 2019 chemistry
U.K. Singhal
A.K. Singhal
Dedicated to my grandparents, parents and teachers
Published by Pearson India Education Services Pvt. Ltd, CIN: U72200TN2005PTC057128
No part of this eBook may be used or reproduced in any manner whatsoever without the publisher’s prior written consent.
This eBook may or may not include all assets that were part of the print version. The publisher reserves the right to remove any material in this eBook at any time.
ISBN: eISBN: 9789353062156 9789353063412
Head Office:15th Floor, Tower-B, World Trade Tower, Plot No. 1, Block-C, Sector-16, Noida 201 301, Uttar Pradesh, India.
Registered Office: 4th Floor, Software Block, Elnet Software City, TS-140, Block 2 & 9, Rajiv Gandhi Salai, Taramani, Chennai 600 113, Tamil Nadu, India.
A Complete Resource Book for JEE Main series is a must-have resource for students preparing for JEE Main examination. There are three separate books on Physics, Chemistry, and Mathematics; the main objective of this series is to strengthen the fundamental concepts and prepare students for various engineering entrance examinations. It provides class-tested course material and numerical applications that will supplement any ready material available as student resource.
To ensure high level of accuracy and practicality, this series has been authored by highly qualified and experienced faculties for all three titles.
About the Book
A Complete Resource Book in Chemistry for JEE Main 2018 is an authentic book for all the aspirants preparing for the joint entrance examination (JEE). This title is designed as per the latest JEE Main syllabus, where the important topics are covered in 34 chapters. It has been structured in user friendly approach such that each chapter begins with topic-wise theory, followed by sufficient solved examples and then practice questions.
The chapter end exercises are structured in line with JEE questions; with ample number of questions on single choice correct question (SCQ), multiple-type correct questions (MCQ), assertion and reasoning, column matching, passage based and integer type questions are included for extensive practice. Previous 15 years’ questions of JEE Main and AIEEE are also added in every chapter. Hints and Solutions at the end of every chapter will help the students to evaluate their concepts and numerical applications. Because of its comprehensive and in-depth approach, it will be especially helpful for those students who prefers self-study than going for any classroom teaching.
Series Features
• Complete coverage of topics along with ample number of solved examples.
• Large variety of practice problems with complete solutions.
• Chapter-wise Previous 15 years’ AIEEE/JEE Main questions.
• Fully solved JEE Main 2017 questions.
• 5 Mock Tests based on JEE Main pattern in the book.
• 5 Free Online Mock Tests as per the recent JEE Main pattern.
Despite of our best efforts, some errors may have crept into the book. Constructive comments and suggestions to further improve the book are welcome and shall be acknowledged gratefully. Kindly share your feedback with me at singhal.atul50@yahoo.com or singhal.atul1974@gmail.com.
Acknowledgements
The contentment and ecstasy that accompany the successful completion of any work would remain essentially incomplete if I fail to mention the people whose constant guidance and support has encouraged me.
I am grateful to all my reverend teachers, especially, late J.K. Mishra, D.K. Rastogi, late A.K. Rastogi and my honourable guide, S.K. Agarwala. Their knowledge and wisdom has continued to assist me to present in this work.
I am thankful to my colleagues and friends, Deepak Bhatia, Vikas Kaushik, A.R. Khan, Vipul Agarwal, Ankit Arora (ASO Motion, Kota), Manoj Singhal, Yogesh Sharma, (Director, AVI), Vijay Arora, (Director, Dronacharya), Rajneesh Shukla (Allen, Kota), Anupam Srivastav, Rajeev Jain (M V N), Sandeep Singhal, Chandan Kumar (Mentor, Patna), P.S. Rana (Vidya Mandir, Faridabad).
I am indebted to my father, B.K. Singhal, mother Usha Singhal, brothers, Amit Singhal and Katar Singh, and sister, Ambika, who have been my motivation at every step. Their never-ending affection has provided me with moral support and encouragement while writing this book.
Last but not the least, I wish to express my deepest gratitude to my wife Urmila and my little, —but witty beyond her years, daughters Khushi and Shanvi who always supported me during my work.
A.K. Singhal
Chemistry Trend Analysis (2007
JEE Mains 2018 PaPEr
1. For 1 molal aqueous solution of the following compounds, which one will show the highest freezing point?
(A) [Co(HO)ClO 23 3 ]. 2
(B) [Co(HO)] Cl 26 3
(C) [Co(HO)Cl] Cl .H 25 22 O
(D) [Co(HO)Cl] Cl 24 22.2HO
2. Hydrogen peroxide oxidizes [Fe(CN)] 6 4 to [Fe(CN )] 6 3 in acidic medium but reduces [Fe(CN)] 6 3 to [Fe(CN)] 6 4 in alkaline medium. The other products formed are, respectively:
(A) HO and (HO OH ) 22 + (B) (H O+ 22OO 2 )andH
(C) (H O) and(HO ) 22 2 ++OOH
(D) HO and (HO O) 22 2 +
3. Which of the following compounds will be suitable for Kjeldahl’s method for nitrogen estimation?
7. The increasing order of basicity of the following compounds is: (I)
(II)
4. Glucose on prolonged heating with HI gives: (A) 6-iodohexanal (B) n-Hexane (C) l-Hexene (D) Hexanoic acid
5. An alkali is titrated against an acid with methyl orange as indicator, which of the following is a correct combination?
