Chemestry for JEE
March 29, 2017 | Author: Sarthak Lalchandani | Category: N/A
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Copyright of this book is reserved by Gujarat Secondary and Higher Secondary Education Board, Gandhinagar. No reproduction of this book in whole or in part, or in any form is permitted without written permission of the Secretary, Gujarat Secondary and Higher Secondary Education Board, Gandhinagar.
Gujarat Secondary and Higher Secondary Education Board, Gandhinagar
QUESTION BANK CHEMISTRY
Price
: ` 70.00
Published by : Secretary Gujarat Secondary and Higher Secondary Education Board, Gandhinagar
I
Contribution 1
Dr. Hasmukh Adhiya (IAS)
Principal Secretary , Education Department Gandhinagar
2
Shri R. R. Varsani (IAS)
Chairman , G.S&H.S.E. Bord, Gandhinagar
3
Shri H. K. Patel (G.A.S)
Dy. Chairman, G.S&H.S.E. Bord, Gandhinagar
4
Shri M. I. Joshi (G.E.S)
Secretary , G.S&H.S.E. Bord, Gandhinagar
Coordination 1
Shri B. K. Patel
O.S.D., G.S&H.S.E. Bord, Gandhinagar
2
Shri D. A.Vankar
Assistant Secretary (Retd.), G.S&H.S.E. Bord, Gandhinagar
5
Shri M. P. Parmar
Assistant Secretary, G.S&H.S.E. Bord, Gandhinagar
Expert Teachers 1.
Shri C. I. Patel (Convenor)
Shri Vidyanagar High School, Ahmedabad
2.
Shri S. B. Gor (Co-Convenor) Ghyanda Girls High School, Ahmedabad
3.
Shri A. I. Patel
4.
Shri V. R. Patel
5.
Shri B. R. Patel
Muktjeevan Vidhyalaya, Ahmedabad
6.
Shri K. K. Purohit
M. K. Higher Sec. School, Ahmedabad
7.
Shri M. B. Patel
New Vidhyavihar for Girls, Ahmedabad
8.
Shri B. A. Nayak
Swaminarayan High School, Ahmedabad
9.
Shri H. M. Patel
Navchetan High School, Ahmedabad
10. Shri S. B. Suthar
R.P.T.P. Science School, Vallabh Vidhyanagar
11. Shri R. N. Patel
R.P.T.P. Science School, Vallabh Vidhyanagar
12. Shri N. N. Shah
Best High School, Ahmedabad
13. Shri J. Y. Mehta 14. Shri I. B. Amlani 15. Smt. M. N. Shethiya 16. Smt. H. N. Nayak 17. Smt. P. S. Thakar
R.P.T.P. Science School, Vallabh Vidhyanagar
18. Shri G. S. Patel 19. Shri M. L. Sharma 20. Shri H. K. Patel
II
P R E FA C E Uptil now , the Students had to appear in various entrance examinations for engineering and medical courses after std-12. The burden of examinations on the side of the students was increasing day-by-day. For alleviating this difficulty faced by the students, from the current year, the Ministry of Human Resource Development , Government of India, has Introduced a system of examination covering whole country. For entrance to engineering colleges, JEE(Main) and JEE(Advanced) examinations will be held by the CBSE. The Government of Gujarat has except the new system and has decided to follow the examinations to be held by the CBSE. Necessary information pertaining to the proposed JEE (Main) and JEE(Advanced) examination is available on CBSE website www.cbse.nic.in and it is requested that the parents and students may visit this website and obtain latest information – guidance and prepare for the proposed examination accordingly. The detailed information about the syllabus of the proposed examination, method of entrances in the examination /centers/ places/cities of the examinations etc. is available on the said website. You are requested to go through the same carefully. The information booklet in Gujarati for JEE( Main) examination booklet has been brought out by the Board for Students and the beneficieries and a copy of this has been already sent to all the schools of the state. You are requested to take full advantage of the same also However, it is very essential to visit the above CBSE website from time to time for the latest information – guidance . An humble effort has been made by the Gujarat secondary and Higher Secondary Education Boards, Gandhinagar for JEE and NEET examinations considering the demands of the students and parents , a question bank has been prepared by the expert teachers of the science stream in the state. The MCQ type Objective questions in this Question Bank will provide best guidance to the students and we hope that it will be helpful for the JEE and NEET examinations. It may please be noted that this “Question Bank” is only for the guidance of the Students and it is not a necessary to believe that questions given in it will be asked in the examinations. This Question Bank is only for the guidance and practice of the Students. We hope that this Question Bank will be useful and guiding for the Students appearing in JEE and NEET entrance examinations. We have taken all the care to make this Question Bank error free, however, if any error or omission is found, you are requested to refer to the text – books.
Date: 02/ 01/ 2013
M.I. Joshi Secretary
III
R.R. Varsani (IAS) Chairman
INDEX UNIT 1
SOME BASIC CONCEPTS IN CHEMISTRY
1
UNIT 2
STATES OF MATTER
20
UNIT 3
STRUCTURE OF ATOM
44
UNIT 4
CHEMICAL BONDING AND ATOMIC STRUCTURE
79
UNIT 5
CHEMICAL THERMODYNAMICS
96
UNIT 6
SOLUTIONS
120
UNIT 7
EQUILIBRIUM
151
UNIT 8
REDOX REACTIONS & ELECTROCHEMISTY
183
UNIT 9
CHEMICAL KINETICS
207
UNIT 10 SURFACE CHEMISTRY
230
UNIT 11 CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES
241
UNIT 12 GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF METALS
263
UNIT 13 HYDROGEN
283
UNIT 14 S - BLOCK ELEMENTS
301
IV
UNIT : 1 SOME BASIC CONCEPTS IN CHEMISTRY Important Points [A] Important formulae : mass Molar mass
1.
No.of moles
2.
No.of moles of gas
3.
No.of moles of Particles
4.
No. of moles of solute Molarity Vol ( L )
5.
Eq. wt. of a salt =
6.
Eq. wt.of element
7.
Volume at STP 22.4
No.of Particles 6.022 1023
M.W.of salt Total + ve charge of metal ion
Atomic Weight Valency
m a nb mn where, a + b are atomic masses and m + n are precentage. Avg. at. mass
8.
n (at mass of element ) % of element in compound = M .W . of compound 100 where, n= No. of atoms of that element
9.
Molarity
10.
Normality
11.
w 1000 MW Wo( g ) Wo = Weight of solvent
12.
Mole fraction ( X )
13.
%W W
14.
ppm
w 1000 M .W . V (ml ) w 1000 E.W . V (ml )
Molality
n nN
W 100 W Wo
weight (vol ) of solute 106 weight (vol ) of solution 1
15.
Molecalar weight = 2 V.D.
16.
Eq.wt of metal
Wt. of metal 1.008 wt of H 2 displaced
17.
Eq.wt of metal
Wt. of metal 11200 Vol of H 2 displaced at STP (mL )
18.
Eq.wt of metal
Wt. of metal 35.5 Wt of Chlorine combined
19.
Eq.wt of metal
Wt. of metal 11200 Vol of Cl2 combined at STP (mL )
20.
Eq.wt of metal
Wt. of metal 8 Wt of oxygen combined
21.
Eq.wt of metal
Wt. of metal 5600 Vol of O2 displaced at STP (mL )
22.
Molority
23.
M 1 V1 M 2 V2 ( Molarity equation)
24.
N1 V1 N 2 V2 ( Normality equation)
25.
n
% W W density 10 Molecular weight
Molecular weight Empirical formula Weight 9 0 ( C ) 32 5
26.
0
27.
K 0 C 273.15
28.
1 L 1 dm 3 , 1 mL 1 cm 3
[B]
Important Facts :
1. 2. 3. 4. 5. 6.
Antoine Lavoisier - Law of conservation of mass Joseph proust - Law of definite proportions John Dalton - Law of Multiple proportions Richter - Law of combining weights. Gay Lussac - Law of combining Volumes. 1 amu = 1.6605 x 10-24 gram
7.
Mass of C atom 1.9926 1023 gram
8. 9. 10.
Avogadro number ( N A ) 6.022 10 23 AZT = Azido thymidine,drug used for aids victims. The limiting reagent is the reagent that is entirely consumed when a reaction goes to completion. Its amount limits the amount of the product formed.
F
12
2
[c]
Precision and Accuracy. The term precision refers for the closeness of the set of values obtained form identical measurements of a quantity. Accuracy refers to the closeness of a single measurement to its true value. Let us take an example to illustrute. this. Three students were asked to determine the mass of a piece of metal where mass is known to be 0.520g. Data obtained by each Student are recorded in table below mesurements in g. 1 2 3 Average Student A 0.521 0.515 .0509 0.515 Student B 0.516 0.515 .0514 0.515 Student C 0.521 0.500 .0520 0.520 The data for student A are neither, precise nor accurate. The data for student B are precise but not accurate. The data for student C are both precise and accurate.
M.C.Q. 1.
2.
Identify the wrong statement in the follwing (AIEEE 2008). (a) CFCs are responsible for ozone layer depletion. (b) Greenhouse effect is responsible for global warming. (c) Ozone layer does not permit I.R. radiation from the sun to reach the earth. (d) Acid rain is mostly because of oxides of ‘N’ and ‘S’. In the reaction
2 Al( s ) 6 HCl( aq ) Al 3 ( aq ) 6Cl ( aq ) 3H 2( g ) (AIEEE 2007)
3.
4. 5.
6.
(a)
6L HC1 ( aq ) is consumed for every 3L, H 2( g ) produced.
(b)
33.6L H 2( g ) is produced regardless of temperature and pressure for every mole of Al that reacts.
(c)
67.2L H 2( g ) at STP, is produced for every mole Al that reacts.
(d)
11.2L H 2( g ) at STP, is produced for every mole HC1 ( aq ) consumed.
Consider a titration of potassium dichromate solution with acidified Mohr’s salt solution using diphenyl amine as indicator. The number of moles of Mohr’s salt required per mole of dichromate is (IIT JEE 2007) (a) 3 (b) 4 (c) 5 (d) 6 Which has maximum number of atoms ? (IIT JEE 2003) (a) 24g of C (12) (b) 56g of Fe(56) (c) 27g of Al (27) (d)108g of Ag (108) What volume of hydrogen gas at 273K and 1 atm pressure will be consumed in obtaining 21.6 g of elemental boron (atomic mass = 10.8) form the reduction of boron trichloride by hydrogen (AIEEE 2003) (a) 89.6L(b) 67.2L (c) 44.8L (d) 22.4L In an organic compound of molar mass 108g/mol, C,H and N atoms are present in 9:1:3.5 by weight Molecular formula can be (AIEEE 2002) (a) C6 H 8 N 2
(b) C6 H10 N
(c) C5 H 6 N 3
(d) C4 H18 N 3 3
7.
Number of atoms in 560 g of Fe (atomic mass = 56) is (a) twice that of 70 g N. (b) half that of 20 g H (c) both (a) and (b) (d) None of these
(AIEEE 2002)
8.
In the standardization of Na2 S 2O3 using K 2 Cr2 O7 by iodometry, the equivalent weight of K 2 Cr2 O7 is (IIT JEE 2001) (a)
Molar mass Molar mass (b) 2 6
Molar mass (d) same as molar mass 3 Mixture X=0.02 mole of [Co(NH3)5SO4] Br and 0.02 mole of[Co(NH3)5Br]SO4 was prepared in 2L of Solution
(c) 9.
1L of mixture X excess AgNO3 Y 1L of mixture X excess BaCl2 Z
10.
Number of mole of Y and Z are (IIT JEE 2003) (a) 0.01, 0.01 (b) 0.02, 0.01 (c) 0.01,0.02 (d) 0.02, 0.02 How many moles of electron weight one kilogram ? (IIT JEE 2002) (a) 6.023 10 23 (b)
1 1031 9.108
11.
1 6.023 108 1054 (d) 9.108 6.023 9.108 An Oxide of metal contains 60% of the metal. What will be the equivalent weight of the metal ?
12.
(a) 12 (b) 40 (c) 24 (d) 48 A container is filled with 2L of water. What will be the volume of water in m3?
(c)
13. 14.
15.
(a) 2 103 (b) 1 103 (c) 2 103 (d) 1 103 The mass of carbon -12 atom considered in the definition of a mole is (a) 0.012Kg (b) 0.12g (c) 120 mg (d) None of these The drug which is used for treating AIDS victims is (a) Azidothymidine (b) Cis- platin (c) Taxol (d) All of these Chose the incorrect statement . (a) The constituents of a compound cannot be separated into simpler substances by physical methods. (b) An element is consists of only one type of particles and these particles may be atoms or molecules. (c) The properties of a compound are same as its constituent elements. (d) Atoms of different elements are different in nature. 4
16.
17. 18. 19.
20.
21.
Which of the following is a pair of physical and chemical property respectively of a substance ? (a) acidity & combustibility (b) colour & density (c) basicity & colour (d) density & acidity. What is the symbol of S.I. unit for the amount of substance ? (a) NA (b) n (c) mole (d) mol What is the symbol of a multiple ‘109’ ? (a) G (b) E (c) n (d) Z Find the correct relation. 5 o o (a) o F 9 (o C ) 32 (b) C ( F 32) 9 5 (c) Both & (a) and (b) (d) Neither (a) nor (b) In chemistry a number is represented in the form N 10n . This method of expressing the number is called scientific notation. What is the value of ‘N’ here. (a) 1 to 10 (b) 0.1 to 9.99 (c) 10 to 100 (d) Any value can be taken What is the correct scientific notation for 0.00016 ?
(a) 1.6 10 4 (b) 16 10 5 (c) 0.16 103 (d)cannot be determined. 22. 23. 24.
25.
26. 27. 28.
29.
How many significant digits are there in 0.25 ? (a) 1 (b) 2 (c) 3 (d) cannot be determined. Which of the following number contains there significant digits ? (a) 0.200 (b) 0.030 (c) 0.0052 (d) 0.002 2+ What is the number of neutrons in Zn ion (Atomic mass namber = 70) (IITJEE 1979) (a) 34 (b) 36 (c) 38 (d) 40 The same amount of Zinc is treated separately with excess of sulphuric acid and excess of sodium hydroxide. What will be the ratio of volumes of hydrogen evolved ? (IITJEE 1979) (a) 1 : 1 (b) 1 : 2 (c) 2 : 1 (d) 9 :4 2.76g of silver carbonate on being strongly heated yields a residue weighing (IITJEE 1979) (a) 2.16g (b) 2.48 g (c) 2.32 g (d) 2.64 g Find the total number of electrons in one molecule of carbon dioxide. (a) 22 (b) 44 (c)66 (d) 88 A gaseous mixture contains oxygen and nitrogen in the ratio of 1 : 4 by weight. Therefore, the ratio of their number of molecules is (a) 1 : 4 (b) 1 : 8 (c) 7 : 32 (d) 3 : 16 Identify the incorrect unit conversion factor. (a)
1cm 3 1mL
(b)
1cm 10mm
(c)
60s 1min
(d) None of these
5
30.
31.
32.
90 g KClO3 on heating gives 2.96g KCl and 1.92g oxygen. Which of the follwing laws is illustrated by this statement ? (a) Law of definite proportion (b) Law of mass conservation (c) Law of multiple proporation (d) Avogadro’s law. Match the following property. A B (i) Law of Multiple proportions. (p) Richter (ii) Law of Combining volumes (q) Proust (iii) Law of Reciprocal proportions. (r) GayLussac (iv) Law of Constant composition. (s) Dalton (a) i - s, ii - p, iii - r, iv - q (b) i - s, ii - r, iii - p, iv - q (c) i - s, ii - r, iii - q, iv - p (d) i - q, ii - r, iii - p, iv - s Two oxides of a metal ‘M’ contain 27.6% and 30.0% of oxygen respectively. If the formula of the first oxide is M3O4 , find that of the second. (a) M 2 O3
33.
34.
35. 36.
37.
38.
(b) M 2 O
(c) MO2
(d) M 3O2
Naturally occuring Boron consists of two isotopes having atomic masses 10.01 and 11.01 respectively. Calculate the percentage of both the isotopes in natural Boron (Atomic mass of natural Boron = 10.81) (a) 20% and 80% (b) 80% and 20% (c) 25% and 75% (d) 75% and 25% Calculate the mass percent of Na and S in sodium sulphate. (a) Na = 16.2%, S = 22.54% (b) Na = 32.39%, S = 11.26% (c) Na = 22.54%, S = 32.39% (d) Na = 32.39%, S = 22.54% Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% oxygen by mass. (a) FeO (b) Fe2O3 (c) Fe3O4 (d) Fe3O2 Calculate the amount of carbon dioxide that can be produced when 1 mole of carbon is burnt in 16 g of dioxygen. (a) 44g (b) 22g (c) 88g (d) 11 g Calculate the concentration of nitric acid in moles per litre which has a density , 1.41 g/mL. %w/w of nitric acid is (a) 15.44M (b) 0.064M (c) 0.077M (d) 12.87M In a reaction : N 2( g ) 3H 2( g ) 2 NH 3( g ) , 2000g N 2 reacts with 1000g H 2 which reactant will left unreacted ? How much ?
39.
(a) N2 , 2428 g
(b) H 2 , 428.6 g
(c) N 2 , 571.4 g
(d) H 2 , 571.4 g
Calculate the number of sulphate ions in 100mL of 0.001M ammonium sulphate solution. (a) 6.022 1019 (b) 6.022 1019 (c) 6.022 1020 (d) 6.022 1020
40.
Calculate the molarity of a solution of ethanol in water in which mole fraction of ethanol is 0.040. (a) 2.31M (b) 0.213M (c) 0.0213M (d) 23.1M 6
41.
Some statements are given below based on the pictures. Identify true and false statements.
(P) (Q) (R) (i) ‘P’ and ‘Q’ both indicates precision and accuracy. (ii) ‘Q’ indicates precision and accuracy white ‘R’ indicates neither precision nor accuracy. (iii) ‘P’ indicates precision but not accuracy. (iv) ‘Q’ indicates both precision and accuracy (a) FTTT 42. 43.
44.
(b) TTTT
(c) TTFT
(d)FTFT
The normality of 0.3M phosphorous acid is (IITJEE 1999) (a) 0.1 (b) 0.9 (c) 0.3 (d) 0.6 An aqueous solution of 6.3g oxalic acid dihydrate is made upto 250 mL. The volume of 0.1 N NaOH required to completely neutralize 10 mL of this solution is (a) 40 mL (b) 20 mL (c) 10 mL (d) 4 mL The pair of the compounds in which both the metals are in the highest possible oxidation state is 3
(a) Fe CN 6 , Co CN 6
3
(b) CrO2 Cl2 , MnO4 3
45.
46.
47.
48.
(c) TiO3 , MnO2 (d) Co CN 6 , MnO3 In the analysis of 0.0500 g sample of feldspar, a mixture of the chiorides of sodium and potassium is obtained, which weighs 0.1180 g. Subsequent treatment of the mixed chlorides with silver nitrate gives 0.2451g of silver chloride. What is the percentange of a sodium oxide and potassium oxide in feldspar ? (a) 10.62% Na2O , 3.58% K 2O
(b) 3.58% Na2O , 10.62% K 2O
(c) 10.62% Na2O , 35.8% K 2O
(d) 35.8% Na2O , 10.62% K 2O
5.5 g of a mixture of FeSO4.7H2O and Fe2 (SO4)39H2O requires 5.4 mL of 0.1N KMnO4 solution for complete oxidation. Calculate the number of mole of Fe2 (SO4)39H2O in the mixture. (a) 0.0095 (b) 0.15 (c) 0.0952 (d) 1.52 A compound contains 28% of nitrogen and 72% of a metal by weight. Three atoms of the metal combine with two atoms of nitrogen. Find the equivalent weight of the metal. (a) 12 (b) 24 (c) 36 (d) 48 The density of a 3M Na2|S2O3 solution is 1.25 g per mL, What is the molalities of Na+ and S2O32-ions ? (a) 3.865, 7.732 (b) 7.732, 3.865 (c) 1.933, 7.732 (c) 7.732, 1.933 7
49.
50.
51. 52.
53.
54.
55. 56.
57. 58. 59.
60.
Haemoglobin present in blood contain 3.72% by mass iron. Calculate the number of iron atoms in 2.0g of haemiglobin. (a) 4.53 X 1026 (b) 4.53 X 1023 (c) 5.95 X 1019 (d) 8 X 1020 How many moles of magnesium phosphate, Mg3(PO4)2 will contain 0.25 mole of oxygen atoms (AEEE 2006) (a) 0.02 (b) 3.125 x 10-2 (c) 1.25 x 10-2 (d) 2.5 x 10-2 The unit J Pa -1 is equivalent to (a) m3 (b) cm3 (c) dm3 (d) none of these -3 The density of Al metal is 2.7 gcm . An irregularly shaped piece of aluminium weighing 40.0g is added to a 100mL graduated cylinder containing 50.0mL of water. upto what height the water level will rise in the cylinder ? (a) 14.8mL (b) 79.6mL (c) 64.8mL (d) 50mL A sample of clay after drying partially was found to contain 50% silica and 7% water.The original sample of clay had 12% water, What is the percentage of silica in the original sample? (a) 50% (b) 5% (c) 43% (d) 47% In which of the following pairs percent compostion of element is not same ? (a) benzene and ethyne (b) But - 2 - ene and Cyclobutane (c) glucose and fructose (d) phenol and ethanol What weight of CuO will be required to provide 200Kg copper (a) 200Kg (b) 79.5Kg (c) 250Kg (d) 100Kg Choose the proper option after studying following statement (T = True, F = False) 1. The percent composition of vinyl chloride and its polymer PVC are same. 2. The perecent compostion of phosphorous trioxide (P2O3)is half than that of its dimer phosphorous hexoxide (P4O6) for each of the elements present in them. (a) T, F (b)F,T (c) T, T (d) F, F Impure sample of ZnS contains 42.34% Zn. What is the percentage of pure ZnS in the smaple ? (a) 67% (b) 63% (c) 58% (d) 37% If the atomic mass of carbon were set at 50 amu, what would be the value of Avog adro’s number ? (a) 5.01 x 1024 (b) 6.022 x 1023 (c) 1.66 x 1024 (d) none of these For which of the following compounds molecular weigh cannot be determined from atomic weights ? (a) Fe4 Fe CN 6
(b) TiO2
(c) TiO1.12
(d) none of these
Which one of the following contains greatest number of oxygen atoms ? (a) 1.0g of O atoms (b) 1.0g of O2 (c) 1.0g of O3 (d) All have same number of atoms
8
61.
Which of the following chemical equation is incorrectly balanced ? (a) Sb2 S3 12 HCl 2H 3 SbCl6 3H 2 S (b) 3IBr 4 NH3 NI3 3NH 4 Br (c) 2 KrF2 2 H 2O 2 Kr 2O2 4 HF (d) PCl3 3 H 2 O H 3 PO3 3 HCl
62.
63. 64.
65. 66.
67.
68.
69.
70.
Match the following Column -I Column -II (i) Cl2O3 (P) basic anhydride (ii) Li2O (Q) acid anhydride (iii) CO2 (R) base (a) (i) - (q) (ii) - (p) (iii) - (q) (b) (i) - (r) (ii) - (q) (iii) - (p) (c) (i) - (p) (ii) - (q) (iii) - (r) (a) (i) - (p) (ii) - (q) (iii) - (p) How many g of NaOH can be obtained by reaction of 1 Kg of Na2CO3 with Ca(OH)2 ? (a) 106 g (b) 850 g (c) 755g (d) 943 g How much calcium oxide (CaO) can be obtained by heating 200 Kg of lime stone theat is 95% pure CaCO3 ? (a) 56Kg (b) 190Kg (c) 170Kg (d) 107Kg Calculate the amount of NaOH required to neutralize 100 mL 0.1M H2SO4. (a) 40g (b) 0.4 g (c) 80 g (d) 0.8 g 3g of an oxide of a metal is converted into chloride and it yielded 5 g of chloride. Find the equivalent weight of the metal. (a) 33.25 (b) 3.325 (c) 12 (d) 20 A compound contains two oxygen atoms, four carbon atoms and number of hy drogen atoms is double of carbon atoms. What is the density of vapour of this compound ? (a) 88 (b) 44 (c) 132 (d) 72 The number of molecules in 100 mL of each of O2,NH3 and CO2 at STP are (a) CO2 O2 NH 3
(b) NH 3 O2 CO2
(c) NH3 = CO2 < O2
(d) NH 3 O2 CO2
Which of the following represents the formula of a compound which contains 26% nitrogen and 74% oxygen ? (a) N2O (b) NO (c) NO2 (d) N2O5 -1 NKg is the unit of (a) momentum (b) velocity (c) Pressure (d) accelaration
9
71.
72.
73.
Which one of the following statements is incorrect ? (a) All elements are homogeneous system (b) Compounds made up of a number of elements are heterogeneous. (c) A mixture is not always heterogeneous (d) Smoke is a heterogeneous mixture. A balanced chemical equation is in accordance with (a) Avogadro’s law. (b) Law of constant proportions (c) Law of conservation of mass (d) Law of gaseous volumes. The atomic weights of two elements X and Y are 20 and 40 respectively. If ‘a’ gm of X contains ‘b’ atoms, how many atoms are present in ‘2a’ gm of Y ? (a) b
74.
75.
76. 77.
78.
79. 80.
(b) a
a (d) ( ) 2
(c) 2b
If the components of air are N2 ,78%, O2, 21% ; Ar, 0.9% and CO2, 0.1% by volume, what will be the molecular weight of air ? (a) 28.9 (b) 32.4 (c) 16.4 (d) 14.5 Calculate the molarity of a solution obtained by mixing 50mL of 0.5M H2SO4 and 75 mL of 0.25M H2SO4. (a) 0.375M (b) 0.35M (c) 0.045M (d) 0.45M Which of the following has the highest normality ? (a) 1M H2SO4 (b) 1M H3PO3 (c) 1M H3PO4 (d) 1M HNO3 In an experiment, 4 gm of M2Ox oxide was reduced to 2.8 gm of the metal. If the atomic mass of the metal is 56 gm/mol, the number of oxygen atoms in the oxide is (AFMC 2010) (a) 1 (b) 2 (c) 3 (d) 4 Match the following Column - I Column - II (i) femto (P) 109 (ii) yotta (q) 10-15 (iii) giga (r) 10-18 (iv) atto (s) 1024 (a) i - q, ii - p, iii - r, iv - s (b) i - s, ii - q, iii - p, iv - r (c) i - q, ii - s, iii - p, iv - r (d) i - r, ii - s, iii - p, iv - q The total number of atoms of all elements present in mole of ammonium dichromate is (a) 19 (b) 6.023 x 1023 (c) 114.47 x 1023 (d) 84 x 1023 0.32 gm of a metal on treatment with an acid gave 112 mL of hydrogen at STP. Calculate the equivalent weight of the metal (a) 58 (b) 32 (c) 11.2 (d) 24 10
81.
82.
83.
84. 85. 86.
87. 88. 89.
90. 91.
92.
For a reaction A + 2B C, the amount of C formed by starting the reaction with 5 moles of A and 8 moles of B is (a) 5 moles (b) 8 moles (c) 16 moles (d) 4 moles 100 mL of PH3 on heating forms P and H2 . The volume change in the reaction is (a) an increase of 50 mL (b) an increase of 100 mL (c) an increase of 150 ml (d) a decrease of 50 mL An organic compound made of C, H, and N contains 20% nitrogen. Its molecular weight is (WBJEE 2009) (a) 70 (b) 140 (c) 100 (d) 65 Volume occupied by one molecule of water (d = 1 gm cm -3) is (a) 9 x 10-23cm3 (b) 6.02 x 10-23cm3 (c) 3 x 10-23cm3 (d) 5.5 x 10-23cm3 Calculate the number of moles in 1m3 gas at STP. (a) 4.46 (b) 44.6 (c) 446 (d) 4460 An ore contains 1.24% of the mineral argentite Ag2S by mass. How many grams of this ore would have to be processed in order to obtain 1.0 g of pure solid silver ? (a) 23.15 g (b) 69.45 g (c) 92.6 g (d) 46.3 g Find the electric charge in couloumb of 9.0 gm of A13+ ions. (a) 9.6 x 104 (b) 6.9 x 104 (c) 2.9 x 105 (d) 4.80 x 10-19 Which of the following is not a homogeneous mixture ? (a) smoke (b) air (c) Brass (d) Aqueous solution of sugar Which of the following has the largest number of atoms ? (a) 0.5g atom of Cu (b) 0.635g of Cu (c) 0.25 moles of Cu atom (d) 1 g of Cu 27 g of A1 (at mass = 27) will react with oxygen equal to (IIT 1978) (a) 24 g (b) 8 g (c) 40 g (d) 10 g Two containers P and Q of equal volumes contain 6 g of O2 and SO2 respectively at 300K and 1 atmosphere. Then (a) No. of molecules in P is less than that in Q (b) No. of molecules in Q is less than that in P (c) No. of molecules in P and Q are same. (d) cannot be determined Which of the following pairs of substances illustrates the law of multiple proportions ? (a) CO and CO2 (b) NaCl and NaBr (c) H2O and D2O (d) MgO and Mg(OH)2 In each of the follwoing questions, two statements are given, one is Assertion (A) and the other is Reason (R). Examine the statements carefully and mark the correct answer according to the instructions given below : (a) If both A and R are correct and R is the correct explanation of A. (b) If both A and R are correct and R is not the correct explanation of A. (c) If A is correct R is wrong. (d) If both A and R are false. 11
93. 94.
A : Normality of 0.1M H2SO4 is 0.2N. R : H2SO4 is a dibasic acid. A : 1 Gram molecule of sulphar also represents 1 gram atom of sulphur. R : Atomicity of sulphur is one.
A : In the equation NH 3 HCl NH 4 Cl , Gay-Lussac’s law is not applicable to NH4Cl. R : NH4Cl is not a gas,. 96. A : Atomic mass of sodium is 23 u. R : An atom of sodium is 23 times heavier than an atom of 12C. 97. A : Pure water, irrespective of its source always contain hydrogen and oxygen in the ratio 1 : 8 by mass. R : Total mass of reactants and products remains constant during physical or chemical change. 98. A : Mass numbers of most of the elements are fractional. R: Mass numbers are obtained by comparing with mass number of 12C. 99. A : The mass of the products formed in a reaction depends upon the limting reactant. R: Limting reactant reacts completely in the reaction. 100. A : Cinnabar is a chemical compound whereas brass is mixture. R : Cinnabar always contain 6.25 times mercury than sulplur by weight. Brass can have any proportion of Cu and Zn. 101. A : 1 L of O2 and 1 L of O3 contains the same number of moles under identical conditions. R : Under identical conditions, 1 L of O2 and 1 L of O3 contain the same number of oxygen atoms. 102. A : The standard unit for expressing atomic mass is amu. R : Now a days amu is represented by ‘u’. 95.
ANSWER KEY 1c
2d
3d
4a
5b
6a
7c
8b
9a
10 d
11 a
12 c
13 a
14 a
15 c
16 d
17 d
18 a 19 d
20 a
21 a 22 b
23 a 24 d
25 a
26 c
27 a
28 c
29 d
30 b
31 b
32 a
33 a 34 d
35 b
36 b
37 a 38 d
39 b
40 a
41 a 42 d
43 a 44 b
45 b
46 a
47 a 48 b
49 c
50 b
51 a
52 c
53 d
54 d
55 c
56 a 57 b
58 d
59 d
60 d
61 c
62 a
63 c
64 d
65 d
66 a 67 b
68 d
69 d
70 d
71 b
72 c
73 a 74 a
75 b
76 c
77 c
78 c
79 c
80 b
81 d
82 a
83 a
84 c
85 b
86 c
87 a
88 a
90 a
91 b
92 a
93 a 94 d
95 a
96 c
97 b
98 d
99 a 100 a 101 c 102 b
12
89 a
SOLUTIONS/HINTS 3.
4.
5.
6.
(d)
6 Fe2 Cr2 O7 2 14 H 6 Fe3 2Cr 3 7 H 2 O
(a)
Number of particles Number of moles
No. of moles of carbon
24 2 12
(b)
2 BC 3 3H 2 2 B 6 HC 2 mol 3 mol 2 mol = 21.6 g
V
nRT 3 0.0821 273 67.2 L P 1
(a)
8.
The redox reaction between potassiumdichromate and Mohr’s salt is :
(b)
mass of carbon =
9 108 72 g 13.5
72 6 12 similarly, no of H and N atoms are 8 and 2 respectively. No.of carbon atoms
During the reaction, Cr2 O7 2 changes to Cr3+ . Hence the change in oxidation number of Cr is 6. 9.
Molar mass 6
(a)
10
Equivalent weight
In 2L solution, there are 0.02 mol Br - ions and 0.02 mole so4 2
2 1 L of mixture X contains 0.01 mol Br – and 0.01 mol SO4 ions. Hence, Y= 0.01 mol Ag Br Z= 0.01 mol BaSO4 (d)
Mass of an electron 9.108 1031 Kg
No. of electron s in 1 Kg
9.108 10
1 9.108 10 31
31
1 6.023 1023 mol 1
108 mol 9.108 6.023
13
11.
(a)
mass of metal = 60 g mass of oxygen = 40 g mass of oxygen = mass of metal 40 g = 60g 8 g = (?) =
25.
8 60 12 40
(a) (i) Zn H 2 SO4 ZnSO4 H 2 (ii) Zn 2 NaOH Na2 ZnO2 H 2 26. (c) Ag 2 CO3( s ) Ag 2 O( s ) CO2( g )
1 mol 0.01 mol
1mol 0.01 mol
There fore mass of residue (Ag2O)
0.01 molarmass of Ag 2O 0.01 232 2.32g
28.
(c)
The ratio by weight =
1 4
1
32.
28 7 32 4 4 32 32 28
Ratio of moles
(a)
Let ‘x’ be the atomic mass of metal ‘M’ In the oixde M3O4, the mass of ‘M’ = 72.4 and that of ‘O’ = 27.6
M 72.4 O27.6 x
16
M 3 O4
72.4 27.6 : 3: 4 x 16
x 56 For second oxide, the mass of ‘M’ = 70 and that of ‘O’ =30
M 70 O30 56
16
M1.25 O1.875 M1 O1.5
OR
M 2 O3
14
33.
37.
(a) Let the % of isotope with atomic mass 10.01 be ‘x’ % of isotope with atomic mass 11.01 = 100-x Avg at mass =
(a)
69% w/w means 100 g nitric acid soution contain 69 g of nitric acid by mass.
moles of HNO3
Vol. of 100 g nitric acid solution
moles per litre
38.
10.01x (100 x)11.01 10.81(Given) 100
(d)
69 1.095 63 100 0.07092 L 1.41
1.095 15.44 0.07092
N 2 3H 2 2 NH 3
28 g 2000 g
6g (?)
2000 6 428.6 g 28 But we are given 1000 g H2 There fore 1000 - 4286 = 571.4 g H2 will left.
39.
40.
3(b) No of moles of (NH4)2SO4 = molarity vol(L) = 0.001 0.1 = 0.0001 No. of SO4 (a)
2
X ETOH
ions 0.0001 6.022 10 23 6.022 1019
n( ETOH ) nETOH n( H 2O )
0.04
n( ETOH ) n( ETOH ) 55.55
n( ETOH ) 2.31
42
(d)
43.
phosphorous acid ( H 3 PO3 ) is a dibasic acid. Its structure is as follows : Normality = basicity x Molarity = 2 x 0.3 = 0.6
(a)
Equivalents of H 2 C 2O 4 . 2H 2 O in 10ml = Equivalents of NaOH 6.3 1, 0000 V 63 250 0.1 40 mL
15
O H
P OH OH
44.
(b) The oxidation states of various metals are : (a) Fe 3, Co 3 (b) Cr 6, Mn 7 (c) Ti 6, Mn 4
45.
(d) Co 3, Mn 6 (b) Suppose amount of NaCl in the mixture = ‘x’ g The amount of KCl in the mixture = (0.118 - x) g NaCl + AgNO3 AgCl + NaNO3 58.5 143.5
143.5 x g.........................(i ) 58.5
x
Similarly AgCl obtained from KCl =
143.5 (0.118 x ) g .........................(ii ) 74.5
But (i) + (ii) = 0.2451 g (Given) Amount of NaCl = 0.0338 g Amount of KCl = 0.0842 g 2NaCl = Na 2 O Now, 117 62 0.03380.0338 % of Na2 O
46.
47.
0.0179 10 0 3.58% ..... 0.5
(a) 5.4 0.1 278 0.150 g 1000
Weight of FeSO4 7 H 2O
Moles of Fe2. (SO4 )3 . 9 H 2O
5.35 0.0095 562
(a)
48.
0.0338 62 0.0179 g 117
Equivalent weight
Atomic Weight Valency
(b)
m
1000 M , 1000d MM s
M Molarity of solution d density of solution
M s Molar mass of solute
1000 3 3.865 1000 1.25 3 158
16
51.
(a)
52.
J Work Nm m3 2 Pa Pr essure Nm (c)
65.
m v
d
V
(d)
m 40 14.8cm3 d 2.7 Water level in cylinder = 50+ 14.8 = 64.8 mL
0.1MH 2 SO4 1000mL solution contains 0.1 mol H 2 SO4 100mL solution contains 0.01 mol H 2 SO4
mass of H 2 SO4 0.01 98 0.98 g 2 NaOH H 2 SO4 Na2 SO4 2 H 2O 2(40) g 98 g (?) 0.98 g 80 0.98 0.8 g 98
66.
(a)
Wt. of metal oxide Eq. wt. of metal Eq. wt. of oxide Wt. of metal Choride Eq. wt. of metal Eq. wt . of Choride 3 E 8 5 E 35.5
67.
E 33.25
(b)
M .F . C4 H 8O2 Molar mass 88
Vapour density
68. 69.
(b)
88 44 2
Equal volumes under identical conditions contain equal no. of molecules
(d)
N 26 O74 N1.85 O4.625 N 2O5 14
70.
(d)
a
16
F ma
f N NKg 1 m Kg 17
73.
(a) No of moles of X
a 20
No.of atoms of X
a N b ( given ) 20 20b a N
2a 40 2a No.of atoms of Y N 40 2 20b N b 40 N (a) No.of moles of Y
74.
Mol. wt. of air
75.
(b)
78 28 21 32 0.9 40 0.1 44 78 21 0.9 0.1
No.of moles of 0.05L H2SO4 = 0.5 x 0.05 = 0.025 No.of moles of 0.075L H2SO4 = 0.25 x 0.075 = 0.01875
Total no. of moles = 0.025 + 0.01875 = 0.04375
Total vol = 0.05L + 0.075L = 0.125L Molarity
79.
80.
0.04375 0.35M 0.125
77.
(c) 1 Mol M2Ox = (2 56 + 16x) gm
Now, (2 56 16 x) gm of oxide 112 gm metal 112 4 4 gm of oxide gm metal 112 16 x
But
112 4 2.8 ( given ) 112 16 x
x = 3
(c) Molecular formula of ammonium dichromate is ( NH 4 )2 Cr2O7 (b) Eq. wt of metal
wt of metal 11200 vol.of H 2 in ml displaced at STP
18
82.
(a)
2 PH 3( g ) 2 P( s ) 3H 2( g ) 2mL
3mL
100ml
(?)
100 3 150ml 2
Increase 50mL
85.
(b) 1m3 1000 L At STP, 22.4 L 1mol
1000 L
86.
1000 44.6 22.4
(c)
Ag 2 S
Ag
ore
248 g
216 g
100 g
(?) 1.0 g 1.148 g
87.
Ag 2 S 1.24 g
(?) 1.148 g 92.58 92.6 g
(a) No. of moles of Al 3
9 0.33 27
No. of Al 3 ions 0.33 6.022 1023 2 1023 Electric charge 3 1.602 1019 10 23 9.6 Coulomb
19
UNIT : 2 STATES OF MATTER Important Points The group of molecules is called matter. Matter is made up of small particles. Matter is in three states, Solid, liquid and gas. The other two states are known as Plasma and Bose, Einstein condensate. The physical state of the matter changes by changing temperature. The physical properties of a subtance are changed by changing its physical state but the chemical properties do not change, sometimes the rate of chemical reaction changes by changing the physical state. During the chemical calculation, it is most essential to have the information about the physical state of substances (reactant or product) and hence it is essential to study the physical state of matter, factors affecting and related some important laws. The deciding factors of the physical state of matter are intermolecular forces, molecular interaction and the effect of thermal energy on the motion of particles. The Dutch scientist van der Waals suggested that the weak forces of attraction exist between the molecules, which cannot be explained by any other chemical attraction is known as van der Waals 0 attractive forces. This force is universal. This force of attraction is exerted upto 4.5 A distance in substance. van der Waals forces depend upon the shape of molecules, number of electrons present in molecules, contact surface of molecules and average intermoleculer distance. The van der Waals forces of attraction are different like (i) Dispersion forces or London forces. (2) Dipole-dipole forces and (3) Dipole-induced dipole forces. Dispersion forces of attraction was first of all proposed by the German scientist Fritz London so it is known as London forces. This type of force of attraction is observed in atoms or molecules, there is a temporary dispersion in electron density that affect the electron density of nearby atom or molecule so the force of attraction is developed and so such effect is called dispersion force. The dipole-dipole forces are observed in permantently dipolar molecules. Such dipolar molecules also have interactive London forces so the cummulative effect of both forces are observed. The dipole-dipole force is stronger than London forces. The dipole-induced dipole forces are observed when dipolar molecules come near to nonpolar molecules. This type of molecules also have London forces and hence the cumulative effect of both forces are observed. The hydrogen bonding is an important intermolecular force. The first elements of groups 5, 6 and 7 due to their high electronegativity combine with hydrogen to form hydride compounds, in which hydrogen bond is observed. There also exists an intermoleculer repulsive forces; and based on that the effect of pressure on solid, liquid and gaseous state explained very easily. The most important factor which decides the physical, state of matter is the effect of thermal energy, on motion of molecules due to this motion of molecules or atoms the energy produced is called thermal energy to keep the molecules near to each other while the thermal energy has tendency to keep the molecules away from each other. By balancing combination of the two opposite factors, the physical state of matter as solid, liquid or gas is decided. Due to weak forces of attraction between molecules of gaseous state have some characteristics. The behaviour of gas is described by the quantitative relation between mass, volume, temperature and pressure and these relations are discovered by experimental observations and such relations are called laws of gases. The relation between pressure and volume of a gas was studied and it is known as Boyle’s law. At constant temperature for a fixed amount gas, pressure (P) varies inversely with its volume (V). Mathematically the Boyle’s law is written as PV = K or P1V1 = P2V2 . The equation d/P = K devised from Boyle’s law where d is the density. The Kelvin temparature is accepted as an SI unit. The relation T = (t + 273.15) K is obtained. On the basis of experimental observations a relation between absolute temperature and volume is obtained, which is known as Charles’
20
V1 V2 V = = K or . The relation between pressure and absoulte T1 T2 T temperature (T) is obtained on the basis of experimental orbservations by scientist Gay Lussac and is P1 P2 P = = K or known as Gay Lussac’s law. Mathematically it is written as . The relation between T1 T2 T volume of a gas and number of molecules was given by Avogadro, which is known as Avogadro’s law. The mathematical form of it is V = K . n. The 00C or 273 K temperature and 1 bar pressure is accepted as a standared value by SI system and hence these values are known as standard temperature and pressure (STP). 1 mole of gas at STP is having volume 22.4 litre and number of molecules equal to 6.022 × 1023 known as molar volume and Avogadro’s number respectively. Combining Boyle’s law and Charles’ P1 V1 P2 V2 PV = = K or law, the relation obtainged is known as combined gas equation. The ideal T1 T2 T gas equation, PV = nRT is also known as equation of state and R is called universal gas constant which has different values in different units. The real gas behaves as ideal gas at high temperature and low pressure and are called ideal gases. The behaviour of real gas is deviated from ideal gas and its study came from the study of effect of pressure and temperature and so the ideal gas equation is written as
law. Mathematically it is written as
æ an 2 ö ç P + 2 ÷ (V - nb) = nRT and this equation is also known as van der Waals equation. The gas can be V ø è liquefied by lowering the temperature and increasing pressure at which gas get liquified is known as critical temperature (TC) and critical pressure (PC) respectively and at critical temperature and critical pressure the volume occupied by 1 mole of gas is called critical volume (VC) and this state is called critical state. The PC, TC and VC values are constant so they are known as critical constants. The liquefication of gas is explained by isotherm. Maxwell and Boltzmann had studied the distriubution of molecules between different possible and plotted graph which is known as Maxwell’s distribution curve. The total pressure of the mixture of two or more than two gases is obtained by the Dalton’s law. Total pressure (P) = pA + pB + pC + pD .... and the partial pressure (p) is calculated from total pressure by equation p1 = X1 × Ptotal. If the % by volume is given then the partial pressure of gas is caculated using equation. Partial pressure p A =
% by volume of gas A × total pressure 100
. The Graham’s law of gaseous diffusion is
1 and using formula the ratio of rate of diffuson of NH3 and HCl gas was obtained practically as d 1.46 + .01. The application of Graham’s law of gaseous diffusion are as given in the text. The Avogadro’s hypothesis is useful to calculate the number of molecules, atoms and total number of atoms in given amount of gas. r µ
The liquid state has its physical properties like fixed volume, fluidity, nonevaporation, vapour pressure, surface tension and viseosity.
21
compressibility, diffusion,
M.C.Q. 1.
What is value of Boyle’s temperature of ethane gas when a= 5.489 dm6 atm mol-2 and b = 0.0638 dm3 atm mol-1 (a) 1048K
2.
3.
(a) Temperature of the gas
(b) Volume of the gas
(c) Number moles of the gas
(d) none of these
6.
7.
a R
a bR
(c)
2a bR
(d) none of these
(b) L mole-1
(c) atm . L mole-1
(d) atm mole-2
A gas expanse through a porons plug and exhibits cooling if its temperature is (a) More than inversion temperature
(b) Less than inversion temperature
(c) Less than critical temperature
(d) Less than Boyle’s temperature
A gas behaves like an ideal gas at (a) High pressure and low temperature
(b) High pressure and high temperature
(c) At low pressure and increasing in volume
(d) Decreasing velocity by lowering temperature
To which of the following gaseous mixture is Dalton’s law not applicable? (b) NH3 + HCl + HBr
(c) O2 + N 2 + CO2 (d) N 2 + H 2 + O2
The degree of dissociation of cl2 at 1500K is 0.45 according to the reaction Cl2 ® 2Cl assumig that both Cl2 and Chlorine atoms behave like ideal gases, calculate the density of the mixture if the pressure of the mixture is 1.5 atm (a) 0.596 g. l-1
9.
(b)
The ratio of van der Waals’ constants a and b has the dimensions of
(a) He + Ne + SO2 8.
(d)290.6K
The Boyle’s temperature for the ideal gases is given by
(a) atm mole-1 5.
(c) 209.6K
The value of universal gas constant R depends upon the
(a) 4.
(b) 104.8K
(b) 0.496 g. l-1
(c) 0.696 g. l-1
(d) 0.396 g. l-1
A gas is kept at 1 atm pressure. To compress it to 1/4th of its initial volume, the pressure to be applied is
1 atm 4 10.The density of a gas at 300K and 1 atm is d pressure remaining constant, at which of the following temperatures will its density become. 0.75 d ?
11.
(a) 1 atm
(b) 2.0 atm
(c) 4.0 atm
(a) 200 C
(b) 300 C
(c) 400K
(d)
(d) 300K
A mixture contains N2O4 and NO2 in the ratio 2 : 1 by volume. The vapour density of the mixture is (a) 45.4
(b) 49.8
(C) 32.6 22
(d) 38.3
12.
At extremely low pressure, the vander waals equation for one mole of a gas may be written as (a) PV =RT + pb
13.
15.
(c) PV = RT
a (d) ( p + )(v - b) = RT v
(b) Vm < 22.4 Litre
(c) Vm = 22.4 L
(d) Vm = 44.8 Litre
The correct order for magnitude of vanderwaals constant b should be (a) C2 H 6 < CO < CO2 < He
(b) CO < C2 H 6 < He < CO2
(c) C2 H 6 < CO2 < CO < He
(d) He < CO < CO2 < C2 H 6
The molecular radius of O2 is 2.88 x 10-10 m calculate the excluded volume per mol of O2 (a) 0.24 dm3
16.
a v
The compressibility of a gas is less than unity at STP Therefore (a) Vm > 22.4 L
14.
(b) PV = RT -
(b) 0.48 dm3
(c) 0.024 dm3
(d) 0.048 dm3
An ideal gas can not be liquefied because (a) The intermolecular force of attraction between the gaseous molecules are negligible. (b) Its critical temperature is very high (c) The vanderwaals constants a and b are very high (d) Of all of these
17.
The values of ‘a’ for the gases O2 , CO2, N2 and CH4 are respectively 1.36, 3.64, 1.39 and 2.253 L2 atm mol-1 which gas can be most easily liquefied ? (a) O2
18.
20.
21.
(c) N2
(d) CH4
(c) Twice that of He
(d) 4 times that of He
The rate of diffusion of H2 is about (a) 1 2 that of He
19.
(b) CO2
(b) 1.4 timesthat of He
Most probable speed, average speed, and RMS (root mean square speed) are related as (a) 1 : 1.224 : 1.128
(b) 1.128 : 1 :1.224
(c) 1 : 1.128 : 1.224
(d) 1.224 : 1.128 : 1.0
At room temperature the mixture of SO2 and O2 gas, compared to theO2 molecule, the SO2 molecule will hit the wall with.. (a) Smaller avarage speed
(b) Greater average speed
(c) Greater kinetic energy
(d) Greater mass
Which has maximum value of mean free path ? (a) CO2
(b) H2
23
(c) O2 (d) N2
22.
4.0 gm ideal gas is filled in a bulb having volme 10 dm3 at a constant temperature T & constant pressure P. If 0.8 gm gas is removed from the bulb to maintain the original pressure at (T + 125)K temperature, what would be the value of T for a gas having molar mass 40 gm mole-1. (a) 500K
23.
(b) 5000C
(c) 773K
(d) 7730C
What would be vlaue of ratio for RMS and average speed of gaseous molecules at a constant temperature ? (a) 1.086 : 1
(b) 1 : 1.086
(c) 2 : 1.086
(d) 1.086 : 2
Hint - 2 2 Pressure and Temperature are constant So : n1T1 = n2T2 W1 W ´ T1 = 2 T2 M1 M2
M1 = M 2
W1T1 = 3.2(T + 125) 4T = 3.2T + 400 \ 0.8T = 400 \T = 500 K
24.
If temperature T2 > T1, which graph of maxwell Botlzmamn distriburibution of molecular speed is correct ? T1 T2
T1
°
N
°
N
(a) O
O
Molecular Speed
°
N
T1
Molecular Speed T2
T2 °
N
(c)
T1
(d) O
25.
T2
(b)
O
Molecular Speed
Molecular Speed
The RMS velocity of an ideal gas at constant pressure varies with density relates as (a) d
(b)
(c) d 2
d 24
(d)
1 d
26.
The ratio of rms (root mean square ) velocities for two different gases is
V1 M2 (a) = M1 V2 27.
µ1 µ2 = (c) M M1 2
m1 m2 = (b) M M1 2
(d)
M2 M1 = m1 m2
At constant temperature, a gas is filled at 1 atm pressure in a closed container. To compress this gas to 1 th of its intial volume, the pressure to be applied is 4
(a) 28.
4 atm 3
(b) 2 atm
1 th 4 atm 4
(b) 5.06 x 105 Pa
(c) 0.505 x 103 Pa
V (L) (22.4L, 273K)
L, (38.8
(b)
) 373K
V (L)
(22.4L, 273K)
(c)
V (L)
(22.4L, 273K)
K) 73 3 , .8 L 8 3 (
T (K)
T (K) (d)
3K) 7 3 , .6L (30
V (L)
(14 .2L , 37 3K)
(22.4L, 273K)
T (K)
T (K)
Helium gas is compressed to half of the volume at 303 K. It should be heated to which temperature for its volume to increase to double of its original volume ? (a) 303K
31.
(d) 1.013 x 105 Pa
The graphs plotted V ® T for one mole of an ideal gas as follows , which graph represent its ideal behaviour at atmospheric pressure ? (a)
30.
1 (d) atm 3
What is the pressure of 380 mm Hg column of a gas in pascal ? (a) 5.05 x 104 Pa
29.
(c)
(b) 606K
(c) 1212K
(d) 300C
When a gas is heated from 298K to 323K at a constant pressure of 1 atm its volume is (a) Increases from V to 1.8 V
(b) Increases from V to 1.08 V
(c) Increases from V to 1.5 V
(d) Increases from V to 2 V
25
Hint 1
380 atm 760 \ 0.987 atm = 105 Pa 380 \ atm = (?) \ 380 mm = 5.05 ´ 104 Pa 760 380mm = 380 torr =
Hint 2 At const pressure according to charles law V1 V2 = =K T1 T2
V (L) ) 73K 3 , 6L (30.
(22.4L, 273K)
22.4 = 0.082 ............( I ) 273 30.6 and = 0.082............( II ) 373
\
T (K) (I) and (II) constant behaves the gas as an ideal Hint - 3 :
According to Charles’ law
V1 V2 = =K T1 T2
V V = 1 at T = 303 K 1 2 1
V T \T2 = 2 1 V2 = 2V1 at T2 = ? V1 2V ´303 = 1 = 303´ 4 = 1212K v1 2
Hint 4 : According to Charles’ law V1 V2 = T1 T2 V1 V = 2 298 323 323 \V = V = 1.08 V 2 298 1 1
32.
At 200 and 760 torr, the sample of air contains 20% O2 & 80% N2 gaseous mixture, find the density of the air (Molarmass O2 = 32 g/mol , N2 = 28 g/mole R =0.082 liter mole-1 K-1) (a) 1.918 gL-1
33.
(b) 2.198 gL-1
(c) 1.198 gL-1
(d) 1.394 gL-1
The ratio of velocities of diffusion of gas A and B is 1 : 4, if the ratio of their masses in a mixtare is 2:3, calculate the ratio of their mole fractions (BIT 1990) (a) 1 : 12
(b) 1 : 24
(c) 1 : 6
26
(d) 4 : 3
34.
Under same conditions of temperature and pressure the volumes of 14g N2 and 36g of O3 are related as : (a) 2VN2 = 3Vo3
35.
38.
(b) 0.2 atm
(c) 3.2 atm
(b) 3200 cal
(c) 4800 cal
(b) It contains 2- covalent bonds
(c) Molecular shape is V- shape
(d) dipole moment of H2O
The compressibility factor of an ideal gas is (b) 1
(c) 2
44.
(b) 67.7%
(c) 52.6%
(d) 25.7%
(b) P
(c) P/2
(d) 4P
The density of nitrogen is maximum at (b) 273K and 2 atm
(c) 546K and 1 atm
(d) 546 K and 2 atm
The temperature at which real gases obey the ideal gas laws over a wide range of pressure is called (a) Critical temperature
(b) Inversion of temperature
(c) Boyle’s temperature
(d) Reduced Temperature
Which of the fallowing gases with have the highest rate of diffusion (a) O2
45.
(c) 52.6g
Two gases A and B having the same temperature T, same pressure P and same volume V are mixed. If the mixture is at the same temperature T and occupied a volume V, the pressure of the mixture is ..
(a) STP 43.
(b) 0.67g
Calculate percentage of NO2 by weight in N2O4 which has vapour density of 36.
(a) 2P 42.
(d)4
A 10 L cylinder of nitrogen at 4.0 atm pressure and 270C – developed a leak. When the leak was repaired 2.36 atm of nitrogen remained in the cylinder at 270C. How many grams of nitrogen escaped?
(a) 27.7% 41.
(d) 1524 cal
(a) fewer electrons than CO2
(a) 18.7g (d) 10.0g 40.
(d) 3.02 atm
The critical temperature of H2O is higher than CO2 because (IIT 1997)
(a) zero 39.
(d) 4VN2 = 3Vo3
The kinetic energy of 4.0 moles of N2 gas at 1270C is (R =2 cal mole-1K-1) (a) 4400 cal
37.
(c) 3VN2 = 4Vo3
Equal masses of Hydrogen and oxygen gases are placed in a closed container, at a pressure of 3.4 atm. The contribution of hydrogen gas to the total pressure is (a) 1.7 atm
36.
(b) 3VN2 = 2Vo3
(b) CO2
(c) NH3
(d) N2
At a given temperature Qx= 39y and My= 2Mx whaere Q and M stand for density and molarmass respectively the ratio of Presures px/py would be (a) 1 4
(c) 6 1
(b) 4 1 27
(d) 1 6
46.
The density of methane at 2.0 atmosphere prassure at 270c is (a) 0.13 gl-1
47.
1 2
(c) 4.0 (He)
(b) 0.1
(c) 0.2
(b) 2
(c)
313 293
(b) 2
(c) 4
1 2
(b) 6.17 x 10-21 J
(c) 6.17 x 10-20J
(d) 7.17 x 10-20 J
(b) 32
(c) 24
(d) 16
(b) 8 9
(c) 1 9
(d)
16 17
The ratio between the root mean square (rms) vlocity of H2 at 50L that of at 800 K is ? (a) 2
56.
(d) none of these
Equal masses of CH4 and H2 are mixed in an empty container at 250C. The fraction of the total pressure exerted by H2 is (a)
55.
(d)0.001
The vapour density of pure ozone would be (a) 48
54.
313 293
(d)
The average kinetic energy of an ideal gas per molecule in SI unit at 250C will be (CBSE 1996) (a) 6.17 x 10-21 KJ
53.
(d) 0.4
Gas eqution PV=nRT is obeyed by (BHV 200) (a) only isothermal process (b) only adiabatic process (c) both a and c
52.
(d) 30.0(C2H6)
If V rms is 30 R 1/2 at 270C calculate the molar mass of the gas in kilogram (DPMT 2005) (a) 1
51.
(b) 32.0 (O2)
As temperature is raise from 200C to 400C, the average kinetic energy of neon atoms changes by a factor of which of the follwoing ? (a)
50.
(d) 26.0 gl-1
The compressibility factor of an ideal gas is (AIIMS - 1997, IIT91997) (a) 0
49.
(c) 1.30 gl-1
The diffusion of methane at a given temperature is twice that of gas x. The molecular weight and the gas is (a) 64.0 SO2
48.
(b) 0.26 gl-1
(b) 4
(c) 1
(d)
1 4
A real gas behave more ideally at (IIT flrelining 1993) (a) Low temperature and low presure
(b) Low temperature and high presure
(c) high temperature and low presure
(d) high temperature and high presure
28
57.
According to the kinetic theory of gases (IIT 1991) (a) The pressure exerted by a gas is proportional to mean square velocity of the molecules. (b) The pressure exerted by the gas is proportional to the root mean square velocity of the molenles. (c) The root mean square velocity is invesly proportional to the temperature (d) The mean transtational K.E of molecule is direlty propostional to the alsolute temperature
58.
The value of vander waals constant a for gses O2, N2, NH3 and CH4 are 1.36, 1.39, 4.37 and 2.253 L2 atm mol-1 respectively the gas which can liquefied most easily will be (a) O2
59.
(b) N2
(b) 0.8L
(d) 55.6L
(c) same
(d) more than double
At constant Volume for a fixed number of a moles of a gas, the pressure of the gas increases with the rise in tempreture due to (IIt 1992) (a) Increase in average molecules speed
(b) Increase rate of collisions amongst
(c) Increase in molecular attraction
(d) Increase in mean free path
Equal mass of methane and oxygen are mixed in an empty container at 2500C the fraction of the total pressure exarted by O2 is (a)
63.
(b) just half
1 2
273 (c) 1 3 ´ 298
(b) 2 3
(d) 1 3
Find the true and false statements from the following on the basis of given graph Fraction of Molecules colliding
62.
(c) 27.8L
If a volume containg gas is compressed to half, how many moles of gas remained in the vessel (a) Just double
61.
(d) CH4
A certain sample of a gas volume 0.24Liter measared at 1.0 atm pressure and 2730C its volume will be (BHU 2005) (a) 0.4L
60.
(c) NH3
0.5 0.4
Maximum probable speed
0.3 0.2 0.1 2
4 6 8 Molecular Speed cm/s
29
10
12
1.
Maxwell and Boltz mann had studied the distribution of molecules between different possible speeds
2.
This graph is known as maxwell’s distribution curve in which kinetic energy and molecular speed of gas is studdied.
3.
The fraction molecues with very high or very low speed is very high
4.
Inecresing the speed fraction also increasing Which becomes maximum and then deereases.
5.
The top portion of curve indicates maximum fraction of molewles and the speed of molecules is called most probable speed which is indicated by µ
6.
On increasing temperature the collison of molecules increases and speed of molecules decreases
(a) TTFTTF 64.
(b) TFTTFT
(c) FTTFFT
(d) TTTTFF
When a real gas behaves as an ideal gas ? (a) Inter molecular attraction among molecules are negligible then (b) At very low pressure and high temperature then (c) When molecular size is very very small and negligible to the volume of container then
(d)
In all the above condition
65.
What is the ratio of diffusion rate of
238
UF6 and
235
UF6 When these gases are diffused under the same
condition of temperature and pressure ? (a) 0.09953 66.
(b) 1.0047
(c) 1.0
(d) 10487
Molecular mass of SO2 gas is 4 times than CH4 therefore (a) Being so2 and CH4 both gases, they diffuse with same rate (b) so2 gas will diffuse 4 times factor than that of CH4 (c) Diffusion of SO2 gas is half than that of CH4 (d) CH4 gas found 4 times faster than SO2
67.
50 ml O2 gas diffused in 80 sec, What time will be required to diffuse same volume of He gas? (a) 22.89 sec
68.
(b) 3.2 gms
(c) 16 gms
(d) 1.6 gms
(b) 6.002 ´ 1023
(c) 7.52 ´ 1023
(d) 3.01 ´ 1023
Molar mass of CaCO3 = 100 gm/mole, how many moleunles are present in 20 gm (aCO3) ? (a) 12.44 ´ 1023
71.
(d) 24.29 sec
How many molecules of 502 gas will be present in 5.6 litre volume of STP ? (a) 1.5 ´ 1023
70.
(c) 92.82 sec
What is the mass (weight) of 6.022 x 1022 oxygen molcules ? (a) 32 gms
69.
(b) 28.29sec
(b) 1.2044 ´ 1023
(c) 1.2044 ´ 1025
(d) 1.2044 ´ 1022
For an ideal gas if pressure is (P), temperature (T) and gas constant is (R) then how many moles of gas 30
will be available in its 10 litre volume ? (a) 10 ´
P RT
(b)
P 1 ´ RT 10
(c) 10 ´
RT P
(d)
RT 1 ´ P 10
STP P1 = 1.5 bar
P2 = 1
V1 = 200 ml
V2 = ?
T1 = 400K
V2 = 273
PV PV 1 1 = 2 2 T1 T2 V2
PV T 1 1 ´ 2 T1 P2
1.5 ´ 200 ´ 273 400 ´ 1
72.
At constant temperature 270C and pressure, 5 litre gas is raised its temperature by 10C, What will be the change in its volume ? th
1 (a) part of its original volume will be increased. 234 th
1 (b) The volume at 300K acquired by a gas will be increased of its part 234
(c) The volume of a gas acquired at O L kmp; will be increased of its
1 234
th
th
1 (d) part of volume of a gas at O0C , will be increased. 234
73.
At 400 K temperature , pressure of 200 ml N2 gas is 1.5 bar, What is its volume at STP? (a) 204.7 ml
74.
(b) 20.47 ml
(c) 402.7 ml
What is correct prediction from the given graph ? H2 Z=
He
PV RT
CH 4
C
CO 2 1.0
31
(d) 40.27 ml
(a) H2 gas is less compressible than an ideal gas as its z > 1.0 (b) CO2 gas is more compressive than the ideal gas as Z > 1.0 (c) CH4 gas is higher compressive its Z < 1 than the ideal gas (d) a and c both 75.
It A,B,C, and D are the graphs plotted for H2, He, CH4 and CO2 Q-75 gases which graph is related for fH2 and CH4 compared to an ideal gases D B
Z=
A
PV RT
C
1.0
(a) D and A 76.
79.
80.
81.
(d) B and C
(b) liquid
(c) gas
(d) none
Which word of the following does not used for states of matter ? (a) Bose - Einstin
78.
(c) A and D
which state of matter whose the intermolecular attractive force do not exist ? (a) solid
77.
(b) A and B
(b) Boyle - Einstein
(c) plasma
(d) solid, gas, liquid
14.2 kg LPG is diffused in a gas cyllinder at 2.5 atm. If 50% lf LPG gas is used up then what will be the pressure of gas will remain in cyllinder ? 14.2 ´ 2.5 (a) 2.5 atm (b) 1.25 atm (c) 5.0 atm (d) 7.1 ´ V 7 -1 -1 . When the unit of R = 8.314 x 10 erg. mole k What are the units of pressure and volume of a gas ? (a) P - dyne cm-2, V-cm3
(b) P - paxal, V-cm3
(c) P - newton m-2, V-cm3
(d) P - atm, V-cm3
unit of R in CGS system is (a) 8.314 ´ 107 erg K -1mole -1
(b) 8.314 JK -1mole -1
(c) 0.082litre atm mole -1 K -1
(d) 1.987 cal mole -1 K -1
What is the value of gas constant R cal.mol-1K-1 ? (a) 0.082
(b) 1.987
(c) 8.314
32
(d) 8.314 x 107
82.
What is not rateted to Z ? (a) z=1 states the ideal behaviour of the gas (b) Gases which has z =1 are an ideal gases (c) z is the ratio of
PV known as compressibility factor nRT
(d) When z < 1 of z > 1, the gases convert into their liqud state 83.
Which equation is the ideal gas equation for the real gases ? (a) PV = nRT (c) ( P =
84.
(b) PV = iRT
am 2 )(V - nb) = CRT v2
86.
(b) molecular interaction
(c) effect of thermal energy on the motion of partcles
(d) Given all.
(d) 25 bar, 15 bar
Whichphysical state is accuired by water in between temperature above than 273 K and below 373 K? (b) liquid
(c) solid
(d) Gas
Which physical state of water is more compressable applying pressure at constant temperature ? (b) water
(c) Vapour
(d) Plasma
Which sabstance can be easily poored form one container into the other at room temperature ? (a) Kerosene
89.
(c) 15 bar, 25 bar
(a) inter molecular forces
(a) Ice 88
(b) 10 bar, 15 bar
Which factor is the deciding factor of physical state of matter ?
(a) plasma 87.
am 2 )(V - nb) = nRT v2
Proportion of O2,U2 and N2 gases are 3 :2: 5, and the Total pressure of this gas mixture container is 50 bar, What are the partial pressure of cl2 and N2 gases respectively? (a) 10 bar, 25 bar
85.
(d) ( P =
(b) Ice
(c) salt
(d) all
Match column I with column II column I
column II
i) Gas
(a) Easily povred form one container to the other
ii) Liquid
(b) difinite shape and fixed volume like a container is acquired
iii) Solid
(c) stats to melt at a definite temperature (d)At constant pressure the increasing in temperature raises Volume effectively
(A) i) - b, ii) - a, iii) - c
(B) i) - a, ii) - b, iii) - c
(C) i) - d, ii) - c, iii) - b
(D) i) - a, ii) - c, iii) - b 33
90.
Observation on physical state of water at 273K up to 373K are given as below, find the Correct option. (A) changing the temperature of water above 273K upto 373K, composition of water is changing gradnally. (B) When temp is changed to rise from 273K on ward the physical state of water changes fom solid - liquid to vapour state. (C) Heating water at 373K temperture, propotion of hydrogen with oxygen is changed (D) Molar mass of water decreases with changing its physical states solid - liquid gaseous an raising the temperature
91.
What is meant by Bose Einstein condensate ? [A] It is the specific state of matter [B] Showing relation E = MC2 for the matter [C] It is an electronic device developed by Bose and Einstein [D] It is an energy of radiation
92.
Which pheno menon will occur when temperature of the matter is changed ? [A] Physical state of matter changes. [B] specific arrangement of molecules in a matter changes [C] chemical properties are not chaging but density is changing [D] Given all
93.
Physical state of matter depend on... [A] Inter molecular forces which keeps moleules near to eachother [B] Thermal energy of kinetic molecules which keeps molecules away from eachother [C] By balncing of combination of two opposite factors is intermolecular forces and thermal energy decide the physical state of matter [D] Given all
94.
95.
What is negative electrical charge on F atom in HF the permanent dipole molecule ? [A] higher than 1.6 x 10-19 C
[A] less than 1.6 x 10-10 C
[C] less than 1.6 x 10-19 C
[D] higher than 1.6 x 10-10 C
State Figure showing dipole - Induced dipole forces in the following ?
(a)
d+ A
d B
+
(b)
34
+d H
d Cl
+
+ H
d Cl
(c)
+ Nonpolar Atom-I
96.
97.
+
+
+d (d) H
Nonpolar Atom-II
d O
+d H
H
d O H
HCl polar molecule comes closer to He molecule, which effect of vander waals forces will be created ? [A] Dipole - induced dipole forces
[B] Dipole induced dipole forces with London forces
[C] London forces
[D] dipole - dipole forces with London forces
Melting point of Rhomic Salphar is higher than phosphorus because ...state [A] size and number of electrons in Rhombic sulphur is more compared to phosphorus [B] sulphur S8 has metallic properties while P4 is nonmetal [C] Inter attratcion forces are lower compared to thermal energy in sulphur than that of in phosphorus [D] Given all
98.
Statement (A) : In liquid state molecules are arranged little for form each other compared to its solid state. hence the effect of pressure is observed in liquid. statement (R) : At 293 K temperature and 1000 bar pressure applied on water than the volume reduced by 4% [A] statement (A) and (R) both correct, statement (R) is the explanation to statement (A) [B] statement (A) and (R) both correct, statement (R) is not the explanation to statement (A) [C] statement (A) is correct, statement (R) is wrong [D] statement (A) is wrong, statement (R) is correct
99.
Common physical states and other two physical states of matter are.... [A] plasma, liquid, Gas are common but solid state and Bose Einstein conden sate are spcial. [B] Bose Einistein condensate and plasma are the other. rtwo states than common physical states gas, liquid and solid. [C] solid, liquid and Gas states are the only physical states : no other state is included. [D] Bose Einistein condensate and plasma are the main rules to decided the common physical states solid, liquid and Gas
100. Which statement is correct in the following. [A] Matter is existing as a individual single moleule [B] A group of matter is called molecule [C] a group of moleules is called matter [D] combination of group of different moleules form the same type of matter
35
101. Which of the following statements is false [A] Matter is made of small particles always exist in solid state [C] matter is solid state possess fixed volume with definite shape [B] matter of the same substance in liquid state has more volume compared to its solid state [D] Beyond solid, liquid and gaseous state, two other physical states are also known as plasma and Bose Einistein condensate. 102. The density of neon will be maximum at (a) NTP
(IIT 1990)
(b) O0C, 2 atm
(c) 2730C, 1 atm
(d) 2730C, 2 atm
103. Pressure of a mixture of 4 g of O2 and 2g of H2 combined in a bulb o f 1 litre at O0C is (AllMs - 2000) (a) 25.215 atm
(b) 31.205 atm
(c) 45.215 atm
(d) 15.210 atm
104. The temperature at which real gases obeys the ideal gas laws over a wide range of pressure in called. (a) critical temperature
(b) proyles temperature
(c) Inversion temperature
(d) Reduced temperature
105. A bottle of NH3 and a square of dry HCL connected through a long tube are opened simultanuulely at both ends, the white ammonia chloride ring first fromed will be (IIT - 1988) (a) At the centre of tuloe
(b) near the hydrogen chloride bottle
(c) near the NH3 bottle
(d) through out the length of the tube
106. At 1000 C and 1 atm, if the density of liquid water is 1.0 g -3 an and that of water vapour is 0.006g cm-3 , then the volume of water moleules in 1 L of steam at this temperature is (IIT - 2000) (a) 6 cm3
(b) 60 cm3
(c) 0.6 cm3
(d) 0.06 cm3
107. Two separate bulbs contains gas A and gas B the density of A is twice as that of gas B. The molecular. mass of gas A is half as that of B If two gases are at same temp, the ratio of the presure of A to that of B is 1 1 (a) 2 (b) (c) 4 (d) 2 4 108. 50 ml of a gas A diffuse throught a membrane in the same time as for this diffiusion of 40 ml of a gas B under identical pressure temperature conditions . If the molecular weight of A = 64, that of B would be (CBSC - 1992) (a) 100
(b) 250
(c) 200
(d) 80
109. Which of the following statement is false ? (BHD - 1994) (a) The product of volume pressure of fixed amound of a gas is independent of temperateure (b) Molecules of differant gases have the same K.E. at a given temperature (c) The gas equation is not valid at high pressure and low temperature (d)The gas costant per moleule is known as Boltzmann Constant. 36
110. Density ratio of O2 and H2 gas is 16 : 1 The ratio of their rms velocities will be (a) 4 : 1
(b) 1 : 16
(c) 1 : 4
(d) 16 : 1
111. at constant temperature and pressure, the rate of diffusion DA and DB of gases A and B having densities PA and PB are related by the expressions (IIt Screening - 1993) 1
é r ù2 (a) DA = ê DB A ú ë rB û
1
é r ù2 (b) DA = ê DB B ú ë rA û
1
é r ù2 (c) DA = ê DB A ú ë rB û
1
é r ù2 (d) DA = ê DB B ú ë rA û
112. If 300ml of gas at 270C is cooled to 70C at constant pressure Its final volume will be (AIIMS - 2000) (a) 135 ml
(b) 540 ml
(c) 350 ml
(d) 280 ml
113. Positive deviation from ideal gas behaviour takes place. because of PV >I nRT PV I nRT
d)
Finite size of the atoms and
PV Vy
(d) Vx = 2Vy
119. What pressure will be exerted in container X when valve -1 is opened (a) 4.1 atm
(b) 8.2 atm
(c) 2.05 atm
(d) 3.84 atm
120. Keeping valve -1 closed, on opening valve 2 between container Zand Y, on expansion of gas how much work will be done by process ? (a) 1.0 litre atm
(b) 14 - litre atm
(c) -14.0 litre atm
(d) zero
121. Opening valve 1 and 2 , on connecting all the three X,Y,Zcontainers with each other, What would be the kineti cenergy of all. gaseous molecules ? (R = 0.082 litre atm /mol.K = 8.314 J/molk) (a) 6842 J
(b) 9974 J
(c) 4988 J
(d)
3832 J
122. Connecting all the three containers by opening valves 1 and 2, if internal pressure of containers are obtained 1.0 atmosphere by lowering temperature of the system, what would be the contribution of partical pressure of H2 gas and N2 gas respectively ? (a) 0.85 atm, 0.15 atm
(b) 0.15atm, 0.85 atm
(c) 0.75 atm, 0.25 atm
(d) 0.25 atm, 0.75 atm
Paragraph - 2 The gases which do not obey general gas equation at all tempratures and pressures are called non ideal or real gases . But such gases show ideal behaviour at low pressure and high temperatures. According to vander waals, the following are two faulty assumptions in kinetic theroy of gases. (1) molecules of gas zero consideraed as point masses and the volume occupied by the gas motecules is neglihigible in comparison to the total volume of gas. 38
(2) It was also assumed that there are no intermolecular attractive forces and the molecules of gas move independently. Hence the vanderwaals equation for non- ideal (real) gases becomes a ö æ ç P + 2 ÷ ( v - b ) = RT for 1 moles and V ø è æ an 2 ö ç P + 2 ÷ ( v - nb ) = nRT for n moles V ø è
here a, and b are vanderwaals constants. a is a measure of intermolecular cular forces in a given gas.It is a measure of icompressibility volume per mole of gas. Deviation of gases from ideal behaviour is studded by plotting graph
PV ( z ) vs P here quantity z is nRT
called compressbility factor. for H2 and He gases
PV = z is always > 1. They show he deviation but for N2,O2CH4andCo2 gases nRT
z is < 1 show -ve deviation at low pressure expected them ideal behaviour value of compressbility factor Z at critiacal always < 1 and real gases show negative deviation as per vander waals equation at critical point
æ a ö ç ÷ (3b) 3 PV 27b 2 ø è c c z= = = Tc æ 8a ö 8 Rç ÷ è 27 Rb ø thus Z is less than 1 at critical point show negative deviation of real gases compare to ideal behaviour here in the above derivation at critical points
a æ 8a ö Tc = ç and Vc = 3b ÷ , Pc = 27b 2 è 27 Rb ø alter natively, constants b=
8 Pc Vc Vc , a = 3Pc Vc 2 , R = 3 3Tc
The units of a : atm L2 mole-2 b : l mole-1 question (1) 39
123. Which statements are correct in the following ? (a)
The real gases donot obey the ideal behaviour at all temperatures and pressures
(b) The gases which donot obey general gas equation are called non - ideal gases (c) Molecules were considered mass and volume less hence They donot occupy volume com pared to total volume in the derivation of vander waals equation (d) vander waals proposed the gas equation for 1 note gas is a ö æ ç P + 2 ÷ ( v - nb ) = RT n ø è
(A) a,b (D)a,d
(B) b,c
(C) c,d
124. (2) question : (2) on what bases the deviation of gases from ideal behaviour is studied ? (a) by plotting graph PV vs T
(b) by plotting graph Z vs T
(c) by plotting graph Z vs P
(d) by plotting graph
(A) a,b 125.
(B) b,c
When a graph Z plotted against P for CH
(a)
(C) c,d 4
PV vs P nRT
(D) a,d
and CO2 gases then the graph obtained as ... (b)
CH 4
CH 4
CO 2
CO 2
Z = 1.0 P
P
Z=1
(c)
(d)
CO 2 CH4
CH 4
Z = 1.0
CO 2 Z = 1.0
P
40
P
126.
What is indicated by the given graph for the behaviour of H2 and CO2 gases correctly ? H2 CO 2
P
Z = 1.0
(a) H2 has value z > 1 show positive deviation from ideal behaviour (b) Value of z =
(c) value of z =
PV for H2 gas is greater than zero but less then 1 nRt
PV for CO2 gas is greater than zero but less than 1 shows its negative deviation of its nRT
ideal behaviour (d) value of Z for H2 is greater than 1 and for co2 it is less then one hence at high pressure co2 gas is more compressible but at low pressure it less compressible than expected from ideal behaviour compressible than expected from ideal behaviour (A) a,b,c
(B) b,c,d
(C) a,c,d
(D) a,b,d
127. At critical point the value of Tc,Pc and Cc interms of vanderwaals equation are respectively. (a)
8a a , , 3b 27 Rb 27b 2
(b)
3a 8a 3b , , 2 8 Rb 27b v
(c)
8a 2 a 3a 8a 3b , , 3nb (d) , , 2 2 27 Rb 27b 8 Rb 27b v
128. Match gases under specified conditions listed in column I with their properties in column II Column I
Column II
1)
H2 (g) (P = 200 atm, T = 273 K)
a) z =
2)
H2 (g) P V= O , T =273 K)
b) Attractive forces are dominant
3)
CO2 (g) ( p = 1 atm , T= 273 K)
c) PV = nRT
4)
real gas SO2 with bigger size of its volume
d) P( V - nb) = nRT
( )
PV ¹1 RT
A) 1 - a, 2 - c, 3 - d, 4- b
B) 1 - b, 2 - d, 3 - c, 4 - a
C) 1 - a, 2 - d, 3 - b, 4 - c
D) 1 - a, 2 - c, 3 - b, 4 - d 41
129. column I 1)
column II aö æ a) ç P + 2 ÷ ( v - b ) = RT v ø è
If force of attraition among the gaseous molecules are negligible
2)
b) PV = RT -
If the volume of the gas molecules
a v
are negligible 3)
At STP
c) PV = RT + PB
4)
At low pressure and high temperature
d) PV = RT .
A) 1) - c, 2) - b, 3) - a, 4)- d
B) 1 - d, 2 - b, 3 - c, 4 - a
C) 1 - c, 2 - a, 3 - b, 4 - d
D)1 - b, 2 - a, 3 - d, 4 - c
130. for a fixed mass of a gas and constant pressure, which of the follownig graphs are releted to V µ T, the charles law ? (a)
(b)
V T
V
T(K)
T(K)
(c)
(d)
V
V
T(°C)
T(K)
(A) b,c,d
(B) a,b,c
(C) a,c,d
(D) a,b,d
131. Which of the following statements is / are correct ? (a) AT high prssure, all real gases are less compressible than ideal gas. (b) H2 he gases are more compressible than ideal gas for all values of pressure æ ö (c) compressiblity factor z = ç nRT ÷ is less than 1 for all real gases at low pressure except H2 and He è ø PV
(d) The compressibility factor z of real gases are quite independent of temperature therefore z is not effected with change in temperature. (A) a,c
(B) b,c
(C) c,d
(D) a,d
Passage A gas Undergoes dissociation as M 4( q ) ® 4 M ( g ) in a closed rigid container having volume 22.4L at 273K If the initialmoles of M4 taken befor dissoliciation is 1 then.
42
132. The total pressure (in attm) after 50% completion of the reaction assuning ideal behaviour is (a) 0.5
(b) 2.5
(c) 2.8
(d) 3.8
133. If the gases are not ideal at the begining total pressure observed is less than 1 atm then (a) z =
PV of M 4 > 1 RT
(b) z =
PV of M 4 < 1 RT
(c) z =
PV of M 4 = 1 RT
(d) z =
PV of M > 1 RT
134. If the gases are not ideal and after 100 % dissociation total pressure is greater than 4 atm, then (a) Compressing of M(q) is easiq then an ideal gas (b) The compression of M(q) is difficult than an ideal gas (c) The compression of m(g) is the same as an ideal gas (d) A gas is non compressible
Answer Key
1 8 15 22 29 36 43 50 57 64 71 78 85 92 99 106 113 120 127
a d a a b c c d d d a a d d b b a d a
2 9 16 23 30 37 44 51 58 65 72 79 86 93 100 107 114 121 128
d c a a c d c c c b d a b d c c c b c
3 10 17 24 31 38 45 52 59 66 73 80 87 94 101 108 115 122 129
b c b b b b c b a c a a c c a c c c a
4 11 18 25 32 39 46 53 60 67 74 81 88 95 102 109 116 123 130
c d b d c a c c a b d b a a b a a a c
43
5 12 19 26 33 40 47 54 61 68 75 82 89 96 103 110 117 124 131
b b c b b a a b a b c d a b a c d c a
6 13 20 27 34 41 48 55 62 69 76 83 90 97 104 111 118 125 132
a b d d b a b c d a c d b a b d c d b
7 14 21 28 35 42 49 56 63 70 77 84 91 98 105 112 119 126 133 134
b d b a c b c c a b b c a a b d a c b b
UNIT : 3 STRUCTURE OF ATOM Important Points Composition of atom Electron (–1 eo ) (1) It was discovered by J.J. Thomson (1897) and is negatively charged particle. (2) Electron is a component particle of cathode rays. (3) Cathode rays were discovered by William Crooke’s & J.J. Thomson (1880). Properties of Cathode rays (i) Cathode rays travel in straight line. (ii) Cathode rays produce mechanical effect, as they can rotate the wheel placed in their path. (iii) Cathode rays consist of negatively charged particles known as electron. (iv)
When cathode rays fall on solids such as
Cu, X rays
are produced.
(v)
The nature of these rays does not depend upon the nature of gas or the cathode material used in discharge tube.
(vi)
The e/m (charge to mass ratio) for cathode rays was found to be the same as that for an e (1.76 10 8 coloumb per gm). Thus, the cathode rays are a stream of electrons.
Proton ( H+, p) (1) Proton was discovered by Goldstein (2) It is a component particle of anode rays. Goldstein (1886) used perforated cathode in the discharge tube and repeated Thomson’s experiment and observed the formation of anode rays. These rays also termed as positive or canal rays. Properties of anode rays (i) Anode rays travel in straight line. (ii) Anode rays are material particles. (iii) Anode rays are positively charged. (iv) Anode rays may get deflected by external magnetic field. (v) Anode rays also affect the photographic plate. (vi) The e/m ratio of these rays is smaller than that of electrons. (vii) Unlike cathode rays, their e/m value is dependent upon the nature of the gas taken in the tube. It is maximum when gas present in the tube is hydrogen. Neutron (on1, n) (1) Neutron was discovered by James Chadwick (1932) according to the following nuclear reaction, 4 Be 9 2 He 4 6 C 12 o n1 or 5 B 11 2 He 4 7 N 14 o n1 (2)
Neutron is an unstable particle. It decays as follows, 1 1 0 0 n 1 H 1 e neutron Proton electon
0 0 antinutrino
44
Name of constant Mass (m)
Charge(e) Specific charge (e/m) Density
Unit Amu Kg Relative
Electron(e–) 0.000546 9.109 × 10–31 1/1837
Proton(p+) 1.00728 1.673 × 10–27 1
Neutron(n) 1.00899 1.675 × 10–27 1
Coulomb (C) Esu Relative C/g Gram / cc
– 1.602 × 10–19 – 4.8 × 10–10 –1 1.76 × 108
+1.602 × 10–19 +4.8 × 10–10 +1 9.58 × 104
Zero Zero Zero Zero
2.17 10
17
1.114 10
14
1.5 10
14
· The atomic mass unit (amu) is 1/12 of the mass of an individual atom of 6 C 12 , i.e. 1.660 10 27 kg . Other non fundamental particles Particle
Symbol
Nature
Positron
e ,1e0 ,
+
Charge esu 10–10 + 4.8029
Neutrino
0
0
Mass (amu)
Discovered by
Anderson (1932)
Anti-proton
p
–
– 4.8029
0.00054 86 < 0.00002 1.00787
Positive mu meson Negative mu meson Positive pi meson Negative pi meson Neutral pi meson
+
+ 4.8029
0.1152
Chamberlain Sugri (1956) and Weighland (1955) Yukawa (1935)
–
– 4.8029
0.1152
Anderson (1937)
+
+ 4.8029
0.1514
–
– 4.8029
0.1514
0
0
0
0.1454
Pauli (1933) and Fermi (1934)
Powell (1947)
Atomic number, Mass number and Atomic species (1)
Atomic number or Nuclear charge (i)
The number of protons present in the nucleus of the atom is called atomic number (Z).
(ii)
It was determined by Moseley as, Where,
X ray’s
a (Z b) or aZ ab
frequency
s 1
Z= atomic number of the metal a & b are constant. (2)
Mass number Mass number (A) = Z + n
45
Z
Different types of atomic species Atomic Similarities species Isotopes (i) Atomic No. (Z) (Soddy) (ii) No. of protons (iii) No. of electrons (iv) Electronic configuration (v) Chemical properties (vi) Position in the periodic table Isobars (i) Mass No. (A) (ii) No. of nucleons
Isotones
Isoelectronic species
No. of neutrons
(i) No. of electrons (ii) Electronic configuration
Differences
Examples
(i) Mass No. (A) (ii) No. of neutrons (iii) Physical properties
1 2 3 (i) 1 H , 1 H , 1 H
16 8
(ii) (iii)
(i) Atomic No. (Z) (ii) No. of protons, electrons and neutrons (iii)Electronic configuration (iv) Chemical properties (v) Position in the perodic table. (i) Atomic No. (ii) Mass No., protons and electrons. (iii) Electronic configuration (iv) Physical and chemical properties (v) Position in the periodic table. At. No., mass No.
(i) (ii)
(i)
18 O, 17 8 O, 8 O
35 17
40 18
40 40 Ar , 19 K , 20 Ca
130 52
30 14
37 Cl , 17 Cl
130 Te, 130 54 Xe, 56 Ba
31 32 Si, 15 P, 16 S
39 40 (ii) 19 K , 20Ca 3 H , 4 He (iii) 1 2
13 14 (iv) 6 C , 7 N
(i)
N 2O, CO2 , CNO (22e )
(ii)
CO, CN , N2 (14e )
2
(iii) H , He, Li , Be (2e ) 3 2 2 (iv) P , S , Cl , Ar , K and Ca (18e )
Electromagnetic radiations (1) Light and other forms of radiant energy propagate without any medium in the space in the form of waves are known as electromagnetic radiations. These waves can be produced by a charged body moving in a magnetic field or a magnet in a electric field. e.g. rays, rays, cosmic rays, ordinary light rays etc. (2) Characteristics (i) All electromagnetic radiations travel with the velocity of light. (ii) These consist of electric and magnetic fields components that oscillate in directions perpendicular to each other and perpendicular to the direction in which the wave is travelling.
46
(3)
A wave is always characterized by the following five characteristics, Crest
Wavelength
Vibrating source Energy
Trough
(i)
Fig. 2.2
Wavelength : The distance between two nearest crests or nearest troughs is called the w avelength. I t is denoted by (lambda) and is measured is terms of centimeter(cm), angstrom(Å), micron( ) or nanometre (nm). 1 Å 10 8 cm 10 10 m ; 1 10 4 cm 10 6 m ; 8 4 7 1nm 10 7 cm 10 9 m ; 1cm 10 Å 10 10 nm
(ii)
Frequency : It is defined as the number of waves which pass through a point in one second. It is denoted by the symbol (nu) distance travelled in one second = velocity =c
(iii)
c
Velocity : It is defined as the distance covered in one second by the wave. It is denoted by the letter ‘c’. All electromagnetic waves travel with the same velocity, i.e., 3 1010 cm / sec . c 3 1010 cm / sec
(iv)
Wave number : This is the reciprocal of wavelength, i.e., the number of wavelengths per centimetre. It is denoted by the symbol (nu bar). It is expressed in cm 1 or m 1 .
1
(v)
Amplitude : It is defined as the height of the crest or depth of the trough of a wave. It is denoted by the letter ‘A’. It determines the intensity of the radiation. The arrangement of various types of electromagnetic radiations in the order of their increasing or decreasing wavelengths or frequencies is known as electromagnetic spectrum. Atomic spectrum - Hydrogen spectrum Atomic spectrum Spectrum is the impression produced on a photographic film when the radiation of particular wavelength is (are) analysed through a prism or diffraction grating. Types of spectrum (1) Emission spectrum: Spectrum produced by the emitted radiation is known as emission spectrum. This spectrum corresponds to the radiation emitted (energy evolved) when an excited electron returns back to the ground state. (i) Continuous spectrum: When sunlight is passed through a prism, it gets dispersed into continuous bands of different colours. If the light of an incandescent object resolved through prism or spectroscope, it also gives continuous spectrum of colours.
47
(ii)
Line spectrum: If the radiation’s obtained by the excitation of a substance are analysed with help of a spectroscope a series of thin bright lines of specific colours are obtained. There is dark space in between two consecutive lines. This type of spectrum is called line spectrum or atomic spectrum.. (2) Absorption spectrum : Spectrum produced by the absorbed radiations is called absorption spectrum. Hydrogen spectrum (1) All these lines of H-spectrum have Lyman, Balmer, Paschen, Barckett, Pfund and Humphrey series. These spectral series were named by the name of scientist discovered them. (2) To evaluate wavelength of various H-lines Ritz introduced the following expression,
1 1 1 R 2 2 z2 c n1 n2
Where R is universal constant known as Rydberg’s constant its value is 109, 678 cm 1 . Plum pudding model of Thomson (1) He suggected that atom is a positively charged sphere having electrons embedded uniformly giving an overall picture of plum pudding. (2) This model failed to explain the line spectrum of an element and the scattering experiment of Rutherford. Rutherford’s nuclear model From the observations of ? ray scattering experiments he concluded that, an atom consists of (i) Nucleus which is small in size but carries the entire mass i.e. contains all the neutrons and protons. (ii) Extra nuclear part which contains electrons. This model was similar to the solar system. (3) Properties of the nucleus (i) Nucleus is a small, heavy, positively charged portion of the atom and located at the centre of the atom. (ii) All the positive charge of atom (i.e. protons) are present in nucleus. (iii) Nucleus contains neutrons and protons, and hence these particles collectively are also referred to as nucleons. (iv) The size of nucleus is measured in Fermi (1 Fermi = 10–13 cm). (v) The radius of nucleus is of the order of 1.5 10 13 cm. to 6.5 10 13 cm. i.e. 1.5 to 6.5 Fermi. Generally the radius of the nucleus ( rn ) is given by the following relation, rn ro ( 1 . 4 10
13
cm ) A 1 / 3
This exhibited that nucleus is 10 5 times small in size as compared to the total size of atom. (vi)
The Volume of the nucleus is about 10 39 cm 3 and that of atom is 10 24 cm 3 , i.e., volume of the nucleus is 10 15 times that of an atom.
(vii) The density of the nucleus is of the order of 10 15 g cm 3 or nucleus is spherical than,
mass number mass of the nucleus 6.023 1023 4 r 3 Density = volume of the nucleus 3 48
10 8
tonnes
cm 3
or
10 12 kg / cc . If
Planck’s quantum theory To explain black body irradiation, Max Planck put forward a theory known as Planck’s quantum theory. (i) The radiant energy which is emitted or absorbed by the black body is not continuous but discontinuous in the form of small discrete packets of energy, each such packet of energy is called a ‘quantum’. In case of light, the quantum of energy is called a ‘photon’. (ii) The energy of each quantum is directly proportional to the frequency ( ) of the radiation, i.e. E
or
E hv
hc
Where, h Planck’s constant = 6.62×10–27 erg. sec. or
6 .62 10 34 Joules sec .
(iii)
The total amount of energy emitted or absorbed by a body will be some whole number quanta. Hence E nh , where n is an integer.. Photoelectric effect (1)
(2)
(3) (4) (5)
When radiations with certain minimum frequency ( 0 ) strike the surface of a metal, the electrons are ejected from the surface of the metal. This phenomenon is called photoelectric effect and the electrons emitted are called photo-electrons. The current constituted by photoelectrons is known as photoelectric current. This minimum frequency is known as threshold frequency. The electrons are ejected only if the radiation striking the surface of the metal has at least a minimum frequency called Threshold frequency. The minimum potential at which the plate photoelectric current becomes zero is called stopping potential. The velocity or kinetic energy of the electron ejected depend upon the frequency of the incident radiation and is independent of its intensity. The number of photoelectrons ejected is proportional to the intensity of incident radiation. Einstein’s photoelectric effect equation According to Einstein, Maximum kinetic energy of the ejected electron = absorbed energy – threshold energy 1 1 1 2 mv max h h 0 hc 2 0
Where, 0 and 0 are threshold frequency and threshold wavelength.
Advantages of Bohr’s theory (i) Calculation of radius of Bohr’s orbit : According to Bohr, radius of nth orbit in which electron moves is
n2 h2 rn 2 2 . 4 me k Z
Where, Orbit number, Mass number Charge on the electron Atomic number of element, k = Coulombic constant Where, n = Orbit number, m = Mass number 9. 1 10 31 kg , e = Charge on the electron 1.6 10 19 Z= Atomic number of element, k = Coulombic constant 9 109 Nm 2c 2 After putting the values of m,e,k,h, we get
rn
49
n2 0.529 Å Z
(iii)
Calculation of velocity of electron Vn
(iv)
Ze 2 2e 2 ZK , Vn nh mr
1/ 2
;
Vn
2.188 10 8 Z cm. sec 1 n
Calculation of energy of electron in Bohr’s orbit Total energy of electron = K.E. + P.E. of electron Substituting of r, gives us
E
2 2 mZ 2e 4 k 2 n2 h2
kZe 2 kZe 2 kZe 2 2r r 2r
Where, n=1, 2, 3……….
Putting the value of m, e, k, h, we get E 21.8 10 12 21. 8 10 19
13.6
Z2 n2
Z2 erg per atom n2
Z2 2
n
J per atom (1J 107 erg)
k.cal / mole
(1 cal = 4.18J)
E 13.6
or
1312 n2
Z2 n
2
eV per atom(1eV 1.6 10-1 9 J )
Z 2kJmol 1
When an electron jumps from an outer orbit (higher energy) to an inner orbit (lower energy)then the energy emitted in form of radiation is given by E En2 En1
2 2k 2 me 4 Z 2 h2
1 1 n2 n 2 2 1
1 1 E 13.6 Z 2 2 2 eV / atom n1 n2
As we know that
E h , c
This can be represented as Where, (6)
R
2 2k 2 me 4 ch
3
and
E 1 2 2 k 2me 4 Z 2 1 1 , 2 3 2 hc ch n1 n2
1 1 1 RZ 2 2 2 n1 n2
; R is known as Rydberg constant. Its value to be used is 109678 cm-1.
Spectral evidence for quantisation (Explanation for hydrogen spectrum on the basis of bohr atomic model) (i) The optical spectrum of hydrogen consists of several series of lines called Lyman, Balmar, Paschen, Brackett, Pfund and Humphrey. (ii) To evaluate wavelength of various H-lines Ritz introduced the following expression,
1 1 1 R 2 2 c n1 n2
Where, R is =
2 2me 4 ch3
Rydberg’s constant
It’s theoritical value = 109,737 cm–1 and It’s experimental value = 109,677. 581cm 1 This remarkable agreement between the theoretical and experimental value was great achievment of the Bohr model. 50
(iii) Comparative study of important spectral series of Hydrogen is shown in following table. S.No. Spectral series (1)
(2)
(3)
(4)
(5)
(6)
(iv)
Transition
Lymen series
Lies in the region Ultraviole t region
Balmer series
Visible region
n1 2
Paschen series
Infra red region
n1 = 3
Brackett series
Infra red region
n1 4
Pfund series
n 2n 2 2 1 22 (n2 n1 ) R
n2 > n1
l max
n1 1
n1 1 and n2 2
n 2 2, 3, 4.... ¥
n 2 3, 4,5.... ¥
n 2 4,5,6.... ¥
4 3R
l max
1 R
lmin
n1 2 and n2 ¥
36 5R n1 3 and n2 4
4 R n1 3 and n2 ¥
l max
l max
144 7R
lmax
16 25 9R
lmin
9 R
lmin
l min
16 R
n1 5
n1 5 and n 2 6
n1 5 and n2 ¥
n 2 6,7,8.... ¥
Humphrey Far series infrared region
n1 6
25 36 l max 11R n1 6 and n 2 7
25 R n1 6 and n2 ¥
l max
36 49 13R
l min
n 22 n22 n12
4 3
9 5 16 7
n1 4 and n2 ¥
Infra red region
n 2 7,8.... ¥
l max
n1 1 and n2 ¥
n1 2 and n2 3
n1 4 and n2 5
n 2 5, 6,7.... ¥
n12 R
lmin
lmin
lmin
36 R
25 9
36 11 49 13
If an electron from nth excited state comes to various energy states, the maximum spectral lines obtained will be
n(n 1) . 2
n principal quantum number..
As n 6 than total number of spectral lines
6(6 1) 30 15. 2 2
Bohr–Sommerfeild’s model It is an extension of Bohr’s model. The electrons in an atom revolve around the nuclei in elliptical orbit. The circular path is a special case of ellipse. Association of elliptical orbits with circular orbit explains the fine line spectrum of atoms. Dual nature of electron (1)
I n 1924, the French physicist, Louis de Broglie suggested that if light has both particle and wave
like nature, the similar duality must be true for matter. Thus an electron, behaves both as a material particle and as a wave. (2)
According to de-broglie, the wavelength associated with a particle of mass m, moving with velocity v is given by the relation
h , mv
where h Planck’s constant.
51
(3)
This was experimentally verified by Davisson and Germer by observing diffraction effects with an electron beam. Let the electron is accelerated with a potential of V than the Kinetic energy is h h 1 mv 2 eV ; m 2 v 2 2eVm mv 2eVm P ; 2k .E m 2eVm 2
(4)
If Bohr’s theory is associated with de-Broglie’s equation then wave length of an electron can be determined in bohr’s orbit and relate it with circumference and multiply with a whole number 2 r n or
Thus (5)
2 r h From de-Broglie equation, . n mv
h 2 r nh or mvr mv n 2
The de-Broglie equation is applicable to all material objects but it has significance only in case of microscopic particles.
Heisenberg’s uncertainty principle This principle states “It is impossible to specify at any given moment both the position and momentum (velocity) of an electron”. h Mathematically it is represented as , x . p 4 Where x uncertainty is position of the particle, p uncertainty in the momentum of the particle Now since p m v So equation becomes, x. mv
h 4
In terms of uncertainty in energy, E . t
or
x v
h 4 m
E and uncertainty in time t, this principle is written as,
h 4
Schrödinger wave equation (1)
Schrodinger wave equation is given by Erwin Schrödinger in 1926 and based on dual nature of electron.
2 2 2 8 2 m 2 2 (E V ) 0 The Schrodinger wave equation is, x 2 y 2 z h Where x, y and z are the 3 space co-ordinates, m mass of electron, h Planck’s constant, E Total energy, V potential energy of electron, amplitude of wave also called as wave function, 2 is mathematical operation to be performed on Y x 2
(2)
The Schrodinger wave equation can also be written as, 8 2 m 2 ( E V ) 0 Where laplacian operator.. h2 52
(3)
Physical significance of and 2 (i)
The wave function represents the amplitude of the electron wave. The amplitude is thus a function of space co-ordinates and time i.e. (x, y, z......times)
(ii)
For a single particle, the square of the wave function 2 at any point is proportional to the probability of finding the particle at that point.
(iii)
If 2 is maximum than probability of finding e– is maximum around nucleus and the place where probability of finding e– is maximum is called electron density, electron cloud or an atomic orbital. It is different from the Bohr’s orbit.
(iv)
The solution of this equation provides a set of number called quantum numbers which describe specific or definite energy state of the electron in atom and information about the shapes and orientations of the most probable distribution of electrons around the nucleus.
It was Erwin Schrodinger who developed a new model of atom in 1920.He incorporated the idea of quantisation, and the conclusions of de-Broglie’s principle and Heisenberg’s principle in his model. In this model, the behaviour of the electron in an atom is described by the mathematical equation
2 2 2 8 m + 2 (E V) known as Schrodinger Wave Equation. ) x 2 y 2 z 2 h (Here x, y and z are three space coordinates, m mass of electron, h Planck’s constant, E Total energy, V Potential energy and 1/J Wave function of electron wave) The above expression can also be expressed as 2
8 m (E V) h2
The permitted solutions of Schrodinger equation are known as wave functions which correspond to a definite energy state called orbital. Thus, the discrete Bohr orbits are replaced by orbital’s i.e., three dimensional geometrical olumes where there is maximum probability of locating the electrons. In simple words, the equation may be interpreted by stating that a body/particle of mass m, potential energy E, velocity v, has wave like characteristics associated with it, with amplitude given by wave function. Probability Distribution In wave mechanics a moving electron is represented by wave function, j. It has on physical significance and refers to the amplitude of electron wave. However, j2 is a significant term and give intensity of electrons. An atomic orbital is a region around the nucleus where there is more probability of intensity of electrons. An orbital is considered as a diffused electron cloud having more density close to the nucleus. The probability of finding an electron in a given volume is understood best in the form of radial probability distribution curves. The probability curves for some orbital’s are given in the figure. The distance of maximum radial probability is radius of an atom. There are two humps for 2p-orbital which means that the 2s electron penetrates a little closer to nucleus. The point at which radial probability becomes zero is known as nodal point. 53
RADIAL PROVABIL ITY
RADIAL PRO VABIL ITY
RADIAL PRO VABILITY
The radial probability plots for some orbitals are shown in the given figure.
DISTANCE
1s
DISTANC E
2s
DISTANCE
2p
ORBITAL WAVE FUNCTIONS AND SHAPES OF ORBITALS According to wave mechanics, atomic orbitals are described bywave functions known as orbital wave functions. These orbital wave functions can be represented by the product of two wave functions, (i) radial wave function and (ii) angular wave function. The radial wave function depends upon distance ‘r’ from the nucleus. On the other hand, angular wave function depends upon the direction given by the angles with respect to different co-ordinate axis. It is found that the wave function for s-orbital is independent of angles and, therefore, s-orbitals do not have angular dependence. Thus, all s-orbitals are spherically symmetrical. However, all other types of orbitals (p, d or f) have angular dependence and, therefore, have directional dependence. Radial Probability Distribution Curves If we draw a graph between radial wave function, R (radial part of wave functionj) and r (distance from nucleus), we obtain graphs as shown below. These graphs are for ls, 2s and 2p-orbitals of hydrogen atom. This type of dependence is known as radial dependence. These plots show radial dependence on only one side of the nucleus. These plots do not have any direct physical significance, but are useful in molecular structure because atomic wave functions are’ needed to build molecular wave functions. It is clear from the graph, that in ls radial wave function, j, is positive everywhere, but for other s orbitals i.e., for 2s or 3sorbitals it is positive in some regions and negative in others. It may be noted that +ve and - ve signs have only relative significance. During superposition (in the formation of molecular orbitals) relative signs play a very important part.
We know that square of the wave function*, R2, represents the probability of finding the electron in a unit volume i.e., probability density. The graphs between square of the radial wave function R2 and r (distance from nucleus) are known as radial probability density graphs. These graphs differ slightly from the earlier graphs as R2 is positive throughout.
54
(“In case R is not real, IRI2 can be taken in place of R2. In such cases IRI2 R. .)
R
2
R
2
R
1s
2
2s
r
2p r
r 2
Graph between R (radialprobabUity density) and r 2
2
2
R dv
R dv
R dv
2s
1s r
r
2p r
Graph between radial probability function R2 dV (or 4pr2R and r2) As R2 represents the probability density, i.e., probability of finding the electron in a unit volume, R2dV gives the probability of finding the given electron in a volume dV. The product R2dV is also known as radial probability distribution function. The graph of R2dV versus r (distance from nucleus) is known as radial probability distribution function graphs. If we observe the radial probability distributive graph of 1S, we find it is quite different from the radial probability density graph. The radial probability density is maximum close to the nucleus, but the radial probability is least. This is due to the fact that volume of the spherical shell (dV) near the nucleus is very small resulting in a small value of radial probability R2dV. At the nucleus (i.e., r 0), dV is zero, hence R2dV as also zero, although R2 is very large at this point. As the distance from nucleus (r) increases, the volume of the shell dV (4 r2dr) increases while R2 decreases. As a result, the radial probability keeps on increasing gradually and reaches a maximum value and then decreases gradually. The maximum in the curve indicates the most probable value and the corresponding distance from the nucleus is called distance of maximum probability (r0). For hydrogen atom in ground state, this has a value of 53 pm. It is important to note that Bohr predicted that the electron will always be at a distance of 53pm from the nucleus for hydrogen atom in ground state. However, according to wave mechanical model the electron is most likely to be found at this distance and there is probability of finding the electron at distances both shorter and longer than this. The radial probability distribution curve for 2s-orbital (n 2, l 0) shows two maxima, a smaller one nearer the nucleus and a bigger one at a larger distance. Comparing the location of the maxima in the 2s orbital, we conclude that an electron in the 2s-orbital has greater probability to stay further away than an electron in the 1s orbital. The radial probability distribution curves of three 2p-orbitals (n 2, l 1) are identical. It shows only one maximum. The distance of maximum probability for a 2p-orbital is slightly less than that for a 2s-orbital. However, in contrast to the curve for 2p-orbital, there is a small additional maximum in the curve for 2sorbital. In other words, 2s-orbital penetrates closer to the nucleus, than a 2p-orbital. Thus, an electron in 2s-orbital has a lower energy than an electron in a 2p-orbital.
55
Some Note worth points 1. 2.
The radius of maximum probability of 1s electron is 0.529Å. The number of regions of maximum probability for 1s, 2p, 3d and 4f is one. For 2s, 3p, 4d and 5f these are two and so on. 3. The small humps indicate that the electron has a tendency to penetrate closer to the nucles. 4. In between the regions of zero electron density called node. More is the number nodes more is the energy of an orbital. 5. In these curves, the first orbital of cash type (1s, 2p, 3d, 4f) has one region of maximum probability and no node. Whereas the first orbital of each type (2s, 3p, 4d, 5f) has two regions of maximum probability and one node so on. Quantum numbers Each orbital in an atom is specified by a set of three quantum numbers (n, l, m) and each electron is designated by a set of four quantum numbers (n, l, m and s). (1) Principle quantum number (n) (i) The maximum number of an electron in an orbit represented by this quantum number as 2n 2 . (ii) It gives the information of orbit K, L, M, N——————. (iii) Angular momentum can also be calculated using principle quantum number (2)
Azimuthal quantum number ( ) (i)
Azimuthal quantum number is also known as angular quantum number. Proposed by Sommerfield and denoted by ‘ ’.
(ii)
It determines the number of sub shells or sublevels to which the electron belongs. Value of
0
1
2
3
Name of subshell
s
p
d
f
Shape of subshell
Spherical
Dumbbell
Double dumbbell
Complex
(iii) (iv)
It tells about the shape of subshells. It also expresses the energies of subshells
(v)
The value of l (n 1) always.
spd f
(increasing energy).
(vii) It represent the orbital angular momentum. Which is equal to
h l (l 1) 2
(viii) The maximum number of electrons in subshell 2(2l 1) (3)
Magnetic quantum number (m) (i) It was proposed by Zeeman and denoted by ‘m’. (ii) It gives the number of permitted orientation of subshells. (iii)
The value of m varies from – to + through zero.
(iv)
It tells about the splitting of spectral lines in the magnetic field i.e. this quantum number proves the Zeeman effect. For a given value of ‘n’ the total value of ’m’ is equal to n2 .
(v)
56
(vi) For a given value of ‘l’ the total value of ‘m’ is equal to (2l 1). (vii) Degenerate orbitals : Orbitals having the same energy are known as degenerate orbitals. e.g. for p subshell p x py pz (4)
Spin quantum numbers (s) (i) It was proposed by Goldshmidt & Ulen Back and denoted by the symbol of ‘s’. (ii)
The value of ' s' is 1/2 and - 1/2, which signifies the spin or rotation or direction of electron on it’s axis during movement. (iii) The spin may be clockwise or anticlockwise. h s (s 1). (iv) It represents the value of spin angular momentum is equal to 2 1 (v) Maximum spin of an atom (spin multiplicity) 2 number of unpaired electrons (vi) This quantum number is not the result of solution of Schrodinger equation as solved for Hatom. Distribution of electrons among the quantum levels
0 0
Designation of orbitals 1s 2s
Number of Orbitals in the subshell 1 1
1 0 1
–1, 0, +1 0 –1, 0, +1
2p 3s 3p
3 1 3
3
2
–2, –1, 0, +1, +2
3d
5
4
0
0
4s
1
4
1
–1, 0, +1
4p
3
4
2
–2, –1, 0, +1, +2
4d
5
4
3
–3, –2, –1, 0, +1, +2, +3
4f
7
n
l
m
1 2
0 0
2 3 3
Shape of orbitals (1)
Shape of ‘s’ orbital (i)
For ‘s’ orbital l=0 & m=0 so ‘s’ orbital have only one unidirectional orientation i.e. the probability of finding the electrons is same in all directions.
(ii)
The size and energy of ‘s’ orbital with increasing ‘n’ will be 1s 2s 3s 4 s.
1S
(iii)
2S
s-orbitals known as radial node or nodal surface. But there is no radial node for 1s orbital since it is starting from the nucleus.
57
(2)
(3)
Shape of ‘p’ orbitals (i) For ‘p’ orbital l=1, & m=+1,0,–1 means there are three ‘p’ orbitals, which is symbolised as p x , p y , p z . (ii)
Shape of ‘p’ orbital is dumb bell in which the two lobes on opposite side separated by the nodal plane.
(iii)
p-orbital has directional properties.
Shape of ‘d’ orbital (i)
For the ‘d’ orbital l =2 then the values of ‘m’ are –2, –1, 0, +1, +2. It shows that the ‘d’ orbitals has five orbitals as
(4)
d xy , d yz , d zx , d x 2 y 2 , d z 2 .
(ii)
Each ‘d’ orbital identical in shape, size and energy.
(iii)
The shape of d orbital is double dumb bell .
(iv)
It has directional properties
Shape of ‘f’ orbital (i)
For the ‘f’ orbital l=3 then the values of ‘m’ are –3, –2, –1,0,+1,+2,+3. It shows that the ‘f’ orbitals have seven orientation as f x ( x2 y 2 ) , f y ( x2 y 2 ) , f z ( x2 y2 ), f xyz , f z 3 , f yz 2 and f xz 2 .
(ii)
The ‘f ‘ orbital is complicated in shape.
Rules for filling of electrons in various orbitals The atom is built up by filling electrons in various orbitals according to the following rules, (1)
Aufbau’s principle This principle states that the electrons are added one by one to the various orbitals in order of their increasing energy starting with the orbital of lowest energy. The increasing order of energy of various orbitals is 1s 2s 2p 3s 3 p 4 s 3d 4 p 5s 4d 5 p 6s 4 f 5d 6 p 7 s 5 f 6d 7 p......... 58
(2)
(n+ ) Rule In neutral isolated atom, the lower the value of (n + ) for an orbital, lower is its energy. However,, if the two different types of orbitals have the same value of (n + ), the orbitals with lower value of has lower energy.
(3)
Pauli’s exclusion principle According to this principle “no two electrons in an atom will have same value of all the four quantum numbers”.
(4)
Hund’s Rule of maximum multiplicity “Electron pairing in and orbitals cannot occur until each orbitals of a given subshell contains one electron each or is singly occupied”. As we now know the Hund’s rule, let us see how the three electrons are arranged in orbitals. The important point ot be remembered is that all the singly occupied orbitals should have electrons with parallel spins i.e in the same direction either-clockwise or anticlockwise . 2px
2py 2pz
2px
2py 2pz
or
Electronic configurations of elements On the basis of the electronic configuration principles the electronic configuration of various elements are given in the following table : The above method of writing the electronic configurations is quite cumbersome. Hence, usually the electronic configuration of the atom of any element is simply represented by the notation.
number of principal
nlx
Number of electrons
symbol of subshell
Some Unexpected Electronic Configuration Some of the exceptions are important though, because they occur with common elements, notably chromium and copper. has 29 electrons. Its excepted electronic configuration is 1s 2 2s 2 2 p 6 3s 2 3 p 6 4 s 2 3d 9 but in reality the configuration is 1s 2 2s 2 2 p6 3s 2 3 p 6 4 s 1 3d10 as this configuration is more stable. Similarly Cr has the configuration of 1s 2 2s 2 sp 6 3s 2 3 p 6 4 s 1 3d 5 instead of 1s 2 2s 2 2 p 6 3s 2 3 p 6 4 s 2 3d 4 . Cu
Factors responsible for the extra stability of half-filled and completely filled subshells, (i) Symmetrical distribution : It is well known fact that symmetry leads to stability. Thus the electronic configuration in which all the orbitals of the same subshell are either completely filled or are exactly half filled are more stable because of symmetrical distribution of electrons. (ii) Exchange energy : The electrons with parallel spins present in the degenerate orbitals tend to exchange their position. The energy released during this exchange is called exchange energy. The number of exchanges that can take place is maximum when the degenerate orbtials (orbitals of same subshell having equal energy) are exactly half-filled or completely. As a result, the exchange energy is maximum and so it the stability. 59
M.C.Q. 1.
What is wrong about anode rays? (A) Their e/m ratio is constant (B) They are deflected by electrical and magnetic field (C) They are produced by ionisation of molecules of the residual gas (D) Their elm ratio depends on nature of residual gas.
2.
When atoms of the gold sheet are bombarded by a beam of á -particles, only a few á-particles get deflected whereas most of them go straight undeflected. This is because (A) The force of attraction on -particles by the oppositely charged electron is not sufficient (B) The nucleus occupies much smaller volume as compared to the volume of atom (C) The force of repulsion on fast moving -particles is very small (D) The neutrons in the nucleus do not have any effect on -particles.
3.
Which of the following is not a characteristic of Planck’s quantum theory of radiations? (A) Radiations are associated with energy (B) Magnitude of energy associated with a quantum is equal to hv (C) Radiation energy is neither emitted nor absorbed continuously (D) A body can emit less or more than a quantum of energy.
4.
Which of the following statements is wrong? The probability of finding the electron in px orbital is (A) Maximum on two opposite sides of the nucleus along x-axis (B) zero at the nucleus (C) same on all the sides around the nucleus (D) zero on the z-axis
5.
In uni electron system, the wave number of any spectral line is directly proportional to (A) the number particles present in the system (B) the velocity of electron undergoing transition (C)
6.
7.
(D) the charge on the nucleus and the ë of light used.
The conclusion that every additional electron enters the orbital with lowest possible energy has been drawn from (A) Pauli’s exclusion principle
(B) Hund’s rule
(C) Aufbau principle
(D) de-Broglie’s equation.
Bohr’s model of atom is not in agreement with (A) Line spectra of hydrogen atom
(B) Pauli’s principle
(C) Planck’s theory
(D) Heisenberg’s principle. 60
8.
Which of the following statements is correct? (A) All electromagnetic radiations do not possess the same velocity (B) Matter waves are associated with electrical and magnetic fields (C) Matter waves and electromagnetic radiations are alike (D) The velocityof matter wave is generally less than that of light
9.
10.
Which experimental observation correctly account for the phenomenon? Experimental observation
Phenomenon
(A) X-rays spectra
Charge on nucleus
(B) -particle scattering
Quantized electron orbit .
(C) Photoelectric effect
The nuclear atom
(D) Emission spectra
Quantizationof energy. .
In the Schrodinger’s wave equation ø represents (A) orbit
11.
12.
(C) wave
(D) radial probability
Which of the following gave the idea of a nucleus of the atom? (A) Oil drop experiment
(B) Davissonand Germer’s experiment
(C) -ray scattering experiment
(D) Austen’s mass spectrogram experiment.
Cathode rays have same charge to mass ratio as (A) -particles
13.
(B) wave function
(B) -rays
(C) Anode rays
(D) Protons
Which of the following statements is/are correct? (A) Isotopes have same number of nucleons (B) Isobars have same number of protons (C) Isotones have same number of neutrons (D) Isobars are atoms of different elements with same isotopic number
14.
According to Bohr’s Model of hydrogen atom (A) Total energy of the electron is quantized (B) Angular momentum of the electron is quantized and given as
l l + 1
(C) Both (A) and (B) (D) None of the above. 15.
Rutherford’s experiment established that (A) inside the atom there are positive centres immersed in sea of electrons (B) nucleus contains protons, neutrons and mesons (C) most of the space in an atom is empty (D) all A, B and C. 61
h 2
16.
Mathematically, Heisenberg’s uncertainty principle can best be explained by (A)
17.
(B)
2
(C) (C)
19.
2
+
2
+
2
2 2
+
2
+
2
2 2
+
2
+
2
( − )
=0
(B)
= 0 (D)= 0(D) (D)
(A)
(B)
(C)
(D)
2
2
+
2 2 2
+
2
+
2 2 2
2
+
( − )
+
2 2
+
=0
( − )
=0
Which of the following expressions imposes the condition of quantization of energy of an electron in an atom? (B) E h v
(C) l
(D) mvr
The total number of electrons in a subshell designated by azimuthal quantum number, l is given as (A) 2 l + 1
21.
( − )
In which of the following electron distributions in ground state, only the Hund’s rule is violated?
(A) E mc2 20.
(D)
The correct Schrodinger’s wave equation for an electron with E as total energy and V as potential energy is (A)
18.
(C)
(B) l2
(C) 4 l + 2
(D) 2l + 2.
According to Bohr’s model of the atom, an electron can revolve around the atomic nucleus in a suitable orbit without emitting energy if its orbit (A) is a perfect circle
(B) is a circle with a large radius
(C) houses a whole number of de-Broglie waves (D) houses odd number of de-Broglie waves. 22.
Which of the following concerning Bohr’s model is false? (A) Predicts that probability of electron near nucleus is more (B) Angular momentum of electron in H atom (C) Introduces the idea of stationary states (D) Explains line spectrum of hydrogen.
23.
The kinetic energy of the electron in the nth orbit of an atom is given by the relation (A) (B) (C) (D) none of these
24.
According to Somerfield’s model, only circular orbit is possible for the electron in (A) K shell
(B) L shell
(C) M shell 62
(D) N shell.
25.
According to Bohr’s atomic model (A) electron on H atom can have only certain values of angular momentum (B) electrons have a particle as well as wave character (C) atomic spectrum of atom should contain only five lines (D) all the above statements are correct. .
26.
The transition of electrons inH atom that will emit maximum energy is
(A) n3 –––––––– n2 27.
(C) n5 –––––––– n4
(D) n6 –––––––– n5
The limiting line in Balmer series will have a frequency of (A) 3.29 × 1015s–1
28.
(B) n4 –––––––– n3 (B) 3.65 × 1014 s–1
(C) 8.22 × 1014s–1
(D) 9.22 × 1014s–1
The wavelength of a spectral line for electron transition is inversely related to (A) Z (nuclear charge)
(B) velocity of electron
(C) number of electrons undergoing transition (D) the energy difference between the energy levels involving transition. 29.
30.
The phenomenon of splitting of spectral lines under the influence of the electric field is known as (A) Photoelectric effect
(B) Stark effect
(C) Zeeman effect
(D) Electromagnetic effect.
If the energy of electron in H atom is given by expression, −
1312 n2
kJ mol-1, then the energy
required to excite the electron from ground state to second orbit is (A) 328 kJ 31.
(B) Balmer lines
34.
(C) Paschen lines
(D) Brackett lines.
12397 × 1010 m ∆E
(C)
12397 ∆E
× 10−10 cm
(D)
12397 ∆E
× 1010 cm
If ionising energy of H atom is 13.6 eV, then the second ionising energy of He should be (A) 13.6eV
(B) 27.2eV
(C) 54.4 eV
(D) Cannot be predicted.
The first line in the Balmer series in the H atom will have the frequency (A) 4.57 × 1014S–1
35.
(D) 1312 kJ.
When electronic transition occurs from higher energy state to a lower energy state with energy difference equal to E electron volts, the wavelength of line emitted is approximately equal to (A) 12397 × 10−10 m (B) ∆E
33.
(C) 984 kJ
In the atomic spectrum of hydrogen, the spectral lines pertaining to electronic transition of n 4 to n 2 refers to (A) Lyman lines
32.
(B) 656 kJ
(B) 3.29 ×1015s–1
(C) 8.22 × 1015S–1
(D) 8.05 × 1013S–1.
The orbital configuration of 24Cr is 3d54s1. The number of unpaired electrons in is (A) 3
(B) 2
(C) 1 63
(D) 4.
36.
How many electrons in 19Khave n 3; l0 ? (A) 1
37.
(D) 10.
(B) 4
(C) 1
(D) 2.
(B) [Ar] 3d6,4s2
(C) [Ar] 3d5, 4s1
(D) [Ar] 3d5,4s2.
(B) 3, 2, -2, + ½
(C) 3, 2, + 1, + ½
(D) 3, 1, + 1, + ½.