BaseAcidEnd point
(A)StrongStrongPink to colourless (B)WeakStrongColourless to pink (C)StrongStrongPinkish red to yellow (D)WeakStrongYellow to pinkish red
6. The predominant form of histamine present in human blood is (pKa, Histidine = 6.0)
(A)
(A) (IV) < (II) < (I) < (III)
(B) (I) < (II) < (III) < (IV)
(C) (II) < (I) < (III) < (IV) (D) (II) < (I) < (IV) < (III)
8. Which of the following lines cor rectly show the temperature dependence of equilibrium constant K, for an exothermic reaction?
9. How long (approximate) should water be electrolyzed by passing through 100 amperes current so that the oxygen released can completely burn 27.66 g of diborane?
10. Consider the following reaction and statements:
[C Br ][Co(NH) Br ]NH o(NH 34)Br23 33 3 +−+ → +
(I) Two isomers are produced if the reactant complex ion is a cis-isomer.
(II) Two isomers are produced if the reactant complex ion is a trans-isomer.
(III) Only one isomer is produced if the reactant complex ion is a trans-isomer.
(IV) Only one isomer is produced if the reactant complex ion is a cis-isomer.
The correct statements are:
(A) (II) and (IV) (B) (I) and (II) (C) (I) and (III) (D) (III) and (IV)
11. Phenol reacts with methyl chlorofor mate in the presence of NaOH to form product A. A reacts with Br2 to for m product B. A and B are respectively:
13. At 518°C the rate of decomposition of a sample of gaseous acetaldehyde initially at a pressure of 363 Torr, was 1.00 Torr s-1 when 5% had reacted and 0.5 Torr s -1 when 33% had reacted. The order of the reaction is:
(A) 0 (B) 2
(C) 3 (4) 1
14. The combustion of benzene (1) gives CO2 (g) and H2O (l). Given that heat of combustion of benzene at constant volume is -3263.9 kJ mol-1 at 25°C; heat of combustion (in kJ mol-1) of benzene at constant pressure will be: (R = 8.314 JK-1 mol-1)
(A) -3267.6 (B) 4152.6 (C) -452.46 (D) 3260
15. The ratio of mass percent of C and H of an organic compound (CxHyOz) is 6 : 1. If one molecule of the above compound (CxHyOz) contains half as much oxygen as required to burn one molecule of compound CxHy completely to CO2 and H2O, then empirical formula of compound CxHyOz is:
(A) C2H4O3 (B) C3H6O3 (C) C2H4O (D) C3H4O2
16. The trans-alkenes are for med by the reduction of alkynes with:
(A) BCl3 and AlCl3 (B) PH3 and BCl3 (C) AlCl3 and SiCl4 (D) PH3 and SiCl4
18. When metal ‘M’ is treated with NaOH, a white gelatinous precipitate ‘X’ is obtained, which is soluble in excess of NaOH. Compound ‘X’ when heated strongly gives an oxide which is used in chromatography as an absorbent. The metal ‘M’ is:
(A) Fe (B) Zn (C) Ca (D) Al
19. According to molecular orbital theory, which of the following will not be a viable molecule?
(A) H 2 2 (B) He 2 2 +
12. An aqueous solution contains an unknown concentration of Ba2+. When 50 mL of a 1 M solution of Na2SO4 is added, BaSO4 just begins to precipitate. The final volume is 500 mL. The solubility product of BaSO4 is 1 × 10-10. What is the original concentration of Ba2+?
(A) 1.0 × 10-10 M (B) 5 × 10-9 M (C) 2 × 10-9 M (D) 1.1 × 10-9 M
(C) He 2 + (D) H 2
20. The major product formed in the following reaction is:
21. Phenol on treatment with CO2 in the presence of NaOH followed by acidification produces compound X as the major product. X on treatment with (CH3CO)2O in the presence of catalytic amount of H2SO4 produces: (A)
25. The compound that does not produce nitrogen gas by the thermal decomposition is:
(A) (NH4)2SO4
(B) Ba(N3)2
(C) (NH4)2Cr2O7
(D) NH4NO2
26. An aqueous solution contains 0.10 M H2S and 0.20 M HCl. If the equilibrium constant for the formation of HS- from H2S is 10 10 7 . × and that of S2 from HSions is 12 10 13 × then the concentration of S2 ions in aqueous solution is:
(A) 510 19 × (B) 510 8 ×
(C) 310 20 × (D) 610 21 ×
27. The oxidation states of Cr in Cr(H O) Cl Cr(C H) 26 36 62 [] [] , Cr(H O) Cl Cr(C H) 26 36 62 [] [] , and KCr(CN) (O)(O )(NH) 22 22 3 [] respectively are:
(A) +3, 0 and +4
(B) +3, +4 and +6
(C) +3, +2 and +4
(D) +3, 0 and +6
22. Which of the following compounds contain(s) no covalent bond(s)?
24. The major product of the following reaction is:
28. The recommended concentration of fluoride ion in drinking water is up to 1 ppm as fluoride ion is required to make teeth enamel harder by converting 3Ca(PO ).Ca(OH) 34 22 [] to:
(A) 3Ca(OH) .CaF 22{}
(B) CaF2 []
(C) 3 2 ()CaFCa(OH) 2 []
(D) 3Ca(PO ). CaF 34 22 []
29. Which of the following salts is the most basic in aqueous solution?
(A) Pb(CHCOO)32 (B) Al(CN) 3
(C) CH COOK 3 (D) FeCl 3
30. Total number of lone pair of electron in I 3 ion is:
(A) 12 (B) 3
(C) 6 (D) 9
Answer keys
Hints and solutions
1. As DTf ∝ mi. Here m is same so decide by using i The complex giving least number of ions will have lowest depression in freezing point and therefore highest freezing point. Hence, option (A) is correct. (Van’t Hoff factor = 1) Hence, the correct option is (A).
2. In acidic medium, HO22 acts as an oxidant as follows
Oxidation number decreases
Oxidation number
In alkaline medium, HO22 acts as reducing agent as follows
Oxidation number increases
Oxidation number decreases
Hence, the correct option is (D).
3. Kjeldahl’s method cannot be not used in the case of nitro, azo compounds and also to the compounds containing nitrogen in the ring, e.g., Pyridine. Hence, here only CH NH 65 2 is suitable for the test.
Hence, the correct option is (C).
4. Glucose on heating with HI gives n-hexane as followed CHO (CHOH)4 (CH2)4 CH3 CH3 CH2OH
Glucose n-hexane Δ HI
Hence, the correct option is (B).
7. Order of basicity in increasing order is as follows
5. As methyl orange is weak organic base. So it is used in the titration of weak base (NH4OH) vs strong acid (HCl)
MeOH Me OH
Unionized form (Yellow)
Ionized form (Red) +− +
In basic medium, equilibrium lies in backward direction and therefore it shows yellow colour.
In acidic medium, equilibrium shifts in forward direction and therefore, colour changes from yellow to red.
Hence, the correct option is (D).
6. NH3 + N
H N
In it, the N-atoms present in the ring will have same pKa values (6.0), while N atom outside the ring will have different pKa value (pKa > 7.4).
Hence, two N-atoms inside the ring will remain in unprotonated from in human blood as their pKa(6.0) < pH of blood (7.4), while the N-atom outside the ring will remain in protonated from because its pKa > pH of blood (7.4).
Hence, the correct option is (A).
Here (III) is most basic because its conjugate acid is stabilized by equivalent resonance. Out of (I) and (IV), (IV) is more basic due to + I effect of -CH3 group (II) is less basic than (I) and (IV) because N atom is sp2(more s%) hybridized. Hence, the cor rect option is (D).
8. DG° = -RT ln K
DH° -TDS° = -RT ln K
° + ° ∆∆ H RT S R ln K
Here ln K vs 1 T graph will be a straight line with slope equal to °∆ H R . As reaction is exothermic, so°∆ H itself will be negative resulting in positive slope. Hence, the cor rect option is (B).
9. BH OB OH O 26 22 3332+ → +
As according to the balanced equation: Here 2766 26 ., gB H i.e., 1 mole BH26 requires 3 mole of O2. Here this oxygen is produced by electrolysis of water as follows.
22 2 4 22 HO HO F →+
As 1 mole O2 is produced by 4 F charge So 3 mole O2 will be produced by 12 F charge
11. The sequence of the reaction is as follows
Hence, the cor rect option is (D).
12. Ba Na SO BaSO Na mL 2 mL,1M 4 ++ + → + 2 450 4 50 2
For BaSO4 Ksp is given as K(BaSO ) = [Ba ][SO] sp 4 2 4 2+− 10 01 10 2 −+=× [Ba]
Now, calculate [Ba+2] in original solution (450 mL) as follows
10-9 × 500 = 450 × M
MM = × =× =× 10500 450 10 9 10 11110 9 99
Hence, the correct option is (D).
13. CHCHOCOCH
PTorr i 3 363 4 = →+
As we know that, rPRn ∝ (1)
Here PR = reactant pressure n = order of reaction
Now rate of reaction is 1.00 torr s-1, when reactant pressure is 363363 5 100 −×
torr = 344.85 torr.
Similarly rate of reaction is 0.5 torr s-1, when reactant pressure is 363363 33 100 −×
torr = 243.21 torr.
So, applying equation (1) 1 05 34485 24321 =
n 2 = (1.418)n ⇒ 21 418 1 n = ∴ n
2
Hence, the correct option is (B).
14. C6H6 (l) + 15 2 O2 (g) → 6CO2 (g) + 3H2O (l)
Dng = 6 15 2 = 3 2
DH = DU + DngRT
DH = ××
3263900 3 2 8314298.J = -3267616 J = -3267.616 kJ
Hence, the correct option is (A).
15. As ratio of mass percent of C and H in CxHyOz is 6 : 1. So, ratio of mole percent of C and H in CxHyOz will be 1 : 2. Hence, x : y = 1 : 2, which is possible in options (A), (B) and (C).