(B) dxy
(C) dyz
(D)
(B) 2n2
(C) (2n+1)
(D)n2.
Which shape is associated with the orbital designated by n 2; / 1 ? (A) spherical
45.
(C) 12
The total number of electrons present in any main energy level can be calculated from (A) (2l + 1)
44.
(B) 8
Which d-orbital does not have four lobes? (A) dx2-y2
43.
(D) 2, 6 and 12.
Which of the following sets of quantum numbers is not possible for 23rd electron of Cr( At. no. 24) ? (A) 3, 2, +2, -½
42.
(C) 2, 6 and 10
The electronic configuration of Mn can be written as (A) [Ar] 4s2
41.
(B) 2, 4 and 6
The number of electrons that can be accommodated in dxy orbital is (A) 10
40.
(D) 3.
In an atom which has 2K, 8L, 8M and 2N electrons in the ground state, the total number of electrons having 1are (A) 20
39.
(C) 4
The maximum number of electrons in s, p and d-subshells are (A) 2 in each
38.
(B) 2
(B) tetrahedral
(C) dumb-bell
(D) pyramidal.
Which of the following statements about quantum numbers is wrong? (A) If the value of l 0, the electron distribution is spherical (B) The shape of the orbital is given by subsidiary quantum number (C) The Zeeman’s effect is explained by magnetic quantum number (D) The spin quantum number gives the orientations of electron cloud.
46.
47.
The two electrons have the following sets of quantum numbers . X: 3, 2, -2, + ½ Y : 3, 0, 0, + ½. What is true of the following? (A) X and Y have same energy
(B) X and Y have unequal energy
(C)X and Y represent same electron
(D) None of the statements is correct.
If the value of azimuthal quantum number of an electron is 2, then which of the following values of magnetic quantum numbers is not permissible, (A) 3
48.
(B) 2
(C) 0
(D) 1.
An isotone of 32Ge76is (i) 32Ge77 (ii) 33As77 (iii) 34Se77 (iv) 34Se78. (A) only (i) & (ii)
(B) only (ii) & (iii)
(C) only (ii) & (iv)
64
(D) (ii), (iii) & (iv)
49.
The fundamental particles which arc responsible for keeping nucleons together is (A) Meson
50.
52.
54.
(D) Electron. .
(B) neutron
(C) alpha particle
(D) proton.
Atoms may be regarded as comprising of protons, electrons and neutrons. If the mass attributed to the neutrons were halved and that attributed to electrons were doubled then atomic mass of would (A) remain approximately the same
(B) be doubled
(C) Approximately be halved
(D) be reduced by approximately 25%.
How many electrons in an atom with atomic number 105 can have (n +l) 8 ? (A) 30
53.
(C) Positron
The positron is as heavy as (A) electron
51.
(B) Antiproton
(B) 17
(C) 15
(D) Unpredictable.
The size of the nucleus is approximately (A) 1/100th of the atom
(B) 1/1000 th of the atom
(C) 1/10000th of the atom
(D) 1/l00000th of the atom.
The line spectrum ot two elements is not identical because . (A) they do not have same number of neutrons (B) they have dissimilar mass number (C) they have different energy level schemes (D) they have different number of valence electrons.
55.
Bohr’s atomic model can explain the spectrum of (A) hydrogen atoms only (B) atoms or ions which are uni electron (C) atoms or ions which have only two electrons (D) hydrogen molecule.
56.
The electronic configuration of a dipositive ion M+2 is 2, 8, 14 and its mass number is 56. The number of neutrons present is (A) 32
57.
(D) 34.
(B)
(C)
(D)
An atom has 2 K, 8 L, 11 M, 2 N electrons, the total number of s-electrons will be (A) 6
59.
(C) 30
An electron of mass m and charge -e moves in circular orbit of radius r around the nucleus of charge + Ze in uni electron system. In C.G.S. system the potential energy of electron is (A)
58.
(B) 42
(B) 8
(C) 10
(D) 4.
In an atom with 2K, 8L, 11M and 2N electrons the number of electrons with m 0 ; s + ½ are (A) 2
(B) 7
(C) 8 65
(D) 16.
60.
For a sub-shell with azimuthal quantum number l, the total values of magnetic quantum number m can be related to l as (A) m ( +2)
61.
62.
(B) m (2 2 + 2)
(B) symmetrical about Y-axis only
(C) spherically symmetrical
(D) unsymmetrical.
The units for equation are −1
−1
2 −1
(C) 2.27 × 10–40 kg ms–1
(D) none
(D)
A near U.V. photon of 300 nm is absorbed by a gas and then remitted as two photons. One photon is red with wavelength760 nm. Hence wavelength of the second photon is (B) 1060nm
(C) 496nm
(D) 300nm
The transition in He+ ion that would have the same wave number as the first Lyman line in hydrogen spectrum is (B) 5––––––––3
(C) 4––––––––2
(D) 6––––––––4
The work function of a metal is 4.2 eV. If radiations of 2000 A0 fall on the metal then the kinetic energy of the fastest photo electron is (B) 16× 1010J
(C) 3.2 × 10–19J
(D) 6.6 × 10–10J
A certain metal when irradiated to light ( 3.2 × 1016 Hz) emits photoelectrons with twice kinetic energy as did photo electrons when the same metal is irradiated by light ( 2.0 × 1016Hz). The of metal (A) 1.2 × 1014 Hz
68.
(C)
(B) 3.33 × 10–43 kg ms–1
(A) 1.6 × 10–19J 67.
−3
2 −2
2
(A) 1.1 × 10–23 kg ms–1
(A) 2––––––––1 66.
(B)
The momentum of a photon of frequency 50 × 1017S–1 is nearly
(A) 460 nm 65.
(D) 2m+1.
(A) symmetrical about X-axis only
2 −1
64.
( −1) 2
An orbital with 0 is
(A) 63.
(C) =
(B) 8× 1015 Hz
(C) 1.2× 1016 Hz
(D) 4× 1012 Hz
The ratio of the radil of the first three Bohr orbit in H atom is (A) 1:
1 1 ∶ 2 3
(B) 1:2:3
(C) 1:4:9
(D) 1:8:27
Hint : rn n2 a0 i.e. r n2 r1 : r2 : r3 12 : 22 : 32 1 : 4 : 9 69.
An electron of a velocity ‘’ is found to have a certain value of de-Broglie wavelength. The velocity to be possessed by the neutron to have the same de-Broglie wavelength is (A)
70.
(B)
(C) 1840
1840
(D)
1840
The momentum of a particle associated with de-Broglie’s wave length of 6 A0 is (A) 1.1 × 10–24kgms–1
(B) 1.1 × 1034kgms–1
(C) 39.6 × 10–34kgms–1
(D) 39.6 × 10–24kgms–1 66
71.
Frequency of matter wave is equal to v frequency, v velocity of particle (A) (K.E.)/2h ℎ
=
72.
v
v=
(B) 2.(KE.)/h ∴
=
Work function is h
1 1000
81.
ℎ
(B) 1.2 × 10–20 J 0
=
ℎ 0
2. . ℎ
1
( K.E. = 2
v2 )
(C) 6 × 10–19 J
(D) 6× 10–12 J
(C) 9.1 × 10–24g
(D)1.008mg
= 6 × 10–19 J
(B) 0.55mg
(C) 1: 8
(C)He+ ion
(B) D atom (B) >13.6eV
(D) 2: 1
(D) Li+2ion
(C) 3.4eV (D) 3.4eV
(B)
1 137
ℎ
(C)
(B)7
1 10
ℎ
(D) same
(C)8
(D)4
In any subshell the maximum number of electrons having same value of spin quantum number is ( + 1)
(B) l + 2
(C) 2l + 1
(D) 4l + 2
For a certain particle, it is found that uncertainty 1840 in velocity is reciprocal of uncertainty in position This implies that (A) Mass of the particle is > unity
(B) Mass of the particle is unity
(C) Mass of the particle < h
(D) Mass of the particle > h / 4
How many electrons in K (Z19) have n 4 and ? (A)1
82.
=
How many electrons in Cu atom have (n + ) 4
(A) 80.
v2 ℎ
Velocity of electron in the first orbit of H-atom is compared to that of velocity of light is approximately
(A) 6 79.
=
The energy required to dislodge electron from excited isolated H atom (IE1 13.6eV) is
(A)
78.
v
∴
For the electronic transition from n 2 to n 1 which one of the following will produce shortest wave length?
(A) 13.6eV 77.
ℎ
(B) 1: 2
(A) H atom 76.
=
In two H atoms A and 13 the electrons move around the nucleus in circular orbits of radius r and 4r respectively. The ratio of the times taken by them to complete one revolution is (A) 1: 4
75.
v
The mass of one mole of electron is (A) 9.1 × 10–28g
74.
∴
(D) l
If threshold wavelength for ejection of electron from metal is 300nm, then work function for photoelectric emission is (A) 1.2 × 10–18 J
73.
v
(C) (K.E./h)
(B)2
(C) 3
(D)4
Suggest two transitions in hydrogen spectrum for which wave number ratio is 108:7 (A) 21, 31
(B) 21, 43
(C) 21, 54 67
(D) 21, 41
83.
The ratio of K.E. to P.E. of an electron in Bohr orbit of H like species is (A) 1 : 2
84.
85.
(B) Inversely proportional to its velocity ,
(C) Independent of its mass and velocity
(D) Unpredictible.
In one joule of energy, the number of photon with wave number equal to is )−1
(B) (ℎ )−1
(B) RH.NA.hc
ℎ 2 . .
(B)
ℎ
(D)0
ℎ 2 . .
(C) both correct
(D) none
ℎ
(B) ½
(C)
ℎ 4
(D)
ℎ 4
(B) 3
(C) 4
(D) infinite
Which of the following electronic transitions require that the greatest amount of energy be absorbed by a hydrogen atom? (B) n2 to n4
(C) n 3 to n 6
(D) n” to n 1
How fast an electron is moving if it has a wavelength equal to the distance it travel in one second? /ℎ
(B)
(C) ℎ/
(D) ℎ/2( . )
If uncertainties in the measurement of position and momentum are equal, the uncertainty in the measurement of velocity is (A) ½
94.
(C) 2.2
Number of waves made by a Bohr electron in one complete revolution in its fourth orbit is
(A) ℎ/
93.
(D) RH.c
If uncertainty in position and velocity are equal then uncertainty in momentum will be
(A) n1 to n2 92.
(C) RH(2hc)
(B) 5.5
(A) 2 91.
(D) ℎ ( )−1
If numerical value of mass and velocity are equal then de-Broglie wave length in terms of K.E. is
(A) ½
90.
)
One unpaired electron causes magnetic moment of 1.1 B.M. The magnetic moment of Fe+2 ion is
(A) 89.
(C) (ℎ
Which of the expression given below gives IE of H atoms in terms of Rydberg’s constant (RH)?
(A) 4.4 88.
(D)– 1 : 1
(A)Directly proportional to its velocity,
(A) RH.hc 87.
(C) 1 : 1
For particles having same KE., the de-broglie wavelength is
(A) (ℎ
86.
(B)– 1 : 2
ℎ
(B)
1 2
ℎ
(C)
1 2
ℎ
(D) none of these
Energy equivalent of 10.00 cm–1 is (A) 19.9 × 10–23 Jper photon
(B) 28.6 × 10–3 kcal mol–1photon
(C)12.0 × 10–3 kJ mol–1photon
(D) all are correct
68
95.
The ratio of energy of the electron in ground state of hydrogen to the electron in first excited state of Be+3 is (A) 1: 4
96.
9 5
(D) 16: 1
(B)
36 5
(C)
(D)
4
5 9
The radius of first Bohr orbit is , then de-Broglie wavelength of electron in 3rd orbit is nearly (A) 2 x
98.
(C) 1: 16
The shortest wavelength of H-atom in Lyman series is x, then longest wavelength in Balmer series of He+ is (A)
97.
(B) 1: 8
(B) 6 x
(C) 9x
(D) x/3
With increasing principle quantum number, the energy difference between adjacent energy levels in H atom (A) decreases
(B) increases
(C) remains constant
(D) decreases for low value of Z and increases for higher value of Z, 99.
The electrons present in K-shell of the atom will differ in (A) principal quantum number
(B) azimuthal quantum number
(C) magnetic quantum number
(D) spin quantum number.
100. The ratio of the ionisation energy of H and Be3+ is.............. (A) 1: 1
(B) 1: 3
(Pb. C.E.T. 1996)
(C) 1 : 9
(D) 1: 16.
101. The maximum number of electrons in a subshell for which l 3 is (Andhra B. Tech 1982) (A) 14
(B) 10
(C) 8
(D) 4.
102. The number of electrons in the M shell of the clement with atomic number 24 is (Andhra B. Tech. 1982) (A) 24
(B) 12
(C) 13
(D) 8.
103. Sodium chloride imparts a yellow colour to the Bunsen flame. This can be interpreted due to the (A) low ionization energy of sodium (B) sublimation of metallic sodium to give yellow vapour (C) emission of excess energy absorbed as a radiation in the visible region (D) photosensitivity of sodium. 104. The exact path of electron 2p-orbital cannot be determined.” The above statement is based upon (Delhi P.M. T. 1981) (A) Hund’s Rule
(B) Bohr’s Rule
(C) Uncertainty principle (D) Aufbau principle.
105. The maximum number of electrons in a subshell is given by the expression (C.B.S.E. 1989) (A) 4l - 2
(B) 4l + 2
(C) 2l + 1 69
(D) 2n2.
106. If r is the radius of first orbit, the radius of nth orbit of the H atom will be (C.B.S.E. 1989) (A) r n2
(B) r n
(D) r2n2
(C) r/n
107. The energy of hydrogen atom in its ground state is -13·6 eV. The energy of the level corresponding to the quantum number n 5 is (A) “0·54 eV
(B) “5·40 eV
(C) “0·85 eV
(D) “2·72 eV.
(C.B.S.E. 1990, M.P.C.E.T. 1999) 108. At 200°C hydrogen molecules have velocity 105 cm sec–1. The de-Broglie wavelength in this case is approximately (A) 2 Å
(B) 1000 Å
(C) 100 Å
(D) 10 Å. (C.B.S.E. 1991)
109. The number of electrons in 3d shell for element with atomic number 26 is (A) 4
(B) 6
(C) 8
(D) 10. (C.E.E. T. Sample Paper 1992)
110. In a set of degenerate orbitals the electrons distribute themselves to retain similar spins as far as possible. This statement is attributed to (A) Pauli’s exclusion principle
(B) Aufbau principle
(C) Hund’s Rule
(D) Slater rules.
(Pb. C.E. T. 1989)
111. The ground state configuration of Fe3+ ion in gaseous state is: (At. No. of Fe 26} (A) [Ar]18 3d3 4s2
(B) [Ar] 18 3d6 4s2
(C) [Ar] 18 3d5
(D) [Ar]18 3d6. (Karnataka C.E.E.T. 1992)
112. If uncertainty in the position of electron is zero, the uncertainty in its momentum would be (A) zero
(B) >
(C) <
(D) infinite. (B.H.U.1992)
(C) 0·053 x 4nm
(D) 0·053 x 20 nm.
113. The radius of second Bohr’s orbit is (A) 0·053 nm
(B) 0·053/4 nm
(B.H. U. 1989) 114. For which of the following sets of quantum -numbers an electron will have the highest energy ? (A) 3,2,1,1/2
(B) 4,2,-1,1/2
(C) 4, 1,0, -1/2
(D) 5,0,0, 1/2. (C.B.S.E. 1994)
115. The uncertainty in the position of an electron (mass 9.1 × 10–28 g) moving with a velocity of 3.0 x 104 cms–1 accurate up to 0.011% will be (A) 1·92 cm
(B) 7·68 cm
(C) 0·175 cm
(D) 3·84cm. (C.B.S.E. 1995)
70
116. The radius of hydrogen atom in the ground state is 0·53 Å, the radius of 3Li2+ in the similar state is (A) 1·06 Å
(B) 0·265Å
(C) 0·17 Å
(D) 0·53 Å. (C.BS.E. 1995)
117. Splitting of spectral lines under the influence of magnetic field is called (A) Stark effect
(B) Zeeman effect
(C) Photoelectric effect
(D) None of these. (A.F.M.C. 1995)
118. The total number of orbitals in a shell with principal quantum number ‘n’ is (B) 2n2
(A) 2n
(C) n2
(D) n + 1. (A.I.I.M.S. 1997)
119. Which of the following expressions gives the de-Brogiie relationship ? (A)
ℎ
=p
(B) l =
ℎ
(C) l =
ℎ
(D) lm =
(M.P.P.M.T. 1998)
120. The uncertainty in the momentum of an electron is 1·0 × 10–5 kg ms–1. The uncertainty in its position will be (h 6·62 x 10–34 kg m2s–1) (A) 1·05 × 10–28 m (B) 1·05 × 10–26 m
(C) 5·27 × 10–30 m
(D) 5·25 ×10–28 m. (Pune A.F.M.C. 1998)
121. If the radius of first Bohr orbit be a0, then the radius of third Bohr orbit would be (A) 3 x a0
(B) 6 x a0
(C) 9 x a0
(D) 1/9 x a0. (M.P.C.E.T. 1998)
122. The first emission line in the atomic spectrum of hydrogen in the Balmer series appears at (A)
9 400
cm–1
(B)
7 144
cm–1
(C)
3 4
cm–1
(D)
5 36
cm–1
(EA.M.C.E.T. 1998) 123. The ratio of specific charge of a proton and an particle is (A) 2 : 1
(B) 1 : 2
(C) 1 : 4
(D) 1 : 1 (M.P.C.E.T 1999)
124. The de-Broglie wavelength of a particle with mass 1g and velocity 100 m/s is (A) 6·63 × 10–33m
(B) 6·63 ×10–34m
(C) 6·63 × 10–35m
(D) 6·65 × 10–35m. (C.B.S.E. 1999)
125. Which of the following sets of quantum numbers belongs to highest energy ? (A) n 4, l 0, m 0, s + 12 (C) n 3, l 1, m 1, s +
1 2
(B) n 3, l 0, m 0, s +
1 2
(D) n 3, l 2, m 1, s +
1 2
(C.P.M.T. 1999)
71
126. If wavelength of photon is 2·2 × 10–11 m, h 6·6 × 10–34 Js, then momentum of photon is (A) 3 × 10–23 kg ms–1
(B) 3·33 × 1022 kg ms–1
(C) 1·452 × 10–44 kgms–1
(D) 6·89 × 1043 kg ms–1. (M.P.C.E.T. 1999)
127. According to Bohr’s theory, the energy required for the transition of H atom from n 6 to n 8 state is (A) equal to energy required for the transitions from n 5 to n 7 state (B) larger than in (A) (C) less than in (A) (D) equal to energy for the transition from n 7 to n 9 state
(Kerala M.E.E. 2000)
128. An electron has kinetic energy of 2·14 × 10 –22J.Its de-Broglie wavelength will be nearly (me 9.1 × 10–31 kg) (A) 9·28 × 10–4
(B) 9·28 × 10–7m
(C) 9·28 × 10–8m
(D) 9·28 × 10–10m. (M.P.C.E.T. 2000)
129. What will be de-Broglie wavelength of an electron moving with a velocity of 1·20 × 105ms–1 ? (A) 6·068 ×10–9
(B) 3·133 × 10–37
(C) 6·626 × 10–9
(D) 6·018 × 10–7 (M.P.CE.T. 2000)
130. The de-Broglie wavelength associated with ball of mass 200 g and moving at a speed of 5 m hour–1 is of the order of (h 6·625 x 10–34 Js) (A) 10–15m
(B) 10–20m
(C) 10–30m
(D) 10–25m (Kerala P.M.T. 2001)
131. The third line of the Balmer series. in the emission spectrum of the hydrogen atom, is due to the transition from the (A) fourth Bohr orbit to the first Bohr orbit (B) fifth Bohr orbit to the second Bohr orbit (C) sixth Bohr orbit to the third Bohr orbit
(D) seventh Bohr orbit to the third Bohr orbit (Kerala P.M.T. 2001)
132. The highest number of unpaired electrons are w present in (A) Fe°
(B) Fe4+
(C) Fe2+
(D.C.E. 2001) (D) Fe3+.
133. Rutherford’s atomic model suggests the existence (A) Atom
(B) Nucleus
(C) -particle
(D) Mesons (A.EM.C. 2001)
134. Which is not true with respect to cathode rays? (A) A stream of electrons
(B) Charged particles
(C) Move with speed as that of light
(D) Can be deflected by magnetic fields (Kerala C.E.T. 2001) 72
135. A element M has an atomic mass 19 and atomic number 9. Its ion is represented by (A) M+
(B) M2+
(C) M–
(D) M2–. (Manipal P.M.T. 2001)
136. Which of the following ions has the maximum magnetic moment ? (A) Mn2+
(B) Fe2+
(C) Ti2+
(D)Cr2+. (A.I.E.E.E. 2002)
137. The value of the energy for the first excited state of hydrogen will be (A) –13·6eV
(B) –3·40eV
(C) –1.51eV
(D) –0·85eV. (M.P.C.E.T. 2002)
138. In hydrogen atom, energy of first excited state is -3·4 eV. Find out the K.E. of the same orbit of hydrogen atom (A) +3·4eV
(B) +6·8eV
(C) –13·6eV
(D) +13·6eV (C.B.S.E. P.M.T. 2002)
139. The energy of the first electron in helium will be (A) –13·6eV
(B) –54·4eV
(C) –5·44eV
(D)zero. (Bihar C.EE. 2002)
140. In a hydrogen atom, if the energy of an electron in the ground state is 13·6 eV, then that in the 2nd excited state is (A) 1·51eV
(B) 3·4eV
(C) 6·04cV
(D) 13·6eV. (A.I.E.E.E. 2002)
141. In the Bohr’s orbit, what is the ratio of total kinetic energy and total energy of the electron (A) –1
(B) –2
(C) 1
(D) +2. (Rajashan P.M.T. 2002)
142. The ratio between kinetic energy and the total f energy of the electrons of hydrogen atom according to Bohr’s model is (A) 2 : 1
(B) 1 : 1
(C) 1 : (–1)
(D) 1 : 2. (Pb. P.M.T. 2002)
143. In Bohr series of lines of hydrogen spectrum, the third line from the red end corresponds to which one of the following inter-orbit jumps of the electron for Bohr orbits in an atom of hydrogen ? (A) 3 ––––– 2
(B) 5 ––––– 2
(C) 4 ––––– 1
(D) 2 ––––– 5. (A.I.E.E.E. 2003)
144. The orbital angular momentum for an electron revolving in an orbit is given by This momentum for an s-electron will be given by 73
( + 1) .
ℎ 2
(A) +
1 2
·
ℎ 2
(B) zero
(C)
ℎ 2
(D) √2
ℎ 2
(A.I.E.E.E. 2003) 145. The atomic number of an element is 35. What is the total number of electrons present in all the p-orbitals of the ground state atom of that element ? (A) 6
(B) 11
(C) 17
(D) 23. (EA.M.C.E.T. 2003)
146. The emission spectrum of hydrogen is found to satisfy the expression for the energy change “E (in Joules) such that “E 2·18 x 10–18 J where n1 1, 2, 3 ........ and n2 2, 3, 4,...... The spectral lines correspond to Paschen series it (A) n1 1 and n2 2, 3, 4
(B) n1 3 and n2 4, 5, 6
(C) n1 1 and n2 3, 4, 5
(D) n1 2 and n2 3, 4, 5 (Kerala Engg. 2003)
147. Among the following series of transition metal ions, the one where all metal ions have same 3d electronic configuration is (A) Ti2+, V3+, Cr4+, Mn5+
(B) Ti3+, V2+, Cr3+, Mn4+
(C) Ti+, V4+, Cr6+, Mn7+
(D) Ti4+, V3+, Cr2+, Mn3+ (C.B.S.E. P.M.T. 2004)
148. For d-electron, the orbital angular momentum is (A) 6 ℎ / 2
(B) 2 ℎ / 2
(C) h / 2
(D) 2h/.
(J & K Med.2004) 149. Time taken for an electron to complete one revolution in the Bohr orbit of hydrogen atom is (A)
4 2
2
ℎ
(B)
ℎ 4 2
(C)
2 2
ℎ2
(D)
ℎ 2
(Kerala P.M.T. 2004) 150. Which of the following sets of quantum numbers is correct for an electron in 4f – orbital ? (A) n 4, 3, m +4, s + ½
(B) n 3, 2, m –2, s + ½
(C) n 4, 3, m + 1, s + ½
(D) n 4, 4, m –4 s - ½ (A.I.E.E.E. 2004)
151. The wavelength of radiation emitted when in a hydrogen atom electron falls from infinity to stationary state 1, would be (Rydberg constant 1·09 × 107m–1) (A) 91nm
(B) 9·1 × 10–8nm
(C) 406 nm
(D) 192 nm (A.I.E.E.E. 2004)
152. The relationship between energy E, of the radiation with a wavelength 8000 Å and the energy of the radiation with a wavelength 16000 Å is (A) E1 6E2,
(B) E1 2E2
(C) E1 4E2
(D) E1 1/2E2 (Kerala Engg. 2005)
74
153. The energy of second Bohr orbit of the hydrogen atom is –328 kJ mol–1, hence the energy of fourth Bohr orbit would be (A) –41 kJ mol–1
(B) –1312 kJ mol–1
(C) –164 kJ mol–1
(D) –82 kJ mol–1 (C.B.S.E. P.M.T. 2005)
154. Which of the following statements in relation to the hydrogen atom is correct? (A) 3s-orbital is lower in energy than 3p-orbital (B) 3p-orbital is lower in energy than 3d-orbital (C) 3s and 3p orbitals arc of lower energy than 3d orbital (D) 3s, 3p and 3d orbitals all have the same energy.
(A.I.E.E.E. 2005)
155. The correct order of number of unpaired electrons in the ions Cu2+, Ni2+, Fe3+ and Cr3+ is (A) Cu2+ > Ni2+ > Cr3+ > Fe3+
(B) Cr3+ > Fe2+ > Ni2+ > Cu2+
(C) Fe3+ > Cr3+ > Cu2+ > Ni2+
(D) Fe3+ > Cr3+ > Ni2+ > Cu2+. (Kerala P.M.T. 2005)
156. The most probable radius (in pm) for finding the electron in He+ is (A) 0·0
(B) 52·9
(C) 26·5
(D) 105·8 (A.I.I.M.S. 2005)
157. Ground state electronic configuration of nitrogen atom can be represented by (A)
(B)
(C)
(D) (I.I.T. 1999)
158. Which one of the following sets of quantum number represeni an impossible arrang- ement ? n
I
ml
ms
(A)
3
2
–2
½
(C)
3
2
–3
½
n
I
m1
ms
(B)
4
0
0
½
(D)
5
3
0
–½. (I.I.T. 1986)
159. The orbital diagram in which Aufbau principle is violated is (A)
(B)
(C)
(D)
75
160. If the speed of electron in Bohr first orbit of hydrogen atom be x, then speed of the electron in 3rd orbit is (A) x/9
(B) x/3
(C) 3x
(D) 9x (I.I.T.1990)
161. If wavelength of photon is 2·2 × 10 – 11m, h 6·6 × 10–34 Js, then momentum of photon is (A) 3 × 10–23 kg/s
(B) 1·452 × 10–44 kg/s (C) 3·33 × 1022 kg/s
(D) 6·89 × 1043 kg/s. (Roorkee 1990)
162. The wave number of first line of Balmer series of hydrogen is 15200 cm–1. The wave number of the first Balmer line of Li1+ ion is (A) 15200 cm–1
(B) 60800 cm–1
(C)76000cm–1
(D) 136,800cm–1. (I.I.T. Screening 1992)
163. Which of the following is violation of Pauli’s exclusion principle ? (A)
(B)
(C)
(D) (I.I.T. Screening 1993)
164. The electrons, identified by quantum number n and (i) n 4, 1 (ii) n 4, 0 (iii) n 3, 2 (iv) n 3, 1 can be placed in order of increasing energy, from the lowest to highest, as (A) (iv) < (ii) < (iii) < (i)
(B) (ii)< (iv) < (i) < (iii)
(C) (i) < (iii) < (ii) < (iv)
(D) (iii) < (i) < (iv) < (ii) (I.I.T. 1999)
165. The wave length associated with a golf ball weighing 200 g and moving at a speed of 5 m/h is of the order (A) 10–10m
(B) 10–20m
(C) 10–30m
(D) 10–40m. (I.I.T. Screening2001)
166. If the nitrogen atom had electronic configuration 1s7, it would have energy lower than that of the normal ground state configuration 1s2 2s2 2p3 because the electrons would be closer to the nucleus. Yet, 1s7 is not observed because it violates. (A) Heisenberg uncertainty principle
(B) Hund’s rule
(C) Pauli’s exclusion principle
(D) Bohr postulates of stationary orbits. (I.I.T. Screening2002)
167. The radius of which of the following orbits is same as that of the first Bohr’s orbit of hydrogen atom ? (A) He+ (n 2)
(B) Li2+ (n 2)
(C) Li2+ (n 3)
(D) Be3+ (n 2) (I.I.T. Screening2004)
76
168. Radial nodes present in 3s and 2p -orbitals are respectively (A) 0, 2
(B) 2, 0
(C) 2, 1
(D) 1,2. (I.I.T. Screening2005)
169. Rutherford’s experiment, which established the nuclear model of the atom, used a beam of (A) - particles, which impinged on a metal foil and got absorbed (B) - rays, which impinged on a metal foil and ejected electrons (C) helium atoms, which impinged on a metal foil and got scattered (D) helium nuclei, which impinged on a metal foil and got scattered
(I.l.T. Screening2002)
ANSWER KEY
1 8 15 22 29 36 43 50 57 64 71 78 85 92 99 106 113 120 127 134 141 148 155 161 168
a d c a b b b a b c d b a a b a c c c c a a c a b
2 9 16 23 30 37 44 51 58 65 72 79 86 93 100 107 114 121 128 135 142 149 156 162
b d c c c c c d b c c c a c d a b c c a c a c a
3 10 17 24 31 38 45 52 59 66 73 80 87 94 101 108 115 122 129 136 143 150 157 163
d b d a b c d b b c b d a d a a c d a c b c c d
4 11 18 25 32 39 46 53 60 67 74 81 88 95 102 109 116 123 130 137 144 151 157 164
c c a a a d b d c b c a c a c b c b c c b a d a 77
5 12 19 26 33 40 47 54 61 68 75 82 89 96 103 110 117 124 131 138 145 152 158 165
c b d a c d a c c c d b a a c c b a b a c b c c
6 13 20 27 34 41 48 55 62 69 76 83 90 97 104 111 118 125 132 139 146 153 159 166
c c c c a a c b a d d b c b c c c d d b b d b c
7 14 21 28 35 42 49 56 63 70 77 84 91 98 105 112 119 126 133 140 147 154 160 167
d c c d a d a b a a b a a a b d b a b a a d b d
HI NT S 2 –2
62. Unit of h = J s, J = kgm s , m = kg,
64. E -= E1 + E2 now 1
2
65. ̅ =
2
−
ℎ
=
1
∴ −1
2
1
1
=
1
1
+
2
1
̅ =
−1
=m
2
−
1
−1
2
̅
=
1
(2)2
2
−
1 2
−1
68. rn = n2 a0 i.e. r ∝ n2 r1 : r2 : r3 = 12 : 22 : 32 = 1 : 4 : 9 ℎ
69. λe = λn
71.
=
83. K .E
ℎ
v=
v
∴
∴
=
v
=
∴
v
=
= ℎ v
∴
1840
=
v2 ℎ
=
2. . ℎ
( K.E. =
1 2
c 1 nhcv now E 1, v x n (hcx) 1 hcx
13.6 Z 2 eV atom1 For hydrogen atom in ground state Z=1 and n=1 for Be+3 ion in first 2 n
excited state Z= 4 and n= 2
RH Z 2 RH 12 I . E RH H n2 12
100. I .Eo
121. Radius of Bohr orbit
n2 a0 Z
122. for Balmer series n1 = 2 and n2 = 3 for first line 138. K.E. of e- in nth orbit = - En = 3.4 e.V 139. En
13.6 2 Z e.V n2
1 2
141. K .E mv 2 P.E
Ze 2 r
Electrostatic force = centrifugal force
v2 )
Ze 2 Ze 2 K .E 1/ 2 P.E 1: 2 2r r P.E 1
85. E hv nh
95. En
ℎ
=
Ze 2 mv 2 P.E mv 2 r2 r
Total energy = K.E + P.E =
1 2 1 K .E mv mv 2 mv 2 1 2 2 Total energy
78
UNIT : 4 CHEMICAL BONDING AND MOLECULAR STRUCTURE Important Points In this unit, the study of chemical bonding and formation of molecule from the atoms are included. The smallest particle of element is atom and the smallest particle in a compound is molecule. The force or the binding that keeps the atoms in the molecule combined during the formation of molecule is called chemical bonding. The concepts like that of Kossel-Lewis, VSEPR principle, valence bond theory, molecular orbital theory have been presented. In chemical bonding, it has more relation with orbitals around the nucleus and especially the valence orbitals. We do not think about the nucleus but we take into consideration the effect due to its positive charge. Scientists Lewis and Kossel have mentioned the approach of chemical bonding. In this, the atom obtains the octet either by losing or by gaining the electron, which is chemically inert. This is called law of octet. Such bonds are called ionic bonds e.g. NaCl. Also, some atoms share electrons with each other and obtain octet structure resulting into stable covalent molecule. e.g. Cl2. To explain the structures of such molecules he mentioned dot and cross symbols and explained the stability of the molecules. Such a bond is called covalent bond. The approach of Kossel Law is explained in detail in the unit. When any bond is formed, the distance between their atoms is called bond length and the angle is called bond angle. As you know the bond lengths of single ( - ) bond, double (=) bond and triple (º) bond are different. The bond angle gives geometrical shapes to molecules viz.1800- linear, 109028' tetrachedral. You will study in detail about covalent bond which can be of three types. (1) Polar covalent bond in which the electron remains dragged towards the more electronegative atom and +d charge on electropositive atom and -d charge on electronegative atoms are shown. As a result the molecule becomes polar. If the electronegativities of the two atoms are same or the difference between them is less, than non-polar bonds, formed by both the atoms sharing the electrons. In coordinate covalent bond, one of the two atoms sharing a pair of electrons and the second atom completes the octet with the help of this gained electron pair. viz. F3B ¬ NH3. Bond ¬ indicates co-ordinate covalent bond. Over and above, bond length, bond angle, bond enthalpy (bond energy) is also an important concept. Shorter the bond length, more will be the stability and so more energy will be required to break it. Thus, the values of enthalpy may be different according to bond formation. The number of bonds is called bond order which we have studied in detail and also the formula to determine it. Born-Haber showed that the enthalpy evolved in formation of compound is the mathematical results of the enthalpies of several reactions of atoms. It is explained in the unit by discussing the formation of compound like NaCl. Sometimes, it so happens that the electron pair instead of being localised on any molecule it localises towards other molecule. Thus, the bonds in the molecule can be shown at different positions in the compounds having same molecular formula. Such structures are called resonance structures and energy associated with the changes of these resonance forms is called resonance energy. This can be studied through the molecules of ozone, carbon dioxide, benzene etc.
79
As we have seen earlier, structures like linear, tetrahedral etc, can be obtained on the basis of bond angle. This study can be used to show the shapes of the molecule by hybridisation of atoms in them, geometrical structures etc. viz. linear BeCl2 - 1800, trigonal BCl3 - 1200 , tetrahedral CH4 109028'. Lewis approach being insufficient to explain the shapes of molecules, Sidgwick and Powell proposed one principle which is known as VSEPR principle which was developed by Nyholm and Gillespie and they proposed certain assumptions. In this it is important to note that when non-bonding electron pairs are there, then they show deviation in geometrical structure and bond angle due to repulsion between electron pairs. e.g. Molecule of water has sp3 hybridisation and so its bond angle must be 109028' but it becomes 1040 30' due to repulsion by two non-bonding electron pairs. Hence, it is called distored tetrachedral. The polarity of bond is a vector quantity. Hence, if a polar bond is formed due to difference in electronegativities but another bond of the same type is formed in its opposite direction, then polar bond will be formed but the resultant polarity of the molecules becomes zero and molecule will be non-polar.
F e.g F
Be
B
F or
F
F
The dipole momentes of polar substances can be calculated for which both the charges +d and
- d and the distance between them is to be utilised. More the value of dipole moment, more will be the polar bond and more will be the ionic bond. One important aspect is that polar substances dissolve only in polar solvents and non-polar substances dissolve in nonpolar solvents. e.g. NaCl will dissolve in water. Napthalene will dissolve in benzene. New hypotheses have been presented affter taking into consideration the limitations of the principles for the approach of covalent bond. Two are main from them : (1) Valence Bond Theory and (2) Molecular Orbital Theory. These concepts are based on quantum mechanics. Heitler and London first of all gave the idea of valence bond theory and it was developed by Pauling and Slater. In the assumptions of valence bond theory the attraction - repulsion forces between positively charged nuclei of two atoms and the electrons arranged in the orbits around them. According to Coulomb's Law if attractive forces are more than repulsive forces then the bond will be formed and molecule will be formed. In this theory, on the basis of the overlapping of valence orbitals different overlaps can be formed. In this type of overlapping the excitation of electrons in valence orbitals can be shown and then formation of molecule by covalent bond with other atoms. viz. In carbon, the electrons of valence orbital 1s2 2s2 2p2 will be excited to give 1s2 2s1 2px1 2py1 2pz1 containing four orbitals with one electron in each and four hydrogen atoms, and hence will give stable molecule like CH4 . The geometrical structure, and bond angle can be expressed from the hybridisation associated with it. In such valence bonds, two types of bonds-s and p are also observed. s bond is a covalent bond; it
80
attains axial overlap of internuclear axis. The stability of this bond is more than that of p bond. In the p-bond the axis of the atomic orbitals undergoing overlapping remains parallel to each other and is perpendicular to internuclear axis. p- bonds are less stable in comparison to s-bonds or they are weaker. Valence bond theory is based on overlapping of valence orbitals. It explains properties like the geometrical shapes, the bond angle etc. very simply but cannot explain magnetic properties. Scientists Mulliken and Hund suggested molecular orbitals like atomic orbitals and proposed molecular orbital theory. Amongst its important points, the idea that atomic orbitals can also form molecular orbitals was taken into consideration. As many atomic orbitals take part in the formation, same number of molecular orbitals, their energy, symmetry etc. were taken into consideration. The formation of these types of atomic orbitals can be shown in the formation of homonuclear molecules like H2, Be2, F2 etc. and heteronuclear molecules like CO, NO etc. Molecular orbitals are formed by linear combination of atomic orbitals-LCAO principle. On the basis of these types of combination two types of molecular orbitals are formed which are known as Bonding Molecular Orbitals (BMO) and Anti-Bonding Molecular Orbitals (ABMO). In the formation of rules these types of BMO and ABMO the principles like Hund's rule of maximum spin, Pauli's exclusion principle, Aufbau principle etc. which are applicable in formation of atomic orbital are also obeyed and maintained. In the unit the molecular orbital diagrams of construction of molecular orbitals from the atomic orbitals for formation of homonuclear molecules from H2 to Ne2 elements as well as for formation of heteronuclear molecules like CO, and NO are shown. From these diagrams, important property like bond order can be calculated. Bond order
=
1 2
{electron in bonding molecular orbitals - electrons in anti - bonding orbital} viz.
ecule bond order
=
1 2
(10 - 4 ) = 3
molecule bond order will be =
1 2
for N2 mol-
i.e. there will be triple bond N º N. In the same way, in NO
[10 - 5] = 2.5. Here, we will note that if the value of bond order
becomes zero, the bond will not be formed e.g. He2. If the value of bond order is integer, the bond will be formed and according to the integer 1, 2, or 3, there will be single (- ), double (=) or triple
(º) bonds respectively. If the value of bond order is fraction, then molecule will attain unstable structure. The molecular orbital theory can explain the magnetic properties e.g. In O2 molecule two unpaired electrons are there and so it is paramagnetic and in N2 molecule, all the electrons are paired and so it is diamagnetic. Thus molecular orbital theory is superior to valence bond theory in this matter. Above this, one important phenomenon is hydrogen bond. The first element of 15, 16, and 17, groups N, O, F being higer electronegative than the other elements of the group it can form covalent mo lecu les like NH3 , H2 O and HF with h yd r ogen . Af terw ar ds th e mo lecu le lik e
H
F
H
F combines with each other H - F
81
molecule through hydrogen bond
H F H F where H ........ (dotted line) indicates hydrogen bond formation. HF,, NH3, H2O possess hydrogen bonds and so their properties are different from those of other elements in the group. Hydrogen bond is of two types (1) Intermolecular and (2) Intramolecular hydrogen bonds. When hydrogen bond is formed between two molecules it is called intermoleculer hydrogen bond e.g. p-chlororphenol and between two groups in the same molecule, it is called intramoleculer hydrogen bond e.g. o-chlorophenol. Intermolecular hydrogen bond is stronger than intramolecular hydrogen bond. The presence of hydrogen bond is the reason for specific properties of the compounds. Viz the retaining of water in the soil, drying of terrylene clothes is faster than that of cotton clothes. After knowing about ionic bond, covalent bond, co-ordinate covalent bond, we shall study the special type of bond present in metals which is called metallic bond. As there are 1, 2 or 3 electrons in the outermost orbit of the metals, they are not able to form covalent bonds. Their ionisation energy is less and attraction of electron towards the nucleus is less. One, two or three electrons are arranged around the nucleus of the atom. Hence, the positively charged nucleus or kernel is there. The electrons around it have attraction towards other nuclei of the atoms in the lattice. Thus, the electron instead of being localised for any one atom, remains delocalised in the whole metal crystal. For this theory Electron Sea model was proposed. In this, the atomic kernel is imagined as floating in the sea, delocalised electrons are arranged around kernel possessing positive charge. Because of this type of metallic bonds, the specific properties of metals, like density, ductility, malleability etc. are different. Co-ordinate covalent bond is a type of covalent bond as seen earlier. The characteristic in it is that from the two atoms undergoing sharing of electrons, only one of the atoms provides a pair of electrons, and is shared by both the atoms. Hence, it is called co-ordinate covalent bond. e.g. In BF3 , three F atoms were bonded with B-atom through three covalent bonds but the octet of B is not complete. Similary in NH3 molecule, three H atoms are bonded with N through three covalent bonds. But N has one non-bonding pair of electrons, which it gives to BF3 molecule and is shared by both the molecules. Hence F3B ¬ NH3 Co-ordinate covalent bond is formed. In this, the molecule which gives pair of electrons is shown by arrow (®) from the molecule which donates it towards the molecule or atom which accepts and shares gained electron pair. You will study more about coordinate covalent bond in the unit of complex salts in Standard-12.
82
M.C.Q. (1)
Which of the following is ionic ? (a) HCl
(2)
(b) CHCl3
(c) IF 5
(d) KI
When molecule is form by chemical bonding between atoms then (a) nucleous of combining atoms are participate (b) valence electrons and inner cell electrons are participate (c) only valence electrons of combining atoms are participate (d) only inner cell electrons of combining atoms are participate
(3)
(4)
Which factor is not responsible for the formation of ionicbond? (a) crystal lattice energy
(b) density
(c) ionisation enthalpy
(d) electron gain enthalpy
According to valence-bond theory which magnetic property oxygen possess ? (a) Paramagnetic
(5)
(b) Lenus Pauling
(c) Hittler and Londan (d) Hund
(b) sp - sp2
(c) sp3 - sp
(d) sp3 -sp 3
Which of the following pair of species is isoelectronic and same structure ? (a) NO3- , SO 3
(8)
(d) Anti Ferromagnetic
In H - C º C - CH = CH2 molecule C 3 - C 2 single bond carbons has which type of hybridization ? (a) sp2 - sp 3
(7)
(c) Diamagnetic
Who was proposed valence-bond theory ? (a) Mulliken
(6)
(b) Ferrimagnetic
(b) SO3, CO 32-
(c) CO32- , ClO3-
(d) NO3-, CO 32-
Which of the following sentence is incorrect for covalent bond ? (a) Strenght of covalent bond depenas upon overlapping at atomic orbitals. (b) Covalent bond is not directional. (c) There is sharing of electrons between atoms bonded by covalent bond (d) Covalent bond is formed between atoms having less difference in their electronegativity.
(9)
Which of the following compound possesses covalent bond ? (a) MgCl2
(b) NaH
(c) BF 3
(d) CsCl
(10) Which of the following molecule possesses polar and nonpolar covalent bond ? (a) NH4Cl
(b) CCl4
(c) H2O 2
(d) HCN
(11) Which of the following compound does not possesses coordinate covalent bond ? (a) CO
(b) SO 2
(c) HNO2 83
(d) HNO3
(12) Which of the following characteristic is not for covalent compound ? (a) They do not possesses particular geometical structure (b) They may be polar or nonpolar (c) Their boiling and melting point is low (d) Generally they are insoluble in water (13) Which of the following possesses ionic and covalent bond ? (a) CO 2
(b) H2SO 4
(C) NH4Cl
(D) NaI
(14) Whhat is Geometrical Structure of ClF3 molecule ? (a) Triogonal bipyramid (b) Corn shpae
(c) sea-saw
(d) T-shape
(15) Which of the following molecule possesses linear structure ? (a) SO2
(b) CO 2
(c) H2O
(d) C2 H 4
(16) Correct structure of SF4 is
(17) Numbers of possible resonating structure of carbonate iong is.... (a) 9
(b) 6
(c) 3
(d) 2
(18) Which of the following molecule has not zero dipol movement ? (a) NF 3
(b) BF 3
(c) CO 2
(d) BeF2
(19) Which of the following molecule possesses highest dipolspace movement ? (a) CCl4
(b) CHCl3
(c) CHCl2
(d) CH3Cl
(20) Which of the following molecule possesses dipol movement ? (a) trans - 1, 2 - dichloro ethene
(b) trans pent - 2 - ene
(c) 2, 2- dimethyl propane
(d) 2, 2, 3, 3- tetra methyl butane
(21) Which of the following molecule has lowest bond space angle ? (a) NH 3
(b) SO 2
(c) H2O
(d) H2S
(c) s(2S)
(d) s * (1S)
(22) Which orbital has highest energy ? (a) s(2Px)
(b) p * (2Py)
84
(23) Which is the paramagnetic species ? -
(a) C N
-
(b) O 2
(c) NO +
(d) CO
(24) Which of the following statement is incorrect when N2 and O2 are convert into N2+ and O2+ respectively ? (a) In O +2 , O - O bond order increases.
(b) In N +2 , N - N bond become weaker..
(c) N +2 become paramagnetic
(d) Increasing dimagnetism in O +2
(25) According to VSEPR theory geomety of which block elements can be explain ? (a) s
(b) p
(c) d
(d) f
(26) Atoms complete octet in valence shell electron during the bond formation. This postulate was proposed by which scientist ? (a) Powel
(b) Lewis
(c) Sigdwick
(d) Mulliken
(27) Crystal formation is which type of reaction ? (a) endothermic and exothermic
(b) endothermic
(c) exothermic
(d) no heat change occurs
(28) Lattice energy of ionic compound depends upon which factor ? (a) Size of ion
(b) Size of ion and charge
(c) charge on ion
(d) Arrangement of ion
(29) Which is correct order for C - O bond length in CO, CO 23 - , CO 2 (a) CO 3-2 < CO 2 < CO
(b) CO 2 < CO 32- < CO
(c) CO < CO 2 < CO 32-
(d) CO < CO 23 - < CO 2
(30) Maximum how many numbers of hydrogen bond can be form by H2O molecule ? (a) 2
(b) 4
(c) 3
(31) In buta 1, 3 - diene (a) only one sp hybridised carbon atom (b) only sp2 hybridised carbon atoms (c) Two sp3 and two sp2 hybridised carbon atoms (d) sp, sp2 and sp3 hybridized carbon atoms
85
(d) 1
(32) Which of the following statement is irrelevant for sigma bond ? (a) strength of sigma bond is not related with overlapping of atomic orbitals. (b) s - bond can form by overlapping of S - P orbitals. (c) s - bond can form by overlapping of end of atomic orbitals of inner center axis. (d) This type of overlapping is also known as axial overlaping (33) In which molecule inter molecular hydrogen bond can be form ? (a) methanol
(b) ethelene glycol
(c) p - nitrophenol
(d) phenol
(34) In which molecule intra molecular hydrogen bond can be form ? (a) o - nitro phenol
(b) aniline
(c) ethylene glycol
(d) all of these
(35) Which of the following pair possesses very strong H - bond ? (a) CH3 COCH3 and CHCl3
(b) HCOOH and CH3 COOH
(c) H2O and H2
(d) SiH4 and SiCl4
(36) Which of the following relation is correct ? (a) Bond order α Bond energy α Bond length α stability (b) Bond orderα
1 α Bond length
1 α stability energy
(c) Bond orderα Bond energy α
1 α stability Bond length
1 α Bond length
1 α stability Bond energy
(d) Bond orderα (37) Molecule : Bondlength :
H2
F2
Cl2
Br2
74pm
144pm
199pm
228pm
Mention more stable molecule from above (a) Cl2
(b) H2
(c) Br2
(38) In water bond angle is 104o 30 because (a) Oxygen atom is sp3 hybridised (b) Repulsion between lone pair election and bonding pair electron (c) Oxygen has high electronegetivity. (d) H2O molecule possesses ''V'' - shape. 86
(d) F2
(39) In which of the following strong H-bond is present ? (a) F - H.....F
(b) O - H.....N
(c) O - H.....O
(d) O - H.....F
(40) Which is correct order for bond dissociation energy in O2 , O +2 , O -2 and O 22(a) O 2 > O +2 > O 22- > O -2
(b) O +2 > O 2 > O -2 > O 22-
(c) O -2 > O 22 - > O +2 > O 2
(d) O -2 > O 22 - > O 2 > O +2
(41) O, P, Q, R elements electronic configuration is given below P = 1s 2 , 2s 2 , 2p2
O = 1s 2 Q = 1s 2 , 2s 2 , 2p
5
R = 1s 2 , 2s2 , 2p 6
Which atom has strong behaviour of electrovalent bond ? (a) O
(b) P
(c) Q
(d) R
(42) In which molecule bond distorsion is more according to VSEPR theory ? (a) SO 2
(b) NH 3
(c) O3
(d) H2O
(c) O 2
(d) F2
(43) Which of the following species is more stable ? (a) O -2
(b) Ne +2
(44) Number of nonbonding electron pair in XeF6, XeF4 and XeF2 respectively (a) 2, 3, 1
(b) 1, 3, 2
(c) 3, 2, 1
(d) 1, 2, 3
(45) On keeping two cube of ice on each other which become one cube which factor is responsible for it ? (a) Van-der waals attraction (b) Hydrogen bond (c) Dipole attraction (d) Covalent bond (46) Determine lattice energy of LiF(S) according to given data. (i) Li(S) ® Li(g)
155.2 KJ mol
(ii) F2(g) ® 2F (g)
æ 75.2 KJΔ ç HD è 2
-
-1
( DsH) ö ÷ ø
(iii) Li(g) ® Li (+g ) + e
520.0 KJ molΔ-1 H( i
(iv) F( g ) + e ® F( g ) 1 (v) Li(s) + F2 ( g ) ® LiF(S) 2
-33.0 KJΔ ( H eg
(a) - 86.7 KJ mol -1
(b) 86.7 KJ mol
-504.1 KJ mol -1
87
)
) -1
(c) -867 KJ mol
( Df H ) -1
(d) 867 KJ mol
-1
(47) Which of the following statment is incorrect for metallic bond ? (a) There is attraction between delocalised electrons and atomic karnel (b) Directionl property is shown by metal (c) Delocalised electron can change their position easily in crystal
S - Orbital
(d) Explanation of metallic bond can be given by 'electron sea model' (48) Why lattice energy of NaCl > KBr ?
(a) When size of negative ion decrease in ionic crystal then lattice energy increases. (b) When volume of positive and negative ion is small than then interionic attraction become more and hence latice energy increases. (c) In ionic crystal when size of positive ion decrease, then lattice energy increases. (d) All of given (49) Number of H - bond form by unpaired electrons of liquid NH3 , H2O and HF respectively are (a) 3, 4, 2
(b) 4, 4, 2
(c) 3, 2, 1
(d) 1, 2, 1
(50) Which of the following pair is not in order for boiling point for 14, 15, 16 and 17 group ? (a) H2O > H2S
(b) HF > HCl
(c) CH4 > SiH 4
(d) NH3 > PH 3
(51) Which of the following compound possesses ionic bond ? (a) CH 4
(b) SiCl4
(c) BF 3
(d) MgCl2
(52) Which of the following relation between BMO and ABMO electrons is correct for stability of diatomic malecule or ion ? (a) Na > N b
(b) Nb > N a
(c) Na + Nb = 0
(d) Na = N b
(53) At what distance van-derwaals attraction exist ? (a) 4.5 ´ 10 -10 m
(b) 0.45nm
(c) 4.5 Ao
(d) Given all
(c) 40 cal. mol-1
(d) 40 Kg cal mol-1
(54) What is bond energy of H-bond ? (a) 40 J mol-1
(b) 40 KgJ mol-1
(55) In which molecule inter molecular H-bond is possible ? (a) CH 3COCH 3
(b) CH 4
(c) SiH4
(d) NH 3
(56) Which of the following characterstic does not possesses by metal ? (a) luminus
(b) ductility
(c) increase in conductance by increase in temperature
(d) malleability
88
(57) On which factor conductance of metals responsible ? (a) ions
(b) delocalised
(c) atomic kernel
(d) number of atoms
(58) Which of the following figure shows electron-sea model ?
(59) According to which group, hydrogen bond is form in protein molecule present in musecls of living organism ? (a) -CO-
(b) -COOR
(c) -CONH-
(d) -COOH
(60) On which factor van-der waalls attraction force does not depend ? (a) numbers of molecules
(b) contact surface area of molecules
(c) shape of molecules (d) numbers of electron in molecules (61) Practicol dipal movement of HCl is 1.03D. If bond length of HCl is 1.275 Ao than what will be the pereentage of ionic nature in HCl ? (a) 7
(b) 17
(c) 43
(d) 21
(62) Which sentence is correct with respect to bond enthalpy ? (a) As bond order is more, then bond dissociation enthalpy is less (b) As atomic volume is more, then bond energy is more. (c) As bond enthalpy is more, then stability of molecule or ion is less. (d) As number of nonbonding election pair on bonded atom then bond enthalpy is less. (63) which of the following orbitals form bonding orbital ?
(a)
(b) S - Orbital P - Orbital
(c) P - Orbital a f - Orbital
(d) S - Orbital P - Orbital
P - Orbital P - Orbital
(64) Mention number of bonding electron pairs and nonbonding electron pairs in NO 3- ion (a) 3, 1
(b) 2, 2
(c) 4, 0
(d) 1, 3
(65) How many numbers of bonding and nonbonding electron pairs in CO2 ? (a) 4, 4
(b) 2, 4
(c) 4, 2 89
(d) 2, 2
(66) Mention proper order of bond length given below. (a) N 2 < N 22- < N 2-
(b) N 22- < N -2 < N 2
(c) N -2 < N 2 < N 22-
(d) N 2 < N 22- < N 2
(c) N2O
(d) Na2O 2
(c) BrF3
(d) SiO23-
(67) Show paramagnetic compound given below. (a) O3
(b) KO 2
(68) Which species possesses pyramidal shape ? (a) OsF2
(b) SO 3
(69) Which of the following does not possesses bond order as CO ? +
(a) N O
-
(b) N O
(c) N2
-
(d) C N
(70) Which rule is violated in the given electronic configuration ?
2P
2P (a) Aufbau
(b) Pauli
(c) Hund
(d) Given all
(71) In which of the following molecule double bond possesses two pispace bond ? (a) S2
(b) O2
(c) C2
(d) H2C = CH 2
(c) SF 6
(d) XeOF4
(72) Mention AB4F2 type molecule. (a) BrF5
(b) XeF4
(73) Which of the following is the correct order for lone pair and bonding pair electrons ? Lp = Lone pair and Bp = Bonding pair (a) Lp - Lp > Lp - Bp > Bp - Bp
(b) Lp - Bp > Lp - Lp > Bp - Bp
(c) Bp - Bp > Lp - Lp > Lp - Bp
(d) Lp - Lp > Bp - Bp > Lp - Bp
(74) Which theory is useful to determine geometrical structure of molecules ? (a) molecular orbital theory
(b) VSEPR theory
(c) Resonance theory (d) Quantam mechanics (75) The one outermost electron present in Na element at (a) one corner of simple cube
(b) eight corner of simple cube
(c) center of simple cube
(d) each corner of simple cube 90
(76) In which molecules / ion have not all the equal bonds ? (a) SF 4
(b) BF4-
(c) XeF4
(d) SiF4
(77) Which of the following has maximum bond angle ? (a) NH 3
(b) CH 4
(c) CO 2
(d) H2O
(78) Which of the following have equal bond order ? (a) O -2
+
(b) CN –
(c) N O
(d) B and C
(79) The type of bond present in CuSO4 .5H2O (a) covalent and co-ordinate covalent
(b) electrovalent and covalent
(c) electrovalent and co-ordinate covalent (d) electrovalent, covalent and co-ordinate covalent (80) Which of the following statement is wrong (a) sp2 hybrid orbitals are formed from two p - atomic and one s - orbitals (b) hybridization is the mixing of atomic orbitals prior to their combing into molecular orbitals (c) d2p2 hybrid orbitals are all at go to one an other (d) d2sp3 hybrid orbitals are directed towords the corners of a regular tetrahedron (81) CO2 is isostructual with (a) SnCl2
(b) HgCl2
(c) C2H 2
(d) SO 2
(82) NH3 has a higher boiling point than expected because (a) its density decreases on freezing (b) with water it forms NH4OH (c) it has strong inter molecular covalent bonds ? (d) it has intermolecular hydrogen bonds. (83) The molecule with zero dipole moment is (a) chloroform
(b) methyl chloride
(c) carbon tetrachloride
(d) methylene chloride
(84) Molecular shaper of SF4 , CF4, XeF4 are (a) the same with 1, 1 and 1 lone pairs of electrons respectively (b) different with 1, 0 and 2 lone pairs of electrons respectively (c) different with 0, 1 and 2 lone pairs of electrons respectively (d) different with 2, 0 and 1 lone pairs of electrons respectively
91
(85) Which of the following has the regular tetrahedral structure ? (a) SF 4
(b) [ Ni (CN ) 4 ] 2-
(c) BF4-
(d) XeF4
(86) In OF2, number of bond pairs and lone pairs of electrons are respectively (a) 2, 8
(b) 2, 6
(c) 2, 9
(d) 2, 10
(87) In O -2 , O2 , O 22- molecular species the total number of antibonding electrons respectively are (a) 7, 6, 8
(b) 1, 0, 2
(c) 6, 6, 6
(d) 8, 6, 8
(88) Match the following and choose the correct Answer Column - I
Column -II
(i) sp3 d 2
(a) Ni [ (CN)4 ] 2-
(ii) sp3 d
(b) SnCl2
(iii) dsp2
(c) IC l4
(iv) sp2
(d) TeCl4
-
(a) i ® d, ii ® a, iii ® c, iv ® b
(b) i ® c, ii ® d, iii ® a, iv ® d
(c) i ® b, ii ® c, iii ® d, iv ® a
(d) i ® a, ii ® b, iii ® c, iv ® d
(89) Among the following compounds, the one that is polar and has the central atom with sp2 hyridisation is (a) HClO2
(b) BF 3
(c) H2CO 3
(d) SiF4
(90) Match the following Set A
Set B
(1) stability of bond
(p) Bond energy
(2) Molecular orbital theory
(q) Bond order
(3) octet rule
(r) Variable Valency
(4) Valence bond theory
(s) Electronic concept of valency
(a) 1 ® q, 2 ® p, r, 3 ® p, 4 ® s
(b) 1 ® p, q, 2 ® p, 3 ® r, 4 ® s
(c) 1 ® p, q, 2 ® r, 3 ® s, 4 ® r
(d) 1 ® p, q, 2 ® q, 3 ® s, 4 ® r
(91) Bond strength increases with (a) Bond length increasing (b) Antibonding eletrons being higher in number (c) Bond order increasing
(d) Bond angle increasing 92
(92) O 22+ will have (a) Bond order lower than O
2
(b) Bond order higher than O2 (c) Bond order lower than H2 (d) Bond order higher than N2 (93) In a molecule number of electrons in BMO are more as compared to ABMO, hence (a) a bond will be formed (b) no bond will be formed (c) information is not sufficient (d) none of the above (94) The bond angle in the ammonium ion is equal to (a) 90o
(b) 104o
(c) 120o
(d) 109o.281
(95) The correct order of dipole moment is (a) CH4 < NF3 < NH3 < H2O
(b) NF3 < CH4 < NH3 < H2O
(c) NH3 < NF3 < CH4 < H2O
(d) H2O < NH3 < NF4 < CH 4
(96) The correct order of the O – O bond length in O2 , H2O2 and O3is (a) O2 > O3 > H2O 2
(b) O3> H2O2 > O 2
(c) H2O2 > O3 > O 2
(d) O2 > H2O 2 > O 3
(97) The bond order of O -2 is (a) 1.0
(b) 1.5
(c) 2.5
(d) 0.5
(c) O -2
(d) N -2
(98) Choose the incorrect statement. (a) s bond is weaker than p bond (b) p bond is weaker than s bond (c) p bond is present along with a s bond (d) s bond can be present alone (99) Which of the following is not paramagnetic ? (a) NO
(b) S2-
(100) Which one of the following compound has sp2 hybridization ? (a) CO 2
(b) SO 2
(c) CO
93
(d) N2O
Answer key 1d 7d 13 c 19 d 25 b 31 b 37 b 43 c 49 d 55 d 61 b 67 b 73 a 79 c 85 c 91 c 97 d
2c 8b 14 d 20 b 26 b 32 a 38 b 44 d 50 c 56 c 62 d 68 a 74 b 80 a 86 a 92 b 98 a
3c 9c 15 b 21 d 27 c 33 c 39 a 45 c 51 d 57 b 63 d 69 b 75 a 81 c 87 a 93 a 99 b
4b 10 c 16 a 22 b 28 b 34 c 40 b 46 a 52 b 58 d 64 c 70 d 76 a 82 d 88 b 94 d 100 b
5c 11 c 17 c 23 b 29 b 35 b 41 c 47 b 53 d 59 c 65 a 71 c 77 c 83 c 89 c 95 a
6b 12 a 18 a 24 c 30 b 36 c 42 d 48 d 54 b 60 a 66 b 72 b 78 d 84 b 90 d 96 c
Hints 1.
Electronegative difference in KI is more
2.
ACC to V.B.T, O2 contain all e- paired. So they are diamagnetic
6.
Here C - C contain double and triple bond
7.
NO 3- and CO 3-2 contains 32 e- so they are iso electric ions Both possess sp2 hybridization
10.
In H2O2 ,
O - H is polar O - O is non-polar
11.
H-O-N = 0
16.
Struchure A is stable
18.
In NF3 , polarity of N
19.
In CH3Cl, dipole moment is due to C-Cl and C-H
F bond and non-bonding e- pair are in opposite direction
94
20.
trans - pent - 2 - ene possess magnetic moment 21.
Compound Bond angle
SO 2 119.5o
29.
CºO
O=C=O
30.
In H2O,
40.
Bond order O +2 = 2.5
H2O 104.5o
H2S 92.5o
NH 3 106.5o
O Contain two lone pair of eH Contain positive change
O2 = 2 O -2 = 1.5 O -22 = 1
42.
In H2O, O contain two lone pair of e- so repulsion is more
46.
D f H o = D sub H o + D D H o + D i H o + D eg H o + D u H o
61.
Theorectical dipole momentum éë1 D = 1 ´ 10 esa cm ùû = 4.8 ´ 10 -10 ´ 1.275 ´ 10 -8
H =q´d
-18
= 6.12 ´10 -10 e.s.u.cm = 6.12 D Pr actical (m) % of ionic character = Theoretical (m)
1.03 ´ 100 6.12 = 16.83 »17 %
=
95
UNIT : 5 - CHEMICAL THERMODYNAMICS Important Points DE =Ep - Er = qv
DE = q - PDV
DE = q + w
w = -nRT loge VV12
DH = DE + PDV
DH = DE + DngRT
Hess Law ΔH steps = ΔH1 + ΔH 2 + ΔH3 Heat absorbed Heat Capacity = Temperature difference
Heat absorbed Specific heat capacity = (Temperature difference) ´ (Weight of subs ta nce in gram) Heat absorbed Molar heat capacity = (Temperature difference) ´ (Molecular weight) Cpν - C = R
Cp Cν
= g (gamma)
w T2 - T1 = = Thermodynamic efficiency q2 T2 q rev DS = SFinal state – SInitial state = T DSsystem + DSsurrounding > 0 …………………. The reaction will occur spontaneously
DSsystem + DSsurrounding < 0 …………………. The reaction will be non spontaneous DSsystem + DSsurrounding = 0 …………………. The reaction will be in equilibrium Entropy change for an ideal gas : In going from initial to final state, the entropy change, DS for an ideal gas is given by the following relations, T2 V2 (i) When T and V are two variables, ΔS = nC v ln . Assuming Cn is constant + nR ln T1 V1 T p2 (ii) When T and p are two variables, ΔS = nCP ln 2 - nR ln . Assuming Cp, is constant T1 p1 (a) Thus, for an isothermal process (T constant), ΔS = nR ln (b) For isobaric process (p constant), ΔS = n Cp ln
V2 or = - nR V1
ln
p2 p1
T2 T1
T2 T1 (d) Entropy change during adiabatic expansion : In such process q = 0 at all stages. Hence DS = 0. Thus, reversible adiabatic processes are called isoentropic process.
(c) For isochoric process (V constant), ΔS = n C v ln
96
V2 DS = R ln V 1
DG = DH –TDS
DfGoreaction = SDfGoProduct – SD fGoReactant æ total s tan dard free energy of ö æ total s tan dard free energy of ö DfGoreaction = ç ÷ ÷ - çç formation of product formation of reac tan t ÷ø è ø è P2 DG = nRT ln P 1 DGo = – 2.303RTlog K
DG = – nFE Cell Kirchhoff’s equation.
ΔH T2 - ΔH T1 T2 - T1
= ΔCP
Kirchhoff’s equation at constant volume,
ΔE T2 - ΔE T1 T2 - T1
= ΔC ν
Effect of Temperature on Spontaneity of Reactions Dr H0
D rS 0
D rG0
Description*
–
+
–
Reaction spontaneous at all temperature
–
–
– (at low T)
Reaction spontaneous at low temperature
–
–
+ (at high T)
Reaction nonspontaneous at high temperature
+
+
+ (at low T)
Reaction nonspontaneous at low temperature
+
+
– (at high T)
Reaction spontaneous at high temperature
+
–
+ (at all T)
Reaction nonspontaneous at all temperature
æ ¶T ö Joule thomson coeffient μ = ç ÷ è ¶p ø H
For cooling m = + ve For Heating m = –ve Neither cooling nor heating m = 0 The temperature at which a real gas shows neither cooling nor heating effect on adiabatic expansion (i.e., m = 0) is called inversion temperature. Hydrogen has highest calorific value. 13.7 Kcal/mol = 57 KJ/mol (be cause of 1cal = 4.2 Joule) Enthalpy of fusion of ice per mole is 6KJ. Order of bond energy in halogen Cl2 > Br2 > F2 > I2. Heat of vapourisation of water per mole is 10.5 KCal. The heat of reaction is independent of the time consumed in the process. æ ¶T ö Joule thomson coeffient μ = ç ÷ è ¶p ø H
97
M.C.Q. 1.
2.
3.
4. 5. 6. 7.
8.
9.
10.
11.
The temperature of the system decreases in an (A) Adiabatic compression (B) Isothermal compression (C) Isothermal expansion (D) Adiabatic expansion If a refrigerator’s door is opened, then we get (A) Room heated (B) Room cooled (C) More amount of heat is passed out (D)No effect on room The cooling in refrigerator is due to (A) Reaction of the refrigerator gas (B) Expansion of ice (C) The expansion of the gas in the refrigerator (D) The work of the compressor The process, in which no heat enters or leaves the system, is termed as (A) Isochoric (B) Isobaric (C) Isothermal (D) Adiabatic Warming ammonium chloride with sodium hydroxide in a test tube is an example of : (A) Closed system (B) Isolated system (C) Open system (D) None of these Out of boiling point (I), entropy (II), pH (III) and e.m.f. of a cell (IV), intensive properties are – (A) I, II (B) I, II, III (C) I, III, IV (D)All the above A thermodynamic state function is (A) one which obeys all the laws of thermodynamics (B) a quantity which is used in measuring thermal changes (C) one which is used in thermo chemistry (D) a quantity whose value depends only on the state of the system. In thermodynamics, a process is called reversible when (A) surroundings and system change into each other (B) there is no boundary between system and surroundings (C) the surroundings are always in equilibrium with the system (D) the system changes into the surroundings spontaneously Which one of the following statement is false– (A) work is a state function (B) temperature is a state function (C) change in the state is completely defined when the initial and final states are specified (D) work appears at the boundary of the system. A mixture of two moles of carbon monoxide and one mole of oxygen, in a closed vessel is ignited to convert the carbon monoxide to carbon dioxide. If DH is the enthalpy change and DE is the change in internal energy, then (A) DH < DE (B) DH > DE (C) DH = DE (D) The relationship depends on the capacity of the vessel At constant T and P, which one of the following statements is correct for the reaction, CO(g) +
1 O2 (g) ® CO2 (g) 2
(A) DH is independent of the physical state of the reactants of that compound (B) DH < DE (C) DH > DE (D) DH = DE 98
(A) The randomness of the universe decreases (B) The randomness of the surroundings decreases (C) Increase is randomness of surroiundings equals the decrease in randomness of system (D) The increase in randomness of the surroundings is greater as compared to the decrease in randomness of the system. 21.
The enthalpy change for a given reaction at 298 K is – x J mol–1 (x being positive). If the reaction occurs spontaneously at 298 K, the entropy change at that temperature (A) can be negative but numerically larger than x/298 (B) can be negative but numerically smaller than x/298 (C) cannot be negative
22.
23.
(D) cannot be positive
Spontaneous adsorption of a gas on a solid surface is exothermic process because (A) enthalpy of the system increases.
(B) entropy increases.
(C) entropy decreases.
(D) free energy change increases.
Identify the correct statement regarding entropy : (A) At absolute zero, the entropy of a perfectly crystalline substance is +ve. (B) At absolute zero, the entropy of a perfectly crystalline substance is zero. (C) At 0°C the entropy of a perfectly crystalline substance is taken to be zero. (D) At absolute zero of temperature the entropy of all crystalline substances is taken to be zero.
24.
Identify the correct statement regarding a spontaneous process : (A) Exothermic processes are always spontaneous. (B) Lowering of energy in the reaction process is the only criterion for spontaneity. (C) For a spontaneous process in an isolated system, the change in entropy is positive. (D) Endothermic processes are never spotaneous.
25. DS will be highest for the reaction
26.
(A) Ca(s) + l/2 O2(g) ® CaO(s)
(B) CaCO3(s) ® CaO (s) + CO2(g)
(C) C(s) + 02(g) ® CO2(g)
(D) N2(g) + O2(g) ® 2NO (g)
The spontaneous flow of heat is always (A) unidirectional from higher temperature to lower temperature (B) from high to low pressure (C) unidirectional from lower temperature to higher temperature (D) from low to high pressure.
27.
Which of the following is zero during adiabatic expansion of the gas (A) DT
28.
(B) DS
(C) DE
(D) All the above
The occurrence of a reaction is impossible if (A) DH is +ve; DS is also + ve but DH < TDS
(B) DH is –ve; DS is also –ve but DH > TDS
(C) DH is – ve; DS is + ve
(D) DH is + ve; DS is – ve 100
29.
30.
31.
32.
33.
34. 35. 36.
37.
Identify the correct statement regarding entropy (A) At 00C, the entropy of a perfectly crystalline substance is taken to be zero (B) At absolute zero of temperature, the entropy of a perfectly crystalline substance is +ve (C) At absolute zero of temperature, the entropy of all crystalline substances is taken to be zero (D) At absolute zero of temperature, the entropy of a perfectly crystalline substance is taken to be zero A container has hydrogen and oxygen mixture in ratio of 4 : 1 by weight, then (A) Internal energy of the mixture decreases (B) Internal energy of the mixture increases (C) Entropy of the mixture increases (D) Entropy of the mixture decreases The second law of thermodynamics says that in cyclic process. (A) Work cannot be converted into heat (B) Heat cannot be converted into work (C) work cannot be completely converted into heat (D) Heat cannot be completely converted into work A heat engine absorbs heat Q1 at temperature T1 and heat Q2 at temperature T2. Work done by the engine is (Q1 + Q2). This data (A) Violates Ist law of thermodynamics (B) Violates Ist law of thermodynamics if Q1 is –ve (C) Violates Ist law of thermodynamics if Q2 is –ve (D) Does not violate Ist law of thermodynamics The molar neutralization heat for and as compared to molar neutralization heat of NaOH and HCl (A) Less (B) More (C) Equal (D) Depends on pressure If the enthalpy of B is greater than of A, the reaction A®B is (A) Endothermic (B) Exothermic (C) Instantaneous (D) Spontaneous Which of the following fuels will have the highest calorific value (kJ/kg) (A) Charcoal (B) Kerosene (C)Wood (D) Dung Which is the best definition of “heat of neutralization” (A) The heat set free when one gram molecule of a base is neutralized by one gram molecule of an acid in dilute solution at a stated temperature (B) The heat absorbed when one gram molecule of an acid is neutralized by one gram molecule of a base in dilute solution at a stated temperature (C) The heat set free or absorbed when one gram atom of an acid is neutralized by one gram atom of a base at a stated temperature (D) The heat set free when one gram equivalent of an acid is neutralized by one gram equivalent of a base in dilute solution at a stated temperature Compounds with high heat of formation are less stable because (A) High temperature is required to synthesise them (B) Molecules of such compounds are distorted (C) It is difficult to synthesis them (D) Energy rich state leads to instability
101
38.
39.
40.
When a gas undergoes adiabatic expansion, it gets cooled due to (A) Loose of kinetic energy (B) Fall in temperature (C) Decrease in velocity (D) Energy used in doing work For a reaction to occur spontaneously (A) (DH – TDS) must be negative (B) (DS + TDS) must be negative (C) DH must be negative (D) DS must be negative If for a given substance melting point is TB and freezing point is TA, then correct variation shown by graph between entropy change and temperature is
(A)
41. 42.
43.
44. 45.
46.
48.
(C)
(D)
A Beckmann thermometer is used to measure (A) High temperature (B) Low temperature (C) Normal temperature (D) All temperature The calorific value of fat is ………. (A) less than carbohydrates and protein (B) less than that of protein but more than carbohydrates (C) less than that of carbohydrates and more than that of protein (D) more than thant of carbohydrates and protein Which of the following processes is accompanied by an increase in entropy ? (A) Normal rubber band to stretched rubber band (B) Normal egg to hard boiled egg (C) Decomposition of N2O5 to N2O to O2 (D) Formation of NH3 for N2H2. Which of the following does not exhibit zero entropy at absolute zero (A) Benzene (B) Glass (C) Pyridine (D) CCl4 The favourable conditions for a spontaneous reaction are (A) TDS > DH, DH = +ve, DS = +ve (B) TDS > DH, DH = +ve, DS = –ve (C) TDS = DH, DH = –ve, DS = –ve (D) TDS = DH, DH = +ve, DS = +ve. A block of ice at –10 °C is slowly heated and converted into steam at 100°C .Which of the following curves represents the phenomenon qualitatively ?
(A)
47.
(B)
(B)
(C)
(D)
On passing CO2 gas in water, its entropy (A) Remains constant (B) Decreases (C) Increases (D) May increase or decrease. When does the reaction occur spontaneously on the basis of the relation DG° = –RT/nK? (A) K = 0 (B) K = 1 (C) K > 1 (D) K < 1 102
49.
In thermodynamics, a process is called reversible when, (A) Surroundings and system change into each other (B) The surroundings are always in equilibrium with the system (C) The system changes into the surroundings spontaneously. (D) There is no boundary between system and surroundings.
50.
Under certain conditions, the value of DG for a hypothetical reaction, X + Y ¾® Z is greater than zero, then – (A) The reaction has tendency to proceed towards Z (B) The reaction has attained equilibrium (C) increase in temperature increases the yield of product Z (D) X and Y predominate in the final mixture
51.
52.
For which of the following processes will energy be absorbed – (A) Separating an electron from an electron
(B) Separating proton from a proton
(C) Separating a neutron from neutron
(D) Separating an electron from neutral atom
For the combustion of 1 mole of liquid benzene at 25ºC, the heat of reaction at constant pressure is given by, C6H 6(l) + 7 O2 (g) ® 6CO2 (g) + 3H2O (l); DH = –780980 cal. What would be the heat of reaction at constant volume? (A) –780090 cal
53.
(B) –780890 cal
(C) –780000cal
(D) –780900 cal
Calculate heat of the following reaction at constant pressure, F2O(g) + H2O(g) ® O2 (g) + 2HF(g) The heats of formation of F2O (g), H2O(g) and HF (g) are 5.5 kcal, –57.8kcal and 64.2 kcal respectively. (A) 76.1 Kcal
54.
(B) 11.9 Kcal
(C) 71.6 Kcal
(D) 91.1 Kcal
Calculate the heat of formation of benzene from the following data, assuming no resonance. Bond energies : C – C = 83 kcal = 140 kcal C – H = 99 kcal Heat of atomisation of C = 170 .9 kcal Heat of atomisation of H = 52.1 kcal (A) 70 Kcal
55.
(B) 75 Kcal
(C) –75 Kcal
(D) –70 Kcal
Calculate DH at 358 K for the reaction Fe2O 3(s) + 3H2(g) ®2Fe (s) + 3H2O(l) Given that, DH298= – 33.29 kJ mole–1 and Cp for Fe2O 3 (s), Fe (s), H2O (l) and H2 (g) are 103.8, 25.1, 75.3 and 28.8 J/K mole. (A) –22.22 KJ/mole
56.
(B) –25. 123 KJ/mole
(C) –28.136 KJ/mole (D) – 30.135 KJ/mole
Ka for CH3COOH at 25ºC is 1.754 × 10–5 . At 50ºC, Ka is 1.633 × 10–5 What will be value of D Sº for the ionisation of CH3COOH? (A) –94.44 J/mole K
(B) –96.66 J/mole K 103
(C) –96.44 J/mole K (D) –90.44 J/mole K
57.
58.
59.
60.
61.
62.
63.
64.
C2H6 (g) + 3.5 O2 (g) ® 2CO2 (g) + 3H2O (g) DSvap (H2O, l) = x1 cal K–1 (boiling point + T1) DHf (H2O, l) = x2 DHf (CO2) = x3 DHf (C2H6) = x4 Hence DH for the reaction is – (A) 2x3 + 3x2 – x4 (B) 2x3 + 3x2 – x4 + 3x1T1 (C) 2x3 + 3x2 – x4 – 3x1T1 (D) x1T1 + X2 + X3 – x4 C (s) + O2 (g) ® CO2, (g); DH = –94.3 kcal/mol CO (g) + O2(g) ® CO2 (g); DH = – 67.4 kcal/mol O2(g) ® 2O (g); DH = 117.4 kcal/mol CO (g) ® C (g) + O(g) ; DH = 230.6 kcal/mol Calculate DH for C (s) ® C (g) in kcal/mol. (A) 171 (B)154 (C)117 (D)145 The difference between DH and DE on a molar basis for the combustion of n–octane at 25°C would be : 25ºC (A) – 13.6 kJ (B) – 1.14 kJ (C) – 11.15 kJ (D) + 11.15 kJ What is the work done against the atmosphere when 25 grams of water vaporizes at 373 K against a constant external pressure of 1 atm ? Assume that steam obeys perfect gas laws. Given that the molar enthalpy of vaporization is 9.72 kcal/mole, what is the change of internal energy in the above process ? (A) 1294.0 cals, 11247 cals (B) 921.4 cals, 11074 cals (C) 1029.4 cals, 12470.6 cals (D) 1129.3 cals, 10207 cals In the reaction CS2 (l) + 3O 2 (g) ¾¾® CO2 (g) + 2SO2 (g) DH = –265 kcal The enthalpies of formation of CO2 and SO2 are both negative and are in the ratio 4 : 3. The enthalpy of formation of CS2 is + 26 kcal/mol. Calculate the enthalpy of formation of SO2. (A) – 90 kcal/mol (B) – 52 kcal/mol (C) – 78 kcal/mol (D) – 71.7 kcal/mol The bond dissociation energy of gaseous H2, Cl2 and HCl are 104, 58 and 103 kcal mol–1 respectively. The enthalpy of formation for HCl gas will be (A) – 44.0 kcal (B) – 22.0 kcal (C) 22.0 kcal (D) 44.0 kcal AB, A2 and B2 are diatomic molecules. If the bond enthalpies of A2, AB & B2 are in the ratio 1 : 1 : 0.5 and enthalpy of formation of AB from A2 and B2 is – 100 kJ/mol–1. What is the bond enthalpy of A2. (A) 400 kJ/mol (B) 200 kJ/mol (C) 100 kJ/mol (D) 300 kJ/mol Given the following data : Substance DH° (kJ/mol) S°(J/mol K) DG° (kJ/mol) FeO(s) – 266.3 57.49 – 245.12 C (Graphite) 0 5.74 0 Fe(s)0 27.28 0 CO(g) – 110.5 197.6 – 137.15 104
Determine at what temperature the following reaction is spontaneous ? FeO(s) + C (Graphite)® Fe(s) + CO(g)
65.
(A) 298 K
(B) 668 K
(C) 966 K
(D) DG° is +ve, hence the reaction will never be spontaneous
One mole of a gas occupying 3 dm3 expands against constant external pressure of 1 atm to a volume of 13 dm3. The work done is – (A) – 10 atm dm3
66.
(C) – 39 atm dm3
(D) – 48 atm dm3
The enthalpy change in the oxidation of glucose is – 2880 kJ mol–1. Twenty five per cent of this energy is available for muscular work. If 100 kJ of muscular work is needed to walk one kilometre, then the maximum distance that a person will be able to walk after eating 120 g of glucose will be (A) 4.8 km
67.
(B) – 20 atm dm3
(B) 2.4 km
(C) 8.4 km
(D) 9.8 km
The heat of formation of liquid methyl alcohol is kilojoule per mole using the following data will be [Heat of vaporisation of liquid methyl alcohol = 38 kJ/mol. Heat of formation of gaseous atoms from the elements in their standard states : H, 218 kJ/mol; C, 715 kJ/mol; O, 249 kJ/mol. Average bond energies : C – H, 415 kJ/mol; C – O, 356 kJ/mol O – H, 463 kJ/mol.] (A) 46.0 kJ/mole
68.
(B) 50.0 kJ/mole
(C) 73.3 kJ/mole
(D) – 266 kJ/mole
10 g of argon gas is compressed isothermally and reversibly at a temperature of 27°C from 10 L to 5 L. q, W, DE and DH for this process are [R = 2.0 cal K–1 mo l–1, log102 = 0.30. [Atomic wt. of Ar = 40.] (A) W = 106.635 cal, q = 103.635 cal, DE ¹ 0 & DH = 0 (B) W = 53.635 cal, q = – 53.635 cal, DE ¹ 0 & DH = 0 (C) W = – 53.635 cal, q = 63.635 cal, DE & DH ¹ 0 (D) W = 103.635 cal, q = – 103.635 cal, DE & DH = 0
69.
Molar heat capacity of water in equilibrium with ice at constant pressure is – (A) zero
70.
(B) infinity (¥)
(C) 40.45 kJ–1 mol–1
(D) 75.48 J K–1 mol–1
Diborane is a potential rocket fuel which undergoes combustion according to the reaction, B2H6 (g) + 3O2 (g) ®B2O3 (s) + 3H2O(g) from the following data, the enthalpy change for the combustion of diborane will be
71.
2B(s) + O2 (g) ®B2O3(s);
DH = – 1273 kJ
H2(g) + O2 (g) ®H2O(l);
DH = – 286 kJ
H2O(l) ®H2O(g) ;
DH = 44 kJ
2B(s) + 2H2 (g) ®B2H6 (g);
DH = 46 kJ
(A) – 2079 kJ mol–1
(C) – 2035 kJ mol–1 (D) – 762 kJ mol–1
(B) – 1091 kJ mol–1
A sample of argon gas at 1 atm pressure and 27°C expands reversibly and adiabatically from 1.25 dm 3 to 2.50 dm3. The enthalpy change in this process will be………. [Cv.m. for argon is 12.48 jK–1 mol–1]. (A) 114.52 J
(B) – 114.52 J
(C) – 57.26 J 105
(D) 57.26 J
72.
73. 74.
75.
76.
Find DG° and DH° for that the reaction CO(g) + O2 (g) ® CO2 (g) at 300 K respectively are, when the standard entropy change is – 0.094 kJ mol–1 K–1. The standard Gibbs free energies of formation for CO2 and CO are – 394.4 and – 137.2 kJ mol–1, respectively. (A) DG° = 257.2 kJ/mol, DH° = 285.4 kJ/mol (B) DG° = 514.4 kJ/mol, DH° = – 570.8 kJ/mol (C) DG° = +514.4 kJ/mol, DH° = 570.8 kJ/mol (D) DG° = – 257.2 kJ/mol, DH° = – 285.4 kJ/mol DH = 30 kJ mol–1, DS = 75 J / k / mol. Find boiling temperature at 1 atm. (A) 400 K (B) 300 K (C) 150 K (D) 425 K Spontaneous adsorption of a gas on a solid surface is exothermic process because (A) enthalpy of the system increases. (B) entropy increases. (C) entropy decreases. (D) free energy change increases. There is 1 mol liquid (molar volume 100 ml) in an adiabatic container initial, pressure being 1 bar Now the pressure is steeply increased to 100 bar, and the volume decreased by 1 ml under constant pressure of 100 bar. Calculate DH and DE. [Given 1 bar = 105 N/m2] (A) DE = 0 J, DH ¹ 0 J (B) DH = 0 J, DE = 10 J (C) DE = 20 J, DH = 890 J (D) DE = 0 J, DH = 10 J The ratio of P to V at any instant is constant and is equal to 1, for a monoatomic ideal gas under going a process. What is the molar heat capacity of the gas (A)
77.
78.
79.
80.
81.
82.
3R 2
(B)
(C)
4R 2
5R 2
(D) 0
The entropy values (in J K–1 mol–1) of H2 (g) = 130.6 Cl2(g) = 223 and HCl(g) = 186.7 at 298 K and 1 atmpressure are given. Then entropy change for the reaction. (A) + 540.3 (B) +727.3 (C) – 166.9 (D) +19.8 A mixture of 2 mole of CO(g) and one mole of O2 in a closed vessel, is ignited to convert the carbon monoxide to carbon dioxide. If DH and DU are enthalpy and internal energy change. Then (A) DH > DU (B) DH < DU (C) DH = DU (D) the relationship depends on the capacity of the vessel. For the reaction of one mole zinc dust with one sulphuric acid in a bomb calorimeter, DU and w correspond to : (A) DU < 0, w = 0 (B) DU < 0, w < 0 (C) DU > 0, w = 0 (D) DU > 0, w > 0 If the enthalpies of formation of Al2O3 and Cr2O3 are – 1596 kJ and – 1134 kJ respectively, then the value of DH for the reaction ; 2Al + Cr2O3 ® 2Cr + Al2O3 is : (A) – 462 kJ (B) – 1365 kJ (C) – 2530 kJ (D) +2530 kJ The internal energy change when a system goes from state A to B is 40 kJ/mole. If the system goes from A to B by a reversible path and returns to state A by an irreversible path what would be the net change in internal energy (A) < 40 kJ (B) Zero (C) 40 kJ (D) > 40 kJ DG° for the reaction x + y ® z is – 4.606 kcal. The value of equilibrium constant of the reaction at 227°C is : (R = 2.0 cal K–1 mol–1 ) (A) 100 (B) 10 (C) 2 (D) 0.01 106
83.
The latent heat of vaporisation of a liquid at 500 K and 1 atm pressure is 10 kcal/mol. What will be the change in internal energy (DE) of 3 moles of liquid at the same temperature? (A) 13.0 kcal
84.
(C) 27.0 kcal
(D) – 27.0 kcal
The work done in ergs for a reversible expansion of one mole of an ideal gas from a volume of 10 litres at 25°C is : (A) 3.43 KJ
85.
(B) – 13.0 kcal
(B) 3.43 Kcal
(C) 3.43 J
(D) 3.43 cal
(C) – 6.20 kcal
(D) 6.20 kcal
Reaction, H2(g) + I2 (g) ® 2HI; DH = 12.40 kcal. According to this, heat of formation of HI will be (A) 12.40 kcal
86.
The heat of combustions of yellow phosphorus and red phosphorus are – 9.91 kJ and – 8.78 kJ respectively. The heat of transition of yellow phosphorus to red phosphorus is : (A) – 18.69 kJ
87.
(C) – 120.6 kcal
(D) +52.8 kcal
(B) – 71030 cal
(C) 70 cal
(B) – 900 kJ
3
–2
(D) – 70 cal 3
(C) 270 kJ
(D) + 900 kJ
The enthalpies of combustion of carbon and carbon monoxide are – 393.5 and – 283 kJ mol–1 respectively. The enthalpy of formation of carbon monoxide per mole is (B) 676.5 kJ
(C) – 676.5 kJ
(D) – 110.5 kJ
If the bond dissociation energies of XY, X2 and Y2 (all diatomic molecules) are in the ratio of 1 : 1 : 0.5 and DHf for the formation of XY is – 200 KJ mol–1. The bond dissociation energy of X2 will be (A) 100 KJ mol–1
92.
(D) – 1.13 kJ
An ideal gas expands in volume from 1 × 10 m to 1 × 10 m at 300 K against a constant pressure of 1 × 105 Nm–2 . The work is :
(A) 110.5 kJ 91.
(B) – 67.6 kcal
–3
(A) – 900 J 90.
(B) +18.69 kJ
The heats of combustion of rhombic and monoclinic sulphur are – 70960 and – 71030 calorie respectively. What will be the heat of conversion of rhombic sulphur to monoclinic sulphur? (A) – 70960 cal
89.
(B) +1.13 kJ
The heat of formation of CO(g) and CO2 (g) are – 26.4 kcal and – 94.0 kcal respectively. The heat of combustion of carbon monoxide will be : (A) + 26.4 kcal
88.
(B) – 12.4 kcal
(B) 200 KJ mol–1
(C) 300 KJ mol–1
(D) 800 KJ mol–1
Consider the reaction, N2(g) + 3H2(g) 2NH3(g); carried out at constant temperature and pressure. If DH and DU are enthalpy change and internal energy change respectively, which of the following expressions is true ? (A) DH = 0
93.
(B) DH = DU
(C) DH < DU
(D) DH > DU
An ideal gas is allowed to expand both reversibly and irreversibly in an isolated system. If Ti is the initial temperature and Tf is the final temperature, which of the following statements is correct ? (A) Tf > Ti for reversible process but Tf = Ti for irreversible process (B) (Tf)rev = (Tf)irrev
(C) Tf = Ti for both reversible and irreversible processes
(D) (Tf)irrev > (Tf)rev 107
94.
Identify the correct statement regarding a spontaneous process : (A) Exothermic processes are always spontaneous. (B) Lowering of energy in the reaction process is the only criterion for spontaneity. (C) For a spontaneous process in an isolated system, the change in entropy is positive. (D) Endothermic processes are never spontaneous. 95. In conversion of lime–stone to lime, CaCO3(s) ® CaO(s) + CO2(g) the values of DH0 and DS 0 are +179.1 kJ mol–1 and 160.2 J/K respectively at 298 K and 1 bar. Assuming that DH° and DS° do not change with temperature, temperature above which conversion of limestone to lime will be spontaneous is : (A) 845 K (B) 1118 K (C) 1008 K (D) 1200 K 96. For a reversible process at T = 300 K, the volume is increased from Vi = 1 L to Vf = 10 L. Calculate DH if the process is isothermal (A) 11.47 kJ (B) 4.98 kJ (C) 0 (D) – 11.47 kJ 97. If at 298 K the bond energies of C–H, C–C, C=C and H–H bonds are respectively 414, 347, 615 and 435 kJ mol–1, the value of enthalpy change for the reaction ; H2C = CH2(g) + H2(g) ® H3C–CH3(g) at 298 K will be – (A) +125 kJ (B) —125 kJ (C) +250 kJ (D) — 250 kJ 98. Considering entropy(s) as thermodynamic parameter, the criterion for the spontaneity of any process is : (A) DS system + DS surroundings > 0 (B) DS system – DS surroundings > 0 (C) DSsystem > 0 only (D) DS surroundings > 0 only 99. Assuming that water vapour is an ideal gas, the internal energy change (DU) when 1 mol of water is vapourisedat 1 bar pressure and 100°C, (Given : Molar enthalpy of vapourization of water at 1 bar and 373 K = 41 kJ mol–1 and R = 8.3 J mol–1 K–1) will be : (A) 37.904 kJ mol–1 (B) 41.00 kJ mol–1 (C) 4.100 kJ mol–1 (D) 3.7904 mol–1 100. The standard enthalpy of formation (DHf°) at 398 K for methane, CH4(g) is 74.8 kJ mol–1. The additional information required to determine the average energy for C – H bond formation would be. (A) the dissociation energy of H2 and enthalpy of sublimation of carbon (B) latent heat of vapourisation of methane (C) the first four ionization energies of carbon and electron gain enthalpy of hydrogen (D) the dissociation energy of hydrogen molecule, H2 101. Standard entropy of X2, Y2 and XY3 are 60, 40 and 50 JK–1 mol–1, respectively. For the reaction,1/2 X2 + 3/2 Y2 ® XY3 DH = – 30 kJ. To be at equilibrium the temperature will be : (A) 500 K (B) 750 K (C) 1000 K (D) 1250 K + 102. On the basis of the following thermochemical data : (DƒGºH (aq) = 0) H2O(l) ¾® H+ (aq) + OH– (aq.) ; DH = 57.32 kJ H2(g) + O2(g) ¾® H 2O(l); DH = –286.20 kJ The value of enthalpy of formation of OH– ion at 25ºC is : (A) –228.88 kJ (B) +228.88 kJ (C) –343.52 kJ (D) –22.88 kJ 108
103. In a fuel cell methanol is used as fuel and oxygen gas is used as an oxidizer. The reaction is CH3OH(l) + 3/2O2 (g) ¾® CO 2(g) + 2H2O(l) At 298 K, standard Gibb’s energies of formation for CH3OH(l), H2O(l) and CO2 (g) are –166.2, –237.2 and –394.4 kJ mol–1 respectively. If standard enthalpy of combustion of methanol is –726kJ mol–1, efficiency of the fuel cell will be : (A) 87% (B) 90% (C) 97% (D) 80% –1 104. The standard enthalpy of formation of NH3 is – 46.0 kJ mol . If the enthalpy of formation of H2 from its atoms is –436 kJ mol–1 and that of N2 is –712 kJ mol–1, the average bond enthalpy of N – H bond in NH3 is (A) – 964 kJ mol– 1 (B) + 352 kJ mol– 1 (C) + 1056 kJ mol– 1 (D) – 1102 kJ mol– 1 105. For a particular reversible reaction at temperature T, DH and DS were found to be both +ve. If Te is the temperature at equilibrium, the reaction would be spontaneous when. (A) Te > T (B) T > Te (C) Te is 5 times T (D) T = Te 106. Choose the correct option about the following sentnences [T= True , F =False] (i) Ice in contact with water constitutes a homogeneous system. (ii) The process is known as isochoric in which the pressure remains constant throughout the change, i.e., dP = 0. (iii) A spontaneous process is reversible in nature. (iv) In an isolated system, one form of energy cannot be converted into another, i.e., internal energy remains constant. [A] FFFF [B] TTTT [C] FTFT [D] FFFT 107. Choose the correct option about the following sentnences [T= True , F =False] (i) Molar heat capacity at constant pressure = Molar heat capacity at constant volume + PDV. (ii) A spontaneous process is accompanied by a decrease in entropy. (iii) DHsub = DHfusion + DHvap. (iv) The standard heat of formation represents the formation of the compound from its elements at 25°C and one atmospheric pressure. (v) Whenever an acid is neutralised by a base, the net reaction is H+ (aq) + OH– (aq) ® H2O(l) ; DH = – 13.7 kcal [A] TFTTF [B] TFTTF [C]TFTFTF [D] TFFFF 108. Match the following : Column I Column II (i) A process carried out infinitesimally slowly (A) Adiabatic (ii) A process in which no heat enters or leaves the system (B) DG = 0 (iii) A process carried out at constant temperature (C) Sublimation (iv) A process in equilibrium (D) DE = 0, DH = 0 (v) A(s) ® A(g) (e) Reversible (vi) Cyclic process (f) Isothermal (A) (i – e, ii – a, iii – f, iv – b, v – c, vi – e) (B) (i – e, ii – a, iii – f, iv – e v – c, vi – d) (C) (i – e, ii – a, iii – f, iv – b, v – c, vi – d) (D) (i – e, ii – a, iii – c, iv – b, v – f, vi – d) 109
109. Match the following Column–I A. Isothermal process B. Adiabatic process C. Isobaric process D. Isochoric process
Column–II P. q = DU Q. w = – PDV R. w = DU S. w = –n RT ln (V2/V1)
(A) A= S B= R C= Q D= P (C) A= R B= S C= Q D= P
(B) A= P B= Q C= R D= S (D) A= S B= R C= P D= Q
Question 110 to 115 are reasoning question choose the correct statement. (A) Statement –1 is true, Statement –2 is true.Statement –2 is the correct explanation for statement –1 (B) Statement –1 is true,Statement –2 is true.Statement –2 is not the correct explanation for statement –1 (C) Statement –1 is true, Statement –2 is False (D) Statement –1 is false Statement –2 is True 110. Statement–1 : The enthalpy of formation of H2O(l) is greater than of H2O (g). Statement–2: Enthalpy change is negative for the condensation reaction H2O (g)® H2O(l) (A) A (B) B (C) C (D) D 111. Statement–1 : Heat of neutralisation of perchloric acid, HClO3, with NaOH is same as that of HCl with NaOH. Statement–2: Both HCl and HClO4 are strong acids. (A) A (B) B (C) C (D) D 112. Statement–1 : When a gas at high pressure expands against vacuum, the work done is maximum. Statement–2: Work done in expansion depends upon the pressure inside the gas and increase in volume. (A) A (B) B (C) C (D) D 113. Statement–1 : In the following reaction :C(s) + O2 (g) ®CO2 (g); DH = DU – RT Statement–2: DH is related to U by the equation, DH = DU – Dng RT (A) A (B) B (C) C (D) D 114. Statement I : The chemical reaction, 3H2(g) + N2(g) ¾® 2NH3 shows decrease in entropy. Statement II: The process passes into equilibrium state when DGT,P becomes zero. (A) A (B) B (C) C (D) D 115. Statement I : Both H and U are state functions. Statement II: Absolute values of H or U can be determined. (A) A (B) B (C) C (D) D 116. 1 mole of NH3 gas at 27° C is expanded adiabatic condition to make volume 8 times (g = 1.33). Final temperature and work done respectively are – (A) 150 K, 900 cal (B) 150 K, 400 cal (C) 250 K, 1000 cal (D) 200 K, 800 cal 110
117. Calculate the work performed when 2 moles of hydrogen expand isothermally and reversibly at 25ºC form 15 to 50 litres. (A) –1438 Cal
(B) –1436 cal
(C) –1348 cal
(D) –1346 cal
118. The DHf0 for CO2(g), CO(g) and H2O (g) are –393.5, –110.5 and –241.8 kJ mol–1 respectively. The standard enthalpy change (in kJ) for the reaction CO2(g) + H2(g) ® CO(g) + H2O (g) is – (A) 524.1
(B) 41.2
(C) – 262.5
(D) – 41.2
119. For reaction carried out in automobiles, what is the value of D H, DS and DG ? 2C8H 18(g) + 25O2(g) 16CO2(g)+ 18H2O(g) (A) +, –, +
(B) –, + , –
(C) –, +, +
(D) +, +, –
120. At 25° temperature equilibrium constant Kp for given reaction is = 1.8 10–7 Then what is the value of DG° ? PCl5 D PCl3 + Cl2 (A)+9197.5
(B) –9197.5
(C) +771
(D) –771.6
121. The conversion A to B is carried out by the following path : Given : DS(A ®C) = 50 e.u. , DS(C ® D) = 30 e.u. , DS(B ® D) = 20 e.u. Where e.u. is entropy unit then DS(A ® B) is (A) +100 e.u.
(B) + 60 e.u.
(C) –100 e.u.
(D) – 60 e.u.
122. For the chemical reaction A + B ¾® P + Q two paths are given in the diagram. Which of the following relationship is correct –
(A) DH1 + DH2 = DH3 + DH 4 (C) DH3 – DH1 = DH4 – DH 2
123. For the reaction,
(B) DH1 + DH2 = DH3 – DH 4 (D) DH1 – DH2 = DH3 + DH 4
+ H — H ¾®
bond energies are given as under –
(i) C—C, 346 kJ/mol (ii) C—H, 413 kJ/mol (iii) H—H, 437 kJ/mol and (iv) C = C, 611 kJ/mol What will be the value of DH 25ºC for the above reaction ? (A) –289 kJ mol–1 (B) – 124 kJ mol–1 (C) + 124 kJ mol–1 (D) +289 kJ mol–1 124. The value of DHfº of U3O 8 is –853.5 KJ mol–1. Also DHº for the reaction 3UO2 + O2 ¾® U3O8, is –76.00 KJ. The value of DHf º of UO2 is approx – (A) –259.17 KJ (B) –310.17 KJ (C) + 259.17 KJ (D) 930.51 KJ 125. The heat procuced by complete neutralisation of 100 ml of HNO3 with 300 ml of decimolar KOH solution is 1.713 kJ. The molarity of HNO3 solution will be – (A) 0.1 (B) 1 (C) 0.3 (D) 0.5 111
Hints (16)
W = 2.303 nRT log
V2 20 = 2.303 ´ 1 ´ 8.314 ´ 107 ´ 298 log V1 10
= 298 ´ 107 ´ 8.314 ´ 2.303 log 2 (20)
As dew formation is spontaneous process therefore entropy or randomness of the universe will increase. As randomeness of the system has decreased but randomness of the surrounding will increase larger so that change is positive.
(21)
It I is because of the fact that for spontaneity, the value of DG = (DH – TDS) should be < 0. If DS is – ve, the value of TDS shall have to be less than DH or the value of DS has to be less than that of x
(24)
In an isolated system, there is no exchange of energy or matter between the system and surrounding. For a spontaneous process in an isolated system, the change in entropy is positive, i.e. DS > 0
(52)
DH = DE + Dng RT Here, Dng = 6 – 7.5 = – 1.5.
(53)
Thus, DE = DH + Dng RT = – 780980 – (–1.5 ) × 2 × 298 = – 780090 calories. F2O (g) + H2O (g) ® O2 (g) + 2HF(g); DH = 76.1 kcal.
(55)
DCp = 2 × 25.1 + 3 × 75.3 – [103.8 + 3× 28.8] = 85.9 J/K mole. We have,
DH 2 - DH1 = DCp T2 - T1
DH 358 - (-33290) = 85.9 358 - 298
(56)
DH358 = –28136 J/mole = –28.136 kJ/mole. (DGº)298 = – 2.303RT log K = – 2.303 × 8.314 × 298 × log (1.754 ×10-5) = 27194 J. (DGº)323 = 2.303 × 8.314 × 323 × log ( 1.633 × 10-5) = 29605 J. DGº = DHº – TDSº 27194 = DHº – 298 DSº 29605 =DHº – 323 DSº DSº = – 96.44J/mol.K
(59)
DH – DE = – 4.5 × 8.315 × 298 J = – 11.15 kJ
(61)
Cs2 (l) + 3O2 (g) ® Co2 (g) + 2SO2 + 2SO2 DH = -265Kcal Let DHf (CO2, g) = 4x and DHf (SO2 , g) = 3x DHreaction = DHf (CO2 ,g) = 2 DHf (SO2 .g) – DHf (CS2 l) – 265 = 4x + 6x –26 x = –23.9 \ DHf (SO2,g) = 3x = -71.7 Kcal / mol. 115
(62)
Given H2 (g)® HCl (g); Cl2 (g) ® 2Cl(g);
DH = 104 kcal
...(1)
DH = 58 kcal
...(2)
HCl (g)® H(g) + Cl(g);
DH = 103 kcal
...(3)
Heat of formation for HCl H2 (g) + Cl2 (g) ®HCl (g); DH = ? Divide equation (1) and (2) by 2, and then add H2 (g) + Cl2 (g) ® H(g) + Cl(g); DH = 81 kcal Subtracting equation (3) from equation (4) HCl (g) ®H(g) + Cl(g) ; – \ (76)
(89) (90)
–
...(4)
DH = 103 kcal
–
...(3)
–
H2 (g) + Cl2 (g) HCl(g) ; DH = – 22.0 kcal Enthalpy of formation of HCl gas = – 22.0 kcal
From first law of Thermodynamics, DE = q + w Þ nCvdT = nCdT – PdV .....(1) Now according to process, P = V and according to ideal gas equation, PV = nRT We have, V2 = nRT nRdT On differentiating, 2VdV = nRdT and PdV = VdV = 2 nRdT So from first equation we have, nCvdT = nCdT – 2 R 4R So, Cv = C – Hence C = 2 2 W = – PDV = – 1 × 105 (1 × 10–2 – 1 × 10–3) = – 1 × 105 × 9 × 10–3 = – 900 J. DH = – 393.5 kJ mol–1
.... (1)
CO(g) + O2(g) ® CO2(g) ; DH = – 283 kJ mol–1 .... (2) On subraction equation (2) from equation (1), we get C(s) + O2(g) ® CO(g) ; DH = – 110.5 kJ mol–1. The enthalpy of formation of carbon monoxide per mole = – 110.5 kJ mol–1. DH = – 393.5 kJ mol–1 (91)
.... (1)
CO(g) + O2(g) ® CO2(g) ; DH = – 283 kJ mol–1 .... (2) Let the bond dissociation energy of XY, X2 and Y2 be x,x and x, KJ/mol respectively, 1 1 X 2 + Y2® XY; DH¦ = – 200 KJ mol–1. 2 2 DHreaction = [(sum of bond dissociation energy of all reactants) – (sum of bond dissociation energy ofproduct)] 1 é1 ù x 0.5x + = ê DH x 2 + DH y2 - DH xy ú = – x = – 200 2 2 2 ë2 û
116
\ x = = 800 KJ mol–1.
Second method
XY ¾¾ ® X (g) + YΔH (g)
=a + kJ / mole
;
……….(i)
X 2 ¾¾ ® 2XΔH
=a+ kJ / mole ;
………. (ii)
Y2 ¾¾ ® 2YΔH
=0.5 + a kJ / mole ;
………. (iii)
1 1 1 1 ´ (ii) + ´ (iii)-(i), gives X 2 + Y2 ¾¾ ® XY ; 2 2 2 2
(92) (93)
(94)
N2 + 3H2 ® 2NH3 DH = DU + DnRT = DU – 2RT.
Dn = 2 – 4 = – 2 \ DH < DU.
In isolated system, the expansion of gas is carried out adiabatically. Since heat exchange between system and surrounding is not possible i.e. q = 0 and secondary wrev is always greater than wirr- therefore for reversible process there must be comparatively higher decreases in internal energy i.e. DU for reversible process will be more negative. Hence, final temperature in reversible process will be smaller than irreversible process. \ (Tf)irrev > (Tf)rev In an isolated system, there is no exchange of energy or matter between the system and surrounding. For a spontaneous process in an isolated system, the change in entropy is positive, i.e. DS > 0. Most of the spontaneous chemical reactions are exothermic. A number of endothermic reaction are spontaneous e.g melting of ice (an endothermic process) is a spontaneous reaction. The two factors which are responsible for the spontaneity of process are (i) tendency to acquire minimum energy (ii) tendency to acquire maximum randomness.
(95)
DGº = DHº – TDSº for a spontaneous process DGo < 0 DHo – TDSo < 0 TDSo > DHo
DH o T> DSo
T>
179.1 ´ 1000 160.2
T > 1117.9 K » 1118 K. (99)
DU = DH – DnRT
(100)
= 41000 – 1 × 8.314 × 373 = 41000 – 3101.122 = 37898.878 J mol–1 = 37.9 kJ mol–1. C + 2H2 ® CH4; DHo = – 74.8 kJ mol–1 In order to calculate average energy for C – H bond formation we should know the followng data.
C(graphite) ® C(g); DH ¦o = enthalpy of sublimation of carbon H2 (g) ® 2H(g) ; DHo bond dissociation energy of H2. 117
(101)
DS° reaction = 50 – 1/2 (60) – 3/2 (40) = –40 JK–1 For reaction to be at equilibrium DG = 0
DH 30000 = = 750 K DS 40 DH = –286.20 kJ
DH – TDS = 0Þ T = (102)
H2(g) + O2(g) ¾® H2O (l) DHr = D Hf (H2O,l) – D Hf (H2 , g) DHf (O2 , g) –286.20 = DHf (H2O (l))
So DHf (H2O, l) = –286.20 KJ/mole H2O (l) ¾¾® H+ (aq) + OH– (aq) DH = 57.32 kJ DHr = DHºf (H+, aq) + DHºf(OH–, aq) – DHºf (H2O, l) 57.32 = 0 + DHºf (OH–, aq) – (–286.20) DHºf (OH–, aq) = 57.32 – 286.20 = –228.88 kJ. (103)
(105)
CH3OH(l) + 3/2 O2 (g) ¾® CO2(g) + 2H2O(l) DGr = DGf (CO2, g) + 2DGf (H2O, (l)) – DGf (CH3OH, (l)) – DGf (O2, g) = –394.4 + 2 (–237.2) – (–166.2) – 0 = – 394.4 – 474.4 + 166.2 = – 868.8 × 166.2 DGr = –702.6 kJ 702.6 ´ 100 = 97%. % efficiency = 726 DG = DH – TDS For spontaneous reaction DG must be negative At equilibrium temperature DG = 0 to maintain the negative value of DG
(117) (120)
T should be greater than Te. V2 50 W = – 2.303 n RT log V = – 2.303 × 2 × 2 × 298 × log = – 1436 calories. 15 1 DG° = – RT ln Kp = – 2.303RT log Kp = – 2.303 1.987 298 log1.8 10–7 = – 2.303 1.987 298 (– 6.7447) = 9197.5 Cal
(131)
Cv =
3 5 RT; Cp = RT for monoatomic gas 2 2
5 7 RT; Cp = RT for diatomic gas 2 2 3 5 5 7 RT + RT RT + RT 2 2 Thus for mixture of 1 mole each, C v = 2 and Cp = 2 2 2 Therefore, Cp / C v = 3RT = 1.5 . 2RT Cv =
118
(132)
78g of benzene on combustion produces heat = – 3264.6 kJ \ 39g will produce =
(137)
-3264.6 = –1632.3 kJ. 2
ΔG = -2.303 RT log K¢ , Here R = 2 cal, T = 300K
K¢ =
10 ´ 15 = 10; 3´ 5
ΔG = -2.303 ´ 2 ´ 300 ´ log10 10
= = -2.303 ´ 2 ´ 300 ´ 1 = -1381.8 cal (140)
For 2 moles of water vapour, Absorbed energy by system is DHvap = 2 x 9720 = 19440 Cal. DH Vap DSvap = T b
19440 = (100 + 273)
= 52.12 Cal. K–1. mole–1 = 52.12 4.184 = 217.6 joule.K–1. mole–1 (142)
DG°= –2.303 RT logk = –2.303 x 1.987 ´ 298 log4 = –1363.7 log 4 = –1363.7 ´ 0.6021 = –821.1 Cal.
(143) (144)
DG° = –nFE°cell = –2 ´ 96500 ´ 1.20J = –231.6 KJ DG = DH – TDS O = 4000 – T × 10
because (DG = 0)
T = 400 k i.e. At 400 k, temperature reaction will be in equilibrium. But at temperature higher than 400 k the value of DG will be negative. Thus at 500 k temperature reaction will be spontaneous. (145)
DGo= DGoP – DGoR = DGoC – [DGoA + DGoB] = – 25 – [–10 –15] = –25 + 25 = 0 DGo= 0
(149)
ΔH =
so
K = 1
13.95 ´ 44 = 278.7 kcal 2.2016 119
UNIT : 6 SOLUTIONS Important Points When two or more than two substance mix and from a uniform or homogeneous mixture, Such a mixture is called solution. Type of solutions The solutions can be found in three states; Solid, Liquid and Gas. The solute and solvent can also be in three states. The physical state of the resulting solution can be decided on the basis of physical state of solute and solvent. Sr No.
Type of solution Solute
1
2
3
Solid solution
Liquid solution
Gaseous solution
Formality ( F ) =
%V V=
Physical state Solvent
Examples Alloy formed from copper and zinc (Brass). Zinc amalgam-Zinc
Solid
Solid
Liquid
Solid
Gas
Solid
dissolved in mercury (Zn/Hg adsorption of H2 gas on Pd.
Solid
Liquid
Liquid
Liquid
Homogeneous mixture of sugar and water. Homogeneous mixture of water and ethanol.
Gas
Liquid
Homogeneous mixture of CO2 gas in water.
Solid
Gas
Liquid
Gas
Homogeneous mixture of camphor in N2 gas. Air containing moisture
Gas
Gas
Mixture of H2 and O2 gas.
1000 ´ mass of solute ( gram )
Formula mass of solute ´ volume of solution ( ml )
100 ´ volume of solute ( ml ) 100 ´ volume of solute = volume of solvent ( ml ) volume of solute + volume of solvent
%W V=
100 ´ mass of solute ( gram ) volume of solution ( ml )
120
parts per million by mass to volume =
amount of solute ( mg )
amount of solution ( litre )
Factors which effect the solubility of gaseous solute in liquid solution formed by homogeneous Mixture of gaseous solute and are given as under. (i)
Nature of gaseous solute and the solvent (ii) Effect of temperature (iii) Effect of pressure Henry’s law : p = KH .X where, KH is Henry’s constant. When solid solute is dissolved in solid solvent is gives solid Solution. The molecules are arranged in two ways: (1) Substituted solid solution (2) Interstitial solid solution Solution-Colligative Properties When solute substances are dissolved in pure solvent, the solutions are obtained. Some properties of solvent change viz. the vapour pressure of a solution prepared from a solvent is less than that of pure solvent, while the boiling point increases and freezing point decreases. The osmotic pressure also changes. The change in these properties depend in number of molecules of solute but not on nature of solute. Such properties are called colligative properties of solution. Raoult’s Law (For Non-volatile Solute) “If dilute and ideal solution is prepared by dissolving non-volatile solute in a volatile solvent, the relative lowering of vapour pressure of the solution is equal to the mole fraction of the dissolved solute.” P10 - P1 0
=
P1
n2 = X2 n1 + n 2
i.e. mole fraction of solute. Where, n1 and n2 are the moles of solvent and solute respectively. |For Very dilute solution n2 O 2 > He (b) N2 > H 2 > O 2 > He (c) O2 > N 2 > H 2 > He (d) O2 > H 2 > N 2 > He
31.
At 293 K temperature, for solubility of all given gases, in water, which gas possesses higher value of KH ? (a) He
(b) N2
(c) H2 126
(d) O 2
32.
At 293 K temperature, for solubility of all given gases, in water, which gas possesses lower value of KH ? (a) Carbon dioxide
33.
(b) formaldehyde
(c) methane
(d) vinyl chloride
At 298 K temperature, for solubility of all given gases, which of the following is the correct ascending order of values of Henry’s constant ? (a) CH4 O 2 > He (b) N2 > H 2 > O 2 > He (c) He > N2 > H2 > O 2 (d) O2 > H2 > N 2 > He
35.
At constant temperature, on the basis of the given graph, which gas possesses higher solubility? (a) A (b) B (c) C (d) D
36.
37.
In which of the following specific condition, CO 2 gas is filled in cold drinks, and in soda water ? (a) at high temperature and high pressure
(b) at low temperature and high pressure
(c) at low temperature and low pressure
(d) at high temperature and low pressure
In which condition, Henry’s law is applicable ? (a) ideal behaviour of gaseous solute at high pressure and low temperature (b) gaseous solute neither associate nor dissociate in solution (c) gaseous solute react with solvent (d) applicable in given all conditions
38.
Now a days, divers uses the cylinder having gaseous mixture contains (a) 2 % O2 and 98 % He
(b) 11.7 % He, 56.2 % N2 and 32.1 % O2
(c) 11.7 % N2, 56.2% He and 32.1% O2
(d) 11.7 % He, 56.2 % O2 and 32.1 % N2
127
39.
Due to which reason, O
2
gas liberates from the blood of tissues of animal bodies
(a) less temperature of tissues
(b) partial pressure of oxygen gas is more in tissues
(c) partial pressure of carbon dioxide is less in tissues (d) partial pressure of oxygen gas is less in tissues 40.
Which of the following is not a substitutional solid solution ? (a) wc
41.
46.
(c) steel
(d) monel metal
(b) decreases
(c) remains constant(d) can’t be predicted
(b) boiling point
(c) freezing point
(d) osmotic pressure
What will be the ratio of any colligative properties of 1.0 m aqueons solutions of Nacl, Na2So 4 ad K4 [Fe(CN)6] (Assume that solute completely (100%) dissociates in the solution) (a) 2:3:4
45.
(b) bronze
Which of the following is a colligative property ? (a) vapour pressure
44.
(d) monel metal
Solute + solvent solution; Δ H > O. what would be the change in solubility of substance on increasing the temperature at equilibrium ? (a) increases
43.
(c) steel
Which of the following is a substitutional solid solution ? (a) wc
42.
(b) brass
(b) 1:2:4
(c) 2:3:5
(d) 1:3:5
At constant temperature, vapour pressure of aqueous solutions of Na2so 4, urea and AlCl3 are equal with the vapour pressure of aqueous solution of 1.2 m kcl solution; then molality of an aqueons of Na 2So 4, urea and AlCl3 are respectively – (a) 3.6 .2.4 m,4.8 m
(b) 0.8 m,2.4 m, 0.6 m
(c) 0.6 m, 3.6 m, 0.8 m
(d) 3.6 m, 1.2 m, 2.4 m
Mention the correct value of y in the Reference of given below (a) 62.5 Torr (b) 37.5 Torr (c) 60 Torr (d) 16.33 Torr
47.
In the reference of given graph, The value of RU – YQ × UV is
Vapour Pressure Mole fraction ® (a) YQ.UV
(b) QU. UW
(c) VW. QU 128
(d) ST.YQ
48.
49.
Ionic substances are completely dissociates in the given solutions, then which of the following solutions possesses highest freezing point ? (a) 0.01m Urea
(b) 0.01m NaCl
(c) 0.01m BaCl2
(d) 0.01m Al2(SO4)3
In Binary ideal solution forms by liquid A and B, at constant temperature, mole-fraction of liquid A in vapour state is 0.4 and its partial vapour pressure is 400 mm, then what will be the partial vapour pressure of B ? (a) 600 mm
50.
(d) CaCl2
(b) 0.2 M K2SO 4
(c) 0.1 M Al2(SO4)3
(d) none of thses
(b) 3.6
(c) 1.2
(d) 0.6
(b) 2.1 m
(c) 3.75 m
(d) 0.6 m
(b) 1 < i < 2
(c) i < 1
(d) i > 2
(b) 0.25 m AlCl3
(c) 0.15m CaCl2
(d) 0.2 m CuSo4
(c) solid
(d) colloidal
Which type of solution, moist air is ? (a) gas
58.
(c) Al2(SO4)3
Which of the following aqueous solution is isotonic with 0.2 m Na4[CoF6] solution ? (Assume that ionic solid substances completely dissociates in the solution) (a) 0.2 m urea
57.
(b) NaCl
Boiling point of the aqueous solution prepared by dissolving 1.5 mole substances in 1000 gm water at 1 atmosphere pressure is 100.5oC; then which of the following alternative is correct for the solution ? (Kb = 0.152oC – kg - mole-1) (a) i = 1
56.
(d) 274 K
At constant temperature, vapour pressure of an aqueous solution of 1.5M NH4NO 3 and xM Al2(SO4)3 are equal; then calculate the molality of an aqueous solution of Al2(SO4)3. (Assume that ionic solid substances completely dissociates in the solution) (a) 0.3 m
55.
(c) 375 K
At constant temperature, osmotic pressure of an aqueous solution of 1.5 M NH4NO 3 and xN Al2(SO4)3 are equal, then mention the value of X. (Assume that ionic solid substances completely dissociates in the solution) (a) 0.1
54.
(b) 271 K
0.2 M aqueous solution of NH4Cl is isotonic with which of the following aqueous solution ? (a) 0.1 M Na3PO 4
53.
(d) 200 mm
Ionic substances are completely dissociates, then aqueous solution of which of the following substances having least freezing point ? (a) glucose
52.
(c) 500 mm
1.0 molal aqueous solution of a substance boils at 100.55 oC ; then, at what approximate temperature, it freezes ? (Kb = 0.51 oC – kg - mole-1 and Kb = 1.86 oC – kg - mole-1) (a) 272 K
51.
(b) 300 mm
(b) liquid
Aqueous solutions are separated by semipermeable membrane. For which pair of the given solution having maximum osmotic pressure ? (Assume that ionic solid substances completely dissociates in the solution) (a) 0.5 m NaCl | 0.1 m Na2SO 4
(b) 0.3 m NaCl | 0.1 m Na2SO 4
(c) 0.5 m NaCl | 0.1 m FeCl3
(d) 0.5 m NaCl | 0.1 m sugar 129
59.
At constant temperature, the vapour pressure of an aqueous solution of Na2SO4 and 0.3 m Na3PO 4 are approximately equal; then, what would be the molality of an aqueous solution of Na2SO 4 ? (Assume that ionic solid substances completely dissociates in the solution) (a) 0.5 m
60.
62.
65.
66.