Now oxygen needed to burn CxHy
CxHy + x y +
4 O2 → xCO2 + y 2 H2O
Now z is half of oxygen atoms needed to burn CxHy here.
2 2 24
Putting on the values of x and y from the given options:
Option (A), x = 2, y = 4 z =+
= 2 4 4 3
Option (B), x = 3, y = 6
=+
= 3 6 4 45 .
Hence, the correct option is (A) (C2H4O3).
16. Na/liq. NH3 reduces alkynes into trans alkene (trans or anti addition).
R C C R C = C R R Na/liq. NH3 trans alkene H H
Hence, the correct option is (D).
17. Here BCl3 and AlCl3 are Lewis acids as both ‘B’ and ‘Al’ has vacant p-orbitals and are e- deficient. SiCl4 is also a Lewis acid as silicon atom has vacant 3d-orbital, so it can accept eHence, the correct option is (A) or (C).
18. Here metal is Al and the reactions are as follows 26 23 22 3 Al HO AlOH H + x +→ () ↓ ()
AlOHNaOHNa Al(OH (Soluble) () () ) 34 +→ []
Hence, the correct option is (D).
19. Species Bond order
(A) H 2 2 B.O. =−[] = 1 2 22 0 (does not exists)
(B) He 2 2 + B.O. =−[] = 1 2 20 1 (exists)
(C) He 2 + B.O. = 1 2 21 05 [] = (exists)
(D) H 2 B.O. =−[] = 1 2 21 05 (exists)
As H 2 has bond order as zero, so it cannot exist. Hence, the correct option is (A).
29. CH3COOK is most basic among the given salts as it gives strong base KOH and weak acid CH3COOH.
Hence, the correct option is (C).
30. I 3 has 3 lone pair e- as follows
Hence, the correct option is (D).
JEE Mains 2017 PaPErs
1. Which of the following compounds will give significant amount of meta product during mono-nitration reaction?
Oxygen (61.4%); carbon (22.9%); hydrogen (10.0%); and nitrogen (2.6%). The weight which a 75 kg person would gain if all 1H atoms are replaced by 2H atoms is:
(A) 15 kg (B) 37.5kg
(C) 7.5 kg (D) 10 kg
8. Which of the following, upon treatment with ter tBuONa followed by addition of bromine water, fails to decolourize the colour of bromine?
2. ∆U is equal to:
(A) Isochoric work (B) Isobaric work (C) Adiabatic work (D) Isothermal work
3. The increasing order of the reactivity of the following halides, for the SN1 reaction is:
(I) CH3CHCH2CH3
Cl
(II) CH3CH2CH2Cl
(III) p—H3CO—C6H4—CH2Cl
(A) (III) < (II) < (I) (B) (II) < (I) < (III) (C) (I) < (III) < (II) (D) (II) < (III) < (I)
4. The radius of the second Bohr orbit for hydrogen atom is:
(Plank’s constant, h = 6.6262 × 10-34 Js; mass of electron = 9.1091 × 10-31 kg; charge of electron, e = 1.60210 × 10-19 C; permittivity of vaccum, ∈0 = 8.854185 × 10-12 kg-1 m -3 A2)
(A) 1.65Å (B) 4.76Å (C) 0.529Å (D) 212Å
5. pKa of a weak acid (HA) and pKb of a weak base (BOH) are 3.2 and 3.4, respectively. The pH of their salt (AB) solution is:
(A) 7.2 (B) 6.9
(C) 7.0 (D) 1.0
6. The formation of which of the following polymers involves hydrolysis reaction?
7. The most abundant elements by mass in the body of a healthy human adult are:
9. In the following reactions, ZnO is respectively acting as a/an:
(a) ZnO + Na2O → Na2ZnO2
(b) ZnO + CO2 → ZnCO3
(A) Base and acid (B) Base and base
(C) Acid and acid (D) Acid and base
10. Both lithium and magnesium display several similar proper ties due to the diagonal relationship; however, the one which is incorrect is:
(A) Both form basic carbonates
(B) Both form soluble bicarbonates
(C) Both form nitrides
(D) Nitrates of both Li and Mg yield NO2 and O2 on heating
11. 3-Methyl-pent-2-ene on reaction with HBr in presence of peroxide forms an addition product. The number of possible stereoisomers for the product is:
(A) Six (B) Zero
(C) Two (D) Four
12. A metal crystallises in a face centred cubic structure. If the edge length of its unit cell is ‘a’, the closest approach between two atoms in metallic crystal will be:
(A) 2a (B) 22a
(C) 2a (D) a 2
13. Two reactions R1 and R2 have identical pre-exponential factors. Activation energy of R1 exceeds that of R2 by 10 kJ mol-1. If k1 and k2 are rate constants for reactions
R1 and R2 respectively at 300 K, then ln(k2/k1) is equal to (R = 8.314 J mol-1K-1):
(A) 8 (B) 12
(C) 6 (D) 4
14. The correct sequence of reagents for the following conversion will be: O CHO CHO HO CH3 CH3 HO
15. The Tyndall effect is observed only when following conditions are satisfied:
(a) The diameter of the dispersed particles is much smaller than the wavelength of the light used.