(b) 1 : 3 : 2
(c) 1 : 2 : 3
(d) 3 : 1 : 2
(a) aqueous solution of sugar
(b) aqueous solution of salt
(c) aqueous solution of CO2
(d) aqueous solution of KNO3
At constant temperature, in a closed vessel, an ideal solution is formed by liquid – A and liquid – B; and mole-fraction of A and B are 0.6 and 0.4 respectively. If vapour pressure of pure liquids are 125.0 and 62.5 mm respectively, then their mole-fraction in vapour state are respectively – (In vessel, no other component is in gaseous form) (b) 0.4 and 0.6
(c) 0.25 and 0.75
(d) 0.75 and 0.25
At constant temperature, two liquids having osmotic pressure p1 and p2 are seperated by semipermeable membrane, then, what will be the osmotic pressure of the system ? (a) p1 + p2
64.
(d) 1.2 m
At constant temperature, solubility of which of the following substances decreases with increase in temperature ?
(a) 0.6 and 0.4 63.
(c) 0.4 m
What will be the ratio of elevation in boiling point of aqueous solution of 1 m sugar, 1 CsCl, and 1 m Na2SO 4 ? (Assume that ionic solid substances completely dissociates in the solution) (a) 3 : 2 : 1
61.
(b) 0.6 m
(b) p1 – p2
(c)
p1 + p 2 2
(d)
π1 - π 2 2
Which of the following pair of solutions forms ideal solution ? (a) Chloro benzene, chloro ethane
(b) benzene-toluene
(c) acetone-chloroform
(d) water, HCL
Which of the following pair forms true solutions ? (a) Hexane, heptane
(b) chloro benzen, bromo benzen
(c) chloro ethane, bromo ethane
(d) phenol, aniline
What state does point s indicate ?
(a) Mole-fraction and partial vapour pressure of both the liquids are same (b) Mole-fraction of the both the liquids are same, but their partial vapour pressures are different (c) Mole-fraction and partial vapour pressures of both the liquids are different (d) Mole-fraction of both the liquids are different, but their partial pressures are same. 130
67.
What would be the elevation in boiling point of 0.1 m NaCl solution ? (Assume that Nacl dissociates completely) (a) kb/10
68.
69.
(b)10 kb
(c) kb/5
(d) k b/20
Which of the following semipermeable membrane is best one ? (a) parchment paper
(b) copper frrocyanide
(c) butter paper
(d) cello phane
At constant temperature, binary ideal solution is formed by two liquids A and B. At equilibrium, mole-fraction of liquid A is 0.7 and in vapour state mole-fraction of A is 0.4 Poa + Pob =90 mm then at the same temperature, what will be the vapour pressure of pure liquid A and B ? (a) 40 mm, 50 mm
(b) 30 mm, 60 mm
(c) 50 mm, 40 mm
(d) 20 mm, 70 mm
70.
At constant temperature, binary ideal solution is formed by two liquids A and B. At equilibrium, mole-fraction of liquid B is 0.4 and vapour state mole-fraction of B is 0.25. PoB=40 mm, then at the same temperature, what will be the vapour pressure of pure liquid ‘A’ ? (a) 80 mm (b) 60 mm (c) 40 mm (d)50 mm
71.
Choose correct alternative for True and False statements for given diagram. (For correct statement T and for wrong statement F) (Assume that ionic solid substances completely dissociates in the solution) (For Correct- Stutement T and for Wrory worry Srut ment F)
(i) Osmotic pressure of the system increase by adding H2O in an aqueous solution of CuSO 4 (ii) The concentration of solution of glucose increases with the passage of time. (iii) Osmotic pressure of the system decreases by adding glucose in the solution of glucose. (iv) The concentration of solution of CuSO 4 increases with the passage of time. (a) FTTF 72.
(b) TFFT
(c) FFTT
(d) TTFF
Choose correct alternative for True and False statements for given figure (For correct statement T and for wrong statement F) (Assume that lomic solid substances completely dissociates in the solution)
(I) concentration of an aqueous solution of FeCl3 increases with the pressure of time. (ii) aqueous solution of FeCl3 gradually turns reddish (iii) concentration of an aqueous solution of KCNS increases with the passage of time (iv) aqueous solution of KCNS remains colourless. (a) FTTF
(b) TFFT
(c) FFTT 131
(d) TTFF
73.
What would be the osmotic pressure of the system at 300 k temperature ? (R=8.314 x 10-2 litre-bar-mole-1 k-1) (Assume that lomic solid substances completely dissociates in an aqueous solution) (a) 25 bar (b) 10.0 membrane bar (c) 24.9 bar (d) 19.95 bar
74.
At constant temperature, 2 litres aqueous solution of each 0.2 M kcl and 0.3 M AlCl3 are in contact with each other by semipermeable membrane. When osmosis stops, then, what mililitre water diffuses from semipermeable membrane to the other side ? (Assume that ionic solids dissociates completely in the aqueous solution) (a) 500 ml
75.
(b) 600 ml
(c) 800 ml
(d) 1000 ml
Choose the correct alternative for the given diagram for correct and wrong statements. (T is for false statement) (Assume that ionic solid substances dissociates completely in the aqueous solution) (I) concentration of solution of NaCl increases with the passage of time (ii) concentration of solution of urea decreases with the passage of time (iii) concentration of solution of NaCl decreases with the passage of time (iv) concentration of solution of urea increases with the passage of time (a) FTTF
76.
(b) 0.1 M NaCl
(c) 0.18 M NaCl
(d) given all
(b) 0.12 M NaCl
(c) 0.1 M NaCl
(d) given all
X M NaCl is isotonic with fluids present in RBC (Red Blood Corpusceles), then what would be the value of x ? (M.w. Of NaCl =58.5 gm/mole) (Assume that ionic solid substances completely dissociates in the solution) (a) 0.15
79.
(d) TTFF
At constant temperature, Which of the following solution is hypotonic in the comparison with fluids in RBC ? (Assume that ionic solid substances completely dissociates in the solution) (a) 0.17 M NaCl
78.
(c) FFTT
Which of the following solution is hypotonic with fluids in RBC ?(Assume that ionic solid substances completely dissociates in the solution) (a) 0.2 M NaCl
77.
(b) TFFT
(b) 0.05
(c) 0.18
(d) 0.78
Which of the following solution is hypotonic in comparison with the solution of 0.4 M glucose? (a) 0.1 M CaCl2
(b) 0.2 M NaCl
(c) 0.15 M FeCl3 132
(d) 0.3 M urea
80.
Which of the following solution is hypotonic in comparison with 0.15 M kCl solution ?(Assume that ionic solid substances completely dissociates in the solution) (a) 0.1 M CaCl2
81.
82.
83.
(a) 5.6% w/v glucose
(b) 2.8% w/v glucose
(c) 1.5% w/v urea
(d) 0.91% w/v urea
(a) 2.02 % w/v glucose
(b) 4.02% w/v glucose
(c) 8.0% w/v urea
(d) both b and c
In Which of the following solution, RBC get burst ? (Molecular wt, CaCl2=111, FeCl3=162.5, glucose=180 and urea=60 gm/mole) (Assume that ionic solid completely dissociates in aqueous solution)
(b) 2% w/v urea
(c) 1.0% w/v FeCl3
(b) 2.7
(c) 3.4
(b) 0.52 m
(c) 0.6 m
w/v
(d)
5.0%
w/v
(d) 3.1 (d) 2.6 m
(b) 0.4
(c) 0.6
(d) given all
A substance associates as trimer in their solution; then, what would be the value of Van’t Hoff factor i ? (b) 0.3
(c) 0.2 (d) 0.25
A substance associates as bimolecule in their solution; then what would be the value of Van’t Hoff factor i ? (a) 0.4 d” i < 1
90.
6.0%
A substances associates in their solution as dimer (or bimolecule), then what will be the value of Van’t hoff factor i ?
(a) 0.4 89.
(d)
CaCl2 ionices 80% in their aqueons solution of 0.2 m CaCl2, then, molality of solution is -
(a) 0.2 88.
(c) 0.09% M FeCl3
FeCl3 ionizes 80% in their aqueous solution, then what will be the value of Vant ‘Hoff factor i ?
(a) 0.48 m 87.
(b) 0.2 MkCl
In Which of the following solution, RBC get shrinks ? (Molecular wt, CaCl2=111, FeCl3=162.5, glucose=180 and urea=60 gm/mole) (Assume that ionic solid completely dissociates in aqueous solution)
(a) 4 86.
(d) 0.12 BaCl2
Which of the following solution is isotonic with fluid of RBC ? (For NaCl, 2=2)
(a) 0.1% w/v CaCl2 glucose 85.
(c) 0.2 M urea
Which of the following solution is isotonic with fluid of RBC ? (For NaCl, i=2)
(a) 1.2% w/v urea glucose 84.
(b) 0.08 M FeCl3
(b) 0.5 d” i < 1
(c) 0 < i < 1
(d) 0.6 d” i < 1
The solute remains as dimer in the m-molal solution; then elevation in boiling point irrelevant with the solution is (a)
mkb 2
(b)
3mkb 5
(c)
133
3mkb 4
(d)
mkb 3
91.
Which of the following ratio is irrelevant formula mass and experimental molecular weight obtained from colligative properties of solution of ionic solid of AB type ? (a) 3 : 2
92.
(b) 5 : 3
(b) 4 : 1
(b) 2 : 5
mkf 2
(b)
mkf 2
(c)
mkf 5
(d)
mkf 8
(b) molarity
(c) molality
(d) normality
(b) Tf -
mkf 3
(c) Tf -
2mkf 3
(d) Tf – 2mkf
(b) – 0.37 C
(c) – 1.45 C
(d) – 0.5 C
Aqueous solution of 0.5 m H2So4 is more concentrated then 0.5 m H2So4 solution; then what will be the possible density of that solution ? (a) 1.07 -
99.
mkf 4
Boiling point of the 0.2 m aqueons solution of a substance is 100.4 oC ; then what would be the freezing point of th solution ? (Kb = 0.513o, K+” = 1.86 (a) – 0.372 C
98.
(d) 1 : 2
A substance associates as trimer in their solution, then what would be the maximum freezing point of their m molal solution is positive ? (a) Tf -
97.
(c) 1 : 4
Which of the following unit of concentration is common in the field of pharmacy ? (a) formality
96.
(d) 5 : 2
A substance associates as trimer in their solution, then which of the following alternative is possible for depression in freezing point of m molal solution ? (a)
95.
(c) 7 : 3
A substance associates as trimer in their solution, then which of the following ratio is irrelevant for real molecular weight and experimental molecular weight obtained from colligative properties of solution ? (a) 2 : 3
94.
(d) 5 : 2
Which of the following ratio of correct molecular wt (formula weight) and experimental molecular weight obtained from colligative properties of solution of ionic solid of AB type is possible ? (a) 5 : 3
93.
(c) 4 : 3
9m m1
(b) 1.06 9m/m1
(c) 1.05 9m/m1
(d) 1.02 9m/m1
Which of the following is irrelevant with the boiling point of an aqueous solution of xm AlCl3 ? (a) Tb + 3 × kb
(b) Tb + 5 × kb
(c) Tb -
7 xkb 2
(d) Tb -
5 × kb 2
100. Which of the following is suitable alternative for density of the solution, when molarity (m) and molality (m) of an aqueous solution of urea is same at fixed temperature ? (molecular wt of urea = 60 gm/mole ? (a) 1 -
3M 50
(b) 1 -
M 25
(c)
134
50 + 3m 50
(d)
25 + 2m 25
101. Choose the correct option for true and false statement. (For true statement ‘T’ and for false statement ‘F’ is used) (i)
solubility of gas in liquid increases with increase in partial pressure of the gas.
(ii)
solubility of gas in liquid increases with increase in temperature
(iii) solubility of gas in liquid isKH is less (iv) solubility of gas in liquid increases, as partial pressure of gas decreases and temperature increases. (a) TTFF
(b) FTTT
(c) TFFF
(d) FFTT
102. Boiling point of an aqueous solution of 0.05m FeCl3 is 100.087 oC; then, what will be the value of Van’t Hoff factor i ? (Kb = 0.513 oC – kg - mole-1) (a) 4
(b) 3.4
(c) 2.5
(d) 2.8
103. Difference of boiling point and freezing point of an aqueous solution of glucose is 104 oC at 1 bar pressure; then what will be the molality of the solution ? (Kb = 0.513o and K+” = 1.86 oC – kg - mole-1) (a) 2.373 m
(b) 1.05 m
(c) 2.151 m
(d) 1.68 m
104. Difference of boiling point and freezing point of 0.2 m acetic acid prepared in benzene is 75.7 oC; then, state the value of Van’t Hoff factor i ? (For benzene, Kb = 2.65 OC – kg – mol–1, Kf = 5.12 oC – kg – mol–1, Tb = 80 oC, Tf = 5.5 oC) (a) 1.44
(b) 0.64
(c) 0.83
(d) 0.77
105. Difference in boiling point and freezing point of 10 kg aqueous solution of urea is 100.2372 OC; then what quantity of urea dissolved in the solution ? (Kb = 0.513o and K+” = 1.86 oC – kg - mole -1) (a) 59.64
(b) 38.946
(c) 51.65
(d) 40.5
106. 500 ml solution of HCl is prepared by dissolving 14.6 gm HCl in water. What will be the molarity of HCl in the solution ? (Molecular weight of HCl = 36.5 gm/mole ) (a) 0.4 M
(b) 0.3 M
(c) 0.8 M
(d) 0.3 M
107. What would be the molality of the solution prepared by dissolving 60 gm NaOH in 1.5 kg water ? (Moleculea weight of NaOH = 40 gm/mole) (a) 0.5 m
(b) 1.0 m
(c) 0.8 m
(d) 0.4m
108. What would be the molarity of 3.0 N H2SO 4 solution ? (a) 6 M
(b) 1.5 M
(c) 3 M
(d) 1 M
109. What quantity of NaOH is needed to prepare 1.2 m, 800 ml NaOH solution ? (a) 3.84
(b) 60
(c) 42
(d) 38.4
110. What would be the molarity and normality of solution prepared by dissolving 19.6 gm H2SO 4 in dissolved water to prepare 800 ml solution ? (Molecular weight of H2SO4 is 98 gm/mole) (a) 0.5 M, 0.25 N
(b) 0.25 M, 0.125 N
(c) 0,25 M, 0.5 N
(d) 0.125 M, 0.25 N 135
111. What amount of H2SO4 required to prepare 2 litre of 0.5 N H2SO4 solution ? (Molecular weight of H2SO4 is 98 gm/mole) (a) 98 gm
(b) 24.5 gm
(c) 49 gm
(d) 73.5 gm
112. What will be the concentration of solution prepared by dissolving 50 gm glucose in 200 gm water ? (a) 25 % w/w
(b) 20 % w/w
(c) 35 % w/w
(d) 15 % w/w
113. What quantity of urea required to prepare 20 % w/w solution having weight 150 gm ? (a) 20 gm
(b) 40 gm
(c) 10 gm
(d) 30 gm
114. What is the normality of an aqueous solution of 0.5 Al2(SO4)3 ? (a) 3 N
(b) 1 N
(c) 1.5 N
(d) 2.5 N
115. What will be the mole-fraction of ethanol in solution, prepared by dissolving 9.2 gm ethanol in 900 gm water ? (Molecular weight of water and ethanol are 18 and 48 gm/mole respectively) (a) 0.04
(b) 0.004
(c) 0.4
(d) 0.0004
116. What will be the mole-fraction of water and NaOH respectively, when 260 gm NaOH dissolved in 1.8 kg water ? (M.W of watll Naoh = 18440) (a) 0.96, 0.4
(b) 0.0962, 0.38
(c) 0.0962, .038
(d) 0.962, 0.0038
117. What would be the mole-fraction of solute in an aqueous solution of a substances heaving strength 4.5 m ? (a) 0.75
(b) 0.075
(c) 0.45
(d) 0.045
118. The density of 98% w/w H2SO 4 solution is 1.8 gm/mole then, molarity of the solution is (a) 20 M
(b) 10 M
(c) 18 M
(d) None of these
119. Molarity and molality of an aqueous solution of H2SO4 are 1.56 (M) and 1.8 (M) respectively; then, waqht will be the density of the solution ? (a) 1.835 gm/ml
(b) 1.55 gm/ml
(c) 1.02 gm/ml
(d) 1.725 gm/ml
120. A solution is preparesd from A, B, C and D mole-fraction of A, B and C are 0.1, 0.2 and 0.4 respectively then, mole-fraction of D is (a) 0.2
(b) 0.1
(c) 0.3
(d) 0.4
121. The density of 4 M H2SO4 solution is 1.992 gm/ml then, what will be the molality of the solution ? (Molecular weight of H2SO4 is 98 gm/mole) (a) 3 M
(b) 3.5 M
(c) 1.2 M
(d) 0.4 M
122. Molarity of 1.2 N aqueous solution of AlCl3 is (a) 3.6 m
(b) 2.4 m
(c) 1.2 m
(d) 0.4 m
123. What will be the molality of the solution prpared using 500 gm of 25 % w/w NaOH and 500 gm of 15 % w/w NaOH solution ? (Molecular weight of NaOH = 40 gm/mole) (a) 12.74 m
(b) 6.25 m
(c) 9 m
136
(d) 5 m
124. What wiil be the molality of solution prepared by taking 25 % w/w NaOH and 15 % w/w NaOH solution ? (Molecular weight of NaOH = 40 gm/mole) (a) 12.74 m
(b) 5.5 m
(c) 9 m
(d) 4 m
125. The density of 2.5 M NaOH solution is 1.15 gm/ml; then, which of the following alternative is correct for molarity and molality ? (a) M > m predicted
(b) M < m
(c) M = m
(d)
can ’t
be
126. Which of the following is correct for an ideal solution ? (a) “H = 0, “V = 0, “S = 0
(b) “H `” 0, “V = 0, “S = 0
(c) “H = 0, “V = 0, “S `” 0
(d) “H = 0, “V `” 0, “S = 0
127. Boiling point of Aqueous solutions of 0.05 m ABC and 0.02 m X2Y3 are same at 1 bar pressure; then, what will be the values of Van’t Hoff factor (i) for solute in both the solutions ? (a) 1.04, 5.1
(b) 1.9, 4.75
(c) 1.2, 3.4
(d) 1.5, 3.7
128. Which of the following substances having concentration of aqueous solution 1% w/w, possesses higher boiling point ? (Molecular weight of Kcl, BaCl2, glucose and Al2(SO4)3 are 74.5, 208, 342 gm’mole respectively) (Assume that inonic solids dissociates completely in their aqueous solution) (a) KCl
(b) BaCl2
(c) glucose
(d) Al2(SO4)3
129. Molecular weight of biomolecules such as protein can be determined by ______ method. (a) osmotic pressure measurement
(b) Depression in freezing point measurement
(c) Elevation in boiling point measurement (d) Vapour pressure measurement 130. In the references of the following graph, UR -Oy = __________. (a) (SU -VW)QU (b) VU(VW – QY) (c) (VW – QY)QU (d) VW(QV – VU) 131. What will be the elevation in boiling point of an aqueous solution of 0.5 m NaCl ? ( i = 1.8) (a)2.04, 5.1
(b) 1.9, 4.57
(c) 1.2, 3.4
(d) 1.4, 3.7
132. When 2 gm phenol id dissolve in 100gm benzene; then depression in freezing point is 0.69 K. If its association is dimeric, then calculate its degree of association (X). Molal depression constant for solvent is 5.12 K - kg - mole-1. (a) 0.0734
(b) 0.374
(c) 0.00734
(d) 0.734
133. At 353 K temperature, the Vapour pressure of pure liquids A and B are 600mm and 800 mm respectively. If mixture of liquids A and B boils at 353 K and 1 bar pressure, then mole proportion of B in percent is (a) 80%
(b) 60%
(c) 20% 137
(d) 40%
134. 90 gm glucose and 120 gm urea dissolved in 1.46 kg aqueous solution, then what will be the boiling point of the solution at 1 bar pressure ? (Kb = 0.512oC -kg – mole-1, molecular weight of glucose and urea are 180 and 60 gm/mole respectively) (a) 100.876oC
(b) 101.024oC
(c) 100.248oC
(d) 100.007oC
135. pH of 0.2M dibasic acid H2A is 1.699; then, what will be its osmotic pressure at T K temperature ? (a) 0.22 RT
(b) 0.02 RT
(c) 0.4 RT
(d) 0.1 RT
136. Boiling point of an aqueous soultion of 0.4m AlCl3 is 100.7oC; then what would be the pressure of ionization of AlCl3 ? Kb – 0.512oC - kg - mole-1. (a) 80.67%
(b) 60.5%
(c) 76.54%
(d) 84.75%
137. The vapour pressure of homogenous mixture of 10 mole of liquid X and 30 mole of liquid Y at constant temperature is 550 mm. In this solution, 10 mole of liquid Y increases, hence, increase in vapour pressure is 10 mm. Then, find the vapor pressure of pure liquid X and Y at that temperature. (a) Pox = 200 mm, Poy = 500 mm
(b) Pox = 400 mm, Poy = 600 mm
(c) Pox = 600 mm, Poy = 300 mm
(d) Pox = 350 mm, Poy = 500 mm
138. What amount of urea dissolved in 1 kg water at constant temperature, so that vapour pressure of the solution reduced by 2% ? ( M.W of urea = 60 gm/mole) (a) 68 gm
(b) 60 gm
(c) 50 gm
(d) 75 gm
139. What would be tne volume of 15% w/v and 5% w/v NaOH solution required to prepare 1 litre aqueous solution of 2M NaOH ? (M.w of Naoh = 40 gram/mole) (a) 300 ml, 700ml
(b) 250 ml, 750ml
(c) 400 ml, 600ml
(d) 280 ml, 720ml
140. At constant temperature, vapour pressure of an aqueous solution of 1.5 kg glucose decreases to 0.98% in comparision with vapour pressure of pure water then, what quantity of glucose in gram dissolved in the solution ? (Molecular weight of glucose = 180 gm/mole) (a) 148.5 gm
(b) 14.85 gm
(c) 125 gm
(d) 135 gm
141. At constant pressure, 0.5 m NaCl aqueous solution is diluted by adding water in it. Which of the following statement is correct in this reference ? (a) Van’t Hoff factor (i) and boiling point of the solution both decreases (b) Van’t Hoff factor (i) and boiling point of the solution both increases (c) Van’t Hoff factor (i) decreases while boiling point of the solution both increases (d) Van’t Hoff factor (i) increases while boiling point of the solution both decreases 142. Boiling point of an aqueous solution of 0.5 m ionic solid substance is 100.5OC; then state the value of i ? (Kb = 0.512oC -kg – mole-1) (a) 1.95
(b) 1.85
(c) 1.25
(d) 0.85
143. Aqueous solution of substance boils at 100.5oC at 1 bar pressure; then at what temperature it freezes ? (Kb = 0.512oC -kg - mole-1, Kf = 1.86oC - kg – mole-1) (a) 11.2oC
(b) 29.84oF
(c) 271.8oK 138
(d) -1.2oC
144. 1.4 m aqueous solution of a weak electrolyte AB2 ionizes 20%, then, state boiling point and freezing point of the solution respectively. (a) 100.86
C, -3.12oC (b) 101oC, -3.65oC
o
(c) 274oC, -3.65oC
(d) 374oC, -3.65oC
145. Solute substance in a 1.4 m aqueous solution associates by 25%, then, find the boiling point and freezing point of solution; where, solute exists as trimer in the solution; thus, n = 3. (Kb = 0.512oC • kg • mole-1, Kf = 1.86oC • kg • mole-1) (a) 100.448oC, -2.28oC
(b) 373.58oK, -3.65oC
(c) 100.59oC, 2.17oC
(d) 213oF, 270.83ok
146. Molecular mass of a weak acid HA is 60gm/mo1 Ifs experimental molecular mass in its 0.7 M aqueous solution obtained from colligative properties is 50gm/mo1. Then calculate ionistion consteint of weak acid HA. (a) 0.023
(b) 0.0085
(c) 0.035
(d) 0.085
147. At constant temperature, the total pressure of a homogeneous mixture of gas-A and gas-B in a closed container collected on water is 2.0 bar. Their ratio of mole frachion is 1.6 If the values of their KH are 2.4 ´ 10+4 bur and 4.8 ´104 bar respectively then calcwate its ration of mole fraction when dissolhed in H2O. (a) 1:2
(b) 2:1
(c) 3:1
(d) 1:3
148. If one of the colligative property of 0.3m aqueous ti sdn. of Na cl and x m aqueows solution of H2So4 then what would be the approximate vulwe of aqueows solution of H2So4 ? (Density of ´ m H2So4 solution = 1.185 gm/ml) (a) 0.464 N
(b) 0.928 N
(c) 0.232 N
(d) 0.53 N
149. At one bar pressure the value of ratio of mole fraction of O2 and N2 gas in air is 1:4 The values of KH of O2 and N2 are 3.3 ´ 107 Torr and 6.60 ´ 107 Torr respectively. Then calculate the value of ratio of mole fractions of O2 and N2 gases will be. (1 Torr = 1mm) (a) 3:1
(b) 2:1
(c) 1:2
(d) 1:3
150. According to Boyle-van’t - Hoff law at a constelnt temperature, osmotic pressure of a solution is directly proportional to its molarity. It means p a C, where C = molarity of solution \ p = Kc Then calcalate the value of k in sI unit at 24o c temperatuve. (R = 8. 314 J/mole k) (a) 24.942 J/mol
(b) 2494.2 J/mol
(c) 0.024942 J/mol
(d) 2.4942 J/mol
151. Boiling point of an aqueous solution of urea at one bar pressure is 373.41 k. Then at a constant temperature, calculate the percentage decrease in vapoure pressure of a solution compared to (kb = 0.512 k.kg. mol-1) (a) 1.42 %
(b) 2.56 %
(c) 4.17 %
(d) 3.44 %
152. Calculate PH of a solution prepared by mixing equal volume of an aqueous solution of HCI having PH = 2 and PH = 5 at 298 k temp. (a) 3.5
(b) 3.0
(c) 7.0 139
(d) 2.3
153. Which of the followingis the correct formula for Rqoult’s law when non-volatile solute is mixed with liquid solveut. where n=mole traction of solute N= mok fraction of solveet, P= vapoure pressure of solution Po = vapoure pressure of pure sovent and DP=Decrease in vapoure pressure. (a)
Dp
Dp n = p N
Dp
n
(b) p0 = N
n
(c) p = n + N
Dp
N
(d) p = n + N
154. Decrease in Freezing point of 75.2 gm pnenol when dissolved in a solvent having kf = 14 k. kg. mole-1. is 7 k. Calculate the percentege of association of phenol if it is forms a dimev in the solutio. (a) 62.5 %
(b) 80.5 %
(c) 70 %
(d) 75 %
155. Which of the following is correct option when kcl is dissolved in H20. ? (a) ∆H = +ve, ∆S = +ve, ∆G = +ve
(b) ∆H = +ve, ∆S = –ve, ∆G = –ve
(c) ∆H = +ve, ∆S = +ve, ∆G = –ve
(d) ∆H = –ve, ∆S = –ve, ∆G = +ve
156. Naphthalene is soluble in ether or benzene because. ? (a) dipole-dipole attraction is equal
(b) London forces are equal
(c) Hydrogen bond
(d) Ionic attraction
157. Four Liauids are given. (i) Water : more polar and capacity to torm H-bond. (ii) Hexanol : moderatly pdar and partial capacity to form H-bond. (iii) Chloro form : moderatly polar and does not capable to form H-bond. (iv) Octane : non polar and does not capable to form H-bond. which of the following pair of liquids mixed with each other in very less proportion. (a) I, IV
(b) I, II
(c) II, III
(d) III, IV
158. Which of the following is applicable for the solubility of gases in liquid. (a) Increases with increase in temperature and pressure. (b) decreases with increase in temp and pressure. (c) Increases with decrease in temp and increase in pressure. (d) Decreases with decrease in temp and increase in pressure. 159. Eoncentration of lead metal in a blood of any person is more than that of 10 microgram. dm-1, than that person is considered as on effect of poision sectrion. Then calculate its concentration in ppb (parts per billion) (a) 1 160. The ratio of
(b) 10
(c) 100
(d) 1000
RT of 6 % w v and 9 % w v is one for both. what would be the value of atomic weight of p
A and B respectively. (AB2 and A2B are electrolytes) (a) 60, 90
(b) 40, 40
(c) 40, 10
(d) 10, 40
161. Decrease in vapour pressure of an aqueous soln. of an electrolyte is 4% what would be the percentage increase in elevation in Boiling point ? (kb = 0.512 k. kg. mol-1) (a) 0.55 %
(b) 0.02 %
(c) 5.5 % 140
(d) 2 %
ANSWER KEY 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
b a d d c a c d a b b c b d b c d a d b a c a c c
26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
d d a b d a b c c d c b b d a a a d c b b c a a b
51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75
c a b d c b a d c c c d b b d d c d a a a c b d d
76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100
142
b a a c c a a a b c b c a b d d a c a a b c d b c
101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125
c b d d a c b b d c c b d a b c b c c c c d b b a
126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150
c d a a c a d a b a a b a a d d a c b d c d a c b
151 152 153 154 155 156 157 158 159 160 161
a d a d c b a c c c a
(70)
(73)
XA = 1─ XB = 1 ─ 0.4 = 0.6 YA = 1 ─ YB = 1 ─ 0.25 = 0.75 pB = p0B × XB pB = PTotal × YB ∴ pB = 40 × 0.4 ∴ 16 = PTotal × 0.25 = 16 mm ∴ PTotal = 64 mm pA = PTotal × YA = 64 × 0.75 = 48 mm Now pA = p0A × XA ∴ 48 = p0A × 0.6 ∴ p0A = 80 mm Whiten × Suppose × m1 H2O is transferred from aqueous solution of KCI to aqueous solution of Alcl3 then the phenomenon osmosis stops. So morality of soluble particles in both solutions becomes equal at this time. M1 = modality of kcl solution M2 = modality of kcl solution when x ml water is reduced. M1 V1 = M2 V2 0.2 × 2000 = M2 ´ (2000 x) \ M2 =
(74)
(75)
(86)
Similarly volume of Alcl3 Solution increases by addition of x m1 H2O. 600 morality of Alcl3 solution = 2000+ X Now morality soluble particles in kcl solution = morality of soluble particles in Alcl3 soln. ∴ n1∙M1 = n2∙M2 2 × 400 4 × 600 ∴ = 2000 ─ X 2000+ X ∴ x = 1000 Liquid present in Red Blood cells is isotonic with 0.91 % wΤv solution of Nacl ∴ morality of soluble particles in 0.91 % wΤv Nacl solution 2 × 1000 × 0.91 = 58.5 × 100 = 2 × 0.1555 = 0.311 M π = MRT where M = effective molarity of solutions kept in contact with semi-permeable membrane. ∴ π = 0.2 M (molarity of soluble particles in Naclsdn) = 3 × 0.2 - 2 × 0.1 =0.4 M. i−1 o dissociation lo = 100 n− 1 Fecl3 ionises 80% in its aqueous solution 80 i−1 ∴ = ∴ i = 3.4 4− 1
100
(87)
400 2000 X
when association of any substance takes place in a solution then degree of associationis < 1. i−1 ∴0 i ≥ n Here n = 2 (given) 1 ∴ 1 > i ≥ means 0.5 ≤ i < 1 2
145
1 −1 n
1−i
∴0<
≤1
1 n
1–
1
∴ 0 < 1 – i ≤ 1– 1
∴ –1< – i ≤ –
n
n
1
∴ 1> i ≥ n Here n = 2 (given) 1 ∴ 1 > i ≥ means 0.5 ≤ i < 1 2
(90)
For association (given) n=2 1 ∴i ≥ 2
mKf
∴ imKf ≥
2 mKf
∴ ∆Tf ≥ (91)
i=
∆T ob
=
∆T cal
2 πob
=
πcal
M M ob
where M = Actual molecular mass of solute Mob = experimental molecular mass of solute. For AB Type ionic compound i ≤ 2 M ∴ ≤2 5:2>2:1 (96)
M ob
The substance forms trimer in a solution due to association. 1 ∴ ≤iCl CH 3COO
(c) OH CH 3COO Cl
133. HSO 4 OH SO 24 H 2O Which is correct about conjugate acid base pair ? (a) HSO 4 is Conjugate acid of base SO 42 (b) HSO 4 is Conjugate base of acid SO 42 (c) SO 24 is Conjugate base of acid HSO 4 (d) None of these 134. Which of following base is weakest (a) NH 4 OH:K b 1.6 10 6
(b) C 6 H 5 NH 2 : K b 3.8 10 10
(c) C 2 H 5 NH 2 : K b 5.6 10 4
(d) C 2 H 7 N : K b 6.3 10 10
166
135. HClO is a weak acid. concentration of H + ion in 0.1 M solution of HClO K a 5 10 -8 will be (a) 7.07 10 5 M
(b) 5 10 9 M
(c) 5 10 7 M
(d) 7 10 4 M
136. Upto what P H must a solution containing a precipitate of Cr OH 3 be adjusted so that all precipitate dissolves (a) upto 4.4 (b) upto 4.1 (c) upto 4.2 (d) upto 4.0 137. NH 4Cl is acidic because (a) On hydrolysis NH 4Cl give weak base NH 4OH and strong acid HCl (b) Nitrogen donates a pair of e (c) It is a salt of weak acid and strong base (d) On hyrdolysis NH 4Cl gives strong base and weak acid 138. 100 ml of 0.04 N HCl aqueous solution is mixed with 100 ml of 0.02 N NaOH solution. The P H of resulting solution is (a) 1.0 (b) 1.7 (c) 2.0 (d) 2.3
ANSWER KEY 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
B C A A C C B A D A A D B A B D C C A C B B D C A
26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
A A D D B D C D C C B A D C D C D D B C A A A B C
51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75
C C A B B D D A D A D B D B D B D D D B B A B A C
76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100
167
C C A B B C C B B B A D C B C A B B A C A C B B C
101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125
D D D A D D B A D B D A A C C D D D C C C A A C A
126 127 128 129 130 131 132 133 134 135 136 137 138
B B C A A D C C B A D A C
Hint : Chemical equilibrium . Hint : 1. Addition of (1) and (2) gives (3) then
C PQ A B BC PQ K AB2 3
K1 K2
K 3 K1 K 2 when the addition of equilibria leads to another equilibria then the product of their equilibria constants gives the equilibria constant of the resultant equilibrium. 3. The value of equilibrium constant does not change in presence of catalyst. 4. Formula : Kp Kc(RT)
Δn (g)
Δn (g) n p - n r
6.
H 2 O is also present in 1 10-8 M HCl solution. So due to self ionisaion of H 2 O , H 2 O H OH
1 10 7 M at 298 K so conc. of H ion in solution increase due to self ionisation of H 2 O Hence P H of HCl solution decreases and its value is less than 7. 9.
N 2(g) O 2(g) 2 NO (g)
P 2NO Kp PN 2 PO2 For reaction, 1 N O (g ) 1 O 2 (g ) N 2 2 2 1
K'P
1
2
2
PO 2 PN 2 PNO
1 1 = Kp 1 2 4 10 4
10. P H 5.4 P H log H 5.4 log H 6.6 3.981 10 6 H
168
1
50 2
11
For a reaction X 2Y initialmol 1 0 at equilibrium mol 1-x 2x Total moles = 1-x+2x = 1+x 2
2x P1 2 PY 4x 2 p1 1 x Kp1 Px 1 x 1 x ------(1) 1 x p1 1 x For a reaction Z P Q Initial 1 0 0 at eqm. 1-x x x Total moles = 1+x
Kp 2
x P2 x p 2 x2 P2 1 x 1 x -------(2) 2 1 x 1 x P2 1 x
Kp1 1 Kp 2 9
4 P1 1 P2 9
P1 1 P1 : P2 1 : 36 P2 36 For salt of weak acid and weak base 1 P H log Ka log kw log K b 2 1 1 1 pKa pkw pk b 2 2 2 1 1 1 4.8 14 4.78 2 2 2 7.01 Equilibrium constant for the reaction
18.
19.
SO 2(g) 1 2 O2(g) SO3(g) KC
1 4.9 10 2
and for reaction 2SO2(g) O2(g) 2SO3(g) 2
1 KC 416.49 2 4.9 10
169
20.
P H 3 Η 3 O 1 103
H O 110 Ka 3
2
3 2
C
21.
0.1
110 5
CO2(g) C(g) 2CO(g) initial pressure 0.5 atm 0 final pressure(0.5-x) 2x Total pressure = 0.5 -x+2x=0.5 +x = 0.8 atm x=0.3 atm. 2
KP
22. 23.
p CO (0.6) 2 1.8 PCO 2 0.2
atm.
According to lowry-bronsted acid base theory in (ii) reaction H 2 PO 4 donates H ion to H 2 O so it acts as an acid. Kp Kc(RT) n(g)
24.
n( g ) 1 Kc 0.0831 457 It means Kp > Kc Water is an amphiprotic solvent as it can accept protons as well as give protons.
25.
Ammonium ion NH 4 is a conjugate acid of NH 3
NH3 H2O NH4 OH Bronsted base Conjugate acid 26.
HSO 4 can act as a bronsted acid as well as bronsted base. HSO 4 H 2 O SO 42 H 3 O Acid HSO 4 H 2 O H 2SO 4 OH
Base 27.
P H 4.7 H 3 O 1.995 10 5 pK w 14 K W 1 10 14
log H3O NowKW H3O OH
OH
110 14 5 10 10 M 5 1.995 10
28.
conjugate base of H 2 PO 4 is HPO 24
29.
H 2 PO 4 H 2 O HPO 42 H 3O Acid conjugate base of OH
OH H 2 O O 2 H 3 O
170
30.
Lewis acid always accepts a pair of e here AlCl 3 accepts a pair of e.
31. 32.
CH 3 3 B accept a pair of e so lewis acid Pressure on equilibrium system increses ,so volume decreses. volume of ice is more than liquid H 2 O so more ice is malted
33.
For endothermic H O reaction, change in temperature affects the equilibrium system and forward reaction takes place . by increasing temp .
34.
K1
and
XeOF4 HF2 K XeOF4 XeO3F2 2 XeF6 H 2 O XeF4 XeF6 XeO3 F2 H 2O K is obtained by XeF4 HF2 K2
K1
K K2
36.
37. 38.
4 6 NO H 2 O Kc NH3 4 O 2 5
K1
(conc.)46(45)
conc1 According to Le -chatelliers principle, if conc. of reactant become doubled, then forward reaction takes place and concentration of product also increases. so equilibrium constant also remains same.
AB4(s) A 4(aq) 4 B (aq) _
K SP A 4 B
4
S(45)4 256S5 1
K 5 S SP 256 39.
2
KSP M2 X 4S3
M S KSP 4 2
40.
1
3
4 1012 4
1
3
1104
CH3COO 2 Ca Ca 2 2CH3COO 0.005 2 0.005 0.01M.
PH 1 log KW logKa log C 2 171
1 log K W 1 log K a 1 log C 2 2 2 1 pkw 1 pka 1 log 1 10 -2 2 2 2
41.
1 2 14 1 2 4.74 1 2 (2) = 7 2.37 1 8.37 change in volume affects number of moks per unit voulme. In a reaction
N2(g) O2(g) 2 NO(g) no. of moles of reactants and product are equal so volume change does not affects the equilibrium. 42.
P H 3 means H 103 M P H 4 means H 104 M
After mixing equal volume total
H 110
-3
10 4 1 2
0.110-4 110 4 2
=
1.110 3 5.5 10 4 2
H 5.5 104 M P H log H log(5.5 10 4 ) 4 0.7404 3.26
43.
P H of HCl should be less than 7. due to self ionisation of H 2 O from acid H 10 8 M from H 2 O H 10 7 M Total H 10 8 10 -7 10 8 (1 10) 11 10 8 M
PH log H log(11108 ) (1.0414 8) = 6.96 H
P 6.96
44.
P of buffer solution H
pK a log
salt Acid
log K a log1 log108 0 8
172
45.
if x is the solubility of AgCl in 0.04 M cacl2 , then
Ag x mol L1
Cl 2 0.04 x 0.08 x 0.08 M KSP of AgCl
Ag Cl
4 1 0 10
46.
P H pka log
5.5 4.5 log
0.08
A g 5 10 9 M
CH3COONa CH3COOH
CH3COONa 0.1
5.5 4.5 logCH 3CooNa 1 logCH 3CooNa 0 CH 3CooNa 1 M 47.
I can accept protons and hence is a base.
50.
N3H N3 H
53. 55.
H 2S can donate proton but can't accept proton. With increase in temperature, ionic product increases. because self ionisation of is endothermic process EDTA is Arrhenius acid as it can give H ions in aqueous solution, bronsted base. as it can accept protons and lewis base because N and O in it can donate lone paris of electrons.
56.
Hydrazoic acid N3 H
57.
KW H OH
58.
mol L1 mol L1 mol 2 L2 Smaller the Kb value, weaker is the base.
59.
On adding NH 3 to water, OH - will increse, Kw H3O OH is constant.
Therefore H3O will decrease 61.
POH P K b log
Salt Base
[Salt] = [Base]
= log(2 10 5 ) log 1 = 0.3010 5.000 0 4.6990 P H 14 p OH 14 4.6990 9.3010
173
62.
P OH pK b log
Salt Base
P OH 14 P H 4.75
4.75 p kb log 1 pk b 4.75 63.
log 1 0
2 BaSO4 (s) Ba (aq) So24(aq)
x mol/lit
x mol/lit
K SP Ba 2 So 24
K SP x 2 1
64 65.
1
x (ksp) 2 (1.5 10 9 ) 2 3.9 10 5 molL1 Hint : As value of KSPis less , solubility is also less.
glycine (NH2CH2COOH) is more acidic than basic. overall ionisation constant K K a1 K a 2 4.5 10 3 1.7 10 10
7.65 10 13
H
K C 7.65 10 13 0.01
0.87 10 7 M P H log(8.7 10 8 ) 7.0605 66.
KSP of BaCo3 Ba 2 CO32
Ba 2 67.
5.110 9 5.110 5 M 110 4
Ba(OH)2 2HCl BaCl2 2H2 O 2 mol HCl neutralize1 mole Ba(OH)2 1 mol HCl neutralize 0.5 mole Ba(OH)2 Ba(OH)2 Ba 2 2OH 1 2
no.of moles of Ba(OH)2 3. 1 2 Ba(OH)2 left 3 0.5 2.5 Ba(OH) 2
2.5 0.05 M 50
or OH 2 0.05 M 0.1 M
174
68. 69.
K W 1.0 1014 Kh 5.65 1010 5 K b 1.7 10 PH
1 K P W PKa log C 2
1 2 14 1 2 4 1 2 log10 2 =7+2-1=8 70.
P H 3 means H 10 3 M 1000 ml juice contains 10 3 mole H ions no.of H ions 10 3 6.022 10 23 in 1000 ml
500 ml juice contains H ions
71.
10 3 6.022 10 23 500 1000
= 3.011 10 20 All alkaline earth metal chlorides MCl 2 on hydrolysis will produce acidic solution MCl 2 H 2 O M(OH) 2 2HCl
because M(OH)2 is a weak base and HCl is a strong acid. but as we go down the group, basic character of hydroxides increses. Hence acidic character decreses. So BaCl2 will have the highest P H . 72.
MX (s) M X K SP S 2 S (K SP )
1
2
(4 10 8 )
MX 2(s) M 2X K SP 2
1
2
2 10 4 M.
K 4S S SP 4 3
1
3
2 105 M.
M3X (s) 3M X 3 KSP KSP = 27S S 27 4
1
4
2.7 1015 27
1
4
2 10 4 1 10 4 2 10 5 MX M 3X MX 2 . 74.
P OH pk b log
Salt weak base
4.7 log
0. 1 5.7 0.01
P H 14 5.7 8.3
175
1104 M
75.
HCOONa is a Salt of weak acid (HCOOH) and Strong base (NaOH) So it is basic. C 6 H 5 NH 3Cl is a Salt of weak base C 6 H 5 NH 2 and strong acid (HCl) so it is acidic. KCN is a Salt of Strong base (KOH) and weak acid (HCN) so it is basic.
76.
AgIO3(s) Ag (aq) IO3 (aq)
K SP S2
S (K SP )
1
2
(1.0 108 )
1
2
1104 mol
S 110-4 283 283 10 4 gm
lit
lit
1000 ml contains 283 10 4 gm of AgIO 3 100 ml contains 28.3 10 4 gm of AgIO 3
77.
Molar concentration of NaOH O H
P
50 10 3 gm -4 40 gm mol 1 10dm 3 1.25 10 M
lo g (1 .2 5 1 0
4
)
0.0969 4.0 3.9031 H
P 14 3.9031 10.0969
78.
0.1 M HCl m eans H 10 1 P H 1
0.1M KOH means OH 10 1 P OH 1 P H 13 0.01 M HCl means H 10 2 P H 2
0.01M KOH means OH 102 POH 2 PH 12 79.
H
ion concentrations are 10 11 , 10 12 and 1013 on mixing equal volumes, H in final solution
10 11 10 12 10 13 10 10 12 110 12 0.110 12 = 3 3 80.
H (A) P pka log
11.110 12 3.7 10 12 M. 3
Salt 4.8 log 0.1 4.8 1 5.8 acid 0.01
H (B) P 4.8 log
0.01 4.8 log 10-1 4.8 1 3.8 0.1
H (C) P 4.8 log
0.1 4.8 0 4.8 0.1
1 1 H (D) P 7 (pka pkb) 7 (4.8 4.8) =7.0 2 2
176
81.
Soda water contains weak acid H 2 CO 3 So its P H 7
83.
P H p K a log
6 5 log log 84.
Salt acid
Salt acid
Salt 1 acid
Salt 10. acid
or
Solubility of Cus (ksp)
1
2
(110
ksp Solubility of Ag 2S 4
) 11016 M 1
3
4 10-45 3 15 110 M 4 1 5
110 At 25 C temp P of H O 7 H 10 M P 6.8 means P 7 H is more than 10 M Solubility of HgS ksp
85.
1
1
1 32 2
4 10-54
2
H
27
M
7
2
H
H
7
Self ionisation of H 2 O is endothermic so by increasing temp H ion increases. 86.
ksp for BaSo 4 Ba 2 SO 42 10 11 0.1 SO 24 SO 24 10 10 M ksp for CaSO 4 Ca 2 SO 24
10 6
0.1
10 5 M SO 24
2
ksp for Ag 2So 4 Ag SO 42
10 5
(0.1)2
10 3 M SO 24
As SO 24 10 10 M in BaSO 4 (least value) it can be precipitated first 87.
A2 B3(s) 2A3 3B2 2x
3x
B
ksp A
3 2
2 3
(2 x) 2 (3 x) 3 4 x 2 27 x 3 108x 5
177
88.
P H P K a log
Salt Acid
4.91 4.76 log log
Salt Acid
Salt 0.15 Acid
Salt antilogof0 .15 1.41 Acid
0.2 50 1000 1.41 V 70.92 ml 0.1 V 1000 89.
pka log(1.8 10 4 ) 3.74
log
90. 91.
Salt P H pka 4.25 3.74 0.51 Acid
Salt antilog of0.51 3.24 Acid
14 11 10 4 1.47 10 kc 6.8 10 Buffer Solution of highest capacity is formed at which
K b kw
P H pka log(1.8 10 4 ) 3.74 93.
Ionic Product of CaF2
10 ksp 10 10 ksp 10 10 ksp 10 10 ksp
in (i) IP Ca 2 F (ii) IP 10 2 (iii) IP 10 5 (iv) IP 10 3 94.
P H pka log
2
12
3 2
8
3 2
11
5 2
13
Salt Salt Acid Acid
= 8 + log 1 = 8 95.
H 2S weak acid H 2SO 4 Strong acid NaCl neutral NaNO 2 basic Hence H3 O will be in the order of
NaNO2 NaCl H 2S H 2SO4
178
ppt. obtain
96.
NaCN is Salt of weak acid (HCN) and Strong base (NaOH) Hence
h
97.
kw 10 14 2.48 10 2 1 k aC 9 (1.3 10 ) 80
Percen tan ge Hydrolysis (2.48 10 2 )100 = 2.48 If x is Solubility of AgCl in 0.04 M CaCl2 , then A g x m ol L 1 C l (0.04 2) x 0.08M
0.08(x) 4 10 10 x 5.0 10 9 M
98.
P H pka
log CH 3COONa 1 CH 3COOH
5.5 4.5 log
CH 3COONa 0.7
4.5 + log CH3COONa 1 log CH 3 COONa 0 CH 3 COONa 7 M
99.
50 ml of0.4 N HCl
0.4 50 0.02 g eq. 1000
0.2 50 0.01g eq. 1000 0.01 g eq of NaOH will Neutalilise 0.01 g eq of HCl HCl left unneutralised = 0.01 g eq vol of Sol. =50+50 =100ml 50 ml of 0.2 N NaOH
HCl
0.01 1000 0.1N 100
or H 0.1M
p H log(0.1) = 1.0 100.
CH 3 COOH
at eq (a 3.4 10 4 )
CH 3COO
3.4 10 4
3.4 10 3 .4 1 0 1 .7 1 0 a 3.4 10 4
H
3.4 10 4
4
4
5
(G iven)
179
a 6.8 10 3
101. A B C D initial 1 1 0 0 conc. Ateqm(10.8) 0.8 0.8 conc.(10.8) 0.2 0.2
Kc
CD 0.8 0.8 16.0 AB 0.2 0.2
Kc
CD 2 2 1 AB 2 2
102. A BCD initial 4 4 conc.
0 0
Ateqm(4 2) 2 2 conc.(4 2) 2
2
103. KC remains same beacause KC is a characteristic constant. PCl5 PCl3 Cl2 2 0 0
104.
2 60 100
2 40 100
moles 1.2 0.8 mol 1.2 0.8 conc. 2 2 lit Kc
2 40 100
0.8 0.8
2
PCl 3 Cl 2 0.4 0.4 0.266 PCl 5 0.6
105. Dissociation of CH3COOH CH3COOH H CH3COO k a1 1.5 103
Dissociation of HCN: HCN H CN for a reaction
k a 2 4.5 10 3
CN CH3COOH CH3COO HCN is
Ka
Ka 1 1.5 10 3 3.33 10 4 Ka 2 4.5 10 10
180
106. A B C D x x 0 0
CD 2x 2x 4 AB x x
kc
2x 2x 107. N 2 O 4 2 NO 2 1 0 (1- ) 2 108. H 2 I 2 2 HI
Total moles = 1- +2 =1+ from equation 2x=3
Initial 4.5 4.5 0 conc. at eqm. (4.5-x)(4.5-x) 2x
x 3 2 1.5
H 2 4.5 1.5 3 Kc
I 2 4.5 1.5 3 109.
K f 1.1 10 2
HI2 3 3 1 H 2 I 2 3 3
K b 1.5 10 3
kf 1.1 10 2 kc 7.33 k b 1.5 10 3 110. initial conc. At eqm. conc.
2 NH3 N 2 3H 2 2 0 0
1
1
3
2 3
KC
111.
N 2 H 2 NH 3
2
1 3 2 22 (1)
3
27
64
P4(S) 5O2(g) P4 O10(s) KC
P4O10 P4(s) O2(g)
5
we know that concentration of a solid component is always taken as a unity
Kc
1 O 2 5
181
112. 2Ag 2O(s) 4 Ag (s) O2(g ) for this reaction Kp Po 2
H2(g) CO2(g) CO(g) H2 O(g)
113.
initial conc. 1 At eqm. (1-x) Kp
129.
PC O PH 2 O PH 2 PC O
1 (1-x)
0 x
0 x
x2 (1 x ) 2
2
q p ApBq (s) PA (aq) q B(aq)
Solubility is PS
qS
mol/lit
K SP (PS) P (qs) q
S( p q ) P P q q 135.
Co 0.1 M Ka 5 10 8
H
Ka Co (5 10 8 0.1)
1
2
(50 10 10 )
7.07 10 5 M.
182
1
2
at equilibrium
log K C =
E cell = -
0.059 C log 1 n C2
T = 298K
E cell =
0.059 log K C n
nFE 0Cell 2.303RT
Faraday's laws of Electrolysis: (i) First law: The amount of products produced at the electrodes by electroysis are directly proportional to the quntity of the electricity passed through the electroytic cell. if w is the mass of the product produced and Q is value of the quntity of electricity passed, than W a Q (ii) Second law : If the different electroytic cells, containing different electrolysis are joined in series and same quntity of electricity os passed through them, than the amount of products obtained at the electrodes are directly proportional to thier equivalent weight. Wa Eq, where W = Mass of product obtained and Eq = Equivalent weifht of product. the modern presentation of Faraday's law was made as follows: "The products, obtained at the electrodes by oxidation and reduction half– reactions have the relation with the moles of the products and stoichiometry of the reaction and the quntity of electricity." " The products, obtained at the electrodes by oxidation and reduction half– reactions have the relation with the moles of the products and stoichiometry of the reaction and the qunantity of electricity." The quantity of electricity passed by 1 more electrons is called one Faraday. 1 Faraday (F) = 1.602 ´10 -19 ´ 6.022 ´ 10 23 electron mole-1
F=
I´ t 96500
= 96487 (@ 96500 ) Coulomb mole-1 (electron)
Efficiency of cell(%) =
Emperimental value of product ´ 100 Theoretical value of product
Gibbs Free Energy and cell Potential ΔG 0 = - nFE 0 Cell
ΔG = Welectrical = - nFE Cell
The resistance of a uniform counductor is directly proportional to its lengh (l) and inversely proportions to its area of cross section (A) R= p
1 where, R = Resitance, A= Area of cross section, l = Lengh, p = proportionality constant. A
Conductivity: The inverse of resistance R is called conductivity G
G=
1 A A 1 l K where K = K = G× R p×l l p A
Specific conductivity K = observed conductivity G x cell constant K ´1000 C Where, K = (Kappa) specific conductivity, c = concentration if solutin in unit of molarity.
Molar conductivity Ù M =
184
Conductivity of strong electrolytes : 1
0 Molar conductivity of a strong ionic electrolyte Ù M = ÙM - AC 2
The value of degree of dissociation for such an electrolyte at the given concentratin will be as below. Degree of dissociation ( a ) = Ka =
Molar conductivity of the solution of a given concentration Ù m 0 Molar conductivity of the solution at infinite diletion Ù m
Ca 2 where, Ka = Dissociation constanto of weak electrolyte, (1 - a )
C = concentration of solution, a = degree of dissociation of weak electrolyte +
Kohlrausch's law " The morar conductivity of an electroyte at infinite dilution λ 0 m is the sum o o if the values of the molar conductivities of positive ion an negative ion present in them l m + and l m+ Ù0 m = v + λ m + v - λ m (where λ 0 m + and λ 0 m - respectively are ionic molar conductivity of positive and negative ion.) (1) Primary cell: The cell which is dead after a long use and which cannot be regenerated i.e. which cannot be reproduced, is called primary cell. e.g. Dry cell. (2) Secondary cell : The cell which can be regenerated or reproduced is called secondary cell. e.g lead storage cell and Ni –Cd storage cell. Dry cell: Reaction at cathode: 2MnO 2 + 2NH +4 + 2e - ® M n O3 + 2NH3 + H 2 O Reaction at anode : Zn( s ) ® Zn( ag ) + 2e The potential of this cell is about 1.5 volt. 2
Mercury cell : Reaction at cathode: HgO + H 2O + 2e - ® Hg ( e ) + 2OH Reaction at anode : Zn( Hg ) + 2OH - ® ZnO ( s ) + H 2O + 2e The complete equation of these reactions is as follows: Zn( Hg ) + MgO ® ZnO ( s ) + Hg (l ) The cell potential of this cell is about 1.35 volt and during the whole life of this cell, no ions an produced in the complate reaction. Lead storage cell. The discharging rections taking place in lead storage cell are as follows : Reaction at cathode: PbO2 ( s ) + 4 H + (aq ) + SO42 - (aq ) + 2e - ® PbSO4 ( s ) + 2 H 2O(l ) Reaction at anode : Pb( s ) + SO42- (aq ) ® PbSO4 ( s ) + 2e The discharging reactions taking place in lead storage cell are as follows: Cathode: PbSO2 ( s ) + 2e - ® Pb( s ) + SO42- (ad ) Anode : PbSO4 ( s ) + 2 H 2O(l ) ® PbO2 ( s ) + 2 H + (aq ) + SO42 (aq ) + 2eThe cell potential of this cell is about 2 volt. Ni–Cd storage cell: Cd ( s ) + 2 Ni (OH ) 3 ( s ) ® CdO ( s ) + 2 Ni (OH ) 2 ( s ) + H 2O(l ) Hydrogen Fuel cell Cathode : O2 ( g ) + 2 H 2O(l ) + 4e - ® 4OH - (aq ) Anode : 2 H 2 ( g ) + 4OH - (aq ) ® 4 H 2O(l ) + 4e Cell reaction : 2 H 2 ( g ) + O2 ( g ) ® 2 H 2O(l ) + 571.7 kj 185
M.C.Q. 1.
2.
Reduction reaction means ________ (a) a process of adding oxygen
(b) a process of removing hydrogen
(c) a process of adding electron
(d) a process of removing electrons
Which substance is oxidizing agent ? (a) a substance donates hydrogen or accepts oxygen (b) a substance donates oxygen or accepts hydrogen (c) a substance experience oxidation (d) a substance donates electron
3.
Which substance is called reducing agent ? (a) a substance donates hydrogen or accepts oxygen (b) a substance accepts hydrogen or donates oxygen (c) a substance expereince reduction
4.
5.
6.
7.
(d) a substance gains electron
Oxidation reaction means _____________ (a) a process of removing electron
(b) a process of adding hydrogen
(c) a process of removel of oxygen
(d) a process of adding electrons
Which of the following is the characterictic of reducing agent ? (a) it experience oxidation.
(b) it experience reduction
(c) it gains electrons
(d) it gives oxygen
Which of the following is the characteristic of oxidizing agent ? (a) it experience oxidation.
(b) it experience reduction
(c) it gains oxygen
(d) it donates electrons
Which of the following statement is true ? (a) there is always reduction occur of oxidizing agent (b) there is always oxidation occur of reducing agent (c) oxidation and reduction are supplimentary processes (d) Given three statements are wrong.
8.
Which of the following statement is wrong ? (a) there is always reduction occur of oxidizing agent (b) there is always oxidation occur of reducing agent (c) oxidation and reduction are supplimentary processes (d) Given three statements are wrong.
9.
Which of the following does not occur, when a rod of Zn metal is dipped in an aqueous solution of CuSO4 ? (a) blue colour of the oxygen fades gradually.
(b) weight of Zn–metal rod decreases.
(c) weight of metal strip of zinc increases. (d) colour of the surface of Zn road become saffron–red. 186
10.
Which of the following observation obtained, when rod of Cu metal is dipped in an aqueous solution of AgNO3? (a) No change in the weight of metal rod of Cu occurs. (b) weight of rod of copper metal decreases (c) solution become bluish gradually (d) colour of the surface of rod of Cu metal does not change
11.
Which substance get oxidized in the reaction : 2Al + Cr2O3 ® Al2O3 + 2Cr ? (a) Al
12.
(b) Cr2O3
(c) Al2O3
(d) Cr
Which substance is a reducting agent in the following reaction ? Reaction : 2Al + Cr2O3 ® Al2O3 + 2Cr (a) Al
13.
(b) Cr2O3
(c) Br
(b) Ca
(d) –1
(d) given all
(c) Cs
(d) O
(c) – 12
(b) –1
(d) + 12
Which of the following oxidation no. does not possess by carbon in any of its compound ? (b) 0
(c) –4
(d) +5
Which of the following oxidation no. does not possess by nitrogen in any of its compound ? 1 3
(b) –3
(c) –4
(d) +5
Which of the following oxidation no. possesses by oxygen in its compounds ? (a) –1
22.
(c) +5
Which of the following oxidation no. does not possess by oxygen in any of its compound ?
(a) 21.
(d) I
Which of the following element always possesses +1 oxidation state in any of its compound ?
(a) –2 20.
(b) +3
(b) Cl
(a) –2 19.
(c) Cl
Oxygen conbines with which of the element by forming chemical bond, then it possesses posi tive oxidation no. ?
(a) F 18.
(d) None of these.
Which of the following oxidation no. does not possess by Cl, Br and I, when they conbines with oxygen forming chemical bond ?
(a) F 17.
(c) Na2S
(b) F
(a) +1 16.
(b) S
Which of following elements does not possess positive oxidation no. in any of its compound ? (a) O
15.
(d) Cr
In the reaction, 2Na + S ® Na2S, which sustance acts as oxidizing agent ? (a) Na
14.
(c) Al2O3
(b) +3
(c) –4
(d) +5
Which type of metal compounds are nomenclate according to stock notation nomenclature method ? (a) Metal compound having fixed oxidation no.
(b) Comppounds of alkali metals.
(c) Metallic compounds having more than one oxidation no. (d) Compounds of non–metal. 187
23.
Molecular fromula of sodium chromate (VI) is _________ (a) Na2Cr2O 7
24.
25.
26.
(b) Na2Cr2O4
(c) Na2CrO4
(d) NaCrO4
What is the name of K2Cr2O7 according to stock notation nomenclature method ? (a) Potassium dichromate (VI)
(b) Potassium chromate (VI)
(c) Potassium dichromate (III)
(d) Potassium dichromate (IV)
What is the name of TiO2, according to stock notation nomenclature method ? (a) Titanium (II) oxide
(b) Titanium oxide (IV)
(c) Titanium (IV) oxide
(d) Titanium (V) oxide
Which of the following is the true for Fe3O4 ? (a) oxidation no. of each Fe atom in Fe3O4 is +3. (b) oxidation no. of each Fe atom in Fe3O4 is +2. (c) The no. of atoms having oxidation no. +2 and +3 is 1:2 in Fe3O4. (d) oxidation no. of each Fe atom in Fe3O4 is + 83
27.