(b) The diameter of the dispersed particle is not much smaller than the wavelength of the light used.
(c) The refractive indices of the dispersed phase and dispersion medium are almost similar in magnitude.
(d) The refractive indices of the dispersed phase and dispersion medium differ g reatly in magnitude.
(A) (a) and (d) (B) (b) and (d) (C) (a) and (c) (D) (b) and (c)
16. Which of the following compounds will behave as a reducing sugar in an aqueous KOH solution?
(A) CH2OH OCOCH3 HO OH HOH2C O
(B) CH2OH HO OH HOH2C O
(C) CH2OH OCH3 HO OH HOH2C O (D) CH2OH OH OH OH HOH2C O
∆° =− r . HkJmol 1 2858
C; Og HO lCHg Og 22 42 () + () → () + () 22
∆° =+ r 1 HkJm ol8903.
Based on the above ther mochemical equation, the value of DrH° at 298 K for the reaction CH gCHg graphite () () () +→ 2 24 will be:
(A) +74.8 kJ mol-1 (B) +144.0 kJ mol-1
(C) -74.8 kJ mol-1 (D) -144.0 kJ mol-1
18. Which of the following reactions is an example of a redox reaction?
(A) XeFO FXeF O 42 26 2 +→ +
(B) XeFPFXeF PF 25 6 +→ []+
(C) XeFH OXeOFHF 62 4 +→ + 2
(D) XeFH OXeO F4HF 62 2 +→ + 2 2
19. The products obtained when chlorine gas reacts with cold and dilute aqueous NaOH are:
(A) ClOand ClO3
(B) ClOand ClO 23
(C) Cl andClO
(D) Cl andClO 2
20. The major product obtained in the following reaction is: Δ BuOK Br H C6H5 C6H5
(A) () () ± CH CH OBuCHCHH 65 t 26 5
(B) CH CH=CHCH 65 65
(C) + () CH CH OBuCHH 65 t 25 ()
(D) () () CH CH OBuCHC H 65 t 26 5
21. Sodium salt of an organic acid ‘X’ produces effervescence with conc. H2SO4.‘X’ reacts with the acidified aqueous CaCl2 solution to give a white precipitate which decolouriszes acidic solution of KMnO4.‘X’ is:
(A) C6H5COONa (B) HCOONa (C) CH3COONa (D) Na 22CO 4
22. Which of the following species is not paramagnetic?
(A) NO (B) CO (C) O2 (D) B2
23. The freezing point of benzene decreases by 0.45°C when 0.2 g of acetic acid is added to 20 g of benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be:
(Kf for benzene = 5.12 K kg mol-1)
(A) 64.6% (B) 80.4% (C) 74.6% (D) 94.6%
24. Which of the following molecules is least resonance stabilized?
(A) (B) O
(C) N (D) O
25. On treatment of 100 ml of 0.1 M solution of COCl3 6H2O with excess AgNO3; 1.2 × 1022 ions are precipitated. The complex is:
1. Nitration is carried out in presence of concentrated HNO3 and concentrated H2SO4. Here, aniline acts as base, in presence of H2SO4 its protonation takes place and anilinium ion is formed.
Anilinium ion is strongly deactivating group and meta directing in nature so it gives meta nitration product in significant amount. (≈ 97%).
2. Using 1st law:
∆U = q + w
For adiabatic process:
q = 0
So, ∆U = w
Hence work involved in adiabatic process is at the expense of change in internal energy of the system.
3. For SN1 reaction, reactivity is decided by ease of dissociation of alkyl halide
R − X R⊕ + X
Higher the stability of R⊕ (carbocation) higher would be reactivity of SN1 reaction.
As stability of cation follows order.
CH3——CH2——CH2
< CH3——CH——CH2 ——CH3 < p—— H3CO——C6H4——CH2
Hence, the correct order is
II < I < III
(A)
4. Radius of nth Bohr orbit in H-atom = 0.53 n Z 2 Å
Radius of second Bohr orbit = 0532 1 2 .( ) × = 2.12 Å
6. Formation of Nylon6 involves hydrolysis of its monomer (caprolactum) in initial state as follows:
Hydrolysis
Caprolactum
H3N——CH2——CH2——
Δ /Polymerization
Nylon6
7. Mass in the body of a healthy human adult has: Oxygen = 61.4%, carbon = 22.9%, Hydrogen = 10.0% and Nitrogen = 2.6%
Total weight of person = 75 kg
Mass due to 1H is =× = 75 10 100 75.kg
1H atoms are replaced by 2H atoms. Hence, mass gain by person = 7.5 kg
8. Br
tert-BuONa
(fails to decolorize the colour of bromine) (no unsaturation)
tert-BuONa
C6H5 Br O
tert-BuONa Br
C6H5
(it decolorizes bromine solution)
(it decolorizes bromine solution due to unsaturation)
tert-BuONa
(it decolorizes bromine solution due to unsaturation)
9. ZnO is an amphoteric oxide but in given reaction it acts as follows:
(a) ZnO + Na2O → Na2ZnO2 acid base salt
(b) ZnO + CO2 → ZnCO3 base acid salt
10. Mg can form basic carbonate like, 56 7 45 2 2 3 2 2 32 23 Mg CH O MgCO Mg OH HO HCO +−++ →⋅ ⋅↓ + ()
While Li can form only carbonate (Li2CO3) not basic carbonate.
11. +H2O2
5 2 3 4
3-methyl pent-2-ene Anti markownikov product (4 stereo isomers possible due to 2 chiral centres as molecule is asymmetric )
12. In FCC unit cell atoms are in contact along face diagonal.
So, 24aR =
Hence, closest distance (2R) = 2 2 2 aa =
13. Using arrhenius equation, KA e Ea RT =⋅ KA e 1 ERT a 1 =⋅ (1) KA e 2 ERT a 2 =⋅ (2)
So, equation (2)/(l) ⇒= K K e 2 1 (E E RT a1 a 2 )
(As Pre-exponential factors of both reactions in same)
Hence, ln ) , K K (E E RT 2 1 aa 1
= = × = 2 10000 8314300 4
14. O O
[Ag(NH3)2]OH Tollens reagent O C
H+/CH3OH (esterification)
15. It is factual.
16. (i) Ester in presence of aqueous KOH solution give SNAE reaction so following reaction takes place.
Aq. KOH Reagent Tollen’s Ring opening
+ Me—C—O
HOCH2 O ⊕ ve silver mirror test
(ii) In above compound, in presence of aq. KOH (SNAE) reaction takes place and a-Hydroxy carbonyl compound is formed which gives ⊕ve Tollen’s test. So this compound behaves as a reducing sugar in an aqueous KOH solution.
17. C; ? r Og HO lCHg Og H H 22 42 f () + () → () + () ∆° = ∆° 22 8903 3935285 80 ∆∆ ∆ rr products rreactant HH H °= °− ° ∑ ∑ ()() 890 31 20 1393 52 2858 ..
∆° () =− =− f CH HkJ/mol 4 8903965 1748
18. XeFO FX eF O 42 26 2 ++ + +→ + 41 6 0
Here, xenon undergoes oxidation while oxygen undergoes reduction. (Redox Process)
NO(15e-) ⇒ One unpaired electron is present in π * molecular orbital. (Paramagnetic)
CO(14e-) ⇒ No unpaired electron is present (Diamagnetic)
O2 (16) ⇒ Two unpaired electrons are present in π * Molecular orbitals. (Paramagnetic)
B2(10) ⇒ Two unpaired electros are present in π bonding molecular orbitals. (Paramagnetic)
23. In benzene acetic acid dimerises as follows: 2 2 CH COOH CH COOH 33()
Percentage degree of association = 94.5%
24. (D) O is nonaromatic (non-planar) and hence least reasonance stabilized.
(A) Benzene is aromatic (6πe -) (B) O furan is aromatic (6πe -)
(C) N pyridine is aromatic (6πe -)
25. Moles of complex Molarity volumeml = × () 1000 = × = 100 01 1000 001 . .mole Moles of ions precipitated with excess of AgNO3 = × × 12 10 60210 22 23 . = 002.moles
C6H5
C6H5
C6H5
C6H5
Number of Cl present in ionization sphere =
Mole of ionprecipitatedwith exessAgNO Mole of complex 3 = 002 0011 2 =
It means 2Cl ions present in ionization sphere. Hence, complex is [CO(H2O)5Cl]Cl2 . H2O
26. DIBAL–H is electrophilic reducing agent which reduces cynide, esters, lactone, amide, carboxylic acid into corresponding aldehyde (partial reduction).
27. NO3 : The maximum limit of nitrate in drinking water can be 50 ppm. Excess nitrate in drinking water can cause diseases like methemoglobinemia, etc.
SO 4 2 : above 500 ppm of SO 4 2 ion in drinking water causes laxative effect otherwise at moderate levels it is normally harmless.
F : Above 2 ppm concentration of F in drinking water causes brown mottling of teeth.
Hence, the concentration given in question of SO 4 2 and NO3 in water is suitable for drinking but the concentration of F (i.e., 10 ppm) makes water unsuitable for drinking purpose.
28. Given chemical equation MCO2HCl2MClH OCO 23 22 +→ ++ 1 gm 0.01186 mol
⇒ from the balanced chemical equation. 1 M 0.01186 =
⇒=M84.3 gm/mol
29.
As Cr+3 is having least reducing potential, so Cr is the best reducing agent here.
30. ions O-2 F- N
+ Mg+2 Atomic number = 8 9 11 12
Number of e = 10 10 10 10
O, F, Na ,Mg 2+ +2 are isoelectronic species here.
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CHAPTER
Basics of Chemistry 1
Chapter Highlights
Matter and its nature, Dalton’s atomic theory, Concept of atom, Molecule, Element, Compound, Precision, Significant figures, SI units, Derived units, Laws of chemical combinations, Mole, Mass, Molecular mass, Equivalent mass, Chemical equation, Stoichiometry of chemical equation and various levels of multiple-choice questions.