Which of the following is the wrong for Fe3O4 ? (a) it is a combination of FeO and Fe2O3. (b) Oxidation no. of each Fe atom in Fe3O4 is not + 83 (c) The no. of atoms havinf oxidation no. +2 and +3 is 1 : 2 in Fe3O4. (d)The proportion of FeO and Fe2O3 is 1 : 2
28.
The no. of peroxy ring in CrO5 is (a) 1
29.
(d) None of these
(b) TiO2
(c) BaO2
(d) SiO2
Which of the following compound possesses super oxide (O21–) ion ? (a) kO2
31.
(c) 3
Which of the following compound possesses peroxide (O22– or [–O–O–]2–) in? (a) kO2
30.
(b) 2
(b) TiO2
(c) BaO2
(d) SiO2
The oxidation no. of sulphur in H2SO3, H2SO4, H2SO5 or sulphuric acid, Caro’s acid (permono sulphuric acid) is respectively________ (a) +4, +6, +7
32.
(d) +5,+6, +6
(b) +7, +2, +6, +5
(c) +6, +2, +6, +5
(b) +5, +6, +6
(c) +4, +6, +6
acid
(d) +6. +3, +7, +5
What is the oxidation No. of Cr in CrO5, K2Cr2O7, K2CrO4 or in chromium potassium dichromate, potassiam chromate respectively (a) +4, +5, +6
34.
(c) +4, +6, +6
The oxidation no. of sulphur in H2S2O8, H2S2O3, H2S2O7, H2S2O6 or marshal (perdisulphuric acid), thiosulphuric acid, oleum, Dithianic acid is respectively (a) +6, +3, +6, +7
33.
(b) +3, +6, +6
p e n t o x id e ,
(d) +6, +6, +6
What is the oxidation no. of phosphorus in H3PO3, H3PO4, H3PO2 or Phosphoric acid, phosphinic acid respectively ?
phosphorus, acid,
(a) +3, +5, +1
(d) +2, +5, +1
(b) +3, +5, +2
(c) +4, +5, +1 188
35.
P O7, H5P3O10, (HPO3)3 or pyrophosphoric acid, penta phosphoric acid, triphosphoric acid respectively ? What is the oxidation no. of phosphorus in H
(a) +4, +5, +3 36.
(b) +3
(c) +7
(d) +5
(b) –1/3, +1, +3, +5
(c) +1, +1, +3, +5
(d) +1/3, +2, +2, +5
(b) +4, 4
(c) +2, +2
(d) +3, +4
(b) +3–x–2y
(c) +3+x+2y
(d) +3+x–y
(b) +2
(c) 0
(d) +3
(b) 0
(c) –5
(d) –4
(b) 3, 2
(c) 2, 2
(d) 4, 3
(b) 15
(c) 7.5
Cr3+ ?
(d) 10
(b) 5
(c) 2
(d) 12
How many mole ferrous (Fe2+) ion oxidized in ferric (Fe3+) ion by the required no. of electrons the oretically to reduced 4 mole Cr2O72– in to Cr3+ ? (a) 8
49.
(d) +5
What moles of Cr2O72– reduced in Cr3+ by the addition of 12 moles of electrons ? (a) 6,
48.
(c) +7
How many moles of elements are added when 2.5 mole Cr2O72– reduced in (a) 12.5
47.
(b) +3
What will be the value of x and y respectively in AlFXOy6– ? (a) 1, 4
46.
(d)+3, –1/3
The value of n in AlFXOyn, if x=2 and y=3 ? (a) –2
45.
(c) +1/3, –1
The value of n in AlFXOyn, if x=1 and y=1 ? (a) +1
44.
(b) +1, –1
The value of n in AlFxOyn is— (a) +3 –x–y
43.
(d) +4, +3, +2, +1
What is the oxidation no. of silicon in zeolite (Na2Al2Si4O12) and tremolite [(Ca2 Mg5(OH)2(Si4O11)2] respectively ? (a) +4, +3
42.
(c) +1, +2, +4, +5
What is the the oxidation no. of nitrogen in N3H, H2N2O2, HNO3 or hydrazoic acid, hyponitrons acid, nitrus acid, nitric acid respectively ? (a) –1, +1, +3, +5
41.
(b)+1, +2, +3, +4
What is the maximum positive oxidation state of chalcogen element in it’s compound ? (a) +6
40.
(d) +4, +5, +5
What is the maximum positive oxidation state of halogen element in any of it’s compound ? (a) +1
39.
(c) +5. +5, +5
What is the oxidation no. of iodine in ICl3, CsI3 respectively ? (a) +3, –1
38.
(b) +6, +5, +5
The oxidation no. of chlorine in HClO, HClO2, HClO3, HClO4 or Hypochlorus acid, chloric acid, chloric acid and perchloric acid respectively are (a) +1, +3, +5, +7
37.
4 2
(b) 24
(c) 48
(d) 12
Theoretically what gram ferrous (Fe2+) ion oxidized in to ferric (Fe3+) ion by passing 2.4125 ´ 105 coulomb electric charge ? (Atomic mass of Fe=56 gram/mole) (a) 70 gram
(b) 140 gram
(c) 14 gram 189
(d) 280 gram
50.
What mole of MnO4– reduced in Mn2+ by the addition of 7.5 mole electrons in MnO4–? (a) 2.5
51.
(d) 20
(b) 0, +1, +3
(c) +1, 0, +3
(d) 0, +2, +1
(b) –2, +3, +3
(c) +2, +3, +4
(d) –2, +3, +4
(b) –3, +3, –2
(c) +3, +3, +3
(d) –2, +3, –2
(b) cyclohexane
(c) benzene
(d) given three compounds
(b) 4, 10, 9
(c) 5, 4, 18
(d) 5, 4, 9
(b) 7
(c) 14
(d) 9
(b) 3
(c) 12
(d) 9
What will be the chage in the oxidation no. (change in no. of electrons), if no. of e– = 10 in balanced eqation given below ? Reaction : aP4 + bNO3– + cH+ ® dPO43– + eNO2 + fH2O (acidic medium) (a) 20
62.
(c) 40
What will be the chage in the oxidation no. (change in no. of electrons) inbalanced eqation given below ? Reaction : 2S + bHNO3 ® CH2SO4 + dNO (acidic medium) (a) 6
61.
(b) 60
In the balanced state of reaction, aBr2 + bOH– + CH2O ® dBrO4– + eHBr (basic medium), if c=9, then what will be the change in oxidation no. (change in no. of electrons). (a) 21
60.
(d) 1.8066 x 1025
What are the values of b, d and f in the balanced state of the following reaction? Reaction : aMnO4–+bAs2O3+ cH2O ® dMn2++eASO43–+fH+(PH B > C
(b) A > C > B
(c) C > B > A
192
(d) C > A > B
81.
Standard oxidation potential of half cells of A/A2+, B/B2+, C/C2+ and D/D2+ are in increasing order, then which of the following statement is correct ? (a) solution of salt of A2+ can be stored in the vessel of metal B. (b) solution of salt of D2+ can’t be stored in the vessel of metal C. (c) solution of salt of D2+ can be stored in the vessel of metal B. (d) given all three statements are wrong.
82.
For which of the following compound, a graph of mular conductivity and (molarity)1/2 is obtained straight line ? (a) CsCl
83.
(b) NH4OH
(b) 3:2:1
(d) –o.11v
(b) less then 2.5 mole
(c) more then 2.5 mole (d) 5 mole
(c) 1:1:2
(d) 1:2:1
What would be the change in PH of the solution, when electrolysis of aqueous solution of CuSO4 is carried out in presence of inert electrodes ? (b) PH decreases
(c) no change in PH
(d) can’t be predicted
Which of the following formula is true to calculate the molar conductivity in conventional symbols ? (a)
^m= 1000×R×l C×A
(b)
G×A ^m = 1000× C× l
1000×G×l C×A
(d)
^m = 1000×C×l R×A
^
(c) m =
89.
(c) –0.33v
(b) 2:2:1
(a) PH increasses 88.
(b) 0.33v
When same electric charge is passed through electrolytic cells containing aqueous solutions of CuSO4, AgNO3 and NiSO4, then what would be prportion of moles of metal obtained at different cathodes respectively ? (a) 2:1:2
87.
(d) 6:3:2
5 faraday electric charge is passed during electrolysis of molten cacl2 solution, then what moles of Ca obtained at cathode experimentally ? (a) 2.5 mole
86.
(c) 6:2:3
Eored for Fe/Fe2+ and Fe2+/Fe3+ half cells are –0.44volt and +0.77volt respectively then what will be the value of Eoox for Fe/Fe3+ half cell ? (a) 0.037v
85.
(d) given all three
Electrolytic cells having molten NaCl, CaCl2 and AlCl3 solutions are connected in series and same electricity is passed then, which of the following ratio of moles of metal obtained at cathode is correct ? (a) 1:2:3
84.
(c) HCOOH
°
°
^KCl- ^ NaCl=23.4 Mho (cm) mole 2
-1
and
°
^
°
NaSr-
^ NaCl=1.8 Mho -(cm) -mole 2
-1
then,
which of the following is the correct order of molar conductivity at intimite dilution in an ascending pattern? (a) NaCl < KCl < NaBr < KBr
(b) NaCl < NaBr < KCl < KBr
(b) KBr < NaCl < NaBr < KCl
(d) KCl < NaCl < NaBr < KBr 193
90.
In an experiment of electroplating, 4 ampere electric current is passed for 2 minutes. Hence, m gram Ag is deposited at cathode. If 6 ampere current is passed for 40 second, then what amount Ag is deposited at cathode ? (a) 2m
91.
(b) 4m
(c) m/2
(d) m/4
Which of the following is correct form of the nernst equation to determine the oxidation potential of Cu/Cu2+ half cell ? RT ×ln [Cu 2+ ] 2F RT ° 2+ ° 2+ ×ln [Cu 2+ ] (b) E Cu|Cu =E Cu|Cu + 2F RT ° 2+ 2+ ×ln [Cu 2+ ] (c) E Cu|Cu =E Cu|Cu + 2F RT 1 ° 2+ 2+ ×ln × (d) E Cu|Cu -E Cu|Cu = 2F [Cu 2+ ] ° 2+ ° 2+ (a) E Cu|Cu =E Cu|Cu -
92.
°
The values of for ^ m A2B, X3Y2 and A3Y are 2.4, 1.5 and 1.8 respectively. Then what will be the °
value of ^ m for XB ?
93.
(a) 1.7 mho – (cm)2 – mole–1
(b) 8.1 mho – (cm)2 – mole–1
(c) 2.1 mho – (cm)2 – mole–1
(d) 0.7 mho – (cm)2 – mole–1
What amount of electric charge required for the reduction of 1 mole Cr2O72– into Cr3+ theoretically ? (a) 6
94.
(b) 3
(c) 1
(d) 4
What does the potential of the electrochemical cell become zero ? (a) Eox of anode and Ered of cathode become equal (b) Eored of anode and Eored of cathode become equal (c) Ered of anode and Ered of cathode become equal (d) Concentration of both the half cell become same
95.
In an electrochemical cell for which of the following alternative shows Ecell = Eocell ? (a) k = 1, then...
(b) cell reaction is in equilibrium, then
(c) concentration of both the half cells become equal, then (d) None of these 96.
For the electrochemical cell, Mg|Mg 2+ (0.04m)||Ag + (C2 ) |Ag ; Ecell = Eocell, then what will be the value of C2 ? (a) 0.04 M
97.
(b) 0.02 M
(c) 0.2 M
(d) 0.0016M
2+ 2+ In an electrochemical cell, Fe|Fe(C ||Cu (0.4m) |Cu , at constant temperature, concentration of both 1)
the solutions increases equally, then cell potential increases. Thus which od the following will be the value of C1 ? (a) 0.8 M
(b) 0.1 M
(c) 0.4 M
194
(d) None of these
98.
Which of the following possesses maximum equivalent conductance at infinite dilution? (a) K+(aq)
99.
(d) Rb+(aq)
(b) 104600 c
(c) 25000 c
(d) 0.26 c
A+ 25° c temperature and 1 bar pressure, oxidation potential of hydrogen half cell is 0.118 volt, then what would be the POH of the solution? (a) 2
101.
(c) Cs+(aq)
The potential of a cell is 2.0 volt. change in free energy (SG) occur for the reaction of cell is –50k.cal then what coulomb electric charge obtained from the cell? (a) 1.08 c
100.
(b) Na+(aq)
(b) 3
(c) 5
(d) 12
2+ 2+ For the electrochemical cell, Zn|Zn (C ||Cu (C |Cu , C2 > C1, then what change 1) 2)
occurs
in cell potential when concentration of solutions of both the half cells increases equally ? (a) increases 102.
(b) decreases
(c) No change occurs (d) Ecell=Eocell
What will be the oxidation potential of Pt | H 2 (g) | H + (aq) half cell at 25oC temperature ? (1 bar)
(a) 0.118 103.
(b) 0.649 v
(P H = 4)
(c) 0.236 v
(d) –0.118v
2+ 2+ ||Cu (C |Cu ; Ecell > EoCell then., what change occurs For the electrochemical cell, Zn|Zn (C 1) 2)
in the value of cell potential, when concentration of both the solution decreased equally ?
104.
105.
(a) Increases
(b) decreases
(c) no change occurs
(d) can’t be predicted
Which equation is suitable to calculate the value of Eocell ? ° ° red+E ° red (a) E cell=E (anode) (cathode)
° ° oxi+E ° oxi (b) E cell=E (cathode) (anode)
° ° oxi-E ° oxi (c) E cell=E (anode) (anode)
° ° oxi+E ° red (d) E cell=E (anode) (cathode)
+ 2+ At 25oc temperature, potential of Mg (2xM ) | Mg and Cu( yM ) | Cu and half cells are 2.365 v and
0.3415 v respectively. What will be the cell potential of the cell formed by these two half cells at 250 c temperature ? (a) 2.0235 v
(b) –2.0235 v
(c) 2.7065 v
195
(d) –2.7065 v
+ (106) Cell potential of the cell formed by connecting half cells Fe |Fe2+ ( x M) and Ag | Ag ( y M) is 1.295 V. if at 250C potential of the | Ag is 0.84 V. then what is the potential of the | at C?
(a) 0.455 V.
(b) – 2.135 V.
(c) 2.135 V.
(d) “ 0.455 V.
0 < ECo 2+ |Co then which value of x is possible? (107) IF ECo 2+ (x M ) |Co
(a) 1.2 M
(b) 0.2 M
(c)1.0 M
(d) can not be decided
(108) If cell potential of standard cell is 0.59 V then equilibrium constant for the cell reaction occuring in the cell at 250 C is _______. (n = 1) (a) 1.0
(b) 10.0
(c) 1010
1 (d) 10
(c) laclanche cell
(d) electrolytic cell
(109) Which of the following cell is different? (a) Daniel cell
(b) lead storage cell
(110) If equilibrium constant of a cell for reaction occuring in the electrochemical cell is 1.9413 ´ 1037 at 250C then what is the std. cell potential of the cell (n=2) (a) 2.2 V.
(b) 1.1 V.
(c) 0.0085 V.
(d) 0.013 V.
(111) On which of the following cell potential of the cell does not depend? (a) temperature
(b) concentration fo the solution of salt bridge
(c) concentration of the solutioin related with cell reaction (d) nature of electrodes (112) What is the value of term (a) 0.007 V.
2.303RT F
in nernst equation at 800C ?
(b) 0.01587 V.
(c) 0.01857 V.
(d) 0.07 V.
(113) Select correct option for the statements given with reference to electrochemical cell (where T= true and F = false) (i) in external circuit e flow from cathode to anode. (ii) in solution electricity conducted through ions. (iii) in external circuit electric current flow from anode to cathode. (iv) anions move from anode to cathode through salt bridge. (a) TFTF
(b) FTFF
(c) FFFT
(d) FTTF
(114) Mention n and Q for the cell reaction taking place in the Pt ∣ Cl2(g,1.0 ܾܽ ∣ ) ݎCl−(C 1 ) ∥ Au3+ (C 2 ) ∣ Au cell. (a) n = 6 , Q =
[Cl − ]6 [Au 3+ ]2 3+
(b) n = 3 , Q = [Au3+ ]2 ⋅[Cl− ]6
(c) n = 3 , Q =[Au ] ⋅[Cl−]3
(d) n = 6 , Q =
1 [Au 3+ ]2 [Cl − ]6
− (115) Mention the oxidation reaction takes place in half cell Pt ∣ H2(g,1.0 ܾܽ ∣ ) ݎOH(aq ). − − (a) H2(g) + 2OH(aq ) ⇌ 2H2 O(l) + 2e − − (c) 2H2 O(l) + 2OH(aq ) ⇌ 3H2(g) + 2e
− − (b) 2OH(aq ) ⇌ H2(g) + O2(g) + 2e − − (d) H2(g) + 2OH(aq ) ⇌ 2H2 O(l) +O2(g) + 2e
(116) Which of the following relation is correct for Faradays 2nd law? where and is quanitity of substance while and are equivalent mass of the substance. (a) m1E1 = m2E2
(b) m1E2 = m2E1 196
(c) m1 + m2 = E1 + E2 (d) E1E2 = m 1m 2
(117) Which of the following cell representation is in correct with reference to cell reaction taking 0 0 place in the cell at 250C ? (EPb 2+|Pb = − 0.13 V. , ESn 2+|Sn = − 0.14 V.) 2+ (a) Sn | Sn2+ (0.1 M) ║ Pb(0.01 M) | Pb 2+ (c) Pb |Pb2+ (0.01 M) ║ Sn(0.1 M) | Sn
2+ (b) Sn | Sn2+ (0.02 M) ║ Pb(0.2 M) | Pb 2+ (d) Sn | Sn2+ (0.05 M) ║ Pb(0.5 M) | Pb
(118) A2+ | A, B2+| B, C2+ | D is increasing order of std. reduction potential then choose correct option for the given state ments. (T = true and F = false ) (i)
rod of metal A dissolve in the solution of metal B2+
(ii) solution of A2+ ion can not be filled in the container of the metal C. 2+ 2+ (iii) reaction D(s) + B(aq 1.0 M) ⇌ D(aq 1.0 M) + B(s) spontaneously in forward direction.
(iv) atoms of metal C can not displace B 2+ ioin from its solution. (a) TFFT
(b) FTTF
(c) FTFT
(d) TFTF
(119) Select correct option in reference to e.m.f. series . (a) from top to bottom tendency to accept e increases . (b) from bottom to top strngth as oxidising agent decreases (c) top to bottom temdency to get oxidised increases . (d) if rod of the metal present at lower positon is the series is dipped in the solution of the metal ion present in the upper part of the solution then rod does not get dissolve 0 (120) What is the value of EMg | Mg 2+ at 250C ? ( EMg |Mg 2+ = 2.36 V. ) ( 0.5 M )
(a) 2.3689 V.
(b) – 2.3689 V.
(c) 2.88 V.
(d) –2.38 V.
(122) If potential of the cell formed by connecting half cell Ag | Ag +( 0.4M) and Al | Al3+ ( 0.1 M) is 0 2.546V then what is the std. cell potential of the cell at 25 C? (a) 1.46 V.
(b) 2.46 V.
(c) 2.64 V.
(d) 2.76 V.
(123) What is the unit of constant 0.059 in Nernst equation? (a) colulomb
(b) volt
(c) faraday
(d) no unit
2+ (124) If n = 3 is taken for the cell Al | Al3+ (0.02 M) ║ Pb(0.2 M) | Pb then what is the value of Q ?
(a) 0.05 M–1
(b) 0.2247 M–1/2
(c) 0.01 M
(d) 0.05 M–1/2
+ (125) For cell Mg | Mg 2+ (0.005 M) ║ Ag (0.04 M) | Ag which of the following option is correct for the value of Q related to n ?
(a) n = 1, Q = 3.125 M1/2
(b) n = 2, Q = 1.7677 M
(c) n = 3, Q = 5.5243 M3/2
(d) n = 4, Q = 0.156 M2
− (126) For the cell Co | Co2+ (0.04 M) | Cl(0.1 M) | Cl2(g,1.0 ܾܽ | ) ݎPt which of the following option is correct for the value of Q related to the value of n?
(a) (b) (c) (d)
n= 1 n=2 n= 3 n= 4
(1) Q = 4.0× 10−4 M 3
(5) Q = 4 M−1
(2) Q = 8 M
(6) Q = 1.6 × 10−7 M6
3 − 2
(3) Q = 8.0× 10−6 M 1
(4) Q = 2 M−2
9 2
3
(7) Q = 0.02 M2 (8) Q = 16 M −2
197
(a) (b) (c) (d)
a – 4, b − 5, c – 2, d − 8 a – 7, b – 5, c – 2, d − 6 a – 7, b – 1, c – 3, d − 6 a – 4, b – 8, c – 7, d − 1
− 3+ Pt | Br2ሺg,1.0 ܾܽ ݎሻ | Br(0.5 (127) For the cell Pt Au value of Q = 20M-4 then what is the M) || Au(0.4 M) | Au appropriate value of n?
(a) n = 6
(b) n = 2
(c) n = 3
(d) n = 4
+ Mg | Mg 2+ (128) If temperature of the cell Mg Ag increases then what changes (0.005 M) || Ag (0.04 M) | Ag observed in the cell potential of the cell?
(a) increases
(b) decreases
(c) do not change
(d) cannot predict
2+ Ni | Ni2+ Cu increases then what changes observed (129) If temperature of the cell Ni (0.05 M) || Cu(0.08 M) | Cu in the cell potential of the cell?
(a) increases
(b)decreases
(c) do not change
(d) cannot predict
2+ (130) If temperature of the cell Fe | Fe2+ (x M) || Cu(0.08 M) | Cu decreases then theoritically cell potential of the cell increases then which value of x is possible?
(a) x = 0.5
(b) x = 0.02
(c) x = 0.05
(d) x = 0.04
2+ (131) If temperature of the cell Zn | Zn2+ (C 1 ) || Sn(C 2 ) | Sn (C1 < C2 ) decreases then theoritically cell
potential what changes are observed in the cell potential of the cell? (a) increases
(b)decreases
(c) do not change
(d) cannot predict
2+ (132) What is the equilibrium constant of the reaction taking place in the cell Zn | Zn2+ (C 1 ) || Cu(C 2 ) | Cu 0 0 at 250C? ( EZn 2+ |Zn = − 0.76 V., ECu 2+ |Cu = 0.34 V.)
(a) 3.3791
(b) 1.9413 ´ 1037
(c) 4.406 ´ 1018
(d) 5.15 ´ 10–38
(133) If for the reaction taking place in one electrochemical cell Kc = 1.3141 ´ 10107 M –1 and n = 2 then for that cell value of E0Cell is ________. (a) 6.32 V.
(b) 0.1186 V.
(c) 0.32 V.
(d) 3.16 V.
(134) If for the reaction taking place in an electrochemical cell Kc = 1.3141 ´ 10307 and n = 6. if standard reduction potential of the cathode is 1.36 V then what is the standard oxidation potential of the anode? (a) 6.32 V.
(b) 1.66 V.
(c) 2.71 V.
(d) 3.16 V.
(135) What is the pH of the solution of HCl if cell potentila of the cell 0 Pt | H2(g,1.0 Zh^) | HCl(X M) || Au3+ (0.001 M) | Au is 1.6655 V ? (E Au 3 +| Au = 1.4 V.)
(a) 3.5
(b) 4.17
(c) 5.5
(d) 4.83
(136) Mention the value of n for the reaction taking place in the cell if its cell potential is 3.16V and equilibrium constant is 1.727 × 10214 M −2 at 250 ܥtemperature. (a) 1
(b) 2
(c) 3
(d) 4
+ 0 (137) If for the cell Mg | Mg 2+ (0.2 M) || H(X M) ∣ H2(g,1.0 ܾܽ| ) ݎPt values of ECell and E Cell are 2.3629 V.. and 2.36 V. respectively at 250C temperature then value of x is ______.
(a) 0.4 M
(b) 0.5 M
(c) 0.1 M
(d) 0.25 M
0 + − is -1.23V and [H+] = 10−7 at 25 C (138) If standard potential of the reaction 2H2 O(݈) ⇌ O2(g,1 bar ) + 4H(aq ) + 4e at 250C temperature then what is the value of potential for the reaction? OR What is the oxidation potential of the pure water?
(a) – 0.41 V.
(b) – 2.62 V.
(c) – 0.817 V. 198
(d) 0.41 V.
(139) If for the reaction
− 0 2H2 O(݈) + 2e− ⇌ H2(g,1 bܽ ) ݎ+ 2OH(aq ,10 −7 M) E = – 0.417 V at 25 ܥ
temperature what is the value of E 0 for the reaction? OR what is the standard reduction potential of the pure water at 250 ܥif potential is – 0.417 V. (a) − 0.83 V. (b) – 2.62 V. (c) – 0.817 V. (d) 0.41 V. 3+ 2+ 3+ 3+ 2+ 3+ (140) If Co | Co , Co | Co , Fe | Fe , Fe | Fe std. oxidation potential are − 0.4167 V., − 1.81 V., 0.0367 V., − 0.77 V. respectively then what is the std. potential of the cell Fe | Fe2+ || Co2+| Co ? (a) 0.72 V. (b) – 0.11 V. (c) 1.0 V. (d) 0.16 V. 2+ (141) If cell potentisl of the cell Pt | H2(g,1.0 bܽ | ) ݎHCl(X M) ║ Cu(0.01 M) | Cu 0.635 V. at 250C 0 then what is the pH of the HCl solution? (ECu 2+ |Cu = 0.34 V.)
(a) 2.0 (b) 5.0 (c) 6.0 (d) 4 2+ (142) If cell potentisl of the cell Mg | Mg (0.01 M) ║ HCl(X M) | H2(g,1.0 Zh^) | Pt 2.09V. at 250C 0 then what is the pH of the HCl solution? (EMg 2+ |Mg = − 2.36 ܸ.)
(a) 6.58 (b) 5.58 (c) 3.58 (d) 4.58 (143) What is the value of E cell and equilibrium constant respectively for the reaction at 250C Fe(s) + 2Ag + Fe2+ (aq ,0.5 M) ⇌ (aq ,0.1 M) + 2Ag (s) ? (a) 1.25 V., 1.04 × 1021 M −1 (b) 1.23 V., 1.507 × 1012 M −1 (c) 1.23 V., 1.264 × 106 M −1 (d) 1.25 V., 1.081 × 1042 M −1 (144) Which reaction would takes place spontaneously in forward direction at 250C if standard oxidation potenitials of Pt | I2(s) | I−, Cu | Cu2+, Cr | Cr 3+, Zn | Zn2+, Sn | Sn2+, Pb | Pb2+, Ag | Ag +, Fe | Fe2+ are 0.54 V., − 0.34 V., 0.74 V., 0.76 V., 0.14 V., 0.13 V., − 0.80 V., 0.77 V. respectively. − 2+ (a) I2(s) + Cu(s) ⇌ 2I(aq ,0.1 M) + Cu(aq ,0.05 M) 3+ 2+ (b) 2Crሺaq ,0.001 Mሻ + 3Zn(s) ⇌ 2Cr(s) + Zn(aq ,0.1 M)
(c) Pb2+ ሺaq ,0.02 Mሻ + Sn(s) ⇌
Pb(s) + Sn2+ (aq ,0.2 M)
3+ (d) 3Ag + ሺaq ,0.1 Mሻ + Fe(s) ⇌ 3Ag (s) + Fe(aq ,0.01 M)
(145) How many gram of Ag will be obtained if 5.0 F quantity of electricity is passed through aqueous solution of AgNO3 ? (a) 270 g (b) 540 g (c) 180 g (d) 135 g (146) If 15 Faraday quantity of electricity is passed through Al3+ (݈) solution then how many gram of Al metal will be obtained ? ( cell efficiency is 80 % .) (At.wt Al = 27 gm mol-1) (a) 135 gm (b) 121.5 gm (c) 108 gm (d) 94.5 gm (147) If 10 Faraday quantity of electricity is passed through molten NaCl during electrolysis 84 L of Cl2 gas is obtained at STP then what is the efficiency of the cell? (a) 80 % (b) 75 % (c) 50 % (d) 90 % (148) During electrolysis oif acqueous solution of NaCl with inert electrodes then at cathode + instead of Na(aq ) reduction of H2 O takes place because ..................... (a) Compare to H2 O std. reduction potential of Na+ (aq ) is more (b) Standard oxidation potential of H2 O is very high (c) of inert electrodes (d) reduction potential of Na+ (aq ) is very less
199
(149) Select correct option for the given statements (where T = true anf F= false ) i) if electrolysis of aqueous solution of CuSO4 is done with inert electrodes then blue colour of solution become faint. ii) if electrolysis of aqueous solution of CuSO4 is done with inert electrodes then pH of the solution increases. iii) if electrolysis of aqueous solution of CuSO4 is done with active electrodes of Cu then weight of anode decreases. iv) if electrolysis of aqueous solution of CuSO4 is done with active electrodes of Cu then precious metals release from cathode. (a) FTFT (b) TFFF (c) TFTT (d) TFTF 3+ (150) On passing necessary quantiy of elecricity through Al(݈) solution 4.5 g Al deposited then + on passing same quantity of electricity through the solution of H(aq ) having sufficient concentration then what volume of hydrogen gas will be obtained at STP.? (at.wt Al = 27 g mol−1 ) (a) 44.4 L (b) 222.4 L (c) 11.2 L (d) 5.6 L (151) Select correct option for the given statements (where T = true anf F= false ) i) if electrolysis of dilute aqueous solution of NaCl is done with inert electrodes then pH of the solution increases. ii) if electrolysis of concentrated aqueous solution of NaCl is done with inert electrodes then at anode chloricne gas is obtained. iii) if electrolysis of dilute aqueous solution of NaCl is done with inert electrodes at cathode hydrogen gas is obtain. iv) if electrolysis of concentrated aqueous solution of NaCl is done with inert electrodes then on adding phenolphthalien to the solution becomes pink. (a) TFTT (b) FTTT (c) FFTT (d) TFTF (152) Select correct option for the given statements (where T = true anf F= false ) i) if electrolysis of dilute aqueous solution of NaCl is done with inert electrodes then pH of the solution increases. ii) if electrolysis of dilute aqueous solution of NaCl is done with inert electrodes then on adding methyl orange to the solution it becomes orange coloured. iii) if electrolysis of concentrated aqueous solution of NaCl is done with inert electrodes pOH of the solution increases. iv) if electrolysis of concentrated aqueous solution of NaCl is done with inert electrodes then the solution obtained turns red litmus to blue. (a) TFTF (b) FTTT (c) FTFT (d) FFTT − 0 0 (153) If at 50 C temperature value of Eox is + 0.83 V for the half cell Pt ∣ H2(g,1.0bܽ ∣ ) ݎOH(aq )
then what is the value of ionic product of water at 500 ? ܥ (a) 1.12× 10−13 (b) 1.102× 10−12 (c) 1.105× 10−12 (d) 1.0× 10−14 (155) What quantity of electricity is required for complerte reduction of all Ag + from 1.0 M AgNO3 is 250 ml aqueous solution ? (a) 2412.5 C. (b) 24125 C. (c) 4825 C. (d) 25250 C. (156) Using 2 g Hg cathode Cd-Hg amalgam is obtained by electrolysis of CdCl2 then how much ampere electric current should be passed for 100 seconds to obtain Cd-Hg having 20% Cd ? (At.wt Cd = 112.5 g mol-1 ) (a) 34.32 A (b) 17.16 A (c) 4.29 A (d) 8.58 A
200
201
202
203
204
205
JEE-UNIT-8 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49
ANSWER KEY c b a a a b d d c c a a b b d a c d d c a c c a c c d b c a c c d a c a d c a b b b c c a b c b a
50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98
c b d d a b b d c a c a b c c b c d d b a d a c d b c d a b c c a d a b d b c b c c a a c d c a c
99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147
b d b c a d c d a c d b b a d d a b a a c a d b b b c c c b a a b b d b c a b c a d c b d a b c b
206
148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196
d d d b c a a b d c a a a a d c d c c b c a c d b d a d c b b b a d c a c b d c b c a c d b c a d
197 198 199 200 201 202 203 204 205 206
d b c a b d b b d a
UNIT - 9 CHEMICAL KINETICS Important Points ñ ñ
ñ
ñ
Chemical Kinetics : The branch of chemistry which deals with the study of the rate of reaction and the factor affecting them. Kinetics - Greek word ‘kinesis’ = movement classification of reaction on the basis of rates: • Very fast reaction : ionic reactions (10-9 sec) • very slow reaction : rusting of iron, radiation from uranium. • Slow reaction : reaction by combining dinitrogen and dihydrogen under certain conditions. Factors Affecting rates of Reaction: (i) The state of substance and the area of surface (ii) concentration of solution. (iii) temperature of system. (iv) Pressure of system (v) Effect of catalyst (vi) Presence of light (If there is any impurity which tries to decrease the rate of reaction then it is calledc a t a l y t i c poison) Rate of Reaction : The rate of reaction is the change in the concentration of any one of the reactants or products per unit time. Average rate of reaction D[R] D[P] =+ Dt Dt Instantaneous Rate of Reaction rav = -
d[R] d[P] =+ as D t ® 0 dt dt Rate of reaction is always positive. The minus sign is used simply to show that the concentration of the reactant is decreasing. Rate determination : In the reactions the stoichiometric coefficients may be different. Rate of reaction can be determined but the determination must be consistent. In chemical kinetics, the following method is accepted. rinst = -
ñ
Rate = •
1 é d[R] ù 1 é d[P] ù =+ ê ê ú VR ë dt û vp ë dt úû
For any reation n1A + n2B ® n3C + n4D Rate = -
1 d [A] 1 d [ B] 1 d [ C ] 1 d [ D ] == = n1 dt n 2 dt n 3 dt n 4 dt
5Br - (aq) + BrO3Rate = -
(aq)
+ + 6H (aq) 3Br2(aq) + 3H 2 O()
+ d é BrO3- ùû 1 d éë Br ùû 1 d éë H ùû 1 d [ Br2 ] =- ë ==+ 5 dt dt 6 dt 3 dt
207
•
ñ
In aqueous solution, there is negligible change in concentration of water and so the change in its concentration is not expressed.
Rate Law : • The presentation of the rate of reaction with reference of concentration of reactants is called rate law. • This rate law in the wide range of concentration of reactants or products is studied and the law that is established is called ‘differential rate equation’ or ‘Rate expression’. Viz
Rate = K [ H 2 ] [ I 2 ]
H 2(g) + I 2(g) 2HI(g)
ñ
Rate constant and order of Reaction : • In most of the reaction carried out, the simple rate equation can be obtained in which rate is proportional to exponents of the concentration of reactant. The exponent is called order of reaction. H 2 + I 2 ® 2HI
Rate a [ H 2 ] [ I 2 ] \ Rate = k [ H 2 ][ I 2 ]
H 2 + Br2 ® 2HBr
Rate a [ H 2 ][ Br2 ]
1
2
\ Rate = K [ H 2 ][ Br2 ] 2 1
2N 2 O5 ® 4NO 2 + O 2
Rate a [ N 2 O5 ] \Rate = K [ N 2 O5 ]
• • •
Where K is rate constant and it is called specific rate constant when concentration of reactant is 1 M. At that time reaction rate = rate constant. It is necessary to note that the order of reaction has no relation with the stoichiometric cofficient of reactant. For the reaction n1A + n2B n3C + n4D • •
Rate = k [A]x[B]y Order of reaction with reference to reactant A is x and with that of reactant B is y. Hence total order of reaction = x + y. Thus the total order of reaction is equal to the sum of exponents of concentration of all reactants.
•
5Br- + BrO3- + 6H+ 3Br2 + 3H2O
•
Rate = K [Br-][BrO3-][H+]2
•
Order of reaction = 1 + 1 + 2 = 4
•
The total order of reaction can be positive, zero or even a fraction
•
Examples of different order.
•
Mo ® N 2(g) + 3H 2(g) (I) 2NH3(g) ¾¾
Rate = k[NH 3 ]o
•
1 (ii) H 2 O 2 ® H 2 O + O 2 2
Rate = k [ H 2 O ] 208
ñ
•
(iii) 2NO 2 + F2 ® 2NO 2 F
Rate = k [ NO 2 ][ F2 ]
•
(iv) 2NO + O 2 ® 2NO2
Rate = k [ NO ] [ O 2 ]
•
(v) CH 3CHO ® CH 4 + CO
Rate = k [ CH 3CHO ]
•
(vi) CO + Cl 2 ® COCl 2
Rate = k [ CO ] [ Cl 2 ]
ñ
1.5
2
1.5
Unit of Rate constant Rate =
ñ
2
dx = k (con) n dt
\K =
dx 1 conc 1 ´ = ´ n dt ( conc ) time ( conc )n
Order of reaction
Unit of K
0
M s-1
Mole litre-1 time-1
1
s-1
time-1
2
M-1 S-1
litre mole-1 time-1
n
M1-n S-1
litre1-n molen-1 time-1
Molecularity : • The number of atoms, ions or molecules of the reactant that take part in the reaction and which experience collision with each other so that the reaction results, it is called molecularity •
Hence, the orders of reactions and molecularities of bimolecular, trimolecular and elementary reaction are same.
•
The possibility of collision of three or more molecules with one another and to result in the reaction is less.
The molecularity more than three is not seen. Molecularity
Order of the reaction
It is equal to the sum of the number of reactant particles taking part in a single step reaction
It is equal to the exponents of the molar concentration of the reactants in the experimentally determined rate equation.
It is always a whole number
It can be fractional
It cannot be zero
It can be zero
Molecularity applies to one step reaction or a single step of reaction which involves several steps for its completion. The molecularity of an overall complex or multistep reaction has no significance or meaning.
Order of reaction applies to a reaction as a wholes, irrespective of the intermediate steps involved for its completion.
It can be obtained from simple balanced equation for the single step reactions.
It cannot be obtained from a simple balanced equation. It is obtained only from experimentally determined rate equation.
It does not help in elucidation of reaction mechanism
It is helpful in the elucidation of reaction mechanism 209
ñ
Zero order reaction : •
The rate of the reaction is proportional to zero exponent of the concentration of reactant.
•
-d [ R ] dt
= k [R ]
\[ R ] = - K.t + [ R ]o
Graph of [R] versus t then straight line. Slope = - k and intercept = [R]o
ñ
First order reaction : •
The rate of the reaction is proportional to exponent one of concentration of reactant.
-
d [R ] dt
kt = n
= k [R ]
[ R ]o [R ]
[ R ] = [ R ]o .e- kt log [ R ]t = -
•
Graph of log[R]t versus t then straight line. Slope = -
t1 = 2
ñ
Determination of order of reaction : (i) (ii) (iii) (iv)
Intial rate method Integrated rate equation method or Graphical method Half life method Ostwald’s isolation method
Ostwald’s Isolation method : •
ñ
0.693 k
Those reaction which are not truly of the first order but under certain conditions become reactions of the first order. Examples : (i) Hydrolysis of ethyl acetate. (ii) Hydrolysis of cane sugar.
Methods :
ñ
K and intercept = log[R]o 2.303
Pseudo – unimolecular or Pseudo first order reaction : •
ñ
K .t + log [ R ]o 2.303
In certain reaction there is involvement of more than one reactants. In this method, the concentration of other reactants in comparison to one reactant is taken in very large proportion. The reaction rate will be indicative with respect to reactant with less concentration because the concentration of other reactants remain almost constant.
Half life method : •
The time taken for half of the reaction to complete.
•
It is very simple method. 210
ñ
•
For zero order reaction tα[R] 1
•
first order reaction t 1 2 is independent of initial concentration
•
second order reaction t 1 2 a [ R ]o
•
nth order reaction t 1 2 a [ R ]o or t 1 2 a
-1
\nk = nA -
n -1
Slope = -
og
Ea RT
Ea 1 . 2.303R T
Graph of log K versus
1 = straight line 1
Ea 2.303R
K2 Ea é 1 1 ù Ea. DT = ê - ú= K1 2.303R ë T1 T2 û 2.303R.T1.T2
Threshold energy : •
Arrhenius equation shows that rate constant increases exponentially with temperature.
•
Increasing temperature from 300 to 310 K, the Kinetic energy increases only by 3% because it is proportional to temperature.
•
The reaction rates have almost doubled by increase in temperature by 10 K.
•
The explanation for this can be given that there must be some pushing energy or threshold energy required for the reaction of molecules.
Arrhenius factors : Ea = NA.E*
K = Ae-Ea/RT
ñ
[ R ]o
Arrhenius equation :
og k = og A -
ñ ñ
1
1- n
K = A.e - Ea /RT
ñ
o
2
where Ea Na E* where A Ea Both A & Ea
= Activation energy = Avogadro number = Kinetic energy = pre-exponential factor or frequency factor = Energy of activation = Arrhenius factors
Theory of collision : •
Max Trauz and William Lewis (1916-18) = Theory of collision.
•
In Chemical reaction, the number of collision per second per unit volume is called collision frequency (z). 211
ñ
•
For biomolecular reaction A + B ® Products
•
Rate = ZAB . e-Ea/RT
•
Rate = P . ZAB . e-Ea/RT
•
ZAB = collision frequency of A and B whose energy is equal to or more than activation energy.
•
P = Probability factor OR steric factor
•
Those collision in which molecules collide with sufficient kinetic energy (threshold energy) and proper direction, resulting into products. Such collision are called effective OR fruitful collision.
Endothermic reaction and Exothermic reaction. •
Minimum potential energy of reactants is less than that of products then the reaction will be the endothermic.
•
Minimum potential energy of reactants is more than that of products then the reaction will be exothermic.
•
Endothermic reation : DH = Hp-Hr = +ve •
•
Exothermic reaction : DH = Hp-Hr = -ve •
•
ñ
DH = Ea-Ear = +ve when Ea > Ear DH = Ea-Ear = -ve when Ea < Ear
Where Ea = activation energy of forward reaction Ear = activation energy of reverse reaction
Effect of catalyst : •
The main function of catalyst is decrease the activation energy, bring energy barrier down and increases the rate of reaction.
•
Equilibrium constant (K) is not changed.
•
Rate of reaction increases.
212
M.C.Q. 1.
2.
Rate of reation is defined as (a) decrease in the concentration of a reactant (b) increase in the concentration of a product (c) change in the concentration of any one of the reactants or products per unit time. (d) all the above three are correct 2A + B ® 3C for the reaction instant rate of reaction is... (a) +
d [ B] 1 d [ C ] 1 d [A] =+ + 2 dt dt 3 dt
(c) + 2 3.
4.
5.
6.
7.
8.
d [A]
= +3
d [ C]
(d) - 2
d [A]
=-
d [ B]
= +3
d [ C]
N2 + 3H2 2NH3 For the reaction the rate of change of concentration for hydrogen is -0.3 x 10-4Ms-1. The rate of change of concentration of ammonia is ... (a) 0.2 ´ 10-4 (b) -0.2 ´ 10-4 (c) 0.1 ´ 10-4 (d) 0.3 ´ 10-4 For the reaction of 4A + B ® 2C + D. Which of the following statement is not correct ? (a) The rate of formation of C and D are equal (b) The rate of formation of D is one half the rate of consumption of A (c) The rate of appearance of C is one half the rate of disappearance of B (d) The rate of disappearance of B is one fourth of the rate of disappearance of A ______ does not affect the rate of reaction. (a) size of the vessel (b) amount of the reactants (c) physical state of reactants (d) DH of reaction In the reaction N2O4(g) ® 2NO2(g) the pressure of N2O4 falls from 0.5 atm to 0.32 atm is 30 minutes, the rate of appearance of NO2(g) is (a) 0.012 atm min-1 (b) 0.024 atm min-1 (c) 0.006 atm min-1 (d) 0.003 atm min-1 In the reaction K1 and K2 are the velocity constants for the forward and backward reaction respectively. The equilibrium constant is
K1
K2
(c) K = K 2
(b) K = K1 ´ K 2
1
(d) none of the above
K1 1
For the reaction A + B + C ® Products, Rate = K [A] 2 [B] 3 [C]. The order of reaction is 11 (c) 5 6 (d) 6 For a reaction pA + qB ® Products. Rate = K[A]m[B]n. Then (a) (p+q) = (m+n) (b) (p+q) ¹ (m+n) (c) (p+q) > (m+n) (d) (p+q) = (m+n) or (p+q) ¹ (m+n)
(a) 3 10.
d [ B]
d [ B] 1 d [A] 1 d [ C] ==+ 2 dt dt 3 dt
dt dt dt dt dt dt The rate of reaction of spontaneous reaction is generally very slow. This is due to the fact that.. (a) the equilibrium constant of the reaction is < 1 (b) the activation energy of the reaction is large (c) the reaction are exothermic (d) the reaction are endothermic
(a) K = 9.
=+
(b) -
(b) 1
213
11.
A + 2B ® C + D For a reaction from following data correct rat law = Mole
12.
13.
14.
15.
16.
17.
18.
19.
liter -1 (B) 0.1 0.2 0.4 0.1
(A) mole lite-1 min-1 1 0.1 6.0 ´ 10-3 2 0.3 7.2 ´ 10-2 3 0.3 2.88 ´ 10-1 4 0.4 2.4 ´ 10-2 (a) Rate = K [A]2[B] (b) Rate = K [A]2[B]2 (c) Rate = K [A][B]2 (d) Rate = K [A][B] In the reaction A + B ® Products, the doubling of [A], increases the reaction rate to four times, but doubling of [B] has no effect on the reaction rate. The rate expression is …. (a) Rate = K [A]2 (b) Rate = K [A] (c) Rate = K [A]2[B]2 (d) Rate = K [A][B] A zero order reaction is one whose rate is independent of …. (a) Reaction vessel volume (b) Concentration of reactants (c) temperature (d) pressure of light The rate constant of a reaction changes when ... (a) pressure is changed (b) concentration of reactants changed (c) temperature is changed (d) a catalyst is added Which of the following is a reaction of zero order ? hJ (a) H 2 + Cl 2 ¾¾ ® 2HCl
(b) 2N 2 O5 ® 4NO 2 + O 2
(c) 2HI ® H 2 + I 2
(d) H 2 + Br2 ® 2HBr
Which of the following is a reaction of fractional order ? (a) 2N 2 O5 ® 4NO 2 + O 2
(b) 2NO 2 + F2 ® 2NO 2 F
(c) H 2 + Br2 ® 2HBr
(d) 2NO + O 2 ® 2NO2
A reaction involving two different reactants can never be a …. (a) bimolecular reaction (b) Unimolecular reaction (c) first order reaction (d) second order reaction For a reaction 3A hProducts, the order of reaction (a) 3 (b) 1, 2 or 3 (c) zero (d) any value between 1 and 3 When concentration of reactant is increased eighteen times the rate becomes two times, the rate of reaction is (b) 1 2 (c) 13 (d) 1 4 The rate determining step in a reaction is A + 2B ® C. Doubling the concentration of B would make the reaction rate... (a) 1
20.
(a) two times
(b) same rate
(c) four times 214
(d) 1 4 times
21.
The rate law of a reaction is rate = K [A]2[B]. On doubling the concentration of both A and B the rate X will become ... (a) x3
22.
23.
(c) 4x2
(b) 8x
(d) 9x
For the reaction CH3COCH3 + I2 + ® Products, the rate is governed by, rate = + K[CH3COCH3] [H ]. The rate order of iodine is = ______. H+
(a) 3 (b) 2 (c) 1 If the order of reaction is zero. It means that
(d) O
(a) rate of reaction is independent of temperature (b) rate of reaction is independent of the concentration of the reacting species (c) the rate of formation of activated complex is zero (d) the rate of decomposition of activated complex is zero 24.
The reactions of higher order are rare because (a) many bady collisions involve very high activation energy (b) many bady collisions have a low probability (c) many bady collisions are not energetically favoured
25.
(d) many bady collisions can take place only in the gaseous phase. 2A +2B ® D + E For the reaction following mechanism has been proposed. A + 2B ® 2C +D (slow)
A + 2C ® E (Fast)
The rate law expression for the reaction is (a) rate = K [A]2[B]2 (c) rate = K [A][B]2 26.
(b) rate = K [A]2[B]2[C] (d) rate = K [A][B]
A2 + B2 ® 2 AB reaction follow the mechanism as given below (i) A2 ® 2A (fast) (ii) A + B2 ® AB + B (slow) (iii) A + B ® AB (fast) the order of overall reaction is (a) 1.5
27.
29.
30.
(c) 0
(d) 1
K3 K1 K2 In the sequence of reaction A ¾¾ ® , B ¾¾ ® , C ¾¾ ® then the rate of determining step of reaction is
(a) A ® B 28.
(b) 2
(b) B ® C
(c) C ® D
For the reaction 2A + B ® Products, reaction rate = K and that of B is halved the rate of reaction will be ... (a) doubled (b) halved (c) unaffected
[A][B]2.
(d) A ® D Concentration of A is doubled (d) four times
In one reaction concentration of reaction A is incereased by 16 times, the rate increases only two times. The order of the reaction would be ... (c) 1 2 (d) 1 4 In the reaction A ® B. When the concentration of A is changed from 0.1 M to 1 M, the rate of reaction increases by a factor of 100. The order of reaction with respect to A is …. (a) 2
(b) 4
(a) 10
(b) 1
(c) 2 215
(d) 3
31.
32. 33. 34. 35.
36. 37.
38. 39.
40.
41.
For the reaction of A + B ® C + D, doubling the concentration of both the reactants increases the reaction rate by 8 times and doubling the initial concentration of only B simply doubles the reaction rate. The rate law for the reaction is (a) r = K [A][B]2 (b) r = K [A][B] (c) r = K [A]½[B] (d) r = K [A]2[B] The unit of rate constant for a zero order reaction is ... (a) litre sec-1 (b) litre mole-1 sec-1 (c) mole litre-1 sec-1 (d) mole sec-1 The rate constant of a reaction has same units as the rate of reaction. The reaction is of ... (a) third order (b) second order (c) first order (d) zero order The rate constant of reaction is 3 ´ 10-3 bar-1 sec-1. The order of reaction is ... (a) 1 (b) 2 (c) 3 (d) 0 The dimensions of the rate constant of a third order reaction involve. (a) only time (b) time and concentration (c) time and square of concentration (d) only concentration -2 3 The rate constant of reaction is 5 ´ 10 litre mole-3 minite-1. The order of reaction is... (a) 1 (b) 2 (c) 3 (d) 4 Which of the following statements is incorrect about the molecularity of a reaction ? (a) Molecularity of a reaction is the number of molecules in the slowest step. (b) Molecularity of a reaction is the number of molecules of the reaction present in the balanced equation. (c) There is no difference between order and molecularity of a reaction. (d) Molecularity is always a positive whole number. For a single step reaction A + 2B ® Products, the molecularity is (a) zero (b) 1 (c) 2 (d) 3 Which of the following statement is false ? (a) For a zero order reaction, the rate changes with temperature. (b) Both order and molecularity of a reaction are always the same. (c) Active mass of 128 g of HI present in a two litre flask is 0.5. (d) For the first order reaction, the rate of reaction halved as the concentration of a reactant halved. If ‘a’ is the initial concentration of the reactant, the time taken for completion of the reaction, it if is of zero order, will be (a) K 2a (b) a 2k (c) a k The reaction 2O3 ® 3O2 proceeds in two steps as follows. (i) O3 O2 + O (fast)
42.
(ii) O + O3 ® 2O2 (slow)
The rate law expression should be... (a) r = K[O3]2 (b) r = K[O3]2[O2]-1 (c) r = K[O3][O2] For reaction of zero order is ... (a) K = [ Ao ]
t
(d) k a
(b) Kt = [ A ] - [ Ao ] 216
(c) Kt = [ A ] - [ Ao ]
(d) r = K[O3]2[O2]1
(d) K =
2.303 [ Ao ] n t [A]
43.
For reaction first order is ... (a) t = K ´ 2.303log
[A] [A] o
(c) [ A ] = [ A ]o .e - Kt 44.
[A] 2.303 log t [A] o
(d) K =
2.303 a log t a+x
For the reaction Zero order (a) t
45.
(b) K =
1
2
a Co 2
(b) t
1
2
a Co
(c) t
1
2
a Co -1
(d) t
1
2
a Co
For reaction first order 0.693 0.693 0.693 0.693 (b) t 1 = (c) t 1 a (d) t 1 a 2 Co 2 2 2 k Co k th Which of the following represents the expression for ¾ life of a first order reaction
(a) t 1 = 46.
(a) 47.
48.
49.
50.
51.
52.
53.
54.
k 4 og 2.303 3
(b)
2.303 3 og k 4
(c)
2.303 og 4 k
(d)
2.303 og 3 k
If initial concentration is doubled, the time for half reaction is also doubled. The order of reaction is ... (a) First (b) Second (c) Third (d) Zero If a is the initial concentration of the reactant, the half life period of the reaction of the nth order is proportional to ... (a) an+1 (b) a1-n (c) an (d) an-1 For the first order reaction, half life is 14 s. The time required for the initial concentration to reduce to 1/8th of its value is ... (a) 28 s (b) 42 s (c) (14)2 s (d) (14)3 s In the first order reaction the concentration of the reactants is reduced to 25% in one hour. The half life period of the reaction is … (a) 120 min (b) 4 hr (c) 30 min (d) 15 min For the First order reaction with half life is 150 seconds, the time taken for the concentration of the reactant to fall from m/10 to m/100 will be approximately (a) 600 s (b) 900 s (c) 500 s (d) 1500 s The half life period of a first order reaction is 15 minutes. The amount of substance left after one hour will be ... 1 1 1 1 (a) (b) (c) (d) 2 4 8 16 For the reaction N2O5 ® 2NO2 + ½O2 t½ = 24 hrs. starting with 10 g of N2O5 how many grams of N2O5 will remain after a period of 96 hours ? (a) 0.63 g (b) 0.5 g (c) 1.77 g (d) 1.25 g In the first order reaction 75% of reactant disappeared in 1.386 h. Calculate the rate constant of reaction. (a) 3.6 ´ 10-3 S-1 (b) 2.8 ´ 10-4 S-1 (c) 17.2 ´ 10-3 S-1 (d) 1.8 ´ 10-3 S-1 217
55. 56. 57.
58.
59.
60.
61. 62.
63.
64.
65.
66.
67.
The minimum amount of energy required for the reacting molecules to undergo reaction is called : (a) potential energy (b) internal energy (c) activation energy (d) threshold energy Increase in the concentration of the reactants leads to the change in (a) heat of reaction (b) threshold energy (c) collision energy (d) activation energy Energy of activation of an exothermic reaction is (a) zero (b) negative (c) positive (d) can not be predicated The chemical reactions in which reactants require high amount of activation energy are generally ................ (a) slow (b) fast (c) instantaneous (d) spontaneous The rate of reaction increases with increase of temperature because ... (a) an increase in the number of activated molecules (b) an increase in the number of collisions (c) lowering of threshold energy (d) activation energy is lowered The activation energy of reaction is equal to (a) Threshold energy + Energy of the products (b) Threshold energy - Energy of the reactants (c) Threshold energy + Energy of the reactants (d) Threshold energy - Energy of the products Collision theory is most satisfactory for ___________ reaction. (a) First order (b) second order (c) Bimolecular (d) Any If Ef and Er are the activation energies of the forward and reverse reactions and the reaction is known to be exothermic then (a) Ef < Er (b) Ef > Er (c) Ef >>> Er (d) Ef = Er Which of the following does not affect the rate of reaction ? (a) size of the vessel (b) physical state of reactants (c) amount of the reactants (d) DH of reaction For a an endothermic reaction, DH represents the enthalpy of reaction. The minimum value for the energy of activation will be ... (a) equal to DH (b) zero (c) more than DH (d) less than DH For an endothermic reaction A ® B. An activation energy of 15 Kcal mole-1 and the enthalpy change of reaction is 5 Kcal mole-1. The activation energy for the reaction B ® A is (a) 10 Kcal mole-1 (b) 20 Kcal mole-1 (c) 15 Kcal mole-1 (d) zero -1 For an exothermic reaction an activation energy of 70 KJ mole and the enthalpy change of reaction is 30 KJ mole-1. The activation energy for the reverse reaction is ... (a) 70 KJ mole-1 (b) 30 KJ mole-1 (c) 40 KJ mole-1 (d) 100 KJ mole-1 The rate constant of the reaction increases by ... (a) increasing the temperature (b) increasing the concentration of reactants (c) carrying out the reaction for longer period (d) adding catalyst 218
68.
Which of the following is the expression for Arrhenius equation ? (a) n
k 2 Ea æ 1 1 ö = ç - ÷ k1 R è T1 T2 ø
(c) k = A.e
- Ea
(b) n k = n A - Ea
RT
(d) All the above
RT
1 helps to calculate T
69.
The Plot of log K vs
70.
(a) Activation energy (b) Rate constant (c) Reaction order (d) Activation energy and frequency factor At 290 K velocity constant of a reaction was found to be 3.2 ´ 10-3. At 300 K, it will be
71. 72.
73.
74.
(a) 1.6 ´ 10-3 (b) 6.4 ´ 10-3 (c) 3.2 ´ 10-4 (d) 3.2 ´ 10-2 The increase in reaction rate as a result of temperature rise from 10 K to 100 K is ... (a) 512 (b) 614 (c) 400 (d) 112 At 300 K rate constant is 0.0231 min-1, for a reaction. Bt at 320 K rate constant is 0.0693 min-1. The activation energy of the reaction is (a) 84 KJ mole-1 (b) 34.84 KJ mole-1 (c) 43.84 KJ mole-1 (d) 30 KJ mole-1 The activation energy of a reaction is 9 Kcal mole-1. The increase in the rate constant when its temperature is raised from 295 to 300 K is approximately (a) 1.289 times (b) 12.89 times (c) 0.1289 times (d) 25% A reactant A forms two products. k1 ® B activation energy E1 (i) A ¾¾ k2 ® C activation energy E2 (ii) A ¾¾ If E2 = 2E1 then K1, and K2 are related as
(a) K 2 = K1.e E1 75.
77. 78.
(b) K 2 = K1.e
E2
RT
(c) K1 = AK 2 .e
E1
RT
(d) K1 = 2K 2 .e
E2
RT
The activation energys of two reaction are E1 and E2 (E1 > E2). If the temperature of the system is increased from T1 to T2, the rate constant of the reaction changes from K1 to K21 in the first reaction and K2 to K21 in second reaction, predict which of the following expression is correct ? k11 k12 > (a) k1 k 2
76.
RT
k11 k12 < (b) k1 k 2
k11 k12 = (c) k1 k 2
k11 k12 = =0 k1 k 2 The rate of reaction 2x + y ® Products. Rate = K[x]2[y]. If x is present in large excess, the order of the reaction is (a) 3 (b) 2 (c) 1 (d) 0
(d)
H+ ® CH3COOH + Et OH. Order of reaction is ... CH3COOEt + H2O ¾¾ (a) 0 (b) 1 (c) 2 (d) 3 In which of the following cases, does the reaction go farthest to completion ? (a) K = 100 (b) K = 10-2 (c) K = 10 (d) K = 1
219
79.
80.
81.
82.
83.
84.
85.
86.
87.
The activation energy of a reaction is zero. The rate constant of the reaction … (a) increase with increase of temperature (b) decrease with increase of temperature (c) decrease with decrease of temperature (d) is nearly independent of temperature Which of the following is the fast reaction ? H2O (a) H 2 + Cl 2 ¾¾¾ ® 2HCl
(b) NO 2 + CO ® NO + CO 2
(c) CH 3CHO ® CH 4 + CO
H2O ® C6 H12 O6 + 6O 2 (d) 6CO 2 + 6H 2 O ¾¾¾
Oxidation of oxalic acid by acidified KMnO4 is an example of autocatalysis. It is due to which of the following ? (a) SO42(b) MnO42(c) Mn2+ (d) K+ What will be the order of the reaction if doubling the concentration of a reactant increases the rate by a factor of 4 and trebling the concentration of the reactant by a factor of 9 ? (a) 1 (b) 2 (c) 3 (d) 0 If the half time for a particular reaction is found to be constant and independent of the initial concentration of the reactants then reaction is of ... (a) 1 (b) 2 (c) 3 (d) 0 o The rate of reaction A + B + C ® Products is given by r = K[A][B] [C]. If A is taken in large excess, the order of the reaction would be (a) 0 (b) 1 (c) 2 (d) nil Rate of chemical reaction can be kept constant... (a) by stirring the components (b) by keeping the temperature constant (c) both of the above (d) none of the above The one which is unimolecular reaction is 1 1 (a) HI ® H 2 + I 2 2 2
1 (b) N 2 O5 ® N 2 O 4 + O 2 2
(c) H 2 + Cl 2 ® 2HCl
(d) PCl3 + Cl 2 ® PCl5
For the reaction H2(g)+ Br2(g) ® 2HBr(g) the experimental data suggests, rate = K[H2][Br2]½. The molecularity and order of reaction respectively for the reaction is (a) 2, 2
88.
89.
90.