BASIC CONCEPTS
MATTER
Any species having mass and occupying space is known as matter. It can exist in the three physical states, namely, solid, liquid and gas.
Matter
Homogeneous mixtures
Elements Compounds Pure substances Mixtures
Heterogeneous mixtures
Figure 1.1 Classification of Matter
Pencil, air, water, justify the physical states and are all composed of matter.
• At the bulk level or macroscopic level, we can further classify matter as mixtures or pure substances.
Mixture
A mixture is composed of two or more substances which are known as its components or constituents (in any ratio).
The components of the mixture can be separated with the help of physical separation methods like filtration, crystallization, distillation.
• A mixture is further classified into two categories— homogeneous and heterogeneous.
• In a homogeneous mixture all the components undergo complete mixing forming a uniform composition as, air or sugar solution.
• In a heterogeneous mixture the composition formed due to the mixing of components is not entirely uniform like in the case of grains mixed with dust etc.
Pure Substance
Pure substances have fixed compositions and their constituents cannot be separated by using simple physical methods of separation.
• A pure substance can be further classified into an element or a compound.
• An element is composed of one type of particle which could either be atoms or molecules. Na, Cu, Ag have only one type of atoms.
• A compound is formed by the combination of two or more atoms or different elements. For example, H2O, CO2
Dalton’s Atomic Theory
An atom is the smallest particle of an element which is neutral in nature, retains all the properties of the element and takes part in a chemical reaction. The word atom was introduced by Dalton (alamos means undivided).
The Dalton’s atomic theory was proposed by Dalton on the basis of laws of chemical combination.
Main assumptions
• Matter (of any type) is composed of atoms.
• An atom is the smallest, fundamental, undivided particle. (Building block material)
• An atom can neither be created nor destroyed.
• Atoms of an element have similar size, energy and properties while atoms of different element differ in these aspects.
• Atoms combine in whole number ratios to form a molecule, therefore, a molecule is the smallest identity that exists individually.
Modern view about atom
According to the modem view:
• An atom is divisible into other smaller particles which are known as subatomic particles. It can also combine in non-whole number ratio as in the case of non-stoichiometric compounds (Berthollide compounds) like Fe0.93O.
• Atoms of same element also differ in mass and mass related properties as in the case of isotopes.
• Chemical reactions involve rearrangement of atoms.
Molecule
The term molecule was introduced by Avogadro. It is the smallest particle (identity) of matter that can exist independently and retains all the properties of the substance. Normally the diameter of the molecules is in the range of 4–20 Å and the molecular mass is between 2–1000.
• In case of macromolecules, the diameter is in the range of 50–250 Å and the molecular weight may be in lakhs
Berzelius Hypothesis
According to the Berzelius hypothesis, “Equal volumes of all the gases contain same number of atoms under the similar conditions of temperature and pressure.”
This hypothesis on application to law of combining volume confirms that atoms are divisible which is in contrary to Dalton’s theory.
PHYSICAL QUANTITIES AND THEIR MEASUREMENTS
In order to describe and interpret the behaviour of chemical species, we require not only chemical properties but also few physical properties. Physical properties are mass, length, temperature time, electric current etc.
Further, to express the measurement of any physical quantity we require its numerical value as well as its unit. Hence, the magnitude of a physical quantity can be given as Magnitude of physical quantity 5 Its numerical value 3 Unit.
Precision and Accuracy
• The measurements are considered accurate when the average value of different measurements is closer to the actual value. An individual measurement is considered more accurate when it differs slightly from the actual value.
• When the values of different measurements are close to each other as well as to the average value, such measurements are called precise.
• In fact, precision is simply the measurement of reproductability of an experiment.
Significant Figures
These are some uncertainties in values during measurement of matter. In order to make accurate measurements we use these figures.
The total number of digits in a number including the last digit with uncertain value is known as the number of significant figures, for example, 14.3256 6 0.0001 has six significant figures.
Rules to determine significant numbers
• All non-zero digits as well as the zeros present between the non-zero digits are significant, for example, 6003 has four significant figures.
• Zeros to the LHS of the first non-zero digit in a given number are not significant figures, for example, 0.00336 has only three significant figures.
• In a number ending with zeros, if the zeros are present at right of the decimal point then these zeros are also significant figures, for example, 33.600 has five significant figures.
• Zeros at the end of a number without a decimal are not counted as significant figures, for example, 12600 has just three significant figures.
• The result of division or multiplication must be reported to the same number of significant figures as possessed by the least precise term, for example, 3.331 3 0.011 5 0.036641 ≈ 0.037.
• The result of subtraction or addition must be reported to the same number of significant figures as possessed by the least precise term, for example, 5.1 1 7.21 1 8.008 5 20.318 ≈ 20.3.