(b) 2, 1½
(c) 1½, 2
(d) 1½, 1½
The rate of reaction for Cl3 C CHO + NO ® CHCl3 + NO + CO is given by equation, rate = K[Cl3 C CHO][NO]. If concentration is expressed in mole litre-1, the unit of K are (a) litre2 mole-2 sec-1 (b) mole litre-1 sec-1 (c) litre mole-1 sec-1 (d) sec-1 For a reaction 2A + B ® Products, the active mass of B is kept constant and that of A is doubled. The rate of reaction will then (a) increase two times (b) increase four times (c) decrease two times (d) decrease four times The conversion of A ® B follows second order kinetics. Doubling the concentration of A will increase the rate of formation B by a factor of (a) ¼ (b) 2 (c) ½ (d) 4 220
91.
Ethyl acetate is hydrolysed in alkaline medium, its order of a reaction and molecularity are respectively (a) 1, 1
92.
(b) 1, 2
(b) log T
The given reaction 2FeCl
3
(a) first order 94.
(c) 1/T
96.
(b) second order
(c) third order
2K1 2 (a) K [ NO 2 ] 2
(b) 2k1 [ NO 2 ] - 2k 2 [ N 2 O 4 ]
(c) 2k1 [ NO 2 ] - k 2 [ N 2 O 4 ]
(d) ( 2k1 - k 2 ) [ NO 2 ]
2
If concentration of reactants is increased by ‘x’, then rate constant K becomes .. k k (a) n (b) (c) k+x (d) k x x The rate constant is given by equation K = p.z.e-Ea/RT which factor should register a decrease for the reaction to proceed more rapidly ? (b) T
(c) Z
(d) P
k ® C. the unit of rate constant is For the reaction A + B ¾¾
(a) sec-1 98.
(d) none of these
K1 In the reverable reaction 2NO2 N2O4, the rate of disappearance of NO2 is equal to K2
(a) E 97.
(d) log 1/T
+SnCl2 ® 2FeCl2 +SnCl4 is an example of ________ reaction
2
95.
(d) 2, 2
According to the Arrhenius equation a straight line is to be obtained by plotting the logarithm of the rate constant of a reaction against ... (a) T
93.
(c) 2, 1
(b) sec-1 mole L-1
(c) sec-1 mole-1 L
(d) sec-1 mole-2 L2
The rate of the gaseous reaction is equal to K[A][B]. The volume of the vessel is suddenly reduced to one forth of the initial volume. The rate of reaction would be ... 1 16 1 8 (b) (c) (d) 16 1 8 1 For reaction Y2 + 2Z ® Product, rate controlling step is Y + ½ Z ® Q. If the concentration of Z is doubled, the rate of reaction will be
(a) 99.
(a) remain the same
(b) become four times
(c) become 1.414 times
(d) become double
100.
The rate law for a reaction given by Rate = K[A]n[B]m. On doubling the concentration of A and halving the concentration of B, the ratio of the new rate of the earlier rate of the reaction will be as 1 (a) m+n (b) n-m (c) 2(n-m) (d) 2 ( m + n )
101.
The time for half lif of a certain reaction A ® Products, is one hour. When the initial concentration of the reactant A is 2 mol L-1 how much time does it take for its concentration to come from 0.50 to 0.25 mole L-1 if it is a zero order reaction ? (a) 0.25 h
(b) 1 h
(c) 4 h
221
(d) 0.5 h
102.
103.
104.
105.
For a first order reaction A ® Products, the concentration of A changes from 0.1 M to 0.025 M in 40 minutes. The rate of reaction when the concentration of A is 0.01 M is (a) 1.73 ´ 10-4 M min-1 (b) 1.73 ´ 10-5 M min-1 (c) 3.47 ´ 10-4 M min-1 (d) 3.47 ´ 10-5 M min-1 In the reaction 2N2O5 ® 4NO2 + O2, initial pressure is 500 atm and rate constant K is 3.38 10-5 sec-1. After 10 minutes the final pressure of N2O5 is (a) 490 atm (b) 250 atm (c) 480 atm (d) 420 atm The half life period of a first order reaction is 6.93 minutes. The time required for the completion of 99% of chemical reaction will be (a) 230.3 min (b) 23.03 min (c) 46.06 min (d) 460.6 min The rate constants K1 and K2 for two different reactions are 1016.e-2000/T and 1015.e-1000/T respectively. The temperature at which K1 = K2 is (a) 1000 K
(b)
2000 k 2.303
(c) 2000 K
(d)
1000 k 2.303
ANSWER KEY 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
c b b a c d a a d d c a b c a c b d c c b d b b c
26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
a c b d c d c d b c d c d b c b b c d a c d b b c
51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75
222
c d a b d c c a a b c a d c a d a d d b a c a c a
76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100
c b a d d c b a b d b b c b d d c c b d a c b c c
101 102 103 104 105
a c a c d
Hints 1.
Defination of the rate of reaction
2.
-
3. 4.
the activation energy of the reaction is large 0.2 ´ 10–4
d [ B] 1 d [A] 1 d [ C] ==+ 2 dt dt 3 dt
d [H2 ] dt Hence 5. 6. 7.
= - 0.3 ´10-4 Ms -1 But Rate = d [ NH 3 ] dt
=-
1 d [H2 ] 1 d [ NH3 ] =+ 3 dt 2 dt
2 d [H2 ] 2 = - -0.3 ´10-4 = 0.2 ´10-4 3 dt 3
(
)
d [ B] d [ D] 1 d [A] 1 d [ C] ==+ =+ 4 dt dt 2 dt dt DH of reaction 0.012 atm min-1
\-
-
d [ N 2O4 ] dt
=+
1 d [ NO 2 ] 2 dt
( 0.32 - 0.50 ) = 0.006 = 1 d [ NO2 ] 30
2
dt
8.
K = K1/K2
9.
1 6
10. 11.
(p+q) = (m+n) or (p+q) ¹ (m+n) Rate = K[A][B]2
Rate = k [ A ] 2 [ A ] 3 [ c ] 1
1
1
\
d [ NO 2 ] dt
= 0.012 atm min -1
1 1 1 11 \ Order of reaction = + + = 2 3 1 6
Keeping [B] constant, [A] is made a 4 times, rate also become 4 times. Hence rate a [A] Keeping [A] constant, [B] is doubled, rate becomes 4 times. Hence rate a [B]2
\ rate = K[A][B]2 12.
rate = K[A]2 [A] doubling, rate becomes four time. Hence rate a [A]2 [B] doubling, no effect on the rate. Hence rate a [B]0
\ rate = K[A]2[B]0 13. 14.
concentration of reactants temperature is changed
15.
hn H 2 + Cl 2 ¾¾ ® 2HCl
16.
H 2 + Br2 ® 2HBr 223
17. 18.
Unimolecular reaction any value between 1 and 3
19.
1 3
20.
four times \rate a [ B]
21.
8x V1 = k [ A ] [ B] = x V2 = k [ 2A ] [ 2B] \V2 = 8x
22. 23. 24. 25.
OO \No I2 in the rate law equation. rate of zero order reaction is independent of the concentration of the reacting species many bady collisions have a low probability rate = K [A][B]2 Rate of reaction for slowest step
26.
1.5 From slowest step rate = k [ B2
n
k 2 é A2 ù 1 = ê ú Or 2 = (8) n \n = k1 ë A1 û 3 2
2
2
From 1st eq. Keq = [ A ] 2 rate = K[B2 ] keq 27. 28.
1
2
][ A ]
\[ A ] = keq 2 . [ A 2 ] 2 1
1
[A2 ]
1
1
1
1
× [A 2 ] 2 = k × keq 2 [A 2 ] 2 [B2 ] = K1 [A 2 ] 2 [B2 ]
C ® D is lowest 2 éBù rate '' = k [ A ][ B] rate '' = k [ 2A ] ê ú ë1û
halved
2
1 2 = k [ A ][ B] 2 1 \x '' = x ' 2
29.
¼
(1) r = k [ A ] (2) 2r = k [16A ] n
n
2r = K [ A ] 16n n
2r K [ A ] 6 1 = \2 = 16n \n = n r 4 K [A] n
30.
2
31.
r = K [A]2[B]
n
cocentration increased = 10 times rate increased = 102 times \ Order = 2 (i) r = k [ A ] [ B] x
y
(ii)8r = k [ 2A ] [ 2B] x
(iii) ¸ (i) @ 2 y = 2\y = 1
(ii) ¸ (i) @ 2 x = 4\x = 2
224
y
(iii) 2r = k [ A ] [ 2B] x
y
32.
mole litre-1 sec-1 rate = K [ R ] , K =
rate
n
[R ]
n
=
M/s n =0 Mn
K = M1- n S-1 \K = M S rate
33.
zero order K =
34.
2
35.
time and square of concentration
K=
K=
rate
[R ]
rate
[R ]
n
n
=
rate
=
[R ]
n
K = rate, when n = 0
bar / s when n = 2 bar n
k = bar -1 S-1
M /S = M1- n S-1when n = 3 n M M /S = M1- n S-1when n = 4 n M
k = liter 3 mole -3 min -1
36.
4 K=
37. 38. 39.
There is no difference between order and molecularity of a reaction. 3 Both order and molecularity of a reaction are always the same.
40.
a
[R ]
n
=
k = litre3 mole -1 min -1
1 k For Zero order reaction t = k [ A ]o - [ A ] But [ A ]o = a
{
}
And when reaction complete [ A ] = 0\t = 41.
a k
r = k[O3]2[O2]-1 From Slowest step r = k [ O3 ][ O ] From eq (i) keq = [ O 2 ][ O ] / [ O3 ]
\[ O ] = Keq [ O3 ] / [ O 2 ]2 \r = k [ O3 ] Keq [ O 2 ] = K1 [ O3 ] [ O 2 ] 2
42.
Kt = [ Ao ] - [ A ]
43.
[ A ] = [ A ]o .e- kt
44.
t 1 2 a Co
45.
t 12 =
46.
2.303 log4 k
-1
0.693 k
t 34 =
2.303 ao 2.303 ao og = og ao 3 k k 4 ao - ao ´ 4
225
47.
Zero For Zero order reaction t 12 a Co
48.
a1- n t 12 a a1- n
49.
42 S
50. 51.
52.
53.
Ao t 12 Ao t 12 Ao ¾¾® ¾¾® \ 3 ´ t 12 2 4 8 T t 12 t 12 30 min 100 % ¾¾ ® 50 ¾¾ ® 25% \ T = 2 ´ t 1 2\t 1 2 = 2 M t 12 M t 12 M t 12 M t 12 M ¾¾® ¾¾® ¾¾® ¾¾® 500S 10 20 40 80 180 t 2 Ao ¾¾ ® 1
\ T @ 3 ´ t 1 2 to 4 ´ t 1 2 @ 450 to 600S 1 a After n + t 1 2 amount left = n 16 2 60 1 1 T = n ´ t 1 2 n = = 4 \ Amount left = a = 15 2 16
0.63 g T = n ´ t 1 2 n =
96 a 10 10 = 4 \ Amount left = n = 4 = = 0.63 24 2 2 16
55. 56. 57. 58. 59. 60. 61. 62. 63. 64.
2.303 a og = 2.8 ´10-4 S-1 1.386 ´ 60 ´ 60 a - 0.75a threshould energy. Collision frequency Positive Activation energy is always Positive Slow an increase in the number of activated molecules. Threshold Energy – Energy of the reactants Bimolecular Ef < Er DH of reaction more than DH
65.
lo Kcal mole -1 DH = Ea - Ea r \+ 5 = 15 - Ea r \Ea r =10
66.
100 Kj mole -1 DH = Ea - Ea r - 30 = 70 - Ea r \Ea r = 100
67. 68. 69. 70. 71.
increasing the temperature. All the above Activation energy and frequency factor. 3.2 ´ 10–4 10k rise, the velocity constant becomes nearly double. 512 Increases of temperature n ´ 10 Increases reaction rate = 29 DT = 100 – 10 = 90 = 9 ´ 10 \ n = 9 \ Increases reaction rate = 29 = 512
54.
2.8 ´10-4 S-1 K =
226
72.
-1 43.84 Kj mole og
og
æ 320 - 300 ö 0.0693 Ea = ç ÷ 0.0231 2.303 ´ 8.3 è 300 ´ 320 ø
og 3 =
73.
74.
Ea æ 20 ö ç ÷ 1.901 è 96000 ø
1.289 times og og
K2 Ea æ T2 - T1 ö = ç ÷ K1 2.303R è T1T2 ø
K2 Ea.DT 9000 ´ 5 = = = 0.1104 K1 2.303R T2 T1 2.303 ´ 2 ´ 300 ´ 295
K2 = 0.1104 K1
K1 = K 2 A.e
E1
Ea = 43.84
K2 = 1.289 K1
K1 = A1.e
RT
- E1
RT
K 2 = K1 ´1.289
K 2 = A 2 .e
- E2
RT
E1 K1 A1 ( E2 - E1 ) / RT 2E - E / RT = ´e = A.e( 1 1 ) = A.e RT K 2 A2
\K1 = K 2 .A.e
75.
K11 K12 > K1 K 2
E1
RT
og
K11 E1 é T2 - T1 ù = ê ú K1 2.303R ë T1T2 û
og
K12 E 2 é T2 - T1 ù = ê ú K 2 2.303R ë T1T2 û
Since E1 > E2 K11 K12 \og ¸ og >1 K1 K2
76. 77. 78. 79.
OR
K11 K12 > K1 K 2
1 The rate is not depend upon the reactant present in excess 1 K = 100 is nearly independent of temperature. Ea = 0 \K = A.e
- Ea
RT
= A.eo = A
80.
hn 6CO 2 + 6H 2 O ¾¾ ® C6 H12 O6 + 6O 2
81.
Mn 2+
82.
2 2n = 4 3n = 9 \n = 2
83. 84. 85.
1 1 none of the above 227
88. 89.
1 N 2 O5 ® N 2 O 4 + O 2 2 1 2, 1 2 litre mole–1 sec–1 increase four times
90.
4 Rate = K [ A ] \rate = K [ 2A ] = 4.K [ A ]
91.
2, 2 CH 3COOC2 H 5 + NaOH ® CH 3COONa + C2 H 5OH
92.
1
86. 87.
2
2
93.
T third order
94.
2K1 [ NO 2 ] - 2K 2 [ N 2 O 4 ]
k1 For 2NO 2 N 2O4 k
2
Rate = \ rate =
2
2
1 d [ NO 2 ] 2 = K1 [ NO 2 ] - K 2 [ N 2 O 4 ] 2 dt
-d [ NO 2 ] dt
= 2K1 [ NO 2 ] - 2K 2 [ N 2 O 4 ] 2
95. 96. 97.
K E Sec–1 mole–1 L \ Second order reaction
98.
16 Volume of the vessel is reduced to one foreth 1
Concentration bocomes 4 ttimes 99.
become 1.414 times
Rate = K [ Y ][ Z] 2 1
\New Rate = 2.k [ Y ][ Z] 2 = 1.414 K [ Y ][ Z] 2 1
1
100.
2(n – m) r = ka b 1
n
æbö r = k ( 2a ) ç ÷ è2ø
m
n
11
m
r11 2n a n b m a - m = = 2n.2- m = 2(n - m) 1 n m r a b
101.
0.25 h For Zero order reaction K =
t=
[ A ]o - [ A ] K
=
[ A ]o 2t
1
2
=
2 =1mol L-1 hr -1 2 ´1
0.50 - 0.25 = 0.25 hr 1
228
102.
3.47 ´10-4 M min -1 K =
2.303 0.1 og = 0.03466 min -1 40 0.025
Rate = K [ A ] = 0.03466 ´ 0.01 = 3.466 ´10-4 M min -1 1
103.
og 104.
K=
490 atm
500 500 = 0.0088 OR = 1.021 OR Pt pt
46.06 min K = t 99% =
pt = 490 atm
0.693 0.693 = = 0.1min -1 1 t2 6.93
2.303 a 2.303 a og = og k a-x 0.1 a - 0.99a
= 23.03 og
105.
2.303 Po 2.303 500 og \3.38 ´10-5 = og t Pt 600 Pt
1000 k 2.303 \10.e
1 = 46.03min 0.01
k1 = k 2 -2000
\n10 -
= 1.e
-1000
-2000
T
= 1015.e
-1000
T
T
2000 1000 =T T
\ 2.303 \T = -
T
\1016.e
2000 1000 =T T
1000 K 2.303
229
UNIT : 10 SURFACE CHEMISTRY Important Points ˆ
The study of chemistry regarding the boundary separating two bulk states or phases is called surface chemistry. This boundary surface is known as interface. It is expressed as hyphen (–) or slash (/). Dissolution, crystallization, catalysis, metallic corrosion are surface phenomena.
ˆ
The surface should be completely pure which can be obtained by vacuum generating method and can be stored also.
ˆ
In this unit, surface phenomena like adsorption, catalysis, colloid and emulsion are studied.
ˆ
In adsorption, the substance which is in solid form and on which other gas or liquid is adsorbed is called adsorbent. The substance that is adsorbed is called adsorbate and the whole phenomenon is called adsorption. The phenomenon opposite to adsorption is called desorption.
ˆ
Absorption is such a phenomenon in which there is homogeneous system viz. any coloured solution but if solid adsorbent like charcoal is added to it then there is decrease in intensity of the colour which is adsorption. The combined phenomenon of adsorption and absorption is called sorption. In adsorption the concentration of adsorbate is more than that in the bulk. More porous the adsorbent more will be adsorption. Adsorption is an exothermic phenomenon.
ˆ
In adsorption, the residual particles on the surface are responsible for the adsorption that is due to difference in forces of attraction.
ˆ
Adsorption is of two types– Physical and Chemical. The points of difference between them are given in the unit.
ˆ
Adsorption is used in many fields as well as in everyday life viz. To wear gas mask in which there is adsorbent to save from the poisonous gas like chlorine. Silica gel is used as adsorbent for keeping the electronic instruments moisture free. In the removal of yellow colour from sugar, the phenomenon of adsorption is used.
ˆ
The factors affecting adsorption are (1) nature of adsorbate (2) nature of adsorbent (3) specific area of adsorbent surface (4) pressure of adsorbed gas (5) temperature. The detailed discussion about each one is included in the unit.
ˆ
At constant temperature the graph of pressure of gas adsorbed or concentration is called adsorption isotherm. There are five different types of isotherms. The study of adsorption isotherm was done by scientist Freundlich and gave the equation pressure (p)) and
1 x x 1 = Kp n or log = log K + log p (for m m n
1 x 1 x = log K + log C (for concentration (C)) where K and = KC n or l log m n m
n are constants. This was an empirical isotherm and had no scientific base.
230
ˆ
Langmuir on the basis of kinetic theory of gases gave isotherm equation. x aC x ap = = , (where a and b are constants) 1 + bC m 1 + bp m
ˆ
The study of Freundlich isotherm can be understood by the study of demonstration experiment in practicals book. There are many uses of adsorption which are described in the unit.
ˆ
Catalysis is also a surface phenomenon. Some chemical reactions are slow. To increase their rates, the substance used in small proportion is called catalyst. This phenomenon is called catalysis. The catalyst is obtained back in original form at the end of the reaction.
ˆ
There are two types of catalysis (1) Homogeneous and (2) Heterogeneous. In homogeneous catalysis, catalyst and the reactant are in one phase e.g. hydrolysis of methyl acetate in presence of H+ (2) In heterogeneous catalysis the catalyst and the reactants are in different phases e.g. Production of sulphuric acid in presence of V2O5 by contact process. There are many uses of homogeneous and heterogeneous catalysis as shown in the unit.
ˆ
The characteristics of catalysis are activity, selectivity that is specific reaction; selection of specific catalyst e.g. zeolite. The zeolite named ZSM-5 is used to obtain gasoline from alcohol. Catalyst increases the rate of reaction but does not affect the equilibrium because it affects equally both the forward and the reverse reaction. Hence, more product is not obtained.
ˆ
Enzymes are proteins and are necessary for biochemical reaction. For every reaction separate and specific enzyme works viz. invertase can transform sugar into glucose and fructose. Urease can decompose urea into ammonia and carbon dioxide.
ˆ
For enzyme catalysis the lock and key model or induced fit model are proposed. As the lock can be opened by suitable key, similarly for a particular reaction suitable enzyme will be required. The enzymes work at the temperature of the body i.e. 298-310 K temperature is considered the best.
ˆ
Colloid chemistry is also a surface phenomenon. The colloidal solution is called sol. There are two components called dispersing phase and dispersion medium in it. The particles are of some particular size so this is a heterogeneous system. Colloids are of two types-Lyophilic and Lyophobic. The colloid which has attraction for solvent (dispersion medium) is used called lyophilic colloid e.g. gum. The colloid which has repulsion towards the solvent (dispersion medium is called lyophobic colloid. If water is as a medium then they are respectively called hydrophilic and hydrophobic. Colloids are of eight types which depend on the dispersing phase and dispersion medium. This is shown in the unit. Multimolecular, macromolecular and associated colloids are also known. In associated colloids molecules come nearer and form an association which is called micelle.
ˆ
The certain temperature, at which the micelle is formed is called Kraft's temperature (TK). Below the critical micelle concentration (CMC), it remains in colloidal state and at higher than that concentration it is changed to solid in the form of precipitates. The formation micelle is obtained in the cleansing action of soap. Molecules like soap are shown as RCOONa, and their ionization form will be RCOO–Na +. From this RCOO–, R part combines with organic impurity and drags inside. It is called tail. The upper charged part COO– attracts dust etc. and removes the dirt. It is called head.
231
ˆ
The methods of preparation of colloids are as follows :
ˆ
Condensation method : In these methods, oxidation, reduction, decomposition etc. types of reactions are associated. In physical methods, excessive cooling is used, and in dispersion methods-mechanical dispersion (use of colloid mill), electrical dispersion (Bredig arc method) and peptization are used.
ˆ
For purification of prepared colloidal solution, a method like dialysis and better method like electrodialysis can be used. On addition of certain electrolytes to colloid solutions, precipitation occurs which is called coagulation. The order of concentration for coagulation for iron sol having positive charge is trivalent > divalent > monovalent for negative ions (anions). Similarly for arsenious sulphide colloid having negative charge, the order of coagulation remains the same but positive ions (cations) are used. Amongst the other methods used for purification of colloids are ultrafiltration and ultracentrifugation.
ˆ
The properties of colloidal solutions-sol are as follows : (1) Colligative properties (2) Optical properties (3) Mechanical and (4) Electrical properties.
ˆ
In colligative properties determination of molecular masses is by osmosis method. In optical properties- Tyndall effect and in mechanical properties Brownian movement and in electrical properties, instrument called electrophoresis is used for determination of electrical charge of colloid. There are two laws given by Hardy and Schulze for the study of coagulation of colloids. The electrolytes having electric charge opposite to that on the colloid are required for coagulation of colloid. For colloid having positive or negative electric charge negative or positive ions of the electrolyte respectively are useful. In concentration, highest concentration of monovalent and less than that concentration of divalent ion and the least concentration of trivalent produces coagulation.
ˆ
Emulsions are also colloids, in which both the dispersion medium and the dispersed phase are in liquid form. They are of two types- Oil / water and water/oil. The examples of water/oil emulsion are cold cream, butter etc; while in oil/water emulsion the examples are milk, vanishing cream etc. There are two methods for the test of emulsions (1) Dye test and (2) Dilution test. Demulsification is the opposite phenomenon.
ˆ
There are many uses of colloids. Its specific uses are rubber plating, sewage disposal, Cottrell smoke precipitator, preparation of nano-substances, medicines, as germicides in metallurgy, construction of roads etc. which are described in detail in the unit.
232
M.C.Q. (1)
On which factors interface depends ? (a) Size of the molecules in the bulk phase. (b) Wieght of the molecules in the bulk phase. (c) Numbers of molucules in the bulk phase. (d) Physical state of molucules in the bulk phase.
(2)
(3)
Which of the following phenomenon in not involved in surface chemistry ? (a) Elecrode reactions
(b) Dissolution
(c) Heterogeneous catalysis
(d) Liquid and its vapour taken in closed vessel.
How much pascal high vacuum is required to achive completely pure surface of metal? (a) 10-8 to 10-9
(4)
(5)
(b) 10-8 to 10-10
(c) 10-6 to 10-9
(d) 10-8 to 10-7
Due to adsorption (a) surface energy increases
(b) surface energy becomes zero
(b) surface energy decreases
(d) no change occurs in surface energy
Which of the following processes is adsortion phenomenon ? (a) Soakd of rayon clothes in coloured solution (b) Contact between silicagel and vapour of water (c) Contact between anhydrous CaCl2 and vapour of water (d) H2S gas in contact to water.
(6)
Which of the following is good adsorbent ? (a) Silica gel
(7)
(9)
(c) Clay
(d) All the given
(c) Indothermic
(d) Exothermic
Which type of process adsorption is ? (a) Isotonic
(8)
(b) Alumina (b) Isochonic
When adsorption phenomenon occurs complete ? (a) H < O,G < O,S < O
(b) G < O,H < O,S > O
(c) H > O,S > O,G < O
(d) G < O,S < O,H > O
Which is correct for adsorption ? (a) Δ H-T Δ S is negative
(b) ΔH is positive
(C) Δ H-T Δ S is positive
(d) TΔS and ΔG=0
(10) Which of the following statement is not true ? (a) The value of adsorption enthalpy of physical adsorption is less than chemical adsorption. (b) Physical adsorption occurs due to van der waals’ forces (c) Chemical adsorption decreases at high temperacture and low pressure. (d) Physical adsorption is reversible. 233
(11) At which temperature chemical adsorption occurs ? (a) At high temperature
(b) At very low temperature
(c) At low temperature
(d) Temperature does not affect.
(12) Whose value is less than zero during adsoption ? (a) D G
(b) Δ H
(c) Δ S
(d)All the given
(13) How molecules of gases are deposited on the surface of solid during physical adsorption ? (a) By electrostatic forces
(b) By chemical forces
(c) By gravitational forces
(d) By van der waals’ forces
(14) What is used to prevante electronic instruments clamaged by the moisture ? (a) Silica gel
(b) Zeolite
(c) Chromatographic plate
(d) All the given
(15) On which factor adsorption of gas on solid adsorbtion depend ? (a) On temprature
(b) On pressure of gas
(c) On nature of adsorbent
(d) All the given
(16) If H2,CH 4,CO2 and NH3 gases are adsorbed by 1 gram charcol at 290 k temperature than deceasing order of their volume is (a) H2>CH4>CO2>NH 3
(b) CO2>NH 3>H 2>CH 4
(c) NH 3>CO2>CH4>H 2
(d) CH4>CO 2>NH3>H 2
(17) Which gaseous molecule has highest value of physical adsorption enthalpy ? (a) H2
(b) N2
(c) H2O
(18) How magnitude of adsorption gas is expressed ? m x (a) (b) m. x (c) x m (19) Which is Freundlich adsorption isotherm equation ? 1 m 1 m n x (a) αp n (b) α p (c) αp n x x m (20) What will be the value of slope after drawing graph of log
(d) He
(d) x+m
(d) None of these x ® log p in Freundlich adsorption m
isotherm? 1 1 1 (a) (b) (c) (d) -k p n a (21) What will be the intercep+ in a graph of Freundlich adsorption isotherm ? 1 1 (a) k (b) log k (c) (d) a n (22) Whose value we can get from intercept in the graph of adsorption isotherm ? 1 (a) a and b (b) n and k (c) (d) log k a 234
(23) Which of the following is not true regarding to Freudlich adsorption isotherm ? (a) This isotherm is applicable in certain limit of pressure (b) Constant k and n change with temprature (c) It shows deviation at low pressure (d) Freundlich isotherm is empirical, there is no theoritical proof of it. 1 (24) If the value of becomes zero in Freundich adsorption isotherm then adsorption is indepen n dent to (a) pressure
(b) temperature
(c) quantity
(d) a and b
1 x (25) If the value of is 1 in Freundlich adsorption isotherm then = ............. n m k (a) (b) kp (c) k (d) none of these p (26) On basis of which theory Langmuir derived isotherm equation ?
(a) Thermodynamics
(b) Kinetic theory of gases
(c) Collosion theory
(d) Wave mechanic theory
(27) Which of the following is Langmuir adsorption isotherm ? x ab x ap m 1 + bp x ap = (a) (b) = (c) = (d) = m 1 + bp m 1 + bp x ab m 1 + bc (28) How will be Langmuir equation at high pressure ? x ap x a x x b (a) = (b) = ap (c) = (d) = m 1 + bp m b m m a m 1 (29) What will be the value of slope in graph of ® according to Langmuir equation ? x p 1 b a (a) (b) (c) (d)k a a b (30) In Indothermic reaction with the increase of temperature adsorption will be (a) constsnt
(b) increase
(c) decrease
(d) none of these
(31) Which adsorbent is used in separation of inert gases by Dewar’s method ? (a) Vanadium pentoxide
(b) Silica gel
(c) Activated charcoal
(d) Allumina
(32) Which of the following are adsorption indicators ? (a) eosin
(b) fluorescin
(c)methelene blue
(d) (a) and (b)
(33) What is not true for catalytic reaction ? (a) Catalyst increases equally both the rate of forward and reverse reactions. (b) Catalyst does not effect to equilibrium constant. (c) Catalyst decreases activation energy (d) Catalyst increases activation energy of chemical equation. (34) Which catalyst forms NH3 and CO2 from urea ? (a) Invertase
(b) Celluase
(c) Urease 235
(d) Pepcine
(35) Which of the following is an example of surface catalysis ? (a) Inversion of sucrose (b) Production of ammonia by Haber’s process (c) Production of H2SO4 by lead chamber process (d) Hydrolysis of ester (36) Which catalyst is used in inversion of sucrose ? (a) Fe(s)
(b) NO (g)
(c) H2SO 4
(d) Cl(g)
(37) Which catalyst is used to obtain methanal from water gas ? (a) Cu
(b) ZnO-Cr2O 3
(c) (a) and (b)
(d) FeO
(38) Which catalyst is used in the decomption of ozone ? (a)Cl2(g)
(b)Cl(g)
(c) O2(g)
(d) all the given
(39) Which catalyst is used to prepare propylene oxide from the reaction between propylene and dioxygen ? (a) Rh-Pd complex
(b) [Rh(CO)2I2]complex (c) Mo(VI)complex (d) Ni-Pd complex
(40) On which factor activity of catalyst depends ? (a) On the strength of chemical adsorption. (b) On the conecentration of products. (c) On the concentration of reactants.
(d) On the physical state of catalyst
(41) By which name this reaction is also known ? reaction: 2 SO2 + O2 ( g ) ® 2 SO3( g ) (a) Shape-selective catalysis
(b) Homogeneous catalysis
(c) Enzyme catalysis
(d) Surface catalysis
(42) CO(g)+H2(g) ® x.What is x ? (a) Methane
(b) Methanal
(c) Formic acid
(d) Formaldehyde
(43) On which factor shape-selective catalysis depends ? (a) Size of reactant malecules
(b) Pore structure of catalyst
(c) Size of product molecules
(d) All the given
(44) How many times reaction rate increase by catalyst ? (a) 1010
to 1020
(b) 102
to 104
(c) 106
to 108
(d) 108
to 1020
(45) What is called colloid system in which dispesing phase and dispersion medium, both are in solid state ? (a) Gel
(b) Emulsion
(c) Sol
(d) Aerosol
(46) What is called colloid system in which dispersing phase is gas and dispersion medium is lquid ? (a) Gel
(b) Areosol
(c) Emulsion 236
(d) Foam
(47) Milk is example of which type of colloid ? (a)Emulsion
(b) Suspension
(c) Gel
(d)Aerosol
(c) Gas in gas
(d) Gas in liquid
(48) Smoke is which type of colloidal system ? (a) Gas in salid
(b) Solid in gas
(49) Which of the following is reversible sol ? (a) Cellulose
(b)
(50) Which type of sol sulphar is ? Fe3+ > Na + > colloid Ba 2+ (a) Multimolecular (c) Associated colloid
(c) Mist
(d) Gelatin +
Na > Ba
2+
> Fe
3+
(b) Micelle (d) Macromolecular colloid
(51) Which of the following is macromolecalar colloid ? (a) Artificial rubber
(b) Protein
(c) Nylon
(d) All the given
(52) What is called to that temperature at which the formation of micelle takes place ? (a) Zero temperature
(b) Kraff temperatue
(c) Kelvin temperature
(d) Absolute temperature
(53) At which condition micelle is formed ? (a)
At concentration higher then critical micelle concentration and lower then kraft temperature.
(b)
At concentration higher then critical micelle concentration and higher then kraft temperature.
(c)
At concentration lower then critical micelle concentration and lower then kraft temperature.
(d)
At concentration lower then critical micelle concentration and higher then kraft temperature.
(54) Which of the following condition is true during the formation of micelle ? (a) DH = -ve, Ds = -ve
(b) DH = +ve, Ds = -ve
(c) DH = +ve, Ds = +ve
(d) DH = -ve, Ds = +ve
(55) What is approximate value of CMC for saap ? (a) 10-9M to 104M
(b) 10-3M to 10-4M (c) 10-9M to 10-14M
(d) 103M to 105M
(56) Which of the following is physical method for the preparation of collodiad sol ? (a) coagulation
(b) peptization
(c) fusion
(d) excessive cooling
(57) Which sol is formed due to hydrolysis of FeCl3 ? (a) FeCl2
(b) Fe(OH)2
(c) Fe2O 3
(d) Fe(OH)3
(58) Which of the following is double deamposition ? (a) SO2+2H2S ® 3S+2H2O
(b) FeCl3+2H2O ® Fe(OH)2+3HCl
(c) As2O3+3H2S ® As 2S3+3H2O
(d) All the given
(59) By which method As2S3 sol can be obtain by the reaction between As2O3andH2S ? (a) Redution
(b) Oxidation
(c) Hydrolysis
(d) Double decomposition
(60) In which of the following method condensation and dispersion are associated ? (a) Excessive cooling (b) Hydrolysis
(c) Bredig’s are 237
(d) Peptization
(61) Which method is is used to obtain sol of gold and silver ? (a) Electric dispersion (b) Peptization
(c) Excessive cooling (d) Mechanical dispersion
(62) Whose menbranes are used in dialysis ? (a) Parchment paper (b) Plastic
(c) Filter paper
(d) Ultrafilter paper
(63) ...............Phenomenon is called reverse to coagulation ? (a) Flocculation
(b) Tyndall
(c) Brownian
(d) Dialysis
(64) Which of the following is correct order of coagulations for the coagulation of As2S3? (a) Fe3+ > Ba 2+ > Na +
(b) Na + > Ba 2+ > Fe3+
(c) Fe3+ > Na 2+ > Ba 2+
(d) Ba 2+ > Na + > Fe3+
(65) Which of the following is correct order of coagulation ions for the coagulation of Fe(OH)3? (a) C1- > SO42- > PO43-
(b) PO43- > SO42- > C1-
(c) SO42- > C1- > PO43-
(d) SO42- > PO43- > C1-
(66) Which is correct order of coagulation of ion necessary for coagulation of calloid sol ? (a) Monovalent ion < divalent ion F
b) F > O > N
c) N > O < F
248
d) O > F > N
14.
The order of ionization energy of K, Ca, & Ba are a) K > Ca > Ba
15.
b) Ca > Ba > K b) Lithium
I.
NH 3
II.
III.
CH + 3
IV. H 3 O + b) i , ii , iv
d) Fluorine
NH – 2
c) i , ii , iii , iv
d) i & iv
c) s– Block
d) f–Block
The element with atomic number 44 belongs a) d– Block
18.
c) Calcium
Pick the iso electronic species from the following
a) ii, iii, iv 17.
d) K > Ba > Ca
The element with zero electron gain enthalpy is a) Argon
16.
c) Ba > K > Ca
b) p–Block
In the third period there are only eight elements because a) It is a short period
b) The 3d orbitals are absent
c) The d orbitals are absent d) When n=3, the maximum number of electrons which can be accommodated are eight 19.
20.
Cho ose the correct electron ic co nfiguration which has the highest difference between first & second ionisation enthalpies. a) 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1
b) 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2
c) 1s 2 2s 2 2p 4
d) 1s 2 2s 2 2p 3
The set of quantum numbers for unpaired electron of an element with atomic number 84 are a) N= 6 , l = 1 , m =
+_
1 , ms =
c) N= 6 , l = 0 , m = 0 , m s = 21.
22.
1/2
1/2
b) N= 5 , l = 3 , m = 0 , m s =
+_
1/2
d) N= 6 , l = 3 , m = 2 , m s =
+_
1/2
The elements with highest ionization enthalpy in a period are a) Alkaline earth metals
b) Halogens
c) Noble gases
d) Lanthanides
Choose the species which is not isoelectronic a) Bo 33–
23.
+_
+_
b) Co 32–
c) No 3 –
d) So 32–
The formation of Mg2+ is as follows I.
Mg (g) ® Mg+ (g) + e –
II. Mg +(g) ® Mg 2+(g) + e –
–737 KJ mo l– 1 –1450 KJ mo l– 1
The energy required in the second steps is higher because a) Mg+ is more electropositive
b) Mg+ has larger size than Mg
c) Mg+ tends to loose only one electron
d) Mg+ has smaller size than Mg
249
24.
The first 2nd and 3rd ionizatio n enthalpies o f galliu m are 579KJmo l–1 ,1979KJmo l–1 & 2962 KJmol–1 even though the iii I.P is highest Ga3+ is the most stable because a) The energy loss is maximum resulting greater stability b) The size of Ga3+ is smallest c) Ga 3+ is most reactive d) It attains a stable configuration
25.
The electronic configuration of M3+ is [ Kr ] 4d 10. Its position in the periodic table is a) Period 4 gp 8
26.
27.
28.
b) Perio d 5 gp 13
c) Perio d 4 gp 18
d) Perio d 5 gp 16
The electronic which will exhibit maximum no. of oxidation states a) 1s 2 2s 2 2p 6 3s 2 3p 5
b) 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 5
c) [Xe] 4f14 5d 6 6s 2
d) [Ar] 4s 2 4p 4
Choose the incorrect order w.r.t properly indicated a) Electro negativity F > Cl > Br
b) Electron affinity Cl > F > Br
c) Oxidizing power F2 > Cl2 > Br2
d) Bond en thalpy F2 > Cl2 > Br2
Choose the correct statement a) As shielding effect increases electro negativity decreases b) As shielding effect increases electro negativity increases c) As ionization potential increases metallic property increases d) As +ve charge on species increases ionic radii increases
29.
30.
The electronic configuration which contain metals, non metals & metalloid is______ a) ns 1 & ns 2
b) ns 2 , ns 2(n–1)d (1–10)
c) ns 2 np 6 & ns
d) ns 2 np 4 & ns 2 np 5
The group in which all the three physical states (solid ,liquid, gas) are observed is a) gp 17
31.
b) Os
c) Fr
d) I
b) 11, 36, 9, 20
c) 20, 36, 11, 9
d) 9, 18, 11, 20
The element with highest electronic affinity belongs to a) Period 1 gp
34.
d) gp 15
Fo ur elem en ts A, B, C & D. D is n on reactive gas. C is a high ly reactive gas, B is a so lid & forms ox ide, A is high reactive solid & used to p rep are Lasagne’ s so lutio n. Choose the correct sequence of possible atomic no. of elements a) 12, 18, 9, 11
33.
c) gp 18
The element which exhibits highest oxidation number is a) Mn
32.
b) gp 14
b) Period 3 gp 17
c) Period 2 gp 17
d) Period 2 gp 16
The atomic no. of B = atomic of A+18 , Statements A & B to a) Same pd & same gp
b) Same pd but different gp
c) Different pd but same gp
d) Different pd and different gp 250
35.
Element B occupies 3rd pd & gp 16 Element C occupies 4 pd & gp 3 The molecular formula of compound formed between B & C is a) B3C 2
36.
b) C 2B 3
c) CB2
d) B2 C
Choose the correct statement w.r.t oxidising property of F a) It is the strongest oxidising agent because it has highest electron gain enthalpy b) It is the strongest oxidising agent due to its small size c) It is the strongest oxidising agent because it has maximum electron negativity d) It is the strongest oxidising agent due to high lattice enthalpy
37.
The name of th e scien tist wh o disco vered the elem ent Unu & its accep ted IUPAC nameis— a) Mendeleev &Mendelinium
b) Seaborg & Seaborgium
c) Mendeleev &Dubinium
d) G.T.Seaborg & Mendelinium
38. Which of the following property does not indicate the periodicity of elements
39.
40.
a) Ionization potential
b) Neutron/ proton Ratio
c) Bonding behaviour
d) Electron negativity
Properties of Li are similar to Mg because a) The size of Li & Mg are different
b) The size by charge ratio is similar
c) The charges are same
d) Both are reactive
From the given set of quantum num bers for th e last electro n o f the ato m, ch oose the elemen t which is a n on metal. The set of Quantu m nu mbers o f A, B, C & D are given below A – n =2,l=1,m =0, C – n=5,l=2,m= a) D
41.
B – n=4,l=0,m=0
2
D – n=6,l=3,m=0
b) C
c) B
d) A
b) Al
c) B
d) Na
Which of the following ions are not isoelectronic with Ar a) Na +
43.
1
Be shows diagonal relationship with a) Mg
42.
+_
+_
b) Ca+2
c) Cl–
The ionisation potential of N > O because a) Ionisation potential increases with decrease in size b) N posses stable half filled p–orbital c) The screening effect in N > O d) O is more electropositive than N 251
d) K +
44.
The physical properties of chromium is most closely related to a) Niobium
45.
46.
47.
d) Calcium
a) [Ar] 3d 10 4s 2 4p 1
b) [Ar] 3d 10 4s 2 4p 4
c) [Ar] 3d 10 4s 2 4p 3n
d) [Ar] 3d 10 4s 2 4p 2
Choose the incorrect statement a)
An element with high electronegativity always has high electron affinity
b)
Electron gain enthalpy is the property of an isolated atom
c)
Electronegativity is the property of a bonded atom
d)
Both electronegativity & electron affinity are equally proportional to nuclear charge & inversely proportional to atomic size
Choose the ox ide which is most basic CuO, MgO, Al2O b) MgO
3
& K2 O
c) CuO
d) Al2 O
3
An elemen t with atomic nu mber 19 will mo st read ily react with the elem ent who se atomic number is a) 18
49.
c) Titanium
The electronic configuration of an element of chalcogen family is
a) K 2 O 48.
b) Tungsten
b) 21
c) 20
d) 17
If graph is drawn between electro n en thalpy & ato mic num ber from 1 to 60, wh ich of the following statement will be true a) Alkali metals are at the maxima & noble gases at the minimum b) Alkali metals are at the minimum & noble gases at the maxima c) Transition elements at maxima d) Maxima & minima are not observed
50.
In a period with increase in atomic number, the metallic character of an element a) Decrease across p d & increases in gp b) increase across pd & decreases in gp c) increase across pd & increases in gp d) Decrease across p d & decreases in gp
51. In elem en t P with electron ic co nfiguration [Ar] 4s1 will com bine with an elem en t of ________ configuration to form a highly soluble ionic solid with high melting point a) [Ar]4s 2 52.
b) [Ne]3s 2 3p 3
c) [Ne] 3s 2 3p 5
d) [Ar] 4s 2 3d 2
In group 14 the lower oxidation state becomes more stable down the group. The reason is a) Inert pair effect
b) Decreases in ionisation potential
c) Metallic character increases
d) Decrease in electron affinity
252
53.
Choose the correct option. Hint T=true F = False I.
In the second period atomic radii of Be is 90pm, F is 64pm, & that of Ne is 160pm
II. Atomic radii decreases from Li to Ne III. The increase in size of Ne is due to presen ce of van derwaals force of attractio n & presence of covalent bond IV. In Ne there is absence of covalent bond therefore the radii is vanderwaals radii V. The order of radii is Metallic > Covalent > Vanderwaals a) TTFTF 54.
I.
Ionisation enthalpy ¥ 1/shielding effect
II.
Ionisation enthalpy ¥ Chemical reactivity
III.
Ionisation enthalpy ¥ 1/Metallic character
IV.
Ionisation enthalpy ¥ Effective nuclear charge b) FFTT
c) TTTF
d) TFTT
I.
C < N < F < C Second ionisation potential
II.
d 5 < p 3 ; d 10< p 6 Half filled order of stability & fully filled orbital’s
III.
Al2O 3 < SiO 2 < P 2O 3 < SO 2 Acid strength
IV.
M 3+ > M2+ > M > M2– Atomic/Ionic radii b) TTTF
c) TTFT
d) TTTT
Choose the correct option I.
Cs + is the most hydrated than other alkali metal
II.
Among the alkali metals, Li has the highest M.P
III.
Li is the strongest reducing agent because of low ionisation enthalpy
IV.
Li is the stro ngest red ucing agen t becau se th e high io nisatio n p otential is compensated by high hydration enthalpy
V.
Li is resemble to Al
a) FTFTF 57.
d) TFFFT
Choose the correct option
a) TFTT 56.
c) TFFTT
Choose correct option
a) TFFT 55.
b) TTTFF
b) TTFTF
c) FFFTF
d) TTTFF
Choose the correct option I.
NaCl < NaI < NaF < NaBr Io nic character
II.
Si < P < C < N
III.
BeCl2 < MgCl2 < CaCl2 < BaCl2 Ion ic character
IV.
Al3+ < Mg2+ < Na + Ionic mobility
a) FTTF
Electron egativity
b) TFFT
c) FTTT 253
d) FFTT
58.
Choose the correct option I.
Transition metals are characterised by variable oxidation state
II.
Elements of IB & IIB are transition elements
III.
Elements of gp1 exhibit only +1 O.S
IV.
Group 17 contains only gases
a) TTFF 59.
The ionisation enthalpy of Be > B
II.
d–Block elements are known as representative elements
III.
Palladium is the only element of fifth period th at h as no electron in fifth energy level
IV.
The second ionisation enthalpy of Al is greater than that of Mg
V.
Among Li, Be, B ,C N ; Li has least value of electron gain enthalpy b) TFFTT
c) TFFFT
d) TFTTT
Choose the correct option I.
The last electron in case of inner transition elements goes to f–orbital
II.
The electron affinity is highest for fluorine
III.
Metallic radius is smaller than covalent radii
IV.
Ar has lesser ionisation enthalpy than K b) TFFF
c) TTTF
d) TTFF
Choose the correct option I.
All halogens exhibit variable oxidation state
II.
s–Block elements do not exhibit variable oxidation state
III.
the most stable oxidation state of Bi is +3
IV.
N exhibits –3, +3 & +5 oxidation state
a) TFTT 62.
d) TTTT
I.
a) TFFT 61.
c) TTTF
Choose the correct option
a) TFTFT 60.
b) TFTF
b) TFFT
c) FTTT
d) FTTF
Choose the correct option I.
O.S o f ‘O’ in OF 2 is –2
II.
Ionisation enthalpy is the minimum amount of energy required to remove an electron from an atom
III.
Screening effect : it is the attraction of electron towards the nuclear
IV.
Half filled orbitals are more stable half fully filled orbitals
a) TTTT
b) FFFF
c) TTFT
254
d) TFFT
Match the following 63.
Set A
Set B
1. Strongest reductant
p) silver
2. Fully filled d–orbital
q) Berkelium
3. Noble metal
r) Co pper
4. Actinide
s) Iodide ion t) Sodium ion
a) 1–s, 2–r, 3–p, 4–q 64.
b) 1–t, 2–r, 3–p, 4–q c) 1–s, 2–t, 3–q, 4–p d) 1–t, 2–s, 3–q, 4–p
Set A
Set B
1. Liquid non metal
p) Lightest metal
2. Metal stored in paraffin
q) Cs
3. Most electropositive metal
r) KOH
4. Strongest alkali
s) K t) group 17 u) CsOH
65.
66.
a) 1–s, 2–q, 3–s, 4–r
b) 1–t, 2–p, 3–s, 4–r
c) 1–t, 2–q, 3–q, 4–r
d) 1–t, 2–p, 3–q, 4–u
Set A
Set B
1. C O
I) Basic oxide
2. CO 2
K) neutral oxide
3. K 2 O
L) Amphoteric oxide
4. Al 2 O 3
M) acidic oxide
5. SiO 2
O) Neutral
a) 1– K, 2 & 5– M, 3– J, 4–L
b) 1, 2 & 5– M, 3– J, 4–L
c) 1–K, 2 – M, 3–L, 4–J
d) 1–J, 2 & 5– M, 3– K, 4–L
Set A
Set B
1. Osmium
p) Hardest metal electric
2. Lead
q) poor conductor of current
3. Tungsten
r) largest size
4. Caesium
s) most reactive solid matter t) highest oxidation state
a) 1– t, 2–q , 3–p , 4–s
b) 1– t, 2–q , 3–p , 4–r
c) 1– t, 2–s, 3–q , 4–t
d) 1– t, 2–q , 3–s, 4–r 255
67.
Set A
Set B
1. Diagonal relationship
q)attraction towards nucleus
2. Shielding effect
r) charge on the nucleus available for other electrons
3. Effective nuclear charge
s) similar polarising power t) Ionisation enthalpy decreases
a) 1–s, 2–t, 3–r 68.
69.
b) 1–t, 2–s, 3–r
c) 1–r, 2–s, 3–t
d) 1–s, 2–r, 3–t
Set A
Set B
1. Br
p) Chalcogen
2. Ba
q) alkali metal
3. Se
r) alkaline earth metal
4. Rb
s) Halogen
a) 1–p, 2–r, 3–s, 4–q
b) 1– s, 2–r, 3–p , 4–q
c) 1–s, 2–r, 3–q, 4–p
d) 1– s, 2–p , 3–r, 4–q
Set A
Set B
1. Hg
p) Liquid non–metal
2. Carbon (Diamond)
q) reacts very violently
3. Bromine
r) reaction endothermic
4. Caesium & F
s) Liquid metal t) extremely high M.P
70.
a) 1–s, 2–r, 3–p, 4–q
b) 1– t, 2–s, 3–p , 4–r
c) 1– s, 2–t, 3–p , 4–q
d) 1– s, 2–t, 3–p , 4–r
Set A
Set B
1. Inner transition elements
p) 3rd perio d
2. Transition
q) s & p Block
3. Typical element
r) d– Block
4. Representative element
s) f–Block t) p–Block
71.
a) 1–r, 2–s, 3–p, 4–q
b) 1– s, 2–r, 3–p , 4–q
c) 1–q, 2–r, 3–s, 4–t
d) 1– s, 2–r, 3–t, 4–q
Set A 1. Be < Al
p) noble gases
2. Aufbau principle
q) p–Block
3. ns 2 np 1–5
r) Diagonal relationship
4. ns np
s) Block deciding rule
2
6
a) 1–r, 2–s, 3–q, 4–p
b) 1– s, 2–r, 3–p , 4–q
c) 1–q, 2–r, 3–q, 4–p
d) 1– r, 2–q , 3–s, 4–p 256
72.
73.
74.
75.
Set A (Atomic no.)
Set B (Position of element)
1. 100
p) d–Block
2. 50
q) s–Block
3. 40
r) lanthanides
4. 11
s) Actinides
a) 1– t, 2–s, 3–p , 4–q
b) 1– r, 2–s, 3–p , 4–q
c) 1– t, 2–p , 3–s, 4–q
d) 1– r, 2–s, 3–q , 4–p
The position of element A, B, C & D are Element
Period
Group
A
4
2
B
3
13
C
3
16
D
4
16
The molecular formula of the oxide of each element in its highest state are a) AO 2, B 3O 2, CO, DO
b) AO, B2 O 3, CO, DO
c) A 2O, B 2O 3, CO 2, DO 2
d) AO, B2 O 3, CO 3, DO 3
With reference of above questio n the oxide which is (i) m ost ion ic (ii) ampho teric (iii) highest M.P (iv) reacts most readily acid only a) AO,BO,AO,AO
b) AO,CO,BO,AO
c) BO, AO, AO, DO
d) DO, AO, BO, CO Assertio n rea son type
75–85 are assertion reason type for each question select the correct cho ice from the following a) Statement 1 is true, statement 2 is true & is correct explanation for statement 1 b) Statement 1 is true, statement 2 is true but is not correct explanation for statement 1 c) Statement 1 is true, statement 2 is false d) Statement 1 is false, statement 2 is true 75. 1.
F atom has lesser electron affinity than atom
2.
The size of F is very small therefore electron electron repulsion is high
76. 1.
The size of X+,X & O – are id entical
2.
The rem oval of electro n decreases th e size while ad ditio n o f e– increases the size.This is primarily due to decreases increases in electronic repulsion 257
95.
96.
Choose the option in which the order is not in accordance to the property indicated. (a) Al3+ á Mg 2+ á Na + á F-
(Increasing ionic size)
(b) Bá Cá Ná O
(Increasing first ionisation enthalpy)
(c) Iá Br á FáCl
(Incresing negative electron gain enthalpy)
(d) Liá Na á K á Rb
(Increasing metallic radius)
Choose the wrong order (a) NH 3 á PH 3 á A 5 H 3 (Acidic) (b) Liá Beá Bá C DiHi (c) Al2 O 3 á MgO á Na 2 O á K 2 O (Basic) (d) Li + á Na + á K + á CS+ (Ionic Radius)
97. Element
I.
Position in perodic table Period Group
A
3
2
B
7
10
C
2
16
D
5
13
the atomic number of B is A) 104
II.
B) 108
C) 110
D) 105
the type and nature of compound form between A & C is A) Sulphide and basic
B) Oxide and amphoteric
C) Sulphide and neutral
D) Oxide and basic
III. Element D is A) Metal (a) 1-B, 2-C, 3-A 98.
B) Metalloid
C) Non–metal
D) Liquid
(b) 1-D, 2-A, 3-C
(c) 1-C, 2-D, 3-B
(d) 1-A, 2-D, 3-A
The properties of elements are given below Element
Property
B
Liquid and forms strongest alkali
C
Non–metal and shining crystal
D
A metal used as catalyst with exceptional electronic configuration
I.
The element B is A) Cs
B) Ga
C) Fr 260
D) Na
ANSWER KEY 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
c c a b c a a c d b d c c b a b a
18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34
b a a c d d d b c d a d a b c b c
35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51
b c d b b d b a b c b a a d b a c
52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68
262
a a d b a c c d b c b a d a b a b
69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85
c b a a d a a d b d a b d b a c b
86 87 88 89 90 91 92 93 94 95 96 97 98 99 100
b d c c a d b c c b b c d a c
UNIT : 12 GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF METALS Important Points Occurrence of elements: Highly electropositive metals of S-block occur in nature in combined state with several anions. The lesser electropositive metal like chloride, silicate, carbonate (p & s block) occur as sulphides; while noble metals like Au, Pt, Ag are dative that is free state. Abundance:
Ten abundant elements in Earth’s crust: Mineral: The chemicalcompounds ormetalcompounds which occur naturally in earth’s crust are known as minerals. Ores: The minerals from which metal can be extracted economically are known as ores. It can be concluded that “all minerals are not ores but all ores are minerals”. Principal Ores of some metals: Metal Ore Al Bauxite Corundum Diaspora Cryolite China Clay Mica Feldspar Cu Malachite Azurite Copper pyrites Copper Glance Fe Iron Pyrites Haematite Magnetite Limonite Siderite Zn Zinc Blend /Sphalerite Calamine Zincite 263
Composition Al2O3.2H2O Al2O3 Al2O3.H2O Na3AlF6 Al2O3.2SiO2.2H2O K2O.3Al2O3.6SiO2 KAlSi3O8 CuCO3.Cu(OH)2 2CuCO3.Cu(OH)2 CuFeS2 Cu2S FeS2 Fe2O3 Fe3O4 Fe2O3.3H2O FeCO3 ZnS ZnCO3 ZnO
Different methods of concentration of ores: 1)
Gravity separation: Principle used: difference in specific gravity.
2)
Magnetic separation: Principle used: difference in magnetic properties.
3)
Froth Flotation: Principle used: differential wetting properties. I.
Substances used: Frothing agent & collecting agent like pine oil and Sodium Ethyl Xanthate respectively.
II.
Froth Stabilizers: Cresols and Aniline.
III. Depressant: Used to prevent certain type of particles from forming the froth. 4)
Leaching: The ore is treated with chemical to make it soluble. It is also used in concentration of noble metals like Gold, Silver, which occur in Free State.
Different methods of reduction Chemical reduction: Using Carbon: Carbon is useful as a reducing agent because it is very cheap and can be used for reduction of several metal oxides. The ore is mixed with coke with flux and heated to high temperature. This is called smelting or pyrometallurgy. Using more reactive metals: For extraction of lesser electropositive metal, Al powder can be used. The process is known Aluminothermite or Thermite. Oxides of Cr, Mn are reduced by Aluminum. Auto reduction: Reducing agent is not required. Cu2S + Cu2O ® 4Cu +SO2 Electrolytic reduction: Highly electropositive metals like Na, Mg, Ca etc cannot be reduced using carbon, carbon monoxide or metal. Therefore these elements are extracted electrolyitcally using fused salts of the metal. It is mainly used for S-block elements and Al. Different methods used in refining The removal of impurities from metals after extraction is known as refining. The refining technique depends upon the nature of impurity and metal and the use to which the metal is put. Refining techniques can be classified into 2 types: (1) Physical
(2) Chemical.
Physical: Method Liquation
Principle or property
Difference in fusibility of metal impurities. Distillation Used for volatile metal. Difference in volatile character. Zone Difference in solubility of refining impurity in molten and solid state.
Process Heated on a sloping hearth. Impurities remain behind and metal flows down. Metal is heated and vapor of pure metal are condensed in a receiver. The circular heater is fitted round a rod of impure metal and is slowly moved down the rod.
265
Metals purified Bi, Sn, Pb, Hg. Zn, Cd,Hg Very high purity viz Ge, Ga
Thermodynamic principles of metallurgy: The basic concepts of thermodynamics are useful in selection of reducing agent for a particular oxide. Free energy of changes during reduction process are important. For a reaction to be spontaneous free energy changes must be negative. Any compound with lower value of standard free energy of formation is more stable. They are converted to oxide prior to reduction. DG = DH - TDS
DG= Change in free energy T= Temperature in Kalvin DS= Enthalpy change
DG 0 = -RTlnK = -2.303 RT log10 K
K= equation constant R= Universal gas constant
If K value is high, it indicted [P] >> [R] & DG 0 becomes more negative Eg.
2Mg(s) + O 2(g) 2MgO (s)-
DG 1000’c = -941KJ/mole
2C (s) + 2CO (g)
DG 1000’c = -439 KJ/mole
MgO + C ® Mg + CO DG = SG p - SG r = -439 + 941 = +502 Since DG is +ve carbon cannot be used as reducing agent 2Zn + O 2 ® 2ZnO
DG 1100’c = -360 KJ mol-1
2C + O 2 ® 2CO
DG 1100’c = -460 KJ mole-1
SG = SG p - SG r = -460 – (- 360) = -100 KJ mol-1 Reduction is possible. Electrochemical Principle Electrode potential DG 0 = - nFE0 This concept is useful in electrolytic reduction. If E0cell is positive the DG 0 is negative. Therefore reaction is spontaneous.
267
Types of iron
preparation
Impurities &Properties
Uses
Pig iron
Obtained from blast furnace
4%C &S ,P
Manufacture of cast iron
Cast iron
Blowing hot air in mixture of pig iron,iron scrap &coke
3%C
Gutter pipes, Lamp posts, casting of articles like stoves etc. It does not rust.
Heating cast iron in reverberatory furnace lined with hematite, limestone is added as flux to remove S,P,Si (impurities)
Purest form of iron.It is extremely tough with high melting point &malleable.