Rounding-off non-significant figures
Rounding-off non-significant figures means dropping of the uncertain or non-significant digits in a number. It is possible as follows:
• If the rightmost digit to be rounded-off is .5, then the preceding number is increased by one, for example, 3.17 is rounded off to 3.2
• If the rightmost digit to be rounded-off is, 5, then the preceding number is kept unchanged, for example, 5.12 is rounded off to 5.1
• If the rightmost digit to be rounded-off is equal to 5, the preceding number is kept as such in case of an even value. However, in case of an odd value it is increased by one, for example, 4.45 is rounded-off to 4.4; 5.35 is rounded off to 5.4
Exponential notation or scientific notation
In case a number ends in zeros that are not to the right of decimal point it is not essential that zeros are significant. For example, 290 has 2 or 3 significant figures and 19500 has 3, 4 or 5 significant figures.
This confusion can be removed when the values are expressed in terms of scientific notations, for example, 19500 can be written as 1.95 3 104 (3 significant figures), 1.950 3 104 (4 significant figures), 1.9500 3 104 (5 significant figures). In this kind of notation, every number can be written as N 3 10n
Here, n 5 Integer,
Table 1.1
SI Units
Measure Unit
Length
Metre (m)
Mass Kilogram (kg)
Time
Second (s)
Temperature Kelvin (K)
Current Ampere (A)
Intensity Candela (Cd)
Amount of Substance Mole (mol)
N 5 Number with non-zero digit to the left of the decimal point.
For example, 0.00069 can be expressed as 6.9 3 10 4 (2 significant figures).
Table
1.2 Derived Units
Measure Derivation
Concentration (C or S) 5 Mass of solute Mass of solution = mol m =mol m 3 3
Volume (V) 5 Length 3 Height 3 Breadth 5 m 3 m 3 m 5 m3
Force (F) 5 Mass 3 Acceleration 5 m 3 a 5 kg m sec 1 5 Newton (N)
Pressure (P) = Force Area = kg msec m 2 2 5 kg m 1 sec 2 5 Pascal (Pa)
Work (W) 5 Force 3 Displacement 5 F 3 d 5 kg m2 sec 2 5 Joule
FACTS TO REMEMBER
Plane angle (Radian, that is, ‘rad’)
Solid angle (Steradian, that is, ‘str’)
Few Prefixes Used for Subsidiary Units
Sub multiples
1 micro (µ) 5 10 6 1 nano (n) 5 10 9
1 femto (f) 5 10 15 1 atto (a) 5 10 18
1 zepto (z) 5 10 21
I yocta (y) 5 10 24
1 Giga (G) 5 109
1 Tetra (T) 5 1012
1 Exa (E) 5 1018
1 Zetta (Z) 5 1021
1 Yotta (Y) 5 1024
1 litre 5 10 3 m3 5 1 dm3
1 atmosphere 5 760 mm or torr 5 101.325 Pa or Nm 2
1 bar 5 105 Nm 2 5 105 Pa
1 calorie 5 4.184 joule
1 eV (electron volt) 5 1.602 3 10 19 joule
1 joule 5 10 7 erg
So, 1 eV 5 1.602 3 10 12 erg
1 cal . 1 J . 1 erg . 1 eV
‘Barn’ is a unit of area to measure the cross section of nucleus.
1 Barn 5 10 28 m2 ≈ 10 24 cm2
LAWS OF CHEMICAL COMBINATIONS
Law of Conservation of Mass
• Law of conservation of mass was proposed by Lavoisier in 1774.
• It was verified by Landolt.
• According to this law, “In a chemical change the total mass of the products is equal to the total mass of the reactants, that is, mass is neither created nor destroyed.” For example, when a solution with calculated weight of AgNO3 and NaCl is mixed, white precipitates of AgCl are formed while NaNO3 remains in solution. The weight of the solution remains the same before and after this experiment.
• It is not applicable to nuclear reactions.
Law of Constant Composition or Law of Definite Proportion
• Law of constant composition was proposed by Proust in 1779.
• It was verified by Star and Richards.
• According to this law, “A chemical compound always contains same elements combined together in same proportion by mass.” For example, NaCl extracted from sea water or achieved from deposits will have 23 g Na and 35.5 g of chlorine in its one mole.
• It is not applicable to non-stoichiometric compounds like Fe0.93 O.
Law of Multiple Proportion
• Law of multiple proportion was proposed by Dalton in 1804.
• It was verified by Berzilius.
• According to this law, “Different weights of an element that combine with a fixed weight of another element bear a simple whole number ratio.” For example, in case of CO, and CO2 weight of oxygen which combines with 12 g of carbon is in 1 : 2 ratio.
• It is applicable when same compound is prepared from different isotopes of an element. For example, H2O, D2O.
Law of Reciprocal Proportion
• Law of reciprocal proportion was proposed by Richter in 1792.
• It was verified by Star.
• According to this law, “When two different elements undergo combination with same weight of a third element, the ratio in which they combine will either be same or some simple multiple of the ratio in which they combine with each other.”
• It is also known as Law of equivalent proportion which states “Elements always combine in terms of their equivalent weight.”
Law of Combining Volume
• Law of combining volume was proposed by Gay-Lussac.
• It applies to gases.
• According to this law, “When gases react with each other they bear a simple whole number ratio with one another as well as the product under conditions of same temperature and pressure.”
AVOGADRO’S LAW
• Avogadro’s law explains law of combining volumes.
• According to this law “Under similar conditions of temperature and pressure equal volume of gases contain equal number of molecules.”