Wrought iron
Hard & brittle
Fe2O3 +C(cast iron)à2Fe +CO Metal from is passed furnace is passed through the rollers to remove impurities
272
Anchors, chains nails etc
Þ
Manufacture of Zn Continued ORE Crushing
Forth flotation if Zn blend
Concentration
ZnS + 3O 2 ® 2 ZnO + 2SO 2
Rosting, Calcination
D ZnCO3(s ) ¾ ¾® ZnO(s ) + CO 2 ( g )
Reduction Chemical with Carbon Δ ZnO(s) + C(s) ¾¾ ® Zn (s) + CO (g)
Leaching
ZnO + H 2SO 4 ¾¾® ZnSO 4 + H 2 O ÏÜع
Heated ZnOwith coal
ZnSO4 impurities like Cd, Fe, Sb & As (i) Zn Dust (ii) Filter Cdppq ZnSO4 + Soluble Salt of Fe, Sb, AS (i) Ca(OH)2
in cylindrical retortØ Zn vapours are cooled and Collected
Impure Zn or spelter Purification
Soluble ZnSO4 Pure ZnSO4 Electrolysis
Liquidation Distillation Electrolytic refining
Pure Zn
273
99.9%
(ii) Filter Pot of hydroxides of Fe, Sb, As
M.C.Q. 1)
The most abundant metal in the earth crust a) Al
2)
5)
10)
b) SiO2
c) FeO
c) Minerals are not ores
d) a & b
d) Ca(OH)2
A metal oxide is reduced with a metal, M1 of pd 3 & Gp13. The process is known as ____ & the metal is b) Thermite & Ga
c) Liquation & Al
d) Thermite & Al
Diaspore & Corundum are ores of ______ & ______ b) Fe & Al
c) Al & Al
d) Si & Al
b) Limonite à Fe2O3
c) Calamine à ZnCO3
Choose the correct option d) a & c
The salt which is most unlikely to occur as mineral is a) Bromide
9)
d) Ore
b) All mineral are ores
a) Siderite à FeCO3 8)
c) Slag
a) All ore are mineral
a) Al & Fe 7)
b) Gangue
Choose the correct statement
a) Pyrometallurgy &Mg 6)
d) Na
An ore after concentration was found to have basic impurities. The flux which can be used is a) CaCO3
4)
c) Ca
The impurities associated are a) Flux
3)
b) Fe
b) Sulphate
c) Oxide
d) Sulphides
During concentration of ore by froth flotation, the ore particles float on surface because: a) Ores are insoluble
b) Sulphides ores are lighter
c) The surface is not wetted by H2O
d) Difference in densities
An ore contains Pbs with impurity Zns. NaCN is added in the froth flotation process ZnS does not form the froth because a) NaCN forms a complex of ZnS on surface of ZnS b) ZnS is not wetted by pine oil c) ZnS is wetted by H2O
11)
The method used for concentration of magnetic ore is a) Gravity separation
12)
b) Froth floatation
c) Magnetic separation
d) Leaching
During froth flotation process the student observed that the froth was disappear after formation. The student added-----_______to the container to overcome the difficulty a) Pine oil
13)
d) NaCN reacts with ZnS ionic compound
b) Cresol
c) Benzene
d) NaCN
Silver ore is related with NaCN(aq) to a) Reduce silver
b) Extract pure silver
c) Refine silver
d) To remove the impurities 274
14)
15)
The principle involved in leaching is a) Difference in volatility
b) Difference in density
c) Difference in solubility
d) Soluble complex formation
Heating of ore in presence of O2 below its melting point is known as a) Roasting
16)
b) Calcinations
c) Smelting
d) b & c
During electrolysis graphite is used as an electrode & not diamond because a) Graphite is cheaper
b) Graphite is soft
c) Graphite posses free electron while diamond doesn’t d) Graphite is non reactive 17)
Group 1 & 2 elements are extracted by a) Thermite process
18)
22)
d) Silver
b) CaO & Cr2O3
c) BaO & U3O8
d) SiO2 & Al2O3
b) Ore of Cu
c) Alloy of Cu
d) Impure copper
a) Increase the conductivity
b) Lower the melting point
c) Increase the mobility of iron
d) All of above
During extraction of metal charcoal powder is sprinkled on top of molten metal. This is useful in preventing b) Formation of alloy
c) Reduction
d) a & b
c) Ultrapure Si
d) Nickel
zone refining is used to obtain ———— metal a) Pure Cu
24)
c) Iron
In Hall Heroult’s process cryolite is added to alumina to
a) Oxidation of metal 23)
b) Manganese
Blister copper is a) Pure copper
21)
d) Cupellation
Several metals are commercially produced by reduction of oxides by carbon. The oxides which can be reduced with carbon are a) ZnO & Fe2O3
20)
c) Bessomerisation
Hydrometallurgy is used in extraction of a) Sodium
19)
b) Electrolytic method
b) Zirconium
Strongly D ® Pure Metal Metal + _______ à Volatile compound ¾¾¾¾ Strongly
This method is known as a) Liquation 25)
c) Zone refining
d) Distillation
c) Ge
d) Ni
c) CuSiO3
d) FeO
______ metal is purified by Mond’s process a) Zr
26)
b) VanArkel b) Ti
The slag obtained during manufacture of Cu is a) CaSiO3
b) FeSiO3
275
27)
28)
29)
The principle used in zone refining a) Fractional distillation
b) Adsorption
c) Fractional crystallisation
d) Chromatographic separation
In the reaction 2MO(s) + C(s) à M(s) + CO2(g) the entropy of the reaction will a) Decreases
b) Increases
c) Remain constant
d) May increases or decreases
In electrolytic refining of copper, anode mud contains a) Earthly impurities
30)
b) Zn & Mn
c) Noble metal
d) Oxides of Cu
Zn is extracted from ZnS. The reducing agent used is _________ & method of refining is ______ a) Coke & Electrolysis
b) Mg & Liquation
c) Coke & Zone refining
d) Coke & fractional distillation
(31 To 40 are match the following sets.) 31)
Set 1
Set 2
1)
Al
p) Haematite
2)
Fe
q) Nuggets
3)
Zn
r) Sphalerite
4)
Ag
s) Feldspar t) Limonite
32)
a) 1-p, 2-t, 3-r, 4-q
b) 1-s, 2-p & t, 3-r, 4-q
c) 1-s, 2-r & t, 3-p, 4-q
d) 1-p, 2-t, 3-r, 4-s
Set 1
Set 2
1)
Pi g iron
p) Hard & brittle
2)
Cast iron
q) Prepared from cast Fe
3)
Wrought iron
r) Prepared by methyl pig iron s) Malleable t) Fe + 4% C
a) 1-t, 2-p & r, 3-q & s
b) 1-t, 2-r , 3- s
c) 1-p, 2-r & t, 3-q & s
d) 1-p, 2-s & r, 3-t & q
33)
Set 1
Set 2
1)
Cr2O3 + Al
p)
electrolysis
2)
Zinc
q)
Bayer’s process
3)
M2O3 + NaOHà
r)
Thermite
D ® M2O3 Soluble ¾¾
s)
Hall Heroult
a) 1-r, 2-p, 3- s
b) 1-r, 2-s, 3- p 276
c) 1-s, 2-p, 3-q
d) 1-r, 2-p, 3-q
34)
35)
Set 1
Set 2
1)
Chromatography
p)
Chemical process
2)
Poling
q)
Difference in solubility
3)
Liquation
r)
Difference in meltingpoint
4)
Zone refining
s)
Low boiling point metal
t)
Adsorption
a) 1-r, 2-p, 3-s, 4-q
b) 1-q, 2-r, 3-p, 4-s
Set 1
Set 2
1)
Earthly impurities
p) Froth flotation
2)
Sulphide ores
q) Magnetic separation
3)
Bauxide
r) Gravity separation
4)
Magnetite
s) Leaching
a) 1-p, 2-r, 3-s, 4-q 36)
1) 5000-8000 K
p)
Pig iron
2) 12700
q)
Molten slag
3) 21700
r)
C + O2 à 2CO
4) 2170
s)
3Fe2 O3 +COà
t)
CaO + SiO2 à CaSiO3
5) > 2170
2Fe3O4 +CO
2
a) 1-s,2 q,3r,4-p
b) 1-s,t ; 2-r,3-q ,4-p
c) 1-r,s ;2-t,3-p,4-q
d) 1-s,2-r,3-q,4-ps
Set 1
Set 2
1)
Bauxite
p)
Bayers
2)
Zinc blend
q)
Blast Furnace
3)
Copper pyrites
r)
Hall- heroult
4)
Haematite
s)
Bessemerisation
t)
Fire Clay cylindrical retort
a) 1-r & p, 2-t, 3-s, 4-q 38)
c) 1-q, 2-r, 3-p, 4-s d) 1-s, 2-p, 3-r, 4-q
Given below are the different temperature reactions & products during extraction of iron in blast furnace
0
37)
b) 1-r, 2-p, 3-s, 4-q
c) 1-t, 2-p, 3-s, 4-q d) 1-p, 2-p, 3-s, 4-q
b) 1-p, 2-t, 3-s, 4-q
Set 1
c) 1-r, 2-s, 3-s, 4-q d) 1-r, 2-t, 3-s, 4-q
Set 2
1)
Zn
p) Auto mobiles
2)
Wrought iron
q) Galvanising
3)
Steel
r) Bell Meta
4)
Copper
s)
Muntz Metal
t)
Anchors
a) 1-s, 2-r, 3-t, 4-q
b) 1-q & r, 2-t, 3-p, 4-r & s
c) 1-q, 2-s, 3-p, 4-r
d) 1-q, 2-t, 3-p, 4-r & s 277
39)
40)
Set 1
Set 2
1)
Electrolysis
p) Wide range of temperature
2)
Zone refining
q) Adsorption
3)
Blast furnace
r)
Electrode potential
4)
Liquation
s)
Noble gas atmosphere
t)
Low meltingpoint
a) 1-r, 2-t, 3-p, 4-t
b) 1-r, 2-s, 3-p, 4-t
Set 1
Set 2
1) Non spontaneous
p) DG= 0
2) spontaneous
q) K < 1
3) Equilibrium
r) DG decreases
c) 1-r, 2-s & t, 3-p, 4-q d) 1-t, 2-s, 3-p, 4-r
s) K > 1 t) a) 1-q, 2-r & s, 3-p & t
b) 1-q, 2-r & s, 3-p
c) 1-t, 2-r & s, 3-p
d) 1-q, 2-r, 3-p
Questions 41 to 51 are assertion reason type
41)
42) 43)
44) 45)
46)
a)
Statement 1 is correct, statement 2 is the correct reason for statement 1
b)
Statement 1 is correct, statement 2 is correct but does not give reason of for statement 2
c)
Statement 1 is correct, statement 2 is false
d)
Statement 1 is false, statement 2 is correct
Statement
1)
Magnesium is mainly extracted by electrolysis of molten electrolyte and not by chemicalmethods.
Statement
2)
The DiH of Mg is very low & hence it is very difficult to reduce Mg2+(aq)àMg(s)
Statement
1)
Cu is obtained by ____ . Then by pdt CO2 obtained
Statement 2)
The by-product obtained is SO2& is used in manufacture of sulphuric acid.
Statement
1)
In Bayer’s process the ore is heated with Conc. Ca(OH)2.
Statement
2)
The ore used in Bayer’s process is atmospheric & is soluble in NaOH therefore ore is concentrated.
Statement
1)
Froth flotation is used for concentration of sulphide ores.
Statement
2)
Cresol is used as a depressant in froth flotation.
Statement
1)
Metal oxide can be easily reduced with carbon
Statement
2)
D ® Melted + CO Metal oxide + C ¾¾
Statement
1)
DG = -nFE0 is applicable to metallurgy
Statement
2)
If cell potential is negative electrolytic reduction of metal ions is possible 278
DG = -x KJ
47) 48) 49)
Statement
1)
Ultra pure silicon is manufactured by vapour phase refining
Statement
2)
Vapour phase refining gives ultra pure metals
Statement
1)
Ni is converted to Ni(CO)4 in Mond process
Statement
2)
Ni(CO)4 is volatiles & the compound can be easily disposable
Statement
1)
Mercury is purified by distillation
Statement
2)
Mercury has low melting point True / false questions (50-58)
50) 1)
Ag & Au are manufacture by hydrometallurgy
2)
Fe can be extracted by electrolytic method
3)
Mg is extracted from aq MgCl2 by passing electric current
4)
Zn is extracted from Zinc blend by using coal or anthracite coal
a) TFFT 51)
b) TFTT
c) FFFT
d) FFTT
c) TTFT
d) FTFT
The metals which can be extracted using carbon/coke 1)
Magnesium
2)
Iron
3)
Potassium
4)
Zinc
a) FTTF
b) TTFF
52) 1)
DG0 = -nFE0
2)
The above equation is the principle used in electrolytic reduction for manufacture of metals
3)
D 2Cu2S(s) + 3O2(g) ¾¾ ®
4)
DG = DH – TDS if free energy change is negative the reaction is spontaneous
a) TFFT
2Cu2O(s) + 2SO2(g) DS is positive in this reaction
b) TTFT
c) TTTT
d) TFTF
c) TFFT
d) TFFF
53) 1)
Kaolinite is an ore of Al
2)
Sphalerite is an ore of sulphide of Cu
3)
Malachite is an oxide ore of copper
4)
sphalerite is iron carbonate
a) TFTT
b) TTFT 279
54) 1)
Ore is heated strongly during calcinations to remove all volatile impurities
2)
Ore is heated with oxygen during roasting to convert sulphide to oxide
3)
Cryolite is added to bauxite in Hall Heroult’s process to increase solubility of bauxite in ____
4)
After leaching of bauxite, it is directly used as an electrolyte
a) TTFF
b) TTFT
c) TTTF
d) TFFT
55) 1.
The reduction reaction by accepting electrons is known as electro nation
2.
In extraction of Gold and silver by process of leaching K > 1
3.
If DH = -1369 KJ/ mol-1, DS= +26 J/mol-1 T = 400K for the reaction A + B à C the reaction is non spontaneous
a) TTT
b) TTF
c) TFT
d) TFF
56) 1.
Chewing of cathode occurs in Hall Heroult’s process
2.
When water is added after digesting boxide, CO2 is bobbled to nutrelise the solution
3.
HCl can not be replaced in (2) because AlCl3 is formed which is highly soluble
a) FTF
b) TTF
c) FFF
d) FTT
57) 1.
Al is used in preparation of parts of airplane and manufacture of alloy alnico.
2.
Copper is used in preparation of tubes of boilers, delta metal and muntz metal.
3.
Copper and aluminum are used in alloys – Duralumin, Aluminum bronze.
4.
Cu & Zn are used in manufacture of German Silver.
a) TFTF
b) TTFT
c) TFFT
d) TFTT
58) 1.
Wet metallurgical process is used for pyrites ores of lower grade.
2.
2Cl- +2H2Oà2OH- +H2 +Cl2 The cell potential is -2.186V.This reaction will take place in forward direction.
3.
Pure Zn metal is called spleter.
4.
The abundance of Al is highest .Its place is third & is about 8.3% by weight.
a) TFFT
b) TTTT
c) FTFT
280
d) FFFT
Linked Comprehension type: 59)
Cassiterite ( SnO 2) and ore of Tin contains FeWO4 & MnWO 4 as impurities as well as small amounts of sulfur. SnO2 is melted with cold, Lime stone and sand in a reveberatory furnace. The impure Tin Sn is purified by liquation I. The ore is concentrated by A) Froth flotation
B) Magnetic separation
C) Leaching
D) a & b
II. The impurities of sulfur can be removed by A) Treatment with NaOH
B) Treatment with HCl
C) Roasting
D) Calcinations
III. If limestone and sand are added during smelting the nature of impurities are A) Acidic and basic
55)
B) Amphotric
C) Acidic
(a) i-B, ii-C, iii-C, iv-B
(b) i-C, ii-D, iii-D, iv-A
(c) i-A, ii-C, iii-B, iv-C
(d) i-A, ii-C, iii-A, iv-B
D) Basic
DG 0f for CuS, CaS, SO2, CS 2 & CuO are -49, -1230, -300.4, +63.6 & - 127.2 KJ/ mol-1 respectively. i. The most stable and unstable compounds are A) SO2 & CaS
B) CaS & CS 2
C) CuS & SO2
D) CaS & SO2
ii. CuS & CaS are reduced with Carbon to give Cu or Ca and CS2. The reaction which will be spontaneous is A) Reduction of CuS because DG is positive B) Reduction of CaS because DG is negative C) Both are spontaneous because DG is positive D) Both are non spontaneous because DG is positive iii. The DG 0 for the reaction 2CuS + 3O2 à 2CuO + 2SO 2 is ________ & the reaction is _________ A) – 756.8KJ & Spontaneous
B) +756.8KJ & non- Spontaneous
C) -378.4KJ & Spontaneous
D) +378.4 KJ & non-Spontaneous
(a) i-B, ii-D, iii-A
(b) i-D, ii-C, iii-B
281
(c) i-B, ii-D, iii-C
(d) i-C, ii-B, iii-D
ANSWER KEY 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
A B B A D C D B C A C b D D A
16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
C B D A D B A C B D B C B C D
31 32 33 34 35 36 37 38 39 40 41 42 43 44 45
282
B A D C B D A D B B A D D C A
46 47 48 49 50 51 52 53 54 55 56 57 58 59 60
C D A A A D B C A B D B A D A
Therefore hydrogen is placed in middle of first period. Isotopes of hydrogen Sl:No: 1 2 3
Name &symbol Protium 1 1 H Deuterium or 12H Tritium or 3 1 H
Atomic number 1
Atomic mass 1
No: of protons 1
No: of neutrons 0
1
2
1
1
1
3
1
2
occurrence Highest 99.9850% 0.015% T: 11 H 1:10
Nuclear stability&t1/2 Stable Stable 12.33yrs Radioactive 3 3 1 H-> 2 He 0 + -1 e
Physical properties Physical properties of isotopes are slightly higher than hydrogen because the mass of isotopes are higher. Chemical properties: The chemical properties are similar because the electronic configuration is same. The rate of reactions differs. Laboratory preparation of hydrogen : Zn +H2SO4 -àZnSO4 +H2 Zn +NaOH -àNa2ZnO2 +H2 (Sodium zincate) Manufacture of Hydrogen: Industrial Preparation: Electrolysis: Electrolyte
Electrodes
a) Acidified water
Pt
b) Aqueous Ba(OH)2
Ni
Hydrogen is liberated at the cathode.
284
Manufacture of Hydrogen
Starting Material
H2 O Steam
Red hot coke
Normal Gas CH4 1270K
Ni Catalysts
CH3 OH (l)
H2O Cu2O 50 bar 673K
CO + H2 Synthesis gas FeCrO 4 as catalyst 673K
H2O
CO2 + H2 Bubble through H2O H2O + CO2 H2CO3 H2 insoluble
CO2 Dissolve
Physical properties: 1.
Colourless, tasteless & odourless
2.
It is diatomic gas, insoluble in water.
3.
It is the lightest element & diamagnetic.
285
Þ
FORMULA : (1) M =
W M ´V 1
M = Molarity W= Weight of solute M1 = Molecular mass V = Volume
(2) N =
W E´V
N = Normality W= Weight of solute E = Equivalent mass V = Volume
gL-1 = N ´ E
(3) (4) (5) (6)
Equivalent weight for H2O2 = 17 For H2O2, N = 2M % w/v = The mass of H2O2 in 100 ml solution.
(7)
Volume = % w v ´ 3.294
(8) %w/v = 3.4 ´ M (9) % Volume = 11.2 ´ M (10) N1V1 = N2V2 (11) éë( gL- )1 ´ ( Volume strength )1 ùû + éë( gL- )2 ´ ( Volume strength )2 ùû +
( )
é gL- ´ ( Volume strength ) ù = ( g L ) ´ ( Volume strength )mix 2û mixture 3 ë
292
M.C.Q. 1.
2.
3.
4. 5. 6. 7.
8. 9.
The element which is the biggest source of energy in future is (a) Monoatomic gas (b) Gaseous non-metal (c) Liquid nonmetal (d) lightest element Dihydrogen is liberated at the anode by electrolysis of : (a) Molten sodium hydride (b) Acidified water (c) Molten sodium chloride (d) Water with Ba(OH)2 The conversion of atomic hydrogen to dihydrogen is : (a) endothermic change (b) Photochemical change (c) exothermic change (d) Nuclear change The isotope of hydrogen with half-life of 12.33 year is : (a) Protium (b) Deuterium (c) Tritium (d) b &c Zinc on reaction with ....... liberates a combastible gas. (a) dil HCl (b) dil KOH (c) H2SO4 (d) a, b, &c Hydrogen gas can be produced from (a) Water gas (b) producer gas (c) coal gas (d) air When Zn pieces are dropped in NaOH solution H2(g) is obtained and soluble ...... is obtamed (a) Na2ZnO3 (b) NaZnO2 (c) Na4ZnO3 (d) Na3ZnO2 673k CO (g) + H 2 O(g) ¾¾¾ [ x ] ® CO 2(g) + H 2(g) . x is ......
(a) Fe (b) Pd (c) FeCrO4 (d) V2O5 H2 can be obtaioned from mixture of CO2&H2 by bubbling the mixture through (a) Water (b) Alkaline Ca2Cl2 (c) Conc H2SO4 (d) Hot Nacl solution
10.
673k 200bar H 2 + A ¾¾¾¾® Alkaline gas . A is [ Fe]
(c) N2
(d) Na
11.
(a) Cl2 (b) O2 The decay product of tritium is : (a) 42 He
(c) 21 H
(d) 23 He
12.
(b) 11 H
The metal Zn, Al, Mg, &Be are placed in different Test tubes. If NaOH is added, the metal which libaate Hydrogen gas are (a) Zn, Al, Mg & Be (b) Zn &Al (c) Mg & Be (d) Zn, Al & Mg
293
13.
The gas used in welding & cutting of metal is a strong ...... (a) Reducing agent (b) Oxidising agent (c) Reducing & oxidising agent (d) Dehydrating agent 14. Hydrogen closely resembles halogens becouse (a) Strong reducing agent (b) diatomic gas (c) it is a colourless gas (d) its is reduction potential is o.oo V 15.
K w = 1.0 ´ 10-14 at 298k because
(a) éëOH - ùû = éë H3O + ùû = (1.0 ´10-14 ) M (b) éëOH - ùû = (1.0 ´10-8 ) M & éë H3O + ùû = (1.0 ´10-6 ) M (c) éëOH - ùû = éë H3O + ùû = (1.0 ´10-7 ) M (d) éëOH - ùû = (1.0 ´10-6 ) M, éë H3O+ ùû = (1.0 ´10-8 ) M 16.
The type of hybridisation of O in H 2O & H 2 H 2(s) is
(a) sp3, sp3 (b) sp2, sp3 (c) sp3, sp2 17. The shape of water moleale is bent and not linear because (a) Bond angle is < 180 (b) sp3 hybridisation (c) Presence of one lone pair of electron (d) sp2 hybridisation 18.
(d) sp3, sp
BH + H 2 O BH 2 + + OH - , H 2 O acts as
(a) Base (b) Reducing (c) acid (d) a &c 19. A metal M belongs to period 3 & group 2 reacts with nitrogen to give compound B. If B is added to water the products are : (a) Mg(OH)2 & NH3
(b) Be(OH)2 & NH3
(c) LiOH & NH3
(d) Ca(OH)2 & NH3
20.
The Only compound whose density in solid state is less if than liguid is (a) Water (b) Sodium hydrocide (c) Nitric acid (d) phosphoras penta chloride 21. Fishes survive in frozen lakes because (a) Ice floats on water (b) Ice acts as an insulator (c) The Solubility of CO 2 in water increaze
(d) a &b 294
22.
Water is most important solvent because it is (a) polar (c) forms H- bond
23.
(b) Non polar (d) a &c
Ice is lighter than water because (a) Density of ice is greater than water (b) The volume of ice is more for given mass of water due to H - bonding (c) Anomalous expansion (d) Oxygen is electronegative & size is large
24.
SiCl 4 + H 2O ® A + HCl A is
(a) Si(OH)4 25.
(b) SiO 2
(c) SiO
(d) SiCl4 × 2H 2O
(c) s - block
(d) p - block & d - block
MH + H 2 O ® MOH + H 2 × M belongs to
(a) p - block 26.
.
(b) d - block
The hydrogen which are used in catalytic reaction are : (a) hydrides of p-block
(b) hydrides of d-block
(c) interstial hydrides (d) b &c 27. Alkali metal do not form interstial hydrides because
28.
29.
30.
31.
(a) alkali metals loose electron readily.
(b) The packing in alkali metals in vay close
(c) Absence of interstitial voids
(d) size is large
The position of the element which forms deficient hydrides. (a) Period 2 group 14
(b) Period 2 group 15
(c) Period 2 group 13
(d) Period 6 group 13
An element forms electron rich hydride. The elctornic configuration of the element is (a) [ He] 2s 2 2p 2
(b) [ He] 2s 2 2p1
(c) [ Ne] 2s 2 2p 2
(d) [ He] 2s 2 2p5
The set if quantum number for valence electron of an element which from election precise of water is (a) n = 4
= 2
(b) n = 2
= 1
(c) n = 3
= 0
(d) n = 2
= 0
The method which can be used for removal of temporary & permanent hardness of water is (a) Decantation
(b) Distillation
(c) Boiling
(d) Filteration 295
32.
Which of the following reacts easily with H2O to form hydrogen (a) HCl
33.
(b)KH
(c) NH 3
(d) B2H 6
It is not advisable to use hard water for washing clothes beacause (a) Precipitate of sodium salt of fatty acid is formed (b) Precipitate of sodium salt of sulphonic acid is formed (c) Precipitate of Magnesium salt of sulphonic acid is formed (d) Precipitate of Magnesium salt of fatty acid is formed
34.
35.
36.
37.
38.
Calgon softens hard water by (a) Precipation of Ca 2+ & Mg 2+ ions
(b) Coagulation of solts
(c) Complexing Ca 2+ & Mg 2+ ions
(d) a & c
clark’s method is used to remove (a) Temporary hardenes
(b) permanent
(c) Hardnes due to soluble SO4-2 of Ca+2, MG+2
(d) Temporary & permanent
Na + D 2 O ® A + B, A & B are ______ & ______respecitively.. (a) NaOH & H 2 O
(b) NaOD & D 2
(c) NaOD & D 2 O
(d) Na2D & D2
The % (mass) of deuterium in heavy water is (a) 18.0
(b) Cannot be predicted
(c) 11.1
(d) 20.0
H 2 O 2 is not used as
(a) Oxidising agent
(b) Redusing agent
(c) Catalyst
(d) Bleaching
Þ
True - False Type
39.
1. H 2 O 2 acts as bleaching agent because of if its oxidising property 2. It is dangerous to used H 2 O 2 is maintenance of environment 3. H 2 O 2 is used is properation of good quality detergents 4. Perhydral is used as a disuifectant (a) TFTT
(b) FTTT
(c) TTFF 296
(d) FFTF
40.
1. A paper with stain of black Pbs, dipped in H 2 O 2 solution turns white 2. The colour of acidified KMnO4 does not diappear when H 2 O 2 is added 3. A basic solution containinig Fe3+ ion turn blue on addition of H 2 O 2 (a) FFF
41.
(b) FTF
(c) TFF
(d) TTT
1. Tritium can be obtaned from natural source. 2. In ionic hydrides the oxidation state of hydrogen is +1 3. The four atom of oxygen in H 2 O 2 are in the same plane 4. Na 2 CO3 removes temporary & permannent hardenes (a) TFTT
42.
(b) FFFT
(c) TFFT
(d) TTTF
Statement S: The position of hydrogen is not fixed, Statement R: Hydrogen resemker alkali metals because of its stable +1 oxidation state. (a) S & R both are correct (b) S is corrct R is correct and explains S. (c) S is incorrect R is correct (d) S is correct R is incorrect Solve the problems from question 43-50
Questions 43-50 - Solve the problems. 43.
44.
45.
Calculate M, N % w/v, gL-1 of 10 Vol H2O2 M N %w/v gL-1 (a) 0.89 1.78 3.036 30.36 (b) 0.78 2.78 2.036 3.036 (c) 0.92 1.95 2.45 3.5 (d) 0.1 0.78 4.0 3.6 30 ml of acidified solution of H2O2 required 30 ml of 0.1NKMnO4 Calculate strength Volume strength & molarity. M gL-1 Volume (a) 0.06 0.7 0.8 (b) 0.12 0.9 0.6 (c) 0.05 1.7 0.56 (d) 0.1 1.0 0.90 374 g of H2O2 is present in 15 lit solution calculate M, N, % w/v & volume strength. M N %w/v Volume gL-1 (a)
0.89
1.2
1.49
6.8
22.0
(b) (c) (d)
1.6
1.3
3.49
7.9
23.1
1.9
1.7
0.49
9.0
22.2
0.733
1.466
2.49
8.2
24.9
297
46.
47.
48.
49.
50.
2.72 g of H 2O 2 is present in 50ml solution Calculate M, N, Strength gL-1 Volume strength H2 O 2 (a) M = 10, N = 2.9, g/l = 53.0, Vol = 17.0 (b) M = 1.6, N = 3.2, g/l = 54.4, Vol = 17.92 (c) M = 0.89, N = 1.8, g/l = 52.9, Vol = 18.1 (d) M = 0.90, N = 1.0, g/l = 5.44, Vol = 16.0 Calculate % w/v, Volume Strength, M & N of a mixture contaning 800 ml of 2.5% w/v, 700 ml of 4.2 w/v & 500 ml 5.3w/v of H2O 2 Solution % W/V g/L M N (a) 3.795 37.95 3.4 2.23 (b) 3.02 3.795 1.116 1.92 (c) 3.795 37.95 1.116 2.23 (d) 2.92 02.89 3.4 1.92 10ml of KMnO 4 Solution is required ti completely oxidise acidic solution of 30ml of 1.5 Volume strength H2O2 Calculate normality of KMnO4 Solution (a) 0.4 (b) 0.65 (c) 0.19 (d) 0.8 500 ml of 5Vol, 400 ml of 10 Volume & 600 ml of 15 volume solution of H2O2 is mixed Calculate volume strength, M, N of resulting solution. Vol % W/V g/l M N (a) 10.33 3.137 31.37 0.92 1.84 (b) 10.33 31.37 3.137 1.84 0.92 (c) 9.95 3.137 32.w 0. 921.84 (d) 9.95 31.37 31.3 1.8 40.92 O2 O2 H 2SO 4 A ¾¾ ® B ¾¾ ® C ¾¾¾ ®D+E
gives green 1st Prepared by Colour flame J.L Thenard A, B, C, D & E are (a) Ba, BaO, BaO2, BaSO 4, H 2O 2 (b) Na, Na2O, Na 2O 2, Na 2SO 4, H 2O (c) Ba, Ba2O 2, BaO 2, BaSO 4, H 2O 2 (d) Ca, CaO, CaO2, CaSO 4, H 2O 2
298
HINTS 43.
Volume = 11.2 ´ M 10 = 11.2 ´ M
N = 2M = 2 ´ 0.89
M = 0.89
N = 1.78
%W / V = 3.4 ´ M = 3.4 ´ 0.89
gL-1 = 3.036
gm/L = 30.36 44.
H2O2 = kMnO4 N1V1 = N2V2
gm/L = N ´ E
N1 ´ 30 = 0.1 ´ 30
gm/L =1.7
V=11.2 ´ M
gm/L = 0.1 ´ 17
= 11.2 ´ 0.05 = 0.56
N1 = 0.1 Molarity of H2O2 Sol. = 0.05 45.
M= =
W M ´V
N = 2M
1
374 34 ´15
2 ´ 0.73
M = 0.733 = 1.466
46.
M= =
W M ´V 1
M= 1.6 47.
= 3.2
Vol =11.2 ´ 0.733
Vol = 8.2
= 3.4 ´ 0.733 % W / V = 2.49 gm/L = 24.9
N = 2M
2.72 34 ´ 0.05 = 2(1.6)
% W / V = 3.4 ´ M
% W / V = 3 .4 ´ M
= 3.4 ´1.6
Vol = 17.92
= 5.44 gm/L = 54.4
2.5 ´ 800 + 700 ´ 4.2 + 500 ´ 5.3 = 2000 ´ % W / V
%W/V = 3.795 gm/L = 37.95 %W/V = 3.4 ´ M 3.795 = 3.4 ´ M
M= 1.116 N = 2.232
299
Vol =11.2 ´1.6
48.
H2O2 kMnO4
Volume =11.2 ´ M
1.5 ´30 = Strength ´10
M = 0.40
V = 4.5 N = 2M= 2´ 0.4 N = 0.8 49.
500 ´ 5 + 400 ´10 + 600 ´15 = Volume ´1500
Volume = 10.33 % W / V = ‘¹ /3.294 %W/V = 3.137 gm/L = 31.37 Volume = 11.2 ´ M M = 0.92 N = 2M N = 1.84
Answer Key 1 2 3 4 5 6 7 8 9 10
d a c c d a b c a c
11 12 13 14 15 16 17 18 19 20
d b a b c a b c a a
21 22 23 24 25 26 27 28 29 30
d d b b c d a c d b
300
31 32 33 34 35 36 37 38 39 40
b b d c a b d c a c
41 42 43 44 45 46 47 48 49 50
b a a c d b c d a a
UNIT : 14 S - Block ELEMENTS Important Points Group-1 (alkali metals) and group-2 (alkaline earth metals) are included in the sblock elements of the periodic table. They are known like this because their oxides and hydroxides are alkaline in nature. Alkali metals possess one and alkaline earth metals possess two s-electrons. They are highly electropositive metals and form monovalent cations (M+) and divalent cations (M2+) respectively. With the increase in atomic number, the physical and chemical properties of alkali metals show regular trend. The atomic and ionic sizes increase on going down in the group and ionization enthalpies decrease in alkali metals. The same type of trend is observed in alkaline earth metals. The first element of each of these two groups, namely, lithium in group-1 and beryllium in group-2 show similarities with the element of the next group viz. Li-Mg and BeAl. This is called diagonal relationship. In fact, the first element of each group shows difference with other elements in the same group i.e their behaviour is anomalous. Alkali elements are bright white, soft metals melting at low temperatures. Li and Na are obtained by electrolysis. They are very active and their compounds are ionic. Their oxides and hydroxides are soluble in water and give strong alkaline solution. Amongst important compounds of sodium are sodium carbonate, sodium hydrogencarbonate, sodium hydroxide, NaOH is produced by Castner Kellner process and sodium carbonate by Solvay ammonia soda process. The chemistry of alkaline earth metals is similar to that of alkali metals. Even then some differences are there, because the atomic and ionic sizes of alkaline earth metals decrease and the charge of the cation increases. Their oxides are less basic than those of alkali metals. Amongst the industrially important compounds of sodium are caustic soda, washing soda and those of calcium are calcium oxide, calcium hydroxide, plaster of Paris, calcium carbonate and Portland cement. The production of cement can be carried out by grinding the mixture of lime stone and clay and heating it in rotary kiln. The clinker obtained is mixed with gypsum (2-3%) which gives fine powder of cement. All these substances have many uses. Monovalent sodium and potassium ions and divalent magnesium and calcium ions are found in larger proportion in biological fluid in a human being. These ions carry out important biological functions like maintenance of ionic equilibrium and nerve impulse conduction which is known as sodium-potassium pump. 301
M.C.Q. (1)
The alkali metals are low melting. Which of the following alkali metal is expected to melt if the room tempreature rises to 300 c ? (a) Na
(2)
(b) K
(c) Rb
(d) Cs
The reducing power of a metal depends on various factors. Suggest the factor which makes Li, the strongest reducing agent in aqueous solution. (a) Sublimation enthalpy (b) Ionisation enthalpy (c) Hydration enthalpy
(3)
Metal carbonates decompose on heating to give metal oxide and carbondioxide. Which of the metal carbonates is most stable thermally ? (a) MgCO3
(4)
(7)
(8)
(d) BaCO3
(b) Ca(OH)2
(c) Sr(OH)2
(d) Ba(OH)2
(b) MgCl2
(c) CaCl2
(d) SrCl2
The order of decreasing ionisation enthalpy in alkali metal is (a) Na > Li > K > Rb
(b) Li > Na > K > Rb
(c) Rb > Na > K > Li
(d) K < Li < Na < Rb
The solubility of metal halides depends on their nature, lattice enthalpy and hydration enthalpy of the individual ions. Among the fluorides of alkali metals, the lowest solubility of LiF in water is due to... (a) Ionic nature of lithium fluoride
(b) High lattice enthalpy
(c) High hydration enthalpy for lithium ion.
(d) Low ionisation enthalpy of lithium atom.
Amphoteric hydroxides react with both alkalies and acids. Which of the following Group - 2 metal hydroxides is soluble in sodium hydroxide ? (a) Be(OH)2
(9)
(c) SrCO3
Some of the group - 2 metal halides are covalent and soluble in organic solvents. Among the following metal halides, the one which is soluble in ethanol is .... (a) BeCl2
(6)
(b) CaCO3
Which of the following metal hydroxide is the least basic ? (a) Mg(OH)2
(5)
(d) Electron - gain enthalpy
(b) Mg(OH)2
(c) Ca(OH)2
(d) Ba(OH)2
In the synthesis of sodium carbonate, the recovery of ammonia is done by treating NH4Cl with Ca(OH)2.The by product obtained in this process is ... (a) NaCl
(b) NaOH
(c) CaCl2
(d) NaHCO3
(10) When sodium is dissolved in liquid ammonia, a solution of deep blue colour is obtained. The colour of the solution is due to ... (a) sodium ion
(b) ammoniated electron
(c) sodium amide
(d) ammoniated sodium ion
(11) By adding gypsum to cement... (a) setting time of cement becomes less.
(b) setting time of cement increases
(c) colour of cement becomes light
(d) shining surface is obtained 302
(12) Dead burnt plaster is ... (a) CaSO4
(b) CaSO 4 × H 2O
1 2
(c) CaSO 4 × H 2O
(d) CaSO 4 × 2H 2O
(13) A substance which gives crimson red flame and breaks on heating to give oxygen and a brown gas is .. (a) Magnesium nitrate
(b) Calcium nitrate
(c) Barium nitrate
(d) Strontium nitrate
(c) Na2CO3
(d) Na 2CO3 × H 2O
(14) The formula of sodash is ... (a) Na 2CO3 × 10H 2O
(b) Na 2CO3 × 2H 2O
(15) Which of the following compounds are readily soluble in water ? (a) BeSO4
(b) MgSO4
(c) BaSO4
(d) both (a) and (b)
(16) Identify the correct formula of halides of alkaline earth metals from the following. (a) BaCl2 • 2H2O
(b) BaCl2× 4H 2 O
(c) CaCl2× 4H 2 O
(d) SrCl2× 4H 2 O
(17) Which of the following statement is true about Ca(OH)2 ? (a) It is used in the preparation of bleaching powder. (b) It is a light blue solid. (c) It does not possess disinfectant property. (d) It is used in the manufacture of cement. (18) Match the elements given in Column - I with the colour they impart to the flame given in Column - II. Column - I
Column - II
(A) Cs
(P) Apple green
(B) Sr
(Q) Brick red
(C) K
(R) Blue
(D) Ca
(S) Crimson red
(E) Ba
(T) Violet
(a) A-P, B-Q, C-S, D-R, E-T
(b) A-Q, B-P, C-R, D-S, E-T
(c) A-R, B-S, C-T, D-Q, E-P
(d) A-S, B-R, C-Q, D-P, E-T
(19) When water is added to compound (A) of calcium, solution of compound (B) is formed. When CO2 is passed into the solution, it turns milky due to the formation of compound (C). If excess of carbon dioxide is passed into the solution, milkiness disappears due to the formation of compound (D). Identify the compound (D). (a) CaO
(b) Ca(OH)2
(c) CaCO3
(d) Ca(HCO3)2
(20) Which alkali metal emits longest wavelength light in Flame test ? (a) Na
(b) K
(c) Cs 303
(d) Li
(21) Which of the following is not known ? (a) KO3
(b) KO4
(c) KO2
(d) K2O2
(22) Which of the following acts as reducing as well as oxidising agent ? (a) NaNO3
(b) Na2O2
(c) Na2O
(d) KNO3
(23) The salt that is added to table salt to make it flow freely in rainy season is ... (a) KCl
(b) KI
(c) Ca3(PO4)2
(d) Na3PO4
(24) Which of the following alkaline earth metal sulphates is least soluble in water ? (a) MgSO4
(b) CaSO4
(c) BaSO4
(d) SrSO4
(25) The hydration energy of Mg2+ is greater than that of .... (a) Al3+ (b) Be2+ (c) Na+ (d) Mg3+ (26) The active constituent of bleaching powder is ... (a) Ca(OCl)2
(b) Ca(OCl)Cl
(c) Ca(C1O2)2
(d) Ca(C1O2)Cl
(27) KO2 is used in oxygen cylinders in space and submarines because it... (a) absorbs CO2 and increases O2 content
(b) eliminates moisture
(c) produces ozone
(d) None of the above
(28) A metal M readily forms water soluble sulphate MSO4, water insoluble hydroxide and oxide MO which becomes inert on heating. The hydroxide is soluble in NaOH. The metal M is... (a) Be
(b) Ca
(c) Mg
(d) Sr
(29) Which of the following is sparingly soluble in water ? (a) NaOH
(b) KOH
(c) LiOH
(d) RbOH
(c) Na
(d) Li
(30) Photo electric effect is maximum in ... (a) Cs
(b) K
(31) Among the following compounds of cement which is present in the highest amount ? (a) Ca2SiO4
(b) Al2O3
(c) Ca3SiO5
(d) Ca3Al2O6
(32) Which pair of the following chlorides do not impart colour to the flame ? (a) BeCl2 and SrCl2
(b) BeCl2 and MgCl2
(c) BaCl2 and CaCl2
(d) MgCl2 and CaCl2
(33) The sequence of hydration enthalpy in following ion is ... (a) Rb + > K + > Cs + > Na + > Li +
(b) Li+ > Na+ > K+ > Rb+ > Cs+
(c) K + >Na + > Rb + > Cs + > Li +
(d) Cs + >Rb + > K + > Na + > Li +
304
(34) In case of alkali metals, the covalent character increases in the order : (a) MI > MBr > MCl < MF
(b) MF < MCl < MBr < MI
(c) MBr < MCl < MI < MF
(d) MF < MBr < MCl < MI
(35) Among the following the least thermally stable is ? (a) K2CO3
(b) Na2CO3
(c) BaCO3
(d) Li2CO3
(36) Which of the following oxides is amphoteric in nature ? (a) MgO
(b) BeO
(c) CaO
(d) BaO
(37) Which of the following characteristics is not related to alkali metals ? (a) Their ions are iso electronic with noble gases. (b) low melting point
(c) low electronegativity
(d) high ionisation energy
(38) Fill in the blanks with proper option given below for the following statement. “All the halides of alkaline earth metals with exception of ‘‘ (a) Barium halide
(b) Strontium halide
’’ are ionic in nature.”
(c) Beryllium halide (d) Calcium halide
(39) K2CO3 can not be prepared by solvay ammonia process because ... (a) K2CO3 is fairly soluble in water.
(b) It has no water of crystallization.
(c) KHCO3 is highly soluble in water.
(d) K2CO3 decomposes in H2O.
(40) The reaction of Cl2 with X gives bleaching powder X is .... (a) CaO
(b) Ca(OH)2
(c) Ca(OCl)2
(d) Ca(O3Cl)2
(41) Which of the following alkaline earth metal sulphates has hydration enthalpy higher than the lattice enthalpy ? (a) SrSO4
(b) MgSO4
(c) CaSO4
(d) BaSO4
(42) A compound (A) on heating gives a colourless gas and a residue that is dissolved in water to obtain (B). Excess of CO2 is bubbled through aqueous solution of B, (C) is formed, which is recovered in the solid form. Solid (C) on gentle heating gives back (A). The compound is ... (a) CaCO3
(b) K2CO3
(c) Na2CO3
(d) CaSO 4 × 2H 2O
(43) For alkaline metal, which of the following trends is incorrect ? (a) Hydration enthalpy : Be > Mg > Ca > Sr (b) Second Ionization enthalpy: Be > Mg > Ca > Sr (c) Density : Sr > Be > Mg > Ca (d) Atomic size : Sr > Ca > Mg > Be (44) Which of the following compounds is most stable ? (a) LiCl
(b) LiI
(c) LiBr 305
(d) LiF
(45) Flame test is not given by ... (a) Be
(b) Sr
(c) K
(d) Ca
(46) The alkaline earth metals forming ionic oxides are ... (a) MgO
(b) BeO
(c) CaO
(d) (a) and (c)
(47) The basic character of the oxides MgO, SrO, K2O, NiO and Cs2O increases in the order : (a) MgO > SrO > K2O > NiO > Cs2O
(b) Cs2O < K2O < MgO < SrO < NiO
(c) NiO < MgO < SrO < K2O < Cs2O
(d) K2O < NiO < MgO < SrO < Cs2O
(48) Which of the following are arranged in increasing order of solubilities ? (a) CaCO3 < KHCO3 < NaHCO3
(b) NaHCO3 < KHCO3 < CaCO3
(c) KHCO3 < NaHCO3 < CaCO3
(d) CaCO3 < NaHCO3 < KHCO3
(49) The compound insoluble in aceticacid is ... (a) Calcium oxide
(b) Calcium carbonate
(c) Calcium oxalate
(d) Calcium hydroxide
(50) Which of the following has the lowest melting point ? (a) LiCl
(b) KCl
(c) NaCl
(d) RbCl
(51) The correct order of decreasing ionic character is ... (a) BeCl2 > MgCl2 > CaCl2 > BaCl2
(b) BeCl2 > MgCl2 > BaCl2 > CaCl2
(c) BeCl2 > BaCl2 > MgCl2 > CaCl2
(d) BaCl2 > CaCl2 > MgCl2 > BeCl2
(52) The highest lattice energy corresponds to ... (a) MgO
(b) CaO
(c) SrO
(d) BaO
(53) How many of the following s -block elements do not give characteristic colours in the flame test ? Li , Be, Ca, Ba, Sr, Mg, Na, K, Ba (a) 3
(b) 4
(c) 2
(d) 5
(54) How many of the following sulphates of metals dissolve in the water ? SrSO4, K2SO4, BeSO4, Li2SO4, MgSO4, BaSO4, Na2SO4, CaSO4, Rb2SO4 (a) 6
(b) 4
(c) 3
(d) 5
(55) How many of the following hydroxides is/are amphoteric in character ? CsOH, LiOH, Ca(OH)2, Be(OH)2, Mg(OH)2, Sr(OH)2, Ba(OH)2, KOH, NaOH. (a) 1
(b) 4
(c) 5
(d) 3
(56) Out of Li, Na, K, Rb and Cs how many of them directly form superoxides on heating with oxygen ? (a) 5
(b) 2
(c) 3 306
(d) 4
(57) How many of the following metals when heated in an atmosphere of N2 gas form nitrides ? Li, Na, K, Rb, Cs, Mg, Ca, Sr, Ba (a) 9
(b) 5
(c) 3
(d) 6
(58) Which of the following is not correct for workfunction of Na+ ions in human body ? (a) An important role in nerve signal transmission. (b) Control of flow of water between cell membrane. (c) For transport of sugar and amino acid in cell. (d) They activate the enzyme. (59) Which of the following is not correct ? heat (a) 2Li 2 O ¾¾¾ ® Li 2 O 2 + 2Li 673k
heat (b) 2K 2 O ¾¾¾ ® K 2 O 2 + 2K 673k
heat ® Na 2O 2 + 2Na (c) 2Na 2O ¾¾¾ 673k
heat ® Rb 2O 2 + 2Rb (d) 2Rb 2O ¾¾¾ 673k
(60) Which of the following has maximum lattice energy ? (a) Li2O
(b) Na2O
(c) MgO
(d) BaO
(61) Which of the following statements is/are correct for “when alkali metals are dissolved in liquid ammonia. We get...” (a) A blue solution in case of dilute alkali metal ammonia solution. (b) If we increase cocentration of metal in ammonia then the blue colour starts changing and finally changes to that of bronze colour. (c) The blue colour of the solution of alkali metal in liquid ammonia is due to excitation of free ammoniated electrons to higher energy levels. (d) All the above are correct. (62) Which of the following pairs of elements possess diagonal relationship ? (a) Li and Mg
(b) Li and Al
(c) Na and Mg
(d) Cs and Ba
(63) Fill in the blanks with suitable option. “The important ingredients of potland cement are dicalcium silicate ‘‘ silicate ... % and tricalcium aluminate ‘‘ ’’ % respectively.” (a) 26 %, 51%, 11%
(b) 51%, 26%, 11%
(c) 11%, 51%, 26%
(d) 26%, 11%, 51%
’’ % , tricalcium
(64) A sample of portaland cement contain 23% SiO2, 3% Al2O3 and 2% Fe2O3 then what would be its silica module (n) ? (a) 3.83
(b) 28
(c) 21.73 307
(d) 4.6
(65) For a good quality of cement, the ratio of silica (SiO2) and alumina (Al2O3) must be between... (a) 3 to 5
(b) 2.5 to 4
(c) 6 to 7.5
(66) Chloro phyll and heamoglobin are complex of ‘‘ (a) Mg2+ and Ca2+
(b) Na+ and K+
(d) 4 to 5.5
’’ and ‘‘‘ (c) Mg2+ and Fe2+
’’ respectivelly.. (d) Cl– and Fe2+
(67) Which of the following is the component of most of the kidney stones ? (a) (COO)2Ca
(b) (COONa)2
(c) (COO)2 Ba
(d) (COO)2Mg
(68) Which of the following metal ions plays an important role in muscle contraction ? (a) K+
(b) Mg2+
(c) Na+
(d) Ca2+
(c) CaF2
(d) CaBr2
(69) White enamel of our teeth is ... (a) Ca3(PO4)2
(b) CaCl2
(70) Two mole of magnesium nitride on reacting with an excess of water gives : (a) One mole of ammonia
(b) Two moles of nitric acid
(c) Four moles of ammonia
(d) Three moles of ammonia
(71) Which of the following metal is used in windows of X - ray tubes ? (a) Be
(b) Mg
(c) Ba
(d) Al
(c) Dolomite
(d) Carnalite
(72) Which of the following is not a Mg ore ? (a) Magnesite
(b) Gypsum
(73) The difference of water molecules in gypsum and plaster of paris is ... (a)
5 2
(b) 2
(c)
1 2
(d) 1
1 2
(74) Which of the following exists in polymeric form ? (a) AlCl3
(b) SiC
(c) BeCl2
(d) B2H6
(75) The electronic cofiguration of metal M is ls2 2s2 2p6 3s2. The formula of its oxide would be ... (a) MO
(b) M2O
(c) M2O3
(d) MO2
(76) The formula of carnalliteis ... (a) KCl MgCl2 2H 2O
(b) K 2 O Al2 O3 6H 2 O
(c) KCl g MgCl2 g 6H 2 O
(d) Na 2 B4 O7 10H 2 O
(77) When NaOH is made, the gas released at the cathode is ... (a) Cl2
(b) H2
(c) O2 308
(d) H2O
(78) Choose proper option for matching column - I and column - II Column - I
Column - II
(A) NaOH
(P) Photo electric cells
(B) Na2CO3
(Q) Coolant in nuclear reactors
(C) Liquid Na
(R) SO2 absorber
(D) Caesium
(S) Detergent
(a) A-R, B-S, C-Q, D-P
(b) A-P, B-Q, C-R, D-S
(c) A-Q, B-P, C-R, D-S
(d) A-S, B-Q, C-P, D-R
(79) In the electrolytic separation of Li, KCl is added to LiCl ... (a) To increase the conductivity of LiCl (b) To lower the fusion temperature of the mixure (c) To decrease the conductivity of LiCl (d) Both (a) and (b) (80) Molecular formula of Glauber’s salt is ... (a) MgSO 4 × 7H 2O
(b) FeSO 4 × 7H 2O
(c) CuSO 4 × 5H 2O
(d) Na 2SO 4 •10 H 2 O
(81) The name “Blue John” is given to which of the following compounds. (a) CaH2
(b) CaF2
(c) Ca3(PO4)2
(d) CaO
(c) Ba
(d) Ag
(82) The wire of flash bulb is made of ... (a) Mg
(b) Cu
(83) 30 gm of Mg and 30 gm of O2 are reacted and the residual mixure contains ... (a) 60 gm of MgO only
(b) 40 gm of MgO and 20 gm of O2
(c) 45 gm of MgO and 15 gm of O2
(d) 50 gm of MgO and 10 gm of O2
(84) Which of the following is not correct ? (a) 4LiNO3(s) ® 2Li2O(s) + 4NO2(s) + O2(g)
(b) 2NaNO3(s) ® 2NaNO2(s) + O2(g) (c) The oxides Li2O and MgO do not give super oxides by combining with more oxygen. (d) Lithium hydrogen carbonate is obtained in solid form. (85) The number and type of bonds between two carbon atoms in calcium carbides are ... (a) one sigma, one pi.
(b) one sigma, two pi.
(c) two sigma, one pi.
(d) two sigma, two pi
(86) Which of the following salts are composed of isoelectronic cations and anions ? (a) NaCl
(b) MgF2
(c) CaS 309
(d) (b) and (c) both
(87) Which are correct statements for Be and Al ? (a) Both are rendered passive by con.HNO3 (b) Both have sp - hybridization in their compounds. (c) Both form acidic oxides (d) Both form hydrides (88) Identify the correct statement ? (a) Gypsum contains a lower percentage of calcium than plaster of paris (b) Gypsum is obtained by heating plaster of paris (c) Plaster of paris can be obtained by hydration of gypsum (d) Plaster of paris is obtained by partial oxidation of gypsum. (89) Which of the following metal is used in the test of elements in organic compounds by Lassigne test ? (a) Li
(b) Na
(c) K
(90) The setting time of dicalcium silicates is ‘‘ (a) 28 days
(b) 1 year
(d) Rb
’’’ (c) 1 week
(d) 24 hour
Assertion - Reason type Questions The questions given below contains statement - 1 (Assertion) and statement - 2 (Reason) Each question has four choices (a), (b),(c) and (d) out of which only one is correct. Choose the correct option as under. (a)
Statement -1 is true, statement - 2 is true Statement - 2 is a correct explanation for statement - 1
(b)
Statement - 1 is true, statement - 2 is true; Statement - 2 is not a correct explanation for statement - 1
(c)
Statement - 1 is true, statement - 2 is false
(d)
Statement - 1 is false, statement - 2 is true.
(91) Statement-1 Alkali metals dissolve in ammonia to give blue solutions. Statement-2 Alkali metals in liquid ammonia give solvated species of the type [M(NH3 ) X ]+ (92) Statement-1 Sodium metal is softer than potassium metal Statement-2 Metalic bonding in Potassium is weaker than inSodium. (93) Statement-1 Be(OH)2 is solublein HCl and NaOH Statement-2 Be(OH)2 is amphoteric in nature 310
(94) Statement-1 Be Forrns [BeF4 ]2 – but Al forrms [AlF6 ]3 – Statement-2 Be doesnot have d–orbitals in the Valence shell but Al has. (95)
Statement - 1 Li
CO3 and Na2CO3 are therally unstable.
2
Statement - 2 Both the carbonates are salts of large cations and large anions. (96) Statement - 1 Metallic character of alkali metals increases on going down a group from top to bottom. Statement - 2 Ionisation enthalpy of alkali metals increases on going down from top to bottom. (97) Statement - 1 Superoxides of alkali metals are diamagnetic. Statement - 2 Superoxides contain the ion O2– which has one unpaired electron. (98) Statement - 1 Alkali metals donot impart colour to the flame. Statement - 2 Their ionization enthalpies are very low. (99) Statement - 1 Sodium cannot be obtained by chemical reduction of its ore. Statement - 2 Sodium is one of the strongest reducing agent. (100) Statement - 1 Beryllium hydroxide becomes soluble in excess alkali forming beryllate ion [Be(OH)4]2Statement - 2 Beryllium ion has greater tendency to form complexes. (101) The half life period of isotopes223Fr is .... (a) 21 hour
(b) 21 second
(c) 21 minute
(d) 21 day
(c) Li
(d) Rb
(102) Sylvine is the mineral of ... (a) K
(b) Na
(103) Which pump is important in biological reaction in human body ? (a) Ca-Mg Pump
(b) K-Fe Pump
(c) Na-K Pump
(d) Fe-Ca Pump
(104) Withrite is which type of salt of Barium ? (a) Carbonate
(b) Sulphate
(c) Chloride
(d) Phosphate
(105) Silastine is the mineral of ... (a) Ca
(b) Ra
(c) Ba
(d) Sr
(106) Which of the following reaction does not occur in solvay ammonia soda process ? (a) (NH 4 ) 2 CO3 + H 2O + CO 2 ® 2NH 4 HCO3 Δ (b) 2KHCO3 ¾¾ ® K 2 CO3 + H 2 O + CO 2 D (c) 2NaHCO3 ¾¾ ® Na 2CO3 + H 2O + CO 2
(d) 2NH 4 Cl + Ca(OH)2 ® 2NH3 + CaCl2 + H 2 O
311
(107) What will be final weight of 286 gm Na2CO3.10H2O by Heating at 373 K ? (a) 206 gm
(b) 162 gm
(c) 186 gm
(d) 124 gm
(108) The order of decreasing polarity in the compounds CaO, CsF, KCl, MgO is (a) CaO, CsF, KCl, MgO
(b) MgO, KCl, CaO, CsF,
(c) KCl, CaO, CsF, MgO
(d) CsF, KCl, CaO, MgO
(109) Which is not an ore of Ca ? (a) Lime stone
(b) Flurospar
(c) Dolomite
(d) Epsomsalt
(110) A certain metal is used to prepare an antacid, this metal accidently catches fire which can not be put out by using CO2 based extiguishers. The metal is (a) C
(b) Ca
(c) Mg
(d) Na
ANSWER KEY (1) (2) (3) (4) (5) (6) (7) (8) (9) (10) (11) (12) (13) (14) (15)
d c d a a b b a c b b a d c d
(16) (17) (18) (19) (20) (21) (22) (23) (24) (25) (26) (27) (28) (29) (30)
a a c d b b c c c c a a a c a
(31) (32) (33) (34) (35) (36) (37) (38) (39) (40) (41) (42) (43) (44) (45)
c b b b d b d c c b b a b d a
(46) (47) (48) (49) (50) (51) (52) (53) (54) (55) (56) (57) (58) (59) (60)
d c d c a a a c a a c b d a c
(61) (62) (63) (64) (65) (66) (67) (68) (69) (70) (71) (72) (73) (74) (75)
312
d a a d b c a d c c a b d c a
(76) (77) (78) (79) (80) (81) (82) (83) (84) (85) (86) (87) (88) (89) (90)
c b a d d b a d d b d a a b a
(91) (92) (93) (94) (95) (96) (97) (98) (99) (100) (101) (102) (103) (104) (105)
b d a a c c d d a b c a b a d
(106) (107) (108) (109) (110)
b d d d c
Hints (1)
Melting point decreases as the strength of metallic bonding decreases with increasing size of the atom.
(2)
Due to small size of the Li+, its hydration enthalpy is the highest and hence Li is the strongest reducing agent.
(3)
Thermal stability of metal carbonates increases as the electropositive character of the metal or the basicity of the metal hydroxide increases from Be(OH)2 to Ba(OH)2, Thus, BaCO3 is the most stable.
(4)
As the ionisation enthalpy increases from Mg ® Ba the M - O bond becomes weaker and weaker down the group, and hence basicity increases down the group. Thus, Mg(OH)2 is least basic.
(6)
Ionisation enthalpy decreases as the atomic size, increases , so, Li > Na > K > Rb
(7)
Due to small size of Li+ and F– ions, lattice enthalpy is much higher than hydration enthaalpy and hence LiF is least soluble among alkali metal Fluorides.
(8)
Be(OH)2 being amphoteric dissolves in NaOH.
(9)
2NH 4 Cl(s) + Ca(OH) 2(aq) ® 2NH3(g) + CaCl2(aq) + H 2 O(1)
D (14) 2Sr(NO3 )2 ¾¾ ® 2SrO + O 2 + 4NO 2 ® brown gas. Sr gives Crimson red flame.
(24) The solubility decreases as we move from CaSO4 to BaSO4. (32) The electrons of Be and Mg are so strongly bonded that they do not get excited in flame. (35) Li2CO3 is not so stable towards heat, being small in size, it is decompose into Li2O and CO2. (41) The hydration enthalpy of Be2+ and Mg2+ ions is more than the lattice enthalpy and so they are soluble in water. D ® CO 2(g) + CaO(s) (42) CaCO3(s) ¾¾
(A) CaO(s) + H 2 O(1) ® Ca(OH) 2(aq)
(B) Ca(OH) 2(aq) + 2CO 2(g) ® Ca(HCO3 ) 2(aq)
(C) D Ca(HCO3 ) 2(aq) ¾¾ ® CaCO3(s) + CO 2(g) + H 2 O(l)
(A) (44) Because of the small size of Li and F, LiF has highest lattice enthalpy and hence most stable. (47) Alkali metal oxides are most basic followed by alkaline earth metal oxides while transition metal oxides are least basic. Amongst alkali and alkaline earth metal oxides, basicity increases down the group. Thus, Cs2O is more basic than K2O and SrO is more basic than MgO. Therefore, the overall order is : NiO < MgO < SrO < K2O < Cs2O 313
(48) The solubility of bicarbonates of alkali metals increases down the group but alkaline earth metal carbonates are insoluble in H2O. (49) CaO, CaCO3 and Ca(OH)2 are all bases and hence must dissolve in acetic acid to form calcium acetate only calcium oxalate does not dissolve in CH3COOH. (53) 2 (Be, Mg) (54) 6 [K2SO4, BeSO4, Li2SO4, MgSO4, Na2SO4, Rb2SO4] (55) 1 [Be(OH)2] (56) 3 [K, Rb, Cs] (57) 5 [Li, Mg, Ca, Sr, Ba] (59) Lithium does not form peroxide. (64) silica mod ule h =
%SiO 2 23 = = 4.6 %Al2 O3 + %Fe2 O3 2 + 3
(70) Mg 3 N 2 + 6H 2O ® 3Mg(OH)2 + 2NH3 1mole
2 moles
2 mole
4 moles
(71) Gypsum = CaSO4 . 2H2O 1 Plaster of paris = CaSO 4 × H 2 O 2
So, difference = 2 –
1 1 =1 2 2
(83) 2Mg + O 2 ® 2MgO 2 ´ 24
32gm
2x24 gm 32gm
2 ´ 40gm
2 ´ 40gm
Here, 48 gm of Mg requires 32 gm of O2 to form 80 gm MgO so, 30 gm of Mg requires 20 gm of O2 to form 50 gm MgO so, 10 gm of O2 is remain. (85) Ca 2+ C- º C- ® 1s and 2p CaC2 : Calcium Carbide (97) correct statement : 1 Superoxides of alkali metals are paramagnetic. (98) correct statement : 1 Alkali metals impart characteristic colour to the bunsen flame. Na 2CO3 × H 2 O(s) 373K (107) Na 2CO3 × 10H 2O(s) ¾¾¾ ® + 9H 2O(g) 286 gm
124 gm
(108) According to Fajan rules, polarity increases as the size of the cation increases. 314
NOTES
315
NOTES
316
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